Category: ISC

ISC Economics Previous Year Question Paper 2015 Solved for Class 12

Maximum Marks: 80
Time allowed: 3 hours

• Candidates are allowed additional 15 minutes for only reading the paper.
• They must NOT start writing during this time.
• Answer Question 1 (Compulsory) from Part I and five questions from Part II.
• The intended marks for questions or parts of questions are given in brackets [ ].

Part – I (20 Marks)
Answer all questions.

Question 1.
Answer briefly each of the following questions (i) to (x): [10 × 2]
(i) Define marginal utility. When can it be negative?
(ii) What is meant by production function?
(iii) Name the market where average revenue is equal to marginal revenue. Give a reason for your answer.
(iv) Give one difference between accounting cost and opportunity cost.
(v) What is the reason for an indeterminate demand curve under Oligopoly?
(vi) What is meant by a propensity to consume?
(vii) Explain discounting bills of exchange as one of the functions of the banks.
(viii) Differentiate between revaluation of currency and appreciation of currency.
(ix) How can gross domestic product at factor cost be obtained from the gross national product at market price?
(x) What is meant by revenue deficit? Explain its implication.
Answer:
(i) Marginal utility is the net addition made to total utility by the consumption of an additional unit. Mu_n = Tu_n – Tu_n-1. when total utility is maximum, marginal utility is zero.

(ii) Production function studies the functional relationship between physical inputs and physical outputs.

(iii) Perfect Competition: Under perfect competition, AR or price remains fixed for the firm. So MR will also be constant. Hence AR = MR.

(iv) Accounting Cost: This is the actual, numeric dollar amount that a firm (company) pays to run their business.
Opportunity Cost: This is the cost of choosing one thing over another.

(v) The main reasons for the indeterminate price and output under oligopoly may be summarised as follows:

• Different Behaviour patterns: In oligopoly due to interdependence, a vast variety of behaviour patterns becomes possible.
• Indeterminate Demand Curve: Another cause of indeterminateness of price and output in an oligopoly is indeterminate demand curve.

(vi) The propensity to consume means the proportions of total income or of in increase an income that consumers tend to spend on goods and services rather than save.

(vii) Discounting of Bills of Exchange
Another important form of bank lending is through discounting or purchasing the bills of exchange. A bill of exchange is drawn by a creditor on the debtor specifying the amount of debt and also the date when it becomes payable. Such bills of exchange are normally issued for a period of 90 days. Thus, creditors cannot get it encashed from the debtors before the maturity period. However, if the creditor needs money before the maturity period (i.e., 90 days), he can get it discounted from the commercial bank. The bank makes payment to the creditor after deducting its commission. When the bill matures, the bank will get payment from the debtors.

(viii) Revaluation means a rise of domestic currency in relation to foreign currency in a fixed exchange rate whereas appreciation implies an increase in the external value of a currency.

Question 2.
(a) Discuss the relationship between the income of the consumer and demand for a commodity with respect to normal goods, inferior goods and necessities. [3]
(b) Differentiate between the extension of demand and an increase in demand, using diagrams. [3]
(c) Explain with the help of a diagram the consumer’s equilibrium through utility approach. [6]
Answer:
(a) Normal good: Demand will increase and the demand curve will shift towards the right.
An inferior good: Demand will decrease and the demand curve will shift towards left.

(b)

(c) In the case of the consumer consumes two or more than two commodities his equilibrium will be determined by the Law of Equi-marginal utility.
According to the Law of Equi-marginal utility, the consumer spends his limited income on different goods in such a way that marginal utility derived by all commodities are equal.
MU_X = M_Y = MU_Z ………
Explanation: Let us now discuss the Law of Equi-marginal Utility with the help of a numerical example: Suppose, the total money income of the consumer = ₹ 5.
Price of good X and Y = ₹ 1 per unit.
So, the consumer can buy a maximum of 5 units of ‘X’ or 5 units of Y. The given table shows the marginal utility which the consumer derives from various units of ‘X’ and ‘Y’.
Table – Consumer’s Equilibrium – 2 Commodities

From the table, it is obvious that the consumer will spend the first rupee on commodity ‘X’, which will provide him utility of 20 utils. The second rupee will be spent on commodity ‘ Y’ to get utility of 16 utils. To reach the equilibrium, a consumer should purchase that combination of both the goods, when
(i) MU of the last rupee spent on each commodity is same; and
(ii) MU falls as consumption increases.
It happens when a consumer buys 3 units of ‘X’ and 2 units of ‘ Y’ because:
MU from last rupee (i.e5th rupee) spent on commodity Y gives the same satisfaction of 12 utils as given by last rupee {i.e., 4th rupee) spent on commodity X; and
MU of each commodity falls as consumption increases.
The total satisfaction of 74 utils will be obtained when a consumer buys 3 utils of ‘X’ and 2 units of ‘ Y’. It reflects the state of the consumer’s equilibrium. If the consumer spends his income in any other order, total satisfaction will be less than 74 utils.

Question 3.
(a) Discuss any two properties of the indifference curve. [3]
(b) Draw diagrams to show the elasticity of demand when it is: [3]
(i) Greater than one
(ii) Less than one
(iii) Unity
(c) Explain the geometric method of calculating the elasticity of supply. [6]
Answer:
(a) (i) An indifference curve (IC) always slopes downward: This property implies that to increase the consumption of X-good, the consumer has to reduce the consumption of Y good so as to remain at the same level of satisfaction.

As shown in the given diagram:
To increase the quantity of ‘X’ good from OX to OX₁, the consumer has to reduce good Y from OY to OY₁.

(ii) Indifference curves are convex to the origin: This property is based on the principle of diminishing marginal rate of substitution. It implies that as the consumer substitutes X for Y, the marginal rate of substitution between them goes on diminishing as shown in the following figure.

AB > CD > EF or Diminishing MRS

(b) (i) Greater than one

(ii) Less than one

(iii) Unity

(c) According to the geometric method, elasticity is measured at a given point on the supply curve.
This method is also known as ‘Arc Method’ or ‘Point Method’.
(i) Highly Elastic Supply (E_s > 1): A supply curve, which passes through the Y-axis and meets the extended X-axis at some point. For example in fig. the supply is highly elastic.
Since LQ is greater than OQ, the elasticity of supply at point A will be greater than one (highly elastic).
In general, we can say that a straight line supply curve passing through the Y-axis or having a negative intercept on X-axis is highly elastic (E_s > 1).

(ii) Unitary Elastic Supply (E_s = 1): If the straight-line supply curve passes through the origin (supply curve SS in fig.), then elasticity of supply will be equal to one. In the diagram, Elasticity of Supply (E_s) = \(\frac { OQ }{ OQ }\) = 1. Hence, the supply is unitary elastic.

(iii) Less Elastic Supply (E_s < 1): If a supply curve meets the X-axis at some point, say, L in Fig., then supply is inelastic. As seen in the fig., E_s = \(\frac { LQ }{ OQ }\) and LQ < OQ. So, E_s < 1, i.e. supply is less elastic.

(iv) Perfectly Elastic Supply: When there is an infinite supply at a particular price and the supply becomes zero with a slight fall in price, then the supply of such a commodity is said to be perfectly elastic. In such a case E_s = Y and the supply curve is a horizontal straight line parallel to the X-axis, as shown in fig.

(v) Perfectly Inelastic Supply: When the supply does not change with change in price, then supply for such a commodity is said to be perfectly inelastic. In such a case, E_s = 0 and the supply curve (SS) is a vertical straight line parallel to the Y-axis as shown in fig.

Question 4.
(a) Show with the help of diagrams, the effect on equilib¬rium price and quantity when
(i) There is a fall in the price of substitute goods.
(ii) There is a rise in the prices of inputs.
(b) The cost function of a firm is given below: [3]

Calculate:
(i) AFC
(ii) AVC
(iii) MC
(c) Explain the law of variable proportions with the help of a diagram.
Answer:
(a) (i) (a) A decrease in the price of a substitute good will disturb the equilibrium point and quantity of the commodity.

(b) Due to a decrease in the price of a substitute good (say coffee) demand for tea will fall because tea has become relatively costly.

(c) A decrease in the price of a substitute good will have a direct impact on the supply (supply will increase) of the good because produces of tea will like to clear their stock as fast as possible.

(d) The new equilibrium can be studied in three different cases.
Case 1: Decrease in Demand = Increase in Supply
When the decrease in demand is proportionately equal to the increase in supply, then leftward shift in demand curve from DD to D₁D₁ is proportionately equal to a rightward shift in supply curve from SS to S₁S₁.
(Fig. A). The new equilibrium is determined at E₁. As the decrease in demand is proportionately equal to the increase in supply, equilibrium quantity remains the same at OQ, but equilibrium price falls from OP to OP₁.

Case 2: Decrease in Demand > Increase in Supply
When a decrease in demand is proportionately more than an increase in supply, then leftward shift in demand curve from DD to D₁D₁ is proportionately more than the rightward shift in supply curve from SS to S₁S₁ (Fig. B). The new equilibrium is determined at E₁. As the decrease in demand is proportionately more than the increase in supply, equilibrium quantity falls from OQ to OQ₁ and equilibrium price falls from OP to OP₁.

Case 3: Decrease in Demand < Increase in Supply
(i) When the decrease in demand is proportionately less than an increase in supply, then leftward shift in demand curve from DD to D₁D₁ is proportionately less than a rightward shift in supply curve from SS to S₁S₁ Fig. C). The new equilibrium is determined at E₁. As the decrease in demand is proportionately less than the increase in supply, equilibrium quantity rises from OQ to OQ₁ whereas, equilibrium price falls from OP to OP₁.

(ii) With a rise in the price of the input, the cost of production will increase and profit will fall. With a fall in profit rate supply will decrease.
A new equilibrium will be established which will be at a higher price, lower quantity demand and lower supply.


(c) Law of Variable proportion
Law of Variable Proportions (LVP) states that as we increase the quantity of only one input keeping other inputs fixed, total product (TP) initially increases at an increasing rate, then at a decreasing rate and finally at a negative rate.
In the given diagram, the quantity of the variable factor has been measured on X-axis and Y-axis measures the total product on Y-axis. The diagram indicates, how the total product and marginal product changes as a result of increases in the quantity of one factor to a fixed quantity of other factors.

The law can be better explained through three stages-
(a) First stage :
Increasing Return to a Factor: In the first stage, every additional variable factor adds more and more to the total output. It means TP increases at an increasing rate and MP of each variable factor rises. Better utilization of fixed factor and increases in the efficiency of a variable factor due to specialization are the major factors responsible for increasing returns. The increasing returns to a factor stage have been shown in the given diagram between O to P. It implies. TP increases at an increasing rate (till point ‘P’) and MP rises till it reaches its maximum point ‘K’. which marks the end of the first phase.

(b) Second stage:
Diminishing returns to a Factor: In the second stage, every additional variable factor adds a lesser and lesser amount of output. It means TP increases at a diminishing rate and MP falls with increase in a variable factor. The breaking of the optimum combination of a fixed and variable factor is the major factor responsible for diminishing returns. The second stage ends at point ‘S’ when MP is zero and TP is maximum (point ‘M’).
2 stage is very crucial as a rational producer will always aim to produce in this phase because TP is maximum and MP of each variable factor is positive.

(c) Third stage:
Negative Returns to a Factor: In the third stage the employment of additional variable factor causes TP to decline. MP now becomes negative. Therefore, this stage is known as negative returns to a factor. Poor coordination between variable and fixed factor is the basic cause for this stage. In fig. the third stage starts after point ‘N’ on MP curve and point ‘M’ on TP curve. MP of each variable factor is negative in the 3 stage. So, no firm would deliberately choose to operate in this stage.

Question 5.
(a) Discuss two features of monopoly. [3]
(b) Show with the help of a diagram, how a perfectly competitive firm earns normal profit in short-run equilibrium. [3]
(c) Explain how a producer can maximise profit by using MR and MC curves. [6]
Answer:
(a) (i) Restriction on Entry and Exit: There exist strong barriers to entry of new firms and exit of existing firms. As a result, a monopoly firm can earn abnormal profits in the long run. These barriers may be due to legal restrictions like licensing or patent rights or due to restrictions created by firms in the form of a cartel.
(ii) Price Discrimination: A monopolist may charge different prices for his product from different sets of consumers at the same time. It is known as ‘Price Discrimination’

(b) Under perfect competition a firm’s, average revenue curve is equal to the marginal revenue curve due to uniform prices for homogeneous goods. In, short-run, a competitive firm can earn supernormal profits, normal profits can suffer losses also. Adjacent diagram shows a perfectly competitive firm earning normal profits. Since the firm is price taker, it has to decide the amount of output it should produce at the given price so as to maximise the profits following the equilibrium conditions.
SMC = MR and AR = AC

In the given diagram OQ is equilibrium output where SMC is equal to MR Since SAC is tangent to P = AR = MR line, the firm covers only its SAC which includes normal profits.

(c) In order to know the position of maximum profit, a firm compares the marginal cost with marginal revenue: So. the first condition of a firm’s equilibrium is that marginal cost must be equal to marginal revenue (MC = MR). It is necessary, but not sufficient condition of equilibrium. A firm may not get maximum profit even when its marginal cost is equal to marginal revenue. So it must fulfil the second condition of equilibrium as well, i.e., marginal cost curve must cut marginal revenue curve from below or the slope of MC curve must be steeper than the slope of MR curve. According to marginal analysis, a firm would, therefore, be in equilibrium when the following two conditions are fulfilled.
1. MC = MR.
2. MC curve cuts the MR curve from below.

Both these conditions of firm’s equilibrium are explained with the help of Fig. In this figure, PP is average revenue (price per unit) as well as marginal revenue curve. It is clear from this figure, that MC curve is cutting MR curve PP at two points ‘A’ and ‘E’. Point ‘A’ cannot indicate the position of equilibrium of the firm as at point A Marginal cost of the firm is still falling or we can say MC is not cutting MR from below.

On the other hand, point E show s that the firm is producing OM units of output. If the firm produces more than OM units of output, its marginal cost (MC) will exceed marginal revenue (MR) and it will have to incur losses. Thus point ‘E’ will represent the equilibrium of the firm. At this point, both the conditions of equilibrium are being fulfilled.
(1) Marginal cost is equal to marginal revenue (MC = MR) and
(2) Marginal cost curve is cutting the marginal revenue curve from below. At point ‘E’ i.e.,’ equilibrium position, the firm is getting maximum profit. In case, the firm produces more or less than OM output, then its profits will be less than the maximum. So the firm, at OM level of output, will have no tendency either to increase or decrease its output from this level. It will, therefore, be in equilibrium at point E.

Question 6.
(a) Find the value of additional investment made by the government, when MPC 05 and the increase in income (ΔY) = ₹ 1000. [3]
(b) What is meant by autonomous consumption? Explain with the help of a diagram. [3]
(c) Explain the concept of deficient demand with the help of aggregate demand and aggregate supply curves. Discuss one physical and one monetary measure to correct it. [6]
Answer:


(c) Deficient demand refers to the situation when aggregate demand (AD) is less than the aggregate supply (AS) corresponding to the full employment level of output in the economy.
The situation of deficient demand arises when planned aggregate expenditure falls short of aggregate supply at the full employment level. It gives rise to the deflationary gap. A deflationary gap is a gap by which actual aggregate demand required to establish full employment equilibrium.
The concepts of deficient demand and deflationary gap are shown in Fig. (a)
(i) In the given diagram, income, output and employment are measured on the X-axis and aggregate demand is measured on the Y-axis.
(ii) Aggregate demand (AD) and aggregate supply (AS) curves intersect at point E, which indicates the full-employment equilibrium.
(iii) Due to a decrease in investment expenditure (ΔI), aggregate demand falls from AD to AD₁. It denotes the situation of deficient demand and the gap between them, i.e., EG is termed as a deflationary gap.
(iv) Point F indicates the underemployment equilibrium.

Physical measure: Reduction in public expenditure.
Monetary Measure: Increase in money supply.

Question 7.
(a) Discuss two qualitative methods of credit control. [3]
(b) Explain any two secondary functions of money. [3]
(c) Discuss the various components of the current account of the balance of payment. [6]
Answer:
(a) (i) Credit Rationing: Rationing of credit is a method by which the Reserve Bank seeks to limit the maximum amount of loans and advances, and also in certain cases fix a ceiling for specific categories of loans and advances.

(ii) Moral Suasion: Moral suasion and credit monitoring arrangement are other methods of credit control. The policy of moral suasion will succeed only if the Reserve Bank is strong enough to influence the commercial banks.

(b) (i) Store of value: In order to be a medium of exchange,rtioney must hold its value over time; that is, it must be a store of value. If money could not be stored for some period of time and still remain valuable in exchange, it would not solve the double coincidence of wants problem and therefore would not be adopted as a medium of exchange. As a store of value, money is not unique; many other stores of value exist, such as land, works of art, and even baseball cards and stamps. Money may not even be the best store of value because it depreciates with inflation. However, still, money is used as a store of value as:
(i) It is more liquid than most other stores of value.
(ii) It is readily accepted everywhere.
(iii) It is easy and economical to store as its storage does not require much space.
(iv) Unit of account: Money also functions as a unit of account, providing a common measure of the value of goods and services being exchanged. Knowing the value of the price of a good, in terms of money, enables both the supplier and the purchaser of the good to make decisions about how much of the good to supply and how much of the good to purchase.
In the absence of the common measure, the seller has to express the value of his good in all other goods. For example, if you want to sell your horse you have to express its value.
1. Horse = 2 cows
1. Horse = 5 bags of wheat
1. Horse = 20 kg of iron

(c) (i) Merchandise: It refers to all such items of exports and imports which are visible and therefore also called Visible Trade’ relating to exports and imports. Current account showing export and import of visible is often referred to balance of trade account.

(ii) Invisible: It refers to all such items which are rendered to rest of the world or received from the rest of the world in the form of services they include the following principal services like travel, transport alias insurance asking and services rendered to the rest of the world are treated like exports and the services received from the rest of the world are treated as imports.

(iii) Transfers: It refers to unilateral transfers i.e., gifts or donations.
These are broadly divided as
(a) official transfers
(b) private transfers.

(iv) Investment income: It refers to income by way of rent, interest and profit. Income earned by our country is shown as receipts, while income earned from the rest of the world from our country is shown as payments.

(v) Compensation of employees: It refers to the income earned or paid to the rest of the world by way of wages and salaries.

Question 8.
(a) Highlight two differences between sales tax and income tax. [3]
(b) What is meant by [3]
(i) Union budget
(ii) State budget
(c) Explain four ways of Redemption of Public Debt. [6]
Answer:
(a)

Sales Tax	Income Tax
(i) It is an indirect tax.	(i) It is a direct tax.
(ii) It is levied on goods and services.	(ii) It is levied on income.

(b) (i) It is the budget prepared by the central government for the country as a whole. This budget is presented in two parts.
(1) Railway budget and
(2) Main Budget.
(ii) It is repaired by state govt, such as one budget of Punjab govt. UP govt. etc.

(c) (1) Sinking Fund Approach: The Government at the regular interval saves a certain amount of money from its budget to meet up the debt obligations and uses this fund for the same when they have accumulated enough money.

(2) Conversion Approach: Conversion of loans is another method of redemption of public debt. It means that an old loan is converted into a new loan. Under this system, 111 high-interest public debt is converted into a low-interest public debt. Prof. Dalton felt that debt conversion actually relaxes the debt burden.

(3) Utilization of Budgetary Surplus: When the Government earns surplus in the budget, it must be utilised for paying the debt. A surplus occurs when public revenue exceeds public expenditure. However, this method is rarely found.

(4) Terminal Annuity: In this method. The government pays off the public debt on a fee basis of a terminal annuity into equal annual instalments including interest along with the principal amount. This is the easiest way of paying off the public debt.

Question 9.
(a) With the help of a diagram, show the circular flow of income in a two-sector model with Savings and Investment. [3]
(A) The growth of the Gross Domestic Product is not a real indicator of economic welfare. Discuss two reasons to justify the given statement. [3]
(c) From the follow ing data, calculate GNP_MP and NNP_FC by Expenditure Method. [6]

(i) Mixed-income of self-employed	450 crores
(ii) Compensation of employees	550 crores
(iii) Private final consumption expenditure	1000 crores
(iv) Net factor income from abroad	-20 crores
(v) Net indirect taxes	150 crores
(vi) Consumption of fixed capital	170 crores
(vii) Net domestic capital formation	380 crores
(viii) Net exports	-30 crores
(ix) Profits	400 crores
(x) Rent	150 crores
(xi) Interest	200 crores
(xii) Government final consumption expenditure	550 crores

Answer:
(a)

In Fig. we have considered a two-sector economy with households and firms. It is assumed that the household sector and producing sector do not save at all. However, in reality, households save a portion of their income for many reasons like precautionary purposes, transaction purposes, etc. On the other way, they borrow money from the capital market for both development and as well as for even non-development purposes. Similarly, firms also save some part of their receipts for the expansion of business, transactional purposes, speculative purposes, precautionary purposes etc. Moreover, the firms also collect loans (i.e., investment) from both banks and non-banking financial institutions to generate capital in their business. Thus, the inclusion of saving (Leakage, i.e., an outflow of income) and investment (Injection, i.e., the inflow of income) in the two-sector economy with the capital market as shown in the Fig. have made the two-sector economy more meaningful and modernised.

(b) 1. Externalities:
(i) Externalities refer to benefits or harms of an activity caused by a firm or an individual, for which they are not paid or penalised.

(ii) Externalities occur outside the market i.e., they affect people not directly involved in the production and or consumption of a good or service. They are also known as spill-over effects. Externalities are not taken into account while calculating GDP.

(iii) For example, setting up industry in a town may add up to domestic product but while calculating GDP the pollution and its impact on society are not taken into account.

(iv) The industrial growth and urbanisation create problems like industrial slumps diseases and crime. The damage to the physical and social environment has an adverse impact on the health and welfare of the community. Thus despite an increase in GDP, no accurate conclusion can be drawn on the welfare level.

2. Distribution of GDP:
(i) GDP shows the total value of goods and services produced in a country. However, it does not exhibit how the GDP is distributed among the people. For example GDP of India is rising but it is not being equally distributed among all the strata of people. The major share of the increase is being grabbed only by the upper-middle class and share of poor of the poorer is decreasing.

(ii) If with every increase in the level of GDP, Distribution is getting more unequal, the welfare level of society may not rise.

(iii) No doubt, GDP is a useful measure of the overall productive activities in a country, yet no welfare implication, on it basis, can be derived, unless it is fairly divided among all the section of the society.

(c) Expenditure Method
GDP_MP = Private final consumption expenditure + Government final consumption expenditure + Gross Domestic Capital Formation + Net Export
1000 + 550 + (380 + 170) + (-30) = ₹ 2,070
NNP_FC = GDP_MP – Depreciation + NFIA – NIT
2070 – 170 + (-20) – 150 = ₹ 1730

ISC Class 12 Economics Previous Year Question Papers

ISC Biotechnology Previous Year Question Paper 2011 Solved for Class 12

Maximum Marks: 80
Time allowed: Three hours

• Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.
• Answer Question 1 (Compulsory) from Part I and five questions from Part II, choosing two questions from Section A, two questions from Section B and one question from either Section A or Section B.
• The intended marks for questions or parts of questions are given in brackets [ ].
• Transactions should be recorded in the answer book.
• All calculations should be shown clearly.
• All working, including rough work, should be done on the same page as, and adjacent to the rest of the answer.

Part-I
(Compulsory)

Question 1
(a) Mention any one significant difference between each of the following : [5]
(i) Nucleotide and nucleoside.
(ii) Plasmids and phages.
(iii) Fluorescence spectrometry and mass spectrometry.
(iv) Primer and primases.
(v) Introns and exons.

(b) Answer the following questions : [5]
(i) What are oil eating bacteria?
(ii) What are single cell proteins?
(iii) Who proposed the Operon concept of gene regulation?
(iv) Explain the term transposons.
(v) Name one polysaccharide found in the cell wall of bacteria and one polysaccharide found in fungi.

(c) Write the full form of the following: [5]
(i) NCBI
(ii) EST
(iii) ROM
(iv) PACE
(v) FMN

(d) Explain briefly: [5]
(i) RNA dependent DNA polymerase
(ii) EMBL
(iii) Isoelectric focussing
(iv) Cosmids
(v) Vitamins
Answer:
(a) (i) Nucleotide: The nucleotide is a condensation product of heterocyclic nitrogen base, a pentose sugar like ribose or deoxyribose, and a phosphate or polyphosphate group.

Nucleoside: Nucleoside consisting of a pentose sugar, usually ribose or deoxyribose, and a nitrogen base-purine or pyrimidine.

(ii) Plasmids: Plasmids are the extra-chromosomal, independent, self replicating, circular, double stranded DNA molecules naturally found in all bacteria and some fungi.

Phages: Bacteriophage (i.e., viruses that infect bacteria) are obligate intracellular parasites that multiply inside bacteria by making use of some or all of the host biosynthetic machinery.

(iii) Fluorescence Spectrometry: Fluorescence spectrophotometry is a technique that assay the state of a biological system by studying the absorption spectra of different radiations and its interactions with fluorescent probe molecules.

Mass Spectrometry: Mass spectrometry (MS) is an analytical technique that involves separation of ions of different mass and energy when fixed magnetic and electric fields are employed.

(iv) Primer: A primer is a short strand of RNA that serves as a starting point for DNA synthesis They are required for DNA replication because DNA polymerases can only add new nucleotides to an existing strand of DNA.

Primases: Primase is an enzyme that catalyzes the synthesis of a short RNA segment (called a primer) complementary to a ssDNA template.

(v) Introns: Introns are non-coding DNA base sequences , which are found between exons, but are not transcribed part of mature mRNA.

Exons: Exons are coding DNA base sequences that are transcribed into wzRNA and finally code for amino acids in the proteins.

(b) (i) Bacteria that decompose and use oil as their source of energy are called as oil eating bacteria e.g., Pseudomonasputida, P.capacia etc. are efficient degraders created through genetic engineering. They need to be established in environment. They (“the oil-eating bacteria”) could digest about two- thirds of the hydrocarbons that would be found in a typical oil spill.

(ii) The term single cell protein (SCP) refers to biomass of dry cells of microorganisms such as yeast, bacteria, mushroom (fungi) and algae which grow on different carbon sources. The name “single cell protein” was used for the first time by the M.I.T. Professor Carol Wilson.

(iii) Jacob and Monod proposed the Operon concept of Gene Expression in bacteria.

(iv) Transposons : Transposons are sequences of DNA that can move around to different positions within the genome of a single cell. Transposons are called “Jumping genes”, and are examples of mobile genetic elements.
Two types of Transposons are possible :

• Retrotrahsposons and
• DNA transposons.

In this process, they may :

• cause mutations
• increase (or decrease) the amount of DNA in the genome.

(v) Peptidoglycans (mucopeptides, glycopeptides, mureins) are the structural elements of bacterial cell walls. Fungi possess cell walls made-up of chitin, polymer of glucosamine,

(c) (i) NCBI: National Centre for Biotechnology Information

(ii) EST : Expressed sequenced tag

(iii) ROM : Read only memory

(iv) PAGE : Polyacrylamide gel electrophoresis

(v) FMN : Flavin mononucleotides.

(d) (i) RNA-dependent DNA polymerase enzyme is used for the synthesis of a complementary DNA strand from RNA molecule as a template. It is produced by HTV and other retroviruses which help them to synthesize DNA from their viral RNA.

(iii) Isoelectric focusing (IEF), also known as electrofocusing, is a technique for separating different molecules by their electric charge differences. It is a type of zone electrophoresis, usually performed on proteins in a gel.

(iv) Cosmids are constructed by uniting some part of the bacterial plasmids with a ‘ori’ gene, an antibiotic selection marker and a cloning site with one or more recently two ‘cos’ sites derived from bacteriophage lambda.

(v) A vitamin is an organic compound required as a nutrient in very small amounts by an organism. They act as co-factor of an enzyme. It cannot be synthesized in sufficient quantities by an organism, and must be obtained from the diet.

Part-II
(Answer any five questions)

Question 2.
(a) Mention two important chemical properties of each of the following: [4]
(i) Monosaccharides
(ii) Proteins .
(b) What are restriction enzymes ? How do thev act ? Give any two examples of restriction enzymes. [4]
(c) What are derived lipids ? Give an example. [2]
Answer:
(a) (i) Chemical properties of monosaccharides :
When a & p isomeric forms of D-glucose are dissolved in water. Their optical rotation changes with time and approaches the final equilibrium value. +53°. This change is known as mutarotation. Mutarotation occurs because of slow conversion of α -D glucose and β-D glucose via open chain form until equilibrium is established giving a constant rotation +53°.

Acetals traditionally derive from the product of the reaction of an aldehyde with excess of alcohol, whereas the name ketal derives from the product of the reaction of a ketone with excess alcohol.

(ii) Chemical properties of proteins :

• Proteins when hydrolyzed by acidic agents, like cone. HC1 yield amino acids in the form of their hydrochlorides.
• Sanger’s reaction : Proteins react with FDNB reagent to produce yellow coloured derivative, DNB amino acid.
• Xanthoproteic test: On boiling proteins with cone. HN03, yellow colour develops due to presence of benzene ring.
• Folin’s test: This is a specific test for tyrosine amino acid, where blue colour develops with phosphomolybdotungstic acid in alkaline solution due to presence of phenol group.

(b) The restriction enzymes are called ‘molecular scissors’. Restriction enzymes are DNA-cutting enzymes present in bacteria. They are obtained from them for use in genetic enginccring-rDNA technology. As sequence are cut within the DNA molecule, they are often called restriction endonucleases.

A restriction enzyme recognizes and cuts DNA only at a particular sequence of nucleotides. For example, the bacterium Hemophilus aegypticus produces an enzyme named Hae III that cuts DNA whenever it identifies the recognition sequence.

A cut is made between the adjacent G and C. HaeIII cleaves both the strands of DNA at the same base pairs producing “blunt” ends.
Examples : EcoRI, HindIII and Bam HI.

(c) Derived lipids are either lipid-like chemicals (e.g., sterols) or derivatives of lipids e.g., terpenes. Prostaglandis and Choline. They are derived from simple and compound lipids by hydrolysis.

Question 3.
(a) Give an account of various enzymes that play a role in the process of DNA replication. [4]
(b) Write short notes on : [4]
(i) Secondary structure of proteins
(ii) Designer oils.
(c) What is meant by synchronization of cell culture. [2]
Answer:
(a) Topoisomerase : Cause single strand breaks and religation.
Helicase: Unwounds a portion of the DNA Double Helix.
RNA Primase : Attaches RNA primers to the replicating strands.
DNA Polymerase delta (δ) : Binds to the 5′ – 3′ strand in order to bring nucleotides and synthesise the daughter leading strand.
DNA Polymerase epsilon (a): Binds to the 3′ – 5′ strand in order to synthesise discontinuous segments starting from different RNA primers.
Exonuclease (DNA Polymerase I): Identifies and removes the RNA Primers.
DNA Ligase : Adds phosphate in the remaining gaps of the phosphate – sugar backbone.
Nucleases : Remove wrong nucleotides from the daughter strand.

(b) (i) Secondary Structure (2° structure): It is development of new stearic relationships amongst the amino acids for protecting their peptide bonds through formation of intrapolypeptide and interpolypeptide hydrogen bonds. Secondary structure is of three types — α-helix, β-pleated and collagen helix. The prefixes α and β signify the first and second types of secondary structure discovered by Pauling and Corey (1951).

α-Helix : The polypeptide chain is spirally coiled, generally in a clockwise or right¬handed fashion (Fig ). There are 3.6 amino acid residues per turn of the spiral. The . spiral is stabilized by straight hydrogen bonds between imide group (-NH-) of one amino acid and carbonyl group (-CO-) of fourth amino acid residue. In this way all the imide and carbonyl groups become hydrogen bonded. R-groups occur towards the outer side of a-helix. a-helix is the final structure in certain fibrous proteins, e.g., keratin (hair, nail, horn), epidermin (skin).

β-Pleated Sheets : Two or more polypeptide chains come together and form a sheet. Condensation is little. However, twisting does occur. The same polypeptide may fold over itself to form two strands for β-pleating. Adjacent polypeptide chains may occur in parallel (e.g., p-keratin) or antiparallel (e.g., silk fibroin). Straight hydrogen bonds occur between imide (-NH-) group of one polypeptide and carbonyl (-CO-) group of adjacent polypeptide. Cross-linkages help in stabilisation of p-pleated sheets.

Collagen Helix : Collagen has a large amount of glycine (25%) and proline (and hydroxyprohne, 25%). It cannot form a-helix due to them. Three of its polypeptide each having about 1000 amino acid residues, come together with each forming an extended left-handed helix. They run parallel, form a right-handed super-helix that is stabilised by hydrogen bonds amongst the three. The triple helix of collagen is often called tropo-collagen. Its one end is stabilised by -S-S- linkages amongst the three chains. Collagen occurs in those tissues where extensibility is limited, e.g., connective tissue, tendons, bones.

(ii) Designer Oil: “Designer oif’ that reduces LDL (“bad”) blood cholesterol levels in humans and increases energy expenditure, which may prevent people from gaining weight. The oil incorporates a phytosterol-based functional food ingredient Phytrol (TM) from Forbes into oil using proprietary technology.

(c) Cell culture synchronisation : Cells in suspension cultures vary greatly in size, shape, DNA, and nuclear content. Moreover, the cell cycle time varies considerably within individual cells. Therefore, cell cultures are mostly asynchronous. It is essential to manipulate the growth conditions of an asynchronous culture in order to achieve a higher degree of synchronisation. A synchronous culture is one in which the majority of cells proceed through each cell cycle phase (Gl, S, G2 and M) simultaneously. Synchronisation can be achieved by following methods.

• Physical methods include selection by volume (size of cell aggregate.)
• Chemical methods include starvation (depriving suspension cultures of an essential growth compound and culture supplying).
• Chemical methods include inhibition (temporarily blocking the progression of events in the cell cycle using a biochemical inhibitor and then releasing the block).

Question 4.
(a) Write the steps in the process of activation of amino acids during the process of translation. [4]
(b) What is the significance of genetic code in protein synthesis ? Mention four important characteristics of the genetic code. [4]
(c) Name the important steps in a single cycle of polymerase chain reaction. [2]
Answer:
(a) The adapter function of the tRNA molecules require the charging of each specific tRNA with its specific amino acid. Since there is no affinity of nucleic acids for specific functional groups of amino acids, this recognition must be carried out by a protein molecule capable of recognizing both a specific tRNA molecule and a specific amino acid.

Activating enzymes called aminoacyl-tRNA synthetases couple each amino acid to its appropriate set of tRNA molecules. There are 20 synthetases for each of the 20 natural amino acids. The reaction involves two steps :

The amino acid-tRNA bond is a high-energy linkage that easily reacts with the amino group of the next amino acid in a protein sequence to form a peptide bond during protein synthesis.

(b) Messenger RNA contains a triplet sequence of nitrogen bases which code for the specific amino acids, used to make polypeptide chains. Each of the sets of three bases is known as a codon or genetic code.

Translation starts with a chain initiation codon (start codon). The most common start codon is AUG which is read as methionine or, in bacteria, as formyl methionine. The three stop codons have been given names : UAG is amber, UGA is opal, sometimes also called umber and UAA is ochre. Stop codons are also called “termination” or “non-sense” codons.

Characteristics of Genetic Code

• Triplet code : Three adjacent nitrogen bases constitute a codon which specifies the placement of one amino acid in a polypeptide.
• Start signal: Polypeptide synthesis is signaled by AUG or methionine codon and GUG — Valine codon. They have dual function.
• Stop signal: Polypeptide chain termination is signaled by three termination codons — UAA, UAG and UGA. They do not specify any amino acid and are hence also called non-sense codon.
• Universal code : The genetic code is applicable universally i.e., the codon specifies the same amino acid from a virus to a tree or human being.
• Non-ambiguous codon : One codon specifies only one amino acid and not any other.

(c) PCR Stages:

• Denaturation step
• Primer Annealing step
• Extension/elongation step (Polymerization)

Question 5.
(a) Write notes on : [4]
(i) Proteomics.
(ii) De-differentiation and Re-differentiation.
(b) Mention the objectives of germplasm conservation using the cell culture technique. What are the limitations of conservation of germplasm using conventional methods? [4]
(c) The base sequence of one strand of DNA is 3′ CATGAC 5′. What will be the base sequence of its : [2]
(i) Complementary DNA strand.
(ii) Complementary RNA strand.
Answer:
(a) (i) Proteomics is the identification, analysis and large scale characterisation of proteome (i.e., the total protein components) expressed by any given cells, tissues and organs under the defined conditions. The major objectives of proteomics are

• to characterise post- transcriptional modifications in protein, and
• to prepare 3D map of cell indicating the exact location of protein.

Proteomics is the direct outcome of advancement made for nucleotide sequencing of different genomes in large scale. This helps to identify various proteins. Generation of information about protein is necessary. Because, protein governs the phenotypic characters of the cells. Merely genome study cannot provide the understanding of mechanism of disease development and various developmental changes occurring in organisms including humans. Moreover, target drugs for many kinds of diseases can be prepared only after understanding the protein modification and protein functions. There are many areas of modem proteomics such as protein expression, protein structure, protein localisation, protein-protein interaction, etc.

(ii) Dedifferentiation: It is a cellular process which occurs in lower life forms such as worms and amphibians in which a partially or terminally differentiated cell reverts to an earlier developmental stage.

Redifferentiation : It is a process by which a group of once differentiated cells return to their original specialised form.

(b) Germplasm Conservation:
The sum total of all the genes present in a crop and its related species constitutes its germplasm; it is ordinarily represented by a collection of various strains and species. Germplasm provides the raw materials (= genes), which the breeder uses to develop commercial crop varieties. Therefore, germplasm is the basic indispensable ingredient of all breeding programmes, and a great emphasis is placed on collection, evaluation and conservation of germplasm.

Limitations of conventional methods:

• Conventionally, germplasm is conserved as seeds stored at ambient temperature, low temperature or ultralow temperature. But many crops produce recalcitrant or short-lived seeds, and in case of clonal crops seeds are not the best material to conserve in view of their genetic heterogeneity and unknown worth.
• Roots and tubers loose viability rapidly and their storage requires large space, low temperature and is expensiv e.
• In addition, materials modified by genetic engineering may sometimes be unstable, and hence may need to be conserved intact for future use.

In such cases, the following approaches of germplasm conservation may be applied:

• freeze preservation,
• slow-growth cultures.
• DNA-clones and
• desiccated somatic embryos/artificial seeds.

(c) DNA -3′ CATGAC 5′

• COMPLIMENTARY DNA: 5′ GTACTG 3′
• COMPLIMENTARY RNA: 5′ GUACUG 3′

Question 6.
(a) Why are enzymes called biocatalysts ? Briefly discuss the mode of enzyme action. [4]
(b) Explain the principle and applications of each of the following biochemical techniques: [4]
(i) Gel permeation.
(ii) Colorimetry.
(c) How are the diseases cystic fibrosis and albinism caused? [2]
Answer:
(a) Catalysts are chemical substances used to speed up chemical reaction without any change of its own. In biological systems, certain biomolecules known as enzymes act in the same manner i.e., they mediate different biochemical reactions but themselves remain unaltered so they are termed as biocatalysts.
Mode of enzyme action : It could be explained by following models :

(b) (ii) Colorimetry
Principle : Colorimetry is the technique of determining the concentration of a chemical in a solution, if it has a colour, is to measure the intensity of the colour and relate the intensity of the colour to the concentration of the solution. In colorimetry, two fundamental laws are applied :

• firstly , a Lambert’s law. relates the amount of light absorbed and the distance it travels through an absorbing medium; and
• Secondly, Beer’s law relates light absorption and the concentration of the absorbing substance.

Application : It is used extensively for identification and determination of concentrations of substances that absorb light (any colour solution). A simple application lies in comparing intensities of radiation transmitted through layers of different thicknesses of two solutions of the same absorbing substance, one with a known concentration, the other unknown.

(c) Cystic fibrosis: A disease that results from a decrease in fluid and salt secretion by a transport protein known as Cystic Fibrosis Transmembrane Regulator (CFTR). As a result of this defect secretion from the pancreas is blocked and heavy dehydrated mucus accumulates in the lungs leading to chronic lung infection.

Albinism: It is a recessive disorder due to non-conversion of typosine into melanin because of gene mutation.

Albinism are of two types :

• Type 1 albinism is caused by defects that affect production of the pigment, melanin.
• Type 2 albinism is due to a defect in the ‘P” gene. People with this type have slight coloring at birth.

Question 7.
(a) Explain how ‘Dolly’ was created. What is its importance in the field of biotechnology? [4]
(b) Differentiate between: [4]
(i) PCR and Gene Cloning.
(ii) Batch Culture and Continuous Culture.
(c) What are polysaccharides ? How are they formed? [2]
Answer:
(a) Dolly, was the first mammal to be cloned by Wilmut et. al (1977). Dolly was the end result of a long research program founded by the British government at the Roslin Institute in Scotland. They used the technique of somatic cell nuclear transfer, where the cell nucleus from an adult udder cell is transferred into an unfertilized oocyte that has had its nucleus removed. The hybrid cell is then stimulated to divide by an electric shock, and the blastocyst that is eventually produced is implanted in a surrogate mother. Dolly was the first clone produced from a cell taken from an adult animal.

The following steps are involved in creation of Dolly :
Isolate donor nucleus : Isolate the nucleus from a somatic (non-reproductive) cell of a udder of adult donor sheep. The nucleus contains all the complete genetic material of the organism. A very small needle and syringe (suction device) is used to penetrate through the cell membrane to hold the nucleus and remove it from the cell.

Get unfertilized eggs: Retrieve some unfertilized egg cells (reproductive) from a female sheep. Many eggs are needed since not all of them will survive the various steps of cloning.

Remove the egg’s nucleus: Remove the egg cell’s nucleus, which contains only one-half of the sheep’s genetic material.

Insert donor nucleus : Insert the nucleus, with all its complete genetic material, isolated from the donor sheep mammal in Step 1 into the cytoplasm of egg cell that has no nuclear material. The egg’s genetic material now contains all traits from the donor adult. This egg is genetically identical to the donor adult cells.

Transfer the egg into womb : Transfer the egg into a receptive female sheep’s womb. Those eggs that survive and implant will continue to develop into embryos. When the offspring is born, it is a clone (genetically identical) of the donor sheep. After cloning was successfully demonstrated through the production of Dolly, many other large mammals have been cloned, including horses and bulls.

(b) (i) PCR:

• Amplification of desired segment of DNA within hours.
• No host cell required.

Gene cloning:

• Amplification of a desirable gene takes more time under lab condition.
• Host cell is required.

(ii) Batch culture:

• In this culture, the same medium and all the cells produced are retained in the culture vessel. Fresh medium is not added.
• The cell number of biomass exhibits a typical sigmoid curve.

Continuous culture:

• In this culture, both cells and used medium are taken out from the continuous culture and replaced by equal volume of fresh medium.
• Here, the cell population is a maintained in a steady state by regularly replacing a portion of used or spent medium.

(c) Polysaccharides are complex polymeric carbohydrate structures, formed by polymerization of monomeric repeating units (either mono-or di-saccharides) joined together by glycosidic bonds.
Example: starch, glycogen, etc. Polysaccharides are formed by enzyme catalyzed condensation reaction between any number of saccharide units.

Starch : It is the nutritional reservoir of plants. They are formed by the polymerisation/condensation of glucose molecules. Two types of starch are possible unbranched starch known as Amylase and branched starch known as Amylopectin. They actually differ in glycosidic linkage.

Glycogen : It is the storage form of glucose in animals. Two chains of glucose molecules joined by an alpha-1, 4-glycosidic bonds are linked by an alpha-1. 6-glycosidic bond to create a branch point.

Question 8.
(a) Briefly explain the concept and application of: [4]
(i) In vitro pollination.
(ii) Protoplast fusion.
(b) Mention the significance of plants obtained by each of the follow ing types of plant tissue culture techniques : [4]
(i) Endosperm culture.
(ii) Anther culture.
(c) State the main achievements of the Human Genome Project. [2]
Answer:
(a) (i) Pollination and fertilization under in vitro conditions offer an opportunity for producing hybrid embry os among plants that cannot be crossed by conventional methods of plant breeding. In nature, Intergeneric or Interspecific hybridization occurs very rarely. This is due-to barriers hindering the growth of the pollen tube on the stigma or style. In such cases, the style or part of it can be excised and pollen grains either placed on the cut surface of ovary or transferred through a hole in the wall of ovary. This technique, called intraovarian pollination, has been successfully applied in such species as Papaver somniferum, Eschscholtiza California, Argemone mexicana.

Ovular Pollination : In this in vitro pollination method, pollen tubes directly penetrate isolated ovules. Interspecific crossing barriers do not occur during penetration of the pollen tubes into the micropyle. Completion of pollen tube penetration is much earlier than the division of the generative cell in sperm cells which may be the primary cause for the failure in fertilization.

Placental Pollination : In this in vitro pollination method, the ovaries are cut into sectors on the day or after the day of stigma receptivity. Each sector contains a placenta with a row of ovules without or with ovary wall. Pollens are abundantly applied on the placenta. In this case also the rate of fertilisation is very slow.

Applications of in vitro Pollination : In plant breeding, the technique of in vitro pollination has lot of potential in at least three different areas,

• overcoming self-incompatibility,
• overcoming cross-incompatibility,
• haploid production through parthenogenesis.

(ii) The technique for protoplast fusion are pritty will defined and highly effective for almost all the systems. Protoplast of desired species/strains are mixed in almost equal proportion; generally they are mixed while still suspended in the enzyme mixture. The protoplast mixture is then subjected to high pH (10.5) and high Ca²⁺ concentration 50 m molL^-1 at 37°C for about 30 min. (high pH Ca²⁺ treatment). This technique is quite suitable for some species while for some other it may be toxic.

Two types of protoplast fusions are :
Spontaneous protoplast fusion: During isolation of protoplasts for culture, when enzymatic degradation of cell walls is affected, some of the protoplasts, lying in close proximity, may undergo fusion to produce homokarvons or homokaryocytes. each with 2-40 nuclei. The occurrence of multinucleate fusion bodies is more frequent, when protoplasts are prepared from actively dividing cells.

This spontaneous fusion, however, is strictly intraspecific. How ever, spontaneous fusion of protoplasts can also be induced by bringing protoplasts into intimate contact through micromanipulators or micropipettes.

There seems to be a correlation between the size of the leaf and the percentage of protoplasts undergoing spontaneous fusion; protoplasts from young leaves are more likely to undergo this fusion.

Induced protoplast fusion: Somatic hybridization is generally used for fusion of protoplasts either from two different species (inter specific fusion) or from two diverse sources belonging to the same species. To achieve this objective, spontaneous fusion may be of no value, and induced fusion requiring a suitable agent (fusogen) is necessary. In animals, inactivated Sendai virus is needed to induce fusion.

In plants, however, the inducing agent first brings the protoplasts together and then causes them to adhere to one another for bringing about fusion. During the last two decades, a variety of treatments have been successfully utilized for fusion of plant protoplasts. These treatments particularly include the following : NaN0₂, high pH with high Ca⁺⁺ ion concentration and polyethylene glycol (PEG).

Applications: Auxin independent growth of hybrids of Nicotiana glauca and A. langsdorffii. Two parental lines cannot produce auxin and thus do not grow on auxin free medium, the hybrid cells produce auxin and are they able to grow and form callus.

(b) (i) Endosperm culture: The endosperm nurses the developing zygotic embryo. For the first time Lamp and Mills (1933) grew7 maize endosperm on nutrient medium 10-20 days after pollination. The endosperm proliferated slightly. In 1949, LaRue succeeded to produce callus from immature endosperm. From India, B.M. Johri and S.S. Bhojwani (1965) at the University of Delhi reported and endosperm culture. Some examples of triploid plants raised from endosperm cultures are : Asparagus officinalis, barley (Hordeum vulgare), rice (Oryza sativa), maize (Zea mays), Prunus persica, Pyrus malus, Citrus gradis, sandle plant (Santalum album).

The triploid plants are self-sterile and usually seedles. This characteristic increases edibility of fruits and desirable in plants such as apple, banana, grape, mulberry, mango, watermelon, etc. These are commercially important edible fruits. The triploids of poplar {Populus tremuloides) have better quality pulpwood. Therefore, it is important to the forest industry,

(ii) Anther culture: When anthers of some plants are cultured on suitable medium to produce haploid plants, it is called anther culture. For the first time S. Guha and P. Maheshwari (1964) produced haploid embryos in vitro from isolated anthers of Datura innoxia. Haploid production has immense use in plant breeding and improvement of crop plants. Haploids provide an easier system for induction of mutation. They can be employed for rapid selection of mutants having traits for disease resistance. The Institute of Crop Breeding and Cultivation (China) has developed the high yielding and blast resistant varieties of rice zhonghua No. 8 and zhonghua No.9 through transfer of desired alien gene.

It is highly useful for the improvement of many crop plants. It is also useful for immediate expression of mutations and quick formation of purelines. This technique was first used in India to produce haploids of Datura. In many plants haploids are also produced by culturing unfertilized ovaries/ovules. In pollen grain culture pollen grains are eseptically removed from the anther and cultured on liquid medium.

(c) The Human Genome Project was an international scientific research project that determined the DNA sequence of the approximately 30,000 — 35,000 genes that make-up the human genome. The greatest achievement of this project is that scientists can find out the structure of human genes and successfully trace the structure of genes. It has been possible to develop the technique regarding the treatment of defective genes only because of the human genome project.

The treatment of the hereditary diseases like cancer has been easier because of this project. DNA interference is a recently developed technique regarding treatment by which it would be possible to treat many incurable diseases. Along with this, the development of human, physical and mental structure etc., are the main achievements of human genome project.

Question 9.
(a) What type of information is obtained from each of the following database. [4]
(i) Taxonomy Browser
(ii) PDB.
(iii) GENSCAN
(iv) PIR.
(b) Explain the method of DNA sequencing using automated DNA sequencing technique. [4]
(c) Write any two examples of each of the following plant hormones commonly used in media preparation: [2]
(i) Auxins.
(ii) Cytokinins.
Answer:
(a) (i) Taxonomy browser: This search tool provides taxonomic information on various species. The Taxonomy database of NCBI has information (including scientific and common names) about all organisms for which some sequence information is available (over 79,000 species).

The server provides genetic information and the taxonomic relationship of the species in question. Taxonomy has links with other servers of NCBI e.g., structure and PubMed.

(ii) PDB (Protein Data Bank) : This database has sequence of those protein whose 3-D structures are known.
Source: NCBI-U.S.A.; EBI, UK.

(iii) Genscan is a notable example of Eukaryotic ab initio gene finders. Genscan is one of the best gene finding algorithms for sequence alignment and gene prediction.

(iv) PIR is a non-redundant annotated protein sequence database, and analytical tools.

(b) Automatic DNA Sequencers: Automatic sequencing machines were developed during 1990’s. It is an improvement of Sanger’s method. In this new method, a different fluorescent dye is tagged to the ddNTP’s. Using this technique, a DNA sequence containing thousands of nucleotides can be determined in a few hours. Each dideoxynucleotide is linked with a fluorescent dye that imparts different colors to all the fragments terminating in that nucleotide. All four labelled ddNTP’s are added to a single capillary tube. It is a refinement of get electrophoresis which separates fastly. DNA fragments of different colors are separated by their size in a single electrophoretic gel.

A current is applied to the gel. The negatively charged DNA strands migrate through the pores of gel towards the positive end. The small sized DNA fragments migrate faster and vice-versa. All fragments of a given length migrate in a single peak. The DNA fragments are illuminated with a laser beam. Then the fluorescent dyes are excited and emit light of specific wavelengths which is recorded by a special ‘recorder’. The DNA sequences are read by determining the sequence of the colours emitted from specific peaks as they pass the detector. This information is fed directly to a computer which determines the sequence. A tracing electrogram of emitted light of the four dyes is generated by the computer. Colour of each dye represents the different nucleotides. Computer converts the data of emitted light in the nucleotide sequences.

(c) (i) IAA (Indole 3-Acetic Acid) and 2, 4-Dichlorophenoxyacetic acid (2, 4-D)
(ii) Kinetin and BAP (Benzyl Amino purine)

ISC Class 12 Biotechnology Previous Year Question Papers

ISC Accounts Previous Year Question Paper 2016 Solved for Class 12

Maximum Marks: 80
Time allowed: Three hours

• Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.
• Answer Question 1 (Compulsory) from Part I and five questions from Part II, choosing two questions from Section A, two questions from Section B and one question from either Section A or Section B.
• The intended marks for questions or parts of questions are given in brackets [ ].
• Transactions should be recorded in the answer book.
• All calculations should be shown clearly.
• All working, including rough work, should be done on the same page as, and adjacent to the rest of the answer.

Section – A
Part – I (12 Marks)
Answer all questions.

Question 1.
Answer briefly each of the following questions : [6 x 2]
(i) What is a contingent liability ? How are contingent liabilities shown in the Balance Sheet of a company prepared as per Schedule III of the Companies Act 2013 ?
(ii) Why is abnormal loss not recorded in the books of a Joint Venture ?
(iii) Give the formula for calculating the outgoing partner’s share in the interim profits of the firm, on the basis of sales made by the firm.
(iv) How is Workmen Compensation Fund, shown in the Balance Sheet of a partnership firm, treated at the time of its dissolution ?
(v) Give any two differences between Reserve Capital and Capital Reserve.
(vi) Give the adjusting entry and closing entry for interest on debentures due to the debenture holders of a company.
Answer:
(i) A liability that might arise on happening of an event in future is known as contingent liability. They are separate shown in the Notes to Accounts’ not in the Balance Sheet.

(ii) Abnormal loss is not recorded in the Joint Venture because the loss automatically gets adjusted as the difference betw een sales and cost of the goods and if insurance claim is received then the loss is minimized.

(iii) \(\frac{\text { Sales for the interim period }}{\text { Total Sales for the last year }} \times \text { profit of the firm } \times \text { share of partner }\)

(iv) It is distributed among the partners in their profit sharing ratio.

(v)

(vi) Interest on debentures a/c Dr.
To Debenture holders a/c
(Being Interest on debentures due)
Statement of Profit and Loss a/c Dr.
To Interest on debentures a/c
(Being interest on debentures transferred to statement of P/L a/c)

Part – II (48 Marks)
Answer any four questions.

Question 2. [12]
Anil and Saji entered into a joint venture to buy and sell old machines. They decided to share profit and losses in the ratio of 3:2.
Anil purchased 200 machines at ₹ 3,000 each and sent them to Saji for sale.
Anil incurred ₹ 26,000 on freight and transit insurance.
Saji took delivery of the machines and incurred ₹ 24,000 as clearing charges and ₹ 6,000 as selling expenses.
Anil drew a bill on Saji for ₹ 4,00,000 which was accepted by Saji. The bill was discounted by Anil for ₹ 3,90,000 with the bank.
Saji was able to sell 190 machines at ₹ 3,850 per machine.
The unsold machines were taken by Anil for his next venture at the original cost plus proportionate non-recurring expenses less 10%.
Saji was entitled to a commission of 2% on the sales made by him.
The co-ventures settled their accounts by means of a bank draft.
It was decided that Anil would maintain a record of all the transactions.
You are required to pass journal entries in the books of Anil.
Answer:

Question 3. [12]
During the year 2014-15, A.B.C. Ltd. issued 10,000 Equity shares of ₹ 50 each at ₹ 55 per share, payable as follows:
On Application ₹ 15
On Allotment ₹ 20 (including premium ₹ 5)
On 1st and Final Call ₹ 20
All the issued shares were subscribed for by the public.
One shareholder holding 500 shares did not pay the amount due on allotment and his shares were immediately forfeited.
Another shareholder holding 100 shares paid the amount of the 1st and Final Call with allotment. After the company had made the 1st and Final Call, 200 of the forfeited shares were reissued as fully called up at ₹ 45 per share.
The share issue expenses were ₹ 7,000 which were written off at the end of the year.
You are required to pass journal entries in the books of the company for the year ending 31st March, 2015.
Answer:

Question 4. [12]
During the year 2014-15, Anderson Ltd. issued 12% Debentures of ₹ 100 each, as per the details given below:
(a) 900 Debentures issued as collateral security to a bank against a loan of ₹ 60,000.
(b) The underwriters were to be paid a commission of ₹ 48,000. 25% of the amount was paid to them in cash and the balance was paid by the issue of Debentures at a discount of 10%, to be redeemed at par.
(c) A machine was purchased for ₹ 2,18,500. The vendor was paid by the issue of Debentures at a premium of 15%, to be redeemed at par.
(d) 5,000 Debentures were issued to the public at 5% premium, to be redeemed at a premium of 5%.
The company wrote off all capital losses arising from the issue of Debentures at the end of the year from its capital profits and if need be from its revenue profits.
You are required to journalize the above transactions in the books of Anderson Ltd.
Answer:

Question 5. [12]
Divya and Pooja are partners in a firm, sharing profits and losses in the ratio of 3:2. On 31st March, 2015, their Balance Sheet was as under:

The partners decided that with effect from 1st April, 2015, they would share profits and losses equally.
For this purpose, they decided that:
(a) Investments to be valued at ₹ 60,000.
(b) Goodwill to be valued at ₹ 24,000.
(c) General Reserve not to be distributed between the partners.
You are required to :
(i) Pass journal entries
(ii) Prepare the revised Balance Sheet of the firm.
Answer:

Question 6.
(a) Pinnacle Instruments Ltd. registered itself with a capital of? 20,00,000 divided into Equity Shares of ₹ 100 each. [9]
On 1st June, 2014, the company issued 5,000 Equity Shares as fully paid to Mila Herbals, as purchase consideration for the purchase of plant and machinery.
The remaining shares were issued to the public at par.
Till the date of the Balance Sheet, the Directors had called from the public,
60% of the nominal value of the shares.
The amount called was received by the company.
You are required to prepare as at 31st March, 2015 :
(i) The Balance Sheet of Pinnacle Instruments Ltd. as per Schedule III of the
Companies Act, 2013.
(ii) Notes to Accounts.
(b) Under which heads and sub-heads will the following items appear in the Balance Sheet of a company as per Schedule IE of the Companies Act, 2013. [3]
(i) Public Deposits
(ii) Calls-in-Advance
(iii) Building under construction
Answer:


(b) (i) Public Deposits → Non-current liabilities → Long term borrowings
(ii) Calls-in-Advance → Current liabilities → Other current liabilities
(iii) Building under construction → Non-current assets → Capital work-in progress

Question 7. [12]
Shankar and Manu are partners in a firm. On 1st April, 2014, their fixed capital accounts showed a balance of ₹ 2,00,000 and ₹ 4,00,000 respectively.
On this date, their current account balances were ₹ 50,000 and ₹ 1,00,000 respectively.
On 1st January, 2015, Shankar introduced additional capital of ₹ 2,00,000 while Manu gave a loan of ₹ 1,50,000 to the firm.
The clauses of their partnership deed provided for :
(a) Interest on capital to be allowed at the rate of 10% per annum.
(b) Interest on drawings to be charged at the rate of 12% per annum.
(c) Profits to be shared by them in the ratio of 3:2.
(d) 10% of the correct net profit to be transferred to General Reserve.
During the financial year 2014-15, both partners withdrew ? 6,000 each at the beginning of every quarter.
The net profit of the firm, before any interest, for the financial year 2014-15 was ₹ 5,00,000.
You are required to prepare for the year 2014 – 15 :
(i) Profit and Loss Appropriation Account.
(u) Partners’Fixed Capital Accounts.
(iii) Partners’ Current Accounts.
(iv) Partner’s Loan Account.
Answer:

Question 8. [12]
Pihu, Geeta and Nita are partners in a firm, sharing profits and losses in the ratio of 3:2:1. On 31st March, 2015, their Balance Sheet was as under :

Nita retires on 1st April, 2015, subject to the following adjustments:
(a) Land and building to be reduced by 10%.
(b) Goodwill to be valued at ₹ 54,000.
(c) Provision for Doubtful Debts to be raised to 10% of the debtors, the excess provision being created from General Reserve. The balance of the General Reserve to be distributed amongst the partners.
(d) Creditors of ₹ 3,000 were paid by Pihu for which she is not to be reimbursed.
(e) The continuing partners to share profits and losses in future in the ratio of 5:4.
(J) Nita to be paid ₹ 29,800 on retirement and the remaining amount in two equal annual installments
together with interest @ 10% per annum on the outstanding balance. The first installment of Nita’s loan to be paid on 31st March, 2016.
You are required to prepare:
(i) Revaluation Account.
(ii) Partners’ Capital Accounts.
(iii) Nita’s Loan Account till it is finally closed.
Answer:

Section – B
(20 Marks)
Answer any two questions

Question 9. [10]
You are required to prepare a Cash-Flow Statement (as per AS-3) for the year 2014-15 from the following Balance Sheets.




Additional Information:
During the year 2014-15 :
(i) A part of the machine was sold for ₹ 21,000 at a profit of ₹ 4,000.
(ii) The company charged ₹ 3,000 as depreciation on its Plant and Machinery.
(iii) New Debentures were issued on 31st March, 2015. at a discount of 10%.
(iv) Interest of ₹ 9,600 was paid on Debentures.
Answer:

Question 10.
(a) What is a Common Size Balance Sheet ? [2]
(b) While preparing a Cash How Statement, identify the following transactions as belonging to Operating Activities, Investing Activities, Financing Activities: [2]
(i) Goodwill written off.
(ii) Interest received by a company on its investments.

Answer:
(a) A common size balance sheet shows the percentage relationship of each assets/liability to total assets/total liabilities including capital.
(b) (i) Operating activities
(ii) Investing activities

Question 11.
(a) Calculate Liquid Ratio from the following (up-to two decimal places): [2]
Current Assets ₹ 1,26,000
Inventories ₹ 2,000
Current Ratio 3:2
(b) From the following Statement of Profit and Loss of Dixon Ltd, for the year 2014 – 15, calculate (up to two decimal places): [8]
(i) Trade Receivables Turnover Ratio
(ii) Inventory Turnover Ratio
(iii) Net Profit Ratio
(iv) Operating Profit Ratio



Answer:

ISC Class 12 Accounts Previous Year Question Papers

ISC Biotechnology Previous Year Question Paper 2016 Solved for Class 12

Maximum Marks: 80
Time allowed: Three hours

• Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.
• Answer Question 1 (Compulsory) from Part I and five questions from Part II, choosing two questions from Section A, two questions from Section B and one question from either Section A or Section B.
• The intended marks for questions or parts of questions are given in brackets [ ].
• Transactions should be recorded in the answer book.
• All calculations should be shown clearly.
• All working, including rough work, should be done on the same page as, and adjacent to the rest of the answer.

Part-I
(Answer all questions)

Question 1.
(a) Mention any one significant difference between each of the following : [5]
(i) Ligases and Helicases
(ii) Introns and Exons
(iii) Gel electrophoresis and Gel permeation
(iv) Sucrose and Starch
(v) Plasmids and Phages

(b) Answer the following questions : [5]
(i) Why is the nutrient medium autoclaved before using it for cell culture?
(ii) Name the enzyme that can synthesize DNA at a high temperature.
(iii) Why are restriction enzymes also called as molecular scissors ?
(iv) Name the nitrogenous bases present in RNA.
(v) Why is Agrobacterium tumifaciens called as the natural genetic engineer of plants ?

(c) Write the full form of each of the following :
(i) YAC
(ii) NCBI
(iii) RAM
(iv) SNP
(v) EMBL

(d) Explain briefly : [5]
(i) Somatic hybridization
(ii) Promoter gene
(iii) Site directed mutagenesis
(iv) DNA probes
(v) Primer
Answer:
(a) (i) Ligases are the enzymes which help in linking up of okazaki (DNA) segments produced on the parent strand with 5′-3′ direction. Hclicases help in unwinding the DNA helix, using ATP hydrolysis as a source of energy.

(ii) Introns are the intervening sequences which do not appear in the mature or processed RNA.
Exons are the coding sequences or expressed sequences that form functional and processed RNA.

(iii) Gel Electrophoresis is a technique by which negatively charged DNA fragments are separated by forcing them to move towards the anode under an electric field through a medium/matrix.

Gel permeation or filtration involves that molecules of different sizes can be separated from each other on the basis of their ability to enter the pores within the beaded gel, followed by passing down a column containing the gel. The technique is used in protein purification.

(iv) Sucrose is a disaccharide formed of glucose and fructose molecules while starch is a polysaccharide molecules formed of large number of glucose molecules arranged in the form of chains.

(v) Plasmids are the extra-chromosomal, self replicating, circular, double stranded DNA molecules present in bacteria.
Phages are the viruses which infect bacteria/cell. lyse it, integrate its DNA into it and replicate with the host chromosome.

(b) (i) The nutrient medium is autoclaved to make it sterilise i.e., free from microbes.

(ii) DNA polymerase isolated from a bacterium Thermus aquations.

(iii) Restriction enzymes are called ‘molecular scissors’ because they make cuts at specific positions/recognition sites within both strands of DNA.

(iv) Adenine, guanine, uracil, cytosine.

(v) Agrobacterium tumefaciens is able to deliver a piece of DNA known as ‘T-DNA’ to transform normal plant cells into tumour cells. By manipulating its Ti plasmid, it has now been modified into a useful cloning vector for delivering gene of our interest into a variety of plants.

(c) (i) YAC = Yeast Artificial Chromosome.

(ii) NCBf = National Centre for Bioinformatics Information.

(iii) RAM = Random Access Memory.

(iv) SNP = Short Nucleotide Polymorphism.

(v) EMBL = European Molecular Biology Laboratory.

(d) (i) Somatic hybrids are the hybrid plants produced through the fusion of protoplasts from two different varieties of plants each haring a desirable character. The hybrid protoplasts can be further grown to form a new plant.

(ii) Promoter gene a gene haring a regulatory sequence of DNA that initiates the expression of a gene.

(iii) Site directed mutagenesis involves induction of specified or desired changes in the base sequence at specified sites of genes, most successfully achieved by overlap extension PCR.
Or
The process of nucleotide changes in the cloned genes by specific in mutagenesis.

(iv) DNA probes are short 15-30 bases long, labelled oligonucleotides (RNA – DNA) used to detect complementary nucleotide sequences after hybridization and the auto radiograms gives many bands of different sizes.

(v) Primer is a short oligonucleotide that hybridizes with the template strand and gives a 3′- OH end at which a DNA polymerase start synthesis of DNA chain.

Part-II
(Answer any five questions)

Question 2.
(a) With reference to amino acids, explain :
(i) Any one physical and any one chemical property of amino acids.
(ii) Essential and non-essential amino acids.
(b) Briefly outline the various steps involved in the gene cloning technique
(c) List any four characteristics of genetic code.
Answer:
(a) They are the building blocks of molecular proteins.
(i) Amino acids are the organic acids (with carboxylic group-COOH) having amino group (-NH₂) generally attached to a-carbon or carbon next to the carboxylic group. Amino acids condenses to produce peptide (-NHCO-) bond.

(ii) Essential amino acids are essential for our body but they are not synthesised inside our body e.g., valine, isoleucine, lysine etc. They are supplemented through diet. Non-essential amino acids are those which are synthesised through transformation and transamination inside our body e.g., serine, alanine etc.

(b) The various steps involved in gene cloning technique are :

• Identification and isolation of desired DNA. .
• Amplification of gene of interest using PCR.
• Fragmentation/cutting of desired DNA and vector DNA by restriction enzymes.
• Ligation/joining of desired DNA fragment into vector using ligase enzyme.
• Transferring the recombinant DNA into host cell/organism by transformation, transfection, electroporation, microinjection (vectorless transfer), particle bombardment gun (Biolistics) (Vectorless transfer) or by Agrobacterium/retrovirus mediated gene transfer..
• Culturing the host cell for obtaining the foreign gene product/recombinant protein.
• Extraction of desired products or downstream processing.

(c) The characteristics of genetic codes are :

• Codes are universal.
• Codes are unambiguous and specific.
• Code is degenerate i.e., some amino acids are coded by more than one codon.
• Codons are read in a continuous fashion i.e., there are no commas or punctuation’s.

Question 3.
(a) Describe the three dimensional structure of DNA as proposed by Watson, Crick and Wilkins. Name a biochemical technique that was used by them to confirm the structure of DNA. [4]
(b) Explain the role of the following enzymes during the process of protein synthesis : [4]
(i) RNA polymerases and amino acyl tRNA synthetase.
(ii) Start codons and end codons.
(c) Why are auxins and cytokinins used in plant tissue culture ? [2]
Answer:
(a) Watson, Crick and Wilkins described the structure of DNA but also indicated how it could be replicated and transfer from one organism to its off spring.
Salient features of double helix structure of DNA include :

• Double helix made of two polynucleotide chains.
• Sugar and phosphate form backbone and N-base project inside.
• The two chains are antiparallel with one chain has the polarity of 5′ → 3′, the other has 3’→ 5.
• Presence of double hydrogen bonds between A=T, and triple bond to between G = C The . purine always comes opposite to pyrimidines to create uniform distance.
• Two chains are coiled in a right handed fashion.
• Pitch (one turn) of helix is 3.4 nm (= 34 A) contains 10 bp with 0.34 nm (3.4 A) gap between adjacent bps. (1 nm = 10-9 m)
• Two chains of DNA are 20 A far apart, due to the pairing of purines with pyrimidines. This distance remains constant.
• The plane of one base pair stacks over the other in double helix to confer stability

Rosalind Franklin confirmed this structure of DNA through X-Ray crystallography.

(b) (i) RNA polymerizes get associated transiently with initiation sigma factor (σ). They do not require a primer and can initiate the synthesis of a new chain on the template strand. It binds to the promoter site to start synthesis. The first nucleotide is labelled as + 1 and is called the transcription start site. It polymerizes ribonucleotides in 5′ → 3′ direction over 3′ → 5′ template DNA.

In prokaryotes, a single RNA polymerase transcribes all m-RNA, r-RNA and t-RNA. However, in case of Eukaryotas there are three different polymerizes i.e., RP+1 for r-RNA, RP-II for m-RNA and RP-III for t-RNA.
Amino acyl t-RNA syntheses couple each amino acid to its appropriate set of t-RNA molecules. There are 20 synthesizes for each of the 20 natural amino acids.

(ii) In prokaryotes, AUG acts as a start/initiation codon as well as codes for methionine. GUG acts as start codon for valine in Eukaryotes. UAA, UAQ UGA function as stop codons and do not code for any amino acid.

(c) Role of auxin :

• It helps in the formation of callus and in the development of xylem along with the promotion of cambial activity.
• It is used for the elongation and enlargement of plant cell and it inhibits the promotion of growths of apical and lateral buds.

Role of cytokinin:

• It stimulates the cell division process.
• It helps in the morphogenesis of plant cell, along with auxins.

Question 4.
(a) Explain how DNA technology has been used to create the following : [4]
(i) Dolly
(ii) Hepatitis B vaccine
(b) Write short notes on : [4]
(i) Batch culture and continuous culture.
(ii) Salinity resistance in crops.
(c) Name any two chemicals used to prepare the gel for gel electrophoresis. [2]
Answer:
(a) (i) Dolly – The first cloned animal
Using nuclear transfer technique, the world’s first mammalian clone – Dolly, was born in February 1996. In 1995, Ian Wilmut and his research group (Scotland) took out udder from a six year old sheep A called clone mother, and put in a special solution. Nucleus of udder cell was taken out and put in a solution. At the same time an unfertilized egg was taken out from another sheep B called egg mother. Nucleus of the egg was removed and nucleated egg was put in a culture medium. The nucleus of udder cell and nucleated egg cell were put together followed by mild electric shock. Consequently nucleus was taken up by the nucleated cell. This cell was incubated onto growth medium then transferred into a surrogate mother. A little lamb Dolly was born in February, 1996.

(ii) It is true that proteins (vaccines) stimulate immune system and cause to secrete specific antibodies. Such specific amino acid sequences in the protein that stimulate immune response are called epitopes. Based on selected epitopes recombinant vaccines may be produced on commercial level which can prove more effective and safer than the conventional vaccines.

Working on these lines, a recombinant Hepatitis B vaccine was produced by cloning the synthetic gene (for the surface antigen of the virus) in yeast cells. This gene expressed well in yeast cells and produced 22 nm particles of hepatitis B virus (HBV) surface antigen (as produced in patients) infected with hepatitis B virus. The recombinant vaccine has high immunogenecity. This product has been marketed as a vaccine for protection against HBV infection.

(b) Tissues and cells cultured in a liquid medium produce a suspension of single cells and cells clumps of few to many cells, are called suspension culture. Suspension cultures are of two types:

(i) Batch culture: In a batch culture, the same medium and all the cells produced are retained in the culture vessel e.g., culture flasks (100-250 ml), fermenters (variable size) etc. The cell number of biomass, of a batch culture exhibits a typical sigmoid curve represented by lag phase (cell number or biomass remain unchanged), log phase (rapid increase in cell number) and finally stationary phase (cell member does not change). The lag-phase lasts about 3-4 cell generation. The stationary phase is forced on the culture.

• due to depletion of nutrients
• and due to accumulation of wastes.

They are maintained by subculturing. They are unsuitable for studies on cell growth and metabolism because of constant change in cell density and nutritional status.

Continuous culture: Here the cell population is maintained in a steady state by regularly replacing a portion of used or spend medium. Such culture systems are either

• closed or
• open system.

In closed system, cells are separated from the used medium, taken out for replacement and added back to the culture so that biomass keeps on increasing. In contrast, both cells and used medium are taken out from open continuous system and replaced by equal volumes of fresh medium.

(ii) A plant which bears a foreign gene for desired function of other organism is called transgenic crop. Using biotechnological approaches stress/salinity tolerant plants can be produced. They secrete stress-related osmolytes such as sugars (fructans and trehalose) sugar alcohols (mannitol) amino acids (betaine, glycine and proline) and other proteins. For example, betaine is highly effective osmolyte that accumulates in some plants during, water stress or high salinity: Betaine is synthesised both in bacteria and plants. A transgenic tobacco was prepared by transferring E.coli bet A. gene through Ti-plasmid. The transgenic tobacco was 80% salt tolerant (i.e. 300 mM) than the normal tobacco.

(c) (i) Agarose gels
(ii) Polyacrylamide gels.

Question 5.
(a) Give the step wise procedure of the Southern Blotting technique. Mention any two differences
between Southern Blotting technique and Northern Blotting technique. [4]
(b) Explain the principle and any two applications of each of the following biochemical techniques:
(i) Ion Exchange Chromatography
(ii) Colorimetry [4]
(c) What is the cause of inborn metabolic disorders? Give any two examples of these disorders. [2]
Answer:
(a) Southern Blotting (Hybridization) Technique : In 1975, Edward M. Southern developed the technique of DNA separation and its hybridization. Therefore, in his honor this technique is known as ‘Southern blotting or Southern hybridization technique’. A specific DNA fragment can be separated and identified in a heterologous population of DNA molecules on the basis of binding of DNA probe with its complementary DNA strand.

The genomic DNA is isolated from the clone and digested with restriction enzymes. The DNA fragments are separated by agarose gel electrophoresis. Different DNA bands are formed on agarose gel which represents DNA fragments of varying sizes. These fragments are transferred from gel to nylon or nitrocellulose membrane. The process of DNA transfer is called ‘blotting’.

A nitrocellulose membrane is put over the gel. Many layers of filter paper are placed over nitrocellulose membrane. This assembly is put in a container having NaOH solution. NaOH denatures DNA and results in formation of single stranded DNA. DNA fragments are transferred from gel to membrane by capillary action.

The DNA fragments are fixed to membrane by using UV radiation or baking at 80°C. The pattern of DNA bands on membrane corresponds to the position of DNA on gel. The membrane is put in solution containing radio labelled DNA probe and incubated for some time. DNA probe hybridizes complementary DNA fragments fixed on membrane. It is gently washed at 12°C and dried. The membrane is exposed through a photographic film. DNA bands formed on photographic film corresponds to the original position of DNA fragments present on agarose gel.

Southern Blotting technique involves separation and identification of a specific DNA fragment. In Northern Blotting technique the RNA is analysed rather than DNA. During southern blotting NaOH denatures DNA to form single stranded DNA which are transferred from gel to nitro cellulose membrane. In northern blotting total RNA molecules are extracted and then mRNA molecules are isolated by using oligo (dT) cellulose Chromatography. RNA samples separated are transferred to a nylon membrane.

(b) (i) Ion Exchange Chromatography Principle: It is based on reversible exchange of ions in solution with ions electrostatically bound to some sort of insoluble support medium. Separation is obtained since different molecules have different degree of interaction with ion exchanger due to difference in their charges, charge densities and distribution of charge on their surface.

Applications:

• A technique for separating two proteins differing by only one charged amino acid.
• It is capable of separating species with very minor differences in their properties.

The ion-exchange may be

• Anion exchanger or
• Cation exchanger.

(ii) Colorimetry:
Principle : It is based on the use of interaction of light energy with coloured solutions of certain molecules as when light passes through a coloured solution, some wavelengths are absorbed more than others. The amount of light absorbed is proportional to the intensity of colour and hence to the concentration of the compound.

Applications:

• Quantitative estimation i.e. concentration in solution.
• Detect and identification of biomolecules

(c) Inborn metabolic disorders are due to change in the gene, for the particular character. Some autosomal genes become recessive and are transmitted to the progeny.
Examples:

1. Albinism
2. Cystic fibrosis
3. Phenylketonuria
4. Sickle-cell anaemia
5. Alkaptonuria

Alkaptonuria: This was one of the first metabolic diseases described by Garrod in 1908. It is an inherited metabolic disorder produced due to deficiency of an oxidase enzyme requiredfor breakdown ofhomogentisic acid (also called alcapton, hence, alkaptonuria is also written as alcaptonuria). Lack of the enzyme is due to the absence of the normal form of gene that controls the synthesis of the enzyme. Hence, homogentisic acid then accumulates in the tissues and is also excreted in the urine. The most commonly affected tissues are cartilages, capsules of joints, ligaments and tendons. The urine of these patients if allowed to stand for some hours in air, turns black due to oxidation of homogentisic acid.

Albinism: It is caused by the absence of the enzyme tyrosinase which is essential for the synthesis of the pigment from dihydroxyphenylalanine. The gene for albinism (a) does not produce the enzyme tyrosinase but its normal allele (A) does. Thus, only homozygous individual (aa) is affected by this disease. Albinos (individuals with albinism) lack dark pigment melanin in the skin, hair and iris. Although albinos have poor vision yet they lead normal life.

Question 6.
(a) Discuss the significance of each of the following techniques used in cell culture technology: [4]
(i) Androgenesis and gynogenesis
(ii) In-vitro pollination
(b) Enumerate any two post transcriptional changes in the mRNA to produce a completely mature mRNA. [4]
(c) State any four achievements of the Human Genome Project. [2]
Answer:
(a) (i) Androgenesis involves production of haploid plants by development of an egg cell
containing male nucleus. The female nucleus is eliminated before fertilization. Gynogenesis includes production of haploid plants by the development of an unfertilized egg cells because of delayed pollination (through use of abortive pollen pre-exposed to ionizing radiations or using an alien pollen). It is found in inter-specific crosses of potato.

(ii) In vitro pollination : A laboratory procedure, in which pollination and fertilisation occur under in vitro condition, offer an opportunity for producing hybrid embryoids among plants that cannot be crossed by conventional method of plant breeding. To overcome interspecific and intergeneric incompatibility cases, the style or part of it can be excised and pollen grains are either placed on the cut surface of ovary or transferred through a hole in the wall of ovary. This technique is called intra-ovarian pollination. It is applied successfully in Papaver somniferum, Argemone mexicana etc. It also involves direct pollination of cultured ovules or excised ovules together with placenta.

(b) Post-transcriptional changes : Modification at 5′ end of primary RNA (Heterogenous RNA or hnRNA) transcript is called capping. During capping, an extra guanosine residue in first added to the terminal nucleoside triphosphate of the primary RNA transcript. Guanosine residue is further modified by addition of methyl groups. Capping prevents the 5′ end of TORNA from being digested by exonucleases. It aids in the transport of m-RNA out of the nucleus and plays an important role in the m-RNA translation.

Addition of a string of adenosine residues, forming a poly (A) tail at 3′ end of the TO-RNA.

The process of removal of those parts of primary transcript, corresponding to introns and joining together of exons, is known as splicing.

(c) Achievement of Human Genome Project:

• Human genome is largely consisting of – 3 × 109 base pairs a lot of repeated sequences.
• 3 billion base pairs were translated into biologically meaningful information by using computers and it led to a new field of bioinformatics.
• Human genome sequencing will enable a new approach to biological research.
• Understanding of function of biological systems increases in public and private sectors due to knowledge from DNA sequences.
• More than 1200 genes are associated with common human disorders such as Cardio- . Vascular diseases, diabetes, (endocrine disease), Alzheimer disease (neurological disorder).
• Efforts for health care are being made to design drugs, genetically modified diets, nutraceuticals etc.

Question 7.
(a) Explain the method used for the construction of a genomic DNA library. Also state how a genomic DNA library differs from a cDNA library. [4]
(b) Discuss any four vector-less methods of transfer of foreign DNA into the host cells. [4]
(c) Define glycosidic bond and peptide bond. [2]
Answer:
(a) DNA Library: DNA library is a collection of DNA fragments of one organism, each carries by a plasmid or virus and cloned in an appropriate host. A DNA probe is used to locate specific DNA sequence in the library. A collection representing the entire genome is called genomic (DNA) library. An assortment of DNA copies of messenger RNA produced by a cell is known as a complimentary DNA (cDNA) library.

Construction of Genomic Library: The process of subdividing genomic DNA into clonable elements and inserting them into host cells is called creating a library.

A complete library, by definition, contains the entire genomic DNA of the source organism and is called as genomic library. A genomic library’ is a set of cloned fragments of genomic DNA.

The process of creating a genomic library includes four steps:

1. In the first step the high molecular weight genomic DNA is separated and subjected to restriction enzy me digestion by using two compatible restriction enzy mes.
2. In the second step, the fragments are then fractionated or separated by using agarose gel electrophoresis to obtain fragments of required size.
3. These fragments are then subjected to alkaline phosphatase treatment to remove the phosphate. In the third step, the dephosphorylated insert is ligated into vector which could be a plasmid, phage or cosmid, depending upon the interest of the researcher.
4. In the last step, the recombinant vector is introduced into the host by electroporation and amplified in host.

In principle, all the DNA from the source organism is inserted into the host, but this is not fully possible as some DNA sequences escape the cloning procedure. Genomic library is a source of genes and DNA sequences. A genomic library is a set of cloned fragments of genomic DNA. Prior information about the genome is not required for library construction for most organisms. In principle, the genomic DNA, after the isolation, is subjected to RE enzyme for digestion to generate inserts.

cDNA libraries V/s Genomic libraries:

• Genomic library is a mixture of fragments of genomic DNA while cDNA obtained from ZM-RNA may cloned to give rise to a cDNA library. Genomic library contains DNA fragments that represent genes as well as those that are not genes. In contrast cDNA library contains only those genes that are expressed in the concerned tissue/organism. In both cases, a mixture of fragments is used for cloning to establish the library.
• Use of cDNA is absolutely essential when the expression of an eukaryotic gene is required in a prokaryote.
• Eukaryotic cDNAs are free from intron sequences.
• As a result of the above, they are smaller in size than the corresponding genes, i.e.. the genes that encoded them.
• A comparison of the cDNA sequence with the corresponding genome sequence permits the delineation of intron/exon boundaries.
• The contents of cDNA libraries from a single organism will vary widely depending on the developmental stage and the cell type used for preparation of the library. In contrast the genomic libraries w ill remain essentially the same irrespective of the developmental stage and the cell type used.
• A cDNA library will be enriched for abundant mRNAs. but may contain only a few or no clones representing rare mRNAs.

(b) Transfection: Transfection is the transfer of foreign DNA into cultured host cells mediated through chemicals. The charged chemical substances such as cationic liposomes, calcium phosphate of DEAE dextran are taken and mixed with DNA molecules. The recipient host cells are overtaxed by this mixture. Consequently the foreign DNA is taken up by the host cells.

Electroporation (Electric Field-mediated Membrane Permeation): In electroporation an electric current at high voltage (about 350 V) is applied in a solution containing foreign DNA and fragile host cells. This creates transient microscopic pores in cell membrane of naked protoplasts. Consequently foreign DNA enters into the protoplast through these pores. The transformed protoplasts are cultured in vitro which regenerate respective cell walls.

Microinjection : In this technique foreign DNA is directly and forcibly injected into the nucleus of animal and plant cells through a glass micropipette containing very fine tip of about 0.5 mm diameter. It resembles with injection needle. In 1982, for the first time Rubin and Spradling introduced Drosophila gene into P-element and microinjected into embryo.

Particle Bombardment Gun (Biolistics): This technique was developed by Stanford in 1987. In this method macroscopic gold or tungsten particles are coated with desired DNA. A plastic micro-carrier containing DNA coated gold/tungsten particles is placed near rupture disc. The particles are bombarded onto target cells by the bombardment apparatus. Consequently foreign DNA is forcibly delivered into the host cells

(c) Glycosidic bond (- O -) is the bond formed between the OH group attached to an anomeric carbon atom of a monosaccharide can easily dehydrate with an -OH group attached to another monosaccharide leading to the formation of a disaccharide.

Peptide bond (-NHCO-) is the bond formed between the carboxylic group (COOH) of one amino acid w ith the amino group (NH2) of another amino acid, with elimination of w ater to form a dipeptide.

Question 8.
(a) Explain the secondary and the quaternary structure of proteins. Mention any two important functions of proteins. [4]
(b) Discuss the method used for DN A sequencing by A utomated DNA sequencing technique. [4]
(c) Give any one example each of in-situ and ex-situ conservation. [2]
Answer:
(a) Secondary Structure (2° structure) of Proteins: It is development of new steric relationships amongst the amino acids for protecting their peptide bonds through formation of intrapolypeptide and interpolypeptide hydrogen bonds. Secondary’ structure is of three types – α-helix β-pleated and collagen helix. The prefixes α and β signify the first and second types of secondary structure discovered by Pauling and Corey (1951).

(i) α-Helix: The polypeptide chain is spirally coiled, generally in a clockwise or right-handed

fashion (Fig.). There are 3.6 amino acid residues per turn of the spiral. The spiral is stabilized by straight hydrogen bonds between imide group (-NH-) of one amino acid and carbonyl group (—CO—) of fourth amino acid residue. In this way all the imide and carbonyl groups become hydrogen bonded. R-groups occur towards the outer side of a-helix. a-helix is the final structure in certain fibrous proteins, e.g., keratin (hair, nail, horn), epidermis (skin).

β-Pleated Sheets : Two or more polypeptide chains come together and form a sheet. Condensation is little. How ever, tw isting does occur. The same polypeptide may fold over itself to form two strands for p-pleating. Adjacent polypeptide chains may occur in parallel (e.g., p-keratin) or anti-parallel (e.g., silk fibrin). Straight hydrogen bonds occur between imide (-NH-) group of one polypeptide and carbonyl (-CO-) group of adjacent polypeptide. Cross-linkages help in stabilization of β-pleated sheets.

Collagen Helix (Fig.) : Collagen has a large amount of glycine (25%) and proline (and hydroxyproline, 25%). It cannot form a-helix due to them. Three of its polypeptide each having about 1000 amino acid residues, come together with each forming an extended left-handed helix. They run parallel, form a right-handed super-helix that is stabilised by hydrogen bonds amongst the three. The triple helix of collagen is often called tropo- collagen. Its one end is stabilised by -S-S- linkages amongst the three chains. Collagen occurs in those tissues where extensibility is limited, e.g., connective tissue, tendons, bones.

Quaternary Structure (4° Structure) of proteins : It is the last or fourth level of protein organisation found in only oligomeric proteins or multimers. The multimeric proteins are formed of two to several polypeptides. The monomers or polypeptide subunits are also called protomers. Protomers may be similar, e.g., two similar polypeptides in enzyme phosphorylase a. It is known as homogeneous quaternary structure. An oligomeric protein having dissimilar subunits shows heterogeneous quaternary structure, e.g., tetrameric haemoglobin with two a (141 amino acids each) and two β (146 amino acids each) polypeptide chains.

Functions of proteins:

• Major group of proteins are enzymes-biocatalyst.
• Myoglobin, a protein found in muscle store oxygen.
• Haemoglobin in RBC’s transports gases in and out of lungs.
• Proteins as structural element; as hairkeratins and bio-membranes.

(b) Automatic DNA Sequencing : In this new method a different fluorescent dye is tagged to the ddNTPs. Using this technique a DNA sequence containing thousands of nucleotides can be determined in a few hours. Each dideoxynucleotide is linked with a fluorescent dye that imparts different colours to all the fragments terminating in that nucleotide. All four labelled ddNTPs are added to a single capillary tube. It is a refinement of gel electrophoresis which separates fastly. DNA fragments of different colours are separated by their respective size in a single electrophoretic gel. A current is applied to the gel. The negatively charged DNA strands migrate through the pores of gel towards the positive end.

The small sized DNA fragments migrate faster and vica versa. All fragments of a given length migrate in a single peak. The DNA fragments are illuminated with a laser beam. Then the fluorescent dyes are excited and emit light of specific wavelengths which is recorded by a special ‘recorder’. The DNA sequences are read by determining the sequence of the colours emitted from specific peaks as they pass the detector. This information is fed directly to a computer which determines the sequence. A tracing electrogram of emitted light of the four dyes is generated by the computer (Fig.). Colour of each dye represents the different nucleotides. Computer converts the data of emitted light in the nucleotide sequences.

(c) In situ conservation refers to ‘on-site’ conservation of plants and animals as such. It is a process of protecting plants or animals in its natural habitat by protecting or cleaning up the habitat, or by defending the species from the predators, e.g., National Park, Sanctuaries and Biosphere Reserve.

Ex situ conservation is ‘off-site conservation’ by protecting an endangered species of plant or animal by removing part of population from a threatened habitat and placing it in a new location such as wild area, zoos or botanical gardens, in vitro gene bank, germplasm-banks, in vivo gene bank.

Question 9.
(a) What is meant by the term genomics ? Mention the difference between structural genomics and functional genomics. [4]
(b) How do the following databases contribute towards managing biological data : [4]
(i) GDB and MGD
(ii) PDB and PIR
(c) Name any two organisms whose genomes have been completely sequenced. [2]
Answer:
(a) Genomics is a scientific discipline of mapping, sequencing and analysing the genome-the complete set of chromosomal and extra-chromosomal genes of an organism. Structural genomics deals with DNA sequencing, sequence assembly, construction of genetic, physical or sequence maps of high resolution of the organism.

Functional Genomics deals with reconstructing genome sequences and to find out the functions they do. It provides novel information about the genome. It helps in understanding of genes and functions of proteins and protein interactions.

(b) (i) GDB (Genome Database): It is the official central repository for genome mapping data created by Human Genome Project. Its central node is located at the hospital for sick children. GDB holds a vast quantity of data submitted by hundreds of investigators. The GDB has many useful genome resource web-links on its resource page.

MGD (Mouse Genome Database) is the primary public mouse genomic catalogue resource. The MGD includes information on mouse genetic markers and nomenclature, molecular segments, phenotypes, comparative mapping data, graphical display of linkage, cytogenetic and physical maps.

(ii) PDB (Protein Data Bank) : This database has the sequence of those proteins, nucleic acid, whose 3-D structures are known by crystallography or NMR spectroscopy.
Source : NCBI-USA; EBI, UK.

PIR (Protein Information Resource): It is an integrated public bioinformatics resource to support genomic, proteomic and system biology, research and scientific studies.

(c) Organisms with completely sequenced genome : Phage λ , HIV, E.coli, Heliobacterpylori, Saccharomyces cerevisiae (yeast), Drosophila melanogaster (fruitfly). (any one organism)

ISC Class 12 Biotechnology Previous Year Question Papers

ISC Accounts Previous Year Question Paper 2011 Solved for Class 12

Maximum Marks: 80
Time allowed: Three hours

• Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.
• Answer Question 1 (Compulsory) from Part I and five questions from Part II, choosing two questions from Section A, two questions from Section B and one question from either Section A or Section B.
• The intended marks for questions or parts of questions are given in brackets [ ].
• Transactions should be recorded in the answer book.
• All calculations should be shown clearly.
• All working, including rough work, should be done on the same page as, and adjacent to the rest of the answer.

Part -1

Question 1.
Answer each of the following questions briefly : [10 x 2]
(i) How will you deal with the following items during the preparation of a cost sheet ?
(a) Salary of Public Relations Officer.
(b) Subscription to technical journals.
(ii) What journal entry will you pass w hen unsold stock is taken over by a Co-venturer assuming that a separate set of books is maintained ?
(iii) Which method of valuation of inventory will you recommend during:
(a) Periods of rising prices.
(b) Periods of falling prices.
(iv) State the entries you will pass in case of transfer from Debtors Ledger to Creditors Ledger under Self Balancing System.
(v) State two differences between average profits and super profits.
(vi) What are the closing entries for interest on calls in arrear account and interest on calls in advance account ?
(vii) State two differences bewtween current ratio and quick ratio.
(viii) List any two types of operating activities.
(ix) Explain the nature of interest on debentures.
(x) How will you deal with a situation when a solvent partner’s capital account reflects a debit balance in the application of Gamer Vs Murray ?
Answer :
(i) (a) The salary of a Public Relations officer is to be considered under the head Selling and Distribution overhead. As PRO is the officer who promotes the companies welfare and goodwill of the company.
(b) Subscription to technical journals will be considered under the head factory overheads. Technical journals arc used by the factory personnel, supervisors and technician

(ii) When unsold stock is taken over by a co-venturer the journal entry to be passed is
Co-Venturer A/c Dr
To Joint Venture A/e stock

(iii) (a) In the condition of rising prices FIFO gives greater profit than LIFO due to difference in the values of closing stock
(b) In the condition of falling prices LIFO gives greater profit than FIFO.

(iv) In the case of transfer under Self Balancing system the journal entries to be passed will be as follows assuming Mr. Singh as the person who is debtor as well as creditor for the common amount.
(a) Singh A/c Dr.
(in the purchase ledger)
To Singh A/c
(Balance of Singh a/c in the purchases ledger transferred to their a/c in the sales ledger)
(b) Purchases ledger adjustment A/c Dr.
(in General ledger)
To General ledger adjustment A c (in purchase ledger)
(Correcting entry of Adjustment Accounts for the transfer)
(c) General ledger adjustment A/c Dr
(in the sales ledger)
To Sales Ledger Adjustment A/c
(in the General ledger)
(Correcting entry of adjustment Accounts for the transfer)

(v)

Average Profits	Super Profits
1. It is (he average of profits of the past few years. 2. The normal rate of return is not relevant in the calculation of the average profits.	1. It is excess of the average profits over the normal profits. 2. The normal rate of return is considered while calculating super profits.

(vi) (i) On Transfer of interest on Calls – in – Arrears A/c to Profit and Loss A/c at the end of the accounting period.
Interest on calls in Arrears A/c Dr
To Profit and Loss A/c (With the amount of interest)
(ii)On transfer to P/L account at the end of accounting period.
Profit and Loss A/c Dr
To interest on Calls-in-advance A/c
(With the account of interest)

(vii)

Current Ratio	Quick Ratio
1. Current ratio relates current assets to current liabilities 2. Ideal ratio 2 : 1	1. Quick ratio relates quick assets to quick liabilities. 2. Ideal ratio 1 : 1

(viii) Two types of operating activities may be stated as :

1. Cash receipts from sale of goods and sen ices.
2. Cash receipts from royalties, fees, commission, etc

(ix) 1. No interest is payable on debentures issued as a collateral security.
2. The balance of interest on debentures account is transferred to the profit and loss account at the end of the year.
3. If the amount of interest accrued and due is not paid, it is known as Interest accrued and Due to interest outstanding. It appears under the head secured loans.

(x) When a solvent partner’s capital account reflects a debit balance in the application of Garner Vs Murray . The partner, having a debit balance, will not haw to bear the loss due to insolvency of a partner.

Question 2. [10]
The following information is available from the records of Singh and Company Limited for the month ended 30^th September 2010 :

Purchases of raw materials — ₹ 1,00,000
Opening stock of finished goods (1000 units) — ₹ 13,600
Direct wages — ₹ 68,000
Factory overhead — ₹ 80% of direct wages
Administrative overhead — ₹ Rs. 2 per unit
Selling and Distribution overhead — ₹ R.s 1.50 per unit
Closing stock of finished goods —₹ 1.800 units
Royalties on production — ₹ 10,000
Sale of scrap of raw materials (normal loss) — ₹ 8,000
The manufacturer sells the product so as to reflect a profit of 25% on sales and 6200 units are sold in the market.
From the above information, you are required to prepare a Cost Sheet showing the total cost for the month ended 30th September 2010.
Note : All calculations are to be made to the nearest rupee and sales are made on the basis of LIFO principle.
Answer:

Part – II

Question 3. [14]
Roger and Suresh sharing profits and losses in the ratio of 3 : 2 jointly agreed to underwrite the subscription of 60,000 equity shares of Rs. 10 each of Parag and Company Limited at a premium of Rs. 3 per share. The underwriting commission is 4% as provided in the Articles. Applications were received from the public only for 40,000 shares and so the underwriters took over the remaining shares. A joint bank account was opened towards which Roger contributed Rs. 55.000 and Suresh Rs. 45,000. A sum of Rs. 7.000 was incurred on various expenses which were paid out of the joint bank account. Parag and Company paid the underwriting commission by cheque. At the close of the venture, the underwriters sold 12.000 shares at the rate of Rs. 15 per share and the rest of the shares were taken up by them at the rate of Rs. 14 per share, in their profit and loss sharing ratio.
Prepare :
(i) Joint Venture Account
(ii) Co-Venturers’ Account
(iii) Joint Bank Account
Answer:


Question 4. [14]
Mr. Khanna maintains his books on sectional balancing ledgers.
Transactions during the month of November 2010 were :


On 01.11.2010, Bought Ledger (Cr) Rs. 30.000 and Sold Ledger (Dr) Rs. 60.000.

From the above particulars, prepare Control Accounts in the relevant ledger.
Answer:
In Nominal Ledger

Question 5. [14]
Given below is the Balance Sheet of Gurmeet and Company Limited as on 31st December 2009 and 31^st December 2010 :


(a) Investments costing Rs. 36,000 were sold for Rs. 30,000 during the year 2010.
(b) New debentures have been issued at the end of the current accounting year.
(c) New investments have been purchased at the end of the current accounting year.
(d) Depreciation charged on machinery during the current accounting year was Rs. 10,000.
From the above information prepare a Cash Flow Statement as per Accounting Standard-3.
Answer:



Question 6. [14]
Ahmed. Bina and Chitra are partners sharing profits and losses in the ratio of 3 : 2 : 1. Their balance sheet as on 31^st March 2010 stood as under :

Ahmed died on 31.03.2010 and the following decisions were taken by the surviving partners according to the partnership deed :
(a) Equipment to be revalued at Rs. 3,50,000 and furniture to appreciate by Rs. 10,000.
(b) A provision of 10% to be created for doubtful debts.
(c) Stock to be revalued at Rs. 83,000.
(d) The goodwill of the firm was valued at Rs. 30,000 on Ahmed’s death.
The firm had a joint life policy of Rs. 90,000. The policy was surrendered and the death claim was realized in full by cheque from the insurance company.
The surviving partners finally agreed that the values of assets and liabilities must remain the same and as such, there must not be any change in their book values as a result of the above mentioned adjustments, excepting the bank balance.
The amount payable to Ahmed was transferred to his executor’s account.
Prepare Partners Capital Account and a Balance Sheet of Bina and Chitra.
Answer:

Question 7. [14]
Sachdeva Tyres and Company Limited issued applications for 100000 equity shares of Rs. 10 each at a premium of Rs. 3 per share. The amount was payable as follows.
(i) On application : Rs. 2 00
(ii) On allotment : Rs. 5 (including premium)
(iii) Balance on the first and the final call.
Applications were received for 150000 shares. Allotment was made pro-rata to all applicants. Sudhir who had applied for 300 shares failed to pay allotment and call money. His shares were forfeited after the first and the final call. Of these. 170 shares w ere reissued to Pramod at Rs. 9 per share fully paid.
Pass the necessary Journal Entries to show the above transactions. Show your w orkings clearly.

Question 8. [14]
(i) Gurung Ltd. took over assets of Rs. 6,00,000 and liabilities of Rs. 60,000 from Batra Ltd. for the purchase consideration of Rs. 5,50,000. It paid the purchase consideration by issuing 8% debentures of Rs. 100 each at 10% premium.
(ii) Gurung Ltd. purchased land from Jaiswal Ltd. for Rs. 4,50,000. The consideration was paid by issuing 5% debentures at a discount of 10%.
(iii) Gurung Ltd. issued 1,000, 6% debentures of Rs. 100 each at a discount of 7% repayable after 5 years at a premium of 10%.
From the above particulars, pass journal entries in the books of Gurung Ltd. to record the transactions.
Answer:

Question 9. [14]
David and Bimal are partners sharing profits and losses in the ratio 3:2. Their Balance Sheet as on 31^st March 2010, was as follows :

They admitted Chander as a new partner on 1.4.2010 and the new profit sharing ratio became 5:3:2. Chander introduced a capital of Rs. 16.000. Chander w as unable to bring any cash for good will and so it was decided to value the goodwill on the basis of his share in the profits and the capital contributed by him. The following revaluations were made at the time of Chander s admission :
(i) Stock had been overvalued by Rs. 750 and furniture by Rs. 500.
(ii) Provision for doubtful debts to be increased by Rs. 100.
(iii) A creditor for Rs. 2,350 was paid off by Bimal privately for which he was not to be reimbursed.
Prepare the Revaluation account, Partner’s capital accounts and a Balance Sheet of the new firm on the date of Chander’s admission. Show your workings clearly.
Answer:

Question 10.
The following figures have been extracted from the books of Arvind and Company Limited
Net sales — Rs. 12,00,000
Net purchases — Rs. 5.00,000
Administrative expenses — Rs. 65.000
Selling and distribution expenses — Rs. 35.000
Gross profit — 20% on sales
Net profit after tax — Rs. 10,00,000
Total assets — Rs. 40,00,000
Equity share capital of Rs. 10 each — Rs. 10,00,000
10% Preference share capital of Rs. 10 each — Rs. 3,00,000
Reserves and surplus — Rs. 2,00,000
8% Debentures — Rs. 8,00,000
Opening debtors — Rs. 1,20,000
Closing debtors — Rs. 80,000
Opening bills receivable — Rs. 60,000
Closing bill receivable — Rs. 40,000
Opening creditors — Rs. 1,30,000
Closing creditors — Rs. 70,000
Closing bill payable — Rs. 50,000
Opening bills payable — Rs. 1,10,000
From the above information, calculate the following :
(a) Total assets to debt ratio
(b) Debt equity ratio
(c) Operating ratio
(d) Operating profit ratio
(e) Earning per share
(f) Debtors turnover ratio
(g) Creditors turnover ratio
Note : All calculations are to be made to two places of decimals.
Answer:

ISC Class 12 Accounts Previous Year Question Papers

ISC Biotechnology  Previous Year Question Paper 2012 Solved for Class 12

Maximum Marks: 80
Time allowed: Three hours

• Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.
• Answer Question 1 (Compulsory) from Part I and five questions from Part II, choosing two questions from Section A, two questions from Section B and one question from either Section A or Section B.
• The intended marks for questions or parts of questions are given in brackets [ ].
• Transactions should be recorded in the answer book.
• All calculations should be shown clearly.
• All working, including rough work, should be done on the same page as, and adjacent to the rest of the answer.

Part-1
(Answer all questions)

Question 1.
(a) Mention any one significant difference between each of the following: [5]
(i) Hybrid and Cybrid.
(ii) DNA polymerase and Taq DNA polymerase.
(iii) Glycosidic bond and Peptide bond,
(iv) Oils and Waxes.
(v) Homopolysaccharide and Heteropolysaccharide.

(b) Answer the following questions : [5]
(i) What is a callus?
(ii) Name the method used for the sterilization of plant hormones and vitamins.
(iii) Why is DNA replication called semi-conservative replication?
(iv) What is a promoter gene?
(v) State two uses of stem cells.

(c) Write the full form of the following :
(i) HGP
(ii) STS
(iii) CSIR
(iv) LAF
(v) SCP

(d) Explain briefly:
(i) Amphipathic property of lipids.
(ii) Replication fork
(iii) Androgenesis
(iv) Transamination.
(v) Active site.
Answer:
(a) (i) Hybrid: Plants produced through the fusion of protoplasts of two different plant species/varieties is called hybrid.

Cybrid: Cybrid or cytoplasmic hybrid are cells or plant containing nucleus of one species but cytoplasm from both the parental species.

(ii) DNA polymerase: A DNA polymerase is an enzyme that catalyzes the polymerization of deoxyribonucleotides into a DNA strand in DNA replication.

Taq DNA polymerase: Taq polymerase, is a thermostable DNA polymerase, originally isolated from bacterium Thermits aquaticus. It is used in polymerase chain reaction.

(iii) Glycosidic bond: A glycosidic bond (C-O-C) is a type of covalent bond that joins a aldose or ketone group of a carbohydrate (sugar) molecule to another group (OH) which may or may not be another carbohydrate.

Peptide bond: A peptide bond (HN – C = 0) is a chemical bond formed between two amino acids when the carboxyl group of one amino acid molecule reacts with the amino group of the other amino acid molecule. thereby releasing a molecule of water.

(iv) Oils: Oils are the esters of unsaturated fatty acid with glycerol. Oils are liquid at room temperature and have low melting point.

Waxes: Waxes are esters of fatty acid other than glycerol. They contain one molecule of long chain fatty acid esterified with one molecule of long chain like cytyl, ceryl or mericvl, mono hydroxy’ alcohol.

(v) Homopolysaccharide: Homopolysaccharides are the complex carbohydrates formed by polymerisation of the monosaccharide monomers.

Heteropolysaccharide: Heteropoly saccharides are the complex carbohy drates w hich are produced by the condensation of more than one type of monosaccharide monomers or their derivatives e.g., chitin, agar.

(b) (i) Callus is a mass of meristematic, undifferentiated cells derived from plant tissue (explants).

(ii) The autoclaving denatures the vitamins and hormones, therefore, the solution of these compounds are sterilized by using Millipore filter paper with pore size of 0.2 micrometer diameter.

(iii) Semi-conservative is that mode of replication in which out of the two DNA strands, one is conserved strand and the other is newly synthesised.

(iv) A promoter gene is a segment of DNA containing the enzyme RNA polymerase that initiates the transcription of the genetic code.

(v) Uses of stem cells :
(1) Bone marrow transplants that are used to treat leukemia (blood cancer).
(2) Treatment for muscular dystrophy.

(c) (i) HGP: Human Genome Project.

(ii) STS: Sequence Tagged Sites.

(iii) CSIR : Council of Scientific and Industrial Research.

(iv) LAF: Laminar Air Flow.

(v) SCP: Single Cell Protein. .

(d) (i) Amphipathic nature : Most membrane lipids are amphipathic, having a non-polar end and a polar end. The amphiphilic nature of some lipids allows them to form structures such as vesicles, liposomes, or membranes in an aqueous environment.

(ii) The point at which the two strands of DNA are separated to allow replication of each strand is known as replication fork.

(iii) Androgenesis is the development of an embryo containing only paternal chromosomes due to failure of the egg to participate in fertilization.

(iv) Transamination is the process of transfer of the a-amino group from the amino acid to an a-keto acid.

(v) Active site is part of an enzyme where substrates bind and undergo a chemical reaction.

Part-II
(Answer any five questions)

Question 2.
(a) Explain the general structure of an amino acid. What do you understand by essential and non-essential amino acids? [4]
(b) What are cloning vectors ? Write the main characteristics of any three types of cloning vectors. [4]
(c) What is a Codon ? Name the start codon and any one end codon. [2]
Answer:
(a) Amino acids are building blocks of macromolecular proteins. They contain amino group and carboxyl group as functional groups.
On the basis of nutritional values amino acids are of two types :
Essential amino acids : These are essential for our body but they are not synthesised inside our body are termed as essential amino acids e.g., valine, leucine, isoleucine, lysine, phenylalanine, methionine, threonine, histidine, arginine.

Non-essential amino acids : They are those which are synthesised through transformation and transamination inside our body e.g., serine, alanine etc.

(b) Cloning vector is a self-replicating DNA molecule that carries foreign DNA insert into a host cell, replicates inside a bacterial (or yeast) cell and amplify to produce many copies of itself and the foreign DNA.

Plasmid: It is an extra chromosomal circular DNA molecule that self replicates inside the bacterial cell and some yeast; cloning limit: 100 to 10,000 base pairs or 0.1-10 kilobases (kb).

Phage : Designed bacteriophage lambda (A.) and Ml 3 : linear DNA molecules, whose region can be replaced with foreign DNA without disrupting its life cycle; cloning limit: 8-20 kb. M13 is a filamentous phage which infects E-coli. Cloning limit: 10 kb.

Cosmids: A constructed extrachromosomal circular DNA molecule that combines features of plasmids and ‘cos’ site of phage ; cloning limit – 45 kb.

Yeast Artificial Chromosomes (YAC) : An artificial chromosome that contains telomeres, origin of replication, a yeast centromere, restriction enzyme site and a selectable marker for identification in yeast cells ; cloning limit: 1 Mb.

(c) Codon is a unit of genetic coding, consists of a series of three adjacent bases (triplet) in one polynucleotide chain of a DNA or RNA molecules, which codes for a particular amino acid during synthesis of proteins in a cell. For example, ATA codes for Leucine.

Start codon : The codon AUG specifies the first amino acid, methionine, in protein synthesis.

End codon : UAG also referred to as amber codon, in mRNA which terminate translation.

Question 3.
(a) Briefly explain the structure of a tRNA molecule. Mention its function during the process of protein synthesis. [4]
(b) Give the stepwise procedure of sequencing of DNA by Sanger’s method. [4]
(c) What is Totipotency ? Give an example of a Totipotent cell. [2]
Answer:
(a) Transfer RNA (tRNA) : It is also called soluble or sRNA. There are over 100 types of tRNAs. Transfer RNA constitutes about 15% of the total RNA. tRNA is the smallest RNA with 70-85 nucleotides and sedimentation coefficient of 4S. The nitrogen bases of several of its nucleotides get modified e.g., pseudouridine (φ), dihydrouridine (DHU), inosine (I). This causes coiling of the otherwise single-stranded tRNA into L-shaped form (three-dimensional, Klug, 1974) or clover-like form (two dimensional, Holley, 1965). About half of the nucleotides are based paired to produce paired stems. Five regions are unpaired of single-stranded—AA-binding site, Tig C loop, DHU loop, extra arm and anticodon loop.

• Anticodon. It is made-up of three nitrogen bases for recognizing and attaching to the codon of wRNA.
• AA-Binding site. It lies at the 3′ end opposite to the anticodon and has CCA — OH group (5′ ends bears G). Amino acid or AA-binding site and anticodon are the two recognition sites of tRNA.
• T φ C loop. It contains pseudouridine. The loop is the site for attaching to ribosomes,
• DHU loop. The loop contains dihydrouridine. It is binding site for aminoacyl synthetase enzyme,
• Extra arm. It is a variable site arm or loop which lies between T ig C loop and anticodon. The exact role of extra arm is not known.

Functions:
tRNA is adapter molecule which is meant for transferring amino acids to ribosomes for synthesis of polypeptides. There are different tRNAs for different amino acids. Some amino acids can be picked up by 2-6 tRNAs. tRNAs carry specific amino acids at particular points during polypeptide synthesis as per codons of mRNA. Codons are recognised by anticodons of tRNAs. Specific amino acids are recognised by particular activating or aminoacyl synthetase enzymes,

They hold peptidyl chains over the mRNAs. The initiator tRNA has the dual function of initiation of protein synthesis as well as bringing in of the first amino acids. There is, however, no tRNA for stop signals.

(b) DNA sequencing: It is the determination of the precise sequence of nucleotides in a sample of DNA.

Sanger dideoxy method: The most popular method for DNA sequencing is called the dideoxy method or Sanger method (named after its inventor, Frederick Sanger, who was awarded the 1980 Nobel prize in chemistry).

The Procedure : The DNA to be sequenced is prepared as a single strand.
This template DNA is supplied with

a mixture of all four normal (deoxy) nucleotides in ample quantities

• dATP
• dGTP
• dCTP
• dTTP

a mixture of all four dideoxynucleotides, each present in limiting quantities and each labeled with a tag.

• that fluoresces a different color :
• dd ATP
• dd GTP
• dd CTP
• dd. TTP

DNA polymerase 1

Because all four normal nucleotides are present, chain elongation proceeds normally until, by chance, DNA polymerase inserts a dideoxy nucleotide (shown as colored letters) instead of the normal deoxvnucleotide (shown as vertical lines). If the ratio of normal nucleotide to the dideoxy versions is high enough, some DNA strands will succeed in adding several hundred nucleotides before insertion of the dideoxy version halts the process.

At the end of the incubation period, the fragments are separated by length from longest to shortest. The resolution is so good that a difference of one nucleotide is enough to separate that strand from the next shorter and next longer strand. Each of the four dideoxynucleotides fluoresces a different color when illuminated by a laser beam and an automatic scanner provides a printout of the sequence.

Limitation : Limitations include non-specific binding of the primer to the DNA, affecting accurate read-out of the DNA sequence, and DNA secondary structures affecting the fidelity of the sequence

(c) Cellular totipotency : Totipotency is the ability of a single cell to divide and produce all the differentiated cells in an organism, including extraembryonic tissues and also the formation of a new organism. Totipotent cells formed during sexual and asexual reproduction include spores and zygotes.

Question 4.
(a) Briefly describe the essential components of the nutrient medium used for the plant tissue culture technique. Also, write the names of any two plant tissue culture media frequently used in the laboratory . [4]
(b) With reference to suspension culture, explain the following : [4]
(i) Achemostat.
(ii) A turbidostat.
(c) What are purines and pyrimidines ? Where are they located in a cell? [2]
Answer:
(a) Nutrient Medium : Virtually all tissue culture media are synthetic or chemically defined; only a few of.them use complex organics, e.g., potato extract, as their normal constituents. A synthetic medium consists of only chemically defined compounds. A variety of recipes have been developed since none of them is suitable for either all plant species or for every purpose. Most of these recipes have been elaborated from those of White (itself evolved from a medium for algae) and Gautheret (based on Knop’s salt solution) The composition is as follow :

Inorganic nutrients: In addition to C, H and O, all nutrient media provide the 12 elements essential for plant growth, viz., N, P, K, Ca, S, Mg (these six are called macronutrients, and are needed in concentrations >0.5 mmol L^-1 or > 0.5 mM), Fe, Zn, Mn Cu, B and Mo (these six are known a micronutrients, and are required in concentrations <0.5 mmol L” ’ ) Generally, iron is provided as iron.EDTAcomplex to keep it available at higher (>5.8) pH. Nitrate is superior to ammonium as the sole N source, but use of NH+ checks the drift of pH towards alkalinity.

Vitamins : For optimum callus growth, the following vitamins are required . inositol, thiamine, pyridoxine and nicotinic acid of which thiamine is essential and the rest are promotory. Pantothenic acid is also known to be promotory but is not included in most of the recipes.

Carbon source : Sucrose (20-50 g L) is the most commonly used carbon source for all cultured plant materials, including even green shoots. In some systems, e.g., monocots, glucose may be superior to sucrose. Plant tissue can utilize other sugars like maltose, galactose, lactose, mannose and even starch but these are rarely used.

Growth regulator: The following growth regulators (GRs) are used in plant tissue culture. Auxins, e.g., IAA (indole-3-acetic acid), IBA (indole-3-butyric acid), NAA (napthalene acetic acid), NOA (naphthoxy acetic acid), 2, 4-D (2,4-dichlorophenoxy acetic acid) etc., are commonly used to support cell division and callus growth (especially 2. 4-D), somatic embryo (SE) induction, rooting, etc. Cytokinins like kinetin (furfurylamino purine), BAP (benzylamino purine), zeatin, 2-ip (isopentenyl adenine), TDZ (thidiazuron, a compound having cytokinin activity) are employed to promote cell division, regeneration of shoots often SE induction and to enhance proliferation and growth of axillary buds. Abscisic acid (ABA) promotes SE and shoot bud regeneration in many species are markedly improves SE maturation. Of the over 20 gibberellins known, GA₃ is almost exclusively used. It promotes shoot elongation and SE germination.

Complex organic additives: In earlier studies, complex additives like yeast extract, coconut milk, casein hydrolysate, com milk, malt extract and tomato juice are used to support plant tissue growth. White’s medium, Murashige and Skoog (MS) are the two common media used for plant tissue culture.

(b) (i) Chemostat :A type of cell culture; a component of medium is in a growth limiting concentration; fresh medium is added at regular interv als and equal volume of culture is withdrawn. But in a chemostat, a chosen nutrient is kept in a concentration so that it is depleted very rapidly to become growth limiting, while other nutrients are still in concentrations higher than required. In such a situation, any addition of the growth-limiting nutrient is reflected in cell growth. Chemostats are ideal for the determination of effects of individual nutrients on cell growth and metabolism.

(ii) Turbidostat: A type of suspension culture; when culture reaches a predetermined cell density, a volume of culture is replaced by fresh medium; works well at growth rates close to the maximum. A continuous culturing method where the turbidity of the culture is kept constant by manipulating the rate at which medium is fed. If the turbidity falls, the feed rate is lowered so that growth can restore the turbidity to its start point. If the turbidity rise the feed rate is increased to dilute the turbidity back to its start point.

(c) Purines and pyrimidines are two of the basic units of the nucleic acids. They are found in a DNA and RNA in a cell Purines is a large sized double ring structure. It contains two bases, i.e., adenine and guanine, Pyrimidines are small sized, single ring structures. It contains three type of bases i.e., thymine, cytosine and uracil.

Question 5.
(a) Explain giving an example how recombinant DNA technology can be used for the formation of
the following: [4]
(i) A vaccine.
(ii) A plant with delayed fruit ripening.
(b) What is osmotic pressure ? Explain any one biochemical technique based on osmotic pressure. [4]
(c) What are dextro-rotatory and laevo-rotatory substances? [2]
Answer:
(a) (i) Recombinant vaccines : Vaccine is produced using recombinant DNA technology. A recombinant vaccine contains protein or a gene encoding a protein of a pathogen origin that is immunogenic A gene coding an immunogenic protein from the pathogen, is isolated, cloned and used for vaccine production The vaccines based on recombinant proteins are also called Sub-Unit Vaccines.

Whole protein vaccine : Hepatitis B vaccine is produced from surface antigens of transgenic yeast by r-DNA technology They can also be produced in genetically engineered microbes, cultured animal cells, possibly in insects and plants.

Recombinant – polypeptide vaccines : In some cases, the immunogenic portion of the protein- recombinant polypeptide is used as vaccine e.g., gene encoding B polypeptide (part of cholera enterotoxin – Ab A2 and B polypeptide) has been cloned and the recombinant B polypeptide produced is being used, in combination with inactivated cholera cells, as an oral vaccine in place of conventional injectable cholera vaccine. Immunogenicity of foot and mouth disease virus coat protein is due to its amino acids 114-160 and also 201 -213. They induce antibodies which neutralize the virus and thereby provide protection against the foot and mouth disease.

Live recombinant vaccine : The most advanced and promising approach in which concerned pathogen gene is introduced into the genome of selected viral / bacterial vector which is suitably attenuated and the live microorganisms are used for vaccination. Vaccinia virus appears to be more promising vector.

DNA vaccines: Recently vaccines based on pathogen naked DNA are being developed. Tue various approaches for DNA vaccines are as follow

• injection of pure DNA (or RNA) preparation into muscles
• reimplantation of autologus cells (cells of the individual to be vaccinated) into which the gene has been transferred and
• particle gun delivery of plasmid DNA which contains the gene in an expression casette e.g., skin cells They elicit humoral immune response and are usually shed off in a few days preventing long term persistence modified cells.

(ii) Delayed fruit ripening
A major problem in fruit marketing is the pre-mature ripening and softening dining transport of fruits. Consequently shelf-life of fruit remains short in the market. During ripening, genes encode the enzyme cellulase and polygalacturonase. Therefore, ripening process can be delayed by interfering the expression of these genes. In the U.S.A., a transgenic tomato named FlavrSavr (flavour saver) was produced where ripening is delayed by , lowering polygalacturonase activity.

A plant growth hormone ethylene is produced during fruit ripening and senescence. It is synthesised from S-adenosylmethionine through an intermediate compound 1-aminocyclopropane-l-carboxylic acid (ACC). There is a large number of bacteria that can degrade ACC. Therefore, bacterial gene (for ACC) deaminase associated with ACC degradation was isolated and introduced into tomato. In transgenic tomato, fruit ripening was delayed because it synthesised lower amount of ethylene (due to inhibition in ACC synthesis) than the normal tomatoes. Such tomatoes and other fruits can be transported to a longer distance without spoilage.

(b) The osmotic pressure is defined to be the pressure required to maintain an equilibrium, with no net movement of solvent. Osmotic pressure is a colligative property, meaning that the osmotic pressure depends on the molar concentration of the solute but not on its identity.

In biochemistry, dialysis is the process of separating molecules in solution by the difference in their rates of diffusion through a semi-permeable membrane, such as dialysis tubing. Typically a solution of several types of molecules is placed into a semi-permeable dialysis bag, such as a cellulose membrane with pores, and the bag is sealed. The sealed dialysis bag is placed in a container of a different solution, or pure water. Molecules small enough to pass through the tubing (often water, salt and other small molecules) tend to move into or out of the dialysis bag, in the direction of decreasing concentration. Larger molecules (often proteins, DNA, or polysaccharides) that have dimensions significantly greater than the pore diameter are retained inside the dialysis bag.

(c) Substances which rotate or deviate the plane of polarized light to the right (clockwise) are called dextrorotatory and are indicated by prefixing (+) or d to their names, e.g., d-glucose or (+) glyceraldehydes; Laevorotatory are the substances which rotate or deviate the plane polarized light to the left and are indicated by prefixing L or (–) to their names; e.g., L-alanine or (–) glyceraldehydes.

Question 6.
(a) Write short notes on : [4]
(i) Single nucleotide polymorphism.
(ii) Bioinformatics databases.
(b) How are biomolecules separated by the following techniques: [4]
(i) Chromatography.
() Centrifugation.
(c) Give two differences between enzymes and inorganic catalysts: [2]
Answer:
(a) (i) Single nucleotide polymorphisms, commonly called SNPs (pronounced “snips”), are the most common type of genetic variation in a nucleotide sequence due to change even in a single base among different individuals. Thus, each SNP represents a difference in a single DNA building block, called a nucleotide. For example, a SNP may replace the nucleotide cytosine (C) with the nucleotide thymine (T) in a certain stretch of DNA. In human genome, SNPs occur at 1.6-3.2 million site. Due to the changes in bases SNPs affect the gene function. DNA fingerprinting of individuals is possible due to these genetic variations in non-coding parts of genome.

(ii) Database : A database is an organized collection of data for one or more multiple uses. One way of classify ing databases involves the type of content, for example : bibliographic, full-text, numeric and image.

Biological databases are libraries of life sciences information, collected from scientific experiments, published literature, high-throughput experiment technolog}, and computational analysis. They contain information from research areas including genomics, proteomics, metabolomics, microarray gene expression, and phylogenetics. Information contained in biological databases includes gene function, structure, localization (both cellular and chromosomal), clinical effects of mutations as well as similarities of biological sequences and structures.
Examples:

• ENA (European Nucleotide Archive) – primary nucleotide data, incorporating EMBL- Bank.
• UniProt-protein databases.
• PDB (Protein Data Bank)-biological macromolecular structure.

(b) (i) Chromatography is a technique of separation of biomolecules involves a sampling mixture containing biomolecules being dissolved in a mobile phase (which may be a gas, a liquid or a supercritical fluid) due to their differential adsorption over an adsorbent medium. The mobile phase is then forced through an immobile, immiscible stationary phase. The phases are chosen such that components of the sample mixture have differing solubilities in each phase.

A component which is quite soluble in the stationary phase will take longer to travel through it than a component which is not very’ soluble in the stationary phase but very soluble in the mobile phase. As a result of these differences in mobilities, sample mixture components will become separated from each other as they travel through the stationary’ phase.

Techniques such as H.P.L.C (High Performance Liquid Chromatography) and GC. (Gas Chromatography) use columns—narrow tubes packed with stationary phase, through which the mobile phase is forced. The sample is transported through the column by continuous addition of mobile phase. This process is called elution. The average rate at which an analyte moves through the column is determined by the time it spends in the mobile phase.

(ii) Centrifugation is a process that involves the use of the centrifugal force for the sedimentation of the components of a mixture with a centrifuge. More-dense components of the mixture migrate away from the axis of the centrifuge, while less-dense components of the mixture migrate towards the axis. Chemists and biologists may increase the effective gravitational force on a test tube so as to move rapidly and completely cause the precipitate (‘’pellet”) to gather on the bottom of the tube. The remaining solution is properly called the “supernate” or ‘ supernatant liquid”. The supernatant liquid is then either quickly decanted from the tube without disturbing the precipitate, or withdrawn with a Pasteur pipette. For example, microcentrifuges are used to process small volumes of biological molecules, cells, or nuclei.

(c) Differences between enzyme and inorganic catalyst are mentioned as follows :

Enzymes	Catalysts
(i) Enzymes are complex organic proteins.	(i) Catalysts are simple inorganic molecules.
(ii) Enzymes catalyses specific types of reactions.	(ii) Catalysts have a wide range.

Question 7.
(a) Differentiate between : [4]
(i) Prokaryotic genome and Eukaryotic genome.
(ii) Somatic embryo and Zygotic embryo.
(b) Explain how a genomic DNA library is formed. How does it differ from cDNA library ? [4]
(c) Name any two inborn metabolic disorders in human beings. Also, write one main symptom for each of them.
Answer:
(a) (i) Prokaryotic Genome:

• Genome are much smaller and simpler.
• Highly repetitive DNA is not found.
• It is without a limiting membrane.
• It is a naked double strand of DNA.

Eukarvotic Geflorne:

• Genome are larger and complex.
•  Occurrence of highly repetitive DNA is found.
• It is bounded by nuclear membrane.
• Double strand of DNA is associated with histone proteins.

(ii) Somatic Embryo: Somatic embryos are formed from plant cells other than egg that are not normally involved in the development of embry os, e.g., callus or explant. No endosperm or seed coat is formed around a somatic embryo.

Zygotic Embryo: Zygotic embryo is formed as a result of double fertilization of the ovule, giving rise to two distinct structures; the zygote and primary endosperm nucleus giving rise to the plant embryo and the endosperm which together go on to develop into a seed.

(b) Construction of Genomic Library : The process of subdividing genomic DNA into clonable elements and inserting them into host cells is called creating a library.

A complete library, by definition, contains the entire genomic DNA of the source organism and is called as genomic library. A genomic library is a set of cloned fragments of genomic DNA. The process of creating a genomic library includes four steps :

In the first step the high molecular weight genomic DNA is separated and subjected to restriction enzy me digestion by using two compatible restriction enzymes.

In the second step, the fragments are then fractionated or separated by using agarose gel electrophoresis to obtain fragments of required size.

These fragments are then subjected to alkaline phosphatase treatment to remove the phosphate. In the third step, the dephosphorylated insert is ligated into vector which could be a plasmid, phage or cosmid, depending upon the interest of the researcher.

In the last step, the recombinant vector is introduced into the host by electroporation and amplified in host. In principle, all the DNA from the source organism is inserted into the host but this is not fully possible as some DNA sequences escape the cloning procedure. Genomic library is a source of genes and DNA sequences. A genomic library is a set of cloned fragments of genomic DNA. Prior information about the genome is not required for library construction for most organisms. In principle, the genomic DNA, after the isolation, is subjected to RE enzyme for digestion to
generate inserts.

cDNA libraries V/S Genomic libraries:

• Genomic library is a mixture of fragments of genomic DNA while cDNA obtained from mRNA may cloned to give rise to a cDNA library.Genomic library’ contains DNA fragments that represent genes as well as those that are not genes. In contrast cDNA library contains only those genes that are expressed in the concerned tissue/organism. In both cases, a mixture of fragments is used for cloning to establish the library.
• Use of cDNA is absolutely essential when the expression of an eukaryotic gene is required in a prokaryote.
• Eukaryotic cDNAs are free from intron sequences.
• As a result of the above, they are smaller in size than the corresponding genes, i.e., the genes that encoded them.
• A comparison of the cDNA sequence with the corresponding genome sequence permits the delineation of intron/exon boundaries.
• The contents of cDNA libraries from a single organism will vary widely depending on the developmental stage and the cell type used for preparation of the library. In contrast, the genomic libraries will remain essentially the same irrespective of the developmental stage and the cell type used.
• A cDNA library will be enriched for abundant mRNAs, but may contain only a few or no clones representing rare mRNAs.

(c) Alkaptonuria : This was one of the first metabolic diseases described by Garrod in 1908. It is an inherited metabolic disorder produced due to deficiency of an oxidase enzyme requiredfor breakdown ofhomogentisic acid (also called alcapton, hence, alkaptonuria is also written as alcaptonuria). Lack of the enzyme is due to the absence of the normal form of gene that controls the synthesis of the enzyme. Hence, homogentisic acid then accumulates in the tissues and is also excreted in the urine. The most commonly affected tissues are cartilages, capsules of joints, ligaments and tendons. The urine of these patients if allowed to stand for some hours in air, turns black due to oxidation of homogentisic acid.

Phenylketonuria (PKU; Foiling. 1934): It is an inborn metabolic disorder in which the homozygous recessive individual lacks the enzyme phenylalanine hydroxylase needed to change phenylalanine (amino acid) to tyrosine (amino acid). Thus, the biochemical abnormality in PKU is an inability to convert phenylalanine into tyrosine leading to hyperphenylalaninemia. Lack of the enzyme is due to the abnormal autosomal recessive gene on chromosome 12. This defective gene is due to substitution. Affected babies are normal at birth but within a few weeks there is rise (30-50 times) in plasma phenylalanine level which impairs brain development. Usually by six months of life severe mental retardation becomes evident.

If these children are not treated about one-third of these children are unable to walk and two-thirds cannot talk. Other symptoms are mental retardation, decreased pigmentation of hair and skin and eczema. Although large amounts of phenylalanine and its metabolites are excreted in the urine and sweat, yet it is believed that excess phenylalanine or its metabolites contribute to the brain damage in PKU. The heterozygous individuals are normal but carriers. It occurs in about 1 in 18000 births among white Europeans. It is very rare in other races.

Question 8.
(a) Give the step-wise procedure of Southern blotting technique. Mention any two important applications of this technique. [4]
(b) What are blunt ends and sticky ends ? How are they formed ? [4]
(c) Name any two industrial enzymes and give their uses. [2]
Answer:
(a) Southern blotting technique : In 1975, Edward M. Southern developed technique of DNA separation and its hybridization. Therefore, in his honor this technique is known as ‘ Southern blotting or Southern hybridization technique’. A specific DNA fragment can be separated and identified in a heterologous population of DNA molecules on the basis of binding of DNA probe with its complementary DNA strand.

The genomic DNA is isolated from the clone and digested with restriction enzymes. The DNA fragments are separated by agarose gel electrophoresis (Fig.). Different DNA bands are formed on agarose gel which represent DNA fragments of varying sizes. These fragments are transferred from gel to nylon or nitrocellulose membrane. The process of DNA transfer is called ‘blotting’.

A nitrocellulose membrane is put over the gel. Many layers of filter paper are placed over nitrocellulose membrane. This assembly is put in a container having NaOH solution. NaOH denatures DNA and results in formation of single stranded DNA. DNA fragments are transferred from gel to membrane by capillary action.

In addition, DNA fragments can also be transferred by vacuum blotting and centrifugation. The DNA fragments are fixed to membrane by using UV radiation or baking at 80°C. The pattern of DNA bands on membrane corresponds to the position of DNA on gel. The membrane is put in solution containing radio labelled DNA probe and incubated for some time. DNA probe hybridizes complementary DNA fragments fixed on membrane. It is gently washed at 12°C and dried.

The membrane is exposed through a photographic film. DNA bands formed on photographic film corresponds to the original position of DNA fragments present on agarose gel.

(b) Restriction enzymes are bacterial proteins that have the ability, to cut both the strands of the DNA molecule at a specific nucleotide sequence. There are hundreds of these restriction enzymes and each can cut the DNA at a specific point and the resulting DNA fragments are all of different lengths. Restriction enzymes can either produce sticky ends or blunt ends.

EcoRI enzyme binds to a region of having specific palindromic’ sequence (where two strands are identical when both are read in the same polarity i.e., in 5’→ 3’ direction). The length of this region is 6 base pairs i.e., hexanucleotide palindrome. It cuts between G and A residues of each strand and produces two single stranded complementary cut ends which are asymmetrical having 5’ overhangs of 4 nucleotides. These ends are called sticky ends or cohesive ends. Because nucleotide bases of this region can pair and stick the DNA fragments again as given below :}

On the other hand, there are some other Type II restriction enzymes which cleave both strands of DNA at the same base pairs but in the centre of recognition sequence, and results in DNA fragments with blunt ends or flush ends. For example Hae111 (isolated from Haemophilus aegypticus, the order of enzyme III), four nucleotide long palindromic sequence and cuts symmetrically the both DNA strands and forms blunt ends as below:

(c) Alpha Amylase : It is widely used enzyme in Food industry and in Laundry detergent.
Papain : It is used in Medicine, Food and Textile industries.

Question 9.
(a) Enlist the main steps in the regeneration of a complete plant from an explant. [4]
(b) Given below is a list of four bio molecules found in a living cell. For each of them, write the class of biomolecules they belong to and their location in a living cell: [4]
(i) Histones.
(iii) Haemoglobin.
(ii) mRNA.
(iv) Glycogen.
(c) Write any two uses of transgenic plants. [2]
Answer:
(a) Steps involved in in vitro regeneration of a complete plants. Regeneration refers to the development of organised structures like roots, shoots, flower buds, somatic embryos (SEs), etc. from cultured cells/tissues; the term organogenesis is also used to describe these events. Root regeneration occurs quite frequently but it is useful only in case of shoots and embryo germination. Only shoot and SE regenerations give rise to complete plants which is essential for applications of tissue culture technology in agriculture and horticulture. Regeneration may occur either directly from the explant or may follow an intervening callus phase.

Basic technique of Plant Tissue Culture: The basic technique of plant tissue culture involves the following steps:

• Preparation and sterilisation of suitable nutrient medium : Suitable nutrient medium as per objective of culture is prepared and transferred into suitable containers. Culture- medium is rich in sucrose, minerals, vitamins, and hormones. Yeast extract, coconut milk are also added. The culture is completely sterilized in an autoclave.
• Selection of explants : Selection of explants such as shoot tip should be done.
• Sterilisation of explants: Surface sterilisation of the explants by disinfectants (e.g., sodium hypochlorite or mercuric chloride) and then washing the explants with sterile distilled water is essential.
• Inoculation: Inoculation (transfer) of the explants into the suitable nutrient medium (which is sterilised by autoclaving to avoid microbial contamination) in culture vessels under sterile conditions is done.
• Incubations : Growing the culture in the growth chamber or plant tissue culture room, having the appropriate physical condition (i.e., artificial light: 16 hours of photoperiod), temperature (- 26°C) and relative humidity (50 – 60%) is required.
• Regeneration : An unorganized mass of cells developing from explants is called callus. The callus gives rise to embryoids which can develop into whole plant if the medium is provided with proper concentration of hormones. This property’ of developing every somatic cell into a Ml fledged plant is called totipotenev. Regeneration of plant, from cultured plant tissues is carried out.
• Hardening: Hardening is gradual exposure of plantlets to an environmental conditions.
• Plantlet transfer : After hardening, plantlets are transferred to the greenhouse or field conditions following acclimatization (hardening) of regenerated plants.

ISC Class 12 Biotechnology Previous Year Question Papers

ISC Economics Previous Year Question Paper 2012 Solved for Class 12

Maximum Marks: 80
Time allowed: 3 hours

• Candidates are allowed additional 15 minutes for only reading the paper.
• They must NOT start writing during this time.
• Answer Question 1 (Compulsory) from Part I and five questions from Part II.
• The intended marks for questions or parts of questions are given in brackets [ ].

Part – I (20 Marks)
Answer All Questions

Question 1.
Answer briefly each of the questions (i) to (xv).
(i) Name and explain the two main branches of economics.
(ii) State the law of equi marginal utility.
(iii) Explain with an example, what kind of a commodity will have an inverse relationship between income and demand.
(iv) Explain the meaning of indivisibility of a factor with an example.
(v) What will be the price elasticity of demand of the points A, B, L and K in the diagram given below:

(vi) State what causes a movement along the supply curve and show it diagrammatically.
(vii) Define marginal cost. With the help of an example, show how marginal cost can be obtained from the total cost.
(viii) Give the modem definition of economic rent.
(ix) Which revenue concept is also called price? Justify your answer by giving a reason.
(x) Distinguish between national income at current prices and national income at constant prices.
(xi) When does the equilibrium quantity in a market remain unchanged with a change in demand? Show it with the help of a diagram.
(xii) What is the significance of freedom of entry and exit of firms under perfect competition?
(xiii) Give one difference between a flexible exchange rate and a fixed exchange rate.
(xiv) What is meant by zero-base budget?
(xv) Explain two merits of direct tax.
Answer:
(ii) Law of Equi-marginal utility states that the consumer in order to maximize his satisfaction should spend his money on two goods in such a manner that the ratio of marginal utility of a commodity to its price becomes equal to the ratio of marginal utility of other commodities to its price. Symbolically,
MU_x / P_x = MU_y/P_y = MU_n/P_n

(iii) Inferior goods have an inverse relationship between their demand and income. For example, the demand for an inferior good like maize may decrease when income increases beyond a particular level because the consumers may substitute it by a superior-good like wheat or rice.

(v) At point ‘A’, the elasticity of demand will be,
The lower segment of the line/upper segment of line = AK/0 = ∞
At point ‘B’ = BK/AB > 1 i.e. e_d > 1
At point ‘L’ = LK /AK < 1 i.e. e_d < 1
At point ‘K’ = 0/AK = 0 i.e. e_d = 0

(vi) Movement along the supply curve is caused due to change in the price of the commodity keeping other factors constant. If the price of that commodity rises, its quantity supply will also rise causing upward movement along the supply curve (i.e. extension of supply) as shown in fig. A. On the other hand, if the price of the commodity falls showing downward movement along the supply curve (i.e. contraction of supply) indicated by downward arrow in fig.B.

(vii) Marginal cost (MC) is an addition made to the total cost (TC) when output is change by one unit i.e.,
MC = ΔTC/Q
MC = TC_n – TC_n-1
For example, the marginal cost of 4th unit is the changed in the total cost when output is increased from three units to four units (i.e. 192 – 162 = 30), as shown in the table given below.

Output Units	TC (₹)	MC
1	20
2	30	30 – 20 = 10
3	50	50 – 30 = 20
4	80	80 – 50 = 30

(ix) Average revenue is revenue earned per unit of the product sold.
\(\mathrm{AR}=\frac{\mathrm{TR}}{\mathrm{Q}}\)
But TR = P × Q
\(\mathrm{AR}=\frac{\mathrm{P} \times \mathrm{Q}}{\mathrm{Q}}=\mathrm{P}\)

(x)

National Income at Current Price (Nominal GDP)	National Price at Constant Price (Real GDP)
Under this GDP is calculated at current prices prevailing in the market for example if we measure India’s National Income of at 2011-12 at the same year’s prices than it is national income at the current price.	Under this GDP is calculated at a base year price. For example, if we measure India’s National income of 2009-10 at 2001-2002 prices than it is national income at a constant price.
This may give a misleading picture of the economic growth of a country because of an increase in National Income maybe because of the increase in price rather than any physical output goods and services.	On the other hand, this gives true picture of economic growth of a country as it is affected by the change in only the physical quantities.
National Income at current price = P1 × Q1 Where P1 = Current Price and Q1 = Current Quantity	National Income at Constant Price = P0 × Q1 Where P0 = Base year Price and Q1 = Current Quantity

(xi) The equilibrium quantity remains unchanged with a change in demand.
When supply curve is perfectly inelastic: When supply curve is perfectly inelastic, a change in demand (increase or decrease) brings about a change in equilibrium price, but the equilibrium quantity remains the same as illustrated in fig.

(xii) Freedom of entry and exit under perfect competition means that new firms are free to enter the industry and existing firms are free to leave the industry if they desire so. This condition ensures that all firms under Perfect competition end up earning only normal profits in the long run. The entry of new firms will increase the total supply by the industry, thus reducing the market price and wiping out supernormal profits. On the other hand, if existing firms are incurring losses, some of them would start leaving the industry, leading to a decrease in supply and a rise in price until the losses are wiped out.

(xiii) The fixed exchange rate is a rate that is fixed and determined by the government of a country and only the government can change it. It is independent of free-market forces of demand and supply. Whereas the flexible exchange rate is that rate which is determined by the demand and supply of different currencies in the foreign exchange market. The government does not intervene in the fixation of the exchange rate.

Part – II
(Answer Any Five Questions)

Question 2.
(a) Calculate the quantity demanded of a commodity when its price increases from ₹ 4 to ₹ 6. The original quantity demanded was 40 units and the price elasticity of demand is 0.5.
(b) Explain how the following phenomena are exceptions to the Law of Demand:
(i) Expectations regarding future prices.
(ii) Conspicuous consumption by a consumer.
(c) Discuss four factors other than price, that affect the demand of a commodity.
Answer:

(b) Expectations regarding future prices: If the price of a commodity is rising today and it is likely to rise more in the future, people will buy more even at the existing price and store it up. They will do this in order to avoid the pinch of higher prices in the future. Similarly, when large fall in the price of a commodity is anticipated, consumers will postpone their purchase even if the prices fall today so as to purchase this commodity at a still lower price in future.

Conspicuous consumption: The law of demand does not apply to status symbol commodities termed as Veblen. These goods are demanded more at higher prices by (rich) consumers to increase their social prestige or as a source of display of wealth or richness. For example, diamond.

(c) Four factors affecting demand are as follows:
Consumers tastes and Preferences: Tastes and preferences depend on social customs, fashion, habits of people etc. which keeps on changing leading to change in customers’ taste and preferences. As a result, the demand for different goods changes. For example, consumers may switch over from cheaper old fashioned goods to costlier ‘mod’ goods.
The income of the consumer: Generally when the income of a consumer goes up, the demand . for a commodity also goes up and when income falls, the demand also falls in case of normal goods.

Price of related goods (i.e. of substitute goods and complementary goods): In the case of complementary goods like car and petrol, the demand for a commodity rises with a fall in the price of complementary good. In the case of substitute goods like tea and coffee, demand for a commodity falls with a fall in the price of other substitutes good.

The expectation of Change in the Price in Future: If the price of a certain commodity is expected to increase in the near future, then people will buy more of that commodity than what they normally buy. There exists a direct relationship between the expectation of change in the prices in future and change in demand in the current period. For example, if the price of petrol is expected to rise in future, its present demand will increase.

Question 3.
(a) Define price elasticity of supply. Draw diagrams when price elasticity of supply is;
(i) Equal to one.
(ii) Greater than one.
(b) Differentiate between returns to variable factor and returns to scale.
(c) Explain with the help of a diagram, the relationship between total product and marginal product.
Answer:
(a) Price elasticity refers to the degree of responsiveness of quantity supplied of a commodity to change in its price. It is calculated as, es = % change in quantity supplied/ % change in the price of a commodity.

(b) Difference between returns to variable factor and returns to scale are as follows:

• The law of Returns to variable factor means the change in physical output when the quantity of one factor is changed, other inputs remain fixed. Returns to scale on the other hand means a change in the physical output when the quantity of all the factors is increased simultaneously at the same proportion.
• The law of variable factors studies the effect of change in one input on the output, whereas Returns to scale studies the effect of change in all inputs on the output.
• The law of variable proportion or Return to a factor is a short-run phenomenon whereas ‘the returns to scale’ is a long-run phenomenon.
• Returns to variable factor take account of the change in factor proportions whereas Returns to scale takes factor proportions to be unchanged.

(c) (i) When MP increases, TP increases at an increasing rate.
(ii) When MP is constant, TP increases at constant rate.

(iii) When MP decreases, TP increases at a diminishing rate
(iv) When MP is zero, TP is maximum.
(v) When MP is negative, TP declines.

Question 4.
(a) Explain diagrammatically how equilibrium price and equilibrium quantity are affected by changes in the demand for a commodity, with the supply remaining constant.
(b) Define the production function. Discuss two criticisms of the Law of Variable Proportions.
(c) How does a perfectly competitive firm earn supernormal profits in the short-run equilibrium? Explain it with the help of a diagram.
Answer:
(a) When demand increases:
An increase in demand is a situation under which demand curve shifts to the right due to other factors i.e., increase in the price of substitutes, decrease in price of complementary goods, increase in income (normal good) etc.

(i) DD is the initial demand curve, crossing the supply curve at point E, the point of initial equilibrium.
(ii) Due to increase in demand, demand curve shifts to the right, from DD to D₁D₁. And at the existing price (OP₁), quantity demanded rises from point E to point F.
(iii) As an immediate impact of the increase in demand, there is excess demand in the market. It is EF (at the existing price).
(iv) Due to the pressure of demand, price of the commodity tends to be higher than the equilibrium price.
(v) Due to the rising price, quantity demanded tends to contract. Contraction of demand occurs from point F towards point K.
(vi) Due to the rising price, quantity supplied tends to extend. The extension of supply occurs from point E towards point K.
The process of extension of supply and contraction of demand continue till new equilibrium K is established, which is at a higher price and higher quantity demanded.

When demand decreases:
A decrease in demand is the situation when the demand curve shifts to the left due to other factors. i.e., decrease in the price of substitute increase in price of complementary goods, a decrease in income etc.
(i) DD is the initial demand curve, crossing the supply curve at point E, the point of initial equilibrium.
(ii) Due to a decrease in demand, demand curve shifts to the left, from DD to D₁D₁. At the existing price (OP₁), quality demanded falls from point E to point F.
(iii) As an immediate impact of the decrease in demand, there is excess supply in the market. It is EF (at the existing price).
(iv) Due to excess supply price of the commodity tends to be lower than the equilibrium price.
(v) Due to lowering price, quantity demanded tends to extend. Extension of demand occurs from point F towards point K.
(vi) Due to lowering price, quantity supply tends to contract. The contraction of supply occurs from point E towards point K.
(vii) The process of extension of demand and contraction of supply continues till new equilibrium K is established which is at the lower price and lower quantity demanded.

(b) A production function shows the maximum quantity of a commodity that can be produced per unit of time with the given amount of inputs when the best production technique available is used.
Two criticism of law of variable proportion is:
Homogeneity of Variable factors: The assumption of homogeneous or equally efficient variable factors is criticized as it is not practical.
Unchanged technologies: Law of variable proportions assumes technology to be given and remain unchanged which makes it unrealistic.

(c) In short-run, a perfectly competitive firm can earn supernormal profits, normal profits and can sustain loss also. A perfectly competitive firm earning SNP is shown alongside in the diagram.

OR shows the AR curve of the firm which is equal to MR. The firm produces equilibrium output OQ at point E where MC cuts MR from below. For OQ output Average Cost (AQ) is QC’ which is less than AR for OQ output which shows that firm earns super-normal profits equal to the shaded area or area of rectangle RECC’ (Total profits). Average profits will be equal to AR-AC i.e., C’Q – EQ = EC’.

Question 5.
(a) Distinguish between monopoly and perfect competition on the basis of:
(i) AR curve.
(ii) Control over the market price.
(b) Explain the following concepts:
(i) Gross profit.
(ii) Transfer Earning of a factor.
(c) Define economic cost. Explain the relationship between total cost, total fixed cost and total variable cost with the help of a diagram.
Answer:
(a) Average Revenue curve: In monopoly competition, AR curve i.e. Demand curve is inelastic so it is a negatively sloping curve, whereas in Perfect competition as AR curve i.e. Demand curve is perfectly elastic to it is a horizontal line parallel to X-axis as shown below.
Control over market price: In monopoly competition, a monopoly firm has full control over the prices of the commodity. A monopolist is Price maker whereas, in Perfect competition, a firm can’t have any influence on the market price, it is considered price taker only.

(c) The economic cost is the sum total of both explicit and implicit cost including normal profits. Economic cost = Explicit cost + Implicit cost (including – normal profits)
Total cost refers to total obligations incurred by the firm in producing any given quantity of output. It is the sum total of all expenditure (on using both fixed and variable factors) incurred in producing a given quantity of output. In short-run total cost comprises of two parts:
(a) Total fixed cost
(b) Total variable cost
i.e., TC = TFC + TVC

Total Fixed Cost: TFC refers to the total cost incurred by the firm on the use of all fixed factors. TFC does not change with the change in output in the short-run production function. It remains the same regardless of the quantity of output. It never becomes zero even at zero levels of output. It is calculated as:
TFC = TC – TVC
where TC = Total cost
TVC = Total variable cost

Total Variable Cost: TVC refers to the total cost incurred by a firm on the use of variable factors. This cost includes payments for raw material, wages, for fuel power etc. These costs vary directly with the level of output, rising as more is produced and falling as less is produced.
TVC = TC – TFC

In the given diagram TC curve is obtained by adding up vertically the TFC curve and TVC curve because TC is the sum of TFC and TVC at every level of output. Since a constant FC is added to the TVC, the shape of the TC curve is same as that of TVC curve. The vertical difference between the TVC curve and the TC curve is the same at levels of output because of TFC. TVC curve starts from zero then increases at a decreasing rate first and at an increasing-rate subsequently with an increase in total output.

Question 6.
(a) Discuss two differences between intermediate goods and final goods.
(b) Distinguish between:
(i) Gross Domestic Product at market price and Net National Product at factor cost.
(ii) Personal income and personal disposable income.
(c) Calculate National Income and Net Domestic Product at market price by Income method from the following data:

(i) Value of output	800 Crores
(ii) Value of intermediate consumption	400 Crores
(iii) Subsidies	10 Crores
(iv) Indirect taxes	60 Crores
(v) Factor income received from abroad	10 Crores
(vi) Factor income paid abroad	20 Crores
(vii) Mixed-income of self-employed	120 Crores
(viii) Rent and royalty	40 Crores
(ix) Interest and profit	20 Crores
(x) Wages and salaries	110 Crores
(xi) Consumption of fixed capital	50 Crores
(xii) Employer’s contribution to social security	10 Crores

Answer:
(b) (i) GDP_mp is the money value of all final goods and services produced in the domestic territory of a country in a year. It includes consumption of fixed capital and indirect taxes. It doesn’t include Net factor Income from abroad whereas NNPfc is the sum total of factor income earned by the normal residents of a country in a year. It excludes depreciation but includes net factor income from abroad. It doesn’t include net direct taxes.

(ii) GDP_mp = NNP_fc – NFIA + Depreciation + Net indirect tax
NNP_fc = GDP_mp – depreciation + NFIA – Net indirect tax
Following is the difference between Personal income and Personal Disposable Income (PDI).
Personal income is the income actually received by persons from all sources in the form of current transfer payments and factor incomes whereas PDI is the personal income remaining with individuals after paying direct personal taxes and other fees and fines to the government.

(c) National Income
NNP_fc = Compensation to employees + Operating surplus + Mixed-income + NFIA.
NNPfc = (x + xii) + (viii + ix) + vii + (v – vi)
= (10 + 110) + (40 + 20) + 120 + (-10)
= ₹ 290 crore.
NDP_mp = NNP_fc – NFLA + Net Indirect Taxes
= 290 – (10 – 20) + (60 – 10)
= 290 + 10 + 50
= ₹ 350 crore.

Question 7.
(a) Explain how the expenditure of the Indian government has risen with reference to:
(i) Increase in developmental activities.
(ii) Increase in population.
(b) How does the fiscal policy of the government control inflation with the following tools:
(i) Public Expenditure.
(ii) Taxation.
(c) Discuss four reasons for the internal borrowing by the government.

Question 8.
(a) Mention two merits and two demerits of international trade. Explain any one merit and anyone demerit of international trade.
(b) How can the government correct an adverse balance of payments through the following measures:
(i) Export promotion.
(ii) Import control.
(c) Explain the Absolute Cost Advantage Theory of international trade with an example.
Answer:
(b) The government can correct the adverse Balance of Payment through the following measures.
Export Promotions: The Government of the country having adverse BOP stimulate exports by reducing export duties, giving subsidies and cash assistance to exporters, providing technical and marketing assistance to export-oriented units, providing facilities like quality control, arranging exhibitions of exportable goods, exempting export goods from taxes, etc. Attempts should be made to attract foreign tourist to encourage tourism. All these measures will increase the volume of exports thereby reducing the deficit in the BOP.
Import Control: To check imports, the Government can impose quota limits may increase import duties or tariffs. This will discourage imports, making BOP favourable.

Question 9.
(a) The following table shows the marginal utility derived from the purchase of books. The price of the book is ₹ 500. Draw a diagram to explain consumer’s equilibrium where MU = P.

Number of Books	M.U.
1	700
2	600
3	500
4	400
5	300

(b) How is the elasticity of demand for a commodity affected by the following factors:
(i) Existence of substitutes of a commodity.
(ii) Nature of a commodity.
(c) The cost function of a firm is given below:

Output	0	1	2	3	4
Total Cost(₹)	60	80	100	111	116

Find:
(i) Total Fixed Cost.
(ii) Total Variable Cost.
(iii) Average Fixed Cost.
(iv) Average Variable Cost.
(v) Marginal Cost.
Answer:
(a) Diagram showing consumer equilibrium where MU_x = P_x is as below. In the diagram number of books consumed are taken on OX axis and MU and Price of the books plotted on Y-axis. As the table shows, with every additional unit of book consumed, MU derived is falling.

According to the consumer equilibrium condition, the consumer can maximize his satisfaction by consuming the units of a given commodity where MU_x = P_x. i.e Marginal utility derived from the consumption of a commodity is equal to the price of that commodity. In the diagram, the equilibrium situation is shown at point E, where prices (₹ 500) is equal to the Marginal utility derived (500) when the consumer consumes 3 units of Book.

(b) The elasticity of demand is affected by many factors two of them are given below:
Existence of substitutes: A commodity will have elastic demand if there are many substitutes available of the commodity, e.g, Pepsi and coca-cola, fruit. A commodity having no substitutes e.g., the salt will have inelastic demand.
Nature of the commodity: Generally the demand for necessities is inelastic and the demand for the luxuries is elastic. This is so because certain goods are essential to life which will be demanded at any price, whereas the goods meant for luxuries can be dispersed easily if they appear to be costly.

(c)

ISC Class 12 Economics Previous Year Question Papers

ISC Economics Previous Year Question Paper 2016 Solved for Class 12

Maximum Marks: 80
Time allowed: 3 hours

• Candidates are allowed additional 15 minutes for only reading the paper.
• They must NOT start writing during this time.
• Answer Question 1 (Compulsory) from Part I and five questions from Part II.
• The intended marks for questions or parts of questions are given in brackets [ ].

Part – I (20 Marks)
Answer all questions.

Question 1.
Answer briefly each of the following questions (i) to (x): [10 × 2]
(i) What is meant by income elasticity of demand?
(ii) Technical advancement leads to cost-saving. With the help of a diagram, explain the effect of technical advancement on the supply curve.
(iii) Explain the meaning of M₁ and M₄ supply of money.
(iv) With the help of a diagram, show how the equilibrium price can remain unchanged even after a rightward shift of the demand curve.
(v) What is meant by the fiscal policy? Name any two instruments of fiscal policy.
(vi) Give two differences between Current Account and Capital Account of Balance of Payment Account.
(vii) What is meant by budget line?
(viii) State whether each of the items given below is included in estimating National Income. Give a reason in each case to justify your answer.
(a) Expenditure on the construction of Express Highway.
(b) Expenditure on the purchase of an old house.
(ix) If the value of MPC is 0-9, find the value of the multiplier.
(x) Explain the meaning of social cost with the help of an example.
Answers:
(i) Income elasticity of demand measures the degree of responsiveness of quantity demanded of a commodity to changes in income of the consumers, other things remaining constant.

(ii) Technical advancement reduces the cost of production and increases profit margin, thereby inducing the producers to produce more and increase the supply, at a given price.

Thus, the supply curve shifts rightward from SS to S₁S₁ due to technical advancement. Quantity supplied increases by the amount Q₁Q₂.

(iii) The supply of money consists of various components. M₁ and M₄ are also components of the money supply.
M₁ = Currency notes and coins with the public (C) + Demand deposits with the banks (DD) + Other deposits with the RBI (OD)
Thus. M₁ = C + DD + OD
and M₄ = M₁ + Saving deposit with post office saving bank (SD) + Total deposits with post office savings organisation (TDP)
Thus, M₄ = M₂ + TDP
where M₂ = M₁ + SD.

(iv) A rightward shift in the demand curve, given the supply curve, the equilibrium price generally increases. However, if the supply curve is horizontal (a perfectly elastic supply of a commodity), then the equilibrium price remains unchanged even after a rightward shift of the demand curve.

The demand curve shifts from D₀D₀ to D₁D₁ but the equilibrium price is OP₀.

(v) Fiscal Policy is defined as the policy under which the government uses the instruments of taxation, public spending and public borrowing to achieve various objectives of economic development, like economic stability, high employment and accelerating economic growth. Two instruments of fiscal policy are:

• Taxation policy
• Public expenditure policy
• Public debt policy.

(vi)

Current Account	Capital Account
(i) It records transactions relating to export and import of goods and services, unilateral transfers and international incomes.	(i) It records transactions relating to the change in assets – both financial and physical.
(ii) These do not change the assets and future liabilities of the country.	(ii) These increases or decreases the assets and future liabilities of the country.
(iii) It is the value of exports minus value of imports, adjusted for international incomes and net transfers.	(iii) It is the record of short term and long term capital transactions, both private and official.

(vii) A budget line is a set of various combinations of two commodities (say, x and y) which a consumer can purchase given his money income and the prices of the two commodities. Here, AB is the budget line.

(viii) (a) It is included. This is because it is a part of the final investment expenditure.
(b) It is not included. This is because National income measures the value of currently produced goods and services Purchase of an old house is already included during the first rime.

(ix) MPC = 0.9.
We know,
MPC + MPS = 1
and Multiplier K = \(\frac { 1 }{ MPS }\)
MPS = 1 – 0.9 = 0.1
and K = \(\frac { 1 }{ 0.1 }\) = 10

(x) Social cost refers to the cost that the society has to bear on account of the production of a commodity. For example, a manufacturing firm produces chemical products, may pour industrial wastes in river water, without proper treatment, thus, saving private costs, but polluting the river water. So, the people taking bath in that river, or fishermen catching fishes from that river, would suffer. This is the social cost of producing chemical products. Several other examples are possible.

Part-II (60 Marks)
Answer any five questions.

Question 2.
(a) Explain any two reasons for the upward slope of the supply curve. [3]
(b) Explain the shape of the following: [3]
(i) Total fixed cost curve.
(ii) Total variable cost curve.
(c) Explain the concept of consumer’s equilibrium with the help of indifference curve analysis. [6]
Answers:
(a) The upward slope of the supply curve can be explained as:
(i) Price level determines the number of profits: Higher the price, larger is the profit that can be earned, ceteris paribus. This gives producers’ incentives to produce more and offer for sale, that is, supply in the market.

(ii) The marginal cost of production: It increases because of the law of diminishing returns, that is, increasing cost with an increase in production. Hence, the producer will be prepared to produce and supply more only at a higher price, so as to cover the higher cost of production.

(iii) An increasing cost industry: In the long run, higher prices may attract new firms in an industry. However, input costs may also increase. So, the new firms would be willing to supply more only at higher prices to cover higher costs of production. So, at higher prices, more firms produce well.

(b) (i) Total fixed costs: Costs involved in employing fixed factors are fixed costs. Rent on land, interest payment on loan capital, salaries paid to permanent employees, etc. are fixed costs. It remains constant at all output levels, even if production is zero.

TFC curve becomes parallel to the output axis because TFC is constant at all levels of Q.

(ii) Total variable costs: Costs incurred by a firm on the use of variable factors are variable costs, like wage cost, cost of raw materials, payment for fuel and power used, etc. It is zero at zero output level.

TVC curve looks like an ‘inverted S’ because it rises at a slow pace, initially and then increases at an increasing rate. This is due to the law of variable proportion.

(c) A consumer attains his equilibrium when he maximises his total utility, given his income and prices of two commodities. We draw the indifference curves and the budget line where the consumer maximises his total utility, subject to budget constraint.

The consumer cannot purchase any combination which lies to the right of budget line AB. Such as combination ‘D’, because it is beyond his budget. However, any combination inside the budget line, say C, is within his reach. But any combination will not give him maximum total utility. Point E is a point, both, on the budget line and on a higher indifference curve (than C or any other point).

At the equilibrium point, the budget line of the consumer should be tangent to a particular indifference curve (here, E). Again, the tangent is the marginal rate of substitution. This is the condition for equilibrium.
The slope of budget line = Slope of the indifference curve
\(\frac{P_{x}}{P_{y}}=M R S_{X Y}\)
Hence, E is the consumer’s equilibrium point with AB as the budget line.

Question 3.
(a) The quantity demanded of a commodity at a price of ₹ 10 per unit is 40 units. Its price elasticity of demand is -2. The price falls by ₹ 2 per unit. Calculate the quantity demanded at the new price. [3]
(b) Discuss two differences between cardinal utility and ordinal utility. [3]
(c) Explain how equilibrium price can be determined with the help of [6]
(i) Demand and supply schedule.
(ii) Demand and supply curves.
Answers:
(a) Given P₀ = ₹ 10
Q₀ = 40
E_d = -2
∆P = – ₹ 2
Therefore, P₁ = ₹ 8
We know,

According to the law of demand, quantity demanded falls with a rise in price and vice-versa, ceteris paribus.
Q₁ = Q₀ + ∆Q = 40 + 16 = 56 units.
Quantity demanded at the new price is 56 units.

(b) Two differences between cardinal and ordinal utility approach:
(i) Cardinal theory expresses utility in quantitative terms as according to this theory, utility is measurable. While the ordinal theory is of the view that utility at best can be ranked or ordered according to preference scales and is thus a qualitative approach.

(ii) Cardinal theory assumes that the marginal utility of money is constant (same whatever is the money income of the consumer). In ordinal theory, it is not necessary to assume constant marginal utility of money.

(ii) Cardinal theory explains price effect only through the substitution effect, ignoring income effect by assuming marginal utility of money to be constant whereas, ordinal theory divides price effect into substitution effect and income effect.

(iv) Cardinal theory fails to explain the case for exceptional demand curves where the demand curve is positively sloped (given good is such an example). However, the ordinal theory is able to explain such demand curves with the help of income effect and the substitution effect.

(v) The assumption of independent utility given up in cardinal theory, it has completely ignored the relation of substitution and complementarity between commodities. This flow is not present in the ordinal theory approach.

(c) (i) A demand schedule is a tabular presentation of combinations of prices and different levels of quantities demanded at those prices while a supply schedule is a tabular presentation of combinations of prices and different levels of quantities supplied at those prices. The table below shows the demand schedule and supply schedule.

Price (per unit)	Quantity Demanded	Quantity Supplied	Market Position
500	30	57	Excess supply
400	40	50	Excess supply
300	45	45	Equilibrium
200	55	35	Excess demand
100	70	20	Excess demand

Other illustrations of such kind are also possible.
Market demand is inversely related to price while market supply is directly related to price. The schedules combined together shows the operation of the law of demand and law of supply.

The equilibrium price is the price at which the consumers are willing to purchase the same quantity which the producers are willing to sell. In the given illustration, at price 500, there is excess supply (by the amount 27 units). As we move down the schedule, we find, at price 200 and below, we have excess demand (by the amount 20 units and 50 units). At price 300, quantity demanded equals quantity supplied and hence, it is the equilibrium price.

When the market is not in equilibrium (say at price 500), producers sell the commodity at a lower price so as to attract buyers, when there is excess supply and hence the market moves towards equilibrium. Similarly, in case of excess demand (say at price 100), the buyers compete against each other for the limited product, leading to a rise in prices. Thus, when the market is in disequilibrium, the actual market price will tend to move and reach equilibrium.

(ii) A demand curve is the graphical presentation of the demand schedule showing various price quantity combinations. While the supply curve is the graphical presentation of the supply schedule. The following diagram combines the market demand crave and the market supply curve.

The equilibrium point is E where the demand curve intersects the supply curve, corresponding to which is the equilibrium price (here P₀).
An excess supply (higher price) reduces price, thus moving towards equilibrium while an excess demand (lower price) leads to a rise in price. Finally, the market always attains equilibrium.

Question 4.
(a) Classify the following into fixed cost and variable cost giving a reason for your answer:
(i) Expenses incurred on raw material
(ii) Interest on capital
(iii) Salaries to permanent employees [3]
(b) Draw the TR and AR curves under perfect competition, with the help of a schedule. [3]
(c) Discuss four differences between perfect competition and monopolistic competition. [6]
Answers:
(a) (i) Variable cost: It is incurred only when there is production.
(ii) Fixed cost: It is incurred on fixed factor capital (in the short run) and remains constant when there is no production.
(iii) Fixed cost: It is payable even when there is no production.

(b) The total revenue (TR) curve, changes in direct proportion to output as the price is constant under perfect competition.
We know,

(c)

Perfect Competition	Monopolistic Competition
(i) Large number of sellers.	(i) Fairly large number of sellers.
(ii) The product is homogenous in nature.	(ii) The product is heterogeneous (differentiated) in nature.
(iii) The buyers have perfect knowledge about the product.	(iii) The buyers have some knowledge about the product.
(iv) There is no market power.	(iv) There is slight market power.
(v) The sellers are a price taker.	(v) The sellers are price maker.
(vi) Selling and advertisement costs do not exist.	(vi) Selling and advertisement costs exist.
(vii) Here, firms earn only normal profits in the long run.	(vii) Here, the firms earn supernormal profits in the long run.

Question 5.
(a) Differentiate between the fixed factor of production and variable factor of production. [3]
(b) Show with the help of a diagram, how a perfectly competitive firm earns supernormal profit in short-run equilibrium. [3]
(c) Explain with the help of a diagram, the relationship between total product and marginal product. [6]
Answers:
(a) The factor of production that cannot be changed with changes in the level of production during the short run, is called a fixed factor. However, the factors of production which changes with change in the level of production both during the short run and long run, are known as variable factors of production.
For instance, during the short run, the plant and machinery are regarded as a fixed factor while raw materials and labour are considered as variable factors.

(b) The price is fixed at P in short-run (AR = MR). The MC passes through the minimum point of SAC curve and cuts MR at R. The firm decides its output at OQ*. Thus the total revenue TR is \(\overrightarrow{\mathrm{P}} \times \mathrm{Q}=\mathrm{O} \overline{\mathrm{P}} \mathrm{R} \mathrm{Q}^{*}\). The total cost (TC) for producing OQ* output = OTSQ*. Thus, the supernormal profit = TR – TC
\(\begin{array}{l}{=\mathrm{O} \overline{\mathrm{P}} \mathrm{RQ}^{*}-\mathrm{OTSQ}^{*}} \\ {=\overline{\mathrm{P}} \mathrm{RTS}}\end{array}\)

(c) The TP curve increases initially, at an increasing rate (till point M) and then at a decreasing rate, reaches a maximum at T and then falls.
When the TP increases at an increasing rate, the marginal product increases. However, when TP increases at a decreasing rate, MP declines. When TP is maximum, MP is zero and becomes negative when TP begins to decline.
This is shown in the diagram.

Question 6.
(a) Explain the primary functions of money. [3]
(b) Explain the following functions of the Reserve Bank of India: [3]
(i) Banker to the government
(ii) Issue of currency notes
(c) Explain the mechanism of credit creation by commercial banks with the help of an example. [6]
Answer:
(a) (i) Medium of exchange: Most important function of money is as a medium of exchange to facilitate transactions. Without money, all transactions would have to be conducted by barter, which involves direct exchange of one good or service for another. Money effectively eliminates the double coincidence of wants problem by serving as a medium of exchange that is accepted in all transactions, by all parties, regardless of whether they desired each other’s goods and services. Use of money has separated the process of sale and purchase.

(ii) Measure of value: Money is the measure of value. It serves as a unit of measurement in terms of which the values of all goods and services are measured and expressed. Expressing the value of the commodity in terms of money is known as its price, which is nothing, but the rate of exchanging a commodity.

(b) (i) Banker to the government: The government maintains an account in the central bank (Reserve Bank of India), just as individuals have in commercial banks. All incomes of the government are deposited and expenses are incurred out of this account. Reserve Bank of India keeps track of all the inflows and outflows. The government also borrows from the central bank (RBI) when necessary.

(ii) Issue of currency notes: The Reserve Bank of India enjoys the monopoly to issue paper notes which are the legal tender and a part of the money supply. In order to prevent the misuse of this power, the RBI has to follow certain principles. It has to keep a certain amount of gold and foreign securities against the issue of notes.

(c) Process of credit creation: It is often said “Loans create deposits and deposits create loans.” The statement can be explained as:
We assume, a situation in an economy where one bank ‘A’ receives a primary cash deposit from a person. After maintaining the Cash Reserve Ratio (CRR) with the central bank, the bank is left with the deposit of say ₹ 1000. Assuming that the Statutory Liquidity Ratio (SLR) is 10% of bank deposits, kept reserved for daily transactions, withdrawal etc. So the loanable fund with bank A becomes ₹ 1000 – ₹ 100 = ₹ 900.

A second person now comes to bank A with a loan requirement. The maximum amount of loan can be ₹ 900. However, bank A instead of providing cash loans will ask the person to open an account with the bank and provide him with the facility of drawing checks to the amount of ₹ 900. The second person after taking the loan, hands over the cheque to a third person who deposits the cheque in his own bank B. Thus a loan of ₹ 900 creates a derivative, deposit of the same amount in bank ‘B’, bank ‘B’ can in-turn lend to a third person a maximum of ₹ 810 (₹ 900 – ₹ 90). In this way in the subsequent stages, the derived deposits in the other banks reduce and the total volume of credit expands. The process continues until the loanable amount becomes negligible. Given SLR, mathematically the commercial banks jointly create a total volume of credit a multiple of the reciprocal of SLR, i.e.,
Volume of total credit = Primary deposit × \(\frac { 1 }{ SLR }\)
\(\frac { 1 }{ SLR }\) is known as the credit multiplier.
Alternatively, total v olume of credit creations = 1000 + 900 + 810 + 729 + …….
This is an infinite GP series = \(\frac { 1000 }{ 1-0.9 }\) = 10,000
Here, the whole banking sector is to be taken into account and good banking habits are to be inculcated among the citizens.

Question 7.
(a) Public expenditure helps in increasing the production of an economy. In this context, discuss any two points of importance of public expenditure. [3]
(b) Explain any two causes of disequilibrium in the balance of payment. [3]
(c) Define fiscal deficit, primary deficit and revenue deficit. Discuss their implications with reference to India. [6]
Answers:
(a) Importance of Public Expenditure on Production of an Economy:
(i) It helps in creating income for various individuals and firms as a result of the purchase of goods and services and factor services from them. Expenditure on defence, development activities etc. creates demand for various goods and services, creating income for those individuals and firms producing them, thus increasing the purchasing power of the people.

(ii) It can help in increasing the production by increasing the efficiency of the people (by spending on education, medical facilities, sanitation etc.).

(iii) It can be used to create human skills through education and training thus increasing the ability to work and produce more.

(iv) It can be used as a means of producing essential raw materials and important inputs in the public sector, thus removing various shortages and ensuring smooth production.

(v) It is helpful in promoting and developing the basic and key industries such as capital goods industries.

(vi) Public expenditure incurred in providing social security schemes increases the purchasing power of low-income groups and hence their ability to work, thus increasing their willingness to work because of secured future assured by these policies.

(b) Causes of Disequilibrium in the Balance of Payment (BOP):
(i) Fall in foreign demand: BOP deficit may arise due to a shift in foreign demand away from domestic goods to foreign products because of changes in preference or lower prices of foreign products. This leads to decreases in exports making the BOP unfavourable.

(ii) Inflationary pressure in the economy: A high rate of inflation at home encourages imports by making it relatively cheaper leading to decrease in the country’s competitiveness in the world market and reduces exports.

(iii) Developmental expenditure: Developing countries are poor. Hence they have to depend upon the developed nations for the supply of machines, technology etc. However, they cannot step up their exports to finance the increased imports, leading to a deficit.

(iv) Increase in the cost structure of export industries: It reduces the volume of exports by reducing the competitiveness of these industries in the world market (it may arise due to higher wages, higher prices of raw materials or inflation). Fall in exports makes the BOP unfavourable.

(v) The decrease in supply: A fall in supply at home say agricultural production due to crop failure say, or labour strike, shortage of raw materials etc. leads to fall in exports and imports may increase to overcome the scarcity resulting an adverse BOP.

(vi) Appreciation in the exchange rate: Appreciation increases the external value of a currency, making imports cheaper and exports dearer, leading to an adverse BOP.

(vii) Increase debt burden: Developing countries import capital largely in the form of portfolio investment, creating a large debt burden. Hence these countries are required to make large payments to the developed countries.

(viii) Demonstration effect: People of underdeveloped countries try to imitate the consumption pattern of developed countries regarding luxuries, leading to an increase in imports and hence a deficit.

(ix) Population pressure: A rapid increase in population in underdeveloped countries increase the demand for consumer goods. Hence exports surplus has fallen resulting in an adverse effect on the BOP.

(x) Political factors: They are also responsible for making BOP unfavourable. Political turmoil and instability in a number of countries-gulf countries, African countries, Afghanistan etc. have an adverse effect on their BOP.

(c) Fiscal deficit is the difference between the total expenditure and the sum of revenue and’ capital receipts excluding borrowings. Thus,
Fiscal deficit = Total budgetary expenditure – Revenue Receipts – Capital Receipts (excluding borrowings)
Implication: It has serious implication for the economy. Government has to borrow to meet this deficit, increasing future liability in the form of interest payment and repayment of loans. Payment of interest increases revenue expenditure, increasing the revenue deficit and thus leading to more borrowings and more interest payments. Hence, it is important to reduce the fiscal deficit for avoiding ‘debt trap’ and smooth functioning of the economy.

Primary deficit refers to the difference between fiscal deficit and interest payments.
Thus, Primary deficit = Fiscal deficit – Interest payments.
Implication: It indicates the real position of the government finances as it excludes the interest burden in respect of loans taken in the past, showing how much the government is borrowing to meets its expenses other than interest payments. It is a measure of the fiscal discipline of the government.

Revenue deficit denotes the difference between the revenue receipts and revenue expenditure. Thus,
Revenue deficit = Revenue Expenditure – Revenue Receipts.
Implication: It indicates the government’s current financial status. In India, the deficit on revenue account is very high. Revenue deficit means dissaving on government account which imposes a burden on the future generations because they have to bear the pinch of the interest burden.

Question 8.
(a) Show the relationship between APC and APS. [3]
(b) Explain the difference between induced investment and autonomous investment. [3]
(c) Discuss the main components of aggregate demand in an economy with the help of a diagram. [6]
Answers:
(a) Since income is either consumed or saved, i.e.,
or \(\frac{\mathrm{C}}{\mathrm{Y}}+\frac{\mathrm{S}}{\mathrm{Y}}=\frac{\mathrm{Y}}{\mathrm{Y}}\)
or APC + APS = 1
i.e., the sum is always unity at all levels of income.

(b) 1. Induced investment is that investment which is undertaken as a result of a change in the level of income whereas Autonomous investment is that type of investment which is not affected by a change in the level of income or output.

2. Induced investment varies directly with a change in income level while autonomous investment is income inelastic.
3. The induced investment shows a positive functional relationship between income and investment Autonomous investment is a horizontal line, parallel to the income axis, indicating some investment at all income levels.

(c) Aggregate demand is generated out of ‘planned’ or ‘desired’ aggregate expenditure in an economy during a particular year. Following are the main components of aggregate demand (AD) in an economy:
(i) Aggregate private consumption expenditure (C): This expenditure creates demand for various consumer goods and services in an economy.
(ii) Aggregate private investment expenditure (I): This expenditure creates demand for various investment goods in an economy.
(iii) Government expenditure (G): The government expenditure in a welfare state creates demand for consumption and investment goods in the economy.
(iv) Net exports (the difference between exports and imports) (X-M): Export income of a country during a particular year denotes the foreign demand for goods and services provided by the domestic country while import payments imply the demand for foreign goods during any particular year.
AD = C + I + G + (X – M)

Here investment is considered autonomous and consumption expenditure consists of two parts – autonomous (\(\overline{\mathrm{C}}\)) and induced (bC, where b = MPC).

Question 9.
(a) Explain the following components of domestic factor income: [3]
(i) Compensation of employees
(ii) Operating surplus
(b) Distinguish between private income and personal income. [3]
(c) From the following data, calculate GNPMP and National Income by using the value-added method: [6]

(i) The gross value of output in the primary sector (at FC)	950 crores
(ii) Gross value of output in the secondary sector (at FC)	470 crores
(iii) Gross value of output in the tertiary sector (at FC)	500 crores
(iv) Value of intermediate goods in the primary sector	360 crores
(v) Value of intermediate goods in the secondary sector	200 crores
(vi) Value of intermediate goods in the tertiary sector	175 crores
(vii) Depreciation	20 crores
(viii) Indirect tax	35 crores
(ix) Subsidy	10 crores
(x) Net factor income from abroad	4 crores

Answers:
(a) (i) Compensation of employees: It is defined as all payments made by producers to their employees in the form of wages and salaries and other payments made in cash and kind and social security benefits in return for labour services. Labour income includes wages and salaries, supplementary labour income and payment in kind.
(ii) Operating surplus: It is the income earned from the ownership and control of capital and is also known as income from property and entrepreneurship. Capital income includes interest, rent, profits, royalties, dividends and other similar incomes.

(b) Private Income = National Income – Income from property and entrepreneurship accruing to the government commercial enterprises and administrative departments – Saving of non-departmental enterprises of the government + Interest on National Debt + Net current transfers from government + Net current transfers from abroad.
Personal Income = Private Income – Undistributed Profits – Corporate Profits Tax – Retained Foreign Earnings

(c) Given:
GVO primary (FC) = 950
GVO secondary (FC) = 470
GVO tertiary (FC) = 500
Value of intermediate goods in primary sector = 360
Value of intermediate goods in secondary sector = 200
Value of intermediate goods in tertiary sector = 175
Depreciation = 20 Indirect tax = 35 Subsidy = 10 NFIA = 4
GVA_FC = Gross value of output in each sector – Intermediate product in each sector
= (950 – 360) + (470 – 200) + (500 – 175)
= ₹ 1185 crore
GNP_MP = GVA_FC + Indirect tax – Subsidy + NFIA
= 1185 + 35 – 10 + 4
= ₹ 1214 crore
National Income (NNP_FC) = GVA_FC – Depreciation + NFIA
= 1185-20 + 4
= ₹ 1169 crore

ISC Class 12 Economics Previous Year Question Papers

ISC Accounts Previous Year Question Paper 2012 Solved for Class 12

Maximum Marks: 80
Time allowed: Three hours

• Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.
• Answer Question 1 (Compulsory) from Part I and five questions from Part II, choosing two questions from Section A, two questions from Section B and one question from either Section A or Section B.
• The intended marks for questions or parts of questions are given in brackets [ ].
• Transactions should be recorded in the answer book.
• All calculations should be shown clearly.
• All working, including rough work, should be done on the same page as, and adjacent to the rest of the answer.

Part -1

Question 1.
Answer each of the following questions briefly: [10 x 2]
(i) State any two uses of Securities Premium as stated in Section 78 of the Companies Act, 1956.
(ii) In a Cost Sheet, how would you treat :
(a) Primary packing material.
(b) Secondary packing material.
(iii) Give two differences between Sacrificing Ratio and Gaining Ratio.
(iv) In case of a Joint Venture business, how is abnormal loss of goods which have been insured, treated in the books of accounts ?
(v) List two instances when a partner ‘s Fixed Capital may change.
(vi) List any two objectives of stock valuation.
(vii) Why is a Journal Ledger Adjustment Account opened ?
(viii) Assuming that the Debt-Equity Ratio of a company is 2 : 1, state whether this ratio would increase, decrease or not change in the following cases :
(a) Issue of new shares for cash.
(b) Repayment of a long-term bank loan.
(ix) What are trade investments ?
(x) The firm with X. Y and Z as partners earned a profit of ? 3,00,000 during the year ended 31st March. 2011. 20% of this profit w as to be transferred to General Reserve. Pass the necessary- journal entry for the same.
Answer:
(i) Two uses of securities premium as stated in Section 78 of the Companies Act, 1956 are :
(a) To issue fully paid bonus shares to the shareholders.
(b) To write off discount on issue of shares and debentures

(iii)

Basis	Sacrificing Ratio	Gaining Ratio
(a) Meaning	It is the ratio in which the old partners surrender a part of their share in favor of new partners.	It is the ratio in which remaining partners acquire the outgoing partners share.
(b) Purpose of calculation	New partner’s share of good-will is divided in sacrificing ratio.	Outgoing partner’s goodwill divided between remaining partners in gaining ratio.

(iv) The total value of abnormal loss is not to be recorded in the Joint Venture Account. However, any claim received from the Insurance Company is to be credited to the Joint Venture Account.

(v) Following are the two instances when a partner’s fixed capital may change :

• When fresh or additional capital has been introduced by a partner.
• When an amount has been withdrawn by a partner to reduce his capital.

(viii) The following would be the effect: It would result in a decrease in the Debt-Equity Ratio.

(ix) Trade investments are long-term investments made by a company in the shares and debentures of another company (not including its subsidiaries).

(x) Profit and Loss Appropriation A/c  Dr.  60,000
To General Reserve A/c  60,000
(Being 20% of profit transferred to General Reserve)

Question 2. [10]
Amit. Paw an and Suresh are partners in a firm, sharing profits in the ratio 2:3:1. Suresh retired on 1st April. 2011. At the time of his retirement:
(a) Goodw ill of the firm w as valued at ₹ 36.000.
(b) The Balance Sheet of the firm showed :
(i) A General Reserve of ₹ 1,20,000.
(ii) A debit balance of ₹ 48,000 in the Profit and Loss Account.
(iii) ₹ 48,000 each in the Joint Life Policy Account and Joint Life Policy Reserve Account. It was decided that the Joint Life Policy would be surrendered on the date of Suresh’s retirement.
Record necessary Journal entries for the above adjustments to be made in the books of the firm on the date of Suresh’s retirement.

Part – II

Question 3. [14]
Paula Fashioners Limited maintains its books under the Sectional Balancing System. From the particulars given below for the year ending 31 st December, 2011, you are required to prepare necessary Control Accounts in the Journal Ledger:

Question 4. [14]
The following extract of costing information relates to a commodity for the year ended 31^st . March. 2007.
1^st April, 2006 :
Raw materials — ₹ 50.000
Finished Products (1,000 tonnes) — ₹ 40,000
Work-in-Progress — ₹12,000

31^st March, 2007:
Raw Materials — ₹ 55,600
Finished Products (2.000 tonnes) — ?
Work-in-Progress — ₹ 40,000

Transactions during the year:
Raw Materials Purchased — ₹ 3,00,000
Direct Wages — ₹ 25,000
Rent. Rates and Insurance of Factory — ₹ 1,00,000
Carriage Inwards — ₹ 3,600
Cost of Factory Supervision — ₹ 20,000
Sale of Finished Products — ₹ 7,50,000
Advertisement and selling expenses @ ₹ 2 per tonne sold.
16,000 tonnes were produced during the year. It was decided to value the closing stock as per FIFO.
Prepare a statement showing:
(a) Value of raw materials used.
(b) Cost of the output for the year.
(c) Value of closing stock.
(d) Profit made during the year.

Question 5.
The Balance Sheet of Cooper and Company as on 31^st  December, 2010 and 31^st December. 2011 arc given below:

Additional Information:
(a) Depreciation charged on building ₹ 10,000.
(b) Depreciation charged on plant ₹ 5,000.
(c) Interest paid on debentures ₹ 7,200 for the year.
(d) Interest paid on public deposit ₹ 9,600 for the year.
From the above information, prepare a Cash Flow Statement as per Accounting Standard-3 for the year ended 31^st December, 2011.
Answer:


Working Notes:
1. Cash credit has been treated as cash equivalent.

Question 6.
In 2010, Ganga Ltd. was registered with an authorized capital of ₹ 1,00,000 in Equity shares of ₹ 10 each. Of these, 4,000 equity shares were issued as fully paid to vendors for the purchase of Plant and Machinery and the remaining 6.000 shares were subscribed for, by the public for cash. During the first year, ? 6 per equity share was called up, on these 6.000 shares, payable ₹ 3 on application. ₹ 1 on allotment and ₹ 2 on the first call. [14]
The amount received in respect of these shares were as follows :
On 5,000 shares, the full amount called.
On 600 shares. ₹ 4 per share.
On 400 shares, ₹ 3 per share.
The company forfeited all those shares on which only ₹ 3 had been received and reissued them at ₹ 4 per share, ₹ 6 called up.
Journalize the transactions in the books of the company and prepare a Calls-in Arrear
Answer:

Question 7.
Anita. Bina and Chitra were in partnership, sharing profits and losses equally. The firm’s Balance Sheet as on 31^st December, 2011 was as follows : [14]

It was decided to dissolve the firm on 31^st December. 2011.
The plant and machinery, debtors and stock were sold by the firm for ₹ 70,000 and the creditors were paid off.
Chitra was declared insolvent and could not meet her liability towards the firm.
From the above, prepare the Realization Account, Partner’s Capital Account and Cash Account assuming that the firm applied the Garner Vs. Murray rule.

Question 8.
From the following information, calculate (up to two decimal places): [14]
(i) Liquid Ratio.
(ii) Current Ratio.
(iii) Proprietary Ratio.
(iv) Working Capital Turnover Ratio.
(v) Gross Profit Ratio.
(vi) Operating Ratio.
(vii) Net Profit Ratio.
Cost of Goods Sold — ₹ 6,00,000
Operating Expenses — ₹ 50,000
Gross Sales — ₹ 8,00,000
Sales Returns — ₹ 10,000
Total Current Assets — ₹ 3,00,000
Total Current Liabilities — ₹ 1,00,000
Total Assets — ₹ 7,00,000
Closing Stock — ₹ 30,000
Prepaid Insurance — ₹ 5,000
Prelim man, Expenses — ₹ 6,000
Share Capital — ₹ 5,60,000
Reserves and Surplus — ₹40,000
Answer:

Question 9.
On 1st January, 2000, Star Ltd. issued 1.000. 12% Debentures of ₹ 100 each at a discount of 5%. repayable as follows : [14]

On 31st December, 2002 — ₹ 20,000
On 31st December, 2003 — ₹ 60,000
On 31 st December, 2004 — ₹ 20,000
The company pays interest on debentures annually.
You are required to:
(a) Pass the Journal entries (including interest) for the year beginning 1st January, 2000 to 31^st December, 2000.
(b) Prepare the ‘Discount on issue of Debenture Account’, till it is finally closed
Answer:


(b) Working Notes : The ratio for writing off discount on issue of debentures based on their redemption is calculated as follows:
1,00,000 : 1,00,000 : 1,00,000 : 80,000 : 20,000 = 5 : 5 : 5 : 4 : 1
Hence, discount to be written off during the first, second and third


As per latest guidelines, discount on issue of debentures is to be written off completely in the year of issue of debentures itself.

Question 10.
The following balances have been extracted from the books of King Furnishings Ltd. as on 31st March. 2011. [14]

You are required to prepare a Balance Sheet of King Furnishings Ltd. as on 31^st March, 2011, in the Horizontal Form, as prescribed under Schedule VI of the Companies Act, 1956.
Answer:

ISC Class 12 Accounts Previous Year Question Papers

ISC Accounts Previous Year Question Paper 2017 Solved for Class 12

Maximum Marks: 80
Time allowed: Three hours

• Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.
• Answer Question 1 (Compulsory) from Part I and five questions from Part II, choosing two questions from Section A, two questions from Section B and one question from either Section A or Section B.
• The intended marks for questions or parts of questions are given in brackets [ ].
• Transactions should be recorded in the answer book.
• All calculations should be shown clearly.
• All working, including rough work, should be done on the same page as, and adjacent to the rest of the answer.

Section-A
Part – I (12 Marks)
Answer all questions.

Question 1. [6×2]
Answer briefly each of the following questions :
(i) Name the account which is prepared to find the profit and loss of a joint venture, if:
(a) One co-venture records all transactions.
(b) All co-ventures record their own transactions.
(ii) What will be the treatment of loan given to a partner by the firm at the time of its dissolution ?
(iii) Give the adjusting entry for interest on capital allowed to a partner, when the firm follows the fixed capital method.
(iv) State, with reason, whether securities premium reserve can be used to write off bad debts.
(v) Give any two differences between a Company s Balance Sheet and a Firm’s Balance Sheet.
(vi) State where will the non-cash transactions be recorded at the time of issue of shares, if all cash transactions are entered in the Cash Book.
Answer:
(i) (a) Joint Co-venture Account
Personal Accounts of other Co-ventures
(b) Memorandum Joint Venture Account
Joint Venture with…. (other Co-venture) Account

(ii) If there is a loan advanced to a partner, the same should be transferred to his capital account thereby reducing the amount of capital repayable to him.

(iii) Interest on Capital A/c Dr.
To Partner’s Current A/c
(Being the interest on capital allowed to partners)
Profit and Loss Appropriation A/c Dr.
To Interest on Capital A/c
(Being the interest on capital transferred to Profit and Loss Appropriation A/c)

(iv) Securities Premium Reserve cannot be used to write off bad debts.
Securities Premium Reserve can be write off for following purposes :

• in paying up unissued shares to be issued as fully paid bonus shares.
• in writing off preliminary expenses.
• for buy – back of shares Under Section 11 A.
• in writing off the expenses etc.

(v) Balance Sheet is prepared as per Schedule III of the Indian Companies Act, 2013. Whereas Balance Sheet is prepared as per Partnership Act, 1932. The details of items of Balance Sheet are to be given in the Notes to Accounts in Company’s Balance Sheet but there is no need to maintain Notes to Accounts in firm’s Balance Sheet.

(vi) Issue of Shares for Consideration other than Cash is shown in the Balance Sheet under the head ‘Share Capital’ and sub-head ‘Subscribed Capital’.

Part – II (48 Marks)
Answer any four questions.

Question 2. [12]
Karan, Ali and Deb are partners in a firm sharing profits and losses in the ratio of 3 : 2 : 1. On 31st March, 2016, their Balance Sheet was as under :

Karan died on 1st July, 2016. An agreement was reached amongst Ali, Deb and Karan’s legal representatives that :
(a) Building be revalued at ₹ 93,500.
(b) Furniture be appreciated by ₹ 10,000.
(c) To write off the Provision for Doubtful Debts since all debtors were good. .
(d) Investments be valued ₹ 38,000.
(e) Goodwill of the firm be valued at ₹ 1,20,000.
(f) Karan’s share of profit to the date of his death, to be calculated on the basis of previous year’s profit which was ₹ 25,000.
(g) Interest on capital to be allowed on Karan’s capital @ 6% per annum.
(h) Amount payable to Karan’s legal representative to be transferred to his legal representative’s loan account.
You are required to :
(i) Pass Journal entries on the date of Karan’s death.
(ii) Prepare the Interim Balance Sheet of the reconstituted firm.
Answer:

Question 3. [12]
Cargo Ltd. invited applications for the issue of 20,000 Equity shares of ? 10 each at a premium of ? 1 per share, payable as follows :
On Application — ₹ 3
On Allotment — The balance (including premium ₹ 1)
Applications were received for 30,000 shares and pro-rata allotment was made to the remaining applicants after refunding application money to 5,000 share applicants.
Nicholas, who was allotted 3,000 shares, failed to pay the allotment money and his shares were forfeited.
Out of these forfeited shares, 1,000 shares were reissued as fully paid-up @ ? 8 per share.
You are required to :
(i) Pass Journal entries in the books of the company.
(ii) Prepare Calls-in-Arrears Account.
(iii) Prepare Share Forfeiture Account.
Answer:

Question 4.
(A) Following balances have been extracted from the books of Universe Ltd. as at 31st March, 2016 :
Particulars
Equity Share Capital (Fully paid shares of ₹ 100 each) — ₹ 4,00,000
Unclaimed Dividend  — ₹ 10,000
Bank Balance  — ₹ 40,000
Security Premium Reserve  — ₹ 75,000
Statement of Profit and Loss (Dr.)  — ₹ 50,000
Tangible Fixed Assets (at cost)  — ₹ 3,50,000
Accumulated Depreciation till date  — ₹ 25,000
Trade Marks  — ₹ 70,000
You are required to prepare as at 31st March, 2016 :
(i) The Balance Sheet of Universe Ltd. as per Schedule in of the Companies Act, 2013.
(ii) Notes of Accounts. . [8]
(B) Chrome Ltd. took over assets of ₹ 6,00,000 and liabilities of ₹ 40,000 of Polymer Ltd. at an . agreed value of ₹ 6,30,000. Chrome Ltd. issued 10% Debentures of ₹ 100 each at a discount of 10% to Polymer Ltd. in full satisfaction of the price. Chrome Ltd. writes off any capital losses incurred during a year, at end of that financial year.
You are required to pass the necessary Journal entries to record the above transactions in the books of Chrome Ltd. [4]
Answer:

Question 5. [12]
Juliet and Rabani are partners in a firm, sharing profits and losses in the ratio of 3 :1. On 31 st March, 2016, their Balance Sheet was as under:

Mike was taken as a partner for 1/4th share, with effect from 1st April, 2016, subject to the following adjustments :
(a) Plant and Machinery was found to be overvalued by ₹ 16,000. It was to be shown in the books at the correct value.
(b) Provision for Doubtful Debts, was to be reduced by ₹ 2,000.
(c) Creditors included an amount of ₹ 2,000 received as commission from Malini. The necessary adjustment was required to be made.
(d) Goodwill of the firm was valued at ₹ 60,000. Mike was to bring in cash, his share of goodwill along with his capital of ₹ 1,00,000.
(e) Capital Accounts of Juliet and Rabani were to be readjusted in the new profit sharing arrangement on the basis of Mike’s capital, any surplus to be adjusted through current account and any deficiency through cash.
You are required to prepare r
(i) Revaluation Account.
(ii) Partners’ Capital Accounts.
(iii) Balance Sheet of the reconstituted firm.
Answer:




Question 6.
(A) Raslii and Runa jointly imdertake to complete the construction of an auditorium for Pascal Ltd. They agreed to share profits and losses in the ratio of 3 : 2.
The contract price was ₹ 8,00,000 of which ₹ 5,00,000 was to be payable to them in cash and the balance in fully paid shares of the company.
A joint bank account was opened in which Rashi contributed ₹ 2,00,000 while Runa contributed ₹ 3,00,000.
The following expenses were incurred to complete the contract :
Salaries and Wages — ₹ 1,25,000
Purchase of material from a supplier on credit — ₹ 2,00,000
Material supplied by Rashi — ₹ 1,00,000
Legal fees paid by Runa — ₹ 85,000
The contract price was duly received after the completion of the project and the accounts of the venture were closed after the supplier was paid ₹ 1,98,000 in lull and final settlement.
Runa took over the shares at ₹ 2,80,000.
Rashi took over the remaining material at ₹ 45,000.
You are required to prepare:
(i) Joint Venture Account.
(ii) Joint Bank Account.
(iii) Shares Account.
(B) Joseph and Leena entered into a Joint venture to sell edible oil. It was decided that Joseph would record all the transactions of the venture.
Joseph supplied 3,000 litres of edible oil costing ₹ 4,50,000 to be sold by Leena, incurring carriage and insurance-in-transit amounting to ₹ 30,000.
20 litres of oil was lost in transit due to leakage which was considered to be normal. Leena incurred ₹ 2,760 as clearing charges and ₹ 2,000 as godown rent. She was entitled to a commission of 2% on the sales made by her.
Leena was able to sell 2,000 litres of oil at ₹ 170 per litre.
The unsold stock was taken over by Joseph at the original cost plus proportionate non-recurring expenses.
You are required to :
(i) Calculate the value of stock taken over by Joseph.
(ii) Pass the relevant Journal entries in the books of Joseph for :
(a) The stock taken over by Joseph.
(b) Commission due to Leena. [4]
Answer:

Question 7.
(A) Mita, Rita and Sandra were partners in a firm, sharing profits and losses in the ratio of 2 : 2 : 1. Mita had personally guaranteed that in any year Sandra’s share of profit, after allowing interest on capital to all the partners @ 5% per annum and charging interest on drawings @ 4% per annum, would not be less than ₹ 10,000.
The capitals of the partners on 1st April, 2015 were :
Mita ₹ 80,000, Rita ₹ 50,000 and Sandra ₹ 30,000.
The net profit for the year ended 31st March, 2016, before allowing or charging any interest amounted to ₹ 40,000.
Mita had withdrawn ₹ 4,000 on 1st April, 2015, while Sandra withdrew ₹ 5,000 during the year.
You are required to prepare the Profit and Loss Appropriation Account for the year 2015-16. [8]
(B) Anita, Asha and Bashir are partners sharing profits and losses in the ratio of 3 : 2 : 1 respectively. From 1st April 2016, they decided to change their profit sharing ratio to 2 : 1: 3. Their partnership deed provides that in the event of any change in the profit changing ratio, the goodwill of the firm should be valued at two years’ purchase of the average super profits for the past three years.
The actual profits and losses for the past three years were :
2015-16 Profit ₹ 40,000
2014-15 Profit ₹ 30,000
2013-14 Loss ₹ 10,000
The average capital employed in the business was ₹ 1,10,000; the rate of interest expected from capital invested was 10%.
You are required to:
(i) Calculate the value of goodwill at the time of change in profit sharing ratio. (Show the workings clearly with the formulae.)
(ii) Pass the Journal entry to record the change. [4]
Answer:




Question 8.
(A) Roshan, Mahesh, Gopi and Jai are partners sharing profits and losses in the ratio of 3 : 3 : 2 : 2. The balances of capital accounts on 1st April, 2015 were : Roshan ₹ 8,00,000, Mahesh ₹ 5,00,000, Gopi ₹ 6,00,000 and Jai ₹ 6,00,000.
After the accounts for the year ended 31st March, 2016 were prepared, it was discovered that interest on capital @ 10% per annum as provided in the partnership deed had not been credited to the partners’ capital accounts before the distribution of profits.
You are required to rectify the error by passing a single adjusting Journal entry. [4]
(B) Mehta and Menon were partners in a firm, sharing profits and losses in the ratio of 7 : 3.
They decided to dissolve their partnership firm on 31st March, 2016. On that date, their books showed the following ledger account balances :
Sundry Creditors ₹ 27,000
Profit and Loss A/c (Dr.) ₹ 8,000
Cash in Hand ₹ 6,000
Bank Loan ₹20,000
Bills Payable ₹ 5,000
Sundry Assets ₹ 1,98,000
Capital A/c
Mehta ₹ 1,12,000
Menon ₹ 48,000

Additional information :
(a) Bills Payable falling due on 31st May, 2016 were retired on the date of dissolution of the firm, at a rebate of 6% per annum.
(b) The bankers accepted the furniture (included in sundry assets) having a book value of ? 18,000 in full settlement of the loan given by them.
(c) Remaining assets were sold for ₹ 1,50,000.
(d) Liability on account of outstanding salary not recorded in the books, amounting to ₹ 15,000 was met.
(e) Menon agreed to take over the responsibility of completing the dissolution work and to bear all expenses of realization at an agreed remuneration of ₹ 2,000. The actual realization expenses were ₹ 1,500 which were paid by the firm on behalf of Menon.
You are required to prepare :
(i) Realization Account.
(ii) Partners’ Capital Accounts. [8]
Answer:

Section – B
(20 Marks)
Answer any two questions

Question 9.
From the information given below, calculate (up to two decimal places):
(i) Operating Ratio.
(ii) Quick Ratio.
(iii) Debt to Equity Ratio.
(iv) Proprietary Ratio.
(v) Working Capital Turnover Ratio.
Particulars
Net revenue from operations — ₹ 12,00,000
Cost of revenue from operation — ₹ 9,00,000
Operating expenses — ₹ 15,000
Inventory — ₹ 20,000
Other Current Assets — ₹ 2,00,000
Current Liabilities — ₹ 75,000
Paid up Share Capital — ₹ 4,00,000
Statement of Profit and Loss (Dr.) — ₹ 47,500
Total Debt — ₹ 2,50,000
Answer:


Question 10. [10]
From the following information of Purity Ltd. calculate:
(i) Cash from Operating Activities
(ii) Cash from Financing Activities

Additional information:
During the year 2015-16:
(a) A piece of furniture costing ₹ 30.000 (accumulated depreciation ₹ 5,000) was sold for ₹ 25,000.
(b) Tax of ₹ 9,000 w as paid.
(c) Interim Dividend of ₹ 4,000 was paid.
(d) The company paid ₹ 3,000 as interest on debentures.
Answer:

Question 11.
(A) What is meant by the term Cash Equivalents as per Accounting Standard 5 ? [2]
(B) The Current Ratio of a company is 2 : 1. State whether the Current Ratio will improve, decline or will not change in the following cases : [2]
(i) Bill Receivable of ₹ 2,000 endorsed to a creditor is dishonoured.
(ii) ₹ 8,000 cash collected from Debtors of ₹ 8,500 in lull and final settlement.

Answer:
(A) Cash and Cash Equivalents are short-term, highly liquid investments which is easily converted into cash. It includes treasury bills, commercial papers, money market funds etc.
(B) (i) It will increase the amount of debtors but reduces the amount of bills receivable. So, current assets will remain same. Hence, there will be no change in the current ratio.
(ii) Cash collected from debtors will increase the bank/cash balance but decrease the amount of debtors. Hence, total current assets remain unchanged. Therefore, there will be no . change in current ratio.

ISC Class 12 Accounts Previous Year Question Papers

ISC Economics Previous Year Question Paper 2013 Solved for Class 12

Maximum Marks: 80
Time allowed: 3 hours

• • Candidates are allowed additional 15 minutes for only reading the paper.

• They must NOT start writing during this time.
• Answer Question 1 (Compulsory) from Part I and five questions from Part II.
• The intended marks for questions or parts of questions are given in brackets [ ].

Part – I (20 Marks)
Answer All Questions

Question 1.
Answer briefly each of the following questions (i) to (xv).
(i) What does zero cross elasticity of demand between two goods imply? Give an example to explain.
(ii) Why is the marginal cost curve U shaped?
(iii) Differentiate between monopoly and monopsony. Give an example for each.
(iv) What is the market period? What is the shape of the supply curve in this period?
(v) Give two assumptions of the law of Variable Proportions.
(vi) Explain the meaning of price ceiling with the help of a diagram.
(vii) Why is the Central Bank considered to be the lender of the last resort?
(viii) What is the Vote-on-account budget?
(ix) Explain how taxation can be used to reduce inequality of income.
(x) What is meant by unlimited legal tender?
(xi) Distinguish between CRR and SLR.
(xii) Calculate the value of multiplier if MPC is equal to MPS.
(xiii) Define GNP at factor cost. How is it different from national income?
(xiv) Explain with the help of an example, how inflation affects the debtors.
(xv) How does an increase in the price of a commodity affect its quantity demanded? Show it with the help of a diagram.
Answer:
(i) Zero cross elasticity of demand between goods implies that two goods are not related to each other. In other words, the change in the price of one commodity (Y) does not affect the demand for another commodity (X).
For example, a change in the price of sugar is not likely to influence the demand for a fan. So their cross elasticity of demand will be zero.

(ii) MC curve is ‘U’ shaped because of the law of variable proportions. As output increases, MC curve negatively slopes (due to increasing returns to the variable factor), reaches the minimum and then slopes positive or upwards due to decreasing returns to the variable factor.

(iii)

Basis	Monopoly	Monopsony
1. Meaning	It is a market structure in there exists only a single seller of a product who is the sole producer of the product which has no close substitutes.	It is a market structure in which there exists only a single buyer of a product in the market.
2. Example	Indian Railways which is owned by the Government of India.	In a village one textile mill purchase cotton from the farmers of that village. So the textile mill is the monopsony firm which is the only buyer of cotton produced by the farmers of the village.

(v) (a) The state of technology is given and remains unchanged.
(b) All the variable factors are equally efficient.

(vi) Price ceiling refers to fixing the maximum price of a commodity at a level lower the equilibrium price.
Let us clear this point by considering the commodity ‘wheat’ and its price determination in fig.
In the diagram, demand curve DD and supply curve SS of wheat intersect each other at point E and, as a result, the equilibrium price of OP is determined.
Suppose, the equilibrium price of OP is very high and many poor people are unable to afford wheat at this price.

As wheat is an essential commodity, the government interferes and fixes the maximum price (known as Price ceiling) at OP1 which is less than the equilibrium price OP.

(vii) Lender of the Last Resort:
The Central Bank acts as the lender of the last resort.
When commercial banks fail to meet their financial requirements from other sources, they can approach the Central Bank which provides them with loan and advances as lender of the last resort.
The Central Bank provides this facility to protect the interest of the depositors and to prevent possible failure of the bank.

(ix) By imposing direct taxes which are progressive in nature, rich people are subjected to a higher rate of taxation as compared to poor people. Higher taxes on luxuries and low taxes on essential commodities also helps in removing inequalities of income. Inequality of income can be further reduced by using the tax proceeds in providing social services which benefit poor people.

(x) Unlimited legal tender is the money, which a person has to accept without any maximum limit. In our country, currency notes of all denominations and coins of 50 paise and higher denominations are unlimited legal tender.

(xi) CRR is the certain percentage of deposits accepted by Commercial Banks which they are required by law to keep with Central Bank in the form of cash reserves. Whereas, Statutory Liquidity Ratio refers to a fixed percentage of the assets of the Commercial Bank which they are required by law to keep with themselves in the form of cash or other liquid assets.

(xii) We know,
MPC + MPS = 1 and K = \(\frac { 1 }{ MPS }\)
When MPC = MPS
MPS + MPS = 1
⇒ 2MPS = 1
⇒ MPS = \(\frac { 1 }{ 2 }\)
⇒ K = \(\frac { 1 }{ MPS }\) = 2

(xiii) GNPFC is the sum total of factor income earned by normal residents of a country, inclusive of depreciation during an accounting year.
GNPFC = NNPFC + Depreciation Whereas, National Income (NNPFC) is the sum total of factor income earned by normal residents of a country during an accounting year.
NNP FC = NDP FC + Net factor income from abroad GNP FC includes depreciation but NNPFC (National Income) does not include depreciation.

(xv) Keeping other things constant an increase in the price of a commodity will inversely affect its quantity demanded. In other words, an increase in the price of say X commodity will reduce or decrease its quantity demanded. The adjacent diagram shows this relationship.

Part – II
(Answer Any Five Questions)

Question 2.
(a) Explain with the help of a well-labelled diagram how a perfectly competitive firm earns normal profit in short-run equilibrium.
(b) Why does the TC curve start from the Y-axis and the TVC curve from the Origin?
(c) Discuss four features of Oligopoly.
Answer:
(a) Under perfect competition a firm’s, average revenue curve is equal to the marginal revenue curve due to uniform prices for homogeneous goods. In short-run, a competitive firm can earn supernormal profits, normal profits can suffer losses also. Figure (A) shows a perfectly competitive firm earning normal profits. Since the firm is price taker, it has to decide the amount of output it should produce at the given price so as to maximise the profits following the equilibrium conditions.
SMC = MR and AR = AC
In the given diagram OQ is equilibrium output where SMC is equal to MR. Since SAC is tangent to P = AR = MR line, the firm covers only its SAC which includes normal profits.

(b) TC (Total cost) curve is the sum total of the total fixed cost and the total variable cost incurred in producing a given amount of output. In other words TC = TFC + TVC. Since Total Fixed Cost (TFC) remains constant regardless of the quantity of output and Total Variable Cost changes with the change in level of output, so at zero levels of output TFC exists but TVC is zero, therefore, TC curve originate not from origin ‘O’ but from the point on OY axis where TFC intersects (refer to the diagram). Since TVC is zero when output is zero, so it starts from the origin as shown in the adjacent diagram.

(c) (i) Few firms: Under oligopoly, there are few large firms each producing a significant portion of the total output. For example, the market for automobiles in India is an oligopolist structure as there are only a few producers of automobiles. The number of firms is so small that action by anyone firm is likely to affect the rival firms. So, every firm keeps a close watch on the activities of rival firms.

(ii) Interdependence: Due to few large firms under oligopoly are interdependent with respect to their price/output policy. Interdependence means that actions of one firm affect the actions of other firms. A firm considers the action and reaction of the rival firms while determining its price and output levels.

(iii) Non-Price Competition: Firms try to avoid price competition for the fear of price war. They use other methods like advertising, better services to customers, etc. to com¬pete with each other. Under oligopoly, firms are in a position to influence the prices. However, they follow the policy of price rigidity. Price rigidity refers to a situation in which price tends to stay fixed irrespective of changes in demand and supply conditions.

(iv) Role of Selling Costs: Due to severe competition and interdependence of the firms, various sales promotion techniques are used. Selling costs are very important and oligopolistic firms spend much on advertisement and customer services in order to promote sales of its product. Advertisement is in full swing under oligopoly, and many times advertisement can become a matter of life-and-death. For example, T.V. commercials war between Coke and Pepsi. It relies more on non-price competition. Therefore, selling costs are more important under oligopoly than under monopolistic competition.

Question 3.
(a) Complete the following table and draw a supply curve for the firm A:

Price per Unit	Supply by firm A	Supply by firm B	Market Supply
2	5	5	?
3	?	10	17
4	9	?	24
5	11	20	?

(b) Explain what happens when the market price is less than the equilibrium price.
(c) Explain the four determinants of supply of a commodity.
Answer:
(a) Complete the following table and draw a supply curve for the firm A:

Price per Unit	Supply by firm A	Supply by firm B	Market Supply
2	5	5	10
3	7	10	17
4	9	15	24
5	11	20	31

(b) (i) If the market price is lower (P2) than equilibrium price (P0), this will create a situation of excess demand.

(ii) The excess demand will put pressure on price. The price would move upward.
(iii) Due to higher price demand will contract and quantity supplied will expand.
(iv) The expansion of quantity supplied and the contraction of demand will continue till the original equilibrium price is restored.

(c) (i) Price of the Commodity: There is a direct relationship between price of a commodity and its supply. Generally, the higher the price, higher the quantity supplied, and lower the price, lower the quantity supplied.

(ii) Price of Related Goods: The supply of a good depends upon the price of related goods. Examples: Consider a firm selling tea. If price of coffee rises in the market, the firm will be willing to sell less tea at its existing price. Or, it will be willing to sell the same quantity only at a higher price.

(iii) Number of Firms: Market supply of a commodity depends upon the number of firms in the market. If there is an increase in the number of firms market supply will increase and if the number of firms decreases market supply will fall.

(iv) The goal of the Firm: If the goal of the firm is to maximise profits, more quantity of the commodity will be offered at a higher price. On the other hand, if the goal of the firm is to maximise sales (or maximise output or employment) more will be supplied even at the same price.

Question 4.
(a) Explain the nature of the AR and MR curves under perfect and imperfect competition.
(b) Explain any one internal and anyone external economy of scale.
(c) How does a producer attain equilibrium under perfect competition through the MR and MC approach?
Answer:
(a) AR and MR curves, under perfect competition, are equal to each other and parallel to OX-axis. The price at AR is constant as it is fixed by industry with every additional sale of unit MR will be equal to AR.


Under imperfect competition, a firm is the price maker having control over the price but it has to lower down its price in order to sell more. So revenue curves (AR and MR) are downward sloping from left to right indicating that more units of output can be sold at a lower price. Therefore AR > MR.

(c) In order to know the position of maximum profit, a firm compares the marginal cost with marginal revenue: So, the first condition of a firm’s equilibrium is that marginal cost must be equal to marginal revenue (MC = MR). It is necessary, but not sufficient condition of equilibrium. A firm may not get maximum profit even when its marginal cost is equal to marginal revenue. So it must fulfil the second condition of equilibrium as well. i.e., the marginal cost curve must cut the marginal revenue curve from below or the slope of the MC curve must be steeper than the slope of the MR curve. According to marginal analysis, a firm would, therefore, be in equilibrium when the following two conditions are fulfilled.
1. MC = MR.
2. MC curve cuts the MR curve from below.

Both these conditions of firm’s equilibrium are explained with the help of Fig. In this figure, PP is average revenue (price per unit) as well as the marginal revenue curve. It is clear from this figure, that MC curve is cutting MR curve PP at two points ‘A’ and ‘E’. Point ‘A’ cannot indicate the position of equilibrium of the firm as at point A Marginal cost of the firm is still falling or we can say MC is not cutting MR from below.

On the other hand, point E shows that the firm is producing OM units of output. If the firm produces more than OM units of output, its marginal cost (MC) will exceed marginal revenue (MR) and it will have to incur losses. Thus point ‘E’ will represent the equilibrium of the firm. At this point, both the conditions of equilibrium are being fulfilled
(1) Marginal cost is equal to marginal revenue (MC = MR) and
(2) The marginal cost curve is cutting a marginal revenue curve from below. At point ‘E’ i.e., equilibrium position, firm is getting maximum profit. In case, the firm produces more or less than OM output, then its profits will be less than the maximum. So the firm, at OM level of output, will have no tendency either to increase or decrease its output from this level. It will, therefore, be in equilibrium at point E.

Question 5.
(a) Explain how the income effect and the substitution effect are the reasons for the downward slope of the demand curve.
(b) Price elasticity of demand for a product is unity. A household buys 50 units of this product when its price is ₹ 10 per unit. If its price rises to ₹ 12 per unit how much quantity of the product will be bought by the household?
(c) A marginal utility schedule of a person is given below. Discuss the law underlying the given schedule:

Pen (units)	1	2	3	4	5
MU (units)	25	20	15	10	5

Answer:
(a) Income effect: It occurs when the price change affects consumer purchasing power and thus lead to a change in quantity demanded. When the price of a commodity falls, the real income of the consumer increases, a part of increased real income may be used to buy more of the commodity under question. Thus, a fall in price of a commodity increases the demand for it. Substitution effect: When the price of a commodity falls and prices of its substitutes remain unchanged, it becomes relatively cheaper in comparison to the other commodities. Thus demand for relatively cheaper commodity increases as consumers normally like to substitute cheaper goods for costlier ones.

(b) Ed = 1, Q = 50, P = ₹ 10, Q1 = x, P1 = ₹ 12

(c) The law underlying the given schedule is the Law of Diminishing Marginal Utility. As per the law of diminishing marginal utility, the level of satisfaction obtained from each additional unit of a commodity consumed keeps on declining as more and more units of a commodity are consumed. This law was formulated by the famous economist. Alfred Marshall. It is based on the following assumptions.

• All the units of a commodity consumed must be same in all respect-in size, colour, quality etc.
• The unit of the good must be the standard unit.
• No change in taste and preference of the consumer.
• No time break in consumption of every successive unit of the commodity.
• No change in the price of a substitute good.

As the schedule illustrates that when the first unit of Pen is consumed, the marginal utility derived is maximum. But with every successive unit pen used, the marginal utility goes on decreasing from 25 utils to 5 utils when 5th unit of pen is used.

Question 6.
(a) Calculate MPC, MPS and APC from the following data:

Income (Y)	Consumption
100	95
110	104

(b) Discuss the fiscal measures used to solve the situation of deficient demand.
(c) Explain how the equilibrium level of income can be determined by aggregate demand and aggregate supply.
Answer:

(b) (1) Increase in government expenditure and investment.
(2) Increase in transfer payments and subsidies.
(3) Reduction in taxes to increase the disposable income of the people.
(4) More use of deficit financing to increase the flow of money.
(5) Repayment of public debts.

(c) According to the modern theory of income and employment determination, in any economy at any given time, income and employment are determined at that level where aggregate demand is equal to aggregate supply.
AD = AS

• In the given figure AS represents Aggregate Supply Curve. It forms 45° angle, signifying’ that each point on it expresses the equal value of income and receipts whereas AD represents Aggregate Demand.
• Both i. e., AD and AS intersect each other at the point which is known as the equilibrium point.
• Point E shows that in equilibrium. position level of employment/output/income is OQ and expected receipts are OP.
• When AD > AS

Now suppose the economy is operating at any point before (left of) the equilibrium point.

As the planned spending (AD) is more than planned output (AS) this means consumers and firms together would be buying more goods than the firms are willing to produce.
As a result, the planned inventory would fall below the desired level.
Firms would be induced to increase production which will lead to an increase in the level of employment/output/income.
The firms will keep on increasing production until AD = AS.

When AS > AD, Under such a situation the economy will be operating beyond E.
As the planned spending is less than planned output (AS) this means consumers and firms together would be buying fewer goods than the firms are willing to produce.
As a result, the planned inventory would rise.
Firms would be induced to decrease production which will lead to decrease in the level of employment/output/income.
The firms will keep on decreasing production till AD = AS.

Question 7.
(a) Explain the following functions of money:
(i) Medium of exchange
(ii) Store of value
(b) Explain bow hank rate and open market operations can be used by the Central Bank to control credit.
(c) How do commercial banks create credit? Explain with the help of an example.
Answer:
(a) Money is an instrument that serves as:

• a medium of exchange
• a measure of value
• a store of value
• a standard for deferred payments.

Functions of Money
1. Medium of exchange: Most important function of money is as a medium of exchange to facilitate transactions. Without money, all transactions would have to be conducted by barter, which involves direct exchange of one good or service for another. Money effectively eliminates the double coincidence of wants problem by serving as a medium of exchange that is accepted in all transactions, by all parties, regardless of whether they desired each other’s goods and services. Use of money has separated the process of sale and purchase.

2. Store of value: In order to be a medium of exchange, money must hold its value over time; that is, it must be a store of value. If money could not be stored for some period of time and still remain valuable in exchange, it would not solve the double coincidence of wants problem and therefore would not be adopted as a medium of exchange. As a store of value, money is not unique; many other stores of value exist, such as land, works of art, and even baseball cards and stamps. Money may not even be the best store of value because it depreciates with inflation. However, still, money is used as a store of value as:

• It is more liquid than most other stores of value.
• It is readily accepted everywhere.
• It is easy and economical to store as its storage does not require much space.

3. Unit of account: Money also functions as a unit of account, providing a common measure of the value of goods and services being exchanged. Knowing the value of the price of a good, in terms of money, enables both the supplier and the purchaser of the good to make decisions about how much of the good to supply and how much of the good to purchase.
In the absence of the common measure, the seller has to express the value of his good in all other goods. For example, if you want to sell your horse you have to express its value.
1. Horse = 2 cows
1. Horse = 5 bags of wheat
1. Horse = 20 kg of iron

4. Standard of deferred payment
(i) Money is accepted as a standard of deferred payments because

• its price remains relatively stable compared to other commodities,
• it has the merit of general acceptability,
• it is more durable compared to other commodities.

It is because of this function of money that there has been a significant expansion of trade.
(ii) Using money as a standard of deferred payments is a direct consequence of the unit of account and store of value functions of money.
(iii) Serving as a standard of deferred payments, money has stimulated the process of capital formation. It is because of this function of money that there has been a considerable growth of the money market as well as the capital market.

(b) (i) Bank rate refers to the rate at which the central bank lends money to commercial banks as the leader of the resort.
(ii) During inflation, the bank rate is increased to suck excess liquidity from the economy or to reduce the money supply.
(iii) During deflection, the bank rate is reduced to increase the money supply.

(c) (i) Buying and selling of government securities in the market is known as open market operations.
(ii) Open market operations have an impact on the lending capacity of the banks.
(iii) It is an important means of controlling the money supply.
(iv) During inflation or excess demand situation the main motive of the Central Bank is to reduce the money supply. To suck excess liquidity from the market the Central Bank sells bonds, government securities and treasury bills.
(v) Due to low money supply, there is a fall in the volume of investment, income and employment resulting in lower demand.
(vi) During deflation, the main motive of the Central bank is to increase the money supply and to increase the money supply the Central Bank buys bonds, government securities and treasury bills.

Question 8.
(a) Explain any two objectives of the fiscal policy in a developing economy.
(b) What are the primary deficit and fiscal deficit in a government budget? What is the implication of the primary deficit on the economy?
(c) Explain cost-push inflation with the help of a diagram.
Answer:
(a) Objectives of fiscal policy are as follows:
Economic Stability: Developing countries face the problem of inflationary rise in prices in the course of economic development reducing the purchasing power of the people through taxes, compulsory savings and public borrowings, increasing public expenditure/investment on production of essential commodities are some of the fiscal measures to contain inflation to bring economic stability.

To achieve full employment: It is the leading objective of fiscal policy. In addition to accelerating economic growth, specific fiscal measures like tax concessions in the use of labour-intensive techniques, subsidisation of labour-intensive products, increased government expenditure on labour-intensive public works like roads, dams: undertaking special employment generation projects etc. will help in increasing the level of employment in developing countries.

(b) Primary deficit is that part of fiscal deficit which indicates borrowing requirement to make up the shortfall in receipts on account of expenditure other than the interest payments. In other words.
Primary Deficit = Fiscal Deficit – Interest payment
Implications of Primary Deficit

• It indicates how much is government borrowing under compulsion to meet expenses other than interest payments.
• A lower or zero primary deficit indicates that the interest payment of earlier loans has forced the government to borrow.

Fiscal Deficit: Refers to the excess of “Total expenditure” over the “sum of Revenue Receipts and non-debt capital receipts. It is calculated as:
Fiscal Deficit = Total Budget Expenditure (Revenue + Capital Exp.) – Revenue Receipts – Non debt Capital Receipt.
It indicates borrowing requirements of the government during the budget year.

Question 9.
(a) Classify the following as final or intermediate goods. Give reasons for your answer.
(i) A car purchased by a company for business purposes.
(ii) Pen or paper purchased by a consumer.
(b) Discuss two reasons why the per capita real income is considered to be a better index of economic welfare than gross domestic product.
(c) Calculate national income and GDP_MP by the income method using the following information:

Items	₹ in Crores
(i) Private final consumption expenditure	1300
(ii) Net factor income earned from abroad	50
(iii) Mixed-income of self-employed	500
(iv) Subsidies	100
(v) Indirect tax	200
(vi) Consumption of fixed capital	1000
(vii) Operating surplus	5000
(viii) Compensation of employees	1500

Answer:
(b) Per capita Real income is considered to be a better index of economic welfare than GDP because of following ing reasons:
(i) Only per capita, real income shows a change in per capita av ailability1 of goods and sen ices to the people of a country which leads to an improvement in the living standard of people as well as economic development.
(ii) Change in Level of Economic Activity: Increase in per capita real income implies an increase in the level of economic activity (in terms of increase in level production not in terms of an increase in market prices).

(c) National Income (NNPFC) = Compensation of Employees + Operating Surplus + Mixed-Income + NFIA
So, NNP_FC = viii + vii + iii + ii
= 1500 + 5000 + 500 + 50
= 7050
GDP_MP = NNP_FC + Depreciation – NFIA + NIT (Net indirect tax)
8100 = 7050 + 1000 – 50 + 100
NNP_FC = ₹ 7050
GDP_MP = ₹ 8100.

ISC Class 12 Economics Previous Year Question Papers

ISC Biotechnology Previous Year Question Paper 2013 Solved for Class 12

Maximum Marks: 80
Time allowed: Three hours

• Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.
• Answer Question 1 (Compulsory) from Part I and five questions from Part II, choosing two questions from Section A, two questions from Section B and one question from either Section A or Section B.
• The intended marks for questions or parts of questions are given in brackets [ ].
• Transactions should be recorded in the answer book.
• All calculations should be shown clearly.
• All working, including rough work, should be done on the same page as, and adjacent to the rest of the answer.

Part-1
(Answer all questions)

Question 1.
(a) Mention any one significant difference between each of the following : [5]
(i) Gene and Genome
(ii) Multi potent cell and Uni potent cell
(iii) Galactose and Glycine
(iv) Batch culture and Continuous culture
(v) Coding region and Non-coding region

(b) Answer the following questions : [5]
(i) Name the enzyme used in PCR. What is the source of this enzyme?
(ii) Why is Bt-cotton resistant to boll worm?
(iii) Mention any two methods of ex-situ conservation of germplasm.
(iv) What is Proteomics?
(v) Glucose and fructose have the same chemical formula (C₆H₁₂0₆), yet they differ in chemical properties. Why ?

(c) Write the Ml form of the following : [5]
(i) GDB
(ii) PIR
(iii) YAC
(iv) NCBI
(v) ddNTP

(d) Explain briefly : [5]
(i) Bacterial Artificial Chromosome
(ii) Vascular differentiation
(iii) Phenylketonuria
(iv) Quartemary protein
(v) Designer oils
Answer:
(a) (i) Gene: Gene is the unit of the genome, consisting of a sequence of DNA that occupies a specific position (locus) on a chromosome and determines a particular characteristic in an organism.

Genome: Genome is the total genetic information or all the genes contained in a haploid set of chromosomes in eukary otes, in a single chromosome in bacteria, or in the DNA or RNA of viruses.

(ii) Multipotent: These cells have the ability to differen-tiate into many of the various type of specialized cell types and can develop into any cell of a particular group or type. e.g., umbilical cord stem cells.

Unipotent: These cells can undergo unlimited reproductive divisions, but can only differentiate into a single type of cell or tissue, e.g., skin cells.

(iii) Galactose: It is a part of disaccharide that is made- up of two sugars. It is found in milk alongwith glucose. Galactose does not occur freely in nature. It is produced in the body during the digestion of disaccharide lactose.

Glycine: Glycine is a neutral amino acid and one of the 20 building blocks of protein. It is a non-essential amino acid, used in purine synthesis, and is a neurotrans¬mitter.

(iv) Batch culture: It is a type of culture in which nutrients are fed continously depending upon the amount consumed without removing growth products.

Continuous culture: It is a open type of culture in which nutrients are supplied from time to time alongwith removal of product in same volume.

(v) Coding region: Coding region (exon) is a part of the DNA that actually codes for a protein.

Non-coding region: Non-coding region (introns) is that part of DNA that does not code directly for a protein.

(b) (i) The enzyme used in PCR is TAQ – DNA polymerase I and the source of this enzyme is Thermus aquaticus.

(ii) Bt-co.tton is an insect-resistant Genetically Modified (GM) variety of cotton seed, which contains a cry gene from Bacillus thuringiensis to kill the bollworm.

(iii) Ex-situ conservation of germplasm refers to maintaining or conserving the germplasm of organism outside their natural seed habitats. Two methods of ex-situ germplasm conservation are seed banks, botanical gardens, zoological parks etc.

(iv) Proteomics is the study of entire complement of proteins, particularly their structures and functions on the large scale. The term “proteomics” was first coined in 1997 and used to make an analog}’ with genomics, the study of the genes. The word “proteome” is derived from “protein’ and “genome”, and this was coined by Marc Wilkins in 1994. It is constantly changing due to intracellular and extracellular factors.

(v) Both Glucose and fructose have the same chemical formula, but they are different because of the different arrangement of the atoms within the molecules. Glucose is an aldose with a -CHO group at position 1 while fructose is a ketose with a -C = O at position 2.

(c) (i) Genome Data Base

(ii) Protein Information Resource

(iii) Yeast Artificial Chromosome.

(iv) National Centre for Biotechnolog} Information

(v) Dideoxynucleoside triphosphate.

(d) (i) Bacterial artificial chromosome (B AC) is a cloning l ector construct, based on a fertility plasmid (or F-plasmid). which is used for transforming and cloning in bacteria, usually E.coli. An ori gene for maintainance of F factor, a selectable marker and many restriction sites for insertion of foreign DNA. The bacterial artificial chromosome’s usual insert size is 300 to 350 kbp.

(ii) Vascular tissues are complex tissues, each consisting of a number of different types of cells. Vascular differentiation refers to the process by which different types cell types arise from precursor cells and become different in structure and function from each other.

(iii) Phenylketonuria is the recessive genetic disorder caused by the absence of the enzyme phenylalanine hydroxylase which catalyzes the conversion of phenylpyruvic acid into hydroxyphcnyl pyruvic acid. It is caused due to mutation of gene.

(iv) Quartemary proteins are the multimeric proteins i.e.. proteins hav ing more than two or more polypeptide chains which are linked to form quartemary structure, e.g.. Haemoglobin.

(v) Designer Oil: “Designer oil” that reduces LDL (‘bad“) blood cholesterol levels in humans and increases energy expenditure which may prevent people from gaining weight. The oil incorporates a phytosterol-based functional food ingredient Phytrol (TM) from Forbes into oil using proprietary technology.

Part-II
(Answer any five questions)

Question 2.
(a) Give a comparative account of DNA and RNA on the basis of their following characteristics: [4]
(i) Chemical composition and structure
(ii) Location and function
(b) Mention the uses of the following in genetic engineering techniques : [4]
(i) Shuttle vectors and Expression vectors
(ii) Restriction endonucleases
(c) What is electroporation ? [2]
Answer:
(a) (i) DNA:

• DNA has 2-Deoxyribose sugar.
• It contains cytosine and thymine as pyrimidine.
• It has a double stranded helix struc-ture.

RNA:

• RNA has ribose sugar.
• It contains cytosine and uracil as pyrimidine.
• It has a single stranded helix.

(ii) DNA:

• DNA occurs in the nucleus chloroplast and mitochondria of cell.
• It controls transmission of hereditary characters.

RNA:

• RNA occurs in cytoplasm of the cell.
• It controls the synthesis of proteins.

(b) (i) Shuttle vectors exist and work and allow DNA to be transferred between both prokaryotes and eukaryotes. The shuttle vector has two origins of replication i.e.. on^E and ori^Euk allowing replication to occur in either system/host. It “shuttles” between two different species. It can be used to perform reverse genetics, e.g.. Yeast episomal plasmid (YEP). Expression vectors allow expressing certain genes directly from their recombinant DNAs A typical expression vector will have a promoter upstream of the DNA containing the sequence to be expressed.

(ii) Restriction endonucleases are enzymes that cleave DNA at specific nucleotide sequences. The sequence recognized is often four to six nucleotides long. For example, the restriction endonucleases Eco RI recognize the sequence. GAATTC.

(c) Electroporation is a mechanical method used to introduce polar molecules into a host cell through the cell membrane. In this procedure, a brief exposure to a high electric voltage pulse temporarily disturbs the phospholipid bilayer, allowing introduction of molecules like DNA to pass into the cell.

Question 3.
(a) What is gene cloning ? Mention the steps involved in this process
(b) Explain the following :
(i) Acidic and basic amino acids
(ii) Phospholipids and glycolipids
(c) State any four objectives of germplasm conservation.
Answer:
(a) The process of formation of similar copies of a desired gene is called gene cloning. Gene cloning is the technique of recombinant DNA technology’ in which a desired gene of interest having a characteristic feature is cloned. Gene cloning involves the replication of DNA fragments by the use of self-replicating vector’s genetic material for its multiplication, expression or integration into host chromosome.

Steps involved in gene cloning :

• In cloning a gene the first step is to isolate the DNA segment from the organism that contains the gene of interest.
• Remove the gene of interest from the DNA, by using restriction enzymes or by PCR.
• Vector is also treated with same restriction enzyme, to cleave it. Vector come to possessing single strand at the ends called stick}’ ends.
• Then the enzy me DNA ligase is used to insert the gene of interest to be cloned into the plasmid. Vector having sicky ends to form recombinant DNA.
• The plasmid or vector acts as a vehicle that transports the desired gene into a host cell, the process is known as transformation.
• Now, these recombinant plasmids are inserted into bacterial host cells, where they replicate to amplify the desired gene, the process is called gene cloning.
• Now the cell can be plated out on an agar medium. The colony of cells containing the desired cloned gene can be identified and isolated.

(b) (i) Amino acids are the basic structural unit of all proteins. A free’ neutral amino acid (a single amino acid) always has an amino group -NH₂. a carboxyl group -COOH, hydrogen -H and a chemical group or side chain -”R”.

Acidic amino acid :
Two amino acids have acidic side chains at neutral pH. These are aspartic acid or aspartate (Asp) and glutamic acid or glutamate (Glu). Their side chains have carboxylic acid groups whose pKa’s are low enough to lose protons, becoming negatively charged in the process. Such amino acids are highly polar.

Basic amino acid :
Three amino acids have three basic side chains at neutral pH. These are arginine (Arg), lysine (Lys). and histidine (His). Their side chains contain nitrogen and resemble ammonia, which is a base. Their pKa’s are high enough that they tend to bind protons, gaining a positive charge in the process.

(ii) Phospholipids are the phosphorylated triglyceride lipids in which one fatty acid is replaced by phosphate group added by phosphorylation. Glvcolipids are the glycosylated lipids in which sugar residue galactose or carbohydrate molecule is added by glycosylation. Phospholipids and glycolipids both are the derivatives of lipids. They form an essential component of cell membrane which plays a role in structure, maintenance and also help in eliciting certain immune reactions.

(c) Objectives of Germplasm Conservation :

• Conservation of rare germplasm arising through somatic hybridization.
• Storage of pollen for enhancing longevity.
• Maintainance of recalcitrant seeds.
• To develop genes for adaptations / endurance to varying, unfavorable biotic/abiotic stresses / environments.
• To develop high yielding varieties.

Question 4.
(a) Why are enzymes temperature sensitive ? Briefly explain the mode of action of enzymes on their substrate. [4]
(b) How is the hormone insulin synthesized, using genetic engineering technique ? State two ways in which this technique is better than the techniques used earlier. [4]
(c) What is a supra-molecular assembly ? [2]
Answer:
(a) Enzymes are temperature sensitive because almost all enzymes are proteins have tertiary structure and only function in a specific range of temperature. Exposing enzymes to high temperature break bonds and can cause them to denature, which alter the shape of the enzyme. Due to change in shape the substrate no longer ‘fits’ inactive site of the enzyme and can no longer function as normal.

Mode of enzyme action : It can be explained by this models :

Lock and key mechanism : This model was proposed by Emil Fisher in 1898. It is also called the template model. According to this model the union of the substrate and the enzyme takes place at the active site, more or less in a manner in which a key fits in a lock and results in formation of an enzyme substrate complex. As the two molecules are involved, this hypothesis is also known as the concept of inter molecular fit. The ES complex is highly unstable and almost immediately this complex breaks to produce the end product of the reaction and regenerate the free enzyme. The ES complex results in the release of energy.

Examples:
Catalase : It catalyzes the decomposition of hydrogen peroxide into water and oxygen.
2H₂O₂ → 2H₂O + O₂
One molecule of catalyses can break 40 million molecules of hydrogen peroxide each second.

The first major medicinal product of genetic engineering is human insulin called Humulin. Insulin is a protein that acts as a hormone to stimulate uptake of blood sugar into tissues, such as the liver and the muscles.
Following are the steps which are involved in insulin synthesis :

• Isolate the gene responsible for producing human insulin protein. The gene is a part of the DNA in a human chromosome.
• Then remove a circular piece of DNA called plasmid from a bacterial cell. Special restriction enzymes are used to cut the plasmid ring open with sticky ends.
• With the plasmid ring open, the gene for insulin is inserted into the plasmid ring and the ring is closed with ligase enzyme forming recombinant DNA. This process is called recombinant technology’.
• The bacterial plasmid DNA now contains the human insulin gene and is inserted into a bacteria.
• Many plasmids with the insulin gene are inserted into many bacterial cells. When the bacterial cells reproduce by dividing, the human insulin gene is also cloned in the newly cloned cells.
• Human insulin protein molecules produced by bacteria are gathered and purified by down stream process by culturing the genetically engineered bacteria, limitless supplies of insulin may be produced.

Two ways in which genetic engineering is better than the technique used earlier:

• Insulin produced by genetic engineering is pure and has no allergic reaction.
• The human insulin is much cheaper when produced by r-DNA technology than was the insulin from cows, as it could be produced much more quickly in greater quantity.

(c) A supra molecular assembly or “super molecule” is a well defined complex of molecules held together by non-covalent bonds. Molecules are combined in the form of sphere or rod. The dimensions of supra molecular assemblies can range from nanometres to micrometers. The process by which a supra molecular assembly forms is called molecular self-assembly.

Question 5.
(a) What is plant tissue culture ? Discuss the organization of a tissue culture laboratory under the following headings: [4]
(i) Media preparation
(ii) Culture room.
(b) Explain any two methods used for the identification of recombinant host cells from the non-recombinant host cells. [4]
(c) Name any four in vivo techniques employed in haploid production. [2]
Answer:
(a) Plant tissue culture is the technique of in vitro maintenance and growth of plant cells, tissues and organs under aseptic conditions on a suitable artificial culture medium contained in small containers under controlled environmental conditions of temperature and light.

(i) Media Preparation Room : An area is required for preparation of media. In such space there should be provision for bench space for chemicals, labware, culture vessels, closures and miscellaneous equipment required for media preparation and dispensing. In this room provision is also made for placing hot plates or stirrers, pH meter, balance, waterbath, burners, oven, autoclave, culture vessel, refrigerator etc.

(ii) Culture Room : All types of cultured plant tissues are incubated under the conditions of well controlled temperature, humidity, illumination and air circulation. The culture room should have light and temperature control system. Generally temperature is maintained at 25±2°C and 20-98% relative humidity and uniform air ventilation. The cultures are grown in diffused light and darkness each for a period of 12 hours.

(b) The introduction of the recombinant DNA in to a suitable host cell is followed by the selection of those cells, which contain the recombinant vectors. There are various selection methods that are based on the expression or non-expression of some of the traits present in the vector or alongwith the cloned gene.

Antibiotic sensitivity : Recombinant plasmid has many traits such as ori recognition site and selectable marker gene. Some of these traits are resistant to certain antibiotics. If the antibiotic resistant gene is present alongwith the cloned gene, it is very easy to select the recombinant transformants directly on a medium supplemented with respective antibiotic.

In most of the cases there are two stages of selection. First is the selection on the basis of . transformed cells i.e., the cells that have taken a plasmid. The second one is to identify the transformed cells that have the recombinant plasmid. The presence of a desired DNA insert can be confirmed either by isolating the recombinant plasmids and digesting it with the same restriction enzyme used for making the recombinant vectors, by PCR, by southern hybridisation with DNA probes, by northern hybridisation with RNA probes and by direct DNA sequencing

lnsertional inactivation : Another method to differentiate between recombinant and non¬recombinant is on the basis of their ability to produce colour.
lnsertional inactivation: In this method, a recombinant DNA is within the coding sequence of an enzyme p-galactosidase. This results into the inactivation of enzyme which is referred to as insertional inactivation.

The bacterial colonies whose plasmids do not have insert, produce blue colour but those with an insert or the recombinant do not produce any colour and are identified as recombinant colonies.

(c) In vivo techniques employed in haploid production are gynogenesis, ovule and rogenesis, genome elimination by distant hybridisation or chemical treatment and semigamy.

Question 6.
(a) Write the principle and any two applications of each of the following biochemical techniques : [4]
(i) Ion – exchange chromatography
(ii) Gel – permeation
(b) What is a genetic code ? Enlist three important properties of genetic code. [4]
(c) What are DNA probes ? [2]
Answer:
(b) (i) Principle of Ion-exchange chromatography : It is defined as the reversible exchange of ions in solution with ions electrostatically bound to some sort of insoluble support medium. Separation is obtained since different molecules have different degree of interaction with the ion-exchanger due to difference in their charges, charge densities and distribution of charge on their surfaces. These interactions can be controlled by varying conditions such as ionic strength and pH.

An ion-exchanger consists of an insoluble matrix to which charged groups have been covalently bound. Ion exchange separations are carried out mainly in columns packed with an ion-exchanger. There are two types of ion-exchanger, namely cation and anion exchangers. Cation exchangers possess negatively charged groups and these will attract positively charged cations. Anion exchangers have positively charged groups that will attract negatively charged anions. After the ion exchange the molecules can be eluted from the matrix by selective desorption. The selective desorption can be achieved by changes in pH and /or ionic concentration or by affinity elution, in which case an ion that has greater affinity for the exchange than has the bound ion is introduced into the system.

Applications:

• Polystyrene and polyphenolic ion exchange resins are more often used to separate srhall molecule such as amino acids, small peptides, nucleotides, N-bases, cyclic nucleotides, organic acids.
• The cellulose ion exchangers are commonly used for proteins, including enzymes, polysaccharides and nucleic acids.

Principle of Gel-permeation chromatography: Gel permeation / filtration chromatography is a separation technique which uses molecular sieves, composed of neutral cross-linked carriers e.g., polymers like agarose, dextrans of different pore sizes. Therefore, it can separate macromolecule of different sizes from one another. Molecules smaller than pore size either the carrier and are retained. They are later eluded (in order of molecular size) and collected. Other names that have been suggested for this technique are : get filtration, molecular or size exclusion chromatography ; or molecular sieve chromatography.

Applications:

• Separation of polysaccharide, enzymes, antibodies and other proteins.
• Separation of non-polar species such as triglycerides in non-aqueous mobile phases.
• Used to analyse the molecular-weight distribution of organic soluble polymer.

The genetic code is called a triplet code, i.e sequence of three nitrogenous bases on m-RNA that specifies the recognition of a particular of a single amino acid. Thus, the information encoded in the sequence of nitrogenous bases must be read in groups of three, (UAC, GGC, UGC).

Three important properties:

1. Triplet code : Three adjacent nitrogen bases constitute a codon which specifies the placement of one amino acid in a polypeptide.
2. Start signal : Polypeptide synthesis is signaled by AUG or methionine codon and GUG — Valine codon. They have dual function.
3. Stop signal: Polypeptide chain termination is signaled by three termination codons — UAA, UAG, and UGA. They do not specify any amino acid and are hence also called non-sense codon.
4. Universal code : The genetic code is applicable universally i.e., the codon specifies the same amino acid from a virus to a tree or human being.
5. Non-ambiguous codon : One codon specifies only one amino acid and not any other.

(c) DNA Probe : It is a solution of radioactive, single-stranded DNA or oligodeoxy nucleotides (a DNA segment of few to several nucleotides). The name probe signifies the fact that this DNA molecule is used to detect and identify the DNA fragment in the gel/membrane that has a sequence complementary to the probe. The probe hybridises with the complementary DNA on the membrane to the greater extent with a low non-specific binding on the membrane. This step is known as hybridisation reaction.

Question 7.
(a) How can the following plants be obtained, using genetic transformation techniques . [4]
(i) Drought and salinity tolerant plants
(ii) Somatic hybrids
(b) Explain the process involved in the transcription of DNA to mRNA. Also, mention any two post transcriptional changes that occur in the mRNA formed. [4]
(c) What are Okazaki fragments ? How are they joined ? [2]
Answer:
(i) Drought tolerance : Water is crucial for all living things. Plants use water as a solvent, a transport medium, an evaporative coolant, physical support, and as a major ingredient for photosynthesis. Without sufficient water, agriculture is impossible. Therefore, drought tolerance is an extremely important agricultural trait.

One way of engineering drought tolerance is by taking genes from plants that are naturally drought tolerant and introducing them to crops. The resurrection plant (Xerophyta viscosa), a native of dry regions of southernmost Africa, possesses a gene for a unique protein in its cell membrane. Experiments have shown that plants given this gene are less prone to stress from drought and excess salinity.

Some genes have been found that control the production of the thin, protective cuticle found on leaves. If crops can be grown with a thickened waxy cuticle, they could be better equipped for dealing with dryness.

Salt tolerance: Irrigation has enabled the transformation of arid regions into some of the world’s most productive agricultural areas. Excess salinity, however, is becoming a major problem for agriculture in dry parts of the world. In several cases, scientists have used biotechnology to develop plants with enhanced tolerance to salty conditions.

Researchers have noticed that plants with high tolerance to salt stress possess naturally high levels of a substance called glycine betaine. Further, plants with intermediate levels of salinity tolerance have intermediate levels, and plants with poor tolerance to salinity have little or none at all. Genetically modified tomatoes with enhanced glycinebetaine production have increased . tolerance to salty conditions.

Another approach to engineering salt tolerance uses a protein that takes excess sodium and diverts it into a cellular compartment where it does not harm the cell. In the lab, this strategy was used to create test plants that were able to flower and produce seeds under extreme salt levels. Commercially available crops with such a modification are still several years away.

(ii) Process, other than the sexual cycle has recently become available for higher plants, which can lead to genetic recombination. This non-conventional genetic procedure involving fusion between isolated somatic protoplasts under in vitro conditions and subsequent development of their product (heterokaryon) to a hybrid plant is known as somatic hybridisation.

Application of Somatic Hybridisation :

• Somatic cell fusion appears to be the only means through w hich two different parental genomes can be recombined among plants that cannot reproduce sexually (asexual or sterile).
• Protoplasts of sexually sterile (haploid, triploid, and aneuploid) plants can be fused to produce fertile diploids and polyploids.
• Somatic cell fusion overcomes sexual incompatibility barriers. In some cases, somatic hybrids between two incompatible plants have also found application in industry or agriculture.
• Somatic cell fusion is useful in the study of cytoplasmic genes and their activities and this information can be applied in plant-breeding experiments.

(b) The process of transcription: Transcription is the process of creating a messenger RNA strand from DNA, performed by the enzyme RNA polymerase, Transcription always occurs in a 5′ → 3′. direction, with polymerase moving 3′ → 5′ along the DNA strand.

Transcription Initiation : There are three steps in transcription :
Initiation : RNA synthesis begins after the RNA polymerase attaches to the DNA and unwinds it. RNA synthesis will always occur on the template strand.

Elongation : RNA polymerase unwinds the DNA double helix and moves downstream and elongates the RNA transcript by adding ribonucleotides in a 5′ → 3′ direction. Each ribonucleotide is added to the growing mRNA strand using the base pairing rules (A binds with T, G binds with C). For each C encountered on the DNA strand a G is inserted in the RNA, for each Q a C and for each T, an A is inserted. Since there is no T in RNA, U is inserted whenever an A is encountered. After RNA polymerase has passed, the DNA restores its double stranded structure.

Termination: When the mRNA is complete, the mRNA is released and the RNA polymerase releases from the DNA.

Two post transcriptional changes that occur in the mRNA formed are:
RNA transcripts eukaryotes are modified or processed, before leaving the nucleus to produce functional wRNA. It is processed in two ways :
(1) 5 ‘ capping : Capping of the pre-mRNA involves the addition of 7-methylguanosine (m7G) to the 5′ end.,
(2) 3′ polyadenylation: The pre-mRNA processing at the 3′ end of the RNA molecule involves cleavage of its 3′ end and then the addition of about 200 adenine residues to form a poly (A) tail. The cleavage and adenylation reactions occur if a polyadenylation signal sequence (5′ – AAUAAA-3′) is located near the 3′ end of the pre-mRNA molecule, followed by another sequence, which is usually (5′ -CCA-3’).

(c) Okazaki fragments are short, newly synthesized DNA fragments produced discontinously in pieces during DNA replication. They are formed on the lagging template strand and are complementary to the lagging template strand. Okazaki fragments are joined together by DNA ligase enzyme.

Question 8.
(a) What is meant by the term genomics ? Write the differences between structural genomics and functional genomics. [4]
(b) Name and explain any four methods of synchronization of cells. [4]
(c) What is meant by Expressed sequence tags ? [2]
Answer:
(a) The word ‘genomics’ has taken root from the term ‘genome’ which is an organism’s total genetic constitution mapping, sequencing and analyzing the genomic information to solve a medical, industrial or biological query. Genomics studies investigate structure and function of genes and do this simultaneously for all the genes in a genome. Genomics is broadly categorized into structural and functional genomics.

Structural Genomics : The structural genomics deals with DNA sequencing, sequence assembly, sequence organisation and management. Basically it is the starting stage of genome analysis i.e,. construction of genetic, physical or sequence maps of high resolution of the organism. The complete DNA sequence of an organism is its ultimate physical map. Due to rapid advancement in DNA technology and completion of several genome sequencing projects for the last few years, the concept of structural genomics has come to a state of transition. Now it also includes systematics and determination of 3D structure of proteins found in living cells. Because proteins in every group of individuals vary and so there would also be variations in genome sequences.

Functional Genomics: It is based on the information of structural genomics the next step is to reconstruct genome sequences and to find out the function that the genes do. This information also lends support to design experiment to find out the functions that specific genome does. The strategy of functional genomics has widened the scope of biological investigations. This strategy is based on systematic study of single gene / protein to all genes/proteins.

Therefore, the large scale experimental methodologies (along with statistically analysed / computed results) characterise the functional genomics. Hence, the functional genomics provide the novel information about the genome. This eases the understanding of genes and function of proteins, and protein interactions.

(b) Cell culture synchronization : Cells in suspension cultures vary greatly in size, shape, DNA, and nuclear content. Moreover, the cell cycle time varies considerably within individual cells. Therefore, cell cultures are mostly asynchronous.

It is essential to manipulate the growth conditions of an asynchronous culture in order to achieve a higher degree of synchronization. A synchronous culture is one in which the majority of cells proceed through each cell cycle phase (G1: S, G2 and M) simultaneously.

Synchronization can be achieved by following methods :

• Physical methods include selection by volume (size of cell aggregate.)
• Chemical methods include starvation (depriving suspension cultures of an essential growth compound and culture supplying).
• Chemical methods include inhibition (temporarily blocking the progression of events in the cell cycle using a biochemical inhibitor and then releasing the block).

(c) An Expressed Sequence Tag or EST is a short sub-sequence of a transcribed cDNA sequence represents a partial gene. They may be used to identify gene transcripts, and are instrumental in gene discovery and gene sequence determination, used in micro-arrays.

Question 9. .
(a) What is Human Genome Project ? Mention its objectives and significant achievements. [4]
(b) Write short notes on : [4]
(i) Locus – link
(ii) Microprocessor
(iii) EMBL
(iv) Taxonomy Browser
(c) What is site-directed mutagenesis ? [2]
Answer:
(a) The Human Genome Project (HGP) : This is an international scientific research project with a primary goal to determine the sequence of chemical base pairs which make-up DNA and to identify the approximately 25,000 genes of the human genome from both a physical and functional standpoint.

Benefits: The work on interpretation of genome data is still in its initial stages. It is anticipated that detailed knowledge of the human genome will provide new avenues for advances in medicine and biotechnology. A number of companies, such as Myriad Genetics started offering easy ways to administer genetic tests to a variety of illnesses, including breast cancer, disorders of homeostasis, cystic fibrosis, liver diseases and many others. Also, the etiologies for cancers, Alzheimer’s disease and other areas of clinical interest are considered likely to benefit from genome information and possibly may lead in the long term to significant advances in their management.

There are also many tangible benefits for biological scientists. For example, a researcher investigating a certain form of cancer may have narrowed down his/her search to a particular gene. By visiting the human genome database on the world wide web, this research can examine what other scientists have written about this gene, including (potentially) the three-dimensional structure of its product, its function(s), its evolutionaty relationships to other human genes, or to genes in mice or yeast or fruit flies, possible detrimental mutations, interactions with other genes, body tissues in which this gene is activated, diseases associated with this gene or other data types.

(b) Locus Link is a National Center for Biotechnology Information (NCBI) online resource. It is designed to link together related information on genetic loci and gene products from several sources.

Microprocessor : A microprocessor or processor is the heart of the computer and it performs all the computational tasks, calculations and data processing etc. inside the computer. Microprocessor is the brain of the computer. In the computers, the most popular type of the processor is the Intel Pentium chip and the Pentium IV is the latest chip by Intel Corporation. The microprocessors can be classified based on the following features.
Instruction Set: It is the set of the instructions that the Microprocessor can execute.
Bandwidth : The number of bits processed by the processor in a single instruction.
Clock Speed : Clock speed is measured in the MHz and it determines that how many instructions a processor can processed.

European Molecular Biology Laboratory (EMBL) : It was established to collect, organise and distribute data on nucleotide sequence and other information rebated to them. Nucleotide Sequence Database (also known as EMBL -Bank) constitutes Europe’s primary nucleotide sequence resource. Main sources for DNA and RNA sequences are direct submission from individual researches, genome sequencing projects and patent applications.

Taxonomy Browser: This search tool provides taxonomic information on various species. The Taxonomy database of NCBI has information (including scientific and common names) about all organisms for which some sequence information is available (over 79,000 species). The server provides genetic information and the taxonomic relationship of the species in question. Taxonomy has links with other servers of NCBI e.g., structure and PubMed.

(c) Site-directed mutagenesis is a molecular biology technique in which a mutation is created at a specific site in the DNA molecule.

ISC Class 12 Biotechnology Previous Year Question Papers