IIT JAM Biology 2015
Previous Year Question Paper with Solution.

1. Which one of the following most accurately describes the process of natural selection?

(a) Selection of one species over a competing species

(b) Selection of individuals that successfully defend themselves against enemies

(c) Selection of individuals that produce more than the average number of offspring

(d) Selection of individuals that are more attractive to the opposite sex

Ans. (*)

Sol. Natural selection is the key mechanism of evolution which explains that organisms adopt to survive and pass their genes to their off springs (reproduction). All the four statements support the theory of natural selection.

2. Identify the statement that is true of operons.

(a) Fine regulation of the expression of individual genes are made possible by operons

(b) Only genes involved in carbohydrate metabolism are present in operons

(c) Feedback inhibition of the biosynthesis of multiple enzymes by a single small molecule is made possible by operons

(d) In the case of inducible operons, the inducer binds to the operator

Ans. (c)

Sol. Operon is present in prokaryote. It is unit of genomic DNA that contains many genes working under the influence of single promotes. Feedback inhibition by a single molecule could be possible by operon.

3. Signaling pathways usually comprise of several intermediate steps that are arranged in the form of a cascade. What is the primary outcome of such an arrangement?

(a) Specificity of signal transduction

(b) Specificity of the cellular response

(c) Amplification of the cellular response

(d) Fine-tuning of the cellular response

Ans. (c)

Sol. Signalling mechanisms involve cascade of reactions in many intermediate steps forming a pathway that allows better functioning of the system as the feedback no longer presents the initial buildup which involves the propagating signal. Amplification of cellular response is also possible by this cascade.

4. Which among the following contain(s) oxygen-rich blood in the human vascular system?

I. Right ventricle

II. Aorta

III. Pulmonary vein

(a) Only I

(b) I and II

(c) I, II and III

(d) II and III

Ans. (d)

Sol. Arteries and veins are the major blood vessels that are connected to heart Aorta is artery that carries blood from left side of heart to body whereas pulmonary vein carries oxygenated blood from lungs to left side of heart which will be pumped to the body.

5. Choose the option that lists the correct sequence of steps involved in gene therapy.

P. Injection of expression vector into patient

Q. Wild-type gene is inserted into expression vector

R. Wild-type gene is isolated and cloned

S. Wild-type gene is transcribed and translated in the patient.

(a) Q, S, P,R

(b) Q, P, R, S

(c) R, P, Q, S

(d) R, Q, P, S

Ans. (d)

Sol. Gene therapy is a useful technique to use desired genes to treat and prevent diseases. It involves the isolation and cloning of the desired gene of interest into expression vector and injecting this vector carrying our desired gene into the patient.

6. Cephalin, a biological surfactant is

(a) choline phosphoglyceride

(b) ethanolamine phosphoglyceride

(c) glycosphingolipid

(d) sphingolipid

Ans. (b)

Sol. Cephalin belonging to phospholipid family is a biological surfactant also known as phosphatidylethanolamine.

It is found in white matter of brain and spinal cord showing hemostatic properties

7. The major product(s) produced by gas phase UV irradiation of 2 pentanone is(are)

(a) acetone and ethene

(b) acetic acid and propionic acid

(c) 2-pentanol

(d) cyclopentane

Ans. (a)


Photoexcited species able to be fragmented. Here, 2-pentanone is a photoexcited species which are fragmented to ethene and acetone.

8. If a projectile lifts off from the surface of the Earth with a speed of 11.2 kms–1, then it can escape from the Earth's gravitational field completely. This is called the escape velocity. If radius of the Earth were 2 times larger and the mass 8 times larger, then the escape velocity (in km s-) would be

(a) 5.6

(b) 11.2

(c) 22.4

(d) 44.8

Ans. (a)

Sol. As we know that

Escape velocity (ve) =

Where, G = Gravitational constant

 M = Mass of planet

  R = Radius of planet

For earth

For other planet


As given,    Mo = 8 Meand Ro = 2Re


From Eqs. (i) and (ii)

9. The speed of an electron (v), in the lowest energy orbit in the Bohr model of the hydrogen atom divided by the speed of light in vacuum (c), is given by (where, m is the mass of the electron, M is the mass of the proton, is the permittivity of free space. a0 is the Bohr radius)





Ans. (d)

Sol. As we know, angular momentum of a revolving electron is given by,


(a) exactly two elements

(b) exactly three elements

(c) exactly four elements

(d) infinitely many elements

Ans. (b)

Sol. Given set P = {x R : (x – 1) (x2 + 1) = 0}

Solution of the equation

(x – 1) (x2 + 1) = 0

x – 1 = 0 or x2 + 1 = 0

  x = 1  or  x2 = – 1

x = = i

  x = 1, i

  P = {1} {  P = {x : x belongs to R}]

Set Q = {x R : x2 – 9x + 2 = 0}

Solution of the equation, x2 – 9x + 2 = 0

11. In a population growing according to the logistic growth model

(a) individuals reproduce according to their physiological capacity

(b) the per capita rate of increase approaches zero as the population nears the carrying capacity

(c) the number of births is always more than the number of deaths

(d) the birth-to-death ratio is not influenced by the carrying capacity

Ans. (b)

Sol. Population growth is studied by logistic and exponential growth models. Logistic model explains the growth phenomena under limited resources which leads to competition among individuals and survival to fittest concept of carrying capacity (k) exists under logistic growth which explains that a given habitat has enough resources to support a max no. of popultation beyond which growth mostly occur.

12. Which part of the genomic DNA contains the sequence corresponding to the 5′ untranslated region (5′ UTR)?

(a) Exon

(b) Intron

(c) Upstream of the transcription start site

(d) Upstream of the promoter

Ans. (a)

Sol. In mRNA, upstream of the initiator codon lies the 5' UTR (untranslated region) which plays important role in regulation of translation process. It contains nucleotide sequence that corresponds to exon of genomic DNA

13. Which one of the following is not true of RNA polymerase II?

(a) It requires a primer to initiate the transcription

(b) It makes an RNA copy of only one strand of a double-stranded DNA at any given time

(c) It does not synthesise rRNA and tRNA

(d) It catalyses the polymerisation of RNA only in the 5′→ 3′ direction

Ans. (a)

Sol. RNA pol II is the most studied RNA polymerase that is used for the transcription to form mRNA No primers are needed for initiation of transcription and the polymerisation occurs in 5'–3' direction. mRNA is formed from only one strand of dsDNA at a given time by RNA pol II.

14. Choose the option that shows the correct pairing of the cellular components in Column I with their corresponding function in Column II.

(a) P-IV, Q-1, R-II, S-III

(b) P-III, Q-IV, R-II, S-I

(c) P-III, Q-I, R-II, S-IV

(d) P-II, Q-I, R-IV, S-III

Ans. (b)

Sol. Dynein is the protein that helps in transforming the chemical energy of ATP into mechanical energy of movement, thus also known as a motor protein. It helps in beating of flagella in bacteria.

Desmosome – They are specialised structures arranged on the lateral sides of plasma membranes used for cell to cell adhesion or attachment of cells.

Endosome – It is found in eukaryotic cells. This is a membrane bound structure involved in endocytic membrane transport pathway that occurs between plasma membrane to lysosome.

Kinesin - This protein is a motors protein in eukaryotes that facilitate movement of organelles. They move along microtubules and consume energy by ATP hydrolysis.

15. An enzyme shows highest activity in the pH range 2.0-3.0. At pH 4.0 and pH 7.0, the enzyme exhibits 50% and 1%, respectively, of its highest activity. Which of the following states of an amino acid residue in the catalytic site is most responsible for its activity profile?

(a) A protonated Asp

(b) A deprotonated Asp

(c) A deprotonated Asn

(d) A protonated Asn

Ans. (a)

Sol. Release of proton by the protonated aspartic acid influences the activity profile of enzyme

16. The specific productivity (qp) of cellulase production by Aspergillus niger follows a linear relationship with the specific growth rate and is of the form qp = where and are constants. Assuming that the values of and are 0.006 and 25, respectively, which type of product formation kinetics is true?

(a) Growth dependent kinetics

(b) Non-growth dependent kinetics

(c) Both growth and non-growth dependent kinetics

(d) Inhibition kinetics

Ans. (b)

Sol. Specific productivity (qp) =

Gluen a = 0.006

= 25

Putting the values we get

qp = 0.006 + 25

Since rate of change of specific productivity with respect to growth rate is constant and independent of growth rate

Production formation kinetics is non-growth dependent.

17. Choose the option that shows the correct pairing of the diseases in Column I with their corresponding causative organisms in Column II.

(a) P-IV, Q-1, R-III, S-II

(b) P-IV, Q-I, R-II, S-III

(c) P-I, Q-IV, R-II, S-M

(d) P-I, Q-IV, R-III, S-II

Ans. (b)

Sol. • Chagas disease – It is caused by protozoan named trypanosoma cruzi spread by insects like kissing bugs triatominae

• Sleeping sickness – It is caused by trypanosoma gambiense and is transmitted by bite of tse – tse fly. It is heavelly prevalent in rural areas.

• Malaria – It is caused by parasitic protozoan named plasmodium falciparum and spread

by mosquit or bites to humans. Symptoms include fever headache

• Plague – It is infections dismease caused by enterobacteria yersinia pestis. It can spread in air, by direct contact by contaminated uncooked food material.

18. Choose the option that shows the correct pairing of the products in Column I with their corresponding microorganisms in Column II.

(a) P-II, Q-III, R-I, S-IV

(b) P-IV, Q-II, R-III, S-I

(c) P-III, Q-IV, R-I, S-II

(d) P-III, Q-I, R-IV, S-II

Ans. (c)

Sol. Citric acid – This acid is found in citrus fruits which acts as a natural preservative for food items and also adds its flavour.

It is produced by aspergillus niger

Polyhydroxy alkonales (PHAs) are linear polyesters produced in nature by bacterial fermentation of sugar as lipids which are produced by recombinant Ralstonia Eutropha to store carbon and energy.

Gentamycin – It is produced by Micromonospora Purpunia and is used to treat Gram –ve infections. It is an aminoglycoside based antibiotic.

Ethanol – It is produced by zymomosas mobilis and is used in alcoholic beverages.

19. Determine the correctness or otherwise of the following Assertion (A) and Reason (R).

Assertion (A) : B-cells secrete antibodies against a virus, while cytotoxic T-cells kill virus-infected cells.

Reason (R) : B-cells confer active immunity, while cytotoxic T-cells confer passive immunity

(a) (A) and (R) are true and (R) is the correct reason for (A)

(b) (A) and (R) are true but (R) is not the correct reason for (A)

(c) (A) is true but (R) is false

(d) (A) is false but (R) is true

Ans. (c)

Sol. B cells that are activated differntiate to form plasma cells that release soluble antibodies and memory cells in blood against pathogen thereby providing passive immunity whereas T cells provide active immunity by killing virus infected cells, alcogesafts, cancer cells etc. Thus essertion is true while reason is false.

20. Which one of the following options shows the correct pairing of the enzyme in Column corresponding application in Column II?

(a) P-IV, Q-I, R-II, S-III

(b) P,IV, Q-III, R-I, S-II

(c) P-IV, Q-I, R-III, S-II

(d) P-I, Q-II, R-IV, S-III

Ans. (a)

Sol. Papain – It is a cysteine protease enzyme present in papaya commonly used in cosmetics, cleaners, heat products, etc.

Bromelain – It is mixture of enzyme found in pineapples that aid protein digestion generally used in baking

Peroxidases – It is used an indicator for adequacy of vegetable blanching due to its high thermal stability and mide distribution. It is also used as an immuno-array marker enzyme.

-amylase – It acts on starch (polysaccharides) producing -maltose by inversion. Animal tissues lack this enzyme but is present in microorganisms might be present in our digestive tract. Optimum temprature for this enzyme is 40 – 50. It is commonly used in beer, flour additive, syrups etc.

21. The rate constant for the reaction O(g) + O3(g) is 8.0 × 10–15 cm3 molecules–1 s–1.

The rate constant in dm3 mol–1 s–1, would be

(a) 4.8 × 10–6

(b) 4.8 × 106

(c) 4.8 × 10–9

(d) 8.0 × 106

Ans. (b)

Sol. n = 8.0 × 10–15 cm3 molecule–1 s–1

1 mol = 6.022 × 1023 molecules

= 8.0 × 10–18 × 6.022 × 1023

= 4.8 × 106 dm3 mol–1 s–1

22. The UV spectrum of 2-butanone and the UV spectrum of methyl vinyl ketone (MVK) are independently recorded and compared. Among the various a , 185 nm, 219 nm, 277 nm and 324 nm, the absorption at a = 324 nm is due to

(a) transition in 2-butanone

(b) transition in MVK

(c) transition in 2-butanone

(d) transition in 2-butanone

Ans. (b)

Sol. 2-butanone and methyl vinyl ketone give transition. In this type of transition, an electron of unshared electron pair present on hetero atom gets excited to (antibonding) orbital. This type of transition requires least amount of energy compared to all the transitions and occurs at lower wavelengths.

2-butanone is not a conjugated system which has absorption at a = 219 nm, while methyl vinyl ketone conjugated system and its absorption at

= 324 nm. Therefore, the absorption at

= 324 nm is due to transition in MVK.

23. The compound meso 2, 3-dibromobutane is obtained by

(a) electrophilic addition of HBr to (E)-1-bromobut-2-ene

(b) electrophilic addition of Br2 to (E)-2-butene

(c) electrophilic addition of Br2 to (Z)-2-butene

(d) nucleophilic addition of Br2 to (Z)-1-bromobut-2-ene

Ans. (c)


24. The major product in the following reaction is





Ans. (c)


25. An archaeological sample (remains of an animal) containing 14C isotope of carbon is found to give 10 beta decays per minute per gram of carbon. It is known that the natural abundance of 14C in organic matter that is in equilibrium with the atmosphere today will give 15 beta decays per minute per gram of carbon. The half life of 14C is known to be 5730 years. The estimated age of the sample (in years) is

(a) 3010

(b) 3350

(c) 3500

(d) 3800

Ans. (b)

Sol. N0 = 15 (Activity/amount of 14C in fresh sample)

Nr = 10 (Activity/amount of loc in dead sample)

t1/2 = 5730 years

= 1.44 × 5730 ln

= 3345.5 = 3350 years

26. The minimum light intensity that the human eye can perceive is 10–10 Wm–2. The area of the opening of our eye (the pupil) is approximately 0.4 cm–2. Consider yellow light with wavelength = 600 nm. The number of photons incident on the retina per second at the minimum intensity for the eye to respond is

(a) 1.5 × 103

(b) 5 × 103

(c) 8 × 103

(d) 1.2 × 104

Ans. (a)

Sol. According to the question given that

l0 = 10–10 W/m2 = 10–10 J/sm2

Area of opening of our eye

A = 0.4 × 104 m2

= 600 nm

= 600 × 10–9 m

Suppose, n numberof photons are incident

= 1.207 × 103

= 1.5 × 103 (closest option)

27. Three NOT gates are connected in series and the output of the last gate is fed back to the input of the first one as shown in the figure. Each gate has a propagation delay of Td = 1 nano second, which means that the gate requires 1 nano second to change the output after the signal arrives at the input. What is the expected output at point A?

(a) Sine wave with a frequency of 666 MHz

(b) Square wave with a frequency of 666 MHz

(c) Random white noise

(d) Square wave with a frequency of 333 MHz

Ans. (d)

Sol. Given delay of each gate,

   Td = 1 ns = 1 × 10–9

Suppose 'O' is applied at input

Time taken to obtain '1' at the output is

T = 3 × 1 ns = 3 ns

Time taken to obtain '0' at the output

T = 3 × 1 ns = 3 ns

In the given circuit either '0' or '1' is obtained.

  Time period of the circuit for an output is 3 ns

  Frequency of the square wave


= 333.33 MHz

It will be square wave of frequency 333 MHz.

28. Let nCr denotes the number . Then, for n = 100, the sum of the series

1 – nCr + nC2nC3 + ... + (–1)r nCr + ... + (–1)n nCn is

(a) 0

(b) 1

(c) 2

(d) 1024

Ans. (a)

Sol. Given, n = 100

1 – nC1 + nC2nC3 + ... + (–1)f nCr + ... + (–1)n nCn

= (1 + nC2 + nC4 + ...) – (nC1 + nC3 + nC5 + ...)

= (nC0 + nC2 + nC4 + ...) – (nC1 + nC3 + nC5 + ...)

= 2n – 1 – 2n – 1 = 0

29. The lengths of two sides of a triangle are 2 units and 3 units and the angle included by these two sides is 60°. The length of the third side of the triangle will be



(c) 4 units

(d) 5 units

Ans. (b)

Sol. According to question,

Let cos A = 60°, b = 2 units, c = 3 units

Find the value of a

   13 – a2 = 6

 a2 = 7

  a = units


30. If A and B are two skew-symmetric matrices, the matrix AB + BA must be

(a) skew-symmetric

(b) symmetric

(c) invertible

(d) not invertible

Ans. (b)

Sol. Given A and B are tow skew-symmetric matrices

A' = – A, B' = – B

Then,   (AB + BA)' = (AB)' + (BA)'

  B' A' + A' B'

= (– B) (– A) + (– A) (– B) [  (AB)' = B' A']

= BA + AB

= AB + BA

  (AB + BA) is a symmetric matrix.

31. In a large wild flower population, assume that no new mutations occur and that no natural selection operates. What factor(s) will affect the frequency of a genotype in this population?

(a) Non-random mating

(b) Gene flow

(c) Outbreeding within the population

(d) Invasion of a new pathogen that kills a large number of individuals in the population

Ans. (a, b, c, d)

Sol. Hardy weinberg principle selectes that gene pool of a given population (which includes thin allele and genotypic frequencies) will remain constant in absence of factors influencing evolution therefore, mutations or natural selection, non random mating, out breeding within population and invariation of new pathogen can influence and disturb this gene pool and lead to evolution.

32. What is (are) the difference(s) between microtubules and microfilaments?

(a) Microtubules are made up of tubulin and microfilaments are made up of intermediate filaments

(b) Microtubules are important for compression resistance and microfilaments bear tension

(c) Microtubules are important for the functions of cilia and flagella and the microfilaments are important for cytoplasmic streaming

(d) Microtubules-muscle contraction; microfilaments-ciliary movement

Ans. (b, c)

Sol. Microtubules and microfilaments are part of cytoskelton of cell that allow the cell to maintain its shape and organelles.

Tubulin is major component of microtubule while actin is major protein of micro filament. Microtubule have role in movement (cillia, flagella) while microfilaments bear tension, help in cytoplasmic streaming, muscle contraction etc.

33. N and N0 represent the number of viable cells at time 't' during sterilisation and at the start sterilisation (t = 0), respectively. Assuming that cell death follows first order kinetics and is the death rate constant, which of the following relationship(s) is/are correct?

(a) N = N0ekt

(b) – ln (N/N0) = kt

(c) N = N, kt2

(d) N – N0 = kt

Ans. (b)

Sol. Given, N = no. of viable cells at time t

N0 = no. of wable cells at start

Following first order kinetics.

34. Which of the following statements about antigen-antibody (Ag-Ab) complexes is (are) True

(a) Hydrogen bonds and van der Waals forces participate in Ag-Ab interactions

(b) lonic bonds and hydrophobic bonds participate in Ag-Ab interactions

(c) The combined strengths of all interactions between a single antigen binding site of an antibody and a single epitope is called activity

(d) Antibody elicited by one antigen can cross react with an unrelated antigen

Ans. (a, b, d)

Sol. Immune complex is formed by binding of epitops of antigen with paratope of antibody through noncovalent interactions like, hydrogen bond, vanderwal forces, electrostatic intractions etc. Avidity explains the accumulative strength of multiple affinities of individual non covalent binding intractions like ligand receptor interaction.

In Ag – Ab complex, cross reactivity is known to occur where an antibody against one Ag can react with an unreacted antigen.

35. The superoxide ion, O2 is produced by the reaction K + O2 KO2. The correct statement pertaining to oxygen and superoxide ion is (are)

(a) oxygen is paramagnetic and has two unpaired electrons

(b) the bond order in oxygen is 2

(c) the bond order in superoxide is 1.5

(d) the superoxide ion is not paramagnetic

Ans. (a, b, c)

Sol. K + O2 KO

 O2 = superoxide

   O2 = 17 electrons

Here, two unpaired electrons are present at so, it is paramagnetic.

36. Among the following compounds, which of these will show two singlets in their 1H-NMR spectrum?

(a) 1, 4-Dichlorobenzene

(b) 1, 2-Dichlorobenzene

(c) Dimethoxymethane

(d) Methylacetate

Ans. (c, d)


37. Among the following pairs of coordination compounds, the pair(s) which represent(s) a case of 'ionisation isomerism' is (are)

(a) [Pt(en)2Cl2]Br2 and [Pt(en)2Br2]Cl2

(b) [Cr(NH3)4 ClBr]NO2 and [Cr(NH3)4ClNO2]Br

(c) [CO(NH3)6][Cr(CN)6] and [Cr(NH3)6][CO(CN)6]

(d) [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br

Ans. (a, b, d)

Sol. In ionisation isomers, there is exchange of ligands occurs between coordination sphere and ionisation sphere. These isomers give different ions when dissolved in water.

So, [Pt(en)2Cl2] Br2 and [Pt(en)2Br2] Cl2 [Cr(NH3)ClBr] NO2 and [Cr(NH3)4ClNO2] Br [Co(NH3)5Br] SO4 and [Co(NH3)5 SO4] Br are the examples of ionisation isomers.

38. The 'strong nuclear force' holds the protons and neutrons (nucleons) together in the nucleus of an atom. It is found that the binding energy per nucleon (for the nucleus of an element), when plotted against the mass number (A) of that element changes very little for 30 < A < 150. The binding energy is lower for A << 30 or A >> 150. This leads us to conclude that

(a) the strong nuclear force must oscillate with distance with a periodicity approximately same as the size of a proton or neutron

(b) the fusion of two elements, both with A << 30 or fission of an element A >> 150 may release energy

(c) the strong nuclear force changes very slowly with distance (i.e. It is long ranged on the scale of the size of nucleus)

(d) the strong nuclear force goes to zero very rapidly with distance (i.e. It is short ranged on the scale of the size of nucleus)

Ans. (b, d)

Sol. The binding energy per nucleon gives a measure or the force which binds the nucleons together inside a nucleus. The binding energy per nucleon is small for lighter nuclei and hence they show nuclear fusion. binding energy per nucleon is smaller for heavier nuclei and hence they show nuclear fission. The strong nuclear force goes to zero very rapidly distance (i.e. it is short ranged forces).

39. Let a and b be two non-zero vectors such that | a + b | = a - b |. Then

(a) a and b are parallel to each other

(b) a and b are perpendicular to each other

(c) a is not a scalar multiple of b

(d) a × b = 0

Ans. (b, c)

Sol. Given a and b be two non-zero vectors such that

 | a + b | = | a – b |

40. Let N be the set of all natural numbers. Consider the relation R on N given by

R = {(m, n) :m-n is divisible by 2). Then,

(a) R is symmetric and transitive

(b) R is symmetric but not transitive

(c) R is reflexive but not symmetric

(d) R is reflexive and transitive

Ans. (a, d)

Sol. Given R = {(m, n) : m – n is divisible by 2}

   (i) Reflexivity

Let a be an arbitrary element of N.

Then, (a – a) = 0, which is divisible by 2.

(ii) Symmetry

Let a, b N such that (a, b) R then

(a, b) R (a – b) is divisible by 2

  –(b – a) is divisible by 2    (b, a) R

  R is symmetric.

(iii) Transitivity

Let a, b, c N such that (a, b) R and (b, c) R

Then, (a, b) R and (b, c) R

  (a – b) is divisible by 2 and (b – c) is divisible by 2

  {(a – b) + (b – c)} is divisible by 2

  (a – c) is divisible by 2

  (a, c) ∈ R

  R is transitive

Thus, R is reflexive, symmetric and transitive

41. The pH of a 0.1 M solution of monosodium succinate ( = 4.19 and = 5.57) is _________ .

Ans. (4.8)

Sol. HOOC – CH2 – CH2 – COONa+ (monosodium succinate)

Ph = 4.8

42. The deactivation rate constant of an enzyme is 0.346 h–1. Assuming that the deactivation process follows first order kinetics, the half-life of the enzyme in minutes is _________ .

Ans. (120 min)

Sol. Given K = 0.346 h–1

t1/2 = ?

= 120 minute.

43. An enzyme preparation containing 10 mg/ml protein shows a specific activity of 50 U/mg The Initial velocity of reaction in a standard 1 ml reaction mixture containing 10 μl of the preparation in ml–1 min–1 is _________ .


Sol. We know that

Initial velocity = specific activity × amount of protein

44. The number of peaks in the 13C-NMR spectrum of CDCl3 is

Ans. (3)

Sol. CDCl3 appears as triplet in 13C-NMR spectra. It comes from splitting with deuterium. The formula for splitting is (2nl + 1), where n is the number of nuclei and l is the spin type. Since, CDCl3 has one deuterium (n = 1), and spin type is 1 (l = 1) then, splitting = 2 × 1 × 1 + 1 = 3,

So, it has 3 peaks.

The CDCl3 signal is 1 : 1 : 1 triplet due to j-coupling of the deuterium which has a spin l = 1 i.e. nucleus having three energy levels.

45. The number of phosphorus-hydrogen bonds in H3PO2 is _________ .

Ans. (2)

Sol. H3PO2

46. A man weighing 70 kg stands on a weighing scale which is placed in an elevator. The elevator is moving up towards its destination floor with a velocity of 1.0 ms–1. As it approaches the destination floor it starts slowing down, such that it comes to rest in 2 seconds. Assuming the acceleration due to gravity, g = 9.8 ms–2, the reading of the weighing scale just before the elevator comes to rest is _________ .

Ans. (721 N)

Sol. First of all, calculate the acceleration of elevator from the given values

Initial velocity u = 1.0 m/s, Final velocity, v = 0, time taken, t = 2 s

From equation of motion

v = u + at

Now, force acting on the man just before elevator stopss is

  Fnet = Mg + Ma

= 70 × 9.8 + 70 + 0.5

= 721 N

721 N will be the reading of scale.

47. A 50-metre tall antenna transmits at 107 MHz (one of the FM radio broadcast frequencies). Calculate the maximum distance from the antenna at which the transmitted signal can be heard. Ignore atmospheric attenuation and give your answer correct to the nearest kilometre only. You are given that the radius of the earth is 6400 km.


Sol. The effective range of the broadcasting antenna of height 'h' is given by

Maximum distance from which signal can be heard,

48. The total number of mappings from the set {1, 2} to the set {3, 4, 5, 6, 7) is _________ .

Ans. (25)

Sol. Given set {1, 2} and {3, 4, 5, 6, 7}

Total number of mapping from set {1, 2} to the set {3, 4, 5, 6, 7} is = 52 = 25

49. The value of the complex number (1 + i)150 + (1 – i)150 is _________ .

Ans. (0)

Sol. Value of complex number (1 + i)150 + (1 – i)150

50. Let the function f : R R be defined by

f(x0 =

If f is continuous at x = 0, then the value of k must be equal to _________ .

Ans. (k = 1)

Sol. Given function f : R —→ R defined by

If f is continuous at x = 0, then

 f(0) = k

Since, f(x) is continuous at x = 0

 LHL = RHL = f(0)

     1 = 1 = k

51. Assume that the gene UNC is essential for the coordinated movement of a nematode and that the wild-type allele U is dominant over the mutant allele u. Similarly, the wild-type allele D of another gene DPY, which is responsible for the normal body length, is dominant over the mutant allele d. Assume the UNC and DPY are on two different chromosomes. If a female of genotype UUDD mates with a male of the genotype uudd, the percentage of the F-2 progeny that will display uncoordinated movement but will have normal body length is _________ .

Ans. (18.75%)

Sol. For movement we know allele U is dominant own allele u

Similarly allele D is dominant own allele d for nominal growth

3 out of 16 progeny shows uncoordinated movement and normal growth

52. The pH of gastric juice in the stomach is 2.0. However, the pH inside the cells that line the stomach is 7.0. For transport of protons from inside the cell to the stomach, the free energy change in kJ mol–1 at 37°C is

[Assume Universal Gas constant R= 8.314 J mol–1 k–1)

Ans. (29.6 kJ/mol)

Sol. DGº = –2.303 RT log Keq

= –5935.6 × –5

= 29,678 J/mol

or 29.6 kJ/mol

53. A solution containing NAD+ and NADH has an optical density of 0.233 at 340 nm and 1.000 at 260 nm. While this solution absorbs at 260 nm, NADH alone absorbs at 340 nm. All measurements are carried out in a 1 cm cuvette. Given the extinction coefficients (see table below), the concentration of the oxidised form of the cofactor in is _________ .


Sol. According to Bear Lambert Law

At 340 nm

54. Drosophila melanogaster is a diploid organism having 8 chromosomes. The number of combinations of chromosomes which are possible in its gametes is _________ .

Ans. (16.0)

Sol. Drosophila melanogaster commonly known as fruit fly is a diploid organism with µ pair of chromosomes. No. of possible compinations will be 2n or 24 = 16.

55. The number of equatorial hydrogens in the following structure is _________ .

Ans. (8 eq H)

Sol. It is a fused rings system.

Chair form of trans decalin is

Hence, there are 8 equatorial hydrogens are present in the trans decalin.

56. The axis of rotation of the earth makes an angle of 66.5° with the plane containing the earth's orbit around the sun (called the plane of the ecliptic). If this angle were 50°, then the area of the earth's surface from which a 'midnight sun' (24 hour daylight) can be observed would change. The ratio of the new area to the previous area is _________ .


Sol. According to Kepler'slaw

So, for two different cases

57. A nuclear power plant generates 1000 Megawatts (MW) of electrical power and used half of its fuel supply in 5 years. The reactor uses 235U with 33% efficiency for the conversion of heat released by nuclear fission to electrical power. Each atom of 235U releases 200 MeV of energy. How many tons of 235U did the reactor start with?

(1 ton = 1000 kg; Avogadro number = 6.023 × 1023 mol–1).

Ans. (11.65 ton)

Sol. 1 atom releases 200 MeV of energy for 33% efficiency, reactor uses

  66 MeV of energy is produced by 1 atom

  2000 × 10× 365 × 24 × 3600 energy will be produced by

= = 5972.72 × 1025 atoms

Now, 6.023 × 1023 atoms of uranium weight 235 g

  5972.72 × 1025 atoms of uranium weight

= = 233038.22 × 102 g

Given that, 1000 × 103 g = 1 ton

   233038.22 × 102 g = × 233937.22 × 102

= 23.30 ton of uranium is required for 10 years.

Since, the reactor uses half of its total fuel capacity. so, it has been started with = 11.65 ton

58. A bat emitting ultrasound at 50 kHz is flying directly towards a solid wall with a speed of 3 ms–1. If the speed of sound in air is 330 ms–1, the frequency of the reflected signal (in kHz) heard by the bat will be _________ .

Ans. (50.91 51 kHz)

Sol. The reflected frequency is given by

= 50.91 = 51 kHz

59. A circle is given by the equation 2x2 + 2y2 + 8x – 20y + 10 = 0. The area of a square whose side equals the radius of the circle is _________ .

Ans. (24 sq units)

Sol. Given equation of circle

2x2 + 2y2 + 8x – 20y + 10 = 0

  2(x2 + y2 + 4x – 10y + 5) = 0

   x2 + y2 + 4x – 10y + 5 = 0 [  2 0] ...(i)

On comparing Eq.(i) with x2 + y2 + 2gx + 2fy + c = 0, we get

Required area of square whose side is equal to the radius of circle = = 24 sq units.

60. The value of the integral dx is _________ .

Ans. (2)


= (1 – 0) – (0 – 1) = 1 + 1 = 2