PYQ 2019 - VedPrep

IIT JAM Biology 2019
Previous Year Question Paper with Solution.

1. The glycosidic linkages in cellulose and amylose are ...................... respectively.

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Amylose is a storage polysaccharide where D-glucose molecules are linked via , 4-glycosidic bond to from a linear structure called amylose. Cellulose is structural polysaccharide where D-glucose molecules are linked via glycosidic bond to form a linear structure called cellulose. The -configuration allows cellulose to form very long, straight chains.

2. A mutation in the chapter locus of lac operon that confers constitutive expression of -galacatosidase is .......................

(a) cis dominant

(b) trans dominant

(d) dominant negative

Ans. (a)

Sol. At operator locus in lac operon repressor binds to blocks the expresion of genes. Mutation in lac operator (O°) does not allow the binding of repressor with operator. Mutation in operator promotes the constitutive expression of Lac Z gene codes for -galactosidase. (O°) is cis dominant because an operator controls oly the genes on same DNA strand (cis). In lac operon, trans acting element is repressor. Dominant negative mutation occurs in the DNA binding sites of the polypeptide.

3. Which one of the points

lies ABOVE the parabola y = 2x² and INSIDE the circle x² + y² = 4?

(a) P

(b) Q

(d) S

Ans. (b)

Sol.

So, Q is the correct answer.

4. Let U = {1, 2, 3, 4, 5}. A subset S is chosen uniformly at random from the non-empty subsets of U. What is the probability that S does NOT have two consecutive elements?

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Let U = {1, 2, 3, 4, 5}

U have 5 elements so, No. of non-empty subset of U will be

U = 2ⁿ – 1 = 2⁵ – 1 = 32 – 1 = 31

Non- empty subset of U = 31

According to question, S does not have two consecutive elements. For this, there are 2 cases.

1. It we remove (2, 4) from subset, we get {1, 3, 5}-Non-consecutive number.

then number of subset (2³).

2. If we remove {1, 3, 4} from subset then we have {2, 5}- Non-cosectuve number.

Here number of subset will be (2²)

So, probability that 'S' does not have two consecutive elements will be

5. Which one of the following figures represents the correct sequence of phases in adult eukaryotic cell cycle?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. G₁ Phase (Gap-1) Corresponds to the interval between mitosis and initiation of DNA replication. Proteins are synthesised in preparation for synthesis phase (S-Phase).

S-Phase G₁ is followed by S-phase during which DNA replication takes place. Completion of DNA synthesis is followed by G₂ phase.

G₂ Phase : Proteins are synthesised in preparation for mitosis.

M-Phase : Mitotic phase corresponding to the separation of daughter chromosomes and usually ending with cell division (cytokinesis).

6. At what pH does poly-Glu in an aqueous solution from -helical structure?

(a) 3

(b) 7

(d) 12

Ans. (a)

Sol. Polyglutamate, a polypeptide made up of only L-Glutamate residues has the alpha helical conformation at pH 3. Polyglutamate is correctly folded when its side chains are neutral and they become neutral at low pH that is 3. When the pH is raised to 7 or more than 7 same charge on adjacent groups in an alpha helix leads to repulsion, destabilisation and unfolding.

7. The dimensions of coefficient of viscosity are ....................

(a) ML^–1T^–1

(b) ML^–1T^–2

(d) ML^–2T^–1

Ans. (a)

Sol. Coefficient of viscosity = Stress/Strain rate

Unit = Pressure/Time^–1

Dimensions = ML^–1T^–2/T^–1

= ML^–1T^–1

8. Match the entries in Group I with the entries in Group II.

(a) P-iv, Q-i, R-ii, S-iii

(b) P-iv, Q-i, R-iii, S-ii

(d) P-ii, Q-iv, R-i, S-iii

Ans. (a)

Sol. (P) Nylon is a polymer made when the appropriate monomers are combined to form long chain. The monomers for nylon 6-6 are adipic acid.

(Q) Natural rubber is polymer of isoprene joined to make long chains.

(R) Starch is polymer of glucose molecule that is most common from of storage of carbohydrates in cells. In glucose, there is 6 carbon atom known as hexose.

(S) Myoglobin is a monomeric protein made up of amino acids. Myoglobin is a protein found in muscles that bind oxygen with its heme group like haemoglobin.

9. The technique that involves impacting samples with electrons is .........................

(a) NMR spectroscopy

(b) ESI mass spectrometry

(d) UV-vis spectroscopy

Ans. (b)

Sol. (a) NMR (Nuclear Magnetic Resonance) is a spectroscopy technique based on absorption of electromagnetic radiation by nuclei of atoms. Experiments are performed on nuclei of atoms not the electrons.

(b) ESI (Electrospray lonisation) is a technique used in mass spctroscopy to produce ions using an electrospray. It is a soft ionisation techniques used for production of gas phase ions of macromolecules.

(c) IR (Infrared) spectroscopy measures the vibration of atoms and determine the functional groups.

(d) UV-Vis-Spectroscopy is a technique measures the absorption of light across the ultraviolet and visible light wavelengths.

10. The orbital angular mementum of hydrogen atom in the ground state is ....................

(a) 0

(b)

(c)

(d) h

Ans. (a)

Sol. The angular momentum/Orbital angular momentum

(L) =

Here l is angular quantum number represented as l subshell

Which was calculated as follows

L =

l = 0

Correct answer is (a)

11.

(a) is 1

(b)

(d) does not exist

Ans. (a)

Sol.

So, the correct answer is 1

12. In how many ways can one write the elements 1, 2, 3, 4 in a sequence x₁, x₂, x₃, x₄ with

(a) 9

(b) 10

(d) 12

Ans. (a)

Sol. 1, 2, 3, 4 elements can be write in nine different manner/ways. Which are as follows:

(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (3, 1), (4, 1), (1, 2, 3), (1, 2, 3, 4) = 9

So, the correct answer is (a)

13.

(a) 2 sec A

(b) 2 cosec A

(d) cosec A

Ans. (b)

Sol.

By taking LCM

{(1 + cosA)² = 2cos A + cos² A}

(Cancelled 1 + cos A)

14. The evolution of eyes in Octopus and in human is an example of ..........................

(a) divergent evolution

(b) convergent evolution

(d) genetic drift

Ans. (b)

Sol. (a) Divergent evolution It occurs when two species evolve differently from a common ancestor, e.g., Darwin's finches.

(b) Convergent evolution It is the process where organisms similar traits to adopt to similar environ,ents or ecologial niches. Camera eyes of the Octopus is very similar to that of humans. Phylogenetic and embryolocial studies suggest that their camera eyes have been acquired independently. They eyes of Octopus and humans have independently evolved after the divergence of two lineages. The eyes of Octopus and humans been described as a typical example of convergent evolution.

(c) Adaptive radiation It is the relatively fast evolution of many species from a single ancestor, e.g. development of mammals after the extinction of dinosaurs.

(d) Genetic drift It is a mechanism of evolution in which allele frequencies of a population change over generation due to chance.

15. Which one of the following modifications occurs both on DNA and protein?

(a) ADP - ribosylation

(b) Methylation

(d) Ubiquitination

Ans. (b)

Sol. Protein methylation "DNA methylation" is an important post-translational modification used in cellular signal transduction pathways. Protein can be methylated at both their N and C-terminal and on side chains nitrogen's of arginine and lysine residues.

ori C contains 11 5'- GATC-3' repeats that are methylated on adenine on both strands. Only fully methylated origins can initiate replication, hemi-methylated cannot initiate DNA replication.

16. Solutions of the following peptides are prepared separately at concentration of 1 mM, Among these four, which one has the highest A₂₈₀?

(a) Ser-Val-Trp-Asp-Phe-Gly-Tyr-Trp-Ala

(b) Gin-Leu-Glu-Phe-Thr-Leu-Asp-Gly-Tyr

(d) His-Pro-Gly-Asp-Val-lleu-Phe-Met-Leu

Ans. (a)

Sol. The optical absorption of proteins is measured at 280 nm. At this wavelength, the absorption of proteins is mainly due to the tryptophan and tyrosine.

(a) Peptide contains both amino acids Tryptophan (Trp) and Tyrosine (Tyr), therefore it shows maximum absorption at 250 nm.

(b) Peptide contains only one tyrosine residue absorbs less.

(d) Peptide does to contain any of these two amino acids tryptophan and tyrosine. If proteins contains no tyrosine or tryptophan residue, A₂₈₀ cannot be used to determine the concentration of the proteins.

17. The free energy required to synthesize a mixed anhydride bond of 1, 3 bisphosphoglycerate is generated by the oxidation of ...........................

(a) an aldehyde to acid

(b) an alcohol of acid

(d) NADH to NAD⁺

Ans. (a)

Sol.

Acetaldehyde Acetic acid

This reaction couples an oxidation step with a phosphorylation step.

Glyceraldehyde -3-phosphate to 1, 3-bisphosphoglycerate.

18. The following reaction is an example of ...........................

(a) enolisation

(b) racemisation

(d) epimerisation

Ans. (c)

Sol. Isomerisation is the process by which cone molecules is transformed into another molecule which has exactly the same atoms but the atoms have a different arrangement. So, the given reaction is an example of isomerisation.

19. Which one of the following parameters changes upon doubling the enzyme concentration?

(a) K_M

(b) V_max

(d) K_eq

Ans. (b)

Sol. According to Michaelis-Menton kinetics, if concentration of substrate is increased more and more, the enzyme molecules work efficiently. If enzyme concentration is double, V_max will be double.

Km It is a value of substate concentration of half maximal velocity . It is independent of enzyme concentration and remains exactly same.

20. Which one of the following statements is a correct description of modes of action of taxol and colchicine?

(a) Taxol causes DNA damage and colchicine prevents microtubule formation

(b) Taxol stabilizes microtubules and colchicine inhibits protein synthesis

(d) Taxol stabilises microtubules and colchicine prevents microtubule formation

Ans. (d)

Sol. In taxol treated cells, microtubules are abundant, appears as thick ordered bundles. In the presence of colchicine, microtubules are absent. Colchicine binds tightly to unpolymerised tubulin and from tubulin colchicine complex. Colchicine is a mitotic inhibitor which inhibits mitosis by desrupting microtubules polymerisation.

21. In a simple microscope, ....................

(a) a lens with negative power is used

(b) the focal length of the lens is less than the least distance for clear vision

(d) magnification depends only on the focal length of the lens

Ans. (b)

Sol. Magnifying power of a simple microscope is

m = 1 +

D – Least distance of distinct vision

F – Fasal length of convex lens

Focal length should be small because smaller the focal length of the lens, greater will be its magnifying power.

22. Which one of the following statements is INCORRECT with respect to bacterial conjugation?

(a) It facilitates transfer of genetic material

(b) It requires flagellum

(d) It can transfer virulence factors

Ans. (b)

Sol. Bacterial conjugation is a sexual mode of genetic transfer from one bacterial cell to another. Pilus is responsible for attachment, transfer of DNa to other cell during bacterial conjugation. Antibiotic resistance genes and virulence factors can be transformed through conjugation.

23. A particle starting from rest is subjected to a constant force. The plot of distance traveled along the direction of the force as a function of time is a/an ......................

(a) straight line

(b) circle

(d) ellipse

Ans. (c)

Sol. By Newton's 2nd law of motion.

F = Force

a = Acceleration

So, constant force results in constant acceleration.

an accelerating body will double the displacement travelled for every unit of time elapsed as shown in graph.

So, the correct answer is (c)

24. Indole Acetic Acid (IAA) is involved in ....................

(a) gravitropism

(b) flowering

(d) senescence

Ans. (a)

Sol. Gravitropism is a growth in response to gravity enables roots to grow downwards into the soil and shoot to grow upwards away from the soil. Indole acetic acid is a natural auxin, responsible for gravitropism.

25. Which one of the following remains uncharged when light waves enter water from air?

(a) Wavelength

(b) Wavenumber1

(d) Intensity

Ans. (c)

Sol. Whenever a wave changes its medium, the frequency of the wave remains constant.

26. According to the kinetic theory of gases, the average energy of diatomic molecule, in an ideal gas depends on ...........................

(a) mass of each atom and the temperature

(b) mass of each atom and the bond length

(d) temperature only

Ans. (d)

Sol. According to the kinetic theory of gas, the average kinetic energy of gas particles is proportional to the absolute temperature of the gas and all gases at the same temperature have the same average kinetic energy.

27. Match the entries in Group I with entries in Group II.

(a) P-ii, Q-i, R-iii, S-iv

(b) P-ii, Q-iii, R-i, S-iv

(d) P-i, Q-iv, R-ii, S-iii

Ans. (b)

Sol. (P) Tuberculosis (TB) is a disease caused by bacteria that are spread through the air form person to person.

(Q) Influenza or Flu is respiratory illness caused by virus. It is highly contagious and is normally spread by cough and sneeze of an infected person.

(R) Malaria is caused by protozoan parasite Plasmodium.

(S) Myasthenia gravis is a rarely acquired autoimmune disorder caused by an antibody mediated blockade of neuromuscular transmission. The autoimmune attack occurs when autoantibodies from against the nicotinic acetylchoine post-synaptic receptros at the neuromuscular junction of skeletal muscles.

28. pK_a increase in the order .....................

(a) HN₃ > NH₃OH⁺ > N₂H₅⁺ > NH₃

(b) NH₃OH⁺ > N₂H₅⁺ > HN₃ > NH₃

(d) HN₃ > N₂H₅⁺ < NH₃ > NH₃OH⁺

Ans. (c)

Sol. All options are incorrect

pKa value for HN₃ is 4.75

pKa value for N₂H₅⁺ is 8.10

pKa value of NH₃ is 35

Correct order should be

NH₃ > N₂H₅⁺ > NH₃OH⁺ > HN₃

This order is not given in options, so all options are incorrect.

29. H₂ reacts with trans (Ph₃P)₂ Ir(CO)Cl to primarily produce ......................

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

It is an oxidative addition reaction when H₂ reacts with trans (Ph₃P)₂ Ir(Co)Cl it forms trans (Ph₃)₂Ir(H₂)(CO)Cl So, the correct answer is (d).

30. Among the following species, the metal centre that has the highest number of unpaired electrons is .............................

(a) VCl₄

(b) Ni(CO)₄

(d) [CdBr₄]2^–

Ans. (a)

Sol. The central metal of

VCl₄, Ni(CO)₄,[AuCl₄]^–, [CdBr₄]^2– have

Vanadium (V) = [Ar]4s²3d³ have 3 unpaired electron.

Nickel (Ni) = [Ar]3d⁸ Ns² have 2 unpaired electron.

Gold (Au) = [Xe]4f¹⁴ 5d¹⁰ 6s¹ have 1 unspired electron

Cadmium (Cd) = [Kr]4d¹⁰5s² have no unpaired electron.

The highest number of unpaired electrons is in Vanadium (VCl₄).

So, the correct answer is (a)

31. Pick the correct statements(s) with respect to the interconversion of the topoisomers of a circularly closed double-stranded DNA.

(a) Only one strand needs to be cut

(b) Both strands have to be cut

(d) ATP is required for interconversion

Ans. (a, b, d)

Sol. DNA topisomerase is a nuclease that breaks a phosphodiester bond in DNA strand.

Type I topoisomerase breaks one strand

Type II topoisomerase breaks both strands of DNA. It is ATP dependent enzyme. it can introduce negative supercoils into DNA using the free energy from ATP hydrolysis.

32. Let U = (1, 2, ...., 15). Let P U consist of all prime numbers. Q U consist of all even numbers and R U consist of all multiples of 3. Let T = P – Q. Then, which of the following is/are CORRECT?

(a)

(b)

(c)

(d)

Ans. (a, c)

Sol. U = {1, 2, .............. 15}

P = {2, 3, 5, 7, 11, 13} Prime numbers

Q = {2, 4, 6, 8, 10, 12, 14} Even numbers

R = {3, 6, 9, 12, 15} Multiples of 3

T = P – Q = {R, 3, 5, 7, 11, 13} – {2, 4, 6, 8, 10, 12, 14}

T = {3, 5, 7, 11, 13}

|T| = 5

Numbers of element in T

So, (a) and (c) are the correct answers.

33.

Which of the following is/are CORRECT?

(a)

(b)

(c)

(d)

Ans. (b, d)

Sol. f(x) = (x – 1) (x – 2) (x – 3) (x – 4)

f(x) is positive when

(i) x < 1 (ii) 2 < x < 3 (iii) x > 4

Here

Here, and have same, signs, i.e., negative (–), and have same sings, i.e., negative (–)

So, the correct answer is (b) and (d).

34. The characteristic oxygen binding profile of haemoglobin show below arises due to the ................

(a) quaternary structure

(b) subunit dissociation

(d) conformational change

Ans. (a, c, d)

Sol. (a) haemoglobin consists of four subunits each with a cofactor called a heme group. Iron is the main component that actually binds to oxygen and each haemoglobin molecule is able to carry four molecules of O₂.

(b) Subunit dissociation of haemoglobin does not contribute in arising oxygen binding profile.

(d) The bonding of the haemoglobin protein and oxygen changes the conformation of the binding site which increase affinity for other oxygen molecules to bind the protein molecule.

Haemoglobin exists in two conformations

a tense or taut (T)

a relaxed (R)

Oxygen binds with nuch higher affinity in the R-state, T-state is conformation of deoxy Hb and stabilised by a greater number of ionic interaction.

The binding of oxygen to haemoglobin subunit in the T-state triggers a change in conformation to the R-state.

35. The advantage(s) of storing chemical energy in the form of starch and not as free glucose is/are that it .................

(a) minimizes diffusion

(b) enables compact storage

(d) protects against chemical reactivity of aldehyde groups

Ans. (a, b, c, d)

Sol. (a) Plants store their food in the form of starch because it is insoluble in water ane cannot be transported in plant cell until it converted into glucose.

(b) Starch consists of two components-Amylose and Amylopectin. Both these structures enable the starch moecules to coi into compact shape.

(d) When conformation of open chain glucose convert in cyclic glucose, carbon number 5 react with CHO group and formed cyclic conformation. Aldehyde group loses its reactivity because it gets bind with carbon, so it protects against chemical reactivity of aldehydes.

36. Which of the following cell types can develop from myeloid lineage?

(a) Macrophages

(b) T-lymphocytes

(d) Erythrocytes

Ans. (a, d)

Sol. From a common myeloid progenitor cell both erythrocytes and macrophages arise.

T-lymphocytes and B-lymphocytes arise from common lymphoid progenitor.

37. Electromagnetic waves

(a) carry energy

(b) carry momentum

(d) do not need a material medium to travel

Ans. (a, b, c, d)

Sol. (a) Electromagnetic waves carry energy. The energy carried by electromagnetic waves is inversely proportional to its wavelength

E = , h = Planck's constant

(b) Electromagnetic waves carry momentum which is proportional to energy.

p =

C Speed at which electromagnetic waves travel in a vacuum, commonly known as 'Speed of light'.

(c) Electromagnetic waves are transverse in nature while travelling in vacuum/free space that a changing magnetic field produces an electric field and vice versa (Maxwell's equations)

(d) Electromagnetic waves do not require a medium to propagate. Medium is required by mechanical waves.

38. Which of the following statement(s) is/are true?

(a) In intrinsic semiconductors, the number of electrons is equal to the number of holes at any temperature

(b) An intrinsic semiconductor changes to an n-type semiconductor upon addition of a trivalent element

(d) In the reverse bias condition, the current in a p-n diode is due to the minority carries

Ans. (a, d)

Sol. (a) In intrinsic semiconductor, the number of holes in valence band is equal to the number of electrons in conduction band. As the temperature increases free electrons and holes get generated equally.

(b) When an intrinsic semiconductor is doped with trivalent element it becomes a P-type semiconductor, P-type means the semiconductor is rich in holes or positive charged ions.

When intrinsic semiconductor is doped with pentavalent element it become N-type semiconductor.

(d) p-n junction diode allows the minority charge carries. Positive charge carries which cross the p-n junction are attracted towards the negative terminal and the negative charged carries attracted towards the positive terminal. p-n junction diode completely blocks the majority charge carriers.

39. BF₃ reacts readily with ...................

(a) C₅H₅N

(b) SnCl₂

(d) (C₅H₅N)—SnCl₂

Ans. (a, d)

Sol. Boron halides such as BF₃ are electron deficient molecules because they do not have an octet electron surrounding the boron atom. They tend to act as Lewis acids by accepting electron pairs from beses to form stable acid base adducts. Such electron donors are pyridine C₅H₅ N and pyridine stannous chloride complex. It will react more readily with

C₅H₅N > (C₅H₅N) — SnCl₂ > SnCl₂.

40. The reaction of (R)-2 bromobutane with CN^– proceeds by .................

(a) retention of configuration

(b) inversion of configuration

(d) formation of (S)-2-methylbutanenitrile

Ans. (b, d)

Sol. Reaction proceed by inversion of configuration. If reaction started with R-(2)-bromobutane then reaction will end up with (S)-(2) methylbutanenitrile or vice-versa.

Substitution reaction results in an inversion of configuration at C-2.

41. C₃-plants utilise 18 molecules of ATP to synthesise one moelcule of glucose from CO₂. How many molecules of ATP equivalents are used by C₄-plants to synthesise one molecule of glucose from CO₂?

Ans. (30 ATP)

Sol. In C₃- plants

for 1 CO₂ molecule -3 ATP molecules are required3 CO₂ molecules (triose phosphate) 9 ATP are required 6 CO₂ molecules (glucose) – 18 ATP are required.

In C₄- plants

1 CO₂ molecules – 5 ATP molecules are requried

3 CO₂ molecules (triose phosphate) – 15 ATP are required

6 CO₂ molecules (glucose) – 30 ATP molecules are required

C₄ pathways consume 30 ATP for the synthesis of one molecule of glucose and C₃ pathway consumes 18 ATP for the synthesis of one molecule of glucose.

42. A 0.1% (w/v)e solution of a protein absorbs 20% of the incident light. What fraction of light is transmitted if the concentration is increased to 0.4%? [Correct to two decimal places]

Ans. (0.4)

Sol. Transmittance (T) = The ratio of the intensity of the transmitted light (l) to the intensity of the incident light (l₀).

Here, l = 80

l₀ = 20

T = 4

Then, the transmittanceconverted into % transmittance (%T)

% T = T × 100% = 4 × 100%

% T = 400%

The transmittance and absorbance are inversely related and the inverse relationship between transmittance and absorbance is not linear, it is logarithmic.

So, A =

Here, A = Absorbance, T = Transmittance

A = – log₁₀ T =

= – log₁₀ %T + log₁₀ 100

= – log₁₀ % T + log₁₀ 10₂

A = 2 – log₁₀ % T

When we put the value of % T in this equation

A = 2 – log₁₀ 400 = 2 – 260

A = 0.60

Absorbance is 0.60

So, transmittance and absorbance are inversely related means – Transmittance = 0.40

So, the correct answer is 0.40.

43. Ley XYZ be an equilateral triangle and let P,Q,R be the midpoints of YZ, XZ and XY, respectively.

Let r =

The value of r is ...........................

Ans. (r = 0.25)

Sol.

Let XY = a

(Theorem : Midpoints of two sides are joined, it is half of third side)

Put the value of Eqs. (ii) and (iii) in equation (i) we get

or r = 0.25

44. Let N be the set of natural numbers and f : NN be defined by

Let fⁿ(x) denote the n-fold composition of f(x) What is the smallest integer n such that fⁿ(13) = 1?

Ans. (9)

Sol.

So, n = 9 is correct answer

45. Heterozygous female fruit flies with gray body and purple eyes were mated with homozygous males with black body and red eyes. The number of off springs obtained and their phenotypes are shown below.

Calculate the recombination frequency

Ans. (0.15)

Sol. Formular for calculating recombination frequency is =

Here, recombinants are Gray body – Red eyes offsprings – 61

Black body – Purple eyes offsprings – 55

= 0.15 × 100 = 15%

46. Proinsulin is an 84 residue polypeptide with six cysteines. How many different disulphide combinations are possible?

Ans. (15)

Sol. A covalent bond is formed between a sulphur atom of two cystiene residue. The bond is called disulphide bond or S-S bond.

According to this equation six cystiene residue are present in proinsulin.

Bond formation takes place between

1 - 2 cys 2 - 3 cys 3- 4 cys

4- 5 cys 5 - 6 cys

(b)

Next disulphide bond takes place between

1 - 3 cys

1 - 4 cys

1 - 5 cys

1 - 6 cys

disulphide bond takes place between

2 - 4 cys

2 - 5 cys = 3 disulphide bonds

2 - 6 cys

(d) Next disulphide bond takes place between

3 - 5 cys

3 - 6 cys

(e) Next disulfphide bond takes place betweeen

4 - 6 — 1 bond

Now, add up all number of possible combinations

5 + 4 + 3 + 2 + 1 = 15

There are 15 combinations of disulphide bond formation.

47. The refractive index of a liquid relative to air is 1.5. Calculate the ratio of the real depth to the apparent depth when the liquid is taken in a beaker.

Ans. (1.5)

Sol. Given – Refractive index = 15

So, according to refractive index formula

Refractive index =

So, the ratio of Dr : Da is 1.5

48. A metallic wire of electrical resistance 40 is bent in the form of a square loop. The resistance between any two diagonally opposite corners is ................. .

Ans. (10)

Sol. Given, Resistance on metallic wire = 40.

When this wire is been form of square the resistance of each side is square is same, so it one fourth of the total resistance.

Now, each side has 10 resistance, so we calculate the two opposite diagonals. So, the two side of square is in series and that series combination is parallel with another series.

So, series combination has 10 + 10 = 20 resistance.

So, (20 || 20) = 10

49. The total number of lone pairs of electrons in NO₂F is ...................

Ans. (8)

Sol. NO₂F

Number of lone pair of electrons in oxygen is 2 + 3 = 5

Number of lone pair of electrons in fluorine is 3.

So, total number of lone pair of electrons is

5 + 3 = 8

50. The total number of multiplet peaks in the ¹HNMR spectrum of 1, 3, 5-tri-isopropylbenzene in CdCl₃ is ............................

Ans. (2)

Sol. There will be 2 peaks

Peak A at 6.78

Peak B at 2.28

Peak A is because of aromatic hydrogens and Peak B is because of propyl hydrogens

So, correct answer is 2.

51. A schematic representation of Restriction Fragment Length Polymorphism (RFLP) analysis of a sample population is shown below. The number of people exhibiting a given pattern is indicated above the lanes.

Calculate the frequency of 6.5 kb allele. [Correct to two decimal places]

Ans. (0.42)

Sol. According to Hardy-Weinberg equilibrium

p² + 2 pq + q² = 1

Given allele is 6.5 kb, we have to calculate q from p.

Here p = 0.65, q = 0.35

From p + q = 1

q = 1 – p = 1 – 0.65 = 0.35

p² + 2pq + q² = 1

(0.65)² + 2(0.65) (0.35) + (0.35)² = 1

0.4225 + 0.455 + 0.1225 = 1

By adding these value it results into 1. So, frequency of given allele = 0.42

52.

Ans. (1)

Sol.

Identify n – 1 in this case

Integration n – 1 in this case

cos 0 = 0

53. Phosphoglucoisomerase catalyses the following reaction

Glu-6 p Fru-6-P

If 0.05% of the original concentration of Glu 6-P remains at equilibrium, then the equilibrium constant of this reaction is ...................

Ans. (1.999)

Sol. Glu — 6P Fru — 6P

Suppose, 100 moles take

At equilibrium 100 – 99.95 reached 99.95 formed = 0.05 remain

equilibrium constant,

54. In a bacterium, a mutation resulted in an incrase of K_s (substrate specific constant) for ammonium from 50 to 5000 without affecting The specific growth rate of the mutant growing on 0.5 mM ammonium in the medium decreases by a factor of ........................

Ans. (10)

Sol. Specific growth rate formula

= Specific growth rate in time^–1

= Maximum specific growth rate in time^–1

S = Concentration of substrate in solution

K_s = Half velocity constant k

According to the question, is unaffected, so put the value of = 1

Initial substrate concentration is 50

K_s increases upto 5000

To calculate of the mutant at 0.5 mM ammonium concentration, convert 50 and 5000 into mM

50 = 0.05 mM

5000 = 5 mM

According to formula

= = 0.010 mM

Convert 0.010 mM into M = 0.010 × 1000 = 10 M

55. The total number of DNA molecules present after 5 cycles of Polymerease Chain Reaction (PCR) starting with 3 molecules of template DNA is .......................

Ans. (96)

Sol. Formula to calculate number of DNA molecule

= a × (2⁵ n)

a = initial number of DNA present

n = number of cycle

According to the question, PCR started with 3 molecules of DNA So, a = 3

Number of cycle = 5

= 3(2)⁵ = 3 × 32 = 96 molecules of DNA are present.

56. Two identical, infinite conducting plates are kept parallel to each other and are separated by a distance d. The uniform charge densities on the plates are . The electric field at a point between the two plates is n , where n is ........................

( is the premitivity of free space)

Ans. (2)

Sol. E = |E₊| + |E_–|

Diversity of electric field for positive charge electric field is directed away from it.

For negative charge electric field is towards the charge.

Electric field due to the infinite charge plate

E = (by one sheet)

But for conducting plates

E =

So, electric field by two conducting plates is

E =

According to formula,

E =

Value of n = 2

57. The concentration of NaCl (in mM) formed at the stoichiometric equivalence point when 10 mL of 0.1 M HCl solution is titrated with 0.2 M NaOH solution is .............................. (as an integer)

Ans. (66.5)

Sol. Reaction used :

Given, NaOH + HCl NaCl(aq) + H₂O(l)

Concentration of HCl (M₁) = 0.1 M

Concentration of NaOH (M₂) = 0.2 M

Volume of HCl (V₁) = 10 mL = 0.01 L

Step 1. To find volume of NaOH at equivalent-point :

M₁V₁ = M₂V₂

V₂ = =

(Volume of NaOH) i.e., V₂ = 0.005 L

Now, find the concentration of NaCl

Let, concentration of NaCl = M₃

Total volume of solution = V₁ + V₂ = (V₃)

= 0.01 + 0.005 = 0.015 L

M₁V₁ + M₂V₂ = M₃V₃

M₃ = =

= = 0.133

Therefore, in nM = 0.133 × 1000 = 133 (mM)

We have two products

mM for NaCl = = 66.5 mM

i.e. between 66-67

58. The standard emf of a cell (in V) involving the reaction, 2Ag⁺ (aq) Ag(s) + Ag²⁺ (aq) at 298 K is .......................... [Correct to two decimal places]

[Given : Ag⁺ (aq) + e Ag(s) ; E° + 0.62 V and Ag²⁺ (aq) + e Ag⁺ (aq); E° = 0.12 V]

Ans. (0.50)

Sol. EMF Electromotive force

E⁰_cell = E⁰_{red cathode} – E⁰_{red anode} ...(i)

2Ag⁺ (aq) Ag(s) + Ag²⁺

Ag²⁺ + e^– Ag⁺ [E° = 0.62 V] ...(ii)

Ag²⁺ + e^–Ag⁺ (aq)

[E° = 0.12 V]

So, put the value of equations (ii) and (iii) in equation (i), we get

E⁰_cell = 0.62 – 0.12

E⁰_cell = 0.50 V

59.

Ans. (126)

Sol.

60. An infinitely long solenoid of radius r and number of turns per unit length n carries a steady current l. The ratio of the magnetic fields at a point on the axis of the solenoid to a point from the axis is ............