1. The antigen binding site of an antibody is present

(a) at the constant region

(b) at the C-terminal

(c) at the variable region

(d) between the constant and the variable region

Ans. (c)

Sol. Antigen binds on the Fab fragment of anti body. Fab consists of constant and variable domain of both heavy and light chain. Variable domains contains CDRs (complementary determining regions) which forms the paratope sites on antibody where antigen will bind. Arm of Y binds an epitope on antigen.

2. Which of the following is not involved in eukaryotic translation?

(a) Ribosome

(b) Spliceosome

(c) mRNA

(d) tRNA

Ans. (b)

Sol. Spliceosome is not involved in prokaryotic translation as it is used in transcription to splice off the introns from pre-mRNA to form mRNA that will go for translation outside the nucleus to cytoplasm and translated with help of translational machinery that includes tRNA and ribosomes.

3. Which of the following statements is correct?

(a) Gram negative bacteria are coloured purple after Gram staining

(b) Gram negative bacteria are commonly more resistant to antibiotics than Gram positive bacteria

(c) Gram negative bacteria cell wall consists of a thick layer of peptidoglycan outside the plasma membrane

(d) Cell wall of Gram negative bacteria does not contain an outer membrane

Ans. (b)

Sol. Lipid layer is present on the peptidoglycan layer in gram-ve bacteria which makes it impermeable for crystal violet stain but application of alcohol dissolves this lipid layer exposing the peptidoglycan layer. Later when safranin is applied cell appears pink

4. The role of enzyme E synthesised by phage X174 during host infection is to

(a) block peptidoglycan synthesis

(b) enhance synthesis of viral +RNA

(c) inhibit lipid metabolism

(d) stimulate dsDNA replication

Ans. (a)

Sol. Enzyme E forms huge pores in host body and inhibit cell wall synthesis. This enzyme targets specific host enzyme Phospho Mur-NAc Pentapeptide Translocase which is an integral membrane protein involved in cell wall synthesis of bacteria.

5. Among CH₄, H₂O, NH₃ and PH₃, the molecule having the smallest percent s character for the covalent bond (X–H) between the central element (X = C, O, N or P) and hydrogen is

(a) CH₄

(b) H₂O

(c) NH₃

(d) PH₃

Ans. (d)

Sol. C—H, O—H, N—H,

n = 2.5 – 2.1, 3.5 –2.1, 3-2.1

Difference =.4 =1.4 =0.9

So, PH₃ has lowest % s character.

6. The result of an electrophoretic separation of a mixture of amino acids X, Y and Z at pH = 5.0 is represented as (Given the isoelectric points of X, Y and Z are 9.87, 3.22 and 5.43 respectively)

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Since x is more electrically charged it will move towards cathode. Net charge is affected by change in pH.

7. cos (x + yx) =

(a) cos (x) cos (yx) – sin (x) sin (yx)

(b) cos (x) cos (yx) + sin (x) sin (yx)

(c) cos (x) sin (yx) – sin (x) cos (yx)

(d) cos (x) sin (yx) + sin (x) cos (yx)

Ans. (a)

Sol. cos(x + yx) = cos x cos yx – sin x sin yx

[  cos(A + B) = cos A cos 8 – sin A sin B]

8. If , then the order of R is

(a) 2 × 3

(b) 3 × 2

(c) 2 × 2

(d) 3 × 3

Ans. (a)

Sol. We can do multiplication of two matrices if A and B are two matrices then AB is defined only if columns of A is equal to rows of B.

Now as in given question order of first matrices is 3 × 2.

So, order of R must be 2 × 3 or 2 × 2,

but as we know the answer of both the matrices, is 3 × 3.

So, R order must be 2 × 3.

9. The average energy of diatomic gaseous molecule at temperature T is k_BT, where k_B is Boltzmann's constant. The average energy of this molecule per degree of freedom is

(a)

(b)

(c)

(d)

Ans. (a)

Sol. The diatomic gas has 5 degree of freedom

Average energy =

So, per degree of freedom energy

10. The refractive index of diamond is 2.419. If the speed of light in vacuum is 3 × 10⁸ m s^–1, then the speed of light in diamond is

(a) 1.240 × 10⁸ ms^–1

(b) 1.352 × 10⁸ ms^–1

(c) 1.521 × 10⁸ ms^–1

(d) 2.433 × 10⁸ ms^–1

Ans. (a)

Sol. Refractive index n = c/v

Where c is the speed of light in a vacuum and v is the speed of light in the medium.

So, v = c/n

Diamond's refractive index 2.42,

11. Which of the following is true of protein synthesis only in prokaryotes'?

(a) Translation and transcription are coupled

(b) The codon AUG codes for the start signal

(c) The tRNA anticodon can bind to two or more different codons

(d) The functional ribosomes contain two subunits constructed of proteins and RNA

Ans. (a)

Sol. Prokaryotes show coupled transcription and translation which means protein synthesis begin even before the mRNA transcription is completed.

12. Match the entries in Group I with that in Group II.

(a) P-4, Q-1, R-3, S-5

(b) P-4, Q-1, R-2, S-3

(c) P-5, Q-4, R-5. S-2

(d) P-5, Q-1, R-2, S-3

Ans. (b)

Sol. Phytase improves Ca²⁺ availability and other minerals in diet

• Cellulase, xylanase, Laccase, Lipases are used in pulp and paper processes.

• Laccase are proteases in fruit used in pulp delignification.

• Bromelain is used in baking where it causes reduction of gluten.

13. If an aldol cleavage of glucose-6-phosphate occurs in glycolysis, it will result in

(a) product of equal carbon chain length

(b) product of unequal carbon chain length

(c) removal of phosphate group

(d) three C₂ compounds

Ans. (b)

Sol. Aldol cleavage of G-6-P is less favored as it forms products of unequal carbon chain length while fructose bisphosphate (FBP) aldol cleavage yields 2(C₃) compounds that are interconvertible and can enter common degrative pathway.

14. The natural geographical distribution of kangaroos is limited to the Australian continent because

(a) abiotic factors determine the distribution

(b) dispersal is limited by accessibility to other continents

(c) kangaroos have not selected habitats in other continents

(d) predators limit the distribution in other continents

Ans. (b)

Sol. Australia is habitat for kangaroos Eg:- We can find musky rat kangaroo in nests on floor of rain forests in North-Eastern Queensland.

Gray kangaroos inhabit forests of Australia and Tasmania.

Antilopine kangaroos are found in Eucalyptus woodlands of Northern Australia.

Tree kangaroo can be found in rainforests of Queensland and on Island of New Guinea.

15. Which of the following is not an example of an adaptive defense mechanism against predation?

(a) Bright colours of bird pollinated flower

(b) Insect that resembles a stick

(c) Nicotine in the tobacco plant

(d) Spines on porcupine

Ans. (a)

Sol. Stick insects camouflage to show appearance of stick, leaf or twig.

Herbivores are selected for their resistance towards metabolites which are produced by plants to decrease herbivores performance Eg.- Nicotine has important role in defense

Spines have been used for defense mechanisms like porcupine body is covered with sharp, stiff spines, keratin coated hairs containing barbs which can penetrate into skin of enemy

16. Match the entries in Group I with that in Group II.

(a) P-4, Q-3, R-1, S-2

(b) P-4, Q-1, R-2, S-3

(c) P-2, Q-i, R-3, S-4

(d) P-1, Q-3, R-4, S-2

Ans. (b)

Sol. Nucleolus is active site for RNA production and source of rRNA

• Spherosomes store and synthesize lipids/oils of the plant. It is a single membrane bound organelle

• -oxidation occurs in peroxisomes which involves breakdown of long chain fatty acids to medium chain fatty acids which are transported to mitochondria and broken down further to form CO₂ and H₂O

• Plasmodesmata are cytoplasmic connections present in plant cells to transport macromolecules between the cell. This cell to cell communication is important for development and growth

17. The nitrogenase of diazotrophs

(a) contains Cu-S center and uses 12 NADH to reduce one N₂

(b) contains one (4Fe-4S) cluster and uses 8 FADH₂ to reduce one N₂

(c) is a complex of Fe-protein and Mo-Fe-protein and uses 16 ATPs to reduce one N₂

(d) is a Mo-Fe-protein and uses 4 ATP and 4 FMNH₂ to reduce one N₂

Ans. (c)

Sol. Reduction of nitrogen to ammonia (NH₃) occurs by enzyme nitrogenase produced by bacteria like cyanobacteria this process of nitrogen fixation involves hydrolysis of 16 ATP. Fe and Mo-Fe are part of nitrogenase structure that have role in electron transfer.

18. During eukaryotic cell division, the amount of DNA doubles

(a) between prophase and anaphase of mitosis

(b) between prophase-I and prophase-II of meiosis

(c) between the G₁ and G₂ phases of the cell cycle

(d) during the M-phase of the cell cycle

Ans. (c)

Sol. Cell cycle composed of interphase and M(mitosis) phase. Interphase involve G₁, S, G₂ phase where G₁ and G₂ are growth phases and S phase in between is the phase for DNA replication After G₂ completion, cell enters M phase, divides and forms 2 new daughter cells by cytokinesis.

19. The correct sequence of the following events in the human female reproductive cycle is

I. Secretion of FSH

II. Growth of corpus luteum

III. Growth of follicle and oogenesis

IV. Ovulation

V. Sudden increase in the levels of LH

(a) I, II, IV, V, III

(b) II, I, III, IV, V

(c) I, III, V, IV, II

(d) I, V, III, IV, II

Ans. (c)

Sol. The correct sequence will be

1. Secretion of FSH

2. Growth of follicle and oogenesis.

3. Sudden increase in the levels of LH.

4. Ovulation.

5. Growth of corpus luteum.

20. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

21. The reaction that produces -bromophenol as the major product is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

22. For an autocatalytic second order reaction R P, the rate law is

[Where v is rate of the reaction and k is the rate constant]

(a) v = k [R]

(b) v = k [R][P]

(c) v = k [R]²

(d) v = k [P]²

Ans. (b)

Sol. In case of auto catalyst, if the product form behave as reactant then only they are considered in rate law.

For example, hydrolysis of ester, acetic acid formed behave as autocatalyst.

In the same way for the auto catalyst reaction, R P.

The rate law   = k [R] [P]

v = k[R][P]

23. In metal-carbonyl complexes, the -back bonding is

(a)

(b)

(c)

(d)

Ans. (c)

Sol. A pair of bonds arises from overlap of filled d-orbitals on the metal with a pair of -antibonding orbitals projecting from the carbon atom of the CO.

So, in metal carboxyl complexes, the -back bonding is d– * types.

24. If u(x) and v(x) are differentiate at x = 0, and if u(0) = 5, u' (Q) = -3, v(0) = -1 and v'(0) = a then the value of at x = 0 is

(a) –20

(b) -7

(c) 6

(d) 13

Ans. (c)

Sol.

As given u(0) = 5, u' (0) = – 3

v(0) = – 1, v' (0) = 2

    = 5 × 2 + (– 1) × (– 3) +

    = 10 + 3 + (3 – 10)

    = 13 – 7 = 6

25. Two dice are thrown simultaneously. The probability that the sum of the numbers obtained is divisible by 7 is

(a) 1/6

(b) 1/36

(c) 0

(d) 1/18

Ans. (a)

Sol. Favourable = (1,6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1) = 6

Total chance = 6 × 6 = 36

Probability = 6 / 36 = 1 / 6

26. If one of the diameters of a circle has end points (2, 0) and (4, 0), then the equation of that circle is

(a) x² – 3x + y² + 5 = 0

(b) x² – 4x + y² + 6 = 0

(c) 2, 1 and 2

(d) x² – 6x + y² + 8 = 0

Ans. (d)

Sol. (x₁, y₁)    (2, 0) and (x₂, y₂)    (4, 0)

Equation of circle

(x – x₁)(x – x₂) + (y – y₁) (y – y₂) = 0

(x – 2) (x – 4) + (y – 0)(y – 0) = 0

x² – 4x – 2x + 8 + y² = 0

x² – 6x + y² + 8 = 0

27. If P = {1, 2, – 1, 3}, Q = {0, 4, 1, 3} and R = {1, 6, 7}, then

(a) {1,2}

(b) {1,3}

(c) {2,1}

(d) {2,3}

Ans. (b)

Sol. P = {1, 2, – 1, 3}

Q = {0, 4, 1, 3}

R = Q R

A = {0, 1, 3, 4, 6, 7)

P A = {1, 3}

28. The position of a particle along the y-axis is y = Pt⁴ + Q. For the equation to be dimensionally consistent, the dimension of P in terms of length [L] and time [T] is

(a) LT^–1

(b) LT^–2

(c) LT^–3

(d) LT^–4

Ans. (d)

Sol. y – Q = Pt⁴

According to the principle of homogenous = = LT^–4

29. Two inductors P and Q having inductance ratio 1 : 2 are connected in parallel in an electric circuit. The energy stored in the inductors P and Q are in the ratio

(a) 1 : 4

(b) 1 : 2

(c) 2 : 1

(d) 4 ; 1

Ans. (c)

Sol. Energy stored in inductor is

U =

Let L₁ = x

L₂ = 2x

30. A body X of mass M moving with velocity v hits a stationary body Y of mass m. If M >> m and X moves with the velocity v', then the velocity of Y after an elastic collision is

(a) 2v

(b) v + v'

(c) v-v'

(d) 2v'

Ans. (b)

Sol. Particle of mass M is moving with velocity v and M is collides with m.

After the collision velocity of mass (m) = v [ M > m]

So, the velocity of Y after collision = (v + v').

31. The cells involved in allergic reactions and containing surface receptors of IgE antibodies and histamine are

(a) basophils

(b) mast cells

(c) monocytes

(d) neutrophils

Ans. (a, b)

Sol. Antigen specific IgE production occurs when IgE binds to F_c receptors on mast cells and basophils which initiates hypersensitivity reaction. Mast cells, basophils, eosinophils have important role in allergic inflammation and innate responses

32. Which of the following is (are) incorrect in the regulation of the trp operon?

(a) It is an example of a negatively controlled repressible operon

(b) The amino acid Trp inactivates the repressor

(c) The amino acid Trp induces the operon

(d) The repressor binds to the operator in the presence of amino acid Trp

Ans. (b, c)

Sol. Tryptophan synthesis in E. coli bacteria occurs by the enzymes coded by trp operon. This operon is a repressible operon showing negative regulation of gene expression. Repressor gets activated in the presence of tryptophan which binds to operator to stop the transcription of genes on operon and vice-versa occurs in absence of tryptophan.

33. Which of the following organs are correctly paired with their function?

(a) Large intestine – Protein digestion

(b) Oral cavity – Starch digestion

(c) Pancreas – Bile production

(d) Small intestine – Fat digestion

Ans. (b, d)

Sol. Enzymes involved in protein digestion are pepsin, chymotrypsin, trypsin present in stomach and small intestine. Salivary gland produce -amylase that help in starch digestion inside mouth. Carbohydrate digestion begins inside the oral cavity. Bile is produced by liver and stored in gall bladder to maintain it's sufficient supply to digest or emulsify fats. Pancreas have both endocrine and exocrine secretions. Pancreatic juices produced by it's exocrine glands help in digestion. Maximum absorption of nutrients and fat digestion occurs in small intestine.

34. The , for homolactic fermentation converting glucose to lactate is –196 kJ mol^–1. If , for the formation of ATP is + 30.5kJ mol^–1, then

(a) homolactic fermentation is 31% energy efficient

(b) the efficiency of energy conservation is 69%

(c) the energy stored in the form of ATP is 31%

(d) the process results in the loss of 31% of energy

Ans. (a, c)

Sol. (a, c)

35. Bacterial plasmid genes of non-chromosomal origin are associated with

(a) providing resistance against antibacterial agents

(b) the degradation of toxic materials

(c) the production of certain toxins

(d) the transfer of genetic material from one cell to another cell

Ans. (a, b, c, d)

Sol. Genes the plasmid are associtated with

• Antibacterial resistance

• Degradation of toxic substances

• Production of toxins

• Transfer of genetic material from one cell to another

36. The elements with atomic numbers 19, 37 and 55

(a) form cubic chloride salts with the coordination number of cation being 6

(b) form ionic fluorides with general formula MF

(c) have lowest density of solids in their respective periods

(d) have lowest ionisation energy in their respective periods

Ans. (b, c, d)

Sol. Atomic number 19, 37 and 55 are alkali metals

• form ionic fluoride with general formula MR

• have lowest density of solids in their respective periods.

• have lowest ionisation energy in their respective periods.

 

37. Fehling's solution

(a) contains a copper complex of tartaric acid

(b) forms a brick-red precipitate with glucose

(c) forms a white precipitate with aldehydes

(d) is used as a test reagent for reducing sugars

Ans. (a, b, d)

Sol. Fehling's solution is a chemical reagent used to differentiate between water-soluble carbohydrate and ketone functional groups and as a far reducing sugars and non-reducing sugars, supplementary to the Tollen's reagent test. The test was developed by German Chemist Hermaner Von Fehling in 1849.

Fehling's solution contains copper (II) ions complexed with tartrate ions in sodium hydroxide solution. Complexing the copper (II) ions with tartrate ions presents precipitations of copper (II) hydroxide.

Fehling's solution is a deep-blue, alkaline solution used to test for the presence of aldhydes (e.g. formaldehyde. HCHO) or other compounds that contain the aldehyde functional group CHO.

The substance to be tested is heated with Fehling's solution, formation of a brick-red precipitate indicate the presence of the aldehyde group. Simple sugar e.g. glucose give a positive test.

38. Which of the following point(s) lies(lie) on the plane 2x + 3y + z = 6?

(a) (0,0, 6)

(b) (0,2.0)

(c) (1.1,1)

(d) (3,0.0)

Ans. (a, b, c, d)

Sol. 2x + 3y + z = 6

using (0, 0, 6)

2 × 0 + 3 × 0 + 6 = 6

using (0, 2, 0)

2 × 0 + 3 × 2 + 0 = 6

Using (1, 1, 1)

2 × 1 + 3 × l + 1 = 6

using (3, 0, 0)

2 × 3 + 3 × 0 + 0 = 6

39. Kinetic theory of an ideal gas is based upon the following assumption(s).

(a) Gases are made of molecules with negligible volume

(b) The gaseous molecules do not possess kinetic energy

(c) The molecules are in constant random motion

(d) Intermolecular forces of attraction are negligible

Ans. (a, c, d)

Sol. A gas is composed of moleculas that are separated by average distances that are much greater than the sizes of the molecules themselves.

The volume occupied by the molecules of the gas is negligible compared to the volume of the gas itself.

• The molecules of an ideal gas exert no attractive forces on each other.

• The molecules are in constant random motion and as material bodies, they obey Newton's laws of motion.

40. The electric field and capacitance of a capacitor in the absence of dielectric material are E and C, respectively. When the capacitor is filled with a dielectric material, the electric field and capacitance of the capacitor becomes E' and C' respectively. Which of the following is(are) correct?

(a) E' > E and C' = C

(b) E' < E and C' > C

(c) E' = E and C' > C

(d) E' > E and C' > D

Ans. (b)

Sol. With presence of dielectric material, capacitance is increased and electric filed is inversely proportional to capacitance.

41. Antigen and antibody interaction is shown by the following scheme

Ab + Ag Ab – Ag

Where Ab is antibody, Ag is antigen and Ab-Ag is antigen-antibody complex. The values of k₁ and k_–1 are 5 ×10^–5 mM^–1 s^–1 and 2 × 10^–7 s^–1, respectively. The dissociation constant for the complex is _________ nM.

Ans. (4)

Sol. k_d (dissociation constant)

Putting the given values

= 4 × 10^–8 = 4 nM

42. The population of a bacterial culture increases from one thousand to one billion in five hours. The doubling time of the culture (correct to 1 decimal place) is _________ min.

Ans. (15.15)

Sol. (15.15) As we know

Doubling time

where t = time interval (hr/min)

b = no. of bacteria after time t = b = 10⁹

B = no. of bacteria at initial level = B = 10³

So putting the values

43. The K_M and v_max of lactate dehydrogenase for conversion of pyruvate to lactate are 1 mM and 5 nMs^–1, respectively. At 0.25 mM pyruvate, the velocity of the reaction catalysed by lactate dehydrogenase is_________ nMs^–1.

Ans. (1)

Sol. According to Michaelis Menten Equation

= 1 × 10^–9 M or 1 nM

44. A linear DNA contains five restriction sites for EcoRI and three restriction sites for BamHI. The number of fragments that will be generated after digestion with EcoRI is _________.

Ans. (6)

Sol. When linear DNA is cut at 5 restriction sites it will produce 6 fragments which are as follows:

5' G A A T C C

3' C T T A A G

5' G

3' C T T A A

5' G A

3' C T T A

5' G A A

3' C T T

5' G A A T

3' C T

5' G A A T C

3' C

45. Total number of singlets observed in the ¹H NMR of the following compound is _________.

Ans. (3)

Sol. The number of singlets observed in the given compounds is 3.

46. The [H⁺] of 0.1 N acetic acid solution is 1.33 × 10^–3.The pH of the solution (correct to two decimal places) is _________ .

Ans. (2.88)

Sol. (2.88) – log(1.33 × 10^–3)

47. The positive root of the equation x⁴ + x² – 2 = 0 is _________ .

Ans. (1)

Sol. x⁴ + 2x² – x² – 2 = 0

x²(x² + 2) – 1(x² + 2) = 0

(x² – 1) (x² + 2) = 0

x = 1, ± 2i  positive root = 1

48.

Ans. (1)

Sol.

49. One gram of radioactive nuclei with a half life of 300 days is kept in an open container. The weight of nuclei remaining after 900 days (correct to 1 decimal place) is _________ mg.

Ans. (125)

Sol. 125

50. Two sources P and Q produce electromagnetic waves with wavelengths , respectively. Source P ejects a photon with a maximum kinetic energy of 4.0 eV from a metal with work function 2.0 eV. The maximum kinetic energy (eV) of a photon ejected by source Q from the same metal is _________ .

Ans. (1ev)

Sol. Wavelength of electromagnetic wave, P =

Wavelength of electromagnetic wave, Q =

Kinetic energy of = 4.9 eV

Kinetic energy = 4 × 1.6 × 10^–19 V

Kinetic energy =

h = 6.62 × 10^–34 Js

c = 3 × 10⁸ m/s

= work function

4 × 1.6 × 10^–19 = 2 × 1.6 × 10^–18

So, from kinetic energy formula,

Kinetic energy =

Q Would be :–

= 3 × 1.6 × 10^–19 – 2 × 1.6 × 10^–19

= 1 × 1.6 × 10^–19

= 1ev

51. The standard oxidation potentials for oxidation of respectively. The standard free energy for oxidation standard conditions (correct to 1 decimal place) i _________ kJ. [Faraday Constant is 96500 C mol^–1]

Ans. (–217.36 kJ)

Sol. (– 217.36)

n = no of electrons

F = faraday constant = 96500 C/mole or 23 Kcal/mol V

= –52 Kcal/mol

or –52 × 4.18 kJ = –217.36 kJ

52. The K_M and v_max of an enzyme are 4 mM inhibitor and 0.1 nM h^–1 respectively. In the presence of 1.5 mM inhibitor, the K'_M and v'_max of the enzyme are 6 nM and 0.1 nH h^–1, respectively. The value of inhibition constant, K₁ (correct to 1 decimal place) is _________ nM.

Ans. (3 mM)

Sol. Since V_max is same before and after inhibition it shows competitive inhibition

Putting the values

k₁ = 3 × 10^–3

or ^k₁ = 3 mM

53. The relation between log₁₀ (MW) [where MW = molecular weight in kDa] of a mixture of protein standards and their retention factors (R_f) obtained from native-PAGE is log₁₀ (MW) = –2R_f + 3. If the measured retention factor for a protein with 180 amino acids is 0.5, then the number of identical monomers in the protein is _________ .

Ans. (5)

Sol. Number of identical monomers in protein = 5

54. In bacteria, a ribosome synthesizes a protein containing 300 amino acids from mRNA in 20 seconds. If the average lifetime of a mRNA is 2 minutes, the number of ribosomes that can translate a single mRNA containing 1350 nucleotides is _________ .

Ans. (4)

Sol. 300 amino acids are synthesized in 20 seconds

This means 300 × 3 codons (900) are translated in 20 seconds

Since average lifetime of mRNA is 2 min (120 seconds is 20 × 6)

this implies it can translate 900 × 6 nucleotides = 5400 in 120 seconds.

So the no. of ribosome that can translate 1350 nucleotide will be

In 20 seconds a ribosome synthesize a protein containing 300 amino acids from mRNA.

300 = = 45 × 60 × 2

Nucleotide = 5400

If the average lifetime of mRNA is 2 min.

The number of ribosomes that can translate mRNA containing 1350 nucleotide    = = 4

55. In 2N H₂SO₄, an organic compound shows fluorescence with quantum yield, = 0.42 and fluorescence rate constant, k_f = 5.25 × 10⁷ s^–1. The observed fluorescence lifetime of it under the same conditions (correct to 1 decimal place) is _________ ns.

Ans. (8)

Sol.

= 0.08 × 10^–9

= 8 × 10^–9 = 8 ns

56. In acidic solution, permanganate ion is reduced by ferrous ion. The number of electrons involved in the reduction of permanganate ion is _________ .

Ans. (5)

Sol. Overall reaction for permanganate ion reduced by ferrous ion in acidic solution

Hence, number of e^– involved is 5

57. If a and b are unit vectors and the angle between them is , then the magnitude of a – b is

Ans. (1)

Sol. Two vectors a and b, we are looking for the difference between them which we can write

d = a – b

Subtracting, two vectors graphically is not obvious, so what we can do is replace the vector b with its negative where we define.

c = – b

So, our first equation becomes

b = a + (– b) = a + c

Above equation can be represented graphically as

Now, we first add a and c in the normal way. Finding the coordinates of a and c from graph.

a = (1, 0)

Now, we add the x and y components of each vector to get our resultant :

So, in polar form, the resultant is

So, magnitude of a – b = 1

58. Using the letters in the word TRICK a new word containing five distinct letters is formed such that T appears in the middle. The number of distinct arrangement is _________ .

Ans. (24)

Sol. At first place, 4 characters can be added. After filling first character any of the remaining 3 characters can be filled in at second position and so hence

= n(4) × n(3) × n(2) × n(1) = 41 = 24

59. An X-ray tube operates at 30 kV. If one electron converts 10% of its energy into a photon at first collision, then the wavelength of the photon (correct to two decimal places) is _________ Å.

[h = 4.14 × 10^–15 eVs^–1, c = 3 × 10³ ms^–1 and e = 1.6 × 10^–19 C]

Ans. (4.14 Å)

Sol. X-ray tube operates at 30 kV = 30000 V

10% of electron energy, i.e. converted in photon

= × 30000

By competition effect

E = eV –

e = charge on electron

V = the accelerating voltage

h = Plank's constant

c = speed of light

Using above formula with values given

1.6 × 10^–19 × × 30000

= 4.14 × 10^–10

= 4.14 A

60. In a mass spectrometer, a deuteron with kinetic energy 17 MeV enters a uniform magnetic field of 2.4 T with its velocity perpendicular to the field.. The deuteron moves in a circular path in the magnetic field. The radius of its path in the magnetic field (correct to two decimal places) is _________ cm. [mass of deuteron is 3.34 × 10^–27 kg, 1 MeV = 1.6 × 10^–13 J and e = 1.6 × 10^–19 C]

Ans. (35)

Sol. Given magnetic field (B) = 2.4 T

    Mass of deuteron (m) = 3.34 × 10^–27 kg

    Kinetic energy (KE) = 17 MeV

= 17 × 1.6 × 10^–13 V

We know that

Radius (r) =

   =

=

=

= 0.3489 m

= 0.3489 × 100

= 34.89 = 35 cm