1. Pure lgG antibody was run on SDS-PAGE under reducing condition. How many bands would you see after staining with Coomassie blue?

(a) 4

(b) 2

(c) 1

(d) 6

Ans. (b)

Sol. Disulphide bonds are broken under reducing conditions which separates light chain and heavy chain 2 light chains will form one band on SDS page and similarly one band will be seen by 2 heavy chains. Hence we obtain 2 bands after staining with coomassie blue.

2. During mitosis, disappearance of the nucleous is hallmark of

(a) metaphase

(b) prophase

(c) anaphase

(d) telophase

Ans. (b)

Sol. Disintegration of nuclear memberane and disappearance of nucleolus are the characteristic features of prophase in mitosis.

3. Eukaryotic cell containing flagella is

(a) cell lining the Fallopian tube

(b) sperm

(c) Paramecium

(d) cell lining the respiratory tract

Ans. (b)

Sol. A sperm cell contains head, neck, middle piece and flagellum which helps in sperm mobility and enable it to ascend in female tract.

4. The circulatory levels of estrogen is derived mainly from

(a) thecal and granulosa cells

(b) gonadotrophs

(c) endometrial epithelia

(d) Leydig cells

Ans. (a)

Sol. Granulosa cells whose estrogen under influence of androstenedione which provided by theca cells. Thus we can say thecal and granulosa cells have important role is estrogen synthesis

5. Radial symmetry is the characteristic feature of which of the following phyla?

(a) Arthropoda

(b) Mollusca

(c) Cnidaria

(d) Chordata

Ans. (c)

Sol. Radial symmetry is observed when body can be divided into 2 identical halves by any plan passing through central axis of body.

Eg. Cnidarians

6. In ecotone, some species become abundant and they are called

(a) sibling species

(b) endemic species

(c) rare species

(d) edge species

Ans. (d)

Sol. Ecotone represents the area of transition between 2 communities where species from both commu- nities reside , thus we observe greater species diversity in this region which explains the edge effect phenomena therefore species living abundantly in ecotone region are known as edge species

7. From the set of 10 numbers {1, 2,..., 10} three numbers are selected at random without replacement. the probability that the sum of these selected numbers is 9, is

(a) 1/40

(b) 1/20

(c) 3/10

(d) 3/80

Ans. (a)

Sol. The possible three numbers whose sum is 9,

= {(1, 2, 6), (1, 3, 5), (2, 3, 4)}

Since three number are chosen from out of 10 numbers. Probability of selecting these numbers whose sum is 9

8. Consider two nuclei with the same mass number A. For which of the following values of A, the fusion reaction is not possible?

(a) 15

(b) 22

(c) 29

(d) 36

Ans. (d)

Sol. For the nuclei having mass number 36, the fusion reaction is not possible because fusion reaction occurs in lighter nuclei. Among all other options, nuclei with mass number 36 is comparatively heavier, thus fusion reaction will not take place.

9. The time-period and the amplitude of an object executing simple harmonic motion, under the restoring force of a spring, are 3.14 s and 0.2 m, respectively. If the mass of the object is 2 kg, then the maximum force (in Newton) exerted by the spring on the object is

(a) 1.6

(b) 3.2

(c) 4.6

(d) 5.2

Ans. (a)

Sol. Given, T = 3.14 s

a = 0.2 m, m = 2 kg

Maximum force exerted by the spring on the object is

= 2 × 4 × 0.2 = 1.6 N

10. The migratory aptitude of (alkyl or aryl) substituents is Baeyer-Villiger oxidation is

(a) methyl < primary < secondary < tertiary

(b) tertiary < secondary < primary < methyl

(c) phenyl < methyl < primary < tertiary

(d) tertiary < primary < methyl < phenyl

Ans. (a)

Sol. The migratory aptitude of alkyl or aryl substituent in Baeyer-Villiger oxidation is methyl < primary < secondary < tertiary. If there is an electron with drawing group on the substituent, then it decreases the rate of migration. According to the carbocation resonance structure of Criegee intermediate, the substituent that can maintain a positive charge, would be most likely to migrate. Tertiary groups are more stable carbocations than secondary groups and secondary groups are more stable than primary. Thus the tertiary > secondary > primary > methyl trend is followed.

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

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

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

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

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

Ans. (b)

Sol. Epilepsy is marked by repeated seizures which are caused by uncontrolled electric discharge from nerve cells in cerebral cortex

• Alzheimer is neurodegenerative disorder marked by loss of memory i.e dementia which is caused by reduced production of acetyl choline is involved in memory and learning.

• Parkinson disease is caused by degeneration of dopaminergic neurons in substantia niagra

• Huntington disease is autosomal dominant neurodegenerative disease caused by loss of motor and cognitive function.

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

Assertion (A) In the process of ATP synthesis in oxidative phosphorylation, ATP synthase is not a part of electron transport chain on inner mitochondrial membrane.

Reason (R) ATP synthase is coupled to electron transport chain through proton motive force.

(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) both (A) and (R) are false

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

Ans. (a)

Sol. ATP synthesis occurs by coupling the ETC with ATP synthase. Electro transport chain (ETC) generates proton motion force (pmf) by transfer of electrons. ATP synthase then works through this pmf to generate ATP in mitochondria so we can say that ATP synthase and ETC are coupled but ATP synthase is not part of ETC

13. Higher levels of glycosylated haemoglobin (HbA1c) indicate

(a) high haemoglobin level

(b) anaemic condition

(c) diabetes

(d) favism

Ans. (c)

Sol. Diabetes is marked by increased levels of HbAlc (glycosylated hemoglobin). Under diabetic con- ditions sugar that gets accumulated becomes glycosylated by combining to hemoglobin. Higher the glucose level in blood higher will be the HbAlc value. HbAlc test is the primary test for diagnosis for diabetes

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

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

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

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

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

Ans. (c)

Sol. Messelson and stahl (1958) prooved that DNA replication is semiconservative

Experiments that gave evidence for origin of life on earth were initially performed by Stanley Miller and HAROLD Urey

ALFRED Harshey and Martha chase (1952) proved that are genetic material is DNA

Chemical method for RNA synthesis was given by Har Gobind Khorana.

A cell for system that could form protein without genetic code was developed by Marshal Nirenberg.

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

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

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

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

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

Ans. (a)

Sol. Bacillus anthracis bacterium causes Anthrax disease. Rubella virus causes Rubella viral infection. Trichophyton is a fungus causing Athlete's foot disease. Leishmaniasis is caused by a protozoan parasite Leishmania

16. In an experiment conducted in the dark, isolated chloroplasts are kept in buffer (pH 4.0) at 4°C until their internal pH is equal to 4.0. Then, they are transferred to a buffer of pH 8.0, and ADP and P_i are added at the same time. Which of the following will happen?

(a) Chloroplasts will be destroyed

(b) Chlorophyll in the chloroplast will release bound Magnesium

(c) Chloroplasts will be intact but no ATP will be produced

(d) Chloroplasts will be intact and ATP will be produced

Ans. (d)

Sol. ATP production doesn't need light but instead needs the proton gradient that will be produced by the pH difference when transferred from pH 4 to pH 8. This gradient serves as the driving force for ATP production.

17. Two mammalian cell lines with doubling times of 12 h and 36 h were cultured with radioactive thymidine for 8 h. The cells were further cultured without the Radioactivity for 72 h. Incorporated radioactivity was measured in equal number of cells in each culture, which revealed that

(a) both the cell lines had the same amount of radioactivity

(b) the fast growing cells had more radioactivity

(c) the slow growing cells had more radioactivity

(d) neither of the cells had any radioactivity

Ans. (c)

Sol. Slow growing cells show more radioactivity than fast growing cells. Radioactivity is measured by thymidine concentration which gets depleted as the cell divides. Since fast growing cells divide 6 times faster than normal cells so they show less radioactivity in comparison to slow dividing cells that replicate 2 times slow than normal cells

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

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

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

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

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

Ans. (b)

Sol. YEAST two hybrid system is used for detection of protein-protein intractions in vivo

• Invitro DNA -protein interaction can be identified using EMSA (electrophoretic mobility shift assay) technique

• Chromatin immunoprecipitation (ChIP) is used to identify DNA- protein interaction in-vivo . We can find desired DNA sequence binding to our known transcription factor.

• Protein structure detection can occur by NMR and x-ray Diffraction techniques NMR work by determining resolution of each atom and X-ray diffraction is detected in single crystals

19. A line L parallel to the vector passes through the point (1, 2, 4) and meets the XY-plane at a point P. The distance between the origin and P is

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Equation of line passing through (1, 2, 4) and parallel to vector is

and equation of XY-plane is

20. Let P(t) denote the population of species at time t. If P(t) is given by the equation = P(1 – P) and if the initial population P(0) = 0.1 million, then the population at t = 1 is

(a)

(b)

(c)

(d)

Ans. (a)

Sol. We have,

 

log p – log (1 – p) = t + C

  log = t + C

t = 0, p = 0.1,

21. Let Z be the set of all integers and f and g are one-one mappings from Z into itself

(a) g(2) = 0

(b) f(3) = 2

(c) g(2) = 1

(d) f(3) = 1

Ans. (a)

Sol. We have, f(g(x)) = g(n + 1) – 1, n is even

g(f(x)) = f(n – 1) – n is odd

Put n = 2

f(g(2)) = g(3) + 1 ...(i)

n = 1

g(f(1)) = f(0) – 1 ...(ii)

  g(3) = f(0) – 1

From Eq. (i) and (ii), we get

f(g(2)) = f(0) – 1 + 1

f(g(2)) = f(0)

   g(2) = 0

22. In a p-n junction, the depletion region has a width of 3 × 10^–7 m and the intensity of electric field in the depletion region is 10⁶ V/m. An electron approaches the junction from the n side with velocity v₁ and enters the p side with velocity v₂. If v₂ = 4 × 10⁵ m/s, then the value of v₁ is

Given data Charge of electron = 1.6 × 10^–19 C; Mass of electron = 9.1 × 10^–31 kg

(a) 3.2 × 10⁵m/s

(b) 4.2 × 10⁵ m/s

(c) 5.2 × 10⁵ m/s

(d) 6.2 × 10⁵ m/s

Ans. (c)

Sol. Given width of depletion region, w = 3 × 10^–7 m

Electric field, E = 10⁶ v/m

   v₂ = 4 × 10⁵ m/s

   v₁ = ?

Electric field in the depletion region is directed from n-side to p-side.

Therefore, the electron faces retarding force in moving from n-side to p-side.

Applying work-energy theorem, we get

    v₁ = 5.15 × 10⁵ = 5.2 × 10⁵ m/s.

23. An -particle and a proton have the same de-Broglie wavelength. Which of the following is also the same for the two particles, if they are moving at non-relativistic speeds?

(a) Frequency

(b) Kinetic energy

(c) Momentum

(d) Speed

Ans. (c)

Sol. Let de-Broglie wavelength of particle be and that of proton be

It is given that,

where,  p_a = momentum of alpha particle and

p_p = momentum of proton.

   p_p = p_a

24. An electron is accelerated from the rest through a potential difference of 400 V. The electron, then enters a uniform magnetic field that is perpendicular to the direction of electrons. The radius of the circular path experienced by the electron is 10 cm. The angular speed of electrons, in radians/sec, is

Given data Charge of electron = 1.6 × 10^–19 C; Mass of electron = 9.1 × 10^–31 Kg

(a) 1.18 × 10⁷

(b) 1.18 × 10⁸

(c) 2.18 × 10⁷

(d) 2.18 × 10⁸

Ans. (b)

Sol. Given, potential difference, = 400 V

Kinetic energy of the electron after accelerated through ΔV is

k = e

where, v = speed of the electron and

m_e = mass of the electron

where, = angular speed of electron and

r = radius of circular path = 10 cm = 10 × 10^–2 m

= 1.185 × 10⁸ = 1.18 × 10⁸ rad/sec

25. Consider two vectors P and Q of equal magnitude. If the magnitude of P + Q is two times larger than that of P – Q, then the angle between them is

(a) 107°

(b) 117°

(c) 127°

(d) 137°

Ans. (c)

Sol. Given |   P | = | Q |

and       | P + Q | = 2 | P – Q |

Now,   (P + Q) . (P + Q) = 4 ( P – Q) . (P – Q)

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

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

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

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

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

Ans. (d)

Sol. Arrhenius equation lnk

Van't Hoff equation lnk

Kirchhoff's law

DrH₂ – DrH₁ = DC_p (T₂ – T₁)

Clausius - Clapeyron equation

27. The major product in the following reaction is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

In presence of strong electron with drawing group, the reactant go for elimination.

Thus, the product will be

28. Of the given isomers of molecules I and II, the meso-form is

CH₃CH(Br)CH(Cl)CH₃ I CH₃CH(Cl)CH(Cl)CH₃ II

(a) (R, R)-isomer of I

(b) (R, S)-isomer of II

(c) (R, S)-isomer of I

(d) (S, S)-isomer of II

Ans. (b)

Sol. For a molecule to be mesoisomer, there must be an asymmetric carbon atom as well as plane of symmetry.

29. Considering the periodic trends of elements, which of the following is not correct?

(a) MgO and Na₂O are basic, and SiO₂ is acidic

(b) Atomic radius decreases in a period from left to right

(c) Order of first ionisation energies: K < Mg < Ca

(d) Order of bond energies: C – C < Si – O < N N

Ans. (c)

Sol. Order of first ionisation energies will be – K > Mg > Ca.

30. Which one is the major product of the following reaction?

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

31. Identify the autoimmune disease among the following.

(a) Type II Diabetes Mellitus

(b) Type I Diabetes Mellitus

(c) Gestational Diabetes

(d) Pernicious Anaemia

Ans. (b,d)

Sol. Diabetes mellites is a type I diabetes which is also an autoimmune disorder as it destroys insulin producing -cells of pancreas.

Pernicious Anaemia is caused by lack of absorption of vitamin B-12 by intestine which leads to decreased production of RBCs. It is also an autoimmune disorder as body attacks the intrinsic factor protein of cells present in lining of stomach that make it.

32. Which of the following statements are true for hydrogen bonds?

Strength of hydrogen bond is

(a) low in a solvent of high dielectric constant

(b) low in a solvent of low dielectric constant

(c) lower in water as compared to organic solvents

(d) higher in water as compared to organic solvents

Ans. (a, d)

Sol. A hydrogen bond is the electrostatic attraction between polar groups that occur when a hydrogen (H) atom binds to highly electronegative atom.

Strength of hydrogen bond is low in a solvent of higher dielectric constant and higher in water as compared to organic solvents.

33. Which of the following statements are true for cellulose?

(a) Cellulose serves a structural role

(b) Cellulose is a branched polysaccharide

(c) Cellulose is a homopolysaccharide composed of linked D-glucose units

(d) Cellulose is a homopolysaccharide composed of linked D-glucose units

Ans. (a, d)

Sol. (a, d) Cellulose is a structural hexosan (glucosan) homopolysacharide having β (1-4) glycosidic linkage forming unbranched linear chains

34. Which of the following are not true for photosynthesis?

(a) Reduction of CO₂ and H₂O

(b) Oxidation of CO₂ and H₂O

(c) Reduction of CO₂ and oxidation of H₂O

(d) Oxidation of CO₂ and reduction of H₂O

Ans. (a, b, d)

Sol. (a, b, d) Both oxidation and reduction process occurs in photosynthesis. H2O is oxidised to produced O₂ and CO₂ is reduced to from sugar molecule.

35. Apoptosis is a controlled process of cell death. The process involves

(a) exposure of phosphatidyl serine on the outer surface of the cell membrane

(b) decreased permeability of the outer mitochondrial membrane

(c) increased lysosomal activity

(d) inter-nucleosomal cleavage of genomic DNA

Ans. (a, d)

Sol. (a, d) Apoptosis is the programmed cell death that involves destruction of cells. Exposure of phosphatidyl serine on the outer leaflet of the membrane acts as a signal for apoptosis which is recognized by phagocytes for its clearance

36. If f(x) = is continuous and differentiable at all points in the interval [0, 2] and f(2) = , 4

(a)

(b)

(c)

(d)

Ans. (a, b)

Sol. We have, f(x) =

f(x) is continuous and differentiable at all points

a + b = c + 1

Also, f(x) is differentiable at x = 1

F'(1^–) = 2a = f'(1⁺) = c

37. In = 60° and S is the mid-point of QR. If QS = PS and PR = 5, then

(a)

(b)

(c)

(d)

Ans. (a,d )

Sol. We have, ln = 60°

 QS = SR = PS

   PR = 5

 ln PQR tan 60° =

  PQ = PR cot 60° =

   PS = PQ =

Area of PSR = PR . RS sin 30°

   

Now,    PS = SR = QS

  S is circumcenter of PQR.

38. The magnetic field in the interior of a long current carrying solenoid can be increased by

(a) increasing the length of solenoid while keeping the number of turns per unit length as constant

(b) increasing the number of turns per unit length

(c) increasing the current in the solenoid

(d) decreasing the length of solenoid while keeping the number of turns per unit length as constant

Ans. (b, c)

Sol. Strength of magnetic field produced inside the solenoid depends on:

(i) Strength of current, i.e., the magnetic field inside the solenoid can be increased by increasing the strength of the current in the solenoid.

(ii) Number of turns of wire, i.e., the magnetic field inside the solenoid can be increased by number of turns of wire in the solenoid.

39. Which of the following statements are correct?

(a) Fluorescence has a much longer decay period than that of phosphorescence

(b) Radiative transition from T₁ to S₀ is phosphorescence

(c) Radiative transition from S₁ to S₀ is fluorescence

(d) Enhancing the life time of the excited state is quenching

Ans. (b, c)

Sol. The radiative transition can occur (a) from the lowest excited singlet S₁ to the ground state S₀ and (b) from the lowest triplet T₁ to ground singlet S₀. The radiative transition between S₁ S₀ + hn leads to fluorescence emission and that from T₁S₀ hν to phosphorescence emission.

40. In the ¹HNMR spectrum of 1-bromopropane (structure shown below), which of the following

(a) Protons a resonate upfield to protons c

(b) Protons b reasonable downfield to protons c

(c) There are two triplets and one quartet in the spectrum

(d) Protons a appear as triplet with high chemical shift in comparison to protons c

Ans. (b, d)

Sol. Figure

C 3 H = triplet, downfield

b Quartet, triplet upfield

a triplet with high chemical shift

41. The net charge on the following peptide at pH 7.0 is ......... .

Val-Asp-Asn-Lys-Ser-lle

Ans. (0)

Sol. To determine the net charge of amino acids at various pH values, we first need to determine the pI for each amino acid. The amino acid carries no net charge when the pH equals to pI. When pH is lower than the pI, the amino acid will carry a net positive charge. When the pH is higher than the pI, then the amino acid carries a net negative charge. In the given sequence (Val - Asp - Asn - Lys - Ser - IIe) Asp contain - 1 charge because it is an acidic amino acid. Whereas Lys is basic amino acid, so it contains +1 charge. All other amino acids are neutral at pH 7. Thus, the given peptide contains '0' net charge.

42. A 152 nm long Watson-Crick double helical DNA (B-DNA) will contain ......... turns.

Ans. (44.7)

Sol. According to HARDY weinberg equilibrium

p + q = 1

Given p (dominant allele frequency) = 0.7

So we can say q = 0.3

Also we know that

p² + q² + 2pq = 1

So genotypic frequency of heterozygote Aa = 2pq

= 2 × 0.7 × 0.3 = 0.42

43. A population is in Hardy-Weinberg equilibrium for a gene with only two alleles ("A" and "a"). If the gene frequency of the allele "A" is 0.7, genotype frequency of heterozygous "Aa" is ......... .

Ans. (0.42)

Sol. According to Hardy-Weinberg equation

p = frequency of dominant allele (represented by A)

q = frequency of recessive allele (represented by a)

For a population in genetic equilibrium

p+ q= 1.0

(the sum of frequencies of both alleles is 100%)

MM  (p + q)² = 1

so,    p² + 2 pq + q² = 1

Here, p² = frequency of AA (homozygous dominant)

    2pq = frequency of Aa (heterozygous)

 q² = frequency of aa (homozygous recessive)

Given p = 0.7

  q = 1.0 – 0.7 = 0.3

 The frequency of heterozygous 'Aa' is

   2pq = 2 × 0.7 × 0.3 = 0.42

44. A receptor binds to its ligand with dissociation constant K_d = 10^–8 M. The concentration of the ligand required to occupy 10% of the receptors would be 10^–x M. The value of x is ......... .

Ans. (9)

Sol. [R] + [L] = [RL]

So

where ([R] = molar concentration of receptor

[L] = molar concentration of ligand

k_d = dissociation constant

As given 10% of receptor is occupied by ligand

so we can say [RL] = 0.1 [R]

k_d = 10^–8 = 0.1 [10^–x]

x = 9

45. The plane x + y + z = 0 intersects the sphere x² + y² + z² = 9 along a circle. If (2, y, z) is a point on the circle, then the value of | y + z | is ......... .

Ans. (2)

Sol. Equation of plane is x + y + z = 0

  Equation of plane is passing through origin (0, 0, 0) and centre and radius of sphere.

x² + y² + z² = 9 is (0, 0, 0) and 3, respectively.

  Centre of circle is (0, 0, 0) (2, y, z) is a point on the circle.

i.e., (2, y, z) is also lie on plane

2 + y + z = 0

y + z = –2    | y + z | = 2

46. In Young's double slit experiment, the slits are separated by a distance of 0.05 mm and the source emits light to two wavelengths 450 and 520 nm.

If the distance between the slit and viewing screen is 2 m, then the separation between 2nd order bright fringes for the two wavelengths is ......... cm.

Ans. (0.56)

Sol. In YDSE, it is given that

Slit separation, d = 0.05 mm = 0.05 × 10^–3 m

Wavelengths,

= 450 nm = 450 × 10^–9 m

= 520 nm = 520 × 10^–9 m

Distance between the slit and the screen is D = 2 m

Distance of nth bright fringe from central maxima is

y_n =

Distance of 2nd order bright fringe due to first wavelength,

y₂₁ = (2)

Similarly, y₂₂ = (2)

Separation between these 2nd order bright fringes

= × 2(520 – 450) × 10^–9

= × 70 m = 0.8 × 7 × 10^–4 × 10 m

= 56 × 10^–4 m = 0.56 cm

47. A jet plane lands on an aircraft carrier at 70 m/s and stops in 3 s. Assuming that the acceleration is constant, the jet plane travels a distance of ......... m before it stops.

Ans. (105)

Sol. Initial speed of jet plane

i.e., u = 70 m/s, t = 3 s, a = 3 and v = 0

v = u + at

0 = 70 + 3a

a =

Now, v² – u² = 2as

S =

Distance = 105 m.

48. The number of signals is ¹³C NMR for the following structure is ......... .

 

Ans. (7)

Sol. The number of signal of ¹³C NMR can be found out by finding the interaction of ¹³C with the neighbour carbon possessing hydrogens.

In this case, there is seven signal in ¹³C NMR.

49. From the database of a clinic it was found that out of 2000 patients who had visited the clinic in a year, 900 had high BP, 900 had high Sugar and 400 had neither high BP nor high Sugar. On a given day, if 20 patients visit the clinic, the expected number of patients who have both high BP and high Sugar is ......... .

Ans. (2)

Sol. Let, n(B) = number of high BP patient

n(S) = number of high sugar patient

n(B S) = n (B) + n(S) – n(B S)

2000 – 400 = 900 + 900 – n(B S)

1600 = 1800 – n(B S)

n(B S) = 1800 – 1600

n(B S) = 200

In a year, number of patient who have suffered from both high BP and high sugar = 200.

In a day = × 20 = × 20 = 2 patient.

50. An enzyme catalyzes the conversion of 4 × 10^–4 M substrate into product at a rate of 20 . If the K_m value for the enzyme is 2 × 10^–4 M, then the value of V_max is .........

Ans. (30)

Sol. According to Michaelis Menten equation

where V is reaction velocity

V_max is maximum velocity

[s] is substrate concentration

K_m is Michaelis constant

As per the question

51. The following polypeptide chain was sequentially treated with dithiothreitol, cyanogen bromide and trypsin.

Phe-Trp-Lys-Tyr-Met-Gly-Ala-Cys-Cys-Pro-Met-Asp-Gly-Arg-Phe-Ala-Gly-Trp

The total number of fragments expected at the end of complete digestion of the polypeptide are .........

(consider that none of the reagents interfere with each other's activities)

Ans. (5)

Sol. Dithiothreitol breaks disulfide bonds of amino-acids like cysteine so we observe no cleavage of polypeptide under it's influence.

CNBr cleaves the Met residues and trypsin cleaves c terminal of arginine and lysine amino acids

So we receive 5 Fragments

52. In maize, the genes for coloured seed and round are dominant over the genes for colourles seed and shrunken seed. Pure breeding strains of the double dominant variety were crossed with the double recessive variety and a test cross of the F1 generation produced the following.

For the above, the distance between the genes for seed colour and seed shape on the chromoscomes would be ......... centimorgan units.

Ans. (3)

Sol. Total no. of off springs = 350 + 396 + 1440 = 800

Recombinant gamete frequency = 14 + 10 = 24

R_f (recombination frequency)

53. A culture of 10⁶ bacteria, with doubling time of 60 min, is grown in a nutrient medium at 37°C. Considering that the nutrients are unlimited, the number of bacteria at the end of 10 h would be ......... × 10⁶.

Ans. (1024)

Sol. Logistic growth occurs under unlimited nutrient condition of medium since doubling time is 60 min so culture will have 2 × 10⁶ cells at end of hour when initially it was 10⁶

So, after 10 h her it will be – (2)¹⁰ × 10⁶

2 – 4 – 8 – 16 – 32 –– 128 – 256 – 512 – 1024 × 10⁶ cells

54. A 50-amino acid residue stretch of a globular protein adopts an extended structure containing a true -helix of 24 residues and b-strand of 26 residues. the total length of the stretch will be ......... nm.

Ans. (11.4)

Sol. 2 helix contains 24 residues

1 turn of -helix contains 3-6 residues

No. of turns having 24 residues

back turn of -helix = 0.54 nm.

6.6 turns = 6.6 × 0.54 = 3.6 nm

It is given that -strand has 26 residues

But we know 1 -strand turns has 2 runiduce

So no. of turns having 26 residuces turns

One turn of strand = 0.6 nm

Length of -strand of 13 turns = 13 × 0.6 = 7.8 nm

Total length of stretch will be length of α-helix + length of β-strand 3.6 nm + 7.8 nm = 11.4 nm.

55. The right limit (x – 3)² (log (x – 3) + cosec (x – 3)²) is .........

Ans. (1)

Sol. (x – 3)²(log(x – 3) + cosec(x – 3)²)

= (3 + h – 3)²(log(3 + h – 3) + cosec(3 + h – 3)²)

= h²[log(h) + cosec²h]

=

= (sin² h log h + 1) = 1 × (0 + 1) = 1

56. If [x] denotes the greatest integer valued function (e.g., [1.16] = 1 and [1.8] = 1) and

dx = L, then L = ......... degrees.

Ans. (30)

Sol.

 

57. A copper wire having a cross-sectional area of 6.62 × 10^–6 m² carries a current 20 A. Assuming that each atom contributes one free electron to the current, the time required by electrons to travel a distance of 1 m is ......... min.

Given data Density of copper = 8.92 g/cm³, molar mass

= 63.5 g/mol and Avogadro number

= 6.02 × 10²³

Ans. (74.77)

Sol. (74.77) Given, A = 6.62 × 10^–6 m², l = 1m, / = 20 A

p = 8.92 g/cm³, M. mass = 63.5 g/mol

 N_A = 6.02 × 10²³, Density =

 

8.92 g/cm³ = [ Area × Length = Volume]

  mass (m) = (8.92 × 6.62 × 10⁶)g

Now,   1e^– = 1 F = 1 mol of Cu = 63.5 g

   (8.92 × 6.62) g of Cu = = 0.9299 F

58. A piece of charcoal, containing 36 g of carbon, found in ancient ruins shows a ¹⁴C activity of 300 decays/min. The tree, from which this charcoal came, has been dead for ......... years.

Given data The ratio of ¹⁴C to ¹²C is 1.3 × 10^–12 in the CO₂ molecules of atmosphere and the half life of ¹⁴C is 5730 yr.

Ans. (4823)

Sol. (4823) ¹⁴C activity would decrease with times. So, in order to estimate the time of death of the tree we need to find the initial activity of the sample.

We have, decay constant

   =

[ T_1/2 = 5730 yr = 5730 × 3.16 × 10⁷ s]

   = 3.83 × 10^–12 S^–1

Number of ¹²C nuclei = Number of mole × N_A

    = × 6.02 × 10²³ = 1.8 × 10²⁴

Since the ratio of ¹⁴C to ¹²C = 13 × 10^–12

Therefore, number of initial ¹⁴C nuclei

N₀(¹⁴C) = 1.3 × 10^–12 × 1.8 × 10²⁴ = 2.34 × 10¹²

Now, we can find the initial activity of sample

R₀ = × N₀

   = 3.83 × 10^–12 × 2.34 × 10¹²

   = 8.96 decays/sec = 538 decays/min

We've relation between present activity (R) and initial activity (R₀) as

= –2.61 × 10¹¹ × (– 0.584)

= 1.524 × 10¹¹ sec

= 4823 years (approx).

59. AT constant pressure, 200 g of water was heated from 10°C to 22°C. The molar heat capacity of H₂O at constant pressure is 75.3 JK^–1 mol^–1. The increase in entropy for this process is ......... JK^–1.

(Consider that molar heat capacity of water is independent of temperature and that water does not expand when heated)

Ans. (34.7)

Sol. (34.7) Given, mass of water = 200 gm

    C_P of water = 753 jK^–1 mol^–1

   T₁ = 10° C

 = (10 + 273) K

   T₁ = 283 K

   T₂ = 22° C

= (22 + 273) K

  T₂ = 295 K

We know that, molecular weight of water = 18 g mol^–1

Number of moles of water (n) = = 11.11111111 mol

Now, we have,

  = 11.11111111 mol × 75.3 JK^–1 mol^–1

  = 34.7 JK^–1

60. The number of optically inactive geometrical isomers of [Pt(NH₃)₂(py)₂Cl₂]²⁺ is ......... .

(where, 'py' is pyridine)

Ans. ([Pt(NH₃)₂Cl₂(Py)₂]²⁺)

Sol. [Pt(NH₃)₂Cl₂(Py)₂]²⁺

Due to presence of plane of symmetry, these four isomers are optically inactive, while fifth isomer is optically active.