General Aptitude

1. Rajiv Gandhi Khel Ratna Award was conferred —————— Mary Kom, a six-time world champion in boxing, recently in a ceremony —————— the Rashtrapati Bhawan (the President's official residence) in New Delhi.

(a) with, at

(b) on, in

(c) on, at

(d) to, at

Ans.

Sol.

2. Despite a string of poor performes, the chances of K. L. Rahul's selection in the team are

(a) slim

(b) bright

(c) obvious

(d) uncertain

Ans.

Sol.

3. Select the world that fits the analolgy :

Cover : Uncover : : Associate : ————

(a) Unassociate

(b) Inassociate

(c) Miasolciate

(d) Dissociate

Ans.

Sol.

4. Hit by flood, the kharif (summer sown) crops in various part of the country have been affected. Officials believe that the loss in production of the kharif crops can be recovered in the output of the rabi (winter sown) crops so that the country can behieve its food-grain production target of 291 million tons in the crop year 2019-20 (July-June). They are hopeful that good rains in July-August will help the soil retain moisture for a longer period, helping winter sown crops such as wheat and pulses during the November-February period.

Which of the following statements can be inferred from the given pasage?

(a) Officials declared that the food-grain production target will be get due to good rains

(b) Officials want the food-gain production target to be met by the November-February period

(c) Officials feel that the food-grain production target cannot be met due to floods

(d) Officials hope that the food-grain production target will be met due to a god rabi produce.

Ans.

Sol.

5. The difference between the sum of the first 2n natural numbers and the sum of the first n odd natural numbers is

(a) n² – n

(b) n² + n

(c) 2n² – n

(d) 2n² + n

Ans.

Sol.

6. Repo rate is the rate at which Reserve Bank of India (RBI) lends commercial banks, and reverse repo rate is the rate at which RBI borrows money from commercial banks.

Which of he following statements can be inferred from the above passage?

(a) Decrease in repo rate will increase cost of borrowing and decrease lending by commercial banks

(b) Increase in repo rate will decrease cost of borrowing and increase lending by commercial banks

(c) Increase in repo rate will decrease cost of borrowing and decrease lending by commercial banks

(d) Decrease in repo rate will decrease cost of borrowing and increase lending by commercial banks

Ans.

Sol.

7. P, Q, R, S, T, U, V, and W are seated a circular table.

I. S is seated opposite to W

II. U is seated at the second place to the right of R.

III. T is seated at the third place to the left of R.

IV. V is a neighbour of S.

Which of the following must be true?

(a) P in neighbour of R

(b) Q is a neighbour of R

(c) P is not seated opposite to Q

(d) R is the left neighbour of S

Ans.

Sol.

8. The distance between Delhi and Agra is 233 km. A car P started travelling from Delhi to Agra and anoter car Q started from Agra to Delhi along te same road 1 hour after the car P started. The two cars crossed each other 75 minutes after the car Q started. Both cars were travelling at constant speed. The speed of car P was 10km/hr more than the speed of car Q. How many kilometers the car Q had travelled when the cars crossed each other?

(a) 66.6

(b) 75.2

(c) 88.2

(d) 116.5

Ans.

Sol.

9. For a matrix M = [m_ij]; i, j = 1, 2, 3, 4 the diagonal elements are all zero and m_ij = –m_ji. The minimum number of elements required to fully specify the matrix is

(a) 0

(b) 6

(c) 12

(d) 16

Ans.

Sol.

10. The profit shares of two comparies P and Q are shown in the figure. If the two companies have invested a fixed and equal amount every year, then the ratio of the total revenue of company P to the total revenue of company Q, during 2013-2017 is ——————.

(a) 15 : 17 (b) 16 : 17 (c) 17 : 15 (d) 17 : 16

Ans.

Sol.

Chemistry

1. An aquesous solution contains a mixture of 10^–8 M NaCl and 10^–8 M HCl.

Choose the correct statement about this solution.

(a) The solution is a buffer with pH less than 7.00.

(b) The solution is a buffer with pH greater than 7.00.

(c) The solution is not a buffer but has its pH less than 7.00.

(d) The solution is not a buffer but has its pH greater than 7.00.

Ans. (c)

Sol. The given mixture is not a buffer as it contains salt of strong acid (NaCl) and strong acid (HCl). Thus, H₃O⁺ ion concentration will be coming from 10^–8 M HCl and H₂O. As its contains HCl so its pH will be just less than 7, because at 10^–8 M concentration of HCl, water will also contribute to the H₃O⁺ ion concentrtion.

2. The coordination complex which has a distorted octahedral structure is

(Given : Atomic numbers of V : 23, Mn : 25, Ni : 28, Cu : 29)

(a) [Ni(H₂O)₆]2+

(b) [Mn(H₂O)₆]2+

(c) [V(H₂O)₆]2+

(d) [Cu(H₂O)₆]2+

Ans. (d)

Sol. The electronic configuration of Cu(II) in complex [Cu(H₂O)₆]²⁺ is [Ar]3d⁹4s⁰ and in octahedral field is : . The asymmetrically electronic configuration in e_g orbitals causes John-Teller distortion. So [Cu(H₂O)₆]²⁺ will have a distorted octahedral structure.

3. In napthalene, the value of the integer "n" according to Huckel's rule of aromaticity is _____.

Ans. (2)

Sol. The structure of naphthalene is

The Huckel rule = (4n + 2)

where n = Number of π-electrons in the aromatic ring.

For naphthalene, n = 10 s

10 = (4n + 2)

^{n= 2}

Correct answer is (2)

4. The azimuthal quantum number (l) of an electron in the dz² orbital of a copper tom (atomic number : 29) is ................... .

Ans. (2)

Sol. The electronic configuration of Cu(Z = 29) is [Ar]4s¹, 3d¹⁰.

The value of azimuthal quantum number = n – 1.

Thus, the value of l for electron in is = 3 – 1 = 2.

5. The standard enthalpy of reaction (in kJ mol^–1) for obtaining three moles of H₂(g) from atomic hydrogen in gas phase is _________. (Given: Standard enthalpy of formation of atomic hydrogen in gas phase is 218 kJ mol^–1)

Ans. (–1308)

Sol. The reaction of H₂ gas formation from atomic hydrogen is

Substituting the values in Eq. (1), we get

6. The correct order of the first ionization energies of He, B, N and O in their corresponding ground state is

(a) He > N > O > B

(b) O > N > B > He

(c) He > B > N > O

(d) N > O > B > He

Ans. (a)

Sol. He is noble gas, so its ionization energy is highest among the given atoms. Nitrogen has completely half-filled p-orbitals (1s², 2s²2p³) in outermost shell, so its ionization energy is higher than oxygen (1s², 2s² 2p⁴) and boron (1s², 2s² 2p¹). Oxygen is more electronegative and have smaller size than boron, so, its ionization energy is more than boron. Hence, the correct order for the ionization energy is He > N > O > B.

7. Based on the molecular orbital theory, which one of the following statements with respect to N₂, N₂⁺, O₂ and O₂⁺ is correct?

(a) Bond orders of N₂ and O₂ are higher than their corresponding cations.

(b) Bond energy of N₂⁺ is higher than their that of N₂, wheres bond energy of O₂⁺ is lower than that of O₂.

(c) The unpaired electrons in N₂⁺ and O₂⁺ are present in and orbitals respectively.

(d) The bond in N₂⁺ is shorter than that in N₂, whereas bond in O₂ is shorter than that in O₂⁺.

Ans. (c)

Sol. According to MOT; electronic configuration of N₂, N₂⁺, O₂ and O₂⁺ are written as follows :

The bond order of N₂ is greater than N₂⁺ and that of O₂ is less than O₂⁺.

In N₂⁺ and O₂⁺, the unpaired electrons is present in and orbitals respectively.

8. Which one of the following statements is incorrect about the diborane molecule?

(a) B-H^t bond is a 2-centre-2-electron bond (H^t: terminal hydrogen)

(b) BH^bB bond is a 3-centre-2-electron bond (H^b: bridged hydrogen)

(c) The bond angle H^tBH^t is 122° (H^t: terminal hydrogen)

(d) The B-H^t bond distance is longer than B-H^b bond distance (H^t: terminal hydrogen, H^b: bridged hydrogen)

Ans. (d)

Sol. The structure of diborane is

The bridged H-atom is sharing its one e^– with both the boron atoms. So, the strength of bridge bond (3c – 2e^–) is poor; thus it is longer as compared to terminal bond (2c – 2e^–).

9. Given below are Newman projections of ethylene glycol and 1,2-difluoroethane about their respective C-C bonds. The most stable conformations (lower energy) of ethylene glycol and 1,2-difluoroethane are

(a) I and III respectively

(b) I and IV respectively

(c) II and III respectively

(d) II and IV respectively

Ans. (a)

Sol. The most stable conformation of ethylene glycol is (I) due to intramolecular H-bonding.

However, size of F atom and H-atom is comparable to each other thus the most stable form of 1,2-difluoroethane is (III).

10. In the reaction given below, choose the condition that gives an anti-Markovnikov's product.

ANti-Markovnikov's product

(a) Peroxde/HCl

(b) Aquous mercuric acetate treatment

(c) Diborane addition

(d) Sulfuric and addition

Ans. (c)

Sol. The addition of diborane follows anti-Markownikoff's product.

Correct answer is (c)

11. Which of the following hexoses will give an osazone that has a different melting point from that of the osazone obtained from D(+) glucose?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. The osazone from D (+) glucose is

The osazone from given hexose is

Thus, osazone obtained from the above hexose have different configuration from D(+) glucose. Hence, its melting point is different from D (+) glucose.

Correct answer is (c)

12. A molecule in solution crystallizes into two different crystal forms with rate of 0.02 s^–1 and 0.13 s^–1. If the crystallization is assumed to be under kinetic control, then the half-life (in seconds, rounded off to one decimal place) of the molecule is _______.

Ans. (4.62 seconds)

Sol. For the two parallel reactions

Substituting values in Eq. (1), we get

Taking natural log of above equation, we get

13. The strandared potential for a cell reaction given below is +0.7V. The standard reaction free energy for this cell is _________ kJ mol^–1.

(Correct up to two decimal places). (Given : Faraday constant, F = 96500 C mol^–1)

Au³⁺(aq) + 3Ag(s) Au(s) + 3Ag⁺(aq)

Ans. (–202.65)

Sol. We know

= –nFE^o_cell ...(1)

Given E^o_cell = +0.7V, F = 96500C

For the given electrochemical reaction

Au³⁺(aq)+ 3Ag(s) Au(s) + 3Ag⁺(aq)

The number of electrons involved (n) = 3

Substituting the values in Eq. (1), we get

= –3 × 96500 × 0.7

= –202.65 kJ mol^–1

14. The activation energy (E_a) estimated for a reaction from the Arrhenius equation is 21 kJ mol^–1. If the frequency factor is assumed to be independent of temperature, then the ratio of the rate constants determined at 298K and 260K is _________ (rounded off to two decimal places).

(Given : Gas constant, R = 8.315 J EC¹ mol^–1)

Ans. (3.45)

Sol. Given, E_a = 21 kJ mol^–1; T₁ = 298K and T₂ = 260K

The ratio of rate constants at two different temperatures is

Substituting the values in Eq. (1), we get

15. At a given pressure, a substance is heated from 2000K to 2600K. If the entropy of the substance is 60 J K^–1 mol^–1 and is assumed to be constant over the given temperature range, then the change in the chemical potential (in kJ mol^–1) of the substance is _________.

Ans. (–36)

Sol. We know,

Chemical potential ...(1)

Substituting the values in Eq. (1), we get

= –60 (2600 – 2000)

= –60 × 600 J mol^–1

= –36000 J mol^–1

= –36.00 kJ mol^–1

Biochemistry

1. Which one of the following hormones initiates a signaling cascade by directly binding to an intra-cellular receptor?

(a) Insulin

(b) Gonadotropin

(c) Progesterone

(d) Epinephrine

Ans. (c)

Sol. Progesterone is the signaling molecule that is non-polar and has ability to cross the membrane quickly. Progesterone hormones has receptors that belong to nuclear family that enter within the cell into the nucleus to bind with DNA for transcription.

2. Which one of the following bonds is not present in ATP?

(a) Phosphoester

(b) Phosphoanhydride

(c) N-Glycosidic

(d) -Glycosidic

Ans. (d)

Sol. Adenosine triphosphate (ATP) provides energy to many processes in living cells. It has ribose sugar, adenine base and three phosphate– and phosphate. The bond between sugar and base is known as N-Glycosidic biond and the bond between -phosphate and sugart is ester bond the known as phosphodiester bond. The phosphoanhydride is energy rich bond, which is present between -phosphate and -phosphate. Hence there is no -glycosidic in the structure of ATP.

3. The reaction involved in the direct conversion of L-phenylalanine to L-tyrosine is

(a) hydroxylation

(b) decarboxylation

(c) transamination

(d) reduction

Ans. (a)

Sol. Hydroxylation reaction is involved in the direct conversion of L-phenylalanine to L-tyrosine using enzyme phenylalanine hydroxylase by the help of co-enzyme that is tetrahydrobiopterin.

4. The human major histocompability complex (MHC) is

(a) Polygenic and monomorphic

(b) Polygenic and polymorphic

(c) Monogenic and polymorphic

(d) Monogenic and monomorphic

Ans. (b)

Sol. The human major histocompatibility complex (MHC) is known as human leucocyte antigen (HLA). This divide into two groups HLA-I and HLA-II, which are further divide into HLA-1,2,3 and HLA-DP, DQ and DR respectively. Hence, it shows that MHC is polygenic. And each gene has multiple alleles in its population hence it shows polymorphism.

5. Har Gobind Khorana and Marshal Nirenberg elucidated the genetic code by using a cell-free protein synthesizing system. It was found that poly(U) and poly(C) result in the synthesis of poly(L-Phe) and poly (L-Pro), respectively, Based on these observations, which one of the following conclusions is correct?

(a) Codon GGG specifies L-Phe and codon AAA specifies L-Pro

(b) Codon CCC specifies L-Phe and codon UUU specifies L-Pro

(c) Codon AAA specifies L-Phe and codon GGG specifies L-Pro

(d) Codon UUU specifies L-Phe and codon CCC specifies L-Pro

Ans. (d)

Sol. Har Gobind Khorana and Marshall Nirenberg elucidated the genetic code by using a cell-free protein synthesizing system. This observation concludes that poly(U) that is codon UUU specifies for L-Phe that shows only U is codon for phenylalanine and poly(C) that is codon CCC specifies for L-Pro shows poly C is codon for proline.

6. Binding of an antibody to its cognate antigen does not involve

(a) Covalent bonds

(b) Electrostatic forces

(c) Van der Waals forces

(d) Hydrogen bonds

Ans. (a)

Sol. Covalent bonds do not involve in binding of an antibody to its cognate antigen, because it is very strong bond and difficult to dissociate. Hence, only electrostatic forces, Van der Waals forces and hydrogen bond are presents in such type of molecular interactions.

7. A globular protein of molecular weight 50 kDa exists as a mixture of monomers and dimers in solution. The most appropriate technique for the separation of these two forms of the protein is

(a) thin layer chromatography

(b) ion exchange chromatography

(c) gel filtration chromatography

(d) paper chromatography

Ans. (c)

Sol. To identify the monomer and dimer protein molecules and their molecular weight that is 50 kDa and 100 kDa respectively, thebest technique is gel filtratino chromatography, also known as molecular exclusion chromatography. It is the only technique that can separate molecule on the basis of molecular weight and it also separates monomers and dimers at its native state.

8. Choose the correct order of molecules according to their ability to diffuse across a lipid bilayer.

(a) CO₂ > H₂O > Glucose > RNA

(b) CO₂ > Glucose > H₂O > RNA

(c) RNA > Glucose > CO₂ > H₂O

(d) H₂O > CO₂ > RNA > Glucose

Ans. (a)

Sol. The correct order of molecules according to their ability to diffuse across a lipid bilayer is CO₂ > H₂O > Glucose > RNA. CO₂ is the non-polar gas with small particles hence, it passes first across the membrane, then H₂O that is small uncharged polar molecule, then glucose that is also uncharged polar but larger than H₂O and shows very less movement. RNA shows negligible movement.

9. When one glucose unit from glycogen gets converted to lactate in the muscle, the net number of ATP molecules produced is ................ .

Ans. (3)

Sol. In the process of glycogenesis, the net gain of ATP is 3. 1 ATP loss during the conversion of fructose-1,6-bisphosphate to fructose-1, 6-bisphosphate by the enzyme fructose-1, 6 bisphosphatase. And, 2 ATP produced ad conversion of 1,3-phosphoglycerate to 3-phosphoglycerate by the enzyme phosphoglycerate kinase. Again 2 more ATP produced at next step, during the conversion of 3-phosphoglycerate to phosphoenolpyruvate acid (PEP).

So, Net gain of ATP = Total ATP – Total loss of ATP

4 – 1 = 3 ATP

10. Considering that the three pK_a's of hisidine are pK₁ = 1.8, pk₂ = 9.2 and pKR = 6.0, its isoelectric point will be ..................... (rounded off to one decimal place).

Ans. (7.6)

Sol. Given,

pK₁ = 1.8, pK₂ = 9.2 and pK_R = 6.0

Since, histidine is a basic amino acid, to find the isoelectric point, we will have to take average of basic pK₂ value and pK_R, that is,

11. One mole of a native protein upon N-terminal analysis yielded one mole each of Asp and Val. Therefore, the protein in its native state exists as a

(a) monomer

(b) homo-dimer

(c) hetero-dimer

(d) tetramer

Ans. (c)

Sol. If one mole of a native protein upon N-terminal analysis yielded one mole each of Asp and Val hence, it shows hetero-dimer with one mole of aspartate and one mole of valine.

12. The prosthetic groups/cofactors involved in both 1e^– and 2e^– transfer in the mitochondrial electron transport chain are:

(a) NAD and NADP .

(b) NAD and FAD

(c) Heme and FMN

(d) Coenzyme Q and FMN

Ans. (d)

Sol. FMN can be converted into FMNH₂ by 2e– and by trans-ferring 1e^– becomes FMNH. Coenzyme Q beciome UQH₂ by 2e^– which is reduced form of ubiquinol and UQH by transferring 1e^– that is semi-ubiquinol.

13. Match the items in Column I with the most appropriate items in Column II and choose the correct option.

(a) P – 2, Q – 1, R – 3, S – 4

(b) P – 4, Q – 1, R – 3, S – 2

(c) P – 1, Q – 2, R – 3, S – 4

(d) P – 4, Q – 1, R – 2, S – 3

Ans. (b)

Sol. Integrin are the cell binding molecules that provide adhesion to endothelium for binding of cells.

Microbial cells are the phagocytic cell found in central nervous system which removes damaged neurons and infections.

Toll-like receptors 7 also known as TLR-7, that innate immune sensor that is responsible for recognizing single stranded RNA.

Dendritic cell (DCs) also known as accessory cells that involve in the presentation of antigens of immune system in mammals.

14. The correct combination of glycosidic linkages present in glycogen is

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Glycogen is found in animals and it is branched like amylopectin. It is formed by mostly alpha 1, 4 glycosidic linkages but branching occurs more frequently than in amylopectin as alpha 1, 6 glycosidic linkages occur about every ten units.

15. Poypeptides are either biosynthesized on the ribosomes using an mRNA template or chemically synthesized by the Merrifield's solid phase method. The correct dierctions of peptide synthesis are

(a) C N direction on the ribosomes and N C direction on the solid phase

(b) N C direction on the ribosomes and C N direction on the solid phase

(c) N C direction in both cases.

(d) C N direction in both cases.

Ans. (b)

Sol. The biosynthesis direction of polypeptides on free ribosome by using mRNA template is N C terminal. And in Merrifield's solid phase method the direction on biosynthesis of protein by chemicals is C N terminal because in this method the N-terminal is blocked and extension is done by C-terminal by chemical coupling.

16. A solution absorbs 20% of the incident light in a cuvette a path length 1.0 cm. The amount of light transmitted by the same solution in a cuvette of 3.0 cm path length is .....................% (rounded off to one decimal place).

Ans. (50)

Sol. Given,

The percentage of incident light absorbed = 20%

Therefore, the percentage of incident light transmitted = 80%

If let us assume that the incident light, I₀ to be 100.

Then, transmitted light, I will be 80.

We know, the formula for transmittance,

We know the relationship between absorbance and transmittance is

Now, we know by Beer-Lambert's law

A =

Absorbance is directly proportional to the path length (l). When we increase the path length from 1 cm to 3 cm that is 3 times, the absorbance would also increase to 3 times that is absorbance would become 0.3.

Now, new transmittance can be find out by suing the same formula

A =

0.3 =

On taking antilog on both sides,

Transmittance percentage can be calculated using,

T = 0.5 × 100 = 50%

17. The second pK_a of phosphoric acid is 6.8. The ratio of Na₂HPO₄ to NaH₂PO₄ required to obtain a buffer of pH 7.0 is .................. (rounded off to two decimal palces).

Ans. (1.58)

Sol. Given,

pK_a = 6.8

pH = 7.0

To find, Na₂HPO₄ : NaH₂PO₄

According to Henderson-Hasselbalch equation

18. A PCR ina 100 µL reaction volume, containing two primers at a concentration of 0.2 µM each, is set up to amplify a 250 base pair DNA fragment. Consider the average molecular weight of one base pair as 660 Da. If the primers are fully consumed by the end of the reaction, the amount of the final PCR product formed ................ µg (rounded off to one decimal place).

Ans. (3.3)

Sol. Given, in PCR

Reaction volume = 100 µL

Concentration of primer = 0.2 µM =

In 100 µL, concentration of primer = = 0.2 × 10^–10 mol

No. of molecules in primer = 0.2 × 10^–10 × 6.023 × 10²³ = 1.204 × 10¹³ molecules

No. of base pair in DNA fragments = 250 bp

Total amount of DNA = 1.204 × 10¹³ × 250 × 660 Dalton

= 198660 × 10¹³ × 1.67 × 10^–24 g [1 dalton = 1.67 × 10^–24 g]

= 331762.2 × 10^–11 g [ 1g = 10⁶ µg]

= 331762.2 × 10^–11 × 10⁶ µg

= 3.3 µg

19. An enzyme obeying Michaelis-Menten kinetics shows a reaction velocity (V) of 10 µmol min^–1 when the substrate concentration [S] equals its K_M. The maimal velocity V_max for this enzyme ................ µmol min^–1 (correct to integer number). (K_M is Michaeis-Menten constant)

Ans. (20)

Sol. We know

When, Michaelis constant (K_M) = Substrate concentration [S]

V_o =

where,

Initial velocity of enzyme V₀ = 10µmol min^–1

Maximum velocity of enzyme, V_max

Putting the values in the equation above, we get

20. The enzyme glucose isomerase catalyzes the inter-conversion of glucose and fructose as shown.

Glucose Fructose

The for this reaction is zero kcal mol^–1. After adding glucose isomerase to a 0.12 glucose solution and allowing the reaction to attain equilibrium, the final concentration of fructose in the reaction mixture will be ................ mM.

Ans. (60)

Sol. Given, = 0

Thus, Product = Subtrate

Hence, the concentration of glucose and fructose will be 0.6 M = 60mM.

Botany

1. Indefinite stamen is a characteristic feature of which of the following plant families?

(a) Malvaceae

(b) Apocynaceae

(c) Poaceae

(d) Brassicaceae

Ans. (a)

Sol. Malvaceae is the family of flowering plant in which there are number of stamens surrounding the pistil and fused together in a column. Examples are cotton, okra, etc.

2. In natural condition, which of the following plants does not exhibit anomalous secondary growth?

(a) Rice

(b) Aloe

(c) Yucca

(d) Dracaena

Ans. (a)

Sol. Anomalous secondary growth is the formation of phloem externally and xylem internally by a single cylindrical cambium. In natural condition, rice does not exhibit anomalous secondary growth due to lack of cambium.

3. In a typical angiosperm under natural condition, primary meristems and usually established during

(a) gametogenesis

(b) embryogenesis

(c) vegetative phase development

(d) secondary growth

Ans. (b)

Sol. Primary meristem is the type of meristem which initials are directly derived from embryonic cells and remain meristematic forever. It gives rise to ground tissues. There are three primary meristem protoderm, ground meristem and procambium.

4. 2-Methoxy-3, 6-dichlorobenzoic acid belongs to which class of plant growth regulators?

(a) Synthetic auxin

(b) Synthetic cytokinin

(c) Strigolactone

(d) Brassinosteroid

Ans. (a)

Sol. Synthetic auxins are chemical compounds that elicity auxin-like functions and have been produced synthetically. For example, naphthalene acetic acid (NAA), 2, 4-dichlorophenoxyacetic acid (2, 4-D), 2, 4, 5-trichlorophenoxyacetic acid (2, 4, 5-T) and 2-methoxy-3, 6-dichlorobenzoic.

5. In a typical green plant, the first stable product of Calvin cycle is

(a) oxaloacetic acid

(b) succinic acid

(c) maleic acid

(d) 3-phosphoglyceric acid

Ans. (d)

Sol. Calvin cycle is light-independent reaction of photosynthesis that converts carbon dioxide and water to organic compounds. Carboxylation is the first step of Calvin cycle, in this step, carbon dioxide is fixed to form a stable organic compound by carboxylation of RuBP takes that is converted to two molecules of three carbon compound called 3-phosphoglyceric acid.

6. Among the following which best describes an organism that lives at the expense of other organism, harmful but usually not killing?

(a) Predator

(b) Symbiotic

(c) Prey

(d) Parasite

Ans. (d)

Sol. Parasite are the organisms that feed on other organsims and may harm them but do not kill the host. They are usually smaller than their host. They may get transferred to the host by contamination of the food or water, casual transfer, contagious transfer (by physical contact), inoculative transfer (by the bite of blood sucking organisms) or vertebrate animals (Zoonotic diseases).

7. The oleo-gum resin asafoetida (hing) is obtained from the cut surface of

(a) stem

(b) root

(c) leaf

(d) fruit

Ans. (a)

Sol. Morphine is the alkaloid of opium that is obtained from latex of poppy plant (Papaver somniferum). They are analgesic and act on CNS to relieve pain. They are called pain killers also.

Quinine extracted from the bark of the Cinchona officinalis tree has been used for a long time to treat malaria.

Atropien is anti-cholinergic or anti-parasympathetic agent that used to cure many diseases such as slow heart rate, nerve agents, etc. it is derived from Hyoscyamus niger.

Vinblastine is a chemotherapy drug that stop the further growth of cancer cells, it is extracted from Catharanthus roseus.

8. 'Bakanae' disease or foolish seedling disease is caused by

(a) Fungus

(b) Bacterium

(c) Virus

(d) Mycoplasma

Ans. (a)

Sol. Bakanae disease or foolish seedling disease is caused by fungus, Gibberella fujikuroi. It mostly infects rice plants.

9. Which of the following chemicals is used for doubling of chromosome numbers during production of doubled haploids in crop plants?

(a) Hygromycin

(b) Kanamycin

(c) Colchicines

(d) Glufosinate

Ans. (c)

Sol. Colchicine is the chemical used for doubling of chromosome numbers during production of 'doubled haploids' in crop plants. This is the formation of embryos from cultured anthers. This turned out to be haploid in nature as they had originated from microspores or pollen grains. If the haploid plants so produced could be diplodized (by using colchicine), then diploid homozygous plants could be produced which would be the same as obtaining pure lines by self-breeding for several generations.

10. An mRNA of a nuclear encoded plant gene, DSH20 has an ORF of 1353 nucleotides. Provided that average molecular weight of amino acid is 110 Dalton (Da), calculate molecular weight of DSH20 protein in kDa (round off to 1 decimal place) is ............... .

Ans. (49.5)

Sol. Given, open reading frame has 1353 nucleotides

We know,

3 nucleotides = 1 amino acid

1353 nucleotides = = 451 amino acids

Last 3 nucleotides code for STOP codon, thus total number of amino acids = 450.

Given, average molecular weight of amino acid = 110 Da.

Molecular weight of 450 amino acids, thus DSH20 protein = 110 × 450 = 49500 Da = 49.5 kDa.

11. Group-I, Group-II and Group-III represent enzyme, product of the enzymatic reaction and metabolic process, respectively.

(a) P-ii-1, Q-iv-3, R-v-2, S-iii-4

(b) P-ii-1, Q-i-3, R-v-4, S-iii-2

(c) P-ii-2, Q-v-3, R-i-4, S-iii-1

(d) P-iii-1, Q-i-3, R-iv-4, S-ii-2

Ans. (b)

Sol. In the first step of glycolysis, phosphoryl group of ATP is transferred to glucose to form glucose-6-phosphate (G6P) in the presence of glucokinase or hexokinase enzyme. This step uses one ATP. This reaction is catalyzed by the enzyme hexokinase (HK) in the presence of magnesium ions.

During TCA cycle, hydration of fumarate to malate is catalyzed by the enzyme fumarase or fumarate hydratase. In this reaction, one molecule of H₂O is required.

During photosynthesis in C₄ plant, the carbon dioxide that enters the mesophyll cells is accepted by a 3-carbon molecule called phosphoenol pyruvate (PEP). The enzyme phosphoenol pyruvate carboxylase (PEPcase) catalyzes the carboxylation of PEP and form an unstable compound oxaloacetic acid (OAA).

Glycolate oxidase is a key enzyme for photorespiration in plants. It catalyzes the conversion of glycolate to glyoxylate to produce hydrogen peroxide in the peroxisome. Hydrogen peroxide is a dangerously strong oxidant which must be immediately split into water and oxygen by the enzyme catalase.

12. Match the following in correct combination between Group I and Group II with reference to the agents that interfere with oxidative phosphorylation.

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

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

(c) P-4, Q-3, R-2, S-5

(d) P-5, Q-2, R-3, S-4

Ans. (a)

Sol. Chemicals such as cyanide, carbon monoxide, malonate, etc., inhibit respiration by binding to metal containing enzymes such as cytochrome oxidase.

Antimycin A, the most potent inhibitors of the mitochondrial respiratory chain, binds to the quinone reduction site of the cytochrome bc₁ complex. It blocks electron transfer complex III from cyt b to cyt c₁.

Aurovertin is an antibiotic that binds to the beta subunits in the F₁ domain and inhibits F₁F₀ ATPase-catalyzed ATP synthesis in preference to ATP hydrolysis.

2, 4-Dinitrophenol is known as the uncoupler of phosphorylation in electron transfer that disppears the proton gradients across bioenergetic membranes like mitochondria.

13. In relation to Agrobacterium mediated genetic engineering in plants, match the following in correct combination.

(a) P-4, Q-3, R-2, S-5

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

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

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

Ans. (c)

Sol. virA is the virulence region in Ti plasmid of Agrobacterium tumefaciens that codes for the receptor of phenolic compounds such as acetosyringone during transfer of T-DNA to plant cells.

virB is the virulence region in Ti plasmid of Agrobacterium tumefaciens that help in conjugation or tube formation.

virD1 possesses DNA topoisomerase that is regulated by the vir A/G.

virG helps in activation of all vir operons on phosphorylation by virA.

14. Match the plant part (Group I) with the product obtained (Group II) and the representative plant species (Group III) in correct combination.

(a) P-2-i, Q-1-iii, R-4-ii, S-3-iv

(b) P-2-i, Q-1-ii, R-4-iv, S-3-iii

(c) P-2-ii, Q-4-iii, R-1-iv, S-3-i

(d) P-4-iii, Q-1-ii, R-2-iv, S-3-i

Ans. (d)

Sol. Cinnamomum zeylanicum is also known as true cinnamon, it is obtained from inner bark of tree cinnamomum. The aroma and flavor of cinnamon is used to make essentail oils and other constituents.

The leaf and leaf buds of Camellia sinensis tree are used to produce tea, it is commonly known as te plant or tea shrub. This plant is known to have high quantity of natural tannins. Tannins are the organic compound that is responsible for the antioxidant activities of black and dark teas.

Crocus sativus is a flowering plant, whose filaments that used to produce saffron. It is also known as saffron crocus.

The mature fruits and seeds of Papaver somniferum is used to obtain many pharmaceutical alkaloids such as codeine, morphine, etc.

15. Select the correct combination by matching the disease, causal organism and the affected plant.

(a) P–iv–1; Q–iii–3; R–i–2; S–ii–4

(b) P–ii–1; Q–i–6; R–iv–4; S–iii–2

(c) P–iii–1; Q–iv–3; R–i–5; S–ii–4

(d) P–iv–1; Q–ii–3; R–iii–2; S–i–5

Ans. (a)

Sol. The stem rust is caused by the fungas Puccinia gramims. It can affect bread wheat, barley, cereal crops, etc.

Wart disease or black scab is a type of disease that is found in potato. It is caused by the chytrid fungus Synchytrium endobioticum.

Cercospora personata is a ascomycetes fungus, that cause plant disease and form leaf spot in many plants such as peanut, ground nut, etc.

Downy mildew is a disease commonly found in grapes. This infects the whole green part of grapes. The main agent of this disease is Plasmopara viticola.

16. Match the following alkaloids with their uses and source plants in correct combination.

(a) P-ii-3, Q-iv-1, R-iii-4, S-i-2

(b) P-ii-1, Q-i-3, R-iv-4, S-iii-2

(c) P-ii-2, Q-iv-1, R-i-4, S-iii-3

(d) P-iii-4, Q-ii-1, R-iv-3, S-i-2

Ans. (b)

Sol. The oleo-gum resin asafoetida (hing) is obtained from the extraction of roots and rhizomes of the plant. It is a medically important plant in India.

17. Match the following ecological terms with their appropriate definitions.

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

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

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

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

Ans. (a)

Sol. Niche is the place of a living organism in the biotic environment and its relations to food and enemies.

Biotas is the living organism (flora and fauna) of a given place ore region.

Trophic level consists of all the organisms in a food web that are the same number of feeding levels away from the original source of energy.

Habitat is the natural environment of an organisms.

18. Arrange the following water reservoirs of earth in decreasing order of the water volume.

P. Streams

Q. Groundwater

R. Glaciers

S. Lakes and island seas

(a) R-Q-S-P

(b) P-Q-R-S

(c) S-P-R-Q

(d) R-P-Q-S

Ans. (a)

Sol. The distribution of water on the Earth's surface is extremely uneven. Only 3% of water on the surface is fresh; the remaining 97% resides in the ocean. Of fresheater, 69% resides in glaciers, 30% underground and less than 1% is located in lakes, rivers and swamps.

19. Selection markers and the corresponding genes used in plant genetic engineering are given below :

Choose the correct combination.

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

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

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

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

Ans. (c)

Sol. Neomycin phosphotransferase II (nptII) is an antibiotic selectable marker, obtained from E. coli. It is resistant to kanamycin.

The hph gene encoding a hygromycin B phosphotransferase has been isolated from an E. coli strain. Its expression confers hygromycin resistance to hph transfected cells.

Plasmid encoded hygromycin B resistance : the sequnce of hygromycin B phosphotransferase gene and its expression in E. coli and Saccharomyces cerevisiae.

Bar gene is resistant to herbicide bialaphos. This gene is cloned from Streptomyces hygroscopicus.

Phosphomannose isomerase (pmi) gene used to catalyze the reversible reaction of interconversion of mannose-6-phosphate and fructose-6-phosphate; so that only cells transferred from pmi can survive.

20. A double homozygous mutant develops green and wrinkled seeds. When it was crossed with a true-breeding plant having yellow and round seeds, all the F1 plants developed yellow and round seeds. After self-fertilization of F₁, the calculated percentage probability of plants with green and wrinkled seeds in the F₂ population (round off to 2 decimal places) is ............... .

Ans. (6.25)

Sol.

Probability of green wrinkled are 1/16.

That is, their percentage probability will be × 100 = 6.25%.

Microbiology

1. The technique of microbial "pure culture" was pioneered by

(a) Edward Jenner

(b) Louis Pasteur

(c) Robert Hooke

(d) Robert Koch

Ans. (d)

Sol. Robert Koch, invented method to purify the bacillus from blood samples and grow pure cultures in 1876, pure cultures allow the pure isolation of a microbe, which is vital in understanding how an individual microbe may contribute to a disease.

Edward Jenner was an English Physician who introduced the vaccine for smallpox.

Louis Pasteur discovered the principles of vaccination, microbial fermentation and pasteurization.

The cell was first discovered by Robert Hooke in 1665 using a microscope.

2. The antibacterial trimethoprim is an inhibitor of

(a) dihydrofolate reductase

(b) dihydropteroate synthetase

(c) N⁵, N¹⁰-methenyl tetrahydrofolate synthetase

(d) serine hydroxymethyl transferase

Ans. (a)

Sol. Folic acid biosynthesis by bacteria is necessary for formation of protein, the final step of this biosynthesis is catalyzed by the dihydrofolate reductase enzyme. There are some drugs that interfere in folic acid synthesis, such as trimethoprim. Trimethoprim is a diaminopyrimidine, which combine with the sulfamethoxazole and forms co-trimoxazole. This co-trimoxazole work as antimicrobial agent that inhibits the dihydrofolate reductase.

3. Choose the correct taxonomical hierarchy among the following.

(a) Species, Genus, Family, Order, Class, Phylum, Domain

(b) Species, Genus, Order, Class, Family, Phylum, Domain

(c) Species, Genus, Order, Family, Class, Phylum, Domain

(d) Species, Genus, Family, Class, Order, Phylum, Domain

Ans. (a)

Sol. Taxonomic hierarchy is the arrangement of taxonimic categories in hierarchical order. The taxonomic categories are arranged in hierarchical in its increasing to decreasing order from domain (kingdom) to species and vice versa as :

Species Genus Family Order Class Phylum Domain

Examples of taxonomical hierarchy

4. Shifting a Saccharomyces cerevisiae culture from fermentative to aerobic respiratory mode will

(a) decrease carbon dioxide production

(b) increase alcohol production

(c) increase glucose consumption

(d) decrease ATP generation per mole of glucose

Ans. (d)

Sol. Fermentation takes place in absence of oxygen and in higher amount of fermentable sugars by yeast and bacteria in theiry cytoplasm. These sugars are converted into ethanol and CO₂ to obtain energy.

Fermentation is the major pathway of energy production in the yeast Saccharomyces cerevisiae, also in aerobic respiration. This process of shifting a Saccharomyces cerevisiae culture from fermentative to aerobic respiratory mode increases glucose consumption for more CO₂ and when glucose decreases, ethanol produced during fermentation is used as a carbon source.

5. Which one of the following diseases is treated by a neuraminidase inhibitor?

(a) Chickenpox

(b) Polio

(c) Influenza

(d) Japanese encephalitis

Ans. (c)

Sol. Neuraminidase inhibitors are drugs that block the function of the viral neuraminidase protein. By blocking this protein enzyme, it stops the release of viruses from the infected host cell and prevents new host cells from being infected. They inhibit the function of neuraminidase of both influenza A and influenza B virus by stopping their reproductive through budding from the host cells.

6. Which one of the following does not provide three-dimensional images?

(a) Atomic force microscopy

(b) Confocal scanning laser microscopy

(c) Differential interference contrast microscopy

(d) Phase-contrast microscopy

Ans. (d)

Sol. Phase-contrast microscopy is a contrast-enhancing optical technique that is often used to produce high-contrast images of transparent specimens, such as live cells in culture.

In rest of all the microscopies mentioned, atomic force microscopy, confocal scanning laser microscopy and differential interference contrast microscopy, three-dimensional image is formed.

7. Which one of the following will increase the resolution of a light microscope?

(a) Decreasing the numerical aperture of the objective lens

(b) Using an objective lens with a longer working distance

(c) Using a medium of higher refractive index

(d) Increasing the wavelength of light

Ans. (c)

Sol. In a light microscope, we can increase the resolution by increasing the refractive index because it slows down the speed of the light that passes through this medium and get the wavelength shorter to give better resolution.

8. Which one of the following conditions favors maximum expression of lac operon genes in E. coli?

(a) Glucose-low, lactose-low, cAMP-high

(b) Glucose-high, lactose-low, cAMP-high

(c) Glucose-low, lactose-high, cAMP-high

(d) Glucose-high, lactose-high, cAMP-low

Ans. (c)

Sol. The lac operon is an inducible operon which induces transcription of structural genes in the presence of lactose or allolactose and hence it is called an inducer. Glucose and galactose cannot induce the lac operon, hence lac operon will be expressed as long as lactose is present. Oxidation of lactose, a disaccharide composed of glucose and galactose, cna provide the cell with metabolic intermediates and energy. There are two proteins lac repressor and catabolite activator protein (CAP) that act as lactose sensor and glucose sensor respectively. The cAMP attaches to CAP, allowing it to bind DNA and help RNA polymerase bind to the promoter, resulting in high levels of transcription.

9. Match the cellular organelle in Group I with its function in Group II.

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

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

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

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

Ans. (b)

Sol. The main function of Golgi apparatus is to carry out the processing of proteins generated in endoplasmic reticulum. It also transports protein to the different parts of cell. It is involved in the protein sorting or protein targeting.

In the cell, the nucleolus, a specialized region in the nucleus is the site of ribosome biogenesis. The nucleolus makes ribosomal subunits from proteins and ribosomal RNA, also known as rRNA. It then sends the subunits out to the rest of the cell where they combine into complete ribosomes.

A major function of the peroxisome is the breakdown of very long chain fatty acids through beta oxidation.

Proteasomes are protein complexes which degrade unneeded or damaged proteins by proteolysis, a chemical reaction that breaks peptide bonds. Enzymes that help such reactions are called proteases.

10. A 250 µL of bacteriophage stock containing 8 × 10⁸ phages mL^–1 is added to 500 µL of E. coli culture containing 4 × 10⁸ cells mL^–1. The multiplicity of infection is ................... .

Ans. (1)

Sol. Given,

Number of bacteriophages = 8 × 10⁸ phages mL^–1

Number of cells = 4 × 10⁸ cells mL^–1

Stock is of 250 µL = 0.25 mL

Now, if 1 mL stock has 8 × 10⁸ phages

0.25 mL stoch has 8 × 10⁸ × 0.25 = 2 × 10⁸ phages per 0.25 mL

Using the formula for multiplication of infection (MOI),

Given,

In 1 mL, number of cells = 4 × 10⁸ cells

In 500 µL or 0.5 mL, number of cells = 4 × 10⁸ × 0.5 = 2 × 10⁸ cells

Now, putting this value in the formula above of MOI, we get,

11. Digestion of an immunoglobulin G (IgG) molecule iwth pepsin will NOT

(a) generate a bivalent antigen binding fragment

(b) generate moovalent antigen binnding fragent

(c) destory the complement binding site

(d) clave the havy chain of IgG molecule

Ans. (d)

Sol. Under anaerobic fermentative growth conditions, one mole of glucose yields 22 g of Streptococcus faecalis or 8.6 g of Zymomonas mobilis. The molar growth yield (YATP) can be calculated as follows:

For S. faecalis:

– Molecular weight of glucose = 180 g/mol

– One mole of glucose yields 22 g of biomass

– Therefore, the molar growth yield (YATP) is 22/180 = 0.122 moles of biomass per mole of glucose.

For Z. mobilis:

– Molecular weight of glucose = 180 g/mol

– One mole of glucose yields 8.6 g of biomass

– Therefore, the molar growth yield (YATP) is 8.6/180 = 0.048 moles of biomass per mole of glucose.

12. Match the process involved in nitrogen or sulfur cycle in Column I with the corresponding microbe in Column II.

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

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

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

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

Ans. (d)

Sol. Most plants absorb nitrogen from the soil in the form of nitrates. For the uptake of nitrates, they must compete with anaerobic bacteria of Pseudomonas species known as denitrifiers. These denitrifiers reduce nitrate to nitrogen, which escape into the atmosphere, through the process called denitrification. Through this process, nitrates are reduced to gaseous nitrogen dioxide, nitrogen oxide, nitrogen monoxide or elemental nitrogen.

NO₃ NO₂ N₂O N₂

The free living nitrogen fixing bacteria are widespread; however, they grow slowly as lare amount of their respiratory energy is required to fix dinitrogen. Some species such as Azobacter, most fix nitrogen under anaerboci conditions or in the presence of low oxygen.

Beggiatoa species accumulate high concentrations of nitrate in their vacuoles and can use the nitrate as an electron acceptor for oxidation of reduced sulfur compounds under anaerobic conditions. These organisms play important roles in the cycling of carbon, sulfur and nitrogen in the marine environments.

2H₂S O₂ 2S + 2H₂O

Rhizobia are diazotrophic bacteria that fix after becoming established inside the root nodules of legumes (Fabaceae). To express genes for nitrogen fixation, rhizobia require a plant host; they cannot independently fix nitrogen.

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

Assertion (A) : Diphtheria exotoxin is an example of A-B type toxin.

Reason (R) : The A component of the toxin is released from the host cell, while the B component inhibits synthesis and kills the host cell.

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

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

(c) Both (A) and (R) are false.

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

Ans. (d)

Sol. The AB toxins are two-component protein complexes secreted by pathogenic bacteria.

"A" component is usually the "active" portion and the "B" component is usually the "binding" portion. B component of exotoxin attaches to host cell receptor.

A-B exotoxin enters the host cell by endocytosis and then A-B exotoxin enclosed in pinched off portion of plasma membrane during pinocytosis. After that, A-B components separate to each other and the component A alters cell function by inhibiting protein synthesis, while the component B is released from the host cell.

14. Which one of the following statements about control of microbial growth is not correct?

(a) Non-ionizing radiation leads to thymine dimers formation in DNA.

(b) Spirochetes and mycoplasma can pass through membrane filters (0.22-0.45 µm).

(c) Use of high concentration of salts and sugars to preserve food is a chemical method of microbial control.

(d) Thermoduric bacteria can survive pasteurization.

Ans. (b)

Sol. (a) Correct : Pyrimidine dimers are molecular lesions formed from thymine or cytosine bases in DNA via photochemical reactions. Ultraviolet light (UV) inducse the formation of covalent linkages between consecutive bases along the nucleotide chain in the vicinity of their carbon-carbon double bonds. Ionizing radiation are clastogens, as they can break the DNA.

(b) Correct : Mycoplasma is very small type of bacterium characterized by its lack of a rigid cell wall. Their diameter is 0.2-0.4 µm. Spirochetes are long slender shape with a diameter of 0.1-0.4 µm. This elasticity allows them to pass through 0.2 µm (bacterial retentive) filters.

(c) Incorrect : Salt or sugar withdraw the available water from within the food to the outside and inserting salt or sugar molecules into the food interior. The result is a reduction of the so called product water activity (a_w), that is necessary for microbial survival and growth. Thus, it is a physical method.

(d) Correct : Thermoduric are bacteria that can tolerate heat. They survive pasteurization and can cause spoilage in dairy products

15. An example of a differential and selective medium in which colonies of Gram-negative bacteria produce large amounts of acidic products and appear green with a metallic sheen is

(a) Blood agar

(b) EMB agar

(c) MacConkey agar

(d) Mannitol salt agar

Ans. (*)

Sol. Culture medium is defined as a medium of nutrients that supports microbial growth. Eosin-methylene blue (EMB) agar is used for both selective and differeential medium. EMB contains the dyes, eosin and methylene blue that inhibit the growth of Gram-positive bacteria. Therefore, EMB is selective for Gram-negative bacteria and provides a color indicator distinguishing between organisms that ferment lactose. When bacterial cells ferment lactose, acid is produced that precipitates the dyes in the medium and the colonies develop a green metallic sheen.

16. Which one of the following is an example of substrate level phosphorylation?

(a) Glucose to Glucose 6-phosphate

(b) Fructose 6-phosphate to Fructose 1, 6-bisphosphate

(c) 1, 3-Bisphosphoglycerate to 3-phosphoglycerte

(d) 2-phosphoglycerate to Phosphoenolypyruvate

Ans. (c)

Sol. Phosphorylation is a process by which ATP is synthesized in chloroplasts and mitochondria. Substrate level phosphorylation is removal of inorganic phosphate. First substrate level of phosphorylation produce 3-phosphoglycerate and ATP by dephosphorylating 1, 3-bisphosphoglycerate via phosphoglycerate kinase producing 3-phosphoglycerate and ATP through first substrate level phosphoryaltion.

17. A bacterial culture containing 3 × 10⁵ live cells were exposed to a newly developed sterilizing agent. After 30 minutes of exposure, 3 live cells remained in culture. The decimal reduction time (in minutes) for the new agent is .............. .

Ans. (6)

Sol. We know the formula for Decimal Reduction Time,

Decimal reduction time =

where,

Final time, t₂ = 30 min

Initial time, t₁ = 0

Number of live cells present initially, log N₀ = log 3 × 10⁵

Number of cells present after exposure, log N₁ = log 3

Putting the values in the formula above, we get,

Decimal reduction time =

18. A bacterial culture has a generation time of 34 minutes. The time taken (in minutes, rounded off to two decimal plates) for the OD₅₅₀ of this exponentially growing culture to increase from 0.25 to 0.85 is ................... .

Assume that OD₅₅₀ has a linear relationship with the cell density.

Ans. (60.18)

Sol. Given,

Generation time, G = 34 min

For OD = 0.25, in 34 minutes it will increase to 0.5.

Thus, the relation set is

Similarly, for OD to reach to 0.85, it there be n number of generations.

2ⁿ =

Taking log on both sides, we get

log 2ⁿ =

For n = 1, time taken = 34 min

For n = 1.77, time taken = 34 × 1.77 = 60.18 min

19. A 100 µL aliquot (10^–4 dilution) of the bacterial culture plated on the nutrient agar gave 4 colonies. The bacterial stock concentration (in million cells/mL, rounded off to one decimal place) is .............. .

Ans. (0.4)

Sol. Aliquot with 10^–4 dilution = 100 µL = 0.1 mL

We know the formula for colony forming unit (CFU) is

Given, No. of cells = 4

Dilution factor = = 10⁴

Volume = 0.1 mL

Thus, in 0.1 mL, number of cells = 4 × 10⁴

In stock solution of 1 L or 1000 mL, number of cells = = 100 million cells

In stock solution of 1 mL, number of cells = = 0.4 million cells.

20. A continuous bacterial culture carried out in a chemostat is set to a flow rate of 40 mL h^–1. The culture volume is equivalent to that of a cubical container having 10 cm sides. The dilution rate (in h^–1, rounded off to two decimal places) of this system is ................... .

Ans. (0.04)

Sol. Flow rate = 40 mL h^–1

Culture volume is equivalent to be cubical container which have side length, a = 10 cm

Volume of cubical container = a³ = 10³ = 1000 cm³ = 1000 mL

Dilution rate, D =

where, Flor rate of dilution, F = 40 mL h^–1

Culture volume, V = 1000 mL

Putting the values in the formula above, we get

Zoology

1. Which one of the following leucocytes is phagocytic and has clear cytoplasm?

(a) Eosinophil

(b) Monocyte

(c) T_H-lymphocyte

(d) Basophil

Ans. (b)

Sol. Monocytes are the type of white blood cells that has phagocytic properties to engulf foreign particles or cell debris. The nucleus of a monocyte is usually kidney-shaped or horseshoe-shaped and the cytoplasm is blue-gray and has a foamy appearance. Cytoplasm of esoniphils, basophils and lymphocytes are filled with granules.

2. Which one ofthe following techniques can be ued for detecting the subcelllular localization of serotonin receptor in intact cells?

(a) Immunoelectron microsopy

(b) SDS-PAGE

(c) Fluorescence in-situ hybridization

(d) Differential centrifugation

Ans. (b)

Sol. Orthologous genes are genes that have the same function in different species and are related by descent from a common ancestor. They can be used to study evolutionary relationships and are identified by comparing the genomes of different species for genes with similar sequences and functions.

3. Which one of the following is not a site for in situ conservation?

(a) Biosphere reserve

(b) Wildlife sanctuary

(c) Zoological garden

(d) Biodiversity hotspot

Ans. (a)

Sol. In situ conservation refers to protecting an endangered species in its natural habitat. In situ approach includes protection of a group of ecosystems through a network of protected areas, such as biodiversity hotspots, national parks, sanctuaries, biosphere reserves and sacred grooves and lakes, it is called on-site conservation. Zoological garden in an example of ex situ conservation that is protecting an endangered species, variety of breed of plant or animal outside its natural habitat.

4. Which one of the following is the precursor molecule for corticosteroids?

(a) Androgen

(b) Estrogen

(c) Pregnenolone

(d) Mineralocorticoids

Ans. (c)

Sol. Pregnenolone or pregn-5-eb-3β-ol-20-one, is an endogenous steroid and precursor/metabolic intermediate in the biosynthesis of most of the steroid hormones, including the progestogens, androgens, estrogens, glucorticoids and mineralocorticoids.

5. Transitional epithelia is found in which one of the following organs?

(a) Liver

(b) Lung

(c) Brain

(d) Urinary bladder

Ans. (d)

Sol. Transitional epithelium is found in organ of urinary system known as urinary bladder. It is a type of compound epithelium having four to six layers. This type of cell can change their shape and structure. It is non-ciliated. It is involved in stretching and relaxation.

6. Visual signal transduction cascade is activated by rhodopsin and involves degradation rather than synthesis of which one of the following second messenger molecules?

(a) cAMP

(b) IP₃

(c) cGMP

(d) DAG

Ans. (c)

Sol. The transcriptin of the light signal in the retinal rod photoreceptor involves the sequential activation of rhodopsin, transducing (G₁) and cGMP phosphodiesterase on the cytoplasmic surface of the disk saccules that fill the rod outer segment. Absorption of light transforms rhodopsin into the activated metarhodopsin II state, which then interacts with G₁ to rapidly catalyze the exchange of GDP for GTP in the nucleotide-binding site of the -subunit. -GTP dissociates and couples to the effector PDE. Activation of PDE results in a decrease in the cytosolic cGMP concentation which in turn leads to closure of cGMP-regualted channels in the plasma membrane, hyperpolarization and neuronal signaling.

7. The genomes of both human and Drosophila code for an amylase that acts on the same substrate. However, the sequence of nucleotides in the genes encoding the two is dissimilar. This is an example of which one of the following types of evolution?

(a) Neutral

(b) Directional

(c) Covergent

(d) Divergent

Ans. (c)

Sol. Since the nucleotide sequence of amylase in human and Drosophila is not similar, but they act on the same substrate, thus it must be an example of convergent evolution. It is defined as the development of similar structures having similar functions in different species as the result of the same kinds of selection pressures.

8. "Round dance" is performed by forager bees to indicate the distance between a food source and their colony. Which one of the following best represents this distance?

(a) 45 meters

(b) 450 meters

(c) 1000 meters

(d) More than 2000 meters

Ans. (a)

Sol. When a food source is very close to the hive (i.e., less than 50 meters away), a forager performs a round dance. She does so by running around in narrow circles, suddenly reversing direction to ther original course. She may repeat the dance several times at the same location or move to another location to repeat the dance. After the round dance has ended, she often distributes food to the bees following her. A round dance, therefore, communicates distance (close to the hive) but no direction.

9. Which one of the following phyla have choanocytes?

(a) Ctenophora

(b) Nematoda

(c) Cnidaria

(d) Porifera

Ans. (d)

Sol. The inner cavity of the Porifera is lined by specialized, flagellated cells called collar cells or choanocytes. It helps phagocytes to engulf food. the species of phylum porifera commonly known as sponges are sessile aquatic organisms with most of them living in marine water than in fresh water.

10. Which one of the following glial cells is not derived from the ectoderm?

(a) Astrocytes

(b) Microglial cells

(c) Oligodendrocytes

(d) Ependyma

Ans. (b)

Sol. Glial cells are the cells of the neural system that perform various supportive functions in neuron. All the glial are derived from ectoderm, except microglial it derived from hemopoietic stem cells. The microglial cells are special tye of macrophages cells that help in maintain the health of central nervous system by remove infection and injured neurons.

11. Tarantulas and mosquitoes both belong to the phylum Artropoda. Which one of the following represents the correct number of legs in them respectively?

(a) 6 and 6

(b) 6 and 8

(c) 8 and 8

(d) 8 and 6

Ans. (*)

Sol.

12. Match the following subcellular organelles in Column I with associated functions in Column II.

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

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

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

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

Ans. (c)

Sol. In th cell, the nucleoli, a specialized region in the nucleus is the site of ribosome biogenesis. Ribosomes are membrane loss organelles present in the cytoplasm that contain ribosomal RNA and protein.

Peroxisomes are membrane bound vesicles that contain a core of oxidative enzymes for peroxide synthesis, such as urate oxidase, D-amino acid oxidase, and -hydroxy acid oxidases. They were named peroxisomes as they act as sites of synthesis and degradation of hydrogen peroxide (H₂O₂).

Glycoprotein biosynthesis occurs in both endoplasmic reticulum and Golgi apparatus.

The Golgi complex can be functionally classified into three interconnected parts : cis, medial and trans Golgi network that functions in identifying and sorting the protein traffic to be sent back to the RER or to proceed ahead of it, by sequentially modifying the protein traffic, mainly by adding carbohydrate moieties enzymatically and to sort the protein traffic to be sent to its correct destination respectively

13. Match the following genetic disorders in Column I with associated typical chromosomal changes mentioned in Column II.

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

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

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

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

Ans. (d)

Sol. The individual which has 47 chromosomes (2n – 1). This condition is called Klinefelter syndrome, which is characterized by mental retardation, underdevelopment of genitalia (leading to sterility) and the presence of feminine physical characteristics (such as breast enlargement). For example : XXX female and XYX male.

The presence of an extra copy of chromosome 21 leads to Down syndrome. There are three copies of a particular chromosome instead of the unusual two copies that is the trisomy of 21^st chromosme. Therefore, the individual would have 47 instead of 46 chromosomes.

Turner syndrome in which 45, X individuals originate from eggs or sperm that lack a sex chromosome or from loss of a sex chromosome in mitosis sometime after fertilization. It is known as the monosomy, in which a single homologous chromosme is lost. The monosomy has 2n – 1 number of chromosomes.

Cri du chat syndrome is genetic disorder which is caused by deltion of genetic material on p arm of chromosome no. 5. Hence, it is also called 5p minus. The infected individual has very high-pitched cry.

14. Match the following components listed in Column I with their respective organs in Column II.

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

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

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

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

Ans. (a)

Sol. In ear, the epithelial membranous laybrinth contains fluid endolymph with high level of potassium ions (K⁺), which play a role in the generation of auditory signals. The membranous labyrinth houses the receptors for hearing and equilibrium.

In the eye, vitreous chamber lies between the lens and the retina. Within the vitreous chamber is the vitreous humor, a transparent jelly-like substance that holds the retina flush against the choroid, giving the retina an even surface for the reception of clear images.

In male sex organ, vas deferens ascends along the posterior border of the caudal epididymis through the spermatic cord and then enters the pelvic cavity. It conveys sperm during sexual arousal from the epididymis toward the urethra by peristaltic contractions of its muscular coat.

In female reproductive cycle, corpus luteum contains the remnants of a mature follicle after ovulation. The corpus luteum produces progesterone, estrogens, relaxin and inhibin until it degenerates into fibrous scar tissue called the corpus albicans.

15. Match the following digestive enzymes in Column I with their respective functions in Column II.

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

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

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

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

Ans. (b)

Sol. Erepin is secereted from the epithelial cells of small intestine and help in the conversion of polypeptides to amino acids.

Steapsin is the lipase enzyme released in pancreatic juice to convert fat into fatty acid and glycerol.

Pepsin is produced in the stomach in inactive form that activated from pepsinogen by pepsin and hydrochloric acid. It help to converts proteins to peptides.

Protein digestion is completed by peptidases in the brush borders. The brush border epithelium also contains enterokinae, which activates trypsinogen of pancreatic juice into trypsin.

16. Which one of the following graphs represents the relationship between ventricular end-diastolic volume and cardiac output in a healthy adult individual rest (solid line) and upon exercise (dotted line)?

(a)

(b)

(c)

(d)

Ans. (a)

Sol. End-diastolic volume (EDV) is the volume of blood in the right and/or left ventricle at end load or filling in (diastole) or the amount of blood in the ventricles just before systole. Cardiac output is the stroke volume times the heart rate and stroke volume is defined as end-diastolic volume minus end-systolic volume.

Frank-starling curve showing the relationship between stroke volume and end diastolic volume. An increase in end diastolic volume causes an increase in stroke volume, thus cardiac output. The curve shifts up in situations such as exercise and down in situations such as heart failure.

17. Match the household insect vectors in Column I with their associated diseases in Column II.

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

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

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

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

Ans. (d)

Sol. Chagas disease is a parasitic disease caused by Triatominae, commonly known as kissing bug (Hemiptera).

Sand fly (Diptera) causes kala azar that carries protozoan parasite called Leishmania. Symptoms can be fever, weakness, enalarged lever, nose bleeding, etc.

Deer fly (Diptera) causes bacterial disease tularemia, commonly konwn as rabbit fever. Symptoms are skin ulcers, fever, etc.

Oriental rat flea (Siphoneptera) also known as rat fly that feed on rodents and carries bubonic plaue. It is transmitted from infected rodent to human through this fly.

18. Match the proteins in Column I with the organs in which they are maximally expressed in Column II.

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

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

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

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

Ans. (c)

Sol. Keratin is fibrous present on the outer layer of skin, it protects the skin from damage. It is also present in hairs, claws, nails, feathers, etc.

Pulmonary surfactant is a mixture of lipids and proteins which is secreted into the alveolar space by epithelial type II cells. The main function of surfactant is to lower the surface tension at the air/liquid interface within the alveoli of the lung.

The pancreas synthesizes three protease enzymes in inactive precursor form. These are trypsinogen, procarboxypeptidase and chymotrypsinogen.

Albumin is a protein made by liver. It helps keep fluid in bloodstream, so it doesnt leak into other tissues. It also carries various substances througout the body, including hormones, vitamins and enzymes.

19. The graph below shows the activity of enzyme pepsin in the presence ofinhibitors aliphatic alcohols (P) or N-acelyl-l-phenylalanine (Q). Which one of the following represents the nature of inhibition by P and Q respectively?

(a) Non-competitive and competitive (b) Competitive and competitive

(c) Non-competitive and uncompetitive (d) Competitive and uncompetitive

Ans. (b)

Sol. P represents competitive inhibitor. For competitve inhibitor, V_max remains unchanged, but K_M increases.

Q represents non-competitve inhibitor. For non-competitive inhibition, V_max is decreased, but K_M remains unchanged in the presence of inhibitor.

20. In Drosophila, the red eye phenotype (W) is dominant over the recessive white eye mutant (w). In a mixed population of red and white eye flies of 10,000 individuals. 3,600 flies were white eyed. The percentage of the heterozygous red eye flies in this population is .................. .

Ans. (48)

Sol. Number of homozygous dominant, white eyed (ww) = 3600

Total number of individuals = 10000

Genotypic frequency of ww, p² = = 0.36

Allelic frequency of w, p = 0.6

We know by Hardy-Weingberg equation,

p + q = 1

q = 1 – 0.6 = 0.4

Frequency of heterozygous population, 2pq = 2 × 0.4 × 0.6 = 0.48

Percentage of heterozygous population = 0.48 × 100 = 48%