General Aptitude

1. The question below consists of a pair of related words followed by four pairs of words. Select the pair that best expresses the relation in the original pair;

Gladiator : Arena

(a) dancer : stage

(b) commuter : train

(c) teacher : classroom

(d) lawyer : courtroom

Ans. (d)

Sol.

2. Choose the most appropriate word from theoptions given below to complete the following sentence:

Under ethical guidelines recently adopted by the Indian Medical Association, human genes are to be manipulated only to correct diseases for which _____ ttreatments are unsatisfactory.

(a) similar

(b) most

(c) uncommon

(d) available

Ans. (d)

Sol.

3. Choose the word from the options given below that is most nearly opposite in meaning to the given word:

Frequency

(a) periodicity

(b) rarity

(c) gradualness

(d) persistency

Ans. (b)

Sol.

4. Choose the most appropriate word from the options given below to complete the following sentence:

It was her view that the country's problems had been ____ by foreign technocrats, so that to invite them to come back would be counter-productive.

(a) identified

(b) ascertained

(c) exacerbated

(d) analyzed

Ans. (c)

Sol.

5. There are two candidates P and Q in an election. During the campaign, 40% of the voters promised to vote for P, and rest for Q. Howerver, on the day of electron 15% of the voters went back on theri promise to vote for P and instead voted for Q. 25% for the voters went back on their promise to vote for Q and instead voted for P. Suppose, P lost by 2 votes, then what was the total number of voters?

(a) 100

(b) 110

(c) 90

(d) 95

Ans. (a)

Sol.

6. The horse has played a little known but very important role in the field of medicine. Horses were injected with toxins of diseases until their blood built up immunities. Then a serium was made from their blood. Serums to fight with diptheria and tetanus were developed this way.

It can be inferred from the passage, that horses were

(a) given immunity to diseases

(b) generally quite immune to diseases

(c) given medicines to fight toxins

(d) given diphtheria and tettanus serums

Ans. (a)

Sol.

7. The sum of n terms of the series 4 + 44 + 444 + ..... is

(a) (4/81) [10^{n + 1} – 9 n – 1]

(b) (4/81) [10^{n – 1} – 9 n –1]

(c) (4/81) [10^{n + 1} – 9n – 10]

(d) (4/81) [10ⁿ – 9n – 10]

Ans. (c)

Sol.

8. Given that f(y) = |y|/y, and q is any non-zero real number, the value of |f(q) – f(–q)| is

(a) 0

(b) –1

(c) 1

(d) 2

Ans. (d)

Sol.

9. Three friends, R, S and T shared toffee from a bowl. R took 1/3^rd of the toffees, but returned four to the bowl. S took 1/4^th of what was left but returned three toffees to the bowl. T took half of the remainder but returned two back into the bowl. If the bowl had 17 toffees left, how many toffees were originally threre is the bowl?

(a) 38

(b) 31

(c) 48

(d) 41

Ans. (c)

Sol.

10. The fuel consumed by a motorcycle during a journey while travelling at various speeds is indicated in the graph below.

The distances covered during four laps of the journey are listed in the table below

From the given data, we can conclude that the fuel consumed per kilometre was least during the lap

(a) P

(b) Q

(c) R

(d) S

Ans. (b)

Sol.

Chemistry

1. Electrophile among the following is

(a) NH₃

(b) SO₃

(c) NO₂

(d) CH C^–

Ans. (c)

Sol. An electrophile is an electron loving specie. An electrophile on virtue of its vacant orbitals, are readily attracted to electron rich centres on the nucleophiles. It may be a positively charged specie or a neutral specie having vacant orbitals.

NO₂ : In nitrogen dioxide, the octet around nitrogen is not complete, due to which he extra unpaired electron contributes to formal charge of +1 or N. Hence, it is attracted towards electron rich species and thus acting as a strong electrophile.

Correct answer is (c)

2. The major product for the following reaction is

(a)

(b)

(c)

(d)

Ans. (c)

Sol. This reaction is an example of Friedel Crafts alkylation reaction, in which electrophile so formed (i.e., primary carbocation) will rearrange to stable one (i.e., tertiary carbocation)

Correct answer is (c)

3. Trouton's rule is obeyed by

(a) hydrogen

(b) methanol

(c) benzene

(d) acetic acid

Ans. (c)

Sol. Trouton's rule gives a relation between a liquid's heat of vaporization and its normal boiling point.

Mathematically, 87 – 88

wherein, H_v = heat of vaporization

T_n = normal boiling point

Trouton's rule is obeyed by benzene, wherein = 87.3; which very well falls within the given range as per the rule.

4. Which one of the following compounds is known as a silanes?

(a) silicon hydrides

(b) silicon halides

(c) silicon hydroxides

(d) silicon oxides

Ans. (a)

Sol. Silanes are a group of compounds having chemical formul Si_nH_{2n + 2}, wherein n = 1, 2, 3 ... Silanes are also called as silicon hydrides.

5. The shape of of PCl₅ is

(a) tetrahedral

(b) square planar

(c) trigonal bipyramidal

(d) square pyramidal

Ans. (c)

Sol.

It exhibits sp³d hybridization and thus trigonal bipyramidal geometry.

6. The correct order of acidity is

(a) C₆H₅COOH < CH₃COOH < C₆H₅OH < C₂H₅OH

(b) CH₃COOH < C₆H₅COOH < C₂H₅OH < C₆H₅OH

(c) C₂H₅OH < C₆H₅OH < C₆H₅COOH < CH₃COOH

(d) C₂H₅OH < C₆H₅OH < CH₃COOH < C₆H₅COOH

Ans. (d)

Sol. Carboxylic acids ions (carboxylate ions) are generally more stable on losing a proton than alkoxide ions, therefore, carboxylic acids are more acidic than alcohols. Further, the aromatic carboxylate ions (benzoate) are more stable than the aliphatic carboxylate ions due to resonance stabilization, therefore, benzoic acid is more acidic than aceitic acid.

Also, the phenoxide ions are more stable than the alkoxide ions due to resonance stabilization of he negative charge over the ring, therefore, phenol is more acidic than ethyl alcohol.

From the above interpretations, the correct acidity order of the following acids is:

C₂H₅OH < C₆H₅OH < CH₃COOH < C₆H₅COOH.

Correct answer is (d)

7. Consider the following equilibrium

SO₂(g) + 1/2 O₂(g) SO₃(g), = –23.5 kCal mol^–1

The formation of SO₃ is favored by

(a) compression and decreasing the temperature

(b) compression and increasing the temperature

(c) expansion and increasing the temperature

(d) expansion and decreasing the temperature

Ans. (a)

Sol. The given rection to form SO₃(g) is n exothermic reaction, which releases 23.5 kcal mol^–1 of heat upon moving in the forward direction. As per the LE Chatelier's principle, an exothermic reaction is favored by decrease in temperature, so as to compensate the heat is given out while moving towards forward direction in the equilibrium reaction.

Further, the effect of pressure upon the position of equilibrium reaction involving gaseous molecules depends upon the number of gaseous molecules depends upon the number of gaseous molecules on either side of equilibrium. As per Le Chatelier's principle, the system tends to oppose the change and remain in equilibrium. Since ther are 1½ molecules on LHS and 1 molecules on RHS an increase in pressure shifts the position of equilibrium towards right side, thereby favoring forward reaction.10

8. A molecule electronic excited state has a life time of 10^–9s, the uncertainty in measuring the frequency (Hz) of the electronic transition is approximately.

(a)

(b)

(c)

(d)

Ans. (a)

Sol. As per the Heisenberg's Uncertinty principle,

And

9. According to the molecular orbital theory, bond order for H₂⁺ species is

(a) 0.5

(b) 1.0

(c) 1.5

(d) 2.0

Ans. (a)

Sol. Bond order can be calculated as :

Bond order =

For species,

Bond order =

10. According to crystal field theory, the electronic configuration of [Ti(H₂O)₆]³⁺ in the ground state is

(a) e¹t₂⁰

(b) t_2g⁰e_g¹

(c) e⁰t₂¹

(d) t_2g¹e_g⁰

Ans. (d)

Sol. In [Ti(H₂O)₆]³⁺, the central metal atom, Ti is in +3 oxidation state. As per the crystal field theory, the d-electron in an octahedral crystal field enters the lower energy orbital to have a ground state electronic configuration as .

11. The ions with lowest and highest radii among O^2–, F^–, Na⁺ and Mg²⁺ are respectively,

(a) Mg²⁺ and O^2–

(b) O^2– and F^–

(c) O^2– and Mg²⁺

(d) Na⁺ and Mg²⁺

Ans. (a)

Sol. All the species O^2–, F^–, Na⁺ and Mg²⁺ are having same number of electrons (10e^–) and thus are isoelectronic. Higher the nuclear charge of isoelectronic species, small shall be the ionic radii. Accordingly, the order of ionic radii of the species is

Common data (Q. 12-13)

The solubility products of FeS, ZnS, CuS and HgS are 1.0 × 10^–19, 4.5 × 10^–24, 4.0 × 10^–38 and 3.0 × 10^–53 respectively.

12. H₂S is passed through an aqueous solution containing all the four metal ions. The metal ion that precipitates first is

(a) Fe²⁺

(b) Zn²⁺

(c) Cu²⁺

(d) Hg²⁺

Ans. (d)

Sol. K_sp(FeS) = [Fe²⁺][S^2–] = 1.0 × 10^–19 ...(1)

K_sp(ZnS) = [Zn²⁺][S^2–] = 4.5 × 10^–24 ...(2)

K_sp(HgS) = [Hg²⁺][S^2–] = 3.0 × 10^–53 ...(3)

K_sp(CuS) = [Cu²⁺][S^2–] = 4.0 × 10^–38 ...(4)

The compound with least solubility product will precipitate first on passing H₂S gas through the mixture metal ions. Since, HgS has least solubility product, therefore, it will precipitate first.

13. The concentration of S^2–, at which FeS begins to precipitaet from the mixture having 0.1M Fe²⁺ is

(a) 1.0 × 10^–17 M

(b) 1.0 × 10^–18 M

(c) 1.0 × 10^–19 M

(d) 1.0 × 10^–20 M

Ans. (b)

Sol. Given, [Fe²⁺] = 0.1M

[S^2–] = ?

Solubility product of FeS, K_sp = 1.0 × 10^–19

FeS  Fe²⁺ + S^2–

K_sp = [Fe²⁺][S^2–]

1.0 × 10^–19 = [0.1][S^2–]

[S^2–] = 1.0 × 10^–18M

Linkes questions (14–15)

Consider the reaction

14. The above reaction is an example of

(a) addition reaction

(b) bimolecular elimination reaction (E₂)

(c) unimolecular substitution reaction (S_N1)

(d) bimolecular substitution reaction (S_N2)

Ans. (c)

Sol. Bromine gets substituted by OH^– in the given reaction. Since, the configuration of the molecules remains the same, that is, retention of configuration. So, the reaction is a two step-reaction wherein carbocation is formed in the first step (since 3° carbon) and hte nucleophile attacks the carbocation. It is a first order reaction, that is, rate of reaction depends only on the stability of hte intermediate carbocation being formed and does not depend on the concentration or strength of the nucleophile.

Hence, above reaction is an example of unimolecular substitution reaction (S_N1).

Correct answer is (c)

15. If the concentration of KOH in the reaction mixture of doubled, the rate of the reaction will be

(a) decreased to one-half

(b) the same

(c) increased by two-times

(d) increased by four-times

Ans. (b)

Sol. Since the reaction is a unimolecular substitution reaction, the rate of reaction is only dependent on the stability of cation and is independent of concentration and strength of he nucleophile.

Rate = k [Carbocation]

Hence, on doubling the concentration of KOH, there shall be no effect on the rate of aforementioned reaction.

Correct answer is (b)

Biochemistry

1. Which one of the following does not inhibit protein biosynthesis?

(a) puromycin

(b) chloramphenicol

(c) cycloheximide

(d) oligomycin

Ans. (d)

Sol. Puromycin is a structural analog of the 3' end of aminoacyl transfer RNA. Like amino acyl tRNA, it binds to the A site of the ribosome peptidyl transferase center. When the A site is occupied by puromycin, peptidyl transferase links the peptide residues of the peptidyl tRNA in the ribosomal P site covalently to puromycin. Since the amide bond cannot be cleaved by the ribosome, no further peptidyl transfer takes place and the peptidyl puromycin complex falls off the ribosome.

Chloramphenicol binds to the 50S ribosomal subunit and blocks peptide bond formation
through inhibition of peptidyl transferase but does not affect the cytosolic protein synthesis in eukaryotes.

Cycloheximide exerts its effects by interfering with the translocation step in protein synthesis (movement of two RNA molecules and mRNA in relation to the ribosome), thus blocking eukaryotic translational elongation.

Oligomycin antibiotic is extracted from Streptomyces bacteria which blocks proton channels (F₀ subunit) due to which ATP synthesis inhibits.

2. The activation of the complement components occurs via three distinct pathways. Which of the following component(s) is specific to the 'Alternate Pathway'?

(a) factor B and D

(b) mannose binding protein

(c) C1qr2s2

(d) C2

Ans. (a)

Sol. In alternative pathway, serum C3 is subject to spontaneous hydrolysis to yield C3a and C3b. The C3b component can bind to foreign surface antigens. The C3b present on the surface of the foreign cells can bind another serum protein called factor B to form complex. Binding to C3b serves as the substrate for an enzymatically active serum protein called factor D. Factor D cleaves the C3b bound factor B, releasing a small fragment (Ba) that diffuses away and generating C3 convertase. Properdin binds C3 convertase and stabilize it.

3. Which one of the following enzymes fixes CO₂ into organic form?

(a) ribulose 50phosphate kinase

(b) ribulose 1,5-bisphosphate carboxylase

(c) pyruvate dehydrogenase

(c) carbonic anhydrase

Ans. (b)

Sol. In Calvin cycle, the first phase, that is carboxylation in which carbon dioxide is incorporated into a five-varbon compound ribulose 1,5 bisphosphate (RuBP). The enzyme that catalyzes this first step is ribulose 1,5 bi-phosphate carboxylase (RuBisCO). It is the most abundant protein in chloroplast and is also said to be the most abundant protein on earth. Its activity is regulated by carbon dioxide, oxygen, magnesium and pH. It is active only when it combines or in presence of carbon dioxide only.

4. Cytochrome C is normally foundin the inner mitochondrial membrane. It is released into the cytoplasm during

(a) apoptosis

(b) necrosis

(c) cell differentiation

(d) cell proliferation

Ans. (a)

Sol. In the intrinsic apoptotic pathway, the functional consequence of pro-apoptotic signaling is mitochondrial membrane perturbation and release of cytochrome c in the cytoplasm, where it forms a complex or apoptosome with apoptotic protease activating factor 1 (APAF-1) and the inactive form of caspase-9. This complex actively hydrolyzes and activates caspase-9 initiating a cascade wherein active caspase-9 activates the executioner caspases-3/6/7, resulting in cell apoptosis.

5. Horseradish peroxidase and alkaline phosphatase are the two enzymes commoly utilized as reagents in ELISA, because these enzymes.

(a) are colored proteins

(b) are very small

(c) have high turnover number

(d) bind to ELISA plates

Ans. (c)

Sol. Alkaline phosphatase (ALP), horseradish peroxidase (HRP) and -d-galactosidase (-DG) are the most frequently used enzymes in ELISA. It is because of their stability, turnover number and lack of interferences. Also, they are detected simply.

6. The polarity of water molecule is due to

(a) Its tetrahedral structure

(b) bonding electrons being attracted more to oxygen

(c) bonding elecrons being attracetd more to hydrogen

(d) its weak electrolytic property

Ans. (b)

Sol. Water is a polar molecule as it contains one oxygen atom which is slightly negative and two hydrogen atoms which are slightly positive in this uneven distribution. In its covalent bonding oxygen attracts more shared electrons. There is attraction between the oppsite charges on molecules, that is, water molecule forms hydrogen bond with other water molecules when highly electronegative oxygen atom of one molecule attracts hydrogen atom of other water molecule which is positive in nature.

7. Cyanide poisoning is due to its direct inhibition of

(a) electron transport chain

(b) fatty acid biosythesis

(c) fatty acid oxidation

(d) nucleic acid biosynthesis

Ans. (a)

Sol. The cyanide ion, CN^–, binds to the iron atom in cytochrome c oxidase in the mitochondria of cells. It acts as an irreversible enzyme inhibitor, preventing cytochrome c oxidase from doing its job, which is to transport electrons to oxygen in the electron transport chain of aerobic cellular respiration.

8. In humans, the largest energy reserve is

(a) liver glycogen

(b) muscle glycogen

(c) blood glucose

(d) adipose tissue triacylglycerol

Ans. (d)

Sol. By far the largest energy reserve in the humen body is adipose tissue triglycerides, and these reserves are an important source of fuel during prolonged endurance exercise. To use this rich source of potential energy during exercise, adipose tissue triglycerides must first be hydrolyzed and the resultant fatty acids delivered to the working muscles.

9. A mixture of four proteins of pIs 11, 7, 5 and 3 are loaded on DEAE anion-exchange column eqquilibrated with low ionic strength buffer of pH 8. Which of the four proteins would be expected to be retained on the column?

(a) protein with pI 11 but not the others

(b) proteins with pIs 11 and 7 but not 5 and 3

(c) proteins with pIs 7, 5 and 3

(d) protein with pI 7 but not the others

Ans. (c)

Sol. On the basis of counter ions, the ion exchangers are of two types :

(i) Cation exchanges: They are negatively charged containing positively charged counter ions (cations).

(ii) Anion exchangers : They are positively charged exchangers contaiing negatively charged counter ions (anion).

At pH 8, for proteins with pIs 3, 5 and 7, the side chains will be negative. Anion exchange resins containing positively charged groups, such as diethyl-aminoethyl groups (DEAE) will bind to negatively charged molecules.

10. Valinomycin, a cyclic peptide antibiotic, facilitates the transport of which one of the following ions?

(a) K⁺

(b) Ca²⁺

(c) Na⁺

(d) H⁺

Ans. (a)

Sol. Valinomycin, an antibiotic isolated from Streptomyces, is an example of ionophores. There are two classes of ionophores, that is, mobile ion carriers and channel formers. Valinomycin is an example of mobile ion carrier. It transports K⁺ down its electrochemical gradient. Its addition makes inner mitochondrial membrane permeable for K⁺. It causes the movement of K⁺ along the concentration gradient from cytosol into the matrix. It decreases the membrane potential component of proton motive force and thus ATP synthesis.

11. Match P, Q, R and S with the appropriate numbers 1 to 6 on the right

P. Basophills 1. Perforin

Q. T-cells 2. Phagocytosis

R. B-cells 3. Albumin

S. Neutrophils 4. Macroglobulin

5. Fc receptors for IgE

6. Plasma cells

(a) P-5, Q-1, R-6, S-2

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

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

(d) P-2, Q-6, R-1, S-3

Ans. (a)

Sol. IgE mediates the immediate hypersensitivity reactions that are responsible for the symptoms of hay fever, asthma, hives and anaphylactic shock. IgE binds to Fc receptors on the membranes of blood basophils and tissue mast cells.

Antigen specific cytotoxic T-lymphocytes (CTL) interact with appropriate target cells and undergo conjugate formation. Formation of a CTL target cell conjugate is a Ca²⁺ dependent, energy requiring step. After attaching to the target cell, the CTL mobilizes its granules directionally toward the target and release the contents of these granules onto the target cell. The major constituents of the granules involved in target cell killing are perforin and granzymes.

The B-lymphocytes mature in the bone marrow and expresses membrane bound antibody. After interacting with antigen, it differentiates into antibody secreting plasma cells and memory cells.

Phagocytosis is a fundamental protective mechanism carried out by these cell types, neutrophils, macrophages, monocytes and dendritic cells.

12. Two purified DNA samples A and B contain equal number of base pairs. Each of these DNA samples has one site each for EcoRI and BamHI restriction enzymes. Complete digestion with both the enzymes yielded 3 DNA bands and 3 DNA bands respectively for A and B upon electrophoresis of the digestion products. Which one of the following explains the observation?

(a) A is circular DNA and B is linear

(b) B is circular DNA and A is linear

(c) A is circular DNA and B could be linear or circular

(d) B is circular DNA and A could be linear and circular.

Ans. (b)

Sol. Circular DNA has no end points. It is circle in nature and is mainly known as DNA plasmid. When this circular DNA is treated with restriction enzyme, it will form a cut and will be seen as a band in electrophoresis.

While linear DNA is straight one having end points when treated with restriction enzyme it breaks and forms two parts, so it forms two bands.

Herein, DNA is treated with two restriction enzymes; thus, if A is linear DNA it yields 3 bands while B is circular DNA as it yields 2 bands.

13. In the following enzyme catalyzed reaction which follows Michaelis-Menten kinetics

K_M is equal to

(a) k_–1/(k₁ k₂)

(b) (k₁ k₂)/k_–1

(c) k₁/(k₂ + k₁)

(a) (k₂ + k₁) / k₁

Ans. (d)

Sol. In the given equation

k₁ is the rate constant for the formation of the enzyme substrate complex, ES, from the enzyme, E, and the substrate S, k_–1 is the rate constant for the reverse reaction, dissociation of the ES complex to free enzyme and substrate, and k₂ is the rate constant for the conversion of the ES complex to product P and the subsequent release of product from the enzyme.

According to the Mischaelis–Menten approach, when the rate of an enzyme catalyzed reaction is measured at varying substrate concentration, the rate depends on the substrate concentration [S]. At a relatively low concentrations of substrate, initial velocity (V) increases almost linearly with an increase in a substrate concentration. The reaction reaches a maximum velocity (V_max) with an increase in substrate concentration and it doesn't increase any further by increasing the concentration of substrate. Concentration of substrate at which reaction velocity is called Michalis constant (K_M). It is the ratio of constants.

14. Match the items in Group I with those in Group II

Group I Group II

P. Progesterone 1. Peptide

Q. Dopamine 2. Fatty acid

R. Vasopressin 3. Carbohydrate

S. Prostaglandin 4. Catecholamine

5. Eicosanoid

6. Steroid

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

(b) P-6, Q-4, R-1, S-5

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

(d) P-6, Q-5, R-1, S-4

Ans. (b)

Sol. Steroids are complex derivatives of triterpens. Each type of steroid is composed of four fused rings called steroids nucleus. Progesterone is a steroid hormone secreted by the corpus luteum. It is responsible for the maintenance of pregnancy hormone.

Catecholamines act as both neurotransmitter and hormones. Epinephrine, dopamine and norepinephrine are collectively known as catecholamines. During stressful situation, these two hormones greatly increase in fight, fright and flight response.

Vasopressin is a peptide that also act as anti-diuretic hormone which increases the permeability of the distal convoluted tubule and is collecting duct of the kidney nephron resulting in less water in the urine. The uring becomes more concentrated as water is conserved. It also causes vasoconstriction, thereby, increasing blood pressure.

Eicosanoids are a family of very potent biological signaling molecules that act as short range messengers. It includes prostanoids and leukotrienes. Prostanoids include the prostaglandins, prostacyclind and thromboxanes. Prostaglandins were first identified in human semen. Almost all mammalian cells except RBC produce prostaglandins.-

15. Three samples of antibodies wereelectrophoresed under denaturing and reducing conditions on a 15% acrylamide gel, followed by stttainning with Coomassie blue dye. Samples 1, 2 an 3 showed two, three and four stainable bands respectively. Which one of the following conclusions can be made from these observations?

(a) sample 1 is IgG, 2 is IgA and 3 is IgM

(b) sample 1 is IgA, 2 is IgM and 3 is IgG

(c) sample 1 is IgG, 2 is IgM and 3 is IgA

(d) sample 1 is IgA, 2 is IgG and 3 is IgM

Ans. (c)

Sol. In a denaturing and reducing conditions (SDS-PAGE), disulfide bonds are borken and an antibody occuring as a monomer gives 2 bands.

In IgG, disulfide bonds hold light chain and heavy chain together in both arms and the heavy chains, The two bands correspond to heavy and light chains.

IgM exists as a pentamer wherein five monomers are joined to each other with disulfide bonds and J chain. It will show three bands, light chain, heavy chain and J chain

IgA is a dimer which contains two antibodies joined by J chain. The functional dimer also consists of secretory component. The dimer IgA give four bands—light chains, heavy chains, J chain and secretory component.

16. Four identical PCR reactions were carried out in tubes named I, II, III and IV. Besides the usual mix of dNTPs, each of the tubes respectively contained ³²P dATP, –³²P dATP, dNTP and rNTP. Which one of the tubes will have radiolabeled PCR product?

(a) tube I

(b) tube II

(c) tube III

(d) tube IV

Ans. (c)

Sol. During elongation of the DNA or in the extension phase in PCR, the incoming dNTP has three phosphates which are called alpha, beta and gamma on the basis of their position from the carbon. The nearest phosphate group is alpha, then beta and the third, farthest phosphate group is gamma. In a reaction, beta and gamma are removed from the dNTP and alpha phosphorous is incorporated in the elongating DNA chain. Thus, the PCR tube 3 with dATP will have radiolabeled PCR product.

17. Match the following

Group I Group II

P. Polynucleotide kinase 1. ATPase

Q. Fluoride 2. GTPase

R. Ras 3. Transketolase

S. Iac operon 4. Enolase

5. 5' end of DNA

6. 3' end of DNA

7. Only positive regulation

8. Positive and negative regulation

(a) P-5, Q-4, R-2, S-8

(b) P-6, Q-3, R-1, S-7

(c) P-4, Q-2, R-1, S-6

(d) P-1, Q-7, R-5, S-3

Ans. (a)

Sol. The polynucleotide kinase utilizes two types of reactions forward reaction and exchange reaction. In the forward reaction, polynucleotide kinase transfers the gamma phosphate from ATP to the 5' end of a polynucleotide. The target polynucleotide lacks a 5' phosphate. It is mainly used for end labeling of DNA molecules.

In the exchange reaction, target polynucleotide that has a 5' phosphate is incubated with an excess of ADP. The removal of water from 2 phosphoglycerate creates a high energy enol phosphate linkage. The enzyme catalyzing this step, enolase, is inhibited by fluoride.

Ras is an intracellular monomeric GTPase switch protein and functions in transducing signals from many different receptor tyrosine kinases. It is a lipid linked protein present at the cytoplasmic face of the membrane.

Lac operon is controlled by both positive and negative regulation. In the absence of an inducer, the lac repressor binds to the lac operator site and prevents transcription of the lac operon. Because binding of a repressor prevents the transcription, it is termed as negative regulation.

18. Collagen, -keratin and tropomyosin have common structural features. They are

P. disulfide bridges to neighboring proteins

Q. repeating sequences of amino acids

R. a high -sheet content

S. superhelical coling

(a) P, Q

(b) Q, R

(c) Q, S

(d) P, R

Ans. (c)

Sol. The fundamental unit of collagen is tropocollagen Tropocollagen consists of three coiled polypeptides called -chains. The -chains are left handed polypeptide helices and have 3.3 amino acid residues per turn. Three chains wind around one another in a characteristic right-handed triple helix. The amino acid sequence in chain is generally repeating tripeptide unit, Gly – X-Y where X is often proline and Y is often 3 or 4 hydroxyproline or 5 hydroxylysine.

Keratins are fibrous proteins present in eukaryotes. They form a large family, with about 30 members being distinguished. They have been classified as either -keratins or -keratins.

The -keratin polypeptide chain is a right handed -helix and rich in hydrophobic amino acid residues Ala, Val, Leu, Ile, Met and Phe. Every keratin polypeptide chain dimerizes to form hetrodimer.

Tropomyosin is a fibrous molecule that consists of two chains, alpha and beta that attach to the F actin in the groove between its filaments.

19. Match the following

Group I Group II

P. Tyrosine hydroxylation 1. Thyroxine

Q. Tyrosine iodination 2. T-cell receptor

R. Tyrosine phosphorylation 3. DOPA

S. Tyrosine oxidation 4. Estradiol receptor

5. Epinephrine

6. Melanin

7. Endorphin

8. Serotonin

(a) P-1, Q-6, R-5, S-4

(b) P-5, Q-7, R-4, S-8

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

(d) P-3, Q-1, R-2, S-6

Ans. (d)

Sol. Dopamine is formed from tyrosine in presence of tyrosine hydroxylase. The thyroid follicles actively accumulate iodine from the blood and secrete it into the colloid. Within the colloid, iodine is oxidized into iodine and attached to a tyrosine residue of thyroglobulin protein.

The attachment of one iodine to tyrosine produces monoiodotyrosine (MIT) the attachment of two iodine produces diiodotyrosine (DIT). When two molecules of DTI are modified and coupled together a molecule of tetraiodothyronine (T4) or thyroxine is produced.

TCR ligation triggers phosphorylation of tyrosine residues in the immunoreceptor tyrosine-based activation motifs of the chain complex.

Melanin is produced through a multistage chemical process known as melanogenesis, where the oxidation of the amino acid tyrosine is followed by polymerization.

20. Scatchard analysis of ligand-receptor interaction yielded the graph shown below. The affinity of the ligand-receptor interaction can be obtained from

(a) Y intercept

(b) X intercept

(c) slope of the time

(d) product of X intercept and Y intercept

Ans. (c)

Sol. The Scatchard plot is a graphical method of analyzing equilibrium ligand binding data. It is used to determine the number of ligand binding sites on a receptor, whether these sites show cooperative interactions, whether more than one class of site exists, and the respective affinities of each site. Ligand binds specifically and non-covalently with the receptor. A receptor may possess one binding site per molecule or mroe than one independent binding sites. Receptor ligand binding is given by the following equation.

R (receptor) + L (ligand) R – L (Receptor – Ligand complex)

The equilibrium constant for this association reaction K_a is

K_a =

K_d = ...(1)

Now, the fractional saturation of sites is

Y = ...(2)

The value of Y ranges from zero, when [RL] = 0 to 1, when [R] = 0

To determine the value of K_d, we can rearrange the Eq. (1) as follows :

[RL] =

Substituting the expression for [RL] in Eq. (2), we obtain

Y =

Y =

By taking the reciprocal, Eq. (3), we get

=

Alternatively, it can be rearranged to

=

A plot of versus Y gives a straight line of slope .

Botany

1. The stalk with which the ovule remains attached to the placenta is called

(a) micropyle

(b) chalaza

(c) funiculus

(d) hilum

Ans. (c)

Sol. The narrow pores or passages left at one end of the ovule by integumentsfor entry of pollen tube into ovule during fertilization are called micropyle.

The swollen basal part of nucellus from which integuments originate is called the chalazal.

Funiculus is a stalk that attaches the ovule to the placenta of the ovary.

The region at which the body of ovules fuses with the funicle is called the hilium.

2. The diploid chromosome number of an organism is 2n = 14, what would be the expected chrosome numbers in a nullisomic?

(a) 12

(b) 13

(c) 15

(d) 16

Ans. (a)

Sol. Nullisomy is when both the pairsof homologous chromosomes are lost. The nullisomics have 2n – 2 number of chromosomes.

Given, diploid chromosome number of an organism, 2n = 14.

In nullisomy, number of chromosome = 2n – 2

Therefore, nullisomy = 2n – 2 = 14 – 2 = 12 chromosomes.

3. The mutagen ethidium bromide acts as a

(a) deaminating agent

(b) alkylating agent

(c) intercalating agent

(d) base analogue

Ans. (c)

Sol. Ethidium bromide binds to DNA molecules by intercalating between adjacent base pairs, causing partial unwinding of the DNA double helix. Bands showing the positions of the different sizes of DNA fragment are clearly visible under ultraviolet irradiation after EtBr staining, as long as sufficient DNA is present.

4. During photorespiration the reactive oxygen species, H₂O₂ is produced in

(a) glyoxysome

(b) Iysosome

(c) peroxisome

(d) dictyosome

Ans. (c)

Sol. Peroxisomes in plant cells are involved in photorespiration by interacting with mitochondria and cytoplasm. They are membrane bound organelles of the cytoplasm that act as sites of metabolic reactions, such as substrate oxidation leading to formation of hydrogen peroxide and subsequent degradation of hydrogen peroxide.

5. One of the defense mechanisms adopted by plants for detoxification of heavy metals is the synthesis of

(a) phytochelatin

(b) calmodulin

(c) tubulin

(d) systemin

Ans. (a)

Sol. Phytochelatins are oligomers of glutathione, produced by the enzyme phytochelatin synthase. They are found in plants, fungi, nematodes and all groups of algae including cyanobacteria. They acts as chelators and are important for heavy metal detoxification.

6. In which one of the following phases of cell cycle the drug colchicine exerts its effect?

(a) G1

(b) G2

(c) S

(d) M

Ans. (d)

Sol. Colchicine is an alkaloid obtained from a plant Colchicum sp. It acts as inhibitor of mitosis by inhibiting spidle formation. As a result, the chromosomes duplicate but remain in the same cell leading to an increase in chromosome number, thereby inducing polyploidy.

7. The transition of water molecule from liquid to glassy state during cryopreservation is termed as

(a) vitrification

(b) hyperhydricity

(c) cryoprotectant

(d) habituation

Ans. (a)

Sol. Vitrification is the process by which water undergoes a phase transition from liquid state to an amorphous glassy state. In this process, glasslike solidification of living cells takes place that completely avoids ice crystal formation during cooling. It avoids ice crystal formation in cryopreserved cells during warming to recover the cells for biological applications.

8. The DNA content of a nucleus can be measured by

(a) ESR spectroscopy

(b) FTIR spectroscopy

(c) flow cytometry

(d) X-ray crystallography

Ans. (c)

Sol. In Flow cytometry technique, a suspension of thousands of chromosomes is made and these chromosomes are stained with a DNA binding fluorochrome. Using a cytometer, the fluorescence of the individual chromosomes is measured and a histogram is generated. Chromosomes of the same size are represented as a peak in the histogram. This is useful for detecting aneuploidy, duplication, etc., in chromosomes.

9. Retrograde signaling involves communication of

(a) nucleus to the chloroplast

(b) endoplasmic reticulum to the nucleus

(c) nucleus to the mitochondria

(d) chloroplast to the nucleus

Ans. (d)

Sol. Retrograde signaling is the process in which plant organelles emit signals to regulate the nuclear genes expression. It acts as a pathway to communicate between chloroplast and nucleus.

10. A photoautotrophic micropropagation system can be established by increasing the

(a) sucrose concentration in the culture medium

(b) CO₂ concentration in the culture medium

(c) agar concentration in the culture medium

(d) NH₄⁺ concentration in the culture medium

Ans. (b)

Sol. In photoautotrophic micropropagation, no exogenous organic component (sugar, vitamins, etc.) is added to the medium. Here, the growth is dependent fully upon photosynthesis. This
system can be established with in vitro environmental control by increasing CO₂ concentration, relative humidity and photosynthetic photon flux that are required for the plant growth and development.

11. Which of the following statements in photosynthesis are correct?

P. The absorptionmaxima for photosystem I (PS I) and PS II are 680 nm and 700 nm, respectively.

Q. Photosynthetic reaction centre contains 300 chlorophyll molecules and the release of one molecule of oxygen requires a minimum of 8 photons.

R. The non-photochemical quenching of excitation energy is enhanced by the presence of zeaxanthin.

S. The photochemical splitting of water occur in PS I.

(a) P, Q

(b) R, S

(c) P, S

(d) Q, R

Ans. (d)

Sol. The reaction center contains chlorophyll a and absorbs light energy of longer wavelength. The reaction centers of PSI and PSII are different and are designated as P700 (far-red) and P680 (red), respectively. The P here stands for pigment and the number refers to the wavelength at which absorption peak is observed. 300 chlorophyll molecules are associated with one reaction center. These are constituents of the antennae. The antennae contain additional accessory pigments such as carotenoids, mainly xanthophylls, including lutein and the related violaxanthinas well as carotene. Some 2500 molecules of chlorophyll was required to produce one molecule of oxygen and that a minimum of eight photon of light must be absorbed in the process.

12. Which of the following statement are true on DNA delivery method during plant transformation?

P. Single stranded nicks are made in T-DNA border repeat by the VirD1, VirD2 and VirD3 protein complex.

Q. virA gene products form the export apparatus on the membrane for the transfer of T-DNA.

R. Gold/Tungsten particles are used as microprojectiles in biolistic method.

S. Acceleration of DNA-coated microprojectiles is carried out with compressed CO₂.

(a) P, S

(b) R, S

(c) P, R

(d) Q, S

Ans. (c)

Sol. Vir B11 with other proteins forms the export apparatus (a type IV secretion system) for delegate of T-DNA.

Compressed He or Xe gas is used for gene gun as they are essentially non-reactive.

13. Match the following plant secondary compounds with their used and source plants.

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

(b) P-3-iv, Q-1-i, R-5-ii, S-6-iii

(c) P-4-iv, Q-3-i, R-1-v, S-2-vi

(d) P-4-iii, Q-2-ii, R-5i, S-6-iv

Ans. (a)

Sol. Commiphora wightii is an arid region plant, highly valued for its medicinally important guggul gum-resin as a source of guggulsterone. It has been used since traditional time for treatment of rheumatoid arthritis.

Shikonin, a naturally occurring naphthoquinone found in the roots of Lithospermum erythrorhizon. It has been used as a red dye for centuries.

Catharanthus roseus is a medicinal plant that produces anti-hypertensive monomeric indole alkaloids, serpentine and ajmalicine and the antitumor dimeric alkaloids, vinblastine and vincristine.

Glycyrrhizin is a saponin like compound that provides the main sweet flavor for Glycyrrhiza glabra, with potential immunomodulating, anti-inflammatory, hepato and neuro-protective and anti-neoplastic activities.

14. Match the gene of interest for various aspects of crop improvement.

Gene insert Aspects of crop improvement

P. bar 1. Tolerance to heavy metals

Q. vip3A 2. Nutritional improvement with increased vitamin A

R. b-Icy 3. Insect resistance

S. gsh-11 4. Herbicide resistance

5. Delayed ripening

6. Resistance to fungal infection

(a) P-4, Q-3, R-5, S-6

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

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

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

Ans. (b)

Sol. A gene which confers resistance to the herbicide bialaphos (bar) has been characterized. The bar gene was originally cloned from Streptomyces hygroscopicus, an organism which produces the tripeptide bialaphos as a secondary metabolite.

Vegetative insecticidal protein (Vip) encoding gene vip3A is a unique class of insecticidal protein. It is part of some transgenic plants for conferring resistance against lepidopteran pests.

To obtaine a functioning provitamin A (beta-carotene) biosynthetic pathway, a single, combined transforming cDNA coding for phytoene synthase (psy) and lycopene beta-cyclase (beta-lcy) both from Narcissus pseudonarcissus and both under the control of the endosperm specific glutelin promoter together with a bacterial phytoene desaturase is introduced in rice endosperm.

Catalyzes the double cyclization reaction which converts lycopene to beta-carotene and neurosporene to beta-zeacarotene.

GshII codes for glutathione transferase and is used to ameliorate heavy metal stress.

15. Match the gene of interest for various aspects of crop improvement.

Plant Protein

P. Rape seed 1. Kafirin

Q. Pea 2. Vicillin

R. Sorghum 3. Gliadin

S. Wheat 4. Napin

5. Zein

6. Patatin

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

(b) P-2, Q-3, R-6, S-1

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

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

Ans. (c)

Sol. The major storage proteins in Brassica napus (rape) seed are cruciferin and napin, which comprise 85-90% of the total proteins.

The major storage protein in legumes including pea is vicilin.

For wheat, the storage protein is gliadin.

Sorghum storage protein is kafrinin.

16. Match the name of the disease wityh the casual organism

Disease Causal organism

P. False smut of rice 1. Plasmopara viticola

Q. Ring rot of potato 2. Collectotrichum falcatum

R. Red rot of sugarcane 3. Corynebacterium sepidonicum

S. Downy mildew of grape 4. Ustalginoidea virens

5. Erwinia amylovora

6. Synchytrium endobioticum

(a) P-1, Q-5, R-2, S-4

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

(c) P-6, Q-2, R-4, S-1

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

Ans. (b)

Sol. False smut of rice disease is caused by Ustilaginoidea virens, it is the perfect sexual stage of Villosiclava virens that reduces both grain yield and quality.

The causal organism for ring rot of potato is Corynebacterium sepidonicum.

Collectotrichium falcatum infects sugarcane crop is the pathogen of red rot disease.

Downy mildew of grape disease is caused by Plasmopara viticola, it is the heterothallic oomycetes that form oospores in leaf litter and soil.

17. Identify the correct statement for phylogenetic systems of classification

P. the most popular phylogenetic systems of clsssification is that of George Bentham and Joseph Dalton Hooke and was published in Genera Platarum

Q. a true phylogenetic systems of classification was proposed by Adolf Engler and was published in Die Naturlichen Pflanzenfamilien

R. the phylogenetic system of classification proposed by John Hutchionson was appeared in The Families of Flowering Plants

S. the origin of dicot from primitive monocot was proposed by Arthur Cronquist in his book Systema Naturae

(a) Q, R

(b) P, Q

(c) R, S

(d) P, S

Ans. (a)

Sol. George Bentham and Sir Joseph Dalton Hooker devised a natural system of classification for seed plants and published it in Genera Plantarum.

Carolus Linnaeus first attempt at classifying plants and animals and published it in 1735 as Systema Naturae.

18. Which of the following statements are true for the plastid genomes?

P. plastid genome is circular in nature with genome size of 120-160 kb.

Q. the plastid ribosomes are with sedimentation coefficient of 80S

R. the gene for the small subunit of ribulose bisphosphate carboxylase (RubisCO) is located in the plastid

S. rRNAs in the plastid genome are arranged in one transcription unit.

(a) P, Q

(b) Q, S

(c) R, S

(d) P, S

Ans. (d)

Sol. The plastid ribosome has sedimentation coefficient of 70S.

There are eight larger subunits of RuBisCO, that are encoded by plastid, while the smaller eight subunits are encoded by nucleus.

19. Identify the correct statements.

P. Specialized parenchymatous cells with tannis and crystals of calcium oxalate are termed as sclereids.

Q. The sieve elements of angiosperms are surrounded by companion cells and are essntial component of phloem loading.

R. The exudation of water byguttatioin occurs through trichomes.

S. The bulliform cells control the unrolling and hydroscopic movement of grass leaves.

(a) P, Q

(b) P, R

(c) Q, S

(d) P, S

Ans. (c)

Sol. Sclereids are typically short cells (oval, spherical or cylindrical) with very thick secondary walls, strongly lignified and provided with numerous simple pits, secondary wall is multilayered. They are dead cells and have a narrow lumen.

Hydathodes are minute pores or vein openings found on the margins and tips of leaves that excrete water in conditions of high atmospheric humidity, that is, in early morning, during night in wet seasons. The process of releasing excess water and the loss of water in the liquid phase is known as guttation.

20. Which of the following statements are incorrect on ecological point of view?

P. Primary succession involving xerosere is initiated in a wet habitat.

Q. Halones commonly found in electronic equipment are one of the active force destroying the protective ozone layer in the stratosphere.

R. Sympatric speciation occurs when the new species evolves in geographic isolation from the parent specis.

S. -diversity is the diversity of species within a habitat or community.

(a) P, Q

(b) P, R

(c) Q, R

(d) Q, S

Ans. (b)

Sol. Xerosere refers to the various stages of succession on a bare rock.

Sympatric (living in same geographical area) speciation refers to speciation occurring due to members of a population occupying different ecological zones in the same geographical area. In other words, they become ecologically specialized so that there is no genetic exchange between populations.

Microbiology

1. Quinolones inhibit bacterial growth by targeting

(a) DNA replication

(b) mRNA translation

(c) RNA polymerase

(d) active transport of nutrients into the cell

Ans. (a)

Sol. Quinolones are the antibiotics which exert their antibacterial effect by preventing bacterial DNA from unwinding and duplicating. They specifically inhibit the ligase activity of the topoisomerase II, gyrase and topoisomerase IV which cuts DNA in order to introduce supercoiling.

2. To select the spontaneously arising histidine auxotrophs in a population, you would use a medium containing

(a) histidine and penicillin

(b) penicillin but no histidine

(c) histidine and Iysozyme

(d) Iysozyme but not histidine

Ans. (a)

Sol. To select for spontaneously arising histidine auxotrophs in a population, we would use histidine and penicillin. Since auxotrophs are uable to synthesize a particular organic compound which is needed for their growth, thus histidine should be present in the medium with penicillin.

3. Which one of the following statements is not associated with contributions of Louis Pasteur?

(a) anthrax is caused by Anthras bacillus

(b) bacteria causing food spoilage come from air

(c) the disease causing organism must be isolated in pure culture

(d) bacteria cause the wine disease

Ans. (c)

Sol. Pasteur started his studies on fermentaion to fight against the diseases of wine. Working ahead with microbes, he found a way to weaken the microbe involved and eventually discovered a solution for anthrax. He proved through a series of experiments that food spoilage was a result of unseen elements in the air.

The statement 'The disease-causing organism must be isolated in pure culture' is one of the Koch's postulates.

4. The active transport of solute in the cell is characterized by

(a) its uptake along the concentration gradient utilizing energy

(b) requirement of a carrier to support transport along the concentration gradient

(c) chemical modification of the solute during its uptake

(d) its uptake against the concentration gradient

Ans. (d)

Sol. Active transport (secretion) is an energy-dependent process that selectively moves a substance against its electrochemical gradient across a cell membrane. It is also known as uphill transport, as it requires the transport of molecules from a low concentration region to a high concentration region.

5. Catabolite repression allows cells to save energy by

(a) inactivating catabolic enzymes

(b) inhibiting symthesis of total RNA

(c) regulating expression of genes required for utilization of less-efficient metabolites

(d) inhibiting translation of mRNAs encoding catabolic enzymes.

Ans. (c)

Sol. Catabolite repression allows micro-organisms to adapt quickly to a preferred (rapidly metabolisable) carbon and energy source first. This is usually achieved through inhibition of synthesis of enzymes involved in catabolism of carbon sources other than the preferred one.

6. A newly emerged variant of Influenza virus can be selectively propagated from the mixed population by addition of

(a) gangcyclovir

(b) tamiflu

(c) interferon gamma

(d) neutralizing antibody

Ans. (b)

Sol. Influenza viruses generate genetic diversity due to the high error rate of their RNA polymerase, often resulting in mixed genotype populations (intra-host variants) within a single infection. This variation helps influenza to rapidly respond to selection pressures, such as those imposed by the immunological host response and antiviral therapy. On the additon of tamiflu (oseltamivir), because of influenza intra-host variation in a transmission chain tamiflu-sensitive and tamiflu-resistant cases may appear. Thus, form the mixed population, only the newly emerged variant of influenza virus that are resistant to antibiotics will propagate.

7. The synthesis of an immunoglobulin in either a secretory or membrane bound form is governed by

(a) allelic exclusion

(b) class switching

(c) different RNA processing

(d) affinity maturation

Ans. (c)

Sol. Differential RNA splicing is a regulated process that results in a single gene coding for multiple proteins. In this case particular exons of a gene may be included or excluded from the final processed mRNA produced from that gene. This process is used in the synthesis of an immunoglobulin in either a secretory or membrane bound form.

8. The cis-trans test can determine wheather a gene codes for

(a) an activator ora repressor

(b) an RNA or a protein

(c) a protein with the same or different amino acids

(d) a diffusible or non-diffusible product

Ans. (c)

Sol. Cis-trans test also known as complementation test is used to test if two mutations of same character occur in the single chromosome (cis position) or in different cistrons in each chromosome of a homologous pair (trans position). Therefore, it can be used to determine if a gene codes for a protein with same or different amino acids.

9. Which of the following are expected to be the abundant inhabitants of a nitrate and sulfate rich soil naturally depleted for oxygen?

(a) Pseudomonas and Azotobacter

(b) Pseudomonas and Desulfovibrio

(c) Azotobacter and Thiobacillus

(d) Nitrosomonas and Nitrobacter

Ans. (b)

Sol. Pseudomonas and Desulfovibrio are expected to be the abundant inhabitants of a nitrate and sulfate rich soil whic his naturally depleted for oxygen.

10. Which one of the following immersion oils would you use to get the best resolution in a light microscope (with 1000X objective)?

(a) an oil with infractive index of 1.6

(b) an oil with infractive index of 1.5

(c) an oil with infractive index of 1.4

(d) an oil with infractive index of 1.3

Ans. (c)

Sol. Immersion oils are used in light microscopy in order to increase the resolving power of a microscope. As a result, both the objective lens and the specimen is immersed in oil of high refractive index, thereby increasing the numerical aperture of the objective lens. Typically, immersion oil with a refractive index of 1.5 is used.

11. Four Hfr strains of E. coli were generated from the same F⁺ strain. The Hfr strains denoted markes in the following order

Strain 1: DQWMT; Strain 2: AXPTM; Strain 3: BNCAX; Strain 4: BDQWM

The order of the markers in the original F⁺ strain is

(a) dqwmtpxacnb

(b) axptmdqwbnc

(c) bncaxptmdqw

(d) bdqwmncaxpt

Ans. (a)

Sol. The order of markers obtained from the Hfr strains that can satisfy the 4 strains obtained on conjugation with F⁺ strain wil be DQWMTPXACNB.

12. Which one of the following forms of the same DNA molecule would bind maximum ethidium bromide?

(a) negatively supercoiled

(b) covalently closed relaxed circle

(c) linear

(d) positively supercoiled

Ans. (c)

Sol. Ethidium binds by inserting itself between the stacked bases in double stranded DNA. The distortion of the sugar phosphate backbone when an intercalating agent binds lengthened the DNA. This changes the properties of DNA considerably. Ethidium bromide that intercalates into circular and supercoiled DNA, cannot stretch it much due to strain, thus, maximum ethidium bromide is likely to bind to the linear DNA.

13. An actively growing culture of E. coli divides in about 20 min. Under laboratory conditions, time taken to replicate the entire genome of this bacterium would be about

(a) 20 min

(b) 40 min

(c) 10 min

(d) 18 min

Ans. (b)

Sol. The replication time for the entire genome of E. coli is about 40 mins, however its doubling time is 20 mins. This is because when a cell is dividing rapidly a new round of replication begins before the current round is completed. This means when a daughter cell is formed, it already has partially replicated.

14. Which of the statements about Corynebacterium diptheriae is not correct?

(a) all strains of C. diphtheriae are producers of diphtheria toxin

(b) diphtheria toxin production can be minimized by high concentration of iron in the medium

(c) diphtheria both inhibits protein synthesis

(d) diphtheria toxin is an A-B toxin secreted as a polypeptide of 62 kDa

Ans. (a)

Sol. Diphtheria toxin is an ADP-ribosyltransferase that inhibits host protein synthesis and is synthesized by toxigenic strains of Corynebacterium diphtheriae that harbor a prophage encoding the toxin.

15. Match the names of investigators in Group I with their contributions in Gruop II

Group I Group II

P. Joseph Lister 1. Role of phagocytosis in infection

Q. John Needham 2. Disproved spontaneous generation

R. Elie Metchinikoff 3. Proved Spontaneous generation

S. Lazaro Spallanzani 4. Use of agar as solidifying agent

5. Use of carbolic acid as disinfectant

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

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

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

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

Ans. (b)

Sol. Joseph Lister is known as the father of antiseptic surgery and is known for using carbolic acid as disinfectant.

John Needham through his experiments showed that there was a life force that produced spontaneous generation.

Elie Metchnikoff is known for his research in immunology. He discovered phagocytosis which turned out to be the major defense mechanism in innate immunity.

Lazzaro Spallanzani disproved spontaneous generation. He challenged and repeated Needham's experiments contradicting his findings.

16. During replication of the E. coli chromosome,Okazaki fragments are produced from

(a) only one of the strands of the circular genome.

(b) both the strands of the circular genome.

(c) one of the strands in one generation and the other strands in the next generation

(d) both the strands of the circular genome provided that the heavy nitrogen (¹⁵N) is present in the medium

Ans. (d)

Sol. Okazaki fragments form because the lagging strand that is being formed have to be formed in segments of 100-200 nucleotides. This is done DNA polymerase making small RNA primers along the lagging strand which are produced much more slowly than the process of DNA synthesis on the leading strand.

17. A new isolate of a facultative anaerobe utilizes either oxygen or pyruvate as terminal electron acceptor. This bacterium was grown either anaerobically with glucose as sole carbon source; or aerobically with lactose as the sole carbon source. Net increase in ATP production (per mole of the carbon source) during the aerobic growth would be

(a) 2-fold

(b) 4-fold

(c) 19-fold

(d) 38-fold

Ans. (c)

Sol. Aerobic metabolism is 19-fold anaerobic metabolism as it produces 38 ATP compared to 2 ATP in anaerobic metabolism.

18. Based on their properties, match the Genera in Group I with those in Group II

Group I Group II

P. Bacillus 1. Sarcina

Q. Neisseria 2. Azotobacter

R. Rhizobium 3. Hyphomicrobium

S. Caulobacter 4. Clostridium

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

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

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

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

Ans. (a)

Sol. Bacillus and Clostridium both are Gram-positive, spore forming bacilli.

Neisseria and Sarcina both are cocci.

Rhizobium and Azotobacter both are nitrogen fixers.

Caulobacter and Hyphomicrobium both are gram-negative, rod-shaped bacteria, which have been found to be closely related.

19. An actively growing culture (20 ml) of E. coli (1 × 10⁵ per ml) was mixed with a total of 100 T4 phage particles, grown further for 40 min and mixed with a few drops of chloroform. Under the conditions used, the generation time of E.coli is 30 min, the infection cycle of phage T4 is 20 min, and the burst size is 100. Assuming that each infection was a successful one, how many plaque forming units would you expect at the end of the experiment?

(a) 10⁴

(b) 10³

(c) 10⁵

(d) 10⁶

Ans. (d)

Sol. Given,

Time, t = 40 min = 0.66h

Generation time, G = 30 min = 0.5h

N₀ = 20 × 10⁵

On using the formula,

log N = appx. 6

So, N = 10⁶

Since, each infection was successful, thus the number of plaques forming units is equal to the number of E. coli cells, i.e., 10⁶.

20. Match the pair of organisms in Group I with their characteristic interactions in Group II

Group I Group II

P. Photoblepharon palpebratus and Vibrio fischeri 1. Mutualism

Q. Pseudomonas and Bdellovibrio 2. Symbiosis

R. Aspergillus and Pseudomonas 3. Antagonism

S. Thiobacillus ferrooxidans and Beijerinckia lacticogenes 4. Parasitism

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

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

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

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

Ans. (a)

Sol. Vibrio fischeri inhabits the light organ of the flash fish Photoblepharon palpebratus in a symbiotic relationship.

Bdellovibrio is a genus of small highly mobile, vibrio shaped gram-negative bacteria which can parasitize and kill other gram-negative bacteria like Psueomonas.

Aspergillus and Pseudomonas exhibit antagonistic behavior where one inhibits the growth of the other.

Thiobacillus ferrooxidans and Beijerinckia exhibit mutualistic relationship and help in faster metal leaching.

Zoology

1. Which one of the following is an example of eumetazoans?

(a) dictyostelium

(b) hydra

(c) sponges

(d) volvox

Ans. (b)

Sol. On the basis of level of organization, kingdom Animalia is divided into Parazoa (cellular level), containing phylum Porifera (sponges) and eumetazoa (tissue/organ/organ system) containing phyla Cnidaria (or Coelenterata), Ctenophora, Platyhelminthes, Aschelminthes (Rotifera, Nematoda), Mollusca, Annelida, Arthropoda, Echinodermata, Hemichordata and Chordata.

Dictyostelium falls under kingdom Protista.

Hydra belongs to phylum Coelenterata of Eumetazoa.

Volvox is an alga that belongs to kingdom Plantae.

2. Which one of the following is characteristic of deuterostomes?

(a) Radially symmetric body

(b) Bilaterally symmetric body

(c) Presence of well-defined digestive system

(d) Formation of anus from blastopore

Ans. (b)

Sol. In deuterostomes, the blastophore becomes the anus. The mouth develops later at the end opposite the blastopore. The mouth and anus are connected by a digestive tract.

3. Extraembryonic tissues are derived from which one of the following?

(a) ectoderm

(b) endoderm

(c) trophoectoderm

(d) mesoderm

Ans. (c)

Sol. During the formation of the blastocyst two distinct cell populations arise : the embryoblast and trophoblast or trophoectoderm.

The embryoblast or inner cell mass, is located internally adn eventually develops into the embryo. It contains some cells that act as stem cells and give rise to other tissues and organs.

The trophoblast is the outer superficial layer of cells that forms the sphere like wall of the blastocyst. It will ultimately develop into extraembryonic outer chorionic sac that surrounds the fetus and the fetal portion of the placenta.

4. Which one of the following type of immune cells is responsible for graft rejection?

(a) B-cells

(b) T-cells

(c) macrophages

(d) eosinophils

Ans. (b)

Sol. The immune response to a transplanted organ consists of both cellular and humoral mechanisms. Although other cell types are also involved, the T-cells are central in the rejection of grafts. The success of an organ or tissue transplant depends on histocompatibility that is, the tissue compatibility between the donor and the recipient. T-cell receptor interacts with a foreign peptide on the surface of an antigen presenting cell (APC), it must recognize both the peptide and the MHC molecule holding that peptide.

5. Which of the following is a main symptom of infection by Wuchereria bancrofti?

(a) swelling of limbs

(b) skin rashes

(c) blindness

(d) brain cyst

Ans. (a)

Sol. Wuchereria bancrofti or filarial worm is the causative agent for wuchereriasis or elephantiasis or filariasis. Enlarged limbs and cracked skin similar to that of elephants are conspicuous features of elephantiasis. The male genitals (scrotum) and mammary glands of the females are also often affected when the worms block the lymph vessels and glands.

6. In insect's tracheal system, the transport of oxygen to the target tissue is done by

(a) fine branches of air tubes extending to almost every cell

(b) a liquid that fills the tracheal tube

(c) a specialized set of cells that produce myglobin

(d) a specialized pigment

Ans. (a)

Sol. In the insect tracheal system, oxygen is transported to the target tissues through a network of fine branches of air tubes called tracheae. These tracheae extend throughout the insect's body and deliver oxygen directly to individual cells, allowing for efficient gas exchange. This respiratory system allows insects to transport oxygen without the need for a circulatory system like that found in vertebrates

7. Which one of the following examples represents an adaptation or a physiological activity that does not minimize the loss of body temperature of animals?

(a) feathers or fur

(b) fat layers in the adipose tissue

(c) shivering

(d) vasodilation

Ans. (d)

Sol. The feather are flexible and light waterproof epidermal structures. Apart from providing waterproof coats and coloration to birds, the feathers in birds are also important for flight and in body heat conservation by insulting them against temperature changes.

Fat layers in the adipose tissue and shivering represents an adaptation or a physiological activity that minimizes the loss of body temperature of animals.

Vasodilation is an increse in the size of the lumen of a blood vessel caused by relaxation of the smooth muscle in the wall of the vessel.

8. Which one of the following hormones is incorrectly paired with its function?

(a) melatonin – biological rhythm

(b) glucagon – increases blood glucose levels

(c) prolactin – stimulates milk secretion

(d) calcitonin – increases blood calcium level

Ans. (d)

Sol. Pineal gland secretes the hormone melatonin. Melatonin is liberated more during darkness than in light and contribute to the setting of the body's biological clock by inducing sleep and helping the body to adjust to jet lag.

Glucagon is produced by the alpha cells, foud in the islet of Langerhan, in the pancreas. It is a peptide hormone that raises blood glucose level.

Prolactin (PRL) secreted by anterior pituitary, together with other hormones, initiates and maintains milk production by the mammary glands.

Parafollicular cells (C cells) in the outer regions of the follicle walls or between follicles, produce calcitonin which helps lower blood calcium levels by inhibiting bone osteoclasts resorption.

9. The term innate behaviour refers to an animal behaviour

(a) that is triggered by an environmental change

(b) that is taught by the parent

(c) that is developmentally fixed

(d) that an organism learns on its own by a hit-and trial approach

Ans. (c)

Sol. Innate behaviors also called instinctive behaviors do not have to be learned or practiced. An animal performs this behavior the first time it is exposed to the proper stimulus. It can be performed in response to a cue without prior experience.

10. Which of the following true about Kreb's cycle?

(a) Kreb's cycle generate NADPH

(b) the enzymes of Kreb's cycle reside in the inter-membrane space of a mitochondria.

(c) it produces ATP, the energy currently of a cell

(d) None of the above

Ans. (c)

Sol. The citric acid cycle or Krebs cycle is carried out in the mitochondria. While both the metabolic pathways use the same substrate, that is, acetyl CoA, the glyoxylate cycle condenses it into a C₄ dicarboxylic acid (succinate) and Krebs cycle leads to oxidation of acetyl CoA into two molecules of CO₂ and produces ATP.

11. A genetic experiment was performed to map the gene(s) for eye colour in a newly-discovered moth species. Sex determination in this moth species: XY - male and XX - female. When blue-eyed males were mated to green-eyed females, all of both male and female progeny had green eyes. When these progeny were mated among themselves, about half of the males of the resulting second generation had blue eyes; however, all females were green-eyed. Which one of the following is consistent with the above data?

(a) Multiple genes control eye colour in this moth species

(b) Gene(s) for eye colour is located on the X chromosome

(c) Gene(s) for eye colour is located on the Y chromosome

(d) Gene(s) for eye colour may not be sex-linked

Ans. (b)

Sol. Given, a genetic experiment was performed to map the gene(s) for eye color in a newly discovered moth species. Sex dermination in this moth species : XY-male and XX-female. When blue-eyed males wee mated to green-eyed females, all of both male and female progeny had green eyes. When these progenies were mated among themselves, about half of the males of the resulting second generation had blue eyes; however, all females were green-eyed. This pattern was linked to the sex chromosomes and was called sex-linked to the sex chromosomes and was called sex-linked inheritance. Male and female offspring from a cross may show different phenotypes if the traits is X-linked. Male inherits its X-chromosome from the mother, while female inherits its X chromosome from both its mother and father. X-linked and autosomal genes assort independently. When genes assort independently, the probabilities associated with the components of the complete genotype are multiplied.

12. In a newly discovered organism, normal development was unaffected when a few blastcimeres were removed from 100-cell stage embryo. However, removal of five cells at the 1000-cell stage abolished the formation of kidney. Which one of the following options most accurately describes the type(s) of specification operating in the development of this organism?

(a) Conditional specification only

(b) Autonomous specification only

(c) Conditional and autonomous specifications

(d) Specification does not occur in this organism

Ans. (a)

Sol. The mode of commitment is sometimes called conditional specification, if the fate of a cell depends upon the conditions in which the cell finds itself. If a blastomere is removed from an early embryo that uses conditional specification, the remaining embryonic cells alter their fates so that the roles of the missing cells can be taken over. This ability of the embryonic cells to change their fates to compensate for the missing parts is called regulation.

13. In which one of the following organisms, it is easiest to distinguish mutations on adjacent base pairs of DNA through genetic recombination experiments?

(a) Bacteriophages

(b) Yeast

(c) Escherichia coli

(d) Bacillus subtilis

Ans. (a)

Sol. Genetic recombination is the exchange of DNA sequence between molecules; in meiosis, it refers to the exchange of genetic material between homologous chromosomes. E. coli are easiest to distinguish mutations on adjacent base pairs of DNAs through genetic recombination experiments. E. coli is used as a tool for genetic engineering in many recombinant experiments as it is simplest and mapped well. It can be used, modified through genetic recombination for a particular pair of DNA.

14. RNA is considered as the first genetic material to have evolved on the earth. Which one of the following properties of RNA is critical for its functioning as the genetic material in the absence of DNA and protein?

(a) The presence of uracil as a base in place of thymine

(b) The RNA is less stable than DNA; therefore RNA has higher probability to evolve as genetic material as compared to DNA

(c) The single stranded RNA has a genotype as well as phenotype

(d) RNA exists in 3 forms while DNA has only one form

Ans. (c)

Sol. The basic constituents of RNA are same for DNA with two differences. In RNA, the pyrimidine base uracil replaces thymine and ribose replaces deoxyribose. RNA cannot form the B-form helix as its additional 2' hydroxyl restricts the arrangement of the sugars in the phosphate backbone. Thus, RNA exists as single stranded structure and exists in 3 forms mRNA, rRNA and tRNA while DNA has only one form, this property of RNA is critical for its functioning as the genetic material in the absence of DNA and protein.

15. The birth control pills contain hormonal formulations that may either arrest the ovulation or prevent the fertilization of egg. Some of the formulations do both. Which one of the following combinations represents a formulation that is likely to affect the process of ovulation and fertilization?

(a) Progesterone and estrogen

(b) Prostaglandin and estrogen

(c) Gonadotrophin and estradiol

(d) Prolactin and estradiol

Ans. (a)

Sol. Oral contraceptive pill are tablets that contain hormones either progestogens or estrogen designed to prevent pregnancy. Some, called combined oral contraceptives (COCs), contain both progestin (hormone with actions similar to progesterone) and estrogens. The primary action of COCs is to inhibit ovulation by suppressing the gonadotropins follicle stimulating hormone and luteinizing hormone, thus preventing the development of a dominant follicle in the ovary. As a result, levels of estrogens do not rise, the mid-cycle LH surge does not occur and ovulation does not take place. Even if ovulation does occur, as it does in some cases, COCs may also block implantation in the uterus and inhibit the transport of ova and sperm in the fallopian tubes.

16. Behavioral studies on animals have shown that there is relationship between mechanism of reproduction and male parental care (protecting eggs or the young ones). In aquatic invertebrates, fishes and amphibians for example, the species that practice internal fertilization rarely show male parental care while a majority of species that practice external fertilization tend to exhibit male parental care. This is likely due to

(a) the male sex in species that practice internal fertilization are unable to defend against the predators

(b) the male sex in species that practice internal fertilization live on female as parasite

(c) the fact that the females of species that practice external fertilization die soon after laying the eggs

(d) the certainty of paternity in species that practice external fertilization and this behavior is reinforced over generation by natural selection

Ans. (d)

Sol. The certainty of paternity is relatively low in most species with internal fertilization because the acts of mating and birth (or mating and egg laying) are separted over time. Certainty of paternity is high when egg laying and mating occur together, as in external fertilization. This may explain why parental care in aquatic invertebrates, fishes, and amphibians, when it occurs at all, is at least as likely to be males as by females.

17. The term biological magnification refers to the increased levels of a toxin seen in successive trophic levels in a food web. Which one of the following options correctly states the reason(s) for the increment of a toxin in the ecosystem?

(a) The toxin is highly toxic to primary producers, relatively less toxic to primary consumers, and non-toxic to secondary consumers. Thus, a higher level of toxin is seen in species representing higher trophic levels

(b) The toxin cannot be degraded by microorganism and consequently persist in the environment for years

(c) The toxin to begin with was not toxic or less toxic, but became more toxic by metabolism in the primary producers

(d) Both (b) and (c)

Ans. (b)

Sol. Biomagnification is the accumulation or increase in the concentration of a substance in living tissue as it moves through a food web (also known as bioaccumulation). These pollutants are non-biodegradable and can be absorbed into living systems. The pollutant of this type, accumulated into the body of an organism, cannot be metabolized or excreted and is passed onto the organisms of the higher trophic level. Heavy metals tend to be stored (accumulating with time) in fatty body tissue. These heavy metals cause significant ecosystem damage and human health problems.

18. From the point of view of the enzymatic reactions, which of the following does not belong here?

(a) Telomerase

(b) Reverse transcriptase

(c) Taq polymerase

(d) Primase

Ans. (d)

Sol. Telomerase enzyme maintains the telomeres at the ends of the chromosomes.

Reverse transcriptase enzyme reads RNA and makes a DNA copy.

DNA polymerase I enzyme from Thermus aquaticus named Taq I or Taq polymerse is generally used to carry out the amplification.

Primase enzyme synthesizes RNA primers.

Reverse transcription, Taq polymerase and primase all of them are used in DNA replication and telomerase does not.

19. Which of the following statements is/are true about juxtacrine signaling?

I. The ligand and the receptor engage in reciprocal signaling

IL Both the ligand and the receptor are membrane associated proteins

III. The ligand gets proteolytically cleaved after binding to the receptor

(a) I only

(b) II only

(c) III only

(d) I, II and III

Ans. (d)

Sol. Juxtacrine signaling is contact dependent. In this, a cell places a specific ligand on the surface of its membrane, and subsequently another cell can bind it with an appropriate cell surface receptor or cell adhesion molecule. Upon ligand binding, the receptor is proteolytically cleaved in a stepwise manner, releasing the intracellular domain into the cytosol where it is translocated to the nucleus. Notch signaling pathway is a significant example for this which is involved in neural development.

20. Which of the following amino acid change (mutation) would MOST adversely affect the structure of an a-helix?

(a) A valine residue changed to an isoleucine residue

(b) A methionine residue changed to a proline residue

(c) An aspartic acid residue changed to a glutamic acid residue

(d) A histidine residue changed to an arganine residue

Ans. (b)

Sol. Methionine is an essential amino acid that is and antioxidant, good source of sulphur and helps inactivate free radicals, it helps in digestion and detoxifies the body of lead and other heavy metals; it prevents build-up of fat in liver and arteries and helps in the treatment of rheumatic fever. Methionine residues contribute to the tertiary structure of folding.