1. F or the complete combustion of graphite and diamond in oxygen individually, the standard enthalpy change values are –393.5 kJ mol–1 and –395.4 kJ mol–1, respectively. Then, the for the conversion of graphite into diamond is

(a) +1.9 kJ mol–1

(b) –1.9 kJ mol–1

(c) +3.8 kJ mol–1

(d) –3.8 kJ mol–1

Ans. (a)

Sol. Given

By reversing the reaction (2), we get

From reaction (1) and (3), we get

Overall reaction: C(graphite) C(diamond);

The enthalpy of the overall reaction is the sum of the enthalpies of reaction (1) and (3)

2. F or a 4s orbital of hydrogen atom, the magnetic quantum number (ml) is

(a) 4

(b) 3

(c) 1

(d) 0

Ans. (d)

Sol. For H-atom, the electronic configuration of hydrogen atom is 1s1. Hence, the number of electrons present in 4s-orbital is zero, that is, 4s0.

Therefore, azimuthal quantum number (l) = 0

Since, ml = –l to +l, thus, ml = 0

3. H ybridization of xenon in XeF2 is

(a) sp

(b) sp2

(c) sp3

(d) sp3d

Ans. (d)

Sol. The hybridization of XeF2 is

Hbd. =

Here, V = Valence electrons of the central atom

M = Number of monovalent atom bonded

C = Cationic charge on the species

A = Anionic charge on the species

Hyb.XeF2 =

The geometry of XeF2 is trigonal bipyramidal and shape is linear.

4. Two equivalents of P react with one equivalent of Q to produce a major product R.

The number of double bonds present in the major product R is ______.

Ans. (11)

Sol. The given reaction si an example of Wittig reaction, in which aldehydes groups reduced to alkene by reacting with Witting reagent (triphenyl phosphonium ylide). The reaction involved is

Correct answer is (11)

5. The total number of possible stereoisomers for the compound with the structural formula CH3CH(OH)CH = CHCH2CH is _____.

Ans. (4)

Sol. The given compound contains one asymmetric carbon (chiral centre) and one double bond, therefore, both optical and geometrical are possible.

Therefore, for one chiral centre, two optical isomers R and S and for one double bond can give two isomers, that is, cis and trans isomers. Thus, total four stereoisomers are possible for the given compound.

6. Among B–H, C–H, N–H and Si–H bonds in BH3, CH4, NH4 and SiH4, respectively, the polarity of the bond which is shown incorrectly is

(a)

(b)

(c)

(d)

Ans. (d)

Sol. The electronegativity of Si is less as compared to H. So, hydrogen gets partial negative charge and Si gets partial positive charge, that is, .

7. Among the following statements,

I. [NiCl4]2– (atomic number of Ni = 28) is diamagnetic

II. Ethylamine is a weaker Lewis base compared to pyridine

III. [NiCl2{P(C6H5)3}2] has two geometrical isomers

IV. Bond angle in H2O is greater than that in H2S.

The correct one is

(a) I

(b) II

(c) III

(d) IV

Ans. (d)

Sol. I. Oxidation state of Ni in [NiCl4]2– is +2 and electronic configuration is [Ar]3d84s0. Thus, it contains unpaired electrons and hence [NiCl4]2– is paramagnetic not diamagnetic.

II. In pyrimidine, nitrogen is sp2 hybridized whereas in CH3–CH2–NH2, nitrogen is sp3 hybridized, therefore, CH3–CH2–NH2 is stronger Lewis base than pyrimidine.

III. Since, [NiCl2{P(C6H5)3}2] has tetrahedral shape, therefore, it does not show geometrical isomers

IV. H2O has bond angle around 104°, as H2S being Drago's compound which have an bond angle around 90°.

8. In [Mn(H2O)6]2+ (atomic number of Mn = 25), the d-d transitions according to crystal field theory (CFT) are

(a) Laporte forbidden and spin forbidden.

(b) Laporte allowed and spin allowed.

(c) Laporte forbidden and spin allowed.

(d) Laporte allowed and spin forbidden.

Ans. (a)

Sol. In [Mn(H2O)6]2+, Mn(II) has electronic configuration [Ar]3d54s0.

So, it is d-d transition, thus = 0, Laporte forbidden and spin multiplicigy changes, 0, spin-forbidden.

9. The major product m in the reaction.

(a)

(b)

(c)

(d)

Ans. (d)

Sol. The reaction involved is

10.

(a)

(b)

(c)

(d)

Ans. (a)

Sol. The reaction is involved is Hofmann elimination, which results in formation of least substituted alkene.

11. The compound, which upon ono-nitration using a mixture of HNO3 and H2SO4, does not give the meta-isomer as the major product, is

(a)

(b)

(c)

(d)

Ans. (c)

Sol. The substituent (–NHCOCH3) is a mild ring activator towards electrophilic substitution reaction. Therefore, upon mono-nitration, acetanilide will give major orthopara product.

Correct answer is (c)

12. T he standard reduction potential (E°) for the conversion of Cr2O72– to Cr3+ at 25°C in an aqueous solution of pH 3.0 is 1.33 V. The concentrations of Cr2O72– and Cr3+ are 1.0 × 10–4 M and 1.0 × 10–3 M, respectively. Then the potential of this half-cell reaction is (Given: Faraday constant = 96500 C mol–1, Gas constant R = 8.314 J K–1 mol–1)

(a) 1.04 V

(b) 0.94 V

(c) 0.84 V

(d) 0.74 V

Ans. (a)

Sol. Given

Eocell = 1.33V

[Cr3+] = 1 × 10–4M

[Cr2O72–] = 1 × 10–3M

At pH = 3

–log[H+] = 3

[H+] = 10–3 M

For the reaction

Cr2O72– + 14H+ + 6e– 2Cr3+ + 7H2O

Number of e– transferred (n) = 6

Using Nernst equation,

Ecell =

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

Ecell =

=

=

= 1.33 – 0.363= 0.936

Ecell 0.94 V

13. T he solubility product (Ksp) of Mg(OH)2 at 25°C is 5.6 × 10–11. Its solubility in water is S × 10–2 g/L–1, where the value of S is _______ (up to two decimal places). (Given: Molecular weight of Mg(OH)2 = 58.3 g mol–1)

Ans. (1.39)

Sol. For Mg(OH)2, the solubility product is

Ksp = S × (2S)2

= 4S3

Where, S is in mol L–1

Ksp = 5.6 × 10–11 (Given)

5.6 × 10–11 = 4S3

S3 =

S = (1.4 × 10–11)1/3 = 0.00024 mol L–1

Solubility in water is

= (0.00024 × 58.3) gL–1

= 0.0139 g L–1 or 1.39 × 10–2 g L–1

Thus, the value of S is 1.39

14. The activation energy (Ea) values for two reactions carried out at 25°C differ by 5.0 kJ mol–1. If the pre-exponential factors (A1 and A2) for these two reactions are of the same magnitude, the ratio of rate constants (k1/k2) is __________ (up to two decimal places). (Given: Gas constant
R = 8.314 J K–1 mol–1)

Ans. (7.54)

Sol. Given,

= 5.0 kJ mol–1

= 5.0 × 103 J mol–1

From Arrhenius equation at two different temperature,

k1 =

k2 =

Since, A1 = A2

15. One mole of helium gas in an isolated system undergoes a reversible isothermal expansion at 25°C from an initial volume of 2.0 liters to a final volume of 10.0 liters. The change in entropy of the surroundings is ______ J K–1 (up to two decimal places). (Given: Gas constant R = 8.314 J K–1 mol–1

Ans. (–13.38)

Sol. Entropy at isothermal conditions for system

= +13.38 J K–1 mol–1

Therefore, entropy for surroundings,

Biochemistry

1. To which one of the following classes of enzymes does chymotrypsin belong?

(a) Oxidoreductase

(b) Hydrolase

(c) Transferase

(d) Isomerase

Ans. (a)

Sol. Chymotrypsin is an enzyme produced in pancreas. This enzyme catalyzes the hydrolysis of the peptide bonds of proteins in the small intestine. Chymotrypsin also catalyzes the hydrolysis of ester bonds.

2. he substrate saturation profile of an enzyme that follows Michaelis-Menten kinetics is depicted in the figure. What is the order of the reaction in the concentration range between 0.8 and 1.4 M?

(a) Zero

(b) Fraction

(c) First

(d) Second

Ans. (a)

Sol. According to the Michaelis–Menten equation, the reaction velocity versus concentration shows the hyperbolic plot. Michaelis–Menten noted two significany features about the kinetics of the reaction:

At low concentration of the substrate (0.2 M–0.6M), velocity is proportional to the substrate concentration and the enzyme catalyzed reaction is a first-order reaction.

At high substrate concentration (0.8 M–1.4M), velocity becomes independent of substrate concentration since rate is no longer dependent on substrate at high concentration and the enzyme catalyzed reaction is of zero-order kinetics.

3. Which one of the following conformations of glucose is most stable?

(a) Boat

(b) Half chair

(c) Chair

(d) Planar

Ans. (c)

Sol. The most stable conformation of glucose (cyclohexane) is chair because it is free of two molecular strains tha is angle strain and torsional strain due to which the bond between molecules cannot break easily.

4. Which one of the following profiles represents the phenomenon of cooperativity?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. The sigmoidal oxygen binding curve of hemoglobin indicates the interactive binding phenomenon termed cooperativity and in given question, figure C shows the sigmoid curve.

5. Which one of the following amino acids is responsible for the intrinsic fluorescence of proteins?

(a) Pro

(b) Met

(c) His

(d) Trp

Ans. (d)

Sol. Intrinsic protein fluorescence is caused by exciting the protein with 280 nm ultraviolet light and observing at approximately 350 nm. The amino acid tryptophan has the strongest fluorescence quantum yield of the amino acids found in proteins. The rest either do not fluoresce or fluoresce weakly. Therefore "Intrinsic Protein Fluorescence" usually refers to the fluorescence emission of the tryptohpan amino acids.

6. The gycosyaltion of the proteins occurs in ................. .

(a) glyoxysomes

(b) lysosomes

(c) Golgi apparatus

(d) plasma membrane

Ans. (c)

Sol. Protein glycosylation is an enzyme-directed chemical reaction that takes place in the ER (endoplasmic reticulum) and the Golgi Apparatus body of the cell. The N-linkage glycosylations happens in both ER and in the Golgi complex while the O-linked glycosylation only occurs in the Golgi complex.

7. Which one of the following properties of the myeloma cells is used in the hybridoma technology to generate monoclonal antibody?

(a) Lack of thymidylate synthase

(b) Over-expression of hypoxanthine-guanine phosphoribosyl transferase

(c) Over-expression of inosine 5'-monophosphate cyclohydrolase

(d) Lack of hypoxanthine-guanine phosphoribosy transferase

Ans. (d)

Sol. Hybridoma technology is a method of forming hybridomas by the fusing of antibody producing B-cel with myeloma cells. These myeloma cells are cancerous B-cells and are HGPRT (Hypoxanthine Guanine Phospho Ribosyl Transferase) negative due to which they are unable to synthesize nucleotides and cannot survive in HAT (Hypoxanthine Aminopterin Thymidine) medium.

8. The movement of protons through the F0F1-ATPase during mitochondrial respiration is required for ................ .

(a) the increase in pH of mitochondrial matrix.

(b) changing the conformation of F0F1-ATPase to expel the ATP.

(c) importing Pi from inter membrane space.

(d) decreasing the affinity of ADP to F0F1-ATPase

Ans. (b)

Sol. According to the current model to ATP synthesis (known as the alternating catalytic model), the transmembrane potential created by proton (H+) cations supplied by the electron transport chain,. drives the H+ cations from the intermembrane space through the membrane via the F0 region of ATP synthase. A portion of the F0 (the ring of c-subunits) rotates as the protons pass through the membrane. The c-ring is tightly attached to the asymmetric central stalk (consisting primarily of the famma subunit), causing it to rotate within the of F1 causing the 3 catalytic nucleotide binding sites to go through a series of conformational changes that lead to ATP synthesis.

9. The number of NADP+ molecules required to completely oxidize one molecue of glucose to CO2 through pentose phosphate pathway is (correct to integer number).

Ans. (12)

Sol. Pentose phosphate pathway is an alternative pathway to glucose oxidation. The sequence of raections is divided into two phases, i.e., oxidative and non-oxidative phases.

In oxidative phase, the glucose-6-phosphate is converted into 6-phosphogluconolacetone in the presence of NADP dependent enzyme. Then 6-phosphogluconolacetone is hydrolyzed into 6-phosphogluconolacetone in the presence of NADP dependent enzyme. Then 6-phosphogluconolacetone is hydrolyzed into 6-phosphogluconate. Then, 6-phosphogluconate undergoes decarboxylation to form ribulose-5-phosphate. The net result of this oxidative phase is generation of 2 moles of NADPH oxidation of one carbon to carbon dioxide.

The second phase is non-oxidative in which ribulose-5-phosphate is siomerized to form ribose-5-phosphate and xylulose-5-phosphate which undergoes transketolase to form glyceraldehyde-5-phosphate which undergoes transketolase to form glyceraldehyde-3-phosphate and sedoheptulose-7-phosphate which undergoes transaldolase to form fructose which undergoes transaldolase to form fructose-6-phosphate and erythose-4-phosphate. The frutose-6-phosphate which undergoes transaldolase to form fructose-6-phosphate and erythose-4-phosphate. The fructose-6-phosphate and glycerabldehyde-3-phosphate gets converted into glucose-6-phosphate.

Overall raection for the pathway is

Glucose-6-phosphate + 12NADP + 7H2O 6CO2 + 12NADPH + 12H+ + Pi

10. Measurement of the absorbance of a solution containing NADH in a path length of 1 cm cuvette at 340 nm shows the value of 0.31. The molar extinction coefficient of NADH is 6200 M–1 cm–1. The concentration of NADH in the solution is ..................... µM (correct to integer number).

Ans. (50)

Sol. Applying Beer's Lambert's law

A =

Given, A (absrobance of solution) = 0.31

(molar coefficient of NADH) = 6200

C (concentration of the NADH solution) = ?

l (path length of the light) = 1 cm

On putting in the formulas above, we get,

0.31 = 6200 × C × 1

0.31 = 6200 C

6200 C = 0.31

C = 0.31 × 1/6200

C = 0.31 × 0.00161

C = 0.00004991 (approx.) = 50 µM.

11. Among the following regents, which one of the combinations of regent will not break the disulphide bonds in the immunogobulin molecules?

P. Reduced glutathione

Q. Dithiothritol

R. Sodium dodecyl sulphate

S. Methionine

(a) R and S

(b) P and R

(c) P and S

(d) Q and R

Ans. (a)

Sol. Dithiothreitol (DTT) and reduced L-glutathione (GSH) are reducing agents. While, the combination of sodium dodecyl sulphate and methionine will not break the disulphide bonds in the immunoglobulin molecules because methionine is a sulfur containing amino acid and thus is not involved in breaking disulphide bond in the Ig molecules.

12. Match the protein elution condition given in Group I with the appropriate chromatography matrices from Group II.

(a) P – 3, Q – 4, R – 1, S –2

(b) P – 2, Q – 4, R – 1, S – 3

(c) P – 1, Q – 2, R – 3, S – 4

(d) P – 4, Q – 2, R – 3, S – 1

Ans. (a)

Sol. On DEAE-sephacryl column is eluted by increasing the concentration of NaCl to 100 mM in buffer.

The supernatant from ammonium sulphate is applied directly to a Phenyl Sepharose column. Elution is achieved with a double linear gradient with decreasing ammonium sulphate and increasing ethylene glycol.

Histidine binds to Ni-NTA and competes with histidine. This can be used at low concentrations (1–2 mM) to inhibitdine. This can be used at low concentration (1–2 mM) to inhibit non-specific binding and at higher concenttations (>20 mM) to elute the His-tagged protein from the Ni-NTA matrix.

Chromatofocusing utilizes ion exchange resins and is typically performed on fast protein liquid chromatography. It elutes species by decreasing the pH of the buffer.

13. Which one of the following is not a neurotransmitter?

(a) Adrenaline

(b) Glutamate

(c) Histamine

(d) Histidine

Ans. (d)

Sol. Neurotramitters are chemical messengers which are synthesized in the presynaptic terminals of neurons and are released on stimulation. The major known neurotransmitters are acetylcholine, norepinephrine, adrenaline, dopamine, serotonin, glycine, glutamate, aspartate and histamine. Histidine is an -amino acid mainly used in biosynthesis of proteins like histamine which is a neurotransmitter synthesized by the process of decarboxylation.

14. The type II hypersensitivity reaction is mainly mediated by ________.

(a) IgE

(b) IgM

(c) IgA

(d) T-cells

Ans. (b)

Sol. Hypersensitivity is an inappropriate immune response that results in tissue damage. It is of four types : Type I, Type II, Type III and Type IV.

Type I hypersensitivity also known as allergic reaction is induced by antigens known as allergens. Type I hypersensitivity reactions are IgE mediated humoral antibody responses.

Type II hypersensitivity is known as cytotoxic reaction as it results in breaking of host cells and is caused by binding to antibody of class IgM or IgG.

Type III hypersensitivity is mediated by immune complex of IgG antibodies with soluble antigens.

Type IV hypersensitivity is mediated by antigenic specific T-cells.

15. Which one of the following reaction mechanisms drives the conversion of low energy 3-phosphoglyceraldehyde to high energy 1, 3-bisphosphogycerate?

(a) Oxidation without anhydride bond formation.

(b) Oxidation coupled with anhydride bond formation.

(c) Substrate level phosphorylation

(d) Formation of carboxylate

Ans. (b)

Sol. In one of the steps in glycolysis process, 3-phosphoglyceraldehyde is converted into 1, 3 biphosphoglycerate in the presence of enzyme glyceraldehyde 3-phosphate dehydrogenase. In this step, there is oxidation of the aldehyde to carboxylic acid by NAD+ and the two phosphate groups are connected by anhydride bond.

16. A polymerase reaction is carried out for 10 cycles in a volume of 1 mL with 5 molecules of template DNA. Assuming that the efficiency of the reaction is 100%, the number of molecules of DNA present in 100 µL at the end of the reaction is ................ (correct to integer number).

Ans. (512)

Sol. After each PCR cycle, number of molecules increases which can be calculated by applying formula, that is,

Mf = Mi × 2n

where,

The initial number of molecules (templates), Mi = 5

The final number of molecules, Mf

The number of cycles performed, n = 10

Volume, V = 1 mL = 100 µL

On substituting values in the formula, we get

Mf = 5 × 210

Mf = 5120 molecules

Thus, in 1000 µL, number of molecules = 5120

In 1 µL, number of molecules = = 5.12

In 100 µL, number of molecules = 5.12 × 100 = 512

17. The secondary structure topology diagram of 400 amino acid long "protein X" is depicted in the figure. The start and end amino acid residue numbers of each -helix are marked. The percentage (correct to integer number) of residues forming -helix is .................. .

Ans. (20)

Sol.

Amino acid residue for first alpha group is 50, for second alpha groups is 24 and for third alpha group is 5, so the total residue becomes 80 (approx.). We have to calculate percentage residue for 400 amino acids by applying formula

That is,

18. An enzyme is following Michaelis–Menten kinetics with substrate S. The fraction of the maximum velocity (Vmax) will be observed with the substrate concentration [S] = 4KM is (correct to one decimal place). (KM is Michaelis–Menten constant).

Ans. (0.8)

Sol. Michaelis–Menten gave mathematical relationship concept between enzyme (E), substrate (S), maximum velocity (Vmax) and Michaelis constant (KM) that is,

It's given

S = 4KM

Putting this in the equation above, we get,

19. The mass spectrum of benzoic acid will generate the fragment as a base peak (100% relative abundance) of m/z (mass to charge ratio) at ..................... (correct to integer number).

Ans. (105)

Sol. Mass of benzoic acid [C6H5COOH] = 122

Now, it can exist in two forms

For monosubstituted aromatics such as for benzoic acid a peak is expected at m/z = 77 corresponding such as for benzoic acid a peak is often observed, and it is present in the spectrum of benzoic acid. However, bond breaking ocuurs more frequently one bond away from the benzene ring (see fragment at 105).

20. The standard free energy values of reactions catalyzed by citrate lyase and citrate synthetase are 670 and –8192 cal mol–1, respectively.

The standard free energy (in cal mol–1) of acetyl-CoA hydrolysis is _________ (correct to integer number).

Ans. (–8862)

Sol. Given,

On adding reaction (i) with (ii), we get

Acetyl CoA + +H2O Acetate + CoA

The standard free energy for the reaction will be

= G1 + G2

On putting the values in the formula above, we get

= –670 + (–8192)

= –8862 cal mol–1

Botany

1. Which of the following genera produces dimorphic seeds that help to broaden the time of germination in a variable habitat?

(a) Xanthium

(b) Pisum

(c) Mangifera

(d) Linum

Ans. (a)

Sol. Mnay species of Xanthium (Family Asteracea) produce dimorphic seeds as an ecological adaptation. Each capitulum produces two seeds per fruit. The upper one is larger, usually non-dormant while the lower one is smaller and exhibits varying degree of innate dormancy. This difference in dormancy helps in maintenance of several generations of population in a field. This contributes greatly to their presence of overwhelming weeds in a field.

2. The genes for micro RNA (miRNA) in plants are usually transcribed by.

(a) RNA polymerase I

(b) RNA polymerase II

(c) RNA polymerase III

(d) RNA polymerase IV

Ans. (b)

Sol. Eukaryotes have multiple types of nuclear RNA polymerase, each responsible for synthesis of a distinct subset of RNA. The miRNA are a class of non-coding RNA that function as a guide in several gene silencing mechanisms. It has been documented that the primary miRNA transcripts (pri-miRNA) has cap and poly A tail which is a characteristics of class II transcripts. Also, on subjecting the miRNA silenced cell lines to -amanitin show decreased level of pri-miRNAs due to inhibition of pol II activity. Also, chromatin immunoprecipitation analyses indicated pol II has physical association with a miRNA promoter. Thus, it's the major polymerase to transcribe mi-RNA.

3. Which of the statements is true for transposable elements Ac and Ds?

(a) Both Ac and Ds are autonomous because they encode their own transposease.

(b) Both Ac and Ds are non-atutonomous because they do encode their own transposease.

(c) Only Ac is autonomous because it encodes its own transposase.

(d) Only Ds is autonomous because it encodes its own transposase.

Ans. (c)

Sol. Transposable elements are the "Jumping genes" that can change their position in the genome. Ac/Ds transposable elements are present in maize. The Ac activator element is autonomous, whereas the Ds dissociation element requires an activator element to transpose. In the presence of an activator (Ac) element, a dissociator (Ds) element can cause chromosome breaks. The nono-autonomous maize Ds transposons can only move in the presence of the autonomous element Ac which comprise a heterogeneous group the share 11-bp terminal inverted repeats (TIRs) and some subterminal repeats but vary greatly in size and composition.

4. Identify the correct statement.

(a) Receptor like kinases play role in gametophytic self incompatibility in Brassicaceae.

(b) Receptor like kinases play role in sporophytic self-incompatibility in Solanaceae.

(c) Ribonucleases play role in sporophytic self incompatibility in Brassicaceae.

(d) Ribonucleases play role in gametophytic self incompatibility in Solanaceae.

Ans. (d)

Sol. Self-incompatibility encompasses a collection of disparate systems that have distinct evolutionary histories and are based on mechanistically different strategies for the inhibition of self-related pollen. Once strategy used by members of the Solanaceae, Rosaceae, Scrophulariaceae and Campanulaceae is directed at inhibiting pollen tubes (Gametophytic incompatibility) after they have grown into the style and is based on the cytotoxic activity of stylar secreted S-RNases, which act inside the pollen tube to inhibit its growth.

Other strategies, such as those used by the Brassicaceae and Papavaraceae are directed at preventing pollen germination or pollen tube ingress into the pistil; in this strategy, self-pollen is inhibited at the stigma surface within minutes of pollen-stigma contact. Self-incopatibility in Brassica is sporophytically controlled by the polymorphic S locus. Two tightly linked polymorphic genes at the S locus, S receptor kinase gene (SRK) and S locus glycoprotein gene (SLG) are specifically expressed and cause incompatibility.

5. Which of the following statement is true for an ecotone?

(a) An ecotone is the synonym of an ecosystem.

(b) An ecotone is an interface zone of two or more ecosystems.

(c) An ecotone is a special feature of land biomass.

(d) An ecotone is exclusively characterized by decreased biodiversity.

Ans. (b)

Sol. An ecotene is a transition area between two biomes. A place where two communities meet and integrate. It may be narrow or wide and it may be local (the zone between a field and forest) ore regional (the transition between forest and grassland ecosystems). An ecotone may appear on the ground as a gradual blending of the two communities across a broad area, or it may manifest itself as a sharp boundary line. An ecotone shows edge effect (changes in population or community structures that occur at the boundary of two or more habitats) and therefore, show greater biodiveristy.

6. Acid rain with a pH of 4.0 is more acidic than the rain with a pH of 6.0 by

(a) 2 times

(b) 10 times

(c) 100 times

(d) 1000 times

Ans. (c)

Sol. The pH of a solution is simply a measure of the concentration of hydrogen ions, H+ calculated using the negative log base 10 of the concentration of the hydronium cations. pH = – log(H3O+).

Difference between pH 4 and pH 6 is two units however, the concentration changes logarithmically. Thus, pH 4 = 10–4 while, pH 6 = 10–6.

Thus, the ratio of pH 4 over pH 6 gives 100. So, acid rain of pH 4 is 100 times more acidic than the acid rain of pH 6.

7. Which of the following plants produces Ylang-yland oil?

(a) Cananga odorata

(b) Carcum copticum

(c) Pandanus odortissimus

(d) Pimenta racemosa

Ans. (a)

Sol. Cananga odorata is also called as the Yalang-Yalng plant; native to Indonesia. The essential oil has a high demand in the perfumery industry is obtained via steam distillation of the flowers.

8. Identify the incorrect statement in connection with polar transport of auxin.

(a) The putative influx carrier AUXI is a cytosolic protein.

(b) Polar auxin transport in root tends to the both acropetal and basipetal in direction.

(c) Naphthyphthalamic acid (NPA) is an inhibitor of polar auxin transport.

(d) AUXI and PIN1 proteins are located in the opposite ends of a cell for polar transport..

Ans. (a)

Sol. Auxin is an important phytohormone, which has polar transport across the plant. The nature of the transport was discovered using inhibitors of poalr transport, like NPA. Polarity of auxin movement is provided by asymmetric localization of auxin carriers (mainly PIN efflux carriers). PIN-FORMED (PIN) and P-GLYCOPROTEIN (PGP) family of protein are major auxin efflux carriers while, AUXIN1/LIKE-AUX1 (AUX/LAX) are major auxin influx carriers. These carriers proteins have transmembrnae spamming domains, via which they adhere to the membranes and facilitate auxin transport.

9. Which of the following stains is used to visualize callose under the microcope?

(a) Alcian blue

(b) Aniline blue

(c) Toluidine blue

(d) Thymol blue

Ans. (b)

Sol. Alcial blue is a cationic dye; usually used to stain positively charged polysaccharides such as glycosaminoglycan of the cartilage.

Aniline blue is used to stain callose, collagen and mason's trichomes. It is also used in fluorescence microscopy, appearing yellow-green when excited by violet light.

Toluidine blue is a basic dye with high affinity for acidic tissue components. It stains nucleic acids blue and polysaccharides purple.

Thymol blue solution is used as a pH indicator.

10. The coding sequence of a gene XLR18 has the single ORF of 783 bp. The approximate molecular weight of the XLR18 protein in KDa is ............... .

Ans. (2.87)

Sol. Since the codons are in triplets, three bases code for a single amino acid thus, 783 bp will be 783/3 = 261 amino acids.

The average weight of an amino acid is 128 Da, but during the formation of a peptide bond there is loss a water molecule thus, average weight of an amino acid in a polypeptide is 110 Da. Thus, mass of the polypeptide XLR18 = 110 × 261 = 28710 Da = 2.87 kDa.

11. Statements given below are either true (T) or false (F). Selected the correct combination.

P. Mitosis occurs exclusively in diploid mother cell.

Q. Mitosis occurs both in diploid and haploid mothe cells.

R. Meiosis occurs exclusively in diploid mother cell.

S. Meiosis occurs both in diploid and haploid mother cells.

(a) P-T, Q-F, R-T, S-F

(b) P-F, Q-T, R-F, S-T

(c) P-T, Q-F, R-F, S-T

(d) P-F, Q-T, R-T, S-F

Ans. (d)

Sol. Mitosis is the equational divison where the chromosomal number reamins same. It occurs in both haploids cells also called fission as well as diploid cells. Whereas, the meiosis is the equational division where the chromosomal number are halved to generate gametes where they fertilize to give back equal number of cells. It occurs only in the diploid cells and not the haploid cells.

12. You are asked to design a genetic construct for high-level expression of a gene encoding the therapeutic protein 18 (TP18) via plasmid transformation. Select the correct set of genetic elements for this construct.

(a) Actin1 promoter TP18 coding sequence Actin1 transcription terminator

(b) Ubiquitin1 promoter TP18 coding sequence Uniquitin1 transcription terminator

(c) rbcS promoter TP18 coding sequenc rbcS transcription terminator

(d) rbcL promoter TP18 coding sequence rbcL transcription terminator

Ans. (d)

Sol. Actin1, Ubiquitin 1, RbcS 1 are the nuclear genomes so the transformation of plastid with these promoters will not yield high levels of expression, whereas, rbcL is encoded by the chloroplast genome and thus, their transformation in the plastid will yield high level of expression.

13. Select the correct combination of the following statements.

P. The cyclic electron transport chain involving PSI results in net production of both ATP and NADPH.

Q. The cyclic electron transport chain involving PSI results in net production ATP.

R. RuBisCO enzyme usually converts RuBP and CO2 into 2-phosphoglycolate and 3-phosphoglycerate.

S. RuBisCO enzyme usually converts RuBP and O2 into 2-phosphoglycolate and 3-phosphoglycerate.

(a) P, Q

(b) R, S

(c) Q, S

(d) P, R

Ans. (c)

Sol. The cyclic flow of electrons that uses photosystem I (P700) and not photosystem II (P680) does not produce any NADPH nor O2. But it does produce ATP and this is known as cyclic phosphorylation.

RuBisCO is an enzyme that has affinity towards both oxygen and carbon dioxide. It has both carboxylation and oxygenation activity. Photorespiration (C-2 cycle), occurs when the CO2 concentration is low, RuBisCO converts RUBP to 2-phosphoglycolate (2-PG) and 3-phosphogycerate.

14. Match the fruit characters with their families and representative plant species.

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

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

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

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

Ans. (b)

Sol.

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

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

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

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

(d) P-4-vi, Q-1-iii, R-3-ii, S-2-v

Ans. (c)

Sol. Black rot is a disease of apples that infects fruit, leaves and bark caused by the fungus Botryosphaeria obtusa.

Loose smut of corn, caused by Ustilago maydis, is easily identified by tumor-like galls.

Fusarium wilt of banana, popularly known as Panama disease, is a lethal fungal disease caused by the soil borne fungus Fusarium oxysporum f. sp. cubense. The fungus enters the plant through the roots and colonizes the xylem vessels thereby blocking the flow of water and nutrients.

Bacterial fruit blotch (BFB) is a serious disease infecting watermelon. The causal organism, Acidovorax avenae subsp. Citrulli can cause significant yield losses depending on the stage of growth in which it infects the crop.

16. Select the correct combination by matching Group I with Group II.

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

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

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

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

Ans. (d)

Sol. Photorespiration is defined as a series of light dependent cyclic reactions in which oxygen combines with RuBP with eventual release of carbon dioxide from the plant resulting in net loss of carbon from the cell. The conversion of glycine to serine in the mitochondria by the enzyme glycine-decarboxylase is its key step, which releases CO2, NH3 and reduces NAD to NADH.

Oxidative decarboxylation of 2-oxoglutarate takes places to form succinyl CoA by the enzyme 2-oxoglutarate dehydrogenase. In this reaction, second CO2 and NADH of TCA cycle are produced.

The nitrogen atom that is transferred to -ketoglutatrate in the transamination reaction is converted into free ammonium ion by oxidative deamination. This reaction is catalyzed by glutamate dehydrogenase.

17. Match the plant alkaloids with their and source species

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

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

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

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

Ans. (c)

Sol. P. Codeine - 4. Anticholinergic - iii. Cola nitida

Q. Caffeine - 2. Stimulant - v. Coptis japonica

R. Scopolamine - 1. Analgesic - vi. Senecio jacobaea

S. Vinblastine - 3. Antineoplastic - ii. Catharanthus roseus

So, the correct match is:

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

18. Identify the correct combination of statements with respect to chemical defense in plants.

P. Pisatin, a phytoalexin produced by Ricinus communis is a constitutive defense compound.

Q. Phaseolus vulgaris produces Phaseolus agglutinin I, which is toxic to the cowpea weevil.

R. A single step non-enzymatic hydrolysis of cyanogenic glycoside releases the toxic hydrocyanic acid (HCN) to protect plant against herbivores and pathogens.

S. Avenacin, a triterpenoid saponin from oat prevents infection by Gaeumannomyces graminis, a major pathogen of cereal roots.

(a) P, Q

(b) Q, S

(c) R, S

(d) P, S

Ans. (b)

Sol. Pisatin is the major phytoalexin produced by the pea (Pisum sativum) plant, the first to be purified and clarified.

Cyanogenic glycosides are a group of nitrile-containing plant secondary compounds that yields cyanide (cyanogenesis) following their enzymatic breakdown with the enzyme beta-glucosidase.

19. In garden pea, dwarf plants with terminal flowers are recessive to tall plants with axial flowers. A true breeding tall plant with axial flowers was crossed with a true-breeding dwarf plant with terminal flowers. The resulting F1 plants were test crossed and the following progeny were obtained :

Tall plants with axial flowers = 320

Dwarf plants with terminal flowers = 318

Tall plants with terminal flowers = 79

Dwarf plants with axial flowers = 83

The map distance between the genes for plant height and flower position ............... cM.

Ans. (20.20)

Sol. Map distance in centimorgan is mapped according to the recombination frequency.

Recombination frequency =

Here, tall plants with terminal flowers and dwarf plants with axial flowers are recombinants.

So, recombinational frequency =

Genetic map distance = 0.2025 × 100 = 20.25 cM.

Malonyl-CoA he substrate of fatty acid synthesis is formed by carboxylating acetyl-CoA using the enzyme acetyl-CoA carboxylase.

20. Two true-breeding snapdragon (Antirrhinum majus) plants, one with red flowers ad another with white flowers were crossed. The F1 plants were all pink flowers. When the F1 plants were selfed, they produced three kinds of F2 plants with red, pink and white flowers in a 1:2:1 ratio. The probability that out of the five plants piced up randomly, two would be pink flowers, two with white flowers and one with red flowers is ——————%.

Ans. (50)

Sol. In the F2 generation of the snapdragon cross, the ratio of red:pink:white flowers is indeed 1:2:1. When you randomly pick five plants, there are different combinations that could fulfill the requirement of having two pink, two white, and one red flower.

Here's a breakdown of the possibilities:

1. PPWWr (pink-pink-white-red)

2. PPWWr

3. PPWWr (pink-pink-white-white)

4. PPWWr

5. PWWWr (pink-white-white-red)

6. PWWWr

Out of the six possible combinations, three match the desired pattern.

So, the probability of randomly picking five plants with two pink, two white, and one red flower is 3 out of 6, which simplifies to 0.5 or 50%. This translates to a percentage, so the correct answer is indeed 50%.

Microbiology

1. David Baltimore's classification of viruses is based on differences in

(a) host cell receptors used by viruses

(b) the pathways required to synthesize virus mRNA

(c) the modes of transmission of viruses

(d) the envelope proteins on the surface of viruses

Ans. (b)

Sol. The Baltimore system of virus classification devised by virologist and Nobel David Baltimore is based on the genomic nature of the viruses. The central theme of Baltimore system of virus classification is that all viruses must synthesize positive-strand mRNAs from their genomes, in order to produce proteins and replicate themselves.

2. Which of the following immune system components can function as an opsonin?

(a) Antibodies

(b) T-cell receptors

(c) Histamines

(d) Interferons

Ans. (a)

Sol. An opsonin is any molecule that enhances phagocytosis by marking an antigen for an immune response or marking dead cells for recycling. Antibodies complementproteins and circulating proteins acts as opsonins.

3. The oral polio vaccine (OPV) consists of

(a) live attenuated virus

(b) killed virus

(c) viral toxin

(d) viral capsid subunit

Ans. (a)

Sol. Oral polio vaccine consists of live attenuated poliovirus strains of each of the serotypes, selected by the ability to mimic the immune response following infection with wild poliovirus but with a significantly reduced incidence of spreading to the central nervous system.

4. Which of the following eukaryotic cellular components carries out intracellular degradation during autophagy?

(a) Nucleus

(b) Golgi bodies

(c) Ribosomes

(d) Lysosomes

Ans. (d)

Sol. Lysosomes are also responsible for autophagy, the gradual turnover of the cell's own components. The first step of autophagy appears to be the enclosure of an organelle (e.g., a mitochondrion) in membrane derived from the ER. The resulting vesicle (an autophagosome) then fuses with a lysosome, and its contents are digested.

5. Analysis of DNA sequences suggest that eukaryotic mitochondrial genomes primarily originated from

(a) fungi

(b) protozoa

(c) algae

(d) bacteria

Ans. (d)

Sol. The mitochondrial genome originated from within the bacterial domain of life. Specifically, among extant bacterial phyla, the -proteobacteria are the closest identified relatives of mitochondria, as indicated, for example, by phylogenetic analyses of both protein-coding genes and ribosomal RNA (rRNA) genes specified by mitochondrial DNA (mtDNA).

6. Binomial nomenclature has NOT yet been adopted for

(a) bacteria

(b) fungi

(c) viruses

(d) protozoa

Ans. (c)

Sol. Binomial nomenclature has not been used for International Committee on Taxonomy of Viruses (ICTV) yet, although it does work well in all other areas of life science. Viruses are mainly classified by phenotypic characteristics, such as morphology, nucleic acid type, mode of replication, host organisms and the type of diseases they cause.

7. Which of the following is NOT an accepted method for sterilization?

(a) Autoclaving

(b) X-rays

(c) Gamma rays

(d) UV rays

Ans. (d)

Sol. A major disadvantage of UV light as a disinfectant is that the radiation is not very penetrating. So, the organism to be killed must be directly exposed to the rays. It may not be effective against all bacterial spores. Another potential problem is that UV light can damage human eyes and prolonged exposure can cause burns and skin cancer in humans. And it may cause damage in human skin cells and permanent damage the eyes.

8. The primary product of nitrogen fixation is

(a) N2

(b) NH4+

(c) NO2–

(d) NO3

Ans. (c)

Sol. Nitrogen fixation is a process by which nitrogen (N2) in the atmosphere is converted primarily into ammonia (NH3). It can be represented by the following equation, in which two moles of ammonia are produced from one mole of nitrogen gas, at the expense of 16 moles of ATP and a supply of electrons and protons (hydrogen ions) :

N2 + 8H+ + 8e– + 16ATP 2NH3 + H2 + 16ADP + 16Pi

9. In humans, the key stages in the life cycle of malarial parasites occur in

(a) red blood cells and the liver

(b) red blood cells and platelets

(c) red blood cells and the pancreas

(d) red blood cells and the gut

Ans. (a)

Sol. The malaria parasite lifecycle involves two hosts. During a blood meal, a malaria-infected female Anopheles mosquito inoculates sporozoites into the human host. Sporozoites infect liver cells and mature into schizonts, which rupture and release merozoites. After this initial replication in the liver, the parasites undergo asexual multiplication in the erythrocytes. Merozoites infect red blood cells. The ring stage trophozoites mature into schizonts, which rupture releasing merozoites. Some parasites differentiate into sexual erythrocytic stages (gametocytes). Blood stage parasites are responsible for the clinical manifestations of the disease.

10. You have a 50 mg mL–1 stock solution of arginine. To prepare 1 liter of growth medium for an arginine auxotroph that requires 70 µg mL–1 of arginine, the volume of this stock solution that should be added is ................... mL (up to 1 decimal point).

Ans. (1.4)

Sol. Given, concenctration, M1 = 50 mg mL–1

Volume to be prepared, V2 = 1 L

Desired concentration, M2 = 70 µg mL–1 = 70 × 10–3 mg mL–1

We know,

M1 × V1 = M2 × V2

50 × V1 = 70 × 10–3 × 1000

V1 = 70/50 = 1.4 mL

11. Accumulating evidence suggest that Domain Archaea is more closely related to Domain Eukarya than to Domain Bacteria. Which of the following properties are shared between eukaryotes and archaea ?

P. Protein biogenesis

Q. Presence of sterol containing membranes

R. Ribosomal subunit structures

S. Adaptation to extreme environmental conditions

T. Fatty acids with ester linkages in the cell membrane

(a) Q, R and T

(b) P, Q, S and T

(c) P and R

(d) R and S

Ans. (c)

Sol. Archaea possess genes and several mtabolic pathways that are more closely related to those of eukaryotes than prokaryotes. All archaea replicate their DNA and synthesize proteins using molecular machines like those of eukaryotes.

12. Match the antimicrobial agents in group I with their category/mode of action in group II.

(a) (i)-(q), (ii)-(s), (iii)-(r), (iv)-(p)

(b) (i)-(s), (ii)-(r), (iii)-(p), (iv)-(q)

(c) (i)-(r), (ii)-(s), (iii)-(q), (iv)-(p)

(d) (i)-(s), (ii)-(r), (iii)-(q), (iv)-(p)

Ans. (c)

Sol. The fluoroquinolones are the only direct inhibitors of DNA synthesis; by binding to the enzyme DNA complex, they stabilize DNA strand breaks created by DNA gyrase and topoisomerase IV.

Amphotoericin B binds with ergosterol, a component of fungal cell membranes, forming pores that cause rapid leakage of monovalent ions and subsequent fungal cell death.

Tetracycline antibiotics are protein synthesis inhibitors. They inhibit the initiation of translation in variety of ways by binding to the 30S ribosomal subunit. They inhibit the binding of aminoacyl tRNA to the mRNA translation complex.

Amoxicillin is in the class of beta-lactam antibiotics. Beta-lactams act by binding to penicillin-binding proteins that inhibit a process called transpeptidation, leading to activation of autolytic enzymes in the bacterial cell wall.

13. Match the microorganisms to their predominant modes of transmission.

(a) (i)-(r), (ii)-(p), (iii)-(s), (iv)-(q)

(b) (i)-(s), (ii)-(q), (iii)-(p), (iv)-(r)

(c) (i)-(q), (ii)-(p), (iii)-(s), (iv)-(r)

(d) (i)-(s), (ii)-(r), (iii)-(p), (iv)-(q)

Ans. (a)

Sol. Bordetella pertussis is highly infectious. It may be spread from person to person by close contact, usually by respiratory aerosols.

Dengue fever is transmitted to humans through the bites of infective female Aedes mosquitoes. When a patient suffering from dengue fever is bitten by a vector mosquito, the mosquito is infected and it may spread the disease by biting other people.

Transmission of Entamoeba histolytica can occur through fecal-oral route (ingestion of food and water, contaminated with feces).

The hepatitis B virus is spread when blood, semen or other body fluid infected with the hepatitis B virus enters the body of a person who is not infected.

14. Match the precursors/intermediates with the corresponding metabolic pathways.

(a) (i)-(q), (ii)-(r), (iii)-(s), (iv)-(p)

(b) (i)-(p), (ii)-(r), (iii)-(s), (iv)-(q)

(c) (i)-(r), (ii)-(p), (iii)-(s), (iv)-(q)

(d) (i)-(r), (ii)-(s), (iii)-(q), (iv)-(p)

Ans. (d)

Sol. In the de novo pathway for purine synthesis, the first purine product is inosine monophosphate (IMP), that serves as a precursor to AMP and GMP.

In L-arginine biosynthesis via L-ornithine, the key intermediate N-acetyl-L-glutamate (NAG) is formed by the enzyme N-acetyltransferase.

Chorismate is an important intermediate that leads to the biosynthesis of several essential metabolites, including the aromatic group amino acids L-phenylalanine, L-tyrosine and L-tryptophan, vitamins E and K, ubiquinone and certain siderophore.

Homocystein is an intermediate in the metabolism of the amino acid methionine. It methylates to methionine.

15. Match the scientists to their area of major contribution

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

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

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

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

Ans. (b)

Sol. Antonie Van Leeuwenhoek discovered "protozoa" the single celled organisms and he called them "animalcules". He also improved the microscope and laid foundation for microbiology.

Carolus Linnaeus is the father of taxonomy, which is the system of classifying and naming organisms. One of his contributions was the development of a hierarchical system of classification of nature.

Sir Alexander Fleming is credited with the discovery of antimicrobial agent penicillin in 1928.

Louis Pasteur demonstrated that microorganisms cause disease and discovered how to make vaccines from weakened or attenuated microbes. He developed the earliest vaccines against fowl cholera, anthrax and rabies.

16. Which of the following combinations would improve the resolution of a microscope?

(i) Increasing the half aperture angle of the objective lens

(ii) Decreasing the wavelength of the illumination source

(iii) Decreasing the numerical aperture of the objective lens

(iv) Decreasing the refractive index of immersion medium

(a) (i) and (ii)

(b) (ii) and (iii)

(c) (ii) and (iv)

(d) (i) and (iii)

Ans. (a)

Sol. The resolution of a microscope is intrinsically linked to the numerical aperture (NA) of the optical components as well as the wavelength of light which is used to examine a specimen.

The higher the value of numerical aperture, the higher the resolution.

The shorter the wavelength, the higher the resolution.

17. Active transport involves the movement of a biomolecule against a concentration gradient across the cell membrane using metabolic energy. If the extracellular concentration of a biomolecule is 0.005M and its intracellular concentration is 0.5M, the least amount of energy that the cell would need to spend to transport this biomolecule from the outside to the inside of the cell is ................... kcal mol–1 (up to 2 decimal points). (Temperature T = 298K and universal gas constant R = 1.98 cal mol–1 K–1)

Ans. (2.717)

Sol. Given,

Concentration outside, Co = 0.005 M

Concentration inside, Ci = 0.5 M

Temperature, T = 298 K

Universal gas constant, R = 1.98 cal mol–1 K–1

Equilibrium gas constant, K = = 0.01

Now, Amount of energy required, = –RT ln K

= – (1.98) (298) ln 0.01

= 2717 cal mol–1 = 2.717 kcal mol–1

18. A continuous cell culture being carried out in a stirred tank reactor is described in terms of its cell mass concentration X and substrate concentration S. The concentration of the substrate in the sterile feed stream is SF = 10 g L–1 and yield coefficient Yx/s = 0.5. The flow rates of the feed stream and the exit stream are equal (F = 5 mL min–1) and constant. If the specific growth rate
the steady state concentration of S is ................... g L–1 (up to 1 decimal point).

Ans. (0.5)

Sol. Given,

Concentration of the substrate in the sterile feed, SF = 10 g L–1

Yield, Yx/s = 0.05

Flow rate, F = 5 mL min–1 = 0.3 L h–1

Volume, V = 3L

S = 0.5 g L–1

19. The initial concentration of cells (N0) growing unrestricted in a culture is 1.0 × 106 cells mL–1. If the specific growth rate (µ) of the cells is 0.1 h–1, the time required for the cell concentration to become 1.0 × 108 cells mL-–1 is ................... hours (up to 2 decimal points)

Ans. (46.05)

Sol. Given,

Initial concentration of cells, N0 = 1.0 × 106 cells mL–1

Final concentration of cells, Nt = 1.0 × 108 cells mL–1

Growth rate, µ = 0.1 h–1

We know,

µ =

Integrating on both sides we get

t =

=

On solving this, we get,

t = 46.05 h

20. The following stoichiometric equation represents the conversion of glucose to lactic acid in a cell:

Glucose + 2Pi + 2ADP 2Lactate + 2ATP +2H2O

If the free energy of conversion of glucose to lactic acid only is = –47000 cal mol–1, the efficiency of energy transfer is _______ % (up to 1 decimal point).

( for ATP hydrolysis is –7.3 kcal/mol).

Ans. (31.06)

Sol. According to the give equation :

1 mole of glucose will hydrolyze 2 moles of ATP

Total free energy, that is, input energy = – 47000 cal mol–1 = – 47 kcal mol–1

Total enregy avaialble = 2 × – 7.3 kcal/mol–1 = – 14.6 kcal mol–1

Therefore,

Efficiency of energy transfer = Output energy/Input energy = 14.6/47 × 100 = 31.06%

Zoology

1. Animals belonging to phylum Echinodermata are closer to chorates than other invertebrate phyla. Which one of the following reasons can account for this relatedness?

(a) Highly evolved nervous system.

(b) Radially symmetric body plan

(c) Deuterostomic development

(d) Well-developed muscles

Ans. (c)

Sol. The echinoderms are different from the other invertebrates as they are deuterostomes rather than protostome. This connects the echinoderms with the chordates. Deuterostomes are one of two distinct evolutionary lines of coelomates that include the echinoderms and chordates. The basic difference between protostomes and deuterostomes arises during embryonic development. In protostomes, the mouth develops from the first indentation of the gastrula and the anus, if present, forms as a second indentation later. In deuterostomes, the anus develops from the first indentation. In deuterostomes, the blastophore becomes the anus. The mouth develops later at the end opposite the blastophore. The mouth and anus are connected by a digestive track.

2. A zoologist recovered some tissue from prserved skin of a woolly mammoth. Further genetic analysis requires DNA isolation and increasing its amount. Which one of the following techniques would be most useful for increasing the amount of DNA?

(a) RELP analysis

(b) Polymerase chain reaction (PCR)

(c) Electroporation

(d) Chromatography

Ans. (b)

Sol. The polymerase chain reaction (PCR) is a technique in which a short region of a DNA molecule is copied many times in vitro in presence of DNA polymerase enzyme. It results in selective amplification of a chosen region of DNA molecule, so it is used for replication of gene or DNA fragment of interest for cloning.

3. In a chemical reaction where the substrate and product are in equilibrium in solution, what will occur if an enzyme is added?

(a) The equilibrium of the reaction will not change.

(b) There will be a decrease in product formed.

(c) Additional substrate will be formed.

(d) The free energy of the system will change.

Ans. (a)

Sol. Some reactions in a cell may be at equilibrium, at least one or often several reactions in a pathway are usually far from equilibrium and therby irreversible. These are the reactions that can be regulated as the metabolic flow through the entire pathway can be increased or decreased by stimulating or inhibiting the activity of the enzymes that catalyze these reactions. When an enzyme is added to a chemical reaction where substrate and product are at equilirbium, there will not be any change in the reaction.

4. Tay-Sachs disease is a human genetic disorder that is associated with defects in which one of the following cellular organelles?

(a) Endoplasmic reticulum

(b) Mitochondria

(c) Golgi apparatus

(d) Lysosome

Ans. (b)

Sol. Tay-Sachs disease is a rare, neurodegenerative disorder in which deficiency of an enzyme (hexosaminidase A) occurs. Lysosomes are the major digestive units in cells. Enzymes within lysosomes break down or digest nutrients, including certain complex carbohydrates and fats. When an enzyme like hexosaminidase A, which are needed to ineffective, they excessively accumulate in the lysosome and are known as gangliosides. This is called abnormal storage. Gangliosides destroy the cell and damaging surrounding tissue.

5. Increase in the existent population of grey peppered moth, Biston betularia, during idustrial revolution in Britain is an example of which ONE of the following evolutionary processes?

(a) Neutral selection

(b) Disruptive selection

(c) Directional selection

(d) Stabilizing selection

Ans. (c)

Sol. The increase in the population of the grey peppered moth, Biston betularia, during the industrial revolution in Britain is an example of (c) directional selection.

Directional selection occurs when a particular extreme phenotype becomes more favorable in a changing environment, leading to a shift in the overall population's characteristics. In the case of the peppered moth, the industrial revolution brought about significant environmental changes, particularly increased pollution from factories.

Before the industrial revolution, the trees and buildings in Britain were covered in light-colored lichens and soot-free, providing a natural camouflage for light-colored peppered moths. However, as pollution from factories increased, the surfaces became darker due to soot deposits, making the light-colored moths more conspicuous to predators against the darker backgrounds.

This change in the environment created a selective pressure in favor of darker-colored moths, as they were better camouflaged against the polluted surfaces. Over time, the proportion of darker-colored moths in the population increased because they had a higher survival rate compared to the lighter-colored moths.

This shift in the population's phenotype toward the darker coloration is an example of directional selection. The selection pressure favors one extreme of the phenotypic range (in this case, darker coloration) due to changes in the environment, leading to a gradual shift in the overall characteristics of the population over generations.

6. Which one of the following is not a characteristic of a cancer cell?

(a) Increase in cell motility

(b) Loss of contact inhibition

(c) Decrease in apoptosis

(d) Uncontrolled meiosis

Ans. (d)

Sol. Increased cell motility is a fundamental characteristic of cancer cells. It is required in order for cells to invade through the basement membrane, represents an initial step in the metastatic cascase and is necessary for cells to move from their primary organ of origin to distant metastatic sites. To divide and grow uncontrollably, a cancer cell not only has to hijack normal cellular growth pathways, but also evade cellular death pathways. Thus, cancer cells and marked by decrease in apoptosis and loss of contact inhibition.

7. Cardiac and cerebral tissues ar derived from the following germ layers respectively

(a) ectoderm and mesoderm

(b) mesoderm and ectoderm

(c) mesoderm and endoderm

(d) endoderm and ectoderm

Ans. (b)

Sol. The primary germ layers (endoderm, mesoderm and ectoderm) are formed and organized in their proper locations during gastrualtion.

Endoderm, the most internal germ layer, forms the lining of the gut and other internal organs.

Mesoderm, the middle germ layer, forms muscle, the skeletal system and the circulatory system.

Ectoderm, the most exterior germ layer, forms skin, brain, the nervous system and other external tissues.

8. An animal's ability to escape from a predator by using the explored knowledge of home area is an example of :

(a) latent learning

(b) insight learning

(c) mimicry

(d) imprinting

Ans. (a)

Sol. Latent learning is ability of an animal to escape by giving them time to create a mental map before a stimulus is introduced from a predator.

9. Bowman's capsuls are present in which one of the following organs/tissues?

(a) Renal cortex

(b) Urinary bladder

(c) Renal medulla

(d) Ureter

Ans. (a)

Sol. At one end of each nephron, in the cortex of the kidney, is a cup-shaped structure called the Bowman's capsule. It surrounds a tuft of capillaries called the glomerulus that carries blood from the renal arteries into the nephron, where plasma is filtered through the capsule. After entering the capsule, the filtered fluid flows along the proximal convoluted tubule to the loop of Henle and then to the distal convoluted tubule and the collecting ducts, which flow into the ureter.

10. Which one of the following is the primary function of lung surfactants?

(a) Remove dust particles from bronchi.

(b) Provide immunity to respiratory tract.

(c) Prevent alveoli from collapsing by decreasing surface tension

(d) Aid in carbon dioxide exchange.

Ans. (c)

Sol. 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. Other functions include, interacting with and subsequent killing of pathogens or preventing their dissemination and modulating immune response.

11. Match the following disorders/diseases listed in Column I to their respective causative agents in Column II.

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

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

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

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

Ans. (b)

Sol. African tick bite fever is caused by Rickettsia sp. The general symptoms may include fever, headache, muscles pains and rashes.

Yellow fever is caused by Favivirus. In mild cases, it causes fever, headache, nausea and vomiting. Serious cases may include fatal heart, liver and kidney conditions.

Microcephaly is caused by Zika virus in which the symptoms vary and include intellectual disability and speech delay. In severe cases, there may be seizures and abnormal muscle functionality.

Sleeping sickness is caused by Trypanosoma gambiense. The initial symptoms are red sore, followed by fever, swollen lymph glands along the back of the neck, shortness of breath, pain the muscles and joints, headaches and irritability. In the later stage, the parasite enters the CNS causing lethargy and unconsciousness.

12. Glucose monomores are joined together by glycosidic linkages to form a cellulose polymer. During this process changes in the free energy, total energy, and entropy respectively are represented correctly by which one of the following options?

(a)

(b)

(c)

(d)

Ans. (d)

Sol. The cellulose polymer chain is derived from D-glucose units, which condense through beta (1-4) glycosidic bonds giving a rigid straight chain having many inter and intramolecular hydrogen bonds among the many hydroxyl groups. When a leaf use carbon dioxide and nutrients to produce cellulose, the randomness and consequently entropy is decreasing . Since the process requires intake of heat thus, enthalpy of formation, is positive. Using , it can be stated that Gibbs free energy is also positive for the process.

13. In Drosophila melanogaster, a mutation in Ultrabithorax, which defines the third segment of the torax or T3 leads to development of four winged flies, as teh halteres develop into a second pair of wings. Which one of the following phenotypes in fly will result from overexpression of Ultrabithorax in the second thoracic segement?

(a) Four winged flies.

(b) Two wings and two halters flies.

(c) Flies with four halters

(d) Flies with two halters.

Ans. (c)

Sol. In Drosophila melangaster, a mutation in Ultrabithorax, which defines the third segment of the thorax or T3 leads to development of four winged flies, as the halteres develop into a second pair of wings. Halters or small wings are present on the metathorax for providing balance during flight. The flies with four halters will result from overexpression of Ultrabithorax in the second thoracic segment.

14. Which one of the following is true in case of respiratory acidosis?

(a) Increased rate of ventilation is a cause of respiratory acidosis.

(b) Blood pH more than 7.

(c) Increased levels of carbon dioxide in blood.

(d) Acidosis can be compensated through reduction of bicarbonate levels in plasma.

Ans. (c)

Sol. When diffusion is slower than normal for example, in emphysema or pulmonary oedema O2 insufficiency (hypoxia) typically occurs before there is significant retention of CO2 (hypercapnia) as a result of which respiratory acidosis occurs in which the hydrogen ions or acid get accumulated thereby reducing the pH.

15. Match the proteins/molecules listed in Column I with the cellular location mentioned in the Column II.

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

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

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

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

Ans. (b)

Sol. The enzyme galactosyltransferase is an adhesion receptor. This enzyme is a resident protein in the Golgi apparatus where it glycosylates proteins.

Cytochrome c oxidase is the last enzyme in the mitochondrial electron transport chain pigments. It accepts an electron from cytochrome c proteins and transfers it to one oxygen molecule to form water.

Clathrin is a protein that plays a major role in the formation of coated vesicles.

The dynamic nature of microtubules results in poles of polymerized and unpolymerized tubulin in the cytosol.

16. In an experiment, nucleus from a Drosophila oocyte was transplanted into the anterior part of another oocyte, at a region opposite to the existing nucleus. Which one of the following phenotypes will be developing egg show?

(a) A ventralized egg with no dorsal appendages.

(b) A dorsalized egg with two dorsal appendages.

(c) A ventralized egg with two dorsal appendages.

(d) A dorsalized egg with four dorsal appendages.

Ans. (d)

Sol. The Drosophila maternal gene gurken participates in a signaling process that occurs between the germ line and the somatic cells (follicle cells) of the ovary. This process is required for correct patterning of the dorsoventral axis of both the egg and the embryo.

Mutations in gurken cause a ventralized phenotype in egg and embryo.

Females containing extra copies of a gurken transgene produce dorsalized eggs and embryos, which is expected if gurken acts as a limiting factor in the dorsoventral patterning process.

17. Match the organism listed in Column I with the features listed in Column II.

(a) I-iii, II-i, III-iv, IV-ii

(b) I-ii, II-iv, III-i, IV-iii

(c) I-iv, II-i, III-ii, IV-iii

(d) I-iv, II-iii, III-ii, IV-i

Ans. (c)

Sol. Tapeworm - Microvilli on Body Surface: Microvilli are tiny, finger-like projections found on the surface of cells. In tapeworms, microvilli increase the surface area of the body, facilitating nutrient absorption from the host's digestive system. This adaptation allows tapeworms, which are parasitic flatworms, to absorb nutrients efficiently.

Jellyfish - Bioluminescence: Bioluminescence is the production and emission of light by living organisms. Jellyfish are known for their ability to emit light through bioluminescence. This light production is typically used for various purposes, such as attracting prey, deterring predators, or communicating with other members of their species.

Trichinella - Viviparous: Viviparity refers to the reproductive method where embryos develop within the mother's body and are then born alive. Trichinella is a genus of parasitic roundworms that exhibit viviparous reproduction. The female Trichinella worm carries and nourishes its developing larvae within its body until they are ready to be released into the host's muscles.

Earthworm - Lateral Heart: Earthworms possess a circulatory system with multiple hearts. The "lateral hearts" are tubular structures found near the front of an earthworm's body segments. These hearts pump blood and distribute it through vessels, helping to circulate oxygen and nutrients to various parts of the worm's body.

Each of these adaptations showcases how different organisms have evolved to thrive in their specific environments and perform vital functions for their survival and reproduction.

18. Which one of the following statements is not part of the classical Darwinian theory of evolution by natural selection?

(a) A trait which is constantly used will get inherited by next generation.

(b) Phenotypic variations exist among the individuals of a population of a species.

(c) Individuals that best fit into a given environment are more likely to survive.

(d) Each population can randomly acquire a distinct and separate suite of variations.

Ans. (a)

Sol. Darwinism is the term given to the explanation offered by Charles Darwin for the origin of species by natural selection. It is also known as Darwinian evolution. The important observations and explanations of Darwinian Theory are :

Species multiply and thus natural populations of organisms increase rapidly as more number of offspring is produced.

Natural selection takes place, as natural resources are limited.

Evolution is gradual, with most populations stable in size, except in the case of seasonal fluctuations.

Species change over time; thus, individuals in a population very extensively in characteristics, with no two individuals being alike.

All organisms have descended from common ancestors and most of the characteristics have been passed from the parent to its offspring.

19. A population of rabbits was determined to have a birth rate of 200 and mortality rate of 50 per year. If the initial population size is 4000 individuals, after 2 years of non-interfered breeding the final population size will be ——————.

Ans. (4550)

Sol. To determine the final population size after 2 years, we need to consider the birth rate and mortality rate.

The birth rate is 200 rabbits per year, and the mortality rate is 50 rabbits per year.

To calculate the final population size, we start with the initial population size of 4000 rabbits and add the births while subtracting the deaths for each year.

After 2 years, the calculation would be: (4000 + (200 – 50)) + (200 – 50) = 4550 rabbits.

Therefore, the final population size after 2 years of non-interfered breeding would be 4550 individuals.

20. In a population which is in Hardy-Weinberg equilibrium, the frequency of occurrence of a disorder caused by recessive allele(q) is 1 in 1100. The frequency of heterozygotes in the population will be ———————. (Give the answer the three decimal places).

Ans. (0.001818)

Sol. In a population in Hardy-Weinberg equilibrium, the frequency of heterozygotes can be calculated using the equation:

2pq = frequency of heterozygotes

Given that the frequency of the recessive allele (q) is 1 in 1100, we can calculate the frequency of the dominant allele (p) as 1 – q.

Let's calculate it step by step:

q = 1/1100 = 0.000909

p = 1 – q = 1 – 0.000909 = 0.999091

Now, we can calculate the frequency of heterozygotes:

2pq = 2 * 0.999091 * 0.000909 0.001818

Therefore, the frequency of heterozygotes in the population is approximately 0.001818