IIT JAM Biology 2018
Previous Year Question Paper with Solution.

1. Which one of the following protozoan parasites belongs to the phylum Apicomplexa?

(a) Toxoplasma gondii

(b) Leishmania donovani

(c) Entamoeba histolytica

(d) Trichomonas vaginalis

Ans. (a)

Sol. Protozoans are divided into four types based on the mode of locomotion. They are :

(a) Amoeboids It has pseudophodia as its locomotory organelle. It includes genera like Amoeba, Entamoeba and Dictyostelium.

(b) Flagellates Flagella is the locomotor organelle. They are divided into two classes depending on their colour

They are

(i) Phytomastigophorea These protozoans contain chlorophyll, e.g., Euglena and Dinoflagellates.

(ii) Zoomastigophorea These protozoans are colourless. Examples include Trichomoan, Hypemastigids and Holomastigotodies, Leishmania.

(c) Ciliates Cillia is the principal locomotory organelle. Paramecium, Stentor, Vorticella and Balantidium are some of its members.

(d) Sporozoans These are non-motile, unicellular, parastitic protists are divided into four-groups which are as follows

(i) Apicomplexa These are spore forming unicellular protists. Examples include Toxoplasma gondii, Plasmodium etc.

(ii) Microsoporidia They are now included in fungi. They have polar filament for penetrating host.

(iii) Ascetosporea They are parasites of animals especially marine invertebrate, e.g., Paramyxids and haplosporids.

(iv) Myxosporidia They are parasites of aquatic animals, e.g., Myxosporea.

2. Which one of the following statements is CORRECT for Mycoplasma?

(a) Their cells are of definite shape

(b) They are resistant to lysis by osmotic shock

(c) Their growth is not inhibited by penicillin

(d) They are nonpathogenic to human.

Ans. (c)

Sol. Mycoplasma is a genus of bacteria which lack cell wall and hence are not of definite shape. Penicillins inhibit bacterial growth by inhibiting cell wall formation, Therefore Mycoplasma remain unaffected as they do not have cell wall.

Several species of mycoplasma are pathogenic to humans. Most common of which is M.pneumoniae which causes pneumonia in human beings.

Mycoplasmas show marked tendency by lysis in low tonicity media as they lack cell wall.

3. Which one of the following organelles is enclosed by a single membrane

(a) Ribosome

(b) Mitochondria

(c) Endoplasmic reticulum

(d) Centrosome

Ans. (c)

Sol. Cell organelles which have double membranes are nucleus, mitochondria, plastids.

Cell organelles having single membrane are SER, RER, Golgi body, central vacuole. ER consists of a network of folded membranes forming sheets tubes or flattened sacs in the cytoplasm

Organelles with no membrane are - Nucleolus, chromatin, ribosomes.

4. Pyramid of energy in a forest ecosystem is

(a) always inverted

(b) dumb-bell shaped

(c) spindle shaped

(d) always upright

Ans. (d)

Sol. An energy pyramid represents the flow of energy in an ecosystem. If follows 10% law, i.e. from first tropic level to second tropic level, only 10% of the total energy is transferred. It is because of this flow that energy pyramid is always upright.

5. In the feedback regulation of an enzyme, the end product binds to the

(a) active site of the enzyme

(b) allosteric site of the enzyme

(c) enzyme-substrate complex

(d) substrate

Ans. (b)

Sol. Feedback inhibition is a form of allosteric regulation in which the final product of a sequence of enzymatic reactions accumulates in abundance. With too much of this product produced, the final product binds to an allosteric site on the first enzyme present in the series of reaction.

6. What is the source of electrons in photosynthesis?

(a) Carbohydrates

(b) Water

(c) CO2

(d) NADH

Ans. (b)

Sol. The source of electrons in green plant and cyanobacterial photosynthesis is water. In photosynthesis, some energy from sun is used to strop electrons from suitable substances such as water producing oxygen gas. The hydrogen freed by splitting of water is used in the creation of two further compounds that serve as short term source of energy.

7. The value of is

(a) 0

(b) 0.75

(c) 1.5

(d) 3

Ans. (c)


Divide the equation by n2

8. Three vectors are as follows

The value of (a + b) . c is

(a) 614

(b) 746

(c) 2

(d) 134

Ans. (a)


9. The logic operation (OR, AND, NOR or NAND) carried out by following circuit is

(a) AND

(b) NOR

(c) OR

(d) NAND

Ans. (c)

Sol. It is combination of NOR and e.g. NOT gate derived from NAND.

Since NOT gate is an electronic circuit that gives an inverted version of the input at its output and at the input of this NOT gate output of NOR gate is present, so result is an 'OR' gate.

10. The reaction of 11-cis-retinal with the lysine residue of a specific protein forms the light-sensitive pigment in the cells of retina. The light-sensitive pigment is an

(a) amide

(b) acid

(c) anhydride

(d) imine

Ans. (d)

Sol. Cis-retinal isomerisation into all trans retinal by light sets oof a series of conformational changes in the opsin eventually leading it to a form called metarhodopsin II which is an imine. In rhodopsin, aldehyde group of retinal is covalently linked to the amino group of a lysine residue on the protein.

11. Viral capsids are made up of morphological subunits called capsomeres. One of the common capsomeres is icosahedral. The icosahedron is a regular polyhedron with

(a) 16 triangular facets and 12 vertices

(b) 20 triangular facets and 12 vertices

(c) 16 triangular facets and 16 vertices

(d) 20 triangular facets and 16 vertices

Ans. (b)

Sol. Viral capsid is made up of regular icoashedral geometry. Regular icosahedral has 30 edges and 20 equilateral triangular faces with five meeting point at each of its 12 vertices. Regular icosahedral provide convex shape to the capsid of the virus.

12. Which of the following feature(s) should be present in a protein to generate strong immune response (antibody production) in an animal?

I. At least one B-cell epitope II. At least one T-cell epitope

III. Proteolytic cleavage site(s)

(a) I only

(b) II and III

(c) I and III

(d) I, II and III

Ans. (d)

Sol. Any protein to be acted as antigen should be able to present itself to immune cells and should be easily available4 for processing. Since immune cells are both B and T lymiophocytes, it will evoke good immune response it it is able to bind to both of them. This is possible if it has both B and T cell epitope. Presence of proteolytic cleavage site (a site which is digested by proteases) will further expose its antigenic properties to the immune system. Hence, all three properties will make it a good antigen.

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

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

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

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

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

Ans. (c)

Sol. Cholera toxin Enterotoxin since it affects the GI tract

Diphtheria toxin Cytotoxin as it enters cell cytoplasm and inhibit protein synthesis.

Lipopolysaccharide Endotoxin as it binds the receptor in monocytes, dendritic cells, macrophages and B-cells which promotes the secretion of pro-inflammatory cytokines and other substances.

Tatanus toxin Neurotoxin as it alters the transmission of neural signals by binding at the presynaptic terminds of the synapse.

14. Proenzyme pepsinogen is secreted from 'P' of gastric mucosa and converted into active enzyme pepsin on exposure to 'Q' secreted from 'R'. Choose the CORRECT combination of P, Q and R.

(a) P - chief cells Q - hydrochloric acid R - oxyntic cells

(b) P - parietal cells Q - enterokinase R - chief cells

(c) P - oxyntic cells Q - hydrochloric acid R - parietal cells

(d) P - peptic cells Q - gastrin R - oxyntic cells

Ans. (a)

Sol. Pepsinogen, a zymogen is secreted by both mucous and chief cells. Once secreted it is activated by hydrocholoric acid secreted by oxynic cells.

15. When bacteria are grown in glucose-depleted media containing high concentration of lactose, expression of lac operon genes is activated by

(a) the binding of lac repressor in the operator site and cAMP-CAP complex in the CAP site

(b) the dissociation of bound lac repressor from the operator site and binding of cAMP-CAP complex in the CAP site

(c) the dissociation of bound lac repressor only from the operator site

(d) the dissociation of both bound lac repressor from operator site and cAMP-CAP complex from CAP site

Ans. (b)

Sol. Lac operon is an inducible operon and is activated only when glucose is absent and lactose is present. For lac operon to be activated, repressor is removed from the operator site once lactose is bound to the repressor protein.

Besides, in the presence of glucose, CAP, required for production of enzymes remains inactive. This CAP is activated by binding of cAMP and cAMP CAP complex led to the transcription of the DNA of the operon. Hence, both dissociation of bound repressor and attachment of cAMP-CAP is required for inducing lac operon.

16. Match the hormones in Group I with their functions in Group II

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

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

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

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

Ans. (b)


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

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

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

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

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

Ans. (c)


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

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

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

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

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

Ans. (d)

Sol. The other name of the different B vitamins are as follows

Vitamin-B1 — Thiamine pyrophosphate

Vitamin B2 — Riboflavin/Flavin mononucleotide

Vitamin B3 — Niacinamide

Vitamin B5 — Panthothenic acid/coenzyme-A

Vitamin B6 — Pyridoxine/pyridoxal phosphate

Vitamin B7 — Biotin

Vitamin B9 — Folates

Vitamin B12 — Cyanocobalamin



(b) x2

(c) 22x

(d) x2x

Ans. (a)


20. The number of three letter words, with or without meaning, which can be formed using letters of the word 'VIRUS' without repetition of letters is

(a) 30

(b) 40

(c) 60

(d) 120

Ans. (c)

Sol. According to fundamental principal of counting.

No. of ways in which first letter can be written is 5

No. of ways in which second letter can be written is 4

No. of ways is which third letter can be written is 3

No. of ways = 5 × 4 × 3 = 60

21. What is the solution of Given C is an arbitrary constant





Ans. (a)


22. The area of an equilateral triangle with sides of length is





Ans. (a)

Sol. Area of equilateral triangle is where 'a' represents its side.

Hence, area have will be

23. Nucleus of a radioactive material can undergo beta decay with half life of 4 minutes. Suppose beta decay starts with 4096 nuclei at t = 0, the number of nuclei left after 20 minutes would be

(a) 1024

(b) 128

(c) 512

(d) 256

Ans. (b)

Sol. The half-life of a ratioactive substance is the time it takes for a given amount of the substance to be reduced by half.

Here t1/2 = 4 min. In t = 20 min

it will reduced by 5 times

So, the amount left will be

24. Which one of the following shows the CORRECT relationship among velocity of light in a medium (v), permittivity of medium and magnetic permeability of medium ?





Ans. (d)

Sol. In a given medium, velocity of light in a medium (v) is inversely proportional to the square root of permittivity and magnetic permeability of the medium.

25. A 30 capacitor is connected to a 240 V, 50 Hz source. If the frequency of the source is changed from 50 Hz to 200 Hz, the capacitive reactance of the capacitor will

(a) increase by a factor of two

(b) increase by a factor of four

(c) decrease by a factor of four

(d) decrease by a factor of two

Ans. (c)

Sol. Since, Reactance of capacitor is inversely proportional to the frequency of the source, it will decrease by a factor of four

26. Match the entries in Group I (Mechanical system) with analogous quantities in Group II (Electrical system)

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

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

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

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

Ans. (d)

Sol. Mechanical-electrical analogies are the representation of mechanical systems as electrical networks. This sort of analogy was given by James C Maxwell in 19th century. These analogies are developed by finding relationships between variables in different domains. According to this analogy, mass is equivalent to inductance, spring constant to reciporocal capacitance, displacement to charge and velocity to current.

27. The achiral molecules among the following (I, II, III and IV) are

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

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

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

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

Ans. (d)

Sol. A molecule is archiral if it is superimposable on its mirror image. Most archiral molecules have a plane of symmetry or a centre of symmetry.

Only in molecule I and IV plane of symmetry is present. In molecule II and III diffeent functional groups are attached making it non-superimoposable.

28. Match the entries in Group I with those in Group II

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

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

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

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

Ans. (a)

Sol. Proline is cyclic amino acid as it contains imine group instead of amino group.

Oxytocin is peptide hormone responsible for ejection of milk.

Aspartame is a dipeptide acting as artificial sweetner.

Penicillin is -lactam antibiotic that interfere in the cell wall formation.

29. Which one of the following statements is CORRECT?

(a) BF3 is a stronger Lewis acid than BI3.

(b) CO and CN are good π-accepting ligands.

(c) cis-Diamminedichloroplatinum (II) has zero dipole moment.

(d) Central atom in BC3 is sp3 hybridized

Ans. (b)

Sol. BF3is weaker lewis acid than Bl3 because empty 'p' orbital of 'B' and orbitals containing lone pair of electrons of 'I' are not of same energy levels due to which the 'P' orbital remains empty in Bl3 whereas in BF3 they are of same energy and thereby undergo back-bonding.

cis-diammine dichloroplatinum (II) is a neutral molecule and cis form always has a dipole moment whereas trans form in such cases, has zero dipole moment.

Central atom in BCl3 is SP2 hybridised as one S and two 'P' orbitals mix here and produce three equivalent orbitals.

Co and CN are good -accepting ligands as they have lone pair of electrons for accepting ligand.

30. In the 1H NMR spectrum, which one of the following compounds will show a triplet?





Ans. (a)

Sol. Proton nuclear magnetic resonance is the applicatiion of nuclear magnetic resonance in NMR spectroscopy with respect to hydrogen-1 nuclei with the molecule of a substance, in order to determine the structure of its molecules. A full H1 atom is called Protium. Since, fig (a) will have hydrogen molecules at three different places, it will show triplet in spectrum anlaysis.

31. Antibody binds to antigen in solution through

(a) ionic interactions

(b) hydrogen bonds

(c) van der Waals interactions

(d) hydrophobic interactions

Ans. (a, b, c, d)

Sol. Antibody binds to antigen in solution by weaker bonds instead of strong covalent bonds. These weak interactions include van der Waals forces, hydrogen or ion-dipole bonds in the shorter distances. Whereas, in long range region, ionic and hydrophobic bonds are important in binding of antigen with antibody.

32. Plasmid mediated antibiotic resistances in bacteria are acquired by

(a) hydrolysis by β-lactamase (penicillin resistance).

(b) expression of aminoglycoside modifying enzyme (kanamycin resistance)

(c) mutation in DNA gyrase (quinolone resistance)

(d) overproduction of dihydrofolate reductase (trimethoprim resistance)

Ans. (a, b)

Sol. Antibiotic resistance can be either plasmid mediated or maintained on the bacterial chromosome. Three fundamental mechanisms of resistance are

(i) Enzymatic degradation of drugs

(ii) Alteration of bacterial proteins that are targets of drugs

(iii) Changes in membrane permeability of bacteria

In first case, penicillin resistance is mediated by production of -lactamase enzyme. Production of antibiotic modifying enzymes and synthesis of antibiotic insensitive strains are other sole mechanisms that confer antibiotic resistance.

Alteration of bacterial targets are the primary resistance mechanisms for antibiotics like trimethoprim. sulphonamides, aminoglycosides, chloramphenicol and quinolone drugs.

Reduced antibiotic penetration is also a resistance mechanism for several other classes of antibiotics including chloramphenicol.

33. Which of the following statements is/are CORRECT for G protein–coupled receptor (GPCR) mediated signaling?

(a) GPCRs contain seven membrane spanning regions

(b) GPCRs are linked to heterotrimeric G protein consisting of subunits.

(c) In the absence of GPCR interacting ligand, subunit of G protein is bound to GTP and complexed with subunits.

(d) In the presence of GPCR interacting ligand, GTP is displaced from subunit of G protein by GDP, GDP bound á subunit dissociates from dimer and activates the effect

Ans. (a, b)

Sol. G-Protein Coupled Receptor (GPCRs) are the largest and most diverse group of membrane receptors in eukaryotes. They consist of single polypeptide that is folded into a globular shape and embedded in a cell's plasma membrane. Seven segments of this moleucle span the entire width of the membrane and are also called seven transmembrane receptors.

G proteins that associate with GPCRs have 3 different subunits , out of which α and β are attached to plasma membrane by lipid anchors.

In the absence of signal, GDP attaches to the α-subunit and the entire GDP-G protein complex binds to the -region near by GPCR. In the presence of signal, GDP is replaced by GTP and it results into dissociation of a protein subunits into two parts viz. GTP- and .

34. Glucose is incubated with enzymes of glycolytic pathway (except pyruvate kinase), gamma 32P-ATP and unlabeled inorganic phosphate. Which of the following products is/are formed?





Ans. (a, b, c, d)

Sol. Glycolytic pathway is as follows

35. In a double stranded DNA, which of the following ratios is/are always equal to 1? A, T, G and C denote the number of bases.

(a) (A+T)/(G+C)

(b) (A+G)/(T+C)

(c) A/G

(d) (G+T)/(A+C)

Ans. (b, d)

Sol. (b,d) In a double-stranded DNA, Chargaff's rules are followed which states that

(i) Purine is always equal to pyrimidine

(ii) Ratio of purine to pyrimidine will be equal to 1.

(iii) A always pair with T and G always pair with C.

Thus, A + G = purine and T + C = pyrimidine

Ratio of (A + G) to (T + C) will be equal to 1.

G + T (where number of T will be equal to number of A)

A + C (where number of C will be equal to number of G).

This ratio (G + T / A + C) is always equal to 1 as it represents number of purine to number of pyrimidines.

36. Consider the equation x3 – 1 = 0. If one of the solutions to this equation is 1, the other solution(s) is/are


(b) i

(c) –i


Ans. (a, d)

Sol. (a,d) x3 –1 = 0

37. Which of the following statements is/are CORRECT regarding self-inductance of a long solenoid having cross sectional area (A), length (l) and having n turns per unit length filled with material of relative permeability ?

(a) It depends on the geometry of solenoid

(b) It does not depend on geometry of solenoid

(c) It depends on cross sectional area of solenoid

(d) It depends on relative permeability of the medium

Ans. (a, c, d)

Sol. (a,c,d) Self-inductance of long solenoid depends on various factors as Given below.

Where L - Coefficient of self induction

l - Length

N - Number of turns per unit length

A - Area of cross section

- Relative permeability of medium

38. If an optician prescribes a corrective lens of power -2.0 D, the required lens

(a) is a concave lens

(b) is a convex lens

(c) has a focal length of +50 cm

(d) has a focal length of -50 cm

Ans. (a, d)


Where D - Diopter (unit for measuring refractive power of lens)

f - Focal length in metres

– Ve - Power indicates concave lens

+ Ve - Power indicates convex lens

– 2 =

f = = – 0.5 m = – 50 cm

39. Which of the following statements is/are CORRECT?

(a) Absorption occurs at all wavelengths if light passes through a given solution.

(b) The efficiency of a photochemical process is often expressed in terms of quantum yield.

(c) The unit of molar extinction coefficient is litre mole-1 cm.

(d) The extent of absorption in a dilute solution would be the same if the concentration is doubled and the path-length of light passing through solution is halved.

Ans. (b, d)

Sol. b,d) Absorption occurs at specific wavelength when light passes through a solution and is given by the equation of Lambert-Beer's law

log = KCl

Where log 5 gives absorption

K - Constant (molar extinction coefficient)

C - Concentration of solution (Mol L–1)

l - Path length (cm)

Unit of K is L mol–1 cm–1

According to second law of photochemistry, efficiency of photochemical process is given by Quantum yield .

40. Which of the following pairs of compounds can be distinguished by iodoform test performed in ammonium hydroxide?

(a) CH3COCH3 and C2H5OH

(b) C2H5OH and CH3OH

(c) CH3COCH3 and C6H5COCH3

(d) C6H5COCH3 and C2H5OH

Ans. (a, d)

Sol. Iodoform test is used to probe the presence of methylketone. This is also the case when testing for specific secondary alcohols containing at least one methyl group in alpha position. It detects the presence of compounds with the structure R COCH3 and alcohols with R CH (OH) CH3. These two structures are present in 'A' and 'D' options.

41. The total number of genetically different types of gametes that will be produced by a heterozygous plant carrying the genotypes AABbCc is _________.

Ans. (4)

Sol. Parent AABbCc

Only 4 gametes are possible.

42. A healthy individual has the cardiac output of 5.5 L and heart rate of 72 beats per minute. The stroke volume of the individual is _________ mL.

Ans. (76)

Sol. Strove volume is the amount of blood pumped by left ventricle per beat


EDV = End systolic volume

43. Both strands of a DNA molecule are labeled with radioactive thymidine and are allowed to duplicate in an environment containing non-radioactive thymidine. The number of DNA molecules that will contain radioactive thymidine after three duplications is _________.

Ans. (2)

Sol. Since, DNA is semiconservative in nature only parental strand will be conserved and nearly synthesised DNA will have unlabeled thymidine.

44. The number of cycles required for complete degradation of Palmitic acid (16 Carbon) by -oxidation is __________.

Ans. (7)

Sol. The process by which fatty acids are oxidised is -oxidation. Carbons are lost as Co-A per cycle of -oxidation. Therefore 7 cycles are required for complete breakdown of a compound with 16 carbons.

45. The value of logn 4–16 is .32. The value of n is _________.

Ans. (2)

Sol. logn 4–16 = – 32

– 16 log4n = – 32

  log4 4 = 2

2n = 4

n = 2

46. The determinant of the matrix is _________.

Ans. (–8)

Sol. (– 8) 1 (6 × 2 – (– 4) – 3 ( 4 + 4) + 0 = 16 – 24 = – 8

47. The equivalent capacitance of following assembly of capacitors is ________ .

Ans. (1)

Sol. Capacitance in parallel is added as C1 = (C1 + C2) (C1 + C2) and in series it is added as

48. The stability of the following carbocation arises from hyperconjugation with _________ number of hydrogen atoms.

Ans. (6)

Sol. Hyperconjugation can occur when lone pairs are formed. Lone pair formation will take place only when H is not directly bounded to carbocation (C+) So, member of 'H' atoms available for hyperconjugation are 6 as 2 methyl groups are attached.

49. Oxidation state of Fe in the complex K3[Fe(CN)5NO] is (+) __________.

Ans. (2)

Sol. Oxidation state of Fe in the given compound is

K3 [Fe (CN)5 NO] + 3 + Fe – 5 + 0 = 0

Fe = 2

50. The mechanism of the following reaction involves the formation of a __________ membered ring.

Ans. (4)

Sol. Since, 3 phenyl groups will be attached to the CH2 of the cyclohexane ring. Resultant compound will be 4 membered ring.

51. The concentration of a purified enzyme is 10 mg/mL. Ten microlitres of the enzyme solution in a total reaction volume of 1 mL catalyses the formation of 20 nanomoles of product in one minute under optimum conditions. The specific activity of the enzyme is _________ unit/mg.

Ans. (0.2)

Sol. It is expressed in µ mol / min / mg. The amount of substrate the enzyme converts per mg protein per unit of time is called specific activity.

Substrate converted = 20 nano moles

Enzyme used = 10 ml in 1 ml

Dilution ratio = 100

Putting in the equation

52. A 100 nucleotide-long single stranded poly-(A) is synthesized from adenosine monophosphate (AMP) at physiological pH. (Atomic mass of C = 12, H = 1, O = 16, P = 31; at physiological pH, Molecular mass of AMP = 345). The molecular mass of the resulting poly-(A) at physiological pH is ___________.

Ans. (32,718)

Sol. The repeated addition of adenosine monophosphate (AMP) at the 3' terminal is called polyadenylation. On adding one AMP to other, one water molecule is also released. Thus for 100 molecules 99 H2O are released.

Molecular mass of 100 AMP's = 100 × 345 = 34500

Molecular mass of 99 H2O = 1782

mass of resultant tail

= 34500 – 1782 = 32,718

53. If a colour-blind woman marries a normal man, the chance that their boy child will be colour-blind is ________%

Ans. (100%)

Sol. (100%) Colourblind woman X Normal man

54. For a 0.1 M aqueous solution of lysine, the pH at which it carries no net charge is ___________. (pKa values for: -carboxylic group = 3.1, -amino group = 8.0, -amino group = 10.8)

Ans. (9.4)

Sol. P1 is the mean value of amino and carboxyl pka's since, it is a dilute solution (0.1 M) of lysine, it will have positive charge overpowering the negative charge. For this reason, we have to ignore pKa of carboxyl group.

55. For a = ________, the following simultaneous equations have an infinite number of solutions:

10+13y = 6

ax +325 =15

Ans. (25)

Sol. Condition for infinite number of solution is

56. If A and B are events such that P(A) = 0.3, P(B) = 0.2 and = 0.45, the value of is ________.

Ans. (0.25)

Sol. (0.25)     P(A) = 0.3

P (B) = 0.2

= 0.45 – 0.20 = 0.25

57. An ultrasound signal of frequency 50 KHz is sent vertically down into a medium. The signal gets reflected from a depth of 25 mm and returns to source 0.00005 seconds after it is emitted. The wavelength of the ultrasound signal in that medium is ________ cm.

Ans. (2)


Where c - Wave speed (m/s)

V - Wavelength (m)

F - Frequency (Hz)

For calculating time, we will divide the given time by 2 as it is giving time for entire journey.

It will cover 25 mm in .00005/2 seconds.

Putting the values in equation

58. The relationship between the applied force F(X) (in Newton) on a body and its displacement X (in metre) is given below. The total amount of work done in moving the body from X = 0 to X = 4 m is ___________ Joule.

Ans. (11)

Sol. Work done = Force × Displacement

Here, work done will be calculated as the area under the graph.

Work done = Area of triangle + Area of rectangle

59. The number of axial C-H bond(s) in the major product (P) of the given reaction is ________.


Ans. (5)

Sol. Product formed will be

In one plane, two C — H axial bonds are present and in second plane 3-CH axial bonds are present. Total 5 axial bonds are present.

60. A first order reaction is 87.5% complete at the end of 30 minutes. The half-life of the reaction is __________ minute(s).

Ans. (10)

Sol. (10) minutes.

Putting this value in t1/2 equation

t1/2 = 10 min.