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601.

In which disease does mosquito transmitted pathogen cause chronic inflammation of lymphatic vessels?(1) Elephantiasis(2) Ascariasis(3) Ringworm disease(4) Amoebiasis

Answer»

Correct option (1) Elephantiasis

Explanation

Elephantiasis is a helminthic disease caused by Wuchereria bancrofti. The infestation is transmitted by female Culex mosquitoes from one individual to the others. The worms live in the lymphatic system. 

Ascariasis is caused by Ascaris lumbricoides. It is an endoparasite of the small intestine of human beings. 

Amoebiasis is caused by Entamoeba histolytica. It lives in the large intestine of humans. Ringworm is a fungal skin disease.

602.

Select the correct match :(1) Alec Jeffreys – Streptococcus pneumoniae(2) Alfred Hershey and  Martha Chase – TMV(3) Matthew Meselson  and F. Stahl– Pisum sativum (4) Francois Jacob and Jacques Monod – Lac operon 

Answer»

Correct Option(4) Francois Jacob and Jacques Monod – Lac operon 

Explanation:

Jacob and Monod (1916) discovered the lac operon. An operon is a part of the genetic material or DNA which acts as a single regulated unit. lt possesses one or more structural genes, an operator gene, a promoter gene, a regulator gene, a repressor gene, and an inducer or corepressor.

Matthew Meselson and F Stahl discovered the semi-conservative mode of DNA replication in E. coli. Alfred Hershey and Martha Chase use T2 Bacteriophage in their experiments to infect E. coli and proved that DNA is the genetic material. Alec Jeffreys (1984) invented the DNA fingerprinting technique. This technique determines nucleotide sequences of certain areas of DNA which are unique to each individual.

603.

Adult human RBCs are enucleated. Which of the following statement(s) is/are most appropriate explanation for this feature ?(a) They do not need to reproduce(b) They are somatic cells(c) They do not metabolize(d) All their internal space is available for oxygen transport(1) only (a)(2) (a), (c) and (d)(b) and (c)(4) only (d)

Answer»

The correct answer is (4)

 In Human RBCs, nucleus get degenerates during maturation and it provide more space for oxygen carrying pigment (Haemoglobin). It lacks many cell organelles including mitochondria so respires anaerobically.

604.

Pick out the mismatched pair(s).I. Wind — Anemophily — MaizeII. Water — Hydrophily — VallisneriaIII. Insects — Entomophily — PoppyIV. Snails — Myrmecophily — LemnaV. Ants — Malacophily — ChrysanthemumChoose the correct option.(a) Only IV(b) Only V(c) IV and V(d) None of these

Answer»

Answer is : (c) IV and V

Pollination by snails is called malacophily and by ants is called myrmecophily, e.g. Anemone nemarosa.

605.

Myelin sheath is produced by :(1) Astrocytes and Schwann cells(2) Oligodendrocytes and Osteoclasts(3) Osteoclasts and Astrocytes(4) Schwann cells and Oligodendrocytes

Answer»

The correct answer is (4)

 Myelin sheath wrapped around the nerve axon. Oligodendrocytes are neuroglial cells which produce myelin sheath in central nervous system while Schwann cell produces myelin sheath in peripheral nervous system.

606.

Coconut fruit is a :(1) Berry(2) Nut(3) Capsule(4) Drupe

Answer»

The correct answer is (3)

Pre-synaptic membrane is involved in the release of neurotransmitter in the chemical synapse. The receptors sites for neurotransmitters are present on postsynaptic membrane of neuron

607.

Attractants and rewards are required for :(1) Entomophily(2) Hydrophily(3) Cleistogamy(4) Anemophily

Answer»

 The correct answer is (1)

Insect pollinated plants provide rewards as edible pollen grain and nectar as usual rewards. In order to materialize and maximize pollination flowers have developed a set of attributes which areaimed at attracting the pollinators called attractants

608.

In which phase of the cell cycle does synthesis of histone proteins take place?(a) Anaphase-I(b) S-phase(c) G1-phase(d) G2-phase

Answer»

Answer is : (b) S-phase

Synthesis of the DNA and the histone proteins takes place during the S-phase (synthetic phase) of interphase in cell cycle.

In this phase, the cell synthesises a replica of its genome by the process of DNA replication. Thus, DNA content becomes doubled.

609.

Arrange the following in correct sequence of basic strength.(a) IV > I > III > II (b) III > I > II > IV (c) II > I > III > IV (d) I > III > II > IV

Answer»

Answer : (d) I > III > II > IV

Piperidine and morpholine both are non-aromatic and N-atom is sp3 hybridised in them, so these are more basic as compared to other two (which are aromatic and have sp2 hybridised N-atom). 

-I-effect of O-atom in morpholine makes it less basic as compared to piperidine. 

Now,

Out of pyridine and pyrolle, pyridine is much stronger base as the lone pair of nitrogen atom in pyridine is not involved in resonance. 

So, 

The order of basic strength is : 

I > III > II > IV.

610.

The ligand N(CH2CH2NH2)3 is :(a) tridentate (b) didentate (c) tetradentate (d) pentadentate

Answer»

Answer : (c) tetradentate 

The ligand N(CH2CH2NH2)3 is tetradentate.

The given ligand have four donor atoms.

They donate four pair of electrons to metal ion.

611.

Figure shows a circuit that contains three identical resistors with resistance R = 9.0 Ω each, two identical inductors with inductance L = 2.0 mH each, and an ideal battery with emf ε = 18 V. The current 'i' through the battery just after the switch closed is:

Answer»

The correct option is (2) 2 A.

Explanation:

When switch in closed, then current will only flow through R2. The surface resistor arm

So, i = i1

Hence, i = i1 = E/R2 = 18/9 = 2 A

612.

Pyruvic acid, the last product of glycolysis is degraded to CO2 and H2O in :– [a] matrix of chloroplasts [b] cytoplasm of mitochondria [c] matix of mitochondria [d] inner membrane of chloroplasts

Answer»

Correct option [c] matix of mitochondria 

613.

The major disadvantage of cosmids is(a) inability to accept more than 40 - 50 kbp of DNA(b) allow the packaging of DNA in phage in vitro(c) also perpetuate in bacteria(d) can produce a complete genome library of 108

Answer»

Answer is : (a) inability to accept more than 40 - 50 kbp of DNA

The major disadvantage of cosmids is their inability to accept more than 40 - 50 kbp of DNA. Cosmids are the vectors, which can accommodate DNA segments upto 45 kbp.

614.

The two important types of secondary structures of proteins are(a) motifs and domains(b) α - helix, β - pleated sheet(c) peptide bond, hydrogen bond(d) R group, amino group

Answer»

Answer is : (b) α - helix, β - pleated sheet

The two important types of secondary structures of proteins are α - helix and β - pleated sheet. β - sheets consist of β - strands connected laterally by atleast two or three backbone H - bonds, forming a generally twisted, pleated sheet. A β - strand is a stretch of polypeptide chain, 3 -10 amino acids long with backbone in an extended conformation.

In α - helix, the chain is spirally coiled, generally in a right - handed manner. This structure is formed through H - bonding in single amino acid chain only.

615.

According to the ‘Blackman’s law of limiting factor’, at a particular time, photosynthesis can be limited by(a) CO2 concentration(b) light(c) Both (a) and (b)(d) Either (a) or (b)

Answer»

Answer is : (d) Either (a) or (b)

According to Blackman’s law, when a process depends on a number of factors, its rate is limited by the pace of the slowest factor. Therefore, at a particular time, photosynthesis can be limited by CO2 concentration or light.

616.

The hepatic portal vein drains blood to liver from:(1) Stomach(2) Kidneys(3) Intestine(4) Heart

Answer»

The correct answer is (3

 In hepatic portal system, hepatic portal vein drains blood to liver from intestine.

617.

An enzyme which has multiple moelcular forms in the same organism is termed as :–[a] coenzyme [b] isoenzyme[c] holoenzyme [d] apoenzyme

Answer»

Correct option [b] isoenzyme

618.

Which of the following statements is correct for ethyne molecule?(a) Ethyne molecule consist of two C - C σ - bonds(b) It has one π - bond(c) It is a non - linear molecule(d) Electron cloud between two C - atoms is cylindrically symmetrical

Answer»

Answer is : (d) Electron cloud between two C - atoms is cylindrically symmetrical

Ethyne molecule is a linear molecule which consists of one C - C σ - bonds, two C - H σ - bonds and two C - C π - bonds. Electron cloud between two C - atoms is cylindrically symmetrical about the internuclear axis.

619.

Consider the following compounds,The correct decreasing order of stability of compounds is(a) I > III > II > IV(b) IV > II > III > I(c) III > II > IV > I(d) IV > I > III > II

Answer»

Answer is :  (b) IV > II > III > I

+M effect of — NH2 > +M of —OH group that disperse the charge of carbocation. Hence, increases the stability —CH3 group shows + I effect, so it will disperse the charge less than —NH2, —OH group whereas —NO2 group shows —M/—I effect due to which the positive charge on the carbocation increases. Hence, stability decreases.

620.

Which of the following is not considered as a part of the tertiary structure of proteins?(a) Hydrogen bonds(b) Electrostatic interactions(c) Hydrophobic effect(d) Disulphide bonds

Answer»

Answer is : (c) Hydrophobic effect

Except hydrophobic effect, all the given are involved in the tertiary structure of proteins.

621.

Molarity of a solution obtained by mixing 800 mL of 0.6 M HCl with 200 mL of 1 M HCl will be(a) 0.4 M(b) 1.6 M(c) 0.68 M(d) 1.68 M

Answer»

Answer is : (c) 0.68 M

M = \(\frac{M_1V_1+M_2V_2}{V_1+V_2}\)

\(=\frac{0.6\times 800+1\times 200}{1000}\)

\(=\frac{480+200}{1000}\)

\(=\frac{680}{1000}\) 

= 0.68 M

622.

Which one is wrongly matched?(a) Gemma cups - Marchantia(b) Biflagellate zoospores - Brown algae(c) Uniflagellate gametes - Polysiphonia(d) Unicellular organism - Chlorella

Answer»

The correct option is(c) Uniflagellate gametes - Polysiphonia

Explanation:

Polysiphonia is a red algae. In it sexual reproduction is of oogamous type. The male sex organ, spermatangium produces non-flagellate male gametes.

In Brown algae, sexual reproduction varies from isogamy, anisogamy to oogamy. In isogamy and anisogamy both the gametes are motile while in oogamy only male gametes are motile. These motile gametes have two unequal laterally attached flagella.

Chlorella is a unicellular organism. It is green algae belonging to class Chlorophyta. In Marchantia, gemma cups are found on its dorsal surface. It contains gammae which help in vegetative propagation.

623.

A plant hormone which allows the seeds to ignore environmental conditions and germinate is(a) abscisic acid(b) ethylene(c) cytokinin(d) gibberellins

Answer»

Answer is : (d) gibberellins

Gibberellins break the dormancy of seeds due to environmental conditions and promote germination. Seed dormancy sets in the seeds to overcome the unfavourable conditions for seedling and germination.

624.

Symptoms alcoholic withdrawal are (a) tranquilizers (b) sedatives (c) opiate (d) stimulants

Answer»

Correct Option (a) tranquilizers

625.

In the processes of ATP synthesis, oligomycin and DCCD act at which place in figrue :(a) A (b) B(c) C (d) D

Answer»

Correct Option (a) A 

626.

Why the bronchioles and lungs in human beings do not collapse even though they lack the cartilaginous rings?(a) Surfactants secreted by the clara cells prevent the collapse(b) The intercostal muscles keep the pleura expanded(c) The elastic cartilage present on their surface prevents them from collapsing(d) None of the above

Answer»

Answer is : (a) Surfactants secreted by the clara cells prevent the collapse

Pulmonary surfactant is a mixture of lipids and proteins which is secreted by the epithelial cells, viz. clara cells. Its main function is to reduce surface tension, thereby preventing the bronchioles and the lungs from collapsing.

627.

When MnO2 is fused with KOH, a purple coloured compound is formed. The compound is(a) Mn2O4(b) KMnO4(c) K2MnO4(d) Mn2O3

Answer»

Answer is : (c) K2MnO4

2MnO2 + 4KOH + O2 \(\rightarrow \underset{\text{Purple green}}{2K_2MnO_4}\) + 2H2O

628.

Along with scales, scutes and bony plates are found in which of the following?(a) Torpedo(b) Salmon(c) Gambusia(d) Hippocampus

Answer»

Answer is : (d) Hippocampus

Hippocampus (seahorse) is a marine bony fish having bony plates and scutes besides scales. It has a brood pouch on the belly of males for incubating eggs.

629.

In plant, apical dominance is a condition where(a) foliar buds are inhibited by gibberellin(b) accessory buds are inhibited by cytokinin(c) axillary buds are inhibited by auxin(d) extra axillary buds are inhibited by abscisic acid

Answer»

Answer is : (c) axillary buds are inhibited by auxin

Apical dominance is a condition where, axillary buds are inhibited by auxin produced by apical meristem. When apical meristem is removed, the axillary bud becomes free from hormone inhibition and like apical meristem, it develops into a stem or flower.

630.

The function of copper ions in copper releasing IUD's is :(1) They inhibit gametogenesis(2) They make uterus unsuitable for implantation(3) They inhibit ovulation(4) They suppress sperm motility and fertilising capacity of sperms

Answer»

(4) Cu2+ interfere in the sperm movement, which suppress the sperm motility and fertilising capacity of sperms.

631.

Which of the following in sewage treatment removes suspended solids ?(1) Secondary treatment(2) Primary treatment(3) Sludge treatment(4) Tertiary treatment

Answer»

The correct answer is (2)

 Primary treatment is a physical process which involves two process, i.e. filtration and sedimentation of big solid waste.

632.

The final proof for DNA as the genetic material came from the experiments of :(1) Hershey and Chase(2) Avery, Mcleod and McCarty(3) Hargobind Khorana(4) Griffith

Answer»

The correct answer is (1)

Hershey and Chase proved that DNA as genetic material. They used bacteriophage for their experiment.

633.

An important characteristic that Hemichordates share with Chordates is :(1) Ventral tubular nerve cord(2) Pharynx with gill slits(3) Pharynx without gill slits(4) Absence of notochord

Answer»

The correct answer is (2)

Pharyngeal gill slits are present in hemichordates and in chordates. Notochord is present in chordates only. Ventral tubular nerve cord is present in non-chordates.

634.

Poison for platinum, a catalyst in contact process of  H2 SO4 is :-(a) S (b) P(c) As(d) C

Answer»

Correct Option (d) C

Explanation :

As acts as poison for Pt in contact process
635.

In mitotic prophase, the sister chromatids are held together by a multi-subunit protein complex called(a) actin(b) cytokinin(c) cohesin(d) tubulin

Answer»

Answer is : (c) cohesin

Cohesin is a multi-subunit protein complex that holds the sister chromatids together after DNA replication, i.e. in prophase and metaphase, during anaphase removal of cohesin leads to the separation of sister chromatids.

636.

Viruses are known as wandering genes because(a) nucleic acid is the only active part of a virus(b) capsid is the outer protective coat(c) nucleic acid is the central core(d) capsid is made up of specific protein

Answer»

Answer is : (a) nucleic acid is the only active part of a virus

Viruses are known as wandering genes because nucleic acid is the only active part of it. The infectivity of virus is due to nucleic acid, while host specificity is determined by the protein coat.

637.

A person performing ‘Yoga’ breathes in as much air as possible. The volume of air inspired is called(a) tidal volume(b) inspiratory capacity(c) vital capacity(d) inspiratory reserve volume

Answer»

Answer is : (b) inspiratory capacity

Total volume of air a person can inspire after a normal expiration is the inspiratory capacity. It includes Tidal Volume (TV) + Inspiratory Reserve Volume (IRV).

638.

In which of the following tissues the benign tumour is enclosed?(a) Connective tissue(b) Epithelial tissue(c) Muscular tissue(d) Nervous tissue

Answer»

Answer is : (a) Connective tissue

Benign tumour does not invade other tissues or spread to other sites. It is usually well-encapsulated in connective tissue

639.

Arrange in the correct order of stability (decreasing order ) for the following molecules:https://www.sarthaks.com/?qa=blob&qa_blobid=2208360274556592691

Answer»

The correct option is (4).

640.

The effective atomic number of cobalt in [Co(NH3)5H2O]3+ is(a) 36(b) 33(c) 24(d) 30

Answer»

Answer is : (a) 36

EAN = (atomic number - oxidation state + 2 x CN)

= 27 - 3 + 2 x 6

= 24 + 12 = 36

641.

Among the given compounds, the most susceptible to nucleophilic attack at the carbonyl group is (a) CH3COOCH3  (b) CH3CONH2 (c) CH3COOCOCH3  (d) CH3COCl

Answer»

Answer given below:

https://bit.ly/2VepkUz

642.

The correct order of nucleophilicity is :–(a) Cl– > Br– > I– > OH– > C2H5O–(b) I– > OH– > Br– > Cl– > C2H5O–(c) I– > Br– > Cl– > OH– > C2H5O–(d) C2H5O– > OH– > Cl– > Br– > I–

Answer»

Correct option (d) C2H5O > OH– > Cl > Br > I

Explanation:

Weaker the acid, stronger is the conjugate base and stronger is nucleophilicity. The order of nucleophilicity is C2H5O > OH > Cl > Br > I

643.

Metal-metal bonding is more frequent in 4d or 5d series than in 3d-serie. This is because of :– (a) Their greater enthalpy of atomization (b) the large size of the orbitals which participates in the metal-metal bond formation (c) their ability to involve ns and (n–1) d-electrons in the bond formation (d) the comparable size of 4d and 5d-series elements

Answer»

Correct option (a) Their greater enthalpy of atomization 

644.

For a body of 50 kg mass, the velocity -time graph is shown in the figure . then force acting on the body is(1) 25 N (2) 50 N (3) 12.5 N (4) 100 N

Answer»

The correct option is (1) 25 N.

Explanation:

Slope under (v-t) curve gives acceleration 

a = 0.5 m/s2

Force = m x a 

= 50 x 0.5

= 25 N

645.

Inbreeding helps in accumulation of superior genes and elimination of undesirable genes is correct or wrong statement

Answer»

It is correct statement.

 Inbreeding helps in accumulation of superior genes and elimination of undesirable genes.

Inbreeding exposes harmful recessive genes that are eliminated by selection. It also helps in accumulation of superior genes and elimination of less desirable genes. Therefore this is selection at each step, increase the productivity of inbred population. Close and continued inbreeding usually reduces fertility and even productivity.

646.

Chromatography is based on the principle of :(a) vaporisation (b) concentration(c) absorption (d) adsorption

Answer»

Answer : (d) adsorption 

Chromatography is based on the principle of differential adsorption of different components over an adsorbent. 

In this method, 

Weakly adsorbed component is eluted first and the strongly adsorbed component is elucted afterwards.

647.

The oxidation state of Ni in tetracarbonyl nickel is :(a) +1 (b) +2 (c) 0 (d) +4

Answer»

Answer : (c) 0

CO is a neutral ligand.

∴Oxidation state of Ni in [Ni(CO)4] is 0.

648.

A resonance air column shows resonance with a tuning fork of frequency 256 Hz at column lengths 32.5 cm and 112.9 cm. The end correction and speed of sound in air are(a) 4.1 cm, 77.7 ms−1(b) 7.7 cm, 411.65 ms−1(c) 5 cm, 224.5 ms−1(d) 6.7 cm, 352.7 ms−1

Answer»

Answer is :  (b) 7.7 cm, 411.65 ms−1

End correction = \(\frac{L_2-3L_1}{2}\)

\(=\frac{112.9-3\times 32.5}{2}\)

\(=\frac{15.4}{2}=7.7\, cm\)

Speed of sound in air,

v = 2v(L2 - L1)

= 2 x 256 x (1.129 - 0.325)

= 411.65 ms-1.

649.

In the complexes [Fe(H2O)6]3+, [Fe(CN)6]3–, [Fe(C2O4)3]3– and [FeCl6]3– more statbility is shown by :–(a) [Fe(H2O)6]3+ (b) [Fe(CN)6]3–(c) [Fe(C2O4)3]3– (d) [FeCl6]3–

Answer»

Correct option (c) [Fe(C2O4)3]3– 

Explanation:

 The central metal ion is Fe3+ and C2O4 2– is negative bi-dentate ligand which forms more stable complex than neutral or monodentate ligand.

650.

In date palm, maturity time as well as size of fruits can be changed by using different pollens. This effect is technically known as(a) xenia(b) metaxenia(c) ruminate(d) mosaic

Answer»

Answer is : (b) metaxenia

Term metaxenia is used for denoting the effect of pollens on structure outside endosperm.

Xenia refers to the effect of pollen on seeds and fruit in a fertilised plant.