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101.

Which of the following is a connecting link as well as a living fossil ?(a) Latimeria (b) Neoilina(c) Euglena (d) Archaeopteryx

Answer»

Correct Option (b) Neoilina

102.

Which of the following theories explains the structure of protoplasm ? (a) Surface-tension theory (b) Colloidal theory (c) Sol-gel theory (d) Viscosity theory

Answer»

Correct Option (b) Colloidal theory 

103.

A carbon compound A forms B with sodium metal and again A forms C with PCl5, but B and C form diethyl ether. Therefore A, and B and C are :–(a) C2H5OH, C2H5ONa, C2H5Cl(b) C2H5Cl, C2H5ONa, C2H5Cl2(c) C2H5OH, C2H6, C2H5Cl2(d) C2H5OH, C2H5Cl, C2H5ONa

Answer»

Correct Option (a) C2H5OH, C2H5ONa, C2H5Cl

Explanation: 

C2H5OH(A) + Na  C2H5ONa(B)

C2H5OH(A) + PCl5  C2H5Cl(C)

C2H5ONa(B) + C2H5Cl (C)  C2H5OC2H5 + NaCl

104.

How many electron(s) in an atom can have n = 3, m = 2?(a) 1 (b) 2 (c) 5 (d) 10

Answer»

Correct option (b) 2 

Explanation:

Only two electrons of an atom have n = 3 and m = 2.

105.

Dipole moment of CH3CH2CH3(I), CH3CH2OH(II) & CH3CH2F(III) is in order :-(a) I < II < III (b) I > II > III(c) I < III < II (d) III < I < II

Answer»

Correct Option (a) I < II < III 

Explanation:

Higher the electronegativity between C—X, higher will be dipole moment .

106.

In pest-resistant legumes,(a) a gene for an enzyme that synthesises a chemical toxic to weevils has been transferred(b) Bt toxins are thousands of times more powerful(c) these crops increase the use of chemical pesticides(d) more resistant to viral attack

Answer»

Answer is : (a) a gene for an enzyme that synthesises a chemical toxic to weevils has been transferred

In pest-resistant legumes, a gene for an enzyme that synthesises a chemical toxic to weevils has been transferred from Bacillus bacteria to the Rhizobium bacteria that live in the root nodules of these plants.

107.

DNA fingerprinting refers to(A) Techniques used for molecular analysis of different specimens of DNA  (B) Techniques used for identification of fingerprints of individuals  (C) Molecular analysis of profiles of DNA samples  (D) Analysis of DNA samples using imprinting devices

Answer»

The correct option is (C) Molecular analysis of profiles of DNA samples.

Explanation:

DNA fingerprinting is a modern technique that compares sets of DNA by locating identical sequences of nucleotides, often for purposes of forensic identification.

108.

Which of the following characteristics represent' Inheritance of blood groups' in humans ?a. Dominanceb. Co-dominancec. Multiple dominanced. Incomplete dominancee. Polygenic inheritance(1) b, c and e (2) a, b and c(3) b, d and e (4) a, c and e

Answer»

Correct option (2) a, b and c

Explanation

Dominance, codominance and multiple alleles are the characteristics that represent 'inheritance of blood groups' in humans. ABO blood groups are determined by the gene I. There are multiple (three) alleles; IA, IB and I0 of this gene. Allele IA and IB are dominant over I0. However, when IA and IB aIIeles are present together, they show codominance.

Therefore, option (2) is correct.

109.

Which of the following classes of Echinodermata does not include free-moving echinoderms?(a) Asteroidea(b) Holothuroidea(c) Echinoidea(d) Crinoidea

Answer»

Answer is : (d) Crinoidea

Crinoidea is a class of stalked and sedentary echinoderms.

110.

Which of the following statements are not disadvantages of hydroponics?I. The cost of these experiments is high.II. It requires skilled people.III. It facilitates production of seasonal vegetables.IV. It avoids soil borne pathogens.Choose the correct option.(a) II and III(b) III and IV(c) I and IV(d) I, II and III

Answer»

Answer is : (d) I, II and III

Hydroponics is a technique of growing plants in a nutrient solution. Hydroponics have been successfully employed as a technique for the commercial production of vegetables such as tomato, seedless cucumber, etc. Thus, this technique facilitates the production of seasonal vegetables. It also avoids soil borne pathogens. However, the cost of these experiments is high and also requires skilled people to perform it.

111.

A person with 44A + XXY chromosome set up has gynecomastia and is Barr body positive. They are symptoms of

Answer»

Klinefelter’s Syndrome is an aneuploid condition marked by the presence of one or more extra chromosomes in a male (44 + XXY/ 44 + XXXY/ 44 + XXXXY). As we know that, to compensate the dosage of X-linked genes in human male and female, one X-chromosome in females is deactivated and is present as darkly stained body near nuclear envelope. The number of Barr bodies is always one less than the total number of X chromosomes. A human male with Klinefelter’s syndrome, 44 + XXY/ 44 + XXXY/ 44 + XXXXY, can have 1, 2 or 3 Barr bodies, depending on the number of X-chromosome. The presence of extra copy of X-chromosome produces some feminine characters in otherwise male individuals. They are sterile feminised males with enlarged breasts (Gynecomastia).

112.

How much time is required for complete decoposition of two moles of water using a current of 2 ampere ?(a) 1.93 x 105 sec(b) 2.93 x 105 sec(c) 0.93 x 105 sec(d) 4.93 x 105 sec

Answer»

Correct Option (a) 1.93 x 105 sec

Explanation

2H2→ 2H2 + H2

1 mole of H2O exchanges 2 moles of electrons, then 2 moles of H2O will exchange 4 moles of electrons. = 4 × 96500 coulomb

=4 x 96500/2sec

= 1.93 x 105 sec.

113.

Among the ligands NH3, en, CN– and CO, the correct order of their increasing field strength, is :–(a) CO &lt; NH3 &lt; en &lt; CN–(b) NH3 &lt; en &lt; CN– &lt; CO(c) CN– &lt; NH3 &lt; CO &lt; en(d) en &lt; CN– &lt; NH3 &lt; CO

Answer»

Correct option (b) NH3 < en < CN < CO

Explanation:

Based on spectrochemical series, ligands arranged in increasing order of crystal field strength are as NH3 < en < CN < CO

114.

Heating an ore in the absence of air below it s melting point is called :–(a) leaching (b) Roasting(c) smelting (d) calcination

Answer»

Correct Option (d) calcination

Explanation : 

Roasting is heating in the presence of air.

115.

Lanthanoids ion which is most likely to be reduced by Cr(+II) is :–(a) Sm (b) Yu(c) Yb (d) All of the above

Answer»

Correct Option (d) All of the above

Explanation

All of them on being reduced will get either stable half-filled orbitals or stable completely filled orbitals.

116.

Each pair forms ideal solution except(a) C2H5Br and C2H5I(b) C6H5Cl and C6H5Br(c) C6H6 and C6H5CH3(d) C2H6I and C2H5OH

Answer»

Answer is : (d) C2H6I and C2H5OH

C2H6I and C2H5OH pair will not form ideal solution. Here, C2H5OH will show H-bonding as well as polarity both.

117.

Which of the following is made up of dead cells?(1) Collenchyma(2) Phellem(3) Phloem(4) Xylem parenchyma

Answer»

The correct answer is (2)

Cork cambium undergoes periclinal division and cuts off thick walled suberised dead cells towards outside i.e. phellem (cork) and it cuts off thin walled living cells i.e., phelloderm on inner side.

118.

A temporary endocrine gland in the human body is :(1) Corpus cardiacum(2) Corpus luteum(3) Corpus allatum(4) Pineal gland

Answer»

The correct answer is (2)

 Corpus luteum is the temporary endocrine gland formed in the ovary after ovulation. It release hormones like progesterone, oestrogen etc.

119.

Which of the following options gives the correct sequence of events during mitosis ?(1) Condensation → nuclear membrane disassembly → arrangement at equator → centromere division → segregation → telophase(2) Condensation → crossing over → nuclear membrane disassembly → segregation → telophase(3) Condensation → arrangement at equator → centromere division → segregation → telophase(4) Condensation ® nuclear membrane disassembly → crossing over → segregation → telophase

Answer»

The correct answer is (1)

The correct sequence of events occur during mitosis would be as follows
(i) DNA condensation occurs so that chromosomes become visible during early to mid-prophase.
(ii) Disassembly of nuclear membrane begins at late prophase or transition to metaphase.
(iii) Chromosomes align at equator during metaphase.
(iv) Centromere division occurs during anaphase forming daughter chromosomes.
(v) During anaphase segregation also occurs in which daughter chromosomes separate and
move to opposite poles.
(vi) Telophase finally leads to formation of two daughter nuclei

120.

The process of separation and purification of expressed protein before marketing is called :(1) Downstream processing(2) Bioprocessing(3) Postproduction processing(4) Upstream processing

Answer»

The correct answer is (1)

The various stages of processing that occur after the completion of fermentation or biosynthetic stage which include separation and purification of product called downstream processing.

121.

Which of the following events does not occur in rough endoplasmic reticulum ?(1) Protein folding(2) Protein glycosylation(3) Cleavage of signal peptide(4) Phospholipid synthesis

Answer»

Correct option (4) Phospholipid synthesis

Explanation

Phospholipid synthesis does not occur in RER. It occurs inside Smooth Endoplasmic

Reticulum ( SER). A signal peptide is a short peptide present at the N-terminus of the newly synthesised proteins. It targets them to the ER and is then cleaved off. RER synthesises proteins. It bears enzymes for modifying polypeptides synthesised by attached ribosomes, e.g. glycosylation.

122.

AGGTATCGCAT is a sequence from the coding strand of a gene. What will be the corresponding sequence of the transcribed mRNA ?(1) AGGUAUCGCAU(2) UGGTUTCGCAT(3) ACCUAUGCGAU(4) UCCAUAGCGUA

Answer»

The correct option is(1) AGGUAUCGCAU

Explanation:

Coding strand is the one that codes for mRNA. It has same nucleotide sequence as that of mRNA except thymine (T) is replaced by uracil (U) in mRNA. Hence, the corresponding sequence of transcribed mRNA by template or non-coding strand (complementary to RNA) is AGGUAUCGCAU.

123.

Many ribosomes may associate with a single mRNA to form multiple copies of a polypeptide simultaneously.Such strings of ribosomes are termed as(1) Polysome(2) Polyhedral bodies(3) Plastidome(4) Nucleosome

Answer»

Correct option (1) Polysome

Explanation

Polysome is a string of ribosomes associated with a single mRNA. polysome helps to produce a number of copies of the same polypeptide. Nucleosome is the unit of eukaryotic DNA that consists of a DNA segment wrapped around a core of eight histone proteins. Nucleosome chain gives a 'beads on string'appearance under electron microscope. plastidome refer to all the plastids of a cell which work as a functional unit.

Polyhedral bodies or carboxysomes are present in several groups of autotrophic bacteria that assimilate inorganic carbon via Calvin cycle, e.g. Cyanobacteria.

124.

Which one of these animals is not a homeotherm?(1) Macropus(2) Chelone(3) Camelus(4) Psittacula

Answer»

The correct option is(2) Chelone

Explanation:

Among the given animals Chelone is not a homeotherm. It is green sea turtle belonging to class-Reptilia which are ectotherms or cold-blooded and their internal body temperature varies according to the ambient environment.

In contrast, Camelus and, Macropus belonging to Class-Mammalia and Psittacula belonging to class-Aves are homeotherms. They can maintain constant body temperature irrespective of the surrounding temperature.

125.

Which of the following animals does not undergo metamorphosis ?(1) Earthworm(2) Tunicate(3) Moth(4) Starfish

Answer»

The correct option is(1) Earthworm

Explanation:

All the given animals except earthworm undergoes metamorphosis. Earthworm exhibits direct development where no larval stage is involved.

Metamorphosis is usually seen in animals exhibiting indirect development, involving a larval stage which Iater transformed into an adult. 

Larval form of moth is caterpillar and that of tunicates is tadpole.

In starfish, bipinnaria larva occurs.

126.

Consider the following complexes : [Ni(CN)4]2- (A) and [Ni(CO)4] (B). Both of them possess similar magnetic behaviour but their geometries are different. The geometries of (A) and (B) are(a) (A) has tetrahedral structure, (B) has square planar structure(b) (A) has tetrahedral structure, (B) has trigonal planar structure(c) (A) has square planar structure, (B) has tetahedral structure(d) (A) has tetrahedral structure, (B) has trigonal pyramidal structure

Answer»

Answer is : (a) (A) has tetrahedral structure, (B) has square planar structure

127.

In the figure shown, m1 = 10 kg, m2 = 6 kg, m3 = 4 kg. If T3 = 40 N, then T2 is(a) 13 N(b) 32 N(c) 25 N(d) 35 N

Answer»

Correct option is : (b) 32 N

Let a be the acceleration of each block, then

T3 = (m1 + m2 + m3)a … (i)

and T2 = (m1 + m2)a … (ii)

From Eqs. (i) and (ii), we get

T2 \(=(\frac{m_1+m_2}{m_1+m_2+m_3})\times T_3\)

\(=(\frac{10+6}{10+6+4})\times 40\)

= 32 N

128.

Two astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will :-(1) Move towards each other.(2) Move away from each other.(3) Will become stationary(4) Keep floating at the same distance between them.

Answer» (1) Both the astronauts are in the condition of weightness. Gravitational force between them pulls towards each other. Hence Astronauts move towards each other under mutual gravitional force.
129.

The correct order of electron gain enthalpy with negative sing F, Cl, Br and I, having atomic number 9, 17, 35 and 53 respectively, is :–(a) I &gt; Br &gt; Cl &gt; F (b) F &gt; Cl &gt; Br &gt; I(c) Cl &gt; F &gt; Br &gt; I (d) Br &gt; Cl &gt; I &gt; F

Answer»

Correct option (c) Cl > F > Br > I 

Explanation:

As we go down the group in Periodic Table, atomic size increases, force of attraction for the added electron decreases, hence electron gain enthalpy decreases.

X(g)+e- → X-(g)

Actual order is Cl > F > Br > I 

The fact that fluorine has a less electron gain enthalpy than chlorine seems to be due to the relatively greater effectiveness of 2p-electron in the small F-atom to repel the additional electron entering the atom than do 3p-electrons in the larger Cl-atom.

130.

Which one of the following does not dissolve in conc. H2SO4 :–(a) CH3 - C ≡C - CH3(b) Ch3 - CH2 - C  ≡ CH(c) CH ≡ CH(d) CH2 ≡ CH2

Answer»

Correct Option (c) CH ≡ CH

Explanation :

If CH ≡ CH were to dissolve in H2SO4 a bisulphite salt of vinyl carbocation H2C = C+H would be formed. The more s-character in the pos itively charged 'C' , less stable is the carbocation and less likely to be formed.

131.

What is the mass of magnesium which completely reacts with 250 cm3 of 1.0 mol/dm3 sulphuric acid :–(a) 6 g (b) 12 g(c) 48 g (d) 96 g

Answer»

Correct option (a) 6 g 

132.

The sulphide ore is converted to oxide before reduction because (a) oxides are easier to reduce (b) sulphide are easily decompose (c) oxide get easily decompose (d) sulphide get oxidised to sulphur

Answer»

Answer Is : (a) oxides are easier to reduce

The sulphide ore is converted to oxide before reduction because oxides are easier to reduce. 

In this way, ore conversion is suitable for reduction.

133.

The separation of primary, secondary and tertiary amines can be done through fractional distillation. Which of the following method will not used to distinguish primary, secondary and tertiary amine? (a) Hinsberg method (b) Hofmann method (c) Liebermann’s nitroso method (d) Victor Meyer’s method

Answer»

Answer is : (d) Victor Meyer’s method

Victor Meyer’s method is used to distinguish between primary, secondary and tertiary alcohols. By identifying the colour produced, the alcohols are identified. Rest all the other methods given are used to distinguish between primary, secondary and tertiary amines.

134.

Given below is a stage of meiosis in an animal cell.Identify the stage being represented by this figure along with structure labelled A.Structure AStage(a)Meiotic spindleTelophase-I(b)CentriolesAnaphase-I(c)CentromereMetaphase-I(d)MicrotubulesTelophase-II

Answer»

Answer is : (b)

The structure A is centriole, which helps in assemblage of meiotic spindle fibres at both ends. The stage of cell division is Anaphase-I. Here the homologous chromosomes separate, with sister chromatids remaining attached to each other.

135.

A T joint is formed by two identical rods A &amp; B each of mass m &amp; length L in the XY plane as shown. Its moment of inertia about axis coinciding with A is :–(a) 2mL2/3(b) mL2/12(c) mL2/6(d) None of these

Answer»

Correct Option (b) mL2/12

Explanation: 

I =  I1 + I1  = I = 0 + ML2/12  = I = ML2/12​​​​​​​

136.

An imcopressible liquid flows through a horizontal tube as shown in figure. Then the velocity V of the fluid is :–(a) 3 m/s (b) 1.5 m/s(c) 1 m/s (d) 2.25 m/s

Answer»

Correct Option  (c) 1 m/s 

Explanation: 

M = m1 + m2

AV1 = AV2 = 1.5 AV

A x 3 = A x 1.5 + 1.5AV = V = 1m/s

137.

The diagram showing the variation of gravitational potential of earth with distance from the centre of eath is :–

Answer»

Correct Option (c)

 

138.

Which of the following is/are true for Leber's hereditary Optic Neurophaty (LHON) I. It is mitochondrial myophathy II. It is characterised by short stature III. It is characterised by acromegaly IV. It is characterised by visual lose (a) All are false (b) II and III (c) I and IV (d) II and IV

Answer»

Correct Option (c) I and IV 

139.

C4 – plants differe from C3 – plants in respect to(a) Number of CO2 molecules used(b) Substrate, which accept the CO2 molecules(c) Number of ATP formed(d) Number of O2 formed

Answer»

Correct Option (b) Substrate, which accept the COmolecules

140.

Match the following columnsColumn IColumn IIA. Macrophytes1. Rooted plants in shallow waterB. Phagotrophs2. Animals ingest foodC. Abiotic components3. TemperatureD. Hydrophytes4. Plants of dry areas5. Plants of aquatic nasitantsCodesABCD(a)5432(b)1234(c)1235d)4321

Answer»

Correct Option  (c)

141.

Which among the following represent central dogma(a) RNA → DNA → Protein(b) Protein → RNA → DNA(c) DNA ⇄RNA  → Protein(d) DNA  → Protein  → RNA

Answer»

Correct Option (c) DNA RNA   Protein

142.

Identify the correct relationship with reference to water potential of a plant cell :(a) Ψw = Ψm+Ψs+Ψp(b) Ψw = Ψm - Ψs - Ψp(c) Ψw = Ψm - Ψs + Ψp(d) Ψw = Ψm + Ψs - Ψp

Answer»

Correct Option (a) Ψw = Ψms+Ψp

 

143.

In bacteria, glucose is utilised first, even if other sources of sugar are available. This happens through a mechanism known as(a) catabolite repression(b) enzyme repression(c) operon repression(d) positive feedback mechanisms

Answer»

Answer is : (a) catabolite repression

Catabolic repression is a system of gene control in some bacterial operons in which glucose is used preferentially and the metabolism of other sugars is repressed in the presence of glucose.

Catabolite repression allows microorganisms to adapt quickly to preferred carbon/energy source (like glucose) first. This is achieved through inhibition of synthesis of enzymes involved in catabolism of carbon sources.

144.

The life history traits of organisms have evolved in relation to(a) Darwinian fitness(b) organisms evolve under selection pressure developed by environmental factors(c) organisms achieve most efficient reproductive strategy(d) All of the above

Answer»

Answer is : (d)

Population of organisms evolves to maximise their reproductive fitness, i.e. Darwinian fitness in the habitat in which they live, under a particular set of selection pressures. Organisms evolve towards most efficient reproductive strategy. Life history traits of organisms have evolved in relation to the constraints imposed by components of habitat in which they live.

145.

In an L - C - R circuit inductance is changed from L to L/2. To keep the same resonance frequency, C should be changed to(a) 2C (b) \(\frac{C}{2}\) (c) 4C (d) \(\frac{C}{4}\)

Answer»

Answer is : (a) 2C 

v = \(\frac{1}{\sqrt{LC}}\)

When L is changed to L/2, 

C must be changed to 2C to keep v same.

146.

An example of colonial alga is :(1) Volvox(2) Ulothrix(3) Spirogyra(4) Chlorella

Answer»

The correct answer is (1)

 Volvox is motile colonial fresh water green alga. It forms spherical colonies.

147.

Anaphase Promoting Complex (APC) is a protein degradation machinery necessary for proper mitosis of animal cells. If APC is defective in a human cell, which of the following is expected to occur ?(1) Chromosomes will be fragmented(2) Chromosomes will not segregate(3) Recombination of chromosome arms will occur(4) Chromosomes will not condense

Answer»

The correct answer is (2)

During anaphase, Anaphase Promoting Complex (APC) is a protein necessary for separation of daughter chromosomes. A defective APC will cause the chromosomes fail to segregate during anaphase.

148.

Which of the following cell organelles is responsible for extracting energy from carbohydrates to form ATP ?(1) Ribosome(2) Chloroplast(3) Mitochondrion(4) Lysosome

Answer»

The correct answer is (3)

The site of aerobic oxidation of carbohydrates in cells to generate ATP are mitochondria.

149.

Out of 'X' pairs of ribs in humans only 'Y' pairs are true ribs. Select the option that correctly represents values of X and Y and provides their explanation: (1) X = 12, Y = 5 True ribs are attached dorsally to vertebral column and sternum on the two ends.(2) X = 24, Y = 7 True ribs are dorsally attached to vertebral column but are free on ventral side.(3) X = 24, Y = 12 True ribs are dorsally attached to vertebral column but are free on ventral side.(4) X = 12, Y = 7 True ribs are attached dorsally to vertebral column and ventrally to the sternum.

Answer»

The correct answer is (4)

In human, 12 pairs of ribs are present out of which 7 pairs of ribs (1st to 7th pair) are dorsally attached to vertebral column and ventrally to the sternum.

150.

Mycorrhizae are the example of:(1) Amensalism(2) Antibiosis(3) Mutualism(4) Fungistasis

Answer»

The correct answer is (3)

 Mycorrhizae is a symbiotic association between fungi and roots of higher plants.