

InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
451. |
Which one of the following cell types is a part of innate immunity?(a) Skin epithelial cells(b) B-cells(c) T-lymphocytes(d) Liver cells |
Answer» Answer is :(a) Skin epithelial cells Innate immunity refers to non-specific defence mechanisms that come into play immediately or within hours of an antigen’s appearance in the body. These mechanisms include physical barriers such as skin epithelial cells, chemicals in the blood and immune system cells that attack foreign cells in the body. |
|
452. |
Which one of the following cell types is a part of innate immunity ?(A) Skin epithelial cells(B) B cells(C) T lymphocytes(D) Liver cells |
Answer» Correct option (A) Skin epithelial cells Explanation: Innate immunity is general defense of body eq. 1. Phagocytosis of invanders by macrophage 2. Restistance of skin to invading micro-organism 3. Destruction of micro-organisms by HCl in digestive juice etc. |
|
453. |
Glycolysis is the breakdown of glucose to pyruvic acid. How many molecules of pyruvic acid are formed from one molecule of glucose ? (A) 1(B) 2(C) 3(D) 4 |
Answer» Correct option (B) 2 Explanation: 2 pyruvic acid molecule are formed from one glucose molecule during glycolysis. |
|
454. |
The number of completely filled shells for the element 16S32 is(a) 1(b) 2(c) 3(d) 4 |
Answer» Answer is : (b) 2 Atomic number of 16S32 is 16. Its electronic configuration using 2n2 rule is 16S = 1s2[Ne], 2s2p6[Ne], 3s2p4[Unfilled] So, number of fully filled orbits or shells is 2. |
|
455. |
Particles used in the Rutherford’s scattering experiment to deduce the structure of atoms(a) had atomic number 2 and were fully ionised(b) had atomic number 2 and were neutral(c) had atomic number 4 and were fully ionised(d) had atomic number 4 and were neutral |
Answer» Answer is : (a) had atomic number 2 and were fully ionised Particles used in Rutherford’s scattering experiment (Geiger-Marsden experiment) are α-particles derived from a tube of radium emanation (or radon). α-particles are helium nuclei 24He, they are fully ionised and have atomic number 2. |
|
456. |
How many ways are there to arrange the letters of the word EDUCATION so that all the following three conditions hold ? - the vowels occur in the same order (EUAIO) - the consonants occur in the same order (DCTN) - no two consonants are next to each other (A) 15(B) 24(C) 72(D) 120 |
Answer» Correct option (A) 15 Explanation: EDUCATION Vowels EUAIO Consonant DCTN = 1 x 6C4 x 1 = 15 |
|
457. |
In his classical experiments on pea plants, Mendel did not use (a) seed shape (b) flower position (c) seed colour (d) pod length |
Answer» Correct option is (d) pod length Mendel did not choose pod length. The seven contrasting traits he took were ● Plant height ● Flower position ● Pod colour ● Pod shape ● Flower colour ● Seed shape ● Seed colour |
|
458. |
In phylum, which group contains the greatest number of species? (a) Class (b) Family (c) Genus (d) Order |
Answer» Correct option is (a) Class The levels of classification from the broadest to the narrowest, i.e. in term of having highest members to the lowest members are kingdom, phylum, class, order, family, genus and species. |
|
459. |
Following the cell cycle scheme given below, what is the probability that a cell would be in M-phase at any given time ?(A) 1/24(B) 1/12(C) 1/6(D) 1/2 |
Answer» Correct option (B) 1/12 Explanation: Total time for cell cycle = 24 hrs. Time for M-phase = 2 hrs So, probability of cell in M-phase at any given time is 2/24 ⇒ 1/12 |
|
460. |
A diploid plant has 14 chromosomes, but its egg cell has 6 chromosomes. which one of the following is the most likely explanation of this ?(A) Non-disjunction in meiosis I and II(B) Non-disjunction in meiosis I (C) Non-disjunction in mitosis(D) Normal meiosis |
Answer» Correct option (B) Non-disjunction in meiosis I Explanation: Egg cell is formed by meiosis. Less number of chromosome indicates non-disjunction in meiosis I. |
|
461. |
The graph shows the relationship between oxygen production in photosynthesis and light intensity for a unicellular green organism in 0.02% sodium hydrogencarbonate solutionThe most likely explanation of the fact that the graph levels off at 180 Jm-2 s-1 is that the system is (a) light limited and carbon dioxide saturated (b) light limited and the temperature is below optimum (c) light saturated and carbon dioxide is unlimited (d) light saturated and the temperature is above optimum |
Answer» Correct option is (d) light saturated and the temperature is above optimum The amount of light given is saturated, not limited. Therefore the answer is (d). The concentration of carbon dioxide is in short supply, hence limiting the rate of photosynthesis. |
|
462. |
Let x,y,z be positive integers such that HCF (x,y,z) = 1 and x2 + y2 = 2z2 . Which of the following statements are true?I. 4 divides x or 4 divides y.II. 3 divides x + y or 3 divides x – y.III. 5 divides z(x2 – y2)(A) I and II only(B) II and III only(C) II only(D) III only |
Answer» Correct option (B) II and III only Explanation: Take combination such as x = 1, y = 7, z = 5 Now, check options (ii) & (iii) statement are correct. |
|
463. |
The respiratory quotient of the reaction given below is2(C51H98O6) + 145O2 → 102CO2 + 90H2O + Energy(a) 0.703(b) 0.725(c) 0.960(d) 1.422 |
Answer» Answer is : (a) 0.703 Respiratory Quotient (RQ) measures the ratio of the volume of carbon dioxide (VC) produced by an organism to the volume of oxygen consumed (VO). The RQ for the given equation is RQ = CO2 produced/O2 consumed = 102/145 = 0.703. RQ = 0.703 |
|
464. |
Which ONE of the following coenzymes is required for the conversion of L-alanine to a racemic mixture of D-and L-alanine ?(A) Pyridoxal-6-phosphate(B) Thiamine pyrophosphate(C) Coenzyme A(D) Flavin adenine dinucleotide |
Answer» Correct option (A) Pyridoxal-6-phosphate Explanation: Information based question |
|
465. |
The number of net ATP molecules produced from 1 glucose molecule during glycolysis is(A) 1(B) 2 (C) 3 (D) 4 |
Answer» Correct option (B) 2 Explanation: In glycolysis net ATP produced = 4 – 2 = 2 ATP |
|
466. |
The source of mammalian hormone ‘relaxin’ is(a) ovary(b) stomach(c) intestine(d) pancreas |
Answer» Answer is : (a) ovary Relaxin hormone is produced by the ovary and the placenta with important effects in the female reproductive system and during pregnancy. In preparation for childbirth, it relaxes the ligaments in the pelvis and softens and widens the cervix. |
|
467. |
Bombyx mori (silkworm) belongs to the order(a) Lepidoptera(b) Diptera(c) Hymenoptera(d) Coleoptera |
Answer» Answer is : (a) Lepidoptera The order of silkworm (Bombyx mori) is Lepidoptera. It is the order of insects that includes butterflies and moths. About 1,80,000 species of the Lepidoptera are described till now. |
|
468. |
I carried 1000 kg of watermelon in summer by train. In the beginning the water content was 99%. By the time I reached the destination, the water content had dropped to 98%. The reduction in the weight of the watermelon was(a) 10 kg(b) 50 kg(c) 100 kg(d) 500 kg |
Answer» Answer is : (d) 500 kg We have, 1000 kg of watermelon in which 99% are water. ∴ 990 kg water and 10 kg rest. Now, x kg watermelon has = 98/100 x water + 2x/100 rest Weight of solid part should remain same ∴ 2x/100 = 10 ⇒ x = 500 ∴ Weight reduction = (1000 - 500) kg = 500 kg |
|
469. |
Correct graph of experimental values of specific heat of a constant volume of hydrogen gas is |
Answer» Correct option is (d) At low temperature, H2 molecule has only translational degrees of freedom ∴ f = 3 Above 600 K molecule be given to vibrate and at above 3000 K molecule dissociates. |
|
470. |
Excess salt inhibits bacterial growth in pickles by(a) endosmosis(b) exosmosis(c) oxidation(d) denaturation |
Answer» Answer is : (b) exosmosis Excess salt inhibits growth in pickles by exosmosis. Salt kills and inhibits the growth of microorganisms by drawing water out of the cells of both the microbe and the food through osmosis (or more specifically exosmosis). Due to hypertonic solution outside the bacterial cell, bacteria will die by plasmolysis. |
|
471. |
The minimum number of plants to be screened to obtain a plant of the genotype AabbCcDd from a cross beteen plants of genotypes AaBbCcDd and AABbCCDd is(A) 8(B) 16(C) 32(D) 64 |
Answer» Correct option (a) 8 Explanation: Number of plants with particular genotype can be calculated by the formula 2n, where n is the number of heterozygous gene in the desired genotype. Therefore, plants with genotype AabbCcDd = 23 = 8. Thus, the correct answer is option A. |
|
472. |
The schematic below describes the status of lac operon in the absence of lactose. Which ONE of the following happens when lactose is present in the cell ?(A) Lactose binds to Pi and stops the transcription of i.(B) Lactose is converted to allolactose, which binds to Plac and results in the displacements of the repressor from O.(C) Lactose is converted to allolactose, which binds to the repressor protein and prevents its interaction with O.(D) Lactose has no effect on the status of the lac operon. |
Answer» Correct option (C) Lactose is converted to allolactose, which binds to the repressor protein and prevents its interaction with O. Explanation: Lac operon is inducible operon here. |
|
473. |
When a pure bred, red flower-producing plant of genotype RR is crossed with a pure bred, white flower-producing plant of genotype rr, all the F1 plants produced pink flowers If all the plants in each generation from F1 to F6 are selfed, what will be the percentage of plants with red and white flowers in the final population consisting of a large number of individuals ? (Consider that flower colour has no effect on reproduction and survival.) (A) 3 – 4(B) 12 –13(C) 49 – 51(D) 97 – 100 |
Answer» Correct option (D) 97 – 100 Explanation: After selfing homozygous indivisiual produces All white & red offspring and heterozygous also produces 50%. Red & White flower. |
|
474. |
The number of positive integers n in the set {2, 3,..., 200} such that 1/n has a terminating decimal expansion is(a) 16(b) 18(c) 40(d) 100 |
Answer» Answer is : (b) 18 We have, n ∈ {2, 3, 4, 5, 6, …, 200} 1/n has terminating decimal of n = 2a x 5b ∴ n = 2, 4, 5, 8, 10, 16, 20, 25, 32, 40, 50, 64, 80, 100, 125, 128, 160, 200 ∴ Total number of n = 18 |
|
475. |
Enzyme X extracted from the digestive system hydrolyses peptide bonds. Which of the following are probable candidates to be enzyme X ?(A) AmylaseB) Lipase (C) Trypsin(D) Maltase |
Answer» Correct option (C) Trypsin Explanation: Enzyme ‘X’ hydrolyses peptide bond so it is a proteolytic enzyme - ⇒ Amylase → Starch digesting enzyme ⇒ Lipase → Fact digesting enzyme ⇒ Trypsin → Protein digesting enzyme ⇒ Maltase → Maltose digesting enzyme (Disaccharides) |
|
476. |
A person with blood group AB has(a) antigen A and B on RBCs and both anti-A and anti-B antibodies in plasma(b) antigen A and B on RBCs, but neither anti-A nor anti-B antibodies in plasma(c) no antigen on RBCs but both anti-A and anti-B antibodies are present in plasma(d) antigen A on RBCs and anti-B antibodies in plasma |
Answer» Answer is : (b) antigen A and B on RBCs, but neither anti-A nor anti-B antibodies in plasma Person with blood group AB have both A and B antigen in the membrane of his red blood cell but lacks both antibodies (a, b) in his plasma. Due to this reason, blood group AB is called universal recipient. |
|
477. |
Let N1 = 255 + 1 and N2 = 165.Then(a) N1 and N2 are co-prime(b) the HCF (Highest Common Factor) of N1 and N2 is 55(c) the HCF of N1 and N2 is 11(d) the HCF of N1 and N2 is 33 |
Answer» Answer is : (d) the HCF of N1 and N2 is 33 It is given that, N2 = 165 = 3 x 5 x 11 and N1 = 255 + 1 As we know that, if n is odd integer then xn + yn is divisible by x + y. So, N1 = 255 + 155 is divisible by 2 + 1 = 3 and N1 = 255 + 155 = (25)11 + (15)11 = (32)11 + (1)11 is divisible by 32 + 1 = 33 ∴ the HCF of N1 and N2 is 33. |
|
478. |
Consider two identical copper spheres A and B. One is placed over a thermally insulating plate, while the other hangs from an insulating thread.Equal amounts of heat are given to the two spheres and temperatures are recorded, then(a) TA = TB (b) TB < TA (c) TB > TA (d) cannot be concluded |
Answer» Correct option is (c) TB > TA Part of heat given to A is used up in doing work (against) gravitational force. So, temperature of B will be slightly higher. |
|
479. |
The polynomial equation x3 – 3ax2 + (27a2 + 9)x + 2016 = 0 has -(A) exactly one real root for any real a(B) three real roots for any real a(C) three real roots for any a 0, and exactly one real root for any a < 0(D) three real roots for any a ≤ 0, and exactly one real root for any a > 0 |
Answer» Correct option (A) exactly one real root for any real a Explanation: f'(x) = 3x2 – 6ax + 27a2 + 9 = 3[x2 – 2ax + 9a2 + 3] = 3((x – a)2 + 8a2 + 3) f'(x) is + ve for x ∈ R so f(x) is monotonic↑ for x ∈ R. |
|
480. |
Gaseous exchange of oxygen and carbon dioxide between alveolar air and capillaries takes place by(a) active transport(b) diffusion(c) carrier-mediated transport(d) imbibition |
Answer» Answer is : (b) diffusion Gaseous exchange occurs at the alveoli in the lungs and takes place by diffusion. The alveoli are surrounded by capillaries so, oxygen and carbon dioxide diffuse between the air in the alveoli and the blood in the capillaries. Diffusion is the movement of gas from an area of high concentration to an area of low concentration. |
|
481. |
Of the periods listed below, which one is the earliest period when ostracoderms, the jawless and finless fishes, appeared?(a) Devonian period(b) Cambrian period(c) Carboniferous period(d) Silurian period |
Answer» Answer is : (b) Cambrian period The class Ostracodermi is represented by the fossil vertebrates of late Cambrian period. The earliest known vertebrates to appear in fossil record were jawless primitive fish-like animals collectively called ostracoderms. These animals resembled the present day cyclostomes (lampreys and hagfishes) in many respects. |
|
482. |
Three successive measurements in an experiment gave the values 10.9, 11.4042 and 11.42. The correct way of reporting the average value is - (A) 11.2080(B) 11.21(C) 11.2 D) 11 |
Answer» Correct option (C) 11.2 Explanation: he correct way of reporting the average value should have exactly the same number of digit after decimal which has least digit after decimal among the data given. |
|