

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.
101. |
The part of the human brain that governs memory and intelligence is :(A) Cerebrum (B) Hypothalamus (C) Medulla (D) Cerebellum |
Answer» The part of the human brain that governs memory and intelligence is Cerebrum. |
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102. |
The heart of an amphibian is usually(a) two-chambered(b) three-chambered(c) four-chambered(d) three and half-chambered |
Answer» Answer is : (b) three-chambered Vertebrate hearts can be categorised by the number of chambers they have, like two-chambered (one atrium and one ventricle) in fishes, three-chambered (two atria and one ventricle) in amphibians and reptiles; and four-chambered (two atria and two ventricles) in birds and mammals. |
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103. |
Which one of the following human cells lacks a nucleus?(a) Neutrophil(b) Neuron(c) Mature erythrocyte(d) Keratinocyte |
Answer» Answer is : (c) Mature erythrocyte Mature human erythrocytes (Red blood cells) lack a nucleus. The absence of a nucleus is an adaptation of the red blood cell for its role. It allows the RBC to contain more haemoglobin and therefore carry more oxygen molecules. It also allows the cell to have its distinctive biconcave shape which aids diffusion. |
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104. |
Which one of the following reactions is catalysed by high-energy ultraviolet radiation in the stratosphere?(a) O2 + O → O3(b) O2 → O + O(c) O3 + O3 → 3O2(d) O + O → O2 |
Answer» Answer is : (b) O2 → O + O In the stratosphere, ozone is created primarily by ultraviolet radiation. When high energy ultraviolet rays strike ordinary oxygen molecules (O2), they split the molecule into two single oxygen atoms, known as atomic oxygen. O2 → O + O A freed oxygen then combines with another oxygen molecule to form a molecule of ozone (O3). |
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105. |
Athletes are often trained at high altitude because(a) training at high altitude increases muscle mass(b) training at high altitude increases the number of red blood cells(c) there is less chance of an injury at high altitude(d) athletes sweat less at high altitude |
Answer» Answer is : (b) training at high altitude increases the number of red blood cells Athletes are often trained at high altitude because the air is ‘thinner’ at high altitudes, means there are fewer oxygen molecules per volume of air. Every breath taken at high altitude delivers less of what working muscles require. To compensate for the decrease in oxygen there occurs more production of red blood cells to aid in oxygen delivery to the muscles. |
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106. |
Consider the isoelectronic ions K+ ,S2- ,Cl- and Ca2+The radii of these ionic species follow the order (a) Ca2+> K+ > Cl- >S2- (b) Cl- > S2- > K+ > Ca2+ (c) S2- >Cl- >K+ >Ca2+ (d) K+ > Ca2+ > S2- > Cl- |
Answer» Correct option is (C) S2- >Cl- >K+ >Ca2+ Isoelectronic species are those species which have same number of electrons. As all the given elements are isoelectronic with each other. Thus, the radii/size of isoelectronic species is inversely proportional to the atomic number, i.e. size \(\propto \frac{1}{Z}\) Thus, the correct order is S2- >Cl- >K+ >Ca2+ |
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107. |
Consider the circuit below. The bulb will light up of :(A) S1 S2 and S3 are all closed.(B) S1 is closed but S2 and S3 are open.(C) S2 and S3 are closed but S1 is open.(D) S1 and S3 are closed but S2 is open. |
Answer» (B) S1 is closed but S2 and S3 are open. |
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108. |
There are 20 units cubes all of whose faces are white, and 44 units cubes all of whose faces are red. They are put together to form a bigger cube (4 x 4 x 4). What is the minimum number of white visible on this larger cube?(a) 20(b) 14(c) 12(d) 8 |
Answer» Answer is : (c) 12 Given, 4 x 4 x 4 cubes is made 64 faces 1 x 1 x 1 cubes. Total cubes = 64, White = 20, Red = 44 To find minimum number of visible white box counting total visible faces of unit cube. Total number of faces of small cube on bigger cube except boundary cubes = 4 x 6 = 24 Counting boundary cubes = 16+8+8 = 32 ∴ Total visible faces = 56 But we have 44 red cube. ∴ Minimum number of white faces cubes which are visible = 56 - 44 = 12 |
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109. |
On each face of a cuboid, the sum of its perimeter and its area is written. Among the six numbers so written, there are three distinct numbers and they are 16, 24 and 31. The volume of the cuboid lies between(A) 7 and 14 (B) 14 and 21 (C) 21 and 28 (D) 28 and 35 |
Answer» Correct option(D) 28 and 35 Explanation: 2(a + b) + ab = 16 ... (1) 2(b + c) + bc = 24 ... (2) 2(c + a) + ca = 31 ... (3) From equation (2), equation (3) =>(a - b) (2 + c) = 7 ... (4) From equation (2) and equation (4) =>4a = 2 + 5b ... (5) Solve equation (1) and (5) b = 2 a = 3, c = 5 Volume = 30 |
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110. |
Among the following, which is an incorrect statement.(a) PH5 and BiCl5 do not exist(b) pπ - dπ bonds are present in SO2(c) SeF4 and CH4 have same shape(d) \(I^+_3\) has bent structure |
Answer» Answer is : (c) SeF4 and CH4 have same shape PH5 does not exist due to very less electronegativity difference between P and H. Hydrogen is slightly more electronegative than phosphorus, thus could not hold significantly the sharing electrons. On the other hand, BiCl5 does not exist due to inert pair effect. This is because on moving down the group, +5 oxidation state becomes less stable while +3 oxidation state become more stable due to inert pair effect. In SO2, pπ-dπ and pπ-pπ both types of bonds are present. SeF4 has sp3 d-hybridisation whereas CH4 has sp3 -hybridisation. Thus, they both have different geometry. \(I^+_3\) has a bent shape due to the presence of 2 lone pairs on central I atom. |
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111. |
The electronegativity of the following elements increases in the order (a) C < N < Si < P (b) N < Si < C < P (c) Si < P < C < N (d) P < Si < N < C |
Answer» Correct option is (c) Si < P < C < N On moving across a period from left to right in periodic table electronegativity increases. This is because across the period the size of an atom decreases. While on moving down the group, electronegativity decreases. Thus, the correct order of electronegativity is Si < P < C < N (1.8) (2.1) (2.5) (3.0) |
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112. |
The number of electrons required to reduce one molecule of oxygen to water during mitochondrial oxidation is(a) 4(b) 3(c) 2(d) 1 |
Answer» Answer is : (a) 4 Four electrons are required to reduce one molecule of oxygen to water during mitochondrial oxidation. O2 + 4e- + 4H+ → 2H2O This process mentioned above takes place during oxidative phosphorylation. It is the metabolic pathway in which cells use enzymes to oxidise nutrients, thereby releasing energy which is used to produce ATP. |
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113. |
Which one of the following conversions does not happen under anaerobic conditions?(a) Glucose to ethanol by Saccharomyces(b) Lactose to lactic acid by Lactobacillus(c) Glucose to CO2 and H2O by Saccharomyces(d) Cellulose to glucose by Cellulomonas |
Answer» Answer is : (c) Glucose to CO2 and H2O by Saccharomyces Conversion of glucose to CO2 and H2O by Saccharomyces is a reaction which takes place in aerobic conditions, i.e. in the presence of oxygen. C6H12O6 + 6O2 → 6CO2 + 6H2O |
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114. |
An amount of 18 g glucose corresponds to(a) 1.8 mole(b) 1 mole(c) 0.18 mole(d) 0.1 mole |
Answer» Answer is : (d) 0.1 mole A mole is the quantity of a substance whose weight in grams is equal to the molecular weight of the substance. 1 mole is equal to 1 moles Glucose, or 180.15588 grams. ∴ 18 g of glucose = x mole × 180 g x mole = 18/180 = 0.1 mole ∴ An amount of 18 g glucose corresponds to 0.1 mole. |
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115. |
Optical activity of DNA is due to its(a) bases(b) sugars(c) phosphates(d) hydrogen bonds |
Answer» Answer is : (b) sugars DNA polymer is made up of nitrogenous base, a sugar and one or more phosphate. Optical activity results due to the molecular asymmetry. The nucleic acid bases have a plane of symmetry. Hence, they do not induce optical activity. Sugars are asymmetric and cause optical activity of DNA. |
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116. |
Desert temperature often varies between 0-50°C. The DNA polymerase isolated from a camel living in the desert will be able to synthesise DNA most efficiently at(a) 0°C(b) 37°C(c) 50°C(d) 25°C |
Answer» Answer is : (b) 37°C Camel belongs to class–Mammalia. Both birds and mammals are homeothermic and have a fixed 37°C body temperature. So, the DNA polymerase isolated from a camel will work efficiently at temperature near its body temperature. |
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117. |
Match the enzymes in Column I with the reactions in Column II. Select the correct combination.Column IColumn IIP.Hydrolasei.Inter-conversion of optical isomersQ.Lyaseii.Oxidation and reduction of two substratesR.Isomeraseiii.Joining of two compoundsS.LigaseivRemoval of a chemical group from a substancevTransfer of a chemical group from one substrate to another(a) P-iv, Q-ii, R-iii, S-i(b) P-v, Q-iv, R-i, S-iii(c) P-iv, Q-i, R-iii, S-v(d) P-i, Q-iv, R-v, S-ii |
Answer» Answer is : (b) P–v, Q–iv, R–i, S–iii – Hydrolases catalyse transfer of a chemical group from one substrate to another. – Lyase catalyses removal of chemical groups from a substrate. – Isomerase catalyses interconversion of optical, geometric or positional isomers. – Ligase catalyses linking together of two compounds. |
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118. |
Similar type of vegetation can be observed, in the same(a) latitude(b) longitude(c) country(d) continent |
Answer» Answer is : (a) latitude Latitude and temperature are related to each other in a way that, as we approaches the equator, the temperature gets warmer and as we approaches the poles, it gets cooler. Since, same vegetation grows in the same climatic zone, therefore similar type of vegetation can be observed in the same latitude. |
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119. |
Similar type of vegetation can be observed, in the same :(A) latitude (B) longitude (C) country (D) continent |
Answer» Similar type of vegetation can be observed, in the same latitude. |
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120. |
Which of the following ecological food chain does not represent an erect pyramid of numbers ?(A) Grass Rodent Snake (B) Tree-Bird-Avian parasite(C) Grass-Deer-Tiger (D) Insect-Chicken-Human |
Answer» Tree-Bird-Avian parasite is ecological food chain does not represent an erect pyramid of numbers. |
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121. |
Which of the following ecological food chains does not represent an erect pyramid of numbers?(a) Grass–Rodent–Snake(b) Tree–Bird–Avian parasite(c) Grass–Deer–Tiger(d) Insect–Chicken–Human |
Answer» Answer is : (b) Tree–Bird–Avian parasite Tree–Bird–Avian parasite does not represent an erect pyramid of number. Instead, it is an inverted pyramid of number. All ecological pyramids of number are erect except in parasitic food chain, where one primary producer supports numerous parasites which further support more hyperparasites. |
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122. |
A nurse measures the blood pressure of a seated patient to be 190 mm of Hg -(A) The blood pressure at the patient's feet is less than 190 mm of Hg(B) The actual pressure is about 0.25 times the atmospheric pressure(C) The blood pressure at the patient's neck is more than 190 mm of Hg(D) The actual pressure is about 1.25 times the atmospheric pressure |
Answer» Correct option (D) The actual pressure is about 1.25 times the atmospheric pressure Explanation: Blood pressure is gauge pressure = 190 mm Hg Atmospheric pressure = 760 mm Hg Actual pressure = 190 + 760 mm Hg = 950 mm Hg = 1.25 x 760 mm Hg |
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123. |
The number of real solutions of the equation 2sin 3x + sin 7x – 3 = 0 which lie in the interval [–2π, 2π] is (A) 1(B) 2(C) 3(D) 4 |
Answer» Correct option (B) 2 Explanation: only possible when sin 3x = 1 & sin 7x = 1 sin 3x = 1 sin 3x = sin (4n + 1) π/2 , n ∈ I 3x = (4n + 1) π/2 ⇒ x = (4n + 1) π/6 sin 7x = sin(4m + 1) π/2, m ∈ I x = (4m + 1) π/14 for common solution (4n + 1) π/6 = (4m + 1) π/14 Solving these 1 = 3m – 7n First solution is m = 5, n = 2 Second solution is m = 12, n = 5 So two solutions are possible |
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124. |
If rice is cooked in a pressure cooker on the Siachen glacier, at sea beach, and on Deccan plain, which of the following is correct about the time taken for cooking rice ?(A) Gets cooked faster on the Siachen glacier(B) Gets cooked faster at sea beach(C) Gets cooked faster on Deccan plain(D) Gets cooked at the same time at all the three places |
Answer» Correct option (D) Gets cooked at the same time at all the three places Explanation: Pressure cooker is used. |
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125. |
Two friends A and B are 30 km apart and they start simultaneously on motorcycles to meet each other. The speed of A is 3 times that of B. The distance between them decreases at the rate of 2 km per minute. Ten minutes after they start, A’s vehicle breaks down and A stops and waits for B to arrive. After how much time (in minutes) A started riding, does B meet A?(a) 15(b) 20(c) 25(d) 30 |
Answer» Answer is : (d) 30 Let the speed of B = x km/h and the speed of A = 3x km/h Distance between A and B = 30 km Given, distance between them decrease at 2 km per minutes. ∴ Distance decrease in one hour = 2 x 60 = 120 km ∴ Total distance travelled by A and B in one hour = (x+3x) km = 4x km ∴ Speed = 120/4 = 30 km/h Hence, speed of B = 30 km/h Speed of A = 90 km/h Distance travelled by A and B after 10 min = 2 x 10 = 20 km So, remaining distance = (30 - 20) = 10 km Time taken by B to distance travelled 10 km = (10/30 x 60) = 20 min Total time taken by A = 20+10 = 30 min |
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126. |
How many positive integers less than 1000 are 6 times the sum of their digits? (a) 0 (b) 1 (c) 2 (d) 3 |
Answer» Answer: B. Number less than 1000 can write abc =100a + 10b +c where a, b, c ∈ {0,1 2 3....... 9 } and a+b+c > 0 The sum of digits of this number is ( a+b+c). . Given, 100a+10b+c= 6 (a+b+c) ∴ 94a+ 4b -5c = 0 Clearly, a > 0. No solution ∴ a = 0 then 4b = 5c This is possible only b = 5 and c = 4 ∴Number is 54. Hence, only one number i.e., 54. |
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127. |
If P(x) be a polynomial with real coefficients such that P(sin2 x) = P(cos2 x), for all x ∈ [0, π/2]. Consider the following statements :I. P(x) is an even function.II. P(x) can be expressed as a polynomial in (2x– 1)2 I. P(x) is a polynomial of even degreeThen.(A) all are false(B) only I and II are true(C) only II and III are true (D) all are true |
Answer» Correct option (C) only II and III are true Explanation: P(sin2 x) = P(cos2 x) P(sin2 x) = P(1 – sin2 x) P(x) = P(1 – x) ∀ x ∈ [0, 1] Differentiable both sides w.r.t. x P'(x) = – P'(1 – x) So P '(x) is symmetric about point x = 1/2 So P '(x) has highest degree odd ⇒ P(x) has highest degree even |
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128. |
A quadrilateral has distinct integer side lengths. If the second-largest side has length 10, then the maximum possible length of the largest side is(a) 25(b) 26(c) 27(d) 28 |
Answer» Answer is : (b) 26 We have, side of quadrilateral has distinct integer second largest size has length 10. Let a = 8, b = 9, c = 10, (All are distinct) We know, in quadrilateral Sum of three sides is greater than fourth side ∴ a+b+c > d ⇒ 8+9+10 > d ⇒ d < 27 ∴ Maximum length of 4th side is 26. |
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129. |
A solid cube and a solid sphere of identical material and equal masses are heated to the same temperature and left to cool in the same surroundings. Then,(a) the cube will cool faster because of its sharp edges(b) the cube will cool faster because it has a larger surface area(c) the sphere will cool faster because it is smooth(d) the sphere will cool faster because it has a larger surface area |
Answer» Answer is : (b) the cube will cool faster because it has a larger surface area Area of cube is more than that of a sphere for same mass and density. Cube also have sharp edges that radiates more effectively than a flat surface. So, rate of cooling for cube is much rapid than sphere. Effect of sharp edges is prominent only at very high temperatures. |
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130. |
A spring balance A reads 2 kg with a block of mass m suspended from it. Another balance B reads 3 kg when a beaker with a liquid is put on its pan. The two balances are now so arranged that the hanging mass m is fully immersed inside the liquid in the beaker as shown in the figure given below.In this situation,(a) the balance A will read 2 kg and B will read 5 kg(b) the balance A will read 2 kg and B will read 3 kg(c) the balance A will read less than 2 kg and B will read between 3 kg and 5 kg(d) the balance A will read less than 2 kg and B will read 3 kg |
Answer» Answer is : (c) the balance A will read less than 2 kg and B will read between 3 kg and 5 kg When block is dipped in water, it displaces some water which exerts buoyant force on block. As a result, reading on scale A will be lower than 2 kg. Due to reaction of block (which is equal to buoyant force), beaker of water is pushed. So, reading of scale B will be more than 3 kg. |
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131. |
The intensity of sound during the festival season increased by 100 times. This could imply a decibel level rise from - (A) 20 to 120 dB(B) 70 to 72 dB(C) 100 to 10000 dB(D) 80 to 100 dB |
Answer» Correct option (D) 80 to 100 dB Explanation: Loudness of sound in decibel dB = 10 log10 (I/I0) when intensity of sound become 100 I then new decibel level = dB' = 10 log10 (100I/I0) dB' – dB = 10 log 10 100 dB' – dB = 20 ∴ decibel rise by 20 dB only one option i.e. 80 to 100 dB match with it |
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132. |
A planet of mass is moving around a star of mass M and radius R in a circular orbit of radius r. The star abruptly shrinks to half its radius without any loss of mass. What change will be there in the orbit of the planet?(a) The planet will escape from the star(b) The radius of the orbit will increase(c) The radius of the orbit will decrease(d) The radius of the orbit will not change |
Answer» Answer is : (d) The radius of the orbit will not change When star shrinks without losing its mass, its gravitational acceleration on its surface increases but there is no change in force exerted by this star on a distant object like a planet. Force of gravitational attraction of star on planet is F = GMm/r2 where, M = mass of star, m = mass of planet and r = orbital radius of planet. As, this force remains same there is no change in orbital radius of planet. |
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133. |
The volume of oxygen at STP required to burn 2.4 g of carbon completely is(a) 1.12 L(b) 8.96 L(c) 2.24 L(d) 4.48 L |
Answer» Answer is : (d) 4.48 L C(s) + O2(g) → CO2(g) 1 mole of carbon reacts completely with 1 mole of oxygen to produce 1 mole of CO2. 12 g of C reacts = 1 mole of O2 2.4 g of C reacts with = 1/12 x 2.4 = 0.2 mole of O2 At STP 1 mole of O2 contains = 22.4 L ∴ 0.2 mole of O2 contains = 22.4 x 0.2 = 4.48 L |
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134. |
The cyclic electron flow during photosynthesis generates(A) NADPH alone.(B) ATP and NADPH.(C) ATP alone.(D) ATP, NADPH and O2. |
Answer» Correct option (C) ATP alone. Explanation: In cyclic photophosphorylation only ATP is formed. |
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135. |
In a mixed culture of slow and fast growing bacteria, penicillin will(A) kill the fast growing bacteria more than the slow growing(B) kill slow growing bacteria more than the fast growing(C) kill both the fast and slow growing bacteria equally(D) will not kill bacteria at all |
Answer» Correct option (A) kill the fast growing bacteria more than the slow growing Explanation: Penicillin inhibit the cell wall formation in bacteria thus it kill rapidly growing bacteria more than the slow growing bacteria. |
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136. |
Consider the linear double stranded DNA shown below. The restriction enzyme sites and the lengths demarcated are shown. This DNA is completely digested with both EcoRI and BamHI restriction enzymes. If the product is analyzed by gel electrophoresis, how many distinct bands would be observed ? (A) 5 (B) 2 (C) 3 (D) 4 |
Answer» Correct option: (c) 3 |
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137. |
A person has 400 million alveoli per lung with an average radius of 0.1 mm for each alveolus. Considering the alveoli are spherical in shape, the total respiratory surface of that person is closest to (A) 500 mm2 (B) 200 mm2 (C) 100 mm2 (D) 1000 mm2 |
Answer» Answer: 1000mm2 Explanation: In lungs, alveoli are tiny sacs which allow oxygen and carbon dioxide to move between the lungs and bloodstream. Average radius of spherical alveoli is 0.1 mm Total number of Alveoli is 400 × 2 = 800 million (since 400 million alveoli in 1 lung) Surface area of a sphere is 4πr2r2 Therefore, total respiratory surface Area = 4πr2r2 × 800 million= 1000mmm2m2 (approx) |
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138. |
Phellogen and phellem respectively denote (a) cork and cork cambium (b) cork cambium and cork (c) secondary cortex and cork (d) cork and secondary cortex |
Answer» Correct option is (b) cork cambium and cork In the dicot stem, the cortical cells get differentiated to give rise to another meristematic tissue, which is called cork cambium or phellogen. On the other side, it forms phellem (cork) and in the inner region it forms secondary cortical cells (phelloderm). |
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139. |
Which one of the following enzymes shows the greatest substrate specificity? (a) Lipase (b) Nuclease (c) Pepsin (d) Sucrose |
Answer» Correct option is (d) Sucrose Sucrose shows the most substrate specificity since it hydrolyses only the disaccharide, sucrose. |
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140. |
An isolated box, equally partitioned contains two ideal gases A and B as shownWhen the partition is removed, the gases mix. The changes in enthalpy (∆H) and entropy (∆S) in the process, respectively, are(a) zero, positive(b) zero, negative(c) positive, zero(d) negative, zero |
Answer» Answer is : (a) zero, positive ∆H for this process = CV∆T = 0 (at constant temperature) ∆S for this process will be positive that is ∆S > 0, the randomness increases the molecules of gases A and B gets intermixed with each other, when the partition is removed. |
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141. |
At 298 K, assuming ideal behaviour, the average kinetic energy of a deuterium molecule is(a) two times that of a hydrogen molecule(b) four times that of a hydrogen molecule(c) half of that of a hydrogen molecule(d) same as that of a hydrogen molecule |
Answer» Answer is : (d) same as that of a hydrogen molecule Average kinetic energy depends upon the temperature and not on the type of gases involved. For any gas, (K.E) avg = 3kT/2 per molecule The (K.E)avg of a deuterium molecule is same as that of hydrogen molecule. |
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142. |
During generation of an action potential, depolarisation is due to (a) K+ efflux (b) Na+ efflux (c) Na+ influx (d) K+ influx |
Answer» Correct option is (c) Na+ influx As the membrane potential is increased, sodium ions channels open, allowing the influx of Na+ ions into the cell. The inward flow of sodium ions increases the concentration of positively charged cations in the cell and causes depolarisation, where the potential of the cell is higher than the cell’s resting potential. |
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143. |
Which homeostatic function of the liver is controlled and monitored in the pancreas? (a) Deamination of amino acids (b) Release of glucose (c) Release of iron (d) Release of toxins |
Answer» Correct option is (b) Release of glucose Glucose is stored in the liver as glycogen. Glycogen can be converted to free glucose by the process of glycogenolysis, which involves the activation of a phosphorylase enzyme by the hormone glucagon. Glucagon is made by the pancreas and is released when the blood sugar levels fall. There release of glucose is a homeostatic function of liver that is controlled and monitored in the pancreas. |
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144. |
Cell division is initiated by (a) centrosome (b) centromere (c) centriole (d) None of these |
Answer» Correct option is (a) centrosome Centrosomes are made up of a pair of centrioles and other proteins. The centrosomes are important for cell division and produce microtubules that separate DNA into two new identical cells. |
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145. |
Match the different types of heart given in Column I with organisms given in Column II. Choose the correct combination.Column IColumn IIP.Neurogenic hearti.HumanQ.Bronchial heartii.King crabR.Pulmonary heartiii.Shark(a) P-ii, Q-iii, R-i(b) P-iii, Q-ii, R-i(c) P-i, Q-iii, R-ii(d) P-ii, Q-i, R-iii |
Answer» Answer is : (a) P-ii, Q-iii, R-i – A neurogenic heart requires nervous input to contract. It is seen in crustaceans like king crab. – Bronchial hearts are myogenic accessory pumps found in coleoid cephalopods like shark that supplement the action of the main, systemic heart. – Pulmonary heart is found in humans where the portion of the circulatory system carries deoxygenated blood away from the right ventricle of the heart to the lungs and returns oxygenated blood to the left atrium and ventricle of the heart. |
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146. |
Given below are the four schematics that describe the dependence of the rate of an enzymatic reaction on temperature. Which of the following combinations is true for thermophilic and psychrophilic organisms?(a) P and P(b) P and S(c) P and R(d) R and R |
Answer» Answer is : (d) R and R Both thermophiles and psychrophiles will show same enzymatic reaction graph. Mostly proteinaceous enzymes are labile to temperature. Thermophiles live at very high temperature while psychrophiles live in the range of −20°C to +10°C. In either case, rising temperature will first raise the rate of reaction but if temperature is still raised continuously, enzymes get denatured, hence reaction rate decreases. |
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147. |
Which of the following is the correct order of size of the given species? (a) I > I- > I+ (b) I+ > I- > I(c) I > I+ > I-(d) I- > I > I+ |
Answer» Correct option is (d) I- > I > I+ Anion is formed by the gain of electron to the neutral atom and cation is formed after the loss of electron from the neutral atom. Hence, cation has smaller size due to increased nuclear charge whereas anion has bigger size than its neutral atom. |
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148. |
Compared to the atmospheric air, the alveolar air has(A) more pO2 and less pCO2(B) less pO2 and pCO2 (C) more pO2 and more pCO2(D) less pO2 and less pCO2 |
Answer» Correct option (B) less pO2 and pCO2 Explanation: Pressure gradient is essential for movement of gases in and out of the body. |
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149. |
Proteins P, Q, and R are associated with intact organellar membrane in a cell. If the intact organellel is treated with a high ionic strength buffer, only protein R remained associated with the membrane fraction. Based on this, one could conclude that(A) P and Q are peripheral membrane proteins. (B) R is a peripheral membrane protein.(C) P and Q are integral membrane bound proteins.(D) P is peripheral and Q is integral membrane protein. |
Answer» Correct option (A) P and Q are peripheral membrane proteins. Explanation: Peripheral proteins can be detached earily |
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150. |
The genetic distance between genes A and B is 10 cm. An organism with Ab combination of the alleles is crossed with the organism with aB combination of alleles. What will be the percentage of the gametes with AB allele combination by an F1 individual ?(A) 1(B) 5(C) 10(D) 50 |
Answer» Correct option (B) 5 Explanation: Recombinants formed = 10% (AB 5% & ab 5%) |
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