KVPY Class-XII (SB/SX)

Result

KVPY Class-XII (SB/SX) - 19-11-2017

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Weekly Quiz Competition

Question 1
1 / -0

The number of 5-tuples (a, b, c, d, e) of positive integers, such that

I. a, b, c, d and e are the measures of angles of a convex pentagon in degrees;
II. a ≤ b ≤ c ≤ d ≤ e; and
III. a, b, c, d and e are in arithmetic progression, is

A
35

B
36

C
37

D
126

Solution

a + b + c + d + e = 540
Let's say first term = a
Common difference = d
n/2[2a + (n - 1)d] = 540
⇒a + 2d = 108

Case-1: d = 0
a = 108
(108, 108, 108, 108, 108)

Case-2: d = 1
(106, 107, 108, 109, 110)
Similarly, it goes up to d = 35.
For d > 35, interior angle > 180^o; which is not possible.
So, from d = 0 to d = 35, total 36 tuples are possible.
Question 2
1 / -0

The distance s travelled by a particle in time t is s = ut - 1/2 gh². The initial velocity of the particle was measured to be u = 1.11 ± 0.01 m/s and the time interval of the experiment was t = 1.01 ± 0.1 s. The acceleration was taken to be g = 9.8 ± 0.1 m/s². With these measurements, the student estimates the total distance travelled. How should the student report the result?

A
1.121 ± 0.1 m

B
1.1 ± 0.1 m

C
1.12 ± 0.07 m

D
1.1 ± 0.07 m

Solution

This question is related to significant numbers as the question asks how the student reports the result.
On analysing the values of u, t and g, the reported result must have two significant numbers.
Question 3
1 / -0

Which of the following statements for air molecules in an air tight container is/are true?

(I) The average speed of molecules is larger than root mean square speed.
(II) Mean free path of molecules is larger than the mean distance between molecules.
(III) Mean free path of molecules increases with temperature.
(IV) The rms speed of nitrogen molecule is smaller than oxygen molecule.

A
Only II

B
II and III

C
II and IV

D
I, II and IV

Solution

According to Kinetic Theory of Gases, for ideal gas, mean free path of molecules is larger than mean distance between molecules.
Question 4
1 / -0

The most abundant metal ion present in the human body is

A
Zn²⁺

B
Ca²⁺

C
Na⁺

D
Fe²⁺

Solution

Calcium is the most abundant metal in the human body, found both in ionic form (necessary for metabolic functions) and in the mineral form in the bones and teeth as hydroxyapatite.
The mineral form of calcium comprises half the weight of our bones and most of our teeth.
Question 5
1 / -0

The complete hydrolysis of XeF₆ results in the formation of

A
XeO₂F₂

B
XeOF₄

C
XeO₃

D
XeO₂

Solution

XeF₆ + 3H₂O → XeO₃ + 6HF
White
explosive
compound
Question 6
1 / -0

Resting membrane potential of a neuron is approximately

A
-70 mV

B
+70 mV

C
-0.7 V

D
+0.7 V

Solution

The resting membrane potential of a neuron is about -70 mV. This means that the inside of the neuron is 70 mV less than the outside. At rest, there are relatively more sodium ions outside the neuron and more potassium ions inside that neuron.
Question 7
1 / -0

Amphimixis is

A
a fusion of pronuclei of male gametes

B
a fusion of pronuclei from male and female gametes

C
a fusion of pronuclei of female gametes

D
the development of a somatic cell into an embryo

Solution

Amphimixis is a type of sexual reproduction, involving the union or fusion of a male and a female gamete.
Question 8
1 / -0

Activation of sympathetic nervous system

A
decreases blood pressure

B
causes pupil contraction

C
increases heart rate

D
causes bronchoconstriction

Solution

The sympathetic nervous system (SNS) releases the hormones (catecholamines - epinephrine and norepinephrine) to accelerate the heart rate.
Question 9
1 / -0

At physiological temperature, sterols in biological membranes

A
increase their fluidity

B
decrease their fluidity

C
increase their permeability to water

D
decrease their permeability to water

Solution

Sterols are essential in all eukaryotic cell membranes. Sterols reduce membrane fluidity and increase membrane rigidity and strength.
Question 10
1 / -0

Which of the following is a heteropolysaccharide?

A
Glycogen

B
Starch

C
Cellulose

D
Hyaluronic acid

Solution

Hyaluronic acid is a polymer of D-Gluconic acid and N-Acetyl D-Glucosamine. So, it is heteropolysaccharide.
Question 11
1 / -0

Bacterial plasmids are genetic entities that

A
are non-transferable to the same bacterial species

B
are capable of independent replication

C
have RNA as genetic material

D
always require integration in the genome for their replication

Solution

Bacterial plasmid is an extra chromosomal genetic material capable of independent replication.
Question 12
1 / -0

Skin prick test on the forearm is conducted to identify the responsible allergen. This is because

A
of the presence of mast cells under the skin

B
the lymphocytes migrate rapidly from the blood to the skin

C
hair follicles can enhance the reaction

D
the neutrophils migrate rapidly from the blood to the skin

Solution

Skin prick test measures specific IgE attached to mast cells in the skin. SPT is usually the first test recommended when an allergy is suspected.
Question 13
1 / -0

Which of the following processes in E.coli does not directly involve RNA?

A
DNA replication

B
Transcription

C
Translation

D
DNA repair

Solution

RNA primers are involved in DNA replication. Transcription and translation directly involve RNA molecules. DNA repair does not involve any RNA molecules.
Question 14
1 / -0

Which of the following statements is incorrect for translation in cytoplasm?

A
One codon codes for only one amino acid.

B
One amino acid may be coded by many codons.

C
More than one amino acids are coded by one specific codon.

D
There are some codons that do not code for any amino acid.

Solution

One specific codon codes for only one amino acid. It cannot code for more than one amino acid.
Question 15
1 / -0

Two homozygous parents harbouring two different alleles of a gene, exhibiting incomplete dominance for flower colour, were used for a genetic experiment. Which of the following statements is incorrect?

A
The F₂ generation will consist of plants of three different flower colours.

B
The genotypic and phenotypic ratios obtained in the F₂ generation will be different.

C
The F₁ generation will be of a different flower colour as compared to both the parents.

D
The genotypic ratio obtained in the F₂ generation will be the same, irrespective of being of complete dominance or incomplete dominance.

Solution

Both genotypic and phenotypic ratios obtained in F₂ generation will be the same. It will be 1 : 2 : 1.
Question 16
1 / -0

Which of the following conditions is essential for a population to be at Hardy-Weinberg equilibrium?

A
Random mating

B
Immigration

C
Emigration

D
Geographical isolation

Solution

Five conditions are required for a population to remain at Hardy-Weinberg equilibrium.

1. A large breeding population
2. Random mating
3. No change in allelic frequency due to mutation
4. No immigration or emigration
5. No natural selection
Question 17
1 / -0

Inbreeding in a population leads to

A
decrease in recessive disorders

B
heterosis

C
increase in homozygosity

D
increase in heterozygosity

Solution

Inbreeding is the production of offspring from the mating or breeding of individuals or organisms that are closely related genetically. It leads to increase in homozygosity in the population.
Question 18
1 / -0

If a pure chlorophyll solution is illuminated with ultraviolet light, then the solution appears

A
green

B
violet

C
red

D
black

Solution

In isolation, some pigments (including chlorophyll) emit light as well as heat after absorbing photons. As excited electrons fall back to the ground state, photons are given off. This afterglow is called fluorescence. If a solution of chlorophyll is isolated from chloroplasts, it will fluoresce in the red-orange part of the spectrum and also give off heat.

Question 19

1 / -0

Botanical names of plants are given in Column - I and the family/order names are given in Column - Il. Choose the appropriate combination from the options given below.

Column - I	Column - Il
(P) Tamarindus indica	(i) Arecaceae
(Q) Cocos nucifera	(ii) Liliaceae
(R) Coichicum automnale	(iii) Solanaceae
(S) Withania somnifera	(iv) Papilionaceae

P - iv, Q - i, R - ii, S - iii

P - iv, Q - ii, R - iii, S - i

P - i, Q - ii, R - iv, S - iii

P - iv, Q - i, R - iii, S - ii

Solution

Column - I	Column - Il
(P) Tamarindus indica	(iv) Papilionaceae
(Q) Cocos nucifera	(i) Arecaceae
(R) Coichicum automnale	(ii) Liliaceae
(S) Withania somnifera	(iii) Solanaceae

Question 20
1 / -0

Nitrogen fixation is inhibited by oxygen. However, in aerobic nitrogen fixing bacteria, nitrogen is fixed in the presence of oxygen. Nitrogenase in such organisms is protected by

A
converting oxygen to form ozone

B
removal of oxygen by leghaemoglobin

C
utilising oxygen for membrane remodelling

D
utilising oxygen for synthesis of pentapeptide chain in peptidoglycan

Solution

In nodules where nitrogen-fixation occurs, the plant tissues contain the oxygen-scavenging molecule (leghaemoglobin). The function of this molecule in nodules is to reduce the amount of free oxygen and to protect the nitrogen-fixing enzyme nitrogenase, which is irreversibly inactivated by oxygen.
Question 21
1 / -0

Frederick Griffith performed an experiment where mice were killed when injected with a mixture of heat killed S-type Streptococcus (HKS) and live R-type Streptococcus (LRS), but not with HKS or LRS separately. The mice were killed because

A
lipids from HKS made LRS virulent

B
RNA from HKS transformed LRS and made it virulent

C
proteins from HKS made LRS virulent

D
DNA from HKS transformed LRS and made it virulent

Solution

Transformation occurs when DNA is taken up by R-strain from dead S-strain.
Question 22
1 / -0

In diabetic patients, the pH of blood plasma can decrease. This leads to acidosis. It occurs because tissues catabolise

A
amino acids, leading to loss of buffering capacity of the blood

B
stored glycogen, leading to the accumulation of pyruvic acid

C
stored fatty acids, leading to the accumulation of beta hydroxybutyric acid and acetoacetic acid

D
nucleic acid pool, leading to decrease in blood pH

Solution

In diabetic patients, excessive breakdown of fatty acids occurs. This leads to the production of acetoacetic acid and β-hydroxy butyric acid which decreases pH of blood.
Question 23
1 / -0

If the number of alveoli in an individual is doubled without changing the total alveolar volume, then the gas exchange capacity of the lungs will

A
increase for both O₂ and CO₂

B
decrease for both O₂ and CO₂

C
remain unaltered for both O₂ and CO₂

D
increase for O₂ and decrease for CO₂

Solution

With increase in the number of alveoli, the surface area will increase. Hence, the gas exchange capacity will increase for both oxygen and carbon dioxide.
Question 24
1 / -0

In an experiment, bacteria were infected with ³²P labelled viruses in the ratio 5 : 1. The culture was rigorously shaken, followed by centrifugation. Radioactivity was

A
lost due to metabolic activity

B
detected in supernatant as inorganic phosphate

C
detected in the supernatant in association with viral capsid

D
detected in bacterial cell pellet

Solution

The experiment mentioned in the question is in accordance with the Hershay and Chase blending experiment. ³²P activity is found in the pellet and ³⁵S activity is found in the supernatant.
Question 25
1 / -0

Let A be a set of all permutations a₁, a₂, ..., a₆ of 1, 2, ..., 6 such that a₁, a₂, ..., a_k is not a permutation of 1, 2, .., k for any k, 1 ≤ k ≤ 5. Then, the number of elements in A is

A
192

B
408

C
312

D
528
Question 26
1 / -0

Let f(x) be a polynomial with integer coefficients satisfying f(1) = 5 and f(2) = 7. The smallest possible positive value of f(12) is

A
5

B
7

C
27

D
15

Solution

f(x) = ax + b
5 = a + b ...(i)
7 = 2a + b ...(ii)
Solving, a = 2 and b = 3
f(x) = 2x + 3
f(12) = 24 + 3
f(12) = 27
Question 27
1 / -0

Consider the following statements regarding the photoelectric effect experiment.

(I) Photoelectrons are emitted as soon as the metal is exposed to light.
(II) There is a minimum frequency below which no photo-current is observed.
(III) The stopping potential is proportional to the frequency of light.
(IV) The photo-current varies linearly with the intensity of light.

Which of the above statements indicate that light consists of quanta (photons) with energy proportional to frequency?

A
I and III only

B
II and III only

C
II, III and IV only

D
I, II and III only

Solution

(I) Photoelectrons are emitted as soon as the metal is exposed to light.
(II) There is a minimum frequency below which no photo-current is observed.
(III) The stopping potential is proportional to the frequency of light.
Question 28
1 / -0

Suppose, non-linked autosomal genes A, B and C control coat colour in an animal, the dominant alleles A, B and C are responsible for dark colour and the recessive alleles a, b and c are responsible for light colour. If a cross between a male of AABBCC genotype and a female of aabbcc genotype produces 640 offsprings in the F₂ generation, then how many of them are likely to be of the parental genotype?

A
10

B
80

C
160

D
640

E
640

Solution

In a trihybrid cross, a total of 64 progeny are produced. Out of these, 8 are of parental genotype. In the problem given, 640 progeny are formed.
Hence, 8/64 × 640 = 80
Question 29
1 / -0

In a population of families having three children each, the percentage of population of families having both boys and girls is

A
10

B
25

C
50

D
75

Solution

Probability of 2 boys and 1 girl = 3/8
Probability of 1 girl and 2 boys = 3/8
Total probability = 3/8 + 3/8 = 6/8 = 75%
Question 30
1 / -0

Matthew Meselson and Franklin Stahl grew E.coli (doubling time is 20 mm) in a medium containing ¹⁵NH₄CI for many generations. Then, it was transferred to a medium containing ¹⁴NH₄CI. After 40 minutes, the cells were harvested and DNA was extracted and subjected to cesium chloride density gradient centrifugation. The proportion of light and hybrid DNA densities will be

A
50% light and 50% hybrid DNA

B
100% light DNA

C
100% hybrid DNA

D
25% light and 75% hybrid DNA

Solution

DNA replication is semiconservative.
Question 31
1 / -0

Which of the following statements is correct in a population interaction between the species X and the species Y?

A
When X benefits and Y is disadvantaged, it is a competition.

B
When both X and Y benefit, it is mutualism.

C
When both X and Y are disadvantaged, it is predation.

D
When both X and Y are disadvantaged, it is parasitism.

Solution

Both species benefit in mutualism.
Question 32
1 / -0

The protein P, the oligosaccharide O and the oligonucleotide N are composed of 100 amino acid residues, 100 hexose residues and 100 nucleotides, respectively. Which of the following orders of molecular weights is correct?

A
P > O > N

B
P > N > O

C
N > O > P

D
O > P >N

Solution

Monomer of protein is amino acid, oligosaccharide is of sugar and oligonucleotide is of nucleotides. Order of its molecular weight is nucleotide > monosaccharide > amino acid.
Question 33
1 / -0

An octapeptide (NH₂-Asn-Glu-Tyr-Lys-Trp-Met-Glu-Gly) is subjected to complete protease and chemical digestion. Choose the incorrect option.

A
Trypsin generates mixtures of dimer and trimer.

B
Trypsin generates tetramers only.

C
Cyanogen bromide generates a hexamer and a dimer.

D
Chymotrypsin generates mixture of dimmer and trimers.

Solution

When an octapeptide (NH2-Asn-Glu-Tyr-Lys-Trp-Met-Glu-Gly) is subjected to complete protease and chemical digestion,the products are formed depending upon the cleavage sites given below:-
For trypsin - After Lys and Arg
For chymotrypsin - After Phe, Trp or Tyr
For cyanogen bromide - After Met

Question 34

1 / -0

Match the enzymes in Column - I with their respective biochemical reactions in Column - Il.

Column - I	Column - Il
(P) Transaminases	(i) Removal of phosphoryl group from a specific amino acid
(Q) Protein kinases	(ii) Removal of -amino group from a specific amino acid
(R) Protein Phosphatases	(iii) Addition of phosphoryl group to a specific amino acid
(S) Dehydrogenases	(iv) Interconversion of optical isomers
	(v) Oxidation and reduction of substrates

P - iv, Q - ii, R - iii, S - v

P - ii, Q - i, R - ii, S - iv

P - ii, Q - iii, R - i, S - v

P - v, Q - ii, R - iii, S - i

Solution

Column - I	Column - Il
(P) Transaminases	(ii) Removal of alpha-amino group from a specific amino acid
(Q) Protein kinases	(iii) Addition of phosphoryl group to a specific amino acid
(R) Protein phosphatases	(i) Removal of phosphoryl group from a specific amino acid
(S) Dehydrogenases	(v) Oxidation and reduction of substrates

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Re-Attempt Weekly Quiz Competition

KVPY Class-XII (SB/SX) - 19-11-2017

Result

KVPY Class-XII (SB/SX) - 19-11-2017

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SHARING IS CARING

Question 1

35

36

37

126

Solution

Question 2

1.121 ± 0.1 m

1.1 ± 0.1 m

1.12 ± 0.07 m

1.1 ± 0.07 m

Solution

Question 3

Only II

II and III

II and IV

I, II and IV

Solution

Question 4

Zn2+

Ca2+

Na+

Fe2+

Solution

Question 5

XeO2F2

XeOF4

XeO3

XeO2

Solution

Question 6

-70 mV

+70 mV

-0.7 V

+0.7 V

Solution

Question 7

a fusion of pronuclei of male gametes

a fusion of pronuclei from male and female gametes

a fusion of pronuclei of female gametes

the development of a somatic cell into an embryo

Solution

Question 8

decreases blood pressure

causes pupil contraction

increases heart rate

causes bronchoconstriction

Solution

Question 9

increase their fluidity

decrease their fluidity

increase their permeability to water

decrease their permeability to water

Solution

Question 10

Glycogen

Starch

Cellulose

Hyaluronic acid

Solution

Question 11

are non-transferable to the same bacterial species

are capable of independent replication

have RNA as genetic material

always require integration in the genome for their replication

Solution

Question 12

of the presence of mast cells under the skin

the lymphocytes migrate rapidly from the blood to the skin

hair follicles can enhance the reaction

the neutrophils migrate rapidly from the blood to the skin

Solution

Zn²⁺

Ca²⁺

Na⁺

Fe²⁺

XeO₂F₂

XeOF₄

XeO₃

XeO₂

The F₂ generation will consist of plants of three different flower colours.

The genotypic and phenotypic ratios obtained in the F₂ generation will be different.

The F₁ generation will be of a different flower colour as compared to both the parents.

The genotypic ratio obtained in the F₂ generation will be the same, irrespective of being of complete dominance or incomplete dominance.

increase for both O₂ and CO₂

decrease for both O₂ and CO₂

remain unaltered for both O₂ and CO₂

increase for O₂ and decrease for CO₂