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KVPY Class-XII (SB/SX) - 2018

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KVPY Class-XII (SB/SX) - 2018
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  • Question 1
    1 / -0

    Let f: [0, 1] → [-1, 1] and g: [-1, 1] → [0, 2] be two functions such that g is injective and g° f: [0, 1] → [0, 2] is surjective. Then,

    Solution

    Let h(x) = g(f(x))
    And h(x) is onto (given)
    ∴ Co-domain of h(x) = Range of h(x)
    Range of h(x) = [0,2] and h(x) = g(f(x) ).
    It means g is giving [0,2] which is also co-domain of g.
    So, g must be onto.
    Now, domain of g = [-1,1] which must be range of f.
    But, co-domain of f = [- 1,1]
    So, f must be onto.

     

  • Question 2
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    A thin metallic disc is rotating with constant angular velocity about a vertical axis that is perpendicular to its plane and passes through its centre. The rotation causes the free electrons in the disc to redistribute.

    Assume that there is no external electric or magnetic field. Then,

    Solution

    Due to contrifugal force, electrons have a tendency to go towards the circumference.

     

  • Question 3
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    The ionic radii of Na+, F-, O2- and N3- follow the order:

    Solution

    For iso-electronic species, the radii are in the following order:
    Cation < Neutral atom < Anion
    Ionic radii increase with the increase in the –ve charge on anion and decrease with the increase in positive charge on the cation. Hence, the decreasing order of the radii is:
    N3- > O2- > F- > Na+

     

  • Question 4
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    Among C, S and P, the element(s) that produce(s) SO2 on reaction with hot conc. H2SO4 is/are

    Solution

    Sulphuric acids reacts with carbon, sulphur and phosphorous as follows:
    C+ 2H2SO4 → CO2 + 2SO2 + 2H2O
    3S + 2H2SO4 → 3SO2 + 2H2O
    P4 + 6H2SO4 → 4H3PO3 + 6SO2

     

  • Question 5
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    The tendency of X in BX3 (X = F, Cl, OMe, NMe) to form a  bond with boron follows the order:

    Solution

    The extent of back bonding in BCl3 is the least because of the energy difference between the 3p orbital of the Cl atom (containing the lone pair) and the empty 2p orbital in Boron. The lone pair in Fluorine, Oxygen and Nitrogen are in the 2p orbitals, but the extent of back bonding depends on the compatibility in the size of the 2p orbitals in these atoms with that of the Boron atom.
    Hence, the correct order for the extent of back bonding is:
    BCl3 < BF3 < B(OMe)3 < B(NMe2)3

     

  • Question 6
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    Consider the following statements about Langmuir isotherm:

    (i) The free gas and adsorbed gas are in dynamic equilibrium.
    (ii) All adsorption sites are equivalent.
    (iii) The initially adsorbed layer can act as a substrate for further adsorption.
    (iv) The ability of a molecule to get adsorbed at a given site is independent of the occupation of neighboring sites.

    The correct statements are:

    Solution

    The statement (iii) is factually incorrect.
    Langumir's isotherm is based on the following assumptions:

    • All sites are equivalent.
    • Each site can hold at most one molecule of the adsorbate (mono-layer coverage only).
    • There are no interactions between adsorbate molecules and the adjacent sites.

    Hence, the initially adsorbed layer cannot act as a substrate for further adsorption.

     

  • Question 7
    1 / -0

    A buffer solution can be prepared by mixing equal volumes of

    Solution

    Mixing equal volumes of NH4OH (0.2 M) and HCl (0.1M) results in the formation of NH4OH and NH4Cl, which is a basic buffer mixture. This will form a perfect buffer because equal quantities of the weak base (NH4OH) and salt of its conjugate acid ( NH4Cl) is present.

     

  • Question 8
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    What is the maximum number of oxygen atoms that a molecule of hemoglobin can bind?

    Solution

    The mammalian hemoglobin molecule can bind (carry) up to four oxygen molecules or 8 oxygen atoms. Hemoglobin is made up of four symmetrical sub-units and four heme groups. Iron associated with heme binds oxygen.

     

  • Question 9
    1 / -0

    Bt toxin, produced by Bacillus thuringiensis, does not kill the producer because the toxin is

    Solution

    Bt toxin is produced by a soil bacterium called Bacillus thuringiensis. This toxin does not kill the bacteria because when it is present in the bacteria, it is in an inactive protoxin form. It becomes active and toxic when it is consumed by insects, such as lepidopterans, coleopterans and dipterans. The activated toxin (delta endotoxins) binds to the epithelial cells in the midgut of an insect and creates pores that cause swelling, eventually killing the insect.

     

  • Question 10
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    An angiosperm was identified with its endosperm of 6n. Assuming that it is a self-pollinating species, which ONE of the following is the correct ploidy of the parent?

    Solution

    The ploidy of endosperm generally is 3n in angiosperms.
    The ploidy of the endosperm given in the question is 6n
    This implies that the two polar nuclei in the central cell are of ploidy 2n each. Similarly, the ploidy of the pollen is 2n. The formation of 6n endosperm is possible only with the above given set of ploidy.
    Hence, n in this case will be 2n.The ploidy of mature plant is double the ploidy of the pollen, i.e. 2 × 2n = 4n.

     

  • Question 11
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    Which ONE of the following statements is TRUE about viruses?

    Solution

    Virus is non-autonomous outside a living host body. As viruses are obligate intracellular pathogens they cannot replicate without the machinery and metabolism of a host cell. Hence, statement 4 is true.
    The genetic material of a virus can either be DNA or RNA. Adenovirus, herpesvirus, and poxvirus are all examples of DNA viruses. Examples of RNA viruses are Ebola virus and polio virus.
    During attachment and penetration into a host cell, the virus attaches itself to a host cell and injects its genetic material into it. Hence,viruses do not possess a protein coat around its genetic material at all stages of their life cycle.

     

  • Question 12
    1 / -0

    Mitochondrial cristae are infoldings of the:

    Solution

    Mitochondrial cristae are folds of the mitochondrial inner membrane that provide an increase in the surface area. This allows a greater space for processes like the electron transport chain and chemiosmosis which help to produce ATP in the final steps of cellular respiration.

     

  • Question 13
    1 / -0

    In biological nitrogen fixation the enzyme nitrogenase converts:

    Solution

    Nitrogenases are enzymes that are produced by certain bacteria, such as cyanobacteria. These enzymes are responsible for the reduction of nitrogen to ammonia. Nitrogenases are the only family of enzymes known to catalyze this reaction, which is a key step in the process of nitrogen fixation.

     

  • Question 14
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    Which ONE of the following properties of normal cell is lost during its transition to cancerous cell?

    Solution

    Contact inhibition is a process of arresting cell growth when cells come in contact with each other. As a result, normal cells stop proliferating when they form a monolayer. If a cell has plenty of available substrate space, it replicates rapidly and moves freely. Contact inhibition is a powerful anticancer mechanism that is lost in cancer cells.

     

  • Question 15
    1 / -0

    Which ONE of the following gases is produced during fermentation by yeast?

    Solution

    Anaerobic respiration (fermentation) involves the breakdown of carbohydrates in the absence of oxygen. In yeasts, fermentation results in the production of ethanol and carbon dioxide – which can be used in food processing.

     

  • Question 16
    1 / -0

    Serine proteases are called so because they-

    Solution

    Serine proteases (or serine endopeptidases) are enzymes that cleave peptide bonds in proteins, in which serine serves as the nucleophilic amino acid at the enzyme's active site. They are found ubiquitously in both eukaryotes and prokaryotes.Serine proteases fall into two broad categories based on their structure: chymotrypsin-like and subtilisin-like.

     

  • Question 17
    1 / -0

    The maximum number of gametes of the pollens produced by a tall pea plant with round, yellow seeds of the genotype TtRrYY, if the three loci are unlinked, would be:

    Solution

    The number of gametes formed is decided by the number of heterozygous alleles present in the given genotype.
    Genotype of the given plant is : TtRrYY
    In the above genotype Rr and Tt are the 2 heterozygous alleles, so here n = 2. (n = number of heterozygous alleles)
    Total number of gamete formed will be: 2n, here n = 2
    Hence, 22 = 4

     

  • Question 18
    1 / -0

    Which ONE of the following statements is TRUE with respect to human ovary?

    Solution

    Estrogen is a steroid hormone that is responsible for the growth and regulation of the female reproductive system and secondary sex characteristics. Estrogen is produced by the granulosa cells of the developing follicle. Progesterone is a steroid hormone that is responsible for preparing the endometrium for uterine implantation of the fertilized egg. If a fertilized egg implants, the corpus luteum secretes progesterone in early pregnancy until the placenta develops and takes over progesterone production for the remaining period of the pregnancy.

     

  • Question 19
    1 / -0

    Which ONE of the following statements is INCORRECT with respect to human antibodies?

    Solution

    Antibodies are synthesised by 'B' cells. Every B cell is specific to a single antigen, but each cell can produce several thousand matching antibodies per second. This prolific production of antibodies is an integral part of the humoral immune response.

     

  • Question 20
    1 / -0

    Concentration (%) of NaCl isotonic to human blood is

    Solution

    0.9% NaCl solution is isotonic for human blood cell. When blood cells come in contact with such a medium, the intracellular and extracellular fluids are in osmotic equilibrium across the cell membrane, and there is no net influx or efflux of water.

     

  • Question 21
    1 / -0

    Which of the following statements is TRUE about the golgi apparatus?

    Solution

    Golgi apparatus is found in animals. Plant cells usually contain smaller arrays of golgi-type vesicles, called dictyosomes. Golgi is responsible for glycosylation of proteins and lipids. Proteins arrive in vesicles, following their assembly in the endoplasmic reticulum. After processing in the golgi apparatus, they are sorted into golgi vesicles, for secretion, storage, or transport to lysosomes.

     

  • Question 22
    1 / -0

    Creutzfeldt Jakob Disease (CJD) is a transmissible disease caused by a:

    Solution

    Creutzfeldt-Jakob Disease (CJD) is caused by an abnormal misfolded protein in the brain, called a prion. Creutzfeldt-Jakob Disease (CJD) gradually destroys brain cells, and causes tiny holes in the brain. People with CJD have ataxia, or difficulty controlling body movements, abnormal gait, speech, and dementia.

     

  • Question 23
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    A researcher found petrified dinosaur faeces. Which of the following is unlikely to be found in this fossil?

    Solution

    Dinosaurs went extinct about 65 million years ago (at the end of the Cretaceous Period), after living on Earth for about 165 million years. The estimated origin of the bamboo is 30 mya in the mid oligocene age.

     

  • Question 24
    1 / -0

    Which of the pairs of amino-acids contains two chiral centres?

    Solution

    All amino acids, except glycine have at least one chiral centre at the α C. Threonine and isoleucine have an additional chiral centre at C. The β carbon of isoleucine is optically active, just as the β carbon of threonine. These two amino acids, isoleucine and threonine, have the fact that they have two chiral centres, in common.

     

  • Question 25
    1 / -0

    In photosynthetic carbon fixation, which of the following reacts with CO2?

    Solution

    Ribulose 1,5 bisphosphate is the 1st CO2 acceptor in C3 cycle. In the carbon fixation stage, carbon dioxide is attached to RuBP by the enzyme rubisco. The resulting 6-carbon product quickly splits into two molecules of a three-carbon compound (3-phosphoglycerate).

     

  • Question 26
    1 / -0

    Match the diseases in Column-I with the routes of infection in Column-II. Choose the CORRECT combination.

    Column-I Column-II
    P. Tuberculosis i. Contaminated food and water
    Q. Dysentry ii. inhalation of aerosol
    R. Filariasis iii. Contact via skin
    S. Syphilis iv. Sexual intercourse
     - v. Mosquito bite
    Solution

    Tuberculosis (TB) can be acquired after inhalation of aerosol droplets containing the bacilli from the cough of infected individuals.
    Dysentery is an inflammatory disease of the intestine, especially of the colon, which always results in severe diarrhea and abdominal pains. It is caused by consuming contaminated food and water.
    Filariasis is a parasitic disease caused by an infection with roundworms of the Filarioidea type. These are spread by blood-feeding diptera, such as black flies and mosquitoes.
    Syphilis is a bacterial infection usually spread by sexual contact. The disease starts as a painless sore — typically on the genitals, rectum or mouth. Syphilis spreads from person to person via skin or mucous membrane contact with these sores.

     

  • Question 27
    1 / -0

    Suppose z is any root of 11z8 + 20 iz7 + 10iz - 22 = 0, where i = √-1. Then, S = |z|2 + |z| +1  satisfies

    Solution

    Rouche's theorem: This theorem states that for any two complex-valued functions f and g are analytic inside some region K with closed contour ∂K, if |g(z)| < |f(z)| on ∂K, then f and f + g have the same number of zeros inside K, where each zero is counted as many times as its multiplicity.
    Using above theorem we conclude that if 11z8 + 20 iz7 + 10iz - 22 = 0, then 1 < |z| < 2
    Since, S = |z|2 + |z| + 1
    The maximum value of S = 22 + 2 + 1 = 7 and
    The minimum value of S = 12 + 1 + 1 = 3
    Hence, 3 < S < 7.

     

  • Question 28
    1 / -0

    Among Ce (4f1 5d1 6s2), Nd (4f4 6s2), Eu (4f7 6s2) and Dy (4f10 6s2), the elements having highest and lowest 3rd ionization energies, respectively, are

    Solution

    Eu (4f76s2) will have the highest value of the third ionization energy as loss of the 3rd electron would disturb the stable electronic configuration of half filled 4f orbital.
    Ce (4f1 5d1 6s2) will have the lowest value of the third ionization energy as the 3rd electron would be removed form the 5d orbital.

     

  • Question 29
    1 / -0

    Tetrapeptide is made of naturally occurring alanine, serine, glycine and valine. If the C-terminal amino acid is alanine and the N-terminal amino acid is chiral, the number of possible sequences of the tetrapeptide is

    Solution

    Note: Glycine is achiral. Therefore, the possible combinations are:
    (A) Valine serine glycine alanine
    (B) Serine valine glycine alanine
    (C) Valine glycine serine alanine
    (D) Serine glycine valine alanine

     

  • Question 30
    1 / -0

    What is the probability that a human individual would receive the entire haploid set of chromosomes from his/her grandfather?

    Solution

    Humans have 46 chromosome in diploid cells, while haploid cells have 23 chromosomes. Every human being receives 1/2 of 46 chromosomes from each parent. Since there is a 50% chance of getting the grandfather's chromosome for each of the 23 in the cell, the probability of getting every one of the 23 from one grandparent is (0.5)23 or (1/2)23.

     

  • Question 31
    1 / -0

    Which ONE of the following primer pairs would amplify the fragment of DNA given below?

    5'-CTAGTCGTCGAT-(N)300-GACTGAGCTGAGCTG-3'
    3'-GATCAGCAGCTA-(N)300-CTGACTCGACTCGAC-5'

    Solution

    5' → 3' is the direction of initiation of replication with the help of primers in PCR. Both forward and reverse primers are used. The sequence of the primer (5'-CTAGTCGTCGAT-3' ) is used to initiate replication in the lower strand and the sequence (5'-CAGCTCAGCTCAGTC-3') is used to initiate replication in the upper strand.

     

  • Question 32
    1 / -0

    How many different blood groups are possible in a diploid species with ABCO blood grouping system involving IAIBICand IOalleles (IOis recessive and others are co-dominant)?

    Solution

    In ABCO blood group system, except for 'O', all alleles are co-dominant. Hence, seven blood groups can be formed. The possible possible blood groups can be A, B, C, O, AB, AC and BC.

     

  • Question 33
    1 / -0

    Within the exponential phase of growth, if the initial surface area and the growth rate of a leaf are 10 mm2 and 0.015 mm2/hour, respectively, the area of the leaf after 4 days would range from

    Solution

    Equation of exponential growth in plants, Wt = W0 × ert
    W0 = 10 mm2
    r = 0.015 mm2/hour
    t = 4 days → 4 × 24 = 96 hours
    wt = 10 × e0.015 × 96 = 42.2 mm2

     

  • Question 34
    1 / -0

    If the acidic, basic and hydrophobic residues of proteins are considered to be red, green and blue in colour, respectively, then a globular protein in aqueous solution would have

    Solution

    Globular proteins are folded such that their tertiary structure consists of the polar, or hydrophilic, amino acids arranged on the outside and the nonpolar, or hydrophobic, amino acids on the inside of the three-dimensional shape. This arrangement is responsible for the solubility of globular proteins in water. Hence, red and green will be on the surface and blue will be at the core.

     

  • Question 35
    1 / -0

    Match the vitamins listed in Column - I with their respective coenzyme form in Column - II. Choose the correct combination.

    Column - I Column - II
    P. Vitamin B1 i. Thiamine pyrophosphate
    Q. Vitamin B2 ii. Flavin adenine dinucleotide
    R. Vitamin B6 iii. Methylcobalamin
    S. Vitamin B12 iv. Coenzyme A
      v. Pyridoxal phosphate
    Solution

    Vitamin B complex is necessary for vital functions and essential for formation of co-enzymes, necessary for enzymatic reactions.
    Thiamine pyrophosphate is the coenzyme of vitamin B1 or thiamine.
    Flavin adenine dinucleotide is a redox-active coenzyme associated with various proteins, which is involved with several important enzymatic reactions in metabolism.
    Vitamin B2, also called riboflavin, is required for FMN/FAD formation.
    Pyridoxal phosphate (PLP, pyridoxal 5'-phosphate), the active form of vitamin B6, is a coenzyme in a variety of enzymatic reactions.
    Methylcobalamin is a cobalamin, a form of vitamin B12. Methylcobalamin has been used as a supplement in patients with vitamin B12 deficiency and in those with diabetes and other neuropathies.

     

  • Question 36
    1 / -0

    Two independent experiments related to photosynthesis were conducted - one with 18O-labelled water (experiment P) and the other with 14C-labelled CO2 (experiment Q). Which ONE of the following options lists the first labelled products in experiments P and Q, respectively?

    (A) P : O2 Q : 3-phosphoglycerate
    (B) P : 3-phosphoglycerate Q : NADPH
    (C) P : O2 Q : ATP
    (D) P : 3-phosphoglycerate Q : 3-phosphoglycerate
    Solution

    By the use of O18 radioisotope it was confirmed that O2 comes from H2O and not from CO2 during light reaction.
    By the use of C14 radioisotope it was confirmed that C in C6H12O6 comes from CO2. 3-PGA is the first labelled product formed in the C3 photosynthetic cycle.

     

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