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.
| 8851. |
How is a mature, functional insulin hormone different from its prohormone form? |
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Answer» Mature functional insulin is obtained by processing of pro-hormone which contains extra peptide called C-peptide. This C-peptide is removed during maturation of proinsulin to insulin |
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| 8852. |
What will be the distribution of phenotypic features in the first generation after a cross between a homozygous female and a heterozygous male for a single locus ?A. `3:1`B. `1: 2: 1`C. `1: 1`D. none of these |
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Answer» Correct Answer - C `F_(1)` or first filial or progeny is the generation iof hybrids produced from a cross between the genetically different individuals called parents. A cross between homozygous and heterzygous parents for a single locus will produce `1:1` ratio of phenotypic features in `F_(1)` generation. |
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| 8853. |
When two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters is independent of the other pair of characters. The statement is independent of the other pair of characters. The statement explains which of the following laws/principles of Mendel ?A. Principle of paired factorsB. Principle of dominanceC. Law of segregationD. Law of independent assortment |
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Answer» Correct Answer - D Cross involving two contrasting character is called dihybrid cross or a two factor cross. The two factors of each trait assort at random and independent of the factors of other traits at the time of meiosis (gametogenesis) and get randomly as well as independently rearranged in the offspring producing both parental and new combination of traits. This forms the basis of the law of independent assortment given by Mendel. |
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| 8854. |
The percentage of ab gamete produced by AaBb parent will beA. `25 %`B. `50 %`C. `75 %`D. `12.5 %` |
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Answer» Correct Answer - A Gametes produced by AaBb parent would be 25 % AB, 25 % aB, 25 % Ab and 25 % ab. |
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| 8855. |
Define the terms Antigen and Antibody. Name any two diagnostic kits based upon them. |
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Answer» An antigen is a foreign substance that elicits the formation of an antibody. Antibody is a protein that is synthesised in response to an antigen. Antigen and antibody show high degree of specificity in binding each other Two diagnostic kits based on antigen-antibody interaction are. a. ELISA for HIV. b. Pregnancy test kits. |
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| 8856. |
How is a mature, functional insulin hormone different from its pro-hormone form? |
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Answer» Mature functional insulin is obtained by processing of pro-hormone which contains extra peptide called C-peptide. This C-peptide is removed during maturation of proinsulin to insulin. |
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| 8857. |
How many types of gametes can be produced by a diploid organism who is heterzygous for 4 loci ?A. 4B. 8C. 16D. 32 |
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Answer» Correct Answer - C Types of gamtes ` = 2^(n)` wher n is number of heterozygous loci. Thus, gametes produced by a diploid organism could be `2^(4) = 16` |
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| 8858. |
ELISA technique is based on the principles of antigen and antibody interaction. Can this technique be used in the molecular diagnosis of a genetic disorder, such as phenylketonuria? |
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Answer» Yes. One can use antibody against the enzyme (that is responsible for the metabolism of phenylalanine) to develop ELISA based diagnostic technique. The patient where the enzyme protein is absent would give negative result in ELISA when compared to normal individual. |
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| 8859. |
Define the following terms:(a) Aestivation(b) Placentation(c) Actinomorphic(d) Zygomorphic(e) Superior ovary(f) Perigynous flower(g) Epipetalous Stamen |
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Answer» (a) Aestivation |
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| 8860. |
Explain with suitable examples the different types of phyllotaxy? |
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Answer» Phyllotaxy refers to the pattern or arrangement of leaves on the stem or branch of a plant. It is of three types, alternate, opposite, and whorled phyllotaxy. |
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| 8861. |
Read the following passage and make notes by drawing and filling the boxes given below: Vertebrate animals can be either warm-blooded or cold-blooded. A cold-blooded animal cannot maintain constant body temperature. The temperature of its body is determined by the outside surroundings. Cold-blooded animals are also called ‘ectothermic, which means outside heat. They are reptiles, amphibians and fishes. Warm-blooded animals are able to regulate their internal temperature. They have fur and feather to keep them warm. They are also called ‘endothermic’ meaning heat inside. They are birds and mammals. |
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Answer» i. warm-blooded ii. cold-blooded iii. ectothermic iv. reptiles v. amphibians vi. fishes vii. endothermic viii. mammals. |
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| 8862. |
What do you understand by the term bio-pesticide? Name and explain the mode of action of a popular bio-pesticide |
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Answer» Biopesticide is a pesticide which is a. not chemical in nature b. more specific in action against the pest c. safer for environment than chemical pesticides A popularly known bio-pesticide is Bt toxin, which is produced by a bacterium called Bacillus thuringiensis. Bt toxin gene has been cloned from this bacterium and expressed in plants. Bt toxin protein when ingested by the insect, gets converted to its active form due to the alkaline pH of the gut. The activated toxin binds to the surface of midgut epithelial cells and create pores that cause cell swelling and lysis and eventually kills the insect. |
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| 8863. |
What will happen to an ecosystem ifa. All producers are removed;b. All organisms of herbivore level are eliminated; andc. All top carnivore population is removed |
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Answer» (a) Reduction in primary productivity. No biomass available for consumption by higher trophic levels/heterotrophs (b) Increase in primary productivity and biomass of producers. Carnivores population will subsequently dwindle due to food shortage. (c) Increase in number of herbivores, Overgrazing by herbivores, Desertification |
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| 8864. |
In an aquarium two herbivorous species of fish are living together and feeding on phytoplanktons. As per the Gausses principle, one of the species is to be eliminated in due course of time, but both are surviving. How? And what possibly happened to both the species? |
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Answer» Each species has a specific position or functional role within the community, called niche. According to the Gausses principle, no two species can live in the same niche. In this case, two herbivorous species are living in the same niche and feeding on phytoplanktons. It may be because of the availability of sufficient phytoplanktons/and or less number of individuals of the fish species. of the two species might have occurred. And though neither of the species have been eliminated, niche overlapping may effect the growth and development of individuals of the species. |
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| 8865. |
What do you understand by the term bio-pesticide? Name and explain the mode of action of a popular bio-pesticide. |
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Answer» Biopesticide is a pesticide which is a. not chemical in nature b. more specific in action against the pest c. safer for environment than chemical pesticides A popularly known bio-pesticide is Bt toxin, which is produced by a bacterium called Bacillus thuringiensis. Bt toxin gene has been cloned from this bacterium and expressed in plants. Bt toxin protein when ingested by the insect, gets converted to its active form due to the alkaline pH of the gut. The activated toxin binds to the surface of midgut epithelial cells and create pores that cause cell swelling and lysis and eventually kills the insect. |
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| 8866. |
In an aquarium, two herbivorous species of fish are living together and feeding on phytoplanktons. As per theGausses principle, one of the species is to be eliminated in due course of time, but both are surviving. How? And what possibly happened to both the species? |
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Answer» Each species has a specific position or functional role within the community, called niche. According to the Gausses principle, no two species can live in the same niche. In this case, two herbivorous species are living in the same niche and feeding on phytoplanktons. It may be because of the availability of sufficient phytoplanktons/and or less number of individuals of the fish species. of the two species might have occurred. And though neither of the species have been eliminated, niche overlapping may effect the growth and development of individuals of the species. |
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| 8867. |
Name the five key tools for accomplishing the tasks of recombinant DNA technology. Also mention the functions of each tool. |
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Answer» i. Restriction endonucleases: for cutting the desired DNA at desired places |
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| 8868. |
Name the five key tools for accomplishing the tasks of recombinant DNA technology. Also mention the functions of each tool. |
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Answer» i. Restriction endonucleases: for cutting the desired DNA at desired places ii. Gel electrophoresis: for separating the desired DNA fragments iii. Ligase enzyme: for creating recombinant DNA molecule. iv. DNA delivery system: like electroporation, microinjection, gene gun method. v. Competant host (usually bacteria / yeast): to take up recombinant DNA |
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| 8869. |
Mating of an organism to double recessive in order to determine whether it is homozygous or heterozygous for a character is calledA. Reciprocal crossB. Back crossC. Dihybrid crossD. Test cross |
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Answer» Correct Answer - D (D) Mating of an organism to double recessive for determining homozygosity or heterozygosity is test cross. |
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| 8870. |
What is the hormone secreted by the growing ovarian follicles? What is the action of that hormone upon the uterus? |
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Answer» The follicles that are growing after menses secrete estrogen. These hormones act upon the uterus stimulating the thickening of the endometrium (the internal mucosa of the uterus). |
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| 8871. |
Blacking of urine when exposed to air a metabolic disorder in human beings. This is due toA. PhenylalanineB. TyrosineC. Homogentisic acidD. Valine replacing glutamic acid |
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Answer» Correct Answer - C (C) Blackening of exposed urine is a metabolic disorder due to homogentisic acid. |
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| 8872. |
What are centrioles? In which type of cell are they present? |
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Answer» Centrioles are tiny cylindrical structures made of nine microtubule triplets. They appear in pairs in the cell. Centrioles participate in the making of the cytoskeleton and of cilia and flagella. In cell division, they play a role in the formation of the aster fibres. Centrioles are structures present in animal cells, in most protists and in some primitive fungi. There are no centrioles in cells of superior plants and in general, it is considered that plant cells do not have centrioles (although this is not entirely correct since some plants have centriole-containing cells). The region where the centrioles are located is called the centrosome of the cell. |
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| 8873. |
What are the main events of the final mitotic period? |
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Answer» The final mitotic phase is telophase. In telophase the following events occur: decondensation of chromosomes, each set located in opposite cell poles; Kotecha formation around each set of chromosomes forming two nuclei; destruction of the mitotic apparatus; reappearing of the nucleoli; beginning of cytokinesis (the division of the cytoplasm to ultimately separate the new cells). |
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| 8874. |
What are the main events of the first mitotic period? |
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Answer» The first mitotic period is prophase. During prophase the following events occur: migration of each centriole pair (centrioles were duplicated in interphase) to opposite cell poles; aster formation around the centriole pairs; formation of the spindle fibres between the two centriole pairs; end of chromosome condensation; the disintegration of the nucleolus; breaking of the karyotheca; dispersion of condensed chromosomes in the cytoplasm; binding of chromosomes to the spindle fibres. |
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| 8875. |
What is the mitotic apparatus? |
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Answer» Mitotic apparatus is the set of aster fibres, radial structures around each centriole pair, plus the spindle fibres, fibres that extend across the cell between the two centriole pairs located in opposite cell poles. The mitotic apparatus appears in prophase and has an important role in the orientation and gripping of chromosomes and other cellular elements causing them to separate and migrate to opposite cell poles. Substances that disallow the formation of the mitotic apparatus, like colchicine, a molecule that binds to tubulin molecules and prevents the synthesis of microtubules interrupt cell division. Colchicine is used to study chromosomes since it paralyzes mitosis when chromosomes are condensed and so are more easily viewed under the microscope. |
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| 8876. |
The picture given below and answer the questions by choosing the most appropriate option:1. Which type of personality assessment is being depicted in the above picture? a. Projective Technique b. Psychometric Tests c. Behavioural Analysis d. Self-report Measures 2. Which of the following is NOT a characteristic of this test? a. It reveals the unconscious mind. b. It can be conducted only on an individual basis. c. Its interpretation is objective. d. The stimuli are unstructured. 3. Identify the name of the test from the options given below. a. Thematic Appreciation Test b. Thematic Apperception Test c. Theatre Apperception Test d. Theatre Appreciation Test 4. Which of the following statements are NOT true of this test? i. In the first phase, called performance proper, the subjects are shown the cards and are asked to tell what they see in each of them. ii. The second phase is called inquiry. iii. Each picture card depicts one or more people in a variety of situations. iv. The subject is asked to tell a story describing the situation presented in the picture.Choose the correct option: a. i, ii b. iii, iv c. i, ii, iii d. ii, iv5. Which of the following is NOT a drawback of this test? a. It requires sophisticated skills and specialised training b. It has problems associated with reliability of scoring c. It has problems associated with validity of interpretations d. It is an indirect measure of assessment.6. Identify the stimuli that are used in such kinds of tests as given in the above picture. a. Stories b. Cartoons c. Ink blots d. Picture cards |
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Answer» 1. a. Projective Technique 2. c. Its interpretation is objective 3. b. Thematic Apperception test 4. a. i,ii 5. d. It is an indirect measure of assessment. 6. d. Picture Cards |
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| 8877. |
The correct order of binary fission in Leishmania is(A) II, III, IV, I(B) I, III, IV, II(C) IV, I, III, II(D) III, I, II, IV |
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Answer» (B) I, III, IV, II |
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| 8878. |
The disposal of urban waste has become a serious concern for the local authorities. Analyze the statement with suitable examples. |
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Answer» The problem of overcrowded, congested and insufficient infrastructure of urban areas results in accumulation of huge urban waste. There are two sources of urban waste. Household or domestic sources and industrial or commercial sources. The mismanagement of urban waste disposal is a serious problem in big cities. Tons of waste come out daily in metropolitan cities and are burnt. The smoke released from the waste pollutes the air. Lack of sewers or other means to dispose of human excretes safely and the inadequacy of garbage collection sources adds to water pollution. The concentration of industrial units in and around urban centres gives rise to a series of environmental problems. Dumping of industrial waste into rivers is the major cause of water pollution. The solid waste generation continues to increase in both absolute and per capita in cities. This improper disposal of solid waste attracts rodents and flies which spread diseases. The thermal plants release a lot of smoke and ash in the air. For example, a plant producing 500mw electricity releases 2000 tons of ash which is difficult to manage. |
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| 8879. |
Why is the frequency of red-green colour blindness is many times higher in males than that in the females? |
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Answer» For becoming colourblind, the female must have the allele for it in both her X-chromosomes; but males develop colourblindness when their solex-chromosome has the allele for it. |
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| 8880. |
Why is the frequency of red-green colour blindness is many timeshigherin males than that in the females? |
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Answer» For becoming colour blind, females must have allele for it in both her X chromosomes. But males develop colour blindness when their sole X chromosome has the allele for it. |
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| 8881. |
A. FlatwormsB. RoundwormsC. EarthwormD. Echinoderms |
| Answer» Round worms is the correct option | |
| 8882. |
Glycosidic linkage at place of branching in starch & glycogen is:(Sorry for the sideways picture) |
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Answer» Correct answer is (A) i.e α(1→6) |
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| 8883. |
Discuss the general characteristics of Group 15 elements with reference to their electronicconfiguration, oxidation state, atomic size, ionisation enthalpy and electronegativity. |
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Answer» (i) Electronic configuration: All the elements in group 15 have 5 valence electrons. Their general electronic configuration is ns2 np3. (iii) Ionization energy and electronegativity First ionization decreases on moving down a group. This is because of increasing atomic |
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| 8884. |
Why does the reactivity of nitrogen differ from phosphorus? |
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Answer» Nitrogen is chemically less reactive. This is because of the high stability of its molecule, |
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| 8885. |
Why does nitrogen show catenation properties less than phosphorus? |
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Answer» Catenation is much more common in phosphorous compounds than in nitrogencompounds. This is because of the relative weakness of the N−N single bond as compared to the P−P single bond. Since nitrogen atom is smaller, there is greater repulsion of electron density of two nitrogen atoms, thereby weakening the N−N single bond. |
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| 8886. |
Assertion : Glucose does not gives 2,4-DNP test Reason : Glucose exists in cyclic hemiacetal form(1) If both assertion and reason are true and reason is the correct explanation of assertion. (2) If both assertion and reason are true but reason is not the correct explanation of assertion. (3) If assertion is true but reason is false. (4) If both assertion and reason are false. |
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Answer» (1) If both assertion and reason are true and reason is the correct explanation of assertion. |
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| 8887. |
Which one of the following reactions is not explained by the open chain Structureof glucose: (a) Formation of pentaacetate of glucose with acetic anhydride. (b) formation of addition product with 2,4 DNP reagent (c) Silver mirror formation with Tollen’s reagent (d) existence of alpha and beta forms of glucose. |
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Answer» Answer is (d) existence of alpha and beta forms of glucose. |
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| 8888. |
In Carius method of estimation of halogen. 0.45 g of an organic compound gave 0.36 g of AgBr. Find out the percentage of bromine in the compound.(Molar masses : AgBr = 188 g mol-1: Br = 80 g mol-1)(A) 34.04%(B) 40.04%(C) 36.03%(D) 38.04% |
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Answer» Correct option is (A) 34.04% Mass of organic compound = 0.45 gm Mass of AgBr obtained = 0.36 gm ∴ Moles of AgBr = \(\frac{0.36}{188}\) ∴ Mass of Bromine = \(\frac{0.36}{188}\) x 80 = 0.1532 gm ∴ % Br in compound = \(\frac{0.1532}{0.45}\) x 100 = 34.04% |
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| 8889. |
Commercially available analogs of 1,25-dihydroxyvitamin D3 (calcitriol) are: a) Doxercalciferol (Hectoral) b) Paricalcitol (Zemplar)c) All of the above d) None of the above |
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Answer» c) All of the above |
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| 8890. |
All of the following antiviral drugs are the analogs of nucleosides, EXCEPT: a) Acyclovir b) Zidovudine c) Saquinavir d) Didanozine |
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Answer» c) Saquinavir |
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| 8891. |
Slip system is (a) Number of slip plane (b) Multiplication of number of slip planes and number of slip direction (c) Number of slip direction (d) None of these |
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Answer» (b) Multiplication of number of slip planes and number of slip direction |
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| 8892. |
The stress required for dislocation to move __________exponentially with the length of the Burgers vector and _________exponentially withthe interplanar spacing of the slip planes. (a) decreases, decreases (b) increases ,increases (c) decreases, increases (d)increases, decreases |
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Answer» The stress required for dislocation to move increases exponentially with the length of the Burgers vector and decreases exponentially withthe interplanar spacing of the slip planes. |
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| 8893. |
Edge dislocations of opposite sign and lying on the same slip plane exert an_________ force on each other(a) attractive (b) repulsive (c) both(d) shear |
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Answer» Edge dislocations of opposite sign and lying on the same slip plane exert an attractive force on each other. |
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| 8894. |
Two edge dislocations of the same sign and lying on the same slip plane exert a ________force on each other (a) attractive (b) repulsive (c) both (d) shear |
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Answer» Two edge dislocations of the same sign and lying on the same slip plane exert a repulsive force on each other . |
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| 8895. |
Copper is ductlile because (a) it is perfect crystal (b) it contain a very high density of dislocation (c)it has glassy structure (d)the stress to move a dislocation in it is low |
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Answer» (d) the stress to move a dislocation in it is low |
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| 8896. |
The strength of a material with no dislocations is _____greater than the strength of a material with a high dislocation density. (a) 20-100 times (b) 100-200 times (c) 200- 300 times (d) 300-400 times |
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Answer» The strength of a material with no dislocations is 20-100 times greater than the strength of a material with a high dislocation density. |
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| 8897. |
Given below are two statements : Statement I : In 'Lassaigne's Test, when both nitrogen and sulphur are present in an organic compound, sodium thiocyanate is formed. Statement II : If both nitrogen and sulphur are present in an organic compound, then the excess of sodium used in sodium fusion will decompose the sodium thiocyanate formed to give NaCN and Na2S.In the light of the above statements, choose the most appropriate answer from the options given below :(A) Both Statement I and Statement II are correct. (B) Both Statement I and Statement II are incorrect.(C) Statement I is correct but Statement II is incorrect.(D) Statement I is incorrect but Statement II is correct. |
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Answer» (A) Both Statement I and Statement II are correct. Both statement I & statement II are correct. |
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| 8898. |
Density of high density polythene is about _______ gm/c.c. (a) 1.18 (b) 1.05 (c) 0.95 (d) 0.99 |
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Answer» Density of high density polythene is about 0.95 gm/c.c. |
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| 8899. |
If 0.1 g of an oranic compound containing phosphorus produces 0.222g of `Mg_(2)P_(2)O_(7)` the percentage of phosphorus present in the compound isA. 31B. 0.2C. 66D. 62 |
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Answer» Correct Answer - D The precentage of P present in the organic compounds is given by Percentage of P `=(62)/(222)xx("weight of" Mg_(2)P_(2)O_(7))/("weight of organic compound")xx100` `=(62)/(222)xx(0.222)/(0.1)xx100=62%` |
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| 8900. |
Density of alumina is nearly______(a) 2g/cc (b) 3 g/cc (c) 4 g/cc (d) 5 g/cc |
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Answer» Density of alumina is nearly g/cc. |
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