Sample questions for CET in GENETICS

GENETICS

1. Which one of the following phenomena of hereditary characters was not known to Mendel ?
(1) Assortment according to probability
(2) Independent assortment
(3) Dominant – recessive traits
(4) Linkage
2. In a monohybrid cross, the phenotypic ratio in the F2 generation will be
(1) 1 : 2 : 1
(2) 3 : 1
(3) 1 : 1
(4) 9 : 3 : 3 : 1
3. A garden pea plant heterozygous for tallness is selfed and 1200 seeds are subsequently germinated. How many seedlings would have the parental phenotype ?
(1) 300
(2) 600
(3) 900
(4) 1200
4. Five out of 20 plants obtained by selfing a red flowered plant were having white flowers. This is an indication that the plant is
(1) Homozygous
(2) Heterozygous
(3) Hemizygous
(4) Homogeneous
5. A pea plant has two genes for height. But, each of its sperm has only one. This illustrates
(1) independent assortment
(2) linked genes
(3) codominance
(4) segregation
6. When a plant heterozygous for tallness is selfed, the F2 generation will have both tall and dwarf plants. This proves the principle of
(1) Dominance
(2) Independent assortment
(3) Segregation
(4) Incomplete dominance
7. To determine heterzyogosity or homozygosity, a plant must be crossed with
(1) Recessive plant
(2) Dominant plant
(3) Homozygous plant
(4) Heterozygous plant
8. A hybrid tall plant of pea can be distinguished from a pure tall pea plant by
(1) Selfing and noting that all plants are short
(2) Selfing and noting that all plants are tall
(3) Crossing with dwarf plant & noting that all plants are tall
(4) Crossing with dwarf plant and noting that 50% are tall and 50% are short
9. In a plant, red flower colour is dominant over white flower colour. A true breeding plant producing red flower is crossed with a pure plant producing white flowers. After selfing the F1 plants, the proportion of plants producing white flowers in the total progeny would be
(1) One third
(2) One fourth
(3) Three fourth
(4) Half
10. If a male who is heterozygous for an autosomal trait mates with a female who is also heterozygous for that trait, what percent of their offsprings are likely to be heterozygous for this trait as well ?
(1) 0
(2) 25
(3) 50
(4) 75
11. When a particular characteristic of an individual like flower colour shows variation among the offspring produced after the individual is selfed, it is said to be
(1) Pure breeding
(2) True breeding
(3) Homozygous
(4) Heterozygous
12. If a heterozygous tall plant is crossed with homozygous dwarf plant, the proportion of the dwarf progeny will be
(1) 25%
(2) 50%
(3) 75%
(4) 100%
13. Two pea plants were subjected to cross pollination. Of the 183 plants produced in the next generation, 94 plants were found to be tall and 89 plants were found to be dwarf. The genotypes of the two parental plants are likely to be
(1) TT and tt
(2) Tt and Tt
(3) Tt and tt
(4) TT and TT
14. The allele for coloured seed coat (S) is dominant over that of white seed coat (s). When flowers of a plant with white seed coat (ss) are pollinated with those of coloured seed coat (Ss), the colour of the coat in the developing seeds will be
(1) White
(2) Coloured
(3) Mosaic
(4) White and coloured in 1 : 1 ratio
15. A monohybrid produces 2 kinds of gametes. A dihybrid produces 4 kinds of gametes. If ‘n’ is the number of characters, how many kinds of gametes are produced by a hybrid ?
(1) 2 x n
(2) 2n
(3) 2/n
(4) n/2
16. Tallness (T) and round shape of seed (R) are dominant over dwarfness (t) and wrinkled shape (r). What will be the ratio of tall plants producing wrinkled seeds in the cross TtRr x TtRr ?
(1) 2/32
(2) 4/32
(3) 6/32
(4) 8/32
17. If a cell of an individual with the genotype AaBb undergoes meiosis, the possible genotypes of the gametes will be
(1) Aa, Bb, AB, ab
(2) AA, BB, aa, bb
(3) AB, Aa, BB, AA
(4) AB, Ab, aB, ab
18. If ‘B’ and ‘b’ are the two alternate forms (alleles) of a gene for a particular trait, which of the following crosses would produce 50% homozygotes and 50% heterozygotes ?
(1) BB x Bb only
(2) Bb x Bb only
(2) Bb x Bb and Bb x bb only
(4) Bb x Bb, BB x Bb and Bb x bb
19. The genetic ratio (phenotypic) of 9 : 3 : 3 : 1 is due to
(1) Incomplete dominance
(2) Independent assortment of genes
(3) Crossing over of chromosomes
(4) Synapsis between homologous chromosomes
20. The F2 generation of a cross produced identical phenotypic and genotypic ratios. It is not an expected Mendelian ratio, and can be attributed to
(1) Independent assortment
(2) Linkage
(3) Incomplete dominance
(4) Principle of dominance
21. In monohybrid crosses, absence of complete dominance is indicated by the F1 plants that exhibit intermediate characters. This can be further confirmed if the phenotypic ratio in F2 is
(1) 3 : 1
(2) 1 : 1
(3) 1 : 2 : 1
(4) 2 : 1 : 1
22. In Mirabilis jalapa, true bred red and white flowered plants are crossed. In F2 generation, they form
(1) 75% red and 25% white flowered plants
(2) 25% red and 75% white flowered plants
(3) 50% red and 50% white flowered plants
(4) 25% red, 50% pink and 25% white flowered plants
23. When two pink flowered four ‘O’ clock (Mirabilis jalapa) plants are selfed, the proportion of the progeny having pink flowers is
(1) 0
(2) ¼
(3) ½
(4) ¾
24. There is a plant in which genes for black flower colour and white flower colour donot show complete dominance or recessiveness. If a plant carrying black flower colour genes is crossed with plant having only white flower colour genes, all the off-springs have grey coloured flowers. If two of these grey flowered plants are crossed, the theoretical progeny ratio would be
(1) All grey
(2) Either all black or all white
(3) 50 % black and 50 % white
(4) 50% grey, 25% black and 25% white
25. In Mirabilis jalapa, RR, Rr and rr determine red, pink and white colours (of flowers) respectively. When F1 hybrid of RR and rr was crossed with dominant parent, the ratio produced is
(1) All red
(2) 2 red : 2 pink
(3) All white
(4) 2 pink : 2 white
26. The molecules on the membranes of RBCs that make up a person’s blood type are
(1) phosphates
(2) lipids
(3) nucleotides
(4) proteins
27. Agglutinogens are present
(1) In the plasma of blood
(2) On the plasma membrane of RBC
(3) In the W B C
(4) In the cytoplasm
28. Agglutinogens are not found in the following blood group;
(1) A
(2) B
(3) AB
(4) O
29. A person is heterozygous for blood group A. His blood group is determined by
(1) Different alleles on the same chromosome
(2) Different alleles on the chromosomes of a homologous pair
(3) Similar alleles on the same chromosome
(4) Similar alleles on the chromosomes of a homologous pair
30. The blood comes under group ‘O’ if:
(1) The blood shows coagulation with both antisera A & B
(2) The blood does not coagulation with antisera A or B
(3) The blood shows coagulation with antiserum A
(4) The blood shows coagulation with antiserum B
31. If both father and mother have blood group `O’, the possible blood group of the children will be
(1) O
(2) A, B, AB, O
(3) A, AB
(4) B, AB
32. If the blood groups of parents are AB and B, what would be the possible blood group of their children ?
(1) A, B and O
(2) A, B and AB
(3) A and B
(4) O
33. If the genotypes for blood groups of parents are IA IB and IB IO, what would be the possible blood groups of their children ?
(1) A, B and AB
(2) A, B and O
(3) A and B
(4) All possible
34. Three children in a family have blood types O, AB and B. What could be the genotypes of their parents ?
(1) IA IB and i i
(2) IA i and IB i
(3) IA IA and IB I i
(4) IB IB and IA IA
35. A child with blood group genotype IAIB is born of a woman with genotype IBIB. The father could not be a man of genotype
(1) IAIB
(2) IBIO
(3) IAIO
(4) IAIA
36. A woman sues a man for the support of her child. She has blood group ‘A’, her child ‘O’ and the man ‘B’. It implies that
(1) The man could be the father of this child
(2) Both woman and man could together produce such a child
(3) They cannot have a child with ‘O’ group
(4) Both (a) and (b) are possible
37. A poor woman and a rich woman delivered babies at the same time in a local nursing home. There was a mix-up. Both the women claimed the male baby as theirs and nobody claimed the female one. The blood group of the disputed female baby is ‘O’, that of poor woman is AB and that of rich woman is A. Who is the actual mother of the female baby ?
(1) Poor woman
(2) Rich woman
(3) Neither of the two
(4) Data insufficient
38. Biologically marriage should be avoided between
(1) Rh+ male and Rh+ female
(2) Rh+ male and Rh- female
(3) Rh- male and Rh+ female
(4) Rh- male and Rh- female
39. When a Rh– mother and Rh+ father produce a Rh– baby,
(1) mother produces antibodies which enter foetal blood causing erythroblastosis foetalis
(2) there will be still birth if the mother is not administered Rhogam antibodies
(3) mother does not produce any antibody and the child will be norml
(4) the baby will always have congenital defects
40. Usually, the recessive character is expressed only when the allele is present in double recessive condition. However, single recessive gene can express itself when the gene is present on
(1) Either an autosome or X – chromosome
(2) The X - chromosome of the male
(3) The X - chromosome of the female
(4) Any autosome
41. A sex linked character is inherited
(1) Only in males
(2) Only in females
(3) In both males and females
(4) Through the autosomes in one sex only
42. Statement A: Colourblindness is more likely to occur in females than in males.
Statement B: The Y chromosome of the male has the genes for distinguishing colour.
(1) Statement A is correct and B is wrong
(2) Statement B is correct and A is wrong
(3) Both the statements A and B wrong
(4) Both the statements A and B correct
43. A person whose father is colour blind marries a lady whose mother is the daughter of a colour blind man. Then,
(1) All the sons are colour blind
(2) Some of the sons are colour blind some are normal
(3) All children are colour blind
(4) All daughters are normal
44. Expression of recessive genes on X - chromosome is definite in males because of
(1) Homozygous nature
(2) Hemizygous nature
(3) Polyzygous nature
(4) Inverted condition
45. When a colour blind man marries the daughter of a colour blind person, in the progeny
(1) 50% of the sons are colour blind
(2) All the daughters are colour blind
(3) All the sons are colour blind
(4) 50% of the daughters & 50% of the sons are colour blind
46. The probability of the daughters of husband and wife having normal vision (their fathers being colour blind) becoming colour blind is
(1) 0%
(2) 25%
(3) 50%
(4) 75%
47. When a haemophilic / colour blind female marries a normal male, in the progeny
(1) Females are normal but carriers and males are haemophilic / colour blind
(2) Females are haemophilic / colour blind and males are normal
(3) Both males and females are haemophilic / colour blind
(4) Both males and females are normal
48. A colour blind boy has two sisters. One of them is normal and the other is colour blind. What can be the possible nature of their parents ?
(1) Colour blind father and normal mother
(2) Colour blind father and carrier mother
(3) Both father and mother are colour blind
(4) Normal father and colour blind mother
49. Holandric genes are the genes present on
(1) The X – chromosome
(2) The Y - chromosome
(3) The autosome
(4) Both X and Y chromosomes
50. Holandric genes are always transmitted to the son
(1) From X chromosome of male
(2) From Y chromosome of male
(3) From X chromosome of female
(4) From X chromosome of either female or male
51. A boy has hairy ears which is Y-linked inheritance. What is the chance that his grandson will inherit the trait from him ?
(1) 0%
(2) 25%
(3) 50%
(4) 100%
52. Haemophilia is
(1) Due to the lack of either clotting factors VIII or IX
(2) Due to recessive gene on X chromosome of male
(3) Failure in coagulation of blood
(4) All these
53. Haemoglobin of patients suffering from sickle cell anemia differs from normal haemoglobin by one
(1) Monosaccharide
(2) Disaccharide
(3) Amino acid
(4) Lipid
54. Sickle cell anaemia is
(1) A sex linked disorder
(2) Iron deficiency disease
(3) An organic disorder
(4) Inherited by a recessive gene on the autosome
55. Statement A: Heterozygous individuals for sickle cell trait of RBC have 50% of their haemoglobin defective.
Statement B: Sickle cell anaemia is due to a defective gene in the sex chromosome.
(1) Both the statements are correct and B is the reason for A
(2) Both the statements are correct and B is not the reason A
(3) Statement A is correct and statement B is wrong
(4) Statement B is correct and statement A is wrong
56. Sickle celled anaemia is due to
(1) Mutation of gene directing the synthesis of a - chain of haemoglobin
(2) Mutation of gene directing the synthesis of b - chain of haemoglobin
(3) Mutation of a gene located on the X – chromosome
(4) Lack of haemoglobin
57. Sickle celled anaemia is caused due to the substitution of
(1) Valine at 6th position of beta - chain of haemoglobin by glutamic acid
(2) Glutamic acid at 6th position of beta - chain of haemoglobin by valine
(3) Glutamic acid at 6th position of alpha - chain of haemoglobin by valine
(4) Valine at 6th position of alpha - chain of haemoglobin by glutamic acid
58. Sickle cell anemia is a disorder caused due to the change in the chemical nature of
(1) Alpha chain of haemoglobin
(2) Beta chain of haemoglobin
(3) Both alpha and beta chains of haemoglobin
(4) Plasma proteins
59. Gynaecomastia is a condition seen in
(1) Down’s syndrome
(2) Turner’s syndrome
(3) Klinefelter’s syndrome
(4) Cri-du-chat syndrome
60. Find out the correct statement
(1) Monosomy and trisomy are the two types of euploidy
(2) Polyploidy is more common in animals than in plants
(3) Polyploids occur due to the failure in complete separation of sets of chromosomes
(4) 2n – 1 condition results in trisomy
61. Statement A: An additional X chromosome in humans results in Klinefelter’s syndrome.
Statement B: Non-disjunction of chromosomes is the chief cause for aneuploidy.
(1) Both the statements are correct and B is the reason for A
(2) Both the statements are correct and B is not the reason A
(3) Statement A is correct and statement B is wrong
(4) Statement B is correct and statement A is wrong
62. The chromosome complement in an individual suffering from Klinefelter’s syndrome is
(1) 44AA + XX
(2) 44AA + XXY
(3) 44AA+XO
(4) 22A+Y
63. A person with Klinefelter’s syndrome is considered a
(1) monosomic
(2) nullisomic
(3) trisomic
(4) triploid
64. One of the following is not synonymous to the rest:
(1) Mongoloid idiocy
(2) Down’s syndrome
(3) 21-trisomy
(4) Turner’s syndrome
65. Down’s syndrome is due to
(1) Crossing over
(2) Sex-linkage
(3) Linkage
(4) Non-disjunction of chromosomes
66. Turner’s syndrome is due to
(1) Monosomic chromosome
(2) Bisomic chromosomes
(3) Trisomic chromosomes
(4) Ploysomic chromosomes
67. Deletion of a part of the 5th chromosome is responsible for
(1) Underdeveloped breasts, sparse pubic hair, small uterus, reduced ovaries, sterility
(2) Small prostate glands, small testes, sparse body hair, long stature due to long limb bones, enlarged breasts
(3) Clogging of blood capillaries, anaemia, reduced blood supply to vital organs, jaundice, muscle cramps, headache, etc.
(4) Very small head, increased inter-orbital distance, severe mental deficiency, moon face, cat cry of new born
68. Cri du chat syndrome is due to
(1) Addition of an autosome
(2) Deletion of a part of an autosome
(3) Addition of a X – chromosome
(4) Deletion of a part of X – chromosome
69. These are the characteristics of Down’s syndrome
(1) Sparse body hair, gynaecomastia, azoospermia, male phenotype
(2) Female phenotype, short stature, absence of menstrual cycle
(3) Microcephaly, hypertelorism, underdeveloped larynx, moon – face
(4) Short stature, broad forehead, epicanthal folds, short phalanges, short flat nose
70. Webbing in the neck, short stature, pigeon chest, underdeveloped breasts, rudimentary ovaries, sterility, etc., are characters of
(1) Down’s syndrome
(2) Klinefelter’s syndrome
(3) Ullrich’s syndrome
(4) Turner’s syndrome

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