Q.1.
on Q459) Which of the following most appropriately describes haemophilia?
-
80%
a) X-linked recessive gene disorder
-
0%
b) Chromosomal disorder
-
20%
c) dominant gene disorder
-
0%
d) Recessive gene disorder
Q.2.
on Q460) A man whose father is a colour-blind, marry a woman, who is a daughter of colour-blind mother. The offspring of this couple will be
-
0%
a) all daughter and sons are colour-blind
-
0%
b) 50% colour-blind and 50% normal son
-
100%
c) carrier normal daughter and colour-blind sons
-
0%
d) colour-blind sons and normal daughter
Q.3.
on Q461) Match the following :-
| Column I; | Column II |
| Genetics | a. Father of genetics |
| W.Bateson | b. To become |
| "Gen" | c. Coined the term genetics |
| Gregor Johann Mendel | d. Study of heredity |
-
0%
a) (1 - d) (2 - c) (3 - a) (4 - b)
-
0%
b) (1 - d) (2 - a) (3 - c) (4 - b)
-
100%
c) (1 - d) (2 - C) (3 - b) (4 - a)
-
0%
d) (1 - d) (2 - b) (3 - C) (4 - a)
Q.4.
on Q462) A normal woman, whose father was colour blind is married with normal man. The sons would be ... ....
-
0%
a) 75% Colour blind
-
0%
b) 50% colour blind
-
100%
c) all normal
-
0%
d) all colourblind
Q.5.
on Q463) Which pair is incorrect?
-
0%
a) Haemophilia - X linked recessive
-
0%
b) baldness - X- linked dominance
-
0%
c) colour blindness- X linked recessive
-
100%
d) antler in deer - sex limited.
Q.6.
on Q464) The karyotype of which among the following genetic disorder shows an additional copy of an allosome:
-
100%
a) Down’s syndrome.
-
0%
b) Klinefelter’s syndrome.
-
0%
c) Turner’s syndrome.
-
0%
d) Edward syndrome.
Q.7.
on Q465) Type of pollination in Pisum sativum is
-
0%
a) Cross pollination
-
0%
b) Self-pollination
-
100%
c) Bud pollination
-
0%
d) Both 2 and 3
Q.8.
on Q466) Which of the following matches is wrong?
-
0%
a) Birds- ZZ (male) - ZW (female)
-
0%
b) Drosophila- XY (male) - XX (female)
-
0%
c) Humans- XY (male), XX (female)
-
100%
d) Cockroach- - XZ (male) - ZZ (female)
Q.9.
on Q467) Which of the following statements are correct? (i) Polyploidy is caused due to failure of cytokinesis after telophase. (ii) polyploidy results into increase in number of autosomes (iii) Turner’s syndrome is an example of polyploidy (iv) polyploidy results into an increase in a whole set of chromosomes
-
0%
a) (i) and (ii)
-
0%
b) (i) and (iii)
-
100%
c) (ii) and (iii)
-
0%
d) (i) and (iv)
Q.10.
on Q468) The monohybrid genotypic ratio ¼TT + ½Tt + ¼tt can be mathematically condensable in the form of:
-
0%
a) (½T + ½T)2
-
100%
b) (½TT + ½tt)2
-
0%
c) (½T + ½t)2
-
0%
d) (½Tt + ½Tt)2
Q.11.
on Q469) Match the following:
| a. Monohybrid phenotypic ratio | 9: 3: 3: 1. |
| b. Monohybrid genotypic ratio | 1: 1: 1: 1. |
| c. Monohybrid test cross ratio | 1: 2: 1. |
| d. Dihybrid phenotypic ratio | 3: 1. |
| e. Dihybrid test cross ratio | 1: 1. |
-
0%
a) a= 4 ; b= 3 ; c= 5 ; d= 1 ; e= 2
-
0%
b) a= 3 ; b= 4 ; c= 5 ; d= 2 ; e= 1
-
0%
c) a= 1 ; b= 2 ; c= 3 ; d= 4 ; e= 5
-
100%
d) a= 4 ; b= 3 ; c= 2 ; d= 1 ; e= 5
Q.12.
on Q470) Muliple alleles are present
-
0%
a) in different chromosomes
-
100%
b) at different loci on chromosomes
-
0%
c) at the same locus on homologous chromosome
-
0%
d) at the non homologous chromosome
Q.13.
on Q471) What is Barr body ?
-
0%
a) One more than the number of X - chromosomes
-
0%
b) One more than the number of Y-chromosomes
-
0%
c) One less than the number of X - chromosomes
-
100%
d) Two less than the number of X – chromosomes
Q.14.
on Q472) Which is very common method to study Human Karyotype ?
-
0%
a) Blood culture method
-
0%
b) Tissue culture method
-
100%
c) Pedigree method
-
0%
d) Statistical method
Q.15.
Asserton (A) :- colchicines is the stimulator of mitosis Reason (R) :- In karyotype chromosomes are arranged in ascending order of size.
-
0%
a) If both (A) and (R) are true and (R) is the correct explanation of (A)
-
100%
b) If both (A) and (R) are true but (R) is not the correct explanation of (A)
-
0%
c) If (A) is true but (R) is false
-
0%
d) If both (A) and (R) are false.
Q.16.
on Q474) In pea plants, yellow seeds are dominant to green, If a heterozygous yellow seeded plant is Corssed with a green seeded plants, what ratio of yellow and green seeded plants, would you expect in F1 generation?
-
0%
a) 9 : 1
-
0%
b) 1 : 3
-
100%
c) 3 : 1
-
0%
d) 50 : 50
Q.17.
on Q475) Test cross in plants or in Drosophilia involves crossing .... ....
-
0%
a) between two genotype with recessive trait
-
100%
b) between two F1 hybrids
-
0%
c) the F1 hybrid with a double recessive genotype
-
0%
d) between two genotypes with dominant traits
Q.18.
on Q476) In a monohybrid cross the F1 ratio of a backcross is .... ....
-
100%
a) 1:1
-
0%
b) 3:1
-
0%
c) 1:2:1
-
0%
d) 9:3:3:3
Q.19.
on Q477) What can the possible blood groups of progeny whose father and mother are of A and B group respectively?
-
0%
a) A and B only
-
0%
b) AB only
-
0%
c) All except O
-
100%
d) A, B, AB and O
Q.20.
on Q478) How many different kinds of gametes will be produced by a plant having the genotype AABbCC?
-
0%
a) Three
-
0%
b) Four
-
100%
c) Nine
-
0%
d) Two
Q.21.
on Q479) The number of linkage group(s) present in Escherichia coli is
-
100%
a) One
-
0%
b) two
-
0%
c) four
-
0%
d) seven
Q.22.
on Q480) In Drosophila crossing over occurs in females but not in males. Genes A and B are 10 map unit apart on chromosome. A female Drosophila with genotype AB/ab and male Drosophila with genotype AB/ab. How many types of gametes are produced by the female and male Drosophila?
-
0%
a) 4 types: 2 types.
-
0%
b) 2 types: 2 types.
-
100%
c) 4 types: 4 types.
-
0%
d) 4 types: 1 type.
Q.23.
on Q484) In case of Pea plant, grey seed colour (G) is dominant to white (g). If the progenies are formed in the following phenotypic ratio, find the genotypes of parents. Grey was crossed with grey, 82 grey and 76 white were formed:
-
0%
a) Gg
-
0%
b) GG
-
100%
c) gg
-
0%
d) All of these
Q.24.
on Q485) Match the representation of diploid cell given in column I with certain type of euploidy/aneuploidy with the numerical representation in column II:
| Column I | Column II |
| a. AABBBCDD | 2n + 2 + 2 |
| b. AABBBBCCDD | 2n – 1 |
| c. AAABBBCCCDDD | 3n |
| d. AABCCDD | 2n + 2 |
| e. AABBBBCCCCDD | 2n + 1 – 1 |
-
0%
a) a= 1 ; b = 2 ; c = 3 ; d = 4 ; e = 5
-
0%
b) a= 5 ; b = 4 ; c = 3 ; d = 2 ; e = 1
-
100%
c) a= 3 ; b = 4 ; c = 5 ; d = 2 ; e = 1
-
0%
d) a= 1 ; b = 4 ; c = 5 ; d = 2 ; e = 3
Q.25.
on Q486) Phenotypic and genotypic ratio are same in:
-
0%
a) Incomplete dominance.
-
0%
b) Mendelian monohybrid test cross.
-
100%
c) Mendelian dihybrid test cross.
-
0%
d) All of the above.
Q.26.
on Q487) Zea mays has 10 pairs of chromosomes . Linkage groups present in it are
-
0%
a) 5
-
0%
b) 10
-
0%
c) 20
-
0%
d) 40
Q.27.
on Q488) What is incorrect statement Gynandromorphs ?
-
0%
a) Individuals who show male characters and female characters
-
0%
b) Individuals who show only female characters
-
0%
c) It happens due to less of X- chromosomes
-
0%
d) It happens due to binacleated eggs
Q.28.
on Q489) What will be expected blood groups in the off spring when there is a cross between AB blood group mother and heterozygous B blood group father ?
-
0%
a) 25 % AB group; 25 % A group; 50 % B group
-
0%
b) 50 % AB group; 25 % A group; 25 % B group
-
0%
c) 25% AB group; 50% O group; 25 % A group
-
0%
d) 25 % O group; 50 % B group; 25 % A group
Q.29.
on Q490) Haemophilia is more commonly seen in human males than in human females because
-
0%
a) a greater proportion of girls die in infancy
-
0%
b) this disease is due to a Y- linked recessive mutation
-
0%
c) this disease is due to an X - linked recessive mutation
-
0%
d) this disease is due to an X- linked dominant mutation
Q.30.
on Q491) How many types and in what ratio the gametes are produced by a dihybrid heterozygous?
-
0%
a) 4 types in the ratio of 9 : 3 : 3 : 1
-
0%
b) 2 types in the ratio of 3 : 1
-
0%
c) 3 types in ratio of 1: 2: 1
-
0%
d) 4 types in the ratio of 1: 1: 1 : 1
Q.31.
on Q492) Sickle cell anaemia has not been eliminated from the African population because:
-
0%
a) it is controlled by recessive genes
-
0%
b) it is not a fatal disease
-
0%
c) it provides immunity against malaria
-
0%
d) it is controlled by a dominant genes
Q.32.
on Q493) The ultimate source of allelic variation is:
-
0%
a) Recombination
-
0%
b) Natural selection
-
0%
c) Mutation
-
0%
d) Drift
Q.33.
on Q494) In a test cross involving F1 dihybrid flies, more parental-type offspring were produced than the recombinant type offspring. This indicates
-
0%
a) chromosomes failed to separate during meiosis
-
0%
b) the two genes are linked and present on the same chromosome
-
0%
c) both of the characters are controlled by more than one gene
-
0%
d) the two genes are located on two different chromosomes
Q.34.
on Q495) Male pattern baldness is a _________trait
-
0%
a) sex-linked
-
0%
b) sex-limited
-
0%
c) sex-influenced
-
0%
d) Y-linked
Q.35.
on Q496) A man of blood group A, marries a woman of blood group B, both of them are heterozygous for blood group, chances of their first child having blood group AB is:
-
0%
a) 25%.
-
0%
b) 50%.
-
0%
c) 75%.
-
0%
d) 100%.
Q.36.
on Q497) Test cross is used to predict the
-
0%
a) Phenotype of test organism
-
0%
b) Sex of test organism
-
0%
c) Genotype of test organism
-
0%
d) Genotypic ratio
Q.37.
on Q498) In Mendel's experiments with garden pea, round seed shape (RR) was dominant over wrinkledSeeds (rr), Yellow cotyledon (YY) was dominant over green cotyledon (yy). What are the expected Phenotypes in the F2 generation of the cross RRYY x rryy?
-
0%
a) Round Seeds with yellow cotyledons, and wrinkled seeds with yellow cotyle dons
-
0%
b) Only round seeds with green cotyledons
-
0%
c) Only wrinkled seeds with yellow cotyledons
-
0%
d) Only wrinkled seeds with green cotyledons
Q.38.
on Q499) First generation after a cross is ___________
-
0%
a) First filial generation
-
0%
b) F1 gneratioon
-
0%
c) Second filial generation
-
0%
d) Both (A) and (B)
Q.39.
on Q500) The chromosomal theory of heredity implies that
-
0%
a) chromosomes are composed of DNA and protein
-
0%
b) genes are composed of chromosomes
-
0%
c) organism cannot live without chromosomes
-
0%
d) genes are located on the chromosomes and are transmitted to the next generation through them.
Q.40.
on Q501) The genotype of a plant showing the dominant phenotype can be determined by
-
0%
a) test cross
-
0%
b) dihybrid crosses
-
0%
c) pedigree analysis
-
0%
d) back cross
Q.41.
on Q502) Given below is a pedigree chart showing the inheritance of a certain sex-linked trait in humans? The Trait traced in the above pedigree chart is
-
0%
a) dominant X-linked
-
0%
b) recessive X-linked
-
0%
c) dominant Y-linked
-
0%
d) recessive Y-linked
Q.42.
on Q503) Number of Barr bodies in Klinefelter’s syndrome:
-
0%
a) 0
-
0%
b) 1
-
0%
c) 2
-
0%
d) 3
Q.43.
on Q504) Furrowed tongue and partially open mouth are characteristics of
-
0%
a) Edward’s syndrome
-
0%
b) Turner’s syndrome
-
0%
c) Down’s syndrome
-
0%
d) Klinefelter’s syndrome.
Q.44.
on Q505) The minimum progeny population size allowing for random union of all kinds of gametes from AaBbCc parents is:
-
0%
a) 9
-
0%
b) 16
-
0%
c) 27
-
0%
d) 64
Q.45.
on Q506) Which of the following is least probable?
-
0%
a) Female child with Sickle cell anaemia
-
0%
b) Male child with haemophilia
-
0%
c) Carrier female for haemophilia
-
0%
d) Female child with haemophilia
Q.46.
on Q507) What does the given pedigree chart shows?
-
0%
a) Sex-linked recessive trait
-
0%
b) Sex-linked dominant trait
-
0%
c) Inheritance of Sickle cell anaemia
-
0%
d) Inheritance of autosomal dominant trait
Q.47.
on Q508) In garden pea, round shape of seeds is dominant over wrinkled shape. A pea plant heterozygous for round shape of seed is selfed and 1600 seeds produced during the cross are subsequently germinated. How many plants would have round seeds?
-
0%
a) 1600.
-
0%
b) 800.
-
0%
c) 400.
-
0%
d) 1200.
Q.48.
on Q509) Match the following:
| a. Coat colour in mouse | Recessive epistasis |
| b. Fruit colour in summer squash | Dominant epistasis. |
| c. Flower colour in sweet pea | Complementary genes |
| d. Fruit shape in Shepherd’s purse | Duplicate genes |
| e. Feather colour in fowl | Inhibitory genes. |
-
0%
a) a= 1 ; b = 2 ; c = 3 ; d= 4 ; e= 5
-
0%
b) a= 2 ; b = 1 ; c = 3 ; d= 4 ; e= 5
-
0%
c) a= 5 ; b = 4 ; c = 3 ; d= 2 ; e= 1
-
0%
d) a= 1 ; b = 5 ; c = 2 ; d= 4 ; e= 3
Q.49.
on Q510) Which one of the followings is caused by a recessive allele of a gene on chromosome?
-
0%
a) Phenylketonuria.
-
0%
b) Tay – Sach’s disease.
-
0%
c) Cystic Fibrosis.
-
0%
d) Huntington’s disease
Q.50.
on Q511) XO-chromosomal abnormality in human beings causes
-
0%
a) Turner's syndrome
-
0%
b) Down's syndrome
-
0%
c) Klinefelter’s syndrome
-
0%
d) none of these.