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Inheritance And Variation Mcq
Quiz 18
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Q.1
on Q459) Which of the following most appropriately describes haemophilia?
0%
a) X-linked recessive gene disorder
0%
b) Chromosomal disorder
0%
c) dominant gene disorder
0%
d) Recessive gene disorder
Explanation
Answer:(a)
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
0%
c) carrier normal daughter and colour-blind sons
0%
d) colour-blind sons and normal daughter
Explanation
Answer:(c)
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)
0%
c) (1 - d) (2 - C) (3 - b) (4 - a)
0%
d) (1 - d) (2 - b) (3 - C) (4 - a)
Explanation
Genetics is the branch of biological sciences which deals with the transmission of characteristics from one generation to another (heredity) and also the action of heredity unit called gene as they bring about the characteristics which they control. The term genetics was coined and used for the first time by W. Bateson (1905). Gregor Johann Mendel appropriately known as father of genetics, proposed the theory of inheritance. Answer:(c)
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
0%
c) all normal
0%
d) all colourblind
Explanation
Answer:(b)
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
0%
d) antler in deer - sex limited.
Explanation
Answer:(b)
Q.6
on Q464) The karyotype of which among the following genetic disorder shows an additional copy of an allosome:
0%
a) Down’s syndrome.
0%
b) Klinefelter’s syndrome.
0%
c) Turner’s syndrome.
0%
d) Edward syndrome.
Explanation
Answer:(b)
Q.7
on Q465) Type of pollination in Pisum sativum is
0%
a) Cross pollination
0%
b) Self-pollination
0%
c) Bud pollination
0%
d) Both 2 and 3
Explanation
Answer:(d)
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)
0%
d) Cockroach- - XZ (male) - ZZ (female)
Explanation
Answer:(d)
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)
0%
c) (ii) and (iii)
0%
d) (i) and (iv)
Explanation
Answer:(d)
Q.10
on Q468) The monohybrid genotypic ratio ¼TT + ½Tt + ¼tt can be mathematically condensable in the form of:
0%
a) (½T + ½T)2
0%
b) (½TT + ½tt)2
0%
c) (½T + ½t)2
0%
d) (½Tt + ½Tt)2
Explanation
The ¼ : ½ : ¼ ratio of TT: Tt: tt is mathematically condensable to the form of the binomial expression (ax +by)2 , that has the gametes bearing genes T or t in equal frequency of ½. The expression is expanded as given below : (½T + ½ t) 2 = (½T + ½t) X (½T + ½t) = ¼ TT + ½Tt + ¼ tt Answer:(c)
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
0%
d) a= 4 ; b= 3 ; c= 2 ; d= 1 ; e= 5
Explanation
Monohybrid genotypic ratio = 1 : 2 : 1. Monohybrid phenotypic ratio = 3 : 1. Monohybrid test cross = 1 : 1. Dihybrid phenotypic ratio = 9 : 3 : 3 : 1. Dihybrid test cross ratio = 1 : 1 : 1 : 1. Answer:(a)
Q.12
on Q470) Muliple alleles are present
0%
a) in different chromosomes
0%
b) at different loci on chromosomes
0%
c) at the same locus on homologous chromosome
0%
d) at the non homologous chromosome
Explanation
Answer:(c)
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
0%
d) Two less than the number of X – chromosomes
Explanation
Answer:(c)
Q.14
on Q472) Which is very common method to study Human Karyotype ?
0%
a) Blood culture method
0%
b) Tissue culture method
0%
c) Pedigree method
0%
d) Statistical method
Explanation
Answer:(a)
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)
0%
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.
Explanation
Answer:(d)
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
0%
c) 3 : 1
0%
d) 50 : 50
Explanation
When heterozygous individual is crossed with recessive individual, 50% of progeny exhibit dominant trait and 50% of progeny exhibit recessive trait. The ratio of yellow seeded plant and green seeded plant will be 1 : 1. Answer:(d)
Q.17
on Q475) Test cross in plants or in Drosophilia involves crossing .... ....
0%
a) between two genotype with recessive trait
0%
b) between two F1 hybrids
0%
c) the F1 hybrid with a double recessive genotype
0%
d) between two genotypes with dominant traits
Explanation
Crossing of F1 individual having dominant phenotype with its homozygous recessive parents is called test cross. The test cross is used to determine whether the individuals exhibiting dominant characters are homozygous or heterozygous. Answer:(c)
Q.18
on Q476) In a monohybrid cross the F1 ratio of a backcross is .... ....
0%
a) 1:1
0%
b) 3:1
0%
c) 1:2:1
0%
d) 9:3:3:3
Explanation
Crossing of F1 hybrid with either of parent. When F1 hybrid is crossed homozygous dominant parent then all progeny express dominant trait. When F1 hybrid is crossed with homozygous recessive parent then 50% of progeny express dominant trait and 50% of progeny express recessive trait. This is also known as test cross. Answer:(a)
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
0%
d) A, B, AB and O
Explanation
Blood type (phenotype)
Genotype
Antigen
Antibodies
A
I
A
I
A
or I
A
I
O
A
b
B
I
B
I
B
or I
B
I
0
B
a
AB
I
A
I
B
Both A and B
Neither a nor b
O
I
O
I
O
Neither A nor B
Both a and b
Answer:(d)
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
0%
c) Nine
0%
d) Two
Explanation
No. of gametes = (2)n where n = no. of heterozygous allele. Here there is only one heterozygous pair. No. of gametes = 21 = 2 Answer:(d)
Q.21
on Q479) The number of linkage group(s) present in Escherichia coli is
0%
a) One
0%
b) two
0%
c) four
0%
d) seven
Explanation
Answer:(a)
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.
0%
c) 4 types: 4 types.
0%
d) 4 types: 1 type.
Explanation
Answer:(a)
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
0%
c) gg
0%
d) All of these
Explanation
The ratio of grey : white is almost 1 : 1. This ratio occurs when one parent is heterozygous and another recessive parent. Genotype of parents is Gg and gg Answer:(a)
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
0%
c) a= 3 ; b = 4 ; c = 5 ; d = 2 ; e = 1
0%
d) a= 1 ; b = 4 ; c = 5 ; d = 2 ; e = 3
Explanation
Answer:(b)
Q.25
on Q486) Phenotypic and genotypic ratio are same in:
0%
a) Incomplete dominance.
0%
b) Mendelian monohybrid test cross.
0%
c) Mendelian dihybrid test cross.
0%
d) All of the above.
Explanation
Phenotypic and genotypic is same in the following: 1) Incomplete dominance 2) Monohybrid test cross 3) Dihybrid test cross Answer:(d)
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
Explanation
Answer:(b)
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
Explanation
Answer:(b)
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
Explanation
Blood type (phenotype)
Genotype
Antigen
Antibodies
A
I
A
I
A
or I
A
I
O
A
b
B
I
B
I
B
or I
B
I
0
B
a
AB
I
A
I
B
Both A and B
Neither a nor b
O
I
O
I
O
Neither A nor B
Both a and b
Answer:(a)
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
Explanation
Answer:(c)
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
Explanation
No. of traits (n)
Experiment
Types of gametes (2
n
)
No. of offsprings (gametes)
2
No. of phenotype (2
n
)
No. of genotype (3
n
)
Phenotypic ratio
Genotypic ratio
2
Dihybrid cross
4
16
4
9
9:3:3:1
2:4:2:1:2:1:1:2:1
Answer:(d)
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
Explanation
Answer:(c)
Q.32
on Q493) The ultimate source of allelic variation is:
0%
a) Recombination
0%
b) Natural selection
0%
c) Mutation
0%
d) Drift
Explanation
Answer:(c)
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
Explanation
If both the characters are controlled by more than one gene then we can't say anything about the recombinant formed or the parental offspring formed as they need crossing over or linkage to occur. Answer:(b)
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
Explanation
Answer:(c)
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%.
Explanation
Blood type (phenotype)
Genotype
Antigen
Antibodies
A
I
A
I
A
or I
A
I
O
A
b
B
I
B
I
B
or I
B
I
0
B
a
AB
I
A
I
B
Both A and B
Neither a nor b
O
I
O
I
O
Neither A nor B
Both a and b
Answer:(a)
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
Explanation
Crossing of F1 individual having dominant phenotype with its homozygous recessive parents is called test cross. The test cross is used to determine whether the individuals exhibiting dominant characters are homozygous or heterozygous. Answer:(c)
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
Explanation
Cross between RRYY and rryy produces four type of phenotype in F2 generation. 1) Round seed with yellow cotyledons 2) Round seed with green cotyledons 3) Wrinkled seed with yellow cotyledons 4) Wrinkled seed with green cotyledons Answer:(a)
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)
Explanation
First generation after a cross between true breeding line is called Filial 1 or F1 generation. It is first hybrid generation. Generation produced by selfing of F1 generation is called second filial or F2 generation. Answer:(d)
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.
Explanation
Answer:(d)
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
Explanation
Crossing of F1 individual having dominant phenotype with its homozygous recessive parents is called test cross. The test cross is used to determine whether the individuals exhibiting dominant characters are homozygous or heterozygous. Answer:(a)
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
Explanation
Answer:(a)
Q.42
on Q503) Number of Barr bodies in Klinefelter’s syndrome:
0%
a) 0
0%
b) 1
0%
c) 2
0%
d) 3
Explanation
Answer:(b)
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.
Explanation
Answer:(c)
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
Explanation
Answer:(d)
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
Explanation
Answer:(d)
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
Explanation
Answer:(c)
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.
Explanation
When heterozygous individual is self-pollinated, 75% of progeny express dominant trait and 25% of progeny express recessive trait. Round seed is dominant over wrinkled seed. Total seed = 1600. No. of round seeds = 75% of 1600 = 1200 Answer:(d)
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
Explanation
Characters
Expected ratios
Example
Monohybrid cross
Phenotypic = 3 : 1
Genotypic = 1 :2 : 1
Pisum sativum
Dihybrid cross
Phenotypic = 9:3:3:1
Pisum sativum
Incomplete dominance
1 : 2 : 1
Mirabilis jalapa
Complementary genes
9 : 7
Lathyrus odoratus
Supplementary genes
9 : 3 : 4
Coat colour of mice
Collaborative supplementary genes
9 : 3 : 3 : 1
Poultry birds – comb pattern
Dominant epistasis
12 : 3 : 1
Fruit colour in
Curcurbita
Recessive epistasis
9 : 3 : 4
Coat colour in mice/pigmentation in onion bulb
Duplicate genes
15 : 1
Fruit shape in
Capsella bursapastoris
Polymeric gene
9 : 6 : 1
Cucurbita pepo
Suppresor gene
13:3
Leaf colour of rice
Answer:(a)
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
Explanation
Answer:(c)
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.
Explanation
Answer:(a)
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