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Inheritance And Variation Mcq
Quiz 6
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Q.1
Gametes of AaBb individual can be ... ... [ AIIMS 1991 ]
0%
a) Aa, Bb
0%
b) AB, ab
0%
c) AB, ab, aB
0%
d) AB, Ab, aB aa
Explanation
There are four possible combinations of gametes for the AaBb parent. Half of the gametes get a dominant A and a dominant B allele; the other half of the gametes get a recessive a and a recessive b allele. Both parents produce 25% each of AB, Ab, aB, and ab. Number of gametes = 2n where n= no. of heterozygous pairs of allele AaBb = 22 = 4. These gametes are AB, Ab, aB, ab. Answer : (d)
Q.2
Which one is not a dominant trait amongst seven pea plants chosen by Mendel... ... [Maipal 2000 ]
0%
a) Flower colour - purple
0%
b) Pod colour - yellow
0%
c) Shape of seed - Round
0%
d) Flower - Axial
Explanation
Traits
Dominant
Recessive
Plant height
Tall (1.2 – 2.0m)(T)
Dwarf (0.25 – 0.5m) (t)
Flower position
Axillary (A)
Terminal (a)
Pod colour
Green (G) or (Y)
Yellow (g) or (y)
Pod shape
Full or inflated (I) or (C)
Constricted (i) or (c)
Flower colour
Violet (V) or (W)
White (v) or (w)
Seed shape
Round (R) or (W)
Wrinkled (r) or (w)
Seed colour
Yellow (Y) or (G)
Green (y) or (g)
Answer : (b)
Q.3
Ratio 1:1:1:1 is obtained from a cross of ... .. [ kerala 2003 ]
0%
a) RRYY × rryy
0%
b) RRYy × rrYy
0%
c) RrYY × Rryy
0%
d) RrYy × rryy
Explanation
Ratio 1 : 1 : 1 : 1 is ratio of dihybrid test cross. In test cross heterozygous hybrid is crossed with pure recessive plant. RrYy × rryy is correct answer. Answer : (d)
Q.4
When yellow round heterozygous Pea plants are self fertilized, the frequency of occurrence of RrYY genotype among the offspring is ... ... [ kerala 2012 ]
0%
a) 9/16
0%
b) 3/16
0%
c) 2/16
0%
d) 6/16
Explanation
Gametes
RY
Ry
rY
ry
RY
RRYY (round, yellow)
RRYy (round,yellow)
RrYY (round, yellow)
RrYy (round, yellow)
Ry
RRYy (round, yellow)
RRyy (round, green)
RrYy (round, yellow)
Rryy (round, green)
rY
RrYY (round, yellow)
RrYy (round, yellow)
rrYY ( wrinkled, yellow)
rrYy (wrinkled, yellow)
ry
RrYy (round, yellow)
Rryy ( round, green)
rrYy ( wrinkled, yellow)
Rryy (wrinkled, green)
RrYY = 2/16 Answer : (c)
Q.5
Which of the following match is correct
0%
a) independent assortment - Segregation of factors
0%
b) Lamarck - Natural selection
0%
c) Hatch and Slack -Chemiosmotic theory
0%
d) Peter Mitchell - Proposed Z-scheme
Explanation
The theory of natural selection was explored by 19th-century naturalist Charles Darwin. Peter D. Mitchell proposed the chemiosmotic hypothesis in 1961. Hill and Bendall proposed the Z scheme. It involves both the photosystem, PS I and PS II. Independent assortment : It states that when two pairs of independent alleles are bought together in F1 hybrid, they show independent dominant effects. in the formation of gametes, law of segregation occurs and the factors assort independently at random. Answer : (a)
Q.6
Hemoglobins of normal and sickle cell patient are subjected to electrophoresis. They will show ... ... [ AMU 2012 ]
0%
a) Same mobility
0%
b) Different mobility
0%
c) No mobility
0%
d) Haemoglobin of patient does not move
Explanation
Because of sickle shape, sickled cells clump together, especially in the smallest blood vessels, the blood becomes more viscous. The result is slowing of the flow, even greater loss of oxygen from the haemoglobin. Thus the mobility of cell will be different. Answer : (b)
Q.7
Which one is correct? ... .. [ Odisha 2002 ]
0%
a) Polygenic character is controlled by multiple genes
0%
b) Numerous intermediate types are found in between the two extremes in polygenic inheritance
0%
c) Height, weight and skin colour are polygenic
0%
d) Polygenic trait is controlled by multiple alleles
Explanation
Quantitative inheritance: It is a type of inheritance controlled by more genes in which the dominant alleles have cumulative effect with each dominant allele expressing a part or unit of trait, the full trait being shown only when all the dominant alleles are present. The genes involved are called polygenes. Quantitative inheritance is therefore, also called polygenic inheritance. Polygenic inheritance does not follow Mendelian ratio. Ratio of F2 generation is 1 : 4 : 6 : 4 : 1. Skin colour, human height and eye colour show polygenic inheritance. Option (d) is incorrect because polygenic triat is controlled by multiple gene not multiple alleles. Answer : (a)
Q.8
In albinism, the absence of which pigment makes the skin and hair light coloured ... ..[ EAMCET 1999 ]
0%
a) Melanin
0%
b) Carotene
0%
c) haemoglobin
0%
d) Chlorophyll
Explanation
Albinism results from inheritance of recessive gene alleles and is known to affect all vertebrates, including humans. It is due to absence or defect of tyrosinase, a copper-containing enzyme involved in the production of melanin. Tyrosinase synthesizes melanin from the amino acid tyrosine.. Absence of this enzymes causes albinism. Answer : (a)
Q.9
A pleiotropic gene is ... ...
0%
a) IA
0%
b) HbS
0%
c) HbA
0%
d) IB
Explanation
Hereditary sickle cell anaemia is an example of pleiotropy. This is an autosome linked recessive trait that can be transmitted from parents to the offspring when both the partners are carrier for the gene (or heterozygous). The disease is controlled by a single pair of allele, HbA and HbS Answer : (b)
Q.10
Heterosis is ... .. [ CPMT 1997 ]
0%
a) Hybrid weakness
0%
b) Hybrid weakness and vigour
0%
c) Hybrid vigour
0%
d) Neither weakness nor vigour of hybrid
Explanation
Hybrid vigour or heterosis is the superiority of hybrid over either of its parents in one or more traits. Answer : (c)
Q.11
The process of removing stamens from floral buds during hybridization experiments is .... .. [ MPPMT 1993 ]
0%
a) Capping
0%
b) Selfing
0%
c) Emasculation
0%
d) Crossing
Explanation
Emasculation is used for the prevention of self-fertilisation. There is the removal of male (pollen) parts of a plant which basically involves the anthers. This helps to perform controlled pollination and breeding purposes. Answer : (c)
Q.12
An individual having different alleles of a gene is ... ... [ Odisha 200 8]
0%
a) heterozygous
0%
b) Homozygous
0%
c) Diploid
0%
d) Mosaic
Explanation
Organisms containing two different alleles or individual containing both dominant and recessive gene of an allelic pair is called heterozygous. Answer : (a)
Q.13
RR ( Red) Antirrhinum is crossed with white (WWW) one. Offspring RW are pink. This is an example of ... ... [ CBSE 1991 ]
0%
a) Dominant - recessive
0%
b) Incomplete dominance
0%
c) Hybrid
0%
d) Supplementary gene
Explanation
There is no dominant-recessive relationship and thus the hybrid is intermediate of both the parent. This phenomenon is called incomplete dominance. Answer : (b)
Q.14
Multiple alleles control inheritance of ... .. [ AMU 2003 ]
0%
a) Colour blindness
0%
b) Sickle cell anaemia
0%
c) Blood groups
0%
d) Phenylketonuria
Explanation
When there is a gene existing in more than two allelic forms, this condition is referred to as multiple allelism. Blood group inheritance is controlled by multiple alleles. Answer : (c)
Q.15
Cross between homozygous black rough ( BBRR) guinea pig and homozygous white smooth guinea pig ( bbrr) produced black and rough animals in F1 generation. presuming the genes to be present on different chromosomes, the percentage of F2 individuals which are heterozygous for both the gene pairs would be ... ... [ Manipur 2005 ]
0%
a) 25%
0%
b) 35%
0%
c) 50%
0%
d) 75%
Explanation
Gametes
BR
Br
bR
br
BR
BBRR
BBRr
BbRR
BbRr
Br
BBRr
BBrr
BbRr
Bbrr
bR
BbRR
BbRr
bbRR
bbRr
br
BbRr
Bbrr
bbRr
bbrr
Since the genes are on different chromosomes, law of independent assortment will be followed BBRR when crossed with bbrr, progeny of genotype BbRr is produced. BbRr × BbRr Heterozygous for both gene = 4/16= 25% Answer : (a)
Q.16
grain colour of wheat is determined by three pairs of polygenes. A cross was made between AABBCC (coloured) and aabbcc (light). In F2 generation what percent of progeny would resemble parents. ... ... [ AIIMS 2008 ]
0%
a) Half
0%
b) Three fourth
0%
c) One third
0%
d) Less than 5%
Explanation
Polygene results in quantitative inheritance which is characterised by occurrence of intermediate forms between the parental type. In case of crossing between AABBCC (dark colour) and aabbcc (light colour) in F2 generation seven phenotypes will obtain with ratio of 1:6:15:20:15:6:1. The total number of progeny is 64, out of which only two will be likely resemble with either parents. Hence, their proportion in F2 generation would be 3.12 i.e., less than 5% Answer : (d)
Q.17
A child with mother of A group and father of AB group, will not have the following blood group. .... .. [ CPMT 2005 ]
0%
a) A
0%
b) B
0%
c) AB
0%
d) O
Explanation
I
A
I
B
I
A
I
A
I
A
(A blood group)
I
A
I
B
(AB blood group)
I
O
I
A
I
O
(A blood group)
I
B
I
O
(B blood group)
O blood group is not possible. Answer : (d)
Q.18
How many contrasting traits of Pea pod were chosen by Mendel? ... ... [ AMU 1998 ]
0%
a) 7
0%
b) 2
0%
c) 4
0%
d) 3
Explanation
Traits
Dominant
Recessive
Plant height
Tall (1.2 – 2.0m)(T)
Dwarf (0.25 – 0.5m) (t)
Flower position
Axillary (A)
Terminal (a)
Pod colour
Green (G) or (Y)
Yellow (g) or (y)
Pod shape
Full or inflated (I) or (C)
Constricted (i) or (c)
Flower colour
Violet (V) or (W)
White (v) or (w)
Seed shape
Round (R) or (W)
Wrinkled (r) or (w)
Seed colour
Yellow (Y) or (G)
Green (y) or (g)
Answer : (a)
Q.19
Emasculation is related to ... .. [ JIPMER 2002 ]
0%
a) Pure line
0%
b) mass selection
0%
c) Clonal selection
0%
d) Hybridization
Explanation
Hybridisation is the process of crossbreeding between genetically dissimilar parents to produce a hybrid. Because garden pea is self-fertilizing, the anthers need to be removed before maturity. The process of removal of anthers is called emasculation. Answer : (d)
Q.20
If Mendel had studied 7 traits using a plant with 12 chromosomes instead of 14, he would have ... .. [ CBSE 1998 ]
0%
a) Discovered the law of independent
0%
b) Not discovered the law of independent assortment
0%
c) Maped the chromosome
0%
d) Amitosis
Explanation
Priniciple of independent assortment is applicable to only those factors or genes which are present on different chromosomes. The seven characters that Mendel choose were present on 14 chromosomes and so they did not show linkage but if present on 12 chromosomes they would have shown linkage and the principle of independent assortment would not have been discovered. Answer : (b)
Q.21
A person with blood group A possesses ... .. [ AFMC 2004 ]
0%
a) Antigen A and antibody B
0%
b) Antigen B and antibody a
0%
c) Antigen A and antibody b
0%
d) No antigen and no antibody
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
O
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 : (c)
Q.22
A dihybrid condition is ... .. [ CBSE 1991 ]
0%
a) tt Rr
0%
b) Tt rr
0%
c) tt rr
0%
d) Tt Rr
Explanation
A phenomenon in which two organisms with two pairs of traits or contrasting characters are crossed is called a dihybrid cross. In the genotype TtRr, T and R are gametes for dominant traits and t and r are gametes for recessive traits. So it is a true dihybrid condition. Answer : (d)
Q.23
Polygenes show ... .. [ CBSE 1996 ]
0%
a) Different phenotypes
0%
b) Different genotypes
0%
c) Both of the above
0%
d) None of the above
Explanation
Polygenes is termed as a gene in which a dominant allele individually produces a slight effect on the phenotype but in the presence of similar other dominant allele controls the quanitative epression of a trait due to cumulative effect. Answer : (a)
Q.24
Out of a population of 800 individuals in F2 generation of cross between yellow round and green wrinkled Pea plants, what would be number of yellow and wrinkled seeds .... .. [ Manipal 2002 ]
0%
a) 800
0%
b) 400
0%
c) 200
0%
d) 150
Explanation
Yellow, round : Yellow, wrinkled, : Green, round : Green, wrinkled 9 : 3 : 3 : 1 Yellow wrinkled seeds = 3/16 × 800 = 150 Answer : (d)
Q.25
Four daughter cells obtained from a single meiosis differ from one another because of .... ... [ BHU 1994 ]
0%
a) Crossing over
0%
b) Independent assortment
0%
c) Change in chromosome number
0%
d) Both A and B
Explanation
During meiosis, position of each chromosome at the midline during metaphase was random, and that there was never a consistent maternal or paternal side of the cell division. Crossing over do take place.Therefore, each chromosome was independent of the other. Thus, when the parent cell separated into gametes, the set of chromosomes in each daughter cell could contain a mixture of the parental traits, but not necessarily the same mixture as in other daughter cells. Answer : (d)
Q.26
ABO blood grouping is determined by three alleles. Possible genotypes and phenotypes are .... .. [ kerala 2012 ]
0%
a) 3, 1
0%
b) 6, 4
0%
c) 4, 6
0%
d) 9, 7
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
O
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
4 phenotype and 6 genotype Answer : (b)
Q.27
Which is correct about traits chosen by Mendel ... .. [ AFMC 2007 ]
0%
a) Terminal pod is dominant
0%
b) Constricted pod is dominant
0%
c) Green coloured pod is dominant
0%
d) Tall plants are recessive
Explanation
Traits
Dominant
Recessive
Plant height
Tall (1.2 – 2.0m)(T)
Dwarf (0.25 – 0.5m) (t)
Flower position
Axillary (A)
Terminal (a)
Pod colour
Green (G) or (Y)
Yellow (g) or (y)
Pod shape
Full or inflated (I) or (C)
Constricted (i) or (c)
Flower colour
Violet (V) or (W)
White (v) or (w)
Seed shape
Round (R) or (W)
Wrinkled (r) or (w)
Seed colour
Yellow (Y) or (G)
Green (y) or (g)
Answer : (c)
Q.28
F2 generation is produced by ... ... [ CPMT 1993]
0%
a) Crossing F1 progeny with one of the parents
0%
b) Selfing the progeny of two identical parents
0%
c) Selfing the parents
0%
d) Recessive cross between individual parents
Explanation
Defination of F2 : The second filial(F2) generation, produced by selfing or intercrossing the F1. Selfing of F1 generation flowers gives F2 generation. Answer : (b)
Q.29
The allele which is unable to express its effect in the presence of another is called ... ... [ CBSE 1991 ]
0%
a) 8
0%
b) 4
0%
c) 2
0%
d) 16
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 : (b)
Q.30
Independent assortment is absent in case of ... .. [ CBSE 2001 ]
0%
a) Genes located on the same chromosome
0%
b) Genes located on homologous chromosomes
0%
c) Genes located on non-homologous chromosomes
0%
d) All the above
Explanation
The law of independent assortment is applicable to only those factors or genes which are present on different chromosomes. Answer : (a)
Q.31
Test cross ratio of 1:1:1:1 proves that ... .. [ Odisha 2011]
0%
a) F1 hybrid produces four different gametes
0%
b) F1 hybride is homozygous
0%
c) Two different progenies are produced by P1 parent
0%
d) Non of the above
Explanation
Gamete
RY
Ry
rY
ry
ry
RrYy
Rryy
rrYy
rryy
ry
RrYy
Rryy
rrYy
rryy
The test cross ratio 1:1:1:1 proves that the F1 hybrid produces four different types of gametes, each one of which is expressed without affecting the other. The four different types of gametes appear side by side. Example: RrYy (round, yellow seed)(F1 generation ) × rryy (Wrinkled, green seed)( recessive plant) F2 generation Round, yellow : Round,green : wrinkled, yellow : wrinkled, green 1 : 1 : 1 : 1 Answer : (a)
Q.32
Genotype F1 individuals can be tested by ... .. [ EAMCET 1996 ]
0%
a) Backcross with homozygous recessive parent
0%
b) Reciprocal crossing
0%
c) Backcrossing with heterozygous parent
0%
d) Backcrossing with homozygous dominants
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.33
B blood group person can donate blood to and receive blood from ... .. [ DPMT 2005 ]
0%
a) B, O and B,O
0%
b) B, AB and B, AB
0%
c) B, AB and B,O
0%
d) AB and A,B
Explanation
Answer : (c)
Q.34
Which of the following genotype does not produce any sugar polymer on the surface of RBCs .. .. [ kerala 2010]
0%
a) IAIA
0%
b) IB i
0%
c) IAIB
0%
d) ii
Explanation
IOIO or ii genotype does not produce any sugar polymer on surface on RBCs. Answer : (d)
Q.35
F2 ratio is chicken came out to be 9 rose comb blacks, 1 single comb black, 3 rose comb white and 3 single comb black. Which are the recessives? [ EAMCET 1996 ]
0%
a) Single comb, white plumage
0%
b) Rose comb, white plumage
0%
c) Single comb, black plumage
0%
d) Rose comb, black plumage
Explanation
Recessive trait is expressed in 6.25% of progeny in F2 generation. Out of 16 progeny, only 1 has single comb, white plumage. Thus, recessive traits are single comb and white plumage. Answer : (a)
Q.36
Phenotypic dihybrid ratio is ... .. [ AFMC 2000 ]
0%
a) 9:3:3:1
0%
b) 15:1
0%
c) 9:6:1
0%
d) 1:2:1
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 : (a)
Q.37
Organisms phenotypically similar but genetically different are ... ..[ EAMCET 1999]
0%
a) Heterozygous
0%
b) Homozygous
0%
c) Monozygous
0%
d) Multizygous
Explanation
Organisms containing two different alleles or individual containing both dominant and recessive gene of an allelic pair is called heterozygous. Answer : (a)
Q.38
A cross between hybrid and its parent is ... ... [ MPPMT 1998 ]
0%
a) Back cross
0%
b) Reciprocal cross
0%
c) Monohybrid cross
0%
d) Dihybrid cross
Explanation
Backcross is a cross of F1 hybrid with either of two parents . Answer : (a)
Q.39
A cross AaBb and aabb yields ... .. [ Pb PMT 1997 ]
0%
a) AaBb and aabb
0%
b) AaBb, Aabb, aaBb and aabb
0%
c) Aabb
0%
d) Aabb
Explanation
Gamete
AB
Ab
aB
ab
ab
AaBb
Aabb
aaBb
Aabb
ab
AaBb
Aabb
aaBb
aabb
AaBb × aabb → AaBb : Aabb : aaBb : aabb = 1 : 1 : 1 : 1 Answer : (b)
Q.40
Match the genetic phenomena with their respective ratios
(a) Inhibitory gene ratio
(1) 9:3:4
(b) Complementary gene ratio
(2) 1:1:1:1
(c) Recessive epistasis ratio
(3) 12:3:1
(d) Dihybrid test cross ratio
(4) 13:3
(e) Dominant epistasis ratio
(5) 9:7
0%
a) a - 5, b - 4, c - 3, d - 2, e - 1
0%
b) a - 4, b - 5, c - 1, d - 2, e - 3
0%
c) a - 1, b - 2, c - 4, d - 3, e - 5
0%
d) a - 5, b - 4, c - 1, d - 2, 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 : (b)
Q.41
Validity of Mendel's law of segregation is established when ... .. [ MPPMT 2004]
0%
a) Two hybrids are crossed
0%
b) A parent is crossed with F1 hybrid
0%
c) Two pure breeding contrasting traits are crossed
0%
d) None of the above
Explanation
Mendel’s law of segregation: In F1 the dominant phenotype appears, the recessive phenotype is not lost but reappears in F2. This suggested that there is no blending of Mendelian factors in F1, but they stay together and only one is expressed. A gamete may carry either the dominant or recessive factors but not both as we find F1 individuals. The concept of segregation is often called Mendel’s first principle. While a 3:1 ratio in F2 generation of a monohybrid cross suggested that segregation of alleles does take place which can be confirmed by test cross. Answer : (a)
Q.42
A man of A - blood group marries a woman of AB blood group. Which type of progeny would indicate that man is heterozygous A? ... . [ CBSE 1993 ]
0%
a) AB
0%
b) A
0%
c) O
0%
d) B
Explanation
I
A
I
O
I
A
I
A
I
A
(A blood group)
I
A
I
O
(A blood group)
I
B
I
A
I
B
(AB blood group)
I
B
I
O
(B blood group)
If a man has heterozygous A blood group, then its genotype would be IAIO. If man has homozygous A blood group, then there will be no progeny of B blood group IA IA IA IAIA (A blood group) IAIO (A blood group) IB IAIB (AB blood group) IAIB (AB blood group) Answer : (d)
Q.43
Offspring produced from a marriage have only O and A blood groups. The possible genotypes of the parents would be ... .. [ KCET 2009]
0%
a) IAIA and IAIB
0%
b) IAIA and IOIO
0%
c) IOIO and IOIO
0%
d) IAIO and IOIO
Explanation
I
A
I
O
I
O
I
A
I
O
(A blood group)
I
O
I
O
(O blood group)
I
O
I
A
I
O
(A blood group)
I
O
I
O
(O blood group)
Genotype for O blood group is IOIO. Genotype for A blood is IAIA or IAIO. Each allele is obtained from each parent. So parents Have possible genotype is IAIO and IOIO. Answer : (d)
Q.44
Blood grouping in human beings is controlled by .... [ DPMT 1996 ]
0%
a) 4 alleles in which A is dominant
0%
b) 3 alleles in which A and B are codominant and i is recessive
0%
c) 3 alleles in which none is dominant
0%
d) 3 alleles in which A is dominant
Explanation
I
O
I
O
I
A
I
A
I
O
(A blood group)
I
A
I
O
(A blood group)
I
B
I
B
I
O
(B blood group)
I
B
I
O
(B blood group)
Inheritance of blood group is an example of multiple alleles. It has three alleles IA, IB and IO or i. In this IA and IB are co-dominant and IO or i is recessive. Answer : (b)
Q.45
Correct reason for Mendel's success was ... .. [ BHU 2001 ]
0%
a) He repeated each experiment several times
0%
b) traits chosen by him had genes far apart so that linkage was absent
0%
c) He kept record of all experiments
0%
d) He used statistical techniques
Explanation
Mendel was lucky in choosing characters which showed complete independent assortment. He only took those traits for his studies which did not show linkage, interaction or incomplete dominance. Answer : (b)
Q.46
Tall read flowered Pea plant crossed to dwarf white flowered plant yields only tall red flowered plants. A test cross shall give a ratio of ... .... [ AMU 1998 ]
0%
a) 1 : 1
0%
b) 3 : 1
0%
c) 1 : 1 : 1 :
0%
d) 1 : 2 : 4 : 6 : 4 : 2 : 1
Explanation
Gametes
TR
Tr
tR
tr
tr
TtRr (tall,red)
Ttrr (tall,white)
ttRr (dwarf,red)
ttrr (dwarf,white)
tr
TtRr (tall,red)
Ttrr (tall,white)
ttRr (dwarf,red)
ttrr (dwarf, white)
Tall plant, red flower (TtRr) × Dwarf plant, white flower (ttrr) → Progeny Tall, red : tall, white : dwarf, red : dwarf,white 1 : 1 : 1 : 1 Answer : (c)
Q.47
Mendel chose contrasting traits in Pea ... .. [ WB 2011 ]
0%
a) Three
0%
b) Two
0%
c) One
0%
d) Seven
Explanation
Mendel experimented on garden pea to study inheritance. He studied the inheritance of seven different pairs of contrasting characters in this plant but considered only one pair at a time. Answer : (d)
Q.48
Red seed colour shows an F2 ratio of 1:4:6:4:it is due to ... .. [ DPMT. 1996 ]
0%
a) Two polygenes
0%
b) Different number of dominant genes
0%
c) Different number of recessive genes
0%
d) Supplementary genes
Explanation
This ratio shows that red seed does not follow mendelian law. It occurs in polygenic inheritance. Answer : (a)
Q.49
Phenotypic and genotypic ratio in F2 generation in incomplete dominance is ... .. [ AMU 2003 ]
0%
a) 1:2:1 and 1:2:1
0%
b) 3:1 and 1:2:1
0%
c) 9:6 and 3:1
0%
d) 9:3:3:1 and 1:2:1:4:1:1:2:1:2:1
Explanation
Phenotypic ratio and genotypic ratio in F2 generation in incomplete dominance is same. The ratio is 1 : 2 : 1 Answer : (a)
Q.50
Gene P and Q are both required in dominant state for normal hearing. A deaf couple has all children with normal hearing. The probable genotype for the couple is ... ...
0%
a) PPqq × ppQQ
0%
b) PPqq × PPqq
0%
c) PpQq × ppqq
0%
d) PPqq × ppQq
Explanation
Gametes
Pq
pQ
PpQq
Gene P and Q both are needed for normal hearing. A deaf couple has all children with normal hearing. This implies that all Children has dominant P and Q in their genotype. We can also conclude that each parent has one dominant trait of two set of allele, i.e., mother carries dominant homozygous PP and father carries Dominant homozygous QQ or vice-versa. PPqq × ppQQ Answer : (a)
0 h : 0 m : 1 s
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