FREE NEET Biology Practice MCQ Test: Principles Of Inheritance And Variation Exercise 1 Questions Answers With Detailed Explanations [PDF]

Principles Of Inheritance And Variation Topic Based

Question : 1

ABO blood grouping is controlled by gene 1 which has three alleles and show co-dominance. There are six genotypes. How many phenotypes in all are possible ?


Answer: (b)

Question : 2

Which mendelian idea is depicted by a cross in which the F1 generation resembles both the parents ?


Answer: (b)

Question : 3

In Drosophila, the sex is determined by


Answer: (b)

According to Bridges, in Drosophila Y-chromosome is heterochromatic so it is not active in sex determination. In Drosophila sex determination takes place by sex index ratio.

Sex index ratio = ${\text"No. of X-chromosomes"}/{\text"No. of set of autosomes"} = X/A$

Gene of femaleness (Sxl- gene) (Sxl = Sex lethal gene) is located on X-chromosome and gene of maleness is located on autosome. Gene of male fertility is located on Y-chromosome and in Drosophila, Y-chromosome plays additional role in spermatogenesis and development of male reproductive organ, so Y-chromosome is essential for the production of fertile male.

Question : 4

A mutation is a


Answer: (d)

Mutation is the sudden inheritable discontinuous variation which appears in an organism due to permanent changes in their genotypes. The term mutation was coined by Hugo de Vries (1901).

Question : 5

The map distance between genes A and B is 3 units, between B and C is 10 units and between C and A is 7 units. The order of the genes in a linkage map constructed on the above data would perhaps be


Answer: (d)

Question : 6

Which of the following statement is not true for two genes that show 50% recombination frequency?


Answer: (a)

The statement in option (b) is not true. It can be corrected as

The tightly linked genes on chromosomes show 100% parental types and 0% recombinants. Two genes that undergo independent assortment indicated by a recombinant frequency of 50% are on non-homologous chromosomes indicated for apart in a single chromosome. As the distance between two genes increases, crossing over frequency increases. This results in the formation of more recombinant gametes and fewer parental gametes. Rest of the statements are true.

Question : 7

A man with blood group 'A' marries a woman with blood group 'B'. What are all the possible blood groups of their offsprings?


Answer: (b)

Question : 8

If a colour blind woman marries a normal visioned man, their sons will be


Answer: (c)

Colour blindness is a recessive sex-linked trait.

principles-of-inheritance-and-variation-class-12-Chapter-5-neet-mcq-test

All sons will be colour blind and all daughters will be carriers.

Question : 9

Mongolian Idiocy due to trisomy in $21^{st}$ chromsome is called


Answer: (c)

Question : 10

Pick out the correct statements.

  1. Haemophilia is a sex-linked recessive disease.
  2. Down's syndrome is due to aneuploidy.
  3. Phenylketonuria is an autosomal recessive gene disorder.
  4. Sickle-cell anaemiais an X - linked recessive gene disorder.


Answer: (c)

Question : 11

The pie chart shows the results of a survey of the incidence of blood groups A, B, AB and O amongst people in Britain. Which of the following conclusions can be deduced from the diagram?

principles-of-inheritance-and-variation-class-12-Chapter-5-neet-mcq-6-586-29

Answer: (d)

5% of the population are of blood group AB, i.e. 1 out of every 20 individuals is AB.

Question : 12

Erythroblastosis foetal is caused when fertilization takes place between gametes of


Answer: (b)

If fertilization takes place between gametes of $Rh^{-}$ female and $Rh^{+}$ male then the resulting foetus' blood is $Rh^{+}$ . The $Rh^{+}$ blood of the foetus stimulates the formation of anti Rh factors in the mother's blood. In second pregnancy (with $Rh^{+}$ foetus), the anti Rh factors of the mother's blood destroy the foetal red blood corpuscles. This is called erythroblastosis foetalis. New born may survive but it is often anaemic. The $Rh^{-}$ child does not suffer.

Question : 13

The types of gametes formed by the genotype RrYy are


Answer: (b)

The formula to determine the number of gametes is

$2^n = 2^{(2)} = 4$

Thus, RrYy would produce 4 gametes of the types RY, Ry, rY, ry.

Question : 14

Two linked genes a and b show 20% recombination. The individuals of a dihybrid cross between ++ /++ × ab/ab shall show gametes


Answer: (a)

Question : 15

The genotypes of a husband and Wife are IAIB and IAi. Among the blood types of their children, how many different genotypes and phenotypes are possible?


Answer: (a)

Husband × Wife = $I^{A} I^{B} × I^{A}i$

Male/Female$I^{A}$$I^{B}$
$I^{A}$$I^{A} I^{A}$$I^{A} I^{B}$
i$I^{A}i$$I^{B}i$

Number of genotypes = 4

Number of phenotypes = 3

$I^{A}I^{A}$ and $I^{A}$i = A

$I^{A}I^{B}$ = AB

$I^{B}i$ = B

Question : 16

In Mendel's experiments with garden pea, round seed shape (RR) was 'dominant over wrinkled seeds (rr), yellow cotyledon (YY) was dominant over green cotyledon (yy). What are the expected phenotypes in the F2 generation of the cross RRYY × rryy ?


Answer: (d)

Question : 17

Test cross in plants or in Drosophila involves crossing


Answer: (a)

Question : 18

Down’s syndrome in humans is due to


Answer: (c)

Down’s syndrome is the chromosomal disorders due to the presence of an additional copy of the chromosome number 21 (trisomy of 21). The affected individual is short statured with small round head, furrowed tongue and partially open mouth and mental development is restarted.

Question : 19

Why is the allele for wrinkled seed shape in garden peas considered recessive ?


Answer: (a)

Allele for wrinkled shape of seed in garden pea plant is considered to be recessive because the trait (character) associated with the allele is not expressed in heterozygotes.

Question : 20

A colour blind mother and normal father would have


Answer: (b)

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