NEET Biology Principles of Inheritance and Variation MCQs Set 26

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Chapter-wise Objective Questions: Principles of Inheritance and Variation

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Multiple Choice Questions

Question 1: A test cross is performed:
(a) by selfing F2 plants
(b) by selfing F1 plants
(c) to determine whether an F1 plant is homozygous or heterozygous
(d) between a homozygous dominant and a homozygous recessive plant
Show Answer & Explanation

Answer: (c) to determine whether an F1 plant is homozygous or heterozygous

Explanation:
1. A test cross crosses an individual showing the dominant phenotype with the homozygous recessive parent.
2. The result tells whether that individual is homozygous or heterozygous.

Teacher's Note:
1. All progeny dominant means the tested plant is homozygous.
2. Progeny in 1 : 1 ratio means the plant is heterozygous.

Question 2: If a cross is made between AA and aa, the nature of the F1 progeny will be:
(a) Genotypically AA, phenotypically a
(b) Genotypically Aa, phenotypically a
(c) Genotypically Aa, phenotypically A
(d) Genotypically aa, phenotypically A
Show Answer & Explanation

Answer: (c) Genotypically Aa, phenotypically A

Explanation:
1. AA × aa gives only Aa.
2. With A dominant, the phenotype is A.

Teacher's Note:
1. Genotype is the allele combination and phenotype is the visible trait.
2. The recessive allele is masked in the heterozygote.

Question 3: A pure tall and a pure dwarf plant were crossed to produce offspring. The offspring were self-crossed. What is the ratio between true-breeding tall and true-breeding dwarf plants in the F2?
(a) 1 : 1
(b) 3 : 1
(c) 2 : 1
(d) 1 : 2 : 1
Show Answer & Explanation

Answer: (a) 1 : 1

Explanation:
1. In F2 the genotypes are 1 TT : 2 Tt : 1 tt.
2. True-breeding (homozygous) tall = TT and true-breeding dwarf = tt, giving 1 : 1.

Teacher's Note:
1. True breeding means homozygous.
2. Do not count the heterozygous Tt plants, which do not breed true.

Question 4: What result would Mendel have got when he self-pollinated a dwarf F2 plant?
(a) All tall plants
(b) Tall and dwarf plants in 3 : 1 ratio
(c) All dwarf plants
(d) Tall and dwarf plants in 1 : 1 ratio
Show Answer & Explanation

Answer: (c) All dwarf plants

Explanation:
1. A dwarf plant is the homozygous recessive tt.
2. Selfing tt gives only tt, so all offspring are dwarf.

Teacher's Note:
1. Recessive phenotype always breeds true.
2. Tall F2 plants, however, may be TT or Tt.

Question 5: What is the dihybrid test cross ratio found by Mendel?
(a) 9 : 3 : 3 : 1
(b) 1 : 1 : 1 : 1
(c) 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1
(d) 3 : 1
Show Answer & Explanation

Answer: (b) 1 : 1 : 1 : 1

Explanation:
1. A dihybrid test cross is TtRr × ttrr.
2. The four gamete types of TtRr unite with only tr, giving four phenotypes in the ratio 1 : 1 : 1 : 1.

Teacher's Note:
1. The test cross ratio equals the gamete ratio of the heterozygote.
2. Equal classes show independent assortment.

Question 6: A cross between a pure tall pea plant with green pods and a dwarf pea plant with yellow pods will produce tall F2 plants, out of 16:
(a) 15
(b) 13
(c) 12
(d) 7
Show Answer & Explanation

Answer: (c) 12

Explanation:
1. Tall plants in F2 = 3/4 of 16 = 12.

Teacher's Note:
1. Pod colour is a second, independent character.
2. So only height decides the number of tall plants.

Question 7: Some of the dominant traits studied by Mendel were:
(a) Round seed shape, constricted pod shape and axial flower position
(b) Yellow seed colour, inflated pod shape and axial flower position
(c) Yellow seed colour, violet flower colour and yellow pod colour
(d) Axial flower position, green pod colour and green seed colour
Show Answer & Explanation

Answer: (b) Yellow seed colour, inflated pod shape and axial flower position

Explanation:
1. Yellow seed, inflated pod and axial flower are all dominant.
2. Constricted pod, yellow pod and green seed are recessive, which rules out the other options.

Teacher's Note:
1. Learn the seven pairs as a table.
2. Each option mixes dominant and recessive traits except one.

Question 8: According to Mendel, "factors" or "genes":
(a) are the units of inheritance
(b) contain the information required to express a particular trait
(c) both of the above
(d) none of the above
Show Answer & Explanation

Answer: (c) both of the above

Explanation:
1. Mendel's factors are units of inheritance and carry the information needed to express a trait.

Teacher's Note:
1. Mendel's factors are now called genes.
2. He proposed them without knowing about chromosomes or DNA.

Question 9: The phenotype of any character will not be affected if the modified allele produces:
(a) a normal enzyme
(b) a non-functional enzyme
(c) no enzyme at all
(d) both 2 and 3
Show Answer & Explanation

Answer: (a) a normal enzyme

Explanation:
1. If the modified allele still produces a normal (equally efficient) enzyme, the phenotype remains unchanged.
2. A non-functional enzyme or no enzyme changes the phenotype.

Teacher's Note:
1. Alleles differ because of changes in DNA sequence.
2. Their effect on the phenotype depends on the product they code for.

Question 10: The recessive characters are:
(a) only expressed in the heterozygous condition
(b) only expressed in the homozygous condition
(c) a blend in the heterozygous condition
(d) always impure
Show Answer & Explanation

Answer: (b) only expressed in the homozygous condition

Explanation:
1. A recessive allele is masked by the dominant allele in the heterozygote.
2. It shows only when both alleles are recessive.

Teacher's Note:
1. A recessive phenotype always has a homozygous recessive genotype.
2. That is why recessive plants breed true.

Question 11: In Mirabilis and Antirrhinum, the appearance of a pink hybrid (Rr) from a cross between a red (RR) and a white (rr) flowered parent indicates:
(a) Incomplete dominance
(b) Segregation
(c) Dominance
(d) Heterosis
Show Answer & Explanation

Answer: (a) Incomplete dominance

Explanation:
1. Neither allele is completely dominant, so the heterozygote shows an intermediate pink colour.

Teacher's Note:
1. This is a blending-type result, but the alleles separate unchanged in the F2.
2. The F2 phenotypic ratio is 1 : 2 : 1.

Question 12: In incomplete dominance, the monohybrid ratio of phenotypes in the F2 generation is:
(a) 1 : 2 : 1
(b) 3 : 1 : 1
(c) 9 : 3 : 3 : 1
(d) 2 : 3 : 1
Show Answer & Explanation

Answer: (a) 1 : 2 : 1

Explanation:
1. In incomplete dominance the F2 phenotypic ratio matches the genotypic ratio.
2. It is therefore 1 : 2 : 1.

Teacher's Note:
1. The heterozygote has its own phenotype.
2. So 3 genotypes give 3 phenotypes.

Question 13: In Mirabilis jalapa, homozygous red-flowered and white-flowered plants are crossed and all F1 plants have pink flowers. In the F2 produced by selfing F1 plants, red, pink and white flowered plants appear in the ratio of:
(a) 1 : 1 : 2
(b) 2 : 1 : 1
(c) 1 : 0 : 1
(d) 1 : 2 : 1
Show Answer & Explanation

Answer: (d) 1 : 2 : 1

Explanation:
1. Selfing Rr gives 1 RR (red) : 2 Rr (pink) : 1 rr (white).

Teacher's Note:
1. Incomplete dominance gives a 1 : 2 : 1 phenotypic ratio.
2. This shows that the alleles do not mix permanently.

Question 14: In a dihybrid cross where one pair of alleles shows incomplete dominance, the F2 genotypic ratio is:
(a) 3 : 6 : 3 : 1 : 2 : 1
(b) 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1
(c) 9 : 3 : 3 : 1
(d) 1 : 2 : 1
Show Answer & Explanation

Answer: (b) 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1

Explanation:
1. Dominance does not change genotypes: (1 : 2 : 1) × (1 : 2 : 1) gives 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1.
2. The ratio 3 : 6 : 3 : 1 : 2 : 1 would be the phenotypic ratio in this case.

Teacher's Note:
1. Genotypic ratio is independent of the type of dominance.
2. Only the phenotypic ratio changes.

Question 15: In which of the following genotype combinations do both gene pairs show a codominant condition (both alleles expressed in the heterozygote)?
(a) HbAHbA, IAIB
(b) HbSHbS, IAIB
(c) HbAHbS, IAIB
(d) HbSHbS, IAIA
Show Answer & Explanation

Answer: (c) HbAHbS, IAIB

Explanation:
1. HbAHbS and IAIB are both heterozygous with both alleles expressed.
2. The other options contain a homozygous pair.

Teacher's Note:
1. Codominance means both alleles of the heterozygote express independently.
2. AB blood group is the classic example.

Question 16: AB blood group shows:
(a) Codominance
(b) Complete dominance
(c) Mixed inheritance
(d) Composite inheritance
Show Answer & Explanation

Answer: (a) Codominance

Explanation:
1. IA and IB are both fully expressed in the AB group.

Teacher's Note:
1. Both sugars A and B appear on the red cell surface.
2. This is codominance.

Question 17: ABO blood group is an example of:
(a) Epistasis
(b) Multiple allelism
(c) Pleiotropism
(d) Complementary genes
Show Answer & Explanation

Answer: (b) Multiple allelism

Explanation:
1. Three alleles (IA, IB, i) of a single gene control ABO blood groups, which is multiple allelism.

Teacher's Note:
1. Multiple alleles are found at the same locus.
2. An individual carries only two of them.

Question 18: A child has blood group O. His parents' blood groups cannot be:
(a) B and O
(b) A and O
(c) AB
(d) A and B
Show Answer & Explanation

Answer: (c) AB

Explanation:
1. An O child is ii, so each parent must carry an i allele.
2. A parent with AB has no i allele and cannot be a parent of an O child.

Teacher's Note:
1. AB parents can have children of A, B or AB only.
2. A and B parents can give an O child if both are heterozygous.

Question 19: If one parent has blood group A and the other parent has blood group B, the offspring may have which blood group?
(a) AB only
(b) O only
(c) B only
(d) A, B, AB, O
Show Answer & Explanation

Answer: (d) A, B, AB, O

Explanation:
1. If both parents are heterozygous (IAi × IBi), they can have children of all four groups A, B, AB and O.

Teacher's Note:
1. The parents' phenotypes alone do not fix the genotypes.
2. The question asks about possible blood groups.

Question 20: A child of O blood group has a father of B blood group. The genotype of the father would be:
(a) IOIO
(b) IBIB
(c) IAIB
(d) IBi
Show Answer & Explanation

Answer: (d) IBi

Explanation:
1. The child is ii, so he received an i allele from his father.
2. A B-group father with an i allele is IBi.

Teacher's Note:
1. IBIB could not give an i allele.
2. The O allele is written as i.

Question 21: A gene that shows its effect on more than one character is:
(a) Polygene
(b) Pleiotropic gene
(c) Multifactor gene
(d) Multiple gene
Show Answer & Explanation

Answer: (b) Pleiotropic gene

Explanation:
1. A pleiotropic gene affects several traits. For example, the gene for phenylketonuria affects mental ability, hair and skin colour.

Teacher's Note:
1. Polygenes are many genes controlling one trait.
2. Pleiotropy is one gene with many effects.

Question 22: Blood grouping in humans is controlled by:
(a) 4 alleles in which IA is dominant
(b) 3 alleles in which IA and IB are dominant
(c) 2 alleles in which none is dominant
(d) 3 alleles in which IA is recessive
Show Answer & Explanation

Answer: (b) 3 alleles in which IA and IB are dominant

Explanation:
1. Three alleles (IA, IB and i) are involved.
2. IA and IB are both dominant to i (and codominant with each other).

Teacher's Note:
1. The i allele is recessive.
2. Six genotypes and four phenotypes result.

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