NEET Biology Principles of Inheritance and Variation MCQs Set 17

Answer: (c) at the same locus on homologous chromosomes

Explanation:
1. Multiple alleles are three or more forms of a gene occupying the same locus on homologous chromosomes.

Teacher's Note:
1. Only two alleles are present in any one diploid individual.
2. The alleles arise by mutation of the same gene.

Answer: (d) Pink

Explanation:
1. Snapdragon shows incomplete dominance, so the heterozygote is pink.

Teacher's Note:
1. The same happens in Mirabilis (four o'clock).
2. The F₂ ratio is 1 red : 2 pink : 1 white.

Answer: (b) pink : white (1 : 1)

Explanation:
1. Rr × rr gives 1 Rr (pink) : 1 rr (white).

Teacher's Note:
1. This is a test cross in an incomplete dominance system.
2. No red plants can appear.

Answer: (b) A, B, AB

Explanation:
1. With AB as one parent, the child gets I^A or I^B from that parent, so group O is not possible.
2. The other parent (A) gives I^A or i, so the children can be A, B or AB.

Teacher's Note:
1. AB parent never has an O child.
2. List the gametes of each parent and combine them.

Answer: (c) I^Bi

Explanation:
1. A child with group B can be I^BI^B or I^Bi. The mother is A (I^AI^A or I^Ai) so she cannot give I^B.
2. The child must therefore have received I^B from the father and i from the mother, making I^Bi.

Teacher's Note:
1. The mother must therefore be I^Ai.
2. Always work out which allele came from which parent.

Answer: (d) O

Explanation:
1. The AB father has no i allele, so he cannot give the O allele.
2. A child with the group O (ii) is therefore impossible.

Teacher's Note:
1. Mother A may give I^A or i.
2. Father AB gives I^A or I^B.

Answer: (d) any of the above

Explanation:
1. If the mother is I^Bi and the father is I^Ai, the children can be A, B, AB or O.
2. So the offspring may have any of the groups A, O or AB.

Teacher's Note:
1. Parents' phenotypes alone do not decide the child's blood group.
2. The question asks what is possible.

Answer: (d) Complementary genes

Explanation:
1. Complementary genes are non-allelic genes that need each other to produce a particular phenotype.

Teacher's Note:
1. Each dominant gene alone cannot give the trait.
2. The F₂ ratio becomes 9 : 7 instead of 9 : 3 : 3 : 1.

Answer: (c) point

Explanation:
1. A single base substitution (GAG to GUG in mRNA) replaces glutamic acid with valine at the sixth position of the β-globin chain. This is a point mutation.

Teacher's Note:
1. A single base change is a point mutation.
2. Frame shift mutations come from insertion or deletion of bases.

Answer: (b) 6 and 4

Explanation:
1. The genotypes are I^AI^A, I^Ai, I^BI^B, I^Bi, I^AI^B and ii (6).
2. The phenotypes are A, B, AB and O (4).

Teacher's Note:
1. Three alleles give 3(3 + 1)/2 = 6 genotypes.
2. Dominance reduces them to 4 phenotypes.

Answer: (a) The mutant haemoglobin molecule undergoes polymerisation under low oxygen tension, causing a change in the shape of RBC

Explanation:
1. The change is in the β-globin chain, not the α-chain.
2. HbS polymerises under low oxygen tension, which makes the red cells sickle shaped.

Teacher's Note:
1. Glutamic acid is replaced by valine at the sixth position of the β-chain.
2. The disease is autosomal and recessive.

Answer: (c) both I^A and I^B express

Explanation:
1. I^A and I^B are codominant, so both antigens appear together.

Teacher's Note:
1. This produces the AB phenotype.
2. Neither allele masks the other.

Answer: (c) One son and one daughter

Explanation:
1. The woman is a carrier, X^CX^c. The man is X^cY.
2. Half the sons and half the daughters are colour blind, so one son and one daughter out of two each.

Teacher's Note:
1. A carrier mother passes the allele to half her children.
2. A colour-blind father gives X^c to all his daughters.

Answer: (a) a holandric character

Explanation:
1. Hypertrichosis (hairy ears) is controlled by a gene on the Y chromosome, so it is a holandric trait.

Teacher's Note:
1. It passes from father to all his sons.
2. Holandric means Y-linked.

Answer: (d) Baldness

Explanation:
1. Haemophilia and colour blindness are X-linked and hypertrichosis is Y-linked.
2. Baldness is a sex-influenced trait on an autosome.

Teacher's Note:
1. Sex-influenced traits are expressed differently in males and females.
2. Baldness is dominant in males and recessive in females.

Answer: (c) Hemizygous

Explanation:
1. A male's X-linked gene has no allele on the Y chromosome, so it is hemizygous.

Teacher's Note:
1. Males are hemizygous for X-linked genes.
2. Hemi means half.

Answer: (c) Sex-influenced character

Explanation:
1. The baldness gene is on an autosome but its expression is influenced by sex hormones, so it is sex influenced.

Teacher's Note:
1. The allele is dominant in males and recessive in females.
2. Hence men are bald more often.

Answer: (c) half of the sons are colour blind

Explanation:
1. The woman is a carrier (X^CX^c) as her father was colour blind.
2. Her sons receive X^C or X^c equally, so half the sons are colour blind.

Teacher's Note:
1. Half the daughters would also be colour blind as the father is X^cY.
2. Always find the mother's genotype from her father.

Answer: (c) 30 map units

Explanation:
1. Recombinants = 1000 − 700 = 300, so recombination frequency = 300/1000 = 30%.
2. One map unit equals 1% recombination, so the distance is 30 map units.

Teacher's Note:
1. Distance between genes is proportional to recombination frequency.
2. Parental type means non-recombinant.

Answer: (a) b, a, c

Explanation:
1. The largest distance (b–c = 28) is the sum of the other two (b–a = 20 and a–c = 8).
2. So a lies between b and c, giving the order b, a, c.

Teacher's Note:
1. The two end genes show the largest crossover value.
2. Check that the shorter distances add up to the longest.

Answer: (d) expressed in males

Explanation:
1. Males have only one X chromosome, so an X-linked recessive gene is expressed in them even when single.

Teacher's Note:
1. In females it is expressed only in the homozygous state.
2. This is why X-linked recessive disorders are more common in males.