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1.In pea plants, smooth (S) peas are dominant over wrinkled (s) peas in the sing

ID: 42982 • Letter: 1

Question

1.In pea plants, smooth (S) peas are dominant over wrinkled (s) peas in the single gene controlling pea texture. A plant with smooth peas is crossed with a plant with wrinkled peas. Of the progeny, 252 are smooth and 247 are wrinkled. What are the genotypes of the parent plants?

2.In humans, having finger hair or not is determined by one gene with two alleles. The allele for finger hair is dominant to the allele for no finger hair. What is the chance a father with no finger hair and a mother with finger hair will have a child that has no finger hair if the mother

Explanation / Answer

1. The given ratio in progeniesis nearly 1:1 which could be found only when the hybrid progeny is test crossed with parent. So the genotype of parent would be : Ss and ss

2. As the father of the mother is not having finger hairs so his genotype would be hh and so the mother must be heterozygous Hh. So the progenies of this heterozygous mother(Hh) and recessive homozygous father(hh) could have Hh(50%) or hh(50%) genotypes. So there are 50 % chances that the progeny will bot have finger hairs.

3. Mother is XAXA means she is normal. In son the X chromosomes comes from mother only.So he will get normal XA chromosome that means the son can not have haemophilia if the mother is normal.

4. Genotype of 75 progenies i.e. purple bodies and long wings could be: AaBB, AABB, AaBb, AABb

Genotype of 25 progenies i.e. yellow bodies and short wings could be: aabb

So the absent genotypes are: AAbb, Aabb, aaBb, aaBB

This is showing that the phenotypic ratio of 9:3:3:1 was modified to 3:1

The inference could be obtained that the progenies surviving would have either :

i) Atleast one "A" and "B" together

ii)Both were absent i.e. recessive homozygous condition.

So A abd B are linked together as absence of any of them causing no survival of the progeny.