Question

GENETICS PROBLEMS

1.  The gene for widow's peak (W) is dominant to the gene for straight hairline (w).  A male homozygote with widow's peak marries a female homozygote with straight hairline, what are the possible genotypic and phenotypic ratios of the F1 and F2 generations that may result from the union?

Phenotype of Parents   ………………………… x …………………………

Genotype of Parents      ………………………… x …………………………

Genes in gametes          ………………………… x …………………………

Genotype of F1 Generation                    …………………………

Phenotype of F1 Generation                   …………………………

Genes in F1 gametes     ………………………… or …………………………

        (Punnett Square)

Genotype of F2 Generation                    ………………. : ……………… : ………………

Phenotype of F2 Generation       ……………………………. : ……………………………

     

2.       The gene for normal vision (C) is dominant to the X-linked recessive gene (c) for colorblindness.

Y chromosomes lack X-linked genes.

a)      If a homozygous woman with normal vision mates with a colorblind man, what is

the probability that they will have a colorblind child?

1.   0%       B.   25%      C.   50%       D.   100%

b)     If a heterozygous woman with normal vision mates with a man with normal

vision, what is the probability that they will have a colorblind son?

1.   0%      B.   25%      C.   50%       D.   100%

Answer 1

Answer 1:phenotypes of parents widow’s peak × straight hair line

                Genotype of parents WW × ww

                Genes in gamates W × w

                Genotype of F1 generation Ww

                Phenotype of F1 generation widow’s peak

                Genes in F1 gamates W or w

     Punnett square(cross between two heterozygotes Ww×Ww)

gamates	W	w
W	WW Widow’s peak	Ww Widow’s peak
w	Ww Widow’s peak	ww Straight hair line

Genotype of F2 generation WW,Ww,ww ( genotypic ratio is 1:2:1)

Phenotype of F2 generation Widow’s peak, straight hair line.(phenotypic ratio is 3:1)

Explanation: Widow’s peak is an autosomal dominant trait hence if W is allele responsible for it, then a person with homozygous(WW) as well as heterozygous (Ww) condition will show the phenotype of Widow’s peak.

The straight hair line trait will be able to express itself only in homozygous (ww)condition.

Answer 2 a: let the genotype of homozygous woman be XCXC

The genotype of colourblind man be XcY

The cross would result in

gamates	XC	XC
Xc	XCXc Carrier female	XCXc Carrier female
Y	XCY Normal male	XCY Normal male

The answer is A ‘0%’

The probability of a colorblind child in this case is 0%

Answer 2b: let the genotype of heterozygous woman be XCXc

The genotype of normal man be XCY

The cross would result in

gamates	XC	Xc
XC	XCXC normal female	XCXc Carrier female
Y	XCY Normal male	XcY colorblind male

The answer is B ‘25%’

The probability of a colorblind son in this case is 25%

Explanation: The character of colorblindness is an X linked recessive trait and males are more prone to show colorblindness as the females have two copies of X chromosome so if one of them is having the defective gene, the gene present on another X chromosome will give the normal vision, so a heterozygote female is carrier female, she can pass on this trait to her son.

For a male if he has the gene for colorblindness then it will be expressed, as there is no copy of the X chromosome

For a female to be colorblind she need to have the defective gene on both the X chromosome