Question #90540

A human female "carrier" who is heterozygous for the recessive, sex-linked trait red color blindness, marries a normal male. a. What proportion of their female progeny will show the trait? b. What proportion of their male progeny will show the trait? c. If one of their daughters marries a normal male, what is the probability that the first son of this marriage will show the trait? d. If one of their daughters marries a normal male, what is the probability that the first child of this marriage will show the trait?

Expert's answer

Legend:

XX - female

XY - male

X_{A} - normal (dominant)

X_{a} - trait (recessive)

X_{A}Xa - normal female heterozygous for the recessive trait

X_{A}Y - normal male

Parents:

X_{A}X_{a} X_{A}Y

Gametes:

X_{A , }Xa , X_{A },_{ }Y

F1 (progeny):

X_{A}X_{A }- female, no trait

X_{A}Y - male, no trait

X_{A}X_{a }- female, no trait

X_{a}Y - male, trait-positive

**Question a: 0%**

**Question b: 50%**

Daughter_1:

X_{A}X_{A} X_{A}Y

Gametes:

X_{A} , X_{A} , Y

F2.1 (progeny):

X_{A}X_{A} - female, no trait

X_{A}Y - male, no trait

Daughter_2:

X_{A}X_{a} X_{A}Y

Gametes:

X_{A} , Xa , X_{A} , Y

F2.2 (progeny):

X_{A}X_{A} - female, no trait

X_{A}Y - male, no trait

X_{A}X_{a} - female, no trait

XaY - male, trait-positive

F2 (total):

2 * X_{A}X_{A} - female, no trait

2 * X_{A}Y - male, no trait

X_{A}X_{a} - female, no trait

X_{a}Y - male, trait-positive

**Question c: 33,3%**

**Question d: 16,7%**

**Answer: **

a: 0% - female with trait

b: 50% - male with trait

c: 33,3% - male with trait

d: 16,7% - child with trait

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