Answer to Question #90540 in Genetics for Rose

Question

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?

Answer 1

Legend:

XX - female

XY - male

X_A - normal (dominant)

X_a - trait (recessive)

X_AXa - normal female heterozygous for the recessive trait

X_AY - normal male

Parents:

X_AX_a X_AY

Gametes:

X_{A ,} Xa , X_A , Y

F1 (progeny):

X_AX_A - female, no trait

X_AY - male, no trait

X_AX_a - female, no trait

X_aY - male, trait-positive

Question a: 0%

Question b: 50%

Daughter_1:

X_AX_A X_AY

Gametes:

X_A , X_A , Y

F2.1 (progeny):

X_AX_A - female, no trait

X_AY - male, no trait

Daughter_2:

X_AX_a X_AY

Gametes:

X_A , Xa , X_A , Y

F2.2 (progeny):

X_AX_A - female, no trait

X_AY - male, no trait

X_AX_a - female, no trait

XaY - male, trait-positive

F2 (total):

2 * X_AX_A - female, no trait

2 * X_AY - male, no trait

X_AX_a - female, no trait

X_aY - 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