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Answer to Question #90540 in Genetics for Rose

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

XA - normal (dominant)

Xa - trait (recessive)

XAXa - normal female heterozygous for the recessive trait

XAY - normal male

Parents:

XAXa XAY

Gametes:

XA , Xa , XA , Y

F1 (progeny):

XAXA - female, no trait

XAY - male, no trait

XAXa - female, no trait

XaY - male, trait-positive

Question a: 0%

Question b: 50%

Daughter_1:

XAXA XAY

Gametes:

XA , XA , Y

F2.1 (progeny):

XAXA - female, no trait

XAY - male, no trait

Daughter_2:

XAXa XAY

Gametes:

XA , Xa , XA , Y

F2.2 (progeny):

XAXA - female, no trait

XAY - male, no trait

XAXa - female, no trait

XaY - male, trait-positive

F2 (total):

2 * XAXA - female, no trait

2 * XAY - male, no trait

XAXa - female, no trait

XaY - 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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