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White (w) is an x-linked recessive and tinman (tn) is an autosomal recessive mutation. What proportion...

White (w) is an x-linked recessive and tinman (tn) is an autosomal recessive mutation. What proportion of white, tinman females (relative to total population) is expected in the F2 starting with a true breeding white female which is wild type for tinman mating with a true breeding male mutant only for tinman.

please do step by step I am so confused. I know the answer is 1/16 but don’t know how??

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Answer #1

Answer:

Female (XwXw; TnTn) x (XWY; tntn) Male ----Parents

F1:

XW tn

Y tn

Xw Tn

XWXw; Tntn (wild color & wildtype tinman—daughter)

XwY; Tntn (white color & wildtype tinman—son)

XWXw; Tntn (F1 female) x (F1 male) XwY; Tntn ---Parents

XWXw x XwY = XWXw (1/4). XWY (1/4). XwXw (1/4) & XwY (1/4)

Tntn x Tntn = TnTn (1/4), Tntn (1/2) & tntn (1/4)

Proportion of white, tinman females = XwXw; tntn = ¼ * ¼ = 1/16

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