Practice with Sex Linkage and Other Cool Stuff - Solutions

1. a) all red eyed, half male

b) 1/4 red eyed female, 1/4 white eyed female, 1/4 red eyed male, 1/4 white eyed male

c) half female, all red eyed; half male, all white eyed.

d) half females are carriers; half males are white-eyed

2. a)    P X^HX^h x X^Hy

            F₁ X^hy

b) no because father is normal so cannot donate X^h

3. a)    P X^NXⁿ x Xⁿy

            F₁ XⁿXⁿ, Xⁿy, X^NXⁿ, X^Ny

b) X^NX^N individual would have to receive X^N from father who would be dead

c) unlikely that X^N would live long enough to breed, also heterozygous individuals would be less common in nature.

4. b - brown; v - vermillion; BV - wt; bv - white

a)        P BBVV x bbvy

            F₁ BbVv, BbVy - all wild type

b)        P BbVv x BbVy

            F₁ BBVV, BBVy, BbVV, BbVy, BBVv, BBvy, BbVv, Bbvy, BbVV, BbVy, bbVV, bbVy, BbVv, Bbvy, bbVv, bbvy

- 9 wild type:3 brown:3 vermillion:1 white

c)        P bbVV x Bbvy

            F₁ BbVv, BbVy, bbVv, bbVy - 2 wild type:2 brown

5. a) 0 - Jake always gives the normal allele

b) P(X^hy) = 1/2 x 1/2 = 1/4

c) 1/2 - Lauren gives the mutant allele half the time

d) P(4 X^hy sons) = 1/4 x 1/4 x 1/4 x 1/4 = 1/16

e) Assuming her mother is normal and not a carrier, she would inherit the X^h allele from her father all the time, making her a carrier.

f) 1/2 chance of being a carrier; 1/2 chance of having hemophilia. She must inherit the X^h from her father so she could not be free of the allele.

6. 1/4 to have a color blind daughter. 1/2 that first son will be color blind. Notice that we are told the child is a son so we do not have to consider the probability of that happening.

7. a) 1/4 (b) 1/2 (c) 1/4 (d) ½

8. a) 1/4

b) no calico males unless nondisjunction occurs resulting in Klinefleter syndrome

9. B — D --- C — A

        10 10 20

10. S — B — F — C

       5.5 2.5 3