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 XHXh x XHy
F1 Xhy
b) no because father is normal so cannot donate Xh
3. a) P XNXn x Xny
F1 XnXn, Xny, XNXn, XNy
b) XNXN individual would have to receive XN from father who would be dead
c) unlikely that XN 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(Xhy) = 1/2 x 1/2 = 1/4
c) 1/2 - Lauren gives the mutant allele half the time
d) P(4 Xhy 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 Xh 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 Xh 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