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.

2. a) X^RX^r, X^ry, X^RX^r, X^ry

b) X^RX^R, X^Ry, X^RX^r, X^ry

c) X^RX^r, X^Ry, X^rX^r, X^ry

3. a) P X^HX^h x X^Hy

F₁ X^hy

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

4. a) P X^NXⁿ x Xⁿy

F₁ XⁿXⁿ, Xⁿy, X^NXⁿ, Xⁿy

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.

5. 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

6. Sounds like a trait caused by a gene on the y chromosomes and so cannot occur in females. In the 1950s, the gene was later found to be a rare autosomal allele.

7. S ---- B ---- F ---- C

       5.5       2.5       3

8. B ---- D ---- C ---- A

       10       10       20

9. a) 0
b) 1/4
c) 1/2
d) 1/16
e) 1/2. Assuming her mother is normal and not a carrier, she would inherit the X^h allele from her father half 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. She could not be free of the allele.

10. 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.

11. a) 1/4
b) 1/2
c) 1/4
d) 1/2

12. a) 1/4
b) no calico males unless nondisjunction occurs resulting in Klinefleter syndrome