1. A woman who is a carrier for hemophilia marries a man who has hemophilia. Tell the following:
Chances of having a child totally normal
Chances of having a boy with hemophilia
Chances of having a girl with hemophilia
Chances of having a girl who is a carrier
They have a girl. What are the chances she is a carrier?

2. A woman who is a carrier for red-green colorblindness marries a man with normal vision. What are the chances they will:
Have a girl that is colorblind?
Have a boy that is colorblind?
Have a girl that is a carrier?
Have a girl that is normal?
Have a boy that has normal vision?
They have a boy. What are the chances he is colorblind?

3. Chromosome mapping Determine the sequence of the genes below.
A – B = C – B = A – C = 15

4. Chromosome mapping
The following distances are determined for several gene pairs.
Determine the sequence of these genes.
What is the distance between k and m?
j – k = j – m = k – l = l – m = 15

5. The sequence is
k l j m or m j l k and the distance between k and m is 21 map units.

6. The following recombination frequencies were determined
The following recombination frequencies were determined. What is the order of these genes on the chromosome?
a – c = 10; a – d = 30; a – e = 6;
b – c = 4
b – d = 16; b – e = 20; c – d = 20
What is the distance between a and b?

7. The order is
e a c b d and the distance between a and b is 14.

8. a. Determine the sequence of the genes below.
b. What is the distance between c and d?
a – b = 8; a – c = 28; a – d = 25;
b – c = 20; b – d = 33

9. d a b c is the order. 53 map units apart.

10. Determine the sequence of the following genes:
P – Q = S – T = P – T = 12
P – S = S – Q = 14
What is the distance between T and Q?

11. P S T Q distance between is T and Q = 5

12. 1. A wild-type fruit fly (heterozygous for gray body color and normal wings was mated with a black fly with vestigial wings. The offspring had the following phenotypic distribution: wild type, 778; black-vestigial, 785; black-normal, 158; gray-vestigial, 162. What is the recombination frequency between these genes for body color and wing type.

13. 2. In another cross, a wild-type fruit fly (heterozygous for gray body color and red eyes) was mated with a black fruit fly with purple eyes. The offspring were as follows: wild-type, 721; black-purple, 751; gray-purple, 49; black-red, 45. (a) What is the recombination frequency between these genes for body color and eye color?

15. A space probe discovers a planet inhabited by creatures who reproduce with the same hereditary patterns as those in humans. Three phenotypic characters are height (T = tall, t = dwarf), hearing appendages (A = antennae, a = no antennae), and nose morphology (S = upturned snout, s = downturned snout). Since the creatures were not "intelligent" Earth scientists were able to do some controlled breeding experiments, using various heterozygotes in testcrosses.
a. For a tall heterozygote with antennae, the offspring were tall-antennae, 46; dwarf-antennae 7; dwarf-no antennae 42; tall-no antennae 5. Calculate the recombination frequencies for both experiment.

17. b. For a heterozygote with antennae and an upturned snout, the offspring were antennae-upturned snout 47; antennae-downturned snout, 2; no antennae-downturned snout, 48: no antennae-upturned snout 3. Calculate the recombination frequencies for both experiments.

19. Using the information from the problem above, a further testcross was done using a heterozygote for height and nose morphology. The offspring were tall-upturned nose, 40; dwarf-upturned nose, 9; dwarf-downturned nose, 42; tall-downturned nose, 9. Calculate the recombination frequency from these data; then use your answer from the problem above to determine the correct sequence of the three linked genes.