A woman who is a carrier for hemophilia A woman who is a carrier for hemophilia marries a man who has hemophilia. Tell marries a man who has hemophilia. Tell the following:the following:

a.a. Chances of having a child totally normalChances of having a child totally normal

b.b. Chances of having a boy with hemophiliaChances of having a boy with hemophilia

c.c. Chances of having a girl with hemophiliaChances of having a girl with hemophilia

d.d. Chances of having a girl who is a carrierChances of having a girl who is a carrier

e.e. They have a girl. What are the chances They have a girl. What are the chances she is a carrier?she is a carrier?

A woman who is a carrier for red-green A woman who is a carrier for red-green colorblindness marries a man with normal colorblindness marries a man with normal vision. What are the chances they will:vision. What are the chances they will:

a.a. Have a girl that is colorblind?Have a girl that is colorblind?

b.b. Have a boy that is colorblind?Have a boy that is colorblind?

c.c. Have a girl that is a carrier?Have a girl that is a carrier?

d.d. Have a girl that is normal?Have a girl that is normal?

e.e. Have a boy that has normal vision?Have a boy that has normal vision?

f.f. They have a boy. What are the chances he is They have a boy. What are the chances he is colorblind?colorblind?

Chromosome mappingChromosome mapping

Determine the sequence of the genes Determine the sequence of the genes below.below.

A – B = 34 C – B = 19 A – C = 15A – B = 34 C – B = 19 A – C = 15

Chromosome mappingChromosome mappingThe following distances are determined for several The following distances are determined for several

gene pairs.gene pairs.

a.a. Determine the sequence of these genes.Determine the sequence of these genes.

b.b. What is the distance between k and m?What is the distance between k and m?

j – k = 12 j – m = 9 k – l = 6 l – m = 15j – k = 12 j – m = 9 k – l = 6 l – m = 15

The sequence is The sequence is

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

The following recombination frequencies The following recombination frequencies were determined. What is the order of were determined. What is the order of these genes on the chromosome?these genes on the chromosome?

a – c = 10; a – d = 30; a – e = 6; a – c = 10; a – d = 30; a – e = 6;

b – c = 4b – c = 4

b – d = 16; b – e = 20; c – d = 20b – d = 16; b – e = 20; c – d = 20

What is the distance between a and b?What is the distance between a and b?

The order is The order is

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

a. Determine the sequence of the genes a. Determine the sequence of the genes below.below.

b. What is the distance between c and d?b. What is the distance between c and d?

a – b = 8; a – c = 28; a – d = 25; a – b = 8; a – c = 28; a – d = 25;

b – c = 20; b – d = 33b – c = 20; b – d = 33

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

Determine the sequence of the following Determine the sequence of the following genes:genes:

P – Q = 17 S – T = 9 P – T = 12 P – Q = 17 S – T = 9 P – T = 12

P – S = 3 S – Q = 14P – S = 3 S – Q = 14

What is the distance between T and Q?What is the distance between T and Q?

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

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.

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?

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.

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.

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.