1) A, B and C are independently assorting genes controlling the production of a black pigment. The genes act in a biochemical pathway as follows:

a) what proportion are colorless?

b) what proportion are red?

Alternatively, red = A_ B_ cc + A_bb + aaBb = (3/4) (3/4) (1/4) + 3/4 (1/4) + 3/4 (1/4)=

9/64 + 12/64 + 12/64 = 33/64

c) what proportion are black?

2) Huntington's chorea is a fatal disease leading to the degeneration of the nervous system. It is found in persons heterozygous for the allele Ht, but has variable expressivity in the sense that a person may develop symptoms as early as childhood or as late as 60 years old, with the mean age of onset at about 40 years.

(a) A man with four children (3 female, 1 male) died of Huntington's chorea. The disease is not known in his widow's family. What are the chances that:

i) All four children carry the disease?

ii) Only the three daughters carry the disease?

iii)Only two children carry the disease?

Apply the sum rule p(HH++) + p (H+H+) + p (H++H) + p(+HH+) + p(+H+H) + p (++HH) = (1/16) (6) = 6/16 = 3/8.

iv) Only one daughter carries the disease?

Apply the sum rule but keep in mind the sexes: Let the first three symbols (H or +) represent females and the last represent the son. p(H+++) + p (+H++) + p(++H+) = 3/16.

(b) In certain families known to carry the Ht allele, the disease is found to "skip" several generations before reappearing. How might this be explained?

3) For this problem you may need the following background information on Drosophila:

cm = carmine is an eye color mutation that causes dull red "carmine" eyes. cm is located at 18.0 mu on the X chromosome

v = vermilion is an eye color mutation that causes bright red "vermilion" eyes.

v is located at 33 mu on the X. The double mutant cm v has orange eyes.

eag = ether-a-go-go causes flies to shake their legs like crazy when under ether anaesthesia. eag is located at 50 mu on the X.

The telomere of the X is located at 0 mu and the centromere at 68 mu.

An interesting X-linked recessive lethal mutation was recently isolated in Drosophila. The new mutation, call it lox, for "lethal on the X", originated in a wild-type genetic background. To map lox females heterozygous for lox were crossed to males whose X chromosome is marked with the recessive mutations cm, v, and eag to generate heterozygous females of genotype: lox +++/ + cm v eag. The heterozygous females were crossed to wild-type males with the following results:

b.) In which interval does lox map? (i.e. Is it between the telomere and cm, between cm and v, between v and eag, or between eag and the centromere?) Explain your reasoning.

lox must map between v and eag.

If lox maps to right of v the cm ++ class can arise only via a triple XO. This would be expected to be a very rare event and indeed is the only zero class among the progeny. All other rare types could originate with only two XOs.

Consider if lox mapped elsewhere:

If left of cm: 1) the ++ eag class would have required two crossovers and should have been less frequent. 2) The cm ++ class should have been very common. 3) the + v eag class should have been rare since it would require a double crossover. 4) The cm + eag class should have been more frequent that the +v+ class since the +v+ class would require 3 XOs to survive.

If left of v: 1) The ++ eag class would have been the least common type because it would require 3 crossovers. 2) The cm ++ class should have been more common because it would require only a single lox-v crossover. 3) The + v eag class should have been more rare because only cm-lox recombinants would survive.

If right of eag: 1) cm v + class would be rare since it would require a double XO to survive. 2) ++ eag class would have been very common. 3) cm ++ class would be more abundant since double XO freq ~ (.17)(.15). 4) + v+ class should have been rarer than cm + eag since it would require 3 XOs to generate the progeny.

c.) Calculate the map position of lox.

Total v-eag recombination events = [+++ ](3)2=6 + [cm v +]110 + [++eag] 46 + [cm+eag] 4 +

[+v+] 10 = 176

v-lox recombinants = 3 + 46 + 4 = 53

lox-eag recombinants = 3 + 110 + 10 = 123

53/176 = 0.301 of the total v-eag crossovers occurred between v and lox.

lox is located at 30.1% of the distance from v-eag = (.301) (17 mu) = 5.1 mu to the right of v at 38.1 mu.

Alternatively 123/176 = .699 of the v-eag crossovers occurred between lox and eag allowing lox to be placed at (.699) (17 mu) = 11.9 mu to the left of eag at 38.1 mu.

4) Explain how balancer chromosomes could be used to stably propagate an autosomal recessive lethal mutation if the balancer chromosome is homozygous viable but carries a recessive sterile allele that eliminates both ovaries and testes. Assume the balancer is marked with a dominant mutation Ha that produces flies with many extra hairs. What phenotypes, and in what proportions, would you expect in the stable stock?

(b) If you want to stably propagate an X-linked lethal mutation with a balancer chromosome containing a recessive sterile allele, would you use a female-specific sterile, a male-specific sterile, or a non sex-specific sterile mutation?

5) Both hemophilia A h(A) and red-shift color blindness (r) are X-linked recessive traits. The h(A) and r genes are separated by 4 cM. In the pedigree shown, what is the probability that the individual male (IV-1) is color blind but not affected by hemophilia?