Question

3. (35 pts) The gene for the cone photoreceptor in the eye and the gene for a blood-clotting factor are both in the X-chromosome. Suppose they are 14 map units apart. Mutation in the photoreceptor gene results in color blindness, and is inherited as an X-linked recessive trait. Mutation in the clotting factor gene results in hemophilia, and is also inherited as an X-linked recessive trait. You have two patients: Sally and Jill. Both patients have a family history of both color blindness and hemophilia. You gather the following data on each patient’s families:

• Sally’s father has both color blindness and hemophilia and her mother is genotypically normal with no family history of either disease.
• Jill’s father is color blind and her mother has hemophilia. There is no history of color blindness on Jill’s mother’s side of the family.
• Sally and Jill are married to genotypically normal males.

a) Which of the following have a chance of having a son who is phenotypically normal? Select all that apply and briefly explain why.

A. Sally’s parents

B. Jill’s parents

C. Sally and her husband  

D. Jill and her husband

b) Who has the greater chance of having a son with both color blindness and hemophilia? Why?

A. Sally

B. Jill

C. Neither. They have the same risk.

c) What is the chance that Sally will produce an egg with the mutation for color blindness but not the mutation for hemophilia? Show your calculations.

A. 86%

B. 50%

C. 43%

D. 14%

E. 7%

d) What is the chance that Jill will produce an egg with the mutation for hemophilia but not the mutation for color blindness? Show your calculations.

A. 86%

B. 50%

C. 43%

D. 14%

E.7%

e) Sally and Jill each have a daughter. The two daughters have undergone genetic testing for the hemophilia mutation. Sally’s daughter tests positive for the hemophilia mutation. Jill's daughter tests negative for the hemophilia mutation. Which daughter has the higher risk of also being a carrier for color blindness? Why?

A. Sally’s daughter

B. Jill’s daughter

C. Neither. They have the same risk.

Answer 1

Picture is attached below :

With Sally's mother and father's genotype, the probability of the girl child being a carrier for colour blindness and hemophilia is 100%. So we conclude Sally is XX^{​​​​​ch.}

with Jill's mother and father's genotype, the probability of girl child being the carrier again is 100%. The genes are on 2 different Xchromosome. So, Jill is X^hX^{​​​​c}.

a)From the picture we can see both Sally's parent and Sally and her husband has a chance of having normal son genotypically. So, answer is A,C

b)A . Sally has greater chance to have son with both hemophilia and colourblindness

d)B. 50%

Jill is X^{​​​​​​h}​​​​​X^​c. Her eggs will be divided into X^{​​​​​​h} & X^{​​​​​​c}.

So the chance is 1/2 = 50%

e)A. Sally's daughter

Since the gene forcolour blindness and hemophilia are 14mapunits apart, the probability of being inherited together is high.