4. (10 points) A young couple (Jane and Joe) consulted a genetic counselor becau

Question

4. (10 points) A young couple (Jane and Joe) consulted a genetic counselor because each of them had a sibling (brother or sister) affected with form of cancer called retinoblastoma. RB is caused by a single autosomal gene, and the disease causing alleles are recessive. Neither the couple nor any of their parents has the disease. In explaining your answers to the following questions please use the symbol B for the dominant, normal RB allele, and b for the recessive, disease causing allele

a) What is the probability that Jane’s mother is a carrier of the RB mutation? (1 point)

b) What is the probability that Joe is a carrier of the RB mutation? (2 points)

c) What is the probability that the first child produced by this couple will have RB? (2 points)

d) If their first child has RB, what is the chance that their second child will have it? (2 points)

e) If they haven’t had any children yet, what is the probability that a child produced by this couple will be a carrier of the RB mutation? (3 points)

Explanation / Answer

a) What is the probability that Jane’s mother is a carrier of the RB mutation? (1 point)

RB is caused by a single autosomal gene, and the disease causing alleles are recessive, two copies of a disease allele are required for an individual to be susceptible to expressing the phenotype. Typically, the parents of an affected individual are not affected but are gene carriers. Since Jane’s sibling is affected, that means he/she obtained each of the mutated b gene from each of the parents.

the probability that Jane’s mother is a carrier of the RB mutation is 100%

b) What is the probability that Joe is a carrier of the RB mutation? (2 points)

Since the parents of Joe are carriers, Bb

Bb x Bb = BB, Bb, Bb, bb

½ or 50% of the offsprings are carriers

So, the probability of Joe being a carrier of the RB mutation is 50%

c) What is the probability that the first child produced by this couple will have RB? (2 points)

Bb x Bb= BB, Bb, Bb, bb

¼ of the children will have RB

But since, Jane and Joe have 50% chance of being carriers

= ½ (Jane carrier) x ½ (Jane carrier) x ¼ (children will have RB) = 1/16

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