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

Question

4. (10 points) A young couple (Jane and Joe) consulted a genetic sounselor.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, andb 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 recessive trait is always expressed in homozygous inividual, that is, when both the alleles of an individual is recessive.

a). Here, the probablity of Jane's mother to be carrier of RB mutation is 100%. Because if any of the sibling of Jane develops the disease, which is true in this case, then both parents are carriers.

b). The probablity that Joe is a carrier of this RB mutation is 50%.

Here both parents are carriers, that is, both have Bb. By crossing this, we get one BB, one bb and two Bb type offspring. To be a carrier, one has to be a heterozygous individual. So the probablity of getting heterozygous is 50%.

c) There is 25% probablity that their first child will have RB disease if both the parents are carriers of the RB mutation. If any of the parent is normal, haing BB gene, then there is zero percent probablity of their child to be affected by the disease.

d). If the first child has RB then it is confirmed that this couple is the carrier of the trait. So, the second child to have the disease is 25%, as explained in B section of this question.

e). Again, If any of the parent normal and a carrier then probablity of child to be a carrier will be 50%. By crossing BB with Bb, we get, two offspring with BB and two with Bb.

But if both the parents are found to be carriers, then the chances of getting a carrier child will be 50% as explained in part b.

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