Title: Consider the attached file a pedigree of a rare autosomal recessive disease Post by: mannip29 on Aug 5, 2013 Consider the attached file a pedigree of a rare autosomal recessive disease. Assume all
people marrying into the pedigree do not carry the abnormal allele. (a) If individuals A and B have a child, what is the probability that the child will have the disease? (b) If individuals C and D have a child, what is the probability that the child will have the disease? (c) If the first child of C * D is normal, what is the probability that their second child will have the disease? (d) If the first child of C * D has the disease, what is the probability that their second child will have the disease? Title: Re: Consider the attached file a pedigree of a rare autosomal recessive disease Post by: Libertarian on Aug 5, 2013 (a) Choosing M for unaffected and m for the disorder, male B must be M/m, and female
A has a 2/ 3 chance of being M/m. T he overall chance of an affected child is 1 * 2/ 3 * 1/ 4 = 1/ 6. (b) I f C' s mother A is heterozygous, C stands a 1/ 2 chance of being heterozygous. D' s mot her must be heterozygous, and D stands a 1/ 2 chance of inheriting that heterozygosity. The overall chance of an affect ed child is 2/ 3 * 1/ 2 * 1 * 1/ 2 * 1/ 4 = 1/ 24. (c) The probability is still 1/ 24. (d) Now that we know individuals C and D must both be M/m, the chance of the second child being m/m is 1/ 4. Title: Re: Consider the attached file a pedigree of a rare autosomal recessive disease Post by: Doctor-2-B on Aug 5, 2013 (a)
Since A is unaffected, her sister is affected, and both her parents are affected, we can determine the genotypes: Parents: both heterozygous Rr Sister: rr If we make a Punnett square for A, we get 1/3 RR, 2/3 Rr, and 0 rr (She's unaffected). Therefore, A has a 2/3 chance of being a carrier. In a mating, she has a 1/2 chance of passing either allele. By the product rule, 2/3*1/2 = 1/3 Since B is unaffected, he can only be Rr (since his mother is affected rr and his father is unaffected). That is, he has a 1/2 chance of passing on an affected r allele. By the product rule, 1/3 * 1/2 = 1/6 (b) Now that we have determined that there is a 2/3 probability for A to be Rr, and we know that her husband will definitely be RR (the question states that no one marrying into the family will carry the affected allele), C has a 2/3 * 1/2 = 1/3 chance to be Rr, otherwise he is RR. D's mother must be Rr because she is unaffected and received an r allele from her own mother. Because D's father is homozygous RR, she has a 1/2 chance of being Rr. When two Rr people mate, there is a 1/4 chance they will have an rr offspring. Multiplying the probabilities, 1/3 * 1/2 * 1/4, we get 1/24 (c) Births are independent events (like the roll of a die), the affectedness/unaffectedness of one offspring will not affect the siblings. Because a normal birth will not distinguish between RR and Rr for either parent, the chance is still 1/24. (d) Here, we have new information. Because C and D are either RR or Rr and they had an affected child, we know that they MUST be Rr (otherwise they would not have been able to produce an rr child). If we make a Punnett square for Rr x Rr, we get 1/4 RR, 1/2 Rr, and 1/4 rr. The 1/4 rr would be the probability of the second child (and all other children they have) being affected. (d) |