2/3 x 1 x 1/4 = 2/12 = 1/6
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1 1. Imagine that you are a genetic counselor, and a couple planning to start a family comes to you for assistance. Charles was married once before, and he and his first wife had a child with cystic fibrosis (a recessive disorder). The brother of his current wife, Elaine, died of cystic fibrosis. What is the probability that Charles and Elaine, who do not have the condition, will have a baby with cystic fibrosis? (Neither Charles nor Elaine s parents have cystic fibrosis.) Answer: Probability that Elaine is heterozygous and probability that Charles is heterozygous and the probability a child born with the disorder from heterozygous parents: 2/3 x 1 x 1/4 = 2/12 = 1/6
2 2. If a man with type AB blood marries a woman with type O blood, what possible blood types would you expect in their children? What fraction would you expect of each type? Answer: Given I A I B x ii, I A I B i i I A i I A i I B i I B i Types possible: A (I A i) and B (I B i) à 2/4 (50%) and 2/4 (50%), respectively 3. Phenylketonuria (PKU) is an inherited disease caused by a recessive allele. If a woman and her husband, who are both carriers, have three children, what is the probability of each of the following? a) All three children are of normal phenotype. b) One or more of the three children have the disease. c) All three children have the disease. d) At least one child is phenotypically normal.
3 Answers: Given Aa x Aa, A a A AA Aa a Aa aa à ¾ normal (AA and Aa) and ¼ with PKU (aa) a) All three children normal: only one condition/combo satisfies the criterion, normal AND normal AND normal ¾ x ¾ x ¾ = 27/64 b) One or more have PKU: 7 conditions/combos satisfy the criteria à any combo of 1 child of the 3 with PKU, any combo of 2 children of the 3 with PKU, or all 3 with PKU:
4 Child 1 Child 2 Child 3 PKU N N à ¼ x ¾ x ¾ = 9/64 N PKU N à ¾ x ¼ x ¾ = 9/64 any 1 of the 3 with PKU N N PKU à ¾ x ¾ x ¼ = 9/64 PKU PKU N à ¼ x ¼ x ¾ = 3/64 PKU N PKU à ¼ x ¾ x ¼ = 3/64 any 2 of the 3 with PKU N PKU PKU à ¾ x ¼ x ¼ = 3/64 PKU PKU PKU à ¼ x ¼ x ¼ = 1/64 all 3 with PKU Summarized: 1 child of 3 à 9/64 x 3 = 27/64 OR 2 children of 3 à 3/64 x 3 = 9/64 OR 3 children of 3 à 1/64 27/64 + 9/64 + 1/64 = 37/64 Rather than having to go through this laborious process, simply think in terms as it relates to the converse of the question posed in part a. If 27/64 is the probability of having all 3 children normal, then 1 27/64 = 37/64 is the remainder probability of at least one child with PKU. It is analogous to asking the question, what is the probability of rolling a 4 on a standard number cube? followed by the question, what is the probability of rolling any number except 4? Probability of rolling 4 = 1/6 Probability of rolling any number except 4 = 1 1/6 = 5/6 c) All three children have PKU: 37/64 only one condition/combo satisfies the criterion, PKU and PKU and PKU ¼ x ¼ x ¼ = 1/64
5 d) At least one child has the normal phenotype: Again, think of this question in terms of the converse of the question asked in part c. Using the answer from c, you can simply solve for the remainder probability of one or more having a normal phenotype by subtracting the probability of all three having PKU (1/64) from all possibilities (1) à 1 1/64 = 63/64 However, for those that want to account for all combos/conditions that should yield the same probability, let s get technical. 7 conditions/combos satisfy the criteria à any combo of 1 child of the 3 normal, any combo of 2 children of the 3 normal, or all 3 normal: Child 1 Child 2 Child 3 PKU PKU N à ¼ x ¼ x ¾ = 3/64 PKU N PKU à ¼ x ¾ x ¼ = 3/64 any 1 of the 3 normal N PKU PKU à ¾ x ¼ x ¼ = 3/64 PKU N N à ¼ x ¾ x ¾ = 9/64 N PKU N à ¾ x ¼ x ¾ = 9/64 any 2 of the 3 normal N N PKU à ¾ x ¾ x ¼ = 9/64 N N N à ¾ x ¾ x ¾ = 27/64 all 3 normal Summarized: 1 child of 3 à 3/64 x 3 = 9/64 OR 2 children of 3 à 9/64 x 3 = 27/64 OR 3 children of 3 à 27/64 9/ / /64 = 63/64 63/64
6 4. In peas, a gene controls flower color with complete dominance such that P = purple and p = white. In an isolated pea patch population, there are 36 purple-flowering plants and 64 white-flowering plants. Assuming Hardy-Weinberg equilibrium within this population, how many of the purple plants are homozygous (PP) and how many are heterozygous (Pp)? Answer: 64/100 = 0.64 = q 2 (the product/square of two recessive alleles frequencies allelic frequency of q x allelic frequency of q); this is the genotypic frequency of homozygous recessive individuals in the population Therefore, square root of q 2 ( q 2 ) = q à 0.64 = 0.8, the allelic frequency of the recessive allele in the population If q = 0.8, then p =? à p = = 0.2, the allelic frequency of the dominant allele in the population If p is the frequency of the dominant allele, then p 2 is the frequency of homozygous dominant individuals in the population (the product/square of two dominant alleles frequencies allelic frequency of p x allelic frequency of p) à p 2 = (0.2) 2 = x 100 = 4 plants homozygous purple 36 4 = 32 plants heterozygous purple or the expression 2pq can be used to find the frequency of heterozygous individuals within the population 2pq = 2(0.2)(0.8) = x 100 = 32 plants heterozygous purple = 4 plants homozygous purple
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