PracticeTest2key - Practice Test 2 PCB3063 Page 1 of 6 1....

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Unformatted text preview: Practice Test 2 PCB3063 Page 1 of 6 1. Consider the following pedigree, which has individuals that are hypersensitive to UV radiation indicated by solid black: a. What is the mode of inheritance for the gene that causes the hypersensitivity? (dominant or recessive) Recessive because affected children can have unaffected parents. b. What is the probability that individual 4 in generation III would have a hypersensitive child if he married a woman that was hypersensitive? Individual 4 in generation III is not hypersensitive, so they must be heterozygous or homozygous for the dominant allele. Call the dominant (non-hypersensitive) allele H and the recessive (hypersensitive) allele h. His parents must both be heterozygotes, so (in general) the probability they would have an HH child is , the probability of a Hh heterozygote is , and the probability of a hh child is . However, individual 4 of generation 3 cannot be hh this means he has a 1/3 probability of being HH and a 2/3 probability of being Hh . If he had a child with a hypersensitive woman, there is a probability of the child being hypersensitive if he is heterozygous ( Hh ). If he is homozygous dominant, the child cannot be hypersensitive. Thus, the probability that individual 4 in generation III would have a hypersensitive child if he married a woman that was hypersensitive is 2/3 times 1/2 = 1/3 . c. What is the probability that individual 6 in generation III would have a hypersensitive child if she married a non-hypersensitive man who has a son from a previous relationship that is hypersensitive? Explain your answer. The genotypic probabilities for individual 6 in generation III are identical to those for individual 4 in generation III (see part b the probabilities are 1/3 probability of being HH and a 2/3 probability of being Hh ). If a non-hypersensitive man had a hypersensitive child from a previous relationship, he must be heterozygous. If individual 6 in generation III is heterozygous, the probability that they will have a hypersensitive child is 1/4. So the probability that individual 4 in generation III would have a hypersensitive child if he married a woman that was hypersensitive is 2/3 times 1/4 = 1/6 . Practice Test 2 PCB3063 Page 2 of 6 2. In a family of five children what is the probability that: a) all are males? 0.5 5 b) three are males and two are females? 0.5 5 Note that this is the probability of any specific birth order (e.g., M M F M F). In previous years I have taught the use of binomial probability formula which would allow you to calculate the probability of any birth order (e.g., three males and two females might be born in the order FFMMM or FMFMM or FMMFM or etc.). You dont have to know how to calculate this....
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This note was uploaded on 04/17/2008 for the course PCB 3063 taught by Professor Marta during the Spring '08 term at University of Florida.

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PracticeTest2key - Practice Test 2 PCB3063 Page 1 of 6 1....

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