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Unformatted text preview: have the allele, and thus develop the disease. b. That his first child will develop Huntington. Use the product rule. 0.5 (chance man has the allele) times 0.5 (chance that he will pass it to his offspring), which equals 0.25 or 25%. c. That one out of three of his children will develop Huntington disease. Using the binomial expansion equation, if the probability of an affected child is 0.25 then the probability of an unaffected child is 0.75. For the binomial expansion equation, n = 3, x = 1, p = 0,25, q = 0.75. P = 3! (0.25 1 )(0.75 3-1 ) 1!(3-1)! The answer is 0.422 or 42.2% d. Draw a pedigree for this family. Ensure that you label each entry with a generation and number (i.e., I-1) and provide a key to your symbols. APPROVED SOLUTION Huntington Possible carrier I-1 I-2 II-1 II-2 III-3 III-2 III-1...
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This note was uploaded on 04/15/2008 for the course CH 388 taught by Professor Eslinger during the Fall '08 term at West Point.
- Fall '08