# q3 - have the allele and thus develop the disease b That...

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APPROVED SOLUTION CH388 (AY08-01) Quiz #3 Name _________________ Lesson 5 Take Home Section ________________ This is a take home quiz, worth 50 points. You must attach a signed documentation of written work cover sheet documenting all sources. This quiz is due at the beginning of class at Lesson 6 (Chromosomes, Mitosis, & Meiosis). Attach additional worksheets as needed. 1. Huntington disease is a rare dominant trait that causes neurodegeneration later in life. A man in his thirties, who already has three children (boy, girl, boy), discovers that his mother has Huntington. What are the following probabilities? a. That the man in his thirties will develop Huntington disease. Construct a Punnett square. Since it is a rare disease, we assume that the mother is a heterozygote and the father is normal. The chances are 50% that the man would

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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.

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q3 - have the allele and thus develop the disease b That...

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