BIOL 121 (2009) Fall Exam 3+ Key

25 if both parents are hets probability mother het 13

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Unformatted text preview: . 3 BIOL 121 NAME: Exam 2 11/04/09 (3 pts) 9) Evolution requires A. Pre- zygotic barriers B. Directional selection C. Genetic variation D. Phenotypic variation E. Assortative mating (3 pts) 10) A new, selectively neutral mutation occurs in one individual of a haploid population (N=10), what is the probability it will increase in frequency to 100%? A. Cannot determine with the information given B. 100% C. 50% D. 20% E. 10% (3 pts) total 30 points 4 BIOL 121 Exam 2 11/04/09 NAME: SHORT ANSWER QUESTIONS ONLY A FEW SENTENCES – AT MOST – ARE NEEDED FOR A COMPLETE ANSWER TO EACH SECTION 11) The married couple shown in the last line of the pedigree below have just discovered that each had an uncle (black squares) with Tay-Sachs disease, a severe and very rare autosomal recessive disorder. The couple wish to have a child, but they would like to know the probability that their child will have Tay-Sachs. i. What is the probability that their first child will have Tay Sachs? (8 pts) Greatgrandparents are heterozygotes Probability left grandfather is heterozygotes = 2/3 – because we know they are not aa Probability right grandmother is heterozygotes = 2/3 Probability parents are heterozygotes = 1/2 if grandparent is het * probability that grandparent is het = 1/3 each Answer = 0.25 if both parents are hets * probability mother het (1/3) * probability father het (1/3) = 2.7% = 1/36 We will also accept Greatgrandparents are heterozygotes Probability left gran...
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