Homework 4 Key

Homework 4 Key - BICD 100 Spring 2008 Page 1 of 4 S. Brody,...

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BICD 100 Page 1 of 4 S. Brody, ext. 42619 Spring 2008 4125 Muir Biology HOMEWORK SET #4 — KEY 30 pts total 1. ( 8 pts total ) a) ( 4 pts ) 2 pq = 2(.8)(.2) = 2 pq = 2(.3)(.7) = .32 for the island population .42 for the immigrant population b) Heterozygote frequency, new population, before mating: ( 2 pts ) = .32(80) + .42(20) 100 = .34 c) Heterozygotes after mating: ( 2 pts ) new frequency of A = .8(80) + .3(20) 80 + 20 = .7 new frequency of a = .2(80) + .7(20) 80 + 20 = .3 2 pq = 2(.7)(.3) = .42 CB:$5CB:$5CB:$5CB:$5CB:$5CB:$5CB:$5CB:$5 2. Frequency of dd = 16/100; frequency d = .4, D = .6, therefore frequency of Dd = .48, DD = .36. ( 8 pts total ) a) Rh + ( × ) Rh can be DD ( × ) dd and Dd ( × ) dd ( 2 pts ) Rh + offspring from Dd ( ! ) dd will be 1 2 (.48) .48 + .36 (The term in the parenthesis normalizes the frequency for the percentage of the parents that are Dd amongst the total Rh + parents.) Similarly from DD ( ! ) dd , we get (1) (.36) .48 + .36 Rh + frequency from all Rh + ( × ) Rh matings will be 1/2(.48) + 1(.36) .48 + .36 = 5/7
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This note was uploaded on 06/07/2008 for the course BICD 100 taught by Professor Nehring during the Spring '08 term at UCSD.

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Homework 4 Key - BICD 100 Spring 2008 Page 1 of 4 S. Brody,...

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