Homework 5 Key

Homework 5 Key - BICD 100 Spring 2008 30 pts total 1. a)...

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BICD 100 Page 1 of 4 S. Brody, ext. 42619 Spring 2008 4125 Muir Biology HOMEWORK SET #5 — KEY 30 pts total 1. ( 10 pts total ) a) ( 4 pts ) b) Origin of extra chromosome: ( 4 pts ) Assume that similar events happen with equal frequency, i.e. MII female leads to equal numbers of bb and aa gametes. Then, looking at Down's offspring, some genotypes are unique to a certain type of nondisjunction. These numbers can then be used (by subtracting from the total) to assign numbers to those genotypes which are not unique to a particular nondisjunction event. genotype total MI female MII MI male MII aab 2 2 aac 2 2 abb 16 11 5 abc 36 11 25 acc 5 5 bbb 7 2 5 bbc 27 2 25 bcc 5 5 = 22 = 8 = 50 = 20 c) Down's from all female nondisjunction: ( 1 pt ) 22 + 8 = 30/100 30/100 × 2 × 10 -3 = 6 × 10 -4
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BICD 100 Page 2 of 4 S. Brody, ext. 42619 Spring 2008 4125 Muir Biology d) Down's from all MII nondisjunction ( 1 pt ) 8 + 20 = 28/100 28/100 × 2 × 10 -3 = 5.6 × 10 -4 2. (
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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 5 Key - BICD 100 Spring 2008 30 pts total 1. a)...

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