Homework 1 Key

Homework 1 Key - F B1 F Br Cinnamon Blue Brown 2 1 1 h 1 pt...

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HOMEWORK SET #1 — KEY 20 pts. total ( 6 pts ) 1. I II III Filled in areas are boring people 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 13 14 15 12 ( 12 answers at 1/4 pt each, 3 pts total ) a) BB Bb bb I-2 II-3 II-4 II-7 II-13 II-15 III-6 III-11 ( 1 pt ) b) BB × BB bb × bb II-13 × II-14 Bb × Bb II-3 × II-4 bb × Bb I-3 × I-4 Bb × BB II-2 × II-1 bb × BB I-1 × I-2 ( 2 pts ) c) chance II-12 × boring fellow 1/16 4 normal children b b B Bb Bb bb bb b ¢ probability of one normal child = 1/2 probability of four normal kids = (1/2) 4 = 1/16
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( 7 pts total ) 2. Bears a) ( 0.3 pt ) F B1 F B1 b) ( 0.3 pt ) F Br F B1 c) ( 0.3 pt ) 1/2 d) ( 0.3 pt ) F Br F Br , zero chance e) ( 0.3 pt ) F c F c and F c F Br f) ( 1.5 pts ) F c F c × F c F c 100% cinnamon F c F Br × F c F c 100% cinnamon F c F Br × F c F Br 3 cinnamon: 1 brown g) ( 3 pts ) F c F ? × F B1 F B1 F c F B1 F Br F B1 × ____________________________________________ _ F 2 F c F Br F c F B1 F B1 F B1
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Unformatted text preview: F B1 F Br __________ Cinnamon Blue Brown 2 : 1 : 1 h) ( 1 pt ) the four from (g), plus: F c F c cinnamon F Br F Br brown ( 7 pts total ) 3. Spinach Spotted is dominant to non-spotted (see cross 5) S = spotted allele s = non-spotted allele ( 3 pts total, each line worth 1/2 pt ) a) Cross # Parents Number and Genotype of Progeny 1 Ss × ss 1/2 Ss 1/2 ss 2 a) SS × SS 1 SS OR b) SS × Ss 1/2 SS 1/2 Ss 3 ss × ss 1 ss 4 SS × ss 1 Ss 5 Ss × Ss 1/4 SS 1/2 Ss 1/4 ss ( 4 pts total ) b) Cross 2 if parents are a) get none ( 1 pt ) b) get 1/2 ( 1 pt ) Cross 4 all of the spotted progeny will produce some non-spotted ( 1 pt ) Cross 5 2/3 of the spotted progeny will produce some non-spotted ( 1 pt )...
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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 1 Key - F B1 F Br Cinnamon Blue Brown 2 1 1 h 1 pt...

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