Question Bank E Answers

Question Bank E Answers - rr ) = 0.49; freq resistant...

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BIO320 Spring 2008 Answer key for question bank E I. Population genetics 1. C freq ( ii ) = 9% = 0.09 = q 2 , so q = 0.3 2.E freq ( aa ) = 0.49 = q 2 , so q = 0.7; p = 1-q = 0.3; freq ( AA ) = p 2 = 0.09 3.D q 2 = 1/1600; q = 1/40; p = 1-q = 39/40; freq (heterozygotes) = 2pq = 78/1600 4.D freq ( BB ) = p2 = 0.09; p = 0.3; q = 0.7; freq (heterozygotes) = 2pq = 0.42 = 42% 5.C freq ( g ) = 0.6 = q; freq ( G ) = p = 1-q = 0.4; freq ( GG ) = p2 = 0.16 6.A (genetic drift) 7.D freq ( b ) in mixed population = 0.8; allele frequencies unchanged by mixing; reach H-W bb ) in any generation = 0.8) 2 = 0.64 8.E freq ( Z ) in mixed population = 0.4, freq ( z ) = 0.6; freq (Zz) constant at 0.48 9.D Freq ( AA + Aa ) = 0.91, so freq ( aa ) = 0.09 = q2; q = 0.3; 2pq = 0.42 10.E Freq ( GG + AG ) = 0.91; freq( AA ) = 0.09 = q 2 ; q = freq ( A ) = 0.3; freq ( G ) = p = 1-q = 0.7 11.C Before insectide selection, freq (
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Unformatted text preview: rr ) = 0.49; freq resistant genotypes ( RR + Rr ) = 0.51; freq ( r) = q = 0.7; freq ( Rr ) = 2pq = 2 (0.3)(0.7) = 0.42. After insecticide; freq ( r) =q = freq ( Rr ) = 0.21/0.51 = 0.412; freq ( r) = q 2 = 0.17 12.D 13.D 14.A 15.E 16.B 17.A 18.E 19.B 20.B 21.A 22.A 23.A 24.C Freq ( A ) = freq ( AA ) + freq ( Aa ) = 0.26 + 0.14 = 0.4; if at H-W equilibrium, freq (Aa) = 2 (0.4)(0.6) = 0.48 > 0.28. 25.B 26.B 27.C 28.C 29. B II. Quantitative genetics 1.A 2. E 3.A 4.A 5.E 6.B 7.C 8.A 9.D 10.C Selection differential S = 25-20 = 5cm; R = h 2 S = 0.4 (5) = 2cm; new fleece length = 22cm. 11.A Selection differential S = 12-10 = 2; R = h 2 S = 0.2 (2) = 0.4lbs; weight of offspring = 10.4lbs. 13.D 14.E (different breed => different genotypes => genotype x environment interaction not predictable) 15.E 16.C 17.B 18.E 19.E 20.C 21.C 22.B 23. A...
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This note was uploaded on 05/29/2008 for the course BIO 320 taught by Professor N/a during the Spring '08 term at SUNY Stony Brook.

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