Chapter 5 - Chapter 5: 5.1 - a. 8, b. 4, c. 1, d. 11, e. 2,...

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Chapter 5: 5.1 - a. 8, b. 4, c. 1, d. 11, e. 2, f. 5, g. 6, h. 3, 1. 10, j. 12, k. 9, l. 7 5.4 - a) 1/4 of the F2 mice will be dancers if the trait is controled by a single gene with complete dominance b) 1/16 of the mice are expected to be dancers being homozygous recessive for two genes c) One gene hypothesis is a better fit with the data 5.5 - Parentals: r s+ / Y (raspberry male) x r+ s / r+ s (sable female) F1 : r s+ / r+ s (wildtype) x r+ s / Y (sable male) F2 : r s+ / r+ s (wildtype), r+ s / r+ s (sable), r s / r+ s (sable), r+ s+ / r+ s+ (wildtype), r+ s / Y (sable), r s+ / Y (raspberry), r s / Y (raspberry sable), r+ s+ / Y (wildtype) distance between raspberry and sable is 11.7 cM (m.u.) 5.7 - a) gametes of the F1: AB and ab (parentals) and Ab and aB (recombinants), the recombinants will make up 40% of the gametes and the parentals will make up 60% F2 phenotypic ratio : 0.59 A_B_ : 0.16 A_bb : 0.16 aaB_ : 0.09 aabb b) parental gametes are Ab and aB (60%) and recombinant gametes are AB and ab (40%),
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This note was uploaded on 04/07/2010 for the course BIO 105 taught by Professor Sullivan during the Summer '08 term at University of California, Santa Cruz.

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Chapter 5 - Chapter 5: 5.1 - a. 8, b. 4, c. 1, d. 11, e. 2,...

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