GenCh2answers

GenCh2answers - 2. Do a testcross (cross to a/a). If the...

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2. Do a testcross (cross to a/a ). If the fly was A/A , all the progeny will be phenotypically A; if the fly was A/a , half the progeny will be A and half will be a. 3. The progeny ratio is approximately 3:1, indicating classic heterozygous-by- heterozygous mating. Since Black ( B ) is dominant to white ( b ): Parents: B/b × B/b Progeny: 3 black:1 white (1 B/B : 2 B/b : 1 b/b ) 7. a. By considering the pedigree (see below), you will discover that the cross in question is T/t × T/t . Therefore, the probability of being a taster is 3 / 4 , and the probability of being a nontaster is 1 / 4 . Also, the probability of having a boy equals the probability of having a girl equals 1 / 2 . (1) p (nontaster girl) = p (nontaster) × p (girl) = 1 / 4 × 1 / 2 = 1 / 8 (2) p (taster girl) = p (taster) × p (girl) = 3 / 4 × 1 / 2 = 3 / 8 (3) p (taster boy) = p (taster) × p (boy) = 3 / 4 × 1 / 2 = 3 / 8 b. p (taster for first two children) = p (taster for first child) × p (taster for second child) = 3 / 4 × 3 / 4 = 9 / 16 13. The results suggest that winged ( A/– ) is dominant to wingless ( a/a ) (cross 2 gives a 3 : 1 ratio). If that is correct, the crosses become Number of progeny plants Pollination Genotypes Winged Wingless winged (selfed) A/A × A/A 91 1* winged (selfed) A/a × A/a 90 30 wingless (selfed) a/a × a/a 4* 80 winged × wingless A/A × a/a 161 0 winged × wingless A/a × a/a 29 31 winged × wingless A/A × a/a 46 0 winged × winged A/A × A/– 44 0 winged × winged A/A × A/– 24 0 The five unusual plants are most likely due either to human error in classification or to contamination. Alternatively, they could result from environmental effects on development. For example, too little water may have prevented the seed pods from becoming winged even though they are genetically winged.
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Chapter Two 11 14. a. The disorder appears to be dominant because all affected individuals have an affected parent. If the trait was recessive, then I-1, II-2, III-1, and III-8 would all have to be carriers (heterozygous for the rare allele). b.
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GenCh2answers - 2. Do a testcross (cross to a/a). If the...

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