Problem Set 2 2010 Answers

Problem Set 2 2010 Answers - Problem Set 2 2010 Answers 2e...

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Problem Set 2 2010 Answers 2 e 3 e 4 a) Trisomies and monomies, (although trisomies are the only case in which the fetus is likely to be viable) as a result of the homologous chromosome pairs failing to segregate. Also the possibility of chromosome breakage, and therefore large chromosomal deletions in the offspring. . d) The easiest would be karyotyping. Large deletions and missing chromosomes would be easily visible. 5) Red x red crosses can give either red or red and pink offspring. Therefore, red in some cases must be a heterozygote, with red dominant to pink. This is supported by cross b, which gives a 3:1 ratio of red to pink. Therefore R = red, and r = pink Similarly, wing x wing can gave a mixture of wing and wingless progeny in cross A and B, indicating that hets have the wings phenotype. In cross A we see a 3:1 segregation of wings to wingless, indicating wing is dominant. In D it's 120:40 = 3:1 again. Therefore W = wings and w = wingless Crosses: a) A) rrWw x rrWw B) RrWw x Rrww C) RRWw x RRww (or one but not both can be Rr) D) RrWw x RrWw E) RrWw x rrww b) The genes are not linked, because the test-cross in E yields a 1:1:1:1 ratio. 6) D=dappled, d=solid; L=long mane, l=short mane. a) Original cross: ddLL x DDll ---> DdLl Chromosomes: dL/dL x Dl/Dl -> dL/Dl b) Testcross: Dl/dL x dl/dl c) Lack of 1:1:1:1 segregation indicates linkage. Parentals:
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Problem Set 2 2010 Answers - Problem Set 2 2010 Answers 2e...

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