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Discussion 1 key only Fall 2009

Discussion 1 key only Fall 2009 - GENETICS(BIO 325 Fall...

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GENETICS (BIO 325) Fall 2009 Discussion 1 key ____________________________________________________________ 1. (b) and (e) 2. Cross the yellow plant with a normal plant. Self the resulting F 1 and look for a consistent, predictable segregation pattern. For example, the presence of a 3 green: 1 yellow segregation ratio would suggest that the yellow phenotype was caused by a recessive mutation. 3. (a) testcross (b) You will observe different segregation in the testcross progeny, depending on the genotype of the red petunia. If the red petunia is R/R, then all testcross progeny will be red; if the red petunia is R/r, then ½ of the testcross progeny will be red (R/r) and ½ will be white (r/r). 4. (a) 36 (b) 36 5. (e) 6. The F 2 genotypic ratio is 1 B/B; 2 B/b; 1 b/b. Considering only the black F 2 , we expect 1 B/B : 2 B/b or two out of every three black guinea pigs to be heterozygous; the fraction is 2/3. 7. P: B/- x b/b Black female white male F 1 : all B/b = all black The female parent could be homozygous (B/B) or heterozygous (B/b) and still be phenotypically black; hence, the symbol B/-.
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