ps2answers - Alleles, Phenotype & Genetic interaction...

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Problem Set #2 Answers: 1. (a). 2 (b). 6 (c). 10 (d). 8 (e). 7 (f). 5 (g). 9 (h). 3 (i). 4 (j). 1 2. Dominance relationships are between alleles of the same gene. Only one gene is involved. Epistasis involves two genes. The alleles of one gene “mask” or “hide” the effects of alleles at another gene. 3. This could be explained by incomplete dominance. RR = red; Rr = pink; rr = white. 4. The long allele (L) is completely dominant to the short allele (l). The flower color trait shows incomplete dominance. PP = purple; Pp = pink; pp = white. 5. The nucleotide sequence of a mutation says nothing about whether it will be dominant or recessive. Dominance and recessiveness depend on the function of the gene and how the phenotype is analyzed. 6 (a). ii (phenotype O) or iI A (phenotype A) or iI B (phenotype B). (b). iI B (phenotype B) or I B I B (phenotype B) or I B I A (phenotype AB). 7 (a). Marble and dotted. (b). spotted and dotted, marbled, and spotted in a 1:2:1 ratio. 8. (a). A Y a Bb Cc X Aa bb Cc (b). Six phenotypes are possible: albino, yellow, brown agouti, black agouti, brown, black. 9. The 9:7 ratio in the F2 indicates that two genes are involved in green pigment formation. We’ll call the two genes A and B. The genotypes of the parental plants that would produce a 9:7 ratio are AA BB and aa bb. A cross between these plants would result in Aa Bb F1 plants. Thus, a test cross of the F1 plants would result in the following genotypes and phenotypes: Aa Bb, aa Bb, Aa bb and aa bb in equal proportions: 1/4 green and 3/4 yellow fruit.
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10. 1/4 would appear to have O type blood, 3/8 have A, and 3/8 have AB. 11 (a). The male would produce bands of 900, 150, 100 and 50 base pairs. The female would produce bands of 500, 400, 200 and 100 base pairs. (b). Since both parents are homozygous for the RFLP marker, their children will contain the top and bottom form. The children will therefore have bands of 900, 500, 400, 200, 150, 100 and 50 base pairs. Since the parents can only make one type of gamete, all the children will have the same banding pattern. 12 (a). The dihybrids have the genotype A y A + Cc , and since the alleles affect coat color, there is probably some epistatic interaction between these genes. Notice that the ratios are not like any that we've discussed for two genes, but looks closest to 9:4:3. Draw a Punnett square and predict the phenotypes. How can you get from the predicted phenotypes to the observed? You need to eliminate 3 from the yellow group, and one from the albino group. Notice from your square that only one genetic combination is consistent with this: those that are homozygous for A y . If you eliminate any mice that are homozygous for A y , you arrive at the correct phenotypic ratios, and this is the only allele that, when eliminated, produces the observed ratios. A y therefore probably encodes an allele that is lethal when homozygous. (b).
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This note was uploaded on 06/03/2008 for the course GENOME 371 taught by Professor Unsure during the Spring '03 term at University of Washington.

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ps2answers - Alleles, Phenotype & Genetic interaction...

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