380F11SolnPS4

380F11SolnPS4 - 01:447:380 Fall 2011 SOLUTIONS TO PROBLEM...

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01:447:380 Fall 2011 SOLUTIONS TO PROBLEM SET 4 Additional Problems 1. Since the genes at A and B loci assort independently, we can treat each one separately. Let’s look at the A’s first…. Aa x Aa ¼ AA, ½ Aa, ¼ aa. Now let’s look at the B’s…. . Bb x Bb ¼ BB, ½ Bb, ¼ bb. Now we can answer the questions using the multiplication rule to combine these probabilities. A) 3/16; p(Ab phenotype) = p(A_ and bb) = ¾ x ¼ B) 82; p(Ab phenotype) = p[(AA genotype and shows A_ phenotype) or (Aa genotype and shows A_ phenotype)] and (bb genotype and shows bb phenotype) = [( ¼ x 0.85) + (1/2 x 0.45)] x (1/4 x 0.75) = 0.082. We are asked for the total number out of 1000, so we must multiply this probability by 1000 to get the final answer. 2. We are told that the starting strains are true-breeding, so we know that the P o must be homozygous. Now we must figure out how many genes contribute to the phenotype observed. If 1 gene is involved, then it’s a monohybrid cross and we should see phenotypic ratios in the F
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This note was uploaded on 10/18/2011 for the course GEN 380 taught by Professor Glodowski during the Spring '11 term at Rutgers.

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380F11SolnPS4 - 01:447:380 Fall 2011 SOLUTIONS TO PROBLEM...

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