Problem 2.35 Part A An organism having the genotype AaBbCcDdEe is self-fertilize

Question

Problem 2.35 Part A An organism having the genotype AaBbCcDdEe is self-fertilized. Assuming the loci assort independently, determine the following proportions: Part A gametes that are expected to carry only dominant alleles. Express your answer using three decimal places. Part B progeny that are expected to have a genotype identical to that of the parent. Express your answer using three decimal places. Part C progeny that are expected to have a phenotype identical to that of the parent. Express your answer using three decimal places.

Explanation / Answer

Part A:-

The total number of gametes is found by formula 2n where n = heterozygous traits. Thus here n= 5, and total number of gametes is 25 = 32. Now there will be only 1 gamete having all dominant alleles having P(ABCDE)= 1/32= 0.031.

Part B:-

Considering each gene as monohybrid cross like Aa * Aa, we will find probalility of Aa (as asked in question) and then one by one finding all required proportions, we will multiply them (because its an INDEPENDENT EVENT).

P[Aa] * P[Bb] * P[Cc] * P[Dd] * P[Ee] = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32 = 0.031

Part C:-

Now to get phenotype same as parent we will consider both AA and Aa genotypes (for each gene) as this will show same dominant trait. But one thing different in this question is we have to apply both AND and OR rule of probability.

P[Aa or AA] * P[Bb or BB] * P[Cc or CC] * P[Dd or DD] * P [Ee or EE] = [ 2/4+1/4] * [ 2/4+1/4] * [ 2/4+1/4] * [ 2/4+1/4] * [ 2/4+1/4] = 3/4 * 3/4 * 3/4 * 3/4 * 3/4 = 243/1024= 0.237

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