Question

Problem 1

(Female) aaBbCcDdEe x AABbCcDdee (Male)

Assume that all the loci assort independently, determine the following proportions:

a)An egg carrying aBcde alleles

b)A sperm carrying aBcde alleles

c)Progeny of genotype AabbCcDdEe or Aabbccddee

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Answer 1

Female Genotype: aaBbCcDdEe
Male Genotype: AABbCcDdee

a) An egg carrying aBcde alleles:

Assuming independent assortment of alleles:

All of the eggs from the female would carry the 'a' allele given that the female only has the 'a' allele. The probability of passing on the B allele is 1/2 as the female has two alleles and the egg could end up with either. Similarly, the probability of the egg carrying the 'c' allele is 1/2, the probability of the egg carrying the 'd' allele is 1/2 and that of the 'e' allele is also 1/2.

Therefore, Proportion of aBcde eggs
= Prob ('a' allele) * Prob ('B' allele) * Prob ('c' allele) * Prob ('d' allele) * Prob ('e' allele)
= 1 * 1/2 * 1/2 * 1/2 * 1/2 = 1/16

Thus, the proportion of aBcde eggs is 1/16.

b) A sperm carrying aBcde alleles:

There would not be any sperm carrying this combination of alleles. This is because the male genotype does not have any 'a' allele. Therefore, the man cannot produce sperm with the a allele, and so the proportion of this allele is 0.

c) Progeny of genotype AabbCcDdEe or Aabbccddee

Female Genotype: aaBbCcDdEe
Male Genotype: AABbCcDdee

Probability of genotype AabbCcDdEe
= Probability of genotype Aa * Probability of genotype bb * Probability of genotype Cc * Probability of genotype Dd * Probability of genotype Ee = 1 * 1/4 * 1/2 * 1/2 * 1/2 = 1/32

In the above case, the man is of the genotype AA, and the woman is of the genotype aa, and therefore, all the progeny will be heterozygous at the A locus. Therefore, the probability is 1. Similarly, at the B locus, the man and woman are both heterozygous and therefore the probability that the progeny is of the genotype bb is 1/4. Similarly, the probability for all the genotypes can be calculated.

Probability of genotype Aabbccddee
= Probability of genotype Aa * Probability of genotype bb * Probability of genotype cc * Probability of genotype dd * Probability of genotype ee = 1 * 1/4 * 1/4 * 1/4 * 1/2 = 1/128

Therefore, the probability of progeny of genotype AabbCcDdEe or Aabbccddee = 1/32 + 1/128 = 5/128