Question

Part A: Say that a person with the genotype “Ee” mates with an individual of the genotype “ee”. Lets also then say that the “E” allele is dominant to that of the “e” allele, what is the probability of them having two offspring, that both coincidently have the “ee” genotype? Part B: Also Lets say 20 % (not the answer but second part to this question) of these individuals have blue eyes and 3% have green eyes. Even if its rare what would be the probability of someone having both blue and green eyes? Can show show how the math works and overall steps if it involves a punnet square? Would appreciate it.

Answer 1

This question requires one to have an understanding of the product rule in genetics. The product rule tells you that the probability of independent events occurring together is the product of their individual probabilities. In our case the phenotype of the second offspring is not dependent on the phenotype of first or vice versa. When a person of the genotype Ee mates with an individual of genotype ee, the genotypes of the F1 progeny (obtained by drawing a Punnet square) will be Ee and ee in a 1:1 ratio. i.e., half of all the offspring will have the genotype Ee and the others will have the genotype ee. Keeping this in mind, the probability of any given child being of the 'ee' genotype is 0.5 (or 1/2).

The probability of two children being of the 'ee' phenotype is the product of their individual probabilities of occurance. So that is p(first child being ee) x p(second child being ee) = 0.5 x 0.5 = 0.25%

Keeping this principle in mind, let us approach the second part of this question. Probability of an individual having blue eyes is given to be 20% and probability of an individual having green eyes is 3%. Since possession of blue or green eyes are independent of each other, we can apply the product rule to find out the probability of an individual having both blue and green eyes.

P(blue and green eyes) = P(blue eyes) X P(green eyes) = 20/100 x 3/100 = 60/10000 = 0.006%.

I hope this helps :)