Question

7. Use the Product Rule and the Sum Rule to determine each of the following probabilities:

• Mendel mated Yellow Round peas (YYRR) to Green Wrinkled peas (yyrr), then performed an F1 selfcross (YyRr x YrRr). What is the probability that an F2 pea chosen at random is heterozygous for both genes?
• From the YrRr x YyRr self cross: What is the probability that an F2 pea chosen at random is heterozygous for the Y gene OR the R gene?
• Cystic Fibrosis is a Mendelian recessive genetic disorder in humans. 5% of the human population is a carrier for CF. What is the probability that two CF carriers will mate?

What is the probability that two CF carriers will mate, AND that they will have an afflicted child?

Answer 1

1)

Probability that F2 pea is heterozygous for both traits
= Prob. that pea is heterozygous for the Yellow gene * Prob. that pea is heterozygous for the Round gene
= 1/2 * 1/2 = 1/4

Since this is a dihybrid cross, 1/4th of the progeny will be heterozygous for both traits. This follows from Monohybrid crosses (Yy x Yy or Rr x Rr), where 1/2 of the progeny is heterozygous.

2)

Probability that the F2 plant is heterozygous for the Y gene or the R gene
= Prob. that the F2 plant is heterozygous for the Y gene but not for the R gene + Prob. that F2 plant is heterozygous for the R gene but not the R gene.
= (1/2 * 1/2) + (1/2 * 1/2) = 1/2

In a self dihybrid cross, of the resulting progeny, 1/2 is heterozygous for a given gene, and 1/2 is not, therefore the probability of each of these scenarios can be calculated and added to get the total probability.

3)

Since 5% of the population is a carrier for CF;

Probability of two carriers mating = Prob(Person 1 is a carrier) * Prob(Person 2 is a carrier)
    = 0.05 * 0.05 = 0.0025

Probability that two people who are carriers mate and have an affected child
   = Probability of two carriers mating * Probability of an affected child from two carrier parents
   = 0.0025 * 0.25 = 0.000625