Question

watch the animation on two - trait crosses and independent assortment, and then answer the question. in humans, photic sneezing (sneezing when exposed to sunlight) and wet ear wax are both the result of dominant alleles, p and e, respectively. if a woman who is unaffected by photic sneezing but is heterozygous for wet ear wax marries a man who is heterozygous for photic sneezing but has dry ear wax, what percentage of their children would likely both sneeze when exposed to sunlight and have dry ear wax?

Explanation:

Step1: Determine parental genotypes

The woman is unaffected by photic sneezing (recessive, pp) and heterozygous for wet ear - wax (Ee), so her genotype is ppEe. The man is heterozygous for photic sneezing (Pp) and has dry ear - wax (recessive, ee), so his genotype is Ppee.

Step2: Determine possible gametes

The woman (ppEe) can produce pE and pe gametes. The man (Ppee) can produce Pe and pe gametes.

Step3: Create a Punnett - square

	pE	pe
pe	ppEe	ppee

Step4: Identify the desired genotype

We want the genotype Ppee (sneeze in sunlight and have dry ear - wax). There is 1 Ppee out of 4 possible genotypes.

Step5: Calculate the percentage

The probability is $\frac{1}{4}=0.25$ or 25%.

Answer:

25%