Final answer:
The probability that two heterozygous parents with an autosomal dominant disorder that has 80% penetrance will produce phenotypically normal offspring is 35%, which includes 25% for homozygous recessive offspring and an additional 10% from the 50% chance of phenotypically normal heterozygous offspring due to incomplete penetrance.
Step-by-step explanation:
To determine the probability of producing phenotypically normal offspring for parents with an autosomal dominant disorder with 80% penetrance, where both parents are heterozygous, we should consider the following:
- First, calculate the probability of inheriting the normal allele from both parents.
- Then, adjust that probability by penetrance to account for the possibility of the disorder not being expressed even if the dominant allele is present.
Each parent, being heterozygous, can provide either the disorder allele (D) or the normal allele (d). The Punnett square predicts that there's a 25% chance for an offspring to be dd (homozygous recessive, and thus phenotypically normal). However, given the 80% penetrance, an offspring with Dd genotype can be phenotypically normal with a 20% chance. The calculation is as follows:
- The chance of being homozygous recessive (dd) is 25%.
- The chance of being heterozygous (Dd) is 50%, but with 20% penetrance, the chance of being phenotypically normal is 50% x 20% = 10%.
- The total probability of being phenotypically normal is 25% (dd) + 10% (20% of Dd) = 35%.
Therefore, the chance that they will produce phenotypically normal offspring is 35%, which does not match any of the options provided; hence, there might be a typographical error in the question or choices given.