Assume that attached (connected) earlobes are a recessive trait with 73% penetrance. For a P generation where one parent is heterozygous and the other is homozygous recessive for the attached earlobe gene, what is the probability that their first child will have attached earlobes?

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BlackTom AI · Accepted Answer

We start by identifying the genetic cross. One parent is heterozygous (Aa) for the attached earlobe gene, and the other is homozygous recessive (aa). For a cross Aa x aa, the offspring have a 50% chance of being Aa and a 50% chance of being aa.

However, the trait is recessive and shows phenotype only when the genotype is aa, and even then only with 73% penetrance. Penetrance means that among individuals with the aa genotype, only 73% actually express the attached earlobe phenotype; the remaining 27% with aa would express the normal earlobe.

Compute the overall probability of an affected child: multiply the probability of obtaining the aa genotype by the penetrance for that genotype. P(aa) = 0.5, and P(phenotype | aa) = 0.73. So P(phenotype) = 0.5 × 0.73 = 0.365.

Interpreting the result in percentage terms: 0.365 × 100 = 36.5%. This value comes from combining the Mendelian 50% genotype expectation with the 73% expression rate of the phenotype among aa individuals.

Now evaluate the other options to see why they don’t match the calculation:
- 73% would imply that all aa individuals express the trait, ignoring the 50% chance of being aa in this cross. It conflates penetrance with the genotype frequency.
- 50% would reflect the genotype outcome (half aa, half Aa) but ignores penetrance; it assumes full expression of the phenotype for aa individuals.
- 18.75% equals 0.5 × 0.375, which might come from misapplying penetrance (e.g., using 0.75 instead of 0.73) or mixing proportions incorrectly; it does not match the correct 0.5 × 0.73 calculation.
- 25% would correspond to 0.5 × 0.5 or to other simplified fractions that don’t incorporate penetrance, so it’s not consistent with the given data.

Thus, the correct approach is to multiply the likelihood of receiving the recessive genotype by the penetrance for that genotype, yielding 36.5% as the overall probability that the first child will have attached earlobes, given the stated cross and penetrance.

BIOL_2004_81774_202509 Homework 1

Assume that attached (connected) earlobes are a recessive trait with 73% penetrance. For a P generation where one parent is heterozygous and the other is homozygous recessive for the attached earlobe gene, what is the probability that their first child will have attached earlobes?

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