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Chapter 3: Problem 25

The autosomal (not X-linked) gene for brachydactyly, short fingers, is dominant to normal finger length. Assume that a female with brachydactyly in the heterozygous condition is married to a man with normal fingers. What is the probability that (a) their first child will have brachydactyly? (b) their first two children will have brachydactyly? (c) their first child will be a brachydactylous girl?

Short Answer

Expert verified
Answer: The probabilities are (a) 1/2 for the first child to have brachydactyly, (b) 1/4 for their first two children to have brachydactyly, and (c) 1/4 for their first child to be a brachydactylous girl.

Step by step solution

01

Determine the genotypes of the parents

The mother has brachydactyly and is heterozygous, which means she has one dominant allele (B) and one recessive allele (b). Her genotype is Bb. The father has normal fingers, which means he has two recessive alleles (b). His genotype is bb.
02

Create a Punnett square

We will create a Punnett square to determine the possible genotypes of their offspring. The Punnett square will be a 2x2 matrix, with the mother's alleles on the top and the father's alleles on the side. \[ \begin{array}{c|cc} & B & b \\ \hline b & Bb & bb \\ b & Bb & bb \end{array} \]
03

Calculate probabilities for part (a)

The probability that their first child will have brachydactyly is equal to the probability of inheriting a dominant allele (B) for brachydactyly. According to the Punnett square, there are two possible outcomes with the genotype Bb and two with the genotype bb. Therefore, the probability that their first child will have brachydactyly (genotype Bb) is: Prob(Bb) = \(\frac{\text{Number of Bb outcomes}}{\text{Total outcomes}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
04

Calculate probabilities for part (b)

The probability that their first two children will have brachydactyly is the product of the probabilities of each child having brachydactyly independently. As we found in Step 3, the probability for each child is \(\frac{1}{2}\). Therefore, the probability that their first two children will have brachydactyly is: Prob(2 children with Bb) = Prob(Bb) × Prob(Bb) = \(\frac{1}{2}\) × \(\frac{1}{2} = \frac{1}{4}\)
05

Calculate probabilities for part (c)

The probability that their first child will be a brachydactylous girl requires us to consider both the child's sex and the presence of brachydactyly. Since the gene for brachydactyly is autosomal (not sex-linked), we can consider that the probability of having a girl is simply \(\frac{1}{2}\) and is independent from having the brachydactyly gene. Thus, the probability that their first child will be a brachydactylous girl is: Prob(brachydactylous girl) = Prob(girl) × Prob(Bb) = \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\)

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