Question

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

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

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.

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. \[ \begin{array}{c|cc} & B & b \\ \hline b & Bb & bb \\ b & Bb & bb \end{array} \]

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) = \(\frac{\text{Number of Bb outcomes}}{\text{Total outcomes}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\)

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) = \(\frac{1}{2}\) × \(\frac{1}{2} = \frac{1}{4}\)

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) = \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\)

Key concepts

Autosomal Dominant

An autosomal dominant trait is one where the presence of just one copy of the dominant allele is enough for the trait to be expressed. Unlike traits linked to sex chromosomes, autosomal traits affect both sexes equally because they are located on non-sex chromosomes.
In the exercise, brachydactyly is described as being controlled by an autosomal dominant allele (denoted as "B").
This means that:

• A person with at least one "B" allele will exhibit brachydactyly.
• If a person has the "Bb" or "BB" genotype, they will have short fingers.
• A person with the "bb" genotype will show the normal finger trait.

Autosomal dominance is significant because it simplifies prediction of inheritance through generations, emphasizing that even one copy of the dominant gene will express the trait.

Brachydactyly

Brachydactyly is a genetic condition characterized by the shortening of the fingers and toes due to abnormal development of the bones. In this context, it is due to a dominant inheritance pattern.
The gene causing brachydactyly is autosomal dominant, meaning it is not linked to the X or Y chromosome.
This condition comes from inheriting a dominant allele from just one parent, which is enough to see this trait in an individual.
Individuals might have:

• Short finger length if they carry the Bb or BB genotype.
• Normal finger length if they carry the bb genotype (no dominant allele present).

In the given exercise, the mother with the "Bb" genotype exhibits brachydactyly and is likely to pass this trait on to her offspring.

Genotype Probability

Genotype probability is a way to predict how likely various genetic outcomes are based on parental genotypes. In the given exercise, a Punnett Square is instrumental in calculating these probabilities.
Here's how it works:

• Genotype Bb: Represents having one dominant and one recessive allele. This genotype is responsible for brachydactyly as it includes the dominant allele, meeting the requirement for expressing a dominant trait.
• Genotype bb: Represents the presence of two recessive alleles, leading to normal finger length.

By examining these genotypes within the Punnett Square, we calculate:

• 50% probability (1/2) for any child to inherit brachydactyly (Bb)
• The probability of independent events (two or more children having the trait) by multiplying their individual probabilities
• Consideration of unrelated probabilities (such as sex of the child) also factors multiplying by respective probabilities

This simple multiplication strategy helps deduce accurate genotype occurrence prediction for offspring.