A man is affected by brachydactyly, an autosomal dominant trait that causes shortening of several fingers. What is the risk that the man's first child will have brachydactyly ?

Question

Accepted Answer

The correct option is 4: 0.5

The correct option is 4, which represents a 50% risk, or 0.5 probability, that the man's first child will inherit brachydactyly. To understand this, we must first analyze the mode of inheritance: **autosomal dominant**. This means the trait is carried on a non-sex chromosome (autosome), and only one copy of the disease-causing allele (let's use 'B') is sufficient to cause the condition. An unaffected individual must have two copies of the normal allele (let's use 'b'), making their genotype **bb**. Since the man is affected, his genotype must be either homozygous dominant (**BB**) or heterozygous (**Bb**). For rare or medically significant autosomal dominant disorders in standard genetics problems, it is almost always assumed that the affected individual is heterozygous (**Bb**). This is because the homozygous dominant condition (BB) is often lethal or extremely rare, and if the man were BB, the risk to the child would be 100%, which is not the correct answer. We must also assume the man's partner is unaffected, making her genotype **bb**. Therefore, the genetic cross is $\text{Bb} \times \text{bb}$. When you create a Punnett square for this cross, the possible offspring genotypes are $\text{Bb}, \text{Bb}, \text{bb}, \text{bb}$. This yields a $1:1$ ratio, or $50%$ chance, for the child to inherit the $\text{Bb}$ genotype and thus be affected by brachydactyly.

The incorrect option is 1: 1.00 (100%)

A $100%$ risk would imply that every child born will have brachydactyly. This outcome would only be possible if the affected father's genotype was homozygous dominant (**BB**) and the cross was $\text{BB} \times \text{bb}$ or $\text{BB} \times \text{Bb}$. In both of these scenarios, all offspring would inherit at least one dominant 'B' allele. However, this is an extremely rare and non-standard assumption for a dominant human disorder like brachydactyly, and it would lead to a different answer than the one provided as correct.

The incorrect option is 2: 0.75 (75%)

A $75%$ risk is the characteristic probability of an offspring inheriting the dominant phenotype when **both parents are heterozygous** for the autosomal dominant trait, meaning the cross would be $\text{Bb} \times \text{Bb}$. This cross yields genotypes $\text{BB}, \text{Bb}, \text{Bb}, \text{bb}$, where $\text{BB}$, $\text{Bb}$, and $\text{Bb}$ all result in the affected phenotype, totaling three out of four possibilities, or $75%$. This is incorrect because the problem does not state the man's partner is also affected, so we must assume she is the common unaffected phenotype, **bb**.

The incorrect option is 3: 0.25 (25%)

A $25%$ risk corresponds to the probability of an offspring inheriting the homozygous recessive genotype (**bb**) from a cross between two heterozygous parents ($\text{Bb} \times \text{Bb}$), which is the chance of having an *unaffected* child in that scenario. This percentage is also the chance of having a homozygous dominant child ($\text{BB}$) in the same cross. Since we are assuming the standard $\text{Bb} \times \text{bb}$ cross, where the risk is $50%$, and the $25%$ value comes from an entirely different genetic mating scenario, this option is factually incorrect for the described parents.