Question

achondroplasia is a common form of dwarfism caused by autosomal dominant mutation in the fibroblast growth factor receptor 3 (fgfr3) gene. if a person with achondroplasia (aa) married and had children with a person of normal height, what is the probability that both their first child and second child would have achondroplasia? multiple choice 1/8 1 1/2

Explanation:

Step1: Determine the genotypes of the parents

The person with achondroplasia has the genotype \(Aa\), and the person of normal height has the genotype \(aa\).

Step2: Use the Punnett square for a single - trait cross

The cross \(Aa\times aa\) gives the following offspring genotypes: \(Aa:aa = 1:1\). The probability of having a child with achondroplasia (\(Aa\)) from a single cross is \(\frac{1}{2}\).

Step3: Use the multiplication rule for independent events

Since the events (having the first child and having the second child) are independent, the probability that both the first and second child have achondroplasia is \(\frac{1}{2}\times\frac{1}{2}\)

Answer:

\(\frac{1}{4}\)

However, looking at the options provided in the image (assuming it's a multiple - choice question where the options are \(\frac{1}{2}\), \(1\), \(\frac{1}{8}\)):
If there was a miscalculation in the problem - understanding (for example, if it was considered as two non - independent events in a wrong way, but based on standard genetics rules of independent fertilization events):
The probability of a single child having achondroplasia from \(Aa\times aa\) is \(\frac{1}{2}\). But if the question was misinterpreted (which is not correct genetically), there is no correct option among the given ones. But if we assume a wrong approach of just using the probability of a single event twice without using the multiplication rule (which is wrong in genetics as these are two independent fertilization events), it still doesn't match the given options. There might be an error in the problem's option set. But if we follow the strict step - by - step of probability for independent events (each child's genotype is an independent event), the correct probability is \(\frac{1}{4}\), but since it's not in the options, and if we assume a wrong approach (which is not valid in genetics) of just taking the probability of a single child (if the question was misphrased as the probability of either child, but no, it's both), there is a problem with the options. But if we consider only the probability of a single child (even though the question says both), \(\frac{1}{2}\) is the probability for a single child, but the question asks for both. So, there is likely a mistake in the problem's option presentation. If we have to choose from the given options based on a wrong interpretation (which is not academically correct), there is no correct option. But if we assume that the question - maker made a mistake and it's about a single child (even though it says both), then \(\frac{1}{2}\) is the answer.