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
1/2
1/8
1
3/4
1/4

Explanation:

Step1: Determine the probability of a single child having achondroplasia

The cross is \(Aa\times aa\). Using the Punnett - square method, the possible genotypes of the offspring are \(Aa\) (achondroplasia) and \(aa\) (normal). The probability of getting \(Aa\) (a child with achondroplasia) is \(\frac{1}{2}\).

Step2: Use the multiplication rule for independent events

Since the events (the first - child's genotype and the second - child's genotype) are independent, we use the multiplication rule. The multiplication rule for independent events \(P(A\cap B)=P(A)\times P(B)\). Here, \(A\) is the event that the first child has achondroplasia and \(B\) is the event that the second child has achondroplasia. So \(P = \frac{1}{2}\times\frac{1}{2}\)

Answer:

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