Question

Duchenne muscular dystrophy is an X-linked recessive disease. A phenotypically normal couple wants to start a family. The woman’s brother has the disease. What is the probability that the couple’s first child will be affected?

Answer 1

Answer:

$\frac{1}{8}$

Explanation:

From the question: Duchenne muscular dystrophy is an X-linked recessive disease.

Now for an X-linked recessive disorder to be affected by a male individual; it only requires the presence of only one copy of the recessive allele of the disease to be present BUT in a female individual. both copies of the recessive allele must be present.

Let the X-linked recessive disease(i.e Duchenne muscular dystrophy) be = (ⁿ)

However, we are told that the couples are normal and are unaffected. Therefore;

the male partner (XY) will be:   $X^NY$

the female partner (XY) will be: $X^NX^N$

Similarly, the question proceeds by telling us that: the woman's brother has the disease.

Definitely, it's possible that this unaffected woman is a carrier of the disease.

So, if she is a carrier; we have her traits to be: $X^NX^n$

NOW, if a cross exist between these couples; we have

$X^NY$     ×        $X^NX^n$

                     $X^N$                         $Y$

$X^N$                $X^NX^N$                  $X^NY$

$X^n$                 $X^NX^n$                   $X^nY$

So, we have offspring as follows:

$X^NX^N$  = normal unaffected female  

$X^NY$     = normal unaffected male

$X^NX^n$   = female carrier for the Duchenne muscular dystrophy disease

$X^nY$      = affected male with the Duchenne muscular dystrophy disease

So, the probability of the child to be affected with the disease = $\frac{1}{4}$

Also, the probability that the first child will be a male or a female = $\frac{1}{2}$

∴

the  probability that the couple’s first child will be affected = $\frac{1}{4}*\frac{1}{2}$

=$\frac{1}{8}$