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ziro4ka [17]
3 years ago
5

Will you pls help with this?

Biology
2 answers:
kipiarov [429]3 years ago
8 0

Answer:

A. Hh

Explanation:

Dvinal [7]3 years ago
7 0

Answer:

i think its C-Aa is that all?

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Duchenne muscular dystrophy is an X-linked recessive disease. A phenotypically normal couple wants to start a family. The woman’
Andreyy89

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}

4 0
3 years ago
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