A man who is heterozygous for the disease Haemophilia marries a woman who is double recessive for Haemophilia. What percentage of their offspring would have the disease?
Let A = the dominant normal allele,
And a = the recessive faulty allele.
The man’s heterozygous gene will be = Aa
And the woman’s double recessive gene will be = aa
Solving for the possible offspring, we have:
Aa + aa = A(aa) + a(aa) = (Aa + Aa) + (aa + aa) = 2Aa + 2aa
Now, out of the four offspring, two will be heterozygous (unlikely to have the disease), and two will be double recessive (sure to have the disease). This means the percentage of offspring likely to have the disease will be: (2/4) X 100 = 50%
Now to the right option…
- Option A is incorrect; certainly not zero percent.
- Option B is also incorrect; not 25% or one-fourth.
- Yes, 50% according to our calculations, C is correct.
- Actually, D is incorrect; not 75% or three-fourth.
You may please note that:
- Double recessive means a gene that makes it highly possible to acquire the disease.
- While double dominant would have meant very highly impossible to have the disease.
- Heterozygous mean healthy but carriers of the disorder.
You may use the questions and answers session to deal further on this topic…
/culled from 2018 JAMB-UTME biology question 32/
A man who is heterozygous for the disease Haemophilia... » QuizTablet
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