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In the European population, about 1 in 2500 people suffers from Cystic Fibrosis, a genetically determined (descared), autosomal disease. Healthy parents have a child suffering from Cystic Fibrosis. The woman remarries a healthy man. What is the chance of a child from this second marriage suffering from Cystic Fibrosis ?

My attempt

http://i.imgur.com/4mhU3Qh.jpg

But the answer given is 1/100

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  • $\begingroup$ Welcome to Biology.SE! Images are not searchable and are problematic for people with visual impairment. Please, always type everything when asking a question on a SE site. $\endgroup$
    – Remi.b
    Commented Nov 8, 2016 at 17:58

1 Answer 1

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Genetics of Cystic fibrosis

Cystic fibrosis is caused by an autosomal recessive allele. We therefore know that the frequency of the homozygote recessive allele is $P_{aa} = \frac{1}{2500}$, where $a$ denotes the recessive allele and $A$ will denote the dominant allele.

Solving the problem

The probability of the second child (from a second husband) has cystic fibrosis is the probability that a second husband is heterozygous ($P_{aA}$) times the probability that this father will transmit the recessive allele $a$ ($\frac{1}{2}$) times the probability that the mother transits the recessive allele $a$ ($\frac{1}{2}$, note that we know the mother is heterozygous as she is healthy but already had a child that suffers from cystic fibrosis).

So the answer is $p_{Aa} \frac{1}{2} \frac{1}{2} = p_{Aa} \frac{1}{4}$, we only need to calculate $p_{Aa}$. Let $x$ denote the frequency of the recessive $a$ allele, then $P_{Aa} = 2 x (1-x)$ at Hardy Weinberg equilibrium. As we know $p_{aa}=\frac{1}{2500}=x^2$, it results that $x=\sqrt{\frac{1}{2500}} = \frac{1}{50}=0.02$. Therefore, $p_{Aa}=2 x (1-x) = 0.0392$. The final answer is $p_{Aa} \frac{1}{4} = \frac{0.0392}{4} = 0.0098 ≈ \frac{1}{100}$.

What you did wrong

You were pretty close to the right answer. In your calculations you multiplied the frequency of the recessive allele $a$ (which I called $x$) by $\frac{1}{4}$ but you should have multiplied the frequency of heterozygote genotypes (which I called $p_{Aa}=2 x (1-x)$) by $\frac{1}{4}$.

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  • $\begingroup$ @Rajdeepdhingra Thanks for letting me know if it answered your question. If you think the question is answered, you can check the answer (top-left of the answer). $\endgroup$
    – Remi.b
    Commented Nov 13, 2016 at 7:30

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