Interesting genetics problem

<p>(not homework help, this is a 2006ish USABO problem)</p>

<p>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:
A. 1 : 25
B. 1 : 50
C. 1 : 100
D. 1 : 625</p>

<p>Now, I actually had to stop and think about this one, which cannot be said for most of the problems I know the answer to. See how long it takes the bio people out there (who haven't seen it before) to get it. I must admit, it does combine both math and bio (and the answer is an approximation) but it is a very good problem, something that I think most AP students wouldn't get.</p>

<p>What’s the answer? :p</p>

<p>What do YOU think is the answer?</p>

<p>I’m on my phone so it’s hard to type out my thought process haha mind pming me answer/thoughts behind it?</p>

<p>Sure. It’s actually pretty simple. :)</p>

<p>Since the disorder is present in the child but not the parent, and the parents are healthy, we know that CF is a recessive disorder. </p>

<p>The new husband is healthy, therefore in order for CF to be present in the newborn, he must be heterozygous at that allele.</p>

<p>Estimated allele frequency for CF sqrt(1/2500) = 1/50</p>

<p>Following are the conditions needed for CF:</p>

<p>Man is heterozygous (1/50)(49/50)(2)
Woman donates the recessive allele (1/2)
Man donates the recessive allele (1/2)</p>

<p>Final tally:</p>

<p>(1/50)(1)(2)(1/2)(1/2) = 1/100</p>

<p>C ?</p>

<p>EDIT: For a more accurate estimation, assume a natural rate of decline for CF after successive generations and find the discrete allelic frequencies for the United States with t = 1.</p>

<p>Hmm tell me if this is correct. Use Hardy Weinberg to get allele frequencies. </p>

<p>Since we know that’s recessive. q^2 = 1/2500 meaning q = 1/50 or 0.02</p>

<p>So p = 0.98</p>

<p>Since we know mother is healthy, yet still had a child with CF she must be a carrier. Likewise with the new father. So we need to determine the probability of the father being a carrier which is 2pq = 2 * 0.02 * 0.98</p>

<p>Probability that two carriers will have a CF baby is 1/4</p>

<p>So (1/4)(2)(1/50)(49/50) ~ 1/100</p>

<p>amirite?</p>

<p>Yep. I’m pretty sure that if you didn’t tell the students at my school to use the Hardy-Weinberg equilibrium to solve it, nobody would get it. Our school is that bad at science.</p>

<p>Wasn’t that hard, was it? I just thought it was interesting for some random reason.</p>

<p>Cool stuff. </p>

<p>Lol, my knowledge of biology consists of what I learned freshman year as well as some random facts from a critical care nursing book I have in my house.</p>

<p>

</p>

<p>I’m the exact opposite. I love the problem solving aspect of Bio but can’t for the sake of my life remember that CF is an autosomal recessive disorder T_T</p>

<p>Not looking forward to the taxonomic sections…</p>

<p>Yea problems involving a simple equation are pretty simple. It’s those minute detail questions that usually get me. (like taxonomy). Hate hate memorizing lol. Sadly it’s what the vast majority of the population thinks biology is -_-.</p>