<p>my PR book showed a simple punnet square, 2 boys 2 girls. one boy was affected one girl was carrier. PR said what was the possiblity of a SON to be affected.</p>
<p>at first i viewed it as a conditional. oh, so it has to be a boy so 1/2=50%</p>
<p>the answer was 25% so i was like....???</p>
<p>a friend told me because the odds of having a boy are 50% and on top of that you 50% chance to be affected if you're a boy, so 50% of 50%= 25%?</p>
<p>whats the correct way/answer?</p>
<p>Your friend is right. Say x is the recessive, or, in your words, “affected” allele, and X is the dominant normal allele, then you have xY crossing Xx. If you set up a punnet square, you get four individuals of Xx, xx, XY, and xY. A boy that is “affected” is xY, which is one out of the four.</p>
<p>The key is to look at the percentage of males affected.</p>
<p>its easier to write it down and do the problem. </p>
<p>Apparently here’s the question [X1 represents the affected allele]</p>
<p>…|…X1 …|…Y </p>
<h2>X1.|.X1X1.|…X1Y </h2>
<p>X…|.X1X…|…XY </p>
<p>so if you look at this “punnet square”, there is a 50% chance of having a CHILD who is affected, and a 25% chance there is a SON who is affected.</p>
<p>Just incase I read the problem wrong:</p>
<p>…|…X …|…Y </p>
<h2>X1.|.X1X.|…X1Y </h2>
<p>X…|.XX…|…XY </p>
<p>XY * X1X = X1X, X1Y, XX, XY</p>
<p>There is still a 25% chance that a MALE will have the trait. I always make mistakes reading like that, but thats the beauty of practice – hopefully you/me/we won’t make the same mistake again</p>
<p>YAY for poorly drawn punnet squares! (sorry it’s on a computer, but hopefully it cleared things up)</p>
<p>OP: That’s strange, I was looking at the same problem a minute ago, and thinking the same thing, but then I remembered that you have the consider the probability of getting a son (50%) and then determine the probability of the son(s) being affected based on the Punnett Square. </p>
<p>So basically - it’s 1/2 X 1/2 = 1/4 = 25%</p>
<p>It’s correct.</p>