The problem is as following:

Primary Ciliary Dyskinesia (PCD) is caused by an autosomal recessive gene. 50% of patients with PCD also have Kartagener Syndrome. Josh has Kartagener Syndrome. What is the chance his brother or sister also have this syndrome?

A  0%
B  6.3%
C  12.5%
D  25%
E  50%

Is this problem actually solvable without knowing the alleles combination of the parents? The answer states C, but perhaps I´m missing something here.

  • 2
    $\begingroup$ You must be expected to assume that both parents are heterozygous. Then there is a 25% chance each offspring would have PCD and half of those offspring would also have Kartenger (ie 12.5%). $\endgroup$ – canadianer Feb 26 '17 at 11:45
  • 2
    $\begingroup$ Why should I expect something like that, is that a rule? $\endgroup$ – Jarry Feb 26 '17 at 13:36
  • $\begingroup$ @Jarry if any parent is homozygous dominant then all of the offsprings will not express recessive allele. $\endgroup$ – JM97 Feb 26 '17 at 13:43
  • $\begingroup$ I didn't say you should expect it… $\endgroup$ – canadianer Feb 26 '17 at 21:22
  • 1
    $\begingroup$ You can only solve this problem if you know that 100% of people with Kartengers also have PCD. The question, at least this portion of it, doesn't say that. You also have to know that the parents are not affected with PCD; the problem doesn't state that either. $\endgroup$ – swbarnes2 Mar 28 '17 at 20:07

Ok, Josh has PCD. That gives three possible genotypes for parents:- ( let the alleles be P and p)

1) Pp x Pp 2) Pp x pp 3) pp x pp

In (1), the probability for next child to be pp is 25% and to have the disease is 12.5%.

In (2) and (3), the probability of any parent to be pp is rare as the p allele is quite rare (discussions with swbarnes2 in comments). So the real probability will be only slightly higher than that of case 1 i.e. 12.5%

  • $\begingroup$ Reduced fertility doesn't mean zero fertility, and they already have a child. $\endgroup$ – swbarnes2 Mar 28 '17 at 20:10
  • $\begingroup$ @swbarnes2 that's the point! They have a child. So the mother even can't have the disease. $\endgroup$ – YAHB Mar 29 '17 at 2:48
  • $\begingroup$ @swbarnes2 I edited the question. Please see if it satisfies the question. Also in such genetic question, it is always yes or no. If there is a "maybe" then its probability will be given. $\endgroup$ – YAHB Mar 29 '17 at 6:49
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    $\begingroup$ In the scenarios with pp parents, the odds of a parent being infertile are 0%. Because they already have a child. So the odds for scenarios 1,2 and 3 are 12.5%, 25%, and 50%. But without knowing the proportion of the p allele in the population, you have no idea how likely each scenario is, so you can't work out the overall percentage. The problem is only solvable if you assume that both parents are carriers, not affecteds. It's a poorly written question, because you can't actually solve it from the info given. You have to make some assumptions. $\endgroup$ – swbarnes2 Mar 29 '17 at 20:33
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    $\begingroup$ The problem is not solvable unless you assume that 1 is what's happening. And when you make that assumption, the answer is present. Realistically, the p allele would be likely be quite rare, so scenarios 2 and 3 would be extremely rare, so the real percentage counting all three would just be a bit higher than 12.5%. $\endgroup$ – swbarnes2 Mar 30 '17 at 17:13

Since the answer is C it means both parents are heterozygous.

There are two possible alternatives:

  1. Both parents are homozygous to the recessive allele
  2. One is homozygous to the recessive allele and the other is heterozygous

In the first case the answer will be 50%, and in the second case it will be 37.5%

So yes, the question is missing crucial information.

  • $\begingroup$ No, in the second case, Pp x pp, 50% of the kids have pp, and 50% of the pp offspring have the syndrome, so 25% total. $\endgroup$ – swbarnes2 Mar 29 '17 at 20:35

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