This isn't exactly homework but I'd like to treat it as such. I'm asked to find the probability that both (III: 2) and (III: 3) are carriers given that (II: 7) and (III: 1) exhibited the recessive trait (affected individuals are marked black in the pedigree below).

Denoting the affected individuals with $aa$ (and so the dominant allele is denoted $A$), I was able to determine the following:

  • Both (II: 1) and (II: 2) are carriers.
  • Both (I: 3) and (I: 4) are carriers.
  • At least one of (I: 1) and (I: 2) must be a carrier.
  • The probability that (II: 6) is a carrier is $\frac{2}{3}$.

The answer is supposed to be $\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}$.

I'm mostly confused as to how to approach this one case at a time. Suppose (II: 6) is indeed a carrier. Then if (II: 5) is homozygous for $A$, then (III: 3) has a $\frac{1}{2}$ probability of being a carrier; if (II: 5) is also a carrier, then (III: 3) has a $\frac{2}{3}$ probability of being a carrier.

Since the event that (II: 5) is homozygous is mutually exclusive from the event that he is heterozygous, would I just add these probabilities (i.e. $\frac{1}{2}+\frac{2}{3}$)? Then multiply by the $\frac{2}{3}$ of (II: 6)? This would give $\frac{7}{9}$, which doesn't seem to get me any closer to the right answer.

A different approach I tried was multiplying the respective probabilities for the (II: 5) events above, which gives $\frac{1}{3}$ and thus one factor of the provided answer. If this is right, and it seems to be, I'm not sure how to apply this to the (III: 2) side of the pedigree.

  • 5
    $\begingroup$ I can probably post an answer later, but here's some a hint in the mean time. In these problems, it is generally assumed that the disease allele is rare. Individuals like II-5, which presumably have no affected relatives, are assumed to be homozygous. $\endgroup$
    – canadianer
    Jan 22, 2015 at 1:33

2 Answers 2

  • II:6 has a $2/3$ probability of being a carrier Aa.
  • II:3 has a $1/2$ probability of being a carrier Aa.
  • Assuming, just like canadianer suggested, that the recessive gene is rare, autosomal, and non-x-linked, this means II:5 and II:4 are both homozygous (AA).

If parents of III:2 are therefore Aa and AA, this gives III:2 the probability of being a carrier of $1/2$. However, we must multiply this value by $1/2$, because III:2's father has a $1/2$ probability of being Aa. Therefore $\frac{1}{2}\frac{1}{2}= 1/4 =$ probability of III:2 being a carrier.

Similarly for III:3, if her parents are Aa and AA, this gives III:3 the probability of being a carrier of $1/2$. However, we must multiple this value by $2/3$, because III:3's mother has a $2/3$ probability of being Aa. Therefore $\frac{1}{2}\frac{2}{3}= 1/3 =$ probability of III:3 being a carrier.

The probability of both III:2 and III:3 being carriers is the product of both their probabilities: $\frac{1}{3}\frac{1}{4}= 1/12$.

  • $\begingroup$ Thanks to everyone that helped. I wasn't aware that making such an assumption was permissible, the textbook makes no mention of that kind of approach. $\endgroup$
    – user170231
    Jan 22, 2015 at 5:21
  • $\begingroup$ Does the same method work if I have to find the probability that neither III-2 nor III-3 are carriers? I'm told the probability is $\dfrac{2}{3}\times\dfrac{3}{4}=\dfrac{1}{2}$ this time. Clearly, the $3/4$ is a stand-alone factor (I can't multiply two fractions to get this product), but I don't see where it originates from. Would I make a different assumption about the uncertain individuals? Aa instead of AA? $\endgroup$
    – user170231
    Jan 22, 2015 at 5:36
  • $\begingroup$ @user170231, where did you get that answer? Was it for either III:2 or III:3 or both being AA? $\endgroup$
    – Anne
    Jan 22, 2015 at 6:07
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    $\begingroup$ @user170231 If it makes it clearer, your new question can be rephrased as "find the probability that both III-2 and III-3 are not carriers?" You already have the probability of each being a carrier: 1/3 and 1/4. The probability of each not being a carrier is then 1-1/3=2/3 and 1-1/4=3/4. $\endgroup$
    – canadianer
    Jan 22, 2015 at 7:24
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    $\begingroup$ To additionally clarify the answer, it also needs to be assumed that one and only one of I:1 and I:2 is a carrier, for the same reasons canadianer gave (the recessive allele is rare). Otherwise, the question cannot be solved without knowing the allele frequencies. $\endgroup$
    – March Ho
    Jan 22, 2015 at 8:51

I can answer your question from the comments. In the second case, we have that one set of parents carry the following probabilities: $AA=1/2$, $Aa = 1/2$, and $AA = 1$. The second set of parents are the following: $AA = 1$, $Aa=2/3$, and $AA = 1/3$.

For the first set, we have the probability of $AA = \frac{1 + 1/2}{2} = \frac{3}{4}$. For the second set, we have $AA = \frac{1 + 1/3}{2} = \frac{2}{3}$. Now the probability of a child being $AA = \frac{3}{4}\frac{2}{3} = \frac{1}{2}$.

  • $\begingroup$ In the first set probability, I understand that $AA=1$ comes from II-4, and the $1/2$ probabilities for $AA$ and $Aa$ come from the possible genotypes for II-3, but which $1/2$ probability contributes to the resultant probability $\frac{1+1/2}{2}$? Is this $[P(AA)_{II-4}+P(Aa)_{II-3}]\times P(AA)_{II-3}$, or $[P(AA)_{II-4}+P(AA)_{II-3}]\times P(Aa)_{II-3}$? I can't seem to make sense of this setup... $\endgroup$
    – user170231
    Jan 22, 2015 at 16:26
  • 1
    $\begingroup$ If we are calculating AA, we take the 1/2 from AA and the 1 from AA giving us a total 3/2 AA out of a total of 2. Then we divide by the total the for probability. $\endgroup$
    – dustin
    Jan 22, 2015 at 16:28
  • $\begingroup$ Oh okay, that makes much more sense now! So the Aa probability isn't directly involved, we're only concerned with the events that the parents are AA. I appreciate the clarification! $\endgroup$
    – user170231
    Jan 22, 2015 at 16:31
  • $\begingroup$ @user170231 it is involved when we divide by the total of 2 = 1 +1/2 +1/2. $\endgroup$
    – dustin
    Jan 22, 2015 at 16:33
  • $\begingroup$ Right, I meant the 1/2 for Aa is involved in the total sample space (denominator) but not the numerator $\endgroup$
    – user170231
    Jan 22, 2015 at 16:36

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