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I recently came across a question that asked for the probability of having a child with an autosomal recessive trait, namely cystic fibrosis. From various parts of the question, we can conclude that the father is definitely a heterozygous carrier and that the mother's parents are both heterozygous carriers. In this situation, the only way to produce a child with cystic fibrosis is if both parents are heterozygous carriers, and this will happen with a probability of $\frac{1}{4}$.

The part that causes confusion is figuring out the probability that the mother is a carrier.

One student reasoned that a cross between two heterozygous parents will produce a 1:2:1 genotypic ratio, so $\frac{1}{2}$ of the progeny will be heterozygous carriers. Therefore, the probability that the mother is a carrier is $\frac{1}{2}$ and the probability of producing a child with cystic fibrosis is $\frac{1}{2}*\frac{1}{4}=\frac{1}{8}$

Another student also referenced the 1:2:1 ratio but reasoned that since the mother is phenotypically normal, that the ratio is actually just 1:2 because there is no chance of being homozygous recessive. Therefore, the probability that the mother is a carrier is $\frac{2}{3}$ and the probability of producing a child with cystic fibrosis is $\frac{2}{3}*\frac{1}{4}=\frac{1}{6}$

These two answers are also found at this other question.

The last solution proposed is the most interesting. This student referenced the Monty Hall Problem, which is indeed somewhat related to this problem (read more about the Monty Hall Problem). Using the same ratio as above, we can conclude that there is a $\frac{1}{4}$ chance that the mother carries no alleles for cystic fibrosis and a $\frac{3}{4}$ chance that she does. The fact that she definitely does not carry two alleles does not change this probability. Therefore, the probability that the mother is a carrier is $\frac{3}{4}$ and the probability of producing a child with cystic fibrosis is $\frac{3}{4}*\frac{1}{4}=\frac{3}{16}$.

Out of these proposed solutions, which one is correct? And why are the others incorrect?

EDIT: To be clear, the solution I initially considered to be correct was the second one.

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Solutions 1 and 2 are not contradictory. 1 is appropriate if you do not know the mother's phenotype, 2 is appropriate if you know she is not affected.

Using the same ratio as above, we can conclude that there is a 1/4 chance that the mother carries no alleles for cystic fibrosis and a 3/4 chance that she does. The fact that she definitely does not carry two alleles does not change this probability.

Yes it does.

Money Hall is about 1) picking, 2) the revealer, knowing where the prize is, reveals something, then 3) you pick again with the new information. That's not really what this question is.

Lets say there are 4 doors; one affected, 2 carrier, and one not a carrier. You do not pick anything (which is why this isn't really a true Monty Hall). If you know the mom's not affected, that's the equivalent of Monty opening the affected door, no mother. Now the odds are clearly 2/3 carrier, 1/3 not, where before they were only 1/2 for carrier. Eliminating a door changes the odds for the others, that's the point of Monty Hall. Eliminating a phenotype does change the odds of the genotype.

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Monty Hall does not apply here because there is no decision point past where information is given: the information is already known that the mother is normal phenotype, and the probability for each child of the mother's parents is independent; in the Monty Hall problem, it is known that exactly 1 of the doors contains the prize. Because this sum is known (that only 1 door has the prize), if you know the contents of one door it changes the probability of the others. That is, the occurrence of the prize/goat is sampled without replacement: the thing behind each door is dependent on what is behind the other doors.

2/3 is the correct probability for the normal phenotype mother to be a carrier. Therefore the probability for a child to be homozygous recessive is (2/3) * (1/2) * (1/2) = 1/6.

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  • $\begingroup$ The part of Monty Hall that appears to relate is that removing a possibility does not mean the remaining possibilities are all equally likely. The way I understand it, the incorrect third solution is saying that removing the homozygous recessive case does not mean the remaining two heterozygous and one homozygous dominant cases are equally likely. $\endgroup$ – Darin Mao Feb 7 at 23:43
  • $\begingroup$ @DarinMao Monty Hall is about adding information about one of the possibilities when you are asking which contains a prize. That is, it is not only known that the probabilities behind the doors are equal, but also that the sum occurrence across all doors is also known (and equal to 1 in the standard Monty Hall problem). Therefore, knowing one door the probability of the other doors changes. That is not true for the mother and her possible sisters, however: her probability is 2/3 regardless of how many sisters she has and regardless of how many of them are or are not carriers or homozygous. $\endgroup$ – Bryan Krause♦ Feb 7 at 23:54

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