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Let's say Jane has Cystic Fibrosis and we know her brother doesn't. What are the chances of her brother's future child having CF? (The chances of Cystic Fibrosis in the general population can be taken as 1 in 25)

My apologies for the very incomplete question, this was all I could remember of the actual question. And the solution went like something along the lines of:

Possibility of Jane's brother's partner having CF: 1 in 25

Possibility of Jane's brother being a carrier: 2 in 3

And this is the part that I am not getting. For me, Jane's brother could only have a 1 in 2 chance of being a carrier (Cc or CC). But the solution went CC, Cc or cC-- this last one, cC, how is that possible? If it's a recessive trait, do we not write the dominant one first? Then how could there be cC and Cc as two separate possibilities?

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And this is the part that I am not getting. For me, Jane's brother could only have a 1 in 2 chance of being a carrier (Cc or CC).

Wrong, because those are not equally likely.

There are 4 equally likely possibilities for any of Jane's siblings:

inherit good allele from Mom, good from Dad,

inherit bad allele from Mom, good from Dad,

inherit good allele from Mom, bad from Dad,

inherit bad allele from Mom, bad from Dad.

We know that the last is not the case for the brother. The first 3 are still equally likely. And in 2/3, brother is a carrier.

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Let me elaborate on swbarnes2’s answer.

The “Possibility of Jane's brother being a carrier” is indeed the trickiest part of the overall question. I have seen that even some teachers of genetics were puzzled when seeing the correct answer (2/3).

It is true that before the brother is born the probability of he being Cc is given by the standard Mendelian rule, which states, for a Cc x Cc mating, ¼ of the offspring will be CC, ¼ cc, and ½ Cc.

But that answer ignores the fact that the brother is already born and (presumably) has an age by which all cc homozygotes have been recognized. Then, we know that he is not cc and can only be CC or Cc, whose probabilities sum up to ¾. It follows that he is Cc with probability ½ / ¾ = 2/3.

This kind of problems are generally tackled by using the Bayes theorem, which is a way to transform a prior probability into a posterior probability. In this case, we know that the prior probability, Pr(Cc), of the brother being a carrier is 50%, and want to calculate the posterior probability, Pr(Cc | not-cc), which is the probability that he is a carrier given that he is not affected. The Bayesian rule has the general form

Pr(A|B) = Pr (B|A) x Pr(A)/P(B),

which in our case reads

Pr(Cc|not-cc) = Pr(not-cc|Cc) x Pr(Cc)/Pr(not-cc).

As the first factor in this equation equals 1 (it is certain that a subject is not cc if he is Cc), it easy to see that the solution corresponds to the above, perhaps more intuitive, calculation. In more complex situations (consider for example if Jane's brother is young, and a certain proportion of cc subjects are diagnosed only late in life) the Bayes rule is useful, as it help decomposing the problem.

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