Two brown-haired people have two children, P and Q. P has blond hair. We therefore believe that each parent is heterozygous and that blond hair is a recessive trait. Q has brown hair. What is the probability that Q is heterozygous?

My answer: 2/3. There are four possibilities Bb, bB, BB,and bb. Before observing the color of Q's hair, we assign equal chance to each. We eliminate bb as a possibility. The chance that it is bB or Bb and not BB is 2/3.

A Biology PhD disagrees with me and says that the probability is 1/2. Who is right?

  • $\begingroup$ now ask that PhD to solve the Monte hall problem. $\endgroup$
    – Memming
    Apr 21, 2013 at 17:18
  • $\begingroup$ @Memming that's en.wikipedia.org/wiki/Monty_Hall_problem $\endgroup$
    – Alan Boyd
    Apr 21, 2013 at 20:22
  • $\begingroup$ I am familiar with the Monty Hall problem. $\endgroup$
    – Hal Canary
    Apr 21, 2013 at 20:24

1 Answer 1


Assumptions: Blonde hair is Homozygous Recessive and that the traits are strictly Mendelian.

The parental generation must be both heterozygotes as at least one child is Blonde (bb). So your cross is Bb x Bb.

Your square is going to look like this:

           _B_            _b_

_B_         BB           Bb

_b_         bB           bb

So of the question is:

What is the probability that Q is heterozygous?

The PhD is INCORRECT. They probably just made a simple Punnet Square in their head and forgot the caveat that you have to exclude "bb" because Q is not Blonde. The possible allele combinations for non-Blonde offspring from a Heterozygotic cross are only BB, bB, Bb.

Therefor the probability that Q is Heterozygous is indeed 2/3 (bB, Bb).

It's not uncommon for PhD students - who are often deep into their own research - to gloss over details and possibly give incorrect answers because the information hasn't been relevant to them for years (most traits aren't Mendelian, strictly recessive or dominant, and rarely involve a single allele), so give them some slack unless they were being a jerk about it. =)

  • $\begingroup$ Just to complete the answer, this essentially is an application of Bayes' theorem $\endgroup$
    – nico
    Apr 21, 2013 at 17:52
  • $\begingroup$ I worked it out that way, too. But adding another layer of mathematics isn't going to prove anything to someone who disagrees with the simple way of seeing things. $\endgroup$
    – Hal Canary
    Apr 21, 2013 at 20:25
  • $\begingroup$ Then you have one more argument: this was probably a high school homework, so it has to use what kids learn about Mendelian heredity, not something deeper and more elaborate. $\endgroup$ Apr 22, 2013 at 11:18

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