A Probability Problem in Heredity

Here’s a fun problem in heredity from the Dover classic Principles of Statistics by M.G. Bulmer. It’s from chapter 2, problem 2.5.

The results of Table 7 on p. 25 (see below) can be explained on the assumption that the genes for flower colour and pollen shape are on the same chromosome but that there is a probability π that one of the genes will be exchanged for the corresponding gene on the other chromosome. If we denote the genes for purple or red flowers by P and p, and the genes for long and round pollen by L and l, then the hybrids from the cross considered will all be of the genotype PL/pl, the notation indicating that the P and L genes are on one chromosome and the p and l genes on the other. When these hybrids are allowed to self-fertilise, there is a chance π that the L and l genes will interchange in one parent, giving Pl/pL; there are therefore really three mating types, PL/pl X PL/pl, Pl/pL X PL/pl and Pl/pL x Pl/pL, which occur with probabilities \( (1 – \pi)^{2} \), \( 2\pi(1 – \pi) \) and \(\pi^{2}\) respectively. Find the probabilities of the four possible phenotypes resulting from the experiment in terms of \( \theta = (1 – \pi)^{2} \).

Here’s the table the problem refers to:

Purple-flowered Red-flowered Total
Long pollen 1528 117 1645
Round pollen 106 381 487
Total 1634 498 2132

The interesting thing about that table is that it violates Mendel’s law of independent assortment. If it obeyed that law, then the probabilities of the resulting phenotypes would be:

Purple-flowered Red-flowered
Long pollen \( \frac{9}{16} \) \( \frac{3}{16} \)
Round pollen \( \frac{3}{16} \) \( \frac{1}{16} \)

We’d expect the purple/long pollen flower to happen about 9/16 = 56% of the time. Instead we see it occurring about 1528/2132 = 72% of the time. As the problem explains this has to do with the way genes are carried on chromosomes. This means we can’t calculate probabilities as you normally would for a dihybrid cross. Therefore it asks us to calculate those probabilities conditional on whether or not the gene for pollen switched chromosomes. Our answer will take the form above, but instead of actual numbers, we’ll express our answer in terms of \( \theta = (1 – \pi)^{2} \). In other words theta is the probability that the gene for pollen did not switch chromosomes.

The hard part of this problem is setting it up. First, we have to recognize that we don’t know which chromosome has the characteristics. The characteristics of mating type PL/pl X PL/pl could both be on the PL chromosomes, in which case the result would be a PL, or a purple/long pollen flower. Or one could be PL and the other pl, which would still be PL, a purple/long pollen flower, since purple and long pollen are dominant characteristics. If both are on the pl chromosome, then the result would be pl, a red/round pollen flower. All told, for the PL/pl X PL/pl mating type we have the following possibilities:

PL/pl X PL/pl mating table (p = \( (1 – \pi)^{2}\) )
PL pl
pl PL pl

For this mating type we get a purple/long pollen flower (PL) three out of four times and a red/round pollen flower (pl) one out of four times. We need to construct similar tables for the other two mating types. But notice the Pl/pL X PL/pl mating type can actually happen in reverse order as PL/pl X Pl/pL, so it needs two tables. Therefore we really need three more tables:

Pl/pL X PL/pl mating table (p = \( \pi(1 – \pi)\) )
PL pl
Pl PL Pl
pL PL pL
PL/pl X Pl/pL mating table (p = \( \pi(1 – \pi)\) )
Pl pL
pl Pl pL
Pl/pL X Pl/pL mating table (p = \( \pi^{2}\) )
Pl pL
Pl Pl PL
pL PL pL

Have to give a quick shout out to http://truben.no/latex/table/ for helping me make those tables! Anyway…so we have 4 tables displaying 16 possible results:

  • 9 out of 16 times you get PL, a purple/long pollen flower
  • 3 out of 16 times you get Pl, a purple/round pollen flower
  • 3 out of 16 times you get pL, a red/long pollen flower
  • 1 out of 16 times you get pl, a red/round pollen flower

If the results were independent, we could just call those probabilities. But they’re not. That’s the whole point of the problem. We have to take into account the probability π of the exchange of genes from one chromosome to the other. Let’s reset the probabilities for the four mating types:

  1. PL/pl X PL/pl = \( (1-\pi)^{2} \)
  2. Pl/pL X PL/pl = \( \pi(1-\pi) \)
  3. PL/pl X pl/pL = \( \pi(1-\pi) \)
  4. Pl/pL X Pl/pL = \( \pi^{2} \)

So the probability of getting pl in the first mating type is \( \frac{1}{4}(1-\pi)^{2} \). Recall the problem asks us to find these probabilities in terms of \( \theta = (1 – \pi)^{2} \), so we can express this as \( \frac{1}{4}\theta \). And there’s one of our answers since pl does not occur in any of the other mating types.

Now we just need to find the other probabilities. To make life easier, let’s go ahead and convert all the probabilities in terms of \( \theta \):

  • \( (1 – \pi)^2 = \theta \)
  • \( (1 – \pi) = \theta^{1/2} \)
  • \( \pi = 1 – \theta^{1/2} \)
  • \( \pi^{2} = (1- \theta^{1/2})^{2} \)

The hardest one to find is PL:

\( PL = \frac{3}{4}\theta + \frac{2}{4}(1 – \theta^{1/2})\theta^{1/2} + \frac{2}{4}(1 – \theta^{1/2})\theta^{1/2} + \frac{2}{4}(1 – \theta^{1/2})^{2} \)
\( PL = \frac{3}{4}\theta + \frac{2}{4}[2(1 – \theta^{1/2})\theta^{1/2} + 1 – \theta^{1/2} – \theta^{1/2} + \theta] \)
\( PL = \frac{3}{4}\theta + \frac{2}{4}[2(\theta^{1/2} – \theta) + 1 – 2\theta^{1/2} + \theta] \)
\( PL = \frac{3}{4}\theta + \frac{2}{4}[2\theta^{1/2} – 2\theta + 1 – 2\theta^{1/2} + \theta] \)
\( PL = \frac{3}{4}\theta + \frac{2}{4}(1 – \theta) \)
\( PL = \frac{3}{4}\theta + \frac{2}{4} – \frac{2}{4}\theta \)
\( PL = \frac{1}{4}\theta + \frac{2}{4} \)
\( PL = \frac{1}{4}(\theta + 2)\)

Next up is pL:

\( pL = \frac{1}{4}(1 – \theta^{1/2})\theta^{1/2} + \frac{1}{4}(1 – \theta^{1/2})\theta^{1/2} + \frac{1}{4}(1 – \theta^{1/2})^{2} \)
\( pL = \frac{1}{4}(\theta^{1/2} – \theta + \theta^{1/2} – \theta + 1 – \theta^{1/2} – \theta^{1/2} + \theta) \)
\( pL = \frac{1}{4}(-2\theta + 1 + \theta) \)
\( pL = \frac{1}{4}(1 – \theta) \)

That just leaves Pl. But if you look at the tables above, you’ll notice Pl appears in the same tables with pL the same number of times. So if we set up an equation to find its probability, we’ll get the same equation we started with when we solved Pl. That means we’ll get the same answer. So we don’t need to solve it. We already know it: \( pL = \frac{1}{4}(1 – \theta) \)

And that finishes the problem. The probabilities of the four possible phenotypes are as follows:

purple-flowered red-flowered
long pollen \( \frac{1}{4}(\theta + 2) \) \( \frac{1}{4}(1 – \theta) \)
round pollen \( \frac{1}{4}(1 – \theta) \) \( \frac{1}{4}\theta \)

One thought on “A Probability Problem in Heredity

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