Genetic drift as a directional factor: biasing effects and a priori predictions

Abstract

The adequacy of Elliott Sober’s analogy between classical mechanics and evolutionary theory—according to which both theories explain via a zero-force law and a set of forces that alter the zero-force state—has been criticized from various points of view. I focus here on McShea and Brandon’s claim that drift shouldn’t be considered a force because it is not directional. I argue that there are a number of different theses that could be meant by this, and show that one of those theses—the idea that drift cannot bias populations to be taken somewhere in the evolutionary space from one generation to the next—is actually false. Not only has this thesis been implicitly assumed in the discussion of the force analogy thus far, but it is also commonly found in a wider range of philosophical and biological texts. I argue that correcting this view, and the usual images associated with it, will thereby bring heuristic benefits that impact the force analogy discussion, but that also go beyond it.

Appendix

Mathematical appendix

In this appendix, a number of precisions are introduced regarding the directional effects mentioned above. I first distinguish between general and particular effects (predictable in every case, or only in a limited subset). The second subsection analyzes if the biases mentioned can be said to have any general direction. And lastly, in the third subsection, McShea and Brandon’s idea that drift’s magnitude can be predicted while its direction cannot is revised.

General and particular directional effects

It is clear that, in each particular case, given enough information, we can predict the probability that a population goes “right” or “left” (i.e. an allele increases or decreases in frequency) in the evolutionary space. But, the discussion above suggests much more: that we can predict that, in general, drift will bias populations to move towards an increase of the most frequent allele, in every generation. This also seems like an intuitive result: If a population with two kinds of things has more of one, then we should expect deviations of the expected value to occur in that one’s direction. The next subsection examines in greater detail whether drift can be said to have this general directional effects.

General directional effects

This subsection examines the generality of the claim that, in every generation, drift will tend to move populations towards an increase of the most frequent allele more than the reverse. Unfortunately, for various reasons, this is not true in general. First, while we can treat gamete sampling as being with replacement, parental sampling is a process without replacement. Now, for this second type of sampling, there are counterexamples to the claim made above. For instance, consider a haploid population with two alleles (A and a) in one locus. If population size at birth is N = 9, the frequency of allele A equals 2/3 (there are 6 A individuals, and so A is the most frequent allele), and the sample size (adult population size) is n = 6, then the expected number of A individuals in the adult population is 4. However, the probability that allele frequency increases (that is, that the sample contains either 5 or 6 A individuals) is smaller than the probability that it decreases (that it contains all 3 a individuals in it)—the probabilities are 0.226 and 0.238 respectively.

The question that arises is, then, if the result holds for sampling processes with replacement (like gamete sampling, or parental sampling when populations are very large). The answer is that this also isn’t the case. Take p = 0.51 and n = 2, for example. What is happening here is that the expected value is not a possible sample value, so a frequency of 0.5 in the sample (i.e. when the sample contains one individual of each kind) will count as a case where the frequency of A decreased. This problem arises because the mean is close to, but not exactly, 2. It might be argued that this is merely an effect of the “incommensurability” between the population and sample sizes, an artifact of the choice of sample size, and that it doesn’t reflect drift’s “intrinsic” direction.

These kinds of cases can be eliminated by putting some mathematical restrictions in place, for example, by establishing that $np \in {\mathbf{\mathbb{Z}}}$. With this restriction, the problem of the general direction can now be analyzed as follows. Call $\hat{p}$ the sample frequency of allele A. What we would need to establish mathematically to prove our general claim is that “If p > 0.5, then $P(\hat{p} > p) > P(\hat{p} < p)$”. This is, of course, equivalent to claiming that “If p > 0.5, then $P(\hat{p} \ge p) > P(\hat{p} \le p)$”. In a sampling process without replacement, this translates to:

$$\sum\limits_{k = np}^{n} {\left( \begin{aligned} n \\ k \\ \end{aligned} \right)} p^{k} (1 - p)^{n - k} > \sum\limits_{k = 0}^{np} {\left( \begin{aligned} n \\ k \\ \end{aligned} \right)} p^{k} (1 - p)^{n - k}$$

Even though this would be an interesting result, even from a purely mathematical point of view, I haven’t found any mention of it in the mathematical (or biological) literature. And, probably due to my limited abilities in mathematics, I haven’t been able to prove it (nor refute it) myself, at least not generally. However, what I have been able to do is to get certain partial results that I think may help illuminate how drift is working in these kind of models.

For example, it is possible to show²⁹ that the result holds for the case where $p = 1 - (1/n)$. This case is interesting because it is one of the most counter-intuitive ones. That is, if in a population of 1000 alleles (gametes) there are 990 A ones, and a sample with replacement is taken of 100 individuals, then what the result claims is that the probability of obtaining 100 A individuals is higher than the sum of the probabilities that 1, 2, 3,…, 98 A alleles are obtained.

Theorem 1

If $p = 1 - \frac{1}{n}$ , then $P(\hat{p} > p) > P(\hat{p} < p)$

Formally, what needs to be established is that:

$$\left( \begin{aligned} n \\ n \\ \end{aligned} \right)p^{n} > \sum\limits_{k = 0}^{n - 2} {\left( \begin{aligned} n \\ k \\ \end{aligned} \right)} p^{k} (1 - p)^{n - k} ,$$

Since $\left( \begin{aligned} n \\ n \\ \end{aligned} \right)$ = 1, and replacing p for the assumed value, this is equivalent to:

$$\left( {1 - \frac{1}{n}} \right)^{n} > \sum\limits_{k = 0}^{n - 2} {\left( \begin{aligned} n \\ k \\ \end{aligned} \right)} \left( {1 - \frac{1}{n}} \right)^{k} \left( {\frac{1}{n}} \right)^{n - k} ,$$

Multiplying both sides by n ⁿ, we get

$$\left( {n - 1} \right)^{n} > \sum\limits_{k = 0}^{n - 2} {\left( \begin{aligned} n \\ k \\ \end{aligned} \right)} \left( {n - 1} \right)^{k} .$$

Next, the following lemma can be established:

Lemma 1

$$\sum\limits_{k = 0}^{n - 2} {(n - 1)^{k} } \left( \begin{aligned} n \\ k \\ \end{aligned} \right) = n^{n} - (n - 1)^{n} - n(n - 1)^{n - 1}$$

Proof of Lemma 1

$$\sum\limits_{k = 0}^{n - 2} {\left( {n - 1} \right)^{k} \left( {\begin{array}{*{20}c} n \\ k \\ \end{array} } \right) = \left( {\sum\limits_{k = 0}^{n} {(n - 1)^{k} \left( {\begin{array}{*{20}c} n \\ k \\ \end{array} } \right)} } \right) - \left( {n - 1} \right)^{n} \left( {\begin{array}{*{20}c} n \\ n \\ \end{array} } \right) - \left( {n - 1} \right)^{n - 1} \left( {\begin{array}{*{20}c} n \\ {n - 1} \\ \end{array} } \right)}$$

The first term in the subtraction can be operated with as follows. Given the binomial theorem, which claims that:

$$\left( {1 + t} \right)^{a} = \left( {\begin{array}{*{20}c} a \\ 0 \\ \end{array} } \right)t^{0} + \left( {\begin{array}{*{20}c} a \\ 1 \\ \end{array} } \right)t^{1} + \left( {\begin{array}{*{20}c} a \\ 2 \\ \end{array} } \right)t^{2} + \left( {\begin{array}{*{20}c} a \\ 3 \\ \end{array} } \right)t^{3}$$

The first term reduces to:

$$\begin{aligned} \sum\limits_{k = 0}^{n} {\left( {n - 1} \right)^{k} \left( {\begin{array}{*{20}c} n \\ k \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} n \\ 0 \\ \end{array} } \right)\left( {n - 1} \right)^{0} + \left( {\begin{array}{*{20}c} n \\ 1 \\ \end{array} } \right)\left( {n - 1} \right)^{1} + \left( {\begin{array}{*{20}c} n \\ 2 \\ \end{array} } \right)\left( {n - 1} \right)^{2} + \left( \cdots \right) + \left( {\begin{array}{*{20}c} n \\ n \\ \end{array} } \right)\left( {n - 1} \right)^{n} } \hfill \\ \quad \quad \quad \quad \quad \quad = \left( {1 + \left( {n - 1} \right)} \right)^{n} = n^{n} \hfill \\ \end{aligned}$$

The third term can also be simplified as follows, which directly gives us the desired result:

$$\left( {\begin{array}{*{20}c} n \\ {n - 1} \\ \end{array} } \right) = \frac{n!}{{\left( {n - 1} \right)!\left( {n - \left( {n - 1} \right)} \right)!}} = \frac{n!}{{\left( {n - 1} \right)!1!}} = n$$

□

Given this lemma, all that remains to be proved is that $\left( {n - 1} \right)^{n} > n^{n} - (n - 1)^{n} - n(n - 1)^{n - 1}$, that is, that $n^{n} < 2(n - 1)^{n} + n(n - 1)^{n - 1}$.

Multiplying both sides by $\frac{1}{{\left( {n - 1} \right)^{n} }}$, we get

$$\left( {1 + \frac{1}{n - 1}} \right)^{n} < 2 + \frac{n}{n - 1} = 2 + \frac{{\left( {n - 1} \right) + 1}}{n - 1} = 3 + \frac{1}{n - 1}$$

or with $m = n - 1,$

$$\left( {1 + \frac{1}{m}} \right)^{m + 1} < 3 + \frac{1}{m},$$

$$\left( {1 + \frac{1}{m}} \right)^{m} < \frac{{3 + \frac{1}{m}}}{{1 + \frac{1}{m}}}$$

Multiplying the term on the right by $\frac{m}{m}$

$$= \frac{3m + 1}{m + 1} = \frac{3m + 3 - 2}{m + 1} = \frac{{3\left( {m + 1} \right) - 2}}{m + 1} = \frac{{3\left( {m + 1} \right)}}{m + 1} - \frac{2}{m + 1} = 3 - \frac{2}{m + 1}$$

What is left to prove is that:

$$\left( {1 + \frac{1}{m}} \right)^{m} < 3 - \frac{2}{m + 1}$$

And this is true, since it can be checked directly for $m \in \{ 2, \ldots ,6\}$, and the left side is bounded by e, while the right side (which represents an increasing function) surpasses e by m = 7. □

A possible demonstration strategy of the general result would be the following. We first show that, for a starting frequency of p = 0.5, $P(\hat{p} > p) = P(\hat{p} < p)$. This is fairly easy to establish. We need to show that:

Theorem 2

$$\sum\limits_{k = 0}^{n/2} {\left( \begin{aligned} n \\ k \\ \end{aligned} \right)} \left(\frac{1}{2}\right)^{k} \left(\frac{1}{2}\right)^{n - k} = \sum\limits_{k = n/2}^{n} {\left( \begin{aligned} n \\ k \\ \end{aligned} \right)} \left(\frac{1}{2}\right)^{k} \left(\frac{1}{2}\right)^{n - k}$$

Proof of Theorem 2

The strategy here is to “pair up” the factors of the sums, such that the probability that k = 0 on the left “pairs” with k = n on the right; k = 1 on the left with k = n − 1 on the right… and, in general, k = r in the left with k = n − r on the right (Fig. 3).

Fig. 3

Pairing of the factors of the sums

Since $\left( {\frac{1}{2}} \right)^{n - k} \left( {\frac{1}{2}} \right)^{n - (n - k)} = \left( {\frac{1}{2}} \right)^{k} \left( {\frac{1}{2}} \right)^{n - k}$, and since

$$\left( \begin{aligned} n \\ k \\ \end{aligned} \right) = \frac{n!}{{k!\left( {n - k} \right)!}} = \frac{n!}{{(n - (n - k))!\left( {n - k} \right)!}} = \left( \begin{aligned} n \\ n - k \\ \end{aligned} \right)$$

then each of the paired factors will be equal to the other. □

Next, I establish the following conjecture: as p increases by “steps” of 1/n, $P(\hat{p} \ge p)$ increases as well. That would give us a sort of induction over p (theorem 2 being the base step, and this conjecture the inductive step). Formally, we need that:

$$\left( \begin{aligned} n \\ n \\ \end{aligned} \right)\left( {1 - \frac{1}{n}} \right)^{n} \left( {\frac{1}{n}} \right)^{0} > \left( \begin{aligned} n \\ n \\ \end{aligned} \right)\left( {1 - \frac{2}{n}} \right)^{n} \left( {\frac{2}{n}} \right)^{0} + \left( \begin{aligned} n \\ n - 1 \\ \end{aligned} \right)\left( {1 - \frac{2}{n}} \right)^{n - 1} \left( {\frac{1}{n}} \right)^{1} > \cdots$$

That is, that

Conjecture 1

$\sum\limits_{k = 0}^{r + 1} {\left( \begin{aligned} n \\ n - k \\ \end{aligned} \right)} \left( {1 - \frac{r + 1}{n}} \right)^{n - k} \left( {\frac{r + 1}{n}} \right)^{k}$ is a decreasing function for r = 0, 1, 2,…, n – 1.

If this could be established it would seem to show something else. Put in the terminology of the coin game image, it would seem to show that the bias of the coin gets bigger and bigger the more one gets closer to one of the absorbing states. However, that would be partially misleading, since proving that $P(\hat{p} \ge p)$ increases when p increases by steps of 1/n is not equivalent to proving that $P(\hat{p} > p)$ increases in the same way. This last statement is, in fact, false. Consider the case where n = 10; for p = 0.9, $P(\hat{p} > p)$ = 0.348, while for p = 0.8, $P(\hat{p} > p)$ = 0.375. It does hold however, in this case, for $P(\hat{p} \ge p)$. This is due to the fact that the increase in the probability of the expected value $(\hat{p} = p)$ is greater than the decrease of the probability of going right.

If the conjecture holds, then this would paint an interesting and complicated picture for the changes in the coins’ bias. That is, the bias would still exist (and the main result needed to claim that drift has a general direction would be established), but the changes in the bias would be somewhat strange: as one moves closer to 100 points, the probability of landing heads or sideways increases. However, at the same time, it could happen that the coin gets closer to 100 points and the bias gets smaller, while this decrease would be compensated by the probability of it landing sideways getting bigger.

In any case, two conclusions from the preceding discussion can be drawn. First, that whether or not drift has a general direction, particular directions can be predicted a priori in particular cases, and this is enough to make the point of the article. And second, that if the conjecture holds, then the general claim can be made, and it is only for the changes in the bias that one gets more complicated results.

The magnitude of drift

Lastly, another interesting point concerns notion of the magnitude of drift, and its differences with the direction. It is generally assumed—e.g. it was by Brandon 2006, p. 324—that the magnitude of drift can be predicted a priori, in a sense that its direction cannot, by considering population sizes.

What the previous considerations show is that the expected magnitude of drift depends not only on the population size, but also on the starting frequencies of the alleles. The last example shows a case where increasing the frequency of an A allele resulted in an increase of $P(\hat{p} = p)$ and in a decrease of both $P(\hat{p} < p)$ and $P(\hat{p} > p)$. That is, a mere increase in the frequency of an allele, without modifying population sizes, made drift “weaker”—it diminished the probability that the sample values deviate from the expected ones. In fact, both magnitude and direction depend upon both factors, population size and initial frequencies.

Additionally, as it happens with the case of the direction, it is not possible to obtain a deterministic prediction about the magnitude of drift simply by specifying the population sizes and the initial frequencies (e.g. a prediction that tells us that p will either increase or decrease by 0.1). At most, specifying that would allow us to say, with a given degree of confidence, that the sample values will occur within a certain interval (the smaller the interval, the less confidence we have). But this prediction is also different than the one given by selection only models—it doesn’t state a particular magnitude, but a range of magnitudes. In sum, it seems that whatever is said about drift’s direction, must also be said about its magnitude.