Drift as constitutive: conclusions from a formal reconstruction of population genetics

Abstract

This article elaborates on McShea and Brandon’s idea that drift is unlike the rest of the evolutionary factors because it is constitutive rather than imposed on the evolutionary process. I show that the way they spelled out this idea renders it inadequate and is the reason why it received some (good) objections. I propose a different way in which their point could be understood, that rests on two general distinctions. The first is a distinction between the underlying mathematical apparatus used to formulate a theory and a concept proposed by that theory. With the aid of a formal reconstruction of a population genetic model, I show that drift belongs to the first category. That is, that drift is constitutive of population genetics in the same sense that multiplication is constitutive in classical mechanics, or that circle is constitutive in Ptolemaic astronomy. The second distinction is between eliminating a concept from a theory and setting its value to zero. I will show that even though drift can be set to zero just like the rest of the evolutionary factors (as others have noted in their criticism of McShea and Brandon), eliminating drift is much harder than eliminating those other factors, since it would require changing the entire mathematical apparatus of standard population genetic theory. I conclude by drawing some other implications from the proposed formal reconstruction.

Appendix: Proofs of theorems

Theorem A1

Let R be the relation $\left\{ {\left\langle {x,\left\{ {x,y} \right\}} \right\rangle /x \in G_{i} \,\& \,\left\{ {x,y} \right\} \in I_{i} } \right\}$. Note that R is a function, since Axiom 3 specifies that R satisfies the uniqueness and existence requisites for functions (each gene has one and only one corresponding individual to which it belongs). Thusly, R can be seen as a G_i → I_i function. Additionally, Axiom 2 implies that each gene is present only once inside each individual (in the g_j ≠ g_k part). All of these tells us that, for each gene, there exists one and only one individual to which it belongs, and which also contains a different gene. In that way, every individual from generation i will be the value assigned by function R to two different genes (arguments) from that generation. Therefore, if |G_i| is the number of genes in generation i, function R establishes |G_i|/2 partitions in its domain G_i. Notice also that R is suryective (every element of the codomain, in this case I_i, is the value of some argument), since, by Axiom 3, no gene can be “loose”. Therefore, the number of partitions also coincides with the number of individuals. Therefore, |I_i| = |G_i|/2, which immediately gives us the desired result.

Theorem A2

For simplicity’s sake, I only prove this for finite populations, and for allele type A (the proof for a is almost identical). By applying the definitions of FreqAT and FreqGT, what needs to be proved is that:

$$\frac{{\left| {\left\{ {g_{k} \in G_{i} /f_{1} \left( {g_{k} } \right) \, = A} \right\}} \right|}}{{\left| {G_{i} } \right|}} \, = \, \frac{{\left( {\left| {\left\{ {i_{k} \in I_{j} /f_{2} \left( {i_{k} } \right) \, = \{ A,A\} } \right\}} \right| \, + \, 1/2\left| {\left\{ {i_{k} \in I_{j} /f_{2} \left( {i_{k} } \right) \, = \{ A,a\} } \right\}} \right|} \right)}}{{\left| {I_{i} } \right|}}.$$

It shall be convenient to call G_i(x) the set $\left\{ {g_{k} \in G_{i} /f_{1} \left( {g_{k} } \right) \, = x} \right\}$ (i.e. the set of genes of type x present in generation i), and I_i({x, y}) the set $\left\{ {i_{k} \in I_{j} /f_{2} \left( {i_{k} } \right) \, = \{ x,y\} } \right\}$ (the set of individuals of genotype {x, y} in generation i). With this terminology, what needs to be proved is that:

$$\frac{{\left| {G_{i} \left( A \right)} \right|}}{{\left| {G_{i} } \right|}} \, = \, \frac{{\left| {I_{i} \left( {\{ A,A\} } \right)} \right| + \raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} \left| {I_{i} \left( {\{ A,a\} } \right)} \right|}}{{\left| {I_{i} } \right|}}$$

Theorem A1 establishes that |G_i| = 2 × |I_i|. Thus, for the previous equality to hold, what needs to happen is that:

$$\begin{aligned} \left| {G_{i} \left( A \right)} \right| & = 2\left( {\left| {I_{i} \left( {\{ A,A\} } \right)} \right| + \raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} \left| {I_{i} \left( {\{ A,a\} } \right)} \right|} \right) \\ & = 2\left| {I_{i} \left( {\{ A,A\} } \right)} \right| + \left| {I_{i} \left( {\{ A,a\} } \right)} \right| \\ \end{aligned}$$

To prove this, I define two new sets, called G_i(A, {A, A}) and G_i(A, {A, a}). These sets will represent the set of A genes present in an {A, A} kind of individual, and in an {A, a} kind of individual. Formally,

$$G_{i} \left( {x, \, \{ x,y\} } \right) = \, \{ g_{k} \in G_{i} /f_{1} (g_{k} ) = x{\text{ and }}\exists g_{j} \in G_{i} {\text{ such that }}f_{1} (g_{j} ) = y\quad \& \quad \{ g_{k} ,g_{j} \} \in I_{i} \}$$

Axioms 3 and 4 imply that $G_{i} (A,\{ A,A\} ) \cap G_{i} (A,\{ A,a\} ) = \varnothing$ and that $G_{i} (A,\{ A,A\} ) \cup G_{i} (A,\{ A,a\} ) = G_{i} (A)$ (the first follows from the fact that no gene is inside more than one individual, while the second from the fact that every gene is inside at least one individual, along with another gene—which must be either of the same or different type). Both of these facts imply that $\left| {G_{i} (A,\{ A,A\} )} \right| + \left| {G_{i} (A,\{ A,a\} )} \right| = \left| {G_{i} (A)} \right|$.

The following two facts, along with the equation just derived, directly give us the desired result.

• 1. $2\left| {I_{i} \left( {\{ A,A\} } \right)} \right| = \left| {G_{i} (A,\{ A,A\} )} \right|$

  This follows from the fact that the members of I_i({A, A}) are pairs of different members of G_i(A,{A, A}), since every member of G_i(A,{A, A}) belongs to an AA individual, along with another G_i(A,{A, A}) member.

• 2. $\left| {I_{i} \left( {\{ A,a\} } \right)} \right| = \left| {G_{i} (A,\{ A,a\} )} \right|$

This follows from the fact that every member of G_i(A,{A, a}) belongs to a (different) Aa individual (i.e. member of I_i({A, a})). This individual also contains a member not belonging to G_i(A,{A, a}, but to G_i(a,{A, a}). Thus, a bijection can be established between both sets, which means that they have the same number of elements.

Theorem A3

This theorem follows from the following two facts. First, that if:

$$\frac{n!}{{x!y!\left( {n - x - y} \right)!}}p^{x} \times q^{y} \times r^{n - x - y}$$

is a multinomial distribution (with p, q, r being the frequencies of three kinds of objects in the population), then the expected value of the frequencies in the sample is exactly p: q: r (i.e. that all kinds of objects maintain their original frequency). Note that the probability assignment for the first sampling process is multinomially distributed.

The second fact is the probabilistic “Law of large numbers”. This law states (with terminology modified to fit the one I am using) that:

$$\mathop {\lim}\limits_{n \to \infty} P\left({\hat{p} - E_{{\hat{p}}} (Sampl(x)) > \epsilon} \right) = 0$$

where n is the size of a sample in a probabilistic sampling process Sampl(x), $\hat{p}$ is the frequency of one kind of object in the sample, $E_{{\hat{p}}} \left( {Sampl\left( x \right)} \right)$ is the expected value of $\hat{p}$ in Sampl(x), and ϵ is an arbitrary number (thus, an arbitrarily small one). In other words, at infinite sample sizes, the probability that deviations from the expected value occur (for the frequency of one kind of object) reduce to zero (for more on this law, see Feller 1971, chapter X). If this is applied to the first sampling process, the result is that actual outcomes should equal their expected outcomes. These two facts together directly imply the desired result.

Theorem A4

The proof of this result uses the same two facts as the proof from above. Since |I_i*| is infinite, by Theorem A3, we get that for every genotype {x, y}:

$$FreqGT\left( {I_{i} *, \, \{ x,y\} } \right) \, = \frac{{w_{x,y} \times FreqGT(I_{i} ,\{ x,y\} )}}{{\bar{w}(i)}}$$

Now, since we assume that all fitnesses are equal, then there exists a number m, such that m = w_A,A = w_A,a = w_a,a. Thus (by definition of $\bar{w}(i)$):

$$\begin{aligned} FreqGT\left( {I_{i} *, \, \{ x,y\} } \right) \, & = \frac{{m \times FreqGT(I_{i} ,\{ x,y\} )}}{{m \times FreqGT(I_{i} ,\{ A,A\} ) + m \times FreqGT(I_{i} ,\{ A,a\} ) + m \times FreqGT(I_{i} ,\{ a,a\} )}} \\ & = \frac{{m \times FreqGT(I_{i} ,\{ x,y\} )}}{{m \times \left( {FreqGT(I_{i} ,\{ A,A\} ) + FreqGT(I_{i} ,\{ A,a\} ) + FreqGT(I_{i} ,\{ a,a\} )} \right)}} \\ \end{aligned}$$

Since the m’s cancel out in the numerator and denominator, and what is left in the denominator (the sum of all the genotype frequencies) equals 1, all of this equals FreqGT(I_i, {x, y}). Thus, after the first sampling process, the frequencies of the genotypes remain identical. Additionally, Theorem A2 implies that allele-type frequencies will also be identical to the originals (since they can be calculated from genotype frequencies, which are themselves identical to the originals). Thus, for every allele-type x, FreqAT($\cup I_{i} *$, x) = FreqAT(G_i, x) (the union set of a set of individuals equals the set of genes present in those individuals).

By using the same two facts as in the demonstration of Theorem A3, and the probability distribution of the second sampling process, we get the desired result.