Further notes on sex ratio evolution

Nasonia vitripennis. Photo: D. Shuker
http://westgroup.biology.ed.ac.uk/BartP/BartP.html

Intuition

A 1:1 sex ratio is generally favoured because individuals of the rarer sex on average obtain more matings, and hence leave more than individuals of the commoner sex. This favours new genes that cause their bearers (mothers) to produce more of the rarer sex, until a 1:1 ratio is again restored. This argument assumes that sons and daughters cost the same for a mother to produce.

As a specific example, consider a wasp that is haplo-diploid (males are haploid, possessing only a single copy of each chromosome, whereas females are diploid). Females control the sex ratio of offspring: all fertilized eggs are female (because they are diploid), and all unfertilized eggs are male (and haploid). Nuclear genes determine this proportion, and natural selection should favour those genes that result in a 1:1 ratio of sons:daughters.

If the above intuitive argument isn’t enough to convince you, here’s a summary of a mathematical proof (the math isn’t required for this course. Read on if you are interested; otherwise skip to the last paragraph).

Proof

The proof is from Box 13.1 of Maynard Smith (Evolutionary Genetics, Oxford University Press, 1989). Suppose that each female in a population of wasps produces on average \(N\) offspring, of which \(m^*\) are sons and \(f^*\) are daughters \((m^* +f^* = N)\). Assume for simplicity that sons and daughters cost the same to produce. Now suppose that a rare dominant gene X, when it occurs in a female, causes her to produce \(m\) sons and \(f\) daughters (the total number of offspring is still the same: \(m+f = N\)). Under what conditions will this new gene increase in frequency?

To determine the answer, first calculate the genotype frequencies: Let the frequency of mutant females X/+ be \(p\) (i.e., one chromosome carries the mutant gene X and the other chromosome carries the wild type +). Since the new gene is rare we can assume that genotype X/X has negligible frequency. Males also carry the gene, although it doesn’t alter their behaviour. They are haploid and can be either X or +; let the frequency of the X males be \(q\). The sum of the fractions of male and female mutants is \[(p + q).\]

Males and females mate at random, and one can list all possible mating combinations and their resulting offspring, to show that in the next generation the sum of the fractions of male and female mutants is now \[(p+q)+Rp,\] where \[R = {1\over 2}({f \over f^*}+{m \over m^*})-1.\]

The new gene thus increases in frequency in the next generation only if \(R>0\) (compare the first two equations). The conditions under which this will happen are revealed by rewriting \(R\) as \[R={(N-2m^*)(m-m^*) \over 2m^*(N-m^*)}\]

If \(m^*<N/2\) (i.e., normal mothers produce fewer than 50% sons), then \(R>0\) and the new gene X increases in frequency only if \(m>m^*\) (i.e., X causes a mother to produce more sons than a normal female). Conversely, if \(m^*>N/2\) (i.e., normal mothers produce more than 50% sons) then the new gene X increases in frequency only if \(m<m^*\) (i.e., X causes a mother to produce fewer sons than a normal female). This proves that the only unbeatable situation (the one in which no mutant can increase in frequency) is when \(m^* = N/2\), that is, when half the offspring are sons and half are daughters.

Cytoplasmic factors subvert this nice balance, as they are inherited only through daughters (male offspring inherit the cytoplasm also but don’t transmit it to their offspring). In the wasp, Nasonia vitripennis, two maternally inherited factors distort the sex ratio. One, a bacteria that lives in the cytoplasm, is called sonkiller as it causes male eggs to die. The other (creature unknown) causes females to fertilize nearly 100% of their eggs, yielding daughters (Nasonia is haplodiploid, and fertilized eggs produce daughters, whereas unfertilized eggs produce sons). Amazingly, there is a third sex-ratio distorter in this wasp, an extra chromosome in the nucleus that is transmitted only through sperm. As you might expect, this factor causes females to produce mainly sons. The mechanism is cruel: The factor is transmitted to eggs through fertilization by sperm, but it then destroys the paternal set of chromosomes, yielding a haploid egg, which is a male. For more details see the list of references for the papers by Werren et al. (1998) and Beukeboom and Werren (1992).