The Plight of Heterozygotes and the Balanced Lethal System of Crested Newts

Heterozygotes have a tough time of it, because no matter how good they are they just can't breed true; heterozyte X heterozygote matings always produce 50% homozygotes. 



Worse, in assortative mating the proportion of heterozygotes will decrease very quickly (and the proportion of homozygotes increase), because the homozygotes will keep to themselves but the heterozygotes will keep producing homozygotes. I suppose in disassortative mating (mating with those different from you) between the two different homozgyotes you will get 100% heterozygotes as a result, but I'm not even sure who heterozygotes would mate with in that situation. I am assuming mating based on genotype (AA with AA, Aa with Aa, aa with aa), ignoring dominance effects that could have, for example, the AA be indistinguishable from the Aa. 

To illustrate assortative mating, I've drawn this (loosely based on a slide from Russell McLaughlin's Population Genetics lectures). To be generous to the heterozygotes, I've started it with 100% heterozygotes for this trait in the population, and you can see that the number of heterozygotes halves each time so it decreases very rapidly. 




Now in the next generation we have 25% of each homozygote, and they stay true and also each get a quarter of the heterozygotes from their parent generation, so they go up to 37.5% frequency. 

(This is all very basic stuff, but it's fun to draw.)

I did a quick Excel graph (I know, the shame, I'm sorry R/ggplot2). This is for one locus and assumes no dominance and complete assortative mating, i.e. individuals of one genotype will always mate with those of their same genotype. You can't see AA because it exactly tracks aa. You can see that the proportion of each homozygote nears (but never quite reaches, though I suppose it would in a non-infinite population) 50% as the proportion of heterozygotes approaches 0%. I hope this is correct but feel free to correct me if not.

This is just how the genetics works. But what if the genotype frequencies have different fitnesses? What if the heterozygote is by far the most fit? Sometimes, this can be resolved by duplicating the gene so that each paralog is one of the alleles, creating a state of permanent heterozygosity to escape the segregational load (Susumu Ohno did say in 1970 that this has risks though, by upsetting the dosage balance of genes, and points to the absence of a duplication of haemoglobin alleles in African populations despite their heterozygote advantage against malaria as evidence for this). 





But what if recombination isn't possible? Crested newts provide a remarkable example of this: 50% of their offspring fail to develop because only heterozygotes for chromosome 1 (1A/1B) are viable. Since only heterozygotes are viable, the parental population will be 100% 1A/1B so 50% of their offspring will be that too and all the 1A/1A and 1B/1B offspring will die. This occurs because 1A and 1B each have a different deleterious recessive allele fixed on a nonrecombining segment (I would reference Grossen et al's 2012 paper in The American Naturalist here, but essentially this entire section is based on that paper so). 

There are various models for how this maladaptive system could have occurred.

• Sims & Sessions: unequal genic exchange between two homologues of an autosomal pair made crossing over impossible in the region, and then inversions and repeat sequences accumulated. Each haplotype, 1A and 1B, would have a different duplication and deletion affecting something essential to embryonic development, and would show extremely strong negative frequency dependence (their fitness would decrease the more of them there were in the population, I think because if there are lots of them they're more likely to be found in a homozygote and thus their vehicle would die), and should die out. 
• Wallace et al: chromosome 1 used to be the sex chromosome, though that is now chromosome 4 (male-heterogametic XX/XY system). 1A and 1B used to be an AA/AB sex determining system, and so BB homozygotes are lethal because B accumulated deleterious mutations in the heterogametic form, I imagine because it was shielded by the A and thus could gain deleterious recessive mutations without selective pressure. The issues Grossen et al raise with this are (a) why are 1A/1A homozygotes lethal in the crested newt then? With an AA/AB system there would have been AAs (b) why the new XY system on chromosome 4 only operated in the formerly heterogametic sex (AB). I guess maybe (b) could be the only thing observed because only the AB survives, but if it's saying it only operates at all there then I don't know. 

Instead of these, Grossen et al propose that chromosomes 1A and 1B are forms of the nonrecombining Y chromosome from an ancestral XX/XY system. The Y would have been shielded and safe to develop deleterious recessive mutations such as deletions, and different deletions caused formation of different haplotypes. They discuss and simulate a whole bunch of different paths this could have taken, incorporating the ancestral XX/XY system and the fact that newt sex is partially determined by temperature (males are more likely to develop when it's warmer, females when it's colder - I had no idea of the mechanism behind this at first but apparently it can be related to differential temperature-regulation of genes and activity of enzymes).

Origin of Y haplotypes - the nonrecombining, degraded Y

Because the Y chromosome does not recombine (at least most of it doesn't - in humans the pseudoautosomal regions do), and because in the XX/XY system it's shielded from the effects of deleterious recessive mutations, it has strange evolutionary patterns. 

• Increased genetic drift - the Y chromosome has an effective population size of 1/4Ne (if you're counting Ne in alleles rather than individuals) because while autosomes are present in two copies in every organism, it is only present in half the individuals of the species and then only in one copy. (Similarly, the X chromosome has 3/4Ne)
• Selective sweeps and background selection - individual mutations don't really get tested because of the lack of recombination, so if the chromosome as a whole is good they'll stay, if not they'll go
• Muller's ratchet - again, no recombination, so deleterious mutations are not lost and there are far more deleterious mutations than positive ones so those build up. 

This results in long blocks of Y chromosome that are passed down as-is leading to a couple of Y haplotypes segregating in the population.  Some interesting examples of these haplotypes are mentioned in the Grossen et al paper, including a guppy species with at least 3 coexisting Y variants which code for 'different male coloration morphs and are thus possibly maintained by frequency-dependent selection occurring through female mate choice'. An experiment that mated sex-reversed (i.e. phenotypic not matching genetic sex) XY female guppies with XY males resulted in 25% YY offspring, which were viable as long as their Y haplotypes were different but show a lethal homozygous phenotype. I wonder do these guppies have a sexual selection or disassortative mating mechanism to avoid mating with a guppy with the same Y as them? Are there enough sex-reversed females for it to have evolved? I figure that for an XY female and XY male mating in a system with three Y haplotypes (assumed equal in frequency), for a given XY female there's a 1/3rd chance that her XY male mate has the same Y haplotype as her, which makes a 1/3*1/4  = 1/12 chance that a particular offspring with any XY male mate will be nonviable. 

Two examples of crosses between an XY male and XY female, the first where they have different Y haplotypes and the second where they don't and 25% of the offspring are nonviable.

The other example they give is of Rana rugosa, a type of wrinkled frog, which has a female-heterogametic ZW/ZZ system:

[D]ifferent populations have fixed different W haplotypes, so that WW individuals are viable when their two W chromosomes stem from different populations but not when they are from the same population...[E]ach haplotype has fixed one or more recessive lethal mutations (e.g., loss of function of some housekeeping genes)

So these Y haplotypes can arise. Once that happens, there's a set of steps they lay out to show how this balanced lethality could come about, which I have diagrammed to the best of my understanding below (they didn't phrase it like this set of states and events).

Top row is female, bottom male. Circled A, B and C refer to events while the A, B, C and D on top refer to states.

We start out in state A, with mm referring to a locus on chromosome 4 that will become important in state D, and X and Y referring to ancestral sex chromosomes that are now chromosome 1, with the Y having haplotypes A and B. (I'm now going to say 'X happened' for brevity but it really means 'the authors propose that X happened). 


In this state, there is no 50% offspring mortality, because Y chromosomes don't occur together - they either don't appear at all, as in the female, or they occur alongside an X, which compensates for any deleterious recessive mutations they carry. 

Event A is a climatic or range change that results in decreased temperature, so that - because newt sex is influenced by temperature, and colder temperatures are feminising - some XY newts are females (State B). These females can then mate with XY males as in the Punnett squares above to produce 50% lethal homozygotes (Ya/Ya or Yb/Yb) and 50% viable heterozygotes (Ya/Yb).  This increase in females produced an imbalanced sex ratio which created a selective advantage for an allele that could increase the number of males, i.e. a masculinising allele, to spread through the population. 

Aside box: Why is an imbalanced sex ratio a problem? 

According to Fisher, the reason most sexually reproducing species have a 1:1 sex ratio is because if the ratio is disturbed, the less common sex will be able to get more mates and thus have an advantage, and so if there's any allele that increases the proportion of that sex, it will be favoured by selection and thus more of the less common sex will be born (assuming it costs the same amount of energy to have either sex, which is not necessarily true e.g. males may need more feeding) until the sex ratio equalises. 

My issue with that is - how common are alleles that make an organism more likely to have offspring of a particular sex? This masculinising one in newts is the first I've heard of, but there may be many especially in quantitative temperature-controlled (fully or partially) systems. 

If the temperature gets even colder (Event B), even some YY newts will develop as females and so since there are enough females the X can be lost from the population by drift, resulting in a fully temperature-controlled sex determination process (State C), such as the YaYb newt developing as female if it's colder, or male if it's warmer. 

Event C was the mutation of a locus on chromosome 4 to a masculinising allele (M), which meant the sex-determination system was no longer (fully) temperature controlled but instead determined partly by the presence or absence of the M allele, with Ya and Yb (the two chromosome 1s), as in state C, not acting as sex chromosomes.

If the X had not been lost by drift, they could have evolved to the much better state (without the 50% segregational load) of MMXX/MMXY, but whenever X was eliminated before the appearance of M, this male-heterogametic mmYaYb/mMYaYb system with 50% segregational load was seen.

The authors explored several different paths in their simulations and also looked at the effect of environmental variability (e.g. if it was very cold but there was a lot of variation the temperature spikes gave chances to produce males and avoid extinction) and population size. The paths were:

• no masculinising mutation on chromosome 4 (i.e. no sex determining chromosomes other than chromosome 1's X/Y) - this meant travelling straight upwards on the diagram below as the temperature decreased.
• early-occurring masculinising mutation on chromosome 4 - in the simulations, this increased in frequency as the temperature dropped until fixation in the population, with first a female-heterogametic system at T = -4 (mMXX/MMXX), then full temperature controlled sex determination (TSD) at T = -6, with X often being fixed by drift in small populations. This presented a problem when temperatures decreased further to T = -8 because this skewed the sex ratio by creating more females as more MMXX newts developed into females, restoring selection for Y except at high environmental variance, where a few MMXX males would still be born and those would have an advantage because MMXY males mating with MMXY females produced some lethal MMYY homozygotes. If X was fixed and Y had been lost, then either the population would go extinct (small N) or survive with female-biased sex ratios.
• late-occurring masculinising mutation on chromosome 4: when mmXY (pale green) was still around, i.e. X had not been lost, M was fixed and the original male-heterogametic system was restored (MMXX/MMXY). However, if X had been lost, i.e. the mmYaYb TSD arrangement had been reached before the introduction of M, the YaYb segregational load system was fixed and a male-heterogametic system now based on the X chromosome emerged, with mmYaYb/mMYaYb (dark green female, blue male). That's the one we see today.

Figure from Grossen et al (2012) showing the different paths and results at different temperatures. For example, the optimal path with no segregational load and fixed MM results in females being the red circle and males being purple in the diagram; the current system  is dark green and blue. Being below and to the left of the line indicates femaleness, with the opposite indicating maleness, so in the ancestral state at T = 0 almost all of the mmXX (yellow) newts will be female except for the few in the dotted circle that go past the dark black line, and similarly with the light green mmXY males. As the temperature increases, more and more of the mmXY and eventually mmYY newts will develop as female, e.g. at T = -6 the dotted line goes straight through the middle of the dark green mmYY circle, so about half of mmYYs will be male and half will be female, which would be a temperature-controlled sex determination (TSD) system (this TSD system would also occur at T -6 with the genotype MMXX). 

So there you go - a very complicated system and one whose answer is highly speculative, but interesting due to its gene-environment interplay.

References

1. Grossen et al (2012). The Balanced Lethal System of Crested Newts: A Ghost of Sex Chromosomes Past? The American Naturalist.

2. Charlesworth & Charlesworth (2000) The Degeneration of Y Chromosomes. Philosophical Transactions of the Royal Society London B. 

(No, these are not properly-formatted references. It's a blog post, these are elleloughranblogspot.ie house rules.)