On a world made up of many small islands, each island cannot support a very large population. Although sometimes individuals from one island make it across the ocean to another, this is usually a rare event. I read once that a genetically diverse population of about 5000 humans would be required to ensure that a population can be sustained without inbreeding, and much less than that means that the genetic diversity is not high enough.

Presume on this island world all land-dwelling species have evolved to have three sexes, such that young share the genetic material of three individuals, not just two. Would sustainable, genetically robust populations require more or fewer individuals in such a set up?

Update: It has been pointed out that perhaps the 5000 number is wrong. I'm not sure of the exact number, and will look into it later. Point is: there is probably a threshold of genetic diversity below which inbreeding is a problem. I'd like to know if that threshold is higher or lower with a third parent.

  • $\begingroup$ How many generations before it is no longer considered an inbreeding problem? $\endgroup$
    – Trevor
    Jun 20, 2019 at 17:40
  • $\begingroup$ I believe that the problem was that eventually the small population would start to produce offspring that are not capable of surviving or reproducing $\endgroup$ Jun 20, 2019 at 17:45
  • $\begingroup$ Island societies interact with each other a lot more than most people think. Polynesia being a good example. And if your islands are really that far apart (or the oceans that treacherous), how on earth did your species evolve in one place then spread out to near total isolation? OR somehow independently evolve not only into 3 sexes but with similar species on various islands? $\endgroup$
    – Cyn
    Jun 22, 2019 at 17:25

5 Answers 5


I'm pretty sure that the minimum population required to avoid inbreeding is drastically less than 5000 people. This question might be helpful in that regard, the selected answer indicates the minimum to be around 160 people I believe. That said, I know that some theories state we would need about 10,000 to effectively represent the ethnic diversity of Earth on an off-world colony, but that is also a lot of speculation and assumption since we don't have any real world examples yet.

For another species with a different reproductive method, the answer could be completely different. If the idea is that reproduction requires 3 parents, then that would drastically increase the required population, eve if we assume all other aspects of reproduction are equivalent. If the genetic distribution between parents is uneven, such as 50% from mother and 25% from each father (using mother to mean carrier parent and father to mean genetic donor), then you would want to increase mothers. If the variation was a 20% mother and 40% each father split, then you would want to increase number of fathers.

Counter to that, if there were more genders but still only 2 required to reproduce, you could end up requiring either a smaller or larger population depending on other factors. How is gender determined? Does it have to be one of the two parent genders, or could it possibly be the third anyway? Is there a higher likelihood of one gender developing over the others? What if 2 genders are mothers and only 1 fathers, or vise versa?

All of that would also have to take into account birthrates, pluralities, and gestation periods. I guess the simple answer is that there's no way to get an exact number, but I think it is safe to assume that an increase in number of required sexual partners will increase minimum population.

Addendum: This could also all be moot, because a population with more than 2 genders is also likely have a higher tolerance for genetic inbreeding, or may not even have a major threshold. Technically speaking we are all inbred since we can all trace our lineage back to a very few number of individuals. For example, everyone with blue eyes is descended from one single person and it is speculated that 1 in every 200 men is descended from Genghis Khan.

  • $\begingroup$ That blue eyed people can trace a common ancestry to a single person (assuming that's true) does NOT mean that they are inbred. The further you go back in time, the more likely it is that any given individually is your ancestor. This has nothing to do with inbreeding, it's just an observation of how family trees spread out. $\endgroup$
    – Harabeck
    Jun 21, 2019 at 16:33
  • $\begingroup$ @Harabeck If all blue-eyed people can trace their ancestry to a single person, then it means that anyone whose parents both had blue eyes is technically, to a minuscule degree, inbred. $\endgroup$ Jun 21, 2019 at 18:53
  • $\begingroup$ @AdmiralJota then everyone is inbred and you've made the term meaningless. $\endgroup$
    – Harabeck
    Jun 21, 2019 at 20:52
  • 1
    $\begingroup$ @Harabeck Well the term technically means being born of two close relatives, but that just brings into question what one considers a 'close relative'. In modern times we generally consider it to mean first cousins or closer relations, although any relations that can be specifically named are often still taboo. That certainly wasn't always the case. Historically it was common for marriages between relatives in cultures that respect bloodlines, especially nobility. And if you want to go back to Adam and Eve (or the evolutionary equivalent) they were probably siblings themselves... $\endgroup$ Jun 23, 2019 at 3:32

The main problem with inbreeding is that a child may get the same recessive gene from both parents, turning it active. Recessive genes are genes that may carry a disorder, but are only active if the child gets the same gene from both parents; otherwise, the healthy version will be the active one. With three genetic parents, the child would have to have the recessive gene from all three parents to activate the disorder, and this is far less likely. In a human population, if 1 in 100 has the bad gene, the chance is 1 in 10,000 that the child will get it from both parents - if there are three parents, the chance will be 1 in a million. This vastly reduces the risk of damaging inbreeding. In fact, the prevalence of the bad gene could be as little as 1 in 22 and still carry a lower risk of harm.

There is a caveat to this: Some hereditary disorders, like hemophilia, are tied to the gender-bearing X chromosome. A male child will get hemophilia if he gets a recessive hemophilia-bearing chromosome from his mother - he can't get a good version from his father, since he gets no X chromosome from him. A girl child only gets hemophilia if she get hemophilia-bearing chromosomes from both her mother and her father. If there are three gender-bearing chromosomes (X, Y, and Z) instead of two, the risk may be greater, or smaller, depending on how gender is determined by the genes.

Not there has to be three gender-bearing chromosomes to get three genders. Say you get an X or a Y chromosome from each parent. The possible combinations are XXX, XXY, XYY, and YYY. It may be that the 'pure' combinations XXX and YYY manifest as the same gender, with XXY and XYY being the two remaining genders. The XXX/YYY gender will then be half as common as either of the other two genders. If an X chromosome carries a recessive disorder, the XYY gender will get it if they receive it from any one of their parents. The XXY gender will need to get it from two parents, and the XXX/YYY gender will need to get it from all three parents. If every 100th X chromosome carries the disorder, one in a 100 XYY's wil have it; one in 10.000 XXY's will have it, and 1 in 2 million XXX/YYY's will have it.

Overall, it hence looks like your three-gender species carries far LESS risk of inbreeding problems than our two-gender species. Depending, of course, on how chromosomes (or the equivalent) works to determine gender. and how common gender-based genetic disorders are.


The other answers are correct that the population size will need to be larger - but they're wrong in their math. It's exponentially more.

Your question is about genetic stochasity in relation to minimum viable population sizes. When tracing/predicting genetic effects, we typically use punnett squares to measure the effects of this. Wikipedia has a good article on inbreeding depression which might be helpful.

For two sexes we use capital-lowercase letters to define traits, such as "Aa", meaning that two creatures both "Aa" mating would produce one of three possible outcomes in their children: AA, Aa, or aa. Using punnett squares we can see there would be a 50% chance of Aa, 25% chance of AA, and 25% chance of aa. The big issue with smaller populations is that "A" or "a" might disappear if AA and AA keep mating (or aa and aa).

Because you're asking about three sexes and there is no third case, I will use "ABC" instead of "Aa?". In this situation there are more than three outcomes: AAA AAB AAC ABA ABB ABC ACA ACB ACC BAA BAB BAC BBA BBB BBC BCA BCB BCC CAA CAB CAC CBA CBB CBC CCA CCB CCC.

That's 27 possible outcomes. Did you see what happened? The math shifted from two dimensions to three. So the punnett square is no longer a square but a cube. The outcomes are no longer 2^2 but 3^3.

What this means is that there will be a far greater risk of genetic inbreeding, so the minimum viable population density will need to increase exponentially. The likelihood of A, B, or C disappearing will increase exponentially. For example, a "C" will disappear if any three creatures in this set of creatures keep mating with each other: AAA, AAB, ABA, ABB, BAA, BAB, BBB. That's far more likely to occur in a three-sex species than a two-sex species.


You increased one side of the equation by 50%, so you must increase the other side of the equation by 50%. So 7500 of the species.

Since you now need an extra parent for every family, and you need 3 children instead of 2 (one to replace each parent) you have therefore increased all the requirements by 50%. Allowing some variation for how inbreeding is calculated, it sounds like the total should go up by 50% as well

  • $\begingroup$ I added an update about the 5000 number. I'm not sure I agree with your logic. Since offspring would now have more genetic material, their children would incorporate far more mixing. Are you saying that additional mixing increases the probability of inbreeding, perhaps? $\endgroup$ Jun 20, 2019 at 17:37
  • $\begingroup$ I am just following a simple formula that says you increased one parent, therefore you need more parents in the pool. Your families would also have to have 3 children instead of 2 to keep the population stable. So if every number is up 50%, then the result should be up 50% as well in my mind $\endgroup$
    – Trevor
    Jun 20, 2019 at 17:42
  • $\begingroup$ Didn't I increase the number of parents? $\endgroup$ Jun 20, 2019 at 17:46
  • $\begingroup$ @MichaelStachowsky Sorry, had 2 things in my mind when I typed that, I edited it and my answer $\endgroup$
    – Trevor
    Jun 20, 2019 at 17:47

What if it still took two individuals to reproduce?

I find it interesting that when we imagine more than two sexes, we assume that some n-way mating is required.

In the world of life on earth, we have examples where there are more than two sexes, such as slime molds. In these cases, it requires only that two individuals of different sexes mate to produce viable offspring. The main difference is that a greater percentage of the population is a potential mate. Unlike what Woody Allen wrote in "Annie Hall", "Bisexuality increases your chance of getting a date on Saturday night," these are heterosexual organisms, but with many sexes. Reviewing articles for this answer the number of sexes varies by species and study from 5 to thousands. The first number I remember was 13, so the odds are 12/13 that a random individual will be appropriate.

A driving factor in gender is protecting mitochondrial bodies from conflict. You can think of the entire cell, and all of reproduction, as a method through which mitochondria close themselves. The mitochondria of males are not passed to the offspring. Only the mitochondria in female cells are cloned. The whole gender we call male is, from the view of the mitochondria, sacrificed -- cut off from propagation -- by the benefits of having the female mitochondria cloned.

If the OP was considering a system like this, how does it affect the calculation?

The calculations haven't considered mitochondrial diseases in humans, which have several sources, including random mutations with each mitochondria, as well as changes in the cell nuclear DNA. Mitochondrial diseases that come from nuclear DNA follow the normal sexual inheritance pattern. Mutational mitochondrial diseases tend to result in early death, so probably impact the replacement rate but not the minimum population size.

If the number of sexes was high enough, the minimum size of the population would be cut in half. If we look at the problem as the number of different genotypes an individual might reproduce:

  • Lower stable limit of reproduction partners: $L$
  • The fraction with whom an individual may reproduce: $F = 1/2$
  • Total population lower stable limit: $P = (1/F) * L$

If each member of a population could reproduce with $12/13$ths of the population:

  • $P = (1/(12/13))*L$ or $(13*L)/12)$

Other answers have assumed $P=5000$, which for the 2-sex system gives $L = 2500$. Substituting $L$ into the 13-sex equation gives $P=5417$, which is a much more efficient population to support in a resource-constrained setting.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .