Genetics question for anyone who's up to it

[dr. sinister]

Here's a hypothetical genetics problem, we assume that simple Mendelian principles apply.

 

There is a population that carries a recessive genetic disorder ‘r’, which is expressed only when an individual inherits two recessive traits, or 'rr'.

 

If there is a starting population of two recessive carriers... "Nr" and "Nr". If they were to mate, show the resultant population frequency for the 1st, 2nd and 3rd generations, assuming that the individuals with the expressed form of the recessive disorder do not survive to reproduce. I.e. in each generation what will be the percentage of those who have "NN", "Nr", "rr".

 

P.s. Don't be a stickler about the wording I kinda just winged it when I wrote up the question. I've worked it out myself, I just want to know if others will get the same result.

 

Thanks.

[Sisyphus]

Assuming no rr survives to breed, every NN and Nr has the same amount of offspring, the mating is random among NNs and Nrs, no individual mates outside its own generation, and the population is large enough that one individual is a negligible percent of the total (i.e., so if there are evenly divided NNs and Nrs, a given NN is just as likely to mate with a NN as a Nr):

 

first generation:

25% NN (1/3 of breeding population)

50% Nr (2/3 of breeding population)

25% rr

 

second generation:

44.4% NN

44.4% Nr

11.1% rr

 

third generation:

56.2% NN

37.5% Nr

6.2% rr

[dr. sinister]

Hmm, I got a different result, but I think the error is mine. In calculating the mating possibilities of generation 2, I assumed that there were only 3 possible combinations. Can you provide the working so I can take a look at it?

 

[Sisyphus]

I did it as follows, for the second generation, and equivalent for the third.

 

Male:*******Female:*****Offspring:******Total:

 

1/3 NN******1/3 NN******1/1 NN********1/9 NN

***********2/3 Nr******1/2 NN********2/18 = 1/9 NN

**********************1/2 Nr********2/18 = 1/9 Nr

2/3 Nr******1/3 NN******1/2 NN********2/18 = 1/9 NN

**********************1/2 Nr********2/18 = 1/9 Nr

***********2/3 Nr******1/4 NN********4/36 = 1/9 NN

**********************1/2 Nr********4/18 = 2/9 Nr

**********************1/4 rr********4/36 = 1/9 rr

 

Adding every possibility up, you get 4/9 NN, 4/9 Nr, and 1/9 rr. Sorry about the crazy formatting. I couldn't get it to leave empty space.

[Mokele]

google "population genetics" or "Hardy-Weinberg" and you'll find everything you need to know.

 

Initially, you have a gene frequency of .5 for the dominant allele (4 alleles on 2 heterozygotes), which declines to 1/(2-p) each generation. Thus the next generation would have a frequency of 2/3rds, the next one of 3/4, then 4/5ths, and so on.

 

There's copious information online about these (see above google searches), and this stuff is actually the foundation of a lot of modern evolutionary biology.

 

Mokele

[dr. sinister]

Thanks, Sisyphus.

 

That was helpful, I'm not sure if it's entirely correct though, it very well may be. I think it maybe the formatting that's misleading, maybe some annotation would clarify your procedure.

 

And Mokele,

 

My purposes are to simply demonstrate the effects of inbreeding in a certain population. Even though human genetic disorders aren't quite so simplistic, I just wanted a way of easily showing the population frequencies as a result of multiple generations of inbreeding.

 

I was having a discussion with someone and they made the claim that inbreeding flushes the autosomal recessive disorder out of the gene pool quicker than outbreeding and I'm pretty confident that this isn't the case.

 

Outbreeding results in as I calculated it a 6% frequency of the Nr genotype by the 3rd generation and entirely eliminates the possibility of the homozygous recessive genotype by the 1st generation.

 

But I wasn't entirely sure about calculating the frequencies of the inbred population. But I think the rule of thumb is a tendency towards homozygous genotypes with population inbreeding.

 

Does anyone have any thoughts on this?

[Sisyphus]

My notation worked as follows: There are two possibilities for the male in each coupling, NN and Nr, with probabilities of 1/3 and 2/3, respectively. The same with females. Thus, by multiplying probabilities together, you get the probability for each specific mating possibility (e.g., 2/9 of all couplings will be Nr male with NN female). Then I took the probabilites of the offspring of each combination, and multiplied them by the probability of the mating possibility to get the portion of the next generation. For example, 1/2 of Nr/NN offspring will be NN, and 1/2 Nr. Thus, 1/9 of the next generation will be made up of NN from a Nr father and a NN mother. I thereby calculate the probability of every single possibility, and then add together identical results to get the total probability.

 

Mokele, are you sure that's still the case if no double-recessive carrier survives to breed? If so, why am I wrong?

[Mokele]

Well, first, the answer I gave doesn't address inbreeding, only selection. The effects of inbreeding are actually very different.

 

There's 5 main factors that influence the gene pool:

1) mutation - adding new genes, good or bad

2) migration - animals leave or enter, taking or bringing in new genes

3) selection - what we just modeled - differential survival or reproduction based on genotype

4) genetic drift - a sort of "random sampling error" caused by small populations

5) inbreeding - mating with close relatives.

 

Sisyphus, we actually got the same answer, it's just expressed differently. I gave the value of p, the total proportion of N alleles in the gene pool, while you gave the value of each of the phenotypes at the first generation. You can convert p to your values by the following method: p is 2/3, and because there's only one other allele (whose frequency is usually called q), p+q=1, and therefore q = 1/3. Now, imagine we just put all the gamates of this population into a big bag, mix well, and pic two at random. The probability of picking two N gamates is p (proportion of N in the gene pool) squared, or 4/9 (what you got). We can get heterozygotes two ways: p then q or q then p, so the probability is 2*p*q, or 4/9 (again, same answer). And recessive homozygotes work like dominants, but this time it's q square, or 1/9th.

 

So we actually got the same answers, just that you figured it out yourself and I used the standard population genetics equations and terms, so it looked different.

 

-----

 

As for inbreeding, that's actually a lot more complicated. The actual effects of inbreeding are that a population will have a disproportionate number of homozygotes (which is why inbred lines of flies or mice are used in lab studies, since they have next to no genetic variation). And selection *can* reduce the number of damaging recessives, thus the frequency in the population, but it runs into a bottleneck: genetic load, as Haldane called it.

 

The principle of genetic load is simple: Stuff isn't just happening at one locus, and in reality, selection, inbreeding, etc affect *all* loci at once. This has lots of effects. First, in a finite population you cannot possibly play out every combination every generation, so traits wind up linked together regardless of selection simply by being on the same animal. Second, selection acts on the animal. If every bad gene only causes a 2% drop in survival rate, you can only realistically have selection on 50 genes.

 

Take your typical inbred animal. It'll be homozygous for bad traits, but homozygous for good ones too, and this will be random with respect to selective properties of the genes. So does the animal live or die? Well, it's got traits going both ways, so it often more or less balances out (though many traits are best in heterozygous form, and thus the net fitness of inbred animals is lower than non-inbred ones). As the inbreeding continues, you have less and less heteozygotes, and, while the net fitness of the population declines, it's more or less the same (just as crappy) for most individuals, thus there's only very limited selection going on. Sure, couple A is more likely to have a kid that's homozygous for damaging mutation X, but couple B is just as likely to have a kid that's homozygous for damaging mutation Y.

 

So basically, under controlled breeding and for animals with one locus (which don't exist), your friend is right, inbreeding lets you flush that trait out faster. However, because inbreeding affects *all* loci, including those you *aren't* flushing out, it'll cause *greater* prevalence of those bad traits. So basically, your friend's misconception is based on the typical way we treat traits separately, without remembering they're part of a whole genome that functions as a unit. Once you take that more realistic approach, you find that any benefits of inbreeding regarding one trait are massively outweighted by the damage done to all other traits in the genome.

 

Mokele

[dr. sinister]

But I don't think inbreeding with even just one locus results in a faster flushing out of the trait. Because by the 3rd generation 37.5% (as Sisyphus has calculated) of the heterozygous recessive trait still exists in the inbred population. While by the 3rd generation in an outbreeding population there is only a 6% (as I've calculated it, I may be incorrect) frequency of the heterozygous trait. Provided that every generation breeds outside of the family.

 

So by that logic, the trait would not be flushed out faster, even when dealing with one locus.

[Mokele]

Sisyphus' and my calculations were for selection, not inbreeding. If you *combine* the two, you technically *could* get a faster response at one locus. Think of it in two steps: 1) inbreeding increases the proportion of homozygotes and decreases the proportion of heterozygotes 2) selection removes those recessive homozygotes. Becuase recessive homozygoes constitute an unusually large proportion of an inbred population relative to an outbred one, more total individuals will die, meaning stronger selection on the inbred population. Also, because the inbred population suffers from a defecit of heterozygotes, the recessive lethal allele cannot "hide" in heterozygotes as effectively as during selection on an outbred population.

 

Of course, while that locus is all neat, the rest of the organism's genome is in such horrible shape that it migth start playing banjo at any minute.

 

Mokele

[dr. sinister]

Right but outbreeding has selection as well and it's more realistic than the selection with he inbred population, since people will always choose to mate with someone outside the family the recessive gene will disperse at a certain frequency through out the population. However, since every individual will choose a mate outside of the family the chance of the recessive allele spreading to successive generations becomes smaller and smaller and eventually by way of probability disappear from the population. So the selection in outbreeding is that each individual finds a mate that is unrelated to them genetically.

 

In that case wouldn't you assume the outbred population to remove the trait faster? Or to a negligible frequency in far less time than an inbred population?

[Mokele]

No, because the smaller the recessive gene frequency in a non-inbred population, the less homozygotes there are, meaning that the genes are "hiding". In contrast, inbreeding leads to homozygosity, so the genes become "exposed" to selection.

 

In any population, the recessive lethal mutations will sometimes be "hidden" from selection in heterozygotes. But in inbred populations, there is an unusually small number of heterozygotes and an unusually large number of both homozygotes for a given gene frequency. Because there are more homozygotes, more individuals carrying the lethal allele die, and less of these alleles are hidden in heterozygote form. This means the frequency of the damaging allele will decrease faster, on account of a greater proportion of those who carry it dying each generation.

 

Mokele

[dr. sinister]

Well that's true the recessive allele can become hidden. But in an non-inbreeding population you get a system of heterozygotes breeding with homozygotes since all the population outside of the family are assumed to be homozygotes. So you would see a similar but quicker result with the outbred population.

[Mokele]

But in an non-inbreeding population you get a system of heterozygotes breeding with homozygotes since all the population outside of the family are assumed to be homozygotes.

 

But if they're not breeding within the family, it's not an inbred system, is it?

[dr. sinister]

But if they're not breeding within the family, it's not an inbred system, is it?

 

Well, I'm talking about the outbreeding population. The inbreeding population would be different.

[Mokele]

Ok, to quickly summarize: You have two populations. Both are under selection, in which there is a lethal recessive allele. In one population, inbreeding occurs.

 

Now, if both populations are of a fixed size, then the inbred population will have more homozygotes of both types (this is actually the definition and how we quantify inbreeding), even though the gene frequency is the same. Because fewer of the recessive lethals are hidden in heterozygotes, more animals die and more copies are removed from the gene pool. Thus th selective pressure on the inbred population is higher.

 

Let's use an example. Both populations have 100 individuals, and 50-50 gene frequencies (which means 200 total genes, 100 of each). The outbred population will have 25 dom homozygotes, 50 heterozygotes, and 25 recessive homozygotes. Those last ones, 25 animals with 50 genes, die. So now you're left with 150 total genes, 100 of the dominant, and 50 of the recessive, which means you've gone from a gen frequency of .5 to .6666 in one generation, a change of .16666.

 

Now, consider a mildly inbred population, which would have 35 dominant homozygotes, 30 heterozygotes, and 35 recessive homozygotes. Note that the initial gene frequency is still 50-50, and that there are 100 animals, 200 genes, and 100 of each allele. Now, those 35 recessive homozygotes die. This leaves us with 130 genes, 100 dominant and only 30 recessive, a new gene frequency of about .77, meaning the change was .27, over half-again as big of a change as with the outbred population.

 

Basically, inbreeding exposes more genes to selection, and therefore allows more rapid loss of the undesirable trait (at the expense of the rest of the genome). That's why it's the #1 choice to top unethical breeders of every species!

 

Mokele

[dr. sinister]

Actually the outbreeding population would not have gene frequencies of,

 

25% 50% 25%

 

Because, the heterozygotes will always mate with a homozygote, heterozygotes will only exist in that one family with the recessive gene.

 

And breeders use inbreeding to express homozygous traits, but not to flush out unwanted traits quicker.

[Mokele]

Actually the outbreeding population would not have gene frequencies of, 25% 50% 25%

 

No, only before selection.

 

Because, the heterozygotes will always mate with a homozygote, heterozygotes will only exist in that one family with the recessive gene.

 

What? This makes no sense. If you're talking about considering only the gene frequencies of one family, then it's pointless unless it's an inbred system because they have nobody else to mate with.

 

In a *population* consisting of many, many individuals of many families (which is where these analyses are actually useful), there is no reason to presuppose that heterozygotes will avoid other heterozygotes because there are plenty of all genotypes from all families.

 

It seems as if you're defining this is a very weird, and not very useful or realistic, way.

 

And breeders use inbreeding to express homozygous traits, but not to flush out unwanted traits quicker.

 

They use it to increase homozygosity for faster selection. Whether it's selection for or again a trait doesn't matter, the point still stands.

 

Mokele