Phallic Question

[Sakkura]

When it comes to X-linked genetic disorders that are often or always fatal at early ages, inbreeding is not much of an issue, since the males are culled. When it comes to autosomal recessive genetic disorders though, inbreeding is highly relevant. So go write a book about the "Austrian lip" disorder that struck the house of Habsburg.

You may want to check this out:
http://en.wikipedia.org/wiki/Haemoph...ropean_royalty

All the inheritance of the hemophilia A allelle in the European royal families was through the women. Most of the male children who received the allelle died at early ages (though some did indeed survive into adulthood).

The Russian royal family inherited the disorder through Princess Alix of Hesse and by Rhine. Thus inbreeding was never involved.

[Rodion Romanovich]

Interesting!

But a mathematical model I made in the last 40 minutes claims you were wrong when you said that inbreeding can only increase the number of homozygotes and never increases the number of heterozygotes as well. I made the following model (please state if any of the biological assumptions are incorrect, but I've gone through the maths once so it's hopefully correct):

Biological assumptions:
The probability of having a recessive allele for disease x is p for the entire population. Since all families aren't exactly equal, some families have higher probabilities for one disease, and lower for another, than have the rest of the population. It's more likely that members of the same family have high probability for the same disease, since they share common ancestors. So within most families, we can find a disease D such that a member of the family has higher probability p of having a recessive allele for D, than people from other families, who have it with probability q. So p > q, but both are smaller than 1 and larger than 0.

It follows that:
(assuming p and q are independent)
Probability of homozygote disease child for inbreeding:
p*p
Probability of either homozygote or heterozygote disease child for inbreeding:
1 - (1-p)*(1-p)
Probability of heterozygote disease child for inbreeding:
1 - (1-p)*(1-p) - p*p

Probability of homozygote disease child without doing inbreeding:
q*p
Probability of either homozygote or heterozygote disease child without doing inbreeding:
1 - (1-q)*(1-p)
Probability of heterozygote disease child without doing inbreeding:
1 - (1-q)*(1-p) - p*q

My claims, along with their proofs:

1. Probability of getting homozygote disease child with inbreeding is thus larger than without inbreeding because:
p > q => p*p > q*q
2. Probability of getting homozygote or heterozygote disease child with inbreeding is also larger than without inbreeding because:
p > q => 1-p < 1-q => 1 - (1-p)*(1-p) > 1 - (1-q)*(1-p)
3. Finally, there are many cases where probability of getting heterozygote also increases with inbreeding compared to no inbreeding, namely, exactly when the following inequality holds:[/B]
1 - (1-p)*(1-p) - p*p > 1 - (1-p)*(1-q) - p*q
Let's look at when the inequality holds:
(1-p)*(1-q) is always greater than (1-p)*(1-p) because p > q by the initial assumption. But it is always the case that p*p > q*p by the same assumption. The inequality holds when the first effect outweighs the second. There are two unknown variables with only one equation, so there no unique solution, but an infinite amount of solutions.

Example:

Since I'm too lazy to investigate the entire solution space, I'll just test a few standard cases, using data from existing diseases:

Say a disease whose average probability over the population (i.e. q in the model above) is 0.01. This is a pretty high probability compared to most genetic diseases, and it can be seen that picking a less common disease would make things worse for your case. Inserting this gives (replacing the > with a = to test where the crossover point lies, when inbreeding becomes worse than non-inbreeding):
1 - (1-p)*(1-p) - p^2 = 1 - (1-p)*(1-0.01) - p*0.01
1 - (1-2*p+p^2) - p^2 = 1 - (1-p-0.01+0.01*p) - p*0.01
1 - 1+2*p-p^2 - p^2 = 1 - 1+p+0.01-0.01*p - p*0.01
p^2 - 0.51*p + 0.005 = 0
This has two solutions:
p = 0.51/2 +- sqrt(0.51^2/4 - 0.005)
p = 0.255 +- 0.245
p1 = 0.5, p2 = 0.01

Conclusion for this example:
So, it turns out that as long as not every family member has as much as 50% chance of this disease, choosing someone with p instead of q probability of the disease (as in inbreeding) will indeed increase the number heterozygotes too, and not just the number of homozygotes.

This is still a highly simplified model (but yours, taking only 1 couple into account, was even more simplified [note: I also take only one couple into account, but by using probabilities simulate doing the same calculation over a large number of couples, and comparing the results for inbred and non-inbred]), but I still think it demonstrates and argues quite clearly why inbreeding indeed causes more heterozygote disease carriers as well. Whereas non-inbreeding behavior decreases the probability of with high probability having one recessive disease allele. The average expected value for P(having a child with some disease) is thus much lower for a non-inbreeding population.

[Sakkura]

@Rodion: Look, I really cba to dig through a long calculation to either find the trivial error or the trivial incorrect assumption. All I will do is cite a basic genetics textbook (here's the website apparently connected to it: http://www.whfreeman.com/iga/):

For both positive assortative mating and inbreeding, the consequence to population structure is the same: there is an increase in homozygosity above the level predicted by the Hardy-Weinberg equilibrium

The sum of probabilities must always be 1, so with increased homozygosity comes decreased heterozygosity (unless/until allelle frequencies change).

[Rodion Romanovich]

@Rodion: Look, I really cba to dig through a long calculation to either find the trivial error or the trivial incorrect assumption. All I will do is cite a basic genetics textbook (here's the website apparently connected to it: http://www.whfreeman.com/iga/):

Your quote doesn't say heterozygote with disease can't increase, it says there's an increase in homozygote in general, which means homozygote without disease can decrease to give room for the increase of both the others. Thus your quote doesn't contradict my post, in fact it just repeats one of the 3 claims I made and proved under the assumption that my biological assumptions hold. Besides, the calculations are done by trivial high school math and should be possible to go through in less than 5 minutes.

The sum of probabilities must always be 1, so with increased homozygosity comes decreased heterozygosity (unless/until allelle frequencies change).

If homozygote for disease increases, heterozygote for disease can increase as well, at the cost of homozygote for no disease decreasing. You're forgetting that we have to look different upon homozygote with disease and homozygote without disease.

Besides, if inbreeding would pay off as you claim, there would be huge consequences to what behavior would be most beneficial, and thus to how humans, and our nearest relatives, would behave. These behaviors are however not observed in our nearest relatives, nor are they very common among human beings:
- it would be common with inbreeding relationships in most animals
- you would be more turned on by your mother and sisters than by non-family members
- only a few in each generation would take part in reproduction
- murdering other individuals in the herd would be nearly harmless, and thus murderous behavior would not be uncommon
- killing the weakest 80% in each generation would be done on a regular basis and would be beneficial for the herd
...basically all the ideologies of the sickest madmen in history would be equal to what would be the most beneficial behavior. So before you claim my argument is wrong, consider that if what you claim is true is true, then you would have to explain why bonobos, common chimps and most humans DON'T exhibit these properties in their behavior, except the occasional human power-hungry sick maniac like Hitler and Stalin (but most scientists agree they became maniacs because of environment rather than genes, for instance both had alcoholic father who beat them as kids, and both grew up during politically unstable times).

[SpawnOfEbil]

Autosomal recessive is not the same as X-linked.

X-linked means that every man who has a faulty X-chromosome has the disease, while females need to have both faulty X chromosomes to have the disease.

Affected males are likely to be too sick to be able to pass on their faulty genes (you missed this out).

[sgsandor]

@ All those who helped answer the questions I apperciate all your help in helping me find the answers to my questions.

@ Any Mod close this thing already, it seems like bad news and I hate to think that I brang that to this community.

[mighty_rome]

It's not your fault sgsandor, your questions were perfectly reasonable. Other people (well, pretty much just one) have caused this thread to get out of hand.

I think it should be closed and deleted, there have been quite a few ugly comments made that are very disrespectful to Jews, Muslims, and many others who read this forum, and we don't need any of this.