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I'm trying to understand, in a simplistic representation of a recessive genetic disease (e.g. Tay Sachs), why wouldn't it "disappear" over time (many generations). Please excuse the use of non-professional terminology below, I hope the point I'm trying to make will be clear.

Let's assume that, for simplicity, people marry at random and each couple has 4 children. Denote the probability of carrying the "bad" gene by p, then a couple of 2 non-carrying people will have 4 healthy children, a mixed couple will have, on average, 2 healthy and 2 carrying children, and a couple of 2 carrying people will have 1 healthy, 2 carrying, and one sick child (which we assume will not have children so we ignore it in further calculations).

Assuming large enough population and random couples and everyone gets married (for simplicity): The probability of a couple of non-carrying: (1-p)^2 The probability of a mixed couple: 2 * p * (1-p) The probability of a 2-carrying couple: p^2

The expected number of healthy children per couple: 4*(1-p)^2 + 2*2*p*(1-p) + p^2 = 4 - 8*p + 4*p^2 + 4*p - 4*p^2 + p^2= 4 - 4*p - p^2 = (2-p)^2 The expected number of carrying: 2*p*(1-p) + 2*p^2 = 2*p - 2*p^2 + 2*p^2 = 2p

Dividing to get the new probability for carrying the gene: 2p / (2-p)^2 Which for small enough p is clearly smaller than p (e.g. for p = 0.05 we get ~0.0263).

If I'm not mistaken this means that the number of those carrying the genes should become exponentially small (in the number of generations).

I realize this is a very simplistic model, marriages are not random, mutations occur etc. but I would expect the basic intuition to hold. What am I missing?

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    $\begingroup$ I haven't read the entirety of the question, but the salient point is that lethal genetic diseases are mostly very rare. Exponential decrease over time and 0 are not the same thing. $\endgroup$ – James Feb 7 '17 at 3:47
  • $\begingroup$ Technically you are correct but if my calculations hold, and we start with p=0.05, we'll be at p=10^-16 within 10 generations which is pretty much the same as 0 $\endgroup$ – nickb Feb 7 '17 at 4:21
  • $\begingroup$ And yet we're not. There may be an error in the assumptions or the calculations. It would be a hard model to get to a good standard. $\endgroup$ – James Feb 7 '17 at 4:24
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Firstly, sorry if this is not so much an answer as a comment, but it is too long to put in the comments section, although I'll try keep it concise...

I can see exactly what you're doing however my math differs at points, and would appreciate you looking over it.

"The expected number of healthy children per couple: 4*(1-p)^2 + 2*2*p*(1-p) + p^2 = 4 - 8*p + 4*p^2 + 4*p - 4*p^2 + p^2= 4 - 4*p - p^2 = (2-p)^2"

Is it not: ... = 4 - 8p + 4p^2 + 4p - 4p^2 + p^2 = 4 - 4p + p^2 = (p-2)^2

Furthermore: "The expected number of carrying: 2*p*(1-p) + 2*p^2 = 2*p - 2*p^2 + 2*p^2 = 2p" - why for the probability of carriers in mixed couples - "2*p*(1-p)" - why have you not multiplied through by two (as with expected healthy children), given that as you said "a mixed couple will have, on average, 2 healthy and 2 carrying children"?

Is it not: 2 *2p(1-p) + 2p^2 = 4p - 4p^2 + 2p^2 = 4p - 2p^2 = 2p(-p + 2)

In this respect I got different expectations. Lastly "Dividing to get the new probability for carrying the gene..." You have done (no. healthy) / (no. carriers). Is the probability for carrying the gene not (no. carriers)/(no. carriers + no. health) - that is the no. carriers as a fraction of the entire population to give a probability?

Using my expectations this would be: [2p(-p+2)] / [2p(-p+2) + (p-2)^2] = [2p(-p+2)] / [-2p^2 + 4p + p^2 - 4p +4] = [2p(-p+2)] / [-p^2 + 4] = [2p(-p+2)] / [(-p + 2)(p + 2)] = 2p / (p+2)

Therefore, with a new probability for carrying of 2p / (p+2), for our "hypothesis" of the diminishing probability of carrying to stand, we need the condition 2p / (p+2) < p to stand, solving for p:

2p < p(p+2) -> 2p < p^2 + 2p -> 0 < p^2 -> p > 0

Since, as a probability, p must be <= 1, and assuming (as we are) that carriers exist and so it must be > 0, p always satisfies this condition - ergo I would conclude that for any value of p - not just small, the probability of carrying will diminish in the next generation. Just as we "hypothesised". However a quick google of "plot y = 2x/(x+2)" would indicate to me that as p approaches 0 it's rate of "diminishing" actually decreases, and it does not get exponentially smaller as you indicated. But solving for y = 0 gives x = 0 so this model would suggest the gene is never actually fully eradicated?! Which of course confuses me, although on a macro scale it might make sense, as you can't have, for example, 0.1 children "carrying", and 99.9 "healthy", only 1 and 99 or 0 and 100.

Thanks for reading this far, any response would be really appreciated, especially given I might be miles off.

Philip.

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  • $\begingroup$ Philip - Thanks a lot! you caught 2 errors (the first of 3 is just a typo replacing + and - and we arrive at the same result). So interestingly for every generation the new probability is 2/(p+2) the previous probability, and the factor indeed gets closer to 1 as p goes to 0. In practice trying to start with p=0.05 (one of every 20 people carrying) I get after 1,000,000 generations that still 2 of every 1,000,000 people are carrying, so even after a very very long time there genes will still be around. Awesome! $\endgroup$ – nickb Mar 2 '17 at 19:51
  • $\begingroup$ New probability is 2p/(p+2) rather than 2/(p+2), and running a quick simulation in Python I get probability falling from 0.2 to 0.000002 after 1,999,960 generations, but yes I agree with your conclusion that they will therefore be around for a long time- which begs the question how did they ever become widespread in the first place?... $\endgroup$ – Philip Hartfield Mar 4 '17 at 11:54

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