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Comment by 6ren | Hacker News Reader

The last section on sex differences is interesting. It explains boys having greater variation in ability than girls by boys having only one X chromosome (XY) while girls have two (XX).

This would be a neat theory, if girls somehow used an average of the two X's... which seems compellingly logical, though the (current) theory is that only one X is used, chosen at random. http://en.wikipedia.org/wiki/Barr_body

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I don't understand how that fits with the existence of X-linked-recessive diseases. (https://en.wikipedia.org/wiki/Genetic_disorder#X-linked_rece...)

These seem to be well-known cases in which you really do get a phenotype that's a function of both X chromosomes.

You're right, it doesn't fit. Now I recall that rates of red-green colour blindness are exactly predicted by whether one or both must be defective (7% for boys, .49% for girls). Both being active but the defective version having no effect would explain the evidence, but doesn't fit the Barr body theory... it would need to know which one to choose (and it wouldn't be "random").

X-chromosome inactivation is actually consistent with X-linked "recessive" conditions, but it's a little tricky, and the wikipedia page doesn't really explain it. Basically, X-linked phenotypes aren't dominant/recessive in the same way as the others. Usually, dominance takes place between chromosome pairs within each cell. However, with X-chromosome inactivation, dominance takes place between cells. For example, women who are labeled "carriers" for colorblindness actually are colorblind in half of the cells in their eyes, but the other half are sufficient to perceive color almost as well as non-carriers.

Hrm, if you're a girl, you get 1 X from Mom and 1 from Dad. So, if Mom's X's are chromosomes a and b, and Dad's is chromosome c, then your X chromosomes will be a,c or b,c with a .5 prob for each. With your suggestion, there's a .5 prob in each case that Dad's X will be used, and so a .5 prob overall that, for a girl, Dad's X will be used.

A guy has to use Mom's X. I think the paper's argument still holds, because there's a larger amount of possibilities for women than men, but I can't take this much further without circular reasoning.

Oh, for the women there's a .25 prob that a is used, and a .25 prob that b is used, and a .5 prob that c is used. Dad's X has the best chance, and his X may have made him a better partner than his peers. Also, a .25 prob for the X that made Mom a better partner. So, a .75 prob overall of getting a great X and a .25 prob of getting an unknown X.

For the dude there's a .5 prob of getting a great X and a .5 prob of getting an unknown X.

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