It said it was fatal to males in utero, and the question asked about live born offspring. Since the males aren’t being born in the first place, I said 50% females and 0% males.

Pedigree = XLD (not all generations affected = X-linked, affects males and females similarly = dominant).

Affected fathers = 100% transmission to daughters, 0% transmission to sons.
Affected mothers = 50% to daughters, 50% transmission to sons. Both parents affected = 100% transmission to daughters (due to father's X chromosome), 50% transmission to sons (due to mother's X chromosome). Both parents affected each transmitting both to daughter (homozygous daughter) = 50% and more severe.

If condition is uniformly fatal to males in utero, then the 50% affected based on transmision (as above) will die in utero, and the 50% not affected will have live births. This means, risk of female being affected = 50% and risk of live-born males being affected = 0%.

https://en.wikipedia.org/wiki/Incontinentia_pigmenti

I knew it had to be X-linked because it said that it was fatal to males who have it in utero, but I had a hard time deciding between dominant and recessive. Ultimately it has to be X-linked dominant because the affected mother in the second generation gave birth to unaffected sons (which can't happen in X-linked recessive). That means the mother in question is heterozygous and her daughters will have a 50% chance of inheriting the disease while her sons have to be unaffected if they live, as previous posters mentioned. I didn't figure this out until way after the test...this one was a doozy. Still not sure that it's XLD because how come nobody in the first gen has it? Guessing it was a spontaneous mutation?

Honest question: Why does it matter what pattern of inheritance it has? For all we know, its multi-factorial/can't be determined.

I feel like the easiest way to answer it is: acknowledging it is "uniformly fatal to males in-utero" = 0% chance for males; and legit just counting the fraction of females in generation III that have the disease = 4/6 ~50%.

Here, the mother is homozygous for the mutation therefore she displays the disease.

We see that,

1. It skips generations
2. Males are more severely affected (die in utero)
3. Females are affected (only when homozygous)

Here we see, that mother can pass defective X chromosome to 50% females & other 50% will be carriers (healthy X chromosome from father, defective from mother) For males, Mother passes defective X chromosome so they die (only 1 X chromosome)

For the males who are living (III 6,11 shown in pedigree), they have been lucky to survive due to mosaicism! ;)