A few days ago I had an MCQ on a test that read:-

Females have _____ % more probability of inheriting an X-Linked Dominant disorder.

A) 50

B) 25

C) 17

D) 1

The answer in the key was C: 17%.

I think they solved the question like this:-

Females could have three genotypes,i.e. $X^D$$X^D$, $X^D$$X^d$, $X^d$$X^d$ Each genotype has a 1/3 probability. Diseased females have a 2/3 probability (66.6666% chance).

Similary males have two genotypes, i.e. $X^D$Y,$X^d$Y. Each has 1/2 probability and therefore 50% chance.

And therefore 66.666%-50%= 16.666% , which is approx. 17%.

But that shouldn't be how it works. I mean, how did they assume that females have an equal chance of inheriting the three genotypes mentioned above.

So what I did was that I took all the possible combinations of parents and found out their offsprings. There were six different combinations of parents.e.g. $X^D$$X^D$x$X^D$Y ; $X^D$$X^d$x$X^D$Y ,etc.

There were a total of 24 offsprings(4 from each pair of parents). 12 were males and 12 were females.

Half of the males(6) had the disease and half didn't. And therefore males had a 50% chance of the inheriting the disease.

9/12 of the females had the disease and 3/12 were normal. Therefore females have a 75% chance of getting the disease.

That means females have 25% more chance of inheriting the disease.

But what should be the correct answer to this question? Am I missing something?

  • 2
    $\begingroup$ This seems like a crappy question if you were given no more information. Men have a 0% chance of inheriting an X-linked allele from their fathers. Women have a 50/50. Both sexes have a 50/50 chance of inheriting a particular allele from their mother. Yes, you are right, you generally do not do a probability problem by listing all the possibilities and presuming they are all equally likely, since that often is not true. $\endgroup$
    – swbarnes2
    Commented Oct 3, 2022 at 18:43

1 Answer 1


I think that the error you're making is that you're assuming that your 12 hypothetical sets of parents are an accurate representation of the population.

You are however correct that the genotypes are not all equally likely, so I don't agree that 17% is right either.

If you assume that that particular set of parents is true, then you've changed the problem such that you can compute a different probability. But you're asking a different question than the prompt you've been given (assuming that no additional information is given, e.g. the parental genotypes or allele frequencies).

If we instead take the actual pool of alleles and sample from them a la Hardy-Weinberg to make up the genotypes, then you instead have a differently posed question. You can't assume that the dominant and recessive alleles are in equal frequency.

I can't claim to understand the question, and I don't agree with the answer you've indicated (17%) for this same reason. XX people have two opportunities to get the dominant allele, vs. only one for males.

Let's call the frequency of the allele in the population $p$. XY people sample one X chromosome, so their probability is simply $p$.

XX people sample two X chromosomes, so their probability of receiving a dominant allele is $p$ for the first X, plus $(1-p) * p$ for the probability of receiving the allele on the second X (conditioning on not receiving it in the first X, thus the $1-p$).

This yields the ratio

$\frac{(1-p)*p + p}{p}$

for the answer to your question.

For high values of $p$ (1), XX people converge to the same probability as XY people, because there are no recessive alleles:

$\frac{(1-1)*1 + 1}{1} = 1$

For low values of $p$, say 0.01, this ratio is close to 2:1 (100% higher).

In your example, where I think implicitly $p=0.5$, we do indeed observe

$\frac{(1-0.5)*0.5 + 0.5}{0.5} = \frac{0.75}{0.5} = 1.5$,

or a 50% higher odds ratio for XX over XY (or alternately $75\%-50\% = 25\%$ higher, as you write). But that is only true for this specific case of $p=0.5$.

There is a value of $p$ in there where 17% will be the excess probability of women vs. men, but it is actually assuming a different $p$. This would be, similar to what you say:

$\frac{(1-p)*p + p}{p} = \frac{2/3}{1/2} \approx \frac{0.67 + 0.67 * (1-0.67)}{.67}$, so they seem to be assuming $p=0.67$ (two in three alleles are the dominant allele in the population).

Maybe I've made a basic math error somewhere here, but I think that the difference between XX and XY probability of inheriting the dominant allele has to depend on $p$ (dominant allele frequency).

  • $\begingroup$ Yes, you are right about the problem in the solution I proposed. I also noticed that just after I had written it down. I think that since we are at a lack of information we have to assume the most ideal conditions. If we assume that the ratio of the dominant to recessive allele in the population is 1, and that the ratio of no. of males to females in the population is 1 too. And that the ratio of homozygous females to heterozygous females is one too. $\endgroup$ Commented Oct 4, 2022 at 3:50
  • $\begingroup$ Since females carry two X chromosomes while males carry one, it becomes evident that the 2/3 of the dominant alleles and 2/3 of the recessive alleles are carried by females, and the rest, i.e. 1/3 of each by males. Now since homozygous to heterozygous ratio for females is one, 2/4 females are heterozygous, 1/4 are homozygous dominant and 1/4 are homozygous recessive. Therefore 3/4 (75%)have the disease. While in males, the chance is obv. 50%. Now, females have a 50% higher chance than that of males. Conversely males a have a 33.333% lower chance than that of females. $\endgroup$ Commented Oct 4, 2022 at 3:57
  • $\begingroup$ @MuhammadFahadIqbalKhan yes, this case agrees with both the result for $p=0.5$ that I give and your result given in the original question (if we ignore issues like whether we take the difference or the ratio between XX and XY). What I think I disagree with is whether the $p=0.5$ case is relevant on its own (or as you put it "most ideal"). $\endgroup$ Commented Oct 4, 2022 at 17:34

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