Hypophosphatemia is an X-linked dominant disorder that causes a type of rickets.

1)
It is an X-liked dominant trait. Let's symbolize XR the dominant (responsible for the disease) and Xr the recessive normal allele.

Males can be either XRY (with the disorder) or XrY (without the disorder)
Females can be XRXR, XRXr (both of these are with the disorder since the allele is dominant), or XrXr without the disorder.

As we said earlier, affected males are XRY. These males passes the XR to all their daughters (obviously, the Y allele cannot be passed to a daughter). So, all daughters will have at least one XR, and they will have the disorder. So, a is correct.

Let's see why the others are wrong:

b) heterozygous females are  XRXr. As we know, sons get the Y allele from their father and the X from their mother. The mother in this case can pass either the XR allele or the Xr allele. So, the sons only have have 50% possibility of inheriting the disease (and be XRY)

c) If the father is normal (XrY, we can assume that since no information is given for the father) then we will have the crossing:
XrY x XRXr
So the daughters will get a normal Xr allele, and have 50% of inheriting from their mother a XR and 50% a Xr. So, there is still the possibility for the daughter to be XrXr and be healthy.

d) Males (XRY) are always passing the Y allele to their sons, so it only depends on the mother whether the son will be healthy or not. So, this is wrong too.

I forgot to mention earlier, an X-linked allele is present on the X chromosome and not on the Y chromosome. So, females have 2 copies of an X-linked allele (XX) while males have only one copy of an X-linked allele XY (Y chromosome doesn't contain the allele)

That's why females can be heterozygous (e.g. XRXr) or homozygous (e.g. XRXR or XrXr). But males can only have one copy of the allele (e.g. XRY or XrY)

Because of that, if a male have one recessive allele (XrY), it is enough to express the recessive trait, since there isn't any dominant allele that can cover the expression of the recessive gene. And, of course, one dominant allele (XRY) expresses the dominant trait as well.


So,
2) XR is dominant, and Xr is recessive.

So, males with red eyes are XRY
Females with white eyes are XrXr

The crossing will be:
XRY x XrXr

Punnett square:
        XR      Y
Xr    XRXr    XrY

So, we have 50% XRXr which are females with red eyes (since they have a dominant allele)
and we have 50% XrY which are males with white eyes (since their sole allele is recessive, the recessive trait is expressed.



Also, other time please use a more appropriate title for your questions, that is relevant to the content of the question.

Thank you so much! Makes a lot more sense when you explained it to me.
I will! this was my first question ever asked wasn't sure how to put the question.

So... which letter is correct?

Post Merge: 9 years ago

I'm thinking A?