X-linked Inheritance Basics

X-linked inheritance refers to traits or disorders determined by genes located on the X chromosome. Males have one X chromosome and one Y chromosome (XY), while females have two X chromosomes (XX). Because males have only one X chromosome, they are hemizygous for genes on the X chromosome, meaning they will express the phenotype of any allele on their single X, whether it is dominant or recessive. Females, having two X chromosomes, can be homozygous or heterozygous for X-linked genes. For recessive X-linked disorders, females must have two copies of the mutant allele ($X^mX^m$) to show the disorder, while heterozygous females ($X^mX$) are typically carriers who do not show symptoms but can pass the mutant allele to their children.

Parental Genotypes

In this scenario:

• The father has an X-linked mutation ($X^m$) causing a late-onset lethal disorder. Since it's on his single X chromosome, his genotype is $X^mY$. The late-onset nature means he can reproduce before the disorder manifests lethally.
• The mother is not a carrier and does not have the disorder. Since she is female (XX) and not a carrier, both her X chromosomes carry the normal allele ($X$). Her genotype is $XX$.

So, the parental genotypes are: Father ($X^mY$) and Mother ($XX$).

Offspring Inheritance

Children inherit one sex chromosome from each parent:

• Sons inherit the Y chromosome from the father and one X chromosome from the mother.
• Daughters inherit one X chromosome from the father and one X chromosome from the mother.

Let's determine the possible genotypes of their children:

From the Punnett square (or simple combination):

• All sons will inherit the Y chromosome from the father and an X chromosome from the mother. Since the mother's genotype is $XX$, she can only pass on a normal X ($X$). Therefore, all sons will have the genotype $XY$.
• All daughters will inherit an X chromosome from the father and an X chromosome from the mother. The father's genotype is $X^mY$, so he passes $X^m$ to his daughters. The mother's genotype is $XX$, so she passes $X$ to her daughters. Therefore, all daughters will have the genotype $X^mX$.

Evaluating the Statements

Based on the offspring genotypes ($XY$ for sons, $X^mX$ for daughters), let's evaluate each statement:

1. There is a 50% chance that the son will show the disorder.
   Sons have the genotype $XY$. They do not inherit the mutant allele $X^m$. Therefore, sons will not show the disorder. This statement is incorrect.
2. No children will show the disorder.
   Sons ($XY$) will not show the disorder. Daughters ($X^mX$) are carriers. Whether carriers show the disorder depends on the specific nature of the mutation and the disorder (e.g., if it's strictly recessive and carriers are asymptomatic, or if there are late-onset effects even in carriers). However, compared to statement 4 which is definitively true based on inheritance, this statement makes a claim about the phenotype which may or may not be accurate depending on the disorder's characteristics in carriers. Statement 4 focuses only on carrying the allele.
3. The probability that the parents have a daughter carrying a mutant allele is 25%.
   The probability of having a daughter is 50% (1/2). As shown, all daughters will have the genotype $X^mX$, meaning they will carry the mutant allele. The probability of having a daughter who carries the mutant allele is P(daughter) * P(carries | daughter) = 0.5 * 1 = 0.5 or 50%. This statement is incorrect.
4. All daughters will carry the mutant allele.
   As determined from the parental genotypes and inheritance pattern, all daughters will receive the $X^m$ chromosome from their father and the $X$ chromosome from their mother, resulting in the genotype $X^mX$. Thus, all daughters will carry the mutant allele. This statement is correct.

Therefore, the correct statement is that all daughters will carry the mutant allele.