Question

the pedigree on the right shows the inheritance pattern for an x - linked recessive disorder. if the female circled in pink married an affected male and had a female child, what is the chance that the child would also be affected? express your answer in a percentage.
%
what is the chance that the child would be a carrier?
%
true or false? the child could be standard.
done

Explanation:

Step1: Determine Genotypes

The circled female is a carrier (XⁿXⁿ, where Xⁿ has the normal allele, Xᵈ has the disorder allele? Wait, no: X - linked recessive, so affected male is XᵈY, carrier female is XⁿXᵈ (wait, no: carrier is XⁿXᵈ? Wait, no: for X - linked recessive, a carrier female has one normal (Xⁿ) and one mutant (Xᵈ) allele: XⁿXᵈ. The affected male is XᵈY.

When they have a female child: the female child gets one X from mother and one X from father.

Mother (carrier): Xⁿ (50%) or Xᵈ (50%)
Father (affected): Xᵈ (100%, since he gives X to daughter)

Step2: Probability of Affected Female Child

A female is affected if she has XᵈXᵈ. So mother gives Xᵈ (50% chance) and father gives Xᵈ (100% chance). So probability = 0.5 * 1 = 0.5? Wait, no: Wait, mother is a carrier: genotype XⁿXᵈ (because in X - linked recessive, carriers are heterozygous, and affected males are XᵈY, unaffected males are XⁿY, affected females are XᵈXᵈ, carriers are XⁿXᵈ, normal females are XⁿXⁿ).

So mother (circled) is a carrier: XⁿXᵈ. Father is affected: XᵈY.

For a female child:
- Mother can pass Xⁿ or Xᵈ (each with 50% chance)
- Father must pass Xᵈ (since he's male, to have a daughter, he gives X; and he has XᵈY, so X is Xᵈ)

So possible genotypes for daughter:
- Xⁿ (from mom) and Xᵈ (from dad): XⁿXᵈ (carrier)
- Xᵈ (from mom) and Xᵈ (from dad): XᵈXᵈ (affected)

So probability of affected (XᵈXᵈ): 50% (since mom has 50% chance to give Xᵈ) → 50%? Wait, no: Wait, mother's X chromosomes: Xⁿ and Xᵈ. So when she has a daughter, she gives one X: either Xⁿ (50%) or Xᵈ (50%). Father gives Xᵈ (100%). So:

- If mom gives Xⁿ: daughter is XⁿXᵈ (carrier)
- If mom gives Xᵈ: daughter is XᵈXᵈ (affected)

So chance of affected: 50%? Wait, no, wait: Wait, the carrier female (XⁿXᵈ) and affected male (XᵈY). Let's do a Punnett square for female offspring (since we're only considering female children, so we look at the X from father (Xᵈ) and X from mother (Xⁿ or Xᵈ)).

Mother's gametes (for X): Xⁿ (50%), Xᵈ (50%)
Father's gametes (for X, since daughter needs X from father): Xᵈ (100%)

So possible female genotypes:
- Xⁿ (mom) + Xᵈ (dad): XⁿXᵈ (carrier)
- Xᵈ (mom) + Xᵈ (dad): XᵈXᵈ (affected)

So probability of affected (XᵈXᵈ): 50% (because mom has 50% chance to give Xᵈ)
Probability of carrier (XⁿXᵈ): 50% (mom gives Xⁿ, 50% chance)

Wait, but wait: the question is "if the female circled in pink married an affected male and had a female child" – so we are only considering female children. So the chance that the female child is affected: 50%? Wait, no, wait: Wait, X - linked recessive: affected male is XᵈY, carrier female is XⁿXᵈ.

When they have a child:

- For a son: gets X from mom (Xⁿ or Xᵈ) and Y from dad. So son's genotype: XⁿY (normal) or XᵈY (affected), each 50%.
- For a daughter: gets X from mom (Xⁿ or Xᵈ) and X from dad (Xᵈ). So daughter's genotype: XⁿXᵈ (carrier) or XᵈXᵈ (affected), each 50%.

So if we are only looking at female children (so we exclude the sons), then among the female children:

- 50% chance to be affected (XᵈXᵈ)
- 50% chance to be carrier (XⁿXᵈ)

Now, the third question: "True or false? The child could be standard." Wait, "standard" probably means normal (not affected, not carrier? No, normal female would be XⁿXⁿ, but the mother is XⁿXᵈ and father is XᵈY. Can the daughter be XⁿXⁿ? No, because father can only give Xᵈ (since he's XᵈY). So the daughter's X from father is Xᵈ, so she must have one Xᵈ. So her genotype is either XⁿXᵈ (carrier) or XᵈXᵈ (affected). So she can't be XⁿXⁿ (normal, standard). So the answer is False.

Wait, let's re - check:

Mother: XⁿXᵈ (carrier)
Father: XᵈY (affected)

Daughter's X from father: Xᵈ (100%)
Daughter's X from mother: Xⁿ (50%) or Xᵈ (50%)

So daughter's genotype:
- XⁿXᵈ (carrier) – 50%
- XᵈXᵈ (affected) – 50%

So no daughter can be XⁿXⁿ (standard, normal), because she gets Xᵈ from father. So the third answer is False.

Step1 (First Question: Affected Female Child):

Probability = (Chance mother gives Xᵈ) * (Chance father gives Xᵈ)
Chance mother gives Xᵈ: 50% = 0.5
Chance father gives Xᵈ: 100% = 1
Probability = 0.5 * 1 = 0.5 = 50%

Step2 (Second Question: Carrier Female Child):

Probability = (Chance mother gives Xⁿ) * (Chance father gives Xᵈ)
Chance mother gives Xⁿ: 50% = 0.5
Chance father gives Xᵈ: 100% = 1
Probability = 0.5 * 1 = 0.5 = 50%

Step3 (Third Question: Standard Child):

As daughter must have Xᵈ from father, she can't be XⁿXⁿ (standard). So answer is False.

Answer:

Affected female child chance: 50%
Carrier female child chance: 50%
Child could be standard: False