Question

here are the two families again, showing the inheritance of the genetic disease you are trying to study. you check back in with the families and find out that veronica is having a child with zayn (it is a big scandal). neither veronica nor zayn has the disease. what is the probability that their child will be homozygous for the dominant allele at the locus you are studying?
1/9
1/6
1/16
4/9
1/4
1/8
2/3
1
1/2
1/3
3/4

Explanation:

Step1: Determine Inheritance Pattern

The pedigree shows affected (black) females with unaffected parents, so the disease is autosomal recessive (aa). Unaffected parents of affected children are carriers (Aa).

Step2: Probability Veronica is Aa

Veronica's parents are carriers (since they have an affected child). Veronica is unaffected, so her genotype is either AA or Aa. Probability she's Aa: Using conditional probability, among non - affected offspring of Aa×Aa (which has genotypes AA:Aa:aa = 1:2:1), the probability of Aa is $\frac{2}{3}$ (since we exclude aa as she's unaffected).

Step3: Probability Zayn is Aa

Same as Veronica, Zayn's parents are carriers (has an affected sister), so Zayn (unaffected) has a $\frac{2}{3}$ chance of being Aa.

Step4: Probability Child is AA

For a child to be AA, both Veronica and Zayn must pass A. But wait, no - wait, actually, first we need to consider the probability that Veronica and Zayn are carriers (Aa) and then the probability of getting AA from their mating. Wait, no, the question is about being homozygous dominant (AA). Let's re - evaluate.

Wait, the disease is autosomal recessive. So normal individuals can be AA or Aa. Let's denote:

Let’s first find the probability that Veronica is AA or Aa, and same for Zayn.

For a family with two carriers (Aa×Aa) having children: the possible genotypes of non - affected children are AA (probability $\frac{1}{3}$) and Aa (probability $\frac{2}{3}$), since the ratio of AA:Aa:aa is 1:2:1, and we exclude aa (affected) for non - affected individuals.

So Veronica: P(AA)=$\frac{1}{3}$, P(Aa)=$\frac{2}{3}$

Zayn: P(AA)=$\frac{1}{3}$, P(Aa)=$\frac{2}{3}$

Now, we want the probability that their child is AA. Let's consider the cases:

Case 1: Veronica is AA (prob $\frac{1}{3}$) and Zayn is AA (prob $\frac{1}{3}$): Child is AA with prob 1. Contribution: $\frac{1}{3}×\frac{1}{3}×1=\frac{1}{9}$

Case 2: Veronica is AA (prob $\frac{1}{3}$) and Zayn is Aa (prob $\frac{2}{3}$): Child is AA with prob $\frac{1}{2}$. Contribution: $\frac{1}{3}×\frac{2}{3}×\frac{1}{2}=\frac{1}{9}$

Case 3: Veronica is Aa (prob $\frac{2}{3}$) and Zayn is AA (prob $\frac{1}{3}$): Child is AA with prob $\frac{1}{2}$. Contribution: $\frac{2}{3}×\frac{1}{3}×\frac{1}{2}=\frac{1}{9}$

Case 4: Veronica is Aa (prob $\frac{2}{3}$) and Zayn is Aa (prob $\frac{2}{3}$): Child is AA with prob $\frac{1}{4}$. Contribution: $\frac{2}{3}×\frac{2}{3}×\frac{1}{4}=\frac{1}{9}$

Now, sum all contributions: $\frac{1}{9}+\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{4}{9}$? Wait, no, that's wrong. Wait, no, actually, the correct way is:

The probability that Veronica is AA is $\frac{1}{3}$, Zayn is AA is $\frac{1}{3}$: child AA: 1.

Veronica AA (1/3) and Zayn Aa (2/3): child AA: 1/2.

Veronica Aa (2/3) and Zayn AA (1/3): child AA: 1/2.

Veronica Aa (2/3) and Zayn Aa (2/3): child AA: 1/4.

But let's calculate the total probability:

P(AA child) = P(Veronica AA)×P(Zayn AA)×P(AA from AA×AA) + P(Veronica AA)×P(Zayn Aa)×P(AA from AA×Aa) + P(Veronica Aa)×P(Zayn AA)×P(AA from Aa×AA) + P(Veronica Aa)×P(Zayn Aa)×P(AA from Aa×Aa)

= ($\frac{1}{3}×\frac{1}{3}×1$) + ($\frac{1}{3}×\frac{2}{3}×\frac{1}{2}$) + ($\frac{2}{3}×\frac{1}{3}×\frac{1}{2}$) + ($\frac{2}{3}×\frac{2}{3}×\frac{1}{4}$)

= $\frac{1}{9}+\frac{1}{9}+\frac{1}{9}+\frac{1}{9}$

= $\frac{4}{9}$

Wait, but let's check again. The key is that both Veronica and Zayn are from families with an affected (aa) sibling, so their parents are Aa×Aa. So the probability that a non - affected child (Veronica or Zayn) is AA is 1/3, Aa is 2/3.

Now, to have a child homozygous dominant (AA), we need to consider the possible genotypes of Veronica and Zayn and the probability of getting AA from their cross.

If Veronica is AA (prob 1/3) and Zayn is AA (prob 1/3): cross AA×AA gives AA (100% chance). So contribution: 1/3 * 1/3 * 1 = 1/9.

If Veronica is AA (1/3) and Zayn is Aa (2/3): cross AA×Aa gives AA (50% chance) and Aa (50% chance). Contribution: 1/3 * 2/3 * 1/2 = 1/9.

If Veronica is Aa (2/3) and Zayn is AA (1/3): cross Aa×AA gives AA (50% chance) and Aa (50% chance). Contribution: 2/3 * 1/3 * 1/2 = 1/9.

If Veronica is Aa (2/3) and Zayn is Aa (2/3): cross Aa×Aa gives AA (25% chance), Aa (50% chance), aa (25% chance). Contribution: 2/3 * 2/3 * 1/4 = 1/9.

Adding these up: 1/9 + 1/9 + 1/9 + 1/9 = 4/9.

Answer:

$\frac{4}{9}$ (corresponding to the option "4/9")