Question

Mr. Thompson has dimples. Mrs. Thompson does NOT have dimples.

They have three children:

Ella – has dimples
Jacob – has dimples
Mila – does NOT have dimples
When Ella grows up, she marries Ethan, who also has dimples.

They have one child, a son named Kai, who does NOT have dimples.

Jacob (from Part 1) grows up and marries Ava, who does NOT have dimples.

They have two children:

Lily – has dimples
Aria – has dimples


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Answer 1

Assumption: dimples are caused by a single gene with a dominant allele D (dimples) and recessive allele d (no dimples); trait shows full penetrance and is autosomal.

Genotype deductions
- Mr. Thompson (dimples): Dd.
- Reason: his wife has no dimples (dd). They produced a child (Mila) without dimples (dd), so he must have contributed a d allele → Dd (he cannot be DD).
- Mrs. Thompson (no dimples): dd (given).
- Ella (dimples): Dd.
- Reason: child of Dd × dd; cannot be DD, so any dimpled child must be Dd.
- Jacob (dimples): Dd (same reasoning as Ella).
- Mila (no dimples): dd (observed).
- Ethan (dimples, married to Ella): Dd.
- Reason: Ella is Dd and their child Kai is dd, so Ethan must have contributed a d; since Ethan shows dimples, he is Dd.
- Kai (no dimples): dd (observed).
- Ava (no dimples): dd (given).
- Lily (dimples): Dd.
- Reason: daughter of Jacob (Dd) × Ava (dd); cannot be DD, so any dimpled child is Dd.
- Aria (dimples): Dd (same reasoning as Lily).

Summary list
- Mr. Thompson: Dd
- Mrs. Thompson: dd
- Ella: Dd
- Jacob: Dd
- Mila: dd
- Ethan: Dd
- Kai: dd
- Ava: dd
- Lily: Dd
- Aria: Dd

Relevant probabilities (under the assumptions)
- Mr. & Mrs. Thompson (Dd × dd): each child 50% Dd (dimples), 50% dd (no dimples).
- Ella × Ethan (Dd × Dd): children 25% DD, 50% Dd, 25% dd. Kai being dd had probability 25%.
- Jacob × Ava (Dd × dd): each child 50% Dd, 50% dd. The chance both daughters are dimpled is (1/2) × (1/2) = 1/4.

If you want, I can show the Punnett squares for each cross.