Question

. Hitchhiker's thumb (H) is dominant to no hitchhiker's thumb (h). A woman who does not have hitchhiker's thumb marries a man who is heterozygous for hitchhiker's thumb. What is the probable genotypic ratio of their children? (SC.912.L.16.1)

*
1 point
0% Hh: 100% hh
50% Hh: 50% hh
75% Hh: 25% hh
100% Hh: 0% hh

Answer 1

To determine the probable genotypic ratio of the children of a woman who does not have hitchhiker's thumb (hh) and a man who is heterozygous for hitchhiker's thumb (Hh), we can use a simple Punnett square.

The woman (hh) can only provide the allele "h", and the man (Hh) can provide either "H" or "h".

Here’s the Punnett square:

```
H h
__________
h | Hh | hh |
__________
h | Hh | hh |
__________
```

From this Punnett square, we can see the possible genotypes of their children:

- **Hh** (heterozygous for hitchhiker's thumb) occurs 2 times.
- **hh** (homozygous for no hitchhiker's thumb) occurs 2 times.

So, the ratio of genotypes is:

- 2 Hh : 2 hh, which simplifies to 1 Hh : 1 hh.

Thus, the genotypic ratio of their children is:

**50% Hh : 50% hh**

So the correct answer is:
**50% Hh: 50% hh**.