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. Given that hitchhiker's thumb is not a sex linked trait, what is the probable genotypic ratio of their children?


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, let's first analyze the genotypes of the parents:

- The woman does not have hitchhiker's thumb (hh).
- The man is heterozygous for hitchhiker's thumb (Hh).

When they have children, we can set up a Punnett square to analyze the possible genotypes of the offspring.

The possible alleles from each parent are:
- From the woman (hh): h, h
- From the man (Hh): H, h

Now we can fill out the Punnett square:

```
H h
----------------
h | Hh | hh |
----------------
h | Hh | hh |
----------------
```

From the Punnett square, we see the following genotypes for the children:
- Hh (heterozygous) = 2 times
- hh (homozygous recessive) = 2 times

Therefore, the genotypic ratio of the children would be:
- 2 Hh : 2 hh, which simplifies to 1 Hh : 1 hh.

This means that the correct genotypic ratio is **50% Hh: 50% hh**.

So, the correct answer is:

**50% Hh: 50% hh**