Answer :
To solve the problem, we need to look at two genetic crosses and determine the likelihood of certain genotypes appearing in the offspring.
### Part 1: Heterozygous Male (Ww) x Homozygous Recessive Female (ww)
1. Parent Genotypes:
- Male: Ww (heterozygous)
- Female: ww (homozygous recessive)
2. Punnett Square for the Cross:
```
| | W | w |
|---|----|----|
| w | Ww | ww |
| w | Ww | ww |
```
3. Offspring Genotypes:
- Ww
- ww
4. Probability of Heterozygous Offspring (Ww):
Out of the four possible combinations:
- Two are Ww (heterozygous)
- Two are ww (homozygous recessive)
Therefore, the probability that an offspring is heterozygous (Ww) is 2 out of 4, which is 0.5, or 50%.
### Part 2: Cross Between Heterozygous (WW) and Homozygous Dominant (WW)
1. Parent Genotypes:
- First individual: WW (homozygous dominant)
- Second individual: Ww (heterozygous)
2. Punnett Square for the Cross:
```
| | W | w |
|---|----|----|
| W | WW | Ww |
| W | WW | Ww |
```
3. Offspring Genotypes:
- WW
- Ww
4. Probability of Homozygous Recessive Offspring (ww):
In this case, none of the offspring can be ww because at least one W will always be present in the combination.
Therefore, the probability of having a homozygous recessive (ww) offspring is 0, or 0%.
In summary, the chance of having a heterozygous offspring (Ww) in the first scenario is 50%, and the chance of having a homozygous recessive offspring (ww) in the second scenario is 0%.
### Part 1: Heterozygous Male (Ww) x Homozygous Recessive Female (ww)
1. Parent Genotypes:
- Male: Ww (heterozygous)
- Female: ww (homozygous recessive)
2. Punnett Square for the Cross:
```
| | W | w |
|---|----|----|
| w | Ww | ww |
| w | Ww | ww |
```
3. Offspring Genotypes:
- Ww
- ww
4. Probability of Heterozygous Offspring (Ww):
Out of the four possible combinations:
- Two are Ww (heterozygous)
- Two are ww (homozygous recessive)
Therefore, the probability that an offspring is heterozygous (Ww) is 2 out of 4, which is 0.5, or 50%.
### Part 2: Cross Between Heterozygous (WW) and Homozygous Dominant (WW)
1. Parent Genotypes:
- First individual: WW (homozygous dominant)
- Second individual: Ww (heterozygous)
2. Punnett Square for the Cross:
```
| | W | w |
|---|----|----|
| W | WW | Ww |
| W | WW | Ww |
```
3. Offspring Genotypes:
- WW
- Ww
4. Probability of Homozygous Recessive Offspring (ww):
In this case, none of the offspring can be ww because at least one W will always be present in the combination.
Therefore, the probability of having a homozygous recessive (ww) offspring is 0, or 0%.
In summary, the chance of having a heterozygous offspring (Ww) in the first scenario is 50%, and the chance of having a homozygous recessive offspring (ww) in the second scenario is 0%.