Answer :
Let's solve this question step by step.
### First Scenario
We have a heterozygous male with the genotype Ww mating with a homozygous recessive female with the genotype ww.
To determine the probability of the offspring being heterozygous (Ww), we can use a Punnett square:
1. Set up the Punnett square:
- Male gametes (Ww): W and w
- Female gametes (ww): w and w
```
Female
w w
-------------------
W | Ww | Ww |
w | ww | ww |
```
2. Fill the Punnett square:
- Top left box: Ww (from W and w)
- Top right box: Ww (from W and w)
- Bottom left box: ww (from w and w)
- Bottom right box: ww (from w and w)
3. Probability Calculation:
- There are two Ww combinations out of four total potential combinations.
- Thus, the probability that an offspring will be heterozygous (Ww) is 2 out of 4, which simplifies to 0.5 (or 50%).
### Second Scenario
Now, let's consider a heterozygous individual (Ww) crossed with a homozygous dominant individual (WW). We need to find the probability of having a homozygous recessive (ww) offspring.
1. Set up the Punnett square:
- Heterozygous individual (Ww): W and w
- Homozygous dominant individual (WW): W and W
```
Female
W W
-------------------
W | WW | WW |
w | Ww | Ww |
```
2. Fill the Punnett square:
- Top left box: WW (from W and W)
- Top right box: WW (from W and W)
- Bottom left box: Ww (from w and W)
- Bottom right box: Ww (from w and W)
3. Probability Calculation:
- In this setup, there are no boxes showing the genotype ww.
- Therefore, the probability of having a homozygous recessive (ww) offspring is 0 out of 4, which is 0%.
### Conclusion
- The chance of the offspring being heterozygous (Ww) in the first scenario is 0.5 (or 50%).
- The probability of having a homozygous recessive (ww) offspring in the second scenario is 0%.
### First Scenario
We have a heterozygous male with the genotype Ww mating with a homozygous recessive female with the genotype ww.
To determine the probability of the offspring being heterozygous (Ww), we can use a Punnett square:
1. Set up the Punnett square:
- Male gametes (Ww): W and w
- Female gametes (ww): w and w
```
Female
w w
-------------------
W | Ww | Ww |
w | ww | ww |
```
2. Fill the Punnett square:
- Top left box: Ww (from W and w)
- Top right box: Ww (from W and w)
- Bottom left box: ww (from w and w)
- Bottom right box: ww (from w and w)
3. Probability Calculation:
- There are two Ww combinations out of four total potential combinations.
- Thus, the probability that an offspring will be heterozygous (Ww) is 2 out of 4, which simplifies to 0.5 (or 50%).
### Second Scenario
Now, let's consider a heterozygous individual (Ww) crossed with a homozygous dominant individual (WW). We need to find the probability of having a homozygous recessive (ww) offspring.
1. Set up the Punnett square:
- Heterozygous individual (Ww): W and w
- Homozygous dominant individual (WW): W and W
```
Female
W W
-------------------
W | WW | WW |
w | Ww | Ww |
```
2. Fill the Punnett square:
- Top left box: WW (from W and W)
- Top right box: WW (from W and W)
- Bottom left box: Ww (from w and W)
- Bottom right box: Ww (from w and W)
3. Probability Calculation:
- In this setup, there are no boxes showing the genotype ww.
- Therefore, the probability of having a homozygous recessive (ww) offspring is 0 out of 4, which is 0%.
### Conclusion
- The chance of the offspring being heterozygous (Ww) in the first scenario is 0.5 (or 50%).
- The probability of having a homozygous recessive (ww) offspring in the second scenario is 0%.
