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Answer :
Certainly! Let's solve the problem step-by-step:
1. First Scenario:
- We are crossing a heterozygous male (Ww) with a homozygous recessive female (ww).
- To determine the probability of having heterozygous (Ww) offspring, we create a Punnett square:
```
Parent 1: Ww
Parent 2: ww
| | W | w |
|---|---|---|
| w | Ww| ww|
| w | Ww| ww|
```
- The possible genotypes of the offspring are: Ww, Ww, ww, and ww.
- Out of these four possible outcomes, two are heterozygous (Ww).
- Therefore, the chance that an offspring will be heterozygous (Ww) is 2 out of 4, which is 0.5 or 50%.
2. Second Scenario:
- This time, we cross a heterozygous (Ww) individual with a homozygous dominant (WW) individual.
- Let's construct another Punnett square for this cross:
```
Parent 1: Ww
Parent 2: WW
| | W | w |
|---|---|---|
| W | WW| Ww|
| W | WW| Ww|
```
- The possible genotypes of the offspring are: WW, Ww, WW, and Ww.
- There are no ww combinations in this scenario, so the probability of having a homozygous recessive (ww) offspring is 0 out of 4, which is 0 or 0%.
In summary, the chance of an offspring being heterozygous in the first scenario is 50%, and the probability of having a homozygous recessive offspring in the second scenario is 0%.
1. First Scenario:
- We are crossing a heterozygous male (Ww) with a homozygous recessive female (ww).
- To determine the probability of having heterozygous (Ww) offspring, we create a Punnett square:
```
Parent 1: Ww
Parent 2: ww
| | W | w |
|---|---|---|
| w | Ww| ww|
| w | Ww| ww|
```
- The possible genotypes of the offspring are: Ww, Ww, ww, and ww.
- Out of these four possible outcomes, two are heterozygous (Ww).
- Therefore, the chance that an offspring will be heterozygous (Ww) is 2 out of 4, which is 0.5 or 50%.
2. Second Scenario:
- This time, we cross a heterozygous (Ww) individual with a homozygous dominant (WW) individual.
- Let's construct another Punnett square for this cross:
```
Parent 1: Ww
Parent 2: WW
| | W | w |
|---|---|---|
| W | WW| Ww|
| W | WW| Ww|
```
- The possible genotypes of the offspring are: WW, Ww, WW, and Ww.
- There are no ww combinations in this scenario, so the probability of having a homozygous recessive (ww) offspring is 0 out of 4, which is 0 or 0%.
In summary, the chance of an offspring being heterozygous in the first scenario is 50%, and the probability of having a homozygous recessive offspring in the second scenario is 0%.
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