Answer :
Certainly! Let's solve the genetic questions step by step.
First scenario: Heterozygous male (Ww) mated with a homozygous recessive female (ww):
1. We have two parent genotypes for this cross:
- Male: Ww (heterozygous)
- Female: ww (homozygous recessive)
2. To find the possible genotypes of the offspring, we set up a Punnett square. The male can pass on either a 'W' or a 'w', and the female can only pass on a 'w' (since she is homozygous recessive).
3. The Punnett square looks like this:
```
| | W | w |
|---|---|---|
| w | Ww | ww |
| w | Ww | ww |
```
4. From this Punnett square, we see the possible offspring:
- Two offspring with genotype Ww (heterozygous)
- Two offspring with genotype ww (homozygous recessive)
5. Therefore, the probability of having a heterozygous Ww offspring is 2 out of 4 possible outcomes, which is 50%.
Second scenario: Heterozygous (W or w) crossed with homozygous dominant (WW):
1. Author's note – There appears to be a typo ("Wh") in the problem, so we assume we work with Ww as it's most logical from the context.
2. Our parent genotypes are:
- One parent: Ww (or W) depending on context interpretation
- Other parent: WW (homozygous dominant)
3. We determine the offspring's genotypes by the Punnett square:
```
| | W | W |
|---|---|---|
| W | WW | WW |
| w | Ww | Ww |
```
4. The resulting offspring genotypes are:
- All offspring are either WW or Ww, as there is no 'w' from the homozygous dominant (WW) parent.
5. Since there are no 'ww' outcomes (homozygous recessive), the probability of having a homozygous recessive offspring is 0%.
To summarize:
- In the first mating scenario, there is a 50% chance of having a heterozygous offspring (Ww).
- In the second scenario, there is a 0% chance of having a homozygous recessive offspring (ww) due to the absence of the recessive allele from the homozygous dominant parent.
First scenario: Heterozygous male (Ww) mated with a homozygous recessive female (ww):
1. We have two parent genotypes for this cross:
- Male: Ww (heterozygous)
- Female: ww (homozygous recessive)
2. To find the possible genotypes of the offspring, we set up a Punnett square. The male can pass on either a 'W' or a 'w', and the female can only pass on a 'w' (since she is homozygous recessive).
3. The Punnett square looks like this:
```
| | W | w |
|---|---|---|
| w | Ww | ww |
| w | Ww | ww |
```
4. From this Punnett square, we see the possible offspring:
- Two offspring with genotype Ww (heterozygous)
- Two offspring with genotype ww (homozygous recessive)
5. Therefore, the probability of having a heterozygous Ww offspring is 2 out of 4 possible outcomes, which is 50%.
Second scenario: Heterozygous (W or w) crossed with homozygous dominant (WW):
1. Author's note – There appears to be a typo ("Wh") in the problem, so we assume we work with Ww as it's most logical from the context.
2. Our parent genotypes are:
- One parent: Ww (or W) depending on context interpretation
- Other parent: WW (homozygous dominant)
3. We determine the offspring's genotypes by the Punnett square:
```
| | W | W |
|---|---|---|
| W | WW | WW |
| w | Ww | Ww |
```
4. The resulting offspring genotypes are:
- All offspring are either WW or Ww, as there is no 'w' from the homozygous dominant (WW) parent.
5. Since there are no 'ww' outcomes (homozygous recessive), the probability of having a homozygous recessive offspring is 0%.
To summarize:
- In the first mating scenario, there is a 50% chance of having a heterozygous offspring (Ww).
- In the second scenario, there is a 0% chance of having a homozygous recessive offspring (ww) due to the absence of the recessive allele from the homozygous dominant parent.
