Answer :
Let's solve the question step-by-step:
1. First Cross: Heterozygous Male (Ww) x Homozygous Recessive Female (ww)
In this genetic cross, we have one parent with the genotype Ww and the other with ww. We can use a Punnett square to determine the probability of different offspring genotypes.
| | W | w |
|------|-----|-----|
| w | Ww | ww |
| w | Ww | ww |
From the Punnett square:
- Two of the four possible offspring (Ww) are heterozygous.
- Two of the four possible offspring (ww) are homozygous recessive.
Therefore, the chance that the offspring will be heterozygous (Ww) is 2 out of 4, which simplifies to a probability of 0.5 or 50%.
2. Second Cross: Heterozygous (Ww) x Homozygous Dominant (WW)
In this cross, we have one parent with the genotype Ww and the other with WW. We use a Punnett square again:
| | W | w |
|------|-----|-----|
| W | WW | Ww |
| W | WW | Ww |
From the Punnett square:
- Two of the four possible offspring (WW) are homozygous dominant.
- Two of the four possible offspring (Ww) are heterozygous.
There are no homozygous recessive offspring (ww) in this case.
Therefore, the probability of having a homozygous recessive offspring (ww) is 0 or 0%.
In summary:
- For the first cross (Ww x ww), the chance of having heterozygous offspring is 0.5 or 50%.
- For the second cross (Ww x WW), the probability of having homozygous recessive offspring is 0.
1. First Cross: Heterozygous Male (Ww) x Homozygous Recessive Female (ww)
In this genetic cross, we have one parent with the genotype Ww and the other with ww. We can use a Punnett square to determine the probability of different offspring genotypes.
| | W | w |
|------|-----|-----|
| w | Ww | ww |
| w | Ww | ww |
From the Punnett square:
- Two of the four possible offspring (Ww) are heterozygous.
- Two of the four possible offspring (ww) are homozygous recessive.
Therefore, the chance that the offspring will be heterozygous (Ww) is 2 out of 4, which simplifies to a probability of 0.5 or 50%.
2. Second Cross: Heterozygous (Ww) x Homozygous Dominant (WW)
In this cross, we have one parent with the genotype Ww and the other with WW. We use a Punnett square again:
| | W | w |
|------|-----|-----|
| W | WW | Ww |
| W | WW | Ww |
From the Punnett square:
- Two of the four possible offspring (WW) are homozygous dominant.
- Two of the four possible offspring (Ww) are heterozygous.
There are no homozygous recessive offspring (ww) in this case.
Therefore, the probability of having a homozygous recessive offspring (ww) is 0 or 0%.
In summary:
- For the first cross (Ww x ww), the chance of having heterozygous offspring is 0.5 or 50%.
- For the second cross (Ww x WW), the probability of having homozygous recessive offspring is 0.
