Answer :
Sure! Let's go through the solution step by step.
First Scenario: Heterozygous Male (Ww) x Homozygous Recessive Female (ww)
1. We have a heterozygous male with genotype Ww and a homozygous recessive female with genotype ww.
2. To find the possible offspring genotypes, let's consider all combinations:
- The male can contribute either allele W or w.
- The female can only contribute allele w (since both her alleles are w).
3. When we create the Punnett square for this cross, the possible offspring genotypes are:
- Ww (if the male contributes W and the female contributes w)
- ww (if the male contributes w and the female contributes w)
4. Thus, the four possible combinations result in:
- Ww (top left of the Punnett square)
- ww (top right of the Punnett square)
- Ww (bottom left of the Punnett square)
- ww (bottom right of the Punnett square)
5. Out of these four possibilities, two are heterozygous (Ww). So, the probability of having a heterozygous offspring is [tex]\( \frac{2}{4} = 0.5 \)[/tex] or 50%.
Second Scenario: Heterozygous (Ww) x Homozygous Dominant (WW)
1. Now let's consider a cross between a heterozygous individual Ww and a homozygous dominant individual WW.
2. For this scenario:
- The heterozygous individual can contribute either W or w.
- The homozygous dominant individual can only contribute W (since both alleles are W).
3. Constructing the Punnett square, the possible offspring genotypes are:
- WW (if the heterozygous contributes W and the homozygous dominant contributes W)
- Ww (if the heterozygous contributes w and the homozygous dominant contributes W for both instances)
4. So, the four possible combinations will be:
- WW
- Ww
- WW
- Ww
5. Since there are no ww (homozygous recessive) offspring in any combinations, the probability of having a homozygous recessive offspring is 0.
This step-by-step explanation aligns with the previously stated outcomes, giving us probabilities of 0.5 for heterozygous offspring in the first scenario and 0 for homozygous recessive offspring in the second scenario.
First Scenario: Heterozygous Male (Ww) x Homozygous Recessive Female (ww)
1. We have a heterozygous male with genotype Ww and a homozygous recessive female with genotype ww.
2. To find the possible offspring genotypes, let's consider all combinations:
- The male can contribute either allele W or w.
- The female can only contribute allele w (since both her alleles are w).
3. When we create the Punnett square for this cross, the possible offspring genotypes are:
- Ww (if the male contributes W and the female contributes w)
- ww (if the male contributes w and the female contributes w)
4. Thus, the four possible combinations result in:
- Ww (top left of the Punnett square)
- ww (top right of the Punnett square)
- Ww (bottom left of the Punnett square)
- ww (bottom right of the Punnett square)
5. Out of these four possibilities, two are heterozygous (Ww). So, the probability of having a heterozygous offspring is [tex]\( \frac{2}{4} = 0.5 \)[/tex] or 50%.
Second Scenario: Heterozygous (Ww) x Homozygous Dominant (WW)
1. Now let's consider a cross between a heterozygous individual Ww and a homozygous dominant individual WW.
2. For this scenario:
- The heterozygous individual can contribute either W or w.
- The homozygous dominant individual can only contribute W (since both alleles are W).
3. Constructing the Punnett square, the possible offspring genotypes are:
- WW (if the heterozygous contributes W and the homozygous dominant contributes W)
- Ww (if the heterozygous contributes w and the homozygous dominant contributes W for both instances)
4. So, the four possible combinations will be:
- WW
- Ww
- WW
- Ww
5. Since there are no ww (homozygous recessive) offspring in any combinations, the probability of having a homozygous recessive offspring is 0.
This step-by-step explanation aligns with the previously stated outcomes, giving us probabilities of 0.5 for heterozygous offspring in the first scenario and 0 for homozygous recessive offspring in the second scenario.
