Answer :
We start by considering the two separate crosses.
[tex]$$\textbf{First Cross: } \text{Heterozygous male } (Ww) \times \text{Homozygous recessive female } (ww)$$[/tex]
1. The heterozygous parent produces gametes with either the [tex]$W$[/tex] or [tex]$w$[/tex] allele, so the possible gametes are [tex]$W$[/tex] and [tex]$w$[/tex].
2. The homozygous recessive parent can only produce gametes carrying the [tex]$w$[/tex] allele.
3. We set up a Punnett square with the male gametes across the top and the female gametes along the side:
[tex]$$
\begin{array}{c|cc}
& W & w \\
\hline
w & Ww & ww \\
w & Ww & ww \\
\end{array}
$$[/tex]
4. In the Punnett square, we see there are 4 possible outcomes. Out of these, 2 outcomes are heterozygous ([tex]$Ww$[/tex]).
5. Therefore, the probability that an offspring is heterozygous is
[tex]$$\frac{2}{4} = 0.5 \text{ (or } 50\%\text{)}.$$[/tex]
[tex]$$\textbf{Second Cross: } \text{Heterozygous individual } (Ww) \times \text{Homozygous dominant } (WW)$$[/tex]
1. The heterozygous parent produces gametes with either the [tex]$W$[/tex] or [tex]$w$[/tex] allele.
2. The homozygous dominant parent will produce only [tex]$W$[/tex] alleles.
3. Using a Punnett square:
[tex]$$
\begin{array}{c|cc}
& W & W \\
\hline
W & WW & WW \\
w & Ww & Ww \\
\end{array}
$$[/tex]
4. All the possible offspring are either [tex]$WW$[/tex] or [tex]$Ww$[/tex].
5. There are no offspring with the homozygous recessive genotype ([tex]$ww$[/tex]), so the probability of a homozygous recessive offspring is
[tex]$$ 0.0.$$[/tex]
In summary:
- The probability of a heterozygous offspring from the first cross is [tex]$0.5$[/tex].
- The probability of a homozygous recessive offspring from the second cross is [tex]$0.0$[/tex].
[tex]$$\textbf{First Cross: } \text{Heterozygous male } (Ww) \times \text{Homozygous recessive female } (ww)$$[/tex]
1. The heterozygous parent produces gametes with either the [tex]$W$[/tex] or [tex]$w$[/tex] allele, so the possible gametes are [tex]$W$[/tex] and [tex]$w$[/tex].
2. The homozygous recessive parent can only produce gametes carrying the [tex]$w$[/tex] allele.
3. We set up a Punnett square with the male gametes across the top and the female gametes along the side:
[tex]$$
\begin{array}{c|cc}
& W & w \\
\hline
w & Ww & ww \\
w & Ww & ww \\
\end{array}
$$[/tex]
4. In the Punnett square, we see there are 4 possible outcomes. Out of these, 2 outcomes are heterozygous ([tex]$Ww$[/tex]).
5. Therefore, the probability that an offspring is heterozygous is
[tex]$$\frac{2}{4} = 0.5 \text{ (or } 50\%\text{)}.$$[/tex]
[tex]$$\textbf{Second Cross: } \text{Heterozygous individual } (Ww) \times \text{Homozygous dominant } (WW)$$[/tex]
1. The heterozygous parent produces gametes with either the [tex]$W$[/tex] or [tex]$w$[/tex] allele.
2. The homozygous dominant parent will produce only [tex]$W$[/tex] alleles.
3. Using a Punnett square:
[tex]$$
\begin{array}{c|cc}
& W & W \\
\hline
W & WW & WW \\
w & Ww & Ww \\
\end{array}
$$[/tex]
4. All the possible offspring are either [tex]$WW$[/tex] or [tex]$Ww$[/tex].
5. There are no offspring with the homozygous recessive genotype ([tex]$ww$[/tex]), so the probability of a homozygous recessive offspring is
[tex]$$ 0.0.$$[/tex]
In summary:
- The probability of a heterozygous offspring from the first cross is [tex]$0.5$[/tex].
- The probability of a homozygous recessive offspring from the second cross is [tex]$0.0$[/tex].
