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Answer :
We first consider the cross between a heterozygous male with genotype $Ww$ and a homozygous recessive female with genotype $ww$. The gametes produced by the male are $W$ and $w$, while the female can only produce $w$. The Punnett square for this cross is:
$$
\begin{array}{c|cc}
& W & w \\
\hline
w & Ww & ww \\
w & Ww & ww \\
\end{array}
$$
In the square, there are four equally likely outcomes. Two of these outcomes are heterozygous ($Ww$). Therefore, the probability that an offspring is heterozygous is:
$$
\frac{2}{4} = 0.5 \quad \text{(or 50\%)}.
$$
Next, consider the cross between a heterozygous individual $Ww$ and a homozygous dominant individual $WW$. The gametes produced by the heterozygous parent are $W$ and $w$, and the homozygous dominant parent produces only $W$. The Punnett square for this cross is:
$$
\begin{array}{c|cc}
& W & W \\
\hline
W & WW & WW \\
w & Ww & Ww \\
\end{array}
$$
Here, all offspring have at least one dominant allele $W$. None of the offspring are homozygous recessive ($ww$). Therefore, the probability of having a homozygous recessive offspring from this cross is:
$$
\frac{0}{4} = 0.
$$
To summarize:
1. For the cross $Ww \times ww$, there is a 50% chance that an offspring will be heterozygous.
2. For the cross $Ww \times WW$, there is a 0% chance that an offspring will be homozygous recessive.
$$
\begin{array}{c|cc}
& W & w \\
\hline
w & Ww & ww \\
w & Ww & ww \\
\end{array}
$$
In the square, there are four equally likely outcomes. Two of these outcomes are heterozygous ($Ww$). Therefore, the probability that an offspring is heterozygous is:
$$
\frac{2}{4} = 0.5 \quad \text{(or 50\%)}.
$$
Next, consider the cross between a heterozygous individual $Ww$ and a homozygous dominant individual $WW$. The gametes produced by the heterozygous parent are $W$ and $w$, and the homozygous dominant parent produces only $W$. The Punnett square for this cross is:
$$
\begin{array}{c|cc}
& W & W \\
\hline
W & WW & WW \\
w & Ww & Ww \\
\end{array}
$$
Here, all offspring have at least one dominant allele $W$. None of the offspring are homozygous recessive ($ww$). Therefore, the probability of having a homozygous recessive offspring from this cross is:
$$
\frac{0}{4} = 0.
$$
To summarize:
1. For the cross $Ww \times ww$, there is a 50% chance that an offspring will be heterozygous.
2. For the cross $Ww \times WW$, there is a 0% chance that an offspring will be homozygous recessive.
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