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Answer :
We begin by analyzing each cross separately.
––––––––––––––––––––––––––––––––––––––––––
**First Cross:**
A heterozygous male with genotype $Ww$ is mated with a homozygous recessive female with genotype $ww$.
1. The possible gametes from the male are $W$ and $w$.
2. The female can only produce gametes with $w$.
We construct the Punnett square:
$$
\begin{array}{c|cc}
& W & w \\
\hline
w & Ww & ww \\
\end{array}
$$
The outcomes are:
- One square gives the heterozygous genotype $Ww$.
- One square gives the homozygous recessive genotype $ww$.
Thus, out of 2 equally likely outcomes, 1 is heterozygous. Therefore, the probability that an offspring is heterozygous is
$$
\frac{1}{2} = 0.5.
$$
––––––––––––––––––––––––––––––––––––––––––
**Second Cross:**
An individual with genotype $WW$ is crossed with another individual with genotype $WW$.
1. Both parents can only produce gametes with $W$.
Constructing a simple Punnett square:
$$
\begin{array}{c|cc}
& W & W \\
\hline
W & WW & WW \\
\end{array}
$$
Since all offspring receive a $W$ from each parent, every offspring has the genotype $WW$, and there is no possibility of obtaining a homozygous recessive ($ww$) offspring. Thus, the probability of having a homozygous recessive offspring is
$$
0.
$$
––––––––––––––––––––––––––––––––––––––––––
**Final Answers:**
- The probability that an offspring will be heterozygous in the first cross is $0.5$.
- The probability of having a homozygous recessive offspring in the second cross is $0$.
These conclusions match the outcomes from the analysis.
––––––––––––––––––––––––––––––––––––––––––
**First Cross:**
A heterozygous male with genotype $Ww$ is mated with a homozygous recessive female with genotype $ww$.
1. The possible gametes from the male are $W$ and $w$.
2. The female can only produce gametes with $w$.
We construct the Punnett square:
$$
\begin{array}{c|cc}
& W & w \\
\hline
w & Ww & ww \\
\end{array}
$$
The outcomes are:
- One square gives the heterozygous genotype $Ww$.
- One square gives the homozygous recessive genotype $ww$.
Thus, out of 2 equally likely outcomes, 1 is heterozygous. Therefore, the probability that an offspring is heterozygous is
$$
\frac{1}{2} = 0.5.
$$
––––––––––––––––––––––––––––––––––––––––––
**Second Cross:**
An individual with genotype $WW$ is crossed with another individual with genotype $WW$.
1. Both parents can only produce gametes with $W$.
Constructing a simple Punnett square:
$$
\begin{array}{c|cc}
& W & W \\
\hline
W & WW & WW \\
\end{array}
$$
Since all offspring receive a $W$ from each parent, every offspring has the genotype $WW$, and there is no possibility of obtaining a homozygous recessive ($ww$) offspring. Thus, the probability of having a homozygous recessive offspring is
$$
0.
$$
––––––––––––––––––––––––––––––––––––––––––
**Final Answers:**
- The probability that an offspring will be heterozygous in the first cross is $0.5$.
- The probability of having a homozygous recessive offspring in the second cross is $0$.
These conclusions match the outcomes from the analysis.
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