Answer :
We start by considering each cross separately.
1. For the cross between a heterozygous male with genotype [tex]$Ww$[/tex] and a homozygous recessive female with genotype [tex]$ww$[/tex], we set up the Punnett square:
[tex]$$
\begin{array}{c|cc}
& w & w \\ \hline
W & Ww & Ww \\
w & ww & ww \\
\end{array}
$$[/tex]
In this Punnett square, there are 4 possible outcomes:
- 2 offspring with the heterozygous genotype [tex]$Ww$[/tex].
- 2 offspring with the homozygous recessive genotype [tex]$ww$[/tex].
Thus, the probability of an offspring being heterozygous is
[tex]$$
\frac{2}{4} = 0.5.
$$[/tex]
2. For the cross between a heterozygous individual with genotype [tex]$Ww$[/tex] and a homozygous dominant individual with genotype [tex]$WW$[/tex], the Punnett square looks like this:
[tex]$$
\begin{array}{c|cc}
& W & W \\ \hline
W & WW & WW \\
w & Ww & Ww \\
\end{array}
$$[/tex]
Here, the offspring are either:
- 2 offspring with the genotype [tex]$WW$[/tex], or
- 2 offspring with the genotype [tex]$Ww$[/tex].
There are no offspring with the homozygous recessive genotype [tex]$ww$[/tex]. Therefore, the probability of obtaining a homozygous recessive offspring is
[tex]$$
\frac{0}{4} = 0.0.
$$[/tex]
So, the final answers are:
- The probability that an offspring will be heterozygous in the first cross is [tex]$0.5$[/tex].
- The probability of having a homozygous recessive offspring in the second cross is [tex]$0.0$[/tex].
1. For the cross between a heterozygous male with genotype [tex]$Ww$[/tex] and a homozygous recessive female with genotype [tex]$ww$[/tex], we set up the Punnett square:
[tex]$$
\begin{array}{c|cc}
& w & w \\ \hline
W & Ww & Ww \\
w & ww & ww \\
\end{array}
$$[/tex]
In this Punnett square, there are 4 possible outcomes:
- 2 offspring with the heterozygous genotype [tex]$Ww$[/tex].
- 2 offspring with the homozygous recessive genotype [tex]$ww$[/tex].
Thus, the probability of an offspring being heterozygous is
[tex]$$
\frac{2}{4} = 0.5.
$$[/tex]
2. For the cross between a heterozygous individual with genotype [tex]$Ww$[/tex] and a homozygous dominant individual with genotype [tex]$WW$[/tex], the Punnett square looks like this:
[tex]$$
\begin{array}{c|cc}
& W & W \\ \hline
W & WW & WW \\
w & Ww & Ww \\
\end{array}
$$[/tex]
Here, the offspring are either:
- 2 offspring with the genotype [tex]$WW$[/tex], or
- 2 offspring with the genotype [tex]$Ww$[/tex].
There are no offspring with the homozygous recessive genotype [tex]$ww$[/tex]. Therefore, the probability of obtaining a homozygous recessive offspring is
[tex]$$
\frac{0}{4} = 0.0.
$$[/tex]
So, the final answers are:
- The probability that an offspring will be heterozygous in the first cross is [tex]$0.5$[/tex].
- The probability of having a homozygous recessive offspring in the second cross is [tex]$0.0$[/tex].
