Answer :
Let's analyze the two crosses one by one.
1. For the first cross, a heterozygous male with genotype [tex]$Ww$[/tex] is mated with a homozygous recessive female with genotype [tex]$ww$[/tex]. The Punnett square is constructed by placing the male gametes ([tex]$W$[/tex] and [tex]$w$[/tex]) along the top and the female gametes ([tex]$w$[/tex] and [tex]$w$[/tex]) along the side:
[tex]$$
\begin{array}{c|cc}
& W & w \\ \hline
w & Ww & ww \\
w & Ww & ww \\
\end{array}
$$[/tex]
In this square there are 4 equally likely outcomes: two are [tex]$Ww$[/tex] (heterozygous) and two are [tex]$ww$[/tex] (homozygous recessive). Therefore, the probability that an offspring is heterozygous ([tex]$Ww$[/tex]) is
[tex]$$
\frac{2}{4} = 0.5.
$$[/tex]
2. In the second cross, a heterozygous individual ([tex]$Ww$[/tex]) is crossed with a homozygous dominant individual ([tex]$WW$[/tex]). The Punnett square for this cross is:
[tex]$$
\begin{array}{c|cc}
& W & W \\ \hline
W & WW & WW \\
w & Ww & Ww \\
\end{array}
$$[/tex]
From this square, there are 4 equally likely outcomes: two outcomes are [tex]$WW$[/tex] (homozygous dominant) and two outcomes are [tex]$Ww$[/tex] (heterozygous). None of these outcomes is homozygous recessive ([tex]$ww$[/tex]). Thus, the probability of obtaining a homozygous recessive offspring is
[tex]$$
0.
$$[/tex]
In summary:
- The chance that an offspring from the first cross is heterozygous is [tex]$0.5$[/tex].
- The probability of having a homozygous recessive offspring in the second cross is [tex]$0$[/tex].
1. For the first cross, a heterozygous male with genotype [tex]$Ww$[/tex] is mated with a homozygous recessive female with genotype [tex]$ww$[/tex]. The Punnett square is constructed by placing the male gametes ([tex]$W$[/tex] and [tex]$w$[/tex]) along the top and the female gametes ([tex]$w$[/tex] and [tex]$w$[/tex]) along the side:
[tex]$$
\begin{array}{c|cc}
& W & w \\ \hline
w & Ww & ww \\
w & Ww & ww \\
\end{array}
$$[/tex]
In this square there are 4 equally likely outcomes: two are [tex]$Ww$[/tex] (heterozygous) and two are [tex]$ww$[/tex] (homozygous recessive). Therefore, the probability that an offspring is heterozygous ([tex]$Ww$[/tex]) is
[tex]$$
\frac{2}{4} = 0.5.
$$[/tex]
2. In the second cross, a heterozygous individual ([tex]$Ww$[/tex]) is crossed with a homozygous dominant individual ([tex]$WW$[/tex]). The Punnett square for this cross is:
[tex]$$
\begin{array}{c|cc}
& W & W \\ \hline
W & WW & WW \\
w & Ww & Ww \\
\end{array}
$$[/tex]
From this square, there are 4 equally likely outcomes: two outcomes are [tex]$WW$[/tex] (homozygous dominant) and two outcomes are [tex]$Ww$[/tex] (heterozygous). None of these outcomes is homozygous recessive ([tex]$ww$[/tex]). Thus, the probability of obtaining a homozygous recessive offspring is
[tex]$$
0.
$$[/tex]
In summary:
- The chance that an offspring from the first cross is heterozygous is [tex]$0.5$[/tex].
- The probability of having a homozygous recessive offspring in the second cross is [tex]$0$[/tex].
