Answer :
Let's work through each cross step-by-step.
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]. We use a Punnett square to show the possible offspring:
[tex]$$
\begin{array}{c|cc}
& w & w \\
\hline
W & Ww & Ww \\
w & ww & ww \\
\end{array}
$$[/tex]
In this grid, there are 4 boxes representing the possible outcomes. Two of these outcomes are heterozygous ([tex]$Ww$[/tex]). Therefore, the probability that an offspring is heterozygous is
[tex]$$
\frac{2}{4} = 0.5 \quad \text{or} \quad 50\%.
$$[/tex]
2. For the second cross, a heterozygous individual with genotype [tex]$Ww$[/tex] is crossed with a homozygous dominant individual with genotype [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]
Here, all possible offspring have either the genotype [tex]$WW$[/tex] or [tex]$Ww$[/tex]. Notice there are no offspring with the homozygous recessive genotype [tex]$ww$[/tex]. That means the probability of having a homozygous recessive offspring is
[tex]$$
\frac{0}{4} = 0 \quad \text{or} \quad 0\%.
$$[/tex]
So, the final answers are:
- The probability that an offspring is heterozygous from the first cross is [tex]$50\%$[/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]. We use a Punnett square to show the possible offspring:
[tex]$$
\begin{array}{c|cc}
& w & w \\
\hline
W & Ww & Ww \\
w & ww & ww \\
\end{array}
$$[/tex]
In this grid, there are 4 boxes representing the possible outcomes. Two of these outcomes are heterozygous ([tex]$Ww$[/tex]). Therefore, the probability that an offspring is heterozygous is
[tex]$$
\frac{2}{4} = 0.5 \quad \text{or} \quad 50\%.
$$[/tex]
2. For the second cross, a heterozygous individual with genotype [tex]$Ww$[/tex] is crossed with a homozygous dominant individual with genotype [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]
Here, all possible offspring have either the genotype [tex]$WW$[/tex] or [tex]$Ww$[/tex]. Notice there are no offspring with the homozygous recessive genotype [tex]$ww$[/tex]. That means the probability of having a homozygous recessive offspring is
[tex]$$
\frac{0}{4} = 0 \quad \text{or} \quad 0\%.
$$[/tex]
So, the final answers are:
- The probability that an offspring is heterozygous from the first cross is [tex]$50\%$[/tex].
- The probability of having a homozygous recessive offspring in the second cross is [tex]$0\%$[/tex].
