Sure, let's go through it step-by-step.
First, we'll address the cross between a heterozygous male with the genotype Ww and a homozygous recessive female with genotype ww.
The Punnett square for Ww (male) x ww (female) is:
[tex]\[
\begin{tabular}{|c|c|c|}
\hline & W & w \\
\hline w & Ww & ww \\
\hline w & Ww & ww \\
\hline
\end{tabular}
\][/tex]
From this Punnett square, we see that there are two possible genotypes for the offspring:
- Ww (heterozygous)
- ww (homozygous recessive)
The probability of each is as follows:
- Ww: 2 out of 4 squares, or 50%
- ww: 2 out of 4 squares, or 50%
Therefore, the chance that the offspring will be heterozygous (Ww) is 50%.
Next, we address the cross between a heterozygous Ww and a homozygous dominant WW.
The Punnett square for Ww (one parent) x WW (other parent) is:
[tex]\[
\begin{tabular}{|c|c|c|}
\hline & W & w \\
\hline W & WW & Ww \\
\hline W & WW & Ww \\
\hline
\end{tabular}
\][/tex]
From this Punnett square, we see that there are also two possible genotypes for the offspring:
- WW (homozygous dominant)
- Ww (heterozygous)
The probability of each is as follows:
- WW: 2 out of 4 squares, or 50%
- Ww: 2 out of 4 squares, or 50%
However, in this case, there is no possibility of producing a homozygous recessive (ww) offspring.
So, the probability of having a homozygous recessive offspring is 0%.
Now let's fill in the blanks:
"If a heterozygous male with the genotype Ww is mated with a homozygous recessive female of genotype ww, there is a 50% chance that [tex]\(\checkmark\)[/tex] of the offspring will be heterozygous.
[tex]\[
\begin{tabular}{|c|c|c|}
\hline & W & w \\
\hline w & Ww & ww \\
\hline w & Ww & ww \\
\hline
\end{tabular}
\][/tex]
If the heterozygous, Ww, is crossed with a homozygous dominant, WW, then the probability of having a homozygous recessive offspring is 0%.
[tex]\[
\begin{tabular}{|c|c|c|}
\hline & W & w \\
\hline W & WW & Ww \\
\hline W & WW & Ww \\
\hline
\end{tabular}
\][/tex]"
