Sure, let's look at each part of the question step by step:
1. First Part: Heterozygous Male (Ww) x Homozygous Recessive Female (ww)
The provided Punnett Square for this cross 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 the possible offspring are:
- Ww (heterozygous)
- Ww (heterozygous)
- ww (homozygous recessive)
- ww (homozygous recessive)
Out of these four possible outcomes, two are heterozygous (Ww) and two are homozygous recessive (ww). Therefore, half or 50% of the offspring will be heterozygous.
So, the correct answer is:
If a heterozygous male with the genotype Ww is mated with a homozygous recessive female of genotype ww, there is a chance that 50% of the offspring will be heterozygous.
2. Second Part: Heterozygous (Ww) x Homozygous Dominant (WW)
The provided Punnett Square for this cross 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 can see the possible offspring are:
- WW (homozygous dominant)
- Ww (heterozygous)
- WW (homozygous dominant)
- Ww (heterozygous)
None of these offspring are homozygous recessive (ww). Therefore, the probability of having a homozygous recessive offspring is 0%.
So, the correct answer is:
If the heterozygous, Ww, is crossed with a homozygous dominant, WW, then the probability of having a homozygous recessive offspring is 0%.
Therefore, the fully completed answer is:
If a heterozygous male with the genotype Ww is mated with a homozygous recessive female of genotype ww, there is a chance that 50% of the offspring will be heterozygous.
If the heterozygous, Ww, is crossed with a homozygous dominant, WW, then the probability of having a homozygous recessive offspring is 0%.
