Let's solve this step by step.
First, consider the mating between a heterozygous male (Ww) and a homozygous recessive female (ww).
Here's the Punnett square for this cross:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & w \\
\hline w & Ww & Ww \\
\hline w & Ww & Ww \\
\hline
\end{array}
\][/tex]
In all four possible resulting combinations (Ww, Ww, Ww, Ww), the offspring will be heterozygous (Ww). Therefore, the chance of the offspring being heterozygous is 100%.
Now consider mating a heterozygous individual (Ww) with a homozygous dominant individual (WW).
Here's the Punnett square for this cross:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & W \\
\hline W & WW & WW \\
\hline w & Ww & Ww \\
\hline
\end{array}
\][/tex]
In this case, the possible resulting combinations are WW, WW, Ww, Ww. All the offspring will either be homozygous dominant (WW) or heterozygous (Ww), so there is no chance of having a homozygous recessive (ww) offspring. Therefore, the probability of having a homozygous recessive offspring is 0%.
In conclusion:
- If a heterozygous male (Ww) is mated with a homozygous recessive female (ww), there is a 100% chance that the offspring will be heterozygous.
- If a heterozygous individual (Ww) is crossed with a homozygous dominant individual (WW), then the probability of having a homozygous recessive offspring is 0%.
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 [tex]\(100\)[/tex]% chance that 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 [tex]\(0\)[/tex].
