Answer :
Let's solve the problem step-by-step.
### First Cross: Ww x ww
In this case, a heterozygous male with the genotype Ww is mated with a homozygous recessive female of genotype ww.
#### Step-by-step:
- The male's gametes can either be W or w.
- The female's gametes can only be w.
Creating a Punnett square:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & w \\
\hline w & Ww & ww \\
\hline w & Ww & ww \\
\hline
\end{array}
\][/tex]
From the Punnett square, we can see the potential genotypes of the offspring: Ww, Ww, ww, and ww.
- There are 2 heterozygous (Ww) combinations out of 4 possible offspring.
- Therefore, the probability that any given offspring will be heterozygous (Ww) is [tex]\(\frac{2}{4} = 0.5\)[/tex] or 50%.
So, there is a 50% chance that the offspring will be heterozygous.
### Second Cross: Ww x WW
In this case, a heterozygous (Ww) individual is crossed with a homozygous dominant (WW) individual.
#### Step-by-step:
- The first parent's gametes can be W or w.
- The second parent's gametes can only be W.
Creating a Punnett square:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & w \\
\hline W & WW & Ww \\
\hline W & WW & Ww \\
\hline
\end{array}
\][/tex]
From the Punnett square, we can see the potential genotypes of the offspring: WW, Ww, WW, and Ww.
- There are 0 homozygous recessive (ww) combinations out of 4 possible offspring.
- Therefore, the probability of having a homozygous recessive (ww) offspring is [tex]\(\frac{0}{4} = 0\)[/tex] or 0%.
So, the probability of having a homozygous recessive offspring is 0%.
### Final Answer
If a heterozygous male with the genotype Ww is mated with a homozygous recessive female of genotype ww, there is a [tex]\( \boxed{50\%} \)[/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]\( \boxed{0\%} \)[/tex].
### First Cross: Ww x ww
In this case, a heterozygous male with the genotype Ww is mated with a homozygous recessive female of genotype ww.
#### Step-by-step:
- The male's gametes can either be W or w.
- The female's gametes can only be w.
Creating a Punnett square:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & w \\
\hline w & Ww & ww \\
\hline w & Ww & ww \\
\hline
\end{array}
\][/tex]
From the Punnett square, we can see the potential genotypes of the offspring: Ww, Ww, ww, and ww.
- There are 2 heterozygous (Ww) combinations out of 4 possible offspring.
- Therefore, the probability that any given offspring will be heterozygous (Ww) is [tex]\(\frac{2}{4} = 0.5\)[/tex] or 50%.
So, there is a 50% chance that the offspring will be heterozygous.
### Second Cross: Ww x WW
In this case, a heterozygous (Ww) individual is crossed with a homozygous dominant (WW) individual.
#### Step-by-step:
- The first parent's gametes can be W or w.
- The second parent's gametes can only be W.
Creating a Punnett square:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & w \\
\hline W & WW & Ww \\
\hline W & WW & Ww \\
\hline
\end{array}
\][/tex]
From the Punnett square, we can see the potential genotypes of the offspring: WW, Ww, WW, and Ww.
- There are 0 homozygous recessive (ww) combinations out of 4 possible offspring.
- Therefore, the probability of having a homozygous recessive (ww) offspring is [tex]\(\frac{0}{4} = 0\)[/tex] or 0%.
So, the probability of having a homozygous recessive offspring is 0%.
### Final Answer
If a heterozygous male with the genotype Ww is mated with a homozygous recessive female of genotype ww, there is a [tex]\( \boxed{50\%} \)[/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]\( \boxed{0\%} \)[/tex].
