Answer :
Sure! Let's work through each part of the question step-by-step.
### Part 1: Crossing Ww (Heterozygous) with ww (Homozygous Recessive)
When a heterozygous male with the genotype Ww is mated with a homozygous recessive female with the genotype ww, we can determine the possible genotypes of the offspring using a Punnett square.
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]
From the Punnett square:
- Two of the offspring (Ww and Ww) are heterozygous.
- Two of the offspring (ww and ww) are homozygous recessive.
So, the probability that an offspring will be heterozygous (Ww) is [tex]\( \frac{2}{4} \)[/tex], which simplifies to 0.5 or 50%.
### Part 2: Crossing Ww (Heterozygous) with WW (Homozygous Dominant)
Now let's look at what happens if we cross a heterozygous individual (Ww) with a homozygous dominant individual (WW).
Using another Punnett square:
[tex]\[
\begin{array}{|c|c|c|}
\hline
& W & w \\
\hline
W & WW & Ww \\
\hline
W & WW & Ww \\
\hline
\end{array}
\][/tex]
From this Punnett square:
- Two of the offspring (WW and WW) are homozygous dominant.
- Two of the offspring (Ww and Ww) are heterozygous.
There are no homozygous recessive (ww) offspring in this cross.
Thus, the probability of having a homozygous recessive (ww) offspring is 0%.
Therefore, for the initial questions:
- The probability that an offspring from the first cross (Ww x ww) will be heterozygous is 0.5 or 50%.
- The probability of having a homozygous recessive offspring in the second cross (Ww x WW) is 0%.
### Part 1: Crossing Ww (Heterozygous) with ww (Homozygous Recessive)
When a heterozygous male with the genotype Ww is mated with a homozygous recessive female with the genotype ww, we can determine the possible genotypes of the offspring using a Punnett square.
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]
From the Punnett square:
- Two of the offspring (Ww and Ww) are heterozygous.
- Two of the offspring (ww and ww) are homozygous recessive.
So, the probability that an offspring will be heterozygous (Ww) is [tex]\( \frac{2}{4} \)[/tex], which simplifies to 0.5 or 50%.
### Part 2: Crossing Ww (Heterozygous) with WW (Homozygous Dominant)
Now let's look at what happens if we cross a heterozygous individual (Ww) with a homozygous dominant individual (WW).
Using another Punnett square:
[tex]\[
\begin{array}{|c|c|c|}
\hline
& W & w \\
\hline
W & WW & Ww \\
\hline
W & WW & Ww \\
\hline
\end{array}
\][/tex]
From this Punnett square:
- Two of the offspring (WW and WW) are homozygous dominant.
- Two of the offspring (Ww and Ww) are heterozygous.
There are no homozygous recessive (ww) offspring in this cross.
Thus, the probability of having a homozygous recessive (ww) offspring is 0%.
Therefore, for the initial questions:
- The probability that an offspring from the first cross (Ww x ww) will be heterozygous is 0.5 or 50%.
- The probability of having a homozygous recessive offspring in the second cross (Ww x WW) is 0%.
