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Answer :
Let's solve the problem step by step:
1. Understanding the First Scenario:
- We have a heterozygous male with the genotype Ww.
- We have a homozygous recessive female with the genotype ww.
When we set up 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]
The offspring genotypes are:
- 2 Ww (heterozygous)
- 2 ww (homozygous recessive)
Therefore, the probability that an offspring will be heterozygous (Ww) is [tex]\( \frac{2}{4} = 0.5 \)[/tex].
2. Understanding the Second Scenario:
- We have a heterozygous individual with the genotype Ww.
- We have a homozygous dominant individual with the genotype WW.
When we set up 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]
The offspring genotypes are:
- 2 WW (homozygous dominant)
- 2 Ww (heterozygous)
In this case, there are no homozygous recessive offspring (ww), so the probability of having a homozygous recessive offspring is [tex]\( \frac{0}{4} = 0.0 \)[/tex].
So, for the questions:
- The probability that the offspring will be heterozygous in the first scenario is 0.5.
- The probability of having a homozygous recessive offspring in the second scenario is 0.0.
1. Understanding the First Scenario:
- We have a heterozygous male with the genotype Ww.
- We have a homozygous recessive female with the genotype ww.
When we set up 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]
The offspring genotypes are:
- 2 Ww (heterozygous)
- 2 ww (homozygous recessive)
Therefore, the probability that an offspring will be heterozygous (Ww) is [tex]\( \frac{2}{4} = 0.5 \)[/tex].
2. Understanding the Second Scenario:
- We have a heterozygous individual with the genotype Ww.
- We have a homozygous dominant individual with the genotype WW.
When we set up 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]
The offspring genotypes are:
- 2 WW (homozygous dominant)
- 2 Ww (heterozygous)
In this case, there are no homozygous recessive offspring (ww), so the probability of having a homozygous recessive offspring is [tex]\( \frac{0}{4} = 0.0 \)[/tex].
So, for the questions:
- The probability that the offspring will be heterozygous in the first scenario is 0.5.
- The probability of having a homozygous recessive offspring in the second scenario is 0.0.
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