Answer :
Certainly! Let's break down the problem step-by-step.
1. Heterozygous (Ww) Male x Homozygous Recessive (ww) Female
We are asked to find the probability of the offspring being heterozygous (Ww).
Considering the provided Punnett square:
\begin{tabular}{|c|c|c|}
\hline & W & w \\
\hline w & Ww & Ww \\
\hline w & Ww & ww \\
\hline
\end{tabular}
From this Punnett square, we can see:
- There are four possible combinations of alleles: Ww, Ww, Ww, ww.
- Three out of these four combinations (Ww, Ww, Ww) result in heterozygous offspring (Ww).
Thus, the probability of the offspring being heterozygous is:
[tex]\[ \frac{3}{4} = 0.75 \][/tex]
Therefore, if a heterozygous male (Ww) is crossed with a homozygous recessive female (ww):
[tex]\[ \text{The probability of the offspring being heterozygous is 0.75}. \][/tex]
2. Heterozygous (Ww) x Homozygous Dominant (WW)
We need to determine the probability of having a homozygous recessive (ww) offspring.
Considering the provided Punnett square:
\begin{tabular}{|c|c|c|}
\hline & W & W \\
\hline W & WW & Ww \\
\hline W & WW & Ww \\
\hline
\end{tabular}
From this Punnett square, we see:
- There are four possible combinations of alleles: WW, Ww, WW, Ww.
- None of these combinations result in homozygous recessive (ww).
Thus, the probability of obtaining a homozygous recessive (ww) offspring is:
[tex]\[ 0 \][/tex]
Therefore, if a heterozygous individual (Ww) is crossed with a homozygous dominant (WW):
[tex]\[ \text{The probability of obtaining a homozygous recessive offspring is 0}. \][/tex]
So, to summarize the correct answers for each statement:
1. If a heterozygous male with the genotype Ww is mated with a homozygous recessive female of genotype ww, there is a [tex]\( \text{chance of } \boxed{0.5} \text{ that the offspring will be heterozygous.} 2. If the heterozygous (Ww) is crossed with a homozygous dominant (WW), then the probability of having a homozygous recessive offspring is \( \boxed{0.0} \)[/tex].
1. Heterozygous (Ww) Male x Homozygous Recessive (ww) Female
We are asked to find the probability of the offspring being heterozygous (Ww).
Considering the provided Punnett square:
\begin{tabular}{|c|c|c|}
\hline & W & w \\
\hline w & Ww & Ww \\
\hline w & Ww & ww \\
\hline
\end{tabular}
From this Punnett square, we can see:
- There are four possible combinations of alleles: Ww, Ww, Ww, ww.
- Three out of these four combinations (Ww, Ww, Ww) result in heterozygous offspring (Ww).
Thus, the probability of the offspring being heterozygous is:
[tex]\[ \frac{3}{4} = 0.75 \][/tex]
Therefore, if a heterozygous male (Ww) is crossed with a homozygous recessive female (ww):
[tex]\[ \text{The probability of the offspring being heterozygous is 0.75}. \][/tex]
2. Heterozygous (Ww) x Homozygous Dominant (WW)
We need to determine the probability of having a homozygous recessive (ww) offspring.
Considering the provided Punnett square:
\begin{tabular}{|c|c|c|}
\hline & W & W \\
\hline W & WW & Ww \\
\hline W & WW & Ww \\
\hline
\end{tabular}
From this Punnett square, we see:
- There are four possible combinations of alleles: WW, Ww, WW, Ww.
- None of these combinations result in homozygous recessive (ww).
Thus, the probability of obtaining a homozygous recessive (ww) offspring is:
[tex]\[ 0 \][/tex]
Therefore, if a heterozygous individual (Ww) is crossed with a homozygous dominant (WW):
[tex]\[ \text{The probability of obtaining a homozygous recessive offspring is 0}. \][/tex]
So, to summarize the correct answers for each statement:
1. If a heterozygous male with the genotype Ww is mated with a homozygous recessive female of genotype ww, there is a [tex]\( \text{chance of } \boxed{0.5} \text{ that the offspring will be heterozygous.} 2. If the heterozygous (Ww) is crossed with a homozygous dominant (WW), then the probability of having a homozygous recessive offspring is \( \boxed{0.0} \)[/tex].