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Answer :
To solve the genetic probability question, let's break down each scenario step by step:
1. Heterozygous Male (Ww) x Homozygous Recessive Female (ww):
- When we mate a heterozygous male (Ww) with a homozygous recessive female (ww), we can create a Punnett square to determine the possible offspring.
- The male (Ww) has alleles W and w.
- The female (ww) has alleles w and w.
The Punnett square looks like this:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & w \\
\hline w & Ww & ww \\
\hline w & Ww & ww \\
\hline
\end{array}
\][/tex]
- From this, you can see the possible offspring genotypes: Ww, ww, Ww, ww.
- Among these, two out of the four possible combinations are heterozygous (Ww).
- This means there is a 2 out of 4 chance, or 50%, for the offspring to be heterozygous.
2. Heterozygous Male (Ww) x Homozygous Dominant Female (WW):
- Now, we're considering a heterozygous male (Ww) crossed with a homozygous dominant female (WW).
- The male (Ww) has alleles W and w.
- The female (WW) has alleles W and W.
The Punnett square looks like this:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & w \\
\hline W & WW & Ww \\
\hline W & WW & Ww \\
\hline
\end{array}
\][/tex]
- From this setup, the possible offspring genotypes are WW, Ww, WW, Ww.
- In this case, none of the offspring are homozygous recessive (ww), since all have at least one dominant allele (W).
- Therefore, the probability of having a homozygous recessive offspring is 0 out of 4, which is 0%.
In summary:
- There is a 50% probability that the offspring from the first mating are heterozygous (Ww).
- There is a 0% probability of having a homozygous recessive (ww) offspring from the second mating.
1. Heterozygous Male (Ww) x Homozygous Recessive Female (ww):
- When we mate a heterozygous male (Ww) with a homozygous recessive female (ww), we can create a Punnett square to determine the possible offspring.
- The male (Ww) has alleles W and w.
- The female (ww) has alleles w and w.
The Punnett square looks like this:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & w \\
\hline w & Ww & ww \\
\hline w & Ww & ww \\
\hline
\end{array}
\][/tex]
- From this, you can see the possible offspring genotypes: Ww, ww, Ww, ww.
- Among these, two out of the four possible combinations are heterozygous (Ww).
- This means there is a 2 out of 4 chance, or 50%, for the offspring to be heterozygous.
2. Heterozygous Male (Ww) x Homozygous Dominant Female (WW):
- Now, we're considering a heterozygous male (Ww) crossed with a homozygous dominant female (WW).
- The male (Ww) has alleles W and w.
- The female (WW) has alleles W and W.
The Punnett square looks like this:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & w \\
\hline W & WW & Ww \\
\hline W & WW & Ww \\
\hline
\end{array}
\][/tex]
- From this setup, the possible offspring genotypes are WW, Ww, WW, Ww.
- In this case, none of the offspring are homozygous recessive (ww), since all have at least one dominant allele (W).
- Therefore, the probability of having a homozygous recessive offspring is 0 out of 4, which is 0%.
In summary:
- There is a 50% probability that the offspring from the first mating are heterozygous (Ww).
- There is a 0% probability of having a homozygous recessive (ww) offspring from the second mating.
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