Answer :
Sure! Let's explore how we find the probabilities step by step:
### Part 1: Heterozygous Male (Ww) and Homozygous Recessive Female (ww)
When a heterozygous male (Ww) is mated with a homozygous recessive female (ww), we can predict the possible offspring using a Punnett square.
1. Genes from each parent:
- The male (Ww) can contribute either a W or a w.
- The female (ww) can only contribute a w.
2. Fill in the Punnett square:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & w \\
\hline
w & Ww & ww \\
\hline
w & Ww & ww \\
\hline
\end{array}
\][/tex]
3. Resulting genotypes:
- Half of the offspring (2 out of 4 squares) will be heterozygous (Ww).
- The other half will be homozygous recessive (ww).
4. Probability of heterozygous offspring:
- The probability of having a heterozygous offspring is [tex]\( \frac{2}{4} = 0.5 \)[/tex].
### Part 2: Heterozygous (Ww) Crossed with Homozygous Dominant (WW)
Now, let's consider a cross between a heterozygous individual (Ww) and a homozygous dominant individual (WW).
1. Genes from each parent:
- The heterozygous individual (Ww) can contribute either a W or a w.
- The homozygous dominant individual (WW) can only contribute a W.
2. Fill in the Punnett square:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & w \\
\hline
W & WW & Ww \\
\hline
W & WW & Ww \\
\hline
\end{array}
\][/tex]
3. Resulting genotypes:
- All offspring (4 out of 4 squares) will have at least one dominant allele, either WW or Ww.
- There are no homozygous recessive offspring (ww).
4. Probability of homozygous recessive offspring:
- The probability of having a homozygous recessive offspring is [tex]\( 0 \)[/tex].
In summary, the probability of having heterozygous offspring in Part 1 is 0.5, and the probability of having homozygous recessive offspring in Part 2 is 0.0.
### Part 1: Heterozygous Male (Ww) and Homozygous Recessive Female (ww)
When a heterozygous male (Ww) is mated with a homozygous recessive female (ww), we can predict the possible offspring using a Punnett square.
1. Genes from each parent:
- The male (Ww) can contribute either a W or a w.
- The female (ww) can only contribute a w.
2. Fill in the Punnett square:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & w \\
\hline
w & Ww & ww \\
\hline
w & Ww & ww \\
\hline
\end{array}
\][/tex]
3. Resulting genotypes:
- Half of the offspring (2 out of 4 squares) will be heterozygous (Ww).
- The other half will be homozygous recessive (ww).
4. Probability of heterozygous offspring:
- The probability of having a heterozygous offspring is [tex]\( \frac{2}{4} = 0.5 \)[/tex].
### Part 2: Heterozygous (Ww) Crossed with Homozygous Dominant (WW)
Now, let's consider a cross between a heterozygous individual (Ww) and a homozygous dominant individual (WW).
1. Genes from each parent:
- The heterozygous individual (Ww) can contribute either a W or a w.
- The homozygous dominant individual (WW) can only contribute a W.
2. Fill in the Punnett square:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & w \\
\hline
W & WW & Ww \\
\hline
W & WW & Ww \\
\hline
\end{array}
\][/tex]
3. Resulting genotypes:
- All offspring (4 out of 4 squares) will have at least one dominant allele, either WW or Ww.
- There are no homozygous recessive offspring (ww).
4. Probability of homozygous recessive offspring:
- The probability of having a homozygous recessive offspring is [tex]\( 0 \)[/tex].
In summary, the probability of having heterozygous offspring in Part 1 is 0.5, and the probability of having homozygous recessive offspring in Part 2 is 0.0.
