Answer :
Sure, let's work through the genetics problem step-by-step to find the probabilities.
First Scenario: Crossing Ww with ww
1. We have a heterozygous male with the genotype Ww, and a homozygous recessive female with the genotype ww.
2. To find the possible outcomes, we use a Punnett square:
- One parent (Ww) can contribute either W or w.
- The other parent (ww) can contribute only w.
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]
3. The possible offspring genotypes from this cross are:
- Ww (heterozygous)
- ww (homozygous recessive)
4. Looking at the offspring, there are 4 possibilities: Ww, Ww, ww, ww.
5. Two out of these four possibilities are Ww (heterozygous).
6. Therefore, the probability that an offspring will be heterozygous is [tex]\( \frac{2}{4} = 0.5 \)[/tex] or 50%.
Second Scenario: Crossing WW with WW
1. In this case, both parents are homozygous dominant with the genotype WW.
2. 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]
3. The possible offspring genotypes are: WW, WW, WW, WW.
4. Since all offspring are WW (homozygous dominant), none can be homozygous recessive (ww).
5. Therefore, the probability of having a homozygous recessive offspring is [tex]\( \frac{0}{4} = 0.0 \)[/tex] or 0%.
These calculations explain why the probabilities are 0.5 for having a heterozygous offspring in the first scenario, and 0 in the second scenario for having a homozygous recessive offspring.
First Scenario: Crossing Ww with ww
1. We have a heterozygous male with the genotype Ww, and a homozygous recessive female with the genotype ww.
2. To find the possible outcomes, we use a Punnett square:
- One parent (Ww) can contribute either W or w.
- The other parent (ww) can contribute only w.
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]
3. The possible offspring genotypes from this cross are:
- Ww (heterozygous)
- ww (homozygous recessive)
4. Looking at the offspring, there are 4 possibilities: Ww, Ww, ww, ww.
5. Two out of these four possibilities are Ww (heterozygous).
6. Therefore, the probability that an offspring will be heterozygous is [tex]\( \frac{2}{4} = 0.5 \)[/tex] or 50%.
Second Scenario: Crossing WW with WW
1. In this case, both parents are homozygous dominant with the genotype WW.
2. 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]
3. The possible offspring genotypes are: WW, WW, WW, WW.
4. Since all offspring are WW (homozygous dominant), none can be homozygous recessive (ww).
5. Therefore, the probability of having a homozygous recessive offspring is [tex]\( \frac{0}{4} = 0.0 \)[/tex] or 0%.
These calculations explain why the probabilities are 0.5 for having a heterozygous offspring in the first scenario, and 0 in the second scenario for having a homozygous recessive offspring.
