Answer :
Let's tackle this problem step by step.
### First Scenario
We are dealing with a heterozygous male (genotype Ww) and a homozygous recessive female (genotype ww). We can use a Punnett square to determine the possible genotypes of the offspring.
Here's what the Punnett square looks like:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & w \\
\hline w & Ww & ww \\
\hline w & Ww & ww \\
\hline
\end{array}
\][/tex]
From the Punnett square, the possible genotypes of the offspring are:
- Ww
- Ww
- ww
- ww
Out of the four possible genotypes, two are heterozygous (Ww) and two are homozygous recessive (ww).
So, the probability that the offspring will be heterozygous (Ww) is:
[tex]\[
\frac{2 \text{ (Ww)}}{4 \text{ (total)}}
\][/tex]
Simplifying, we get:
[tex]\[
\frac{2}{4} = 0.5
\][/tex]
So, there is a 0.5 (or 50%) chance that the offspring will be heterozygous.
### Second Scenario
In this scenario, a homozygous dominant (WW) is crossed with another homozygous dominant (WW). Let's figure out the possible genotypes of the offspring using another Punnett square:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & W \\
\hline W & WW & WW \\
\hline W & WW & WW \\
\hline
\end{array}
\][/tex]
From this Punnett square, the possible genotypes of the offspring are:
- WW
- WW
- WW
- WW
In this case, all the offspring are homozygous dominant (WW). There are no homozygous recessive (ww) offspring.
So, the probability that the offspring will be homozygous recessive (ww) is:
[tex]\[
0 \text{ out of } 4 = 0
\][/tex]
Therefore, the chance of having a homozygous recessive offspring is 0 (or 0%).
### Conclusion
Here are the final answers:
- There is a 0.5 chance that the offspring will be heterozygous when a heterozygous male (Ww) is mated with a homozygous recessive female (ww).
- There is a 0 chance of having a homozygous recessive offspring when two homozygous dominant (WW) individuals are crossed.
### First Scenario
We are dealing with a heterozygous male (genotype Ww) and a homozygous recessive female (genotype ww). We can use a Punnett square to determine the possible genotypes of the offspring.
Here's what the Punnett square looks like:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & w \\
\hline w & Ww & ww \\
\hline w & Ww & ww \\
\hline
\end{array}
\][/tex]
From the Punnett square, the possible genotypes of the offspring are:
- Ww
- Ww
- ww
- ww
Out of the four possible genotypes, two are heterozygous (Ww) and two are homozygous recessive (ww).
So, the probability that the offspring will be heterozygous (Ww) is:
[tex]\[
\frac{2 \text{ (Ww)}}{4 \text{ (total)}}
\][/tex]
Simplifying, we get:
[tex]\[
\frac{2}{4} = 0.5
\][/tex]
So, there is a 0.5 (or 50%) chance that the offspring will be heterozygous.
### Second Scenario
In this scenario, a homozygous dominant (WW) is crossed with another homozygous dominant (WW). Let's figure out the possible genotypes of the offspring using another Punnett square:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & W \\
\hline W & WW & WW \\
\hline W & WW & WW \\
\hline
\end{array}
\][/tex]
From this Punnett square, the possible genotypes of the offspring are:
- WW
- WW
- WW
- WW
In this case, all the offspring are homozygous dominant (WW). There are no homozygous recessive (ww) offspring.
So, the probability that the offspring will be homozygous recessive (ww) is:
[tex]\[
0 \text{ out of } 4 = 0
\][/tex]
Therefore, the chance of having a homozygous recessive offspring is 0 (or 0%).
### Conclusion
Here are the final answers:
- There is a 0.5 chance that the offspring will be heterozygous when a heterozygous male (Ww) is mated with a homozygous recessive female (ww).
- There is a 0 chance of having a homozygous recessive offspring when two homozygous dominant (WW) individuals are crossed.
