Answer :
Certainly! Let's break down the genetic crosses and figure out the probabilities:
### Problem 1:
Parents' Genotypes:
- Heterozygous male: Ww
- Homozygous recessive female: ww
Possible Offspring Genotypes:
To determine the potential genotypes of the offspring, we can use a Punnett square. Here’s how it works:
- One parent's alleles (Ww) are written across the top.
- The other parent's alleles (ww) are written along the side.
This results in the following Punnett square:
[tex]\[
\begin{array}{c|c|c}
& W & w \\
\hline
w & Ww & ww \\
\hline
w & Ww & ww \\
\end{array}
\][/tex]
Possible offspring genotypes are:
- 2 Ww (heterozygous)
- 2 ww (homozygous recessive)
Probability of Heterozygous Offspring:
- Out of 4 possible genotypes, 2 are Ww.
- Therefore, the chance of offspring being heterozygous (Ww) is [tex]\( \frac{2}{4} \times 100\% = 50\% \)[/tex].
### Problem 2:
Parents' Genotypes:
- Heterozygous: Ww
- Homozygous dominant: WW
Potential Offspring Genotypes:
Let's perform another Punnett square:
- One parent's alleles (Ww) are placed at the top.
- The other parent's alleles (WW) are placed along the side.
This yields:
[tex]\[
\begin{array}{c|c|c}
& W & w \\
\hline
W & WW & Ww \\
\hline
W & WW & Ww \\
\end{array}
\][/tex]
Potential offspring genotypes:
- 2 WW (homozygous dominant)
- 2 Ww (heterozygous)
Probability of Homozygous Recessive Offspring:
- There are no ww genotypes among the possibilities.
- Thus, the chance of having a homozygous recessive offspring (ww) is [tex]\( 0\% \)[/tex].
In summary, for the first cross, there is a 50% chance for heterozygous offspring, and for the second cross, there is a 0% chance of having a homozygous recessive offspring.
### Problem 1:
Parents' Genotypes:
- Heterozygous male: Ww
- Homozygous recessive female: ww
Possible Offspring Genotypes:
To determine the potential genotypes of the offspring, we can use a Punnett square. Here’s how it works:
- One parent's alleles (Ww) are written across the top.
- The other parent's alleles (ww) are written along the side.
This results in the following Punnett square:
[tex]\[
\begin{array}{c|c|c}
& W & w \\
\hline
w & Ww & ww \\
\hline
w & Ww & ww \\
\end{array}
\][/tex]
Possible offspring genotypes are:
- 2 Ww (heterozygous)
- 2 ww (homozygous recessive)
Probability of Heterozygous Offspring:
- Out of 4 possible genotypes, 2 are Ww.
- Therefore, the chance of offspring being heterozygous (Ww) is [tex]\( \frac{2}{4} \times 100\% = 50\% \)[/tex].
### Problem 2:
Parents' Genotypes:
- Heterozygous: Ww
- Homozygous dominant: WW
Potential Offspring Genotypes:
Let's perform another Punnett square:
- One parent's alleles (Ww) are placed at the top.
- The other parent's alleles (WW) are placed along the side.
This yields:
[tex]\[
\begin{array}{c|c|c}
& W & w \\
\hline
W & WW & Ww \\
\hline
W & WW & Ww \\
\end{array}
\][/tex]
Potential offspring genotypes:
- 2 WW (homozygous dominant)
- 2 Ww (heterozygous)
Probability of Homozygous Recessive Offspring:
- There are no ww genotypes among the possibilities.
- Thus, the chance of having a homozygous recessive offspring (ww) is [tex]\( 0\% \)[/tex].
In summary, for the first cross, there is a 50% chance for heterozygous offspring, and for the second cross, there is a 0% chance of having a homozygous recessive offspring.
