Answer :
Certainly! Let's go through the genetic cross scenarios step-by-step:
First Scenario:
- We have a heterozygous male with the genotype Ww and a homozygous recessive female with the genotype ww.
- We use a Punnett square to determine the possible genotypes of their offspring.
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & w \\
\hline w & Ww & ww \\
\hline w & Ww & ww \\
\hline
\end{array}
\][/tex]
- Looking at the Punnett square:
- Two squares show the genotype Ww (heterozygous).
- Two squares show the genotype ww (homozygous recessive).
- Therefore, out of the four possible outcomes, two are heterozygous (Ww).
- The probability that an offspring will be heterozygous (Ww) is:
[tex]\( \frac{2}{4} = \frac{1}{2} \)[/tex] or 50%.
Second Scenario:
- We have a heterozygous individual Ww crossed with a homozygous dominant individual WW.
- Again, using a Punnett square:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & w \\
\hline W & WW & Ww \\
\hline W & WW & Ww \\
\hline
\end{array}
\][/tex]
- In this Punnett square:
- Two squares show the genotype WW (homozygous dominant).
- Two squares show the genotype Ww (heterozygous).
- None of the squares shows ww (homozygous recessive).
- Therefore, the probability of having a homozygous recessive offspring (ww) is:
[tex]\( 0 \)[/tex] or 0%.
So, from the given scenarios:
1. The probability of having heterozygous offspring from the first cross is 50% or [tex]\( \frac{1}{2} \)[/tex].
2. The probability of having a homozygous recessive offspring from the second cross is 0%.
First Scenario:
- We have a heterozygous male with the genotype Ww and a homozygous recessive female with the genotype ww.
- We use a Punnett square to determine the possible genotypes of their offspring.
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & w \\
\hline w & Ww & ww \\
\hline w & Ww & ww \\
\hline
\end{array}
\][/tex]
- Looking at the Punnett square:
- Two squares show the genotype Ww (heterozygous).
- Two squares show the genotype ww (homozygous recessive).
- Therefore, out of the four possible outcomes, two are heterozygous (Ww).
- The probability that an offspring will be heterozygous (Ww) is:
[tex]\( \frac{2}{4} = \frac{1}{2} \)[/tex] or 50%.
Second Scenario:
- We have a heterozygous individual Ww crossed with a homozygous dominant individual WW.
- Again, using a Punnett square:
[tex]\[
\begin{array}{|c|c|c|}
\hline & W & w \\
\hline W & WW & Ww \\
\hline W & WW & Ww \\
\hline
\end{array}
\][/tex]
- In this Punnett square:
- Two squares show the genotype WW (homozygous dominant).
- Two squares show the genotype Ww (heterozygous).
- None of the squares shows ww (homozygous recessive).
- Therefore, the probability of having a homozygous recessive offspring (ww) is:
[tex]\( 0 \)[/tex] or 0%.
So, from the given scenarios:
1. The probability of having heterozygous offspring from the first cross is 50% or [tex]\( \frac{1}{2} \)[/tex].
2. The probability of having a homozygous recessive offspring from the second cross is 0%.
