Sure! Let's break down the genetic cross results for the given scenarios:
Scenario 1:
In the first scenario, we are crossing a heterozygous male with the genotype [tex]\( Ww \)[/tex] with a homozygous recessive female with the genotype [tex]\( ww \)[/tex].
The Punnett square for this cross looks like this:
[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, we can see the potential genotypes of the offspring:
- Two offspring are [tex]\( Ww \)[/tex] (heterozygous).
- Two offspring are [tex]\( ww \)[/tex] (homozygous recessive).
There are a total of 4 possible genotypes. Out of these, 2 are heterozygous [tex]\( Ww \)[/tex]. Therefore, the probability that an offspring will be heterozygous [tex]\( Ww \)[/tex] is:
[tex]\[
\frac{2}{4} = 0.5
\][/tex]
Thus, there is a 0.5 (or 50%) chance that the offspring will be heterozygous.
Scenario 2:
In the second scenario, we are crossing a heterozygous [tex]\( Ww \)[/tex] individual with a homozygous dominant [tex]\( WW \)[/tex] individual.
The Punnett square for this cross looks like this:
[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, we can see the potential genotypes of the offspring:
- Two offspring are [tex]\( WW \)[/tex] (homozygous dominant).
- Two offspring are [tex]\( Ww \)[/tex] (heterozygous).
There are a total of 4 possible genotypes. None of these genotypes are homozygous recessive [tex]\( ww \)[/tex]. Thus, the probability of having a homozygous recessive [tex]\( ww \)[/tex] offspring is:
[tex]\[
0
\][/tex]
Therefore, the probability of having a homozygous recessive offspring in this cross is 0 (or 0%).
To summarize:
1. There is a 0.5 chance that the offspring will be heterozygous in the first cross.
2. The probability of having a homozygous recessive offspring in the second cross is 0.
