Answer :
Let's analyze both crosses step by step.
1. For the first cross, a heterozygous male with genotype [tex]$Ww$[/tex] is mated with a homozygous recessive female with genotype [tex]$ww$[/tex].
- The male produces two types of gametes: one carrying the allele [tex]$W$[/tex] and the other carrying the allele [tex]$w$[/tex].
- The female produces only one type of gamete because she is homozygous recessive: each gamete carries the allele [tex]$w$[/tex].
When we set up a Punnett square using these gametes, we have:
[tex]$$
\begin{array}{c|cc}
& W & w \\
\hline
w & Ww & ww \\
w & Ww & ww \\
\end{array}
$$[/tex]
In the resulting offspring, two out of four (or [tex]$50\%$[/tex]) are heterozygous ([tex]$Ww$[/tex]).
2. For the second cross, a heterozygous individual with genotype [tex]$Ww$[/tex] is crossed with a homozygous dominant individual with genotype [tex]$WW$[/tex].
- The heterozygous individual produces gametes carrying either the allele [tex]$W$[/tex] or the allele [tex]$w$[/tex].
- The homozygous dominant individual can only produce gametes carrying the allele [tex]$W$[/tex].
The Punnett square for this cross is:
[tex]$$
\begin{array}{c|cc}
& W & w \\
\hline
W & WW & Ww \\
W & WW & Ww \\
\end{array}
$$[/tex]
Here, none of the offspring are homozygous recessive ([tex]$ww$[/tex]). This means that the probability of having a homozygous recessive offspring is [tex]$0\%$[/tex].
Thus, the final answers are:
- [tex]$50\%$[/tex] of the offspring will be heterozygous ([tex]$Ww$[/tex]) in the first cross.
- The probability of having a homozygous recessive ([tex]$ww$[/tex]) offspring in the second cross is [tex]$0\%$[/tex].
1. For the first cross, a heterozygous male with genotype [tex]$Ww$[/tex] is mated with a homozygous recessive female with genotype [tex]$ww$[/tex].
- The male produces two types of gametes: one carrying the allele [tex]$W$[/tex] and the other carrying the allele [tex]$w$[/tex].
- The female produces only one type of gamete because she is homozygous recessive: each gamete carries the allele [tex]$w$[/tex].
When we set up a Punnett square using these gametes, we have:
[tex]$$
\begin{array}{c|cc}
& W & w \\
\hline
w & Ww & ww \\
w & Ww & ww \\
\end{array}
$$[/tex]
In the resulting offspring, two out of four (or [tex]$50\%$[/tex]) are heterozygous ([tex]$Ww$[/tex]).
2. For the second cross, a heterozygous individual with genotype [tex]$Ww$[/tex] is crossed with a homozygous dominant individual with genotype [tex]$WW$[/tex].
- The heterozygous individual produces gametes carrying either the allele [tex]$W$[/tex] or the allele [tex]$w$[/tex].
- The homozygous dominant individual can only produce gametes carrying the allele [tex]$W$[/tex].
The Punnett square for this cross is:
[tex]$$
\begin{array}{c|cc}
& W & w \\
\hline
W & WW & Ww \\
W & WW & Ww \\
\end{array}
$$[/tex]
Here, none of the offspring are homozygous recessive ([tex]$ww$[/tex]). This means that the probability of having a homozygous recessive offspring is [tex]$0\%$[/tex].
Thus, the final answers are:
- [tex]$50\%$[/tex] of the offspring will be heterozygous ([tex]$Ww$[/tex]) in the first cross.
- The probability of having a homozygous recessive ([tex]$ww$[/tex]) offspring in the second cross is [tex]$0\%$[/tex].
