Answer :
We begin by examining each cross separately.
[tex]$$\textbf{First cross: } Ww \times ww$$[/tex]
1. The heterozygous male ([tex]$Ww$[/tex]) produces two types of gametes: one with allele [tex]$W$[/tex] and one with allele [tex]$w$[/tex].
2. The homozygous recessive female ([tex]$ww$[/tex]) produces only gametes with allele [tex]$w$[/tex].
3. Forming the Punnett square:
[tex]$$
\begin{array}{c|cc}
& W & w \\
\hline
w & Ww & ww \\
\end{array}
$$[/tex]
4. From the square, there is one heterozygous genotype ([tex]$Ww$[/tex]) and one homozygous recessive genotype ([tex]$ww$[/tex]). Thus, the probability that an offspring is heterozygous is
[tex]$$
\frac{1}{2} \text{ or } 50\%.
$$[/tex]
---
[tex]$$\textbf{Second cross: } Ww \times WW$$[/tex]
1. The heterozygous individual ([tex]$Ww$[/tex]) again produces gametes [tex]$W$[/tex] and [tex]$w$[/tex].
2. The homozygous dominant individual ([tex]$WW$[/tex]) produces only gametes with allele [tex]$W$[/tex].
3. Construct the Punnett square:
[tex]$$
\begin{array}{c|cc}
& W & w \\
\hline
W & WW & Ww \\
\end{array}
$$[/tex]
4. The offspring produced are either homozygous dominant ([tex]$WW$[/tex]) or heterozygous ([tex]$Ww$[/tex]). There is no possibility for a homozygous recessive genotype ([tex]$ww$[/tex]). Therefore, the probability of having a homozygous recessive offspring is
[tex]$$
0.
$$[/tex]
---
\textbf{Final answers:}
- The probability that an offspring is heterozygous in the first cross is [tex]$\frac{1}{2}$[/tex] (50%).
- The probability of having a homozygous recessive offspring in the second cross is [tex]$0$[/tex].
[tex]$$\textbf{First cross: } Ww \times ww$$[/tex]
1. The heterozygous male ([tex]$Ww$[/tex]) produces two types of gametes: one with allele [tex]$W$[/tex] and one with allele [tex]$w$[/tex].
2. The homozygous recessive female ([tex]$ww$[/tex]) produces only gametes with allele [tex]$w$[/tex].
3. Forming the Punnett square:
[tex]$$
\begin{array}{c|cc}
& W & w \\
\hline
w & Ww & ww \\
\end{array}
$$[/tex]
4. From the square, there is one heterozygous genotype ([tex]$Ww$[/tex]) and one homozygous recessive genotype ([tex]$ww$[/tex]). Thus, the probability that an offspring is heterozygous is
[tex]$$
\frac{1}{2} \text{ or } 50\%.
$$[/tex]
---
[tex]$$\textbf{Second cross: } Ww \times WW$$[/tex]
1. The heterozygous individual ([tex]$Ww$[/tex]) again produces gametes [tex]$W$[/tex] and [tex]$w$[/tex].
2. The homozygous dominant individual ([tex]$WW$[/tex]) produces only gametes with allele [tex]$W$[/tex].
3. Construct the Punnett square:
[tex]$$
\begin{array}{c|cc}
& W & w \\
\hline
W & WW & Ww \\
\end{array}
$$[/tex]
4. The offspring produced are either homozygous dominant ([tex]$WW$[/tex]) or heterozygous ([tex]$Ww$[/tex]). There is no possibility for a homozygous recessive genotype ([tex]$ww$[/tex]). Therefore, the probability of having a homozygous recessive offspring is
[tex]$$
0.
$$[/tex]
---
\textbf{Final answers:}
- The probability that an offspring is heterozygous in the first cross is [tex]$\frac{1}{2}$[/tex] (50%).
- The probability of having a homozygous recessive offspring in the second cross is [tex]$0$[/tex].
