Answer :
We begin by analyzing each cross separately.
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Cross 1: Heterozygous Male ([tex]$Ww$[/tex]) × Homozygous Recessive Female ([tex]$ww$[/tex])
1. The heterozygous male has genotype [tex]$Ww$[/tex], so his possible gametes are [tex]$W$[/tex] and [tex]$w$[/tex].
2. The homozygous recessive female has genotype [tex]$ww$[/tex], so her only possible gamete is [tex]$w$[/tex].
Setting up the Punnett square:
[tex]$$
\begin{array}{c|cc}
& W & w \\ \hline
w & Ww & ww \\
w & Ww & ww \\
\end{array}
$$[/tex]
3. From the Punnett square, we see that out of the four possible combinations, two are heterozygous ([tex]$Ww$[/tex]).
4. This tells us that [tex]$\dfrac{2}{4} = \dfrac{1}{2}$[/tex] of the offspring, which is 50%, will be heterozygous.
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Cross 2: Heterozygous ([tex]$Ww$[/tex]) × Homozygous Dominant ([tex]$WW$[/tex])
1. The heterozygous individual ([tex]$Ww$[/tex]) produces gametes [tex]$W$[/tex] and [tex]$w$[/tex].
2. The homozygous dominant individual ([tex]$WW$[/tex]) produces only [tex]$W$[/tex] gametes.
Setting up the Punnett square:
[tex]$$
\begin{array}{c|cc}
& W & W \\ \hline
W & WW & WW \\
w & Ww & Ww \\
\end{array}
$$[/tex]
3. All offspring will have at least one dominant [tex]$W$[/tex] allele. There are no offspring with the genotype [tex]$ww$[/tex], which would be homozygous recessive.
4. Thus, 0% of the offspring will be homozygous recessive.
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Final Answers:
- 50% of the offspring from the cross between a heterozygous male ([tex]$Ww$[/tex]) and a homozygous recessive female ([tex]$ww$[/tex]) will be heterozygous.
- 0% of the offspring from the cross between a heterozygous individual ([tex]$Ww$[/tex]) and a homozygous dominant individual ([tex]$WW$[/tex]) will be homozygous recessive.
--------------------------------------------------
Cross 1: Heterozygous Male ([tex]$Ww$[/tex]) × Homozygous Recessive Female ([tex]$ww$[/tex])
1. The heterozygous male has genotype [tex]$Ww$[/tex], so his possible gametes are [tex]$W$[/tex] and [tex]$w$[/tex].
2. The homozygous recessive female has genotype [tex]$ww$[/tex], so her only possible gamete is [tex]$w$[/tex].
Setting up the Punnett square:
[tex]$$
\begin{array}{c|cc}
& W & w \\ \hline
w & Ww & ww \\
w & Ww & ww \\
\end{array}
$$[/tex]
3. From the Punnett square, we see that out of the four possible combinations, two are heterozygous ([tex]$Ww$[/tex]).
4. This tells us that [tex]$\dfrac{2}{4} = \dfrac{1}{2}$[/tex] of the offspring, which is 50%, will be heterozygous.
--------------------------------------------------
Cross 2: Heterozygous ([tex]$Ww$[/tex]) × Homozygous Dominant ([tex]$WW$[/tex])
1. The heterozygous individual ([tex]$Ww$[/tex]) produces gametes [tex]$W$[/tex] and [tex]$w$[/tex].
2. The homozygous dominant individual ([tex]$WW$[/tex]) produces only [tex]$W$[/tex] gametes.
Setting up the Punnett square:
[tex]$$
\begin{array}{c|cc}
& W & W \\ \hline
W & WW & WW \\
w & Ww & Ww \\
\end{array}
$$[/tex]
3. All offspring will have at least one dominant [tex]$W$[/tex] allele. There are no offspring with the genotype [tex]$ww$[/tex], which would be homozygous recessive.
4. Thus, 0% of the offspring will be homozygous recessive.
--------------------------------------------------
Final Answers:
- 50% of the offspring from the cross between a heterozygous male ([tex]$Ww$[/tex]) and a homozygous recessive female ([tex]$ww$[/tex]) will be heterozygous.
- 0% of the offspring from the cross between a heterozygous individual ([tex]$Ww$[/tex]) and a homozygous dominant individual ([tex]$WW$[/tex]) will be homozygous recessive.
