Answer :
Let’s analyze each cross step by step.
Step 1. Cross Between a Heterozygous Male and a Homozygous Recessive Female
The male has the genotype
[tex]$$Ww,$$[/tex]
and the female has the genotype
[tex]$$ww.$$[/tex]
The male can contribute either the [tex]$W$[/tex] allele or the [tex]$w$[/tex] allele with equal probability, while the female can only contribute the [tex]$w$[/tex] allele. We set up a Punnett square as follows:
[tex]$$
\begin{array}{c|cc}
& w & w \\
\hline
W & Ww & Ww \\
w & ww & ww \\
\end{array}
$$[/tex]
Each box represents a possible combination of alleles in the offspring. Here, two of the four outcomes are heterozygous ([tex]$Ww$[/tex]) and two are homozygous recessive ([tex]$ww$[/tex]).
Thus, the probability that an offspring will be heterozygous is
[tex]$$\frac{2}{4} \times 100\% = 50\%.$$[/tex]
Step 2. Cross Between a Homozygous Dominant and Another Homozygous Dominant
For the second situation, both parents have the genotype
[tex]$$WW.$$[/tex]
Since each parent can only contribute the [tex]$W$[/tex] allele, all offspring will receive a [tex]$W$[/tex] allele from each parent. The Punnett square for this cross is:
[tex]$$
\begin{array}{c|cc}
& W & W \\
\hline
W & WW & WW \\
W & WW & WW \\
\end{array}
$$[/tex]
Every box in the Punnett square shows the genotype [tex]$WW$[/tex]. This means that no offspring will be homozygous recessive ([tex]$ww$[/tex]).
Thus, the probability of having a homozygous recessive offspring is
[tex]$$0\%.$$[/tex]
Final Answer:
1. The probability that an offspring will be heterozygous from the first cross is [tex]$\boxed{50\%}.$[/tex]
2. The probability of having a homozygous recessive offspring from the second cross is [tex]$\boxed{0\%}.$[/tex]
Step 1. Cross Between a Heterozygous Male and a Homozygous Recessive Female
The male has the genotype
[tex]$$Ww,$$[/tex]
and the female has the genotype
[tex]$$ww.$$[/tex]
The male can contribute either the [tex]$W$[/tex] allele or the [tex]$w$[/tex] allele with equal probability, while the female can only contribute the [tex]$w$[/tex] allele. We set up a Punnett square as follows:
[tex]$$
\begin{array}{c|cc}
& w & w \\
\hline
W & Ww & Ww \\
w & ww & ww \\
\end{array}
$$[/tex]
Each box represents a possible combination of alleles in the offspring. Here, two of the four outcomes are heterozygous ([tex]$Ww$[/tex]) and two are homozygous recessive ([tex]$ww$[/tex]).
Thus, the probability that an offspring will be heterozygous is
[tex]$$\frac{2}{4} \times 100\% = 50\%.$$[/tex]
Step 2. Cross Between a Homozygous Dominant and Another Homozygous Dominant
For the second situation, both parents have the genotype
[tex]$$WW.$$[/tex]
Since each parent can only contribute the [tex]$W$[/tex] allele, all offspring will receive a [tex]$W$[/tex] allele from each parent. The Punnett square for this cross is:
[tex]$$
\begin{array}{c|cc}
& W & W \\
\hline
W & WW & WW \\
W & WW & WW \\
\end{array}
$$[/tex]
Every box in the Punnett square shows the genotype [tex]$WW$[/tex]. This means that no offspring will be homozygous recessive ([tex]$ww$[/tex]).
Thus, the probability of having a homozygous recessive offspring is
[tex]$$0\%.$$[/tex]
Final Answer:
1. The probability that an offspring will be heterozygous from the first cross is [tex]$\boxed{50\%}.$[/tex]
2. The probability of having a homozygous recessive offspring from the second cross is [tex]$\boxed{0\%}.$[/tex]
