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1. If a heterozygous male with the genotype [tex]WW[/tex] is mated with a homozygous recessive female of genotype [tex]ww[/tex], there is a chance that [tex]\square[/tex] of the offspring will be heterozygous.



\[

\begin{array}{|c|c|c|}

\hline

& W & w \\

\hline

w & Ww & ww \\

\hline

w & Ww & ww \\

\hline

\end{array}

\]



2. If the heterozygous [tex]Ww[/tex] is crossed with a homozygous dominant [tex]WW[/tex], then the probability of having a homozygous recessive offspring is [tex]\square[/tex].



\[

\begin{array}{|c|c|c|}

\hline

& W & w \\

\hline

W & WW & Ww \\

\hline

W & WW & Ww \\

\hline

\end{array}

\]

Answer :

- First cross (Ww x ww) yields 50% heterozygous offspring.
- Second cross (Ww x WW) yields 0% homozygous recessive offspring.
- The probability of heterozygous offspring in the first cross is $\boxed{50\%}$.
- The probability of homozygous recessive offspring in the second cross is $\boxed{0\%}$.

### Explanation
1. Problem Analysis
We are given two separate genetic crosses and asked to determine the probabilities of specific offspring genotypes for each. The first cross involves a heterozygous male (Ww) and a homozygous recessive female (ww). The second cross involves a heterozygous individual (Ww) and a homozygous dominant individual (WW).

2. First Cross: Probability of Heterozygous Offspring
For the first cross (Ww x ww), we can use the Punnett square to determine the possible offspring genotypes. The Punnett square shows that there are two possible genotypes: Ww and ww. Two out of the four offspring are Ww (heterozygous). Therefore, the probability of heterozygous offspring is $\frac{2}{4} = \frac{1}{2} = 50\%$.

3. Second Cross: Probability of Homozygous Recessive Offspring
For the second cross (Ww x WW), we again use the Punnett square to determine the possible offspring genotypes. The Punnett square shows that there are two possible genotypes: WW and Ww. None of the offspring are ww (homozygous recessive). Therefore, the probability of homozygous recessive offspring is $\frac{0}{4} = 0\%$.

4. Final Answer
Therefore, the probability of heterozygous offspring in the first cross is 50%, and the probability of homozygous recessive offspring in the second cross is 0%.

### Examples
Understanding genetic probabilities is crucial in breeding programs, whether for agriculture or pets. For example, a farmer might want to know the likelihood of certain traits appearing in their crops, or a breeder might want to predict the coat color of kittens. These probabilities help make informed decisions about which individuals to breed to achieve desired outcomes.