High School

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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 $\square$ 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 $\square$.



\[

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

\hline

& W & W \\

\hline

W & WW & Ww \\

\hline

w & Ww & ww \\

\hline

\end{array}

\]

Answer :

- Cross Ww x ww yields $\frac{1}{2}$ heterozygous offspring.
- Cross Ww x WW yields $0$ homozygous recessive offspring.
- The probability of heterozygous offspring in the first cross is $\frac{1}{2}$.
- The probability of homozygous recessive offspring in the second cross is $0$.
$\boxed{0}$

### Explanation
1. Problem Analysis
We are given two separate genetic crosses and asked to determine the probabilities of specific offspring genotypes. First, we analyze the cross between a heterozygous male (Ww) and a homozygous recessive female (ww). Then, we analyze the cross between a heterozygous male (Ww) and a homozygous dominant male (WW).

2. Ww x ww Cross
For the first cross (Ww x ww), we construct a Punnett square to visualize the possible offspring genotypes:

| | W | w |
| :---- | :-- | :-- |
| w | Ww | ww |
| w | Ww | ww |

From the Punnett square, we can see that there are two possible genotypes for the offspring: Ww (heterozygous) and ww (homozygous recessive). Two out of the four possible outcomes are heterozygous (Ww). Therefore, the probability of heterozygous offspring is $\frac{2}{4} = \frac{1}{2}$.

3. Ww x WW Cross
For the second cross (Ww x WW), we construct a Punnett square to visualize the possible offspring genotypes:

| | W | w |
| :---- | :-- | :-- |
| W | WW | Ww |
| W | WW | Ww |

From the Punnett square, we can see that there are two possible genotypes for the offspring: WW (homozygous dominant) and Ww (heterozygous). None of the offspring have the ww (homozygous recessive) genotype. Therefore, the probability of homozygous recessive offspring is $\frac{0}{4} = 0$.

4. Final Answer
Therefore, for the first cross (Ww x ww), the probability of heterozygous offspring is $\frac{1}{2}$, and for the second cross (Ww x WW), the probability of homozygous recessive offspring is $0$.

### Examples
Understanding the probability of offspring genotypes is crucial in agriculture for predicting traits in crops or livestock. For example, a farmer might want to know the likelihood of certain disease-resistant genes being passed on to the next generation of plants.

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