Hardy-Weinberg Equilibrium Problems worksheet with detailed solutions.

Here are the step-by-step solutions for the three Hardy-Weinberg problems shown in your document.

Problem 1

Given:
* Frequency of dominant allele ($A$) = $p = 0.19$
* Frequency of recessive allele ($a$) = $q = 0.81$

(a) Calculate the percentage of heterozygous individuals.
1. Heterozygous individuals have the genotype $Aa$. In the Hardy-Weinberg equation, this is represented by the term $2pq$.
2. Plug in the values: $2 \times 0.19 \times 0.81$.
3. Calculation: $0.38 \times 0.81 = 0.3078$.
4. Convert to percentage: $0.3078 \times 100 = 30.78\%$.
*(Note: The document rounds this to 31%, but 30.78% is more precise).*

(b) Calculate the percentage of homozygous recessives.
1. Homozygous recessive individuals have the genotype $aa$. This is represented by the term $q^2$.
2. Plug in the value for $q$: $0.81^2$.
3. Calculation: $0.81 \times 0.81 = 0.6561$.
4. Convert to percentage: $0.6561 \times 100 = 65.61\%$.
*(Note: The document rounds this to 66%).*

---

Problem 2

Given:
* Total sheep = 900
* White sheep (Dominant phenotype $W\_$) = 891
* Black sheep (Recessive phenotype $ww$) = 9

Calculate allelic frequencies ($p$ and $q$).
1. Start with the recessive group because we know their exact genotype ($ww$).
2. Find the frequency of the recessive genotype ($q^2$):
$$q^2 = \frac{\text{Number of black sheep}}{\text{Total sheep}} = \frac{9}{900} = 0.01$$
3. Find the frequency of the recessive allele ($q$) by taking the square root of $q^2$:
$$q = \sqrt{0.01} = 0.1$$
4. Find the frequency of the dominant allele ($p$). Since $p + q = 1$:
$$p = 1 - 0.1 = 0.9$$

Result:
* Frequency of $W$ ($p$) = 0.9
* Frequency of $w$ ($q$) = 0.1

---

Problem 3

Given:
* Frequency of recessive homozygote genotype ($q^2$) = 0.09

Calculate the percentage of individuals homozygous for the dominant allele.
1. First, find $q$ by taking the square root of the given frequency:
$$q = \sqrt{0.09} = 0.30$$
2. Next, find $p$ using the equation $p + q = 1$:
$$p = 1 - 0.30 = 0.70$$
3. The question asks for "homozygous for the dominant allele" ($AA$), which is represented by $p^2$.
4. Calculate $p^2$:
$$0.70^2 = 0.49$$
5. Convert to percentage: $0.49 \times 100 = 49\%$.

──────────────────────────────────────

Final Answer:
1(a) 30.78% (or approx. 31%)
1(b) 65.61% (or approx. 66%)
2 $p = 0.9$, $q = 0.1$
3 49%