Hardy-Weinberg Practice Problems worksheet with step-by-step questions on allele frequencies and population genetics.

Let's solve each problem step by step using the Hardy-Weinberg principle. The Hardy-Weinberg equations are:

$$
p^2 + 2pq + q^2 = 1 \quad \text{(genotype frequencies)}
$$
$$
p + q = 1 \quad \text{(allele frequencies)}
$$

Where:
- $ p $ = frequency of dominant allele
- $ q $ = frequency of recessive allele
- $ p^2 $ = frequency of homozygous dominant individuals
- $ 2pq $ = frequency of heterozygous individuals
- $ q^2 $ = frequency of homozygous recessive individuals

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Problem 1: Lethal Recessive Condition

> A lethal recessive condition is responsible for the death of 1 out of every 20,000 babies born in South America.

This means that 1 in 20,000 newborns is homozygous recessive (aa) and dies due to the condition. So:

#### a. What is the frequency of the recessive allele in the population?

We know:
- $ q^2 = \frac{1}{20,000} = 0.00005 $

Now solve for $ q $:

$$
q = \sqrt{0.00005} = \sqrt{5 \times 10^{-5}} \approx 0.00707
$$

So, the frequency of the recessive allele (q) ≈ 0.00707

✔ Answer a: $ q \approx 0.00707 $

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#### b. What is the frequency of the dominant allele in the population?

Use $ p + q = 1 $

$$
p = 1 - q = 1 - 0.00707 = 0.99293
$$

✔ Answer b: $ p \approx 0.99293 $

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#### c. What is the frequency of homozygous recessive individuals in the population?

This is $ q^2 $, which we already used:

$$
q^2 = \frac{1}{20,000} = 0.00005
$$

✔ Answer c: $ q^2 = 0.00005 $

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#### d. How many people, out of 1,000 individuals in this population, would carry the recessive allele?

"Carry the recessive allele" means either heterozygous (Aa) or homozygous recessive (aa). But since aa individuals die early, we assume they are not present in the adult population. However, the question asks about carriers, which typically refers to heterozygotes (since they carry the allele but don’t express the disease).

So we want the number of heterozygous individuals (Aa), whose frequency is $ 2pq $.

Calculate:

$$
2pq = 2 \times 0.99293 \times 0.00707 \approx 2 \times 0.00702 \approx 0.01404
$$

So, frequency of carriers ≈ 0.01404

In 1,000 individuals:

$$
0.01404 \times 1000 = 14.04 \approx 14 \text{ people}
$$

✔ Answer d: Approximately 14 people out of 1,000 carry the recessive allele.

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Problem 2: Snapdragon Flowers with Incomplete Dominance

> In snapdragons, R = red flowers (dominant), r = white flowers (recessive).
> Incomplete dominance → pink flowers = heterozygous (Rr).
> Population: 100 snapdragons
> 50 have pink flowers (Rr)
> 25 have white flowers (rr)
> So remaining 25 must be red (RR)

Let’s find the frequencies.

Total = 100

- RR (red): 25 → frequency = $ \frac{25}{100} = 0.25 $
- Rr (pink): 50 → frequency = $ \frac{50}{100} = 0.50 $
- rr (white): 25 → frequency = $ \frac{25}{100} = 0.25 $

Now use these to find allele frequencies.

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#### a. What is the frequency of the dominant allele in the population?

Dominant allele = R

Each RR contributes 2 R alleles → $ 25 \times 2 = 50 $ R alleles
Each Rr contributes 1 R allele → $ 50 \times 1 = 50 $ R alleles
Each rr contributes 0 R alleles

Total R alleles = $ 50 + 50 = 100 $

Total alleles in population = $ 100 \text{ individuals} \times 2 = 200 $

So:

$$
p = \frac{100}{200} = 0.5
$$

✔ Answer a: $ p = 0.5 $

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#### b. What is the frequency of the recessive allele in the population?

Recessive allele = r

Each rr contributes 2 r alleles → $ 25 \times 2 = 50 $ r alleles
Each Rr contributes 1 r allele → $ 50 \times 1 = 50 $ r alleles
RR contributes 0

Total r alleles = $ 50 + 50 = 100 $

So:

$$
q = \frac{100}{200} = 0.5
$$

✔ Answer b: $ q = 0.5 $

Alternatively: $ q = 1 - p = 1 - 0.5 = 0.5 $

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#### c. What is the frequency of heterozygous individuals?

From data: