Hardy Weinberg Problem Set: Calculate allele and genotype frequencies using a mouse population with dark and light fur phenotypes.

Let’s solve this step by step.

We are told:
- There are 20 mice total (count them in the image — 4 rows × 5 = 20).
- Phenotypes: dark (D) and light (d).
- Light is recessive → so only genotype dd shows light phenotype.
- Dark is dominant → genotypes DD or Dd show dark phenotype.

First, count how many mice are light-colored (recessive phenotype):

Looking at the image (we’ll count carefully):

Row 1: positions 3, 5 → 2 light
Row 2: positions 1, 3, 5 → 3 light
Row 3: position 4 → 1 light
Row 4: positions 2, 4 → 2 light

Total light (dd) = 2 + 3 + 1 + 2 = 8

So, number of individuals with recessive phenotype (q²) = 8/20 = 0.4

Wait — let me double-check that count.

Actually, let’s list them row by row to be sure:

Row 1 (left to right):
1. dark
2. dark
3. light ←
4. dark
5. light ← → 2 light

Row 2:
1. light ←
2. dark
3. light ←
4. dark
5. light ← → 3 light

Row 3:
1. dark
2. dark
3. dark
4. light ←
5. dark → 1 light

Row 4:
1. dark
2. light ←
3. dark
4. light ←
5. dark → 2 light

Total light = 2 + 3 + 1 + 2 = 8

Yes, correct.

So:

A. Frequency of individuals that display the recessive trait (light) = q² = 8/20 = 0.4

B. Frequency of recessive allele (q) = √(q²) = √0.4 ≈ 0.6325 (we’ll keep more decimals for accuracy)

But wait — Hardy-Weinberg assumes random mating, no selection, etc. But here we’re just calculating from observed data.

C. Frequency of dominant allele (p) = 1 - q = 1 - √0.4 ≈ 1 - 0.6325 = 0.3675

D. Percentage of homozygous dominant (DD) = p² = (√0.4 is q, so p = 1 - q)

Actually, better to compute numerically:

q² = 0.4 → q = √0.4 = √(2/5) = approximately 0.632455532

Then p = 1 - q ≈ 1 - 0.632455532 = 0.367544468

Then p² = (0.367544468)² ≈ 0.1351

So percentage = 0.1351 × 100 ≈ 13.51%

E. Percentage of heterozygotes (Dd) = 2pq = 2 × p × q ≈ 2 × 0.3675 × 0.6325 ≈ ?

Calculate:

0.3675 × 0.6325 ≈ let's compute:

0.3675 × 0.6 = 0.2205
0.3675 × 0.0325 ≈ 0.01194375
Total ≈ 0.23244375

Then 2 × that ≈ 0.4648875 → 46.49%

F. Suggest a reason for the number of d alleles being high.

Since q² = 0.4, meaning 40% of population is dd, and q ≈ 0.63, which is fairly high.

Possible reasons: maybe the environment favors light color? Or maybe there was recent migration bringing in more d alleles? Or perhaps genetic drift? Or maybe the population is not in Hardy-Weinberg equilibrium because of selection against dark mice?

But since it says “suggest a reason”, we can say: Perhaps the light-colored mice have a survival advantage in their environment (e.g., camouflage), leading to higher frequency of the recessive allele.

Alternatively, maybe the population recently experienced a bottleneck or founder effect that increased the d allele frequency.

But simplest: natural selection favoring the recessive phenotype.

Now, let’s write answers clearly.

Note: The problem says “Calculate” for A–E, and “Suggest” for F.

Also, note: In part E, it says “the percentage of mice in the population that are heterozygous, Dd.” We calculated using Hardy-Weinberg formula, but actually, if we had genotype counts, we could calculate directly. However, we don’t know how many are DD vs Dd among the dark ones — we only know phenotypes.

Ah! Important point: We cannot directly observe genotypes — only phenotypes. So for parts D and E, we must use Hardy-Weinberg assumptions to estimate genotype frequencies from phenotype frequencies.

That’s standard for these problems unless otherwise specified.

So our approach is correct.

But let’s verify total adds up:

q² = 0.4 (dd)
p² ≈ 0.1351 (DD)
2pq ≈ 0.4649 (Dd)
Sum: 0.4 + 0.1351 + 0.4649 = 1.0000 → good.

Percentages:
dd: 40%
DD: ~13.51%
Dd: ~46.49%

Now, for final answers:

A. q² = 8/20 = 0.4
B. q = √0.4 ≈ 0.632 (or exact form √(2/5))
C. p = 1 - q ≈ 0.368
D. p² ≈ 0.135 → 13.5%
E. 2pq ≈ 0.465 → 46.5%
F. Reason: e.g., Natural selection may favor the light-colored mice, increasing the frequency of the recessive allele.

But let’s use precise values.

Actually, since 8 out of 20 are dd, q² = 0.4 exactly.

q = √0.4 = √(4/10) = 2/√10 = √10 / 5 ≈ 3.16227766 / 5 ≈ 0.632455532

p = 1 - q ≈ 0.367544468

p² = (1 - √0.4)^2 = 1 - 2√0.4 + 0.4 = 1.4 - 2√0.4

But better to compute numerically.

For school level, they might expect rounded answers.

Typically, two decimal places for frequencies, one for percentages? Let’s see.

In many textbooks, they round to two or three decimals.

I’ll go with:

A. 0.4
B. 0.63
C. 0.37
D. 14% (since 0.1351 rounds to 14%)
Wait, 0.1351 is closer to 13.5%, but sometimes they want whole percent.

Check calculation again.

p = 1 - sqrt(0.4) = 1 - 0.632455532 = 0.367544468

p² = 0.367544468^2 = let's calculate:

0.3675 * 0.3675 = ?

0.36*0.36=0.1296
0.36*0.0075=0.0027
0.0075*0.36=0.0027
0.0075*0.0075≈0.00005625
Better: (0.3675)^2 = 0.3675 * 0.3675

Compute: 3675 * 3675 / 10^8

Or: 0.3675 * 0.3675

= (0.37 - 0.0025)^2 = 0.37^2 - 2*0.37*0.0025 + (0.0025)^2 = 0.1369 - 0.00185 + 0.00000625 ≈ 0.13505625

So p² ≈ 0.1351 → 13.51%

Similarly, 2pq = 2 * 0.3675 * 0.6325 ≈ 2 * 0.2324 = 0.4648 → 46.48%

For reporting, perhaps:

A. 0.4
B. 0.63
C. 0.37
D. 13.5%
E. 46.5%
F. [reason]

Some might prefer fractions.

q² = 8/20 = 2/5 = 0.4
q = √(2/5) = √10 / 5
But probably decimal is fine.

Now for F: "Suggest a reason for the number of d alleles in the population."

Since q is high (0.63), meaning d allele is common.

Reasons could include:

- The recessive trait (light color) provides a selective advantage (e.g., better camouflage in sandy environment).
- Genetic drift in a small population.
- Recent immigration of mice carrying the d allele.
- Mutation pressure (less likely).

Most straightforward: natural selection favoring the light phenotype.

Final Answer section should have all answers labeled.

Let me compile:

A. 0.4
B. 0.63
C. 0.37
D. 13.5%
E. 46.5%
F. The high frequency of the d allele may be due to natural selection favoring the light-colored mice, as they might be better camouflaged in their environment, leading to higher survival and reproduction rates.

To be precise, let's use more accurate rounding.

In some curricula, they expect:

q = sqrt(0.4) = 0.632 → often rounded to 0.63
p = 1 - 0.632 = 0.368 → rounded to 0.37
p² = 0.368² = 0.135424 → 13.5%
2pq = 2*0.368*0.632 = 2*0.232576 = 0.465152 → 46.5%

Yes.

For F, any reasonable biological reason is acceptable.

Now, write Final Answer.