Hardy-Weinberg equilibrium worksheet for biology students, illustrating gene expression in San Mariano frogs and calculating allele frequencies in a population.

Explanation:
Let’s go step by step through each question.

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1. What is the definition of microevolution of a population?
Microevolution means a change in allele frequencies in a population over time.
In this case, it's specifically about how the frequency of the *A* and *a* alleles changes from one generation to the next — especially under selection (like freezing tolerance).

So the answer is:
A change in allele frequencies in a population over time.

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2. What are the Hardy-Weinberg equations for genotype frequencies?
The Hardy-Weinberg principle says that if no evolution is happening (no selection, mutation, migration, drift, and random mating), then genotype frequencies stay constant and follow these formulas:

- Frequency of AA = $ p^2 $
- Frequency of Aa = $ 2pq $
- Frequency of aa = $ q^2 $

Where:
- $ p $ = frequency of dominant allele *A*
- $ q $ = frequency of recessive allele *a*
- And $ p + q = 1 $

So the equations are:
$ p^2 $ for AA, $ 2pq $ for Aa, $ q^2 $ for aa.

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3. Calculate the allele frequency for the “A” allele (p), assuming the population is currently in H-W equilibrium.

We’re told:
- There are 19 individuals total.
- 18 have the functional *A* protein → they are either AA or Aa
- 1 does not have the functional protein → must be aa (since only *aa* lacks the protein; *A* is dominant for function)

So:
- Number of aa = 1
- So genotype frequency of *aa* = $ \frac{1}{19} $

Under H-W equilibrium:
Frequency of *aa* = $ q^2 = \frac{1}{19} $

So:
$ q = \sqrt{\frac{1}{19}} \approx \sqrt{0.05263} \approx 0.2294 $

Then:
$ p = 1 - q = 1 - 0.2294 = 0.7706 $

But wait — let’s double-check using counts instead of frequencies, to avoid rounding error.

Each individual has 2 alleles → total alleles = $ 19 \times 2 = 38 $

- The 1 *aa* individual contributes 2 a alleles.
- The remaining 18 individuals are either AA or Aa — but we don’t know how many of each.

However, the problem says: *“assuming the population is currently in H-W equilibrium”*. That means genotype frequencies follow $ p^2 $, $ 2pq $, $ q^2 $. So we can use the observed *aa* count to estimate $ q^2 $, then get $ q $, then $ p $.

Yes — that’s the standard approach when assuming H-W.

So:
- Observed frequency of *aa* = $ \frac{1}{19} $
- So $ q^2 = \frac{1}{19} $
- $ q = \sqrt{1/19} = \frac{1}{\sqrt{19}} \approx \frac{1}{4.3589} \approx 0.2294 $
- $ p = 1 - q \approx 0.7706 $

Rounded to 2 decimal places: p ≈ 0.77

But the problem might expect an exact expression or a decimal. Let’s keep more precision for now.

Alternatively, maybe we can find exact allele count if we assume H-W holds exactly with integer individuals — but 19 is small, so exact fit may not be possible. The standard method is to use observed *aa* frequency = $ q^2 $, solve for $ q $, then $ p $.

So:
$ q = \sqrt{1/19} $
$ p = 1 - \sqrt{1/19} $

But likely they want a decimal. Let’s compute accurately:

$ \sqrt{19} = 4.35889894354 $
So $ q = 1 / 4.35889894354 = 0.22941573387 $
Then $ p = 1 - 0.22941573387 = 0.77058426613 $

Rounded to 2 decimals: 0.77
To 3 decimals: 0.771

But let’s see part 4 — it says “Calculate the allele frequencies for ‘A’ allele (p), assuming the population is currently in H-W equilibrium. p = ___ , q = ___”

So they want both p and q.

We’ll write:
- $ q = \sqrt{1/19} \approx 0.23 $
- $ p = 1 - q \approx 0.77 $

But maybe they expect us to use the exact count method differently? Let’s think again:

If the population is in H-W equilibrium, then expected number of *aa* = $ q^2 \cdot N $, and we observe 1 out of 19. So yes, $ q^2 = 1/19 $. That’s correct.

So answer for #3:
p ≈ 0.77, q ≈ 0.23
(Or more precisely: p = 1 − √(1/19), q = √(1/19))

But since the problem gives blank with “p = ___, q = ___”, and earlier parts use decimals (like q=0.9), likely they want decimals rounded to 2 places.

Let’s hold that.

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4. Calculate the allele frequencies for “A” allele (p), assuming the population is currently in H-W equilibrium.
This is same as #3 — probably a repeat or formatting issue. The blanks are:
p = ___, q = ___

From above:
- Observed *aa* = 1/19 → $ q^2 = 1/19 $
- So $ q = \sqrt{1/19} \approx 0.2294 $
- $ p = 1 - 0.2294 = 0.7706 $

Rounded to two decimal places:
p = 0.77, q = 0.23

Check: $ p + q = 0.77 + 0.23 = 1.00 $ ✔
And $ q^2 = 0.23^2 = 0.0529 $, and 1/19 ≈ 0.05263 — very close. Good enough for this level.

Some teachers accept 0.77 and 0.23.

Alternatively, if they want more precise: p = 0.771, q = 0.229 — but given the context, two decimals is fine.

Let me verify once more:
Total alleles = 38
If q = 0.2294, then number of *a* alleles = 38 × 0.2294 ≈ 8.72 → not integer, but that’s okay — allele frequency is a proportion, not required to give integer counts in small samples.

So final answers:

1. A change in allele frequencies in a population over time.
2. $ p^2 $, $ 2pq $, $ q^2 $
3. p ≈ 0.77
4. p = 0.77, q = 0.23

But the problem lists 4 questions, and #3 and #4 seem overlapping. Looking back:

> 3. Calculate the allele frequency for the “A” allele (p), assuming the population is currently in H-W equilibrium. In this instance, you must solve “p” before “q”!
> 4. Calculate the allele frequencies for “A” allele (p), assuming the population is currently in H-W equilibrium. p = ___, q = ___

Probably #3 wants just p (with reasoning), and #4 wants both filled in.

So for #3: p = 1 − √(1/19) ≈ 0.77
For #4: p = 0.77, q = 0.23

Since the instruction says “ONLY THEN provide the final answer”, and the format asks for Final Answer at the end — and the user likely expects the numeric answers for blanks — I’ll give the final numeric answers as requested in #4 (since it has blanks).

Final Answer:
p = 0.77, q = 0.23