1. Some Population Genetic Analysis to Get Us Started

In the last tutorial, we counted allele frequencies of a small population of mice, some of which were albino, and others with normal coloration. Because albinism is caused by a recessive allele, a, we know that the only mice that will show the albino phenotype are the ones with the genotype aa. Pigmented mice can have either the genotype AA or Aa.

• The “a” allele
• The “A” allele
• Homozygous recessive individuals (genotype “aa”).
• Heterozygotes (genotype “Aa”)
• Homozygous dominant individuals (genotype “AA”).

To help you in this analysis, note that if the total population size is 200, that means that there are 400 alleles in this slice of the mouse gene pool (each mouse has two alleles, one from each parent).

Note that it might be easier to do this analysis on paper. If you want to print out a paper version of this gene pool, here’s a link to a PDF version of the table below.

1.a. Population Genetics Analysis Task 1

1.b. Population Genetics Analysis Task 2

2. Population Genetic Analysis is about counting the Ps and the Qs

In 1908, the mathematician G.H. Hardy and the obstetrician Wilhelm Weinberg independently published papers describing the key principles of population genetics. Their work mathematically describes the relationships between the alleles, genotypes, and certain phenotypes in a population’s gene pool. Note, however, that these relationships only apply to a population that’s in equilibrium. In an evolutionary context, equilibrium means roughly the same as non-evolving. In other words, these are gene pools in which allele frequencies are staying the same over time. In later tutorials in this series, we’ll explore the conditions that allow gene pools to remain in equilibrium, and (more interestingly) the conditions that cause gene pools to change.

In the notation that’s typically used to represent Hardy and Weinberg’s work, p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele.

As you’ve discovered above, p + q = 1. In plain English, that means that the frequency of the dominant allele + the frequency of the recessive allele = 100% of the alleles for that gene in the population.

Now, that probably seems pretty obvious. But let’s take that expression, and square both sides:

(p + q)(p+q) = 1 * 1

What you get is

p² + 2pq +q² = 1

That might not seem like a big deal, but this expression is astoundingly useful. That’s because, in this ideal, non-evolving, population:

• p² is the frequency of individuals who are homozygous dominant (such as AA, in our population above)
• 2pq is the frequency of heterozygotes (such as Aa)
• q² is the frequency of homozygous recessives (such as aa).

The frequency of homozygous recessives is the key. That’s because the homozygous recessive phenotype has a known genotype. If you see an individual with a recessive phenotype, you know that individual’s genotype. For example, think of the mice in the sample population above. Because the albino allele (a) is recessive, any albino mouse has to have the genotype “aa.” If you count the frequency of albino mice, you know one of the expressions in the Hardy-Weinberg equation (q²). From there, you can figure out q (just by taking the square root of q²). And from there, you can figure out many other things about that population’s genetic structure (at least, for that gene).

Again, let’s turn back to our population of 200 mice above. You discovered, through counting, that the frequency of albino mice (homozygous recessive mice with the “aa” genotype) was 0.49.

• That means that q² = 0.49.
• If you want to know q (the frequency of the recessive allele) just take the square root of 0.49. That’s 0.7
• Since p + q = 1, that means that the frequency of “A” has to be 1.0 – 0.7, which equals 0.3.
• Since the frequency of heterozygotes is 2pq, you can calculate that 2 * 0.3 * 0.7 = 0.42 (which means that heterozygotes should make up 42% of the population).
• And, you can also calculate that the expected frequency of homozygous dominant individuals should be 0.3², or 0.09.

That’s a pretty high yield from two equations! But let’s spend a few moments with flashcards to get these symbols and equations memorized.

3. Use Cross-Multiplication Tables!

When solving population genetics problems involving the Hardy-Weinberg equation, you’ll make your life much easier (and your test scores much higher) if you use cross-multiplication tables. It’s MUCH simpler than solving for any of the variables in the equation p² + 2pq + q² = 1. Let me show you how these work.

Using the symbols of the Hardy-Weinberg equation, here’s what a cross-multiplication table looks like:

So let’s walk through the steps of how you’d use this to solve population genetics problems.

STEP 1: Draw a Punnett square.

STEP 2: Place the alleles you’re analyzing on the sides and inner squares as if this were a monohybrid cross (Aa x Aa)

STEP 3: Now, as a proof of concept, let’s add the values for p and q from our sample mouse gene pool above.

STEP 4: Notice that if you cross multiply, you can find the values of p², q², and pq.

And, of course, if you had any value, you can simply use the functions on your calculator to find all the rest. For example, let’s say that by observing a population, you discovered that the frequency of homozygous recessives is 4%, or 0.04. The frequency of homozygous recessives is q², so just plug that into the bottom right corner, where q² goes.

Knowing that aa is 0.04, you take the square root of that to get the frequency of q.

Now you can figure out p, because 1 – q = p (in this case, 1 – 0.2 = 0.8)

And now cross multiply to figure out the rest.

This means that from one value (the frequency of homozygous recessives, which was 0.04), we’ve been able to determine

• the frequency of the recessive allele (20%)
• the frequency of the dominant allele (80%)
• the frequency of heterozygotes (32%)
• the frequency of homozygous dominant (64%)

4. Solving Hardy-Weinberg Problems Using Cross-Multiplication Tables

These practice problems were inspired (or copied) from various websites including:

• http://www.k-state.edu/parasitology/biology198/hardwein.html
• http://biogirl757.weebly.com/uploads/3/9/6/8/39680816/hardy_weinberg_problem_set_key.pdf

What’s Next?

Proceed to Topics 7.4 -7.5, Part 3: The Hardy-Weinberg Principle (the next tutorial in AP Bio Topics 7.4 – 7.5, Population Genetics)