What are the differences in the two possible methods to calculate the Hardy-Weinberg Equation?

Question

I am more than flabbergasted by calculations of Hardy-Weinberg equilibria. The formula's theory assumes a binomal distribution of allele frequencies in a population, and hence allows the comparison of observed phenotype frequencies to an ideal model distribution.

$$ p^2 + 2pq + q^2 = 1 $$

with

$p^2$ = Frequency of homozygous phenotype PP

$2pq$ = Frequency of heterozygous phenotype PQ

$q^2$ = Frequency of homozygous phenotype QQ

There are two different methods of working with the HW equation that I have come across, and they do not seem to be compatible; used on the same data sets they yield wildly different results. I feel like I'm not seeing the forest for the trees, to be honest.

Is either of these methods (technically) wrong? How do I decide which one to use for each data set (i.e. is it valid to use method #1 on 3-allele-data sets)? Or do I just use whatever method is easiest? Doesn't this lead to discrepancies in the results? This seems like a big thing, especially when comparing real-world observation data to assumed HW distributions.

Example data set (EST/California) and results of HW calculations. I'm using observed phenotype in the place of the original data's observed genotype.

Phenotype     n  |   HW1   HW2p   HW2q
PP           37  |  33.9   37.1   28.8
PQ           20  |  25.0   23.0   28.2
QQ            7  |   4.5    3.8    7.0

---

Calculation method #1

As suggested by above-linked data set source and Beebee/Rowe's Introduction to Molecular Ecology, alleles are simply added up to get values for p and q respectively:

Phenotype    n  |   P    Q
PP          37  |  74    0   (37 * PP = 74 P)
PQ          20  |  20   20   (20 * PQ = 20 P + 20 Q)
QQ          7   |   0   14   ( 7 * QQ = 14 Q)
Sums        64  |  94   34

From this we can directly infer p/q: $p = 94/(94+34) =0.73$ and $q = 34/(94+34) = 0.27$; with the condition $p + q = 1$ remaining true.

Using these values for p and q in the HW equation yields: $$ p^2 + 2pq + q^2 = 1 \\ (0.73)^2 + 2(0.73)(0.27) + (0.27)^2 = 1 $$ And a population-wide phenotype distribution of

Phenotype      n
PP         33.92    p^2 * 64
PQ         24.96    2pq * 64
QQ          4.48    q^2 * 64

---

Calculation method #2

The second method of utilising the HW equation is a lot more direct, and just about every text book I could find uses this method to solve 3+ allele problems (as, for example, blood types); Wikipedia commends its usefulness in calculating heterozygote frequencies of genetic diseases in large populations.
In this calculation frequencies are directly inferred as per the definition of the HW equation. This means $p^2$ is equal to the frequency of the phenotype PP (see above). $$ p^2 = \frac{37}{64} \\ p = {\sqrt 0.578} \\ p \approx 0.76 $$ $$ q = 1 - p = 0.24 $$ Leading to a HW distribution of $$ (0.76)^2 + 2(0.76)(0.24) + (0.24)^2 = 1 $$

With a new ideal phenotype distribution:

Phenotype      n
PP          37.1    p^2 * 64
PQ          23.0    2pq * 64
QQ           3.8    q^2 * 64

Inverting the calculation by starting with $q^2 = 7/64$ leads to even more diverging results:

Phenotype      n
PP          28.8    p^2 * 64
PQ          28.2    2pq * 64
QQ           7.0    q^2 * 64

Using both $p^2$ and $q^2$ inevitably violates the condition $(p+q)^2 = 1$.

Answer 1

Okay, so the Hardy-Weinberg Equilibrium is just that: an equilbrium.

• In physics, a system (some object) is at equilibrium when no net energy crosses its boundary with the surroundings.
  • no energy change means no change in motion (the state of the object) due to outside influence on the system.
• In chemistry, a system (the beaker/reaction container/cytosol) is at equilibrium when no change in the concentration of the chemical species inside of it occurs over time
  • no chemical makeup change means the (molecular) state of the atoms in the system is free of energetic influence from the outside world.
• In the H-W Equilibrium, a system (the gene pool) is at rest when no change in the frequency(concentration) of its alleles occurs over time.
  • no change in the frequency of alleles means that there is no outside influence on the state of the gene pool.

Just like in basic physics, we have to define our system. If our system is the gene pool, then even the behaviors of the organisms in whose cells those alleles are found are outside of the system. The system is only the alleles aka gene pool.

The most important part about understanding the H-W Equilibrium is that it only applies when those conditions of the gene pool being free from outside influence hold. In fact, in the link you gave, the introduction has it in the second sentence (don't feel bad, I missed it the first few times, too—too focused on applying the algebra to understand what it was really saying):

The model has five basic assumptions:

1) the population is large (i.e., there is no genetic drift)

2) there is no gene flow between populations, from migration or transfer of gametes

3) mutations are negligible

4) individuals are mating randomly

5) natural selection is not operating on the population.

Let's break them down/group them:

• 2: Obvious in terms of equilibrium == system is in isolation from surroundings
• 4 & 5: Selection pressure (4: sexual selection, or 5: natural selection) is not pushing on the system.
• 1 & 3: Random effects are not pushing on the system (1 = statistical sampling error that can push small populations, and 3 = mutation that pushes the system like adding a new chemical species to a system at rest.)

---

With that solid grounding in the theory, let's move on to your questions about the algebra. Remember, the equilibrium only applies when the gene pool is not being pushed around by external factors. Which means that the physical copies of alleles (gametes) should pair randomly with the possible gametes at every mating. If those assumptions didn't hold true, then the math of the equilibrium wouldn't hold up. If you run the numbers on a system at equilbrium, barring small amounts of sampling error (reducible with a bigger n), the H-W Model's predictions will be accurate. All the methods will yield similar results.

However, if the predicted values don't match the actual with one method, you can assume that other methods will also yield inaccurate predictions, and that the two different calculation methods' predictions will vary widely from each other, largely because of the assumptions you are making (Method 2: 1 = p+q), or aren't making (Method 2).

Answer 2

You are throwing numbers into an equation without understanding it.

If you know the numbers for pp, pq and qq, you don't use HWE to recalculate them. So your last statement in Calculation 1 is nonsense. What you've done there is calculated the proportions for a population in HWE where the p=0.73, but you were already given the actual distribution in the population, so pretending like your theoretical calculation trumps the given data is nonsense.

Calculation 2 only works perfectly if your population is exactly in HWE. Yours is a little off, that's why it won't give you quite the same p and q's as calculating it properly, by actually counting up the p alleles in the homozygotes and heterozygotes.