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Do you like math?
Sometimes when people are starting out in biology, for some reason they assume there
is no math in biology.
But then, they soon find out…they were wrong.
My math teachers were telling the truth when they said math is in everything.
Chi squares, osmotic pressure calculations, standard curves with gel electrophoresis,
Punnett square ratios…those all have math.
But one of my favorite topics that involves biology and math, and is also absolutely fascinating,
is the Hardy Weinberg Equilibrium.
It got its name from a mathematician and a physician, and it states that a population’s
alleles and genotype frequencies are constant unless there is some type of evolutionary
force acting upon them.
A reminder from our ecology videos, a population in this case is a group of organisms that
are all the same species and can breed with each other and have fertile offspring.
However, there can be variety among a population as they are not clones of each other.
But based on Hardy Weinberg equilibrium, the population’s allele and genotype frequencies
would remain constant if no evolutionary force acts upon them.
If we have these frogs here, assume they are all the same species in this example, but
there are some slight differences.
In our simplified example, some of the frogs are lighter green and some are darker green.
We’re going to use a simple genotype in this example.
All frogs in this population will either be GG, Gg, or gg.
There is also an allele frequency in our population.
We’re going to say here that 60% of a frequency of 0.6 of the alleles are G. 40% or the frequency
of 0.4 of the alleles are g.
Notice the percentages add up too 100%, and the frequencies add up to 1.
Now to be in Hardy Weinberg equilibrium, we have to have 5 assumptions.
Again, you can only be in the equilibrium if no evolutionary force is acting upon it.
1) No selection.
No natural selection is acting upon these frogs.
That means neither dark green nor light green will have any impact on reproductive fitness.
2) No mutation.
Baby frogs inherit genes from their parents, and there are never mutations.
3) No migration.
Frogs can’t come in.
Frogs can’t go out.
4) Large population.
There are a lot of frogs.
It turns out small populations are more vulnerable to genetic drift by the way, see our genetic
5) Random mating.
The frogs mate without any specific choice.
All of these five assumptions must be kept in order for Hardy Weinberg Equilibrium to
So, in real life, does this generally happen?
For example, in real life, maybe in a certain environment, predators can more easily see
the light green frogs so perhaps they are eaten more and have less reproductive fitness.
So if Hardy Weinberg Equilibrium is unrealistic in nature, why does the Hardy Weinberg Equilibrium
That’s where the math comes in.
The Hardy Weinberg Equilibrium gives you this baseline to compare how an evolving population
could compare to one that remains constant without evolutionary forces acting upon it.
And now, let’s explore the math.
So there are two equations in Hardy Weinberg Equilibrium that we will focus on.
We’re going to start with the first one: p + q = 1
In this equation, p= the dominant allele frequency in the population and q= the recessive allele
frequency in the population.
By the way, Hardy Weinberg Equilibrium doesn’t mean that p has to equal q.
And the dominant allele frequency in any population doesn’t have to be a larger number than
the recessive allele frequency in a population; that’s a misconception because dominant
alleles aren’t always the more common allele.
The equation does say that the dominant allele frequency and the recessive allele frequency
have to equal 1.
So in my example here, if I say 60% of the alleles are G. 40% of the alleles are g.
0.6 is p and 0.4 is q.
So this equation p + q = 1 is for those allele frequencies.
But what if I wanted to know the genotype frequencies?
Meaning, I wanted to know the frequency of frogs that are homozygous dominant which is GG,
heterozygous which is Gg, or homozygous recessive which is gg.
Then I can use this other Hardy Weinberg Equilibrium equation: p2 +2pq + q2 =1
Just like with the previous equation, I like to write out what these symbols stand for.
So p2 is the homozygous dominant frequency, the frequency of GG in this case.
2pq is the heterozygous frequency, so Gg frequency in this case.
q2 is the homozygous recessive frequency so gg frequency in this case.
Let’s plug those previous p and q values in.
P2 would then be .36.
2pq would be 0.48.
q2= 0.16 Cool huh?
Now again, that is if it was in Hardy Weinberg equilibrium.
The only thing is that when calculating Hardy Weinberg Equilibrium problems, you don’t
always have the p and q values.
So we’re going to just do an example.
New population of frogs.
So, forget the previous frequencies, this is a new frog population in a new frog land.
But we’ll still use the same allele letters.
So here is the information you get for this new population.
There are 500 frogs and of those, 375 frogs are dark green.
The rest are light green.
With that information, please solve all genotype frequencies and allele frequencies if in Hardy
Step 1) Determine first whether you’re going to work with the first equation or
the second equation.
Since I’m working with individuals here that have genotypes, I’m going to work with
the second equation.
Step 2) Figure out a value you can determine.
So 375 frogs are dark green out of 500.
That would mean 125 frogs are light green since there are 500 frogs total.
But I can’t use whole numbers like that as both of these equations ultimately are
equal to 1.
I need frequencies for these equations.
Now I could say 375 dark green frogs out of 500 total frogs is = 0.75.
But all I know is that would be the frequency of dark green frogs.
The problem is dark green frogs could have genotype GG or they could have Gg.
I can’t assume they’re one or the other, so I shouldn’t use that value.
The recessive genotype, resulting in a light green trait, is safe to use though because
I know that light green frogs are the genotype gg.
They can’t be anything else.
So 125 frogs out of 500 frogs = 0.25.
That means, from that equation, the value q2=0.25
Step 3) Take a value you solved from the previous step and calculate from there.
So if I know that q2=0.25, then I could go ahead and solve for q.
If q2=0.25, I can determine q if I take the square root of 0.25.
Therefore q= 0.5.
That’s the allele frequency for the recessive allele g.
If I know the q value, I can find out the p value using the first equation!
Since p + q= 1, now I know that p=0.5.
That’s the allele frequency for the dominant allele G.
Now I have p and I have q, and I can take care of everything else!
I can use the second equation to determine the homozygous dominant frequency, the heterozygous
frequency, and I already knew the homozygous recessive frequency from the beginning.
p2= .25. for the genotype frequency GG. 2pq= 0.5 for the genotype frequency Gg.
q2= 0.25 for the genotype frequency gg.
Phew that’s a lot.
Finally, some last tips for solving Hardy Weinberg equations:
1) I used nice numbers.
Meaning, I didn’t need a calculator for my examples.
It’s not always like that.
You might need a calculator for the numbers you’re given, and you might need to round.
2) When you get your final results, always check that the values you calculated do = 1
for both equations.
This can really help you check your work.
3) Be careful not to assume too much with given info.
Remember when I was given a value of 375 frogs out of 500 being dark green, I didn’t necessarily
know whether they were the GG or Gg genotype.
Instead I chose to work with the recessive genotype since I knew light green frogs would
be gg. 4) Practice.
And practice some more.
There are a lot of practice problems you can find online.
And remember, this skill that can be very useful for determining the extent of evolutionary
forces because you have this base comparison of Hardy Weinberg Equilibrium to compare it to.
Well, that’s it for the Amoeba Sisters, and we remind you to stay curious.