Back in the early 1900s,
a couple of scientists named
G. H. Hardy and Wilhelm Weinberg
started thinking that Darwin's theory of natural selection,
survival of the fittest,
and Mendel's ideas about inheritance of genes
could be combined to help us understand
how populations evolve.
They were interested in understanding...
one, what is the frequency of specific alleles in a population?
So, what fraction of the total number of alleles
are dominant, capital B,
or recessive, lowercase b?
And two, as the organisms mate
and have offspring over many generations,
whether the frequency of those alleles
changes over time.
So, if you start out with 60% of the alleles being capital B
and 40% being lowercase b.
After the people mate,
are there still 60% capital B and 40% lowercase b?
Notice also that the allele frequencies
-- 0.6 and 0.4 --
add up to 1.
This is a key aspect of learning how to solve
Hardy-Weinberg equilibrium problems,
so keep it in mind for later.
Hardy and Weinberg postulated
that if the allele frequencies
did not change over time,
then the population was in equilibrium.
We would say that such a population
is in Hardy-Weinberg equilibrium.
In contrast, if the allele frequencies
do change over time,
the population is not in Hardy-Weinberg equilibrium.
For a population to be in Hardy-Weinberg equilibrium
-- so, for the allele frequencies to not change over time --
there are five criteria that must be met.
These are, number one,
the population must be large;
number two, no mutations occur;
number three, no migration into or out of the population,
also called gene flow;
number four, mating must be random;
and number five, no natural selection can occur.
If these five criteria are met,
the population is in Hardy-Weinberg equilibrium.
Evolution is not occurring at that locus,
and the allele frequencies
will remain the same over time.
If you calculate allele frequencies
and find that they have changed over time,
you can say that the population is not in Hardy-Weinberg equilibrium,
meaning that it has evolved.
Let's go through two related examples
of Hardy-Weinberg equilibrium problems,
and learn both how to calculate allele and genotype frequencies,
as well as to start thinking about why
a population may or may not be evolving.
Example number one...
you are studying the coat color locus
for a population of 100 squirrels
living in a forest along the coast of California.
The dominant allele, capital B, gives black fur,
while lowercase b, the recessive allele,
gives white fur and is recessive.
In population genetics, we call the frequency of the dominant allele "p"
and the frequency of the recessive allele "q".
The sum of these allele frequencies
-- all the alleles in the population --
must equal 1.
Now, let's relate this to the genotypes,
which is the probability of having an individual
that is homozygous dominant,
that is, having one allele to be capital B
and the other, also, a capital B,
is p times p.
That equals p^2.
The probability of being heterozygous,
having one capital B
and one lowercase b,
is 2 times p times q.
And the probability of being homozygous recessive
is q times q,
which equals q^2.
The sum of the genotype frequencies
must also equal 1,
which is this equation:
p^2 + 2pq + q^2 = 1.
Now, back to your squirrels.
You find that 20 of the 100 squirrels have white fur,
which corresponds to 0.2.
White fur is homozygous recessive,
so we know that these squirrels have the same genotype:
little b little b.
Since we know that q^2 is the genotype frequency
of homozygous recessive
-- little b little b --
we know that 0.2 equals q^2.
We can solve for q
by taking the square root of 0.2,
which gives us q equals 0.45.
Since p plus q equals 1,
we can easily solve for p,
which is 0.55.
These squirrels breed and, in a second generation,
you go back to the forest
and want to determine if the population
is in Hardy-Weinberg equilibrium.
You count and see that 40 of the 200 squirrels
in this generation have white fur.
This means that the frequency
of the recessive genotype,
q^2, is 0.2.
We can now easily solve for q,
the recessive allele frequency,
so q equals 0.45.
We can easily solve for p,
which is 0.55.
Based on these equations,
let's see if our population is in Hardy-Weinberg equilibrium.
We started with p equals 0.55
and q equals 0.45
in the first generation.
And we calculated that p equals 0.55
and q equals 0.45
in the second generation.
So, the allele frequencies did not change over time.
This means we can accept the Hardy-Weinberg equilibrium null hypothesis
and say that, yes, this population
is in Hardy-Weinberg equilibrium
and has not undergone evolution.
Example number two...
you are also studying a population of squirrels
that live on a nearby beach.
In the first generation,
they had p equals 0.55
and q equals 0.45.
In the second generation,
you find that the frequency of the dominant allele,
p, is 0.7,
and the frequency of the recessive allele,
q, is 0.3.
This population of squirrels on the beach
is not in Hardy-Weinberg equilibrium
since the allele frequencies
did change over generations.
So, we reject the null hypothesis
and can say that this population evolved.
Now, what factors caused this population
to undergo evolution?
To answer this, think about the five criteria
for Hardy-Weinberg equilibrium,
and let's come up with some ideas
as to which of these criteria were not met.
Number three: no migration
into and out of the population.
Maybe the water level went down,
so some white squirrels living on the coast
could come from shore
and start living on the island.
Number five: no natural selection can occur.
Maybe the white squirrels
stood out more against the sand
than the black squirrels,
so they were more noticeable to predators
and got eaten.
Now, let's do a third example
to see how we can use the Hardy-Weinberg equilibrium idea
to solve population genetic problems.
You can use the Hardy-Weinberg equilibrium
to solve population genetic problems
if you assume that the population
is in Hardy-Weinberg equilibrium.
Let's say that you are now studying eye color
in a population of squirrels.
You have either brown eyes or blue eyes,
and you know that blue is recessive.
You look at all of your squirrels
and find that 4% of the population has blue eyes.
How can you figure out,
A, the frequency of the brown eye allele,
and B, what percentage of the population
is heterozygous for this trait?
Assuming the population follows Hardy-Weinberg equilibrium,
let's solve A and B.
To solve A, let's start with the information
that 4% of the population has blue eyes.
This is telling you the frequency
of the recessive genotype: blue eyes.
So, we know that 0.04 equals q^2.
We can easily solve for q,
the frequency of the recessive allele,
which is 0.2.
Therefore, given that p plus q equals 1,
we know that p equals 0.8.
This is the allele frequency of brown eyes.
To solve B,
let's use the equation p^2 + 2pq + q^2 = 1.
The frequency of the heterozygous genotype is 2pq,
so 2 times 0.8 times 0.2
equals 0.32.
To get the percentage,
we can multiply by 100.
So, 32% of the population of squirrels
is heterozygous at the eye color locus.
Overall, Hardy-Weinberg equilibrium
is a way to apply Mendelian genetics
-- the idea that traits like coat and eye color in squirrels
get inherited --
to large populations,
and to figure out whether that trait is undergoing evolution.
Through this lesson, you have learned
how to calculate genotype and allele frequencies
over several generations
and use these calculations to determine
if a population is or is not
in Hardy-Weinberg equilibrium.
If the population is in Hardy-Weinberg equilibrium,
you can use Hardy-Weinberg equilibrium
to solve population genetics problems.
On the other hand, if the population
is not in Hardy-Weinberg equilibrium,
we can start thinking about what selective pressures
might be in place
that are causing evolution
and anticipate how these selective pressures
may shape future generations.
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