>> Hello and welcome to the Penguin Prof Channel.
Today's episode is about population genetics
and Hardy-Weinberg problems.
So we're going to get right into it.
By the end of this video, you will be able
to solve the typical Hardy-Weinberg problem
like this one.
This is the one we're going to do.
There's going to be a population of individuals.
They're going to vary in a particular phenotype
and you're going to calculate the allelic frequencies
and the genotype frequencies.
So stay tuned.
I am assuming some knowledge here.
If these principles are not familiar to you, you're going
to want to check out my videos on Mendelian genetics
and solving genetics problems.
I will put the links in the down bar below.
You will also need a calculator with a square root function
to solve these problems.
Now you might assume that the Hardy-Weinberg Principle came
about from two people named Hardy and Weinberg
who worked together on this.
And that's actually not true.
I think it's important to put their work into perspective.
On a timeline here, we're going to start with Darwin.
Now Darwin had a lot of great ideas about natural selection
and changing through time, but like all of his contemporaries,
he believed that offspring were a blended version
of their parents.
Now his work preceded Mendel.
Mendel's famous publication, his experiments with his pea plants
in which he showed that what actually gets inherited
are particles.
That offspring are not a blended version of their parents.
Now you'd think this was a big deal, but in fact,
in Mendel's time it wasn't.
His paper was not viewed as anything more
than a paper on pea propagation.
His paper was cited three times over the next 30 years.
It was, however, rediscovered
by two very important people Hugo de Vries and Carl Correns,
who were Dutch botanists and they were two
of the earliest geneticists.
De Vries introduced the concept of the gene,
he called them pangenes.
And they rediscovered Mendel's work, so we're going
to put Mendel back on this timeline right here, in 1900,
which is kind of cool considering he actually died
in 1884.
So, now Mendelian genetics gets really hot.
Everything is about these particles,
there's no more blending.
And we've got William Bateson and Reginald Punnett.
Bateson was the first to use the term genetics.
He and Punnett cofounded genetic linkage,
as well as the Journal of Genetics in 1910.
Now in 1908 during a lecture,
Punnett was asked why recessive phenotypes persist over time.
So, for example, if brown eyes are dominant over blue eyes,
why is it that over time, you don't end
up with a whole population of brown-eyed individuals?
Punnett was kind of stumped by this
and he didn't know the answer, but luckily he played cricket
with somebody who did.
G.H. Hardy was a mathematician,
actually he was a pure mathematician,
so he wasn't really interested in applied mathematics.
But because Punnett was a good friend, he decided to help
and he wrote a paper and in June of 1908,
he very eloquently explained some
of the mathematics behind the inheritance of alleles
and genotypes in population and it was met with great happiness
and it was called Hardy's Law.
And it was all good until 1943 when it was discovered
that somebody kind of beat Hardy to the punch
and a German physician by the name
of Weinberg published a paper that was more eloquent,
that was more clear,
it was overall just better, in January of 1908.
So Hardy's Law became no longer Hardy's Law, now he's got
to share it with Weinberg.
So it became the Hardy-Weinberg Principle.
Turns out though that somebody beat both of them,
the American geneticist, William Castle wrote
about these principles in 1903.
So in truth, this should be called the Castle-Weinberg-Hardy
Principle, but that's a mouthful.
So what is all this about?
This is all about frequencies.
And we are interested in the frequencies
of a particular allele or a genotype in a population.
There's tons of applications for this,
just one example would be determining the frequency
of disease causing alleles in a population.
So, for example, it's very important to know the frequency
of the allele that causes cystic fibrosis.
What the Hardy-Weinberg Principle states is the
following, that the frequency of alleles and genotypes
in a population will remain constant over time
in the absence of other evolutionary influences.
So we're going to make some assumptions here.
And there are a lot
of assumptions, as you can see here.
And some of them might strike you as being almost impossible.
You would be right.
In the real world, many of these assumptions are not going
to be true.
So if you're asking yourself, well, then what's the point?
That's a good question.
But in practice, it is the violation
of Hardy-Weinberg assumptions
that are sometimes the most important.
So in other words, knowing how the real values deviate
from the expected ones actually tells you a lot.
Okay, so the nuts and bolts of how we're going
to do this, here we go.
The frequency of the dominant allele,
we're going to use the letter P to designate that
and the frequency of the recessive allele,
we're going to use Q for that.
So, the first of my two helpful hints,
the sum of all possible outcomes must equal one.
If you want lots of examples of this, you want to check
out that solving genetics problems video and we're talking
about the inheritance of a particular allele here.
But this is also true for frequencies.
So if you've got one dominant allele and one recessive allele,
you're talking about the frequencies.
If you add those frequencies together, they must equal one
because you've got to have one or the other
if there only two alleles involved.
Well, what's a genotype?
A genotype is two alleles.
So if you're interested in genotype frequency,
then you've got allele number one, p plus q equals one
and allele number two, p plus q equals one.
Now a genotype is allele number one and allele number two.
That means multiply.
So if both of those are equal to one --
oh, does this look familiar?
Yes, it does.
You can use the foil rule, multiply those out
and you get p squared plus 2pq plus q squared equals 1.
And this is going to be the key for Hardy-Weinberg equilibria.
And the reason is p squared is going to give you the frequency
of homozygous dominant genotype, 2pq is the frequency
of the heterozygous genotype, q squared is the frequency
of the homozygous recessive genotype and that is it.
If you happen to have a nasty question
with three alleles involved, don't worry,
it's exactly the same.
It's a little bit more involved, but it's the same principle.
So that's the good news.
Okay, we're going to stick with two alleles though
and the best way to get a handle
on all this is just to do a problem.
So we're going to have a population of penguins
and we're going to look at one phenotype in particular,
the color of their feet.
So we're going to have a little -- oh, aren't they cute --
a little population of penguins and they're going
to have two different foot colors.
And we're going to use big Y for the yellow foot allele,
so individuals that are homozygous dominant
and heterozygous will express the yellow foot phenotype.
And we're going to use a lowercase y
for the blue foot phenotype, so only individuals
who are homozygous recessive will express blue feet.
Typical Hardy-Weinberg question, we'll ask you
to calculate the frequencies of both of those alleles,
as well as the frequencies
of all three genotypes in the population.
So the second helpful hint is that the key
to the Hardy-Weinberg problems lies
in the homozygous recessive individuals.
And the reason is because those individuals, you can count,
you can see them, right.
You can't see the difference between the homozygous dominant
and the heterozygous individuals.
But you can see the homozygous recessive ones.
So that's going to be the key.
Here's our sample problem.
We're going to have a population of 1000 penguins.
We're going to have 12 of them with blue feet.
So that leaves you with 988 with yellow feet.
Find the frequency of the blue and yellow allele,
as well as the frequencies of all three genotypes.
So, the four steps to solving Hardy-Weinberg problems,
I know they look scary, but we're going
to take them one by one.
Step number one, assign the alleles.
The frequency of the dominant allele, in this case,
the yellow foot color, big Y, the frequency of that is going
to be p. The frequency of the blue color allele,
the recessive allele, that's q. So we're going to fill
out a little table, I like to make a little table looks
like this.
So big Y is going to be p and little y is going
to be q. That was easy.
That's step one.
Step two is to calculate q by taking the square root
of the number of homozygous recessive individuals.
Pay attention to this.
The number of penguins with blue feet is not q,
its q squared remember, because they need two copies
of that recessive allele in order to express that phenotype.
Students make mistakes here all the time.
The number of homozygous recessive individuals is q
squared, not q. So you can get q very easily,
but you got to take the square root of the number
of homozygous recessive individuals.
So what the problem tells you is that there are 12 of them,
12 out of 1000, that's .012, that's q squared.
To get q you got to take the square root of that number,
.1095, we're going to round that to .11 --
that's q. So we're going to fill that in right here.
Now the good news is once you have q, you can easily get p
because p plus q must equal 1.
So back on this table --
by the way, you notice I'm rounding everything
to two significant figures.
So to get p, I just subtract .11 from the number 1.
So I get the allele frequency of the dominant allele to be 0.89.
Once I have p and q,
I can calculate all the other genotype frequencies.
So I'm going to get the frequency
of homozygous dominant individuals,
that's going to be p squared.
That's .89 squared.
I'm going to get for the heterozygous individuals, 2pq,
so in this case, that's 2 times .89 times .11 and that's .20.
And now I filled out the table completely and that's it.
That's really all you need.
So you can fill out this entire table by knowing only the number
of homozygous recessive individuals.
This is really the key and if you keep this in mind,
it shouldn't be too bad.
Just one more little visual here.
Some students of mine find this very helpful if --
if they forget the equation of they forget what it means.
If you cross two heterozygous individuals, so in other words,
cross a p and a q against a p and a q, then you'll get this.
And that should help you to make some sense of it.
Okay, if you want to try another one, of course you do.
Here's some mice that come in two different coat colors.
Same thing, all I tell you is the number of mice
that express the homozygous recessive phenotype.
So here's a little table for you to fill out and give it a go.
I'm going to put the answer in the description bar
and also a link to the step-by-step solutions.
As always, I hope that was helpful.
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