The Hardy-Weinberg Principle - Online Tutor, Practice Problems & Exam Prep

The Hardy-Weinberg equation predicts genotype frequencies in a diploid population with two alleles, represented as p2 + 2pq + q2 = 1. It assumes random mating and no evolution. By calculating allele frequencies p + q = 1, students can derive genotype frequencies. This model serves as a null hypothesis for testing population equilibrium, indicating potential evolutionary changes if actual frequencies deviate from predictions.

Introduction to Hardy-Weinberg

Hello, folks, and welcome to what is definitely my favorite equation in all of science, and that is the Hardy-Weinberg equation. Alright. The Hardy-Weinberg equation, we're going to say predicts genotype frequencies for a gene with 2 alleles in a diploid population. And we can see the equation is written out down here, but we're gonna work up to it so we understand what it means. Alright.

Now before we go on, I just want to say this is something that if you're at all nervous about using math in biology class, this is where it's going to come up. Right? So we're going to get an equation, and there are going to be things that there we're going to have to solve for. So here, we're going to see where that equation comes from, and we will practice using it a bunch coming up. I also think, though, that this is just a pretty profound idea, at least in my mind.

We're going to be able to predict genotype frequencies for a whole population, given just a little bit of information. That means that you don't need to know individual matings that are going on now out in the wild. If you know just a little bit of starting information, you can predict what that entire population is going to look like in terms of genotype frequencies. Now to do that, you have to make some assumptions. The two assumptions that we have to make here are random mating and no changes to the allele frequency.

Remember, a change to allele frequency, we call evolution. So this is for populations that are not evolving. Now you might say those seem kind of like big assumptions, and they are. We're gonna go into these in a lot more detail coming up, and we're even gonna break them up into sort of more finer distinctions, specific assumptions going forward. But it's also a lot kinda like in physics class when you learn about motion.

You start by ignoring friction. Right? It doesn't perfectly model what's happening out there in the world, but it still has a lot of predictive power and really helps you understand how populations work. Alright. So we're going to say that if a population does match this prediction, populations that match this prediction are said to be in Hardy Weinberg equilibrium.

And that's a term that you probably should be familiar with. Alright. So let's see how this works. We're gonna start by thinking about allele frequencies. So remember, we have \( p \) that's equal to the frequency of the first allele, which we're gonna denote with a big A.

We have \( q \) as the frequency of the second allele, which we're gonna denote with a little a. And because there are only 2 alleles, we have that \( p+q = 1 \). Alright. Well, let's see how we can use those to predict these expected genotypes. Alright.

We're gonna start by putting all our alleles in the gene pool. So again, we're not worried about individual matings here. We're imagining all the alleles from the population taken out of individuals and basically put in one big bag, and we're gonna shake it up. And then because we are assuming random mating, we're just gonna sort of go in and pick out 2 random alleles and pull them out, and that's gonna represent an individual. So to do that, we're gonna figure out what is the chance of being each genotype, either a big A big A homozygote, a heterozygote, or a little a little a homozygote.

Alright. Well, if we have all those alleles in the same bag, when we reach in, what's the chance of being that big A big A homozygote? Well, the chances well, the chance that you pull out a big A first, and that chance is \( p \). Right? The frequency of that first allele.

So it's gonna be \( p \). And then what's the chance the second allele that you pull out is also a big A? Well, Well, that's gonna be \( p \) as well. So the chance that you're a big A big A homozygote is \( p^2 \) or \( p \times p \), In other words, the frequency of that big allele squared.

Alright. Well, we can now just jump down the frequency of the little a little a homozygote should be pretty obvious now as well. Well, what's the chance that you reach into that big that bag and you pull out a little a? Well, that's the frequency of the little a, which is \( q \). What's the chance that the second allele that you pull out is also a little a?

It's gonna be the frequency of that little a again. So it's gonna be \( q^2 \) or \( q \times q \). Alright. What's then the chance of being the big A little a heterozygote? Alright.

Well, the chances well, we're gonna reach into the bag. What's the chance that we pull out that big A? Well, that's gonna be \( p \). And then what's the chance after we pull out that big A that we pull out a little a? It's gonna be \( p \times q \).

Right? But there's actually another way that we could get a heterozygote. We could have pulled the little a out first. So we're gonna have to add this to the frequency or the chance that we pull out the little a first, and that was \( q \) times the chance that we pull the big A out second. Right?

That's another way to make a heterozygote. So, \( q \times p \). Alright. Now \( p \times q + q \times p \), we can just rewrite that. We have 2 p's and 2 q's, so it's just gonna be \( 2pq \).

2pq. Alright. So in other words, what's the chance of being a big A big A homozygote? Well, that's gonna be the frequency of those big A big A homozygotes in a population. So what's the frequency of those big A big A homozygotes?

\( P^2 \). What's the frequency of the heterozygotes? \( 2pq \). What's the frequency of the little a little a homozygotes? We expect that to be \( q^2 \).

Alright. That brings us to our Hardy-Weinberg equation here. So we have these three, things that we just calculated here, \( p^2 \), \( 2pq \), plus \( q^2 \). Now that represents the frequency of each genotype, but because that's all the possible genotypes, well, that equals 100% of the individuals in the population. So \( p^2 + 2pq + q^2 \) is going to equal 1.

Alright. So now you can see right? We have these two equations, \( p + q = 1 \) and \( p^2 + 2pq + q^2 = 1 \). You can see how we can come up with some math problems to solve here. Either we're going to have to solve for \( p \) or \( q \), or the frequency of certain genotypes.

And to do that, well, a lot of the math is going to be some simple plug-and-chug. Some of it's going to be some very basic algebra, but we're going to spend some time going through those and seeing how we can calculate those things in different problems that you might see. Again, we'll practice that coming up, and I'll see you there.

The Hardy-Weinberg Principle Example 1

In this example, we're going to be using Hardy Weinberg and then graphing our results. It says here the table below gives the allele frequencies. Remember that's p and q for 4 populations that we've named populations a, b, c, and d. Importantly, these populations are in Hardy Weinberg. Alright.

So based on the information, we want to fill in the values for p2, q2, and 2pq. Then using the graph, we want to plot the genotype frequencies for populations a, b, c, and d. So let's get going. We see here we have allele frequencies p and q. Going down in the columns and going across, we have these populations a, b, c, and d. We want to figure out the genotype frequencies. That's p2 or the frequency of that first homozygote, 2pq or the frequency of the heterozygote, and q2 for the frequency of that second homozygote. So we can do this really quickly, right? P equals 0.

Well, what’s p2 equal then? Well, 0 squared is 0. 2pq is going to be 0 times 1 times 2 or 2 times 1 times 0 is going to be 0. And q2, well, we're just going to take 1 squared and that equals 1. Alright.

For our next one, population b p2, well, we're going to take 0.2 square that, that equals 0.04. 2pq is going to be 0.2 times 0.8 times 2 is going to be 0.32. And q2 0.8 squared that equals 0.64. All right, so now for population c, well 0.4 squared for p2 that equals 0.16, 2pq 2 times 0.4 times 0.6 equals 0.48, and q2 0.6 squared equals 0.36 and our final one here population d, they're at 0.5 and 0.5. So p2 equals 0.5 Squared which is 0.25.

That's also what q equals. So that's 0.25 and 2 times 0.5 times 0.5 equals 0.5. Alright. So we've filled in our genotype frequencies, our predicted genotype frequencies for those populations. So now let's graph it.

Well, on our y axis, it says genotype frequencies. So that's what we just calculated, that p22pq and q2. And we are going from 0 up to 1. And on the x axis, we have allele frequencies and it has 0 and q plotted right next to each other. So p goes from 0 down to 1, and in those same populations, remember because p plus q equals 1, q goes from 1 down to 0.

Alright. So let's start graphing. I'm going to graph, all the p2, then all the 2pqs, then all the q2. For population a, when p equals 0 and q equals 1, we're talking about right here on the graph was p2 equals 0. So dot goes there.

Alright. The next one is when p equals 0.2 and q equals 0.8. So we're talking about right here and there q2 p squared sorry equals 0.04. So real low on the graph there. Now when p equals 0.4 and q equals 0.6, we're talking about right here on the graph and that's going to be p2 equals 0.16, which is gonna be right about there.

Oops, I do it in the wrong spot right about there. Is that not? There we go. Finally, when, for population d, when p and q both equal 0.5,

In the Hardy-Weinberg equation, what does the term 2pq represent?

• A. The probability of being a homozygote for the first allele.
• B. The probability of being a homozygote for the second allele.
• C. The probability of being a heterozygote.
• D. The combined probability of all genotypes.

Which of the following statements about the Hardy-Weinberg equation are true?

I) For a gene with two alleles, q² represents the frequency of one of the homozygotes.

II) If p and q both equal 0.5, you expect 50% of the population to be heterozygous.

III) The p² and q² terms will be equal as they both represent the probability of being a homozygote.

• A. I & II only.
• B. II & III only.
• C. I & III only.
• D. I, II, & III.

Predicting Allele or Genotype Frequencies Using Hardy-Weinberg

We said that with the Hardy Weinberg equation, it's likely that you're going to be asked to do some math problems. And I understand that can be a little intimidating for some folks doing math in biology class, but here we're going to go over the very basic approach to how to do this, and then we'll look at some specific problem types that you're likely to see. So here, we're just going to talk about how you can predict allele or genotype frequencies using Hardy Weinberg very generally. So we'll start just by saying that Hardy Weinberg can be used to predict allele or genotype frequencies. Now when we introduced Hardy Weinberg, we said that it is used to predict genotype frequencies.

But if you're given those genotype frequencies, you can also calculate the allele frequencies just by working backward. Right now, a really big thing in this type of problem is that we need to make those assumptions we need to talk about. We need to assume that the population is in Hardy Weinberg equilibrium. And those two assumptions that cause a population to be in equilibrium, remember that's that there's random mating in the population and that no evolution is occurring. Now when you see a problem like this, like we're going to go over here, it should say some version of assume the population is in Hardy Weinberg.

Because you need to make that assumption to do the types of problems we're going to do. Alright. So when you see a problem like that, we're going to use these following just sort of generalized steps. And these steps are very generalized, but again, we'll go through some real problems coming up. Alright.

Remember, first off, when you see a problem like that, just remember your equations. Well, we have an equation for allele frequencies. We have p+q, frequency of the first allele plus the frequency of the second allele. Together, those should equal 1 or 100% of the alleles in the population because we're always doing problems with only 2 alleles. Alright.

Then we have our genotype frequencies, and this is the Hardy Weinberg equation. p2 + 2pq + q2 = 1 Remember, p2, that's the frequency of the first homozygote. 2pq, the frequency of the heterozygote. And q2 is the frequency of the second homozygote, and they add to 1 because that's all the possible genotypes in this population.

Alright. Well, once you remember your equations, just figure out which one, one of those variables are you given. What do you know? Maybe you're given p, maybe you're given q, maybe you're given some sort of combination. You're given, like, p2, the frequency of those homozygotes.

Alright? So just figure out what do you know, and then identify which variable or variables the problem is asking for. Is it asking for p? Is it asking for q? Is it asking for something like 2pq, the number of heterozygous?

And then we're just going to solve for the missing variables. Right? And that's just going to take some very simple arithmetic or maybe some very basic algebra in some cases. Alright. The last thing to remember, right, we're talking about alleles, and genotype frequencies, but remember, genotypes cause phenotypes.

So sometimes you will be given a phenotype or asked to report your answer in terms of phenotypes. So if that's the case, just remember your simple dominance relationships. Remember, big A, big A, and big A little a genotypes. Those have that dominant allele in them. So those genotypes will display the dominant trait.

And little a little a genotype, they have the recessive allele, so they will display the recessive trait. And if you're asked to use phenotypes or report it in phenotypes, you just need to do that little conversion either at the beginning or at the end. Again, we'll practice that and see that coming up. I'll see you there.

A. Calculating Genotype Frequency from Allele Frequency

Probably the most straightforward type of problem you're going to get using Hardy Weinberg is calculating genotype frequency from allele frequency. And because this is the first of a few different problem types we're going to be looking at, we have this labeled here. That genotype frequencies equation. Alright? So we have over here our sort of, you know, real straightforward simple steps to solve.

Remember, we're going to remember our equations. We'll figure out what we know, figure out what the problem is asking for, and then we're just going to solve. So let's give it a try. We hear the example we have says, find the predicted genotype frequencies if the frequency of the A allele is 0.2 and the frequency of the little a allele is 0.8. Alright.

So what do we know? Well, the frequency of the A and the little a alleles, that's p and q. So I'm just going to write those down as p and q. \(p = 0.2\) and \(q = 0.8\). Alright.

What do we want to know? What's it asking for? It's asking for the genotype frequencies. So we're looking for that \(p^2\), \(2pq\), and \(q^2\). Alright?

So let's write that down. So first, we're looking for the frequency of the big A big A homozygote, that's going to equal \(p^2\). We're looking for the frequency of the heterozygotes, that's going to equal \(2pq\). And we're looking for the frequency of the little a little a homozygotes, that's going to equal \(q^2\). Alright, well we have the variables, we know what we need to calculate, we can just plug and chug.

So let's do it. \(p^2\), that equals \(0.2^2\), right? You can use your calculator, just do 0.2 times 0.2 is going to equal 0.04. \(2pq\), well \(2pq\) equals 2 times 0.2 times 0.8. Alright, well 0.2 times 0.8, so we're going to do now 2 times, that comes out to 0.16.

\(2pq \times 0.16\) is 0.32 and \(q^2\). Well, \(q^2\) is just going to equal \(0.8^2\). \(0.8^2\) again, you can use your calculator just you know, do 0.8 times 0.8 that equals 0.64. Alright, so with that I solve the problem, right? What's the frequency of the big A big A homozygote?

That is going to be 0.04. What's the frequency of the heterozygote? It's going to be 0.32. And what's the frequency of the little a little a homozygote? It's going to be 0.64.

Alright. Now because I often make a lot of silly math mistakes, I always like to just check myself because remember the Hardy Weinberg equation says that these should add to 1. So let's do that. I'm just going to do \(0.04 + 0.32 + 0.64\). Does that equal 1?

Well, \(0.04 + 0.32\) equals 0.36 plus 0.64. Sure enough, that equals 1.00. I didn't make any silly math mistakes. I got my genotype frequencies. We solve the problem.

Alright. Again, we're going to try some other problem types. You should check them out.

B. Calculating Allele Frequency from a Genotype Frequency

Another type of problem you may be asked to do using the Hardy-Weinberg equation is calculating allele frequency from genotype frequency. Now remember, the Hardy-Weinberg equation uses allele frequencies to calculate genotype frequencies. So here, we just need to sort of work backward, and that's our first note. It just says work backward from genotype to find the allele frequency, and we have an example that we can work through to see how this works. The example says, predict the allele frequency for both alleles if the population is 1% little a little a homozygous.

Alright. So steps to solve, very general here. Remember our equations. Remember, we have p+q=1. And we also for genotype frequencies or that Hardy-Weinberg equation, we have p2+2pq+q2=1.

Alright. We've remembered those now. Identify what is given. Well, here it says that 1% of the population is little a, little a, homozygote. So in those equations, which variable or combinations of variables, what part of those equations represents the little a, little a homozygotes?

Well, remember that's q2. So q2=0.01. Alright. Now what is the problem asking for? That's what we gotta do next.

Well, the problem is asking for P and Q. So now let's solve. Well solving for Q should be pretty straightforward here, right? If q2 = 0.01, solve for that, all we do is we take the square root of both sides of the equation. That gives us q=0.01. In other words, the frequency of the little a allele in this gene pool is 0.1 or 10%.

But now we need to solve for P, and we have this little equation here, p+q=1. We can solve for P using that, right? So we can write up here, p+q=1.

And I know what Q is, so I can substitute that in. p+0.1=1. Now you can probably do that in your head, but let's be formal about it. We'll subtract 0.1 from both sides of the equation. That means that p=1-0.1=0.9.

So p=0.9. That's what we were looking for. We were looking for the frequency of both alleles. So p=0.9 and q=0.1. Those are my answers.

In other words, the frequency of the big A allele is 0.9 or 90% of the alleles in that gene pool, and the frequency of the little a allele is 0.1 or 10% of the alleles in that gene pool. Now if you're looking for fun on a Friday night, you can go back to your Hardy-Weinberg equation. You could even now calculate the number or the frequency of the big A homozygotes. You could calculate the heterozygotes. We're not going to do that here.

But again, you could do that, and some problems may ask for that information. Alright. We're going to look at another problem type coming up. You don't want to miss it. So I'll see you there.

C. Calculating Allele Frequency from Phenotype Frequency

We now want to look at problems that calculate allele frequency from a phenotype frequency. And because this is the third type of problem we're looking at using the Hardy-Weinberg equation, we have this labeled as c. Now, this can be a little more confusing because we're dealing with a phenotype frequency, and in some cases, there's more than one genotype that can cause the same phenotype. So we're going to start off here just saying that when given a phenotype frequency, the first thing you want to do is try to convert that to a genotype frequency in some way. Alright.

And this is most confusing in cases of simple dominance because in those cases, you have two genotypes that cause the same phenotype. So for simple dominance, we're going to say here, remember to get the dominant trait, well, you have two genotypes that can cause it. You have the big A big A homozygote and the heterozygous. And then for the recessive trait, you just have one genotype, just that little a little a homozygous. Okay.

So how do we get the frequency of the dominant trait? Well, big A big A, we know we can calculate that the frequency of that using \(p^2\). And the frequency of the heterozygote, we know we can calculate using \(2pq\). And of course, the frequency of the recessive trait there, that's just \(q^2\) because that's the frequency of that little a little a homozygote. Okay.

So, that dominant trait we're going to have to combine that \(p^2\) and \(2pq\) to figure out what that is. Alright. So knowing that, let's keep that in mind and let's check out our example. It says in a population, 64% of individuals display the dominant trait. Alright.

Then we want to know, though, what percentage of the population do you predict to be heterozygous? Alright. This is going to take some steps to get there, but we can do it. So first thing we want to do, remember our equations. Right?

We have two equations we can use, \(p + q = 1\), and we also have the genotype frequencies, that Hardy-Weinberg equation, \(p^2 + 2pq + q^2 = 1\). Alright. Now let's identify what this problem gives us. Well, it tells us that 64% of individuals display the dominant trait. Alright?

So that I want to write down. That's what we know. So I'm going to write down 0.64. That's the frequency we're dealing with. That's equal to the dominant trait, but we said here, what's the dominant trait?

Well, that's going to be \(p^2 + 2pq\). So I'm going to write that here. Point 64 equals \(p^2 + 2pq\). Alright. Now we want to figure out well identify what variable or variables the problems ask for.

What is this problem asking for? It's asking for the percentage of the population that's going to be heterozygous. We calculate that using \(2pq\). So I'm going to write that all the way over here because this is going to be my answer once I can do that little equation there. But I have work to do it.

Alright. Now this seems like it should be pretty easy. Right? I feel like this is something I should just be able to solve, but this is one equation with two variables in it, p and q, and you can't solve something like that just on its own. We need more information or more equations.

Luckily, we have that. Alright. So let's think how we can start breaking this down though. So when I look at this \(p^2 + 2pq\) that's part of my Hardy-Weinberg equation. Right?

So I'm just going to write out the whole equation here. I have \(p^2 + 2pq + q^2\), that frequency of the recessive genotype, and that is going to equal 1. Alright. But I know what \(p^2 + 2pq\) equals. Alright.

Here I'm just going to move down the page a little bit here, so we can see it all a little better. So I know what this equals. So I can substitute that in. Right? So \(p^2 + 2pq\), that equals 0.64.

So 0.64 plus \(q^2\) equals 1. Alright. That I know how to solve. Okay? So to solve this, well, I just need to first subtract 0.64 from both sides of the equation.

So minus 0.64 and minus 0.64. That gives me well, \(q^2\) is going to equal well, 1 minus 0.64 is going to be 0.36. Well, now I can get q. Right? All I'm going to do here is take the square root of both sides of the equation.

That's going to give me q which is equal to the square root of 0.36, which is 0.6. Alright. So to get to \(pq\), I'm going to need p and q. So I'm going to write those up here because this is what I need to figure out next. \(p\) equals and \(q\) equals, but I know \(q\).

\(q\) equals 0.6. So how can I get \(p\)? Well, we have an equation for that. Right? \(P+q = 1\).

So I'm going to write that now. \(P+q = 1\). But I know what \(q\) is, so I can substitute that in. \(P + 0.6 = 1\). Right now to get \(p\) by itself, I'm going to subtract 0.6 from both sides of the equation.

Minus 0.6, minus 0.6. That gives me \(p = 1 - 0.6\) is 0.4. So \(p = 0.4\). Alright. Now I have \(p\) and I have \(q\), so I can do my final part of the equation here.

I can figure this out. \(2pq\). Well, that's going to be equal to 2 times 0.6 times 0.4. Right? And that's going to be equal to 2 times 0.6 times 0.4 is 0.24 and that equals 0.48.

Alright. So 0.48, that's my answer. That's the frequency of the heterozygotes in this population. Now just note the question asked for it in a percent, so 0.48, we can write that just as 48%. 48% of the individuals in this population, I expect to be heterozygous if 64% of the individuals display the dominant trait.

Alright. Now a question like this, it takes a lot of steps to get there, but none of those steps were really that difficult. Right? We just got to keep plugging and chugging until we get to \(p\) and \(q\). And once you get down to \(p\) and \(q\), you can plug those back into the Hardy-Weinberg equation, and figure out all your different genotypes.

Alright. We're going to practice all these more coming up. Give them a try.

Imagine a population of cats where one gene codes for eye color. The E allele codes for yellow eyes and is dominant. The e allele codes for green eyes and is recessive. The frequency of the E allele is 0.6, and the frequency of the e allele is 0.4. Assuming the population is in Hardy-Weinberg equilibrium, what percent of the population will have green eyes?

• A. 36%
• B. 24%
• C. 16%
• D. 48%

Sickle cell anemia is a recessive trait. In Nigeria, it is estimated that 2.4% of the population is born with sickle cell anemia. Heterozygotes for the sickle cell allele are said to have the sickle cell trait and have few sickle cell symptoms. Assuming that the population is in Hardy Weinberg equilibrium for the sickle cell gene, what percentage of the population would you expect to have the sickle cell trait?

• A. ~71.4%
• B. ~13%
• C. ~15.5%
• D. ~26%

PTC is a chemical that elicits a bitter taste response in some people. The ability to taste PTC depends on the presence of the PTC-tasting allele. Individuals with at least one copy of the PTC-tasting allele taste PTC as bitter. It is estimated that about 70% of the population taste PTC. Assuming that the population is in Hardy-Weinberg equilibrium for this trait, what percent of the population would you expect to be heterozygous for this trait?

• A. ~42%
• B. ~49%
• C. ~45%
• D. ~54%

Testing if a Population is in Hardy-Weinberg Equilibrium

Another type of problem that you may see is testing if a population is in Hardy-Weinberg Equilibrium. So here, you're going to be given data from an actual population, and you want to know if it matches the Hardy-Weinberg expectation. So we're going to broadly say here to start, just to test if a population is in Hardy-Weinberg equilibrium, we need to compare those actual genotype frequencies to the Hardy-Weinberg expectations. Now, the reason you may want to do that is that Hardy-Weinberg equilibrium is sometimes used as a null model. In experimental design or statistics, a null model is the thing you compare your data to.

A null model is sort of what you would expect to happen if everything else is equal, if nothing interesting is going on. So that means that if the population is not in Hardy-Weinberg, if those two things don't match, well, we can assume one of two things. Because remember, for things to be in Hardy-Weinberg, we had some pretty big assumptions going in. So we can assume that at least one of those assumptions is broken. We can assume that there is either non-random mating in the population or that some type of evolution is going on.

Maybe there is natural selection on this gene, and it's removed alleles from the population. So in that way, we can look at populations and test whether these things are actually happening out there in the world. Now, to do this, we're given a sample from a particular population, and we see that it has 250 big A big A homozygotes.

We see that it has 100 heterozygotes and 150 little a little a homozygotes. And our question is, is this population in Hardy-Weinberg equilibrium? Well, we have these steps to solve here. They're pretty straightforward, and it's a decent amount of math, but we've actually done pretty much all this math before, so we should know how to do it.

Step 1, we are going to calculate allele frequencies. We need to calculate p and q, but we've already practiced that. Step 2, we need to plug p and q into the Hardy-Weinberg equation.

That gives us those expectations, and that gives us our p2, our 2pq, and our q2. And then, well, we need to compare it to the original. But the original here actually counts individuals. So we need to figure out what our expected counts would be. So we are going to multiply the expected frequencies by the actual number of individuals in that sample.

This will give us something to compare, and our final thing here is just to compare it to the original data. Now if you're really doing this, you'd need to use statistics and see if it's statistically significant. But in an introductory biology class, usually, they'll just make the difference big enough that you don't need statistics. It's very obvious if it is different.

Let's try this out. For our sample, the first thing we need to do is calculate the allele frequency. Again, we know how to do this:

We calculate p. We're going to take 2 times the number of big A big A homozygotes plus the number of heterozygotes and divide all of that by 2 times the total number of individuals. We calculate that 2 times 250 (number of bigA bigA) + 100 (heterozygotes) equates to 600. Dividing 600 by 1000 (total alleles) gives p = 0.6.

We already know what q equals, since p + q = 1. But again, to double-check, q calculation would be the number of heterozygotes (100) plus 2 times the number of little a homozygotes (150 each), divided by twice the total individuals (1000). This calculation confirms that q = 0.4.

Now, moving on to the genotype frequencies:

• p2 for bigA bigA will be 0.36
• 2pq for the heterozygotes will be 0.48
• q2 for little a little a will be 0.16

Now, we multiply these frequencies by the total sample size (500) to get the expected number of individuals:

• 180 expected big A big A homozygotes
• 240 expected heterozygotes
• 80 expected little a little a homozygotes

Now comparing these to the original (250 big A big A, 100 heterozygotes, 150 little a little a), we can see the population is not in Hardy-Weinberg equilibrium as the expected frequencies do not match the observed ones, indicating either non-random mating or some type of evolutionary process affecting allele frequencies.

We'll practice this more coming up. Check it out.

In a population of glass frogs (Hyalinobatrachium munozorum) you collect 200 individuals. Of those, you determine that 160 are homozygous for the H allele, 10 are homozygous for the h allele, and 30 are heterozygous. Is this population in Hardy-Weinberg equilibrium? If not, why not?

• A. Yes. The population is in Hardy-Weinberg equilibrium.
• B. No. There are more heterozygotes than predicted by the Hardy-Weinberg equation.
• C. No. There are fewer HH homozygotes than predicted by the Hardy-Weinberg equation.
• D. No. There are more HH homozygotes than predicted by the Hardy-Weinberg equation.

Jorge and Claire are given the following data on fur density in red pandas at zoos in the United States. Is the following population in Hardy-Weinberg equilibrium?

160 homozygous for the F allele, 137 heterozygotes, and 29 homozygous for the f allele.

• A. Yes, the population is in Hardy-Weinberg equilibrium.
• B. No, we would expect there to be more heterozygotes in the population.
• C. No, we would expect there to be more ff homozygotes in the population.
• D. No, we would expect there to be more FF homozygotes in the population.