Patrick Roisen

M.Ed., Stanford University
Winner of multiple teaching awards

Patrick has been teaching AP Biology for 14 years and is the winner of multiple teaching awards.

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Hardy Weinberg Equilibrium

Patrick Roisen
Patrick Roisen

M.Ed., Stanford University
Winner of multiple teaching awards

Patrick has been teaching AP Biology for 14 years and is the winner of multiple teaching awards.


To be honest, back when I was in college and even when I first started teaching Biology, whenever I hit the Hardy Weinberg questions, I thought big whoop who cares. It’s just a couple of equations that describe a population genetics. And they only work if things don’t change and how often does that happen? It wasn’t until a colleague of mine was getting all excited about teaching Hardy Weinberg, that I found that maybe it is a big warp and maybe I should care.

You see in science, whenever you’re studying something you always want a control group, to compare to your experimental group. And if you are studying something like human evolutions, it’s really hard to get funding, to build yourself a giant second earth. Put your control humans on, compared to this earth's humans. Instead, using some basic statistical math, you can just mathematically create a control group.

It’s kind of like if a friend of yours. If you think there is something funny going on with his coin, well you don’t need to sit there and flip it 100 times and then compare that Result to 50 other coins that you also flipped 100 times. Instead you could just say, well assuming there’s nothing weird with the way he’s flipping it, or weird with the weight or shape of the coin., it should be 50-50, it’s the same thing.

I assume if you’re watching this video, you’ve done some genetic problems. And you know just how useful Punnett squares for predicting the outcome, when you have two different organisms breeding. Well the Hardy Weinberg equations are essentially Punnett squares for an entire population.

Before I go much further we’re going to spend a little bit time defining the terms of the Hardy Weinberg equations and making sure you know what those mean. Next as a statistical tool, Hardy Weinberg is based on a few assumptions. And if it’s predictions don’t much reality, then it indicates that one or more of those assumptions have been broken. The last thing we’ll do is, we’ll go through a couple of the standard Hardy Weinberg equations problem sets that you’ll see on the actual on the actual AP exam.

I’ve already mentioned one of the basic concepts of Hardy Weinberg, which is a population. A population is a group of interbreeding organisms in a particular area. So what are some of the other terms? Well there is this other concept called a gene pool. You got to understand a gene pool is an abstract idea. You can’t go swimming it. What a gene pool is just the sum total of all of the genes in a population. Let’s take a look at an example.

In this population, we have six individuals. And we’re going to be looking at the tongue rolling gene, the ability to do this. That’s a dominant trait. So we can see here that we have two people who are homozygous dominant for the tongue rolling ability. Three individuals who are heterozygous, and then one individual in our population of six, who is homozygous recessive i.e. a non tongue-roller.

So how big is our gene pool? How many copies of this gene do we have? Well we know that every individual has two copies of every gene, that’s Mendel’s first law. So two times six is 12. So we have a gene pool of six genes here. So what’s the frequency of that big R within our gene pool? Let’s take a look.

So I could write big R is one, two and I do some math I count up. And I see seven out of my 12 possible alleles is the big R. I type that into my calculator and get roughly that’s equal to 0.58. In Hardy Weinberg terms they call that the frequency of the dominant allele. And they use the variable ‘p’ to represent whatever the dominant allele is.

What if I was trying to figure out the frequency of this recessive little r, or in Hardy Weinberg terms, ‘q’? Well I could sit there and I could count out the number of the little r’s divide that by twelve. Or I can rip up my incredible math skills and I can say, "Well I know p plus q that frequency of the recessive allele that’s a grand total of one."

So I just plug in my numbers 0.58+q= 1. Subtract 0.58 from both side and I get q equals 0.42. Not so tough. Now what if I was trying to, well what does this mean? It means that the chances of me randomly spinning around and pointing at somebody, and that person happens to have at least big R is 0.58. What’s the chance of that person having two big R’s? Well, if you remember from your math classes, I will just take the probability of it happening once and multiply by the probability of that happening again. Essentially 0.58 times 0.58 or more simply p.

Hey wait a minute, let’s take a look at that p+q=1 thing, again. So p+q=1 let’s see. if I square both sides of this equations I get p² plus 2pq plus q² and 1² is 1.So p² is the chances of somebody being homozygous dominant, well that makes sense. The chance of getting a p and p. q² is the frequency of the homozygous recessive individuals. And then that leaves 2pq to be those individuals who are heterozygous. So that makes sense, p times q, why 2. Well you could get the big R from your mommy and little r from daddy, or the other way around. So there’s two ways to get a dominant and recessive and trait. So that’s why we have that 2 there, and is not just because of the math, it's because it works and it represents reality.

Now I’ve been flinging a bunch of terms out there at you. So let’s take a quick look and make sure that everybody is on the same page with this.

So p is the frequency of the dominant allele in the gene pool. q is the frequency of the recessive allele in the gene pool. So these are describing the abstract concept of the gene pool. P² is the frequency of homozygous dominant people or organisms in the population. So here we’re talking about actual numbers of actual creatures. 2pq is the frequency of heterozygous individuals in the population. While q² is the frequency of homozygous recessive individuals.

So that is basically how Hardy Weinberg works. Now we say this is how it should it out. Let’s take a quick look at our original gene pool and let's plug in our numbers and see if this works.

Well we have here P², here is p. So 0.58² that's p² equals, I do some quick math and I get 0.34. And if you remember, that’s roughly 1/3, that’s 2 of our 6, hey yeah that works. What about 2pq? Well 2pq equals 2 times 0.58 times 0.42. You do some math and that’s simply 0.49. Again I’m rounding a little bit interestingly. And if I look, that’s rough 1/2 and 3 out of my 6 hey it matches.

Now you see how I’m plugging in the numbers on that, I'm just skipping to the end? Ways to get points on essays on Hardy Weinberg. And they like to ask these sorts of questions. Show the math, show the equation and it’ll help increase your score. Let’s do that last one q² and again see if it matches our predictions. q² is simply 0.42². Again I do some simple math and I get 0.18. And 0.18 that’s roughly a 6 and hey 1/6 is my population marks.

So yes this population is in Hardy Weinberg genetic equilibrium. So I’d say that the population was in Hardy Weinberg equilibrium. And that means that from one generation to the next, those gene frequencies shouldn’t change. But there is some assumptions that’s based on. So let’s take a closer look at what are the assumptions of the Hardy Weinberg equations.

First, since it’s a statistical method, it has to be assumed that we’re talking a large population. So if you predict from the Hardy Weinberg equations that you should have 64% of your population is homozygous dominant. But it turns out only say 47% are, then that is an indication that you’re either talking about a small population. Or at some point in the past, the population size greatly get reduced and that caused a tweaking of the predicted outcome.

And this is kind of like if you’re doing that thing with your friend’s coin, and you’re flipping the coins and you’re only flipping three times you know. If you only flip it like I said three times, you can’t get 1.5 heads and 1.5 tails, it just doesn’t work that way.

The thing next is no migration in and out of the population. If some individuals come in, who are all homozygous dominant, then they’re going to skew the results. So that’s another thing that Hardy Weinberg assumes isn’t happening.

Similarly no mutation. You can’t have the things changing. Because if a big R changes into a little r or vice versa again that’s skews the results. Just like if you’re flipping the coin, and instead of getting head or getting tail, the heads turns into tail or a into a foot that just screws things up.

Next up, random mating. Again this is statistical method. And it’s based on the idea that these combination are coming together randomly. And so, if you’ve been able to eliminate the first three assumptions, then that might clue in that the skewing of your Hardy Weinberg predictions is due to non random mating.

The next big one is natural selection. And this like I said is a biggy. Natural selection is giving advantage or disadvantage to one allele or the other. If say having little r was a big advantage, then the one individuals who’s homozygous recessive, he’s going to have great success in providing children for the next generation. So you’ll see a great increase in the numbers of little r’s in the gene pool.

So that’s the assumptions of the Hardy Weinberg equations. And again if everything matches our assumptions, then yes you’re in equilibrium and the frequencies will stay say one generation to the next. But if they aren’t matching out predictions, then this is the nice thing about Hardy Weinberg. Because it give scientist a direction. Rather than just saying something seems wrong, instead they can say, something is wrong. It doesn’t much my numbers, now I have direction in which to go, to solve the problem. To figure out what’s the cause.

So how will Hardy Weinberg show up on the actual exam? Before I get into it, I really recommend that you go online, and check out the official version of the AP lab that deals with Hardy Weinberg. Because that will give you a good example of the kind of questions they may show you and some of the contexts. Also in your bonus materials folder, I’ve also included a worksheet that I gave to my students.

Now because you’re getting a chance to work on it at home I’ve made the numbers pretty hard. On the AP biology test, and in fact on most teacher's test they don’t let you use a calculator. And that’s actually a good thing because that means the numbers that they’re going to give you are pretty easy. So if you don’t know them already, I’d spent a little bit of time just brushing up on your common squares and common square roots. And it’ll make spotting the right answer much faster and much easier.

So let’s take a look at one of the first kinds of questions that they’re going to show you. So here for example, if the frequency of little r, the non rolling alleles is 0.6, then what’s the frequency of the heterozygote? Or the heterozygous individuals in the population?

Well I recall little r is, that’s q. So if I know q equals 0.6, and this is one of the tricks for maxing out your score on the AP exam, the essay portion. State everything. So q equals 0.6, well then I write my equation p plus q equals 1. Again, I just got myself a point by stating one of the equations. I plug in my numbers. And again, show your work. Imagine you’re doing one of your first geometry proofs, and your teacher's sitting there with a ruler ready to get you if you skip something.

So I plug it in, so p plus 0.6 equals 1. So I solve and I get p equals 0.4. So far so good, so now p is 0.4, q is 0.6. So heterozygous individuals they’re 2pq. Plug in my numbers 2 times 0.6 times 0.4, sorry I reversed p and q. I do some math and I get 0.48. So far so good.

Now this was one of the easier questions, because they gave you what q was. What they really like to do is trick you. Let’s take a look at the standard trick question. And I can tell you it’s a standard trick question because I use it all the time on my test. 0.91 of the population can roll their tongue, which is a dominant trait. What’s the frequency of the homozygous dominant individuals?

Now people will sit there and go, "I know it’s 0.91." They’re trying to soccer you in. Because they see 0.91 they say they can roll their tongue, it’s an easy question. No it isn’t. Remember this is the potion of the population that has at least one big R. This includes both the homozygous dominants and the heterozygotes. So the trick is find the q’s, find the q² individuals.

If 100% of the population is 1, then 1 equals p², plus 2pq, plus q² right. So let me plug in. I know that P² plus 2pq is 0.91 plus q². That tells me q² is 0.09. Now I got that, now I take my square root of both sides and I get q equals 0.3. I’ve got q, now p plus q equals 1. Plug it in, go over here. And so I get p equals I do some math 0.7. Now I just say p² my homozygous dominant, is 0.7² or 0.49 and there we go. That’s it, that’s not that hard.

So the trick is, read what they’re giving you, if they say the frequency of the gene is then they’re giving you p or q. The frequency of the alleles. If they say some fraction of the population, then you start thinking they’re trying to trick me. Read very carefully. Are they saying 91% of the population do this dominant thing? They’re trying to sneak it past you. Look for the freaky ones, look for homozygous recessives. Find them, and then you can get q. Once you get q then you can get p, and then you can plug it in to figure out whatever genotype that you want. And that’s it.

Outside of the whole square root thing, none of the math is actually all that hard. And in fact, in the most recent question that involved Hardy Weinberg, they actually gave you p, and they gave you a q. So don’t get too wordy about it.

And remember, the Hardy Weinberg equation is all about describing the gene frequencies, and genotype within a population. Now, it’s based on some assumptions, things like known natural selection or a large sample size, or large population. So if all those assumptions are met, then you don’t expect to see any changes from one generation to the next. If you do see changes, or you see that the predictions don’t match the outcome, then it’s because one or more of those assumptions have been broken.

Go through the end of the chapter questions in your textbook, do some of the problems, and that will give you the practice that you need to do well in the AP exam. And if you can handle all of the questions on my worksheet, then you’re pretty good. You’ve mastered Hardy Weinberg. Just be careful I made them tough and I’ll make you actually to think.

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