Hardy-Weinberg calculations for A-level Biology

Hardy-weinberg calculations for A-level-biology
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When you’re doing Hardy-Weinberg calculations for A-level Biology you need to make sure you fully understand what the two equations mean and can match up the information you’ve given to the equations. This is what we’ll cover in this tutorial

The two equations of the Hardy-Weinberg principle

You have two equations to work with.

Equation 1. p+ q = 1.0

This equation describes the frequencies of the two alleles. The frequency of the dominant allele is p and the frequency of the recessive allele is q.

Equation 2. p2 + 2pq + q2 = 1.0

This equation describes the frequencies of the three possible genotypes: p2 = the frequency of the homozygous dominant genotype, 2pq = the frequency of the heterozygous genotype and q2 = the frequency of the homozygous recessive genotype.

You will have seen these equations before, but did you pick up on the fact that the first equation relates to the alleles and the second relates to genotypes? This means that if you want to calculate something about the alleles you will get the answer by using the first equation, while if you want to know something about one of the genotypes (or phenotypes, as we’ll see soon) you will get your answer from the second equation.

Regardless of what you want to calculate, it’s likely that you have to use both equations at some point. For example, if you want to calculate the frequency of an allele, but you’re given some information about the frequency of people with a recessive disease, you’ll have to start with the second equation, and finish with the first. Let’s look at this in more detail.

Hardy weinberg calculations for A-level biology
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Making sense of the info you’re given

Before you tackle a question you need to make sure you understand the information you’ve been given. Remember, both equations use frequencies, so this is what you’re given you can input the data straight in the equation and you’ll get your answer straight back out (after a bit of problem solving).

It’s likely that the question you get will be a bit more complicated that that. First of all, you might not be given frequencies, and you might have to work quite hard to figure out how the information you have fits with the equations.

Let’s look at an example

Hardy-Weinberg calculations for A-level Biology example 1:

Sickle cell anaemia is a recessive disease. 1 in 500 African Americans has sickle cell anaemia. How many carriers of sickle cell anaemia are there per 500 African Americans?

This is one of the types of questions that requires the most work. Here’s how I would approach this question:

1. Work out what I actually need to calculate and I can get that using one of the equations of the Hardy-Weinberg principle. I can see that I’ve been asked to work out the number of carriers. Carriers of recessive diseases have the heterozygous genotype so I need to calculate 2pq. This won’t give me the final answer though: I’ve been asked to calculate the number of carriers per 500 people, 2pq is just the frequency (e.g. how many people per 1). To convert from frequency to number of carriers per 500 I just need to do 2pq x 500.

I know know what information I need to get, and how to convert that into a final answer, so the next step is to decide how to start

2. Work out how the information I have fits with the 2 equations and find a starting point. I’ve been told that 1 in 500 people have sickle cell anaemia. Since this is a recessive disease it means that 1 in 500 people have the homozygous recessive genotype i.e. are q2 . I can therefore work out q2 by doing 1/500 = 0.002.

3. Problem solve your way through it. We now have a variable that fits with one of the Hardy-Weinberg equations (q2)and we know what we need to calculate (2pq), so now we need to use out problem solving skills to get from one to another. Here are the steps we need to take:

q2 = 1/500 

    = 0.002

Then use this to calculate q:

q = √q

q = √0.002

   = 0.00447

Then calculate p:

p = 1.0 – q

  = 1.0 – 0.996 = 0.004

You can now calculate 2pq:

2pq = 2 x 0.00447 x 0.996

       = 0.00890

Finally, calculate the number of carriers in the population:

Number of carriers = frequency x population size

        = 0.00890 x 500

      = 4.45 

      = 4 carriers in the population

Common mistakes

The key to doing these calculations is to make sure you follow steps 1 and 2 and take time to think really carefully about what you’ve been given and how it fits with the equations. In my experience the main reason that students struggle with these questions is that they don’t do this, so don’t really know what the info they have means and therefore don’t know what to do with it. Another common mistake is not taking a problem solving approach and just trying to do what they’ve done before without thinking about whether that’s relevant. For example, they end up square rooting an allele frequency because they’ve done other questions that involved a square root, even though this doesn’t make sense based on the equations. Mistakes with squaring/square rooting are a particular problem so remember to double check your working.

Your turn to practice

Here’s an example for you to practice with. Follow the 3 steps above to tackle this question. It’s different to the first example, so you won’t be able to copy the exact steps, but remember the point is that you’ll need to use your problem solving skills anyway. When you’ve had a go, scroll down to see the solution.

Hardy-Weinberg calculations for A-level Biology example 2

Neurofibromatosis is caused by a dominant allele. In the United Kingdom the incidence of neurofibromatosis is 1 in 4,500. What is the frequency of the allele for Neurofibromatosis in the United Kingdom?

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Solution

Step 1: The question asked you to calculate the frequency of the allele for Neurofibromatosis. This is the dominant allele p. 
You can calculate p by using the equation p + q = 1.0 and this will be the final step you need to do to get your answer.

Step 2: You are told that 1 person in 4,500 people have Neurofibromatosis. As this is a dominant disease, 1 person in 4,500 is either homozygous dominant (p2) or heterozygous (2pq). Therefore you can use this information to get p2 + 2pq.

Step 3: Problem solve your way through it (yes, this one requires quite a lot of work).

p2 + 2pq = 1/4500 = 0.000222

Now you have this you can calculate q2:

q2 = 1-(p2 + 2pq)

    = 1 – 0.000222

    = 0.99978

Now you can calculate q:

q = √q2

   = √0.99978

  = 0.99989

Finally you can calculate p:

p = 1- q = 1- 0.99989

= 0.00011. This is your final answer

You’ll notice I kept 5 significant figures in throughout. If in doubt keep more significant figures than you think you need – you’re less likely to go wrong than if you round too early.

What now

The examples I’ve shown you here are about as hard as it gets with these calculations, so if you’re confused don’t be too worried. It takes quite a bit of practice to get to the point where you can confidently work out what to do, so go do as many questions as you can find and remember to follow the three steps outlined in this tutorial.

Coming soon: Hardy Weinberg calculations for A-level Biology cheat sheet.