# Homework 2 Hardy Weinberg Problems And Answers

The Hardy-Weinberg formulas allow scientists to determine whether evolution has occurred. Any changes in the gene frequencies in the population over time can be detected. The law essentially states that if no evolution is occurring, then an equilibrium of allele frequencies will remain in effect in each succeeding generation of sexually reproducing individuals. In order for equilibrium to remain in effect (i.e. that no evolution is occurring) then the following five conditions must be met:

Obviously, the Hardy-Weinberg equilibrium cannot exist in real life. Some or all of these types of forces all act on living populations at various times and evolution at some level occurs in all living organisms. The Hardy-Weinberg formulas allow us to detect some allele frequencies that change from generation to generation, thus allowing a simplified method of determining that evolution is occurring. There are two formulas that must be memorized:

p = frequency of the dominant allele in the population

q = frequency of the recessive allele in the population

p^{2} = percentage of homozygous dominant individuals

q^{2} = percentage of homozygous recessive individuals

2pq = percentage of heterozygous individuals

Individuals that have aptitude for math find that working with the above formulas is ridiculously easy. However, for individuals who are unfamiliar with algebra, it takes some practice working problems before you get the hang of it. Below I have provided a series of practice problems that you may wish to try out. Note that I have rounded off some of the numbers in some problems to the second decimal place:

You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:

- The frequency of the "aa" genotype.
- The frequency of the "a" allele.
- The frequency of the "A" allele.
- The frequencies of the genotypes "AA" and "Aa."
- The frequencies of the two possible phenotypes if "A" is completely dominant over "a."

Sickle-cell anemia is an interesting genetic disease. Normal homozygous individials (SS) have normal blood cells that are easily infected with the malarial parasite. Thus, many of these individuals become very ill from the parasite and many die. Individuals homozygous for the sickle-cell trait (ss) have red blood cells that readily collapse when deoxygenated. Although malaria cannot grow in these red blood cells, individuals often die because of the genetic defect. However, individuals with the heterozygous condition (Ss) have some sickling of red blood cells, but generally not enough to cause mortality. In addition, malaria cannot survive well within these "partially defective" red blood cells. Thus, heterozygotes tend to survive better than either of the homozygous conditions. If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cell gene?

There are 100 students in a class. Ninety-six did well in the course whereas four blew it totally and received a grade of F. Sorry. In the highly unlikely event that these traits are genetic rather than environmental, if these traits involve dominant and recessive alleles, and if the four (4%) represent the frequency of the homozygous recessive condition, please calculate the following:

- The frequency of the recessive allele.
- The frequency of the dominant allele.
- The frequency of heterozygous individuals.

Within a population of butterflies, the color brown (B) is dominant over the color white (b). And, 40% of all butterflies are white. Given this simple information, which is something that is very likely to be on an exam, calculate the following:

- The percentage of butterflies in the population that are heterozygous.
- The frequency of homozygous dominant individuals.

A rather large population of Biology instructors have 396 red-sided individuals and 557 tan-sided individuals. Assume that red is totally recessive. Please calculate the following:

- The allele frequencies of each allele.
- The expected genotype frequencies.
- The number of heterozygous individuals that you would predict to be in this population.
- The expected phenotype frequencies.
- Conditions happen to be really good this year for breeding and next year there are 1,245 young "potential" Biology instructors. Assuming that all of the Hardy-Weinberg conditions are met, how many of these would you expect to be red-sided and how many tan-sided?

A very large population of randomly-mating laboratory mice contains 35% white mice. White coloring is caused by the double recessive genotype, "aa". Calculate allelic and genotypic frequencies for this population.

After graduation, you and 19 of your closest friends (lets say 10 males and 10 females) charter a plane to go on a round-the-world tour. Unfortunately, you all crash land (safely) on a deserted island. No one finds you and you start a new population totally isolated from the rest of the world. Two of your friends carry (i.e. are heterozygous for) the recessive cystic fibrosis allele (c). Assuming that the frequency of this allele does not change as the population grows, what will be the incidence of cystic fibrosis on your island?

You sample 1,000 individuals from a large population for the MN blood group, which can easily be measured since co-dominance is involved (i.e., you can detect the heterozygotes). They are typed accordingly:

Using the data provide above, calculate the following:

- The frequency of each allele in the population.
- Supposing the matings are random, the frequencies of the matings.
- The probability of each genotype resulting from each potential cross.

Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the Caucasian population of the United States. Please calculate the following.

- The frequency of the recessive allele in the population.
- The frequency of the dominant allele in the population.
- The percentage of heterozygous individuals (carriers) in the population.

In a given population, only the "A" and "B" alleles are present in the ABO system; there are no individuals with type "O" blood or with O alleles in this particular population. If 200 people have type A blood, 75 have type AB blood, and 25 have type B blood, what are the alleleic frequencies of this population (i.e., what are p and q)?

The ability to taste PTC is due to a single dominate allele "T". You sampled 215 individuals in biology, and determined that 150 could detect the bitter taste of PTC and 65 could not. Calculate all of the potential frequencies.

What allelic frequency will generate twice as many recessive homozygotes as heterozygotes?

Here is a tutorial to perfectly understand Hardy-Weinberg Rule! If you feel like you just need a brief reminder, you can skip the text until the section **In short...** and try out the exercices just to check your understanding.

## Terms you should know a priori

I will not define the following terms, so make sure you understand them

## What is Hardy-Weinberg rule?

Hardy-Weinberg rule (HWr) describes a relationship between allele frequency and genotype frequencies. I will present the maths later.

## Alleles and Genotypes frequency

Let's assume we have a bi-allelic (two alleles) locus. The possible alleles are called and .

**Notation for genotype frequencies**

Note that I will use a slightly unusual notation on purpose.

Let $f_{AA}$ be the frequency of the homozygous individual caring the allele on both haplotype (both chromosome). Let $f_{BB}$ be the frequency other homozygote. Let $F_{AB}$ be the frequency of heterozygote which inherited the allele from the mother and the allele from the father. Finally $F_{BA}$ is the frequency of heterozygotes which inherited the allele from the mother and the allele from the father.

Most of the time people write $f_{AB}$ (or similar notation) to designate all heterozygotes irrespective of whether the allele come from the father or the mother. In this answer I make the difference of which parent gave which allele. I will instead note $f_{het}$ the frequency of all heterozygotes. Therefore $f_{het} = f_{AB} + f_{BA}$. Similarly, I will note $f_{hom}$ the frequency of all homozygotes. Therefore, $f_{hom} = f_{AA} + f_{BB}$.

**Notation for allele frequency**

Unlike above, I will use a class notation here.

Let the frequency of the allele be $p$ and the frequency of the allele be $q$.

**All genotype frequencies must sum to 1**

Of course, if you add up the frequency of all types of individuals in a population, you should reach 1 (or 100% if you prefer). Per consequence,

$$f_{AA} + f_{AB} + f_{BA} + f_{BB} = f_{hom} + f_{het} = 1 $$

is our first important result.

*Problem 1*

There is 20% of heterozygote in the population, what fraction of the population is homozygote?

Answer: (hover to see the answer)

$f_{hom}=0.8$ (or $f_{hom}=$80%)

*Problem 2*

There are 10% of individuals and 15% of individuals in the population, what fraction of individuals are heterozygotes

Answer:

$f_{het} = 0.75$ (or $f_{het} =$ 75% )

**All allele frequencies must sum to one**

The same is true for the allele frequency, they must sum to one. This is even easier than before. Using above notation it means

$$p + q = 1$$

You will note that an obvious consequence of this result is that $1-p = q$ is also true. This is the second important result.

*Problem 3*

The frequency of alleles which are the allele is 0.3. What is the frequency of the allele?

Answer:

$q = 0.7$

## Hardy-Weinberg equilibrium

Let's now talk about HWr.

**HWr assumptions**

HWr makes a number of assumptions that I will not detail them here. Please have a look at the post Assumptions of Hardy-Weinberg rule for more information.

**HWr exercice without explanations**

Here are exercices that one may consider they come too early in my explanations. Don't worry too much if you can't solve them but please take the time to try solving them. Please do not try to apply known formulas about HWr, only use your logic and the above formulas (allele frequencies must sum to 1 and genotype freuqencies must sum to 1)

*Problem 4*

In a population, one observes that 20% of the population is homozygote for the allele and that there are no heterozygous individuals at all. What is the frequency of the allele?

Answer:

$f_{A} = 0.2$

Below is a drawing of the population (assuming N=20 individuals) to help understand the answer.

AA | AA | AA | AA | BB | BB | BB | BB | BB | BB | BB | BB | BB | BB | BB | BB | BB | BB | BB | BB

*Problem 5*

In a population, one observes that all alleles are present in heterozygote only. The frequency of heterozygotes is 0.1. What is the frequency of of the allele?

Answer:

$f_{A} = 0.95$

The hard part here is to first calculate $f_{B} = 0.05$. Let's draw the population (assuming only N=20 individuals) and that will likely help you out.

AA | AB | AA | AA | AA | AA | AA | AA | AA | AA | AA | AB | AA | AA | AA | AA | AA | AA | AA | AA

**HWr formula**

As said above, HWr describes a relationship between allele frequency and genotype frequency

Without much explanations, here are these relationships

$$f_{AA} = p^2$$

$$f_{AB} = pq$$

$$f_{BA} = pq$$

$$f_{BB} = q^2$$

Therefore,

$$f_{hom} = p^2 + q^2$$

$$f_{het} = 2pq$$

To build your intuition, you should try to consider for yourself extreme cases where one allele is fixed (fixed=reached a frequency of one). For example what are the expected genotype frequencies when

- $p=1$ and $q=0$?
- $p=q=0.5$?
- $p=0$ and $q=1$?

**How HWr is often expressed**

Mixing up the idea of the above relationships and the fact that all genotype frequencies must sum up to one, one can also write

$$p^2 + 2pq + q^2 = 1$$

It is also common to add the fact that all allele frequencies must sum up to one as well and write (replace $q$ by $1-p$)

$$p^2 + 2p(1-p) + (1-p)^2 = 1$$

**Why do these relationship make sense?**

*Problem 6*

Imagine you are drawing alleles in a population of alleles like you would draw cards from a deck of cards! You draw a single allele. If the frequency of allele is $p$, what is the probability of drawing an allele?

Answer:

$p$

Similarly the probability of picking a allele is $q=1-p$.

*Problem 7*

Ok, now. Assuming that the fact that you already drew an allele does not change the allele frequency in the population (because the population is very large), what is the probability of drawing two alleles in a row?

Answer:

It is $p$ for the first draw and $p$ for the second draw. The overall probability is therefore $p \cdot p = p^2$

Here we go! You just solved why $f_{AA} = p^2$. You can apply the exact same logic to find out $f_{BB} = q^2$.

Now the probability to first draw a allele and then a allele is $pq$. Therefore, $f_{AB} = pq$. The probability to first draw a allele and then a allele is $qp=pq$ as well. Therefore, $f_{BA} = pq$.

**Exercice**

*Problem 8*

In a population with two alleles for a certain locus, A and B, the allele frequency of A is 0.7. What is the frequency of heterozygotes if the population is in Hardy-Weinberg equilibrium?

(You will note I renamed the alleles to match my above notation)

Try to solve this problem yourself now!

Answer:

It is a given that $p=0.7$. The question is what does $f_{het}$ equals to?

Let's start by calculating $q$.

$q=1-p=0.3$.

Then, let's calculate the the genotype frequencies.

$f_{AA} = p^2 = 0.49$.

$f_{BB} = q^2 = 0.09$.

$f_{het} = 2pq = 0.42$.

Here we go! We've got the answer. Let's make sure the answer makes sense by summing up the genotype frequencies to ensure we get 1.

$f_{hom} = f_{AA} + f_{BB} = 0.58$

$f_{hom} + f_{het} = 0.42 + 0.58 = 1$

All good!

*Problem 9*

At a bi-allelic locus, the allele is dominant over the allele . The phenotype associated to the dominant allele is present in 10% of the individuals of the population. What is the frequency of the allele ?

Answer:

The genotypes coding for the phenotype associated with the dominant alleles are and . Therefore, $f_{AA} + f_{Aa} = 0.1$ and therefore, $f_{aa} = 0.9$. The frequency of the allele is $p = \sqrt{f_{aa}} = \sqrt{0.9} ≈ 0.95$. To conclude the frequency of the allele is $q = 1 - p = 1 - \sqrt{0.9} ≈ 0.05$

*Problem 10*

You will find a HW problem involving selection at **this post**.

*Problem 11*

You will find a HW problem involving a population of triploids individuals at **this post**.

*Problem 12*

Your will find a HW problem for sex-linked loci at **this post**

## In short...

Here is a summary for a bi-allelic locus

*Allele freqs sum up to one*

$$p + q = 1$$

*Genotype freqs sum up to one*

$$f_{AA} + f_{AB} + f_{BA} + f_{BB} = f_{hom} + f_{het} = 1 $$

You will note that most often, when people write $f_{AB}$, they just mean $f_{het}$ and therefore don't make the distinction between $f_{AB}$ and $f_{BA}$.

*Hardy-Weinberg rule*

Note that Hardy-Weinberg rule holds only under a number of assumptions (such as random mating, panmictic population, no selection, ...) that I have not detailed here (see Assumptions of Hardy-Weinberg rule for more information).

$$f_{AA} = p^2$$

$$f_{AB} = pq$$

$$f_{BA} = pq$$

$$f_{BB} = q^2$$

Therefore,

$$f_{hom} = p^2 + q^2$$

$$f_{het} = 2pq$$

*Putting all togther*

The whole thing put together is often expressed as

$$p^2 + 2p(1-p) + (1-p)^2 = 1$$