# Re-establishment of Hardy Weinberg Equilibrium After Selection

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I have been given this question:

In a population of 2500 mice, light-colored fur is dominant to dark-colored fur. 2400 mice originally have light-colored fur. However, an owl species moves into their environment, preying mainly upon the light-colored mice as they are easier to spot when hunting at night. When the owls finally leave, 80% of the dark-colored mice are left, and only 10% of the light-colored mice. After many generations, Hardy-Weinberg equilibrium is re-established. Approximately what percentage of mice now have dark-colored fur?

A) 10%

B) 15%

C) 20%

D) 25%

E) 30%

F) 35%

My thought process goes as such:

When the owls leave, the groups are 240 light mice and 80 dark mice.

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

q is recessive allele, p is dominant allele.

$$0.5^2 + 2*0.5*0.5 + 0.5^2 = 1$$

Apply fitness:

$$0.8*(0.5^2) + 0.1*(2*0.5*0.5) + 0.1*(0.5^2) = 0.275$$

(0.8 and 0.1 correlate to the fitness of dark and light mice, respectively)

Now divide everything by 0.275 to figure out new allele frequencies:

$$0.727 + 0.182 + 0.091 = 1$$

This means that 0.727 = 73% is the new frequency for homozygous recessive (dark) mice.

However, the correct answer is B) 15%

Any idea what I did wrong/how the correct answer is achieved?

Before the owls arrive, 100 out of the 2500 mice are dark-colored. This ratio provides the frequency of the recessive allele

\$\$q^2 = frac{100}{2500}\$\$

So \$q\$ is 10/50 or 0.2. This means that the frequency of the dominant allele \$p\$ is 1 - 0.2 or 0.8.

Knowing these frequencies, it is possible to calculate the number of each allele

\$\$AA = (0.8)^2 imes 2500 = 1600\$\$ \$\$Aa = 2 (0.8) (0.2) imes 2500 = 800\$\$ \$\$aa = (0.2)^2 imes 2500 = 100\$\$

Of the light-colored mice, 2/3 are AA and 1/3 are Aa. The same proportions hold after the owls leave, but the total humber has been reduced. The dark-colored mice are all aa.

\$\$AA = frac{2}{3} imes 240 = 160\$\$ \$\$Aa = frac{1}{3} imes 240 = 80\$\$ \$\$aa = 80\$\$

From these values, the total number of each allele may be calculated

\$\$A = 2 imes 160 + 80 = 400\$\$ \$\$a = 80 + 2 imes 80 = 240\$\$

These numbers provide the frequency of each allele post-selection

\$\$p = frac{400}{640} approx 0.63\$\$ \$\$q = frac{240}{640} approx 0.37\$\$

Applying Hardy Weinberg for the new equilibrium

\$\$aa approx (0.37)^2 approx 0.14 \$\$

which is, "approximately" 15%.

## Re-establishment of Hardy Weinberg Equilibrium After Selection - Biology

Allele frequencies (or percentages, if you prefer) in a population will remain in Hardy-Weinberg Equilibrium (HWE) from generation to generation if the following assumptions are met:

1. Natural selection is not occurring
2. Migration (Gene Flow) is not occurring
3. Mutation is not occurring
4. Genetic Drift is not occurring (drift is less likely in populations of large size)
5. Mating occurs at random

Although these assumptions are rarely true in the natural world, they allow us to calculate an expected allele frequency. Significant differences between the observed and expected frequencies indicate that "something" (i.e. one or more of the above) is going on, and therefore tell us that "microevolution" is occurring.

Calculating Expected Allele and Genotype frequencies:

In the simplest possible situation we have a single gene with only two alleles. These alleles might be A and a, or A1 and A2. Let's say that A or A1= tall, and a or A2= short. Don't worry for now whether the alleles are dominant and recessive or co-dominant. They will have frequencies p and q in a population. (Because there are only two possibilities and they have to add up to 100%, p + q = 1.)

If we know the allele frequencies, we can predict the genotype frequencies. The expected genotype frequencies of the two alleles are calculated as shown. This ought to look familiar: it's our old friend the Punnet's Square. Allele A or A1 has a frequency of p, and allele a or A2 has a frequency of q. Multiply the allele frequencies to the get the probability of each genotype.

p q A1 p p 2 pq A2 q pq q 2

In other words, p 2 + pq + pq + q 2 = 1, or 100%. The expected frequencies of the genotypes are therefore:

 Genotype Expected Frequency AA or A1A1 p * p = p 2 Aa or A1A2 pq + pq (or 2pq) aa or A2A2 q * q = q 2

Let's take a look at some graphs of this to make it a little easier to see. For values of p from 0 to 1, in intervals of 0.1, here's what we get:

Red represents the frequency of the AA or A1A1 genotype, green is the Aa or A1A2 genotype, and blue is the aa or A2A2 genotype.

All of the above has to do with the allele and genotype frequencies we would expect to see. Next, let's look at the real world situation so we can compare.

Calculating Observed Allele and Genotype Frequencies:

In a real world population, we can only see phenotypes, not genotypes or alleles. However, in a population of genotypes AA, Aa and aa, the observed frequency of allele A equals the sum of all of the AA genotype plus half of Aa genotype (the A half). The observed frequency of allele a is therefore half of the Aa individuals (the a half) plus all of aa individuals . If you know one value, you can of course just subtract it from 1 (100%) to get the value of the other. In other words, the observed frequency of A = 100%(AA) + 50%(Aa) and a = 50%(Aa) and 100%(aa)

 Phenotype Genotype Makeup Frequency Tall AA 100% A p 2 Tall A a 50% A and 50% a 2pq Short aa 100% a q 2
or
 Phenotype Genotype Makeup Frequency Tall A1A1 100% A1 p 2 Medium A1 A2 50% A1 and 50% A2 2pq Short A2 A2 100% A2 q 2

Tip: If the alleles are codominant, each phenotype is distinct (you can distinguish between tall, medium and short) and your job is easier. If the alleles are dominant and recessive, we can't visually tell the homozygous AA from the heterozygous Aa genotypes (both are tall), so it's best to start with the homozygous recessive (short) aa individuals. Count up the aa types and you have the observed q 2 . Then, take the square root of q 2 to get q, and then subtract q from 1 to get p. Square p to get p 2 and multiply 2*p*q to get the observed heterozygous Aa genotype frequency.

If observed and expected genotype frequencies are significantly different , the population is out of HWE.

 Genotype Frequencies AA Aa aa Observed Expected Difference
or
 Genotype Frequencies A1A1 A1A2 A2A2 Observed Expected Difference

Question: Why might observed and expected phenotype frequencies differ? Imagine the following scenarios where natural selection is at work. Situation one favors only one tail of the distribution. Perhaps the tallest, perhaps the shortest, but not both. This is directional selection. Now imagine that both tails of the distribution are selected against, and only the middle is favored. This is called stabilizing selection. Next imagine that the extremes on both ends are favored. This is called disruptive selection. In each of these scenarios, what would happen over time?

Before (dotted line) and after (yellow shaded area) directional selection, stabilizing selection, and disruptive selection.

One common misconception is that dominant alleles will rise in frequency and recessive alleles will decline in frequency over time. In reality, allele frequencies will not change from one generation to the next if the assumptions listed above are not violated. A good example of this is human ABO blood type. Type O blood is recessive but it remains the most common.

In the hwe.xlsx Excel Spreadsheet, there are three examples to help make this more concrete.

Example 1 : Allele A is dominant and allele a is recessive. Set the original frequencies of p (allele A) and q (allele a) at 0.6 and 0.4 in Generation 1. These are highlighted in blue. All other numbers are calculated from these two original data points. The frequency of genotype AA is determined by squaring the allele frequency A. The frequency of genotype Aa is determined by multiplying 2 times the frequency of A times the frequency of a. The frequency of aa is determined by squaring a. Try changing p and q to other values, ensuring only that p and q always equal 1. Does it make any difference in the results?

Example 2 : Alleles A 1 and A 2 are co-dominant. In this case, A 1 is at a frequency of 0.25 and A 2 is at a frequency of 0.75.

Example 3 : Alleles A and a are dominant and recessive. Note that allele A is at very low frequency despite being dominant. Does it increase in frequency?

The second sometimes confusing thing about HWE is that after all of the examples above, you may wonder if it is possible for the observed and expected frequencies to differ. Here's an example where they do:

In a population of snails, shell color is coded for by a single gene. The alleles A 1 and A 2 are co-dominant. The genotype A 1 A 1 makes an orange shell. The genotype A 1 A 2 makes a yellow shell. The genotype A 2 A 2 makes a black shell. 1% of the snails are orange, 98% are yellow, and 1% of the snails are black.

Observed frequency of A 1 allele = 0.01 + 0.5(.98) = 0.50 = 50%

p 2 = Expected frequency of A 1 A 1 = 0.25

2pq = Expected frequency of A 1 A 2 = 0.50

q 2 = Expected frequency of A 2 A 2 = 0.25

 Phenotype Orange Yellow Black Genotype A 1 A 1 A 1 A 2 A 2 A 2 Observed 1% 98% 1% Expected 25% 50% 25% Difference -24% +48% -24%

There are significantly fewer orange and black snails than expected, and significantly more yellow snails than expected. It appears that this is a case of stabilizing selection, since both tails appear to be strongly selected against.

## Practice Question

In plants, violet flower color (V) is dominant over white (v). If p = 0.8 and q = 0.2 in a population of 500 plants, how many individuals would you expect to be homozygous dominant (VV), heterozygous (Vv), and homozygous recessive (vv)? How many plants would you expect to have violet flowers, and how many would have white flowers?

[practice-area rows=&rdquo2&Prime][/practice-area]
[hidden-answer a=&rdquo84391&Prime]The expected distribution is 320 VV, 160Vv, and 20 vv plants. Plants with VV or Vv genotypes would have violet flowers, and plants with the vv genotype would have white flowers, so a total of 480 plants would be expected to have violet flowers, and 20 plants would have white flowers.[/hidden-answer]

In theory, if a population is at equilibrium&mdashthat is, there are no evolutionary forces acting upon it&mdashgeneration after generation would have the same gene pool and genetic structure, and these equations would all hold true all of the time. Of course, even Hardy and Weinberg recognized that no natural population is immune to evolution. Populations in nature are constantly changing in genetic makeup due to drift, mutation, possibly migration, and selection. As a result, the only way to determine the exact distribution of phenotypes in a population is to go out and count them. But the Hardy-Weinberg principle gives scientists a mathematical baseline of a non-evolving population to which they can compare evolving populations and thereby infer what evolutionary forces might be at play. If the frequencies of alleles or genotypes deviate from the value expected from the Hardy-Weinberg equation, then the population is evolving.

Use this online calculator to determine the genetic structure of a population.

## Re-establishment of Hardy Weinberg Equilibrium After Selection - Biology

Knowing that evolution is a change in the frequency of alleles in a population over a period of time, the Hardy-Weinberg equilibrium was tested (Lapiana, 1994). Using Hardy-Weinberg’s equations the statistical data of a scenario in which the recessive gene resulted in a negative phenotype was analyzed. In populations where negative phenotypes are carried, the frequency of that allele rapidly decreases. Using beads and random selection over ten generations it was concluded that natural selection removes negative genotypes in a short period of time.

In this lab we learned about Hardy-Weinberg equilibrium and equation which helps us estimate the frequency of the alleles, that is p 2 +2pq+q 2 =1. One represents 100%, p represents the dominant alleles,, q represents the recessive allele while 2pq are the heterozygous alleles, for example Ff . We also had a hypothesis for this lab which was, “If the Hardy-Weinberg equilibrium is tested, the frequency of alleles will remain constant.”

The basic idea of the experiment was to predict and test the frequency of the alleles “F” and “f”. We did this using beads, the white beads were the dominant alleles and the red ones were recessive. Also, to simulate natural selection we were asked to select against the red beads after which we got our results, which will be discussed further in detail about in the data/results section.

This experiment was conducted in the New Tech High at Coppell Biology Lab on December 5, 2015 and December 7, 2015. The materials include white beads symbolizing one dominant fur allele and red beads symbolizing one recessive no fur allele. We had a total of ten trials signifying each new generation of alleles being paired to create one bunny. After all the alleles are matched up each round we would take out the homozygous recessive allele bunnies because they died off. In our data you can see that after each generation we put how many heterozygous dominant (FF), homozygous dominant (Ff), and homozygous recessive individuals (ff) were made.

The results showed that the individuals with ff alleles decreased over several generations due to natural selection. Every generation the number of individuals with the alleles Ff and ff decreased. The number of individuals with FF alleles increased over time. The population of individuals with ff alleles were extinct after the 8th generation.

We can conclude that the Hardy-Weinberg Law was observed throughout this lab. Our hypothesis was proven incorrect, as we had stated that the amount of recessive alleles would stay relatively constant. However, the data supports the statement that as alleles are passed through generations, the amount of recessive alleles decreased, since the homozygous recessive individuals died off due to natural selection and survival of the fittest. Also, the amount of total individuals in the population decreased from 50 in generation 1 to 26 in generation 10. The population started with 50 F alleles and 50 f alleles, and by the end there were only 4 f alleles left. If the four recessive alleles were pair together to make two homozygous dominant individuals, the population would remain at a 100% homozygous dominant gene frequency and would remain unchanged unless breeding from individuals outside the population occurred.

Using a simulation, we tested natural selection through the lens of the Hardy-Weinberg equilibrium. After collecting our data we recognised that our hypothesis was incorrect. Instead of the frequency of recessive alleles remaining constant, the negative recessive phenotype was almost completely dissolved in 10 generations. The majority of the error in our experiment occurred during the random selection of the beads. Using beads that are congruent would eliminate the majority of the human error. The statistical analysis of our data was accomplished using the Hardy-Weinberg equations. We conclude that natural selection follows the principles of the Hardy-Weinberg equilibrium.

## MrBorden's Biology Rattler Site Room 664

Learning Objective:Construct an explanation based on evidence that the process of evolution primarily results from four factors: (1) the potential for a species to increase in number, (2) the heritable genetic variation of individuals in a species due to mutation and sexual reproduction, (3) competition for limited resources, and (4) the proliferation of those organisms that are better able to survive and reproduce in the environment

Evaluate the evidence for the role of group behavior on individual and species’ chances to survive and reproduce.

QFD : “I am thankful for all of those who said NO to me. Its because of them I’m doing it myself.” – Albert Einstein

?FD :name 3 adaptations made by mammals over the last 500 years

Learning Objective:Construct an explanation based on evidence that the process of evolution primarily results from four factors: (1) the potential for a species to increase in number, (2) the heritable genetic variation of individuals in a species due to mutation and sexual reproduction, (3) competition for limited resources, and (4) the proliferation of those organisms that are better able to survive and reproduce in the environment

Evaluate the evidence for the role of group behavior on individual and species’ chances to survive and reproduce.

1) handout on natural selection

QFD : “Go where you are celebrated – not tolerated. If they can’t see the real value of you, it’s time for a new start.” – Unknown

?FD : describe the difference between divergent and convergent evolution

Learning Objective: Construct an explanation based on evidence that the process of evolution primarily results from four factors: (1) the potential for a species to increase in number, (2) the heritable genetic variation of individuals in a species due to mutation and sexual reproduction, (3) competition for limited resources, and (4) the proliferation of those organisms that are better able to survive and reproduce in the environment

Evaluate the evidence for the role of group behavior on individual and species’ chances to survive and reproduce.

1) Hardy Weinberg problems

QFD : Don’t say you don’t have enough time. You have exactly the same number of hours per day that were given to Helen Keller, Pasteur, Michaelangelo, Mother Teresea, Leonardo da Vinci, Thomas Jefferson, Albert Einstein, etc…

?FD : Describe how genes play a role in evolution of species, be specific

Learning Objective:Construct an explanation based on evidence that the process of evolution primarily results from four factors: (1) the potential for a species to increase in number, (2) the heritable genetic variation of individuals in a species due to mutation and sexual reproduction, (3) competition for limited resources, and (4) the proliferation of those organisms that are better able to survive and reproduce in the environment

Evaluate the evidence for the role of group behavior on individual and species’ chances to survive and reproduce.

QFD : “I am thankful for all of those who said NO to me. Its because of them I’m doing it myself.” – Albert Einstein

?FD :name 3 adaptations made by mammals over the last 500 years

Learning Objective:Construct an explanation based on evidence that the process of evolution primarily results from four factors: (1) the potential for a species to increase in number, (2) the heritable genetic variation of individuals in a species due to mutation and sexual reproduction, (3) competition for limited resources, and (4) the proliferation of those organisms that are better able to survive and reproduce in the environment

Evaluate the evidence for the role of group behavior on individual and species’ chances to survive and reproduce.

1 The effects of natural selection may be countered by

2 In relation to natural selection, evolution is the

3 Scientists generally agree that heterozygous advantage is

4 The occurrence of large or small beak sizes among seed crackers in the absence of medium-sized beaks is an example of

5 Sickle-cell trait in humans is a classic example of ____________________.

A) how mutations can lead only to tragic outcomes

B) why outbreeding is important

C) the superior fitness seen in heterozygotes

D) how every organism is an integrated gene complex

6 A person with sickle cell trait, having one S allele and one normal, will be resistant to malaria and eventually develop sickle cell anemia.

7 For a woman living in central Africa, which genotype would be the most advantageous to have?

A) homozygous for the sickle cell allele

B) heterozygous for the sickle cell allele

C) homozygous for the normal hemoglobin allele

D) it doesn’t matter all are equally advantageous

8 Mating with relatives is called

9 The random loss of alleles in a population is called

10 Which of the following factors is most likely to contribute to gene flow between populations?

11 In the Hardy-Weinberg equation, the term 2pq represents the frequency of the

12 A scientist measures the circumference of acorns in a population of oak trees and discovers that the most common circumference is 2 cm. What would you expect the most common circumference(s) to be after 10 generations of stabilizing selection?

B) greater than 2 cm or less than 2 cm

C) greater than 2 cm and less than 2 cm

D) can’t tell from the information given

13 Refer to question 16, but this time answer what you would expect after 10 generations of disruptive selection.

B) greater than 2 cm or less than 2 cm

C) greater than 2 cm and less than 2 cm

D) can’t tell from the information given

14 Refer to question 16, but this time answer what you would expect after 10 generations of directional selection.

B) greater than 2 cm or less than 2 cm

C) greater than 2 cm and less than 2 cm

D) can’t tell from the information given

15 Why is genetic polymorphism important to evolution?

A) individual variability provides the raw material for natural selection to act on

B) genes cannot mutate unless they are polymorphic

C) only heterozygous individuals are selected in natural populations

D) the Hardy-Weinberg equilibrium is less likely to be disturbed in polymorphic populations

E) none of the above genetic polymorphism is not important to evolution

16 In a population of wildflowers, the frequency of the allele for red flowers was 0.8. What was the frequency of the white allele, the only other allele for flower color?

17 Referring to question 16, what is the frequency of homozygous red flower plants in the population?

18 Referring to question 16, what is the frequency of homozygous white flower plants in the population?

19 Referring to question 16, what is the frequency of plants in the population that are heterozygous for flower color?

20 What is the ultimate source of genetic variability?

21 The movement of new genes into a population as a result of migration or hybridization is called

22 A virus killed most of the seals in the North Sea (e.g., dropped the population from 8000 to 800). In an effort to help preserve the species, scientists caught 20 seals and used them to start a new population in the northwest Pacific Ocean. Which of the following factors would most likely have the least impact in this new population?

A) increases the rate of mutation

B) increases the proportion of homozygous individuals in a population

24 Evolution by natural selection works best on a population having no variation.

25 According to Darwin’s theory of evolution, evolution occurs through natural selection operating on populations in ecosystems.

26 The theory of population genetics and how evolution occurs includes all but which one of the following

B) The size of the population is small.

C) There is no influx of genes from other populations.

D) No genotype has selective advantage over another.

27 Using the Hardy-Weinberg Principle, which expression represents the frequency of the homozygous recessive genotype?

28 Which statement most accurately reflects what population geneticists refer to as “fitness”?

A) Fitness is the measure of an organism’s adaptability to various habitats.

B) Fitness reflects the number of mates each individual of the population selects.

C) Fitness refers to the relative health of each individual in the population.

D) Fitness is a measure of the contribution of a genotype to the gene pool of the next generation.

29 Mutation is a relatively unimportant source of variation and is not the foundation for evolution.

30 The fact that the majority of human newborns weigh around 7 pounds is reflective of ________________.

D) None of these is correct.

31 Organisms that are least likely to experience extinction over the long term are most likely to be found in _______________.

A) areas inhabited by humans

32 Which one of the following would cause the Hardy-Weinberg principle to be inaccurate?

A) The size of the population is very large.

B) Individuals mate with one another at random.

C) Natural selection is present.

D) There is no source of new copies of alleles from outside the population.

E) None of the answers is correct.

33 Which one of the following populations would most quickly lead to two groups with few shared traits?

A) a population with disruptive selection

B) a population with directional selection

C) a population with stabilizing selection

D) a population with no selection

34 Mutations tend to have little effect on the allele frequency in a population.

35 The effects of genetic drift are most apparent in small populations.

early fossil trilobite

## Re-establishment of Hardy Weinberg Equilibrium After Selection - Biology

What will happen in the next generation? To answer this question, we will use the Hardy-Weinberg principle, which applies basic rules of probability to a population to make predictions about the next generation. The Hardy-Weinberg principle predicts that allelic frequencies remain constant from one generation to the next, or remain in EQUILIBRIUM, if we assume certain conditions (which we will discuss below).

For example, if the allelic frequencies of alleles A and a in the initial population were p = 0.8 and q = 0.2, the allelic frequencies in the next generation will remain p = 0.8 and q = 0.2. The conditions for Hardy-Weinberg equilibrium are rarely (if ever) encountered in nature, but they are fundamental to understanding population genetics. When a population deviates from Hardy-Weinberg predictions, it is evidence that at least one of the conditions in not being met. Scientists can then determine why allelic frequencies are changing, and thus how evolution is acting on the population.

The conditions for Hardy-Weinberg equilibrium:

1. Population is infinitely large -&ndash or large enough to minimize the effect of genetic drift, which is change in allele frequencies due entirely to random chance (and not selection).
2. No selection occurs - so all the individuals in the population have an equal chance of surviving and reproducing.
3. Mating is random &ndash so that an individual is equally likely to mate with any potential mate in the population, regardless of genotype or phenotype.
4. No migration - so no alleles enter or leave the population.
5. No mutation - so allelic characteristics do not change

Because mating is random (Condition 3, above), we can think of these diploid individuals as simply mixing their gametes. We do not need to consider the parental origin of a given gamete (i.e. if it comes from a heterozygous or homozygous parent), but simply the proportion of alleles in the population. For example, for the population mentioned previously with p value of 0.8 and q value of 0.2, we can think of a bag of mixed gametes with 80% of which are A and 20% are a .

Therefore, on the paternal side (the sperm) we have the given proportions of the two alleles (0.8 of allele A and 0.2 of allele a ) freely mixing with the eggs (the maternal contributions), which have the alleles in the same proportions (0.8 of A and 0.2 of a ).

The probability of an A sperm meeting an A egg is 0.8 x 0.8 = 0.64. The probability of an A sperm meeting an a egg is 0.8 x 0.2 = 0.16. The probability of an a sperm meeting an A egg is 0.8 x 0.2 = 0.16. The probability of an a sperm meeting an a egg is 0.2 x 0.2 = 0.04.

Therefore in the following generation, we would expect to have the following proportion of genotypes:

That is, if there were a thousand offspring, there would be:

This in turn translates to 1600 A alleles (640 + 640 + 320), and 400 a alleles (320 + 40 + 40). 1600/2000 = 0.8 and 400/2000 = 0.2 that is, the allele frequencies are the same as in the parental generation.

To generalize: if the allele frequencies are p and q, then at Hardy-Weinberg Equilibrium you will have (p + q) X (p + q) = p 2 + 2pq + q 2 as the distribution of the genotypes.

• The frequency of AA individual will be p 2 .
• The frequency of Aa individuals will be 2pq.
• The frequency of aa individuals will be q 2 .

Furthermore, the frequency of A alleles will be p 2 + pq (equal to the frequency of AA individuals plus half the frequency of Aa individuals). Since p + q =1, then q = 1 - p. The frequency of A alleles is p 2 + pq, which equals p 2 + p (1 &mdash p) = p 2 + p &mdash p 2 = p that is, p stays the same from one generation to the next. The same can be shown for q.

So we see that with random mating, no selection, no migration or mutation, and a population large enough that the effects of random chance are negligible, the proportion of alleles in a population stays the same from generation to generation.

Let&rsquos test your knowledge of this topic:
In a population that is in Hardy-Weinberg equilibrium, the frequency of the dominant allele A is 0.40. What is the frequency of individuals with each of the three allele combinations, AA , Aa and aa ?

Frequency of AA individuals: _______
Frequency of Aa individuals: _______
Frequency of aa individuals: _______

Since the frequency of allele A is 0.4, the frequency of allele a is 1 &ndash 0.4 = 0.6.

When the population is in Hardy-Weinberg equilibrium:

The frequency of AA individuals is (0.4)(0.4) = 0.16
The frequency of Aa individuals is 2(0.6)(0.4) = 0.48
The frequency of aa individuals is (0.6)(0.6) = 0.36

(Note: a good way to check if your answers are correct is to verify that the values add up to 1.)

## Deviation from the Hardy-Weinberg Equilibrium

The Hardy-Weinberg Equilibrium can be used as a null hypothesis, compared to values from a real population, to describe statistically significant deviations from the Equilibrium. If the deivation is significant, then the gene frequencies are changing and thus, evolution is occurring.

### Example 3:

In a population of Zebra Mussels (Dreissena polymorpha), 1,469 individuals have the CC form of the enzyme phosphoglucose isomerase (PGI), an enzyme critical in the second step of glycolysis. Another 138 have the Cc form of PGI, and another 5 individuals have the cc form of the enzyme. All of these forms of the enzyme are fully functional.

#### Problem:

Is this population in Hardy-Weinberg Equilibrium?? [In other words, the null hypothesis is that the population is at H-W equilibrium.]

#### Solution:

1. p = frequency of dominant allele (C) = 2(CC) + 1(Cc) / 2n
= 2 x 1,469 + 138 / 2 (1,469 + 138 + 5) = 3,076 / 3,224 = 0.954
2. q = 1 - p = 0.046
3. For this example, HW expectations are:
• Exp(CC) = p 2 n = 0.9542 x 1,612 = 1,467.4
• Exp(Cc) = 2pqn = 2 x 0.954 x 0.046 x 1,612 = 141.2
• Exp(cc) = q 2 n = 0.0462 x 1,612 = 3.4

Determine deviation of actual from HW expectations:

The Paleontological Research Institution and its Museum of the Earth
1259 Trumansburg Road &bull Ithaca, NY 14850 USA
phone: 607-273-6623 &bull fax: 607-273-6620

## How to use chi-squared to test for Hardy-Weinberg equilibrium

This post demonstrates the use of chi-squared to test for Hardy-Weinberg equilibrium. There is a question on a recent (February 2020) AP Biology practice test that required this calculation. The question is a secure item, so the exact question will not be discussed here. There is a previous post on this blog explaining how to test for evolution using the null hypothesis and chi-squared.

mv2.png/v1/fit/w_300,h_300,al_c,q_5/file.png" />

For our examples, we'll use the fictional species featured in many of the evolution simulations. The population demonstrates incomplete dominance for color. There are two alleles red and blue. Heterozygotes have a purple phenotype.

mv2.png/v1/fit/w_235,h_77,al_c,q_5/file.png" />

Chi-squared is a statistical test used to determine if observed data (o) is equivalent to expected data (e). A population is at Hardy-Weinberg equilibrium for a gene if five conditions are met random mating, no mutation, no gene flow, no natural selection, and large population size. Under these circumstances, the allele frequencies for a population are expected to remain consistent (equilibrium) over time. The H-W equations are expected to estimate genotype and allele frequencies for a population that is at equilibrium. The equations may not accurately predict the frequencies if the population is not at equilibrium (for example, if selection is occurring). However, it is possible that, even with the presence of an evolutionary force, a population may still demonstrate the expected H-W data.

mv2.png/v1/fit/w_300,h_300,al_c,q_5/file.png" />

In the case of a trait showing incomplete dominance, the heterozygotes are distinct from the homozygous dominant individuals, which allows the genotype and allele frequencies to be calculated directly (without the H-W equations). This direct calculation can be compared to values based on H-W calculations to determine if the population is at H-W equilibrium.

For the first example, we'll use a simple data set (not generated by a simulation). In this case, there are 50 total individuals in the population 10 are red, 10 are purple, and 30 are blue. These are the observed values for the chi-squared analysis.

## Why is the Hardy Weinberg Equation used?

Here are some of the Uses & Applications of the Hardy Weinberg Equation. Let’s Know :

1. This is used to calculate the genetic variation of a population at equilibrium.

2. This is used to determine the allele and genotype frequencies in a population.

3. This is used to determine the number of individuals in a population with the same allele and genotype in the particular locus of the chromosome.

4. This equilibrium is used in population genetical studies as it indicates the unchanging frequency of alleles and genotypes in a stable, idealized population which is not evolving.

5. This can also be used in determining the complete dominance of the genotype when the two alleles in the genotype cannot be distinguished.

6. This can also be used to calculate the frequency of heterozygous carriers of harmful recessive genes.

7. This is also helpful in calculating the frequency of the genotype with more than two alleles. In such cases, the equation changes a bit. Just for three alleles, the equation will be (p+q+r) 2 =1, for four alleles the equation will be (p+q+r+s) 2 =1, and so on.

8. This equilibrium can also be used for calculating the gene frequencies in case of sex-linked loci in both males and females.

9. This can also be used to determine the linkage disequilibrium if linkage equilibrium is not reached in a gene pool (i.e. the total genes and their alleles in a population). Linkage equilibrium is the state of genetic equilibrium in a population that can be reached due to genetic exchange by recombination with two or more alleles at one locus and another locus on the same chromosome with two or more alleles.

10. Hardy-Weinberg principle provides a mathematical baseline for studying non-evolving populations by comparing them with the evolving populations and thereby to come with various analysis, predictions, and conclusions about what evolutionary forces might be at play in the evolution of the population.

## The Hardy-Weinberg Theorem

Godfrey Hardy was an English mathematician. Wilhelm Weinberg was a German doctor. Each worked alone to come up with the founding principle of population genetics. Today, that principle is called the Hardy-Weinberg theorem. It shows that allele frequencies do not change in a population if certain conditions are met. Such a population is said to be in Hardy-Weinberg equilibrium. The conditions for equilibrium are:

1. No new mutations are occurring. Therefore, no new alleles are being created.
2. There is no migration. In other words, no one is moving into or out of the population.
3. The population is very large.
4. Mating is at random in the population. This means that individuals do not choose mates based on genotype.
5. There is no natural selection. Thus, all members of the population have an equal chance of reproducing and passing their genes to the next generation.

When all these conditions are met, allele frequencies stay the same. Genotype frequencies also remain constant. In addition, genotype frequencies can be expressed in terms of allele frequencies, as the Table below shows. For a further explanation of this theorem, see Solving Hardy Weinberg Problems at http://www.youtube.com/watch?v=xPkOAnK20kw.

Hardy and Weinberg used mathematics to describe an equilibrium population (p = frequency of A, q = frequency of a, so p + q = 1): p 2 + 2pq + q 2 = 1. Using the genotype frequencies shown in Table below, if p = 0.4, what is the frequency of the AA genotype?

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