Name: ______

AP Biology

Population Genetics and Evolution Lab

Introduction

In 1908, G.H.Hardy and W. Weinberg independently suggested a scheme whereby evolution could be viewed as changes in frequency of alleles in a population of organisms. In this scheme, if A and a are alleles for a particular gene locus and each diploid individual has two such loci, then p can be designated as the frequency of the Aallele and q as the frequency of the a allele.

For example, in a population of 100 individuals (each with two loci) in which 40% of the alleles are A, p would be 0.40. The rest of the alleles (60%) would be would be a and q would be equal to 0.60. Therefore,

p + q = 1.0

These are referred to as allele frequencies. The frequency of the possible diploid combinations of these alleles (AA, Aa, aa) is expressed as:

p2 +2pq +q2 = 1.0

Hardy and Weinberg also argued that if five conditions are met, the population's alleles and genotype frequencies will remain constant from generation to generation. These conditions are as follows:

  1. No mutations must occur so that new alleles do not enter the population.
  2. No gene flow can occur (i.e. no migration of individuals into, or out of, the population).
  3. Random mating must occur (i.e. individuals must pair by chance)
  4. The population must be large so that no genetic drift (random chance) can cause the allele frequencies to change.
  5. No selection can occur so that certain alleles are not selected for, or against.

It is important to remember that the Hardy-Weinberg Theorem describes a hypothetical, non-evolving population that is in equilibrium. Most populations will not meet these five conditions.If a population's allelic frequencies are changing, then the population is undergoing evolution.

So of what value is such anequation? It provides a yardstick by which changes in allelic frequencies can be measured. The simulations in this lab are intended to demonstrate how the equation is used.

Four Hardy-Weinberg Case Studies

Case 1 (Test of an Ideal Hardy-Weinberg Community)

The entire class will represent a breeding population, so find a large open space for its simulation. In order to ensure random mating, choose another student at random. In this simulation, we will assume that gender and genotype are irrelevant to mate selection.

The class will simulate a population of randomly mating individuals with an initial frequency of 0.5 for the dominant allele (A) and 0.5 for the recessive allele (a)as well as genotype frequencies of 0.25AA, 0.50Aa, and 0.25aa. Each member of the class will receive four cards. Two cards will have the A allele and two cards will have the a allele. The four cards will represent the four possible haploid gametes produced throughthe process of meiosis. Each "parent" will be contributing a haploid set of chromosomes to the next generation.

Procedure

  1. Turn the four cards over so the letters are not showing, shuffle them, and take the card on top to contribute to the production of the first offspring. Your partner should do the same. Put the cards together. The two cards represent the alleles of the first offspring. One of you should record the genotype of this offspring in the Case 1 section on the next page. Each student pair must produce two offspring, so all four cards must be reshuffled and the process repeated to produce a second offspring.
  2. The other partner should then record the genotype of the second offspring in the Case 1 section on the next page. Using the genotypes produced from the matings, you and your partner will mate again using the genotypes of the two offspring. That is, student 1 assumes the genotype of the first offspring, and student 2 assumes the genotype of the second offspring.
  3. Each student should obtain, if necessary, new cards representing their alleles in his or her respective gametes after the process of meiosis. For example, student 1 becomes the genotype Aa and obtains cards A,A,a,a; student 2 becomes aa and obtains cards a,a,a,a. Each participant should randomly seek out another person with whom to mate in order to produce offspring of the next generation. You should follow the same mating procedure as for the first generation, being sure you record your new genotype after each generation in the Case 1 section.
  4. Class data will be collected after each generation for five generations. Your teacher will collect class data after each generation by asking you to raise your hand to report your genotype.

Case 1 Individual Data / Initial / F1 / F2 / F3 / F4 / F5
My Genotype / aa / aa / Aa / Aa / Aa
Case 1 Class Data / Surviving Genotypes / Surviving Alleles
Generation # / AA / Aa / aa / Total Individuals / A / a / Total Alleles
Parental / 9 / 10 / 7 / 26 / 28 / 24 / 52
F1 / 7 / 10 / 9 / 26 / 24 / 28 / 52
F2 / 5 / 15 / 6 / 26 / 25 / 27 / 52
F3 / 6 / 15 / 5 / 26 / 27 / 25 / 52
F4 / 7 / 14 / 5 / 26 / 28 / 24 / 52
F5 / 8 / 13 / 5 / 26 / 29 / 23 / 52

Complete the table below. For the population (class data), what are the theoretical allele andgenotype frequencies in the initial parental generation? Based on the Hardy-Weinberg theorem, what would the theoretical allele and genotype frequencies be for the 5thgeneration? What are the actual allele and genotype frequencies at the end of the 5thgeneration?

Case 1
Data Analysis / Frequency of Genotypes / Frequency of Alleles
Generation# / p2
(AA) / 2pq
(Aa) / q2
(aa) / p
(A) / q
(a)
Parental (H-W Theoretical) / 0.35 / 0.39 / 0.26 / 0.50 / 0.50
F5 (H-W Theoretical) / 0.25 / 0.50 / 0.25 / 0.50 / 0.50
F5 (Actual)

Case 1 Analysis Questions

  1. What does the Hardy-Weinberg equation predict for the new p and q?Do the results obtained in the simulation agree with the expected results based on the Hardy-Weinberg Theorem?
  1. Perform a Chi-Square analysis of the Case 1 F5 data (H-W theoretical vs. actual) to determine if the difference between the observed and expected values occurred simply by chance. In order to do this, you will need to convert the F5 H-W theoretical frequencies to numbers of individuals and numbers of alleles by multiplying the frequency by the total number of individuals/alleles and rounding to the nearest whole number.

F5 Genotypes
(# of Individuals) / F5 Alleles
(# of alleles)
p2
(AA) / 2pq
(Aa) / q2
(aa) / Total / A / a / Total
Observed -(Actual)
Expected -(H-W Theoretical)
Difference - (O-E)
Difference Squared - (O-E)2
(O-E)2/E
2 = Σ (O-E2)/E
Degrees of Freedom

Accept the null hypothesis Reject the null hypothesis

Degrees of Freedom / Probability
0.90 / 0.50 / 0.25 / 0.10 / 0.05 / 0.01
1 / 0.016 / 0.46 / 1.32 / 2.71 / 3.84 / 6.64
2 / 0.21 / 1.39 / 2.77 / 4.61 / 5.99 / 9.21
3 / 0.58 / 2.37 / 4.11 / 6.25 / 7.82 / 11.35
4 / 1.06 / 3.36 / 5.39 / 7.78 / 9.49 / 13.28
5 / 1.61 / 4.35 / 6.63 / 9.24 / 11.07 / 15.09
  1. Based on the above analysis, should you accept or reject the null hypothesis (i.e., the difference between the observed and expected values occurred simply by chance)? Explain.
  1. What major assumption(s) for a population in Hardy-Weinberg equilibrium were not strictly followed in this simulation?

Case 2 (Selection)

In this case you will modify the simulation to make it more realistic. In the natural environment, not all genotypes have the same rate of survival; that is, the environment might favor some genotypes while selecting against others. An example is the human condition sickle-celled anemia. It is a condition caused by a mutation on one allele, in which a homozygous recessive does not survive to reproduce.

For this simulation you will assume that the homozygous recessive individuals never survive. Heterozygous and homozygous dominant individuals always survive.

Procedure

  1. The procedure is similar to that for Case 1. Start again with your initial genotype, and produce your "offspring" as in Case 1. This time, however, there is one important difference. Every time your offspring is homozygous recessive (aa) it does not reproduce. Since we want to maintain a constant population size, the same two parents must try again until they produce two surviving offspring. You may need to get new allele cards from the pool.
  2. Proceed through five generations, selecting against the homozygous offspring 100% of the time.
  3. Add up the genotype frequencies that exist in the population and calculate the new p and q frequencies in the same way as you did in Case 1.

Case 2 Individual Data / Initial / F1 / F2 / F3 / F4 / F5
My Genotype / Aa / AA / AA / AA / Aa / AA
Case 2 Class Data / Surviving Genotypes / Surviving Alleles
Generation # / AA / Aa / aa / Total Individuals / A / a / Total Alleles
Parental / 0 / 26 / 0 / 26 / 26 / 26 / 52
F1 / 11 / 15 / 0 / 26 / 37 / 15 / 52
F2 / 10 / 16 / 0 / 26 / 36 / 16 / 52
F3 / 9 / 17 / 0 / 26 / 35 / 17 / 52
F4 / 14 / 12 / 0 / 26 / 40 / 12 / 52
F5 / 14 / 12 / 0 / 26 / 40 / 12 / 52
Case 2
Data Analysis / Frequency of Genotypes / Frequency of Alleles
Generation # / p2
(AA) / 2pq
(Aa) / q2
(aa) / p
(A) / q
(a)
Parental (H-W Theoretical) / 0.00 / 1.00 / 0.00 / 0.50 / 0.50
F5 (H-W Theoretical) / 0.25 / 0.50 / 0.25 / 0.50 / 0.50
F5 (Actual) / 0.00

Case 2 Analysis Questions

  1. How do the new fifth generation frequencies of p and q compare to the fifth generation frequencies in Case 1? What accounts for this difference?
  1. How has the allelic frequency of the population changed?
  1. What major assumption(s) were not strictly followed in this simulation for a population in Hardy-Weinberg equilibrium?
  1. Predict what would happen to the frequencies of p and q if you simulated another 5 generations.
  1. Since homozygous recessives are strongly selected against, would you expect therecessive (a) allele to be completely removed from the population? In other words, in a large population, would it be possible to completely eliminate a deleterious recessive allele? Explain.
  1. Explain how lethal alleles frequently are able to remain in a population, even though the homozygous recessive individuals die before reproductive age.Give an example of this situation in real populations.
  1. The allele causing Huntington Disease has no apparent redeeming features, yet it is remaining constant in the population.Using at least two complete sentences, explain why.

Case 3 (Heterozygote Advantage)

From Case 2, it is easy to see what happens to the lethal recessive allele in a population. However, data from human populations sometimes show an unexpected high frequency of a deleterious allele in some populations. Sometimes there is a slight advantage to being heterozygous for a trait rather than homozygous dominant. So the situation is now more complicated: homozygous recessives are still strongly selected against and do not survive to reproduce, but now, in addition, homozygous dominants have a lower reproductive rate than heterozygotes. We will incorporate this fact into our simulation.

Procedure

  1. Keep everything the same as in Case 2 (all homozygous recessives do not survive to reproduce), but now if your offspring is AA, you must flip a coin. If the coin lands heads up, the offspring does not survive; if the coin lands tails up, the offspring does survive. As in Case 1 & 2, after successfully reproducing, you become your surviving offspring and mate at random with another individual in the population.
  2. This case will be run for 10 generations. Class data will also be collected for analysis.

Case 3 Individual Data / Initial / F1 / F2 / F3 / F4 / F5 / F6 / F7 / F8 / F9 / F10
My Genotype / Aa / Aa / Aa / Aa / Aa / AA / AA / AA / Aa / Aa / AA
Case 3 Class Data / Surviving Genotypes / Surviving Alleles
Generation # / AA / Aa / Aa / Total Individuals / A / a / Total Alleles
Parental / 0 / 26 / 0 / 26 / 26 / 26 / 52
F1 / 3 / 23 / 0 / 26 / 29 / 23 / 52
F2 / 6 / 20 / 0 / 26 / 32 / 20 / 52
F3 / 8 / 18 / 0 / 26 / 34 / 18 / 52
F4 / 9 / 17 / 0 / 26 / 35 / 17 / 52
F5 / 12 / 14 / 0 / 26 / 38 / 14 / 52
F6 / 11 / 15 / 0 / 26 / 37 / 15 / 52
F7 / 12 / 14 / 0 / 26 / 38 / 14 / 52
F8 / 7 / 19 / 0 / 26 / 33 / 19 / 52
F9 / 9 / 17 / 0 / 26 / 35 / 17 / 52
F10 / 14 / 12 / 0 / 26 / 40 / 12 / 52
Case 3
Data Analysis / Frequency of Genotypes / Frequency of Alleles
Generation # / p2
(AA) / 2pq
(Aa) / q2
(aa) / p
(A) / q
(a)
Parental (H-W Theoretical) / 0.00 / 1.00 / 0.00 / 0.50 / 0.50
F10 (H-W Theoretical) / 0.25 / 0.50 / 0.25 / 0.50 / 0.50
F10 (Actual) / 0.00

Case 3 Analysis Questions

  1. Explain how the changes in p and q frequencies in Case 3 (Heterozygote Advantage) compare with the frequencies in Case 1(H-W Equilibrium) andCase 2 (Selection)?
  1. Do you think the recessive allele would ever be eliminated in Case 3? Explain.
  1. What is the impact of heterozygote advantage on genetic variation in a population? Explain.
  1. Describe a real-life example of heterozygote advantage.

Case 4 (Genetic Drift)

Remember that even though natural selection is creating adaptive change, it is not the only force molding a population. Equally important are the forces of random chance that can cause changes over time in a population even though they are not adaptive. We will simulate this by creating smaller population the classroom.

Procedure

  1. The class will be divided into several smaller sub-populations. These populations will remainisolated from each other and cannot interbreed.
  1. Keep the mating process the same as in Case 1 (all individuals survive and reproduce). Asbefore, after successfully reproducing, you become your surviving offspring and mate atrandom with another individual, but only in your sub-population. Record the genotype ofyour offspring in the Data Table below.

Case 4 Individual Data / Initial / F1 / F2 / F3 / F4 / F5
My Genotype / Aa / AA / Aa / aa / Aa / Aa
  1. On the next page, record class data — and the data from other sub-populations — for data analysis.

Case 4 Class Data / Surviving Genotypes / Surviving Alleles
Generation # / AA / Aa / aa / Total Individuals / A / a / Total Alleles
Parental / 0 / 26 / 0 / 26 / 26 / 26 / 52
F1 / 5 / 14 / 7 / 26 / 24 / 28 / 52
F2 / 7 / 15 / 4 / 26 / 29 / 23 / 52
F3 / 8 / 14 / 4 / 26 / 30 / 22 / 52
F4 / 7 / 13 / 6 / 26 / 27 / 25 / 52
F5 / 3 / 16 / 7 / 26 / 22 / 30 / 52
Group / 1 / 2 / 3 / 1 / 2 / 3 / 1 / 2 / 3 / 1 / 2 / 3 / 1 / 2 / 3 / 1 / 2 / 3 / 1 / 2 / 3
F5 / 0 / 0 / 3 / 7 / 4 / 5 / 3 / 4 / 0 / 10 / 8 / 8 / 7 / 4 / 11 / 13 / 12 / 5 / 20 / 16 / 16
Case 4
Data Analysis / Frequency of Genotypes / Frequency of Alleles
Generation # / p2
(AA) / 2pq
(Aa) / q2
(aa) / p
(A) / q
(a)
Parental (H-W Theoretical) / 0.00 / 1.00 / 0.00 / 0.50 / 0.50
F5 (H-W Theoretical) / 0.25 / 0.50 / 0.25 / 0.50 / 0.50
F5 (Actual)
Group 1 F5 (Actual)
Group 2 F5 (Actual)
Group 3 F5 (Actual)

Case 4 Analysis Questions

  1. Compare the data in the different sub-populations with the F5 data for the entire class.How do they differ from the expected results based on the Hardy-Weinberg Theorem?
  1. What do these results indicate about the importance of population size as an evolutionary force? How is thisissue significant in conservation biology and endangered species conservation?
  1. Briefly describe a real-life example of genetic drift.