Grebe courtship: http://www.youtube.com/watch?v=jgl7OlSY2ZM&feature=related
Bird callers: http://video.aol.com/video-detail/piedmont-hs-bird-callers-appear-on-letterman/1266515351
AP Biology Lab 10
Population Genetics and Evolution
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 A allele 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 would be (60%) would be a and q would be equal to 0.60. p + q = 1 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 5 conditions are met, the population's alleles and genotype frequencies will
remain constant from generation to generation. These conditions are as follows:
 The breeding population is large. (Reduces the problem of genetic drift.)
 Mating is random. (Individuals show no preference for a particular mating type.)
 There is no mutation of the alleles.
 No differential migration occurs. (No immigration or emigration.)
 There is no selection. (All genotypes have an equal chance of surviving and reproducing.)
The Hardy-Weinberg equation describes an existing situation. Of what value is such a rule? It provides a way by which
changes in allelic frequencies can be measured. If a population's allelic frequencies change, it is undergoing evolution.
Part I: Estimating Allele Frequencies for a Specific Trait
Using the class as a sample population, the allele frequency of a gene controlling the ability to roll the tongue can be
estimated. Rolling the tongue is evidence of the presence of a dominant allele in either a homozygous (AA) or heterozygous
(Aa) condition. The inability to roll the tongue is dependent on the presence of the two recessive alleles (aa). To estimate the
frequency of the tongue-rolling allele in the population, one must find p. To find p, one must first determine q (the frequency of
the non rolling allele).
Procedure
1. Determine whether you are a roller or a non roller.
2. A decimal number representing the frequency of rollers (p2+2pq) should be calculated by dividing the number of rollers in
the class by the total number of students in the class. A decimal number representing the frequency of the non rollers (q2) can
be obtained by dividing the number of non rollers by the total number of students. You should then record these numbers in
Table 1.
3. Use the Hardy-Weinberg equation to determine the frequencies (p and q) of the two alleles. The frequency q can be
calculated by taking the square root of q 2. Once q has been determined, p can be determined because 1-q=p. Record these
values in Table 1 for the class and also calculate and record values of p and q for the North American population.
Table 1 Phenotypic Proportions of Rollers and Non Rollers and Frequencies of the Determining Alleles
Allele Frequency
Based on the H-W
Equation
Phenotypes
Rollers (p2+2pq)
Class
Population
North American
Population
(#)
(%)
0.55
Non Rollers(q2)
(#)
p
q
(%)
0.45
Data
1. Fill in the missing data in the table above based on class information (i.e., number of rollers and non-rollers in
the class, percentage of rollers and non-rollers in the class, and allele frequency in the class and in the North
American Population).
2. Calculate the percentage of heterozygous rollers (2pq) in your class. What is it?
3. Calculate the percentage of the North American population that is heterozygous for the roller allele. What is it?
Part II: Case Studies (Fondly Referred to as “Grebe, Grebe”)
You as a class will become a population of randomly mating grebes. Each of you will receive your initial grebe
genotype cards (AA, Aa or aa) at the beginning of the simulation. Grebes are aquatic birds similar to ducks. You
will play the role of a grebe for the remainder of today’s activity, so we will begin with the information you need for
successful “grebing”. In order to insure random mating, you must be completely promiscuous. Choose any of the
students in the class and confidently approach them: they will not refuse. You will indicate your intentions using
the classic line, “Grebe, Grebe”, to which your future (albeit brief) mate will reply, “Grebe, Grebe”. You will then
grebe by placing your two allele cards behind your back and randomly shuffling them. Each parent contributes a
haploid set of chromosomes to the next generation. Each couple must have two offspring. The first two cards
displayed will become the genotype of one student. Repeat by shuffling the cards behind the back again and
display two more cards. These will represent the genotype of the second student.
To maintain a constant population size, after each grebing session, the parent genotype dies, and you assume
the genotype of one of the offspring, and your partner assumes the other offspring’s genotype. To do this, you
may need to visit the central allele supply and pick up a new card. After each grebing session, be sure to record
your new genotype in the appropriate slot on the data table. Thank your partner and proceed to a new individual.
Follow the exact same procedures, being sure to record your new genotype after each generation. After five
generations, the class total for AA, Aa and aa will be recorded.
Case I: Hardy-Weinberg Equilibrium
The starting class population will have genotype frequencies of 0.25AA, 0.5Aa and 0.25aa (in other words, 25% of
the class will receive two “A” cards, 50% of the class will receive one “A” and one “a” card, and 25% will receive
two “a” cards). Record your initial genotype on the data page.
Number of A alleles present at the fifth generation
Number of offspring with genotype AA = ___
Number of A alleles = ___
Number of offspring with genotype Aa = ___
Number of A alleles = ___
Total A alleles = ___
p = frequency of A
p = Total number of A alleles/Total number of alleles in the population
p = ___
Number of a alleles present at the fifth generation
Number of offspring with genotype aa = ___
Number of a alleles = ___
Number of offspring with genotype Aa = ___
Number of a alleles = ___
Total a alleles = ___
q = frequency of a
q = Total number of a alleles/Total number of alleles in the population
q = ___
Questions
1. What does the Hardy-Weinberg equation predict for the new p and q you found at the end of five generations?
2. Do the results you obtained in this simulation agree? If not, why not?
3. What major assumption(s) 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.
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 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.
Proceed through five generations, selecting against the homozygous offspring 100% of the time. Then add up the
genotype frequencies that exist in the population and calculate the new p and q frequencies in the same way as it
was done in Case 1.
Number of A alleles present at the fifth generation
Number of offspring with genotype AA = ___
Number of A alleles = ___
Number of offspring with genotype Aa = ___
Number of A alleles = ___
Total A alleles = ___
p = frequency of A
p = Total number of A alleles/Total number of alleles in the population
p = ___
Number of a alleles present at the fifth generation
Number of offspring with genotype aa = ___
Number of a alleles = ___
Number of offspring with genotype Aa = ___
Number of a alleles = ___
Total a alleles = ___
q = frequency of a
q = Total number of a alleles/Total number of alleles in the population
q = ___
Questions
4. How do the new frequencies of p and q compare to the initial frequencies? How has the allelic frequency of the
population changed?
5. Predict what would happen to the frequencies of p and q if you simulated another 5 generations. Explain your
prediction.
6. In a large population, would it be possible to completely eliminate a deleterious recessive allele? Explain.
Case III: Heterozygote Advantage
Data from many human populations show an unexpectedly high rate of people that are heterozygous for the
sickle cell trait. These people are protected from malaria to a certain extent due to the presence of some
abnormal red blood cells. We will incorporate this fact into our simulation. In this round, keep everything the same
as in Case II where aa dies all the time. However, in this case, if your offspring is AA, flip a coin. Heads the
offspring lives, tails the offspring dies of malaria and the same parents must mate again until they have a viable
offspring (Aa). After five generations, total the genotypes for the class and record AA, Aa and aa. Now, keep your
most recent genotype and go through five more generations, recording the results. Finally, again start with the
most recent genotype for another five generations. Record all class totals for AA, Aa and aa.
Number of A alleles present at the fifth generation
Number of offspring with genotype AA = ___
Number of A alleles = ___
Number of offspring with genotype Aa = ___
Number of A alleles = ___
Total A alleles = ___
p = frequency of A
p = Total number of A alleles/Total number of alleles in the population
p = ___
Number of a alleles present at the fifth generation
Number of offspring with genotype aa = ___
Number of a alleles = ___
Number of offspring with genotype Aa = ___
Number of a alleles = ___
Total a alleles = ___
q = frequency of a
q = Total number of a alleles/Total number of alleles in the population
q = ___
Questions
7. How do the new frequencies of p and q compare to the initial frequencies in Case 1?
8. How has the allelic frequency of the population changed?
9. What were some general trends you saw throughout this activity?
Data
Case 1 (Hardy-Weinberg Equilibrium)
Case 2 (Selection)
Initial Class Frequencies:
Initial Class Genotypes:
AA ________ Aa________ aa_________
AA ________ Aa________ aa_________
My initial genotype: _______________
My initial genotype: _______________
F1 Genotype ______
F1 Genotype ______
F2 Genotype ______
F2 Genotype ______
F3 Genotype ______
F3 Genotype ______
F4 Genotype ______
F4 Genotype ______
F5 Genotype ______
F5 Genotype ______
Final Class Genotypes:
Final Class Genotypes:
AA ________ Aa________ aa_________
AA ________ Aa________ aa_________
Final Class Frequencies
Final Class Frequencies
p _________ q __________
p _________ q ________
Case 3 (Heterozygote Advantage)
Initial Class Genotypes:
AA ________ Aa________ aa_________
My initial genotype: _______________
Genotypes by generation:
F1 Genotype ______
F6 _______
F11_______
F2 Genotype ______
F7 _______
F12 _______
F3 Genotype ______
F8 _______
F13 _______
F4 Genotype ______
F9 _______
F14 _______
F5 Genotype ______
F10 _______
F15 _______
Final Class Genotypes (after 5, 10 and 15 generations):
AA ________ Aa________ aa_________
Final Class Frequencies (after 5, 10 and 15 generations) =
p _________ q _________