1. Basic Genetics of Platypapyrus foursuitii
Platypapyrus foursuitii is a diploid organism. One feature that makes the species particularly amenable for genetic studies is that their chromosomal material takes the form of playing cards and can easily be handled like cards. Each card represents alleles in the gene pool, and two cards together represent the genotype of an individual. A person can hold any number of different individual genotypes, depending on the sample size you want. For a class of 25, you can have a population size of 50 by giving each student four cards. Sample sizes much less than 50 can result in significant fluctuations due to sampling error. For 50 individuals, you will need 100 cards or two decks.
2. A one-locus, two-allele model (50 individual organisms in the population)
These instructions assume 50 individuals in the population, each individual containing two cards. If you have 25 students, you can give each one two pairs to work with. If you have some other number, distribute accordingly. For example, for a class or 20, give all 20 students four cards and then give 10 of the students two more cards. If you have large numbers, you can expand the number of cards accordingly.
For a one locus, two allele model, use red and black colors as alleles of the coat color genes. Ignore all the suits - if it's red (heart or diamond) it's allele R, and if it's black (spades or clubs) it's allele B. There are three possible genotypes to expect in such a case: RR, RB and BB. If you want, you can make up phenotypes to go with these genotypes, but the important point is to watch what the genes do, so phenotypes might cloud the issue.
Sort the cards from three decks into two piles - the black cards and the red cards. Eliminate the Jokers. There are 52 cards per deck, so there will be a total of 78 red cards and 78 black ones. Create 20 individuals homozygous for the B allele by making 20 pairs of black cards. Create 30 individuals homozygous for the R allele by making 30 pairs of red cards.
The individual cards make up the "gene pool" of Platypapyrus foursuitii. This gene pool contains 40 black cards and 60 red cards, so the frequency of the black allele is 40/100 = 0.4, and the frequency of the red allele is 60/100 = 0.6. Because of the way you set the population up, all of the individuals are homozygous, either BB or RR. The frequencies of diploid genotypes are 0.6 for RR (20/50), 0.4 for BB (30/50) and 0 for RB. You may want to calculate these values in class, but it is a good idea to have thought through these and all subsequent calculations before class.
4. The class exercise
Explain the biology and genetics of Platypapyrus foursuitii to the students and give out the pairs of cards that represent each individual. Tell them to keep the pairs together just as they are now until they receive further directions. Point out that all the organisms are homozygous, and have the class calculate the frequency of each gene and the frequency of each diploid genotype.
Now, the population is ready to undergo a round of random mating. Here's how to do that. "Random mating" means that mating is without regard to the genotype of the individuals. So in this case, it is important that individuals not know about alleles of potential mates. Mating in this species is reciprocal - each time two organisms mate they trade one card. The students get up and mill around the room, carrying their cards with them and keeping the cards in pairs. Carry the cards so that the color cannot be seen by other students. To assure adequate mixing of the population, have the students say "hello" when they encounter another student. The fourth time they say "hello" to someone they can trade cards. That means giving one card of a pair to the other person and receiving a card from the other person in return. When a card of a pair has been replaced by mating, that pair should not be mated again in this round of mating. Students continue to mill around until all of their individuals have been mated.
Everyone returns to their seats and the class calculates the frequency of genes and genotypes once again. Since no cards have either entered or left the gene pool, the frequency of R will still be 0.6, and the frequency of B will be 0.4. But the frequency of genotypes will have changed. Determine the number of RR, RB and BB. Since there is a total of 50 individuals, the frequency of each type is fairly easily determined by counting the number of that type and dividing by 50. The exact numbers will vary somewhat each time you do it but will hover around RR=0.36; RB= 0.48; BB=0.16. Discuss why the numbers have changed from the initial figures.
The level of this discussion will depend on the class. Here are possible topics in ascending order of mathematical sophistication:
There were no RB in that initial generation, but with random mating some of the RR will trade with the BB to form heterozygotes. That increase in RB will have to be at the expense of RR and BB since those are the only possibilities.
The chance of two gametes getting together during random mating depends on the frequency of each gene in the population. Since there are more RR than BB to start with, there will be more chances of forming RR than of forming BB, so there are fewer BB than RR after random mating.
The chance of two independent events happening at the same time is the product of their individual probabilities. The chance of having a B is 0.4, so the chance of putting two B alleles together during random mating is 0.4 X 0.4 or 0.16. The chance of having an R is 0.6, so the chance of putting 2 R alleles together is 0.6 X 0.6 or 0.36. All that remains are RB, so that's 1-(0.16+0.36) = 1- 0.52 = 0.48.
A slightly more sophisticated way to understand the frequency of RB is to consider that one way of making RB is to have an R and receive a B. That will happen with a frequency of 0.6 X 0.4 = 0.24. On the other hand, you could also have a B and receive an R. That will happen with a frequency of 0.4 X 0.6 = 0.24. Thus the total frequency for RB should be 0.48.
Why is the frequency actually obtained not exactly 0.36, 0.48 and 0.16? This allows a discussion of sampling error. If we used more individuals each generation, the numbers should come out close to the expected values more frequently.
Note that no algebraic consideration of this problem has occurred yet. That can come later after the students have a feel for the issues if you think it should come at all.
Now go through a second round of random mating. But before doing so, ask the students to predict the frequencies of genes and genotypes they expect to see in the next generation. (Again, since no addition or subtraction of cards occurs, there should be 0.6 R and 0.4 B as before. And because cards are being paired at random, the genotypes should come out close to 0.36 RR, 0.48 RB and 0.16 BB.) After random mating, calculate the frequencies again.
The Hardy-Weinberg Equilibrium is a description of a gene pool on which no forces are acting. If no forces come into play, there should be no change in gene frequencies, and the genotype frequencies will also be the same every generation since they are determined by the frequencies of the genes. This point has been demonstrated without recourse to p's and q's and the usual Hardy-Weinberg terminology. That can now be introduced to students who will have an idea where it is all going. At this point, if appropriate, you can introduce Hardy-Weinberg language.
4. The simulation in Hardy-Weinberg terms
Assume a sexually reproducing population that is very large and mating at random and in which there is no migration (no cards enter or leave the original collection), no mutation (no cards change color during the experiment), and no selection (there is no disadvantage to being either red or black. The chance of being included in the next generation is simply a function of your frequency.).
Let the frequency of one allele (let's say red in our case) be p and the frequency of the other allele (black) be q. Since there are only two different alleles, p+q=1.
After random mating, the frequency of RR will be p2; of RB will be 2pq; and of BB will be q2. Since these are the only possible genotypes, p2 + 2pq + q2 = 1. This will be true generation after generation as long as the assumptions hold. If the frequencies do not approximate those expected by the Hardy-Weinberg Equilibrium, then one or more of the assumptions must be incorrect. This is of great value to population geneticists since it tells them that there is some interesting research to do finding out what assumptions are incorrect.
The effect of sample size
One important assumption of the Hardy-Weinberg Equilibrium is that the population is very large, so large that there is no sampling error. This is a condition of the equilibrium that is frequently not met, and in small populations there may be considerable deviations from expected values. In a population of this exercise, the size is already fairly small, and you will sometimes get some fairly noticeable changes in genotype frequencies from one generation to the next. But because this model is set up with an invariant frequency of haploid gametes, that can always correct itself in the next generation.
But try this: Go through a round of random mating followed by calculation of gene and genotype frequencies. Then split the group into two smaller populations. They don't have to be populations of equal size. In fact, the effect will often be more noticeable if you split into one small population (8 or 10 individuals) and one with all the rest. Now calculate the gene frequencies for each new population. Are they the same? Often they will not be. It is even possible, although not very frequent, that the smaller population will be made up of only one color.
This kind of change in gene frequency due only to sampling error in small populations is called "genetic drift," and the effect of splitting off a small population with significantly different gene frequencies is called the "founder effect." Once a new founder population is formed, if it grows again to a larger population less subject to sampling error, the new population can again maintain the new gene frequencies because of Hardy-Weinberg Equilibrium.
Selection and mutation
As this simulation is set up, it is difficult to simulate selection without being misleading. That is because the energy for natural selection comes from the production of excess offspring. There are no excess offspring in this model - each generation produces exactly enough offspring to replace the parents. If you select against one of the genes by preventing it from reproducing, the population will decrease in size.