10-16-06, Lab 8 - Population Genetics and Evolution - Hardy-Weinberg Lab Report

Vocabulary: Hardy, Weinberg, Hardy-Weinberg law of genetic equilibrium, loci, DNA, histone, chromosome, allele, allele frequency, homozygous, homozygous recessive, genotype, gene pool, allelic frequencies, microevolution, natural selection,

Write a pre-lab to address the purpose of the lab, how it relates to what we are talking about. We finished the unit on evolution. This is about evolution, relate it to our unit on evolution. Alleles are variations in a gene pool. Changes in gene frequency are identifiable by Hardy-Weinberg. There are dominant and recessive alleles and it's not always predictable. The gene that we are working with, in part 1, the gene that we are going to use is connective or detached ear lobes. If there is a bend at all, then it is detached. Connective are recessive. Attached are dominant. Bryan's earlobe data:

Connective - 10

Detached - 7

Total population: 17

Figure out the totals for this room. These are phenotypes. Phenotypes are the physical or outward expression of an allele variation.

Class connective: 5

Class detached: 7

Total population of class room: 12

Connective: 15

Detached: 14

Total: 29

100% of the population is 29 and we are trying to find the frequencies of alleles . p^2 represents the homozygous dominant alleles. Q^2 represents the homozygous recessive alleles. In any population, when we deal with phenotypes, phenotypes, there is only one allele variation that we know for sure is homozygous and that is q^2. q^2 represents ressessive and recessive and we see that with connected earlobes. We know that people with connective earlobes have two recessive genes. Q^2 is equal to what n our population?

Q^2 = 15 people, you do not square 15. You have 15 people. What is 15 the percentage of? How do we figure out what percentage of the population that q^2 is equal to? Divide 15 by 29. It's .52. Take the square root of .52 because that's our percentage and we are trying to get `q'. So, q is equal to .72. How do we figure out p from that? 1-q is equal to p and is .28.

Q^2 = 15 people

Q^2 = .52

Q = square root of .52

According to Hardy-Weinberg, 72% of the population are recessive alleles (from our population data). You can see that our numbers do not necessarily reflect that. Our numbers are 15 and 14, even though we have 72% recessive alleles. What's q^2?

Now that we know p and q we can plug it into our formula, p^2 + 2pq + q^2 = 1. So we have .08 + 2(.28)(.72) + .52 = 1

2pq = .40 … we know that p^2 and q^2 and 2pq have to equal 100%. If we know q^2 and p^2 and add those two together, subtract from 1, and we get our .40 number there.

P^2 is homozygous dominant, q^2 is homozygous recessive, and 2pq is heterozygous dominant. They are equivalences—and we know that the three must equal 100% when added up.

Connective earlobes are recessive. The attached earlobes are 14 and it's mixed. We are figuring percentages. The first thing that we have to do is we have to figure what percentage of the total population is represented by the recessive population. All of those 15 folks are q^2. All of the members that are recessive are called q^2 because they have two recessive alleles—q^2 = 15/29.

14 = p^2 + 2pq and we'd have to know what those are.

Hardy-Weinberg will be on the AP exam. First, find Q. Q is the only thing that you can figure out from most given population information. To find Q, get the square root of the percentage which is in this case .52 (as the percentage - 52% of the total population).

P^2 is the homozygous, and 2pq is heterozygous (one big, one little). The number of q^2 represents the percent of the population with the two alleles in the population. Take the square root of q^2 and that's .72. We know that all populations have a `total number of members” - we know that p+q = 1. So plug in .72+p = 1 and then we have p = 1 - .72 and then we can find out that p = .28

P^2 = .08

8% of those people observed are homozygous dominant. 40% of those people are heterozygous dominant, and 52% of those people are recessive (homozygous recessive, that is).

Hardy-Weinberg's formula there allows us to predict the distribution of the alleles on larger scales. Our sample was small and random - but a larger sample size would be useful.

The population is said to be evolving when the percentages change. That's what this is all about.

There are two alleles in your population - a dominant A and a recessive a. All of you are heterozygous. There are the same number of alleles for dominance as recessiveness. Each of you has 2 dominant and 2 recessive alleles.

P^2 + 2pq + q^2 = 1

What will it look like ? What are our beginning frequencies?

Frequencies are determined by the number of alleles in the population. Our beginning frequencies are going to be 25% with homozygous dominant, 50% of the population is going to be heterozygous dominant, and 25% of the population is going to be homozygous recessive. The reason for that is because all of the alleles are equal. There are not more dominant alleles than recessive alleles.

AA = .25

Aa = .50

Lowercase .aa = .25

Initial genotype is Aa

You need to go through five generations. Each time that you match up for a mating pair, make sure you know which offspring and there's 2 offspring produced for each partnership. Do not mate with the same person. We have to spread this population out. Hardy Weinberg said that we can only do this if it is a non-random mating. Get up and for a long time, have babies.

Day two (Tuesday). Let's get started. You have your initial genotype (Aa). Did a first round of breeding that gave you F₁ genotype. During meiosis, those two alleles (Aa) have to be separated. The Aa is the initial genotype of the parent. During meiosis, the DNA content is replicated. Replication produces four physical alleles. Those four cards that you have in your hand represent four alleles. These four alleles (Aa, Aa) have the opportunity to be scattered into four different haploid gametes (sex cells) and when they combine they produce a new diploid organism. Suppose that your parent generation is aa, and little a little a replicates, then it will only make two more little a's. What's the only allele that you are going to scatter? Little “a”. We are leaving that step out if you have a homozygous allele combination as resulting from one of your offspring. If your genotype is AA then you can give up a big A, and if it's aa then only one `a'. If it's big a, little a, then you have to shuffle all four cards because we don't know what the outcome is.

Read the procedure on page 92. Select one card from your genotype, partner does the same - then another is done.

This was the data from 10/16 and 10/17

Case 1

F₁ generation

AA (1), Aa (9), aa(2)

F₂ generation

AA (2) homozygous dominant, Aa (7) heterozygous, aa (3) homozygous recessive

F₃ generation

AA (2), aa (4), Aa (6)

F₄ generation

AA (4), Aa (8), aa (0)

F₅ generation

AA (4), Aa (6), aa (2)

My data was: Aa, AA, AA, AA, Aa, aa

Apply Hardy-Weinberg to each of those five generations. Determine if there is a trend developing. There is a very interesting trend between F₃ and F₄ because there was no homozygous recessive.

The next scenario (case 2), this talks about selection. The last one that we did was a perfect Hardy-Weinberg equilibrium (random, no selection, etc.). Case two says that the environment may favor some genotypes. Sickle-cell anemia is an example in human populations. West African genetics study on malaria. Those individuals with the sickle-cell gene (very painful) - most of those children die before they are ten years old. There are two selective forces - homozygous recessive (sickle cell is selecting against this). The heterozygous members are those that survive. The reproductive class in West Africa where malaria is prominent is heterozygous dominant for the sickle-cell trait.

Do five genotypes. The only thing that you need to remember is “don't breed with somebody in another generation”.

My data for case two. F₁ genotype - AA, F₂ genotype - Aa, F₃ genotype - Aa, F₄ genotype - Aa, F₅ genotype - AA.

Case 2

F₅ - homozygous dominant (AA) is (5). That means Aa is 7 (for heterozygous). That shows us a significant gene frequency change from the initial gene frequency. Why haven't we completely eliminated the recessive allele? It's like a carrier gene in that heterozygous trait. The human gene for color blindness is carried from generation to generation and suddenly show up in a male because there is no matching gene to go with it—it's carried on the female chromosome.

Case 3

There is a selection against homozygous dominant. It is not as lethal as the sickle-cell gene. Everytime you get homozygous dominant (AA), you are going to flip a coin. You flip a coin, and if it's heads, your AA lives, and if tails, your AA dies and you have to do it again, and you flip a coin each time you get AA - and if it dies then do it again. Stop after five generations. Spread out in the room. Get up and move around.

My data: Aa, Aa, Aa, Aa, AA (F₁-F₅)

F₅ - AA (4), Aa (8), aa - 0. We are doing five more generations. Whatever genotype that you were in the F₅ generation is what you start with. As a part of your lab report, you will figure out p and q and frequencies at a later time for this. We still need to do case four.

Let's do another five generations. The next set of generations: Aa, Aa, AA, Aa, AA.

F₁₀ - Aa (heterozygous) (7) and AA is (5).

In case four, we have isolated populations. Get in groups of four. Go through five generations just like we did in case one - start with Aa and the initial frequency is (.25, .5, .25) and nothing dies and you will do five generations of those. Share around. There are at least three different combinations. Do five generations and see the differences.

My data: Aa, Aa, Aa, aa, Aa

First group: 2 AA, 2 Aa, 0 aa,

second group: 2 AA, 1 Aa, 1 aa,

third group: 2 AA, 1 Aa, 1 aa

You will be comparing the F₅ of each.

An introduction to the lab report

The Hardy-Weinberg theorem involved the use of