Biology 102 Lab 6: Evolutionary Processes Microevolution

Objectives:

To observe a population in Hardy Weinberg equilibrium and demonstrate how selection pressures can change allele frequencies and lead to microevolution and evaluate how certain aspects in nature can influence evolution, for this lab show all your calculations. These calculations may include in the word document or if clearly and concisely labeled and clear and easy to read may be added as a scanned document (not photo)

Evolution is a population phenomenon. Although individuals in a population do not evolve, the population composition changes over time because of several evolutionary factors. The change in gene frequencies in a population resulting from mutations, selection, migrations, and genetic drift is known as microevolution. The mechanism of microevolution is well understood and are in fact used to increase the productivity of crops and livestock. Macroevolutions are the changes that are associated with the origin of a species is less understood.

Microevolution

Models of microevolution have been developed since 1858 when Charles Darwin published his version of the theory of evolution. Darwin’s work was monumental in that is established the role of natural selection in evolution. However, Darwin worked 50 years before Mendel’s idea about genetic mechanism became widely known. Darwin was unaware of the basic concepts you know now about genes, inheritance, DNA, and physiology. Researcher in the last 75 years have established the hereditary mechanism and population data that supports Darwin’s theory.

A population of organism can be considered as the expression of the gene pool, or the collection of genes in that population. In diploid organisms, the genes in the gene pool occur as pairs in each individual and the individual may be homozygous or heterozygous at a particular locus. Population genetics, the basis of microevolution, deals with the frequency of genes and genotypes in a population and attempts to quantify the influences of mutation, selection and other factors on genes and genotypic frequencies.

As example might help to clarify this point. Assume that a population of randomly mating diploid organism has two alleles that can occur at a particular locus (A and a) in a large population (over 100 individuals) we would expect to find three genotypes in the population (AA, Aa, aa) The frequency of these genotypes would depend on the frequencies of the alleles A and a in the gene pool. If the allele A occurs at a high frequency in the population, you would expect to find the genotypes AA and Aa at higher frequencies. The laws of probability can be used to determine the frequency of genes in a population or to predict genotypic frequencies if the gene frequencies are known. The multiplication law of probability is most often used. It states that the frequency or probability of two independent events occurring together is equal to the arithmetic product of their individual relative frequencies or probabilities. It is easy to apply this law once you have figured out how this language relates to biological populations.

If a large population is randomly mating with no selection or migration occurring, al gametes have an equal chance of combining with other gametes of the opposite sex. Since gametes are haploid, the relative frequency of certain gametes will be equal to the relative frequency of the alleles in the gene pool. Note that the relative frequency is a percentage or its decimal equivalent. Therefore, the probability of producing offspring with the genotype AA will be equal to the product of the relative frequencies of the A and A gametes. (each gamete is independent since it is produced by a different parent, and the genotype AA is the result of those gametes coming together at fertilization). The relative frequency of the genotype Aa will be equal to the relative frequency of A gametes in the population multiplied by the relative frequency of a gametes and the relative frequency of genotype aa will be equal to the frequency of a bearing gametes times the frequency of a bearing gametes. The difficulty is to determine the relative frequency of each kind of gamete.

Usually, relative gene frequencies are estimated by determining the frequency of each genotype in the population. If the phenotype corresponding to the aa genotype differs from the phenotype of AA and Aa as in the case when the a is recessive or blends, the relative frequency of aa can be determined simply by counting all the phenotypes and expressing their occurrence as a percentage of the total population. Once determining the square root of the aa genotype relative frequency

Relative frequency aa= (relative frequency a) x (relative frequency a) or (relative frequency a)2

Therefore, the relative frequency of a can also be expressed as the

Square root of the relative frequency of aa

Again, this equation only holds true when there is no selection or migration occurring.

If only two alleles are involved, as in our example, then obviously the relative frequencies of A and a must add up to 100%.

Relative frequency of A+ relative frequency of a= 1

Relative frequency of A= 1-the relative frequency of a

Thus, it is possible given the relative frequency of a, to find that of A by subtraction. From there, it is a simple process to calculate the relative frequencies of genotype AA and Aa even if they are not observable in the population. The standard notation used by most biologist for this model is as follows.

If

P=relative frequency of A

Q=relative frequency of a

And p+q=1

Then the relative frequencies of the genotypes can be expressed as

AA=pxp or p2

Aa=pxq or pq

aA=qxp or pq

aa=qxq or q2

Therefore, the frequency of genotypes in a population can be expressed in the binomial expansion equations.

P2+2pq+q2= 1 or (p+q)2=1 This is the Hardy Weinberg Equation

This equation accounts for all possible combinations of two factors (in this case 2 alleles and their frequencies)

Population Genetics Simulation

The Hardy Weinberg law tells us that the relative frequencies of genes do not change in large populations unless there is an influence such as selection, mutation, gene flow or differential mating. If quantitative estimates of the effects of these influences are available, they can be included as factors that change gene frequencies from generation to generation, allowing us to predict what the future holds for the population. A factor that must be considered, however is random deviation for expected results such fluctuations, known as genetic drift, will be explained later.

Population genetic experiments with higher diploid organism may take several weeks to several years to perform, even for organism with short life cycles such as fruit flies. This makes it difficult to illustrate evolutionary mechanism with live organism during a single lab period. It is however possible to simulate a population of organism using beans of different colours to represent different genotypes and than draw them two at a tie to simulate mating’s. Since the random drawing of beans gives you the genotypes in the mating, you can forecast the composition of the next generation using Mendelian genetics. Keep in mind that this is a simulation based on randomness, any genotype can mate with any other genotype, resulting in random combination of gametes and alleles. Furthermore, since beans are not added or removed from the total population and since they do not spontaneously change colours, the summation assumes that no selection, migration, or mutation is occurring. This fulfills the requirements of the Hardy Weinberg Law

To simulate the population in the laboratory a coffee can containing 2000 beans of different colours will be used. The can contains 49% red beans, 42% brown beans and 9% white. These percentages will correspond to the genotypes in a hypothetical population in which there are 2 alleles at a given locus.

Colour Relative frequency (%) Genotype

White 9 aa

Speckled 42 Aa

Red 49 AA

The frequencies of alleles in the population can be determined in one of two ways, by counting or by calculation. For the counting method, assuming that there are 2000 individuals in a population (you physically counted 2000 individuals) the relative frequency in the above table must be multiplied by 2000 to get the number of each genotype.

To calculate gene frequencies, the following accounting should be done. Since the AA genotype has two alleles and the heterozygote (Aa) has one than

(number of AA individuals) x2= 2000X0.49x2=1960

(number of Aa individuals) x1=2000x0.42x1=840

Therefore, the number of A alleles in a population gene pool =2800

How many a allele are in the population gene pool listed above?

The relative frequency of the A allele in the gene pool can also be expressed as

%A= amount of A/number of A +number of a X100

Using the above equation, calculate the relative frequency of the A allele in the sample population (show your work)

What is the equation you would use to calculate the relative frequency of the a allele in the gene pool?

Using that equation calculate the relative frequency of a allele in your sample population.

The second method for determining the frequencies of A and a for your population is by calculation.

Calculate the relative frequencies of alleles A and a for your population using the above example and the Hardy Weinberg Equation to see f they are the same as by the counting methods. Remember that p2is the relative frequency of AA. What is q2equal to?

Solve the following. Show your work.

P=relative frequency of A= (square root of p2)

Q=relative frequency of a= (square root of q2)

P= ____ in bean simulation

Q=_____ in bean simulation

Are the relative frequencies of A and a the same by the counting and calculation methods? Explain why or why not?

The table below, is a representation of our can of beans and sampling. To obtain samples the can is mixed and rotated to randomly disperse the beans within it. Reaching into the can without looking 2 beans were drawn from the can at a time to simulate a random mating. This was completed 20 times, while returning the beans to the can after each draw. Complete the table.

Mating pairs (colourxcolour) genotypes Total mating

Red X red 8

Red x brown 1

Red x white 2

Brown x brown 4

Brown x white 4

White x white 1

Transfer these results into the table below and fill in the expected genotypic ratios (medelian genetics) from each cross for the next generation. The first cross is filled in for you.

Mating Number of times Genotypic ratio in offspring

AA x AA 8 4AA

AAXAa

AAxaa

AaXAa

Aaxaa

aaxaa

Assume that each mating produces 4 offspring. To determine the number of offspring of each genotype produced by each mating type, use the following equation.

Number of each genotype produced by each mating type = (number of mating’s) x (expected genotypic ratio as a fraction) x4

Fill in the table below.

Mating Number of each genotype

AA Aa Aa

AA x AA

AA x Aa

AA x aa

Aa x Aa

Aa x aa

Aa xaa

Tally up the number of offspring of each genotype to determine the total number of offspring of each genotype in the next generation. What are the frequencies of each genotype in the next generation? Record your results in the table below.

Genotype Number of offspring % of total

AA

Aa

aa

Now count the total number of A and a alleles in this generation (for example AA has 2 A alleles)

Number of A alleles=

Number of a alleles=

Compare the relative frequencies of the alleles in the population as follows.

Frequency of A= amount of A/number of A + number of a=

Frequency of a=amount of a/number of A +number of a=

Calculate the frequency again this time using the Hardy Weinberg equation and how does this compare to the answer obtained by counting.

Remember since p+q=1, when there are only two alleles, you can find p using the following equations

Q2=frequency of aa

Q=square root of q2

P=1-q

Using the genotypic frequencies of the parental population, calculate the Hardy Weinberg value for p by this method.

Do all three techniques yield the same frequencies for the new generation? Provide proof.

Note: if they do not check your math and consider the effects of small sample size. Out of the large population only 20 mating’s. When mating populations are small, random sampling of the gene pool does not always occur. By change you may have drawn all mating pairs of the same type of mean retherm than drawing them in proportion to the occurrence. Random deviations from the expected frequencies are called “genetic drift “and most likely will influence your results.

To determine the effects of sample size (size of the mating population) on results, this example was completed an additional 18 times (this is to estimate if each student in the class completed this task of random sampling from the can of beans) and the results are compiled in the table below.

Student Number of each genotype

AA Aa aa

1 38 26 16

2 40 32 8

3 40 32 8

4 36 34 10

5 40 36 4

6 42 30 8

7 38 34 8

8 40 34 6

9 40 32 8

10 38 36 6

11 36 36 8

12 40 38 2

13 40 36 4

14 38 34 8

15 40 34 6

16 42 36 2

17 38 36 6

18 40 28 12

Calculate the gene frequencies for the new generation in this larger sample.

Frequency of A=

Frequency of a=

Compare these new results to the Hardy Weinberg predictions, are these values closer to the predicted values when compared to earlier looking at values from the single sampling? And if so why?

Since the simulation assumes random mating, no differential reproduction, no selection, no mutation, and no migration, what does this result illustrate about evolution.

If changes in gene frequency occurred in this simulation are, they due to random processes (genetic drift) or directed processes?

The previous example simulation demonstrated the Hardy Weinberg law that gene frequencies do not change from generation to generation when there is random mating in a large population and no other factors are operating to change gene frequencies. It is apparent than, that evolution will not occur under these types of conditions and that the Hardy Weinberg law provides a standard against which natural populations can be measured. IF gene frequencies are measured in two successive generations and are compared to the Hardy Weinberg equations, it is possible to determine whether the population is evolving: that is straying from random toward directed combinations. To simulate what would happen if a selective force were operating on a population and differential reproduction occurred, we are going o use the bean population again and draw 20 mating pairs, tallying the mating in the table board assuming that each student completed this in class.

Mating pairs (colour x colour) genotypes Total mating from 20 pairs Sum of class totals

Red x red AA x AA 6 102

Red x brown AA x Aa 7 123

Red x white AA x aa 3 53

Brown x brown Aa x Aa 2 50

Brown x white Aa x aa 1 23

White x white Aa x aa 1 9

Transfer the results to the following table and fill in the expected genotypic ratios for the offspring from each cross (the first one is completed for you

Mating pairs (colour x colour) genotypes Number of times Genotypic ratio in offspring

Red x red AA x AA 102 4 AA

Red x brown

Red x white

Brown x brown

Brown x white

White x white

Assume that each mating produces 4 offspring, determine the number of offspring of each genotype produced by each mating type and record your results in the table below.

Mating pair genotypes Number of AA Number of Aa Number of aa

AA x AA 408

AA x Aa

AA x aa

Aa x Aa

Aa x aa

aa x aa

Now let us assume that a selection factor is operating on the population, such as a disease in the environment. Genotype aa can mate with any other genotype, but half of the zygotes of genotype aa die during development because of the disease. For fertilizations results from each mating. But four offspring are not necessarily produced because of the disease. In the table below summarize the number of viable offspring of each genotype produced from the cross in question 22. Remember that half of the aa offspring die from the aa x aa cross, aa x Aa and Aa x Aa mating.

Genotype Number of offspring (viable) Percent of total

AA

Aa

aa

What are the frequencies of alleles A and a in this new living generation?

Frequency of A=

Frequency of a=

From your work in the previous examples, you know the gene frequencies in the original population. Therefore, you can use the H-W equation to calculate the frequencies of the genotypes that should be observed in the absence of selection and differential reproduction in the new generation.

Using the H-W equation, the relative frequencies are

AA= p2=

Aa= 2pq=

Aa=q2 =

What should the frequencies of A and a be according to the H-W equation

Frequency of A=

Frequency of a=

Compare the gene frequencies obtained by calculation to those obtained from the selection simulation. Why are they different?

Does this illustrate how evolution takes place? How?

If selection against the homozygous recessive genotype were to continue, why wouldn’t the allele eventually disappear from the population?

If selection against the aa were to cease and no other factors changed what would happen to the frequencies of A and a in future generations?

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