Biology Homework. Read the Lab 12 procedures and watch the online Hardy-Weinberg video (https://youtu.be/xPkOAnK20kw and posted on BlackBoard), then complete this assignment prior to lab.
1. The Hardy-Weinberg Theorem states…
2. What are the five key assumptions that are necessary for the H-W Theorem to be valid?
3. Write the Hardy-Weinberg equation:
4. Dominant allele “R” has a frequency (p) of 0.45 in a particular gene pool. Calculate the following showing all your work and using the proper variables for each value (e.g. p, q, p2, q2, 2pg).
a. The frequency of allele “r” in that same gene pool?
b. The proportion of the population that has the genotype RR.
c. The proportion of the population that has the genotype Rr.
d. The proportion of the population that has the genotype rr.
5. If 17% of a population displays the recessive trait for Disease B, what are the frequencies of the recessive allele “b” and the dominant allele “B” in the gene pool?
6. You perform an experiment where you allow a large population of fruit flies to mate randomly. The parental generation had 30% homozygous recessive genotypes. The F1 generation consisted of 100 flies, 40 of which displayed the recessive trait. Calculate the expected values for each phenotype assuming Hardy-Weinberg equilibrium, then fill in the table below and use the Chi-Square test instructions document (posted online) to compare your calculated X2 value with the tabulated X2 value for a P-value of 0.05.
| |
# of dominant phenotype individuals |
# of recessive phenotype individuals |
| Observed value (o) |
|
|
| Expected value (e) |
|
|
| Deviation (o – e) = d |
|
|
| d2 |
|
|
| d2/e |
|
|
| Calculated Chi-square (X2) = Σd2/e |
|
| Degrees of Freedom |
|
| Tabulated X2 value at P=0.05
(from X2 instructions document) |
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a. According to your analysis above, are the observed proportion of genotypes in the F1 generation the same, or significantly different, than those expected according to the H-W theorem?
b. If you allowed your F1 generation to mate, what would you expect the frequency of the recessive allele (q) to be in the F2 generation, assuming the H-W theorem applies?
Lab 12: Population Genetics I: Hardy-Weinberg Theorem
| OBJECTIVES
After completing this exercise, you should be able to:
1) Explain Hardy‑Weinberg equilibrium in terms of allelic and genotypic frequencies and relate these to the expression (p + q)2 = p2 + 2pq + q2 = 1 .
2) Describe the conditions necessary to maintain Hardy‑Weinberg equilibrium.
3) Use the marble model to demonstrate Hardy-Weinberg equilibrium and conditions for evolution.
4) Test hypotheses concerning the effects of evolutionary change (migration, mutation, genetic drift by either bottleneck or founder effect, and natural selection) using a computer model.
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Introduction
Charles Darwin’s unique contribution to biology was not that he “discovered evolution” but, rather, that he proposed a mechanism for evolutionary change ‑ natural selection, the differential survival and reproduction of individuals in a population. In On the Origin of Species, published in 1859, Darwin described natural selection and provided abundant and convincing evidence in support of evolution, the change in populations over time. Evolution was accepted as a theory with great explanatory power supported by a large and diverse body of evidence. However, at the turn of the century, geneticists and naturalists still disagreed about the role of natural selection and the importance of small variations in natural populations. How could these variations provide a selective advantage that would result in evolutionary change? It was not until evolution and genetics became reconciled with the advent of population genetics that natural selection became widely accepted.
Ayala (1982) defines evolution as “changes in the genetic constitution of populations.” A population is defined as a group of organisms of the same species that occur in the same area and interbreed or share a common gene pool. A gene pool is all the alleles at all gene loci of all individuals in the population. The population is considered the basic unit of evolution. Populations evolve, individuals do not. Can you explain this statement in terms of the process of natural selection?
In 1908, English mathematician G. H. Hardy and German physician W. Weinberg independently developed models of population genetics that showed that the process of heredity by itself did not affect the genetic structure of a population. The Hardy‑Weinberg theorem states that the frequency of alleles in the population will remain the same regardless of the starting frequencies. Furthermore, the equilibrium genotypic frequencies will be established after one generation of random mating. This theorem is valid only if certain conditions are met:
1. The population is very large.
2. Matings are random.
3. There are no net changes in the gene pool due to mutation; that is, mutation from “A” to “a” must be equal to mutation from “a” to “A”.
4. There is no migration of individuals into and out of the population.
5. There is no selection ‑ all genotypes are equal in reproductive success.
It is estimated, for example, that before the Industrial Revolution in Great Britain, more than 90% of the peppered moths were light colored, while less than 10% were dark. Under Hardy‑Weinberg equilibrium, these proportions would be maintained in each generation for large, random breeding populations with no change in the mutation rate and migration rate, as long as the environment was relatively stable. The process of heredity would not change the frequency of the two forms of the moth. Later in this laboratory, you will investigate what happened to these moths as the environment changed following the Industrial Revolution.
Basically, the Hardy‑Weinberg theorem provides a baseline model in which gene frequencies do not change and evolution does not occur . By testing the fundamental hypothesis of the Hardy‑Weinberg theorem, evolutionists have investigated the roles of mutation, migration, population size, nonrandom mating, and natural selection in effecting evolutionary change in natural populations. Although some populations maintain genetic equilibrium, these exceptions are intriguing to scientists.
Use of the Hardy‑Weinberg Theorem
The Hardy‑Weinberg theorem provides a mathematical formula for calculating the frequencies of alleles (e.g. “A” or “a”) and genotypes (e.g. “AA”, Aa” or “aa”) in populations. If we begin with a population with two alleles at a single gene locus ‑ a dominant allele, “A”, and a recessive allele, “a”‑ then the frequency of the dominant allele is p , and the frequency of the recessive allele is q . Therefore, p + q = 1 . If the frequency of one allele, p, is known for a population, the frequency of the other allele, q, can be determined by using the formula q = 1 ‑ p .
During sexual reproduction, the frequency of each type of gamete produced is equal to the frequency of the alleles in the population. If the gametes combine at random, then the probability of randomly combining an “A” allele with another “A” allele to produce an “AA” genotype in the next generation is p x p = p2 , according to the product rule – the probability of Event 1 AND Event 2 is equal to the product of their individual probabilities. Likewise the probability of “a” combining with “a” to form “aa” is q x q = q2 . The heterozygote “Aa” can be obtained two ways, with either parent providing a dominant allele and the other a recessive allele. According to the sum rule – the probability of Event 1 OR Event 2 is equal to the sum of their individual probabilities. Therefore, the probability of combining allele “A” from parent 1 and “a” from parent 2 is equal to p x q (or pq ), while the probability of the opposite (combining allele “a” from parent 1 and “A” from parent 2) is also equal to pq , therefore the sum of the two possibilities is 2pq . These genotypic frequencies can be obtained by multiplying p + q by p + q , in other words (p + q)2 . The general equation then becomes
(p + q)2 = p2 + 2pq + q2 = 1
Hardy-Weinberg Equation: p2 + 2pq + q2 = 1
To summarize:
For allele frequencies:
p = frequency of “A” allele
q = frequency of “a” allele
For genotype frequencies:
p2 = frequency of AA genotype
2pq = frequency of Aa genotype
q2 = frequency of aa genotype
Follow the steps again in this example.
1. If alternate alleles of a gene, “A” and “a”, occur at equal frequencies, p and q, then during sexual reproduction, 0.5 of all gametes will carry “A” and 0.5 will carry “a”.
2. Then p = q = 0.5.
3. Once allelic frequencies are known for a population, the genotypic makeup of the next generation can be predicted from the general equation. In this case,
p2 + 2(pq) + q2 = 1
0.52 + 2(0.5 x 0.5) + 0.52 = 1
0.25 + 0.50 + 0.25 = 1 (all genes in the population)
This represents the results of random mating as shown in Figure 12.1.
4. The genotypic frequencies in the population are specifically
p2 = frequency of AA = 0.25
2pq = frequency of Aa = 0.50
q2 = frequency of aa = 0.25
5. The allelic frequencies remain p = q = 0.5.
In actual populations the frequencies of alleles are not usually equal. For example, 4% of a population might be albinos (a recessive trait). In other words q2 = 0.04 and the frequency of the albino allele could be calculated as the square root of 0.04.
1. Albino individuals = q2 = 0.04 (genotypic frequency); therefore, q = √0.04 = 0.2 (allelic frequency).
2. Since p + q = 1 , the frequency of p is (1 ‑ q), or 0.8. So 4% of the population are albinos (genotypic frequency = 0.04), and 20% of the alleles in the gene pool are for albinism (allele frequency of “a” = 0.2). The other 80% of alleles are for normal pigmentation (allelic frequency of “A” = 0.8). Note: you could not determine the frequency of A by taking the square root of the frequency of all normally pigmented individuals because you cannot distinguish the heterozygote (2pq) and the homozygote (p2) for this trait. Therefore you must use the p + q = 1 equation.
3. The genotypic frequencies of the next generation now can be predicted from the general Hardy‑Weinberg theorem. First determine the results of random mating by completing Figure 12.2, filling in all the missing probabilities based on the data from above (q = 0.2, p = 0.8).
What will be the genotypic frequencies from generation to generation, provided that alleles p and q remain in genetic equilibrium?
AA = Aa = aa =
The genetic equilibrium will continue indefinitely if the conditions of the Hardy‑Weinberg theorem are met. How often in nature do you think these conditions are met? Although natural populations may seldom meet all the conditions, Hardy‑Weinberg equilibrium serves as a valuable model (a null hypothesis) from which we can predict/detect genetic changes in populations as a result of natural selection or other factors. This allows us to understand quantitatively and in genetic language how evolution operates at the population level.
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Figure 12.1. Random mating in a population at Hardy‑Weinberg equilibrium. The combination of alleles in randomly mating gametes maintains the allelic and genotypic frequency generation after generation. The gene pool of the population remains constant, and the populations do not evolve. |
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Figure 12.2. Random mating for a population at Hardy‑Weinberg equilibrium. Complete the mating combinations for albinism and normal pigmentation.
AA = p2 =
aA = pq =
Aa = pq =
aa = q2 = |
Testing Hardy‑Weinberg Equilibrium
Using a Marble Model
Materials
Box containing 100 marbles of two colors
Introduction
Working in pairs, you will test Hardy‑Weinberg equilibrium by simulating a population using colored marbles. The box of marbles represents the gene pool for the population. Each marble should be regarded as a single gamete, the two colors representing different alleles of a single gene. Each box should contain 100 marbles of the two colors in the proportions specified by the instructor. Record in the space provided below the color of the marbles and the initial frequencies for your gene pool.
A = (color) allelic frequency
a = (color) allelic frequency
(p + q)2 = p2 + 2(pq) + q2 = 1
(0.5 + 0.5)2 = 0.52 + 2(0.5 x 0.5) + 0.52 = 1
1 = 0.25 + 0.50 + 0.25 = 1 (all genes in the population)
1. There are 100 alleles in your box, how many diploid individuals are represented in this population?
2. What would be the color combination of the marbles needed to produce a homozygous dominant individual?
3. What would be the color combination of the marbles needed to produce a homozygous recessive individual?
4. What would be the color combination of the marbles needed to produce a heterozygous individual?
Hypothesis
State the Hardy‑Weinberg theorem in the space provided. This will be your hypothesis (it is sort of a null hypothesis…assuming no selection, etc.)
Predictions
Predict the genotypic frequencies of the population in future generations (if/then). Deductive thinking.
PART I Procedure-
1. Without looking, randomly remove two marbles from the box. These two marbles represent one diploid individual in the next generation. In the table to the right, record a tally of the diploid genotype (AA, Aa, or aa) of the individual formed from these two gametes.
2. Return the marbles to the box and shake the box to reinstate the gene pool. By replacing the marbles each time, the size of the gene pool remains constant and the probability of selecting any allele should remain equal to its frequency. This procedure is called sampling with replacement.
3. Repeat steps 1 and 2 (select two marbles, record the genotype of the new individual, and return the marbles to the box) until you have recorded the genotypes for 50 individuals who will form the next generation of the population.
| AA individuals |
Aa individuals |
aa individuals |
|
|
|
|
PART I Results-
1. Before calculating the results of your experiment , determine the expected frequencies of genotypes and alleles for the population. To do this, use the original allelic frequencies for the population provided by the instructor. (Recall that the frequency of A = p, and the frequency of a = q.) Calculate the expected genotypic frequencies using the Hardy‑Weinberg equation p2 + 2pq + q2 = 1 . The number of individuals expected for each genotype can be calculated by multiplying 50 (total population size) by the expected frequencies. Record these results in Table 12.1.
Table 12.1 Expected Genotypic and Allelic Frequencies for the Next Generation Produced by the Marble Model
| Parent
Populations |
EXPECTED New
Populations |
| Allelic
Frequency |
Genotypic Number (# individuals) and Frequency (proportion) |
Allelic
Frequency |
| A
|
a |
AA
# =
Freq.=
|
Aa
# =
Freq.=
|
aa
# =
Freq.=
|
A |
a |
2. Next, using the results of your experiment , calculate the observed frequencies in the new population created as you removed marbles from the box. Record the number of diploid individuals for each genotype in Table 12.2, and calculate the frequencies for the three genotypes (AA, Aa, aa). Add the numbers of each allele, and calculate the allelic frequencies for A and a. These values are the observed frequencies in the new population. Genotypic frequencies and allelic frequencies should each equal 1.
Table 12.2 Observed Genotypic and Allelic Frequencies for the Next Generation Produced by the Marble Model.
| Parent
Populations |
OBSERVED New
Populations |
| Allelic
Frequency |
Genotypic Number (# of individuals) and Frequency (proportion) |
Allelic
Frequency |
| A
|
a |
AA
# =
Freq.=
|
Aa
# =
Freq.=
|
aa
# =
Freq.=
|
A |
a |
3. To compare your observed results with those expected, you can use a chi-square test of the genotype frequencies . Table 12.3 will assist in the calculation of the chi-square test. Fill out this table, but then use excel to perform a chi-squared test and calculate an actual p-value.
Table 12.3 Chi-Square of Results from the Marble Model
| |
# of AA individuals |
# of Aa individuals |
# of aa individuals |
| Observed value (o) |
|
|
|
| Expected value (e) |
|
|
|
| Deviation (o – e) = d |
|
|
|
| d2 |
|
|
|
| d2/e |
|
|
|
| Calculated Chi-square value (X2) = Σd2/e |
|
Degrees of freedom (# of possible genotypes – 1) = __________, P-value from excel chi-square test = ____________
Are your observed genotypic frequencies the same, or significantly different, than what is expected according to H-W Equilibrium? Why?
PART II Procedures-
1. Follow the same procedures as in PART I, except this time keep your eyes open as you select the marbles. Decide on one of the two colors to have a “selective advantage”, then make your selections favoring choosing that color. (You don’t need to choose them every time, but don’t choose the marbles randomly.)
2. Record your tallies for the new population below, then calculate your “observed frequencies” and record them in Table 12.4, below.
| AA individuals |
Aa individuals |
Aa individuals |
|
|
|
|
3. Disregard the fact that you did not select the marbles randomly, and use your original “expected frequencies” from Table 12.1 (according to H-W Equilibrium) to run a chi-square test on the genotype frequencies, using excel .
PART II- Results
Table 12.4 Observed Genotypic and Allelic Frequencies for the Next Generation Produced by the (non-random) Marble Model.
| Parent
Populations |
OBSERVED New
Populations |
| Allelic
Frequency |
Genotypic Number (# of individuals) and Frequency (proportion) |
Allelic
Frequency |
| A
|
a |
AA
# =
Freq.=
|
Aa
# =
Freq.=
|
aa
# =
Freq.=
|
A |
a |
Chi-square test p-value:
Are your observed genotypic frequencies the same, or significantly different, than what is expected according to H-W Equilibrium? Why?
Post-lab 5 Questions
(detach and turn in next week along with your abstract)
1. In your PART I newly produced generation, what proportion of your population was…
a. Homozygous dominant?
b. Homozygous recessive?
c. Heterozygous?
2. Do your results in PART I match your predictions for a population at Hardy-Weinberg equilibrium?
3. If you continued the PART I simulation for 25 generations…
a. What would you expect to happen to the frequencies of each allele?
b. Would that population be evolving? Explain your response.
4. Consider each of the conditions for the Hardy‑Weinberg model. Does your model meet each of those conditions?
Consider what you did in PART II:
5. How did the simulation in PART II differ from your methods in PART I?
6. Which condition necessary for H-W Equilibrium was violated?
7. What did your chi-square test allow you to identify about the genotype frequencies in the new population in PART II?
8. Is the population in PART II evolving? Explain your response.
9. How could this method be useful when studying population genetics of real organisms?
10. What might be some of the difficulties you would encounter while using this in a real population genetics study?
For example, if the “A” allele has a frequency of 0.75 in a gene pool, then p = 0.75, and the frequency of “a” can be calculated as q = 0.25.
If “A” has a frequency of p = 0.75 the probability of getting an “AA” genotype is:
p2 = 0.752 = 0.5625
Likewise for “a”, with q = 0.25, probability of “aa” is:
q2 = 0.252 = 0.0625
Prob. of “Aa” or “aA” is:
2pq = 2(0.75 x 0.25) = 0.375
NOTE:
p2 + 2pq + q2 = 1
0.5625+0.375+0.0625 =1
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Bio 112 Bignami & Olave Spring 2016
Biology Homework
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