Evolution Lab
This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA
Overview
Download & View Evolution Lab as PDF for free.
More details
- Words: 2,189
- Pages: 9
Introduction:
The Hardy-Weinburg theorem states that the gene pool of a nonevolving
population remains constant over generations. According to this theorem, the Mendelian system has no tendency to alter allele frequencies and they should remain constant forever. However, in reality, some other factor always intervenes and therefore HardyWeinburg equilibrium cannot occur in nature. The system works similar to that of a deck of cards: no matter how many times the deck is reshuffled, the deck itself remains the same. The equation for Hardy-Weinburg equilibrium uses the letter p to represent the frequency of one allele and the letter q to represent the frequency of the other allele. Accordingly, p + q = 1, meaning that the combined frequencies of all possible alleles must add up to 100% for that population locus. The equation calculating frequencies of alleles in the gene pool if frequencies of genotypes are known is: p2 + 2pq + q2 = 1. This theory is important because it explains how Mendelian inheritance preserves genetic variation from one generation to the next. For a population to be in Hardy-Weinburg equilibrium, it must satisfy five conditions. These conditions are, first, a very large population size. Second, no migration can occur. Third, no net mutations will be accounted for. Fourth, random mating is required. And finally, no natural selection can occur, meaning all genotypes must have an equal chance of survival and reproduction. The Hardy-Weinburg equation describes the expected norm. If all five conditions are met, no change should occur within the allele/genotype frequencies of a population. The equation provides a norm by which evolutionary change can be measured.
The purpose of the following laboratory exercise was to not only understand the aforementioned law of genetic equilibrium, but also to study the relationship between evolution and changes in allele frequency using a sample population. Other objectives include calculating allele and genotype frequencies using the H-W theorem, discuss the effect of natural selection on allelic frequencies, and explain/predict effect on allelic frequencies of selection against homozygous recessive. The null hypotheses were assumed for the laboratory. These hypotheses state that no relationship will be found between evolution and changes in allele frequency and Hardy-Weinburg equilibrium theorem will not give an appropriate comparative measure of evolution. Also, the ability to taste PTC cannot be used to determine allele frequencies for a classroom or population of people. Protocol:
In part one of the laboratory exercise, allele frequencies for a specific trait within a sample population (consisting, in this case, of a classroom of students) was estimated. To do this, special PTC (phenylthiocarbamide) chemical papers were obtained. Short strips of PTC paper were removed and handed out to students. The students placed the strips on the tip of their tongues. If the student was a PTC taster, he/she would sense a bitter taste. The number of tasters/nontasters was recorded, and then the decimal number representing the frequency of tasters (p2 + 2pq) was calculated, as was the frequency of nontasters (q2). The Hardy-Weinburg equation was used to determine the frequencies of the two alleles, and the results were recorded. In part two, each student was given four cards. Two of the four cards were labeled with the capital letter A on one side of the card, and the remaining two cards were labeled with a lowercase a. It was assumed that the class was a population of randomly
mating heterozygous individuals, each with the genotype Aa. Each student found a partner and they both shuffled their cards face down so that the letter was not shown. The top card of each of the partners’ pile was contributed to create Partner 1’s offspring genotype. The cards were placed back in their respective piles, reshuffled, and the top card was again contributed, this time to create Partner 2’s offspring genotype. Each of the genotypes was recorded. The two partners assumed the genotype of their offspring by creating a hand of four cards of the appropriate genotype: for AA, a hand of four capital A’s, for aa, a hand of all lowercase a’s, and Aa, two A’s and two a’s, just as before. The two individuals then randomly selected a different (or possibly the same) partner to make another generation of offspring with the same card-shuffling procedure. Class data was collected. The process was repeated for five generations, each time randomly selecting a mate, assuming offspring’s genotype, and recording class data. The third part of the exercise was intended to demonstrate the probability of recessive genotypes characterized by genetic diseases surviving to reproductive maturity. The procedure followed was similar to that in the preceding part of the exercise. The initial genotype (Aa) was assumed by all students. With a partner, each student determined the genotype of his/her offspring and the offspring of his/her partner. This time, however, every time an aa genotype was created, the offspring theoretically dies. The set of partners who produced the fatal aa genotype tried again until they created a surviving genotype. This process was repeated for five generations, each time selecting against the recessive offspring. After each generation was created, the offspring of each partner was recorded, as well as class data.
Data:
The following table lists the percentages of phenylthiocarbamide tasters based on the chemical paper test performed in part one of the exercise. The ability to taste PTC is evidence of a presence of a dominant allele (either homo- or heterozygous). The inability to taste PTC suggests a homozygous recessive allele. The results in Table 1 list the percent of tasters and nontasters in the class and in the North American population overall.
Table 1: Phenotypic Proportions of Tasters and Nontasters and Frequencies of the Determining Alleles
Phenotypes
Allele Frequency Based on the HW Equation p q
% Tasters (p2 + 2 pq)
% Nontasters (q2)
Class Population
60%
40%
0.3676
0.6325
North American Population
55%
45%
0.3292
0.6708
According to the chart, and through calculations using the formulas p2 + 2pq + q2 and l - q = p, the percentage of heterozygous (2pq) tasters was approximately 46.5%. The percentage of heterozygous PTC tasters in North America was found to be about 44.2%, so therefore the class average is relatively normal compared to the population of the entire continent. The average of the class was only around 2% higher. In the next table, Table 2, the class results are listed for Part 2 of the exercise in which partners used genotype cards to create offspring for five generations. These results are the total numbers of offspring with an AA, Aa, or aa genotype in the class of sixteen students.
Table 2: Data of Part 2 of Laboratory Exercise Showing Class Totals for Each Genotype
AA
1
Offspring’s Genotype (AA, Aa, or aa) Aa
4
6
6
2
Aa
2
8
6
3
Aa
3
7
6
4
Aa
5
8
3
5
Aa
6
5
5
Generation
Class Totals for Each Genotype Aa
aa
The genotype frequencies for the sample class population can be computed using the equation: Frequency of (AA/Aa/aa)= Total of (AA/Aa/aa)/Total AA + Total Aa + Total aa. For the frequency of AA, the fraction divides 20/80. The frequency of Aa is 34/80. The frequency of aa is 26/80. The theoretical genotype frequencies of the beginning population where p and q = 0.5 are: p2 (AA) = .25, 2 pq (Aa) = .5, and q2 (aa) = .25. However, after five generations of mating in the sample classroom population, the frequencies were slightly different. For p2 (AA), the frequency was .25; for 2pq (Aa), the frequency was .425; for q2 (aa), the frequency was .325. Data 1: Calculations for A and a Alleles in Fifth Generation
Number of offspring with genotype AA =6x2= 12 A alleles Number of offspring with genotype Aa = 5x1 = 5 A alleles Total A alleles: 17 Total # alleles in population= 32 P= .53125 Number of offspring with genotype aa = 5x2 = 10 a alleles Number of offspring with genotype Aa = 5x1 =5 a alleles Total a alleles: 15 Total # alleles in population = 32 Q = .46875
In the final table, Table 3, the survival rate of genetic diseases is calculated. Both the hetero- and homozygous dominant genotypes will survive in a mock environment, however a recessive genotype will not. There is no column for aa recessive genotype because it did not survive into the first generation or to reproductive maturity. Table 3: Chart of Genotypes Excluding that Which is Affected by Genetic Disease
1
Offspring’s Genotype (AA, Aa, or aa) Aa
2
Generation
Class Totals for Each Genotype AA Aa 5
11
AA
8
8
3
Aa
6
10
4
Aa
5
11
5
Aa
4
12
Data: Calculations for Alleles Present at the Fifth Generation for Table 3
Number of offspring with AA alleles 4x2= 8 Number of offspring with Aa alleles x1 = 12 Total A alleles = 20 Total # of alleles in population: 32 P = .625 Number of offspring with genotype aa x2= 0 Number of offspring with genotype Aa = 12 Total a alleles = 12 Total # of alleles in population: 32 Q = .375 In this part of the laboratory, it becomes apparent that through natural selection, individuals with the genotype aa are eliminated, causing a decline in the number of a alleles in this case. Therefore, the p frequencies in Part 3 are higher than those in Part 2 of the lab exercise (in which the values were 0.5 for both p and q) and the
q frequencies are lower than the q frequency in Part 2. This shows a trend toward A dominance. If a similar process of selection proceeded for another five generations, the frequency of q will continue to decrease, but it will not reach zero because the heterozygous Aa will remain. Likewise, the frequency of A will continue to increase. The difference in allele frequencies between Part 2 and Part 3 after five generations is relatively apparent with the given data. In Part 2, the frequencies for A and a are very closely related, with the a alleles at a slight loss. In Part 3, however, the frequencies are quite different; for p, the frequency was .625 and q, a mere .375. This is a significant difference. Q, or recessive a, had a much lower frequency in Part 3 and P, or A, had a higher frequency. Conclusion:
In this lab, the relationship between evolution and changes in allele
frequencies were tested and observed. Using the Hardy-Weinberg equation, it was determined how to calculate the frequencies of alleles and genotypes in the gene pool of a population. For the Hardy-Weinberg theory to apply, the population must be in equilibrium. This has five requirements; a large population, no migration, no selection, no mutations, and random mating. Since only a classroom of students was used as a mock population, the first of the five requirements listed was not met, therefore random mating could not be ensured. In Part 2, all students started out as heterozygous, Aa. After five “mating” simulations, the number of heterozygotes, dominant homozygotes, and recessive homozygotes were counted. The Hardy-Weinberg theory predicts the p and q frequencies to be .5 and .5. However, these results were not obtained. This is normal, because Hardy-Weinberg Equilibrium does not occur in nature, nor was the experiment performed under ideal conditions. Again, the small population size and not-so-random
mating was probably the source of error here. In Part 3, the homozygous recessive genotype assumed a deadly genetic disease and could not live to reproductive maturity. This allowed for only homozygous dominant and heterozygous to survive. This created a drastic difference in frequency; p was higher than q because the frequency of q in the population would decrease due to natural selection. In Part 1, it became apparent that PTC was tasted by a majority of the population in not only the classroom, but also in North America. This, in combination with the other data collected and discussed relating to Parts 2 and 3, leads to the rejection of the null hypotheses. The Hardy-Weinberg Equilibrium indeed proved an appropriate measure of evolutionary change in all parts of this experiment. Also, there was an evident relationship between evolution and changes in allele frequency in, especially, Part 3 of the exercise. The results of this lab were relatively accurate in the sense that the procedure was performed as expected, but the population size of the classroom of students was much too small. However, this was probably expected on the creation of the lab, in order to show that Hardy-Weinberg equilibrium does not, and should not, occur in nature. It is simply a norm by which evolutionary change can be measured. The results of the lab exercises were conclusive and promoted a firm conviction on the rejection of the null hypotheses. Bibliography:
Campbell, Neil. Reece, Jane. 2002. Biology: Sixth Edition. Addison and Wesley Longman, Inc. California. The College Board and ETS. Advanced Placement Biology Laboratory Manual
for
Students. New York: 1991.
Measuring Allelic Frequency and Evolutionary Change Using the Hardy-Weinberg Theorem and PTC Tasting
Tessa Rodes April 7, 2009
The Hardy-Weinburg theorem states that the gene pool of a nonevolving
population remains constant over generations. According to this theorem, the Mendelian system has no tendency to alter allele frequencies and they should remain constant forever. However, in reality, some other factor always intervenes and therefore HardyWeinburg equilibrium cannot occur in nature. The system works similar to that of a deck of cards: no matter how many times the deck is reshuffled, the deck itself remains the same. The equation for Hardy-Weinburg equilibrium uses the letter p to represent the frequency of one allele and the letter q to represent the frequency of the other allele. Accordingly, p + q = 1, meaning that the combined frequencies of all possible alleles must add up to 100% for that population locus. The equation calculating frequencies of alleles in the gene pool if frequencies of genotypes are known is: p2 + 2pq + q2 = 1. This theory is important because it explains how Mendelian inheritance preserves genetic variation from one generation to the next. For a population to be in Hardy-Weinburg equilibrium, it must satisfy five conditions. These conditions are, first, a very large population size. Second, no migration can occur. Third, no net mutations will be accounted for. Fourth, random mating is required. And finally, no natural selection can occur, meaning all genotypes must have an equal chance of survival and reproduction. The Hardy-Weinburg equation describes the expected norm. If all five conditions are met, no change should occur within the allele/genotype frequencies of a population. The equation provides a norm by which evolutionary change can be measured.
The purpose of the following laboratory exercise was to not only understand the aforementioned law of genetic equilibrium, but also to study the relationship between evolution and changes in allele frequency using a sample population. Other objectives include calculating allele and genotype frequencies using the H-W theorem, discuss the effect of natural selection on allelic frequencies, and explain/predict effect on allelic frequencies of selection against homozygous recessive. The null hypotheses were assumed for the laboratory. These hypotheses state that no relationship will be found between evolution and changes in allele frequency and Hardy-Weinburg equilibrium theorem will not give an appropriate comparative measure of evolution. Also, the ability to taste PTC cannot be used to determine allele frequencies for a classroom or population of people. Protocol:
In part one of the laboratory exercise, allele frequencies for a specific trait within a sample population (consisting, in this case, of a classroom of students) was estimated. To do this, special PTC (phenylthiocarbamide) chemical papers were obtained. Short strips of PTC paper were removed and handed out to students. The students placed the strips on the tip of their tongues. If the student was a PTC taster, he/she would sense a bitter taste. The number of tasters/nontasters was recorded, and then the decimal number representing the frequency of tasters (p2 + 2pq) was calculated, as was the frequency of nontasters (q2). The Hardy-Weinburg equation was used to determine the frequencies of the two alleles, and the results were recorded. In part two, each student was given four cards. Two of the four cards were labeled with the capital letter A on one side of the card, and the remaining two cards were labeled with a lowercase a. It was assumed that the class was a population of randomly
mating heterozygous individuals, each with the genotype Aa. Each student found a partner and they both shuffled their cards face down so that the letter was not shown. The top card of each of the partners’ pile was contributed to create Partner 1’s offspring genotype. The cards were placed back in their respective piles, reshuffled, and the top card was again contributed, this time to create Partner 2’s offspring genotype. Each of the genotypes was recorded. The two partners assumed the genotype of their offspring by creating a hand of four cards of the appropriate genotype: for AA, a hand of four capital A’s, for aa, a hand of all lowercase a’s, and Aa, two A’s and two a’s, just as before. The two individuals then randomly selected a different (or possibly the same) partner to make another generation of offspring with the same card-shuffling procedure. Class data was collected. The process was repeated for five generations, each time randomly selecting a mate, assuming offspring’s genotype, and recording class data. The third part of the exercise was intended to demonstrate the probability of recessive genotypes characterized by genetic diseases surviving to reproductive maturity. The procedure followed was similar to that in the preceding part of the exercise. The initial genotype (Aa) was assumed by all students. With a partner, each student determined the genotype of his/her offspring and the offspring of his/her partner. This time, however, every time an aa genotype was created, the offspring theoretically dies. The set of partners who produced the fatal aa genotype tried again until they created a surviving genotype. This process was repeated for five generations, each time selecting against the recessive offspring. After each generation was created, the offspring of each partner was recorded, as well as class data.
Data:
The following table lists the percentages of phenylthiocarbamide tasters based on the chemical paper test performed in part one of the exercise. The ability to taste PTC is evidence of a presence of a dominant allele (either homo- or heterozygous). The inability to taste PTC suggests a homozygous recessive allele. The results in Table 1 list the percent of tasters and nontasters in the class and in the North American population overall.
Table 1: Phenotypic Proportions of Tasters and Nontasters and Frequencies of the Determining Alleles
Phenotypes
Allele Frequency Based on the HW Equation p q
% Tasters (p2 + 2 pq)
% Nontasters (q2)
Class Population
60%
40%
0.3676
0.6325
North American Population
55%
45%
0.3292
0.6708
According to the chart, and through calculations using the formulas p2 + 2pq + q2 and l - q = p, the percentage of heterozygous (2pq) tasters was approximately 46.5%. The percentage of heterozygous PTC tasters in North America was found to be about 44.2%, so therefore the class average is relatively normal compared to the population of the entire continent. The average of the class was only around 2% higher. In the next table, Table 2, the class results are listed for Part 2 of the exercise in which partners used genotype cards to create offspring for five generations. These results are the total numbers of offspring with an AA, Aa, or aa genotype in the class of sixteen students.
Table 2: Data of Part 2 of Laboratory Exercise Showing Class Totals for Each Genotype
AA
1
Offspring’s Genotype (AA, Aa, or aa) Aa
4
6
6
2
Aa
2
8
6
3
Aa
3
7
6
4
Aa
5
8
3
5
Aa
6
5
5
Generation
Class Totals for Each Genotype Aa
aa
The genotype frequencies for the sample class population can be computed using the equation: Frequency of (AA/Aa/aa)= Total of (AA/Aa/aa)/Total AA + Total Aa + Total aa. For the frequency of AA, the fraction divides 20/80. The frequency of Aa is 34/80. The frequency of aa is 26/80. The theoretical genotype frequencies of the beginning population where p and q = 0.5 are: p2 (AA) = .25, 2 pq (Aa) = .5, and q2 (aa) = .25. However, after five generations of mating in the sample classroom population, the frequencies were slightly different. For p2 (AA), the frequency was .25; for 2pq (Aa), the frequency was .425; for q2 (aa), the frequency was .325. Data 1: Calculations for A and a Alleles in Fifth Generation
Number of offspring with genotype AA =6x2= 12 A alleles Number of offspring with genotype Aa = 5x1 = 5 A alleles Total A alleles: 17 Total # alleles in population= 32 P= .53125 Number of offspring with genotype aa = 5x2 = 10 a alleles Number of offspring with genotype Aa = 5x1 =5 a alleles Total a alleles: 15 Total # alleles in population = 32 Q = .46875
In the final table, Table 3, the survival rate of genetic diseases is calculated. Both the hetero- and homozygous dominant genotypes will survive in a mock environment, however a recessive genotype will not. There is no column for aa recessive genotype because it did not survive into the first generation or to reproductive maturity. Table 3: Chart of Genotypes Excluding that Which is Affected by Genetic Disease
1
Offspring’s Genotype (AA, Aa, or aa) Aa
2
Generation
Class Totals for Each Genotype AA Aa 5
11
AA
8
8
3
Aa
6
10
4
Aa
5
11
5
Aa
4
12
Data: Calculations for Alleles Present at the Fifth Generation for Table 3
Number of offspring with AA alleles 4x2= 8 Number of offspring with Aa alleles x1 = 12 Total A alleles = 20 Total # of alleles in population: 32 P = .625 Number of offspring with genotype aa x2= 0 Number of offspring with genotype Aa = 12 Total a alleles = 12 Total # of alleles in population: 32 Q = .375 In this part of the laboratory, it becomes apparent that through natural selection, individuals with the genotype aa are eliminated, causing a decline in the number of a alleles in this case. Therefore, the p frequencies in Part 3 are higher than those in Part 2 of the lab exercise (in which the values were 0.5 for both p and q) and the
q frequencies are lower than the q frequency in Part 2. This shows a trend toward A dominance. If a similar process of selection proceeded for another five generations, the frequency of q will continue to decrease, but it will not reach zero because the heterozygous Aa will remain. Likewise, the frequency of A will continue to increase. The difference in allele frequencies between Part 2 and Part 3 after five generations is relatively apparent with the given data. In Part 2, the frequencies for A and a are very closely related, with the a alleles at a slight loss. In Part 3, however, the frequencies are quite different; for p, the frequency was .625 and q, a mere .375. This is a significant difference. Q, or recessive a, had a much lower frequency in Part 3 and P, or A, had a higher frequency. Conclusion:
In this lab, the relationship between evolution and changes in allele
frequencies were tested and observed. Using the Hardy-Weinberg equation, it was determined how to calculate the frequencies of alleles and genotypes in the gene pool of a population. For the Hardy-Weinberg theory to apply, the population must be in equilibrium. This has five requirements; a large population, no migration, no selection, no mutations, and random mating. Since only a classroom of students was used as a mock population, the first of the five requirements listed was not met, therefore random mating could not be ensured. In Part 2, all students started out as heterozygous, Aa. After five “mating” simulations, the number of heterozygotes, dominant homozygotes, and recessive homozygotes were counted. The Hardy-Weinberg theory predicts the p and q frequencies to be .5 and .5. However, these results were not obtained. This is normal, because Hardy-Weinberg Equilibrium does not occur in nature, nor was the experiment performed under ideal conditions. Again, the small population size and not-so-random
mating was probably the source of error here. In Part 3, the homozygous recessive genotype assumed a deadly genetic disease and could not live to reproductive maturity. This allowed for only homozygous dominant and heterozygous to survive. This created a drastic difference in frequency; p was higher than q because the frequency of q in the population would decrease due to natural selection. In Part 1, it became apparent that PTC was tasted by a majority of the population in not only the classroom, but also in North America. This, in combination with the other data collected and discussed relating to Parts 2 and 3, leads to the rejection of the null hypotheses. The Hardy-Weinberg Equilibrium indeed proved an appropriate measure of evolutionary change in all parts of this experiment. Also, there was an evident relationship between evolution and changes in allele frequency in, especially, Part 3 of the exercise. The results of this lab were relatively accurate in the sense that the procedure was performed as expected, but the population size of the classroom of students was much too small. However, this was probably expected on the creation of the lab, in order to show that Hardy-Weinberg equilibrium does not, and should not, occur in nature. It is simply a norm by which evolutionary change can be measured. The results of the lab exercises were conclusive and promoted a firm conviction on the rejection of the null hypotheses. Bibliography:
Campbell, Neil. Reece, Jane. 2002. Biology: Sixth Edition. Addison and Wesley Longman, Inc. California. The College Board and ETS. Advanced Placement Biology Laboratory Manual
for
Students. New York: 1991.
Measuring Allelic Frequency and Evolutionary Change Using the Hardy-Weinberg Theorem and PTC Tasting
Tessa Rodes April 7, 2009
