Evolution Lab
Pre- Lab: Strawfish
Background:
In the strawfish pre-lab, 20 blue and 20 yellow straws were placed in a brown paper bag, and then at random 2 straws were pulled out and set together until there were no more straws in the bag. Two blue straws symbolized the alleles for a blue fish, two yellow straws symbolized the alleles for a yellow fish, and one yellow and one blue straw symbolized the alleles for a green fish. In test #2 we tested for preferential predation. The predators preferred the yellow. The straws were pulled out of the bag in twos. After all the straws were out of the bag, the amount of surviving fish and the number of surviving alleles were recorded. For each following generation, the predators “ate” half of the population of yellow fish. Those straws were taken out of the bag and put to the side. Then the next generation was recorded.
In test #4 we tested for heterozygote advantage. The green fish were able to camouflage themselves in the algae, and because the yellow and blue fish were easier to find, predators preferred them. The same procedure was followed for pulling out the straws in pairs but instead of the predators “eating” only the yellow fish, they ate every other blue fish and every other yellow fish. Those straws were taken out of the bag for the next generations.
In the strawfish pre-lab, 20 blue and 20 yellow straws were placed in a brown paper bag, and then at random 2 straws were pulled out and set together until there were no more straws in the bag. Two blue straws symbolized the alleles for a blue fish, two yellow straws symbolized the alleles for a yellow fish, and one yellow and one blue straw symbolized the alleles for a green fish. In test #2 we tested for preferential predation. The predators preferred the yellow. The straws were pulled out of the bag in twos. After all the straws were out of the bag, the amount of surviving fish and the number of surviving alleles were recorded. For each following generation, the predators “ate” half of the population of yellow fish. Those straws were taken out of the bag and put to the side. Then the next generation was recorded.
In test #4 we tested for heterozygote advantage. The green fish were able to camouflage themselves in the algae, and because the yellow and blue fish were easier to find, predators preferred them. The same procedure was followed for pulling out the straws in pairs but instead of the predators “eating” only the yellow fish, they ate every other blue fish and every other yellow fish. Those straws were taken out of the bag for the next generations.
TEST #2
Test #4
2. Did the allele frequency change much over the 4 generations in this test? Explain.
The allele frequency did change over the 4 generations. In test #2, the blue allele frequency was constant but the yellow allele frequency decreased because the predators were eating half of the yellow fish of each generation. However in test #4, the allele frequency decreased at the same rate for both the yellow and blue because the predators were eating half of both of the fish populations in each trial. There was an equal rate of removal of the fish from the population.
The allele frequency did change over the 4 generations. In test #2, the blue allele frequency was constant but the yellow allele frequency decreased because the predators were eating half of the yellow fish of each generation. However in test #4, the allele frequency decreased at the same rate for both the yellow and blue because the predators were eating half of both of the fish populations in each trial. There was an equal rate of removal of the fish from the population.
Hardy-Weinberg Equation:
p² + 2pq + q² = 1
p + q = 1
p² + 2pq + q² = 1
p + q = 1
p = frequency of the dominant allele
q = frequency of the recessive allele
q = frequency of the recessive allele
Conditions for Hardy Weinberg:
· No mutations
· Random mating
· No gene flow
· No natural selection
· Large population size
· No mutations
· Random mating
· No gene flow
· No natural selection
· Large population size
AP Lab 8: Population Genetics and Evolution Lab
Case 1: Test of an Ideal Hardy-Weinberg Population
The class represents a breeding population. Each student will choose another student at random to simulate random mating, The class is a population of randomly mating heterozygous individuals with an initial gene frequency of p=.5 and q=.5.
The initial genotype frequencies are AA=.25, Aa=.50, and aa=.25.
Every student will be given 2 cards that act as gametes. One card is the dominant A while the other is the recessive a. When two students "mate," each will randomly select one of their two initial cards. The combination is the genotype of the succeeding generation. The students then assume the genotype of the produced generation and continue "mating" with other students, following the same procedure each round. Record each generation in a data table.
For the second case, all of the aa children die. Therefore, when mating, if an aa is produced, continue "breeding" until either an Aa or AA is produced. Record each generation in a data table.
The class represents a breeding population. Each student will choose another student at random to simulate random mating, The class is a population of randomly mating heterozygous individuals with an initial gene frequency of p=.5 and q=.5.
The initial genotype frequencies are AA=.25, Aa=.50, and aa=.25.
Every student will be given 2 cards that act as gametes. One card is the dominant A while the other is the recessive a. When two students "mate," each will randomly select one of their two initial cards. The combination is the genotype of the succeeding generation. The students then assume the genotype of the produced generation and continue "mating" with other students, following the same procedure each round. Record each generation in a data table.
For the second case, all of the aa children die. Therefore, when mating, if an aa is produced, continue "breeding" until either an Aa or AA is produced. Record each generation in a data table.
Results/Data:
Hardy Weinberg Calculations :
Why did our case 1 data not match the initial Hardy-Weinberg?
For the Hardy-Weinburg principle to be effective, the population must coincide with five conditions. This means that there would be a large population size, random mating, no mutations, no gene flow, and no natural selection. Although in our class there were only some of those conditions. There were no mutations, population or gene flow and random mating. Also the population size was relatively small; there are 20 or so students in the class. This is why the Hardy-Weinburg principle could not be completely accurate.
For the Hardy-Weinburg principle could have been more applicable and our results would have matched more accurately if we had "mated" with other classes in the school. The results of Case 1 would then match the initial frequencies more.
Sources of Error:
A major source of error in the lab was the reporting of data. In case 1, 289 alleles were reported. 289 is not a multiple of twelve, meaning the number of alleles was not accurate. The amount of alleles every student was supposed to record and report was either wrong or not counted correctly. Then in case 2 the number of alleles reported was 334, which is also not a multiple of twelve. These reports effected the Hardy-Weinburg calculations. Had the source of error been fixed after case 1, then maybe the results would have matched the initial frequencies better.
For the Hardy-Weinburg principle to be effective, the population must coincide with five conditions. This means that there would be a large population size, random mating, no mutations, no gene flow, and no natural selection. Although in our class there were only some of those conditions. There were no mutations, population or gene flow and random mating. Also the population size was relatively small; there are 20 or so students in the class. This is why the Hardy-Weinburg principle could not be completely accurate.
For the Hardy-Weinburg principle could have been more applicable and our results would have matched more accurately if we had "mated" with other classes in the school. The results of Case 1 would then match the initial frequencies more.
Sources of Error:
A major source of error in the lab was the reporting of data. In case 1, 289 alleles were reported. 289 is not a multiple of twelve, meaning the number of alleles was not accurate. The amount of alleles every student was supposed to record and report was either wrong or not counted correctly. Then in case 2 the number of alleles reported was 334, which is also not a multiple of twelve. These reports effected the Hardy-Weinburg calculations. Had the source of error been fixed after case 1, then maybe the results would have matched the initial frequencies better.
Lab Repost:
![Picture](/uploads/1/7/3/7/17376319/7469712.jpg?740)
Hardy Weinberg Equation:
p² + 2pq + q² = 1
p + q = 1
p = frequency of the “A” dominant allele
q = frequency of the “a” recessive allele
Control Hardy Weinberg Equations:
7(2) + 6 = 20
20/40 = .5
p=.5
q=.5
.5² +2(.5) (.5) +.5²=1
Trial #2-4 Hardy Weinberg Equations:
6+4+2 =12
4+4+4=12
6+4+2=12
34+24+16=72
24+12 = 36
36/72
p=.5
24+12 = 36
36/72
q=.5
.5² +2(.5) (.5) +.5²=1
p² + 2pq + q² = 1
p + q = 1
p = frequency of the “A” dominant allele
q = frequency of the “a” recessive allele
Control Hardy Weinberg Equations:
7(2) + 6 = 20
20/40 = .5
p=.5
q=.5
.5² +2(.5) (.5) +.5²=1
Trial #2-4 Hardy Weinberg Equations:
6+4+2 =12
4+4+4=12
6+4+2=12
34+24+16=72
24+12 = 36
36/72
p=.5
24+12 = 36
36/72
q=.5
.5² +2(.5) (.5) +.5²=1