Genetics Lab
Strawfish
In the experiment about the "strawfish", twenty blue and twenty yellow straws were placed into a paper bag. These pieces of straw were a representation of "strawfish" alleles and each color had incomplete dominance. Two of these straws were drawn out of the bag to represent one strawfish, so when two blue straw alleles were drawn the fish was blue with the genotype BB. If there were two yellow straw alleles drawn, the fish had yellow traits with the genotype YY. If one blue straw allele and one yellow straw allele was drawn, the fish was a heterozygous green with a genotype BY or YB. The strawfish experiment was a model of what happens to traits in reproduction due to allele count. Each test represented environmental factors of natural selection and they explained how some traits or even some species could result in change or extinction.
In the first test, there were 4 generations, the first being the control without predation. After the control, the number of each color of fish and their alleles was counted to see which fish were left in each generation. In this test, every other blue fish was consumed by a predator and only the surviving fish alleles were returned to the bag. The second trial resulted in a green majority and the blue and yellow fish had equal populations. In the third trial, the number of green and yellow fish were equal while the blue population decreased significantly. In the final trial, there was only one blue fish remaining showing that the blue fish would disappear completely. The yellow fish's population remained nearly constant in number compared to the other three trials while the green fish remained with the greatest number of fish. Throughout the test, the green fish survived the longest in each trial while the blue nearly became extinct due to predation and lack of reproduction. Because green was the color of fish that lasted the longest, a lot of alleles were lost and diversity of the colors decreased. However, because the yellow fish seemed to stay constant in its population over the four generations, the green fish will eventually disappear as well. This is due to the predators adapting to the resources of food. In time, they will settle for the fish with the closest genotype and phenotype.
In the first test, there were 4 generations, the first being the control without predation. After the control, the number of each color of fish and their alleles was counted to see which fish were left in each generation. In this test, every other blue fish was consumed by a predator and only the surviving fish alleles were returned to the bag. The second trial resulted in a green majority and the blue and yellow fish had equal populations. In the third trial, the number of green and yellow fish were equal while the blue population decreased significantly. In the final trial, there was only one blue fish remaining showing that the blue fish would disappear completely. The yellow fish's population remained nearly constant in number compared to the other three trials while the green fish remained with the greatest number of fish. Throughout the test, the green fish survived the longest in each trial while the blue nearly became extinct due to predation and lack of reproduction. Because green was the color of fish that lasted the longest, a lot of alleles were lost and diversity of the colors decreased. However, because the yellow fish seemed to stay constant in its population over the four generations, the green fish will eventually disappear as well. This is due to the predators adapting to the resources of food. In time, they will settle for the fish with the closest genotype and phenotype.
Data Table for Test #1: Preferential Predation (Predators prefer blue fish)
Graph Set for Test #1
Hardy-Weinburg Equation
The Hardy-Weinburg equation is a test that determines the evolution of populations by tracking the possible genotypes or genetic makeup in a population from generation to generation. The equation is p^2 + 2pq + q^2 = 1 where 'p' is the frequency of the homozygous dominant allele (AA), 'q' is the frequency of the homozygous recessive allele (aa), and the 'pq' part of the equation represents the heterozygotes' alleles (Aa or aA). Because there are only two alleles, the frequency of one plus the frequency of the other must equal a whole so p + q = 1. The phenotypes or observable traits of the populations can only narrow the possibility of origin to one type of homozygous or a heterozygous pair of alleles. This equation will determine whether or not the population is in the Hardy-Weinburg equilibrium. The population must fit the equilibrium otherwise the equation will not accurately analyze the evolution of the population. Other rules that the population must follow are no mutations, random mating, no natural selection acting upon the population, a significantly large population size, and no gene flow. It it uncommon that a population follows all of these rules at one time but as time progresses, the population will reach the desired equilibrium.
Cases 1 and 2
In this activity, each student was given two cards representing their gametes. One had a "A" and the other had an "a" producing a heterozygous genotype (Aa or aA). Two parents give one gamete to their child, so each student exchanged with another contributing one card or gamete to the offspring. The genotype offspring acquired became the student's new genotype and then continued to mate with another student by trading cards or gametes. This process was repeated several times.
My initial genotype (my dear representative Bailey's anyway) was Heterozygous "Aa." Below are the generations of genotypes and their possible phenotypes.
F1: AA - complete dominant traits and alleles
F2: AA - complete dominant traits and alleles
F3: AA - complete dominant traits and alleles
F4: AA - complete dominant traits and alleles
F5: Aa - possible dominant or recessive phenotype and heterozygous alleles.
In Case 1, the class had 289 alleles; 174 dominant "A" alleles and 115 recessive "a" alleles. The "p" of Case 1 equals 0.602 and the "q" is 0.398.
Plug the variables in to the H-W equation-- 0.3624 + 0.4792 + 0.1584 = 1.
The initial frequency of the class for homozygous dominant alleles (AA) = 0.25, for heterozygous (Aa) = 0.5, and homozygous recessive (aa) = 0.25. The actual frequencies for Case 1 were AA = 0.3624, Aa = 0.4792, and aa = 0.1584. The reason for differences in the initial and actual frequencies of class data stem from the class size being too small thus not even fitting the rules of the Hardy-Weinberg equation. Because the genotype proportions changed through the "generations," the population successfully evolved and the result is a higher number of homozygous dominant gametes and a decreased number of homozygous recessive gametes.
In Case 2, a variable of death was added to the homozygous recessive children. This caused the students to mate until reproducing a child that could survive.
My initial genotype was again Aa.
F1: Aa - possible dominant or recessive phenotype and heterozygous alleles.
F2: Aa - possible dominant or recessive phenotype and heterozygous alleles.
F3: Aa - possible dominant or recessive phenotype and heterozygous alleles.
F4: Aa - possible dominant or recessive phenotype and heterozygous alleles.
F5: Aa - possible dominant or recessive phenotype and heterozygous alleles.
For this case, the initial class frequency was A = 0.25, Aa = 0.5, and aa = 0.25. The actual gamete count for the class was 334 total with 220 "A" alleles and 114 "a" alleles. 'p" equals 0.659 and "q" equals 0.341. After these variables were added to the equation, the making the actual class frequencies became AA = 0.434, Aa = 0.45, and aa = 0.116.
The death of the homozygous recessive children caused a great impact on the number of "aa" organisms as well as impacted the number of heterozygotes. If the class had been larger, the recessive alleles would have survived because of the existence of heterozygotes. However, even though the genotype is affected by these deaths, the phenotype would not change. The phenotypes or observable traits of the homozygous recessive would be invisible to the heterozygotes. The actual recessive gametes would not disappear, only the physical traits. This is a basic principle of evolution and explains why several organisms of different species have similar genotypes but different phenotypes.
My initial genotype (my dear representative Bailey's anyway) was Heterozygous "Aa." Below are the generations of genotypes and their possible phenotypes.
F1: AA - complete dominant traits and alleles
F2: AA - complete dominant traits and alleles
F3: AA - complete dominant traits and alleles
F4: AA - complete dominant traits and alleles
F5: Aa - possible dominant or recessive phenotype and heterozygous alleles.
In Case 1, the class had 289 alleles; 174 dominant "A" alleles and 115 recessive "a" alleles. The "p" of Case 1 equals 0.602 and the "q" is 0.398.
Plug the variables in to the H-W equation-- 0.3624 + 0.4792 + 0.1584 = 1.
The initial frequency of the class for homozygous dominant alleles (AA) = 0.25, for heterozygous (Aa) = 0.5, and homozygous recessive (aa) = 0.25. The actual frequencies for Case 1 were AA = 0.3624, Aa = 0.4792, and aa = 0.1584. The reason for differences in the initial and actual frequencies of class data stem from the class size being too small thus not even fitting the rules of the Hardy-Weinberg equation. Because the genotype proportions changed through the "generations," the population successfully evolved and the result is a higher number of homozygous dominant gametes and a decreased number of homozygous recessive gametes.
In Case 2, a variable of death was added to the homozygous recessive children. This caused the students to mate until reproducing a child that could survive.
My initial genotype was again Aa.
F1: Aa - possible dominant or recessive phenotype and heterozygous alleles.
F2: Aa - possible dominant or recessive phenotype and heterozygous alleles.
F3: Aa - possible dominant or recessive phenotype and heterozygous alleles.
F4: Aa - possible dominant or recessive phenotype and heterozygous alleles.
F5: Aa - possible dominant or recessive phenotype and heterozygous alleles.
For this case, the initial class frequency was A = 0.25, Aa = 0.5, and aa = 0.25. The actual gamete count for the class was 334 total with 220 "A" alleles and 114 "a" alleles. 'p" equals 0.659 and "q" equals 0.341. After these variables were added to the equation, the making the actual class frequencies became AA = 0.434, Aa = 0.45, and aa = 0.116.
The death of the homozygous recessive children caused a great impact on the number of "aa" organisms as well as impacted the number of heterozygotes. If the class had been larger, the recessive alleles would have survived because of the existence of heterozygotes. However, even though the genotype is affected by these deaths, the phenotype would not change. The phenotypes or observable traits of the homozygous recessive would be invisible to the heterozygotes. The actual recessive gametes would not disappear, only the physical traits. This is a basic principle of evolution and explains why several organisms of different species have similar genotypes but different phenotypes.
Sources of Error
Possible sources of error in Cases 1 and 2 include:
- Individuals not correctly counting their initial genotype or total alleles which gave odd totals for the class.
- People switching cards to mate with the same person repeatedly which decreases diversity and counteracts the randomness of reproducing.
- Rounding the p and q values.