Natural Selection, Chi-square & Hardy-Weinberg Calculations

Transcription

1 BIOL 0 LAB 5 Natural Selection, Chi-square & Hardy-Weinberg Calculations Variability exists in all natural populations. For a wide variety of reasons, some phenotypes (visible characters) or genotypes exploit the environment more efficiently than others do. These phenotypes leave proportionately more offspring than their counterparts. If this phenotypic characteristic is heritable, offspring will resemble their parents, and the population will eventually consist mostly of individuals of the successful phenotypes. This weeding out of the less fit phenotypes is the process of natural selection, the accompanying change in the population is evolution, and the features that ultimately come to characterize the species are then viewed as adaptations to the environment. Since the key to evolutionary success is the production of fertile offspring, it requires numerous generations to demonstrate change through time. A model system then may provide a simulation of the process. In this model system, a plastic fork/spoon and a cup represent a predator. Two seeds represent different phenotypes within a prey species. Picking up seeds with a fork from the tabletop and putting them in the cup can simulate a feeding frenzy. Seeds that remain after the frenzy are the survivors and are the ones that can reproduce to reestablish the population. [An alternative model is to use the thumb and index or middle finger on one hand to pick up the seeds and deposit them in the cup]. Work in groups of three or four. For each feeding frenzy have two predators and one timer. You should rotate the tasks. To start, count out 100 of each seed type (corn and green lentil), mix them and spread them evenly on the entire table clear of everything except the seeds. During a feeding frenzy the two predators should pick up as many seeds as possible in a 5-second interval and place them in a cup. These are the prey that have been removed from the population, it is the survivors that breed according to their survivorship and reestablish the population at the equilibrium population size of 400. Example: Proportion Starting Number Surviving Adjusted Prey Population Eaten Survivors Population Population Green 00-0 = = = = 176 Total=

2 The adjusted population then serves as the starting population for the next generation and next 5-second feeding frenzy. The procedure needs to be followed up to 5 generations. Merely observing changes through time are insufficient evidence that change has occurred. We thus need to provide statistical evidence that can be provided using a simple chi-square test. Your Assignment (Due next Lab Period: 1 points) (1) propose an appropriate hypothesis for this experiment (1 point) () Conduct a Chi-square analysis on the starting population for Generation 0 vs. the adjusted population of Generation 5. ( points) (3) Calculate both the allele and expected genotype frequencies for all 5 generations (in all cases use the adjusted population for each generation). (5 points) (4) Finally, refer back to the genetics lab where you tabulated your phenotype and genotype for the list of monozygotic traits. Recall that you also tabulated the # in the class that were homozygous recessive for each trait. Use this data to calculate both allele and genotype frequencies for all of these traits. (4 points)

3 PREY STARTING NUMBER EATEN SURVIVORS PROPORTION SURVIVING GEN ADJUSTED GEN 1 GEN GEN 3 GEN 4 GEN 5 Assume that the green lentil is the homozygous recessive phenotype in this population(aa) Hardy-Weinberg Calculations Let A = p and a = q Thus, allele frequencies in a given population can be expressed as p + q = 1 Let AA = p Let Aa = pq Let aa = q Thus, genotype frequencies in a population can be expressed as p + pq + q = 1 3

4 Chi-square A relatively simple yet powerful test often used in science is the Chi-square analysis. Essentially, this is the statistical assessment of how well actual data fits an expected pattern. For example, suppose a scientist raises 100 plants and, because of the laws of genetics, predicts a 3:1 ratio of yellow to blue flowers (75 yellow, 5 blue). The actual ratio was 84 yellow: 16 blue. The statistical question is: Are the "observed" frequencies (84 to 16) significantly different from the "expected" frequencies (75 to 5)? Testing this question using the Chi-square method proceeds as follows: X =! ( O " E) E Where X is the Chi-square value, O is the observed frequency, E is the expected frequency, and is the symbol for summation over k categories. In this example, there are two categories: yellow flower color and blue flower color. In order to determine the significance of the X value, you must also determine the number of degrees of freedom for the problem. Degrees of freedom (df) refer to how many values have to be known to know all of the values in a problem. In this case, the degrees of freedom is k-1, which means that if you know all but one of the values for the categories, you will know the other (i.e., if you know that out of 100 plants, 84 are yellow, then 16 must be blue if that's the only other flower color - k =, so - 1 = 1 df). So, calculations of X may be summarized as follows: Category (Flower Color) Yellow Blue n Observed (O) Expected (E) 75 5 df = k-1 = -1 = 1 X ( O =! " E) = E ( 84! 75) 75 + ( 16! 5) 5 4

5 = = = 4.30 X = 4.3 calculated This value is then compared to a tabled value of X. By convention, this tabled value is usually set at 0.05 (95% level of significance) and the df are determined by the number of categories. Therefore, from the table (Table 5.1) at 0.05 and 1 df.: X = 3.84 table Because 4.3 is greater than 3.84, we determine that the numbers we found for flower color are significantly different than what is expected. The observed frequencies (84 and 16) are significantly different than the expected frequencies (75 and 5) at 0.05 level and 1 df. In other words, we are 95% certain that values of 84 yellow and 16 blue flowers came from a sample other than one which has 75 yellow and 5 green flowers. _ Table 5.1. Chi-Square Values. P Values D.F