BIOL 2060 Study Guide - Midterm Guide: Statistical Hypothesis Testing, Carolinian Forest, Binomial Distribution

Introduction: A Chi-square test is used to compare observed data with expected data according to a hypothesis. For instance, if you were crossbreeding 2 heterozygous pea plants, you would expect to see a 3:1 phenotypic ratio in the offspring. In this case, if you were to breed 400 pea plants, you would expect to see 300 plants showing the dominant trait and 100 showing the recessive trait. But what happens if you observe only 260 plants with the dominant trait and 140 plants with the recessive trait? Does this mean something is wrong with Mendelian genetics or is this difference in expected results just due to chance (random sampling error)? These are the questions that can be answered using Chi-square statistics. The results of this statistical test is used to either reject or accept (fail to reject) the null hypothesis. The null hypothesis states there is no significant difference between the observed results and the expected results. This means that if the null hypothesis is accepted, the difference in observed and expected results was just a matter of chance and so the observed results basically "fit" with what was expected. Degrees of freedom (df) = number of independent outcomes (Y) being compared less 1 df = Y-1 At the 95% confidence interval we are 95% confident that there is a significant difference between the observed and expected results, therefore rejecting the null hypothesis. Probability Value - Is the decimal value determined from the X2 table and is the probability of accepting the null hypothesis. A 0.05 probability value equates to a 95% confidence interval.

The Chi-squared test formula is: Example: If we cross two pea plants that are heterozygous yellow pods, we would expect a 3:1 phenotypic ratio. So let's say we actually did the cross and got 280 plants with green pods and 120 plants with yellow pods. Question: Is this a 3:1 phenotypic ratio? This is the value of Chi-squared Test. We have a total of 400 plants and we expect a 300 green:100 yellow phenotypic ratio If the calculated Chi-squared value is less than the critical value listed in the Chi-squared table, then we accept the null hypothesis. This means that there is no significant difference between the observed and the expected values. Our degrees of freedom (df) = 2 outcomes - 1, or df = 1. Now we go the X2 table below and using the df = 1 and probability value of 0.05, our critical value is 3.84. Since our calculated X2 value is 5.33, and is larger than the critical value, we reject the null hypothesis and can say (at 95% confidence) that there is a significant difference between our observed and expected values.

The parent generation is yellowed podded and green podded pea plants. You cross a yellow podded pea plant with a green podded pea plant and you get 100% yellow podded plants in the F1 Generation (Phenotypic ratio 4 : 0, yellow to green). What will be the expected phenotypic ratio when you allow the F1 generation to reproduce?

Fill out the Punnett square.

If we actually did the cross and got 1150 yellow and 350 green. Would this be a consistent with what was expected?

Learning Outcomes Questions

1. Why would you run a Chi-squared test?

2. Determine the degrees of Freedom of the phenotypic ratio for this genetic cross.

a. 1

b. 2

c. 3

d. 4

e. 5

3. Using the data given, what is the result of your Chi-squared analysis? x2= ___.

4. Using the results of your Chi-squared analysis, do we fail to reject or reject the null hypothesis?

You and your hypothetical lab partner each develop an experimental plan to test this hypothesis. The two experimental plans are presented below. Evaluate each plan for its pros and cons in the experimental design using the bullet points below. Both experimental plans have some good elements and areas that could be improved.

Experiment A) Mature 2 inch sections (sprigs) of Elodea will be used to measure oxygen production under different light conditions. 5 sprigs will be tested. One will be kept in the dark (no light). The others will be exposed to a light filtered with cellophane. One sprig’s light will be filtered through clear cellophane to receive all wavelengths of light, and the other sprigs’ light will be filtered through red, blue, and green cellophane, respectively. The no-light plant will be kept in a dark cabinet, while the other plants will be together on a counter. Each plant sprig will be placed in equal sized upside-down test tubes in 100ml of tap water so that oxygen bubbles produced by the plant will be trapped at the top of the test tube. A ruler will be used to measure the length of the oxygen pocket collected at the top of the test tube. The experiment will continue until one sprig produces 5 cm of oxygen in the tube.

Experiment B) 3 cm sections (sprigs) of Elodea of various ages will be used to measure oxygen production under different light conditions. 15 sprigs will be divided into 3 groups of 5 sprigs each. One group will receive natural sunlight, one group will be exposed to a blue light bulb under a box to block natural light, and the final group will be exposed to a red light bulb under a similar box to block natural light. Each group of plant sprigs will be trapped under a clear plastic funnel that is covered by an upside-down test tube to collect oxygen bubbles produced by all the sprigs in the group. All groups will be in 100ml of water of the same temperature and pH. The experiment will continue for 24 hours, and then the volume of oxygen gas produced will be measured by multiplying the length of the oxygen pocket by the circle described by the test tube, ?r². The average volume produced by all groups will be graphed and compared to determine which condition produced the most total volume of oxygen.

Consider the following questions to evaluate experimental plans A and B (DO NOT ANSWER THESE QESTIONS)

What is the independent variable suggested by the hypothesis? Do both plans test the independent variable equally well?

What is the dependent variable in each experiment? Which experiment’s dependent variable best matches the prediction in the hypothesis?

What are the experimental and control groups for each experiment? Which experiment gives the best opportunity to assess the effect of the independent variable?

How does each experiment account for standardized variables? Which experiment do you think is the better controlled experiment?

How is sample size addressed in each experiment? Which would you find to be more convincing?

Does either experiment describe any statistical analyses to be performed on the data?

ANSWER THIS QUESTION: BELOW

Considering your analysis of each of the questions above,write an improved experimental design to test the hypothesis “If blue light is more effective at promoting photosynthesis, then more oxygen gas (O₂) will be produced when plants are exposed to blue-filtered light compared to other wavelengths.”You may use elements of the above experiments

Hemoglobin and Fitness Instructions Directions: Neutral Evolution

1. Obtain 20 beans of two different colors (e.g., white and red). Count out 16 white and 4 red beans. The white beans represent the Hn allele and the red beans represent the Hs allele. This is the genetic makeup of your starting population. (Note: You can use any objects that can readily be categorized into two groups, such as coins, colored rocks, or paper clips.)

2.Calculate the frequency of both alleles [f(Hn) and f(Hs)] and record them in Table 1. In our experiment frequency is a measure of how many copies of a given allele exist in the gene pool (i.e., a proportion). Use decimal values. 


3.Arrange the beans into pairs. These pairs represent the genotype of each of 10 individuals in the population. Record the number of individuals with each genotype [f(Hn Hn), f(Hn Hs), and f(HsHs)] in Table 1. 


4.Now imagine that the individuals are living and reproducing with each individual reproducing at the same rate (i.e., all individuals produce two copies of each of their alleles into the next generation). Obtain enough beans to represent the next generation— the offspring generation—and then let the parental generation “die”. 


5.Calculate the frequency of each allele in the offspring generation and record it in Table 1. 


Answer the questions that follow in Table 1. 


Table 1

Answer the following questions to help you understand the exercise:

What happened to the frequency of the common allele? 


What happened to the frequency of the rare allele? 


What happened to the frequency of the common and rare alleles when the starting frequencies were different from yours

What happens to allele frequencies from one generation to the next if there are no evolutionary forces acting on the population?