BIO 204 Lecture Notes - Lecture 7: Chi-Squared Test, Isopoda, Null Hypothesis
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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?
To determine if our data is consistent with expected results. | ||
a To determine if our data is consistent with expected results. b To determine if our data exactly matches the expected results. | ||
c To determine the expected results. | ||
d | To compare the phenotypic ratios to the genotypic ratios. |
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= ___.
a. | 2.22 | |
b | 2.71 | |
c | 4.36 | |
d | 187.78 | |
e | 448.27 |
4. Using the results of your Chi-squared analysis, do we fail to reject or reject the null hypothesis?
a. | Fail to reject the null | |
b. | Reject the null | |
c. | It cannot be determined from the data given |
Need help with evolution questions:
1.) A haplotype is best defined as the ________________.
haploid genotypes of all the gametes produced by a diploid individual | |
ABO blood type conferred by an individual gamete |
genotype of either the paternal or maternal chromosomal complement |
multilocus genotype of a chromosome or gamete |
2.) Which of the following statements regarding linkage disequilibrium is true?
Exists when D is less than zero, but not when it is greater than zero. |
Is reduced by sexual reproduction. |
Is increased by crossing-over during meiosis. |
Is increased by any random sampling error that happens to create or destroy certain chromosome genotypes but not others. |
Is reduced by selection that favors certain combinations of genotypes. 3.) Selection on multilocus genotypes in random-mating populations leads to linkage disequilibrium when _______________.
|
Experiment 1 Fermentation by Yeast Experiment Inventory Labware (4) 250 mL Beakers (1) 100 mL Graduated Cylinder (1) Test Tube Rack (5) Fermentation Tubes = (10) Test Tubes (5 plastic and 5 glass; see Figure 4) (1) Measuring Spoon (4) Pipettes (1) Ruler Note: You must provide the materials listed in *red. EXPERIMENT 1: FERMENTATION BY YEAST Yeast cells produce ethanol, C2 H6 O, and carbon dioxide, CO2 , during alcoholic fermentation. In this experiment, you will measure the production of CO2 to determine the rate of fermentation in the presence of different carbohydrates with fermentation tubes. Note: Regular table sugar is sucrose, a disaccharide, which is made up of glucose and fructose. Glucose is a monosaccharide. Figure 4: Fermentation tubes. Note how the smaller, plastic test tube is inverted into the larger glass tube. You will create five fermentation tubes in this experiment. PROCEDURE 1. In this experiment, you will mix yeast with sugar, Equal®, and Splenda®. Before you begin, develop a hypothesis predicting what will happen when the sugar/sweeteners are mixed with yeast. Will fermentation occur? Why or why not? Record your hypothesis in the post-lab questions. 2. Use the permanent marker to label three 250 mL beakers as Equal®, Splenda®, and Sugar. 3. Empty the Equal®, Splenda®, and Sugar packets into the corresponding beakers. 4. Fill the Equal® and Splenda® beakers to the 100 mL mark with warm tap water. 5. Fill the Sugar beaker to the 200 mL mark with warm tap water. 6. Mix each beaker thoroughly by pipetting the solution up and down several times. Use a new pipette to mix each solution. Each beaker now contains a 1% solution. Set these aside for later use. 7. Completely fill one of the smaller plastic tubes with tap water and invert the larger glass tube over it. Push the small tube up into the larger tube until the top connects with the bottom of the inverted tube. Invert the fermentation tube (Figure 4) so that the larger tube is upright (there should be a small bubble at the top of the internal tube). Note: Repeat Step 7 several times as practice. Strive for the smallest bubble possible. When you feel comfortable with this technique, empty the test tube(s) and proceed to Step 8. CAUTION: Do not try to force the plastic test tube into the glass test tube. This might cause your glass test tube to break, causing you injury. If your plastic test tubes do not fit easily, please call eScience Labs for replacement glass tubes. If you are able to set up at least two fermentation tubes, continue with the experiment, but know that you will have to perform steps 12-15 in multiple steps. 8. Use the permanent marker to label the fourth 250 mL beaker as Yeast. 9. Fill this beaker with 175 mL of warm tap water. It should be between 30 and 40o C (warm to the touch). 10.Open the yeast package, and use the measuring spoon to measure and pour 1 tsp. yeast into the beaker. Pipette the solution up and down until all of the yeast is mixed homogenously into the solution. Note: Make sure the yeast solution remains homogenous before each test tube is filled in the proceeding steps. The yeast density is fairly high, and the yeast may settle to the bottom of the beaker if it rests for an extended period of time. 11. Use the permanent marker to label the big glass and small plastic test tubes as 1, 2, 3, 4, and 5. 12.Use the 100 mL graduated cylinder to measure and pour 15 mL of the following solutions into the corresponding small plastic test tubes: Tube 1: 1% Glucose Solution Tube 2: 1% Sucrose Solution Tube 3: 1% Equal® Solution Tube 4: 1% Splenda® Solution Tube 5: 1% Sugar Solution Note: Thoroughly rinse the graduated cylinder between each measurement. 13.Fill the remaining volume in each small tube to the top with the yeast solution. 14.Slide the corresponding larger tube over the small tube and invert it as practiced in Step 7. This will mix the yeast and sugar/sweetener solutions. 15.Place the fermentation tubes in the test tube rack, and use a ruler to measure (in millimeters) the initial air space in the rounded bottom of the internal tube. Record these values in the Table 1. 16.Allow the test tubes to sit in a warm place (approximately 30 °C) for two hours. Placement suggestions include: a sunny window sill, atop (not in!) a warm oven heated to approximately 85 °C (185 °F on an oven setting), or under a very bright (warm) light. 17.At the end of the fermentation period, use your ruler to measure (in millimeters) the final gas height (total air space) in each tube. Record this data in Table 1. 18.Calculate the difference between the initial and final gas height in each tube. Record this data in Table 1.
EXPERIMENT 1: FERMENTATION BY YEAST
Result Tables
Table 1: Yeast Fermentation Data
Tube | Initial Gas Height (mm) | Final Gas Height (mm) | Net Change (mm) |
---|---|---|---|
1 | |||
2 | |||
3 | |||
4 | |||
5 |
Post-Lab Questions
Include your hypothesis from Step 1 here. Be sure to include at least one piece of scientific reasoning in your hypothesis to support your predictions.
Did you notice a difference in the rate of respiration between the various sugars? Did the artificial sugar provide a good starting material for fermentation?
Was anaerobic fermentation occurring? How do you know (use scientific reasoning)?
If you observed respiration, identify the gas that was produced. Suggest two methods you could use for positively identifying this gas.
Hypothesize why some of the sugar or sweetener solutions were not metabolized, while others were. Research the chemical formula of Equal® and Splenda® and explain how it would affect yeast respiration.
How do the results of this experiment relate to the role yeast plays in baking?
What would you expect to see if the yeast cell metabolism slowed down? How could this be done?
Indicate sources of error and suggest improvement (for example, what types of controls could be added?).