173 Exam 2 Example Questions

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Exam 2 Example Questions 1. DIETARY FIBER AND GASES IN THE BREATH In some countries, dietary fiber is added to some soft drinks to help with constipation, or as a diet aid. Two types of soluble dietary fiber that are the most popular in soft drinks are polydextrose and oligosaccharide. It is unknown if these dietary fibers in commercial soft drinks is fermented by gut microbes. You are asked to investigate four brands of soft drinks to determine whether they are fermented by gut bacteria. You investigate the following soft drinks: Fibertini – contains 5g polydextrose Fiber & Go – contains 3g polydextrose and 0.6g oligosaccharide Fiberlicious – contains 3g oligosaccharides Yum Yum Fiber – contains 7g polydextrose To measure fermentation of the soft drinks, you collect breath samples of 45 healthy subjects (23 women and 22 men, between ages 20 – 40 years old) with no history of gastrointestinal issues or antibiotic use for at least 6 months prior to the study. Soft drinks were randomly assigned to each subject and their breath was sampled before each soft drink was consumed and every 15 min thereafter for 6 hours. Breath samples were collected using Quintron Breath Sampling Kits and the levels of hydrogen (H 2 ), methane (CH 4 ) and carbon dioxide (CO 2 ) in each sample were measured using a Quintron Breath Analyzer. (a) How does the fermentation of dietary fiber in the gut result in gases in the breath? In your response, be sure to describe how each of the breath gases (H 2 , CH 4 , and CO2) are created during the process. (2 points) - Dietary fiber enters the gut and is broken down by primary degraders resulting in intermediate products plus H2 and CO2 (0.5 point) - The intermediate products are broken down further into butyrate by butyrate producers. This results in the release of more H2 and CO2 (0.5 point) - H2 and CO2 combine to form CH4 or some people have methane producing microbes present in their gut (0.5 point) - H2, CO2, and CH4 gases enter the blood stream through the intestinal wall and travel to the lungs where they are exchanged and exhaled. (0.5 point) (b) Based on the information provided above, state your hypothesis for this experiment. (1 point) Dietary fiber in soft drinks is fermented by gut microbes. (c) What would be an appropriate negative control for this experiment? (1 point) Tap water or a substance without fiber (no effect or no increase in fermentation) (d) What might be a good positive control for this experiment? (1 point) 1

A dietary fiber that is known to be fermented by gut microbes, or a substance that will result in increased gases in the breath. A probiotic is also an acceptable answer You collect data on your 45 subjects and report the mean concentration of H 2 and CH 4 in breath samples over time following the consumption of each soft drink. Your results are graphed below, where open circles are the positive control; closed circles are Fibertini; open triangles are Fiber & Go; open squares are Fiberlicious; and closed squares are Yum Yum Fiber. (e)Based on the figures above, approximately how long after soft drink consumption did it take for hydrogen to start to increase in the breath? (1 point) About 60 – 120 min (1 – 2 hours); also accepted 40 – 50 min (f) Based on your results, what dietary fiber (oligosaccharide or polydextrose) produces more H 2 and CH 4 in the breath? How do you know? (1 point) Oligosaccharide because the two soft drinks that contained this type of fiber (Fiberlicious and Fiber & Go) had the highest H2 and CH4 in the breath compared to the other soft drinks. 2

(g) You notice that the breath sampling data from 6 of your subjects had consistently very low H2 and CH4 for every reading, regardless of the soft drink being tested. You remember that these subjects all have healthy GI tracts and no recent antibiotic intake, so their gut microbiomes are presumed to be healthy. How else might you explain these results, AND what can be done to fix this problem? (2 points) The subjects are probably collecting their breath samples at the start of their breath instead of in the middle of their exhale (1 point). This results in the recently inhaled air being expelled into the tube, which is high in CO2 and low in H2 and CH4. A breath sampled in the middle of the exhale includes air from deeper in the lungs that has undergone gas exchange with the blood (1 point). Any answer related to health of the gut microbiome in these subjects will not be accepted. The question prompt clearly states this is not a concern. (h) When planning a controlled experiment, it is important to keep all variables constant, except for the variable that is being studied (i.e., fiber in soft drinks). List TWO variables in this study that may confound the results. Be specific. (1 point) Possible answers must include experimental setup variables, NOT experimental errors. (0.5 point each): - The amount of dietary fiber in each soft drink is inconsistent. This may affect the fermentation results. - The subjects had a large age range of 20 years. Different age groups may have different gut microbiomes, which may affect the results. - Males and females included in the study - Different soft drink were randomly assigned to different individuals that might have different microbiomes. - Different levels of fiber in the regular diet - Breath samples were not fasting so some participants may have eaten more recently than others. Experimental errors will not be accepted (i.e., errors consuming drinks, breath sampling errors, etc.) 2. CRIME SCENE You are a forensic scientist and you have just arrived at a crime scene in which a murder has taken place. To investigate the crime, your first task is to collect blood samples in the hope that the murderer has left traces of their DNA. After sampling, you go to the lab to isolate and analyze the sample of DNA you collected. DNA profiling is a way of analyzing differences among individuals at the DNA level using molecular techniques such as PCR and gel electrophoresis. While most of our DNA is the same in every person, there are some regions that are unique to each individual and these regions are called Short Tandem Repeats (STRs). These are long stretches of small repeated sequences. The number of repeats in any of these sequences differs among individuals due to mutations that accumulate through generations. The unique combination of these variable regions makes up the DNA profile. After isolating the DNA from the blood samples collected at the crime scene and the blood collected from possible suspects, you use PCR to amplify the STR sequences of each sample. You add 2µl of sample DNA to 23µl 3

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premixed PCR cocktail in a sterile PCR tube. You also prepare positive (known sample of DNA) and negative control (sterile water) tubes. You place the PCR tubes in the thermocycler for 30 cycles. Each cycle includes 1 minute at 95°C, 1 minute at 52°C, and 1 minute at 72°C. (a) Explain what is happening at each of these temperatures in the PCR reaction. (1 point) 1 minute at 95°C = denature DNA; 1 minute at 52°C = primers anneal; 1 minute at 72°C = strands elongate (b) The primers you selected for your PCR reactions have melting temperatures at 51°C and 58°C, respectively. Will your primer set still be able to successfully amplify the DNA in your PCR reaction? Explain. (2 points) No, the temperature melting of the two primers should be within 5°C of each other and be above the annealing temperature (52C). This set has Tm that are too far apart and primer 1 has too low of a melting temperature for this reaction. Different primers need to be selected. (c) After appropriate amplification, the PCR product needs to be purified so that the solution only contains the isolated STR sequences. What other components are present in the PCR reaction tube that need to be removed? List at least FOUR. (2 points) Answers may include four of the following (0.5 each): Excess primers, dNTPs, longer DNA fragments, the original template strand, buffer, salt, Taq polymerase (d) You run your purified PCR samples on a gel to analyze the number and size of the STR sequences. After running the gel you notice that your negative control has three distinct bands. What do these results mean? Why is this considered a ‘failed’ PCR reaction and what should you do next? (2 points) There should not be any bands in the negative control. Your PCR master mix was likely contaminated with foreign DNA (1 point). The PCR reaction will have to be re-ran (1 point) (e) After finally running a successful PCR reaction, you obtain the gel results shown to the right. The lanes include a ladder (L), your victims blood (V), Evidence sample 1 (E1), Evidence sample 2 (E2), samples collected from two suspects (S1 and S2), and a positive control. Based on the DNA profiles presented in the gel, which suspect murdered the victim? (1 point) Suspect 1 4

Figure 1. Lace bug herbivory mediated interactions between S. altissima and a native (focal) plant species. COMPETITION BETWEEN INVASIVE AND NATIVE PLANTS Invasive species are organisms that have been transported by human activities into a region in which they do not normally occur. These invaders compete with native wildlife and these interactions can be influenced by biotic and abiotic factors in the environment. For example, the invasive plants may serve as a refuge for herbivores, resulting in increased herbivory on native plants. Ecologists study these competitive interactions to understand how to manage and minimize the impact of invasive species. Solidago altissima (Tall goldenrod) is from the aster/daisy family (Asteraceae) and is normally found in the United States. However, S. altissima has been introduced in Japan where it competes with other native Asteraceae plant species. As an ecologist working for the U.S. Fish and Wildlife Service, you want to compare the competitive effects of S. altissima in the U.S. and Japan. More specifically, you want to know how competition with S. altissima affects herbivory of native plants by lace bugs in Japan compared to the US. (1) Is the competition between S. altissima and native plant species exploitative or interference competition? (0.5 point) Exploitative (2) Name TWO abiotic factors that may influence the competition between S. altissima and native plant species. (0.5 point) Any two of these: Space (above or below ground), water, nutrients, light Using common garden experiments, you evaluated lace bug damage on native (focal) plants with and without S. altissima in Japan and the U.S. The abundance of each species is the same in all gardens. Figure 1 shows the experimental setup and your predicted results. Arrows indicate competition between plants, with a larger arrow indicating stronger competition. (3) Based on the information in Figure 1, how do you predict competition with S. altissima will affect herbivory of focal species in Japan? (0.5 point) In Japan, the presence of S. altissima will increase herbivory of focal plants. (4) Based on the information in Figure 1, how do you predict competition with S. altissima will affect herbivory of focal species in the U.S.? (0.5 point) In the U.S., lace bug damage will be the same whether S. altissima is present (mixed culture) or absent (monoculture). 5

You run the garden experiments in Shiga, Japan (Figure 2a) and Kansas, U.S. (Figure 2b) using a variety of different focal plant species and you obtain the results shown in Figure 2. (5) Based on the results in Figure 2, how does competition with S. altissima affect herbivory of focal plants in Japan compared to the U.S.? (1 point) In Japan, the presence of S. altissima increased herbivory of focal plants. (0.5 point) In U.S., the presence of S. altissima actually decreased herbivory of focal plants. (0.5 point) (6) Do the results support your predictions from part c & d? Explain. (1 point) Student answers may vary depending on predictions above. 0.5 point for yes/no, and 0.5 for explanation Sort of, we predicted that herbivory in the U.S. would be the same in the presence and absence of S. altissima. However, the results show that the presence of S. altissima actually decreased herbivory of native plants in the U.S (Fig. 2B). All three focal species had decreased herbivory by lace bugs when S. altissima was present. Students may also respond with “no” and explain that the first prediction was supported but the second one was not. This should get full credit (7) Using what you know about evolution, why do we observe stronger competitive effects on focal plants in regions where a competitor is invasive (i.e., Japan) compared to its native range (i.e., United States)? How might this change over time? (2 points) These invaders compete more strongly with native wildlife because native wildlife do not share an evolutionary history with the invader and, therefore, are less adapted to competition with the invader (1 point). Over time, focal species will adapt to the invasive species through natural selection. Individual focal plants that are able to compete better with the invasive species will survive and reproduce at a higher rate and eventually these traits will become dominant in the population (1 point). You decide to take your research a step further to examine how the abundance of S. altissima affects the fitness of other plant species in Japan. You run an 8-week de Wit replacement series experiment. You hypothesize that the abundance of S. altissima will have a negative effect on the fitness of focal species. (8) Define fitness and explain how competition can decrease fitness in two competing plant species. (1.5 points) Fitness is an individual’s overall ability to survive and reproduce . (0.5 point) 6

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Resources are necessary for an individual to survive, and if these resources are limited due to competition with another species, this can reduce the survival and reproduction of a species. (1 point) (9) You are given five garden beds to setup your de Wit experiment. In each box, indicate the number of plants for each species that would need to be planted in each bed to perform an appropriate de Wit replacement series. Assume you will plant a total of 60 plants per garden bed. (2.5 points) 0.25 point for each value (each bed is worth 0.5 point) The combinations should be 0/60, 15/45, 30/30, 45/15, 60/60. This represents 0, 25, 50, 75, 100% proportions of each species. Half credit if student puts percentages instead of numbers. At the end of the 8-week experiment, you measure plant fitness by harvesting the plants and measuring their survival, productivity (weight in kg) and reproduction (number of seeds produced). You synthesize your data and draw a de Wit series graph that shows the effect of S. altissima competition on focal plant productivity. (10) Based on the de Wit graph only, does the abundance of S. altissima have a negative effect on the fitness of focal species? Briefly explain. (1 point) Yes, the data line is below the 1:1 line indicating interspecific competition. (11) After running statistical analyses on your data, your results show that the number of seeds produced by focal plants decreased as S. altissima abundance increased (p = 0.001). However, the results also showed that the survival of focal plants actually increased as S. altissima abundance increased (p = 0.054). Assuming a significance threshold of α = 0.05, do these results support your hypothesis from above? Explain . (2 point) Yes, these results support the hypothesis (1 point) because a decrease in the number of seeds would be a decrease in reproduction, which is required for fitness. The increase in survival is not significant; therefore, we can conclude that fitness may still be decreasing. (1 point) 7 Bed 5 0 S. altissima 60 Focal species Bed 4 15 S. altissima 45 Focal species Bed 3 30 S. altissima 30 Focal species Bed 2 45 S. altissima 15 Focal species Bed 1 60 S. altissima 0 Focal species

173 Exam 2 Example Questions

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