Unit 3 Review (1)

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AP Bio Review Unit 3: Cellular Energetics Multiple Choice Practice Total Score: 31/35 FRQ Score: (14/15) 1. The enzyme trypsin aids in protein digestion in the small intestine. The relative activity of trypsin at different pH values is shown in Figure 1. Which of the following statements best explains the activity levels of trypsin shown in Figure 1? a. The small intestine releases inhibitor molecules that block the activity of trypsin unless it is at its optimum pH. b. The number of effective collisions between trypsin and its substrate increase at higher pH values. c. As pH values increase, the substrate concentration decreases, leading to an eventual decline in the rate of the trypsin-catalyzed reaction. d. At extremely low pH values, trypsin is denatured and cannot function efficiently 2. It is estimated that oxygen production first evolved in photosynthetic prokaryotes approximately 2.7 billion years ago. The first photosynthetic prokaryotes are presumed to be similar to today’s cyanobacteria. Which of the following best supports the claim that photosynthetic prokaryotes were responsible for the oxygen in Earth’s atmosphere? a. The light reactions of photosynthesis split carbon dioxide into carbon and oxygen. b. The light reactions of photosynthesis split water into hydrogen ions and oxygen. c. The Calvin cycle splits glucose into carbon, hydrogen, and oxygen. d. The Calvin cycle splits water into hydrogen ions and oxygen. 3. Which of the following questions is most relevant to understanding the Calvin cycle? a. How does chlorophyll capture light? b. How is ATP used in the formation of 3-carbon carbohydrates? c. How is NADP + reduced to NADPH? d. How is ATP produced in chemiosmosis? Use the following information to answer questions 4 & 5: A student placed 20 tobacco seeds of the same species on moist paper towels in each of

two petri dishes. Dish A was wrapped completely in an opaque cover to exclude all light. Dish B was not wrapped. The dishes were placed equidistant from a light source set to a cycle of 14 hours of light and 10 hours of dark. All other conditions were the same for both dishes. The dishes were examined after 7 days and the opaque cover was permanently removed from dish A. Both dishes were returned to the light and examined again at 14 days. The following data were obtained. 4. According to the results of this experiment, germination of tobacco seeds during the first week is a. increased by exposure to light b. unaffected by light intensity c. prevented by paper towels d. accelerated in green-leaved seedlings 5. Additional observations were made on day 21, and no yellow-leaved seedlings were found alive in either dish. This is most likely because a. yellow-leaved seedlings were unable to absorb water from the paper towels b. taller green-leaved seedlings blocked the light and prevented photosynthesis c. yellow-leaved seedlings were unable to convert light energy to chemical energy d. a higher rate of respiration in yellow-leaved seedlings depleted their stored nutrients Use the following information to answer questions 6 – 10: Photosynthetic activity can be measured using chloroplasts suspended in a buffered solution containing DCPIP, a dye that can accept electrons from the electron transport chain of photosynthesis. Transfer of electrons to DCPIP decreases the relative absorbance of a specific wavelength of light (605 nm) by a solution that contains the dye. A buffered solution containing chloroplasts and DCPIP was divided equally among six identical samples. The samples were placed at various distances from a lamp, and then all samples were exposed to white light from the lamp for 60 minutes at room temperature. Sample 3 was wrapped in foil to prevent any light from reaching the solution. At 20-minute intervals, the photosynthetic activity in each sample was determined by measuring the relative absorbance of 605 nm light. The results of the experiment are provided below. Relative Absorbance of 605 nm Light (arbitrary units) Sample Distance from Lamp (cm) 0 min 20 min 40 min 60 min

1 15 0.89 0.61 0.34 0.04 2 30 0.90 0.67 0.41 0.14 3* 30 0.88 0.87 0.86 0.87 4 45 0.86 0.69 0.47 0.26 5 60 0.92 0.75 0.59 0.41 6 75 0.88 0.79 0.71 0.58 * Wrapped in foil 6. Which of the following provides the best indication that light is required for the activation of electron transfer reactions in chloroplasts? a. Calculating the rate of change of the absorbance for sample 1 b. Comparing the observed results for sample 2 and sample 3 c. Repeating the entire experimental procedure at night d. Including multiple trials for all the samples 7. Which of the following can be reasonably concluded from the experimental results? a. Chloroplasts must be suspended in a buffer solution to function properly. b. The optimal temperature for activation of electron transfer is 25°C. c. DCPIP inhibits biochemical reactions in suspended chloroplasts. d. Light from a lamp can substitute for sunlight in stimulating chloroplast processes 8. If an additional sample containing the chloroplast/DCPIP solution was placed at a distance of 90 cm from the lamp, which of the following predictions would be most consistent with the experimental results? a. The concentration of DCPIP in the solution will increase exponentially. b. The absorbance at 60 minutes will be roughly equal to 1.4. c. The change in absorbance over time in the solution will be less than that of the other samples. d. The temperature of the solution will exceed 75°C. 9. Which of the following descriptions of photosynthesis best explains the results of the experiment? a. Availability of electrons for transfer to DCPIP depends on light energy. b. Movement of DCPIP across chloroplast membranes occurs in less than 60 minutes. c. Chlorophyll molecules degrade rapidly in the presence of DCPIP. d. DCPIP can only be used to measure photosynthetic activity at low light levels. 10.Which of the following scientific questions could be investigated using a similar experimental setup? a. How much carbon dioxide is required by a plant cell to produce one molecule of glucose? b. What wavelength of light best activates electron transfer reactions in chloroplasts? c. Which molecule in chloroplasts accepts activated electrons from DCPIP during photosynthesis? d. Are the same genes that are expressed in chloroplasts also expressed in mitochondria? 11.What most likely causes the trends in oxygen concentration shown in the graph below?

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a. The water becomes colder at night and thus holds more oxygen. b. Respiration in most organisms increases at night. c. More organisms are respiring at night than during the day. d. Photosynthesis produces more oxygen than is consumed by respiration during the day. 12.The chemical reaction for photosynthesis is 6 CO 2 + 12 H 2 O + light energy ? C 6 H 12 O 6 + 6 O 2 + 6 H 2 O If the input water is labeled with a radioactive isotope of oxygen, 18O, then the oxygen gas released as the reaction proceeds is also labeled with 18O. Which of the following is the most likely explanation? a. During the light reactions of photosynthesis, water is split, the hydrogen atoms combine with the CO2, and oxygen gas is released. b. During the light reactions of photosynthesis, water is split, removing electrons and protons, and oxygen gas is released. c. During the Calvin cycle, water is split, regenerating NADPH from NADP+, and oxygen gas is released. d. During the Calvin cycle, water is split, the hydrogen atoms are added to intermediates of sugar synthesis, and oxygen gas is released. 13.Students in a class measured the mass of various living organisms. They then kept the organisms in the dark for 24 hours before remeasuring them. None of the organisms were provided with nutrients during the 24-hour period. The data are as follows. Organism Starting Mass (g) Final Mass (g) Elodea (submerged aquatic plant) 15.10 14.01 Goldfish 10.10 9.84 Sea anemone 25.60 24.98 Which of the following is the best explanation for the pattern of change in mass of the organisms over time? c. Water loss due to evaporation b. The law of conservation of matter d. Cellular respiration d. Growth and reproduction Use the following information to answer questions 14 & 15: An experiment to measure the rate of respiration in crickets and mice at 10°C and 25°C was performed using a respirometer, an apparatus that measures changes in gas volume. Respiration was measured in mL of O 2 consumed per gram of organism over several five- minute trials and the following data were obtained.

Organism Temperature (°C) Average respiration (mL O 2 /g/min) Mouse 10 0.0518 Mouse 25 0.0321 Cricket 10 0.0013 Cricket 25 0.0038 14. During aerobic cellular respiration, oxygen gas is consumed at the same rate as carbon dioxide gas is produced. In order to provide accurate volumetric measurements of oxygen gas consumption, the experimental setup should include which of the following? a. A substance that removes carbon dioxide gasc. A glucose reserve b. A plant to produce oxygen D. A valve to release excess water 15.According to the data, the mice at 10°C demonstrated greater oxygen consumption per gram of tissue than did the mice at 25°C. This is most likely explained by which of the following statements? a. The mice at 10°C had a higher rate of ATP production than the mice at 25°C. b. The mice at 10°C had a lower metabolic rate than the mice at 25°C. c. The mice at 25°C weighed less than the mice at 10°C. d. The mice at 25°C were more active than the mice at 10°C. 16.Where on an enzyme does the substrate bind? a. active site b. allosteric site c. C-terminus d. N-terminus 17.Describe the effect on the free energy of doubling the amount of enzyme. (enzymes have no effect on free energy. They decrease the activation energy, but it does not change free energy, free energy depends on the concentration of reactants and products, but enzymes do not change either of them) a. double the free energy b. half the free energy c. triple the free energy d. no effect on free energy 18.Which structure is unaffected by denaturation? a. Primary b. Secondary c. Tertiary d. Quaternary 19.Which environmental conditions can cause denaturation of the enzyme? a. slight increase in temperature b. extreme increase in temperature c. decrease in temperature d. extreme increase in temperature and pH changes 20.Which of the following will increase the rate of the reaction? a. increase product b. increase substrate c. increase inhibitor d. increase enzyme 21.Describe why an increase in temperature increases the rate of the reaction. a. increase in temperature increases the speed/kinetic energy of particles b. increase in temperature activates more enzyme to increase the rate c. increase in temperature results in increase in substrate available d. increase in temperature allows for a tighter fit in the active site

22.Describe the difference between competitive and noncompetitive inhibitors. (competitive inhibition can be overcome by adding enough substrate, however the binding itself is irreversible, while in noncompetitive inhibitors the binding is reversible because it changes the shape of the site where the enzyme binds and it itself binds in the allosteric site) a. noncompetitive binds to active site and competitive binds to other site b. competitive is irreversible and noncompetitive is reversible binding c. competitive is irreversible and noncompetitive is reversible binding d. noncompetitive is irreversible and competitive is reversible binding 23.In order to maintain order and power cellular processes, a. energy input must exceed energy loss b. energy input equals energy loss c. energy loss must exceed energy input d. energy loss equals energy input 24.What is the process of an endergonic reaction being fueled by an exergonic reaction called? a. second law of thermodynamics b. energy coupling c. first law of thermodynamics d. endergonic/exergonic switch 25.Where did photosynthesis first evolve? a. prokaryotic organisms b. alien organisms c. eukaryotic organisms d. abiotic processes 26.What are the products of the light-dependent reactions of photosynthesis? a. ATP & NADH (glycolysis- cellular respiration) b. CO 2 & H 2 O (cellular respiration) c. NADPH & ATP (light-dependent) d. Organic Compounds & O 2 (cellular respiration) 27.What is the function of the electron transport chain between Photosystem II and Photosystem I? a. establish electrochemical gradient of protons b. to funnel electrons into photosystem I for the Calvin cycle c. transport electrons to oxygen to make water d. to synthesize ATP using ATP synthase 28.Where does the Calvin cycle take place? a. thylakoid membrane (light dependent reactions) b. cristae (ETC, oxidative phosphorylation) c. stroma (calvin cycle) d. matrix (krebs cycle) 29.Where does the electron transport chain take place? a. chloroplasts b. prokaryotic plasma membrane c. mitochondria d. all of the choices 30.Which of the following are electron carriers for the electron transport chain in cellular respiration? a. NADPH b. FADH 2 c. NADH d. NADH & FADH 2 31.Which process releases energy in glucose to form ATP, NADH, and pyruvate? a. Glycolysis b. Oxidative Phosphorylation c. Krebs Cycle d. Fermentation 32.What is oxidation? a. loss of electrons b. loss of water

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c. gain of electrons d. gain of water 33.In the electron transport chain of cellular respiration, where is the electrochemical gradient of protons generated? a. cytosol b. cristae c. matrix d. intermembrane space 34.Which process does NOT require oxygen? a. Glycolysis b. Oxidative Phosphorylation c. Krebs Cycle d. Aerobic respiration 35. Describe the pathway of electrons in the light reactions. a. H 2 O ? PSI ? ETC ? PSII ? NADPH b. H 2 O ? PSII ? ETC ? PSI ? NADPH c. H 2 O ? PSII ? ETC ? PSI ? NADH d. H 2 O ? PSI ? ETC ? PSII ? NADH

Unit 3 Free Response Practice (Answer in blue and correction in red) (14/15) 1. Microcystis aeruginosis is a freshwater photosynthetic cyanobacterium. When temperatures increase and nutrients are readily available in its pond habitat, M. aeruginosis undergoes rapid cell division and forms an extremely large, visible mass of cells called an algal bloom. M. aeruginosis has a short life span and is decomposed by aerobic bacteria and fungi. Identify the metabolic pathway and the organism that is primarily responsible for the change in oxygen level in the pond between time I and II AND between times III and IV. Time 1-2: metabolic pathway is photosynthesis; the organism that is responsible for change in oxygen level is cyanobacteria. Time 3-4: metabolic pathway is cellular respiration; the organism that is responsible for change in oxygen level is bacteria and fungi. (2/2) 2. Many species of bacteria grow in the mouths of animals and can form biofilms on teeth (plaque). Within plaque, the outer layers contain high levels of oxygen and the layers closest to the tooth contain low levels of oxygen. The surface of the tooth is covered in a hard layer of enamel, which can be dissolved under acidic conditions. When the enamel breaks down, the bacteria in plaque can extract nutrients from the tooth and cause cavities. Certain types of bacteria (e.g. Streptococcus mutans ) thrive in the innermost anaerobic layers of the plaque and are associated with cavities. Other types of bacteria ( Streptococcus sanguinis ) compete with S. mutans but are unable to thrive in acidic environments. (a) Identify the biochemical pathway S. mutans uses for metabolizing sugar and describe how the pathway contributes to the low pH in the inner layers of plaque. S. mutans use fermentation and this pathway produces acids like lactic acid because they live in an anaerobic environment. The pH drops during

fermentation because it releases organic acids. (2/2) (b) Normal tooth brushing effectively removes much of the plaque from the flat surfaces of teeth, but cannot reach the surfaces between teeth. Many commercial toothpastes contain alkaline components, which raise the pH of the mouth. Predict how the population sizes of S. mutans AND S. sanguinis in the bacterial community in the plaque between the teeth are likely to change when these toothpastes are used. S mutans will decrease because they can only survive in environments with low pHs, acidic environments. However, S. sanguinius will increase because they can survive in more areas than the S mutans, non acidic environments. (1) 3. An absorption spectrum indicates the relative amount of light absorbed across a range of wavelengths. The graphs below represent the absorption spectra of individual pigments isolated from two different organisms. One of the pigments is chlorophyll a, commonly found in green plants. The other pigment is bacteriorhodopsin, commonly found in purple photosynthetic bacteria. The table below shows the approximate ranges of wavelengths of different colors in the visible light spectrum. Color Wavelength (nm) Violet 380 – 450 Blue 450 – 475 Cyan 475 – 495 Green 495 – 570 Yellow 570 – 590 Orange 590 – 620 Red 620 - 750 (a) Identify the pigment (chlorophyll a or bacteriorhodopsin) used to generate the absorption spectrum in each of the graphs above. Explain and justify your answer. Graph 1: The pigment used is bacteriorhodopsin; this organism appears to be

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violet because it shows a low absorption rate for violet, 380-450 nm. Instead it absorbs the green light range and reflects the violet. Graph 2: The pigment used is chlorophyll a; this organism appears green because it is absorbing the red and blue wavelengths while reflecting the green wavelengths. Green plants do not absorb green wavelengths and instead reflect them. (3/3) (b) In an experiment, identical organisms containing the pigment from Graph II as the predominant light-capturing pigment are separated into three groups. The organisms in each group are illuminated with light of a single wavelength (650 nm for the first group, 550 nm for the second group, and 430 nm for the third group). The three light sources are of equal intensity, and all organisms are illuminated for equal lengths of time. Predict the relative rate of photosynthesis in each of the three groups. Justify your predictions. Group 1 will have the intermediate rate because the level of absorption of the 650 nm is between 550 nm and 430 nm so the energy available for photosynthesis will also be intermediate. Group 2 will have the lowest rate because there is the least amount of absorption at 550 nm, therefore there will be less energy available to drive photosynthesis. Group 3 will have the fastest rate because there is the greatest amount of absorption at 430 nm so there will be the most energy available to drive photosynthesis. (5/5) (c) Bacteriorhodopsin has been found in aquatic organisms whose ancestors existed before the ancestors of plants evolved in the same environment. Propose a possible evolutionary history of plants that could have resulted in a predominant photosynthetic system that uses only some of the colors of the visible light spectrum. The aquatic organisms were absorbing green light more than any other color so over time, natural selection would act and favor the organisms who absorbed the wavelengths that were available to them, rather than favoring the organisms that were not taking in wavelengths of light that were not as abundant. (unabsorbed wavelengths of light were available resources that could be exploited, absorbing light at lower wavelengths could cause damage (short wavelength= high energy), and higher wavelength absorption would not give enough energy. Genetic drift could have eliminated pigments that absorb certain wavelengths of light or mutations altered the pigments used by the organisms) (1/2)

Unit 3 Review (1)

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