2017 Ap Biology Frq Answers

Deconstructing the 2017 AP Biology Free Response Questions: A Comprehensive Guide

The AP Biology exam is a significant hurdle for many high school students, and the free-response questions (FRQs) often prove to be the most challenging part. This comprehensive guide will dissect the 2017 AP Biology FRQs, providing detailed answers, explanations, and valuable insights into the concepts tested. Understanding these questions and answers is crucial for current students preparing for the AP Biology exam and for anyone wanting a deeper understanding of fundamental biological principles. We'll explore each question thoroughly, highlighting key concepts and demonstrating how to effectively answer these complex questions.

Question 1: Enzyme Activity and Regulation

This question focused on enzyme activity, specifically the impact of environmental factors and regulatory mechanisms. Students were presented with data on enzyme activity under varying conditions.

Part A: This section asked about the effect of temperature and pH on enzyme activity. A successful response would clearly describe the optimal temperature and pH for the enzyme, explaining the underlying reasons for the observed changes in activity at different temperatures and pH levels. The answer should mention concepts like enzyme denaturation at extreme temperatures and how changes in pH can alter the enzyme's three-dimensional structure and active site, impacting its ability to bind to the substrate.

Example Answer Part A: The data shows that the enzyme exhibits optimal activity at a temperature of approximately 37°C and a pH of approximately 7. At temperatures significantly above or below 37°C, enzyme activity decreases. This is likely due to denaturation, where the enzyme's three-dimensional structure is disrupted, altering the shape of the active site and reducing its ability to bind to the substrate. Similarly, deviations from a pH of 7 cause changes in the enzyme's charge distribution, potentially affecting its conformation and substrate binding.

Part B: This part dealt with the concept of enzyme inhibition. Students were asked to explain the effects of a specific inhibitor on enzyme activity, distinguishing between competitive and non-competitive inhibition. A strong answer would correctly identify the type of inhibition and explain the mechanism by which the inhibitor affects enzyme function. This includes describing how the inhibitor interacts with the enzyme (e.g., binding to the active site or an allosteric site) and its impact on the enzyme's ability to bind to the substrate and catalyze the reaction.

Example Answer Part B: If the inhibitor binds to the active site, preventing substrate binding, it is competitive inhibition. If it binds to an allosteric site, causing a conformational change in the enzyme and reducing its activity, it is non-competitive inhibition. The data should be used to support the identification of the inhibition type, explaining how the inhibitor's presence affects the maximum reaction rate (Vmax) and the Michaelis constant (Km). For example, a competitive inhibitor will increase Km but not change Vmax, while a non-competitive inhibitor will decrease Vmax but not necessarily change Km.

Part C: This section tested the understanding of enzyme regulation. Students were asked to discuss different ways that cells regulate enzyme activity. A comprehensive response would address several mechanisms including allosteric regulation, feedback inhibition, and covalent modification (e.g., phosphorylation). Each mechanism should be explained with specific examples.

Example Answer Part C: Cells regulate enzyme activity through various mechanisms. Allosteric regulation involves binding of a molecule to a site other than the active site, inducing a conformational change that affects enzyme activity. Feedback inhibition is a type of allosteric regulation where the end product of a metabolic pathway inhibits an enzyme earlier in the pathway. Covalent modification, such as phosphorylation, involves the addition or removal of a phosphate group, which can activate or deactivate an enzyme.

Question 2: Cellular Respiration and Energy Production

This question focused on cellular respiration and the processes involved in energy production within cells.

Part A: This part dealt with the different stages of cellular respiration (glycolysis, pyruvate oxidation, Krebs cycle, and oxidative phosphorylation). Students were asked to describe the location and main products of each stage. A complete answer would accurately identify the cellular location for each process (cytoplasm for glycolysis, mitochondrial matrix for pyruvate oxidation and the Krebs cycle, and inner mitochondrial membrane for oxidative phosphorylation) and list the key products generated in each stage (ATP, NADH, FADH2, CO2).

Example Answer Part A: Glycolysis occurs in the cytoplasm and produces 2 ATP, 2 NADH, and 2 pyruvate molecules. Pyruvate oxidation takes place in the mitochondrial matrix and produces acetyl-CoA, NADH, and CO2. The Krebs cycle, also in the mitochondrial matrix, generates ATP, NADH, FADH2, and CO2. Oxidative phosphorylation, occurring across the inner mitochondrial membrane, utilizes the electron transport chain and chemiosmosis to produce a large amount of ATP.

Part B: This section tested the understanding of the role of oxygen in cellular respiration. Students needed to explain the importance of oxygen as the final electron acceptor in the electron transport chain and the consequences of its absence. A strong answer would describe how the electron transport chain functions to generate a proton gradient, which drives ATP synthesis through chemiosmosis. It would also explain that in the absence of oxygen, the electron transport chain would halt, leading to a significant reduction in ATP production and a shift towards fermentation.

Example Answer Part B: Oxygen acts as the final electron acceptor in the electron transport chain. Electrons are passed down a series of protein complexes, releasing energy used to pump protons across the inner mitochondrial membrane, creating a proton gradient. This gradient drives ATP synthesis through chemiosmosis. Without oxygen, the electron transport chain stops, resulting in a dramatic decrease in ATP production. Cells resort to anaerobic respiration (fermentation) to generate a small amount of ATP.

Part C: This part asked students to describe how ATP is generated during cellular respiration, emphasizing the role of chemiosmosis and the proton gradient. A thorough answer would describe the process of oxidative phosphorylation, including the electron transport chain and ATP synthase. It would explain how the movement of protons down their concentration gradient through ATP synthase drives the synthesis of ATP.

Example Answer Part C: The majority of ATP generated during cellular respiration comes from oxidative phosphorylation. Electrons from NADH and FADH2 are passed through the electron transport chain, releasing energy used to pump protons (H+) from the mitochondrial matrix into the intermembrane space. This creates a proton gradient across the inner mitochondrial membrane. Protons flow back into the matrix through ATP synthase, a molecular turbine, which uses the energy from this flow to synthesize ATP from ADP and inorganic phosphate (Pi). This process is called chemiosmosis.

Question 3: Plant Structure and Function

This question investigated different aspects of plant structure and function.

Part A: This section focused on the process of transpiration in plants. Students needed to explain how transpiration contributes to water movement in plants and the factors that influence the rate of transpiration. A comprehensive answer would describe the cohesion-tension theory, explaining how water molecules are pulled up through the xylem due to transpiration at the leaves. It would also discuss environmental factors that affect transpiration rates, such as temperature, humidity, wind speed, and light intensity.

Example Answer Part A: Transpiration, the evaporation of water from leaves, is a major driving force for water movement in plants. According to the cohesion-tension theory, water molecules are cohesive and adhere to the xylem walls, forming a continuous column of water. Transpiration creates tension, pulling water upwards from the roots. Higher temperatures, lower humidity, higher wind speed, and greater light intensity increase transpiration rates, as they increase the rate of evaporation from the leaves.

Part B: This part dealt with the structure and function of different plant tissues. Students were asked to describe the structural adaptations of different tissues (e.g., xylem, phloem, and mesophyll) that contribute to their functions in the plant. A successful answer would clearly describe the structure of each tissue and explain how its structure relates to its function. For example, the xylem's thick-walled cells provide structural support, while the phloem's sieve tubes facilitate sugar transport. The mesophyll cells with their numerous chloroplasts are essential for photosynthesis.

Example Answer Part B: The xylem, composed of tracheids and vessel elements with lignified walls, efficiently transports water and provides structural support. The phloem, containing sieve tubes and companion cells, transports sugars (photosynthates) throughout the plant. The mesophyll cells, packed with chloroplasts, are the primary sites of photosynthesis. The arrangement of these cells in leaves maximizes light absorption and gas exchange.

Part C: This section tested the understanding of the adaptations of plants to different environments. Students were asked to describe how specific plant adaptations contribute to their survival in arid environments (deserts). A good response would mention adaptations such as reduced leaf surface area, thick cuticles, deep root systems, CAM photosynthesis, and water storage tissues. The answer should explain how these adaptations reduce water loss and increase water uptake in water-scarce environments.

Example Answer Part C: Plants adapted to arid environments exhibit several features to minimize water loss and maximize water uptake. Reduced leaf surface area (e.g., spines instead of leaves) decreases the surface area for transpiration. Thick cuticles reduce water loss through evaporation. Deep root systems access groundwater. CAM photosynthesis minimizes water loss by opening stomata at night and fixing carbon dioxide at night. Succulent tissues store water for periods of drought.

Question 4: Animal Structure and Function

This question explored various aspects of animal structure and function.

Part A: This section focused on the process of gas exchange in animals. Students were asked to describe the mechanisms of gas exchange in the lungs of mammals. This involves explaining how oxygen is taken up and carbon dioxide is released. A thorough answer would include the processes of inhalation and exhalation, describing the role of the diaphragm and intercostal muscles in changing lung volume and pressure. It would also mention the structure of alveoli and their role in gas exchange, emphasizing the importance of a large surface area and thin respiratory membranes.

Example Answer Part A: Gas exchange in mammalian lungs occurs through inhalation and exhalation. The diaphragm contracts and flattens, and intercostal muscles contract, increasing the volume of the thoracic cavity and reducing pressure. This draws air into the lungs. Oxygen diffuses from alveoli (tiny air sacs with a large surface area and thin walls) into the capillaries surrounding them, while carbon dioxide diffuses from the capillaries into the alveoli. During exhalation, the diaphragm relaxes and the intercostal muscles relax, decreasing the volume of the thoracic cavity and increasing pressure, forcing air out of the lungs.

Part B: This part dealt with the transport of oxygen and carbon dioxide in the blood. Students were asked to describe the mechanisms involved in the transport of these gases, mentioning the role of hemoglobin. A successful response would explain that oxygen binds to hemoglobin in red blood cells, forming oxyhemoglobin, which increases oxygen-carrying capacity. Carbon dioxide is transported in several forms: dissolved in plasma, bound to hemoglobin (carbaminohemoglobin), and as bicarbonate ions (HCO3-). The reaction between CO2 and water, catalyzed by carbonic anhydrase, is crucial in the formation of bicarbonate ions.

Example Answer Part B: Oxygen is transported primarily bound to hemoglobin in red blood cells. Hemoglobin's four subunits each bind to an oxygen molecule, significantly increasing the blood's oxygen-carrying capacity. The binding of oxygen is cooperative, meaning the binding of one oxygen molecule increases the affinity for subsequent oxygen molecules. Carbon dioxide is transported in three forms: dissolved in plasma, bound to hemoglobin (carbaminohemoglobin), and as bicarbonate ions (HCO3-). Carbonic anhydrase catalyzes the reaction between CO2 and water to form carbonic acid (H2CO3), which dissociates into H+ and HCO3-. The bicarbonate ions are transported in the plasma.

Part C: This section tested the understanding of homeostatic mechanisms in the body. Students were asked to explain how the body maintains blood glucose levels. A strong answer would describe the roles of insulin and glucagon in regulating blood glucose, explaining how these hormones affect glucose uptake by cells and glycogen storage in the liver and muscles.

Example Answer Part C: Blood glucose levels are maintained through a negative feedback loop involving insulin and glucagon. When blood glucose levels rise after a meal, the pancreas releases insulin. Insulin stimulates glucose uptake by cells, particularly muscle and liver cells, and promotes glycogen synthesis (glycogenesis) in the liver and muscles, lowering blood glucose levels. When blood glucose levels fall, the pancreas releases glucagon. Glucagon stimulates glycogen breakdown (glycogenolysis) in the liver, releasing glucose into the bloodstream and raising blood glucose levels. This system keeps blood glucose within a relatively narrow range.

Conclusion

Mastering the AP Biology FRQs requires a deep understanding of core biological concepts, the ability to apply this knowledge to novel situations, and effective communication skills. By carefully reviewing these 2017 FRQs and their detailed explanations, students can gain valuable insights into the types of questions they might encounter on the exam and develop strategies for crafting thorough and well-supported answers. Remember to practice with various FRQs from previous years to further strengthen your understanding and improve your test-taking skills. Good luck!