1The First Law of Thermodynamics, when applied to biological systems, states that energy can be:
concept of thermodynamics and application in biological system
Easy
A.Completely destroyed
B.Used up entirely with no byproducts
C.Transformed from one form to another
D.Created from nothing
Correct Answer: Transformed from one form to another
Explanation:
The First Law of Thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, only converted from one form to another, such as light energy being converted to chemical energy in photosynthesis.
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2In thermodynamics, what does 'entropy' measure?
concept of thermodynamics and application in biological system
Easy
A.The total energy of a system
B.The heat released by a reaction
C.The degree of disorder or randomness in a system
D.The speed of a chemical reaction
Correct Answer: The degree of disorder or randomness in a system
Explanation:
Entropy is a fundamental concept in thermodynamics that quantifies the amount of randomness or disorder within a system. According to the Second Law of Thermodynamics, the entropy of an isolated system always tends to increase.
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3Living organisms are considered which type of thermodynamic system because they exchange both energy and matter with their surroundings?
concept of thermodynamics and application in biological system
Easy
A.Adiabatic system
B.Closed system
C.Isolated system
D.Open system
Correct Answer: Open system
Explanation:
An open system can exchange both energy (like heat and work) and matter (like food and waste) with its environment. Living organisms constantly do this to maintain their structure and function.
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4An exothermic reaction is a chemical reaction that:
exothermic and endothermic reactions
Easy
A.Releases heat into the surroundings
B.Does not involve any change in heat
C.Requires a catalyst to start
D.Absorbs heat from the surroundings
Correct Answer: Releases heat into the surroundings
Explanation:
The prefix 'exo-' means 'out'. Exothermic reactions release energy, usually in the form of heat, causing the temperature of the surroundings to rise.
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5If a chemical reaction causes its container to feel cold to the touch, the reaction is most likely:
exothermic and endothermic reactions
Easy
A.Endothermic
B.Isothermic
C.Exothermic
D.Neutral
Correct Answer: Endothermic
Explanation:
The prefix 'endo-' means 'in'. An endothermic reaction absorbs heat from its surroundings to proceed. This absorption of heat makes the container feel cold.
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6What is the sign of the change in Gibbs Free Energy () for a spontaneous, exergonic reaction?
endergonic and exergonic reactions
Easy
A.Negative ()
B.It varies and cannot be determined
C.Zero ()
D.Positive ()
Correct Answer: Negative ()
Explanation:
Exergonic reactions release free energy, meaning the products have less free energy than the reactants. This results in a negative value for the change in Gibbs Free Energy () and indicates the reaction can occur spontaneously.
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7An endergonic reaction is a reaction that:
endergonic and exergonic reactions
Easy
A.Releases energy
B.Requires a net input of energy to proceed
C.Has a negative
D.Occurs spontaneously
Correct Answer: Requires a net input of energy to proceed
Explanation:
Endergonic reactions are non-spontaneous and require an input of free energy from the surroundings to occur. They have a positive change in Gibbs Free Energy ().
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8The breakdown of ATP (adenosine triphosphate) into ADP (adenosine diphosphate) and phosphate is a classic example of which type of reaction in cells?
endergonic and exergonic reactions
Easy
A.Synthesis reaction
B.Endergonic reaction
C.Photosynthetic reaction
D.Exergonic reaction
Correct Answer: Exergonic reaction
Explanation:
The hydrolysis of ATP is a highly exergonic reaction, releasing a significant amount of free energy. Cells couple this reaction to endergonic reactions to power cellular work.
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9In which organelle does photosynthesis primarily take place in plant cells?
photosynthesis
Easy
A.Mitochondrion
B.Chloroplast
C.Ribosome
D.Nucleus
Correct Answer: Chloroplast
Explanation:
Chloroplasts are the specialized organelles within plant and algal cells that contain chlorophyll and are the site of photosynthesis.
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10What are the three main inputs (reactants) required for the process of photosynthesis?
photosynthesis
Easy
A.Nitrogen, carbon dioxide, and oxygen
B.Sugar, oxygen, and sunlight
C.Carbon dioxide, water, and sunlight
D.Oxygen, sugar, and water
Correct Answer: Carbon dioxide, water, and sunlight
Explanation:
Photosynthesis uses energy from sunlight to convert carbon dioxide () and water () into glucose (a sugar) and oxygen.
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11What is the name of the primary pigment used by plants to capture light energy for photosynthesis?
photosynthesis
Easy
A.Hemoglobin
B.Carotene
C.Melanin
D.Chlorophyll
Correct Answer: Chlorophyll
Explanation:
Chlorophyll is the green pigment found in chloroplasts that absorbs light energy, primarily in the blue and red parts of the spectrum, which drives the photosynthetic process.
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12What is a major gaseous product released during photosynthesis?
photosynthesis
Easy
A.Methane
B.Oxygen
C.Carbon dioxide
D.Nitrogen
Correct Answer: Oxygen
Explanation:
During the light-dependent reactions of photosynthesis, water molecules are split, which releases oxygen gas () as a byproduct.
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13Where in the cell does the metabolic process of glycolysis occur?
glycolysis
Easy
A.Nucleus
B.Cytoplasm (or cytosol)
C.Chloroplast
D.Mitochondrial matrix
Correct Answer: Cytoplasm (or cytosol)
Explanation:
Glycolysis is the initial stage of cellular respiration and it takes place in the cytoplasm of virtually all cells, both prokaryotic and eukaryotic.
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14Glycolysis literally means the 'splitting' of what molecule?
glycolysis
Easy
A.Glucose
B.ATP
C.Pyruvate
D.Water
Correct Answer: Glucose
Explanation:
The term glycolysis comes from 'glyco' (sugar) and 'lysis' (splitting). It is the metabolic pathway that breaks down one molecule of glucose (a 6-carbon sugar) into two molecules of pyruvate (a 3-carbon compound).
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15What is the net gain of ATP molecules produced directly from one molecule of glucose during glycolysis?
glycolysis
Easy
A.2
B.36
C.0
D.4
Correct Answer: 2
Explanation:
Although 4 ATP molecules are produced during glycolysis, 2 ATP are consumed in the initial investment phase, resulting in a net gain of 2 ATP.
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16What is the primary end product of glycolysis, which can then enter the Krebs cycle after further processing?
glycolysis
Easy
A.Pyruvate
B.Acetyl-CoA
C.Lactic acid
D.Citric acid
Correct Answer: Pyruvate
Explanation:
Glycolysis breaks a six-carbon glucose molecule into two three-carbon molecules of pyruvate. Pyruvate is a key crossroads molecule in metabolism.
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17In eukaryotic cells, where does the Krebs cycle (also known as the citric acid cycle) take place?
Krebs cycle
Easy
A.Cytoplasm
B.Chloroplast stroma
C.Outer mitochondrial membrane
D.Mitochondrial matrix
Correct Answer: Mitochondrial matrix
Explanation:
The Krebs cycle is a central part of cellular respiration and occurs in the mitochondrial matrix, the innermost compartment of the mitochondria.
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18The Krebs cycle is a key stage in which overall metabolic process?
Krebs cycle
Easy
A.DNA replication
B.Fermentation
C.Cellular Respiration
D.Photosynthesis
Correct Answer: Cellular Respiration
Explanation:
Cellular respiration is the process of breaking down glucose to produce ATP. It consists of three main stages: glycolysis, the Krebs cycle, and the electron transport chain.
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19Before entering the Krebs cycle, pyruvate is first converted into which two-carbon molecule?
Krebs cycle
Easy
A.Glucose
B.Acetyl-CoA
C.Citric acid
D.Lactic acid
Correct Answer: Acetyl-CoA
Explanation:
Pyruvate from glycolysis is transported into the mitochondria and undergoes oxidative decarboxylation to form Acetyl-CoA, which is the molecule that actually enters the Krebs cycle.
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20Which of the following are major products of the Krebs cycle that carry high-energy electrons to the electron transport chain?
Krebs cycle
Easy
A.Carbon dioxide and Water
B.NADH and FADH₂
C.Pyruvate and Oxygen
D.ATP and Glucose
Correct Answer: NADH and FADH₂
Explanation:
A primary function of the Krebs cycle is to harvest high-energy electrons from carbon compounds. These electrons are transferred to the electron carriers NAD+ and FAD to form NADH and FADH₂, which then donate the electrons to the next stage of respiration.
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21A living cell maintains a high degree of internal order and complexity. How is this consistent with the Second Law of Thermodynamics, which states that the entropy of an isolated system always increases?
concept of thermodynamics and application in biological system
Medium
A.The cell increases the entropy of its surroundings by releasing heat, making the total entropy change of the universe positive.
B.The cell converts entropy into energy to maintain order.
C.Biological processes are an exception to the Second Law of Thermodynamics.
D.The cell is a closed system, so the Second Law does not apply.
Correct Answer: The cell increases the entropy of its surroundings by releasing heat, making the total entropy change of the universe positive.
Explanation:
Cells are open systems. To maintain their internal order (low entropy), they take in energy and release waste products and heat into the environment. This release of heat increases the entropy of the surroundings more than the cell's internal entropy decreases, ensuring the total entropy of the universe increases, thus satisfying the Second Law.
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22The hydrolysis of ATP to ADP and inorganic phosphate (Pi) has a of approximately -30.5 kJ/mol. A cell needs to synthesize molecule X from molecule Y, which is an endergonic reaction with a of +15.0 kJ/mol. How can the cell accomplish this?
endergonic and exergonic reactions
Medium
A.By coupling the synthesis of X to another endergonic reaction.
B.By increasing the concentration of molecule Y to shift the equilibrium.
C.By coupling the synthesis of X to the hydrolysis of ATP, resulting in a net negative .
D.The synthesis of X will proceed spontaneously, but very slowly.
Correct Answer: By coupling the synthesis of X to the hydrolysis of ATP, resulting in a net negative .
Explanation:
By coupling the endergonic synthesis of X (+15.0 kJ/mol) with the exergonic hydrolysis of ATP (-30.5 kJ/mol), the overall coupled reaction has a net of -15.5 kJ/mol. A negative indicates the overall process is spontaneous and can proceed.
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23If a plant's chloroplasts are treated with a chemical that specifically inhibits the function of Photosystem I (PSI) but not Photosystem II (PSII), which of the following would be the most immediate consequence?
photosynthesis
Medium
A.ATP synthesis could continue to some extent, but NADPH production would halt.
B.ATP synthesis would stop, but NADPH production would continue.
C.Oxygen production would cease immediately.
D.Carbon fixation in the Calvin cycle would be unaffected.
Correct Answer: ATP synthesis could continue to some extent, but NADPH production would halt.
Explanation:
In the light-dependent reactions, Photosystem II (PSII) contributes to the proton gradient for ATP synthesis, while Photosystem I (PSI) is primarily responsible for reducing NADP+ to NADPH. If PSI is inhibited, the final step of linear electron flow is blocked, halting NADPH production. However, electrons from PSII can still contribute to the proton gradient, allowing some ATP synthesis to continue.
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24The 'investment phase' of glycolysis involves the consumption of ATP. What is the primary purpose of phosphorylating glucose and fructose-6-phosphate during these initial steps?
glycolysis
Medium
A.To immediately produce NADH for the electron transport chain.
B.To trap the sugar molecules inside the cell and increase their reactivity for subsequent cleavage.
C.To directly generate a proton gradient for chemiosmosis.
D.To signal the cell that glucose levels are high.
Correct Answer: To trap the sugar molecules inside the cell and increase their reactivity for subsequent cleavage.
Explanation:
Adding phosphate groups makes the sugars negatively charged, preventing them from diffusing back across the nonpolar cell membrane. This phosphorylation also 'activates' the sugars, making them less stable and more reactive for the later steps of glycolysis, such as the cleavage of fructose-1,6-bisphosphate.
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25What is the primary role of oxaloacetate in the Krebs cycle?
Krebs cycle
Medium
A.It is the final product of the cycle that exits to form glucose.
B.It combines with acetyl-CoA to form citrate and is regenerated at the end of the cycle.
C.It is a high-energy electron carrier similar to NADH.
D.It is the source of carbon atoms that are released as CO2.
Correct Answer: It combines with acetyl-CoA to form citrate and is regenerated at the end of the cycle.
Explanation:
The Krebs cycle begins when the 2-carbon acetyl-CoA combines with the 4-carbon molecule oxaloacetate to form the 6-carbon molecule citrate. Through a series of reactions, two carbons are lost as CO2, and oxaloacetate is regenerated, allowing the cycle to continue with the next molecule of acetyl-CoA.
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26A particular biochemical reaction is endothermic () but proceeds spontaneously within a cell at body temperature. According to the Gibbs free energy equation (), what must be true for this to occur?
concept of thermodynamics and application in biological system
Medium
A.The reaction must be coupled to ATP hydrolysis.
B.The reaction must cause a large increase in entropy () such that the term overcomes the positive .
C.The temperature (T) must be very low.
D.The reaction must result in a significant decrease in entropy ().
Correct Answer: The reaction must cause a large increase in entropy () such that the term overcomes the positive .
Explanation:
For a reaction to be spontaneous, its Gibbs free energy change () must be negative. In the equation , if the reaction is endothermic ( is positive), the only way for to be negative is if the entropy change () is positive and large enough, so that the term is a large negative number that outweighs the positive .
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27The combustion of glucose in a bomb calorimeter is a highly exothermic reaction. In a cell, aerobic respiration is also exothermic. How does the cell harness the energy from this process without being damaged by a massive, single release of heat?
exothermic and endothermic reactions
Medium
A.The cell releases the energy in a series of small, controlled steps, capturing it in chemical bonds of ATP.
B.The cell uses enzymes to make the reaction endothermic instead of exothermic.
C.The cell converts all the heat energy directly into useful work.
D.The cell is an isothermal system, so no heat is actually released.
Correct Answer: The cell releases the energy in a series of small, controlled steps, capturing it in chemical bonds of ATP.
Explanation:
Cellular respiration breaks down glucose through a long series of enzyme-catalyzed reactions. This stepwise oxidation allows the energy to be released gradually. A significant portion of this energy is captured in the high-energy phosphate bonds of ATP molecules, with the rest released as heat, which prevents cell damage.
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28Using an isotope of oxygen, O, scientists have traced the path of atoms in photosynthesis. If a plant is supplied with water (HO) and normal carbon dioxide (CO), where would you expect to find the O isotope at the end of the process?
photosynthesis
Medium
A.In the ATP molecules synthesized.
B.In the molecular oxygen (O) released as a byproduct.
C.In the glucose (CHO) produced.
D.In the carbon dioxide released during photorespiration.
Correct Answer: In the molecular oxygen (O) released as a byproduct.
Explanation:
During the light-dependent reactions, water molecules are split in a process called photolysis to provide electrons for the electron transport chain. This process releases electrons, protons (H+), and molecular oxygen. Therefore, the oxygen atoms from the water molecule (HO) are the ones that form the gaseous oxygen (O) released.
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29The enzyme phosphofructokinase-1 (PFK-1) is a key regulator of glycolysis. It is allosterically inhibited by high levels of a specific molecule. Which of the following molecules is most likely to act as an inhibitor of PFK-1?
glycolysis
Medium
A.Glucose-6-phosphate
B.ATP
C.ADP
D.Pyruvate
Correct Answer: ATP
Explanation:
PFK-1 catalyzes an early, committed step in glycolysis. High levels of ATP signal that the cell has an abundance of energy. To prevent the wasteful breakdown of glucose when energy is not needed, ATP binds to an allosteric site on PFK-1, inhibiting its activity and slowing down the entire glycolytic pathway. Conversely, high levels of ADP or AMP would activate it.
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30Starting from one molecule of glucose, how many molecules of FADH are produced by the Krebs cycle?
Krebs cycle
Medium
A.6
B.4
C.1
D.2
Correct Answer: 2
Explanation:
Glycolysis breaks one molecule of glucose (6 carbons) into two molecules of pyruvate (3 carbons). Each pyruvate is converted to one molecule of acetyl-CoA, which then enters the Krebs cycle. Therefore, one molecule of glucose leads to two 'turns' of the Krebs cycle. Since each turn produces one molecule of FADH, two turns produce two molecules of FADH.
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31A reversible reaction A <=> B has a standard free energy change () of +5.7 kJ/mol. What does this indicate about the reaction under standard conditions?
endergonic and exergonic reactions
Medium
A.The reaction is exergonic and the concentration of B will be greater than A at equilibrium.
B.The reaction is endergonic and the concentration of A will be greater than B at equilibrium.
C.The reaction is at equilibrium, with equal concentrations of A and B.
D.The reaction will proceed rapidly toward B.
Correct Answer: The reaction is endergonic and the concentration of A will be greater than B at equilibrium.
Explanation:
A positive signifies an endergonic reaction. The equilibrium constant K_eq is related to by . A positive means K_eq is less than 1. Since K_eq = [B]/[A], a K_eq < 1 indicates that at equilibrium, the concentration of the reactant [A] will be greater than the concentration of the product [B].
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32C4 plants like corn are more efficient at photosynthesis in hot, dry climates than C3 plants like rice. What is the key biochemical adaptation in C4 plants that accounts for this difference?
photosynthesis
Medium
A.They use a different pigment instead of chlorophyll to capture light.
B.Their light-dependent reactions produce more ATP and NADPH per photon.
C.They perform the Calvin cycle only at night to prevent water loss.
D.They fix CO into a 4-carbon compound to concentrate it for the Calvin cycle, minimizing photorespiration.
Correct Answer: They fix CO into a 4-carbon compound to concentrate it for the Calvin cycle, minimizing photorespiration.
Explanation:
In hot, dry conditions, plants close their stomata, which reduces CO intake and favors photorespiration in C3 plants. C4 plants have a spatial separation of processes. They initially fix CO into a 4-carbon acid in mesophyll cells, then transport this acid to bundle-sheath cells where CO is released at a high concentration, favoring the Calvin cycle over photorespiration.
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33The Krebs cycle is considered amphibolic, meaning it has both catabolic and anabolic roles. Which of the following is an example of an anabolic function of the Krebs cycle?
Krebs cycle
Medium
A.The complete oxidation of acetyl-CoA to CO.
B.The reduction of NAD and FAD to NADH and FADH.
C.The withdrawal of intermediates like -ketoglutarate and oxaloacetate for amino acid synthesis.
D.The production of ATP via substrate-level phosphorylation.
Correct Answer: The withdrawal of intermediates like -ketoglutarate and oxaloacetate for amino acid synthesis.
Explanation:
While its primary role is catabolic (breaking down acetyl-CoA), the Krebs cycle's intermediates are also starting points for biosynthetic (anabolic) pathways. For example, -ketoglutarate can be converted to the amino acid glutamate, and oxaloacetate can be converted to aspartate. This dual function makes the cycle amphibolic.
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34In the absence of oxygen, muscle cells undergo lactic acid fermentation. What is the primary purpose of converting pyruvate to lactate in this process?
glycolysis
Medium
A.To regenerate NAD from NADH, allowing glycolysis to continue.
B.To produce additional ATP.
C.To lower the pH of the cell, which activates glycolytic enzymes.
D.To directly shuttle lactate into the Krebs cycle.
Correct Answer: To regenerate NAD from NADH, allowing glycolysis to continue.
Explanation:
Glycolysis requires a constant supply of NAD to act as an electron acceptor. During anaerobic fermentation, the electron transport chain is unavailable to re-oxidize NADH. The reduction of pyruvate to lactate oxidizes NADH back to NAD, providing the necessary coenzyme for glycolysis to continue producing a small amount of ATP.
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35Enzymes are biological catalysts that dramatically increase the rate of biochemical reactions. How do they achieve this from a thermodynamic perspective?
concept of thermodynamics and application in biological system
Medium
A.They decrease the overall Gibbs free energy change () of the reaction, making it more exergonic.
B.They lower the activation energy () of the reaction without changing the overall .
C.They change the equilibrium constant (K_eq) to favor the products.
D.They increase the overall Gibbs free energy change () of the reaction, providing more energy.
Correct Answer: They lower the activation energy () of the reaction without changing the overall .
Explanation:
Enzymes do not alter the thermodynamics (the initial and final energy states) of a reaction; the overall remains the same. Instead, they provide an alternative reaction pathway with a lower activation energy barrier. By lowering the energy required to reach the transition state, they significantly speed up the rate at which the reaction reaches equilibrium.
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36A spontaneous reaction occurs in a test tube, and you observe that the test tube becomes cold. Which of the following correctly describes this reaction?
exergonic and endergonic reactions
Medium
A.Exothermic and Exergonic
B.Endothermic and Exergonic
C.Exothermic and Endergonic
D.Endothermic and Endergonic
Correct Answer: Endothermic and Exergonic
Explanation:
'Spontaneous' means the reaction is exergonic (). The observation that the test tube gets cold means the reaction is absorbing heat from its surroundings, which is the definition of an endothermic reaction (). This is possible if the reaction causes a large increase in entropy (), making the term in the Gibbs equation large and negative enough to overcome the positive .
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37During glycolysis, ATP is produced by substrate-level phosphorylation. Which of the following statements best describes this process?
glycolysis
Medium
A.ATP synthase uses rotational energy to bind ADP and Pi.
B.Light energy is used to phosphorylate ADP.
C.ATP is generated by the movement of protons across a membrane down a concentration gradient.
D.A phosphate group is transferred from a high-energy intermediate substrate directly to ADP.
Correct Answer: A phosphate group is transferred from a high-energy intermediate substrate directly to ADP.
Explanation:
Substrate-level phosphorylation is a direct method of ATP synthesis where an enzyme transfers a phosphate group from a phosphorylated substrate molecule (like 1,3-bisphosphoglycerate or phosphoenolpyruvate) directly to an ADP molecule. This is distinct from oxidative phosphorylation, which involves a proton gradient and ATP synthase.
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38What is the critical linking step that connects the products of glycolysis with the Krebs cycle, and where does it occur in a eukaryotic cell?
Krebs cycle
Medium
A.The conversion of citrate to isocitrate in the mitochondrial matrix.
B.The conversion of pyruvate to lactate in the cytoplasm.
C.The conversion of pyruvate to acetyl-CoA in the mitochondrial matrix.
D.The conversion of glucose to pyruvate in the cytoplasm.
Correct Answer: The conversion of pyruvate to acetyl-CoA in the mitochondrial matrix.
Explanation:
Pyruvate, the end product of glycolysis, is transported into the mitochondrial matrix. There, the enzyme complex pyruvate dehydrogenase catalyzes its conversion into acetyl-CoA. This reaction also produces NADH and CO. Acetyl-CoA is the molecule that then enters the Krebs cycle.
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39The primary function of the Calvin cycle in photosynthesis is to:
photosynthesis
Medium
A.Produce ATP and NADPH for use in the light-dependent reactions.
B.Absorb light energy and convert it into chemical energy.
C.Use ATP and NADPH to synthesize sugar (G3P) from carbon dioxide.
D.Split water molecules to release oxygen and electrons.
Correct Answer: Use ATP and NADPH to synthesize sugar (G3P) from carbon dioxide.
Explanation:
The Calvin cycle constitutes the 'synthesis' part of photosynthesis. It uses the chemical energy (ATP) and reducing power (NADPH) produced during the light-dependent reactions to fix atmospheric carbon dioxide into organic molecules, ultimately producing glyceraldehyde-3-phosphate (G3P), a 3-carbon sugar.
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40Why are many metabolic pathways, like glycolysis, composed of numerous small steps, with several reactions being highly exergonic (large negative )?
endergonic and exergonic reactions
Medium
A.To ensure each step is reversible, allowing the pathway to run backwards easily.
B.To minimize the number of enzymes required for the entire process.
C.These highly exergonic steps act as 'control points' and make the overall pathway irreversible, giving it a clear direction.
D.To make the overall pathway endergonic and store energy.
Correct Answer: These highly exergonic steps act as 'control points' and make the overall pathway irreversible, giving it a clear direction.
Explanation:
While many steps in a pathway may be near equilibrium, a few key reactions have a large negative . These steps are essentially irreversible under cellular conditions. They act as regulatory points and ensure that the entire pathway proceeds in one direction, preventing a futile cycle.
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41A living organism is an open system that maintains a highly ordered state, which appears to contradict the Second Law of Thermodynamics (entropy tends to increase). How is this local decrease in entropy primarily achieved and sustained?
concept of thermodynamics and application in biological system
Hard
A.By reaching a state of thermodynamic equilibrium with the environment, minimizing energy flux.
B.By coupling internal ordering processes to external reactions that increase the total entropy of the universe (organism + environment) by an even greater amount.
C.By utilizing enzymes to make energetically unfavorable reactions favorable, thereby violating the Second Law locally.
D.By creating energy from nutrients to directly reverse the entropy increase within the system.
Correct Answer: By coupling internal ordering processes to external reactions that increase the total entropy of the universe (organism + environment) by an even greater amount.
Explanation:
Living systems are not at equilibrium and do not violate the Second Law. They maintain order by taking in energy-rich molecules (like glucose) and breaking them down, releasing heat and simple, low-energy molecules (like CO₂ and H₂O) into the environment. This process drastically increases the entropy of the surroundings, more than compensating for the decrease in entropy within the organism itself. Thus, the total entropy of the universe increases.
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42A metabolic reaction has a change in enthalpy () of +20 kJ/mol and a change in entropy () of +100 J/mol·K. At what temperature will this reaction transition from being non-spontaneous (endergonic) to spontaneous (exergonic)?
endergonic and exergonic reactions
Hard
A.Above 200 K
B.The reaction is always exergonic regardless of temperature.
C.The reaction is always endergonic regardless of temperature.
D.Below 200 K
Correct Answer: Above 200 K
Explanation:
A reaction becomes spontaneous when the Gibbs free energy () is negative. The equation is . For to be zero (the transition point), . First, convert units to be consistent: . Then, solve for T: . Since the entropy term () is negative, increasing the temperature above 200 K will make increasingly negative, driving the reaction to be spontaneous.
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43If the methyl carbon () of an acetyl-CoA molecule is radiolabeled with C and enters the Krebs cycle, after one complete turn of the cycle, where will the radiolabel primarily reside?
Krebs cycle
Hard
A.Released as CO₂
B.In the succinate molecule, which is symmetric.
C.In the oxaloacetate molecule, distributed between the C2 and C3 positions.
D.In the -ketoglutarate molecule.
Correct Answer: In the oxaloacetate molecule, distributed between the C2 and C3 positions.
Explanation:
The methyl carbon of acetyl-CoA becomes C4 of citrate. Through the initial steps, it becomes the C4 of isocitrate and then the C3 of -ketoglutarate. The two carbons lost as CO₂ in the first turn originate from the oxaloacetate that started the cycle. After conversion to the symmetric succinate molecule, the label is randomized. By the time oxaloacetate is regenerated, the C label is distributed within its carbon skeleton, not released as CO₂ until subsequent turns.
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44The standard free energy change () for the aldolase reaction in glycolysis (Fructose-1,6-bisphosphate DHAP + G3P) is highly positive (+23.8 kJ/mol). How does the cell ensure this reaction proceeds in the forward direction?
glycolysis
Hard
A.Aldolase is allosterically activated by high concentrations of ATP.
B.The rapid consumption of the products (DHAP and G3P) by subsequent enzymes keeps their concentrations extremely low, pulling the reaction forward.
C.The high concentration of the substrate (Fructose-1,6-bisphosphate) is sufficient to overcome the large positive .
D.The reaction is coupled directly to ATP hydrolysis.
Correct Answer: The rapid consumption of the products (DHAP and G3P) by subsequent enzymes keeps their concentrations extremely low, pulling the reaction forward.
Explanation:
This is a classic example of Le Châtelier's principle in biochemistry. While the standard free energy change () is positive, the actual free energy change () inside the cell is near zero or slightly negative. This is because the next enzymes in the pathway, triose phosphate isomerase and G3P dehydrogenase, immediately use up the products (DHAP and G3P). This keeps the product concentration far below equilibrium, which, according to the equation , makes the actual favorable for the forward reaction.
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45A C4 plant and a C3 plant are grown in a sealed chamber with a low CO₂ concentration and a high O₂ concentration under bright light. Which of the following outcomes is most likely?
photosynthesis
Hard
A.The C4 plant will have a significantly higher net photosynthetic rate due to the suppression of photorespiration.
B.The C3 plant will have a higher photosynthetic rate because it does not expend extra ATP to concentrate CO₂.
C.Both plants will have nearly identical photosynthetic rates as light is the limiting factor.
D.Both plants will switch entirely to cyclic photophosphorylation and cease carbon fixation.
Correct Answer: The C4 plant will have a significantly higher net photosynthetic rate due to the suppression of photorespiration.
Explanation:
The conditions described (low CO₂, high O₂) strongly favor photorespiration in C3 plants, where RuBisCO inefficiently binds O₂ instead of CO₂. C4 plants have a mechanism (the C4 pathway using PEP carboxylase) to concentrate CO₂ in bundle sheath cells where the Calvin cycle occurs. This effectively outcompetes O₂ for RuBisCO's active site, suppressing photorespiration and allowing the C4 plant to maintain a much higher rate of net photosynthesis under these specific stress conditions.
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46A cell is treated with a drug that specifically inhibits isocitrate dehydrogenase. Which of the following metabolic shifts would be the most immediate and pronounced consequence?
Krebs cycle
Hard
A.Depletion of all Krebs cycle intermediates and a halt in glycolysis.
B.Accumulation of citrate and isocitrate, and a decrease in the cellular NADH/NAD⁺ ratio.
C.A rapid shift to anaerobic fermentation, even in the presence of oxygen.
D.Increased production of ATP via substrate-level phosphorylation within the Krebs cycle.
Correct Answer: Accumulation of citrate and isocitrate, and a decrease in the cellular NADH/NAD⁺ ratio.
Explanation:
Inhibiting isocitrate dehydrogenase creates a bottleneck in the Krebs cycle. The substrates for this enzyme, isocitrate and its precursor citrate, will accumulate. Since this is a major NAD⁺-reducing step, its inhibition will lead to less NADH being produced by the cycle. This will decrease the overall cellular NADH/NAD⁺ ratio, signaling a lower energy state to other pathways.
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47ATP hydrolysis has a of approximately -30.5 kJ/mol, but the actual free energy change () in a cell can be as high as -50 kJ/mol. What is the primary reason for this discrepancy?
concept of thermodynamics and application in biological system
Hard
A.The concentrations of ATP, ADP, and inorganic phosphate (Pi) in the cell are far from the standard state concentration of 1 M.
B.The temperature inside a cell is significantly higher than the standard state temperature (298 K).
C.Enzymes like ATPases alter the equilibrium constant of the reaction, making it more exergonic.
D.The standard free energy value does not account for the energy released as heat.
Correct Answer: The concentrations of ATP, ADP, and inorganic phosphate (Pi) in the cell are far from the standard state concentration of 1 M.
Explanation:
The actual free energy change () is dependent on the reaction quotient, Q, as shown by the equation . In a living cell, the concentration of ATP is kept very high relative to the concentrations of its products, ADP and Pi. This makes the ratio Q = [ADP][Pi]/[ATP] very small, causing the term to be large and negative, thus making the actual much more negative (more exergonic) than the standard .
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48An inhibitor, dinitrophenol (DNP), can dissipate the proton gradient across the thylakoid membrane. If DNP is added to functional chloroplasts undergoing photosynthesis, what would be the effect on ATP synthesis and NADPH production?
photosynthesis
Hard
A.ATP synthesis would halt, but NADPH production would continue for some time.
B.ATP synthesis would increase, and NADPH production would decrease.
C.NADPH production would halt, but ATP synthesis would continue for some time.
D.Both ATP synthesis and NADPH production would halt immediately.
Correct Answer: ATP synthesis would halt, but NADPH production would continue for some time.
Explanation:
ATP synthesis in chloroplasts is driven by chemiosmosis, which requires a proton gradient across the thylakoid membrane. DNP eliminates this gradient, so ATP synthase can no longer function, halting ATP production. However, NADPH production is the result of the linear electron transport chain where electrons from Photosystem I are passed to NADP⁺. This process is not directly dependent on the proton gradient (though the gradient is formed by it), so it can continue as long as there is light and an electron source.
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49Arsenate () is a chemical analog of phosphate () and can substitute for it in the reaction catalyzed by glyceraldehyde-3-phosphate dehydrogenase. However, the resulting 1-arseno-3-phosphoglycerate is unstable and immediately hydrolyzes. What is the net ATP yield from glycolysis per molecule of glucose in the presence of arsenate?
glycolysis
Hard
A.1 ATP
B.0 ATP
C.2 ATP
D.-2 ATP
Correct Answer: 0 ATP
Explanation:
In normal glycolysis, the reaction catalyzed by glyceraldehyde-3-phosphate dehydrogenase is followed by the phosphoglycerate kinase step, which generates 2 ATP (one for each 3-carbon unit) via substrate-level phosphorylation. Arsenate uncouples these two steps. By forming an unstable arseno-intermediate that hydrolyzes, it bypasses the substrate-level phosphorylation step. Therefore, the 2 ATP that would have been generated at this stage are lost. Since 2 ATP are invested at the beginning of glycolysis and the final substrate-level phosphorylation step (pyruvate kinase) still produces 2 ATP, the net yield becomes 2 - 2 = 0 ATP.
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50The dissolution of ammonium nitrate () in water is a highly endothermic process (), yet it occurs spontaneously, making it useful in cold packs. What thermodynamic property must be true for this spontaneous, endothermic process?
exothermic and endothermic reactions
Hard
A.The process must be coupled to an exothermic reaction.
B.The change in entropy () must be large and positive.
C.The change in entropy () must be negative.
D.The change in Gibbs free energy () must be positive.
Correct Answer: The change in entropy () must be large and positive.
Explanation:
For a process to be spontaneous, the change in Gibbs free energy () must be negative. The relationship is . If the process is endothermic ( is positive), the only way for to be negative is if the entropy term, , is large and negative. This requires that the change in entropy, , must be large and positive. In this case, the dissolution of the solid crystal into ions in solution represents a massive increase in disorder (entropy), which overcomes the positive enthalpy change.
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51Anaplerotic reactions replenish Krebs cycle intermediates. If a cell is undergoing rapid synthesis of amino acids such as glutamate and glutamine, which enzyme would be most critical for maintaining the function of the Krebs cycle?
Krebs cycle
Hard
A.PEP carboxykinase, which converts oxaloacetate to phosphoenolpyruvate.
B.Succinyl-CoA synthetase, which converts succinyl-CoA to succinate.
C.Malic enzyme, which converts malate to pyruvate.
D.Pyruvate carboxylase, which converts pyruvate to oxaloacetate.
Correct Answer: Pyruvate carboxylase, which converts pyruvate to oxaloacetate.
Explanation:
Glutamate and glutamine are synthesized from the Krebs cycle intermediate -ketoglutarate. If the cell is rapidly siphoning off -ketoglutarate for amino acid synthesis, the cycle will be depleted of its intermediates. Pyruvate carboxylase catalyzes the most important anaplerotic reaction, converting pyruvate directly into oxaloacetate. This replenishes the cycle's carbon skeleton, allowing it to continue functioning despite the removal of intermediates for biosynthesis.
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52Phosphofructokinase-1 (PFK-1) is a key regulatory enzyme in glycolysis. A cancer cell line exhibits a high glycolytic rate (the Warburg effect) even in the presence of abundant oxygen and high ATP levels. This phenotype could be explained by a mutation in PFK-1 that...
glycolysis
Hard
A.eliminates its binding site for its activator, AMP.
B.increases its affinity for its substrate, fructose-6-phosphate.
C.makes it highly sensitive to allosteric inhibition by ATP.
D.decreases its affinity for its allosteric inhibitor, citrate.
Correct Answer: decreases its affinity for its allosteric inhibitor, citrate.
Explanation:
Normally, high levels of ATP and citrate (which signals that the Krebs cycle is saturated) allosterically inhibit PFK-1, slowing down glycolysis. In cancer cells exhibiting the Warburg effect, glycolysis runs unchecked. A mutation that makes PFK-1 insensitive to its inhibitors (like ATP or citrate) would cause the enzyme to remain active even when cellular energy is high, leading to a constitutively high glycolytic rate. Decreasing its affinity for citrate would remove a key negative feedback signal.
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53In cyclic photophosphorylation, electrons are shunted from ferredoxin back to the cytochrome b6f complex instead of being used to reduce NADP⁺. What is the primary purpose and outcome of this alternative pathway?
photosynthesis
Hard
A.To protect the photosystems from photodamage in low-light conditions.
B.To exclusively produce NADPH when cellular ATP levels are already high.
C.To produce O₂ without the need for water splitting at Photosystem II.
D.To produce additional ATP without producing NADPH, balancing the ATP/NADPH ratio needed for the Calvin cycle.
Correct Answer: To produce additional ATP without producing NADPH, balancing the ATP/NADPH ratio needed for the Calvin cycle.
Explanation:
The Calvin cycle requires more ATP (3 per CO₂) than NADPH (2 per CO₂). Linear electron flow produces ATP and NADPH in roughly equal amounts. Cyclic photophosphorylation provides a mechanism to produce only ATP. By cycling electrons from PSI back through the cytochrome complex, more protons are pumped into the thylakoid lumen, driving ATP synthase without involving PSII or producing NADPH. This allows the cell to adjust the ATP/NADPH output to meet the specific demands of carbon fixation.
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54A metabolic pathway consists of three sequential, enzyme-catalyzed reactions: A B ( kJ/mol), B C ( kJ/mol), and C D ( kJ/mol). Which statement accurately describes this pathway from a thermodynamic standpoint?
concept of thermodynamics and application in biological system
Hard
A.The overall pathway is endergonic and requires continuous ATP input at every step to proceed.
B.The overall pathway is exergonic, and the conversion of B to C can drive the thermodynamically unfavorable steps.
C.The pathway cannot proceed spontaneously because the first step is highly endergonic.
D.The pathway is at equilibrium, with no net flux from A to D.
Correct Answer: The overall pathway is exergonic, and the conversion of B to C can drive the thermodynamically unfavorable steps.
Explanation:
The overall standard free energy change for a pathway is the sum of the changes for the individual steps: kJ/mol. Since the total is negative, the overall pathway is exergonic and spontaneous under standard conditions. The highly exergonic second step (B C) is thermodynamically 'coupled' to the other steps by keeping the concentration of its reactant (B) low and its product (C) high relative to their equilibrium values, effectively pulling the first reaction forward and pushing the third one.
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55Succinate dehydrogenase is unique among the enzymes of the Krebs cycle because it is also part of the electron transport chain (Complex II). What is a direct consequence of this dual role?
Krebs cycle
Hard
A.It is the primary site of regulation for the entire Krebs cycle, responding directly to the cellular ATP/ADP ratio.
B.The oxidation of succinate to fumarate yields FADH₂ instead of NADH, resulting in fewer ATP molecules being produced per molecule of succinate oxidized.
C.It is the only step in the Krebs cycle that directly consumes ATP.
D.It directly pumps protons across the inner mitochondrial membrane, contributing to the proton-motive force.
Correct Answer: The oxidation of succinate to fumarate yields FADH₂ instead of NADH, resulting in fewer ATP molecules being produced per molecule of succinate oxidized.
Explanation:
The standard reduction potential of the FAD/FADH₂ couple is higher than that of NAD⁺/NADH. The energy released by oxidizing succinate is not sufficient to reduce NAD⁺. Instead, FAD is used as the electron acceptor. Since FADH₂ donates its electrons to Complex II, bypassing Complex I (the first proton pump), its electrons contribute less to the proton gradient than electrons from NADH. Consequently, FADH₂ oxidation yields approximately 1.5 ATP, whereas NADH yields about 2.5 ATP.
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56In the absence of oxygen, human muscle cells convert pyruvate to lactate. What is the primary thermodynamic and kinetic reason for this reaction?
glycolysis
Hard
A.To regenerate NAD⁺ from NADH, which is essential for the glyceraldehyde-3-phosphate dehydrogenase step to continue, thus allowing continued ATP production via glycolysis.
B.To lower the intracellular pH, which activates key glycolytic enzymes.
C.To directly shuttle lactate into the Krebs cycle as an alternative fuel.
D.To produce additional ATP through the conversion of pyruvate to lactate.
Correct Answer: To regenerate NAD⁺ from NADH, which is essential for the glyceraldehyde-3-phosphate dehydrogenase step to continue, thus allowing continued ATP production via glycolysis.
Explanation:
Glycolysis requires a continuous supply of NAD⁺ to act as an electron acceptor in the glyceraldehyde-3-phosphate dehydrogenase reaction. In aerobic conditions, NADH is re-oxidized to NAD⁺ by the electron transport chain. In anaerobic conditions, this is not possible. Lactic acid fermentation solves this problem by using NADH to reduce pyruvate to lactate, thereby regenerating the NAD⁺ needed to keep glycolysis running and producing a small but rapid supply of ATP.
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57Reaction coupling is a fundamental principle in metabolism. The synthesis of glutamine from glutamate and ammonia is endergonic ( kJ/mol). How does the cell make this reaction proceed?
endergonic and exergonic reactions
Hard
A.By coupling it to the hydrolysis of one molecule of ATP to ADP, making the net reaction exergonic.
B.By performing the reaction at a very high temperature to overcome the energy barrier.
C.By increasing the concentration of glutamate to extremely high levels, forcing the reaction forward.
D.By using an enzyme that changes the reaction's equilibrium constant to favor glutamine.
Correct Answer: By coupling it to the hydrolysis of one molecule of ATP to ADP, making the net reaction exergonic.
Explanation:
The cell couples the endergonic glutamine synthesis to the highly exergonic hydrolysis of ATP ( to -50 kJ/mol). The overall coupled reaction proceeds in two steps, often involving a phosphorylated intermediate. The large negative free energy change from ATP hydrolysis is harnessed to drive the unfavorable synthesis reaction, resulting in a net negative for the combined process, making it spontaneous.
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58The primary carboxylation reaction in C3 photosynthesis is catalyzed by RuBisCO. This enzyme also has a competing oxygenase activity. Which statement most accurately analyzes the thermodynamics of these two competing reactions?
photosynthesis
Hard
A.Both carboxylation and oxygenation are energetically neutral, with the direction determined solely by the relative concentrations of CO₂ and O₂.
B.Carboxylation is endergonic and requires direct coupling to ATP hydrolysis at the active site of RuBisCO.
C.The carboxylation reaction is highly exergonic and essentially irreversible, while the oxygenase reaction (photorespiration) is energetically costly, consuming ATP and reducing power.
D.Oxygenation is more exergonic than carboxylation, which is why it is the preferred reaction at high temperatures.
Correct Answer: The carboxylation reaction is highly exergonic and essentially irreversible, while the oxygenase reaction (photorespiration) is energetically costly, consuming ATP and reducing power.
Explanation:
The addition of CO₂ to RuBP is a highly exergonic step ( kJ/mol), which commits the carbon to the Calvin cycle. In contrast, the oxygenase activity initiates the photorespiratory pathway, a salvage pathway that consumes O₂ and produces CO₂. This salvage pathway requires the expenditure of both ATP and NADPH to convert the products back into a usable form (3-PGA), representing a net energy loss for the cell.
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59A patient is diagnosed with a severe deficiency in the E3 component (dihydrolipoyl dehydrogenase) of the pyruvate dehydrogenase complex (PDC). This enzyme is also a component of the -ketoglutarate dehydrogenase complex. Which fuel source could their cells still metabolize most effectively to produce ATP?
Krebs cycle
Hard
A.Lactate
B.Fatty acids
C.Alanine
D.Glucose
Correct Answer: Fatty acids
Explanation:
A defective E3 component cripples both the conversion of pyruvate to acetyl-CoA and the conversion of -ketoglutarate to succinyl-CoA. This effectively shuts down ATP production from glucose, lactate, and amino acids like alanine (which convert to pyruvate). However, fatty acid beta-oxidation produces acetyl-CoA directly, which can enter the Krebs cycle. While the cycle itself is impaired at the -ketoglutarate step, the acetyl-CoA can still be oxidized to citrate and isocitrate, generating some NADH. More importantly, the FADH₂ and NADH generated during beta-oxidation itself can feed into the electron transport chain, providing a significant source of ATP independent of the defective enzymes.
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60A biological system at a non-equilibrium steady state has a continuous flux of matter and energy. How does the rate of entropy production () in such a system compare to a system at equilibrium?
concept of thermodynamics and application in biological system
Hard
A. and is minimized for the steady state, whereas at equilibrium, .
B. and is maximized for the steady state, whereas at equilibrium, is minimized but still positive.
C. for the steady state (as it maintains order), whereas at equilibrium, .
D. is constant and equal for both the steady state and equilibrium.
Correct Answer: and is minimized for the steady state, whereas at equilibrium, .
Explanation:
A system at equilibrium has no net fluxes and thus produces no entropy (). A living system, being in a non-equilibrium steady state, constantly undergoes irreversible processes (like metabolism) and thus has a positive rate of entropy production (). According to the principles of non-equilibrium thermodynamics (as proposed by Prigogine), for stable steady states near equilibrium, the system will arrange itself to have the minimum possible rate of entropy production for the given constraints.