Biochemistry Quiz: Thermodynamics In Biochemistry

What this quiz covers

This quiz focuses on Thermodynamics In Biochemistry, giving you a quick way to practice the rules, question types, and explanations that matter most for Biochemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The reaction has a very high activation energy, which prevents it from proceeding at a measurable rate without a catalyst. (correct answer)
2. The large negative ΔG°' indicates that the reverse reaction is favored, preventing the net breakdown of glucose.
3. The process has a large positive entropy change, which slows down the reaction kinetics under standard conditions.
4. The reaction is actually at equilibrium under these conditions, so there is no net change in the amount of glucose.

Explanation: This question highlights the crucial distinction between thermodynamics (spontaneity, ΔG) and kinetics (reaction rate). A large negative ΔG indicates that a reaction is thermodynamically favorable, but it provides no information about how fast the reaction will occur. The rate is determined by the activation energy (Ea). The oxidation of glucose has a very high activation energy barrier, meaning it is kinetically stable. Enzymes in biological systems provide an alternative reaction pathway with a lower activation energy, allowing the reaction to proceed rapidly.

Question 2

1. The decrease in the system's entropy is exactly balanced by an increase in the system's enthalpy, keeping ΔG negative.
2. The process is spontaneous because the universe tends toward lower energy, and the folded protein is a lower enthalpy state.
3. The process is driven by an external energy source, such as ATP hydrolysis, which is coupled to the folding reaction.
4. The second law of thermodynamics requires that the total entropy of the universe (system + surroundings) increases for a spontaneous process. (correct answer)

Explanation: When you encounter questions about protein folding thermodynamics, remember that spontaneity depends on the complete thermodynamic picture, not just what happens to the protein itself. The key insight is distinguishing between the system (the protein) and the universe (system + surroundings). While protein folding does decrease the conformational entropy of the polypeptide chain ($\Delta S_{system} < 0$), this doesn't prevent spontaneity because the second law of thermodynamics only requires that $\Delta S_{universe} > 0$ for spontaneous processes. Answer D correctly identifies this principle. During folding, the protein releases energy as it forms favorable interactions (hydrogen bonds, van der Waals forces, etc.). This energy increases the molecular motion in the surrounding water molecules, creating a large positive $\Delta S_{surroundings}$ that more than compensates for the negative $\Delta S_{system}$. The result: $\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} > 0$. Answer A incorrectly suggests enthalpy changes balance entropy changes within the system alone—but spontaneity requires considering the surroundings too. Answer B oversimplifies by focusing only on enthalpy, ignoring entropy's role entirely. Answer C wrongly implies protein folding needs external energy coupling like ATP hydrolysis, when in fact it's thermodynamically favorable on its own. Remember this pattern: when a biological process seems thermodynamically paradoxical (like decreasing entropy being spontaneous), look beyond the immediate system to the complete universe. The surroundings often provide the thermodynamic driving force through energy dissipation.

Question 3

1. The reaction is directly coupled to the hydrolysis of an ATP molecule within the aldolase active site, making the net process exergonic.
2. The subsequent reactions in the glycolytic pathway rapidly consume the products, keeping their concentrations extremely low. (correct answer)
3. The enzyme aldolase significantly lowers the ΔG°' of the reaction, changing it from a positive to a negative value.
4. Cellular temperatures are much higher than standard conditions, which provides the necessary energy to overcome the positive ΔG°'.

Explanation: Even if a reaction has a positive ΔG°', it can proceed in the forward direction if the actual free energy change (ΔG) is negative. ΔG is influenced by the concentrations of reactants and products (ΔG = ΔG°' + RTln(Q)). In glycolysis, the products of the aldolase reaction (dihydroxyacetone phosphate and glyceraldehyde-3-phosphate) are immediately used in the next steps. This rapid consumption keeps their concentrations very low, making the mass-action ratio Q << 1. This results in a large, negative RTln(Q) term that overcomes the positive ΔG°', making the actual ΔG negative.

Question 4

1. The binding displaces many ordered water molecules from the binding site, and the primary interactions are hydrophobic.
2. The ligand and protein form strong hydrogen bonds and ionic interactions, but the ligand loses significant rotational freedom upon binding. (correct answer)
3. The binding process is entirely driven by the hydrophobic effect, with minimal formation of specific polar contacts in the binding pocket.
4. The binding causes a large-scale disordering of the protein structure, while forming a few weak van der Waals contacts with the ligand.

Explanation: A favorable (negative) ΔH is characteristic of bond formation, such as strong hydrogen bonds and ionic interactions (salt bridges), which release energy. An unfavorable (negative) ΔS indicates that the system has become more ordered. This occurs when two molecules (ligand and protein) combine to form one complex, and the ligand loses its translational and rotational freedom. This combination of ΔH < 0 and ΔS < 0 is classic for binding events dominated by specific, strong polar interactions.

Question 5

1. The net ΔG is +33 kJ/mol, and the coupled reaction remains non-spontaneous and will not proceed without further energy input.
2. The net ΔG is -33 kJ/mol, meaning the coupled reaction is spontaneous and will proceed in the forward direction. (correct answer)
3. The net ΔG is -75 kJ/mol, making the reaction strongly spontaneous, as the free energies are always additive in magnitude.
4. The net ΔG cannot be determined without knowing the standard free energy change (ΔG°') for the hydrolysis of ATP.

Explanation: The free energy changes of coupled reactions are additive. The net ΔG for the process is the sum of the ΔG for the endergonic reaction and the ΔG for the exergonic (driving) reaction. Net ΔG = ΔG(X→Y) + ΔG(ATP hydrolysis) = (+21 kJ/mol) + (-54 kJ/mol) = -33 kJ/mol. Since the net ΔG is negative, the overall coupled reaction is thermodynamically favorable (spontaneous) and will proceed.

Question 6

1. A large, positive standard free energy change (ΔG°'), indicating that the process requires energy input from ATP.
2. A standard free energy change (ΔG°') near zero, indicating that the process is near equilibrium.
3. A rapid rate of electron flow, because the large potential difference overcomes the activation energy barriers.
4. A large, negative standard free energy change (ΔG°'), indicating that the process is highly spontaneous. (correct answer)

Explanation: When you encounter questions about electron transport and thermodynamics, focus on the fundamental relationship between reduction potential and free energy. These concepts are directly linked through the equation $\Delta G°' = -nF\Delta E°'$, where n is the number of electrons transferred and F is Faraday's constant. The negative sign in this equation is crucial. When the electron transport chain has a large positive standard reduction potential difference ($\Delta E°'$), this automatically creates a large negative standard free energy change ($\Delta G°'$). A negative $\Delta G°'$ indicates a thermodynamically favorable, spontaneous process that releases energy - exactly what we observe in cellular respiration. Answer D correctly identifies this relationship. The large positive $\Delta E°'$ translates directly to a large negative $\Delta G°'$, making electron transport highly spontaneous and energy-releasing. Answer A reverses the relationship, incorrectly suggesting positive $\Delta G°'$ (energy-requiring). This would occur with a negative $\Delta E°'$, not positive. Answer B suggests the process is near equilibrium, which would require $\Delta E°'$ close to zero - contradicting the "large positive" value given. Answer C confuses thermodynamics with kinetics. While $\Delta E°'$ tells us about energy favorability, it says nothing about reaction rates, which depend on activation barriers and enzyme activity. Remember this key relationship: positive $\Delta E°'$ always means negative $\Delta G°'$. When you see "large positive reduction potential," immediately think "highly spontaneous process."

Question 7

1. A large negative enthalpy change (ΔH) resulting from the formation of strong covalent bonds between adjacent phospholipid molecules.
2. A large positive entropy change (ΔS) of the system as the phospholipids move from a disordered to an ordered state.
3. A large positive entropy change (ΔS) of the surrounding water molecules, which is the dominant factor in the overall free energy calculation. (correct answer)
4. A large negative enthalpy change (ΔH) due to favorable van der Waals interactions, which is sufficient to overcome the entropic penalty.

Explanation: This process is driven by the hydrophobic effect. While the phospholipids themselves become more ordered (a negative ΔS for the system), they sequester their hydrophobic tails away from water. This releases the highly ordered water molecules that previously formed solvation shells ('cages') around the individual lipid tails. The resulting increase in the disorder of the water (a large positive ΔS for the surroundings) is the dominant thermodynamic contribution, making the overall ΔG negative.

Question 8

1. The reaction will proceed in the reverse direction (F6P → G6P) because the standard free energy change is positive.
2. The reaction requires coupling to ATP hydrolysis to become favorable because its standard free energy change is positive.
3. The reaction will proceed in the forward direction (G6P → F6P) because the low product-to-reactant ratio makes the actual free energy change (ΔG) negative. (correct answer)
4. The reaction is at equilibrium within the cell because the actual free energy change (ΔG) is approximately zero under these concentration conditions.

Explanation: The actual free energy change (ΔG) determines the spontaneity of a reaction under cellular conditions, not the standard free energy change (ΔG°'). The relationship is given by ΔG = ΔG°' + RTln(Q), where Q is the mass-action ratio, [F6P]/[G6P]. Here, Q = 14/83 ≈ 0.169. Since Q < 1, ln(Q) is a negative value. The calculation is ΔG = 1700 J/mol + (8.315 J/mol·K)(310 K)ln(0.169) ≈ 1700 - 4588 = -2888 J/mol, or -2.9 kJ/mol. Because ΔG is negative, the reaction is spontaneous in the forward direction.

Question 9

1. A massive amount of strain energy is stored within the terminal phosphoanhydride bond, and breaking this 'high-energy bond' releases this energy.
2. The products, ADP and inorganic phosphate, have significantly greater resonance stabilization and less electrostatic repulsion than the reactant ATP. (correct answer)
3. The reaction produces two molecules from one, leading to a large increase in enthalpy (ΔH > 0) that drives the process forward.
4. The terminal phosphate group of ATP is inherently unstable and spontaneously detaches without enzymatic catalysis, releasing energy as heat.

Explanation: The term 'high-energy bond' is a common misconception. The favorability of ATP hydrolysis comes from the fact that the products (ADP and Pi) are more stable than the reactant (ATP). This increased stability is due to several factors, including: 1) relief of electrostatic repulsion between the adjacent negative charges on the phosphate groups, 2) greater resonance stabilization of the phosphate products, and 3) better solvation of the products by water.

Question 10

1. At equilibrium, the concentration of reactants will be significantly greater than the concentration of products.
2. The reaction will proceed very rapidly toward equilibrium because the free energy change is large and negative.
3. At equilibrium, the concentration of products will be significantly greater than the concentration of reactants. (correct answer)
4. The reaction is irreversible under all cellular conditions and will proceed entirely to products, leaving no reactants.

Explanation: The standard free energy change (ΔG°') is related to the equilibrium constant (K'eq) by the equation ΔG°' = -RTln(K'eq). A large, negative ΔG°' corresponds to a large K'eq (K'eq > 1). A large K'eq indicates that at equilibrium, the ratio of products to reactants is very high. Therefore, the concentration of products will be significantly greater than the concentration of reactants. Thermodynamics (ΔG) provides no information about the rate of a reaction.

Question 11

1. The process will become more spontaneous because the favorable TΔS term will increase in magnitude. (correct answer)
2. The process will become less spontaneous because the unfavorable ΔH term becomes more significant at higher temperatures.
3. The spontaneity of the process will not be affected by temperature, as it depends only on the signs of ΔH and ΔS.
4. The process will become more spontaneous because the activation energy barrier will be lowered at higher temperatures.

Explanation: The spontaneity of a process is determined by the Gibbs free energy change, ΔG = ΔH - TΔS. In this case, the enthalpy change (ΔH) is positive (unfavorable), and the entropy change (ΔS) is positive (favorable). The TΔS term is therefore negative (favorable). As temperature (T) increases, the magnitude of the favorable -TΔS term increases. This will eventually overcome the unfavorable positive ΔH, making ΔG more negative and thus making the process more spontaneous.

Question 12

1. The cell is in a state of chemical equilibrium, where the rate of ATP synthesis exactly matches the rate of ATP hydrolysis.
2. The cell is in a low-energy state, as indicated by the system being far from its equilibrium point.
3. The cell is in a dynamic steady state, which is maintained by a constant input of energy from catabolism. (correct answer)
4. The cell has reached maximum entropy, and the high potential of ATP hydrolysis is a result of this disorder.

Explanation: Living organisms are open systems that are not at equilibrium. They exist in a dynamic steady state (or homeostasis), where the concentrations of metabolites are kept relatively constant but are not at their equilibrium values. This non-equilibrium state is maintained by a continuous flux of energy and matter (e.g., from the catabolism of nutrients). If a cell were to reach equilibrium, ΔG for all its reactions would be zero, meaning no net work could be done, which is the definition of death.

Question 13

1. Why the overall pathway of glycolysis is reversible under cellular conditions.
2. How the energy from a high-phosphoryl-transfer-potential compound can drive the formation of ATP. (correct answer)
3. Why pyruvate kinase is an allosterically inhibited enzyme in the glycolytic pathway.
4. How the reaction can proceed without a significant change in the overall entropy of the system.

Explanation: This is an example of substrate-level phosphorylation. PEP has a very high phosphoryl-transfer potential, indicated by its large, negative ΔG°' of hydrolysis. This means the transfer of its phosphate group to a recipient is highly favorable. The energy released from the 'hydrolysis' of PEP (-61.9 kJ/mol) is much greater than the energy required to synthesize ATP (+30.5 kJ/mol). Coupling these two processes results in a highly exergonic net reaction (ΔG°' = -61.9 + 30.5 = -31.4 kJ/mol), effectively using the chemical energy in PEP to create ATP.

Question 14

1. The overall ΔG°' is -11.5 kJ/mol; the pathway is spontaneous only if the first step is coupled to ATP hydrolysis.
2. The overall ΔG°' is -21.5 kJ/mol; the pathway is spontaneous under standard conditions despite an initial endergonic step. (correct answer)
3. The overall ΔG°' is the ΔG°' of the rate-limiting step (-18.5 kJ/mol); the pathway is spontaneous overall.
4. The overall ΔG°' cannot be calculated because the first step is endergonic, which would block the entire pathway.

Explanation: The standard free energy changes for sequential reactions in a pathway are additive. The overall ΔG°' for the conversion of A to D is the sum of the ΔG°' values for each step. Overall ΔG°' = (+2.0 kJ/mol) + (-18.5 kJ/mol) + (-5.0 kJ/mol) = -21.5 kJ/mol. Because the overall standard free energy change is negative, the complete pathway is thermodynamically favorable under standard conditions, even though the first step is individually unfavorable.

Question 15

1. 250%, because the energy released is much greater than the energy required for the transport process.
2. 60%, because a significant portion of the energy from ATP hydrolysis is lost as heat.
3. 100%, because the energy from ATP hydrolysis is perfectly coupled to the transport of the solute.
4. 40%, calculated as the useful energy captured divided by the total energy available from the driving reaction. (correct answer)

Explanation: When you encounter questions about coupled biochemical reactions, focus on thermodynamic efficiency—how much of the available energy is actually captured for useful work versus lost as heat. Thermodynamic efficiency is calculated as the ratio of useful energy captured to total energy available from the driving reaction. Here, the transport process requires +18 kJ/mol of work, while ATP hydrolysis provides -45 kJ/mol of energy. The efficiency is therefore: $\frac{18 \text{ kJ/mol}}{45 \text{ kJ/mol}} \times 100\% = 40\%$ This means 40% of the ATP's energy goes toward transport work, while 60% is dissipated as heat—typical for biological processes. Answer A incorrectly suggests 250% efficiency, which violates thermodynamic laws. You cannot extract more energy from a reaction than it provides—efficiencies above 100% are physically impossible. Answer B states the correct percentage (60%) but misidentifies what it represents; 60% is the energy lost as heat, not the efficiency. Answer C claims perfect coupling (100% efficiency), which never occurs in real biological systems due to entropy and heat dissipation. Answer D correctly identifies that efficiency equals useful energy divided by total available energy, yielding 40%. Study tip: For coupled reaction problems, always calculate efficiency as (energy required for work)/(energy available from driving reaction) × 100%. Remember that biological processes are never 100% efficient—some energy is always lost as heat due to the second law of thermodynamics.

Question 16

1. The overall reaction is thermodynamically favorable with $\Delta G°' = -8$ kJ/mol, demonstrating energy coupling (correct answer)
2. The overall reaction is unfavorable because the first step has a positive $\Delta G°'$ value
3. The reactions cannot be coupled because they involve different chemical species as products
4. The overall $\Delta G°'$ is +32 kJ/mol because unfavorable reactions amplify when coupled

Explanation: For coupled reactions sharing a common intermediate, the overall $\Delta G°'$ is the sum of individual $\Delta G°'$ values: (+12) + (-20) = -8 kJ/mol. Since the overall $\Delta G°'$ is negative, the coupled process is thermodynamically favorable, even though the first step alone is unfavorable. This demonstrates energy coupling. Choice B incorrectly focuses only on the first step. Choice C misunderstands coupling through shared intermediates. Choice D incorrectly adds the absolute values instead of the algebraic sum.

Question 17

1. The enzyme changes the thermodynamics by lowering the activation energy and making $\Delta G$ negative
2. The actual $\Delta G$ is negative due to non-standard concentrations of reactants and products in the cell (correct answer)
3. The positive $\Delta G°'$ value must be incorrect since spontaneous reactions always have negative standard free energies
4. The reaction is being driven by direct ATP hydrolysis occurring simultaneously at the enzyme active site

Explanation: $\Delta G°'$ refers to standard conditions (1 M concentrations), while actual cellular $\Delta G = \Delta G°' + RT\ln(Q)$ where Q is the reaction quotient. If reactant concentrations are high and product concentrations are low compared to standard conditions, the $RT\ln(Q)$ term can be sufficiently negative to make the overall $\Delta G$ negative despite a positive $\Delta G°'$. Choice A incorrectly states that enzymes change thermodynamics. Choice C is wrong because many spontaneous cellular reactions have positive $\Delta G°'$ values. Choice D assumes ATP coupling without evidence.

Question 18

1. $\Delta H < 0$ and $\Delta S < 0$, because both enthalpy and entropy oppose the reaction at low temperatures
2. $\Delta H < 0$ and $\Delta S > 0$, because the reaction becomes more favorable at higher temperatures
3. $\Delta H > 0$ and $\Delta S < 0$, because the positive $\Delta G$ values indicate an endothermic reaction
4. $\Delta H > 0$ and $\Delta S > 0$, because $\Delta G$ decreases linearly with increasing temperature (correct answer)

Explanation: When you encounter temperature-dependent free energy data, you need to apply the Gibbs-Helmholtz equation: $\Delta G = \Delta H - T\Delta S$. This relationship reveals how enthalpy and entropy changes affect reaction spontaneity at different temperatures. Looking at the data, $\Delta G$ becomes more negative (more favorable) as temperature increases, changing from +2.1 kJ/mol at 280 K to -1.5 kJ/mol at 340 K. This temperature dependence tells us about the signs of $\Delta H$ and $\Delta S$. Since $\Delta G$ decreases as temperature increases, the $-T\Delta S$ term must be becoming more negative faster than any positive contribution from $\Delta H$. This happens when $\Delta S > 0$ (entropy increases). For the reaction to transition from nonspontaneous to spontaneous at higher temperatures, we need $\Delta H > 0$ and $\Delta S > 0$. At low temperatures, the positive $\Delta H$ dominates, but at high temperatures, the $-T\Delta S$ term (which becomes more negative) eventually overcomes the positive enthalpy term. Answer A is wrong because negative enthalpy and entropy wouldn't explain the temperature trend. Answer B incorrectly suggests $\Delta H < 0$; if enthalpy were negative, the reaction would be favorable at all temperatures with positive entropy. Answer C misinterprets the data by assuming $\Delta S < 0$; negative entropy would make the reaction less favorable at higher temperatures, opposite to what we observe. Remember: when $\Delta G$ becomes more negative with increasing temperature, look for positive $\Delta H$ and $\Delta S$ - the entropy term eventually dominates at high temperatures.

Question 19

1. $\Delta G_2 = +18 + (8.314 × 310 × \ln(0.01)) = -11,800$ kJ/mol, an unrealistic value due to incorrect units
2. $\Delta G_2 = +18 + (0.008314 × 310 × \ln(0.01)) = -6.1$ kJ/mol, so the step becomes thermodynamically favorable
3. $\Delta G_2 = +18 - (0.008314 × 310 × \ln(0.01)) = +29.9$ kJ/mol, making the unfavorable step even more unfavorable
4. $\Delta G_2 = +18 + (0.008314 × 310 × \ln(0.01)) = +6.1$ kJ/mol, so Step 2 remains unfavorable (correct answer)

Explanation: When you encounter questions about free energy changes under physiological conditions, you need to distinguish between standard conditions ($\Delta G°'$) and actual cellular conditions using the relationship $\Delta G = \Delta G°' + RT \ln Q$. For Step 2, you have $\Delta G°'_2 = +18$ kJ/mol, temperature T = 37°C = 310 K, and reaction quotient Q₂ = 0.01. The gas constant R must be in kJ/(mol·K), so R = 0.008314 kJ/(mol·K). Calculating: $\Delta G_2 = +18 + (0.008314 × 310 × \ln(0.01)) = +18 + (2.58 × (-4.61)) = +18 + (-11.9) = +6.1$ kJ/mol. Answer A uses incorrect units for R (8.314 instead of 0.008314), leading to an impossibly large negative value. Answer B makes a sign error in the final calculation—it shows the RT ln Q term as negative when it should be added to the positive result. Answer C incorrectly subtracts the RT ln Q term instead of adding it; since ln(0.01) is negative, subtracting a negative becomes addition, making the reaction even more unfavorable. Answer D correctly applies the equation and shows that while the RT ln Q term makes the reaction less unfavorable than under standard conditions, Step 2 remains thermodynamically unfavorable ($\Delta G > 0$). Study tip: Always convert R to the correct units (0.008314 kJ/(mol·K) for energy in kJ) and remember that ln Q can be negative when Q < 1, which makes reactions more favorable than standard conditions predict.

Question 20

1. The favorability remains unchanged because $\Delta G$ is independent of temperature for protein folding reactions
2. Unfolding becomes less favorable because the higher temperature increases the enthalpic penalty for breaking interactions
3. Unfolding becomes more favorable because $\Delta G$ decreases to approximately +1 kJ/mol at the higher temperature (correct answer)
4. Unfolding becomes thermodynamically favorable with $\Delta G = -15$ kJ/mol due to the large entropic contribution at high temperature

Explanation: When you encounter thermodynamics problems involving protein folding at different temperatures, remember that $\Delta G$ depends on temperature through the relationship $\Delta G = \Delta H - T\Delta S$. You'll need to first calculate the entropy change, then apply it at the new temperature. From the given data at 25°C (298 K): $\Delta S = \frac{\Delta H - \Delta G}{T} = \frac{180 - 25}{298} = +0.520$ kJ/(mol·K). This positive entropy change makes sense because unfolding increases molecular disorder. At 65°C (338 K), using the same $\Delta H$ and $\Delta S$: $\Delta G = 180 - (338)(0.520) = 180 - 176 = +4$ kJ/mol. This is approximately +1 kJ/mol, confirming that unfolding becomes more favorable (though still thermodynamically unfavorable) at higher temperature. Choice A is incorrect because $\Delta G$ absolutely depends on temperature—this is fundamental thermodynamics. Choice B misunderstands the temperature effect; while $\Delta H$ represents the enthalpic penalty, it remains constant, and temperature affects the entropic term $T\Delta S$. Choice D contains a calculation error—the correct $\Delta G$ is about +1 kJ/mol, not -15 kJ/mol, so unfolding remains thermodynamically unfavorable. The key insight is that higher temperatures amplify the entropic contribution ($T\Delta S$), making processes that increase disorder (like protein unfolding) relatively more favorable. Always calculate $\Delta S$ from initial conditions first, then apply $\Delta G = \Delta H - T\Delta S$ at the new temperature.