Principles of Bioenergetics and Thermodynamics (1D)
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MCAT Biological and Biochemical Foundations of Living Systems › Principles of Bioenergetics and Thermodynamics (1D)
A metabolic reaction releases heat to the surroundings ($\Delta H<0$) but is observed to be nonspontaneous in the forward direction under the tested conditions ($\Delta G>0$). Which conclusion is most consistent with $\Delta G=\Delta H-T\Delta S$?
$\Delta G$ and $\Delta H$ must always have the same sign, so the observation is impossible
$\Delta S$ must be positive because exothermic reactions always increase disorder
$\Delta S$ must be sufficiently negative so that $-T\Delta S$ outweighs the negative $\Delta H$
Nonspontaneity implies the reaction has a high activation energy, not an unfavorable $\Delta G$
Explanation
This question tests understanding of bioenergetics and thermodynamics principles. Nonspontaneity (ΔG > 0) can occur with exothermic ΔH < 0 if ΔS is sufficiently negative, making -TΔS positive and dominant. The reaction releases heat but is nonspontaneous. Choice D is correct as negative ΔS can outweigh negative ΔH. Choice C fails by claiming ΔG and ΔH must match signs, ignoring entropy. In crystallization, negative ΔS opposes exothermic ordering. Analyze by solving for ΔS range where ΔG > 0 despite ΔH < 0.
A cell drives an endergonic biosynthetic reaction by rapidly removing its product via sequestration into a vesicle. Which prediction is most consistent with the effect on $\Delta G$ for the biosynthetic reaction (with $\Delta G = \Delta G^\circ' + RT\ln Q$)?
Removing product increases $Q$, making $RT\ln Q$ more positive and driving $\Delta G$ downward
Removing product affects rate but cannot affect $\Delta G$ because free energy depends only on temperature
Removing product decreases $Q$, making $RT\ln Q$ more negative and driving $\Delta G$ downward
Removing product changes $\Delta G^\circ'$ by altering the enzyme’s active site
Explanation
This question tests understanding of bioenergetics and thermodynamics principles. Removing product decreases Q, making RT ln Q more negative and thus ΔG more favorable for endergonic reactions. The cell sequesters product in vesicles. Choice D is correct as lower Q drives ΔG downward via the logarithmic term. Choice B errs by claiming increased Q, misdefining the effect. In biosynthesis like fatty acid synthesis, product removal pulls pathways. Simulate by calculating ΔG before and after product depletion.
A researcher compares two catalysts for the same reaction and finds both yield the same equilibrium composition, but Catalyst 1 reaches equilibrium faster. Which conclusion is most consistent with thermodynamics?
Catalyst 1 must increase $K_{eq}$, shifting equilibrium more toward products
Catalyst 1 increases entropy of the system, which is why equilibrium is reached sooner
Catalyst 1 likely lowers activation energy more, increasing rate without changing $\Delta G^\circ$ or $K_{eq}$
Catalyst 1 must make $\Delta G^\circ$ more negative, increasing the driving force
Explanation
This question tests understanding of bioenergetics and thermodynamics principles. Catalysts speed reactions by lowering Ea but do not change equilibrium composition or ΔG°. Both catalysts yield same equilibrium, but one is faster. Choice C is correct as Catalyst 1 likely lowers Ea more, accelerating without affecting thermodynamics. Choice B fails by saying it increases Keq, confusing catalysis with equilibrium shift. In industrial processes, select catalysts for rate, not yield change. Compare by measuring time to equilibrium and final compositions.
For a reaction at 298 K, $K_{eq}=10^{-2}$. Use $\Delta G^\circ = -RT\ln K_{eq}$ with $R=8.314\ \text{J·mol}^{-1}\text{·K}^{-1}$. Which statement about $\Delta G^\circ$ is most consistent with these data?
$\Delta G^\circ$ cannot be determined without $\Delta H^\circ$
$\Delta G^\circ$ equals zero because $K_{eq}$ is dimensionless
$\Delta G^\circ$ is positive because $\ln(10^{-2})$ is negative
$\Delta G^\circ$ is negative because $K_{eq}$ is less than 1
Explanation
This question tests understanding of bioenergetics and thermodynamics principles. ΔG° is positive when Keq < 1 because ΔG° = -RT ln Keq, and ln(Keq < 1) is negative, making - (negative) positive. For Keq = $10^{-2}$, ΔG° > 0. Choice B is correct as $ln(10^{-2}$) negative leads to positive ΔG°. Choice A is wrong, reversing the sign relationship. In redox reactions, positive ΔG° indicates unfavorable electron transfer. Compute ΔG° from Keq to classify reactions as reactant- or product-favored.
In a reconstituted vesicle system, ATP synthase is supplied ADP and $P_i$ but no proton gradient. A light-driven proton pump is then activated to generate a gradient. Which outcome would be expected based on thermodynamic coupling?
ATP synthesis occurs only if $\Delta H$ for ADP+$P_i$ is negative
ATP synthesis is unchanged because ATP synthase only lowers activation energy
ATP synthesis decreases because gradients always oppose chemical work
ATP synthesis increases because the proton gradient provides free energy to drive an otherwise unfavorable phosphorylation
Explanation
This question tests understanding of bioenergetics and thermodynamics principles. ATP synthesis is endergonic and driven by the proton gradient's energy in chemiosmotic coupling. Activating the proton pump generates the gradient in the vesicle system. Choice A is correct as the gradient provides free energy to make phosphorylation favorable. Choice B errs by claiming gradients oppose work, reversing the coupling mechanism. In mitochondria, inhibit electron transport to see ATP drop. Test coupling by measuring ATP production with imposed gradients.
A reaction $A \rightleftharpoons B$ has $\Delta G^\circ = +5.7\ \text{kJ/mol}$ at 298 K. Use $\Delta G^\circ = -RT\ln K_{eq}$ with $R=8.314\ \text{J·mol}^{-1}\text{·K}^{-1}$. Which prediction is most consistent with the principle relating $\Delta G^\circ$ and equilibrium?
$K_{eq}<1$, so A predominates at equilibrium
The reaction is spontaneous in the forward direction at any concentrations because $\Delta G^\circ$ is defined at equilibrium
$K_{eq}=1$ because the sign of $\Delta G^\circ$ does not affect equilibrium
$K_{eq}>1$, so B predominates at equilibrium
Explanation
This question tests understanding of bioenergetics and thermodynamics principles. The standard free energy change (ΔG°) relates to the equilibrium constant via ΔG° = -RT ln Keq, where positive ΔG° indicates Keq < 1 and reactant predominance at equilibrium. For the reaction A ⇌ B, ΔG° = +5.7 kJ/mol implies Keq < 1. Choice C is correct because it logically follows that A predominates at equilibrium due to the positive ΔG°. Choice A fails by mistakenly reversing the relationship, claiming Keq > 1, which highlights an error in sign interpretation. To apply this principle elsewhere, consider ATP hydrolysis where negative ΔG° predicts product favorability. Calculate Keq from ΔG° in metabolic reactions to predict equilibrium compositions.
A reaction in a cell has $\Delta G^\circ' = +2\ \text{kJ/mol}$. The cell maintains product concentration far below substrate concentration such that $Q=10^{-4}$ at 310 K. Use $\Delta G = \Delta G^\circ' + RT\ln Q$ with $R=8.314\ \text{J·mol}^{-1}\text{·K}^{-1}$. Which prediction is most consistent with reaction spontaneity in vivo?
Forward reaction is spontaneous only if temperature decreases, since $RT\ln Q$ becomes positive
Forward reaction cannot be spontaneous because $\Delta G^\circ'$ is positive
Forward reaction can be spontaneous despite positive $\Delta G^\circ'$ because $\ln Q$ is negative
Forward reaction is spontaneous only if an enzyme is added to change $\Delta G^\circ'$
Explanation
This question tests understanding of bioenergetics and thermodynamics principles. Even with positive $\Delta G^\circ'$, a reaction can be spontaneous if $Q \ll K_\text{eq}$, making $RT \ln Q$ sufficiently negative in $\Delta G = \Delta G^\circ' + RT \ln Q$. Here, $\Delta G^\circ' = +2$ kJ/mol but $Q = 10^{-4}$ at 310 K yields negative $\Delta G$. Choice D is correct as the negative $\ln Q$ term drives forward spontaneity. Choice B is wrong, claiming positive $\Delta G^\circ'$ prevents spontaneity, overlooking nonstandard conditions. Apply to gluconeogenesis where product removal pulls endergonic steps. Use concentration ratios to compute $\Delta G$ in metabolic flux analysis.
A reaction has $\Delta H<0$ and $\Delta S<0$. Which outcome would be expected for spontaneity as temperature increases, based on $\Delta G=\Delta H-T\Delta S$?
The reaction becomes spontaneous only if $\Delta S$ becomes positive at higher temperature
Spontaneity decreases with temperature because $-T\Delta S$ becomes more positive
Spontaneity increases with temperature because $-T\Delta S$ becomes more negative
Spontaneity is independent of temperature because $\Delta H$ is negative
Explanation
This question tests understanding of bioenergetics and thermodynamics principles. For reactions with ΔH < 0 and ΔS < 0, increasing temperature makes -TΔS more positive, potentially rendering ΔG positive. The reaction's spontaneity decreases with temperature. Choice B is correct as -TΔS becomes less negative (more positive), reducing spontaneity. Choice A fails by reversing the effect, claiming increased spontaneity. In protein denaturation, high temperature overcomes negative ΔS. Predict temperature dependence by evaluating ΔG at varying T.
Two reactions are measured at 298 K: Reaction 1 has $\Delta G^\circ=-12\ \text{kJ/mol}$; Reaction 2 has $\Delta G^\circ=+12\ \text{kJ/mol}$. A cell couples them by sharing an intermediate so they proceed together 1:1. Which conclusion about the coupled process is most consistent with Gibbs free energy additivity?
The coupled process must be nonspontaneous because one step has positive $\Delta G^\circ$
The coupled process has $\Delta G^\circ=0$ and can be driven forward by maintaining nonstandard concentrations
Coupling changes each reaction’s $K_{eq}$ so that both become strongly product-favored
The coupled process has $\Delta H^\circ=0$, so it is at equilibrium
Explanation
This question tests understanding of bioenergetics and thermodynamics principles. Free energy changes are additive in coupled reactions, allowing an exergonic step to drive an endergonic one if net ΔG < 0, though ΔG° = 0 here means equilibrium at standard conditions. The coupled process has ΔG° = -12 + 12 = 0 kJ/mol. Choice D is correct because at ΔG° = 0, nonstandard concentrations can drive the process via RT ln Q. Choice B errs by claiming a positive step makes the whole nonspontaneous, ignoring additivity. In glycolysis, coupled steps maintain flux despite some positive ΔG°. Compute net ΔG° for pathways to assess overall equilibrium.
A researcher measures $\Delta H = +10\ \text{kJ/mol}$ and $\Delta S = +60\ \text{J·mol}^{-1}\text{·K}^{-1}$ for a binding process at 298 K. Use $\Delta G = \Delta H - T\Delta S$. Which outcome would be expected based on these thermodynamic properties?
Binding is spontaneous because the positive entropy term can outweigh the positive enthalpy
Binding is nonspontaneous because any positive $\Delta H$ makes $\Delta G$ positive
Binding is spontaneous only if $\Delta H$ is negative, regardless of $\Delta S$
Binding is at equilibrium because $\Delta S$ is positive
Explanation
This question tests understanding of bioenergetics and thermodynamics principles. Spontaneity of binding is determined by $\Delta G = \Delta H - T \Delta S$, where a positive $\Delta S$ can drive a process even if $\Delta H$ is positive. For this binding, $\Delta H = +10 \text{ kJ/mol}$ and $\Delta S = +60 \text{ J/mol·K}$ at 298 K yield negative $\Delta G$. Choice C is correct because the entropy term outweighs the enthalpy, making binding spontaneous. Choice B errs by stating positive $\Delta H$ always makes $\Delta G$ positive, ignoring the entropy contribution. In protein folding, hydrophobic effects provide positive $\Delta S$ to drive structure formation. Calculate $\Delta G$ for ligand binding at different temperatures to assess entropy-enthalpy compensation.