Enzymes & Bioenergetics Basics

Question 1

(a) [4]

• First Law: energy is conserved; cells convert energy forms (chemical ↔ chemical/mechanical) but do not create it. (1)
• Second Law: total entropy of universe increases in any spontaneous process. (1)
• A cell decreases its own (system) entropy by building order, but releases heat and disordered products (CO₂, H₂O) to surroundings. (1)
• The increase in surroundings' entropy exceeds the decrease in system entropy, so $\Delta S_{universe} > 0$ — no violation. (1)

(b) [4]

• Glutamine synthesis has $\Delta G'^{\circ} = +14 > 0$ → endergonic. (1)
• Coupled: $\Delta G = +14 + (-50) = -34\ \text{kJ mol}^{-1}$. (2)
• Negative → spontaneous (exergonic overall). (1)

(c) [6]

• Rearrange: $K'_{eq} = e^{-\Delta G'^{\circ}/RT}$. (1)
• $RT = 8.314 \times 310 = 2577.3\ \text{J mol}^{-1}$. (1)
• Exponent: $-14000/2577.3 = -5.432$. (2)
• $K'_{eq} = e^{-5.432} \approx 4.4\times10^{-3}$. (1)
• $K'_{eq} \ll 1$ → at standard conditions equilibrium favours reactants; reaction does not proceed forward appreciably without energy input. (1)

(d) [6]

• $E_a$ = minimum energy needed to reach the transition state / activated complex from reactants. (2)
• Enzyme lowers $E_a$ by stabilising the transition state. (1)
• In $k = Ae^{-E_a/RT}$, smaller $E_a$ makes the exponent less negative → larger $k$ → faster rate. (2)
• $\Delta G$ depends only on initial and final states, not the path, so it is unchanged; enzyme alters kinetics not thermodynamics; $K_{eq}$ unchanged. (1)

Question 2

(a) [8]

• Rate of ES formation: $k_1[E][S]$. (1)
• Rate of ES breakdown: $(k_{-1}+k_2)[ES]$. (1)
• Steady-state assumption: $d[ES]/dt = 0$, so $k_1[E][S]=(k_{-1}+k_2)[ES]$. (1)
• Define $K_M = (k_{-1}+k_2)/k_1$, so $[ES]=[E][S]/K_M$. (1)
• Conservation: $[E]_T=[E]+[ES]$ → $[E]=[E]_T-[ES]$. Substitute: $[ES] = \dfrac{[E]_T[S]}{K_M+[S]}$. (2)
• $v_0 = k_2[ES]$, and $V_{\max}=k_2[E]_T$, giving $v_0 = \dfrac{V_{\max}[S]}{K_M+[S]}$. (2)

(b) [4]

• Set $[S]=K_M$: $v_0 = \dfrac{V_{\max}K_M}{K_M+K_M}=\dfrac{V_{\max}K_M}{2K_M}=\tfrac12 V_{\max}$. (2)
• $K_M$ = substrate concentration giving half-maximal velocity; a measure of enzyme–substrate affinity (low $K_M$ = high affinity). (2)

(c) [4]

• $v_0 = \dfrac{120\times 12}{4+12} = \dfrac{1440}{16} = 90\ \mu\text{mol min}^{-1}$. (4)

(d) [8]

• Competitive inhibitor binds active site; raises apparent $K_M$; $V_{\max}$ unchanged (overcome by excess substrate). (2)
• Non-competitive inhibitor binds elsewhere (allosteric); lowers apparent $V_{\max}$; $K_M$ unchanged. (2)
• Function (4):

def v0(S, Vmax, KM, I, Ki, mode):
    if mode == "competitive":
        KMapp = KM * (1 + I/Ki)
        return Vmax * S / (KMapp + S)
    elif mode == "non-competitive":
        Vapp = Vmax / (1 + I/Ki)
        return Vapp * S / (KM + S)

Question 3

(a) [5]

• Lock-and-key: rigid active site complementary to substrate; only exact fit binds. (2)
• Induced-fit: active site flexible; substrate binding induces conformational change to fit and strain bonds. (2)
• Induced-fit better explains: broad specificity / conformational changes seen by X-ray (e.g. hexokinase closing over glucose). (1)

(b) [6]

• Allosteric regulation: effector binds a site distinct from active site, altering enzyme conformation and activity (activator or inhibitor). (2)
• Feedback inhibition: end product of a pathway allosterically inhibits an early (committed-step) enzyme, preventing overproduction — e.g. isoleucine inhibits threonine deaminase in the Thr→Ile pathway. (2)
• Difference: feedback inhibitor binds an allosteric site (not competing at active site) and is structurally unrelated to substrate; competitive inhibitor mimics substrate and binds active site directly. (2)

(c) [5]

• Temperature curve: rises with T (increased kinetic energy/collisions) to optimum ~37 °C, then falls sharply as heat denatures the enzyme (H-bonds break, tertiary structure lost). (2)
• pH curve: bell-shaped peak at pH 7; either side, altered ionisation of active-site/substrate groups reduces binding and catalysis. (2)
• Reversible vs irreversible: mild deviations reduce rate reversibly (charge changes); extreme heat/pH cause irreversible denaturation. (1)

[
  {"claim":"Coupled dG of glutamine synthesis with ATP is -34 kJ/mol","code":"result = (14 + (-50)) == -34"},
  {"claim":"Keq for +14 kJ/mol at 310K is approx 4.4e-3","code":"import sympy as sp; K = sp.exp(sp.Rational(-14000,1)/(sp.Float(8.314)*310)); result = abs(float(K)-0.00443) < 1e-4"},
  {"claim":"v0 at S=12, Vmax=120, KM=4 equals 90","code":"result = 120*12/(4+12) == 90"},
  {"claim":"At S=KM, v0 = Vmax/2","code":"Vmax,KM = sp.symbols('Vmax KM', positive=True); v = Vmax*KM/(KM+KM); result = sp.simplify(v - Vmax/2) == 0"}
]