Energy, Enzymes and ATP
General Biology · Section 3.1 · 17 study cards
Thermodynamics as biologists use it, free energy and coupling, ATP as the energy currency, and how enzymes and inhibitors work.
Practice this set → Spaced repetition, card by card. No account needed.
Method
Separate the two questions: which way, and how fast
Almost every mistake in this unit comes from mixing these up. Thermodynamics answers only which way a reaction is favoured to go and how much free energy it can yield, and the sign of ΔG carries that answer. Kinetics answers how fast it gets there, and activation energy carries that answer. Enzymes act entirely on the second question and never on the first. Before answering any question in this unit, decide which of the two it is about.
Work free-energy problems in a fixed order
- Write the reaction of interest with its ΔG, and note the sign. Positive means it will not go on its own.
- Write the reaction you intend to couple to it, normally ATP hydrolysis at about −7.3 kcal per mole.
- Add the two ΔG values. Free energy changes are additive along a pathway.
- If the sum is negative, the coupled process is thermodynamically possible; say so explicitly.
- Add the kinetic caveat: possible is not the same as fast, so name the enzyme or state that one is required.
Treat the enzyme as a shape, not as a symbol
When a question changes temperature, pH, or adds an inhibitor or a metal ion, ask what happens to the three-dimensional shape of the active site. Heat past the optimum and extremes of pH break the weak bonds holding the fold, so the site no longer fits the substrate. A competitive inhibitor leaves the shape intact but occupies the site. A noncompetitive or allosteric effector leaves the site empty but distorts it. Framing all four cases as one question about shape means you only have to remember one idea.
Use the saturation curve to predict experiments
Sketch rate against substrate concentration whenever an experiment is described. The curve rises and then flattens. Adding substrate helps only on the rising part; on the flat part only more enzyme helps. This single sketch answers most questions about inhibitor type as well, since the diagnostic is whether extra substrate restores the original maximum rate.
Definitions and theorems
Worked example
The first step of glycolysis converts glucose to glucose-6-phosphate. Written with free phosphate, the reaction has a ΔG of about +3.4 kcal per mole. Hydrolysis of ATP to ADP and Pᵢ has a ΔG of about −7.3 kcal per mole. Explain, with a calculation, how the cell carries out the first step, and state why hexokinase is still required.
Classify the target reaction. A ΔG of +3.4 kcal per mole is positive, so adding free phosphate to glucose is endergonic and will not proceed to any useful extent on its own.
Identify the energy source. ATP hydrolysis at −7.3 kcal per mole is exergonic and larger in magnitude than the requirement, so it is a candidate for coupling.
Add the free energy changes: (+3.4) + (−7.3) = −3.9 kcal per mole. Free energy changes are additive, so the coupled reaction overall has a negative ΔG.
State the conclusion from the sign. A negative overall ΔG means the coupled reaction is exergonic and thermodynamically favourable, so glucose-6-phosphate can be formed.
Describe the mechanism, not just the arithmetic. The coupling works because the phosphate is transferred directly from ATP to glucose on the enzyme, rather than ATP being hydrolysed separately. Two separate reactions in the same beaker would simply lose the energy as heat.
Answer the last part. ΔG says nothing about rate: the reaction is favourable but the activation energy is far too high for it to happen fast enough at body temperature. Hexokinase lowers that activation energy and holds ATP and glucose in the correct orientation, and it does not change the −3.9 value at all.
Common mistakes
- Saying ATP has high-energy bonds that store energy. Bond breaking always costs energy. ATP hydrolysis is exergonic because ADP and free phosphate together are more stable than ATP and water, through charge repulsion relief and resonance stabilisation. Phrase every answer as a comparison of reactants with products.
- Claiming an enzyme makes a reaction more favourable. Enzymes change only the activation energy. Answers that say an enzyme changes ΔG, shifts the equilibrium, or makes an endergonic reaction spontaneous lose the mark even when the rest is correct.
- Reading spontaneous as fast. Spontaneous means only that ΔG is negative. A spontaneous reaction can take centuries. If a question asks why a favourable reaction is not happening, the answer is activation energy, not thermodynamics.
- Getting the inhibition test backwards. Remember which one extra substrate defeats: competitive inhibitors sit in the active site and can be outnumbered, so more substrate restores the maximum rate. Noncompetitive inhibitors distort the site from elsewhere, so more substrate changes nothing.
- Treating denaturation as reversible slowing. Past the temperature optimum an enzyme is not merely sluggish; its fold has collapsed and activity is usually lost for good. Cooling the tube back down does not generally restore the rate, which is why the curve is asymmetric.
Practice it
Reading the method is not the same as being able to recall it under pressure. This set drills 17 cards one at a time and schedules each card separately, so the ones you keep missing come back sooner.