AOL 1 : Energy & Enzymes

Define and differentiate the terms metabolism, anabolism, and catabolism.

• Metabolism

  • Metabolism is the sum of all chemical reactions that occur in a cell or organism.

  • These reactions are organized into metabolic pathways, in which the product of one reaction becomes the substrate of the next.

  • Metabolism has two halves: catabolism and anabolism.

• Catabolism

  • Catabolic pathways break down large, complex molecules into smaller, simpler ones.

  • They generally release energy, much of which is captured in ATP.

  • Example: cellular respiration breaks down glucose into CO₂ and H₂O.

• Anabolism

  • Anabolic pathways build large, complex molecules from smaller building blocks.

  • They generally require an input of energy, often supplied by ATP.

  • Example: photosynthesis builds sugar from CO₂, and cells build proteins from amino acids.

• How they relate

  • Catabolism supplies the energy and small building blocks that anabolism uses, so the two are linked through ATP and shared intermediates.

Define the terms kinetic energy and potential energy.

• Kinetic energy

  • Kinetic energy is the energy of motion.

  • Examples: a moving object, the random motion of molecules (thermal energy or heat), and flowing electrons.

  • Faster molecular motion means more thermal energy and a higher temperature.

• Potential energy

  • Potential energy is stored energy that is due to an object’s position or structure.

  • Examples: water behind a dam, a stretched spring, and a concentration or charge gradient across a membrane.

  • Chemical energy is a form of potential energy stored in the arrangement of atoms and bonds in molecules. Examples are glucose and ATP.

• Interconversion

  • Energy can change between these forms.

  • Example: when a cell breaks down glucose, the chemical potential energy in its bonds is converted into the kinetic energy of heat and into the chemical potential energy of ATP.

Explain the first and second laws of thermodynamics.

• First law (conservation of energy)

  • Energy cannot be created or destroyed. It can only be transferred or transformed from one form to another.

  • The total energy of the universe is constant.

  • Example: a cell does not make energy. It converts the chemical energy in food into ATP, movement, and heat.

• Second law

  • Every energy transfer or transformation increases the entropy (disorder) of the universe.

  • Some energy is always converted to heat, which increases the random motion of molecules and is not available to do work.

  • Consequently, no energy conversion is 100% efficient.

• Applying the laws to living things

  • A cell is highly ordered and has low entropy, but it does not violate the second law. It maintains order by taking in energy and releasing heat and disordered waste products (such as CO₂ and H₂O) to its surroundings.

  • The entropy of the universe (cell plus surroundings) still increases.

Define and relate the terms free energy, enthalpy, entropy, exergonic, and endergonic.

• Enthalpy (H)

  • Enthalpy is the total heat content of a system, which reflects the energy stored in its bonds.

  • ΔH is the change in enthalpy. It is negative when a reaction releases heat (exothermic) and positive when it absorbs heat.

• Entropy (S)

  • Entropy is a measure of disorder or the number of ways a system’s energy and particles can be arranged.

  • ΔS is positive when disorder increases (for example, one large molecule breaking into many small ones).

• Free energy (G)

  • Free energy is the energy in a system that is available to do work at constant temperature and pressure.

  • The change in free energy is given by the equation ΔG = ΔH − TΔS, where T is the absolute temperature in kelvin.

• Exergonic and endergonic reactions

  • Exergonic reactions have a negative ΔG (ΔG < 0). They release free energy and are thermodynamically favorable (spontaneous).

  • Endergonic reactions have a positive ΔG (ΔG > 0). They require an input of free energy and are not spontaneous.

  • At ΔG = 0, the reaction is at equilibrium.

• How they relate

  • A reaction is favorable when it releases heat (negative ΔH) and/or increases disorder (positive ΔS). These two factors combine in ΔG to decide whether a reaction is exergonic or endergonic.

  • “Spontaneous” refers only to thermodynamic favorability, not to speed.

Discuss how an endergonic reaction can be “coupled” to an exergonic reaction.

• The idea of coupling

  • A cell can drive an endergonic (unfavorable) reaction by linking it to an exergonic (favorable) reaction so that the overall process is exergonic.

  • Free energy changes are additive, so the overall reaction is favorable if the sum of the two ΔG values is negative.

• Requirements

  • The reactions must share a common intermediate or be carried out by the same enzyme, so that energy is actually transferred.

  • The exergonic reaction must release at least as much free energy as the endergonic reaction requires.

• The most common example: ATP

  • ATP hydrolysis (ATP + H₂O → ADP + Pᵢ) is strongly exergonic, with ΔG°′ ≈ −7.3 kcal/mol (−30.5 kJ/mol).

  • In a typical coupled reaction, an enzyme transfers the phosphate group from ATP to a reactant, which forms a phosphorylated intermediate at higher energy that can then react more easily.

  • Example: in glutamine synthesis, ATP phosphorylates glutamate, and the activated intermediate then reacts with ammonia. The overall reaction is favorable.

• Calculation example

  • An endergonic reaction with ΔG = +4 kcal/mol coupled to ATP hydrolysis with ΔG = −7.3 kcal/mol gives an overall ΔG of −3.3 kcal/mol, which is favorable.

Define and relate the terms substrate, active site, activation energy, transition state, and reaction rate.

• Substrate

  • The substrate is the reactant that an enzyme acts on.

• Active site

  • The active site is the specific region of the enzyme, usually a pocket or groove, where the substrate binds and the reaction occurs.

  • Its shape and chemical properties match the substrate, which gives the enzyme specificity.

• Transition state

  • The transition state is the unstable, high-energy intermediate arrangement of atoms that exists briefly as reactants are converted into products.

  • It sits at the peak of the energy diagram.

• Activation energy (Eₐ)

  • The activation energy is the energy barrier, which is the difference in free energy between the reactants and the transition state.

  • Even a favorable (exergonic) reaction needs this input to start.

• Reaction rate

  • The reaction rate is how fast the reactants are converted to products.

  • It depends on the activation energy. A lower barrier means more molecules can reach the transition state, so the rate is faster.

• How they relate

  • The substrate binds the active site, and the reaction proceeds through the transition state. The size of the activation energy barrier sets the reaction rate.

  • Activation energy is a kinetic property (how fast), while ΔG is a thermodynamic property (whether favorable).

Describe in general terms how enzymes are able to alter the rate of chemical reactions.

• Key principle

  • Enzymes are biological catalysts, mostly proteins. They speed up reactions by lowering the activation energy.

  • They do not change the ΔG of the reaction or the equilibrium, and they are not used up.

  • Enzymes speed up the reaction in both directions.

• Sequence of events

  1. The substrate binds to the active site, forming an enzyme–substrate complex.

  2. The active site stabilizes the transition state, often by changing shape slightly (induced fit), which lowers the activation energy.

  3. The reaction occurs, and the products form at the active site.

  4. The products are released, and the enzyme is free to bind another substrate.

• Ways the active site lowers activation energy

  • It holds substrates in the correct orientation and close proximity.

  • It puts strain on substrate bonds.

  • It provides a favorable microenvironment, such as a local pH or charge.

  • It uses amino acid side chains or cofactors to take part directly in the chemistry.

• Example

  • Catalase breaks down hydrogen peroxide (H₂O₂) into water and oxygen, and it speeds the reaction by a factor of millions.

Explain how small molecules (e.g., drugs or regulatory molecules) are able to increase or decrease the activity of an enzyme.

• Inhibitors (decrease activity)

  • Competitive inhibitors

    • They resemble the substrate and bind the active site, which blocks the substrate.

    • Their effect can be overcome by adding more substrate.

  • Noncompetitive (allosteric) inhibitors

    • They bind at a different site (the allosteric site) and change the enzyme’s shape so that the active site works poorly.

    • They are not overcome by more substrate.

  • Irreversible inhibitors

    • They bind covalently and permanently disable the enzyme. Example: penicillin inhibits a bacterial cell wall enzyme.

• Activators (increase activity)

  • Allosteric activators bind to an allosteric site and stabilize the active shape of the enzyme, which increases its activity.

  • Some enzymes need cofactors or coenzymes (such as metal ions or vitamin-derived molecules) before they become active.

• Why this works

  • Enzyme activity depends on the 3D shape of the protein. Binding of a small molecule can change that shape, which raises or lowers the enzyme’s ability to bind substrate or carry out catalysis.

• Biological and medical examples

  • Drugs such as statins inhibit enzymes (HMG-CoA reductase) to lower cholesterol.

  • Cells use regulatory molecules, such as the end product of a pathway, to switch enzymes on or off.

Predict whether a reaction will be exergonic or endergonic based on ΔH and ΔS.

• The equation

  • ΔG = ΔH − TΔS, with T in kelvin (always positive).

  • A reaction is exergonic if ΔG < 0 and endergonic if ΔG > 0.

• Four cases

  1. ΔH negative, ΔS positive: ΔG is always negative, so the reaction is exergonic at all temperatures. It releases heat and increases disorder.

  2. ΔH positive, ΔS negative: ΔG is always positive, so the reaction is endergonic at all temperatures.

  3. ΔH negative, ΔS negative: the reaction is exergonic only at low temperatures, where the −TΔS term is small.

  4. ΔH positive, ΔS positive: the reaction is exergonic only at high temperatures, where the entropy term outweighs the positive ΔH.

• Example

  • Breaking down a large polymer into monomers has a positive ΔS, because the products are more disordered. If ΔH is also negative, the reaction is clearly exergonic.

• Tip

  • Check the signs of ΔH and ΔS first. Only the mixed-sign cases (3 and 4) depend on temperature.

Determine whether two reactions can be successfully coupled based on their changes in free energy.

• The rule

  • Two reactions can be coupled when the sum of their ΔG values is negative.

  • The reactions must also share a common intermediate or be linked by an enzyme.

• Step-by-step approach

  1. Identify the ΔG of each reaction, using the correct sign and noting which direction the reaction runs.

  2. Add the two values.

  3. If the total is less than zero, coupling is successful. If the total is zero or positive, coupling will not drive the reaction forward.

• Examples

  • Reaction A: ΔG = +5 kcal/mol, and reaction B (ATP hydrolysis): ΔG = −7.3 kcal/mol. The sum is −2.3 kcal/mol, so coupling works.

  • Reaction A: ΔG = +10 kcal/mol, and reaction B: ΔG = −7.3 kcal/mol. The sum is +2.7 kcal/mol, so coupling does not work.

• Notes

  • An exergonic reaction can drive an endergonic reaction only if it releases more free energy than the endergonic reaction requires.

  • If a reaction is run in reverse, change the sign of its ΔG.

Evaluate how changing cellular conditions will change the free energy of a reversible reaction.

• Key equation

  • ΔG = ΔG°′ + RT ln Q, where Q is the reaction quotient, which equals [products] / [reactants] at any given moment.

  • ΔG°′ is the standard free energy change, and ΔG is the actual value under cellular conditions.

• Effects of changing concentrations

  • Adding more reactant or removing product makes Q smaller, so ΔG becomes more negative and the forward reaction is more favorable.

  • Adding more product or removing reactant makes Q larger, so ΔG becomes more positive and the forward reaction is less favorable.

  • At equilibrium, Q equals K and ΔG = 0, so there is no net change.

• Biological examples

  • Cells keep the ATP/ADP ratio high, far from equilibrium, so ATP hydrolysis has a much more negative ΔG in the cell than the standard value.

  • In a metabolic pathway, if the next reaction rapidly uses up the product, it keeps Q low and pulls the reaction forward.

• Other conditions

  • Temperature changes ΔG through the TΔS term, and pH changes the concentration of H⁺ (a reactant or product in many reactions).

Interpret graphs of free energy vs. progress of reaction to determine the thermodynamic favorability, likelihood of coupling, and relative rates of reactions for different chemical reactions.

• What the graph shows

  • The y-axis is free energy and the x-axis is the progress of the reaction (reactants on the left, products on the right).

• Thermodynamic favorability

  • Compare the starting (reactant) level with the ending (product) level.

  • If the products are lower than the reactants, ΔG is negative and the reaction is exergonic (favorable).

  • If the products are higher, ΔG is positive and the reaction is endergonic.

  • ΔG is the vertical difference between the two levels.

• Likelihood of coupling

  • A reaction with a large drop in free energy can drive one with a smaller rise.

  • Compare the sizes of the drop and the rise. If the drop is larger, the sum of the ΔG values is negative, and coupling is possible.

• Relative rates

  • The reaction rate depends on the height of the peak (the activation energy), which is the difference between the reactants and the top of the barrier.

  • A lower peak means a faster reaction. A higher peak means a slower reaction.

  • Favorability and rate are independent. A very favorable reaction can still be slow if its peak is high.

• Reading tip

  • Use the start and end levels for ΔG (thermodynamics), and the peak height for the rate (kinetics).

Interpret data to assess changes in enzyme function and predict the effects (i.e., activating or inhibiting) of regulatory molecules.

• Step-by-step approach

  1. Identify the control condition (enzyme without the regulatory molecule) and the experimental condition (with the molecule).

  2. Compare the reaction rate (or product formed over time) between the two conditions.

  3. If the rate is lower with the molecule, it is an inhibitor. If it is higher, it is an activator.

  4. Assess how it works by checking the effect of raising substrate concentration.

• Distinguishing inhibitor types from data

  • If high substrate overcomes the inhibition, so that the same maximum rate is reached, the inhibitor is competitive.

  • If high substrate does not restore the maximum rate, the inhibitor is noncompetitive (allosteric).

• Other clues

  • No change in rate means the molecule has no regulatory effect on that enzyme.

  • A rate that depends on concentration of the molecule shows how strong the effect is.

• Example

  • If adding molecule X cuts product formation in half and adding extra substrate does not recover it, X is probably a noncompetitive inhibitor.

Evaluate the effect of regulatory molecules (e.g., feedback inhibitors) on metabolic pathways.

• Metabolic pathway basics

  • In a pathway (A → B → C → D), each step is catalyzed by a different enzyme.

  • Regulating early or committed steps controls the whole pathway.

• Feedback inhibition

  • In feedback (end-product) inhibition, the final product of a pathway inhibits an enzyme early in the pathway, usually by binding an allosteric site.

  • When the product accumulates, the pathway slows. When the product is used up, inhibition is released and the pathway speeds up.

  • This is a negative feedback loop that prevents waste of energy and materials.

• Effects of regulatory molecules on the pathway

  • Inhibiting the early enzyme reduces flow through the pathway. The intermediates and the final product decrease.

  • Inhibiting a later enzyme causes the substrate of that step to build up while the products after it decrease.

  • Activating an enzyme increases flux through the steps that follow it.

• Example

  • In bacteria, the amino acid isoleucine inhibits the first enzyme in its own synthesis pathway, so the cell stops making it when enough is present.

Relate the pH scale to the concentration of hydrogen ions (aka H+ or “protons”) in a solution.

• Definition

  • pH = −log₁₀[H⁺], where [H⁺] is the concentration of hydrogen ions in moles per liter.

• Interpreting the scale

  • pH 7 is neutral, with [H⁺] = 10⁻⁷ M (pure water).

  • pH below 7 is acidic, with a higher [H⁺].

  • pH above 7 is basic (alkaline), with a lower [H⁺].

• Key relationships

  • The scale is logarithmic, so each 1-unit change in pH is a 10-fold change in [H⁺].

  • A lower pH means more H⁺.

  • Example: a solution at pH 3 has 10,000 times more H⁺ than a solution at pH 7.

• Acids and bases

  • An acid is a substance that donates H⁺ (raising [H⁺] and lowering pH).

  • A base is a substance that accepts H⁺ or releases OH⁻ (lowering [H⁺] and raising pH).

• Biological relevance

  • Most cells keep the cytosol near pH 7.2, and many enzymes work only in a narrow pH range.

  • Buffers resist changes in pH.

Evaluate environmental effects (e.g., pH, temperature) on enzyme catalyzed reactions rates.

• Temperature

  • As temperature rises from low values, the rate increases, because molecules move faster and collide more often.

  • Each enzyme has an optimum temperature where the rate is highest (about 37 °C for many human enzymes).

  • Above the optimum, the enzyme denatures: its hydrogen bonds and other weak interactions break, the 3D shape is lost, the active site is destroyed, and the rate drops sharply.

  • At very low temperatures, the enzyme is not denatured, but the rate is low.

• pH

  • Each enzyme has an optimum pH, and activity falls on either side of it.

  • Changes in pH alter the charge of amino acid side chains, which disrupts ionic bonds and hydrogen bonds, changes the shape of the protein, and affects substrate binding.

  • Extreme pH can denature the enzyme.

  • Examples: pepsin works best at about pH 2 in the stomach, and trypsin works best at about pH 8 in the small intestine.

• Other factors

  • Substrate concentration increases the rate until the enzyme is saturated.

  • Enzyme concentration increases the rate in proportion, if substrate is plentiful.

  • Salt concentration can also affect protein shape.

• Graph shape

  • Plots of rate versus temperature or pH are typically bell-shaped, with a peak at the optimum.

Relate anabolic and catabolic pathways to overall changes in free energy.

• Catabolic pathways

  • They break large molecules into small ones, an overall exergonic process with negative ΔG.

  • They release free energy, and part of it is captured by making ATP and electron carriers (such as NADH).

  • Entropy generally increases as products become smaller and more numerous.

• Anabolic pathways

  • They build complex molecules from simple ones, an overall endergonic process with positive ΔG.

  • They require an input of free energy, supplied by coupling to ATP hydrolysis or the use of electron carriers.

• How they connect

  • The free energy released by catabolism is used to make ATP, and the ATP is used to drive anabolism.

  • Energy is not recycled perfectly, because some is lost as heat (second law of thermodynamics), so cells need a constant energy supply.

• Example

  • Cellular respiration (catabolic): glucose + O₂ → CO₂ + H₂O has ΔG° of about −686 kcal/mol.

  • Photosynthesis (anabolic): CO₂ + H₂O → glucose + O₂ has ΔG° of about +686 kcal/mol and is powered by light energy.

Interpret free energy diagrams to predict the free energy changes involved in a reaction, the kinetics of a reaction, and the impact of adding an enzyme to a reaction.

• Free energy change (ΔG)

  • Find the difference between the product level and the reactant level.

  • Products lower means ΔG negative (exergonic). Products higher means ΔG positive (endergonic).

• Kinetics

  • The activation energy is the difference between the reactant level and the peak (transition state).

  • A lower peak means a faster reaction. A higher peak means a slower reaction.

  • A reaction can be thermodynamically favorable and still be very slow if the barrier is high.

• Impact of adding an enzyme

  • The peak is lowered, so the activation energy decreases and the reaction rate increases.

  • The reactant and product levels stay the same, so ΔG does not change.

  • The equilibrium is the same, but it is reached faster.

  • The enzyme lowers the barrier for both the forward and reverse reactions.

• How to describe the enzyme curve

  • On the diagram, the enzyme-catalyzed pathway is usually drawn as a lower peak (sometimes with several smaller peaks) between the same start and end points.

Interpret enzyme kinetics graphs to determine the relative rate of an enzyme catalyzed reaction under different reaction conditions.

• Rate versus substrate concentration graph

  • At low [substrate], the rate rises steeply and nearly linearly, because many active sites are free.

  • At high [substrate], the curve levels off at Vmax, the maximum rate, because all enzyme active sites are occupied (saturation).

  • Km is the substrate concentration at half of Vmax. A lower Km means the enzyme binds its substrate more tightly (higher affinity).

• Comparing conditions

  • Draw a vertical line at a given [substrate] and compare the heights of the curves. A higher curve means a faster rate under that condition.

  • More enzyme raises Vmax.

• Effects of inhibitors on the graph

  • Competitive inhibitor: Vmax stays the same, but more substrate is needed to reach it, so Km increases and the curve shifts to the right.

  • Noncompetitive inhibitor: Vmax decreases, and the curve plateaus lower.

• Other graphs

  • Rate versus temperature or pH curves are bell-shaped. The rate at a given value is read from the curve, and the peak is the optimum.

  • Product versus time graphs have a steeper slope for a faster reaction, and the slope flattens as substrate runs out.

Predict the impact of different drugs on enzyme activity based on varying reaction conditions.

• Step-by-step approach

  1. Identify the type of drug: competitive inhibitor, noncompetitive (allosteric) inhibitor, irreversible inhibitor, or activator.

  2. Consider the conditions, especially the substrate concentration, and decide how the drug’s mechanism responds to them.

  3. Predict the effect on rate, Vmax, and Km.

• Competitive inhibitor

  • It competes with the substrate for the active site.

  • At low substrate, inhibition is strong. At high substrate, inhibition is reduced or overcome, because substrate outcompetes the drug.

  • Vmax is unchanged and Km increases.

• Noncompetitive (allosteric) inhibitor

  • It binds at a separate site and lowers the enzyme’s activity, regardless of substrate concentration.

  • Adding substrate does not overcome the inhibition.

  • Vmax decreases.

• Irreversible inhibitor

  • It permanently disables enzyme molecules, so the effective amount of active enzyme decreases and Vmax drops. More substrate does not help.

• Allosteric activator

  • It increases the rate, especially at lower substrate concentrations, by stabilizing the active form of the enzyme.

• Effect of other conditions

  • Raising the drug concentration increases the inhibition up to saturation.

  • Changes in pH or temperature that alter enzyme shape can change how well the drug binds.

• Example

  • If a drug’s inhibition disappears when substrate is increased 100-fold, the drug is competitive.