Ten lessons linking molecular state, assay calculations, enzyme kinetics and energy accounting, with two interactive visuals each.
Introductory biochemical reasoning, not a full major curriculum. Clinical diagnosis, drug advice, exhaustive pathways and real-cell ATP-yield predictions are outside scope. No jurisdiction-specific certification standard is claimed.
General chemistry, logarithms, molar units, basic algebra and familiarity with proteins and reactions.
Course outline
pH moves the proton balance
Use a one-site acid equilibrium to calculate protonation fractions.
Charge belongs to each site
Calculate a hypothetical molecule charge from explicit ionization states.
A folded shape is not a sequence
Separate levels of protein structure and count peptide bonds.
Absorbance needs a path length
Use Beer-Lambert concentration calculations with stated assay assumptions.
Saturation is not equilibrium
Use Michaelis-Menten and distinguish steady state from equilibrium.
One inhibitor, two different questions
Predict competitive-inhibition effects using the stated kinetic model.
Turnover needs an enzyme denominator
Calculate kcat and keep catalytic efficiency distinct from capacity.
Favorable is not fast
Use reaction free energy and concentration dependence.
Coupling needs a shared chemical path
Add free-energy changes only for a physically coupled net process.
A gradient is a link, not a magic ATP count
Trace energy conversion and calculate an explicitly toy proton budget.
Sources and curriculum note
Sources fetched October 4, 2026. These are archived courses, not a current universal syllabus. All teaching scenarios are explicitly hypothetical.
Read every lesson below. The interactive reader above contains the same explanations, with visual tools and quizzes.
1. pH moves the proton balance
Learning goal: Use a one-site acid equilibrium to calculate protonation fractions.
For a single acid site HA ⇌ H⁺+A⁻, pKa describes the acid dissociation equilibrium. In a dilute ideal model, pH=pKa+log₁₀([A⁻]/[HA]). At pH=pKa, the two forms are equally abundant. Above pKa the deprotonated form dominates; below pKa the protonated form dominates. It is a balance, not a sudden switch.
Let r=10^(pH-pKa). The deprotonated fraction is r/(1+r); the protonated fraction is 1/(1+r). A ratio of ten is not a fraction of ten or exactly one. The simple concentration relation assumes activities can be approximated by concentrations. Protein environments can shift pKa, and multiple interacting sites need a more careful model.
In an explicitly hypothetical one-site acid with pKa 6 at pH 7, r=10 and deprotonated fraction is 10/11≈0.90909. The chosen pKa is a toy input, not a claimed value for a named amino acid. Use the formula and given assumptions rather than inventing an actual molecule or predicting a medical effect.
Official source grounding: B0, B3. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
If HA has one relative unit, A⁻ has ten, totaling eleven.
Fraction is 10/11≈0.90909; remaining fraction 1/11≈0.09091.
Practice problem and solution
New hypothetical one-site acid, pKa=6.8, total concentration 200 μM, pH 7.4. Treat the site as ideal and isolated. Enter the deprotonated concentration in μM, to 2 decimals.
The ratio is 10^(7.4-6.8)=10^0.6=3.981. The deprotonated fraction is 3.981/4.981=0.7993. Concentration is 0.7993·200=159.85 μM.
Mental model: pH changes a protonation balance, not an on/off switch.
Common trap: Treating a ratio as a fraction.
2. Charge belongs to each site
Learning goal: Calculate a hypothetical molecule charge from explicit ionization states.
A molecule can carry both positive and negative sites. A zwitterion has opposite charges within one molecule and may have net zero charge; it is not a molecule without charged groups. For free amino acids, the amino and carboxyl groups can carry opposite charges. Side chains add further acid-base possibilities.
Use explicit charge conventions. An acid site HA is neutral here and A⁻ has charge -1. A basic site BH⁺ has charge +1 and B is neutral. Protonating a site raises its charge by one unit under these conventions. Free amino-acid terminal groups are not the same as peptide-bonded backbone groups; do not count every residue as a free amino acid.
For an original hypothetical three-site molecule, two deprotonated acid sites contribute -1 each and one protonated basic site contributes +1. Net charge is -1. This is a bookkeeping exercise with stated states, not a claim that real groups are completely ionized at a particular pH. Fractional mean charge would require equilibrium fractions instead.
Official source grounding: B0, B1. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Hypothetical molecule has two A⁻ acid sites and one BH⁺ basic site. Find net charge.
First acid contributes -1; second acid contributes -1.
The protonated base contributes +1.
Sum -1-1+1=-1 charge unit.
Charged sites remain even if another state combination gives net zero.
Practice problem and solution
New hypothetical molecule with three independent acidic sites (pKa 3, each charge -1 when deprotonated) and two independent basic sites (pKa 8, each charge +1 when protonated), at pH 6. Enter the expected net charge using fractional occupancy, to 4 decimals.
Acid site deprotonated fraction: 10^3/(1+10^3)=0.999001, so three sites give -2.9970. Basic site protonated fraction: 1/(1+10^(-2))=0.990099, so two sites give +1.9802. Net=-2.9970+1.9802=-1.0168.
Mental model: Net charge sums site charges.
Common trap: Counting peptide residues as free amino acids.
3. A folded shape is not a sequence
Learning goal: Separate levels of protein structure and count peptide bonds.
Primary structure is the amino-acid sequence of a polypeptide. Secondary structure describes local backbone organization such as helices and sheets. Tertiary structure is the overall fold of one chain. Quaternary structure describes organization of multiple polypeptide chains. These are descriptive levels, not four compulsory stages every protein must pass through.
The peptide bond has partial double-bond character that constrains backbone geometry. Secondary motifs use organized backbone interactions; side chains contribute to the overall fold. Disrupting folding does not automatically cleave the covalent peptide backbone. State which bonds or interactions a hypothetical treatment changes before claiming that sequence was lost.
A hypothetical complex with two separate linear chains of four residues each has three peptide bonds per chain, six total. Association into a complex does not create a peptide bond between chains unless explicitly stated. It supplies a quaternary arrangement; each chain still has its own primary sequence. This is an invented complex, not a named real protein.
Official source grounding: B1. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Count peptide bonds in a hypothetical complex with two separate linear chains of four residues.
A linear chain joins neighboring residues with one bond per join.
Four residues have three joins.
Two separate such chains have 2·3=6 peptide bonds.
Noncovalent subunit association adds zero peptide bonds under the stated assumption.
Practice problem and solution
New hypothetical complex: three separate linear chains of 150, 90 and 60 residues, held together by one disulfide link between chains. A protease then cleaves 4 peptide bonds in the 150-residue chain. Enter the peptide bonds remaining.
Bonds before cleavage: 149+89+59=297. The disulfide link is not a peptide bond, so it adds none. After cleaving 4, 297-4=293 peptide bonds remain.
Mental model: Sequence, fold and subunits answer different questions.
Common trap: Using total residues minus one across separate chains.
4. Absorbance needs a path length
Learning goal: Use Beer-Lambert concentration calculations with stated assay assumptions.
In a suitable linear absorbance regime, Beer-Lambert law is A=εlc. A is dimensionless, ε can be in M⁻¹cm⁻¹, l in cm, and c in mol/L. Solve c=A/(εl). Protein extinction depends on composition and conditions, so a numerical ε must be supplied or verified for the particular assay rather than guessed.
Correct the absorbance for a matched blank and account for dilution. A diluted sample concentration is not the stock concentration. The simple model assumes suitable optical behavior and a known path length; scattering, mixtures and nonlinear response can defeat a naive calculation. A single absorbance reading does not prove purity.
In an explicitly hypothetical assay, blank-corrected A=0.30, ε=15000 M⁻¹cm⁻¹ and l=1 cm give c=0.000020 M=20 μM. A fivefold dilution means stock concentration 100 μM. These are invented teaching inputs, not a real protein measurement or a universal protein extinction coefficient.
Official source grounding: B0. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Hypothetical A=0.30 after blank correction, ε=15000 M⁻¹cm⁻¹, l=1 cm; sample was diluted fivefold. Find stock concentration.
Diluted c=0.30/(15000·1)=0.000020 M.
Convert M to μM: 0.000020·10⁶=20 μM.
Stock is five times diluted concentration: 100 μM.
The result assumes the supplied extinction coefficient and linear optical model apply.
Practice problem and solution
New hypothetical blank-corrected measurement: raw A=0.36, blank A=0.04, ε=9000 M⁻¹cm⁻¹, path length 0.5 cm, sample diluted 1:20 (dilution factor 20) before reading. Enter the original stock concentration in μM, to 1 decimal.
Corrected A=0.32. c=0.32/(9000·0.5)=7.111×10⁻⁵ M in the cuvette. Times 20 gives 1.4222×10⁻³ M, which is 1422.2 μM.
Mental model: Absorbance estimates concentration with a calibrated model.
Common trap: Forgetting path length or dilution.
5. Saturation is not equilibrium
Learning goal: Use Michaelis-Menten and distinguish steady state from equilibrium.
For a simple single-substrate initial-rate model, v=Vmax[S]/(Km+[S]). Under the usual quasi-steady-state approximation, enzyme-substrate complex forms and disappears at nearly equal rates after the early transient. That does not mean the overall reaction is at thermodynamic equilibrium: product can still accumulate.
Vmax is kcat times total enzyme concentration for this model. Km is the substrate concentration at half Vmax. Km is generally a combination of rate constants, not automatically a binding dissociation constant; it approaches Kd only under an additional rapid-equilibrium-type rate assumption. Avoid using the word affinity as an unconditional definition.
In a hypothetical assay Vmax=100 μM/min, Km=2 mM, at [S]=2 mM the rate is 50 μM/min; at [S]=8 mM it is 80 μM/min. Increasing substrate does not increase the rate indefinitely. The bar cases are outputs of the toy formula, not observed enzyme measurements. Assume initial rates, substrate excess, and no cooperative behavior.
Official source grounding: B2. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Hypothetical Vmax=100 μM/min, Km=2 mM. Find v at S=8 mM.
Use the same units for S and Km.
S/(Km+S)=8/(2+8)=0.8.
v=100·0.8=80 μM/min.
At S=Km, fraction is 1/2, but that half-capacity condition is not equilibrium.
Practice problem and solution
New hypothetical enzyme in the simple Michaelis-Menten model with Vmax=100 μM/min. At S=2 mM the measured rate is 40 μM/min. Find Km, then enter the predicted rate at S=9 mM in μM/min.
40=100·2/(Km+2) gives Km+2=5, so Km=3 mM. At 9 mM: v=100·9/(3+9)=75 μM/min.
Mental model: Steady-state kinetics does not imply equilibrium.
Common trap: Defining Km unconditionally as affinity.
6. One inhibitor, two different questions
Learning goal: Predict competitive-inhibition effects using the stated kinetic model.
For ideal reversible competitive inhibition, v=Vmax[S]/(αKm+[S]) with α=1+[I]/Ki. The apparent Km increases to αKm while Vmax is unchanged. At very high substrate the inhibition is overcome in this ideal model. A rate at one substrate concentration still falls; unchanged Vmax does not mean unchanged rate.
An inhibition pattern is not a complete structural proof of a binding site. The competitive model describes mutually exclusive effective substrate and inhibitor occupancy under its assumptions. Mixed, uncompetitive, irreversible and cooperative behavior require other models. Do not classify all inhibitors using only one lower rate reading.
In a hypothetical assay Vmax=100 μM/min, Km=2 mM, α=3 and S=2 mM, v=25 μM/min rather than the uninhibited 50. Saturating capacity stays 100 while the substrate needed for half-capacity becomes 6 mM. The bars compare toy model outputs, not actual drug results, and no treatment recommendation is implied.
Official source grounding: B2. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Hypothetical competitive inhibitor with α=3, Vmax=100 μM/min, Km=2 mM, S=2 mM. Find v.
Apparent Km=αKm=3·2=6 mM.
v=100·2/(6+2)=200/8=25 μM/min.
Without inhibitor, v=100·2/(2+2)=50 μM/min.
The ideal limiting Vmax remains 100, although the fixed-substrate rate halves.
Practice problem and solution
New hypothetical competitive inhibitor in the ideal model: Vmax=80 μM/min, Km=4 mM, I/Ki=3. Find the substrate concentration in mM that gives a rate of 40 μM/min with the inhibitor present.
α=1+3=4, so apparent Km=16 mM. The rate is Vmax/2 when S equals the apparent Km, so S=16 mM. Without the inhibitor, 4 mM would give half Vmax.
Mental model: Unchanged capacity does not mean unchanged rate.
Common trap: Treating one rate change as proof of mechanism.
7. Turnover needs an enzyme denominator
Learning goal: Calculate kcat and keep catalytic efficiency distinct from capacity.
In the simple saturating model, Vmax=kcat[E]total. Therefore kcat=Vmax/[E]total and has inverse-time units. Vmax depends on how much enzyme is present; it is not an intrinsic turnover comparison unless enzyme amounts match. Use active-site concentration consistently with the definition of kcat.
At low substrate, v≈(kcat/Km)[E]total[S]. The ratio kcat/Km describes catalytic efficiency in that low-substrate regime and has concentration⁻¹time⁻¹ units. Higher kcat alone does not guarantee higher low-substrate rate if Km also changes. Saturating and substrate-limited questions require different comparisons.
In a hypothetical assay Vmax=0.60 μM/s and active-site concentration 0.020 μM, kcat=30 s⁻¹. If enzyme amount doubles without changing kinetic parameters, Vmax doubles to 1.20 μM/s but kcat remains 30. These inputs are toy values, not measurements of a real catalyst or claims about diffusion limits.
Official source grounding: B2. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Hypothetical Vmax=0.60 μM/s and active sites 0.020 μM. Find kcat.
Use matching μM units in numerator and denominator.
kcat=0.60/0.020=30 s⁻¹.
This is turnover per active site per second at saturation under the model.
Doubling enzyme doubles Vmax, not the individual site turnover.
Practice problem and solution
New hypothetical assay: total enzyme 0.10 μM, of which 60% is active, gives Vmax=2.4 μM/s. Find kcat from active sites, then enter the predicted Vmax in μM/s for 0.25 μM total enzyme with the same active fraction.
Active sites=0.06 μM, so kcat=2.4/0.06=40 s⁻¹. New active sites=0.25·0.6=0.15 μM, so Vmax=40·0.15=6.0 μM/s.
Mental model: Capacity and per-site turnover are different.
Common trap: Comparing Vmax without matching enzyme concentration.
8. Favorable is not fast
Learning goal: Use reaction free energy and concentration dependence.
Thermodynamic favorability is distinct from reaction rate. At fixed conditions, ΔG<0 favors the forward direction, ΔG>0 favors reverse, and ΔG=0 describes equilibrium. A favorable reaction may remain slow because of its activation barrier. Catalytic pathways can change rates without changing the reaction free-energy difference or equilibrium constant.
The relation ΔG=ΔG°′+RT ln Q uses a dimensionless reaction quotient and compatible units. The biochemical prime marks a specified transformed standard state, commonly fixed pH; it is not the actual mixture concentration. Logarithm here is natural log, not log base ten. Concentrations used in an ideal approximation are divided by standard concentration so Q is dimensionless.
In a hypothetical transformed reaction with ΔG°′=5 kJ/mol, R=0.008314 kJ/(mol K), T=300 K and Q=0.1, ΔG≈-0.74311 kJ/mol. A positive standard value can therefore coexist with forward-favorable actual conditions. These are invented reaction parameters, not a named real biochemical reaction or a claim about cellular concentrations.
Official source grounding: B3. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Mental model: Favorability and speed answer different questions.
Common trap: Treating ΔG°′ as the actual value for every mixture.
9. Coupling needs a shared chemical path
Learning goal: Add free-energy changes only for a physically coupled net process.
Free-energy changes add when reactions are added to form a net reaction under compatible conditions. Cancel a shared intermediate and retain stoichiometric coefficients. A favorable summed ΔG identifies a possible net direction, but simply placing an unrelated exergonic reaction nearby does not guarantee useful coupling.
Biochemical coupling requires a mechanism that links the transformations, often through a shared intermediate or transfer event. ATP participates in many such transformations, but a broken bond alone does not release free energy in isolation: the whole reactant-to-product change matters. Keep thermodynamic accounting separate from claims that a mechanism has been established.
In an explicitly hypothetical two-step scheme A→B costs +12 kJ/mol, while B+C→D releases -20 kJ/mol. The shared B cancels to give A+C→D with ΔG=-8 kJ/mol per net event. The toy energy values are not real ATP numbers. If the second process did not consume B or otherwise couple to the first, arithmetic alone would not prove that A transforms.
Official source grounding: B3. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Hypothetical A→B has +12 kJ/mol and B+C→D has -20 kJ/mol. Find net reaction and ΔG.
Add reactants A+B+C and products B+D.
Cancel one B on each side, leaving A+C→D.
Add compatible values: 12+(-20)=-8 kJ/mol.
The proposed shared-intermediate scheme links the steps; its favorable net value does not specify how fast it runs.
Practice problem and solution
New hypothetical sequence of shared-intermediate steps: A→B +21 kJ/mol, B→C -34 kJ/mol, C+D→E +8 kJ/mol. Enter the net ΔG in kJ/mol for three complete passes of the sequence.
B and C appear as product then reactant, so they cancel. One pass is A+D→E with ΔG=21-34+8=-5. Three passes give -15 kJ/mol.
Mental model: Coupling requires a link and compatible thermodynamic accounting.
Common trap: Assuming an unrelated favorable reaction drives any unfavorable one.
10. A gradient is a link, not a magic ATP count
Learning goal: Trace energy conversion and calculate an explicitly toy proton budget.
In respiration, electron transport can generate a proton electrochemical gradient across a membrane, and proton movement through ATP-synthesizing machinery can drive ATP formation. Chemical and electrical parts both contribute to that gradient. A mechanism diagram is an energy-dependency map, not a literal drawing of membrane proteins.
A gradient requires sufficiently separated compartments and a pathway that couples proton return to synthesis. Proton leakage bypasses productive coupling. It can reduce ATP yield without requiring every electron-transfer event to stop. Actual ATP yields depend on machinery and transport costs; do not turn a toy stoichiometry into a universal ATP-per-carrier rule.
In a hypothetical sealed two-compartment model, a budget of 12 proton equivalents is available and the stipulated cost is 4 equivalents per ATP, including all modeled costs. Ideal output is 3 ATP units. If 4 equivalents leak away first, the remaining 8 support 2 ATP units. Those numbers are deliberately invented for accounting and are not measured mitochondrial stoichiometry.
Official source grounding: B5, B6. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Hypothetical proton budget: 12 equivalents, cost 4 per ATP including all costs, leak 4 equivalents. Calculate yield.
This is a toy accounting result, not a real ATP yield or biological stoichiometry.
Practice problem and solution
New hypothetical toy model: 36 proton equivalents are available, 40% leak back without use, and each ATP costs a stipulated 4.5 equivalents. Enter the ideal ATP units made, to 1 decimal.
Leak is 0.4·36=14.4, so useful budget is 21.6. 21.6/4.5=4.8 ATP units. These numbers are toy assumptions, not real mitochondrial values.
Mental model: Follow the coupled pathway and state yield assumptions.
Common trap: Reporting a toy proton cost as a real universal ATP ratio.