# ADAT practice questions: Biochemistry and physiology

Ten original ADAT practice questions on biochemistry and physiology, each answered on this page with a rationale and a source.

Last updated: 2026-09-18.

## Question 1

A molecule initially contains adenine linked to ribose and one phosphate attached to that sugar. A specified reaction removes only the phosphate; the base–sugar connection and the sugar's 2′ hydroxyl remain unchanged. Which description of the remaining base–sugar molecule follows?

- A. A nucleotide, because retaining the base preserves its nucleotide identity.
- B. A deoxyribonucleoside, because phosphate removal removes the 2′ oxygen.
- C. A free nitrogenous base, because the phosphate was the base–sugar connection.
- D. A ribonucleoside, because the base and ribose remain without a phosphate.

**Answer D:** A ribonucleoside, because the base and ribose remain without a phosphate.

The starting description is a ribonucleotide. Removing its only phosphate while preserving the adenine–ribose connection leaves a ribonucleoside, specifically adenosine. The stipulated 2′ hydroxyl remains; phosphate removal and replacement of a 2′ hydroxyl by hydrogen are different changes. Why the other choices do not fit: A: A base is present in both nucleosides and nucleotides. A nucleotide also contains phosphate; the specified remaining molecule has none. B: The stem explicitly preserves the 2′ hydroxyl. Deoxyribose's 2′ hydrogen cannot be inferred from phosphate removal. C: The base–sugar connection is explicitly preserved. A phosphate on the sugar is not that connection, so the result is not a free base alone.

Source: [Alberts et al., Molecular Biology of the Cell, 4th ed., 2002, [Chemical Components of a Cell](](https://www.ncbi.nlm.nih.gov/books/NBK26883/)

## Question 2

A fictional protein consists of two associated polypeptide chains. After a treatment, the chains are separate. Measurements establish unchanged amino-acid sequences, intact peptide backbones and preserved local α-helical segments. No full three-dimensional structure or activity measurement is available after treatment. Which conclusion is best supported?

- A. The treatment cleaved the primary structure where the two chains had associated.
- B. The quaternary association changed, while preservation of complete function is unestablished.
- C. The preserved α helices establish that each chain retained its entire tertiary structure.
- D. The intact backbones establish that the protein's organization did not change.

**Answer B:** The quaternary association changed, while preservation of complete function is unestablished.

Quaternary organization concerns association between polypeptide chains, and that association has changed. Intact sequences and peptide backbones establish preservation of primary connectivity, while local α helices provide limited secondary-structure information. Neither observation establishes the entire tertiary conformation or post-treatment function. Why the other choices do not fit: A: Separating two associated chains does not itself cleave either peptide backbone; the stem directly establishes intact backbones and unchanged sequences. C: Local secondary-structure preservation does not determine the complete three-dimensional arrangement of a chain. The necessary tertiary-structure measurement is absent. D: Primary connectivity is one organizational level. The measured loss of association between chains is a real change at another level.

Source: [Alberts et al., same edition, [Shape and Structure of Proteins](](https://www.ncbi.nlm.nih.gov/books/NBK26830/)

## Question 3

Two closed reaction mixtures have the same reactants, products, temperature and other conditions. One contains a catalyst that is regenerated overall and supplies a route with a lower activation barrier. The catalyst does not couple the reaction to another reaction or alter reactant/product free energies. Which prediction follows?

- A. The equilibrium constant increases because the products become lower in free energy.
- B. The equilibrium constant decreases because the catalyst binds the transition state.
- C. Equilibrium is approached faster, but the equilibrium constant remains unchanged.
- D. The forward reaction accelerates while the reverse reaction is thermodynamically forbidden.

**Answer C:** Equilibrium is approached faster, but the equilibrium constant remains unchanged.

Under the specified same-reaction, same-condition comparison, the catalyst alters the kinetic route and increases the rate of approach to equilibrium. It does not change the reactant–product free-energy difference or the equilibrium constant. Lowering a barrier is distinct from moving a thermodynamic endpoint. Why the other choices do not fit: A: The stem excludes an altered product free energy. A lower activation barrier does not imply a larger equilibrium constant. B: Stabilizing a transition-state route does not by itself lower the equilibrium constant, which concerns reactant/product equilibrium rather than the height of that barrier. D: Catalysis does not make the reverse reaction thermodynamically impossible. Faster approach to the same equilibrium is compatible with forward and reverse reaction pathways.

Source: [MIT 5.07SC, Fall 2013, [Lectures 5–6](](https://ocw.mit.edu/courses/5-07sc-biological-chemistry-i-fall-2013/e9540e994623cb307d2e08dcd3de9d72_MIT5_07SCF13_Lec5_6.pdf)

## Question 4

A fictional assay is stipulated to follow v₀=Vmax[S]/(Km+[S]) exactly, with one substrate, noncooperative steady-state initial rates, stable active enzyme, substrate excess, negligible depletion/product/reverse flux and no limiting additional reactant or cofactor. pH and temperature are fixed, and each rate uses the same 25 nM concentration of catalytically competent active sites. At [S]=6 µM, v₀=24 µM/min; at [S]=18 µM, v₀=40 µM/min. Which pair of Km and kcat is compatible with both rates and the active-site concentration?

- A. Km=9 µM; kcat=2,400 s⁻¹.
- B. Km=6 µM; kcat=40 s⁻¹.
- C. Km=9 µM; kcat=40 s⁻¹.
- D. Km=9 µM; kcat=0.040 s⁻¹.

**Answer C:** Km=9 µM; kcat=40 s⁻¹.

The first rate gives Vmax=4(Km+6); the second gives Vmax=(20/9)(Km+18). Equating them yields 36Km+216=20Km+360, so Km=9 µM and Vmax=60 µM/min. Convert 25 nM to 0.025 µM: kcat=60/0.025=2,400 min⁻¹=40 s⁻¹. The parameters reproduce both supplied rates: 60×6/15=24 and 60×18/27=40. Why the other choices do not fit: A: The computed turnover is 2,400 per minute. Reporting that number per second omits the division by 60. B: A measured substrate concentration is not automatically Km. Km 6 with kcat 40 and 0.025 µM active sites gives Vmax 60 and predicts 30 µM/min at substrate 6 µM, contradicting the given 24. D: This value results from treating 25 nM as 25 µM before dividing by 60. The active-site concentration is 0.025 µM, a thousandfold smaller, so turnover per site is a thousandfold larger.

Source: [MIT 5.07SC, Fall 2013, [Lectures 7–8](](https://ocw.mit.edu/courses/5-07sc-biological-chemistry-i-fall-2013/b4fd2e5ca7f7bfcffd8f98ca6afff53d_MIT5_07SCF13_Lec7_8.pdf)

## Question 5

A fictional enzyme's uninhibited parameters are Vmax 120 µM/min and Km 20 µM. Separate experiments with reversible inhibitors P and Q fit the classical single-substrate reversible full-inhibition model exactly under controlled initial-rate conditions: P gives Vmax,app 30 µM/min and Km,app 10 µM; Q gives Vmax,app 60 µM/min and Km,app 30 µM. Active-enzyme concentration is unchanged; tight binding, irreversible/time-dependent loss, aggregation and cooperativity are excluded. Within the stated model, which interpretation is supported?

- A. P and Q both have general mixed patterns; their apparent Km changes have opposite directions.
- B. P is uncompetitive because its Vmax and Km both fall; Q has a general mixed pattern.
- C. P is pure noncompetitive because its Vmax falls; Q is competitive because its Km rises.
- D. P is competitive because its Km falls; Q is uncompetitive because its Vmax falls.

**Answer A:** P and Q both have general mixed patterns; their apparent Km changes have opposite directions.

For P, Vmax falls fourfold while Km falls only twofold; the unequal changes fit general mixed inhibition with fES 4 and fE 2. For Q, Vmax falls twofold while Km rises 1.5-fold, giving fES 2 and fE 3. Both factors exceed 1 and differ within each case. General mixed inhibition can lower or raise apparent Km. These fitted patterns do not establish a unique molecular binding location. Why the other choices do not fit: B: In the stated uncompetitive model, Vmax and Km decrease by the same factor. P's fourfold and twofold decreases are unequal, so merely observing two decreases is insufficient. C: Pure noncompetitive inhibition preserves Km, which P does not. Competitive inhibition preserves Vmax, which Q does not. Each half of this option omits a required comparison. D: Competitive inhibition in this model raises apparent Km without changing Vmax; P does neither. Uncompetitive inhibition lowers Km with Vmax by the same factor; Q instead raises Km.

Source: [Strelow et al., [Mechanism of Action Assays for Enzymes](](https://www.ncbi.nlm.nih.gov/books/NBK92001/)

## Question 6

Two structural records describe the same small molecule before and after a controlled change in solution conditions: CH3–C(=O)–NH–CH2–C(=O)–OH becomes CH3–C(=O)–NH–CH2–C(=O)–O−. The drawings specify the displayed protonation states; all unmarked atoms and connections are unchanged. A nearby binding site retains a positive charge, and separation and solvent conditions for the contact are otherwise matched. Which interpretation follows from the actual connectivity and charges?

- A. The terminal group remains formally neutral, so its interaction with the positive site can be classified only as hydrogen bonding.
- B. The terminal group becomes a carboxylate, and the adjacent amide C–N connection is cleaved during that change.
- C. The terminal group becomes a carboxylate, adding a charge-based attraction to the positive site while the amide remains intact.
- D. The terminal group becomes an aldehyde, adding a carbonyl dipole while the amide remains intact.

**Answer C:** The terminal group becomes a carboxylate, adding a charge-based attraction to the positive site while the amide remains intact.

The terminal C(=O)–OH has lost its displayed proton and is now C(=O)–O−, a carboxylate. The negative group can attract the stipulated positive site electrostatically. The separate C(=O)–NH connection is still present: it is an amide, not the bond removed in this protonation change. This establishes a local interaction contribution under the matched conditions, not the whole molecule's binding affinity or its predominant state at an unstated pH. Why the other choices do not fit: A: The explicit minus sign is a formal negative charge, not merely a partial charge within a polar neutral bond. Hydrogen bonding can coexist with other interactions, but it does not erase the charge-based attraction in this comparison. B: Carboxylate identification is correct, but the displayed carbonyl C–N bond is unchanged. Removing the carboxyl proton is different from hydrolyzing an amide. D: An aldehyde has H directly attached to its carbonyl carbon. The terminal carbon here remains attached to two oxygens; losing the hydroxyl proton does not create an aldehyde.

Source: [Alberts et al., Molecular Biology of the Cell, 4th ed](https://www.ncbi.nlm.nih.gov/books/NBK26883/)

## Question 7

Use cofactor broadly for a required nonprotein enzyme helper and coenzyme for an organic helper. Two purified systems have established requirements: system Z needs a bound Zn2+ ion for its peptide-hydrolysis reaction; system N needs NAD+ and forms NADH while its substrate is oxidized. Removing the relevant helper stops each reaction, and replacing that helper restores it under otherwise matched conditions. Which classification and role match both systems?

- A. Zn2+ is an inorganic cofactor; NAD+ is an organic coenzyme accepting reducing equivalents in the specified reaction.
- B. Zn2+ is an inorganic cofactor; NAD+ is an organic coenzyme carrying an acetyl group in the specified reaction.
- C. Zn2+ is an inorganic coenzyme; NAD+ is an organic coenzyme accepting reducing equivalents in the specified reaction.
- D. Zn2+ is an inorganic coenzyme; NAD+ is an organic coenzyme carrying an acetyl group in the specified reaction.

**Answer A:** Zn2+ is an inorganic cofactor; NAD+ is an organic coenzyme accepting reducing equivalents in the specified reaction.

The ion is an inorganic helper and therefore a cofactor, not an organic coenzyme under the stated convention. NAD+ is an organic helper whose conversion to NADH accompanies acceptance of reducing equivalents as the substrate is oxidized. Acetyl-group carriage instead describes an acetyl-CoA relationship. Restoration demonstrates the stipulated helper requirement; it supplies no nutritional dose or treatment inference. Why the other choices do not fit: B: The zinc classification fits. The NAD+/NADH change, however, is a redox relationship; it does not transfer an acetyl group onto NAD as the explanation claims. C: The NAD role fits, but being required for catalysis does not make a zinc ion organic. The declared coenzyme convention excludes this inorganic ion. D: It both misclassifies zinc as an organic-helper category and substitutes acyl carriage for the specified NAD redox role.

Source: [Alberts et al., Molecular Biology of the Cell, 4th ed](https://www.ncbi.nlm.nih.gov/books/NBK26911/)

## Question 8

A fictional transformed reaction has ΔG°′=+2.40 kJ/mol at 300 K. Temperature, pressure, pH, pMg and ionic strength remain fixed. Its dimensionless Q′ is 0.20; use R=0.008314 kJ/mol/K and ΔG′=ΔG°′+RT lnQ′. No kinetic measurements are supplied. Which conclusion is supported?

- A. The forward reaction is unfavorable because the standard transformed value is positive.
- B. The actual forward change is favorable, but the reaction rate is not established.
- C. The actual forward change is favorable and must be faster than any reaction with positive ΔG′.
- D. The reaction is at equilibrium because Q′ is below 1.

**Answer B:** The actual forward change is favorable, but the reaction rate is not established.

RT ln(0.20) is approximately −4.014 kJ/mol, so actual ΔG′ is approximately −1.614 kJ/mol. The negative sign favors forward change at this composition. No rate follows from that sign; activation barriers and kinetic conditions are not supplied. Why the other choices do not fit: A: This uses the standard value while ignoring the composition term. The actual value is negative at Q′ 0.20. C: The favorable sign does not rank reaction rates. A favorable conversion may still be slow, and no comparative kinetic data are supplied. D: Equilibrium requires ΔG′=0 and Q′=K′, not merely Q′<1. Here the calculated actual free-energy change is negative.

Source: [Alberts et al., Molecular Biology of the Cell, 4th ed](https://www.ncbi.nlm.nih.gov/books/NBK26838/)

## Question 9

A fictional stoichiometric ledger follows two free glucose molecules. One goes through glycolysis and both of its pyruvates are converted to lactate. The other goes through glycolysis, two pyruvate-dehydrogenase reactions and two complete TCA turns. Count NADH formed minus NADH consumed by just these listed reactions; do not include respiratory reoxidation, shuttles or other uses. Which pair is correct for net NADH generation and total CO2 release?

- A. 12 NADH and 6 CO2.
- B. 10 NADH and 4 CO2.
- C. 10 NADH and 6 CO2.
- D. 8 NADH and 6 CO2.

**Answer C:** 10 NADH and 6 CO2.

The lactate-directed glucose forms 2 NADH in glycolysis and consumes 2 during lactate formation, giving 0 net NADH and 0 CO2. The oxidative branch forms 2 cytosolic NADH in glycolysis,2 matrix NADH at PDH and 6 at TCA:10 total. PDH releases 2 CO2 and TCA releases 4:6 total. This is a pathway ledger, not an assertion that NADH accumulates unchanged in a living cell or that newest acetyl carbons are the first-turn CO2. Why the other choices do not fit: A: Twelve counts NADH formation from both glycolyses plus PDH/TCA but fails to subtract the 2 NADH consumed in lactate formation. B: Four CO2 counts the two TCA turns but omits the 2 released at PDH. Glycolysis and lactate formation add no CO2 here. D: Eight is the matrix NADH formation on the oxidative branch. The 2 cytosolic NADH from that branch also belong in the requested ledger; only the other branch consumes its glycolytic NADH through lactate formation.

Source: [Alberts et al., Molecular Biology of the Cell, 4th ed](https://www.ncbi.nlm.nih.gov/books/NBK26838/)

## Question 10

Two fictional preparations have the same endpoint glucose-6-phosphate concentration. Preparation R has a specified additional immediate need for NADPH but little need for new ribose. Enzymes for both oxidative and nonoxidative pentose-phosphate reactions are available. Which proposed interpretation best matches the product requirement and the limit of the concentration measurement?

- A. Oxidative PPP can supply NADPH; equal endpoint G6P concentrations establish equal total G6P turnover.
- B. Glycogen synthesis supplies NADPH; equal endpoint G6P concentrations leave its production and consumption rates unresolved.
- C. Nonoxidative PPP rearrangements supply NADPH; equal endpoint G6P concentrations leave its production and consumption rates unresolved.
- D. Oxidative PPP can supply NADPH and nonoxidative reactions can redistribute pentose carbon; endpoint G6P does not establish turnover.

**Answer D:** Oxidative PPP can supply NADPH and nonoxidative reactions can redistribute pentose carbon; endpoint G6P does not establish turnover.

Oxidative PPP chemistry generates NADPH. When pentose need is relatively low, nonoxidative carbon rearrangements can connect excess pentose carbon with glycolytic intermediates. That is a route consistent with the stated requirement, not a measured flux. Equal G6P pool sizes can coexist with different production and consumption rates. Why the other choices do not fit: A: The NADPH source is correct, but an endpoint pool size does not identify its input and output rates. Equal concentration does not establish equal total turnover. B: The measurement limit is correct, but glycogen synthesis assembles glucose residues rather than generating the NADPH supplied by oxidative PPP chemistry. C: The measurement limit is correct, but nonoxidative PPP reactions rearrange carbon skeletons. Assigning NADPH generation to those rearrangements confuses their role with the oxidative branch.

Source: [Alberts et al., Molecular Biology of the Cell, 4th ed](https://www.ncbi.nlm.nih.gov/books/NBK26838/)

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Official reference: [ADA 2026 ADAT Candidate Guide](https://www.ada.org/-/media/project/ada-organization/ada/ada-org/files/education/2026_adat_candidate_guide.pdf). Original exam-style questions written for study, never recalled exam content. Independent educational preparation, not affiliated with or endorsed by the American Dental Association. Practice results are not calibrated ADAT scores.