Practice Chemical & Physical Foundations questions for the Medical College Admission Test. Every question includes a full explanation of why the correct answer is right and why the tempting distractors are wrong.
10 example questions with full explanations. Use the interactive practice above to work through the complete set.
Question 1hard
During coronary angiography, a physician observes that an atherosclerotic plaque has reduced the radius of a coronary artery from 3.0 mm to 1.5 mm over a length of 2.0 cm. Assuming laminar flow and constant viscosity of blood (η = 0.003 Pa·s), by what factor does the resistance to flow change in the stenotic segment compared to the normal segment, and how does this affect flow rate if the pressure gradient is held constant?
- A.Resistance increases by a factor of 4; flow rate decreases to 1/4 of its original value
- B.Resistance increases by a factor of 16; flow rate decreases to 1/16 of its original value✓
- C.Resistance increases by a factor of 8; flow rate decreases to 1/8 of its original value
- D.Resistance increases by a factor of 2; flow rate decreases to 1/2 of its original value because cross-sectional area is halved
Correct answer: B
Poiseuille's law states that resistance R = 8ηL/(πr⁴), so resistance is inversely proportional to the fourth power of the radius. When radius is halved (from 3.0 mm to 1.5 mm), R_new/R_old = (r_old/r_new)⁴ = (3.0/1.5)⁴ = 2⁴ = 16. Since flow rate Q = ΔP/R (analogous to Ohm's law), at constant pressure gradient, Q decreases by the same factor of 16. Option A (factor of 4) reflects only the second power of radius, a common error confusing Poiseuille's law with Ohm's law applied to cross-sectional area. Option C (factor of 8) incorrectly applies a cubic relationship. Option D confuses area (r²) with the r⁴ dependence in Poiseuille's law. The r⁴ dependence makes small changes in vessel radius clinically devastating for blood flow.
Question 2hard
A galvanic cell is constructed using a zinc electrode in 1.0 M Zn²⁺ solution and a copper electrode in a solution where [Cu²⁺] = 0.0010 M. Given that E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V, and using the Nernst equation at 25°C (RT/F = 0.0257 V), what is the cell potential under these non-standard conditions?
- A.1.10 V; the non-standard [Cu²⁺] has no effect because zinc concentration is at standard conditions
- B.0.92 V; the Nernst equation reduces the cell potential because the low [Cu²⁺] disfavors the forward reaction at the cathode
- C.1.28 V; the low [Cu²⁺] concentration increases the driving force for copper deposition, raising the cell potential
- D.1.01 V; applying the Nernst equation to the full cell reaction with Q = [Zn²⁺]/[Cu²⁺] gives E_cell = 1.10 − (0.0257/2) × ln(1000) ≈ 1.10 − 0.089 ≈ 1.01 V✓
Correct answer: D
The standard cell potential E°_cell = E°_cathode − E°_anode = 0.34 − (−0.76) = 1.10 V. The cell reaction is Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), so Q = [Zn²⁺]/[Cu²⁺] = 1.0/0.0010 = 1000. The Nernst equation gives E_cell = E°_cell − (RT/nF) × ln(Q) = 1.10 − (0.0257/2) × ln(1000) = 1.10 − (0.01285)(6.908) = 1.10 − 0.0888 ≈ 1.01 V. Option A incorrectly ignores the effect of [Cu²⁺] on the cathode half-reaction potential. Option B gives 0.92 V, which would require an even lower [Cu²⁺] or an arithmetic error. Option C incorrectly reasons that low [Cu²⁺] increases driving force (it actually reduces it, since Q > 1 means less favorable conditions compared to standard). The key insight is that Q = [products]/[reactants] includes only aqueous species, and since Q > 1, E_cell < E°_cell.
Question 3hard
An anesthesiologist uses a pulse oximeter that operates by shining red light (660 nm) and near-infrared light (940 nm) through a patient's fingertip. The device exploits differences in light absorption between oxyhemoglobin (HbO2) and deoxyhemoglobin (Hb). A medical student notes that the oximeter also detects the pulsatile component of the signal to distinguish arterial blood from venous blood and tissue. If the speed of light in the oximeter's optical fiber is 2.0 × 10⁸ m/s instead of 3.0 × 10⁸ m/s (vacuum), what is the index of refraction of the fiber, and what is the frequency of the 660 nm light INSIDE the fiber?
- A.n = 1.5; frequency inside the fiber = 3.0 × 10¹⁴ Hz, because frequency changes when light enters a denser medium
- B.n = 0.67; frequency inside the fiber = 4.5 × 10¹⁴ Hz, because the fiber is less optically dense than vacuum
- C.n = 1.5; frequency inside the fiber = 4.5 × 10¹⁴ Hz, the same as in vacuum, because frequency is invariant across media✓
- D.n = 1.5; frequency inside the fiber = 2.0 × 10¹⁴ Hz, because the wavelength shortens proportionally and frequency must decrease to conserve energy
Correct answer: C
The index of refraction n = c/v = (3.0 × 10⁸ m/s)/(2.0 × 10⁸ m/s) = 1.5, which is physically realistic for optical fiber (n > 1). When light enters a medium, its frequency remains constant — it is the wavelength that changes (λ_medium = λ_vacuum/n). The frequency in vacuum: f = c/λ = (3.0 × 10⁸ m/s)/(660 × 10⁻⁹ m) = 4.55 × 10¹⁴ Hz ≈ 4.5 × 10¹⁴ Hz, and this frequency is unchanged inside the fiber. Option A incorrectly states that frequency changes — a fundamental misconception. Option B gives n < 1, which is physically impossible for a passive optical medium in classical optics (only possible in metamaterials, not tested here). Option D incorrectly claims frequency decreases to conserve energy, confusing photon energy (E = hf, which is determined by frequency alone and remains constant) with the wave speed relationship.
Question 4medium
A researcher studying enzyme kinetics dissolves a weak acid drug (HA) with a pKa of 6.2 in a buffer solution maintained at pH 7.2. What is the ratio of the conjugate base (A⁻) to the acid (HA) in this solution, and what fraction of the drug exists in the ionized (A⁻) form?
- A.A⁻/HA = 1:10; approximately 9% ionized
- B.A⁻/HA = 1:1; approximately 50% ionized
- C.A⁻/HA = 100:1; approximately 99% ionized
- D.A⁻/HA = 10:1; approximately 91% ionized✓
Correct answer: D
Using the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]), we get 7.2 = 6.2 + log([A⁻]/[HA]), so log([A⁻]/[HA]) = 1.0, meaning [A⁻]/[HA] = 10:1. The fraction ionized is 10/(10+1) = 10/11 ≈ 91%. Option A reverses the ratio, which would apply if the pH were one unit below the pKa. Option B would only be correct if pH = pKa. Option C would require the pH to be 2 units above the pKa (i.e., pH 8.2).
Question 5medium
A patient's arterial blood has a PCO₂ of 40 mmHg and pH of 7.40. In the bicarbonate buffer system, CO₂(g) + H₂O(l) ⇌ H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq). If the patient hyperventilates and PCO₂ drops to 20 mmHg, which of the following best predicts the direction of the equilibrium shift and its effect on blood pH?
- A.Equilibrium shifts left, consuming HCO₃⁻ and H⁺, raising blood pH (respiratory alkalosis)✓
- B.Equilibrium shifts right, producing more H⁺ and HCO₃⁻, lowering blood pH (respiratory acidosis)
- C.Equilibrium shifts left, consuming HCO₃⁻ and H⁺, lowering blood pH (respiratory acidosis)
- D.Equilibrium is unaffected because CO₂ is a gas and does not participate in the aqueous equilibrium
Correct answer: A
By Le Chatelier's principle, reducing the concentration of a reactant (CO₂) causes the equilibrium to shift to the left to replace it. The leftward shift consumes H⁺ and HCO₃⁻, decreasing the concentration of free protons and thus raising the blood pH above 7.40 — this is called respiratory alkalosis. Option B is incorrect because a rightward shift would occur if CO₂ were added, not removed. Option C incorrectly pairs a leftward shift with lower pH; consuming H⁺ raises, not lowers, the pH. Option D is wrong because dissolved CO₂ in the blood is in direct equilibrium with carbonic acid and actively participates in the aqueous buffer system.
Question 6easy
Which of the following functional groups is present in a compound with the general structure R-CO-NH-R'?
- A.Ester
- B.Amide✓
- C.Carboxylic acid
- D.Ketone
Correct answer: B
An amide contains a carbonyl group (C=O) directly bonded to a nitrogen atom, giving the general structure R-CO-NH-R' (or R-CO-NR'R''). This is distinct from an ester (R-CO-O-R'), a carboxylic acid (R-COOH), and a ketone (R-CO-R'), none of which contain nitrogen bonded to the carbonyl carbon. Amides are particularly important in biochemistry as the peptide bond linking amino acids is an amide linkage.
Question 7easy
A reaction has ΔH = −80 kJ/mol and ΔS = −200 J/(mol·K). At 25°C (298 K), what is the sign of ΔG, and is the reaction spontaneous?
- A.ΔG is positive; the reaction is non-spontaneous
- B.ΔG is zero; the reaction is at equilibrium
- C.ΔG is negative; the reaction is spontaneous✓
- D.ΔG is negative; the reaction is non-spontaneous
Correct answer: C
Using ΔG = ΔH − TΔS: ΔG = −80,000 J/mol − (298 K)(−200 J/mol·K) = −80,000 + 59,600 = −20,400 J/mol ≈ −20.4 kJ/mol. Because ΔG is negative, the reaction is spontaneous at 298 K.Ccommon error is to assume that a negative ΔS always makes a reaction non-spontaneous; it only opposes spontaneity, and at low enough temperatures the negative ΔH can dominate, as it does here.
Question 8easy
An electron in a hydrogen atom transitions from the n = 3 energy level to the n = 1 energy level. Which of the following best describes what occurs during this transition?
- A.A photon is absorbed, and the electron gains energy
- B.A photon is absorbed, and the electron loses energy
- C.No photon is involved; only kinetic energy changes
- D.A photon is emitted, and the electron loses energy✓
Correct answer: D
When an electron falls from a higher energy level (n = 3) to a lower energy level (n = 1), it releases energy in the form of an emitted photon whose energy equals the difference between the two levels (E = hf). Absorption of a photon would cause the electron to move to a higher energy level, not a lower one. The energy of the emitted photon corresponds to the Lyman series (UV region) for hydrogen.
Question 9easy
A 2.0 kg block is pushed 5.0 m along a frictionless horizontal surface by a constant horizontal force of 10 N. How much work is done on the block by this force?
- A.50 J✓
- B.4 J
- C.10 J
- D.100 J
Correct answer: A
Work is calculated using W = Fd cosθ. Since the force is horizontal and the displacement is also horizontal, θ = 0° and cosθ = 1. Therefore W = (10 N)(5.0 m)(1) = 50 J. The mass of the block (2.0 kg) is not needed for this calculation.Bcommon error is multiplying mass by distance instead of force by distance.
Question 10easy
Which type of intermolecular force is primarily responsible for the relatively high boiling point of water compared to other small molecules of similar molar mass?
- A.London dispersion forces
- B.Hydrogen bonding✓
- C.Dipole-dipole interactions
- D.Ionic bonding
Correct answer: B
Water's unusually high boiling point (100°C) relative to its small molar mass (18 g/mol) is due to extensive hydrogen bonding between molecules. Hydrogen bonds form when hydrogen is covalently bonded to a highly electronegative atom (O, N, or F) and is attracted to a lone pair on another electronegative atom. While water also exhibits London dispersion and dipole-dipole forces, these are weaker and cannot account for the anomalously high boiling point. Ionic bonding is not an intermolecular force in liquid water.