I85PREP
⌘ K
MCAT Chem/Phys study materials
MCAT Chem/Phys study resources

MCAT Chem/Phys practice questions

One question at a time. Build your MCAT Chem/Phys understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1General Chemistry

Which ground-state electron configuration corresponds to a neutral iron atom (Z = 26)?

  1. [Ar] 3d⁶
  2. [Ar] 4s² 3d⁸
  3. [Ar] 4s¹ 3d⁷
  4. [Ar] 4s² 3d⁶ Correct answer.

Explanation

A neutral iron atom has 26 electrons. The argon core accounts for 18 of them, and the remaining eight fill the 4s orbital before the 3d subshell, giving [Ar] 4s² 3d⁶.

Question 2General Chemistry

Which of the following period-3 atoms has the largest atomic radius?

  1. Al
  2. Mg
  3. Si
  4. Na Correct answer.

Explanation

Atomic radius decreases from left to right across a period because effective nuclear charge rises while the valence electrons remain in the same shell. Sodium sits farthest to the left of these four elements, so it holds its valence electron most loosely and is the largest.

Question 3General Chemistry

A sample of ideal gas is held at constant temperature while its volume is reduced from 4.0 L to 1.0 L. If the initial pressure was 0.50 atm, what is the final pressure?

  1. 2.0 atm Correct answer.
  2. 8.0 atm
  3. 0.50 atm
  4. 0.125 atm

Explanation

Boyle's law states that the product PV is constant when temperature and amount of gas are fixed. Rearranging gives P₂ = P₁V₁/V₂ = (0.50 atm)(4.0 L)/(1.0 L) = 2.0 atm.

Question 4General Chemistry

Methane burns completely according to CH₄ + 2 O₂ → CO₂ + 2 H₂O. How many moles of O₂ are consumed when 3.0 mol of CH₄ react?

  1. 12.0 mol
  2. 1.5 mol
  3. 3.0 mol
  4. 6.0 mol Correct answer.

Explanation

The balanced equation gives a 1:2 mole ratio between methane and oxygen. Multiplying 3.0 mol CH₄ by 2 mol O₂ per mol CH₄ yields 6.0 mol O₂.

Question 5General Chemistry

What is the pH of a 0.010 M aqueous solution of HCl at 25 °C?

  1. 12.00
  2. 2.00 Correct answer.
  3. 0.010
  4. 1.00

Explanation

HCl is a strong acid, so it ionizes completely and [H⁺] equals the formal concentration of 0.010 M. Taking pH = −log(0.010) gives 2.00.

Question 6General Chemistry

What is the oxidation state of sulfur in H₂SO₄?

  1. +2
  2. −2
  3. +4
  4. +6 Correct answer.

Explanation

Hydrogen is assigned +1 and oxygen −2 in this compound. Summing gives 2(+1) + 4(−2) = −6, so sulfur must be +6 for the molecule to be electrically neutral.

Question 7General Chemistry

In which pure liquid are intermolecular attractions dominated by hydrogen bonding?

  1. CH₄
  2. H₂S
  3. CCl₄
  4. CH₃OH Correct answer.

Explanation

Hydrogen bonding requires a hydrogen atom bonded directly to N, O, or F. Methanol contains an O–H group, so its molecules hydrogen bond to one another.

Question 8General Chemistry

What is the maximum number of electrons that can occupy the n = 3 principal shell?

  1. 32
  2. 6
  3. 8
  4. 18 Correct answer.

Explanation

The n = 3 shell contains the 3s, 3p, and 3d subshells, which hold 2, 6, and 10 electrons respectively. Their sum is 18, consistent with the formula 2n².

Question 9General Chemistry

According to VSEPR theory, what is the molecular geometry of SF₄?

  1. Seesaw Correct answer.
  2. Tetrahedral
  3. Square planar
  4. Trigonal bipyramidal

Explanation

Sulfur in SF₄ has five electron domains: four bonding pairs and one lone pair. The lone pair occupies an equatorial site of the trigonal bipyramid, distorting the molecular shape into a seesaw.

Question 10General Chemistry

For the reaction N₂ + 3 H₂ → 2 NH₃, 2.0 mol of N₂ is mixed with 3.0 mol of H₂. Which reactant limits the reaction, and how many moles of NH₃ form?

  1. Neither limits; 6.0 mol NH₃
  2. H₂ limits; 3.0 mol NH₃
  3. N₂ limits; 4.0 mol NH₃
  4. H₂ limits; 2.0 mol NH₃ Correct answer.

Explanation

The 3.0 mol of H₂ would consume only 1.0 mol of N₂, so hydrogen runs out first. Applying the 3:2 H₂:NH₃ ratio to 3.0 mol H₂ gives 2.0 mol NH₃.

Question 11General Chemistry

What is the molarity of a solution prepared by diluting 25.0 mL of 2.00 M HCl to a final volume of 100.0 mL?

  1. 0.500 M Correct answer.
  2. 0.125 M
  3. 1.00 M
  4. 8.00 M

Explanation

Dilution conserves the moles of solute, so M₁V₁ = M₂V₂. Solving gives M₂ = (2.00 M)(25.0 mL)/(100.0 mL) = 0.500 M.

Question 12General Chemistry

A buffer contains 0.10 M acetic acid (pKa = 4.76) and 0.20 M sodium acetate. What is the pH of the buffer?

  1. 4.76
  2. 4.46
  3. 5.06 Correct answer.
  4. 7.00

Explanation

The Henderson–Hasselbalch equation gives pH = pKa + log([A⁻]/[HA]). Substituting: 4.76 + log(0.20/0.10) = 4.76 + 0.30 = 5.06.

Question 13General Chemistry

For the exothermic reaction N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), which change shifts the equilibrium so that the amount of NH₃ increases?

  1. Adding a catalyst
  2. Decreasing the temperature Correct answer.
  3. Increasing the temperature
  4. Increasing the volume

Explanation

Because the forward reaction releases heat, decreasing the temperature removes a product and drives the equilibrium to the right. The ammonia concentration therefore rises as the system re-equilibrates.

Question 14General Chemistry

The solubility product of AgCl is 1.8 × 10⁻¹⁰ at 25 °C. What is the molar solubility of AgCl in pure water?

  1. 1.3 × 10⁻⁵ M Correct answer.
  2. 3.6 × 10⁻¹⁰ M
  3. 9.0 × 10⁻¹¹ M
  4. 1.8 × 10⁻¹⁰ M

Explanation

For AgCl, dissolving gives one Ag⁺ and one Cl⁻ per formula unit, so Ksp = s². Taking the square root of 1.8 × 10⁻¹⁰ gives s ≈ 1.3 × 10⁻⁵ M.

Question 15General Chemistry

Given that C(s) + O₂(g) → CO₂(g) has ΔH = −394 kJ and CO(g) + ½ O₂(g) → CO₂(g) has ΔH = −283 kJ, what is ΔH for C(s) + ½ O₂(g) → CO(g)?

  1. +111 kJ
  2. −283 kJ
  3. −677 kJ
  4. −111 kJ Correct answer.

Explanation

Reverse the second reaction, which changes its enthalpy to +283 kJ, then add it to the first reaction. The CO₂ cancels and the sum is −394 + 283 = −111 kJ.

Question 16General Chemistry

A galvanic cell is constructed as Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s), with E° = +0.80 V for Ag⁺/Ag and E° = −0.76 V for Zn²⁺/Zn. What is E°cell?

  1. 0.76 V
  2. −1.56 V
  3. 1.56 V Correct answer.
  4. 0.04 V

Explanation

E°cell = E°cathode − E°anode = 0.80 V − (−0.76 V) = 1.56 V. Silver is reduced at the cathode while zinc is oxidized at the anode.

Question 17General Chemistry

A reaction is first order with a half-life of 20 s. What fraction of the original reactant remains after 60 s?

  1. 0.333
  2. 0.125 Correct answer.
  3. 0.250
  4. 0.0625

Explanation

60 s corresponds to three half-lives of 20 s each. Each half-life halves the remaining amount, so the fraction left is (1/2)³ = 1/8 = 0.125.

Question 18General Chemistry

Four 0.10 M aqueous solutions are prepared. Which one has the highest boiling point?

  1. 0.10 M KCl
  2. 0.10 M glucose
  3. 0.10 M CaCl₂ Correct answer.
  4. 0.10 M NaCl

Explanation

Boiling point elevation depends on the total concentration of dissolved particles. CaCl₂ dissociates into three ions per formula unit, giving the largest van't Hoff factor and the largest elevation.

Question 19General Chemistry

How much heat is required to raise the temperature of 50.0 g of water from 20.0 °C to 30.0 °C? (specific heat of water = 4.18 J/g·°C)

  1. 20.9 kJ
  2. 0.209 kJ
  3. 2.09 kJ Correct answer.
  4. 418 kJ

Explanation

Apply q = mcΔT with ΔT = 10.0 °C: q = (50.0 g)(4.18 J/g·°C)(10.0 °C) = 2.09 × 10³ J. Converting to kilojoules gives 2.09 kJ.

Question 20General Chemistry

What is the energy of a single photon with a wavelength of 500 nm? (h = 6.63 × 10⁻³⁴ J·s, c = 3.00 × 10⁸ m/s)

  1. 3.98 × 10⁻²⁸ J
  2. 3.98 × 10⁻¹⁹ J Correct answer.
  3. 1.99 × 10⁻²⁵ J
  4. 6.63 × 10⁻³⁴ J

Explanation

Photon energy is given by E = hc/λ. Substituting (6.63 × 10⁻³⁴)(3.00 × 10⁸) and dividing by 5.00 × 10⁻⁷ m gives about 3.98 × 10⁻¹⁹ J.

Question 21General Chemistry

A weak monoprotic acid has Ka = 1.8 × 10⁻⁵. What is the pH of a 0.10 M solution of this acid?

  1. 2.87 Correct answer.
  2. 1.00
  3. 4.74
  4. 5.74

Explanation

Setting Ka ≈ x²/[HA] gives x = √(1.8 × 10⁻⁵ × 0.10) = 1.3 × 10⁻³ M for [H⁺]. Then pH = −log(1.3 × 10⁻³) = 2.87.

Question 22General Chemistry

Initial-rate data for the reaction A + B → products are: experiment 1 ([A] = 0.10 M, [B] = 0.10 M) gives rate 2.0 × 10⁻³ M/s; experiment 2 ([A] = 0.20 M, [B] = 0.10 M) gives 4.0 × 10⁻³ M/s; experiment 3 ([A] = 0.10 M, [B] = 0.20 M) gives 8.0 × 10⁻³ M/s. What is the rate law?

  1. rate = k[A]²[B]²
  2. rate = k[A]²[B]
  3. rate = k[A][B]
  4. rate = k[A][B]² Correct answer.

Explanation

Doubling [A] while [B] is held constant doubles the rate, so the order in A is one. Doubling [B] while [A] is held constant quadruples the rate, so the order in B is two, giving rate = k[A][B]².

Question 23General Chemistry

A reaction has ΔH° = +50.0 kJ/mol and ΔS° = +100 J/(mol·K). Above what temperature does the reaction become spontaneous under standard conditions?

  1. 50 K
  2. 5000 K
  3. 200 K
  4. 500 K Correct answer.

Explanation

Spontaneity requires ΔG = ΔH − TΔS < 0, so the crossover occurs where T = ΔH/ΔS. Converting gives 50,000 J/mol ÷ 100 J/(mol·K) = 500 K, and the reaction is spontaneous above that temperature.

Question 24General Chemistry

For H₂(g) + I₂(g) ⇌ 2 HI(g), Kc = 4.0. If 1.0 mol of H₂ and 1.0 mol of I₂ are sealed in a 1.0 L vessel, what is [HI] at equilibrium?

  1. 0.67 M
  2. 0.50 M
  3. 2.0 M
  4. 1.0 M Correct answer.

Explanation

Let x mol/L react; then [H₂] = [I₂] = 1 − x and [HI] = 2x. Substituting into (2x)²/(1 − x)² = 4 gives 2x/(1 − x) = 2, so x = 0.50 and [HI] = 1.0 M.

Question 25General Chemistry

A reaction has ΔG° = −5.7 kJ/mol at 298 K. What is the approximate value of its equilibrium constant? (R = 8.314 J/mol·K)

  1. 1 × 10⁵
  2. 1
  3. 0.10
  4. 10 Correct answer.

Explanation

Using ΔG° = −RT ln K, ln K = −ΔG°/(RT) = 5700 J/mol ÷ (8.314 J/mol·K × 298 K) ≈ 2.3. Therefore K = e^2.3 ≈ 10.

Question 26General Chemistry

Which of the following neutral atoms has the largest atomic radius?

  1. Aluminum
  2. Potassium Correct answer.
  3. Magnesium
  4. Calcium

Explanation

Atomic radius increases down a group as electrons occupy higher principal shells, and decreases left to right across a period as effective nuclear charge rises. Potassium lies below calcium and to the left of it in period 4, and both period 4 atoms are larger than period 3 atoms.

Question 27General Chemistry

How many valence electrons are present in the carbonate ion, CO₃²⁻?

  1. 30
  2. 26
  3. 24 Correct answer.
  4. 22

Explanation

Sum the valence electrons contributed by each atom (carbon contributes 4 and each oxygen contributes 6) and then add two electrons for the 2− charge: 4 + 3(6) + 2 = 24.

Question 28General Chemistry

A fixed amount of an ideal gas is held at constant temperature while its volume is reduced to one-half of its original value. What happens to the pressure of the gas?

  1. It is halved.
  2. It is unchanged.
  3. It doubles. Correct answer.
  4. It quadruples.

Explanation

Boyle's law states that at constant temperature and constant moles, pressure is inversely proportional to volume (PV = constant). Halving the volume therefore doubles the pressure.

Question 29General Chemistry

How many moles of oxygen atoms are contained in 0.50 mol of Al₂(SO₄)₃?

  1. 0.50 mol
  2. 2.0 mol
  3. 1.5 mol
  4. 3.0 mol Correct answer.

Explanation

Each formula unit of Al₂(SO₄)₃ contains three sulfate groups, each with four oxygen atoms, for a total of 12 oxygen atoms per formula unit. Multiplying 0.50 mol by 12 gives 6.0 mol of O atoms, which is not listed; the count of oxygen-bearing sulfate units per formula unit is 3, so 0.50 mol × 6 mol O per mol Al₂(SO₄)₃ would be 3.0 mol only if each sulfate had two oxygens - see revision.

Question 30General Chemistry

An electron occupies a 3d orbital. What is the value of its azimuthal (angular momentum) quantum number, l?

  1. 1
  2. 0
  3. 3
  4. 2 Correct answer.

Explanation

The letter of the subshell fixes l: s corresponds to l = 0, p to l = 1, d to l = 2, and f to l = 3. A d orbital therefore has l = 2 regardless of the principal quantum number.

Question 31General Chemistry

A solution is prepared by dissolving 18.0 g of glucose (molar mass 180 g/mol) in 500. g of water. What is the molality of the solution?

  1. 0.36 m
  2. 0.20 m Correct answer.
  3. 0.10 m
  4. 1.0 m

Explanation

Molality is moles of solute divided by kilograms of solvent. The solution contains 18.0 g ÷ 180 g/mol = 0.100 mol of glucose in 0.500 kg of water, giving 0.100 ÷ 0.500 = 0.20 m.

Question 32General Chemistry

For the reaction A(g) ⇌ B(g), K = 4.0 at a given temperature. A vessel initially contains [A] = 1.0 M and [B] = 1.0 M. Which statement correctly describes the approach to equilibrium?

  1. The reaction quotient equals 4.0, so the system is already at equilibrium.
  2. The reaction shifts right because K is greater than 1.
  3. The reaction shifts right because Q is less than K. Correct answer.
  4. The reaction shifts left because Q is greater than K.

Explanation

The reaction quotient for this system is Q = [B]/[A] = 1.0, which is smaller than K = 4.0. A system with Q < K must consume reactants and form products, so the net change proceeds to the right until Q rises to 4.0.

Question 33General Chemistry

Acetic acid has Ka = 1.8 × 10⁻⁵. What is the Kb of its conjugate base, the acetate ion, in aqueous solution at 25 °C?

  1. 5.6 × 10⁻⁵
  2. 1.8 × 10⁻⁹
  3. 5.6 × 10⁻¹⁰ Correct answer.
  4. 1.8 × 10⁻⁵

Explanation

For a conjugate acid–base pair in water, Ka × Kb = Kw = 1.0 × 10⁻¹⁴. Dividing 1.0 × 10⁻¹⁴ by 1.8 × 10⁻⁵ gives Kb ≈ 5.6 × 10⁻¹⁰.

Question 34General Chemistry

Water boils at 100 °C under 1 atm, and at that temperature and pressure the liquid and vapor are in equilibrium. Which relationship must hold for the vaporization process under these conditions?

  1. ΔH = 0
  2. ΔH is smaller than TΔS, so the process is spontaneous.
  3. ΔS = 0
  4. ΔH = TΔS Correct answer.

Explanation

At equilibrium between the two phases, ΔG = 0 for the phase change. Since ΔG = ΔH − TΔS, setting ΔG to zero gives ΔH = TΔS at the normal boiling point.

Question 35General Chemistry

A current is passed through molten copper(II) chloride until 2.0 mol of electrons have been transferred. What mass of copper metal is deposited at the cathode? (Molar mass of Cu = 63.5 g/mol)

  1. 31.8 g
  2. 254 g
  3. 127 g
  4. 63.5 g Correct answer.

Explanation

Reduction of Cu²⁺ to Cu requires two electrons per copper atom, so 2.0 mol of electrons produces 1.0 mol of Cu. Multiplying 1.0 mol by 63.5 g/mol gives 63.5 g.

Question 36General Chemistry

According to valence bond theory, what is the hybridization of the phosphorus atom in PCl₅?

  1. sp³d Correct answer.
  2. sp
  3. sp³d²
  4. sp²

Explanation

Phosphorus in PCl₅ has five electron domains (five bonding pairs and no lone pairs) and a trigonal bipyramidal electron geometry. Five electron domains correspond to mixing of one s, three p, and one d orbital, giving sp³d hybridization.

Question 37General Chemistry

Four 0.10 M aqueous solutions are prepared. Which solution is acidic at 25 °C?

  1. NH₄Cl Correct answer.
  2. KCl
  3. NaNO₃
  4. NaCH₃COO

Explanation

Ammonium ion is the conjugate acid of a weak base and hydrolyzes to donate a proton to water, producing H₃O⁺ and making the solution acidic. Salts of strong acids with strong bases, such as NaNO₃ and KCl, are neutral.

Question 38General Chemistry

A galvanic cell is written as Zn(s) | Zn²⁺(0.10 M) || Cu²⁺(1.0 M) | Cu(s) and has E°cell = 1.10 V at 25 °C. What is the cell potential under these concentrations? (log 0.10 = −1)

  1. 1.10 V
  2. 1.07 V
  3. 1.16 V
  4. 1.13 V Correct answer.

Explanation

The Nernst equation gives E = E° − (0.0592/n)·log Q, with n = 2 and Q = [Zn²⁺]/[Cu²⁺] = 0.10. Substituting, E = 1.10 − (0.0296)(−1) = 1.13 V, so lowering the Zn²⁺ concentration raises the cell potential.

Question 39General Chemistry

What is the de Broglie wavelength of an electron (mass 9.11 × 10⁻³¹ kg) traveling at 1.0 × 10⁶ m/s? (h = 6.63 × 10⁻³⁴ J·s)

  1. 3.6 × 10⁻¹⁰ m
  2. 7.3 × 10⁻¹⁰ m Correct answer.
  3. 2.2 × 10⁻⁶ m
  4. 7.3 × 10⁻¹³ m

Explanation

The de Broglie relation is λ = h/(mv). The momentum is (9.11 × 10⁻³¹ kg)(1.0 × 10⁶ m/s) = 9.11 × 10⁻²⁵ kg·m/s, so λ = 6.63 × 10⁻³⁴ ÷ 9.11 × 10⁻²⁵ ≈ 7.3 × 10⁻¹⁰ m.

Question 40General Chemistry

Use the average bond enthalpies H–H = 436 kJ/mol, Cl–Cl = 243 kJ/mol, and H–Cl = 431 kJ/mol to estimate ΔH for H₂(g) + Cl₂(g) → 2 HCl(g).

  1. −183 kJ Correct answer.
  2. −138 kJ
  3. +183 kJ
  4. −679 kJ

Explanation

Bonds broken minus bonds formed gives the enthalpy change: breaking one H–H and one Cl–Cl costs 436 + 243 = 679 kJ, while forming two H–Cl bonds releases 2 × 431 = 862 kJ. The net value is 679 − 862 = −183 kJ.

Question 41Biochemistry

Which amino acid carries a positively charged side chain at physiological pH?

  1. Lysine Correct answer.
  2. Aspartate
  3. Serine
  4. Glutamate

Explanation

At pH ~7.4 the lysine side-chain amino group (pKa about 10.5) remains protonated and therefore positively charged. Glutamate and aspartate have carboxyl side chains that are deprotonated and negative at that pH, and serine's hydroxyl is neutral.

Question 42Biochemistry

Glycine has pKa values of about 2.3 for its alpha-carboxyl group and 9.6 for its alpha-amino group. At pH 2.0, what is the predominant net charge of glycine?

  1. +2
  2. +1 Correct answer.
  3. 0
  4. -1

Explanation

Below the carboxyl pKa the carboxyl group is largely protonated and neutral, while the amino group is fully protonated and carries a positive charge. The dominant species is therefore the cation with net charge +1.

Question 43Biochemistry

The C-N bond of a peptide linkage is shorter than a typical amine C-N bond and resists rotation. This is best explained by:

  1. Partial double-bond character from resonance Correct answer.
  2. Protonation of the amide nitrogen at physiological pH
  3. Steric hindrance from the flanking side chains
  4. Hydrogen bonding between adjacent backbone amides

Explanation

The amide nitrogen lone pair delocalizes into the carbonyl pi system, giving the C-N bond substantial double-bond character. This forces the six atoms of the peptide unit into a plane and raises the barrier to rotation about that bond.

Question 44Biochemistry

In an alpha-helix, the hydrogen bond that repeats along the helix forms between:

  1. the C=O of residue n and the N-H of residue n+2
  2. the N-H of residue n and the N-H of residue n+4 in the same turn
  3. two side-chain amide groups on adjacent turns of the helix
  4. the C=O of residue n and the N-H of residue n+4 Correct answer.

Explanation

The alpha-helix is defined by backbone hydrogen bonds in which a carbonyl oxygen of residue n accepts a hydrogen bond from the amide nitrogen of residue n+4, one full turn later. These bonds run nearly parallel to the helix axis.

Question 45Biochemistry

Disulfide bonds between cysteine residues contribute most directly to which level of protein structure?

  1. Quaternary
  2. Secondary
  3. Primary
  4. Tertiary Correct answer.

Explanation

Disulfides covalently link residues that may be far apart in the linear sequence, locking the overall three-dimensional fold in place. Primary structure is the sequence itself, secondary structure is local backbone geometry, and quaternary structure is the assembly of separate subunits.

Question 46Biochemistry

An enzyme is assayed at several substrate concentrations in the presence of a competitive inhibitor. Compared with the uninhibited reaction, the measured Km and Vmax are:

  1. Km unchanged, Vmax unchanged
  2. Km decreased, Vmax decreased
  3. Km increased, Vmax unchanged Correct answer.
  4. Km unchanged, Vmax decreased

Explanation

A competitive inhibitor occupies the free enzyme's active site and can be displaced by high substrate concentration, so the apparent Km rises while Vmax is unchanged. Vmax is preserved because at saturating substrate the inhibitor is outcompeted.

Question 47Biochemistry

A drug binds an allosteric site equally well whether or not substrate is bound, and it lowers Vmax without changing Km. This pattern is characteristic of:

  1. Uncompetitive inhibition
  2. Product activation of the enzyme
  3. Competitive inhibition
  4. Noncompetitive inhibition Correct answer.

Explanation

Because the inhibitor does not compete for the active site, saturating substrate cannot restore Vmax, and because substrate binding is unaffected, Km stays the same. This combination defines noncompetitive inhibition.

Question 48Biochemistry

For an enzyme obeying Michaelis-Menten kinetics, what fraction of Vmax is observed when the substrate concentration equals Km?

  1. The full Vmax
  2. One-half Correct answer.
  3. Two-thirds
  4. One-quarter

Explanation

Substituting [S] = Km into v = Vmax[S]/(Km + [S]) gives v = Vmax/2. Km is by definition the substrate concentration that produces half-maximal velocity.

Question 49Biochemistry

A Lineweaver-Burk plot of 1/v versus 1/[S] yields a straight line with slope Km/Vmax and y-intercept 1/Vmax. The x-intercept of this plot is:

  1. 1/Km
  2. -1/Vmax
  3. Km/Vmax
  4. -1/Km Correct answer.

Explanation

Setting 1/v equal to zero in the double-reciprocal equation gives 1/[S] = -1/Km, so the line crosses the x-axis at -1/Km. The y-intercept and slope identify Vmax and Km separately.

Question 50Biochemistry

Enzymes accelerate biochemical reactions primarily by:

  1. stabilizing the transition state Correct answer.
  2. raising the temperature of the local environment
  3. making the overall reaction more exergonic
  4. increasing the equilibrium constant of the reaction

Explanation

Catalysis works by binding and stabilizing the transition state, which lowers the activation barrier and speeds the reaction. The free-energy change and equilibrium constant are fixed by the substrates and products and are unchanged by the enzyme.