Reason About Scientific Principles, Theories, and Models - Chem & Physics

A biochemist compares two weak acids HA and HB at 25°C. Solutions are prepared with equal formal concentrations (same total acid added) and no added salt. The measured pH of HA solution is lower than the pH of HB solution. Based on Brønsted–Lowry acid–base theory and equilibrium reasoning, what conclusion can be drawn?
Answer: HA has a larger $K_a$ (smaller p $K_a$ ) than HB, producing more H $^+$ at equilibrium. This question tests Brønsted–Lowry acid-base theory and equilibrium constants for weak acids. The acid dissociation constant $K_a$ quantifies acid strength, with larger $K_a$ (smaller pK_a) indicating stronger acids that produce more H⁺ at equilibrium. For equal concentrations, HA with lower pH than HB must have larger $K_a$ , as it dissociates more, yielding higher [H⁺]. Therefore, choice A logically follows from equilibrium considerations. Choice B is incorrect because smaller $K_a$ would produce higher pH, not lower. A strategy for similar questions is to approximate [H⁺] ≈ √( $K_a$ C) for weak acids and compare pH values. Always check if the weak acid approximation ( $K_a$ << C) is valid.
A lipid vesicle is prepared with an initial internal solution of 200 mM sucrose (impermeant) and is placed in an external solution of 200 mM sucrose plus 100 mM urea (urea is membrane-permeable). Assume constant temperature and that water is permeable. Using the osmotic pressure model ( $\pi = iMRT$ ) and the concept of effective osmoles, what will most likely happen to vesicle volume shortly after transfer and then at long times?
Answer: Shrink initially, then return toward original volume as urea equilibrates across the membrane. This question tests understanding of osmotic pressure and membrane permeability using the van 't Hoff equation. Initially, the external solution has higher total osmolarity (300 mOsm vs 200 mOsm internal), creating an osmotic gradient that drives water outward, causing the vesicle to shrink. However, urea is membrane-permeable and will equilibrate across the membrane over time, eventually reaching 100 mM on both sides. At equilibrium, both compartments will have 200 mM sucrose, but the internal will also have 100 mM urea, making both sides 300 mOsm total. Since the effective osmotic pressure depends only on impermeant solutes at equilibrium, and sucrose is equal on both sides, the vesicle returns toward its original volume. Choice A correctly predicts initial shrinkage followed by volume recovery. Choices B and D incorrectly predict swelling, while C fails to account for the initial osmotic imbalance.
An investigator compares diffusion of a small neutral drug across a planar lipid membrane using Fick's first law, $J = -D\,\Delta C/\Delta x$ , where $J$ is flux, $D$ is the diffusion coefficient, and $\Delta x$ is membrane thickness. Two membranes have identical composition and area, but membrane 2 is twice as thick as membrane 1; the concentration difference across each membrane is held constant. What conclusion can be drawn from Fick's law about the relative fluxes?
Answer: Membrane 2 has half the flux of membrane 1 This question tests understanding of Fick's first law of diffusion and how membrane thickness affects molecular flux. Fick's first law states that flux J = -D(ΔC/Δx), where D is the diffusion coefficient, ΔC is the concentration difference, and Δx is the membrane thickness. When membrane 2 has twice the thickness of membrane 1 (Δx₂ = 2Δx₁) while all other parameters remain constant, the flux through membrane 2 becomes J₂ = -D(ΔC/2Δx₁) = J₁/2. The flux is inversely proportional to membrane thickness, so doubling the thickness halves the flux. Choice C is incorrect because it ignores the inverse relationship between flux and thickness explicitly shown in Fick's law. When applying Fick's law, remember that flux decreases as the diffusion distance increases, which explains why thinner membranes facilitate faster transport.
A metal microelectrode is inserted into tissue, and a redox couple at the surface is modeled by standard reduction potentials. The measured cell potential is approximated by $E_{cell} = E_{cathode} - E_{anode}$ under the same conditions. If the cathode half-reaction becomes less favorable (its reduction potential decreases) while the anode is unchanged, which prediction is most consistent with the model?
Answer: $E_{cell}$ decreases This question tests understanding of electrochemical cell potentials. The cell potential is calculated as Ecell = Ecathode - Eanode, where both are reduction potentials. If the cathode reduction potential decreases (becomes less positive or more negative) while the anode potential remains constant, the difference Ecathode - Eanode decreases, making Ecell smaller. This represents a less favorable overall cell reaction. For example, if Ecathode changes from +0.8 V to +0.6 V while Eanode stays at +0.2 V, then Ecell changes from 0.6 V to 0.4 V. Choice C is incorrect because cell potential depends on both electrodes, not just one. To analyze electrochemical cells, remember that a more positive cell potential indicates a more spontaneous reaction.
A researcher models an action potential upstroke as rapid opening of voltage-gated Na $^+$ channels, approximating Na $^+$ movement as current through a resistor: $I = V/R$ . If channel opening decreases the effective membrane resistance to Na $^+$ by a factor of 5 at the same driving voltage, which prediction is most consistent with Ohm's law?
Answer: Na $^+$ current increases by a factor of 5 This question tests the application of Ohm's law to ion channel conductance. Ohm's law states I = V/R, showing that current is inversely proportional to resistance for a given voltage. When Na+ channels open, the membrane resistance to Na+ decreases by a factor of 5 (R_new = R_old/5). At the same driving voltage, the current becomes I_new = V/(R_old/5) = 5V/R_old = 5I_old. This five-fold increase in current represents the rapid Na+ influx during the action potential upstroke. Choice C is incorrect because Ohm's law explicitly shows that current depends on both voltage and resistance. To analyze ion channel effects, remember that opening channels decreases resistance and increases current proportionally.
A researcher uses a simple capacitor model for a cell membrane: a lipid bilayer of thickness $d$ and area $A$ separates charge on two conducting solutions. Capacitance is approximated by $C = \varepsilon A/d$ , where $\varepsilon$ is the permittivity of the membrane. If cholesterol incorporation increases membrane thickness by 20% with negligible change in $A$ and $\varepsilon$ , which prediction is most consistent with the model?
Answer: Capacitance decreases by about 17% because $C \propto 1/d$ This question tests the application of the parallel-plate capacitor model to biological membranes. The capacitance formula C = εA/d shows that capacitance is inversely proportional to membrane thickness. When cholesterol increases thickness by 20% (d_new = 1.2d_old), the new capacitance becomes C_new = εA/(1.2d) = C_old/1.2 ≈ 0.833C_old. This represents a decrease of approximately 17% (1 - 0.833 = 0.167). The physical basis is that thicker dielectric materials require more voltage to establish the same electric field, reducing charge storage capacity. Choice B is incorrect because it reverses the relationship between capacitance and thickness. To solve capacitor problems, remember that capacitance decreases as plate separation increases, which is why myelination (increasing membrane thickness) reduces axonal capacitance.
A lab calibrates a pH electrode by measuring the potential difference across a glass membrane. The electrode response is modeled by the Nernst equation: $E = E^\circ - \frac{RT}{F}\ln\left(\frac{a_{\mathrm{H}^+\,(in)}}{a_{\mathrm{H}^+\,(out)}}\right)$ . During an experiment at constant temperature, the external solution becomes 10-fold more acidic while the internal reference remains constant. Which prediction is most consistent with the Nernst model?
Answer: The magnitude of $E$ changes by an amount proportional to $\ln(10)$ This question tests the application of the Nernst equation to predict potential changes in pH electrodes. The Nernst equation relates electrode potential to ion activity: E = E° - (RT/F)ln( $a_H$ +,in/ $a_H$ +,out). When the external solution becomes 10-fold more acidic, [H+]_out increases by a factor of 10, making the activity ratio $a_H$ +,in/ $a_H$ +,out decrease by a factor of 10. Taking the natural logarithm of 1/10 gives ln(0.1) = -ln(10), so the potential change is ΔE = +(RT/F)ln(10). At room temperature, this equals approximately +59 mV per 10-fold change in H+ concentration. Choice B is incorrect because E° is indeed constant, but the logarithmic term changes with ion activities. To solve Nernst equation problems, focus on how changes in ion concentrations affect the logarithmic term, remembering that ln(10) ≈ 2.303.
A model membrane is made of phospholipids with negatively charged headgroups. A peptide contains two variants: Variant X has Lys residues on one face (positively charged at physiological pH), while Variant Y replaces those Lys with Gln (polar but uncharged). In buffered saline at pH 7.4, which prediction is most consistent with electrostatic interactions as the dominant initial driver of membrane association?
Answer: Variant X associates more strongly because opposite charges attract, increasing adsorption to the membrane surface. This question tests understanding of electrostatic interactions between charged biomolecules and membranes. The membrane has negatively charged phospholipid headgroups, while Variant X has positively charged lysine residues (at pH 7.4, lysine's amino group is protonated). Opposite charges attract according to Coulomb's law, so Variant X will associate more strongly with the membrane through favorable electrostatic interactions. Variant Y with uncharged glutamine residues lacks these attractive electrostatic forces and will associate less strongly. Answer D incorrectly states that like charges repel, misidentifying the charge interaction between positive lysine and negative headgroups. When analyzing protein-membrane interactions, identify the charges on both components at the given pH and apply the principle that opposite charges attract while like charges repel.
A nanopore sequencing device detects ions moving through a pore under an applied voltage. The pore is narrowed, decreasing cross-sectional area while keeping the same voltage drop across the pore and the same electrolyte conductivity. Using the model $R = \rho L/A$ and $I = V/R$ , which prediction is most consistent with narrowing the pore?
Answer: Current decreases because resistance increases as area decreases. This question tests Ohm's law and resistance in conductive paths. Resistance R = ρ L / A increases as A decreases, so at constant V, current I = V/R decreases. Narrowing the pore increases R, decreasing I. Thus, choice A is consistent. Choice B is incorrect because smaller A increases R, not decreases. For similar problems, compute R ratios from geometry. Verify if conductivity ρ remains constant with pore size.
A 100 mL sample of a protein solution is placed in a rigid calorimeter (constant volume). During a folding transition, the solution releases 2.0 kJ of heat to the surroundings. No pressure–volume work is performed. Using the first law of thermodynamics with the sign convention that heat released by the system is $q<0$ , which statement is most consistent with the change in internal energy $\Delta U$ of the system during folding?
Answer: $\Delta U < 0$ because $q<0$ and $w=0$ at constant volume here. This question tests understanding of the first law of thermodynamics in a constant volume process. The first law states ΔU = q + w, where ΔU is the change in internal energy, q is heat, and w is work. At constant volume, no pressure-volume work can be performed (w = 0), so ΔU = q. Since the system releases 2.0 kJ of heat to the surroundings, q = -2.0 kJ (negative because heat flows out of the system). Therefore, ΔU = -2.0 kJ < 0, indicating that the internal energy of the system decreases during protein folding. Answer A incorrectly assumes that creating ordered structure increases internal energy, confusing entropy with energy. For constant volume processes, always remember that ΔU = q when no other forms of work are present.
A solution of a globular protein is heated slowly at constant pressure. Differential scanning calorimetry shows an endothermic unfolding transition. Based on thermodynamic reasoning for phase/structural transitions, which conclusion can be drawn about the enthalpy change for unfolding at the transition?
Answer: Unfolding has $\Delta H > 0$ at the transition, consistent with heat absorption to disrupt stabilizing interactions. This question tests thermodynamic interpretation of calorimetry data for protein unfolding. An endothermic peak in DSC indicates heat absorption (ΔH >0) during unfolding, as energy is required to break stabilizing interactions. Thus, choice A is consistent with the transition. Choice B is incorrect because endothermic means heat absorption, not release. For similar problems, relate peak direction to process enthalpy. Verify baseline subtraction and scan rate effects on peak shape.
A student models a peptide bond as having partial double-bond character due to resonance between C=O and C–N. Which prediction is most consistent with this resonance model for the geometry around the peptide bond in proteins?
Answer: Rotation about the C–N bond is restricted, favoring a planar arrangement of atoms around the peptide bond. This question tests resonance models in organic chemistry for peptide bonds. Resonance gives the C–N bond partial double-bond character, restricting rotation and favoring planarity. Thus, choice A aligns with the model. Choice B is incorrect because resonance stabilizes specific conformations, not all. A strategy is to draw resonance structures and predict bond orders. Check if steric factors or solvents affect planarity in proteins.
A researcher models oxygen binding to hemoglobin using a Hill-type cooperative model. Under condition 1, the Hill coefficient is $n_H \approx 2.8$ ; under condition 2 (after adding 2,3-BPG), $n_H$ remains similar but the $P_{50}$ increases. Which prediction is most consistent with this model interpretation?
Answer: Cooperativity is similar, while oxygen affinity decreases (right shift). This question tests interpretation of Hill coefficient and cooperativity in oxygen binding models. The Hill coefficient (nH) quantifies cooperativity - values greater than 1 indicate positive cooperativity, with higher values indicating stronger cooperativity. P50 is the partial pressure at which hemoglobin is 50% saturated, indicating oxygen affinity (lower P50 = higher affinity). When 2,3-BPG is added, the Hill coefficient remains similar (~2.8), meaning cooperativity is maintained, but P50 increases, indicating decreased oxygen affinity (right shift of the binding curve). This is physiologically important as 2,3-BPG helps hemoglobin release oxygen to tissues. Choice B correctly identifies that cooperativity remains similar while oxygen affinity decreases. Choice A is incorrect because the similar nH values indicate cooperativity is maintained. When analyzing cooperative binding, distinguish between changes in cooperativity (nH) and changes in affinity (P50 or K50).
A sample of CO $_2$ in a closed container is modeled as an ideal gas. The temperature is increased from 300 K to 330 K while volume and moles are held constant. Which prediction is most consistent with the ideal gas law $PV = nRT$ regarding the pressure change?
Answer: Pressure increases by a factor of $330/300$ This question tests the application of the ideal gas law to predict pressure changes with temperature. The ideal gas law PV = nRT can be rearranged to P = nRT/V, showing that pressure is directly proportional to temperature when n and V are constant. When temperature increases from 300 K to 330 K, the pressure increases by the same ratio: P_new = P_old × (330/300) = 1.1 × P_old. This 10% increase in temperature produces a 10% increase in pressure. Choice B is incorrect because it confuses molecular kinetic energy with bulk pressure; faster-moving particles create more pressure, not less. To solve ideal gas problems with constant volume, use the proportionality P₁/T₁ = P₂/T₂.
A microfluidic channel carries an electrolyte solution. The researcher applies an electric field along the channel and models ion drift with $v_d = \mu E$ , where $\mu$ is mobility. If the electric field magnitude is doubled while temperature and solution composition remain constant, which prediction is most consistent with this drift model for a given ion?
Answer: The drift velocity doubles because $v_d$ is proportional to $E$ at constant $\mu$ . This question tests the drift velocity model in electrophoresis. The equation $v_d$ = μ E shows drift velocity proportional to electric field E at constant mobility μ. Doubling E doubles $v_d$ . Therefore, choice A follows logically. Choice B is incorrect because higher field increases velocity, not reduces mobility. A transferable strategy is to isolate variables in the linear relation. Check if ohmic assumptions hold at high fields.
A peptide contains a histidine side chain modeled as a weak base with $\mathrm{p}K_a \approx 6.0$ for the conjugate acid. The peptide is moved from cytosol (pH 7.4) to lysosome (pH 5.0). Which prediction is most consistent with acid–base equilibrium for the histidine side chain?
Answer: Histidine is more protonated in the lysosome than in the cytosol This question tests the application of acid-base equilibria to amino acid ionization. With pKa ≈ 6.0 for the histidine side chain, we can use the Henderson-Hasselbalch equation to predict protonation states. In the cytosol (pH 7.4), since pH > pKa, the histidine is predominantly deprotonated (neutral). In the lysosome (pH 5.0), since pH < pKa, the histidine is predominantly protonated (positively charged). The shift from pH 7.4 to 5.0 crosses the pKa, causing a dramatic increase in protonation. Choice B is incorrect because it reverses the effect: lower pH means higher [H+], which drives protonation, not deprotonation. To predict ionization states, compare the environmental pH to the pKa: when pH < pKa, the protonated form predominates.
An MRI contrast agent contains a paramagnetic ion that increases the relaxation rate of nearby water protons. The effect is modeled qualitatively as a local magnetic field perturbation that enhances dephasing. Which prediction is most consistent with increasing the concentration of the paramagnetic agent in tissue?
Answer: Proton relaxation becomes faster (shorter relaxation times) because more local field perturbations increase dephasing interactions. This question tests the mechanism of paramagnetic relaxation enhancement in MRI. Paramagnetic ions create local magnetic field fluctuations, accelerating spin dephasing and shortening relaxation times (faster relaxation). Higher concentration increases perturbations, speeding relaxation. Therefore, choice A is consistent with the model. Choice B is incorrect because paramagnetics enhance dephasing, not prevent it. For similar questions, recall T1 and T2 dependencies on field inhomogeneities. Verify if the agent affects T1 or T2 predominantly based on context.
A lab compares the osmotic pressure of two dilute aqueous solutions at the same temperature: Solution X contains 0.10 M glucose (non-electrolyte), Solution Y contains 0.10 M NaCl (assume ideal dissociation to Na $^+$ and Cl $^-$ ). Using the van 't Hoff model $\Pi = iMRT$ , which prediction is most consistent?
Answer: Solution Y has higher osmotic pressure because its van 't Hoff factor $i$ is larger. This question tests the van 't Hoff equation for osmotic pressure. Osmotic pressure Π = i M R T, where i=1 for glucose and i=2 for NaCl (dissociating into two ions). At same M and T, NaCl has higher Π due to larger i. Thus, choice A is correct. Choice C is incorrect because i differs, affecting particle count. For similar questions, calculate effective particle concentration iM. Verify ideal dissociation and dilute conditions.
A drug is a weak acid (HA) with $pK_a = 4.0$ . It is placed in two aqueous compartments at $25^\circ\text{C}$ : compartment X at $\text{pH}=2.0$ and compartment Y at $\text{pH}=7.0$ , separated by a membrane permeable to neutral HA but poorly permeable to $\text{A}^-$ . Using the Henderson–Hasselbalch model, which prediction is most consistent with ion trapping at equilibrium?
Answer: More total drug accumulates in Y because HA deprotonates to $\text{A}^-$ and is trapped. This question tests application of the Henderson-Hasselbalch equation to predict ion trapping across membranes. The Henderson-Hasselbalch equation shows that at pH 2 (below pKa 4), the drug exists predominantly as neutral HA, while at pH 7 (above pKa), it exists predominantly as ionized A-. Since only neutral HA can cross the membrane, drug molecules that enter compartment Y as HA will deprotonate to A- and become trapped because A- cannot cross back. This creates a concentration gradient that drives more HA from X to Y, resulting in drug accumulation in the high pH compartment. Choice B correctly identifies that more drug accumulates in Y because HA deprotonates to A- and is trapped. Choice A is incorrect because at pH 2, the drug is mostly protonated (HA), not ionized. Ion trapping is a key principle in drug distribution: weak acids accumulate in basic compartments, while weak bases accumulate in acidic compartments.
A researcher models O $_2$ transport in blood using Henry's law for dissolved gas: $C = k_H P$ . At 37°C, plasma is equilibrated with O $_2$ at two different partial pressures while hemoglobin is absent (protein removed). Which conclusion can be drawn from Henry's law regarding the relationship between dissolved O $_2$ concentration and O $_2$ partial pressure under these conditions?
Answer: Dissolved O $_2$ concentration is proportional to O $_2$ partial pressure, assuming constant temperature. This question tests reasoning about Henry's law for gas solubility in liquids. Henry's law states that the concentration of a dissolved gas (C) is proportional to its partial pressure (P) above the liquid, with C = $k_H$ P, where $k_H$ is the Henry's constant depending on temperature and solvent. Here, in plasma without hemoglobin, dissolved O₂ follows Henry's law directly, so concentration increases linearly with partial pressure at constant temperature. Thus, choice A is correct as it accurately reflects this proportionality. Choice B is incorrect because gas solubility actually increases with pressure, not decreases, per Henry's law. For similar questions, plot C vs. P to visualize linearity and check if other factors like temperature or binding proteins are absent. Confirm saturation limits are not reached, as Henry's law applies below saturation.

Start this quiz