MCAT® Exam

A. They are spectator ions and do not participate in the net reaction.

In a net ionic equation, ions that do not participate in the actual chemical change (i.e., they remain dissolved in solution before and after the reaction) are called spectator ions. Nitrate (NO₃⁻) and sodium (Na⁺) ions are present on both sides of the full ionic equation and do not form a precipitate or gas, nor do they act as reactants in the formation of AgCl(s).

A. The initial concentrations of hydrogen peroxide and iodide ions.

To determine the initial rate of a reaction, the initial concentrations of reactants and catalysts are paramount. These values directly influence how quickly the reaction begins, and changes in them can significantly alter the observed rate. Other factors like temperature also play a role, but among the given options, initial concentrations are most critical for defining the starting conditions.

C. At the pI, the protein has no net charge, reducing solubility and promoting aggregation.

The isoelectric point (pI) is the pH at which a molecule, such as a protein, carries no net electrical charge. When a protein has no net charge, the repulsive forces between individual protein molecules (due to like charges) are minimized. This reduction in electrostatic repulsion allows hydrophobic interactions and van der Waals forces to dominate, leading to increased protein-protein interactions, aggregation, and ultimately precipitation out of solution, as its solubility is at a minimum.

D. Protein synthesis

Ribosomes, composed of ribosomal RNA (rRNA) and proteins, are the cellular machinery responsible for protein synthesis (translation). In bacteria, ribosomes consist of a 30S and a 50S subunit that combine to form a 70S ribosome. An antibiotic targeting the 50S subunit would directly interfere with the assembly or function of the bacterial ribosome, thereby inhibiting protein synthesis.

B. 5' to 3'

DNA polymerase can only add nucleotides to the 3' end of a growing DNA strand. This fundamental property dictates that synthesis proceeds in the 5' to 3' direction along the newly forming strand. This leads to the continuous synthesis on the leading strand and discontinuous synthesis (Okazaki fragments) on the lagging strand.

C. Respiratory acidosis

The normal arterial blood pH is typically 7.35-7.45. A pH of 7.25 indicates acidosis. Elevated PCO2 suggests a respiratory component, as CO2 forms carbonic acid in the blood. Since bicarbonate (the metabolic component) is normal, the primary cause of the acidosis is the respiratory system. Therefore, this is respiratory acidosis.

D. Facilitated diffusion

The key observations are: transport increases with concentration gradient (suggesting passive transport), it eventually plateaus (indicating saturation of a carrier mechanism), and it does not require direct ATP hydrolysis (ruling out primary active transport). Facilitated diffusion uses a carrier protein or channel, which can become saturated, and moves molecules down their concentration gradient without direct energy input.

C. Visible light has a longer wavelength than X-rays.

All electromagnetic radiation, including visible light and X-rays, travels at the speed of light in a vacuum (approximately 3 x 10^8 m/s), so option B is incorrect. Energy (E) is directly proportional to frequency (f) and inversely proportional to wavelength (λ) (E = hf = hc/λ). X-rays are on the higher energy end of the electromagnetic spectrum, meaning they have higher frequency and shorter wavelengths compared to visible light. Therefore, visible light has a longer wavelength and lower frequency and energy than X-rays. This makes option C correct.

D. The magnetic field strength will be doubled.

The magnetic field strength (B) inside a solenoid (a common form of electromagnet) is directly proportional to the current (I) flowing through its coil. The formula for the magnetic field inside a long solenoid is B = μ₀nI, where μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current. If the current (I) is doubled, and n remains constant, then B will also double.

B. Blue-green

The color of a transition metal complex is determined by the wavelengths of visible light that it *does not* absorb, but instead reflects or transmits. If a complex absorbs red light, its perceived color will be the complementary color to red. On the color wheel, the complementary color to red is blue-green (or cyan).

B. Nonsense mutation

A nonsense mutation is a point mutation that results in a premature stop codon, leading to a truncated and often non-functional protein. The scenario explicitly states a 'premature stop codon' due to a single nucleotide substitution.

B. Add an aqueous solution of NaHCO₃ (sodium bicarbonate) to selectively deprotonate benzoic acid.

Acid-base extraction relies on selectively converting one component of a mixture into its charged (ionic) form, which is soluble in the aqueous layer, while the other component remains uncharged and soluble in the organic layer. Benzoic acid is a carboxylic acid (weak acid, pKa 4.2). Aniline is a weak base (pKb 9.4, so its conjugate acid has pKa 14-9.4 = 4.6). To separate them, we need to selectively ionize one. Sodium bicarbonate (NaHCO₃) is a weak base (conjugate acid H₂CO₃ has pKa1 6.4, so NaHCO₃ acts as a base with pKb ≈ 7.6). Its conjugate acid (H₂CO₃) has a pKa (6.4) that is greater than benzoic acid's pKa (4.2) but less than aniline's conjugate acid pKa (4.6). Therefore, NaHCO₃ is strong enough to deprotonate benzoic acid (making it water-soluble as benzoate) but not strong enough to protonate aniline significantly to make it water-soluble. This allows for selective separation. Options A and B would likely convert both or neither effectively for selective separation. Option D would not work as aniline is a base and would need an acid to be protonated.

B. 1000 bp

PCR amplifies a specific region of DNA defined by the forward and reverse primers. The 'target gene' length of 1000 bp refers to the region between the primer binding sites. Successful PCR amplification will produce copies of this target region, so the product size will be the length of the target gene itself.

D. Plasmids

In prokaryotes, circular double-stranded DNA is typical for the main chromosome. Smaller, extrachromosomal circular DNA molecules that can replicate independently are known as plasmids. These often carry genes for antibiotic resistance or other advantageous traits.

B. Acid chloride (RCOCl)

The electrophilicity of a carbonyl carbon is enhanced by electron-withdrawing groups and diminished by electron-donating groups. The ability of the leaving group (or substituent on the carbonyl carbon) to donate electrons via resonance or induction affects this. Halides (like Cl) are highly electronegative and poor electron donors via resonance, making them strongly electron-withdrawing. Amides have a nitrogen atom that can strongly donate electrons via resonance, making the carbonyl carbon less electrophilic. Esters have an oxygen atom that can also donate electrons via resonance, but less effectively than nitrogen. Carboxylic acids fall in between, with the -OH group having some resonance donation. Acid chlorides, with the highly electronegative chlorine atom, have the least resonance stabilization of the carbonyl and the strongest inductive withdrawal, making their carbonyl carbon the most electrophilic and thus most reactive towards nucleophiles.

C. Oxygen (O)

Oxygen forms a diatomic molecule (O2) with a double bond. A double bond consists of one sigma bond and one pi bond, resulting in a bond order of 2. Nitrogen forms a triple bond (bond order 3), while fluorine forms a single bond (bond order 1). Carbon typically forms networks or more complex molecules, not simple diatomic molecules with a bond order of 2.

C. Both ethyl and methyl groups are in equatorial positions.

For cis-1,2-disubstituted cyclohexanes, one substituent must be axial and the other equatorial in the same chair conformation. However, the question asks for the *most stable* chair conformation. Larger groups prefer to be in the equatorial position to minimize steric strain (1,3-diaxial interactions). In cis-1-ethyl-2-methylcyclohexane, since the groups are cis, they must be on the same side of the ring. If one is axial, the other must be equatorial. To achieve the most stable conformation where both groups are equatorial, a ring flip would be required if they started axial/equatorial. However, due to the cis relationship on adjacent carbons, if one is equatorial, the other must be axial. The statement 'both ethyl and methyl groups are in equatorial positions' for cis-1,2 is incorrect as this would imply a trans relationship. Let's re-evaluate. Cis-1,2 means one up, one down. So, if C1 is 'up' (equatorial), then C2 must be 'up' (axial) to be cis. If C1 is 'up' (axial), then C2 must be 'up' (equatorial) to be cis. Therefore, for cis-1,2, one group must be axial and the other equatorial. The *more stable* conformation will have the *larger* group in the equatorial position. Ethyl is larger than methyl. So, the ethyl group should be equatorial and the methyl group axial. Upon re-reading the question and options, option B is incorrect for a cis-1,2 disubstituted cyclohexane. Let's re-examine the options considering the nature of cis-1,2. If we consider the example of cis-1,2-dimethylcyclohexane, one methyl is axial, and the other is equatorial. The more stable chair conformation will place the BULKIER group in the equatorial position. Ethyl is bulkier than methyl. So, the ethyl group should be equatorial, and the methyl group should be axial. This corresponds to option D. My initial reasoning for the explanation was flawed, let me correct it and the option selection.

C. Beta cells

Insulin is a hormone produced by the beta cells of the pancreatic islets. Its primary role is to lower blood glucose levels. A deficiency in insulin production, as suggested by the symptoms and high blood glucose, points directly to a problem with beta cells.

A. The -OH group on the lowest chiral carbon (C5) is on the right.

D-glucose is defined by the configuration of its lowest chiral carbon. In a Fischer projection, if the -OH group on this carbon (C5 for hexoses) is on the right, it is a D-sugar. If it's on the left, it's an L-sugar. Glucose is an aldohexose, meaning it has an aldehyde group (not a ketone) and six carbons. While glucose does form a cyclic hemiacetal in solution, a Fischer projection is a linear representation.

C. Acid-base extraction

Carboxylic acids are acidic and can be deprotonated by a base (like NaHCO₃ or NaOH) to form a water-soluble carboxylate salt. Aldehydes are generally neutral and not deprotonated by weak bases. This difference in acid-base properties allows for separation using acid-base extraction, where the carboxylic acid moves into the aqueous layer as a salt, while the aldehyde remains in the organic layer.

A. Covalent bonding

Reversible binding in biological systems, such as drug-receptor interactions, primarily involves non-covalent forces. These include hydrogen bonding, ionic interactions (salt bridges), and Van der Waals forces (London dispersion forces, dipole-dipole). Covalent bonding, by definition, involves the sharing of electrons and is typically much stronger and less readily reversible under physiological conditions. While some drugs form irreversible covalent bonds, the question specifies 'reversible binding'.

C. 0.004 s⁻¹

For a first-order reaction, the integrated rate law is ln[A]t - ln[A]0 = -kt, or ln([A]t/[A]0) = -kt. Initial concentration [A]0 = 1.0 M. Final concentration [A]t = 1.0 M - 0.20 M = 0.80 M. Time (t) = 50 s. Substitute these values: ln(0.80 / 1.0) = -k * 50 s. ln(0.80) = -k * 50 s. -0.223 = -k * 50 s. k = 0.223 / 50 s = 0.00446 s⁻¹. Rounding to one significant figure for the given options, the closest is 0.004 s⁻¹.

A. 100 MBq

The half-life is 6 hours. The total time elapsed is 18 hours. Number of half-lives = Total time / Half-life = 18 hours / 6 hours = 3 half-lives. After 1 half-life: 800 MBq / 2 = 400 MBq. After 2 half-lives: 400 MBq / 2 = 200 MBq. After 3 half-lives: 200 MBq / 2 = 100 MBq.

A. No change in Vmax and an increase in Km.

Competitive inhibitors bind to the enzyme's active site, competing with the substrate. This effectively increases the apparent Km because more substrate is needed to reach half Vmax. However, if enough substrate is added, the inhibitor can be outcompeted, allowing the enzyme to reach its original Vmax.

B. The reaction rate will approach V_max and become largely independent of further substrate increase.

In Michaelis-Menten kinetics, when substrate concentration is significantly higher than K_M, the enzyme becomes saturated with substrate. At this point, the enzyme is working at its maximum capacity (V_max), and further increases in substrate concentration will not significantly increase the reaction rate because all active sites are continuously occupied.