1. Biological and Biochemical Foundations of Living Systems
Biochemistry: Macromolecules, Enzymes, Metabolism, Molecular Biology
Biology: Cell Structure and Function, Genetics, Microbiology, Physiology
Organic Chemistry: Functional Groups, Reaction Mechanisms
General Chemistry: Atomic Structure, Chemical Reactions, Solutions
2. Chemical and Physical Foundations of Biological Systems
General Chemistry: Periodic Table, Stoichiometry, Thermodynamics
Organic Chemistry: Structure and Properties
Physics: Mechanics, Fluids, Waves, Electricity and Magnetism
Biochemistry: Biomolecules, Enzyme Kinetics
3. Psychological, Social, and Biological Foundations of Behavior
Psychology: Cognition, Perception, Learning, Memory
Sociology: Social Structures, Inequality, Demographics
Biology: Nervous System, Endocrine System, Behavior
4. Critical Analysis and Reasoning Skills (CARS)
Reading comprehension strategies
Passage analysis
Logical reasoning
Sample Topic: Enzyme Kinetics (Biochemistry)
MCQ 1:
Which of the following changes would increase the maximum velocity (Vmax) of an
enzyme-catalyzed reaction?
A) Increasing the substrate concentration beyond saturation
B) Increasing the enzyme concentration
C) Adding a competitive inhibitor
D) Increasing the temperature beyond the enzyme’s optimal point
,Correct Answer: B) Increasing the enzyme concentration
Rationale:
A: Increasing substrate concentration beyond saturation does not increase Vmax since the
enzyme is already working at full capacity.
B: Increasing enzyme concentration provides more active sites, which increases Vmax.
C: A competitive inhibitor affects the apparent affinity (Km) but does not change Vmax.
D: Increasing temperature beyond optimal denatures the enzyme, decreasing Vmax.
MCQ 2:
An enzyme exhibits Michaelis-Menten kinetics. If the substrate concentration is equal to
Km, what is the velocity of the reaction relative to Vmax?
A) 0
B) 0.25 Vmax
C) 0.5 Vmax
D) Vmax
Correct Answer: C) 0.5 Vmax
Rationale:
At substrate concentration [S] = Km, the enzyme is operating at half its maximum velocity by
definition of Km in Michaelis-Menten kinetics.
MCQ 3:
A non-competitive inhibitor affects enzyme kinetics by:
A) Increasing Km without affecting Vmax
B) Decreasing both Km and Vmax
C) Decreasing Vmax without affecting Km
D) Increasing both Km and Vmax
Correct Answer: C) Decreasing Vmax without affecting Km
,Rationale:
Non-competitive inhibitors bind to an allosteric site, reducing enzyme activity regardless
of substrate concentration. This lowers Vmax.
Km remains unchanged because substrate binding is not affected.
50 deep MCAT-style questions (with answers + concise rationales) for Biological &
Biochemical Foundations of Living Systems.
Biological & Biochemical Foundations — 50 MCQs with answer +
rationale
1. Which bond links monosaccharides in a polysaccharide?A) Peptide bond
B) Glycosidic bond
C) Phosphodiester bond
D) Ester bond
Answer: B) Glycosidic bond
Rationale: Monosaccharides are joined by glycosidic (O-glycosidic) bonds; peptide
bonds join amino acids, phosphodiester bonds join nucleotides.
2. An enzyme shows increased activity when AMP is bound to an allosteric site. AMP
is acting as a:
A) Competitive inhibitor
B) Uncompetitive inhibitor
C) Allosteric activator (positive effector)
D) Cofactor
Answer: C) Allosteric activator (positive effector)
Rationale: Binding to an allosteric site that increases activity is an allosteric activator;
cofactors are usually required for catalysis but not described here.
3. Which process produces the most ATP per glucose under aerobic conditions?
A) Glycolysis alone
B) Krebs cycle and oxidative phosphorylation
C) Fermentation
D) Pentose phosphate pathway
Answer: B) Krebs cycle and oxidative phosphorylation
Rationale: Complete aerobic oxidation via TCA + oxidative phosphorylation yields the
most ATP (~30–32 ATP/glucose).
4. A mutation that changes an amino acid from a polar to nonpolar residue in an
enzyme’s active site will most likely:
A) Have no effect on activity
B) Prevent substrate binding if hydrogen bonds were critical
C) Increase Km by stabilizing enzyme–substrate complex
D) Convert enzyme to allosteric enzyme
, Answer: B) Prevent substrate binding if hydrogen bonds were critical
Rationale: Replacing polar residues that form hydrogen bonds can disrupt substrate
binding and enzyme activity.
5. Which of these vitamins is a precursor for NAD+?
A) Vitamin C
B) Niacin (B3)
C) Thiamine (B1)
D) Riboflavin (B2)
Answer: B) Niacin (B3)
Rationale: Niacin is precursor for NAD+/NADP+; riboflavin is precursor for
FAD/FMN.
6. During aerobic respiration, the final electron acceptor is:
A) NAD+
B) FAD
C) Oxygen
D) Pyruvate
Answer: C) Oxygen
Rationale: Oxygen accepts electrons at the end of the electron transport chain to form
water.
7. The presence of a 5′ cap and poly-A tail on eukaryotic mRNA functions to:
A) Increase transcription rate
B) Stabilize mRNA and aid in translation initiation
C) Direct mRNA to mitochondria
D) Promote alternative splicing
Answer: B) Stabilize mRNA and aid in translation initiation
Rationale: 5′ cap and 3′ poly-A tail protect mRNA from degradation and enhance
translation.
8. If an enzyme follows Michaelis-Menten kinetics, doubling enzyme concentration
will:
A) Double Vmax and double Km
B) Double Vmax and leave Km unchanged
C) Leave Vmax unchanged and double Km
D) Halve Vmax
Answer: B) Double Vmax and leave Km unchanged
Rationale: Vmax is proportional to enzyme concentration; Km is an intrinsic property of
the enzyme–substrate interaction.
9. Which cellular organelle is the primary site of fatty acid β-oxidation in eukaryotic
cells?
A) Mitochondrion
B) Endoplasmic reticulum
C) Golgi apparatus
D) Peroxisome
Answer: A) Mitochondrion
Rationale: Most β-oxidation occurs in mitochondria; peroxisomes handle very long-
chain fatty acids initially in some cells.