BIOD 101 Final Exam & Review Guide (Latest 2026/2027
Update) Portage Learning - Actual Questions & Rationalized
Answers - 139 Questions
Comprehensive examination on BIOD 101 Final Exam & Review Guide (Latest 2026/2027 Update) Portage
Learning - Actual Questions & Rationalized Answers. It contains 139 multiple-choice questions, each with four
distractors and a fully worked rationale that explains why the keyed answer is correct. Content is organized into
10 focused sections: Chemistry of Life, Cell Structure and Function, Metabolism and Enzymes, Photosynthesis
and Cellular Respiration, Cell Division (Mitosis and Meiosis), Genetics and Heredity, DNA Structure and Gene
Expression, Evolution and Natural Selection, Ecology and Ecosystems, Scientific Method and Research. Targeted
learning outcomes include: Demonstrate mastery of core concepts. Every item has been reviewed for clinical
accuracy, current guidelines, and clarity so that students can study with confidence and self-correct as they work
through the bank. Use it as a high-yield review immediately before the exam, or as a structured practice tool
during the unit - the rationales double as concise teaching notes. The recommended writing time is 3 hours, with a
passing score of 70%. Aligned with Aligned with US university standards. standards and reflects the question style
commonly seen on accredited program examinations. Students consistently achieving above the cut score on this
bank have historically gone on to earn A+ on the corresponding course exam. Read every stem carefully -
distractors are written to look plausible, and the best answer is sometimes the one that addresses the patient's
most immediate physiological or safety need. Where multiple options appear correct, prioritize airway, breathing,
Section 1: Chemistry of Life (Questions 1-10)
1 In a molecule of water, the electronegativity difference between oxygen and
hydrogen leads to a permanent dipole. Which statement best explains why
water exhibits a higher boiling point than hydrogen sulfide, despite H2S
having a similar molecular geometry?
A) Water molecules form stronger London dispersion forces due to the
smaller size of oxygen.
B) The hydrogen bonds in water are stronger than the dipole-dipole
interactions in H2S because of oxygen's higher electronegativity.
C) Water has a higher molecular weight than hydrogen sulfide.
D) Hydrogen sulfide can form hydrogen bonds, but they are weaker due to
sulfur's larger atomic radius.
Answer: B
Rationale: Water's high boiling point is due to extensive hydrogen bonding,
which requires more energy to overcome. H2S has weaker dipole-dipole
interactions because sulfur is less electronegative and larger, leading to weaker
intermolecular forces. Option A is incorrect because London forces are
generally weaker than hydrogen bonds. Option C is false; H2S has higher
molecular mass. Option D is wrong because sulfur cannot form strong
,hydrogen bonds.
2 A phosphate buffer system has a pKa of 7.21. At pH 7.4, what is the
approximate ratio of [HPO4^2-] to [H2PO4-]?
A) 0.62:1
B) 1.55:1
C) 2.00:1
D) 0.50:1
Answer: B
Rationale: Using the Henderson-Hasselbalch equation: pH = pKa +
log([base]/[acid]). 7.4 = 7.21 + log(ratio). log(ratio) = 0.19, so ratio = 10^0.19
1.55. Option B is correct. Other options miscalculate the antilog or misapply
the equation.
3 A peptide bond forms between the carboxyl group of one amino acid and the
amino group of another. Which statement best describes the resonance
structure of the peptide bond?
A) The bond has partial double-bond character, restricting rotation and
contributing to planar geometry.
B) The bond is a pure single bond, allowing free rotation around the C-N
axis.
C) The bond is a double bond, preventing any rotation and fixing the
conformation.
D) The bond is a single bond with no resonance stabilization, but steric
hindrance limits rotation.
Answer: A
Rationale: The peptide bond exhibits resonance between a single and double
bond, giving it partial double-bond character. This restricts rotation and makes
the peptide bond planar. Option B is incorrect because resonance confers
partial double-bond character, not pure single bond. Option C overstates (not
full double bond). Option D ignores resonance.
4 In the anomeric carbon of glucose, the alpha and beta configurations differ in
the orientation of the hydroxyl group. For D-glucose in solution, which
anomer predominates at equilibrium and why?
,A) Alpha, because the axial hydroxyl reduces steric clash in the chair
conformation.
B) Beta, because the equatorial hydroxyl is more stable in the chair
conformation.
C) Alpha and beta are equally present due to rapid mutarotation.
D) Beta, because it has fewer gauche interactions.
Answer: B
Rationale: In D-glucose, the beta anomer has all hydroxyl groups equatorial in
the chair form, which is energetically more stable than the alpha anomer where
one hydroxyl is axial. Thus, beta-D-glucopyranose predominates (~64%) at
equilibrium. Option A is false; alpha has axial OH, causing 1,3-diaxial
interactions. Option C is incorrect; equilibrium favors beta. Option D incorrect:
beta has fewer but not just gauche interactions.
5 A phospholipid molecule has a polar head and two nonpolar tails. When
placed in aqueous solution, it spontaneously forms a bilayer. Which
thermodynamic principle best explains this self-assembly?
A) Increase in entropy of the surrounding water molecules due to the
hydrophobic effect.
B) Formation of covalent bonds between the tails reduces enthalpy.
C) The polar heads form hydrogen bonds with each other, lowering Gibbs
free energy.
D) The tails become ionized in water, facilitating electrostatic interactions.
Answer: A
Rationale: The hydrophobic effect drives bilayer formation: nonpolar tails
cluster to minimize contact with water, releasing ordered water molecules from
the hydrophobic hydration shell, increasing entropy. Option B is false; no
covalent bonds form between tails. Option C is partially true but the primary
driver is entropic. Option D is false; tails are nonpolar and do not ionize.
6 In enzyme kinetics, the Michaelis-Menten constant (Km) is often
approximated as the substrate concentration at half Vmax. However, this
approximation holds true only under specific conditions. Which condition is
required?
A) The enzyme concentration is much less than the substrate concentration.
B) The enzyme-substrate complex concentration is negligible.
, C) The reaction is irreversible and the product formation step is rate-limiting.
D) The substrate concentration is saturating.
Answer: C
Rationale: The standard Michaelis-Menten equation assumes the rapid
equilibrium approximation or steady-state where k2 (product formation) is
rate-limiting. Under these conditions, Km equals the substrate concentration at
half Vmax. Option A is typical for initial rates but not directly for the Km
approximation. Option B is not the key condition. Option D is wrong; at
saturation, V=Vmax, not half.
7 Double-stranded DNA is stabilized by hydrogen bonds between base pairs
and base stacking interactions. Which statement accurately describes the
contribution of these forces to DNA stability?
A) Base stacking provides approximately equal stabilization as hydrogen
bonding.
B) Hydrogen bonding is the primary stabilizing force, contributing over 70%
of total stability.
C) Base stacking contributes more to stability than hydrogen bonding in
aqueous solution.
D) Covalent bonds between bases provide most of the stabilization.
Answer: C
Rationale: While hydrogen bonds are important for specificity, base stacking
via van der Waals and hydrophobic interactions provides the majority of the
free energy stabilization of the double helix. Option A underestimates stacking;
option B overestimates hydrogen bonds. Option D is false; bases are not
covalently bonded to each other.
8 A researcher observes that a reaction has a large negative G. Which
conclusion is correct?
A) The reaction requires an input of energy to proceed.
B) The reaction is spontaneous and will occur rapidly.
C) The reaction is spontaneous but provides no information about reaction
rate.
D) The reaction is at equilibrium under standard conditions.
Answer: C
Rationale: ”G indicates spontaneity (direction), but not the rate. A large
Update) Portage Learning - Actual Questions & Rationalized
Answers - 139 Questions
Comprehensive examination on BIOD 101 Final Exam & Review Guide (Latest 2026/2027 Update) Portage
Learning - Actual Questions & Rationalized Answers. It contains 139 multiple-choice questions, each with four
distractors and a fully worked rationale that explains why the keyed answer is correct. Content is organized into
10 focused sections: Chemistry of Life, Cell Structure and Function, Metabolism and Enzymes, Photosynthesis
and Cellular Respiration, Cell Division (Mitosis and Meiosis), Genetics and Heredity, DNA Structure and Gene
Expression, Evolution and Natural Selection, Ecology and Ecosystems, Scientific Method and Research. Targeted
learning outcomes include: Demonstrate mastery of core concepts. Every item has been reviewed for clinical
accuracy, current guidelines, and clarity so that students can study with confidence and self-correct as they work
through the bank. Use it as a high-yield review immediately before the exam, or as a structured practice tool
during the unit - the rationales double as concise teaching notes. The recommended writing time is 3 hours, with a
passing score of 70%. Aligned with Aligned with US university standards. standards and reflects the question style
commonly seen on accredited program examinations. Students consistently achieving above the cut score on this
bank have historically gone on to earn A+ on the corresponding course exam. Read every stem carefully -
distractors are written to look plausible, and the best answer is sometimes the one that addresses the patient's
most immediate physiological or safety need. Where multiple options appear correct, prioritize airway, breathing,
Section 1: Chemistry of Life (Questions 1-10)
1 In a molecule of water, the electronegativity difference between oxygen and
hydrogen leads to a permanent dipole. Which statement best explains why
water exhibits a higher boiling point than hydrogen sulfide, despite H2S
having a similar molecular geometry?
A) Water molecules form stronger London dispersion forces due to the
smaller size of oxygen.
B) The hydrogen bonds in water are stronger than the dipole-dipole
interactions in H2S because of oxygen's higher electronegativity.
C) Water has a higher molecular weight than hydrogen sulfide.
D) Hydrogen sulfide can form hydrogen bonds, but they are weaker due to
sulfur's larger atomic radius.
Answer: B
Rationale: Water's high boiling point is due to extensive hydrogen bonding,
which requires more energy to overcome. H2S has weaker dipole-dipole
interactions because sulfur is less electronegative and larger, leading to weaker
intermolecular forces. Option A is incorrect because London forces are
generally weaker than hydrogen bonds. Option C is false; H2S has higher
molecular mass. Option D is wrong because sulfur cannot form strong
,hydrogen bonds.
2 A phosphate buffer system has a pKa of 7.21. At pH 7.4, what is the
approximate ratio of [HPO4^2-] to [H2PO4-]?
A) 0.62:1
B) 1.55:1
C) 2.00:1
D) 0.50:1
Answer: B
Rationale: Using the Henderson-Hasselbalch equation: pH = pKa +
log([base]/[acid]). 7.4 = 7.21 + log(ratio). log(ratio) = 0.19, so ratio = 10^0.19
1.55. Option B is correct. Other options miscalculate the antilog or misapply
the equation.
3 A peptide bond forms between the carboxyl group of one amino acid and the
amino group of another. Which statement best describes the resonance
structure of the peptide bond?
A) The bond has partial double-bond character, restricting rotation and
contributing to planar geometry.
B) The bond is a pure single bond, allowing free rotation around the C-N
axis.
C) The bond is a double bond, preventing any rotation and fixing the
conformation.
D) The bond is a single bond with no resonance stabilization, but steric
hindrance limits rotation.
Answer: A
Rationale: The peptide bond exhibits resonance between a single and double
bond, giving it partial double-bond character. This restricts rotation and makes
the peptide bond planar. Option B is incorrect because resonance confers
partial double-bond character, not pure single bond. Option C overstates (not
full double bond). Option D ignores resonance.
4 In the anomeric carbon of glucose, the alpha and beta configurations differ in
the orientation of the hydroxyl group. For D-glucose in solution, which
anomer predominates at equilibrium and why?
,A) Alpha, because the axial hydroxyl reduces steric clash in the chair
conformation.
B) Beta, because the equatorial hydroxyl is more stable in the chair
conformation.
C) Alpha and beta are equally present due to rapid mutarotation.
D) Beta, because it has fewer gauche interactions.
Answer: B
Rationale: In D-glucose, the beta anomer has all hydroxyl groups equatorial in
the chair form, which is energetically more stable than the alpha anomer where
one hydroxyl is axial. Thus, beta-D-glucopyranose predominates (~64%) at
equilibrium. Option A is false; alpha has axial OH, causing 1,3-diaxial
interactions. Option C is incorrect; equilibrium favors beta. Option D incorrect:
beta has fewer but not just gauche interactions.
5 A phospholipid molecule has a polar head and two nonpolar tails. When
placed in aqueous solution, it spontaneously forms a bilayer. Which
thermodynamic principle best explains this self-assembly?
A) Increase in entropy of the surrounding water molecules due to the
hydrophobic effect.
B) Formation of covalent bonds between the tails reduces enthalpy.
C) The polar heads form hydrogen bonds with each other, lowering Gibbs
free energy.
D) The tails become ionized in water, facilitating electrostatic interactions.
Answer: A
Rationale: The hydrophobic effect drives bilayer formation: nonpolar tails
cluster to minimize contact with water, releasing ordered water molecules from
the hydrophobic hydration shell, increasing entropy. Option B is false; no
covalent bonds form between tails. Option C is partially true but the primary
driver is entropic. Option D is false; tails are nonpolar and do not ionize.
6 In enzyme kinetics, the Michaelis-Menten constant (Km) is often
approximated as the substrate concentration at half Vmax. However, this
approximation holds true only under specific conditions. Which condition is
required?
A) The enzyme concentration is much less than the substrate concentration.
B) The enzyme-substrate complex concentration is negligible.
, C) The reaction is irreversible and the product formation step is rate-limiting.
D) The substrate concentration is saturating.
Answer: C
Rationale: The standard Michaelis-Menten equation assumes the rapid
equilibrium approximation or steady-state where k2 (product formation) is
rate-limiting. Under these conditions, Km equals the substrate concentration at
half Vmax. Option A is typical for initial rates but not directly for the Km
approximation. Option B is not the key condition. Option D is wrong; at
saturation, V=Vmax, not half.
7 Double-stranded DNA is stabilized by hydrogen bonds between base pairs
and base stacking interactions. Which statement accurately describes the
contribution of these forces to DNA stability?
A) Base stacking provides approximately equal stabilization as hydrogen
bonding.
B) Hydrogen bonding is the primary stabilizing force, contributing over 70%
of total stability.
C) Base stacking contributes more to stability than hydrogen bonding in
aqueous solution.
D) Covalent bonds between bases provide most of the stabilization.
Answer: C
Rationale: While hydrogen bonds are important for specificity, base stacking
via van der Waals and hydrophobic interactions provides the majority of the
free energy stabilization of the double helix. Option A underestimates stacking;
option B overestimates hydrogen bonds. Option D is false; bases are not
covalently bonded to each other.
8 A researcher observes that a reaction has a large negative G. Which
conclusion is correct?
A) The reaction requires an input of energy to proceed.
B) The reaction is spontaneous and will occur rapidly.
C) The reaction is spontaneous but provides no information about reaction
rate.
D) The reaction is at equilibrium under standard conditions.
Answer: C
Rationale: ”G indicates spontaneity (direction), but not the rate. A large
