BIOCHEM 210 FINAL EXAM 2026/2027 | Comprehensive
Biochemistry Review | Latest Q&A Verified | Portage
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MODULE 1: FOUNDATIONS
Q1: Calculate the pH of a solution containing 2.5 × 10⁻⁵ M HCl at 25°C.
A. 2.40
B. 4.60 [CORRECT]
C. 5.40
D. 9.40
Correct Answer: B
Rationale: HCl is a strong acid that dissociates completely, so [H⁺] = 2.5 × 10⁻⁵ M. Using
the formula pH = -log[H⁺]: pH = -log(2.5 × 10⁻⁵) = -(log 2.5 + log 10⁻⁵) = -(0.40 - 5.00) =
4.60. Option A incorrectly calculates -log(2.5 × 10⁻⁵) as 2.40 (sign error). Option C
transposes digits (5.40 instead of 4.60). Option D calculates pOH instead of pH (14 -
4.60 = 9.40), confusing acid and base calculations.
Q2: Which functional group is characterized by a carbonyl group (C=O) bonded to a
hydroxyl group (-OH)?
A. Aldehyde
B. Ketone
C. Carboxylic acid [CORRECT]
D. Ester
Correct Answer: C
Rationale: Carboxylic acids contain the -COOH functional group, combining a carbonyl
(C=O) with a hydroxyl (-OH) on the same carbon. Aldehydes (A) have -CHO (carbonyl
with H). Ketones (B) have carbonyl between two carbons (R-CO-R'). Esters (D) have
-COOR' (carbonyl with OR' group). The carboxyl group is ionizable (pKa ~4-5), making
carboxylic acids acidic and critical for amino acid chemistry (Module 4).
,Q3: Using the Henderson-Hasselbalch equation, what is the pH of a buffer solution
containing 0.10 M acetic acid (pKa = 4.76) and 0.30 M sodium acetate?
A. 4.26
B. 4.76
C. 5.26 [CORRECT]
D. 5.76
Correct Answer: C
Rationale: Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]) = 4.76 + log(0.30/0.10) =
4.76 + log(3) = 4.76 + 0.48 = 5.26. The buffer pH is higher than pKa because the
conjugate base (acetate) exceeds the weak acid (acetic acid). Option A subtracts
instead of adding log(3). Option B would require equal concentrations. Option D uses
log(10) instead of log(3).
Q4: Which property of water is MOST critical for its role as the universal biological
solvent?
A. Low heat capacity
B. High polarity and hydrogen bonding capability [CORRECT]
C. Low surface tension
D. Non-polar character
Correct Answer: B
Rationale: Water's high polarity (electronegativity difference O-H) and extensive
hydrogen bonding enable dissolution of ionic and polar compounds, stabilization of
biomolecular structures, and participation in metabolic reactions. Water has high (not
low) heat capacity (A is incorrect), enabling temperature regulation. Water has high (not
low) surface tension (C is incorrect). Water is polar (D is incorrect). These properties
derive from water's bent molecular geometry and electronegativity differences.
Q5: Which type of isomerism involves compounds with identical molecular formulas but
different spatial arrangements that are non-superimposable mirror images?
A. Structural isomerism
B. Geometric isomerism
C. Enantiomerism (optical isomerism) [CORRECT]
,D. Conformational isomerism
Correct Answer: C
Rationale: Enantiomers are stereoisomers that are non-superimposable mirror images
(like left and right hands), characterized by chirality (typically tetrahedral carbon with
four different substituents). They rotate plane-polarized light equally but in opposite
directions. Structural isomers (A) differ in bond connectivity. Geometric isomers (B,
cis/trans) differ in arrangement around double bonds or rings but aren't mirror images.
Conformational isomers (D) interconvert by bond rotation and aren't distinct isolable
compounds.
Q6: In the context of acid-base chemistry, what is the relationship between pKa and acid
strength?
A. Higher pKa indicates stronger acid
B. Lower pKa indicates stronger acid [CORRECT]
C. pKa is unrelated to acid strength
D. pKa equals acid strength only at pH 7
Correct Answer: B
Rationale: pKa = -log Ka, so lower pKa corresponds to higher Ka (acid dissociation
constant), indicating greater acid strength (more complete dissociation). Strong acids
like HCl have pKa < 0; weak acids like acetic acid have pKa ~4.76. This logarithmic
relationship is inverse—each pKa unit represents a 10-fold difference in acid strength.
Understanding pKa is essential for predicting ionization states of amino acid side
chains and drug molecules at physiological pH.
MODULE 2: ORGANIC CHEMISTRY
Q7: Which hybridization state corresponds to a carbon atom forming four single bonds
in a tetrahedral geometry?
A. sp
B. sp²
C. sp³ [CORRECT]
D. dsp³
, Correct Answer: C
Rationale: sp³ hybridization (mixing one 2s and three 2p orbitals) creates four
equivalent orbitals with 109.5° tetrahedral angles, as seen in alkanes and saturated
carbons. sp (A) creates linear geometry (180°, two bonds, e.g., alkynes). sp² (B) creates
trigonal planar geometry (120°, three bonds, e.g., alkenes, carbonyls). dsp³ (D) involves
d-orbitals not used by carbon in organic chemistry. Hybridization determines molecular
geometry, bond angles, and reactivity.
Q8: Which class of biomolecules is characterized by being insoluble in water, soluble in
nonpolar solvents, and serving primarily for energy storage and membrane structure?
A. Carbohydrates
B. Proteins
C. Lipids [CORRECT]
D. Nucleic acids
Correct Answer: C
Rationale: Lipids are defined by solubility properties (hydrophobic/hydrophilic
amphipathic nature) rather than shared structural features. They include fatty acids,
triacylglycerols (energy storage), phospholipids (membranes), steroids (signaling), and
waxes. Carbohydrates (A) are typically water-soluble. Proteins (B) have variable
solubility based on amino acid composition. Nucleic acids (D) are water-soluble
polyanions. Lipid solubility derives from hydrocarbon chains that cannot form hydrogen
bonds with water.
Q9: Which functional group is present in primary amines but NOT in secondary amines?
A. Nitrogen atom
B. Two carbon groups attached to nitrogen
C. Two hydrogen atoms attached to nitrogen [CORRECT]
D. Lone pair of electrons on nitrogen
Correct Answer: C
Rationale: Primary amines have the general formula R-NH₂ (two H atoms on N);
secondary amines have R₂NH (one H on N); tertiary amines have R₃N (no H on N). All
Biochemistry Review | Latest Q&A Verified | Portage
Learning | Pass Guaranteed - A+ Graded
MODULE 1: FOUNDATIONS
Q1: Calculate the pH of a solution containing 2.5 × 10⁻⁵ M HCl at 25°C.
A. 2.40
B. 4.60 [CORRECT]
C. 5.40
D. 9.40
Correct Answer: B
Rationale: HCl is a strong acid that dissociates completely, so [H⁺] = 2.5 × 10⁻⁵ M. Using
the formula pH = -log[H⁺]: pH = -log(2.5 × 10⁻⁵) = -(log 2.5 + log 10⁻⁵) = -(0.40 - 5.00) =
4.60. Option A incorrectly calculates -log(2.5 × 10⁻⁵) as 2.40 (sign error). Option C
transposes digits (5.40 instead of 4.60). Option D calculates pOH instead of pH (14 -
4.60 = 9.40), confusing acid and base calculations.
Q2: Which functional group is characterized by a carbonyl group (C=O) bonded to a
hydroxyl group (-OH)?
A. Aldehyde
B. Ketone
C. Carboxylic acid [CORRECT]
D. Ester
Correct Answer: C
Rationale: Carboxylic acids contain the -COOH functional group, combining a carbonyl
(C=O) with a hydroxyl (-OH) on the same carbon. Aldehydes (A) have -CHO (carbonyl
with H). Ketones (B) have carbonyl between two carbons (R-CO-R'). Esters (D) have
-COOR' (carbonyl with OR' group). The carboxyl group is ionizable (pKa ~4-5), making
carboxylic acids acidic and critical for amino acid chemistry (Module 4).
,Q3: Using the Henderson-Hasselbalch equation, what is the pH of a buffer solution
containing 0.10 M acetic acid (pKa = 4.76) and 0.30 M sodium acetate?
A. 4.26
B. 4.76
C. 5.26 [CORRECT]
D. 5.76
Correct Answer: C
Rationale: Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]) = 4.76 + log(0.30/0.10) =
4.76 + log(3) = 4.76 + 0.48 = 5.26. The buffer pH is higher than pKa because the
conjugate base (acetate) exceeds the weak acid (acetic acid). Option A subtracts
instead of adding log(3). Option B would require equal concentrations. Option D uses
log(10) instead of log(3).
Q4: Which property of water is MOST critical for its role as the universal biological
solvent?
A. Low heat capacity
B. High polarity and hydrogen bonding capability [CORRECT]
C. Low surface tension
D. Non-polar character
Correct Answer: B
Rationale: Water's high polarity (electronegativity difference O-H) and extensive
hydrogen bonding enable dissolution of ionic and polar compounds, stabilization of
biomolecular structures, and participation in metabolic reactions. Water has high (not
low) heat capacity (A is incorrect), enabling temperature regulation. Water has high (not
low) surface tension (C is incorrect). Water is polar (D is incorrect). These properties
derive from water's bent molecular geometry and electronegativity differences.
Q5: Which type of isomerism involves compounds with identical molecular formulas but
different spatial arrangements that are non-superimposable mirror images?
A. Structural isomerism
B. Geometric isomerism
C. Enantiomerism (optical isomerism) [CORRECT]
,D. Conformational isomerism
Correct Answer: C
Rationale: Enantiomers are stereoisomers that are non-superimposable mirror images
(like left and right hands), characterized by chirality (typically tetrahedral carbon with
four different substituents). They rotate plane-polarized light equally but in opposite
directions. Structural isomers (A) differ in bond connectivity. Geometric isomers (B,
cis/trans) differ in arrangement around double bonds or rings but aren't mirror images.
Conformational isomers (D) interconvert by bond rotation and aren't distinct isolable
compounds.
Q6: In the context of acid-base chemistry, what is the relationship between pKa and acid
strength?
A. Higher pKa indicates stronger acid
B. Lower pKa indicates stronger acid [CORRECT]
C. pKa is unrelated to acid strength
D. pKa equals acid strength only at pH 7
Correct Answer: B
Rationale: pKa = -log Ka, so lower pKa corresponds to higher Ka (acid dissociation
constant), indicating greater acid strength (more complete dissociation). Strong acids
like HCl have pKa < 0; weak acids like acetic acid have pKa ~4.76. This logarithmic
relationship is inverse—each pKa unit represents a 10-fold difference in acid strength.
Understanding pKa is essential for predicting ionization states of amino acid side
chains and drug molecules at physiological pH.
MODULE 2: ORGANIC CHEMISTRY
Q7: Which hybridization state corresponds to a carbon atom forming four single bonds
in a tetrahedral geometry?
A. sp
B. sp²
C. sp³ [CORRECT]
D. dsp³
, Correct Answer: C
Rationale: sp³ hybridization (mixing one 2s and three 2p orbitals) creates four
equivalent orbitals with 109.5° tetrahedral angles, as seen in alkanes and saturated
carbons. sp (A) creates linear geometry (180°, two bonds, e.g., alkynes). sp² (B) creates
trigonal planar geometry (120°, three bonds, e.g., alkenes, carbonyls). dsp³ (D) involves
d-orbitals not used by carbon in organic chemistry. Hybridization determines molecular
geometry, bond angles, and reactivity.
Q8: Which class of biomolecules is characterized by being insoluble in water, soluble in
nonpolar solvents, and serving primarily for energy storage and membrane structure?
A. Carbohydrates
B. Proteins
C. Lipids [CORRECT]
D. Nucleic acids
Correct Answer: C
Rationale: Lipids are defined by solubility properties (hydrophobic/hydrophilic
amphipathic nature) rather than shared structural features. They include fatty acids,
triacylglycerols (energy storage), phospholipids (membranes), steroids (signaling), and
waxes. Carbohydrates (A) are typically water-soluble. Proteins (B) have variable
solubility based on amino acid composition. Nucleic acids (D) are water-soluble
polyanions. Lipid solubility derives from hydrocarbon chains that cannot form hydrogen
bonds with water.
Q9: Which functional group is present in primary amines but NOT in secondary amines?
A. Nitrogen atom
B. Two carbon groups attached to nitrogen
C. Two hydrogen atoms attached to nitrogen [CORRECT]
D. Lone pair of electrons on nitrogen
Correct Answer: C
Rationale: Primary amines have the general formula R-NH₂ (two H atoms on N);
secondary amines have R₂NH (one H on N); tertiary amines have R₃N (no H on N). All
