Portage Learning BioChem 210 Final Exam: Ultimate Prep Bundle (2025/2026
Updated) – 300+ Verified Q&A & Metabolic Pathway Guides
This all-inclusive study guide is meticulously designed for students taking the Portage
Learning BioChem 210 final exam. It features over 300 verified practice questions with
detailed rationales covering everything from the properties of carbon to the complex
steps of the urea cycle and oxidative phosphorylation. The guide includes visual
breakdowns of metabolic pathways and clear explanations for identifying functional
groups and protein structures. Fully updated for the 2025/2026 academic year, this
resource ensures you have the core knowledge and calculation skills needed to secure
an A on your proctored final.
Which functional group is characterized by a carbon double-bonded to an oxygen
and single-bonded to a hydroxyl group (-COOH)?
A. Ketone
B. Aldehyde
C. Carboxylic Acid
D. Ester
Answer: C. Carboxylic Acid
Rationale: Carboxylic acids are central to amino acids and fatty acids. Esters (D)
would have an -OR group instead of -OH.
At a pH above its pKa, will an acidic functional group (like a carboxyl group) be
protonated or deprotonated?
A. Protonated (neutral)
B. Deprotonated (negative)
C. Protonated (positive)
D. Deprotonated (neutral)
Answer: B. Deprotonated (negative)
Rationale: When pH > pKa, the environment is basic enough to "strip" the proton (H+)
from the acid, leaving it with a negative charge.
Which of the following amino acids is classified as non-polar (hydrophobic)?
A. Lysine
B. Aspartate
,C. Valine
D. Serine
Answer: C. Valine
Rationale: Valine has an aliphatic isopropyl side chain. Serine is polar, Lysine is basic
(+), and Aspartate is acidic (-).
What type of bond stabilizes the primary structure of a protein?
A. Hydrogen bonds
B. Ionic bonds
C. Peptide bonds
D. Disulfide bridges
Answer: C. Peptide bonds
Rationale: Primary structure is the linear sequence of amino acids held together by
covalent peptide (amide) bonds.
In an alpha-helix, the hydrogen bonds occur between:
A. R-groups of distant amino acids.
B. The carbonyl oxygen and the amide nitrogen four residues apart.
C. Two different polypeptide chains.
D. Sulfur atoms in Cysteine.
Answer: B. The carbonyl oxygen and the amide nitrogen four residues apart.
Rationale: This specific spacing creates the tight, right-handed coil characteristic of the
alpha-helix.
A non-competitive inhibitor binds to:
A. The active site, blocking the substrate.
B. The substrate itself.
C. An allosteric site, changing the enzyme's shape.
D. The product of the reaction.
Answer: C. An allosteric site, changing the enzyme's shape.
Rationale: Because it doesn't compete for the active site, it lowers Vmax but does not
change the Km.
What happens to Vmax and Km in the presence of a competitive inhibitor?
A. Vmax decreases, Km increases.
B. Vmax stays the same, Km increases.
,C. Vmax decreases, Km stays the same.
D. Both stay the same.
Answer: B. Vmax stays the same, Km increases.
Rationale: You can overcome competitive inhibition by adding more substrate (reaching
Vmax), but it takes a higher concentration to get there (higher Km).
Hemoglobin’s affinity for oxygen increases as more oxygen molecules bind. This is
known as:
A. Competitive inhibition
B. Cooperativity
C. Denaturation
D. Feedback inhibition
Answer: B. Cooperativity
Rationale: Hemoglobin is an allosteric protein; the binding of one O2 shifts the subunits
from a "T" (tense) state to an "R" (relaxed) state.
Which molecule acts as a "high-energy" electron carrier in its reduced form during
catabolism?
A. NAD+
B. NADH
C. ADP
D. FAD
Answer: B. NADH
Rationale: NADH carries two high-energy electrons and one proton to the Electron
Transport Chain.
Glycolysis takes place in which part of the cell?
A. Mitochondria
B. Nucleus
C. Cytosol
D. Endoplasmic Reticulum
Answer: C. Cytosol
Rationale: All enzymes for the 10 steps of glycolysis are located in the cytoplasm.
What is the net yield of ATP and NADH from one molecule of glucose during glycolysis?
A. 4 ATP, 2 NADH
, B. 2 ATP, 2 NADH
C. 2 ATP, 4 NADH
D. 32 ATP, 10 NADH
Answer: B. 2 ATP, 2 NADH
Rationale: Glycolysis uses 2 ATP and produces 4 ATP, resulting in a net gain of 2 ATP.
Which enzyme catalyzes the "first committed step" of glycolysis?
A. Hexokinase
B. Pyruvate Kinase
C. Phosphofructokinase-1 (PFK-1)
D. Aldolase
Answer: C. Phosphofructokinase-1 (PFK-1)
Rationale: PFK-1 is the key regulatory enzyme. Once fructose-1,6-bisphosphate is
formed, the molecule must finish glycolysis.
Under anaerobic conditions, pyruvate is converted to ________ in muscle cells to
regenerate NAD+.
A. Acetyl-CoA
B. Ethanol
C. Lactate
D. Oxaloacetate
Answer: C. Lactate
Rationale: Lactate dehydrogenase reduces pyruvate to lactate, allowing glycolysis to
continue in the absence of oxygen.
The conversion of pyruvate to Acetyl-CoA is catalyzed by the Pyruvate Dehydrogenase
Complex. This reaction produces:
A. 1 CO2 and 1 NADH
B. 2 ATP
C. 1 FADH2
D. 1 Glucose
Answer: A. 1 CO2 and 1 NADH
Rationale: This is an oxidative decarboxylation that bridges glycolysis to the Citric Acid
Cycle.
Updated) – 300+ Verified Q&A & Metabolic Pathway Guides
This all-inclusive study guide is meticulously designed for students taking the Portage
Learning BioChem 210 final exam. It features over 300 verified practice questions with
detailed rationales covering everything from the properties of carbon to the complex
steps of the urea cycle and oxidative phosphorylation. The guide includes visual
breakdowns of metabolic pathways and clear explanations for identifying functional
groups and protein structures. Fully updated for the 2025/2026 academic year, this
resource ensures you have the core knowledge and calculation skills needed to secure
an A on your proctored final.
Which functional group is characterized by a carbon double-bonded to an oxygen
and single-bonded to a hydroxyl group (-COOH)?
A. Ketone
B. Aldehyde
C. Carboxylic Acid
D. Ester
Answer: C. Carboxylic Acid
Rationale: Carboxylic acids are central to amino acids and fatty acids. Esters (D)
would have an -OR group instead of -OH.
At a pH above its pKa, will an acidic functional group (like a carboxyl group) be
protonated or deprotonated?
A. Protonated (neutral)
B. Deprotonated (negative)
C. Protonated (positive)
D. Deprotonated (neutral)
Answer: B. Deprotonated (negative)
Rationale: When pH > pKa, the environment is basic enough to "strip" the proton (H+)
from the acid, leaving it with a negative charge.
Which of the following amino acids is classified as non-polar (hydrophobic)?
A. Lysine
B. Aspartate
,C. Valine
D. Serine
Answer: C. Valine
Rationale: Valine has an aliphatic isopropyl side chain. Serine is polar, Lysine is basic
(+), and Aspartate is acidic (-).
What type of bond stabilizes the primary structure of a protein?
A. Hydrogen bonds
B. Ionic bonds
C. Peptide bonds
D. Disulfide bridges
Answer: C. Peptide bonds
Rationale: Primary structure is the linear sequence of amino acids held together by
covalent peptide (amide) bonds.
In an alpha-helix, the hydrogen bonds occur between:
A. R-groups of distant amino acids.
B. The carbonyl oxygen and the amide nitrogen four residues apart.
C. Two different polypeptide chains.
D. Sulfur atoms in Cysteine.
Answer: B. The carbonyl oxygen and the amide nitrogen four residues apart.
Rationale: This specific spacing creates the tight, right-handed coil characteristic of the
alpha-helix.
A non-competitive inhibitor binds to:
A. The active site, blocking the substrate.
B. The substrate itself.
C. An allosteric site, changing the enzyme's shape.
D. The product of the reaction.
Answer: C. An allosteric site, changing the enzyme's shape.
Rationale: Because it doesn't compete for the active site, it lowers Vmax but does not
change the Km.
What happens to Vmax and Km in the presence of a competitive inhibitor?
A. Vmax decreases, Km increases.
B. Vmax stays the same, Km increases.
,C. Vmax decreases, Km stays the same.
D. Both stay the same.
Answer: B. Vmax stays the same, Km increases.
Rationale: You can overcome competitive inhibition by adding more substrate (reaching
Vmax), but it takes a higher concentration to get there (higher Km).
Hemoglobin’s affinity for oxygen increases as more oxygen molecules bind. This is
known as:
A. Competitive inhibition
B. Cooperativity
C. Denaturation
D. Feedback inhibition
Answer: B. Cooperativity
Rationale: Hemoglobin is an allosteric protein; the binding of one O2 shifts the subunits
from a "T" (tense) state to an "R" (relaxed) state.
Which molecule acts as a "high-energy" electron carrier in its reduced form during
catabolism?
A. NAD+
B. NADH
C. ADP
D. FAD
Answer: B. NADH
Rationale: NADH carries two high-energy electrons and one proton to the Electron
Transport Chain.
Glycolysis takes place in which part of the cell?
A. Mitochondria
B. Nucleus
C. Cytosol
D. Endoplasmic Reticulum
Answer: C. Cytosol
Rationale: All enzymes for the 10 steps of glycolysis are located in the cytoplasm.
What is the net yield of ATP and NADH from one molecule of glucose during glycolysis?
A. 4 ATP, 2 NADH
, B. 2 ATP, 2 NADH
C. 2 ATP, 4 NADH
D. 32 ATP, 10 NADH
Answer: B. 2 ATP, 2 NADH
Rationale: Glycolysis uses 2 ATP and produces 4 ATP, resulting in a net gain of 2 ATP.
Which enzyme catalyzes the "first committed step" of glycolysis?
A. Hexokinase
B. Pyruvate Kinase
C. Phosphofructokinase-1 (PFK-1)
D. Aldolase
Answer: C. Phosphofructokinase-1 (PFK-1)
Rationale: PFK-1 is the key regulatory enzyme. Once fructose-1,6-bisphosphate is
formed, the molecule must finish glycolysis.
Under anaerobic conditions, pyruvate is converted to ________ in muscle cells to
regenerate NAD+.
A. Acetyl-CoA
B. Ethanol
C. Lactate
D. Oxaloacetate
Answer: C. Lactate
Rationale: Lactate dehydrogenase reduces pyruvate to lactate, allowing glycolysis to
continue in the absence of oxygen.
The conversion of pyruvate to Acetyl-CoA is catalyzed by the Pyruvate Dehydrogenase
Complex. This reaction produces:
A. 1 CO2 and 1 NADH
B. 2 ATP
C. 1 FADH2
D. 1 Glucose
Answer: A. 1 CO2 and 1 NADH
Rationale: This is an oxidative decarboxylation that bridges glycolysis to the Citric Acid
Cycle.
