Uoft 2025 Fall BCH210 Final exam Questions With Complete Solutions
Course
BCH210
1. Enzyme Kinetics (Michaelis–Menten)
Question:
An enzyme has a Vmax=120 μmol/minV_{max} = 120 \, \mu mol/minVmax=120μmol/min and
Km=5 mMK_m = 5 \, mMKm=5mM.
Calculate the reaction velocity when substrate concentration is [S]=5 mM[S] = 5 \,
mM[S]=5mM.
Solution:
Use Michaelis–Menten equation:
v=Vmax[S]Km+[S]v = \frac{V_{max}[S]}{K_m + [S]}v=Km+[S]Vmax[S]
v=120×55+5=60010=60 μmol/minv = \frac{120 \times 5}{5 + 5} = \frac{600}{10} = 60 \, \mu
mol/minv=5+5120×5=10600=60μmol/min
Answer:
✅ v=60 μmol/minv = 60 \, \mu mol/minv=60μmol/min
2. Competitive Inhibition
Question:
How does a competitive inhibitor affect KmK_mKm and VmaxV_{max}Vmax?
Solution:
Competitive inhibitor binds active site
Can be overcome by increasing substrate
Effects:
KmK_mKm increases (lower apparent affinity)
VmaxV_{max}Vmax unchanged
Answer:
✅ Km↑K_m ↑Km↑, VmaxV_{max}Vmax unchanged
3. Protein Structure
Question:
Describe the forces stabilizing tertiary structure of proteins.
,Solution:
Tertiary structure is stabilized by:
Hydrophobic interactions
Hydrogen bonds
Ionic (salt bridges)
Disulfide bonds (covalent)
Van der Waals forces
Answer:
✅ Combination of non-covalent + covalent interactions maintains 3D folding
4. Hemoglobin Oxygen Binding
Question:
Explain why hemoglobin shows a sigmoidal oxygen binding curve.
Solution:
Due to cooperative binding
Binding of one O₂ increases affinity for next
T → R conformational shift
Answer:
✅ Sigmoidal curve arises from positive cooperativity
5. Glycolysis Regulation
Question:
Name the rate-limiting enzyme of glycolysis and how it is regulated.
Solution:
Enzyme: Phosphofructokinase-1 (PFK-1)
Activated by:
o AMP
o Fructose-2,6-bisphosphate
Inhibited by:
, o ATP
o Citrate
Answer:
✅ PFK-1 controls glycolysis via energy charge signals
6. ATP Yield Calculation
Question:
Calculate total ATP produced from 1 glucose molecule under aerobic conditions.
Solution:
Step ATP Yield
Glycolysis 2 ATP + 2 NADH
Pyruvate → Acetyl-CoA 2 NADH
TCA Cycle 2 ATP + 6 NADH + 2 FADH₂
Convert:
NADH = 2.5 ATP
FADH₂ = 1.5 ATP
ATP=2+(10×2.5)+(2×1.5)=2+25+3=30ATP = 2 + (10 \times 2.5) + (2 \times 1.5) = 2 + 25 + 3 =
30ATP=2+(10×2.5)+(2×1.5)=2+25+3=30
Answer:
✅ ~30–32 ATP per glucose
7. DNA Replication
Question:
Why is DNA synthesis always in the 5’ → 3’ direction?
Solution:
DNA polymerase adds nucleotides to 3’-OH group
Incoming nucleotide provides energy via triphosphate hydrolysis
Ensures proofreading capability
Course
BCH210
1. Enzyme Kinetics (Michaelis–Menten)
Question:
An enzyme has a Vmax=120 μmol/minV_{max} = 120 \, \mu mol/minVmax=120μmol/min and
Km=5 mMK_m = 5 \, mMKm=5mM.
Calculate the reaction velocity when substrate concentration is [S]=5 mM[S] = 5 \,
mM[S]=5mM.
Solution:
Use Michaelis–Menten equation:
v=Vmax[S]Km+[S]v = \frac{V_{max}[S]}{K_m + [S]}v=Km+[S]Vmax[S]
v=120×55+5=60010=60 μmol/minv = \frac{120 \times 5}{5 + 5} = \frac{600}{10} = 60 \, \mu
mol/minv=5+5120×5=10600=60μmol/min
Answer:
✅ v=60 μmol/minv = 60 \, \mu mol/minv=60μmol/min
2. Competitive Inhibition
Question:
How does a competitive inhibitor affect KmK_mKm and VmaxV_{max}Vmax?
Solution:
Competitive inhibitor binds active site
Can be overcome by increasing substrate
Effects:
KmK_mKm increases (lower apparent affinity)
VmaxV_{max}Vmax unchanged
Answer:
✅ Km↑K_m ↑Km↑, VmaxV_{max}Vmax unchanged
3. Protein Structure
Question:
Describe the forces stabilizing tertiary structure of proteins.
,Solution:
Tertiary structure is stabilized by:
Hydrophobic interactions
Hydrogen bonds
Ionic (salt bridges)
Disulfide bonds (covalent)
Van der Waals forces
Answer:
✅ Combination of non-covalent + covalent interactions maintains 3D folding
4. Hemoglobin Oxygen Binding
Question:
Explain why hemoglobin shows a sigmoidal oxygen binding curve.
Solution:
Due to cooperative binding
Binding of one O₂ increases affinity for next
T → R conformational shift
Answer:
✅ Sigmoidal curve arises from positive cooperativity
5. Glycolysis Regulation
Question:
Name the rate-limiting enzyme of glycolysis and how it is regulated.
Solution:
Enzyme: Phosphofructokinase-1 (PFK-1)
Activated by:
o AMP
o Fructose-2,6-bisphosphate
Inhibited by:
, o ATP
o Citrate
Answer:
✅ PFK-1 controls glycolysis via energy charge signals
6. ATP Yield Calculation
Question:
Calculate total ATP produced from 1 glucose molecule under aerobic conditions.
Solution:
Step ATP Yield
Glycolysis 2 ATP + 2 NADH
Pyruvate → Acetyl-CoA 2 NADH
TCA Cycle 2 ATP + 6 NADH + 2 FADH₂
Convert:
NADH = 2.5 ATP
FADH₂ = 1.5 ATP
ATP=2+(10×2.5)+(2×1.5)=2+25+3=30ATP = 2 + (10 \times 2.5) + (2 \times 1.5) = 2 + 25 + 3 =
30ATP=2+(10×2.5)+(2×1.5)=2+25+3=30
Answer:
✅ ~30–32 ATP per glucose
7. DNA Replication
Question:
Why is DNA synthesis always in the 5’ → 3’ direction?
Solution:
DNA polymerase adds nucleotides to 3’-OH group
Incoming nucleotide provides energy via triphosphate hydrolysis
Ensures proofreading capability
