PORTAGE LEARNING BIOCHEM MODULE 4
EXAM FINAL PAPER 2026 FULL SOLUTION
QUESTIONS AND ANSWERS 100% CORRECT
◉ Hemoglobin Binding Curve.
Answer: 4 subunits present in hemoglobin that can be either T or R -
state. Cooperative binding leads to a sigmoidal curve.
◉ Binding Cooperativity.
Answer: When one subunit of hemoglobin changes from T to R-state
the other sites are more likely to change to R-state as well. Leads to
sigmoidal graph.
◉ Homotropic Regulation of Binding.
Answer: Where a regulatory molecule is also the enzyme's substrate.
◉ Heterotropic Regulation of Binding.
Answer: Where an allosteric regulator is present that is not the
enzyme's substrate.
◉ Hill Plot.
,Answer: Turns sigmoid into straight lines. Slope = n (# of binding
sites). Allows measurement of binding sites that are cooperative.
◉ pH and Binding Affinity (Bohr Affect).
Answer: As [H+] increases, Histidine group in hemoglobin becomes
more protonated and protein shifts to T-state. O2 binding affinity
decreases.
◉ CO2 binding in Hemoglobin.
Answer: Forms carbonic acid that shifts hemoglobin to T-state. O2
binding affinity decreases. Used in the peripheral tissues.
◉ BPG (2,3-bisphosphoglycerate).
Answer: Greatly reduces hemoglobin's affinity for O2 by binding
allosterically. Stabilizes T-state. Transfer of O2 can improve because
increased delivery in tissues can outweigh decreased binding in the
lungs.
◉ Michaelis-Menton Equation.
Answer: V0 = (Vmax[S]) / (Km + [S])
◉ Km in Michaelis-Menton.
Answer: Km = [S] when V0 = 0.5(Vmax)
,◉ Michaelis-Menton Graph.
Answer:
◉ Lineweaver-Burke Graph.
Answer: Slope = Km/Vmax
Y-intercept = 1/Vmax
X-intercept = - 1/Km
◉ Lineweaver-Burke Equation.
Answer: Found by taking the reciprocal of the Michaelis-Menton
Equation.
◉ Kcat.
Answer: Rate-limiting step in any enzyme-catalyzed reaction at
saturation. Known as the "turn-over number". Kcat = Vmax/Et
◉ Chymotripsin.
Answer: Cleaves proteins on C-terminal endof Phe, Trp, and Tyr
◉ Competitive Inhibition Graph.
Answer: Slope changes by factor of α. Slope becomes αKm/Vmax.
X-intercept becomes 1/αKm
, Y-intercept does not change.
Vmax does not change.
◉ Uncompetitive Inhibition Graph.
Answer: Does not change slope.
Changes Km and Vmax.
Results in vertical shift up and down.
Y-intercept becomes α'/Vmax
X-intercept becomes -α'/Km
◉ Mixed Inhibition Graph.
Answer: Allosteric inhibitor that binds either E or ES.
Pivot point is between X-intercept and Y-intercept.
◉ Non-Competitive Inhibition Graph.
Answer: Form of mixed inhibition where the pivot point is on the x-
axis. Only happens when K1 is equal to K1'.
◉ Ionophore.
Answer: Hydrophobic molecule that binds to ions and carries them
through cell membranes. Disrupts concentration gradients.
EXAM FINAL PAPER 2026 FULL SOLUTION
QUESTIONS AND ANSWERS 100% CORRECT
◉ Hemoglobin Binding Curve.
Answer: 4 subunits present in hemoglobin that can be either T or R -
state. Cooperative binding leads to a sigmoidal curve.
◉ Binding Cooperativity.
Answer: When one subunit of hemoglobin changes from T to R-state
the other sites are more likely to change to R-state as well. Leads to
sigmoidal graph.
◉ Homotropic Regulation of Binding.
Answer: Where a regulatory molecule is also the enzyme's substrate.
◉ Heterotropic Regulation of Binding.
Answer: Where an allosteric regulator is present that is not the
enzyme's substrate.
◉ Hill Plot.
,Answer: Turns sigmoid into straight lines. Slope = n (# of binding
sites). Allows measurement of binding sites that are cooperative.
◉ pH and Binding Affinity (Bohr Affect).
Answer: As [H+] increases, Histidine group in hemoglobin becomes
more protonated and protein shifts to T-state. O2 binding affinity
decreases.
◉ CO2 binding in Hemoglobin.
Answer: Forms carbonic acid that shifts hemoglobin to T-state. O2
binding affinity decreases. Used in the peripheral tissues.
◉ BPG (2,3-bisphosphoglycerate).
Answer: Greatly reduces hemoglobin's affinity for O2 by binding
allosterically. Stabilizes T-state. Transfer of O2 can improve because
increased delivery in tissues can outweigh decreased binding in the
lungs.
◉ Michaelis-Menton Equation.
Answer: V0 = (Vmax[S]) / (Km + [S])
◉ Km in Michaelis-Menton.
Answer: Km = [S] when V0 = 0.5(Vmax)
,◉ Michaelis-Menton Graph.
Answer:
◉ Lineweaver-Burke Graph.
Answer: Slope = Km/Vmax
Y-intercept = 1/Vmax
X-intercept = - 1/Km
◉ Lineweaver-Burke Equation.
Answer: Found by taking the reciprocal of the Michaelis-Menton
Equation.
◉ Kcat.
Answer: Rate-limiting step in any enzyme-catalyzed reaction at
saturation. Known as the "turn-over number". Kcat = Vmax/Et
◉ Chymotripsin.
Answer: Cleaves proteins on C-terminal endof Phe, Trp, and Tyr
◉ Competitive Inhibition Graph.
Answer: Slope changes by factor of α. Slope becomes αKm/Vmax.
X-intercept becomes 1/αKm
, Y-intercept does not change.
Vmax does not change.
◉ Uncompetitive Inhibition Graph.
Answer: Does not change slope.
Changes Km and Vmax.
Results in vertical shift up and down.
Y-intercept becomes α'/Vmax
X-intercept becomes -α'/Km
◉ Mixed Inhibition Graph.
Answer: Allosteric inhibitor that binds either E or ES.
Pivot point is between X-intercept and Y-intercept.
◉ Non-Competitive Inhibition Graph.
Answer: Form of mixed inhibition where the pivot point is on the x-
axis. Only happens when K1 is equal to K1'.
◉ Ionophore.
Answer: Hydrophobic molecule that binds to ions and carries them
through cell membranes. Disrupts concentration gradients.
