BIOCHEMISTRY C 785 READINESS CHECK 1B
TEST PAPER 2026 COMPREHENSIVE
SOLUTIONS PRACTICE
◉ Temperature Denaturation of Protein. Answer: Midpoint of
reaction is Tm.
◉ Cooperative Protein Folding. Answer: Folding transition is sharp.
More reversible.
◉ Folding Funnel. Answer: Shows 3D version of 2D energy states.
Lowest energy is stable protein. Rough funnel is less cooperative.
◉ Protein-Protein Interfaces. Answer: "Core" and "fringe" of the
interfaces. Core is more hydrophobic and is on the inside when
interfaced. Fringe is more hydrophilic.
◉ π-π Ring Stacking. Answer: Weird interaction where aromatic
rings stack on each other in positive interaction.
◉ σ-hole. Answer: Methyl group has area of diminished electron
density in center; attracts electronegative groups
,◉ Fe Binding of O2. Answer: Fe2+ binds to O2 reversible. Fe3+ has
an additional + charge and binds to O2 irreversibly. Fe3+ rusts in O2
rich environments.
◉ Ka for Binding. Answer: Ka = [PL] / [P][L]
◉ ϴ-value in Binding. Answer: ϴ = (bound / total)x100%
ϴ = [L] / ([L] + 1/Ka)
◉ Kd for binding. Answer: Kd = [L] when 50% bound to protein.
Kd = 1/Ka
◉ High-Spin Fe. Answer: Electrons are "spread out" and result in
larger atom.
◉ Low-Spin Fe. Answer: Electrons are less "spread out" and are
compacted by electron rich porphyrin ring.
◉ T-State. Answer: Heme is in high-spin state. H2O is bound to
heme.
◉ R-State. Answer: Heme is in low-spin state. O2 is bound to heme.
,◉ O2 Binding Event. Answer: O2 binds to T-state and changes the
heme to R-state. Causes a 0.4A movement of the iron.
◉ 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
TEST PAPER 2026 COMPREHENSIVE
SOLUTIONS PRACTICE
◉ Temperature Denaturation of Protein. Answer: Midpoint of
reaction is Tm.
◉ Cooperative Protein Folding. Answer: Folding transition is sharp.
More reversible.
◉ Folding Funnel. Answer: Shows 3D version of 2D energy states.
Lowest energy is stable protein. Rough funnel is less cooperative.
◉ Protein-Protein Interfaces. Answer: "Core" and "fringe" of the
interfaces. Core is more hydrophobic and is on the inside when
interfaced. Fringe is more hydrophilic.
◉ π-π Ring Stacking. Answer: Weird interaction where aromatic
rings stack on each other in positive interaction.
◉ σ-hole. Answer: Methyl group has area of diminished electron
density in center; attracts electronegative groups
,◉ Fe Binding of O2. Answer: Fe2+ binds to O2 reversible. Fe3+ has
an additional + charge and binds to O2 irreversibly. Fe3+ rusts in O2
rich environments.
◉ Ka for Binding. Answer: Ka = [PL] / [P][L]
◉ ϴ-value in Binding. Answer: ϴ = (bound / total)x100%
ϴ = [L] / ([L] + 1/Ka)
◉ Kd for binding. Answer: Kd = [L] when 50% bound to protein.
Kd = 1/Ka
◉ High-Spin Fe. Answer: Electrons are "spread out" and result in
larger atom.
◉ Low-Spin Fe. Answer: Electrons are less "spread out" and are
compacted by electron rich porphyrin ring.
◉ T-State. Answer: Heme is in high-spin state. H2O is bound to
heme.
◉ R-State. Answer: Heme is in low-spin state. O2 is bound to heme.
,◉ O2 Binding Event. Answer: O2 binds to T-state and changes the
heme to R-state. Causes a 0.4A movement of the iron.
◉ 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
