BIOCHEMISTRY - LEHNINGER 6E REVIEW 2lightcat-files.s3.amazonaws.com/packets/biochemistry-17...Bohr...

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BIOCHEMISTRY - LEHNINGER 6E

REVIEW 2

CONCEPT: BINDING AFFINITY ● Protein-ligand binding is reversible, like a chemical equilibrium □ [S] – substrate concentration □ [E] – enzyme concentration

● Ligands bind to proteins via the same noncovalent forces that dictate protein structure

□ H-bonding, ionic interaction, hydrophobic interaction, and van der Waals forces □ Allows interactions between ligand and protein to be transient● Ligand binding graph

□ θ – fraction of protein’s ligand binding sites bound to ligand □ [Ligand] – ligand concentration, the units are arbitrary □ Kd (dissociation constant) – concentration of ligand at which ½ the binding sites are occupied

- A measure of affinity of a protein for a ligand


REVIEW 2

CONCEPT: BINDING AFFINITY PRACTICE: Match the dissociation constants to the lines on the graph. Name Kd (M) A 2 x 10-6 B 1 x 10-7 C 1 x 10-6 D 4 x 10-8 E 9 x 10-7

PRACTICE: Which protein binds more tightly to the ligand? What is the dissociation constant of protein 1, and the association constant of protein 2?

[Ligand] (nM) θ of Protein 1 θ of Protein 2 0.5 0.2 0.05 1 0.5 0.2 2 0.8 0.5 3 0.9 0.8


REVIEW 2

CONCEPT: O2 TRANSPORTATION ● Hemoglobin (Hb) carries nearly all O2 (99%)

□ Hemoglobin has 4 subunits: each subunit can bind 1 O2 molecule □ Binding is ____________________, subsequent O2 is easier to bind, due to interactions between the subunits

● Bohr effect – hemoglobin's binding affinity oxygen depends on the concentration of CO2 □ Inverse relationship, so high CO2 and low pH (high acidity) à less oxygen binding □ Shifts that lower pH favor the T state □ Shifts that raise pH favor the R state

● 2,3 Diphosphoglycerate (DPG) – a byproduct of glycolysis in erythrocytes, also called 2,3 bisphosphoglycerate (BPG)

□ Increases the amount of oxygen hemoglobin can unload by making hemoglobin go into the T state □ Shifts O2-Hb curve to the ____________________ (increases disassociation of oxygen from Hb)

Left Shift (Hb has high O2 affinity= low disassociation)

Right Shift (Hb has low O2 affinity= high disassociation)

____PCO2 ____PCO2 ____Acidity (high pH) ____Acidity (low pH) ____DPG ____DPG ____Temperature ____Temperature


REVIEW 2

CONCEPT: ENZYMES ● Enzymes control the rates of reaction

□ Do not affect net ∆G, lower Ea, increasing the reaction rate □ Do not affect the equilibrium constant, just reach equilibrium faster

● 5. Isomerase – shifts a functional group within the same molecule ● 2. Transferase (kinase, deaminase) – transfers functional groups from one molecule to another ● 3. Hydrolase (lipase, protease, nuclease) – hydrolyzes a bond ● 1. Oxidoreductase (dehydrogenase, hydroxylase, etc) – transfers electrons from reductant to oxidant ● 4. Lyase (carboxylase, aldolase) – makes a double bond ● 6. Ligase – joins two substrates


REVIEW 2

CONCEPT: ENZYME KINETICS ● Michaelis-Menten Model of enzyme kinetics – Vo = □ Vo (initial velocity) – y axis □ [S] (substrate concentration) – x axis

□ Vmax – maximum velocity of the enzyme (only gets here when substrate concentration reaches saturation) □ KM (Michaelis constant) – concentration of the substrate when half of the binding sites on the enzyme are filled

- Substrate concentration at which enzyme is at ½ Vmax (Vo = ½ Vmax) - KM = [S] at ½Vmax

● Reaction rate = ∆[S]/min, can use rate of disappearance of substrate, or rate of formation of product

□ KM does not depend on enzyme concentration, it is an intrinsic property of the enzyme □ Vmax varies with enzyme concentration, it is an extrinsic property of the enzyme

● The Michaelis constant (KM) can be used to compare enzymes and determine their affinity for a substrate. □ Relative KM is inversely proportional to substrate affinity □ High KM means low affinity for substrate because it takes a great concentration to reach ½ capacity □ Low KM means high affinity for substrate because it takes a lesser concentration to reach ½ capacity ● Enzymes are limited by the rate of diffusion


REVIEW 2

CONCEPT: ENZYME KINETICS ● Lineweaver-Burk equation – 1/Vo = (KM/Vmax)(1/[S]) + (1/Vmax) y = mx+b ● Lineweaver-Burk Plots: □ y-intercept – □ x-intercept – □ slope – ● Kcat – number of substrate molecules converted to product by single enzyme per unit time □ Called the catalytic constant or turnover number

□ Kcat = Vmax/[enzyme], units are in s-1 ● Kcat/KM for many enzymes is close to the diffusion-controlled limit


REVIEW 2

CONCEPT: ENZYME KINETICS PRACTICE: Use the following data to answer the following questions.

[Substrate] (mM) Vo (mM/sec) [Enzyme] (mM) 0.5 0.04 50

0.67 0.05 50 1 0.067 50 2 0.1 50

1. Graph a Michaelis-Menten plot 2. Graph a Lineweaver-Burk plot 3. Determine KM 4. Determine Vmax 5. Determine Kcat


REVIEW 2

CONCEPT: ENZYME KINETICS ● Competitive inhibition – the inhibitor and substrate compete for the active site

□ Competitive inhibition increases the amount of substrate needed for a reaction to reach Vmax □ Affects KM because it competes with the substrate, therefore causes increase in KM

□ Vmax remains the same (NOT changed) □ Saturation with __________________ can overcome competitive inhibition

□ Isosbestic point at x=0 ● α – “effect of inhibitor”, α = 1 + [I]/Ki

[l] – concentration of inhibitor Ki – binding affinity for inhibitor

α = 1 means no inhibitor, as alpha goes up inhibitor concentration increases ● Uncompetitive inhibition –inhibitor binds enzyme-substrate complex □ Prevents release of substrate □ Decreases KM and Vmax

● α’ (alpha prime) = 1 + [l]/Ki’ (Ki prime) □ Parallel lines


REVIEW 2

CONCEPT: ENZYME KINETICS ● Mixed inhibition – inhibitor can bind to enzyme or enzyme-substrate complex, but doesn’t have the same affinity for both □ Can increase or decrease KM depending on if its affinity for enzyme vs. enzyme-substrate complex □ Always decreases Vmax

TYPE OF INHIBITION X-AXIS INTERCEPT Y-AXIS INTERCEPT None -1/KM 1/Vmax Competitive -1/αKM 1/Vmax Uncompetitive -α’/KM α’/Vmax Mixed -α’/αKM α’/Vmax

● Noncompetitive inhibition:


REVIEW 2

CONCEPT: ENZYME KINETICS PRACTICE: Use the following data to answer the following questions.

[Substrate] (mM) Vo (mM/sec) Vo w/ inhibitor (mM/sec) 0.5 0.04 0.0286 0.67 0.05 0.0333 1 0.067 0.04 2 0.1 0.05

1. What type of inhibitor is this? 2. What is the Ki ([I]=2mM)?


REVIEW 2

BIOCHEMISTRY - LEHNINGER 6E REVIEW 2lightcat-files.s3.amazonaws.com/packets/biochemistry-17...Bohr...

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