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Lecture5-Enzyme activity

Lecture5-Enzyme activity. Enzymes: have an active site where substrates bind and are converted to products Active site: constituted from diverse regions of the polypeptide chain and is flexible

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Lecture5-Enzyme activity

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  1. Lecture5-Enzyme activity Enzymes: have an active site where substrates bind and are converted to products Active site: constituted from diverse regions of the polypeptide chain and is flexible Enzyme Denaturation: are denatured at high pH and temperature Hydrogen, electrostatic and hydrophobic bonds are disrupted Specific Activity: moles of substrate converted to product per unit time (sec or min) per mg of protein Turnover Number: moles of substrate converted to product per mole of the active site of the enzyme

  2. Substrate binding to the active site Lock and key fit Induced fit

  3. COENZYMES • Many enzymes require a coenzyme or cofactor for activity • Apoenzyme + Coenzyme  Holoenzyme • (inactive) (active) • Coenzymes are derived from vitamins and act as co-substrates and are converted into products • Cofactors are metal ions such as Cu, Mg, Mn, Fe: are not usually converted to products • Coenzymes and cofactors alter the conformation around the active site of the enzyme

  4. Vitamins: water-soluble

  5. ACTIVATION ENERGY • For Product formation, input of energy (activation energy) is required • S------> S* ----------> P • (activated Complex) • Activation Energy: • energy in Kcal/mole required to convert one mole of substrate to the activated complex • Enzymes lower activation energy • The magnitude of decrease in activation energy is identical for forward and backward reactions

  6. Free Energy Changes Gs = Gs0 + 2.303 RT log [S] Gp = Gp0 + 2.303RT log [P] ∆G = ∆G0 + 2.303 RT log [P]/[S] ∆G= Free energy change of the reaction ∆G0 = Standard free energy change for the reaction R= cal/degree Kelvin and is equal to ~2.0 T = degrees Kelvin = 273 + degree Centigrade At equilibrium: ∆G = 0 so 0= ∆G0 + 2.303 RT log [P]e/[S]e = ∆G0 + 2.303RT log Keq ∆G0 = -2.303RT log Keq

  7. FREE ENERGY • ΔG • is free energy of the reaction • is = ΔGo + 2.303RT log [P]/[S] • is equal to ΔGo when [P]=[S]= 1 M • is not changed when enzyme is present • has negative value for a spontaneous reaction • is the sole determinant whether the reaction will proceed in the direction written

  8. STANDARD FREE ENERGY CHANGE • ΔGo • is equal to ΔG when [P] = [S] =1M • values can be used to determine Keq • is equal to zero when Keq is 1; is negative when Keq >1 and is positive when Keq is less than 1 • values for two reactions are additive if there is a common intermediate

  9. CALCULATE Keq FROM KNOWN ΔGo at 27 degree centigrade • If ΔG0 = -7 Kcal/mole; then • -7x1000 cal/mole = -2.303RT logKeq • - 7000 = -2.303 x2x300 logKeq • 7000/1400=5 = logKeq • Therefore, Keq = antilog5 = 105

  10. Standard Free Energy Changes in Glycolysis

  11. Enzymes play important roles in normal homeostasis • Example of respiration

  12. Lecture 5-Learning Objectives • General properties of enzymes: conformation, stability • Active site, specific activity, turnover number • Coenzyme-factors-their origin and role • Activation energy-how it is related to rates of enzymatic reactions • Change in free energy of reaction (ΔG’)-driving force for the reaction • Standard free energy change (ΔG0’)---its usefulness in calculating equilibrium constant (Keq)

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