Lecture5-Enzyme activity-Meisenberg and Simmons pp39-49

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Structure of HIV RT
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Lecture5-KumarEnzyme 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

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

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Vitamins water-soluble
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ACTIVATION ENERGY

• For Product formation, input of energy
  (activation energy) is required
• S------gt S ----------gt 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

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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 Pe/Se ?G0 2.303RT
  log Keq
• ?G0 -2.303RT log Keq

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FREE ENERGY

• ?G
• is free energy of the reaction
• is ?Go 2.303RT log P/S
• is equal to ?Go when PS 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

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STANDARD FREE ENERGY CHANGE

• ?Go
• is equal to ?G when P S 1M
• values can be used to determine Keq
• is equal to zero when Keq is 1 is negative when
  Keq gt1 and is positive when Keq is less than 1
• values for two reactions are additive if there is
  a common intermediate

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CALCULATE Keq FROM KNOWN ?Go at 27 degree
centigrade

• If ?G0 -7 Kcal/mole then
• -7x1000 cal/mole -2.303RT logKeq
• - 7000 -2.303 x2x300 logKeq
• 7000/14005 logKeq
• Therefore, Keq antilog5 105

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Standard Free Energy Changes in Glycolysis
Reaction deltaG0 (Kcal/mole)
G ATP ? G-6-P ADP H -4
G-6-P ?? F-6-P 0.4
F-6-P ATP ? F1,6-BP ADP H -3.4
F 1,6-BP ? DHAP G-3-P 5.7
DHAP ? G-3-P 1.8
G-3-P Pi NAD? 1,3-DPG NADH H 1.5
1,3-DPG ADP? ATP 3-PG -4.5
3-PG ?? 2-PG 1.1
2-PG ? PEP H2O 0.4
PEP ADP H? Pyr ATP -7.5
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Enzymes play important roles in normal homeostasis

• Example of respiration

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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 (?G)-driving
  force for the reaction
• Standard free energy change (?G0)---its
  usefulness in calculating equilibrium constant
  (Keq)