ATP Synthase Lecture # 14

What is ATP Synthase? ATP (adenosine triphosphate) – the molecule of energy for cells, drives many biochemical reactions like muscle contraction, DNA and protein synthesis, etc ATP Synthase- enzyme that catalyses ATP synthesis and hydrolysis http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/A/ATP.html

Where is it found? A membrane enzyme: Found in: bacterial plasma membranes The thylacoid membrane in chloroplasts The inner mitochondrial membrane of eukaryotic cells http://www.biologie.uni-osnabrueck.de/biophysik/Feniouk/images/Rb_capsulatus.jpg

Parts of ATP Synthase Consists of two domains F0 and F1 http://www.sigmaaldrich.com/Area_of_Interest/Biochemicals/Enzyme_Explorer/Key_Resources/Metabolic_Pathways/ATP_Synthase.html

The Domains Hydrophobic F0 domain sits in the membrane - performs proton translocation Hydrophillic F1 portion protrudes from membrane - performs ATP synthesis/hydrolysis 3 alternating alpha and beta subunits http://nobelprize.org/chemistry/laureates/1997/illpres/boyer-walker.html

What does it do? It makes ATP from ADP and inorganic phosphate (Pi) ADP + Pi + ATP The overall equation is:     ADP3- + HPO42- + H+ + nH+outside membrane(+ charge)     ATP4- + H2O + nH+inside membrane (- charge) energy from protons diffusing across membrane with the gradient

How does it function? First a proton gradient is established Protons collect on one side of the membrane Then the protons flow through a channel in the enzyme causing the protein subunits to rotate http://www.sp.uconn.edu/~terry/images/anim/ATPmito.html

The “stalk” rotates in 120° increments causes the units in the F1 domain to contract and expand The structural changes facilitate the binding of ADP and Pi to make ATP Each subunit goes through 3 stages Open State – releases any ATP Loose State – ADP and Pi molecules enter the subunit Tight State – the subunit contracts to bind molecules and make ATP

Animations http://www.stolaf.edu/people/giannini/flashanimat/metabolism/atpsyn1.swf Protons cross membrane through the ATP synthase enzyme http://www.stolaf.edu/people/giannini/flashanimat/metabolism/atpsyn2.swf Rotary motion of ATP synthase powers the synthesis of ATP

Interesting Facts Contains 22722 atoms 23211 bonds connected as 2987 amino acid groups 120 helix units and 94 sheet units Generates over 100 kg of ATP daily (in humans) One of the oldest enzymes-appeared earlier then photosynthetic or respiratory enzymes Smallest rotary machine known Picture http://webct.uga.edu/public/FRES1010CG/BCMB401098ATPSyn2.gif/

Our Model

Enzyme Activity and units

Enzyme units Amounts of enzymes can either be expressed as molar amounts, as with any other chemical, or measured in terms of activity, in enzyme units. The enzyme unit (U) is a unit for the amount of a particular enzyme used for its activity. “One U is the amount of the enzyme that catalyzes 1 micro mole of substrate per minute”. The conditions also have to be specified: one usually takes as STP

Enzyme Activity Unit of enzyme activity: Enzyme activity = moles of substrate converted per unit time = rate × reaction volume Used to measure total units of activity in a given volume of solution. Specific activity: It is the amount of product formed by an enzyme in a given amount of time under given conditions per mg of total protein. = rate × reaction volume / mass of total protein. Molecular activity: Used to compare activities of different enzymes.

Enzyme Activity Classical units: Unit of enzyme activity: mmol substrate transformed/min = unit Specific activity: mmol substrate/min-mg E = unit/mg E Molecular activity: mmol substrate/min- mmol E = units/mmol E

Enzyme Activity New international units: Unit of enzyme activity: mol substrate/sec = katal Specific activity: mol substrate/sec-kg E = katal/kg E Molecular activity: mol substrate/sec-mol E = katal/mol E

Example 1 The rate of an enzyme catalyzed reaction is 35 μmol/min at [S] = 10-4 M, (KM = 2 x 10-5). Calculate the velocity at [S] = 2 x 10-6 M. Work the problem.

Example 1 Answer The rate of an enzyme catalyzed reaction is 35 μmol/min at [S] = 10-4 M, (KM = 2 x 10-5). Calculate the velocity at [S] = 2 x 10-6 M. First calculate VM using the Michaelis-Menton eqn: VM [S] VM (10-4) VM (10-4) v = -----------, so: 35 = ------------------ = -------------- KM + [S] 2 x 10-5 + 10-4 1.2 x 10-4 VM = 1.2(35) = 42 mmol/min; then calculate v: 42 (2 x 10-6) 84 x 10-6 v = ------------------------ = ------------ = 3.8 mmol/min 2 x 10-5 + 2 x 10-6 22 x 10-6

Example 2 An enzyme (1.84 μgm, MW = 36800) catalyzes a reaction in presence of excess substrate at a rate of 4.2 μmol substrate/min. What is the TON in min-1 ? What is the TON in sec-1 ? Work the problem.

Example 2 Answer An enzyme (1.84 μgm, MW = 36800) catalyzes a reaction in presence of excess substrate at a rate of 4.2 μmol substrate/min. What is the TON ? 1.84 μgm μ mol E = ------------------------- = 5 x 10-5 μmol E 36800 μgm/ μmol 4.2 μmol/min TON = ------------------ = 84000 min-1 5 x 10-5 μmol

Example 2 Answer What is the value of this TON (84000 min-1) in units of sec-1 ? 84000 min-1 1 sec-1 TON E = ------------------ x ---------- = 1400 sec-1 60 min-1

Example 3 Ten micrograms of carbonic anhydrase (MW = 30000) in the presence of excess substrate exhibits a reaction rate of 6.82 x 103 μmol/min. At [S] = 0.012 M the rate is 3.41 x 103 μmol/min. a. What is Vm ? b. What is KM ? Work these.

Example 3 The rate in presence of excess substrate is Vmax so: Vmax = 6.82 x 103 μmol/min. b. At [S] = 0.012 M the rate is 3.41 x 103 μmol/min which is ½ Vmax so: KM = 0.012 M. This may also be determined using the Michaelis-Menton equation.