1. Chapter 3

3. * Likelihood that a product will form when a substrate molecule enters the active site * Sometimes, with low enzyme affinity, a substrate will enter an active site but instead of making a product, substrate will leave the active site before a reaction takes place and a product forms

4. * The rate at which an enzyme can convert a substrate molecule into product per second * Relies heavily on amount of collisions between enzyme and substrate molecule * Maximum number of chemical conversions of substrate molecules per second that a single catalytic site will execute for a given enzyme concentration * “Kcat” in formulas

5. * Vmax  theoretical maximum rate of an enzyme ; tells us how well an enzyme performs * At Vmax  ALL enzyme molecules are bound to substrate molecule * Enzyme is SATURATED with substrate * How to Measure Vmax: * Reaction rate is measured at different substrate concentrations while keeping enzyme concentration constant * As substrate concentration INCREASES  reaction rate rises until it reaches maximum rate (Vmax) * Vmax is where your graph levels off… * Find the y-value for the plateau…this is Vmax

6. * Plotted against substrate concentrations * Produces ASYMPTONIC curve * In practice: * Curve never truly flattens out * In theory: * Curve flattens out when there is an infinite substrate concentration * Impossible to read a value of Vmax off a graph…so we must use another graph….

7. * Create a double reciprocal plot (aka Lineweaver-Burke plot) * Plot inverse of substrate concentration on x-axis (1/[S]) * Plot inverse of velocity on y-axis (1/velocity) * Benefits: * 1/infinity substrate concentration = ZERO * Zero can be plotted on your graph * Vmax can be accurately found off of graph * Resulting line is a straight line (not curve) * Accurately determine Vmax * Accurately determine Km (Michaelis-Menton Constant)

8. * Find 1/Vmax * Point where the line crosses/intersects the y-axis * Point of intersection on y-axis is where 1/[S] is equal to zero (when [S] is infinity) * Once 1/Vmax is determined, Vmax can be calculated * Km (Michaelis Menten Constant) can also be determined from the double reciprocal plot

9. * Km (abbreviation) * Measure of the affinity of an enzyme for its substrate * The substrate concentration at which an enzyme works at HALF its maximum rate (1/2 Vmax) * At this point, HALF the active sites of the enzymes are occupied by the substrate * Higher affinity of the enzyme for the substrate  lower concentration is needed for Km Higher affinity= lower Michaelis-Menton constant = reaction will occur much more quickly (towards its Vmax) Lower affinity = higher Micahelis-Menton constant = reaction will occur much more slowly Low Km = High affinity of enzyme for substrate = more product High Km = Low affinity of enzyme for substrate = less product

10. * Use double reciprocal plot * -1/Km is the point where the line for the graph intersects x-axis * Use the value of -1/Km to calculate Km * Value of Km can vary depending on many factors * Identity of substrate * Temperature * pH * Presence of particular ions * Overall ion concentration * Presence of poisons * Presence of pollutants * Presence of inhibitors

11. * Vmax * Info about maximum rate of reaction that is possible * Km * Measures affinity of the enzyme for the substrate * Higher affinity = more likely product will be formed when substrate enters active site * High affinity produces LOW Km value * Low affinity produces HIGH Km value… * What would give you a HIGH Km value (low affinity for substrate to go to active site)? * INHIBITORS!!!!

13. * The smaller the 1/Km value is  larger the Km value is * The larger the 1/Km value is  smaller Km value is * The smaller the 1/Vmax value is  larger the Vmax value is * The larger the 1/Vmax value is  smaller Vmax value is Large Vmax  means you need to add lots a substrate to make reaction happen faster Small Vmax  means only a little substrate is need to make reaction happen fast Large Km value  LOW enzyme affinity (some reason, the substrates do NOT want to bind to active site) Low Km Value  HIGH enzyme affinity (substrates immediately bind to active sit of enzymes…only a small substrate concentration is needed to saturate your enzymes’ active sites)

20. * Steps: * Create table recording “v” (velocity) vs. substrate concentration [S] * Plot rate of reaction “v” (velocity) vs. substrate concentration [S] * V  y-axis * [S]  x-axis * Can’t tell Vmax from graph…. * Revisit table and now fill in additional columns, adding 1/v and 1/[S] by doing simple calculations * Now, make a new graph * Plot 1/v on y-axis and 1/[S] on x-Axis * Plot values from table on this new graph * Draw straight line through points….ALL the way through y-axis and onto negative side of x-axis * Your y-intercept is the 1/Vmax when the 1/[S] is 0 (which, do the math, would equal INFINITY!!!! Endless substrate, yay!) * Your x-intercept is the 1/km * Plug in these values into the above equations, solve for Km and Vmax, now you have determined Vmax for your original graph!!! [S]/ arbitrary unit 1/[S]/ arbitrary unit v/ arbitrary unit 1/v / arbitrary unit