*(2.3) Enzymes: Enzyme Kinetics

T/F: The Km value of a system can be altered by changing the concentration of enzyme or substrate.

False - The Km value is an intrinsic property of the enzyme-substrate system and cannot be altered by changing enzyme/substrate concentrations.

T/F: Although hemoglobin is a cooperative enzyme, it does not display a sigmoidal binding curve because it acts as a transport protein rather than an enzyme.

False - although it is true that hemoglobin functions as a transport protein rather than an enzyme, is cooperative and therefore displays a sigmoidal curve when graphed on a Michaelis Menten plot.

Cooperatively can also be quantified using a numerical value called _________________. Elaborate on how this metric is used to indicate the nature of binding by the molecule.

Hill's coefficient If HC > 1, positively cooperative binding is occuring, such that after one ligand is bound the affinity of the enzyme for further ligand(s) increases. If HC < , negatively cooperative binding is occurring, such that after one ligand is bound the affinity of the enzyme for further ligand(s) decreases. If HC = 1, the enzyme does not exhibit cooperative binding.

In terms of enzyme kinetics: The rate of an enzymatic reaction cannot increase once the system has reached _______________.

Saturation

Biochemistry Concept Check 2.3 (pg. 52)

Solutions on pg. 65

1) What is a Lineweaver-Burk plot? 2) The actual data are represented by the portion of the graph to the right of the y-axis. Considering this, why is the line extrapolated into the upper left quadrant? 3) The intercept of the line with the y-axis gives the value of _________, while the intercept of the line with the x-axis gives the value of ________. 4) When are LB plots especially useful and why?

The Lineweaver-Burk plot is a double reciprocal graph of the Michaelis Menten equation - the same data graphed in this way yields a straight line. 2) The line is extrapolated into the upper left quadrant to determine it's intercept with the x axis 3) The intercept of the line with the y-axis gives the value of (1/Vmax), while the intercept of the line with the x-axis gives the value of (-1/Km). 4) The LB plot is especially useful when determining the type of inhibition that an enzyme is experiencing because Vmax and Km can be compared without estimation.

Elaborate on the Michaelis-Menten model of enzyme kinetics (i.e., give the reaction w/variables)

The rate of reaction (v) depends on the concentration of both the enzyme [E] and the substrate [S]. The enzyme and substrate react at rate k1 to form the enzyme-substrate complex (ES). From here the enzyme substrate complex can either dissociate at rate k-1 or turn into enzyme + product at rate kcat (k2 in image).

1) T/F: Enzymes showing cooperative kinetics are often regulatory enzymes in pathways. 2) T/F: Cooperative enzymes are not subject to activation and inhibition.

True (e.g. phosphofructokinase-1 in glycolysis) 2) False - Cooperative enzymes are subject to activation and inhibition, both competitively and through allosteric sites.

How do changes in substrate concentration affect the reaction rate (enzyme velocity): a) When substrate concentration is less than Km b) When substrate concentration is above Km (as it approaches Vmax)

a) When substrate concentration is less than Km, changes in substrate concentration will have a high impact (increase) on the reaction rate. b) When substrate concentration is above Km, changes in substrate concentration will have a lower impact (increase) on the reaction rate - this impact becomes less and less as rate continues to approach Vmax.

With regard to the Michaelis-Menten model of enzyme kinetics: As the amount of substrate increases, the enzyme is able to increase its rate of reaction until it reaches a maximum enzymatic reaction rate (Vmax); once Vmax is reached, adding more substrate will not increase the rate of reaction. 1) Demonstrate this graphically in a plot of substrate concentration vs reaction velocity.

see image

1) What is an enzyme's catalytic efficiency? 2) A (smaller/larger) Kcat will result in a higher catalytic efficiency. 3) A (smaller/larger) Km will result in a higher catalytic efficiency.

1) An enzyme's catalytic efficiency is the ratio of Kcat/Km 2) A large Kcat (high turnover) will result in a higher catalytic efficiency 3) A small Km (high substrate affinity) will result in a higher catalytic efficiency

1) Explain the idea of enzyme cooperativity. 2) What kind of curve do cooperative enzymes show when graphed on a Michaelis Menten plot (v vs. [S]) ?

1) Cooperative enzymes have multiple subunits and multiple active sites. Subunits and enzymes may exist in one of two states: 1. A low affinity tense state (T) 2. A high affinity relaxed stat (R) Binding of the substrate encourages the transition of other subunits from the T state to the R state, which increases the likelihood of substrate binding by these other subunits. Conversely, loss of substrate can encourage the transition from the R state to the T state, and promote dissociation of substrate from the remaining subunits. 2) They show a sigmoidal (S-shaped) curve due to cooperatively among substrate binding sites (see image)

1) Outside of environmental conditions, what are the two main factors that influence enzyme kinetics? 2) This is described by the ____________ model of kinetics.

1) In addition to environmental conditions, the rate of the reaction depends on the concentration of both the enzyme and the substrate. 2) Michaelis-Menten model of enzyme kinetics.

1) Define Km. 2) How are Km values useful in comparing enzymes?

1) Km is the Michaelis Constant. It is the substrate concentration at which half of the enzyme's active sites are full (when v = 1/2 of vmax) 2) We can use Km values to assess an enzyme's affinity for it's substrate. An enzyme with a lower Km value reflects a high affinity for it's substrate. Conversely, a high Km value reflects a low affinity fo the enzyme for it's substrate.

1) When the enzyme velocity is half of Vmax, the Michaelis constant (Km) will be equal to ____________. 2) Show this algebraically.

1) Substrate concentration [S] 2) See image

1) With regard to enzyme kinetics, what is denoted by the term Vmax? 2) How do you increase Vmax?

1) Vmax is an enzyme's maximum velocity - the maximum rate that the enzyme reaches when the concentration of substrate is equal to or greater than the concentration of enzyme (i.e., it is saturated.) 2) The only way to increase Vmax is by increasing the enzyme concentration, meaning the cell must induce the expression of the gene encoding the enzyme.

Regarding Michaelis Menten enzyme kinetics: 1) What do the variables Vmax and Kcat represent? 2) What are the units for each of them? 3) How are they related to one another? (give equation) 4) Most enzymes have Kcat values between _______ and ________.

1) Vmax represents maximum enzyme velocity; Kcat, or turnover number, measures the number of substrate molecules "turned over" or converted to product, per enzyme molecule per second. 2) Vmax = moles of enzyme per second (mol/s); Kcat = (s^-1) 3) the turnover number equation: Vmax = [E]Kcat 4) 101 and 103

1) When you have a high enzyme concentration relative to substrate, what would you expect in terms of the kinetics of the enzymatic reactions taking place? 2) How would you expect the rate of the reaction change if you slowly added more substrate? 3) How would you expect the rate of the reaction to change once you've added so much substrate that there is an excess of substrate relative to the concentration of enzyme?

1) You would expect to quickly form products (i.e., reach equilibrium quickly) because there are so many active sites available. 2) As you slowly added more substrate, the rate of the reaction will increase - That is, more reactions will be taking place in the same amount of time (increased output). This is because there are plenty of available active sites to perform catalysis on the additional substrate. 3) You would expect the maximum rate at the point where the enzyme and substrate concentrations are equal. At that point, unlike before, adding more substrate (making the substrate concentration exceed the enzyme concentration) will not cause the rate to increase - it will remain the same. This is because, although you are adding more substrate molecules, there are no more available active sites to accept them - the system has reached saturation.

1) When comparing two enzymes, the one with the (higher/lower) Km has the lower affinity for its substrate. 2) Explain.

1) higher 2) The enzyme with the higher Km has a lower affinity for it's substrate due to the fact that It requires a higher concentration of substrate to reach a the point of half-saturation. In other words, BECAUSE the affinity is lower the enzyme binds less tightly to it's substrate and as a result is able to release it more rapidly. Because it "turns over" substrate more rapidly, it requires more substrate for it to reach a point where half of the active sites are in use (i.e. half-saturation).

1) For a given concentration of enzyme, the Michaelis-Menten relationship generally graphs as a ______________. 2) Explain what this means conceptually in terms of MM enzyme kinetics.

1) hyperbola. 2) This can be seen in the graph - When substrate concentration is less than Km, changes in substrate concentration will greatly affect the reaction rate. However, at high concentrations exceeding Km, the reaction rate increases much more slowly. As the reaction rate infinitely approaches Vmax, the rate becomes independent (unaffected) by substrate concentration.

1) Give the Michaelis Menten equation and define all variables. 2) What does this equation function to do, conceptually? 3) What assumptions must be made in order to use this equation?

1) see image 2) This equation relates the velocity of the enzyme to substrate concentration [S] 3) The concentration of enzyme must be kept constant

1) Restate the Michaelis Menton equation using Kcat instead of Vmax 2) At very low substrate concentrations, where Km>>[S], this derived equation can be further simplified - show how.

1) see image (top equation) 2) See image (bottom equation)