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35 Cards in this Set
- Front
- Back
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What does a small Km tell you about the affinity of an enzyme for the substrate
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small Km means high affinity, it takes a low concentration of enzyme to half saturate the enzyme
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What does a large Km tell you about the affinity of an enzyme for its substrate
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Large Km means low affinity because a high concentration of substrate is needed to half saturate the enzyme
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What is the law of mass action and how does it relate to reaction rates? Is it universally applicable to all reactions?
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Law of Mass action= the observed rate of reaction is proportional to the concentration of the substrates and a proportionality constant K, Vo=k[A][B]
Note that this law is based on Vo being proportional to the collisions between the substrates and therefore only applies to elementary reactions (first order for all reactants) |
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What are the two assuptions made when using the MM equation
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1. initial rate-[S] is the amount added and is constant, no P yet and therefore no reverse rxn
2. Steady state-[ES] is nearly constant |
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What is the significance of the initial rate assumption when using the MM equation?
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-states that [S]=amount added and is constant
-no reverse reaction This is significant because we need to know the substrate concentration in order to calculate the reaction rate |
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What is the significance of the steady state assumption when using the MM equation
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-states that [ES] is nearly constant
-this is necessary becuse this must remain constant if we are going to use our K's to substitue for a value that we can use in the equation |
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what is the MM equation?
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Vo=(Vmax[S])/(km +[S])
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How is a change in enzyme concentration reflected in the MM equation
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The Vmax term depends on the enzyme cocentration, Vmax=k2[E]
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How can the MM equation be simplified when [S]<<km. Describe what's occuring in this situation
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If [S] is much less than km, we know that the enzyme is not close to being saturated. In this case, we can simplify the equation to Vo=(Vmax[S])/km, in which case the rate is linearly proportional to [S]
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How can the MM equation be simplified when [S]>>km
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In this case the enzyme is fully saturated and they only thing limited the rate is the intrinsic properties of the enzyme. The equation can be simplified to V=Vmax[S]/[S], or V=Vmax and the rate is independent of [S].
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What is km and what are it's units
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Km is the substrate concentration at which 1/2 V max is achieved. ITs units are concentration
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what equation is used to construct the lineweaver burke plot
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1/Vo=(km/Vmax[S]) + 1/Vmax
or if you think in terms of y=mx+b, 1/vVo= (Km/Vmax)(1/[S]) + (1/Vmax) |
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What enzyme properties can be determined from each of the following parts of an LB plot?
Slope, Yint, Xint |
slope=Km/Vmax
Yint= 1/Vmax Xint= -1/Km |
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What is the significance of a log dose response curve
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-allows you to measure velocity over a wide range of substrate concentrations
-inflection point is EC50 (50% affective) |
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What does a competitive inhibitor bind to?
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enzyme only (NOT ES) at the same site as the substrate
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What affect does a competitive inhibitor have on the apparent Km of an enzyme on a LB plot
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The km appears to increase.
It appears as if more substrate is need to achieve 1/2 saturation. This is because the substrate must compete with the inhibitor for the active site |
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What affect does a competitive inhibitor have on the Vmax of an enzyme on a LB plot
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The vmax does not change.
If you were able to add enough substrate to totally outcompete the inhibitor, the enzyme could still function at maximal capacity. The competitive inhibitor does not affect the enzyme's ability to catalyze the reaction, it only changes how much substrate can reach the active site |
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What affect does a competitive inhibitor have on the slope of an enzyme on a LB plot
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As more inhibitor is added, the slope (Km/Vmax) increases. This reflects the fact that the apparent Km increases but the Vmax remains unchanged
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What affect does a competitive inhibitor have on the x intercept of an enzyme on a LB plot
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The x intercept (-1/Km) moves towards the Y axis (becomes less negative).
This reflects the fact that the apparent Km increases. Km increases because a higher substrate concentration is needed to half saturate the enzyme beause the inhibitor is binding the same site. (Careful, don't get confused w/ the fraction and the negative sign) |
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What affect does a competitive inhibitor have on the Y intercept of an enzyme on a LB plot
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The Y intercept (1/Vmax) does not change.
Competitive inhibitors do not change the enzyme's ability to operate at maximal capacity if it is able to become saturted with substrate. The inhibitor inhibits saturation, not function. |
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what does a non-competitive inhibitor bind
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equally to E and ES
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What affect does a non-competitive inhibitor have on the apparent Km of an enzyme on a LB plot
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The apparent Km is commonly unchanged because the inhibitor is not affecting substrate binding, just the ability of the enzyme to catalyse the reaction
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What affect does a non-competitive inhibitor have on the Vmax of an enzyme on a LB plot
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The Vmax is decreased because the inhibitor is preventing the enzyme from properly catalyzing the reaction
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What affect does a non-competitive inhibitor have on the slope of an enzyme on a LB plot
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The slope (km/vmax) increases because Km is unchanged (usually) but Vmax decreases
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What affect does a non-competitive inhibitor have on the x intercet of an enzyme on a LB plot
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no change if Km is unchanged. The inhibitor is not affecting the affinity of the enzyme for the substrate (substrate can bind normally) so Km is not affected
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What affect does a non-competitive inhibitor have on the Y-int of an enzyme on a LB plot
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The Y intecept (1/Vmax) increases because Vmax decreases
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What does an uncompetitive inhibitor bind to
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ES only
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What affect does an uncompetitive inhibitor have on the apparent Km of an enzyme on a LB plot
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decreased, Think in terms of Le Chatellier's principle, the inhibitor binds only to the ES complex and decreases it's concentration. This drives the E+S equalibrium to shift towards making more ES so it appears as if the enzyme has a higher affinity for the substrate
In graphical terms, the x intercept moves farther away from 0 meaing that the value is 1/a bigger number. 1/a biger number is a smaller number. |
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What affect does an uncompetitive inhibitor have on the Vmax of an enzyme on a LB plot
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The Vmax will decrease because the ESI complex is catalytically inactive, the [E] appears to decrease and Vmax=k2[E]
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What affect does an uncompetitive inhibitor have on the slope of an enzyme on a LB plot
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Slope (Km/Vmax) is unchanged since both Km and Vmax decrease, all lines will be parallel
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What affect does an uncompetitive inhibitor have on the x intercept of an enzyme on a LB plot
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The x intercept (-1/km) will move away from the Y axis because the Km is decreased so the fraction is getting more negative.
Km appears to decrease because ES is pulled away as ESI so the E+S<=>ES equilibrium shifts right making it appears as if the enzyme has a higer affintiy (smaller Km) |
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What affect does an uncompetitive inhibitor have on the Yint (Vmax) of an enzyme on a LB plot
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The Y int (1/vmax) increases because V max decreases
Vmax decreases because the enzyme is stuck in ESI can can't make E+P E+S <=>ED<=>E+P ES<=>ESI think in terms of lechatlier's principle |
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what would a plot of Vmax vs. [E] look like for an irreversible inhibitor compared to a control
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The slopes would be identical since the active enzyme is unchanged. The X intercept for the inhibited enzyme would be larger because when the cocentration of enzyme is low, all of it is rendered nonfunctional by the inhibitor
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T/F allosteric enzymes that show cooperativity obey Michaelis-Menton Kinetics
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False
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T/F substrate binding always faciltes further binding
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false, the Kohsland model for sequential allosteric behavior demostrates that the allosteric effect can be positive or negative
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