TBR 7 - Metabolic Components Flashcards
A molecule with (less/more) energy is more stable.
Less. Looking at the reaction chart on page 109, note B, the product, is more stable because it has less energy.
True or false. At low temperatures, fewer molecules have enough energy to surpass the transition state of a reaction.
True. In biological molecules, raising the temperature is usually not a possibility. For many organisms, the temp must be in a certain range.
Describe what transition state analogs are and what purpose they serve in medicine.
TSA are molecules that look like the transition state of molecules belonging to a particular ezymatic reaction. TSA are excellent inhibitors of the catalytic process.
Describe what enzymes are and their role in physiology.
Enzymes are catalysts which are often employed to lower the transition state’s activation energy.
Do enzymes alter the equilibrium of a reaction? What do enzymes alter in reactions?
A catalyst will NOT alter the equilibrium of a reaction. A catalyst will alter the RATE of the reaction.
True or false? An enzyme accelerates the forward reaction as well as the reverse reaction by precisely the same factor.
True.
Describe HOW enzymes speed up reactions.
EZ contain specific regions called active sites to which SPECIFIC substrate molecule will bind. EZ also stabilize the TS and they carry out acid-base catalysis by precisely positioning the catalytic groups of certain AAs ( ser, glu, lys, etc.) found within the active site pocket
Define Vmax.
Vmax = the maximal rate/velocity of a reaction. This reaction rate can be obtained when ALL the enzyme’s active sites are saturated with substrate. Vmax = K3[Etotal]
Give the Michaelis-Menten Equation.
V = (Vmax*[S])/([S] + Km)
If you plot the velocity of a reaction as a function of substrate concentration, we will get a _____ curve.
Hyperbolic curve.
What is Km?
Km = the substrate concentration at which the reaction rate is half its maximal value (Vmax/2).
E + S ES –> E + P, what happens when k2»»»k3?
The EZ-substrate complex (ES) will have a tendency to dissociate to E and S rather than form E and P.
E + S ES –> E + P, what is the Km when k2»»»k3?
Km = K2/K1. It is ONLY in this situation that Km is a measure of the binding strength of the ES complex.
A ___ Km value indicates ____ binding of the ES complex while a ____ Km value indicates a strong binding of the ES complex.
A high Km value indicates a weak binding of the ES complex while a low Km value indicates a strong binding of the ES complex.
Not all cases reactions that involve binding of substrate to protein result in a hyperbolic curve. Allosteric enzymes have a _____ curve.
Allosteric enzymes have a non-hyperbolic, more sigmoidal curve when plotted as velocity versus substrate concentration.
A lineweaver/double reciprocal plot allows us to estimate Vmax and Km. How do we do this?
If we take the best fit line, the y-axis intercept is 1/Vmax. The x-axis intercept will be at -1/Km.
On a Lineweaver and Burk plot, where are HIGH [S] values located.
If 1/[S] is approaching zero, then it must mean that [S] is approaching infinity.
Describe the turnover number.
When an EZ is completely saturated with substrate, then the # of substrate molecules which are converted to product per unit time is referred to as the turnover number. Kcat = Vmax/[total EZ concentration]
Name the two major types of enzyme inhibition.
Reversible and irreversible.
Describe the two types of reversible inhibition.
A competitive inhibitor will compete with the substrate for the active site on the enzyme. The non-competitive inhibitor will bind to another portion of the enzyme. Allosteric inhibition is a subset of this.
How does a competitive inhibitor affect a reaction?
It decreases the rate of catalysis of the enzyme.
How does a competitive inhibitor affect the Vmax of a reaction?
A “CI” can be overcome at high [S]. Initially we have a lower Vmax, but as we approach an infinite [S], the concentration of the “CI” becomes negligible.
How does a competitive inhibitor affect the Km of a reaction.
It results in an apparent change in Km. The Km of a competitively inhibited enzyme appears to shift to a higher value. The word apparent is used because each EZ has a characteristic Km value for a given substrate.
Describe the lineweaver burk plot for competitive inhibition.
Vmax remains constant. There are apparent Km changes. A competitive inhibitor will increase the slope of the lines found in the plot. Intersection of lines at the y-axis.
