SA enzymes Flashcards
(a) Vmax is about 700. In a plot of V vs. [S], the asymptote is Vmax. Simple inspection of the data
shows the approach to Vmax—the rate increases by only 1 unit when [S] increases fivefold.
(b) Km is about 8 μM, the [S] at which the velocity is half-maximal. Because Vmax is about 700,
1/2 Vmax is about 350. The [S] at that rate is about 8 μM.
Why is the Lineweaver-Burk (double reciprocal) plot (see Box 6, p. 206) more useful than the standard V vs. [S] plot in determining kinetic constants for an enzyme? (Your answer should probably show typical plots.)
The plot of V vs. [S] is hyperbolic; maximum velocity is never achieved experimentally, because it is impossible to do experiments at infinitely high [S]. The Lineweaver-Burk transformation of the Michaelis-Menten equation produces a linear plot that can be extrapolated to infinite [S] (where 1/[S] becomes zero), allowing a determination of Vmax.
Km = about 2 mM (the concentration of S needed to achieve one-half of Vmax, which is about 500). The total enzyme present is producing about 500 μmol of product per minute. Because the turnover number is 5,000/min, the amount of enzyme present must be 0.1 μmol; 1 μmol of enzyme would produce 5,000 μmol product/min.
Methanol (wood alcohol) is highly toxic because it is converted to formaldehyde in a reaction catalyzed by the enzyme alcohol dehydrogenase:
NAD+ + methanol → NADH + H+ + formaldehyde
Part of the medical treatment for methanol poisoning is to administer ethanol (ethyl alcohol) in amounts large enough to cause intoxication under normal circumstances. Explain this in terms of what you know about examples of enzymatic reactions.
Ethanol is a structural analog of methanol, and competes with methanol for the binding site of alcohol dehydrogenase, slowing the conversion of methanol to formaldehyde, and allowing its clearance by the kidneys. The effect of ethanol is that of a competitive inhibitor.
a) 100; b) 0.0002; c) 66.7; d) 40; e) competitive
On the enzyme hexokinase, ATP reacts with glucose to produce glucose 6-phosphate and ADP five orders of magnitude faster than ATP reacts with H2O to form phosphate and ADP. The intrinsic chemical reactivity of the —OH group in water is about the same as that of the glucose molecule, and water can certainly fit into the active site. Explain this rate differential in two sentences or less.
The binding of glucose to hexokinase induces a conformation change that brings the amino acid residues that facilitate the phosphoryl transfer into position in the active site. Binding of water alone does not induce this conformational change.
Why is a transition-state analog not necessarily the same as a competitive inhibitor?
The structure of a competitive inhibitor may be similar to the structure of the free substrate. Similar structure will mean that the competitive inhibitor can associate with the enzyme at the active site, effectively blocking the normal substrate from binding. A transition-state analog, however, is similar in structure to the transition-state of the reaction catalyzed by the enzyme. Often a transition- state analog will bind tightly to an enzyme, and is not easily competed away by substrate.
The scheme S → T → U → V → W → X → Y represents a hypothetical pathway for the metabolic synthesis of compound Y. The pathway is regulated by feedback inhibition. Indicate where the inhibition is most likely to occur and what the likely inhibitor is.
What is a zymogen (proenzyme)? Explain briefly with an example.
A zymogen is an inactive form of an enzyme that is activated by one or more proteolytic cleavages in its sequence. Chymotrypsinogen, trypsinogen, and proelastase are all zymogens, becoming chymotrypsin, trypsin, and elastase, respectively, after proper cleavage.
