Biochemistry

Exam I

11 February 2004

1. Acetylcholinesterase is inhibited by various organophosphates, including haloxon and nitrophenylhalon:

The rate of loss of activity of the enzyme at a given inhibitor concentration is apparently 1^st order. With NPH, the second inhibitor shown, apparent 1^st order rate constants at 0.3, 0.9 and 3 mM are 0.15, 0.46 and 1.5 per min. What would the 1^st order rate be at 10 mM inhibitor? What kind of inhibition is this? Explain how these compounds inhibit this enzyme. If the enzyme is separated from the inhibitor and incubated in buffer at room temperature, activity returns. This is also an apparently 1^st order reaction. Whether the inactive enzyme was produced with haloxon or NPH, the rate constant for reactivation is 0.017 per min. Suggest an explanation for these observations.

Apparently the 1^st order rate constant depends linearly upon the [NPH]. This would give a value of 5 per min at 10 mM. This could be, in a sense, a competitive inhibition. It is an inactivation, caused by a covalent linkage between inhibitor and active site. We saw the same effect with diisopropylfluorophosphate and the serine proteases. Activity returns when the inhibitor is removed. However, the inhibitor has covalently modified the enzyme, so a covalent change must occur to restore activity. This is a 1^st order reaction, or a solvolysis, with the phosphate group left by the inhibitor is transferred to a water molecule. The rate of reactivation is the same for both inhibitors, because both leave exactly the same group covalently bound to the active serine, and so it reacts with water at exactly the same rate, no matter how it was formed.

2. Consider the thermodynamics of protein folding. Certain solvents will cause a protein (for example, RNase) to unfold; replacing the solvent with a buffered aqueous solution will cause it to refold. If we think of it as a simple two state system, folded and unfolded protein, what will affect the equilibrium between the two states? Suggest reasons why some solvent changes shift the equilibrium toward the unfolded state, while in water, the equilibrium favors the folded state.

The equilibrium state is determined by the difference in free energy between the two systems: the folded and unfolded dissolved protein. These systems include the solvents. Since the question refers to thermodynamics, we need a discussion of the DH and DS between the two. What governs the DH? Bonding interactions, mostly. Especially H-bonding, in water. What governs DS? The ordering of solvent and/or protein. The folded protein is obviously more highly ordered than the unfolded. The ordering of solvent is a little less obvious, but has to do with the solvation of nonpolar groups by the polar water. These are the kind of issues that needed to be discussed.

3. We discussed the likely mechanism by which RNase catalyzes RNA hydrolysis. We discussed evidence supporting the catalytic mechanisms of lysozyme and the serine proteases. What kind of evidence would support the catalytic mechanism of RNase that we suggested? Be specific.

The mechanism of RNase involves two histidines acting as general acids/bases. It also includes ionic interactions with the phosphate of the substrate. Since protons are moving around a lot in this mechanism, you might be tempted to try to trace their movement with labeling. That is not possible in a protic solvent, because all ionizable protons equilibrate rapidly in such a solvent and become equally labeled. The traditional approach would have been covalent modification. These histidines are reactive; probably more so than other histidines. They could be modified with an alkylating reagent. We talked about TPCK with the chymotrypsin. It would not be a good choice here, because it does not resemble the substrate, but a general histidine reagent would work fine. We can always isolate the histidines at the active site by doing the modification with substrate present and absent. Amino acids that are protected by substrate are probably at the active site. So, if a histidine modifying reagent labeled the suspect histidines in the absence, but not the presence, of substrate, and if this modification led to inactivation, this would be consistent with the mechanism. Similar things could be done with other active site residues, especially those involved in binding the phosphate.

A more modern approach would be site directed mutagenesis to change the suspect histidines, on at a time, and together, into other amino acid residues. The whole range could be tested, but them most important ones would be those unable to act as a proton donor/acceptor. Perhaps a threonine, or a phenylalanine.

Other approaches are also possible. Some suggested using a DNA substrate; the pH profile also is suggestive.

4. a. In anemia, the Hb content of the blood is decreased. If we compare a person with 50% of the normal [Hb], to a person poisoned by CO, with 50% of the Hb sites occupied by CO, the anemic individual is much less impaired than the CO intoxicated one. Explain. (CO has 200X more affinity for Hb than O₂.)

Both individuals have the same absolute number of O₂ binding sites available. However, the CO intoxicated person has those sites distributed in Hb molecules that have at least some CO bound. The binding of CO, as with any ligand at the heme, favors the R form. Effectively, then, the CO intoxicated person is trying to deliver O₂ to the tissues with a Hb whose cooperativity is highly impaired, and whose O₂ affinity is high, and unchanging. O₂ binding is not the problem here, O₂ release is. Because of CO’s much higher affinity for Hb, it is released only slowly.

b. A mutant Hb has a cysteine replacing Tyr-145 of the b chain. X-ray crystallography indicates that this cys makes a disulfide bond with Cys-93 of the same chain, affecting the position of the C-terminus. How will this affect the following Hb properties? The value of L, oxygen affinity, Bohr effect, Hill constant, BPG affinity.

We are told that this mutation and new S-S bond affects the position of the C-terminus. The C-terminus is locked into salt bridges in the T form of Hb. It is rather mobile in the R form. This difference favors the T form. Will these salt bridges be the same in the mutant? We are not directly told. Since the C-terminus has changed position, I would guess not. This would, in effect, destabilize the T form relative to the R form. This decreases the value of L, would tend to increase the [R], and therefore the O₂ affinity. Greater (or lesser) binding of O₂, though, does not necessarily change the Bohr effect. What does change it is the lack of the salt bridge. A large chunk of the Bohr effect is due to the breaking of the salt bridge in going from T to R. This lowers the pK_a of basic groups in the salt bridges, which allows them to then release some of their protons. Since we are assuming the salt bridge isn’t there, it is unlikely there will be any lowering of the pK_a. What about the Hill constant? The value of the Hill constant depends on the existence of two conformational isomers, with no intervening structures. It is possible that n will change in this mutant, but it would have to be because of subtle structural changes; in other words, the destabilization of T might have the effect of making other states, less accessible before, now more available. Or the T state might be completely eliminated. If [T] has gone down, the apparent BPG affinity will also decrease.

5. Asp-52 is critical in the functioning of lysozyme. Predict the likely effects of changing this residue to a glutamate; or an alanine. When lysozyme is incubated with (NAG)₄, the slow formation of (NAG)₆ and (NAG)₂ is observed. Explain. How would these mutants be likely to affect this reaction?

This asp is critical because it seems to stabilize the transition state, and it forms a covalent bond to the C1 of part of the substrate. Glu has similar chemical properties and might work almost as well. It is not the same length, which could alter the position enough to significantly reduce the activity of the enzyme, but it would be hard to be sure in advance. Alanine, on the other hand, is an asp with the carboxyl group chopped off. The ala will be unable to stabilize the + charge in the transition state, or to form a covalent bond when the substrate bond is broken. We would expect that enzyme to be almost inactive.

Lysozyme will hydrolyze (NAG)₄, although very slowly. This hydrolysis is catalyzed by the lysozyme forming a covalent bond to the NAG that is on the C1 side of the cleaved bond. In other words, (NAG)₄ must bind to sites B-E or C-F of the lysozyme for hydrolysis to occur. A-D won’t work. Since (NAG)₂ is produced, but not free NAG, we must conclude that it binds C-F. Now the disaccharide in E&F will be released, but the other remains covalently bound. (NAG)₄ now binds in E&F (two sugars remain unbound), and the aspartyl-(NAG)₂ transfers the (NAG)₂ to the (NAG)₄ to make (NAG)₆. Although it is not impossible to imagine the reverse reaction from non-enzyme bound (NAG)₂ and (NAG)₄, it is thermodynamically so unfavorable that the rate is probably too slow to account for the observation. The covalent intermediate between enzyme and (NAG)₂ is a much better glucosyl donor than the free disaccharide or tetrasaccharide.

6. Two enzymes involved in glycolysis are NAD dependent dehydrogenases: glyceraldehyde 3-phosphate dehydrogenase (GAPDH) and lactate dehydrogenase (LDH). Suggest ways to use what you know about alcohol dehydrogenase to determine and verify the stereospecificity of these two enzymes. You have CH₃CD₂OH available to you.

We can use ADH, NAD and CH₃CD₂OH to make R-NADD. We know the stereospecifity of ADH. Now we can use R-NADD with 1,3-bisphosphoglycerate and GAPDH. If the product glyceraldehyde has D, we know that GAPDH has the same stereospecificity as ADH with regard to NAD; that the H^- adds to the re face. The same can be done with LDH. LDH also produces a stereocenter in its product, lactate. Determining its stereospecificity with regard to lactate would require the ability to make pure lactate enantiomers, or to determine the configuration of the lactate produced.

7. In a Dixon plot, 1/v is plotted against [I] at various [S]. How can this be used to determine the K_I?

We start with the Michaelis-Menten equation, v=V_max[S]/(K_M + [S]). The unmixed inhibitor equations alter this by multiplying either K_M or [S] in the denominator by 1 + [I]/K_I. So, for example, with competitive inhibition, 1/v = K_M/(V_max[S]) x (1 + [I]/K_i) + 1/V_max. This can be rearranged into the equation of a line, where 1/v can be considered the y-coordinate, and [I] the x-coordinate. If we take measurements at various [I], with 3 different [S], we can get 3 straight lines, whose slopes are K_M/(V_max[S]K_I). 3 equations, 3 unknowns, we can solve. Same can be done when inhibitor binds only to the ES complex, with slightly different results.