Chapter 8

Show that the half-life for a first-order reaction is inversely proportional to the rate constant, and determine the constant of proportionality.

An enzyme contains an active site aspartic acid with a \(\mathrm{p} K_{\mathrm{a}}=5.0\), which acts as a general acid catalyst. On the accompanying template, draw the curve of enzyme activity (reaction rate) versus \(\mathrm{pH}\) for the enzyme (assume that the protein is stably folded between \(\mathrm{pH} 2-12\) and that the active site Asp is the only ionizable residue involved in catalysis). Briefly explain the shape of your curve.

The folding and unfolding rate constants for a myoglobin mutant have been determined. The unfolding rate constant \(k_{\mathrm{F} \rightarrow \mathrm{U}}=3.62 \times 10^{-5} \mathrm{~s}^{-1}\) and the folding rate constant \(k_{\mathrm{U} \rightarrow \mathrm{F}}=255 \mathrm{~s}^{-1}\), where \(\mathrm{F}\) is the folded protein and \(U\) is the unfolded (denatured) protein. For wild-type myoglobin, \(\Delta G_{\mathrm{F} \rightarrow \mathrm{U}}^{\mathrm{or}}=+37.4 \mathrm{~kJ} / \mathrm{mol}\). Which myoglobin is more thermodynamically stable, the mutant or the wild-type?

In some reactions, in which a protein molecule is binding to a specific site on DNA, a rate greater than that predicted by the diffusion limit is observed. Suggest an explanation. [Hint: The protein molecule can also bind weakly and nonspecifically to any DNA site.]

Would you expect an "enzyme" designed to bind to its target substrate as tightly as it binds the reaction transition state to show a rate enhancement over the uncatalyzed reaction? In other words, would such a protein actually be a catalyst? Explain why or why not.

The initial rate for an enzyme-catalyzed reaction has been determined at a number of substrate concentrations. Data are as follows: \begin{tabular}{cc} {\([\mathrm{S}](\mu \mathrm{mol} / \mathrm{L})\)} & \(v\left[(\mu \mathrm{mol} / \mathrm{L}) \mathrm{min}^{-1}\right]\) \\ \hline 5 & 22 \\ 10 & 39 \\ 20 & 65 \\ 50 & 102 \\ 100 & 120 \\ 200 & 135 \\ \hline \end{tabular} (a) Estimate \(V_{\max }\) and \(K_{\mathrm{M}}\) from a direct graph of \(v\) versus [S]. Do you find difficulties in getting clear answers? (b) Now use a Lineweaver-Burk plot to analyze the same data. Does this work better?

The catalytic efficiency of many enzymes depends on \(\mathrm{pH}\). Chymotrypsin shows a maximum value of \(k_{\text {cat }} / K_{\mathrm{M}}\) at \(\mathrm{pH}\) 8. Detailed analysis shows that \(k_{\text {cat }}\) increases rapidly between \(\mathrm{pH} 6\) and 7 and remains constant at higher \(\mathrm{pH}\). \(K_{\mathrm{M}}\) also increases rapidly between \(\mathrm{pH} 8\) and 10 . Suggest explanations for these observations.

The following data describe the catalysis of cleavage of peptide bonds in small peptides by the enzyme elastase. \begin{tabular}{ccc} Substrate & \(\boldsymbol{K}_{\mathbf{M}}(\mathbf{m} \mathbf{M})\) & \(\boldsymbol{k}_{\text {cat }}\left(\mathbf{s}^{-1}\right)\) \\ \hline PAPA \(\downarrow \mathrm{G}\) & \(4.0\) & 26 \\ PAPA \(\downarrow \mathrm{A}\) & \(1.5\) & 37 \\ PAPA \(\downarrow \mathrm{F}\) & \(0.64\) & 18 \\ \hline \end{tabular} The arrow indicates the peptide bond cleaved in each case. (a) If a mixture of these three substrates was presented to elastase with the concentration of each peptide equal to \(0.5 \mathrm{mM}\), which would be digested most rapidly? Which most slowly? (Assume enzyme is present in excess.) (b) On the basis of these data, suggest what features of amino acid sequence dictate the specificity of proteolytic cleavage by elastase. (c) Elastase is closely related to chymotrypsin. Suggest two kinds of amino acid residues you might expect to find in or near the active site.

Initial rate data for an enzyme that obeys Michaelis-Menten kinetics are shown in the following table. When the enzyme concentration is 3 nmol \(\mathrm{ml}^{-1}\), a Lineweaver-Burk plot of this data gives a line with a \(y\)-intercept of \(0.00426\left(\mu \mathrm{mol}^{-1} \mathrm{ml} \mathrm{s}\right)\). \begin{tabular}{cc} {\([\mathbf{S}] \boldsymbol{\mu} \mathbf{M}\)} & \(\boldsymbol{v}_{\mathbf{0}}\left(\boldsymbol{\mu} \mathbf{m o l} \mathbf{m l}^{-1} \mathbf{s}^{-1}\right)\) \\ \hline 320 & 169 \\ 160 & 132 \\ \(80.0\) & \(92.0\) \\ \(40.0\) & \(57.2\) \\ \(20.0\) & \(32.6\) \\ \(10.0\) & \(17.5\) \\ \hline \end{tabular} (a) Calculate \(k_{\text {cat }}\) for the reaction. (b) Calculate \(K_{\mathrm{M}}\) for the enzyme. (c) When the reactions in part (b) are repeated in the presence of \(12 \mu \mathrm{M}\) of an uncompetitive inhibitor, the \(y\)-intercept of the Lineweaver-Burk plot is \(0.352\left(\mu \mathrm{mol}^{-1} \mathrm{ml} \mathrm{s}\right)\). Calculate \(K^{\prime}{\underline{\phantom{xx}}}_{\mathrm{I}}\) for this inhibitor.

\- The inhibitory effect of an uncompetitive inhibitor is greater at high \([\mathrm{S}]\) than at low [S]. Explain this observation.

In kinetics experiments, the hydrolysis of the substrate sialic acid by neuraminidase appears to obey Michaelis-Menten kinetics. Neuraminidase activity is critical for viral infectivity; thus, this enzyme is the target of much work by pharmaceutical companies to develop a drug to treat influenza virus infection. The drug "Tamiflu" is a competitive inhibitor of neuraminidase. Initial rate data collected at \(\mathrm{pH}=6.15,37^{\circ} \mathrm{C}\) with \(0.021 \mu \mathrm{M}\) neuraminidase and \(25.0 \mu \mathrm{M}\) sialic acid gives a Lineweaver-Burk plot with a slope of \(51.2 \mathrm{~s}\). (a) Recall from Problem 23 that the \(k_{\text {cat }}\) for neuraminidase at \(\mathrm{pH}=6.15\), \(37{ }^{\circ} \mathrm{C}\) is \(26.8 \mathrm{~s}^{-1}\). Calculate \(K_{\mathrm{M}}\) for the hydrolysis of sialic acid. (b) When the reactions in part (a) are repeated in the presence of \(0.040 \mu \mathrm{M}\) of Tamiflu, the slope of the Lineweaver-Burk plot is \(198.8 \mathrm{~s}\). Calculate the value of \(K_{\mathrm{I}}\) for Tamiflu.