Exam 4 - 2021

1.A. What is the "order" of the reaction : $V=k[A][B]$$V = k[A][B]$ ?

• $2$$2$

1.B. What are enzyme allosteric effectors and how do they affect enzyme activity?

• Small chemicals that bind to multisubunit enzymes

  • inducing cooperative conformational changes

    • that then increase or decrease the rate of enzymatic catalysis

1.C. What is the likely optimum pH for an enzyme that has His acting as a general acid / base catalyst in the active site?

• centered around the $p{K}_a$$pK_a$ of histidines ionizable side chain ( $6.0$$6.0$ ) so that the side chain can fluctuate between being protonated or deprotonated depending on if pH is above or below 6.0.

1.D. What are enzyme cofactors, and what functions do they provide that are essential for enzyme activity?

• non-protein substance , typically vitamins and metal ions
• they help activate the enzyme by changing shape , balancing negative charge , acting as a ligand binding sink.

1.E. Sketch a graph that clearly shows the steady state approximation, which is the basis for Michaelis-Menten calculations (make sure to label all the lines/ axes).

2. A competitive inhibitor yielded the following equations from a Lineweaver-Burk plot:

   • No Inhibitor : $y=0.4843x+0.1951$$y = 0.4843x + 0.1951$
   • 3 mM Inhibitor : $y=0.7550x+0.1969$$y = 0.7550x + 0.1969$
   • 5 mM Inhibitor : $y=1.0062x+0.1862$$y = 1.0062x + 0.1862$

What is the apparent $K_M$$K_M$ at each inhibitor concentration and what is $K_I$$K_I$ for this data?

• $K_M^{app}=\alpha K_M$$K_M^{app} = \alpha\ K_M$

• $\alpha =1+\frac{[Inhibitor]}{K_I}$$\alpha = 1 + \frac{[Inhibitor]}{K_I}$

• $\alpha =\frac{K_m^{app}}{K_m}$$\alpha = \frac{K_m^{app}}{K_m}$

• Y-Intercept = $\frac{1}{V_{max}}$$\frac{1}{V_{max}}$

  • $V_{max}=\frac{1}{\text{Y-intercept}}$$V_{max} = \frac{1}{\text{Y-intercept}}$

• X-Intercept = $-\frac{1}{K_M}$$-\frac{1}{K_M}$

  • $K_M=-\frac{1}{\text{X-intercept}}$$K_M = - \frac{1}{\text{X-intercept}}$

• No Inhibitor one lets us find $K_M$$K_M$ :

  • Set the equation to zero and solve for x-intercept =

    • $0=0.4843x+0.1951$$0 = 0.4843x + 0.1951$
    • $\frac{-0.1951}{0.4843}=x$$\frac{-0.1951}{0.4843} = x$
    • $x=-0.40284947346685934$$x = -0.40284947346685934$

  • Find $K_M$$K_M$

    • $K_M=-\frac{1}{-0.40284947346685934}=2.482316760635572$$K_M = - \frac{1}{-0.40284947346685934} = 2.482316760635572$

  • Now we can solve for $K_M^{app}$$K_M^{app}$ at each concentration of inhibitor

    • $3mM$$3\ mM$ :

      • $0=0.7550x+0.1969$$0 = 0.7550x + 0.1969$
      • $\frac{-0.1969}{0.7550}=x$$\frac{-0.1969}{0.7550} = x$
      • $x=-0.260794701986755$$x = -0.260794701986755$
      • $K_M^{app}=-\frac{1}{-0.260794701986755}=3.834433722701879mM$$K_M^{app} = - \frac{1}{-0.260794701986755} = 3.834433722701879\ mM$

    • $5mM$$5\ mM$ :

      • $0=1.0062x+0.1862$$0 = 1.0062x + 0.1862$
      • $\frac{-0.1862}{1.0062}=x$$\frac{-0.1862}{1.0062} = x$
      • $x=-0.18505267342476645$$x = -0.18505267342476645$
      • $K_M^{app}=-\frac{1}{-0.18505267342476645}=5.403866809881848mM$$K_M^{app} = - \frac{1}{-0.18505267342476645} = 5.403866809881848\ mM$

  • Now we can solve for $K_I$$K_I$ at each concentration of inhibitor

    • $3mM$$3\ mM$ :

      • $\alpha =\frac{3.834433722701879mM}{2.482316760635572mM}=1.5446996062340212$$\alpha = \frac{3.834433722701879\ mM}{2.482316760635572\ mM} = 1.5446996062340212$
      • $K_I=\frac{3mM}{\left(1.5446996062340212\right)-1}=5.507622854258314mM$$K_I = \frac{3\ mM}{\left(\ 1.5446996062340212 \ \right) - 1} = 5.507622854258314\ mM$

    • $5mM$$5\ mM$ :

      • $\alpha =\frac{5.403866809881848mM}{2.482316760635572mM}=2.1769448990459392$$\alpha = \frac{5.403866809881848\ mM}{2.482316760635572\ mM} = 2.1769448990459392$
      • $K_I=\frac{5mM}{\left(2.1769448990459392\right)-1}=4.248287242718945mM$$K_I = \frac{5\ mM}{\left(\ 2.1769448990459392 \ \right) - 1} = 4.248287242718945\ mM$

3. $\mathrm{\Delta }\mathrm{\Delta }{G}^{‡}$$\Delta \Delta G^{\ddagger}$ for an enzymatic reaction at $25°C$$25°C$ is $13kJ\cdot mo{l}^{-1}$$13\ kJ \cdot mol^{-1}$ . Calculate the rate enhancement compared to the nonenzymatic reaction $(R=8.134J\cdot mo{l}^{-1}\cdot K^{-1})$$\left(\ R = 8.134\ J \cdot mol^{-1} \cdot K^{-1} \ \right)$

   $$\text{Rate Enhancement}=e^{\frac{\mathrm{\Delta }{\mathrm{\Delta }}^{‡}}{R\ast T}}$$
   $$\frac{13kJ}{mol}\ast \frac{1000J}{1kJ}=\frac{13,000J}{mol}$$
   $$\text{Rate Enhancement}=e^{\left(\frac{\frac{13,000J}{mol}}{\frac{8.314J}{mol\cdot K}\ast 298K}\right)}=e^{(\frac{13,000J}{mol}\ast \frac{mol}{8.314J}\ast \frac{1}{298})}=e^{5.247072537145237}=190.0092085702474$$

4. The specificity pocket of trypsin is show in the figure below. If the trypsin enzyme were mutated to have a Lys at position 189 (instead of serine) how would you predict the Km values for the following substrate peptides would change?

   • Substrate Peptide A – has Arg at position N-1 ( position next to scissile bond , N )
   • Substrate Peptide B – has a Glu at position N-1 ( position next to scissile bond , N )

   • Please note you do not need to determine the actual $K_m$$K_m$ values , just the change relative to wild type trypsin.

   

   • normally it has a negative charge

     • if we replace it with a positive charge , the $K_M$$K_M$ for substrate $A$$A$ would go up ,

       • $K_m$$K_m$ for substrate $B$$B$ decreases

    

5. Determine the velocity of the reaction : $\mathrm{X}+\mathrm{Y}⟶\mathrm{Z}$$\ce{X + Y->Z}$ when sample concentrations are $[X]=3\mu M$$[X] = 3\ \mu M$ and $[Y]=5\mu M$$[Y] = 5\ \mu M$. The $k$$k$ is $400M^{-1}\cdot s^{-1}$$400\ M^{-1} \cdot s^{-1}$

   $$\text{rate}=k\ast [X]\ast [Y]$$
   $$\text{rate}=\left(\frac{4.0\ast {10}^{2}M}{seconds}\right)\ast (3.0\ast {10}^{-6}M)\ast (5\ast {10}^{-6}M)=\frac{6\ast {10}^{-9}M}{1second}$$

    

6. The kinetic parameters for two substrates that are acted on by an enzyme are given below. Which is a better substrate for this enzyme ( as measured by efficiency of the enzyme ) ? Justify your answer.

   Substrate	$K_M$$K_M$ ( M )	$k_{cat}$$k_{cat}$ ( sec-1 )
   A	$1.1$$1.1$	$4\ast {10}^7$$4 * 10^7$
   B	$4.7\ast {10}^{-4}$$4.7 * 10^{-4}$	$4\ast {10}^3$$4 * 10^3$

   • Substrate $A$$A$ has the highest catalytic efficiency constant $\left(k_{cat}\right)$$\left(\ k_{cat} \ \right)$

   • Substrate $B$$B$ has the best binding constant $K_M$$K_M$

   • If we only care about efficiency , then Substrate $A$$A$ is the best

      

7. When experiments were conducted to study a newly discovered enzyme, the following observations were made: the enzyme has a very broad pH range in which it is at its optimum catalytic activity, ranging from pH = 6.5 to 8.5. However a competitive inhibitor is effective only at pH 6.5 , not 8.5. Explain this observation.

   • enzyme is active in 6.5 to 8.5

   • competitive inhibitor can only bind at 6.5

   • at pH of 8.5 , the competitive inhibitor is deprotonating , and then no longer able to interact with the enzyme

     • unlikely enzyme would be deprotonating