Lysozyme

As we saw in lecture two of the Metabolism series, lysozyme cleaves its substrate N-acetyl glucosamine (NAG) between the fourth and fifth residues of the sugar molecule.  In the model to the right, the two subunits of the lysozyme molecule are depicted in wireframe mode with NAG represented in purple. 

Where are the catalytic residues of lysozyme (Glutamate 35 and Aspartate 53) located with respect to the cleavage site? Click on the button below to see them represented in  spacefilled mode. You can see just how good a fit the substrate is within the enzymes active site by clicking on the button below.  The substrate is thought to be held in place by making more than a dozen hydrogen bonds with the enzyme.Glutamate 35 and Aspartate 53 are shown in cyan.  

Question

The lactone whose structure is pictured below resembles the structure of NAG and is is a potent inhibitor of lysozyme.   How do you think this occurs at the molecular level?  

If we were to add this inhibitor to a solution of NAG and carry out a kinetic study of the reaction, what do you think will happen to the value  K_M (Michaelis constant) ?  What will happen to the maximum velocity of the reaction (Vmax) ? 

(Please try to reflect on the problem and scroll down for the solution only when you have attempted the question). 

 

 

 

 

 

 

 

Solution

The structure of the lactone closely resembles the substrate (NAG) and therefore readily competes with NAG for access to the active site of lysozyme.  It is therefore a competitive inhibitor and will increase the value of K_M  (remember this value tells us about the affinity of the enzyme for substrate).  In the presence of such an inhibitor, more NAG is needed to achieve Vmax (remember the plot of [S] vs velocity?), hence K_M increases.  In contrast, Vmax will be unchanged as with enough substrate present, the enzyme can still reach the levels of Vmax achieved in the absence of inhibitor (the inhibitor is out-competed).  K_M and Vmax are typically determined by Lineweaver-Burk analysis (practical MDL1-3) and the effects of a competitive inhibitor on such a plot is shown below.

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