Chapter 9 Possible Essay Questions

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  1. 1) What are the precursor steps that the catalytic triad go through before encountering substrate?
    a.       The catalytic triad consists of a serine residue hydrogen bonded to the imidazole ring of a histidine residue. The –NH group of the histidine residue is, in turn, bonded to an aspartate residue. Before proteolysis can occur, the catalytic triad must create a powerful nucleophile that is capable of attacking the very unreactive carbonyl carbon atom. So, to form this nucleophile upon binding of the substrate, the histidine positions the side chain of serine and polarizes it to facilitate deprotonation. Histidine will then take a proton from serine, creating an alkoxide residue, which is stronger than a hydroxide ion. The aspartate will then stabilize the histidine and position it so that it can better accept the proton. Now that the alkoxide is formed, it can attack the carbonyl carbon. 
  2. 1)      Explain peptide hydrolysis of chymotrypsin.
    a.       Peptide hydrolysis proceeds in eight steps. The first step is substrate binding. Upon substrate binding, the alkoxide of the side chain of serine will attack the carbonyl carbon atom, creating a tetrahedral intermediate with a negative charge on oxygen. To neutralize the negative charge, a structure located in the polypeptide backbone known as the oxyanion hole will interact with the oxygen, more specifically, the –NH of an asparagine residue. After the formation of the tetrahedral intermediate, the histidine will donate a proton to the amino group formed by cleavage of the peptide bond, causing the release of the amine component of the sbstrate and the formation of the acyl-enzyme intermediate. After the acylation is complete, deacylation will occur. Water will come in and bind at the site that the amine group has just departed from. Histidine will strip a hydrogen away from the water, creating a hydroxyl group that will attack the carbonyl carbon atom, creating yet another tetrahedral intermediate with a negative charge around the oxygen. Similarly to the first time, the oxyanion hole will help neutralize this charge. The tetrahedron will eventually collapse, forming the carboxylic acid product. Release of this product resets the enzyme for acylation again. 
  3. 1)      How do chymotrypsin, trypis, and elastase differ?
    a.       Chymotrypsin cuts peptide bonds after residues with large hydrophobic side chains. Trypsin will cleave peptide bonds next to lysine or arginine. Elastase will cleave the bonds next to small hydrophobic side chains. The differences in their cleavage is due to the binding pocket that they possess. For example, whereas chymotrypsin has a binding pocket with plenty of space to allow large hydrophobic amino acids to go in, trypsin has an aspartate located on the bottom of the pocket, causing it to attract lysine and arginines, which are positively charged. Furthermore, elastase contains larger valine molecules, which are too large to allow the entry of large hydrophobic side chains. Therefore, they will bind small hydrophobic side chains.  
  4. 1)      Explain how they tested whether the catalytic triad was the mechanism? 
    The way in which they tested this was through site-directed mutagenesis. By replacing one of the residues of the catalytic triad with an alanine, they were able to determine the significance of each residue to peptide bond cleavage. The conversion of serine to alanine caused a great decrease in function of the enzyme. It dropped to one-millionth of its function. Converting histidine caused a dramatic decrease in activity as well. Converting the aspartate caused a functioning of the molecule to a lesser degree, but not as detrimental as the other two. The change of all of them to alanine caused a decrease similar to changing just the serine or histidine. What this demonstrated was the importance of the catalytic triad and, more specifically, the serine-histidine pair
  5. 1)      What are the alternatives to peptide-bond hydrolysis other than serine proteases?
    a.       Alternatives to peptide-bond hydrolysis are cysteine proteases, aspartyl proteases, and metalloproteases. All function to form a nucleophile that will attack the carbonyl carbon. However, their methods are slightly altered. Cysteine proteases function similarly to serine proteases except that, instead of employing a serine residue with an oxygen to attack the carbonyl carbon, it uses a sulfur to attack the carbonyl carbon. After the histidine activates the cysteine residue, the sulfur can attack. Because this sulfur is more reactive than oxygen, the aspartate is not necessary for cysteine proteases. They function with simply a cysteine and a histidine. Aspartyl residues are different in that they employ two aspartate residues—one deprotonated and one protonated. The aspartate residue in its deprotonated form will position the water molecule and activate it to prepare it for deprotonation. The protonated aspartate residue will polarize the carbonyl carbon atom, making it more susceptible to attack. Finally, metalloproteases will function through the use of a metal ion that will form the nucleophile by activating the water molecule. All three function to 1) produce the nucleophile, 2) polarize the carbonyl carbon, and 3) stabilize the intermediate
  6. 1)      What is the simple mechanism for carbon dioxide hydration? 
    a.       The simple mechanism of carbon dioxide hydration is theproduction of the hydroxyl group, binding of the substrate, attack of the carbon dioxide, forming HCO3-, which leaves the enzyme, preparing it for another round of carbon dioxide hydration. 
  7. 1)      What is special about the structure of carbonic anhydrases?
    a.       Carbonic anhydrases possess a zinc molecule that facilitates the formation of the nucleophile of the pH. Zinc-bound water acts has a drop in pKa from 15.7 to 7, allowing deprotonation at neutral pH and creating a large number of hydroxide ions that can react effectively with the carbon dioxide that binds to the carbonic anhydrase. Zinc is bound to four or more ligands. Three of these sites are occupied by histidine residues. The fourth is bound to water, which will activate it. 
  8. 1)      What is the benefit of using metal ions?
    a.       They have positive charges, can exist in numerous forms, and and they can form strong, yet kinetically labile, bonds. 
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Chapter 9 Possible Essay Questions
2014-10-16 20:26:51
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