Chemistry 532: Key to Exam 2

1. (25 points) Suppose you are studying the inactivation of enzyme X by an active site-directed alkylating agent. Enzyme X is a homodimer, i.e. it is composed of two identical protein subunits.

You measure the concentration of enzyme X in solution using the Bradford dye-binding protein assay. This protein assay involves the binding of the dye Commassie blue to proteins, resulting in a spectral shift in the dye which is measured by spectrophotometry. You first make a standard curve by measuring the spectral signal versus the microgram of standard protein, bovine serum albumin (BSA), and show that the spectral signal is linearly proportional to the microgram amount of BSA. You then measure the spectral signal for a sample of your enzyme X and use the standard curve made with BSA to estimate the microgram amount of enzyme X. Since you know the amino acid sequence of enzyme X and hence the molecular weight of dimeric enzyme X, you convert this microgram estimate into moles of dimeric enzyme X and now you have an estimate of the molar concentration of dimeric enzyme X in solution. You then add sub-stoichiometric amounts of active site-directed alkylating agent to this solution of enzyme X and you measure the percent remaining enzymatic assay of enzyme X. You make the remarkably discovery that 1 mole of inactivator is sufficient to drop the enzymatic activity of 1 mole of dimeric enzyme X from 100% to 0%. You were expecting that it would take 2 moles of inactivator to fully inactivate 1 mole of dimeric enzyme X since the dimer has two active sites. Thus you are starting to think of some possibilities such as:

1) inactivation of one active site of the dimer leads to a conformational change that prevents the other active site of the dimer from reacting with inactivator;

2) both active sites, one from each monomer, face each other in the dimer and together they form a single active site;

3) both active sites are near each other in space and they function as separate active sites, but when the inactivator alkylates one of the active sites it is big enough to protrude into the other active site and block access of substrate to both active sites even though it is covalently attached only to one active site.

You may assume that your enzyme assay is faithful, i.e. the % enzymatic activity is proportional to the % of active enzyme. From the data presented above, can you really be sure that 1 mole inactivator is able to reduce the enzymatic activity to 0% of 1 mole of dimeric enzyme X? Explain. If you feel that the conclusion of the above experiment is not certain, provide a description of a re-designed experiment which would allow you to more convincingly test the issue of how many moles of inactivator are required to fully inactivate 1 mole of dimeric enzyme X.

Answer: The potential problem is with the dye-binding protein assay. Although the assay gives a linear response for spectral signal vs ug of BSA or vs ug of enzyme X, the slope of this curve for the two proteins is probably different. Thus you cannot use a standard curve prepared from BSA to get the ug amount of enzyme X. One needs to measure the concentration of protein X by a better method. One can take the amino acid sequence of enzyme X and calculate its extinction coeff. at 280 nm quite accurately, then measure the OD280 of a solution of pure enzyme X to estimate its molar concentration. Even this method is not perfect, if 50% of your enzyme X is denatured, it will still give the OD280 nm signal but then only 1 mole of inactivator per dimer of X will be needed to fully inactivate the enzyme.

2. (10 points) Suppose you design an enzyme inhibitor Y that is very potent at inhibiting the enzyme in vitro. You now want to determine if Y can block the action of the enzyme in the cell. Suppose the target enzyme is in the cytosol of the cell. Suppose you determine that Y cannot pass through the plasma membrane of the cell. But you consider the possibility that Y can get into the cell via endocytosis. By endocytosis I mean the plasma membrane invaginates and pinches off to form a vesicle in the cytosol of the cell. The aqueous compartment of this vesicle comes from the extracellular aqueous compartment so you are sure that your inhibitor Y in the extracellular fluid will be able to get into the aqueous compartment of the vesicle which pinched off from the plasma membrane. After this vesicle is formed, it fuses with the ER. So in this way inhibitor Y is delivered inside of the cell. Is it necessarily true that inhibitor Y will gain access to the cytosolic enzyme target? Explain.

Answer: Remember the lumen of the ER is not connected to the cytosol. So your inhibitor will have to cross the ER membrane to get to the cytosol. So it is not necessarily true that your inhibitor will gain access to the cytosolic enzyme target.

3. (15 points) Suppose you have a gene for your protein ligated into an expression plasmid via the restriction sites Bam HI and Xba I as shown below:

Only the top strand DNA sequence is shown: expression plasmid

CAGGAAGGATCCATGAAAGCGGGAATC- - - - - - - - - - - AAGGTCGGGGAGTAATCTAGAGGCATA

BamHI site Translation start site Translation stop site XbaI site

The dotted line in the above sequence is the remaining portion of the coding sequence of your gene.

Suppose you want to change the C-terminal Lysine of your protein to a Glutamate (i.e. you want to change the C-terminal Glutamate codon GAG to that for Lysine AAG). Your strategy is as follows. You will digest the expression plasmid with both Bam HI and Xba I to release the insert with your gene. Then into the linearized expression vector (which you can purify away from the released gene fragment on an Agarose gel) you will ligate in a PCR fragment that contains the mutant version of your gene (with the C-terminal Glutamate codon replaced with the Lysine codon). Thus you want to generate a PCR fragment that can be digested with Bam HI and Xba I to generate sticky ends that can be ligated into the Bam HI/XbaI-digested expression vector. You want the DNA sequence of this PCR fragment to be identical to the fragment containing the wild-type (non-mutant) gene released from the expression plasmid by Bam HI/Xba I digestion, except that the C-terminal Glutamate codon GAG is changed to AAG for Lysine. Design a pair of PCR primers that you can use for this task. The target for the PCR reaction will be the expression plasmid containing the wild-type gene (shown above). The PCR primers should be 24 nucleotides long (including 4 nucleotides for the clamp and 20 nucleotides of non-clamp sequence) and should have a G/C content close to 50%. For your answer, write down the nucleotide sequence of these primers, both of them written in the conventional 5’-to-3’ direction (5’ end on the left, 3’ end on the right).

Answer: below is the sequence above with its complement drawn EXCEPT THAT THE GAG CODON WAS CHANGED TO AAG:

CAGGAAGGATCCATGAAAGCGGGAATC- - - - - - - - - - - AAGGTCGGGAAGTAATCTAGAGGCATA

GTCCTTCCTAGGTACTTTCGCCCTTAG TTCCAGCCCTTCATTAGATCTCCGTAT

PRIMER 1: AGCTGGATCCATGAAAGCGGGAAT

PRIMER 2: AGCTTCTAGATTACTTCCCGACCT

THE UNDERSCORE SEQUENCE IN THE PRIMERS IS THE CLAMP, ANY 4 NUCLEOTIDES WILL SUFFICE. NOTE THAT PRIMER 2 HAS THE MUTANT CODON IN IT SO THAT THE PCR FRAGMENT WILL HAVE A C-TERMINAL LYSINE. THE PRIMER 2 WILL MATCH THE TARGET WITH 19 OUT OF 20 NUCLEOTIDES (1 MIS-MATCH FOR THE MUTATION) THIS IS GOOD ENOUGH TO PRIMER PCR ESPECIALLY SINCE YOUR TARGET IS A SINGLE PLASMID, NOT A DNA LIBRARY.

4. (15 points) It appears that one of the primary roles of immunophilin proteins such as FKBP is to mediate the conformational switching of signaling proteins such as calcineurin. It is unknown whether the proline cis-trans isomerization activity of the immunophilins is essential for their general function. What are the general features of the proline cis-trans isomerization reaction that makes it an attractive candidate for a reaction that could regulate conformational switching.

Answer: There are two features. One is that the rate of isomerization is relatively slow and, thus, spontaneous conformational switching would not occur rapidly, which allows for control of the state of the protein by another protein. The second is that the relative energies of the two isomers are similar and, thus, a large thermodynamic driving force is not required to switch between the two state of the protein.

5. (10 points) Labeling of proteins with NHS-esters sometimes forms esters between serine, theonine, and tyrosine sidechains and the labeling acyl group. The esters can be easily removed by subsequent treatment of the labeled protein with hydroxylamine or hydrazine. Draw the chemical structure for the product of the reaction between serine and a generic NHS-ester (i.e., RCOONHS) and illustrate how hydroxylamine removes the ester.

6. (10 points) Explain why TCEP is preferable to other reagents that reduce disulfides (i.e., DTT, beta-mercaptoethanol, etc.) when trying to label a protein with a cysteine-reactive conjugate.

Answer: TCEP does not react with cysteine-reactive conjugates because it is a phosphine-based reducing agent. The other reagents are thiol based and these will react with cysteine-reactive reagents and, thus, they have to be removed prior to labeling.

7. (15 points) The BC8 antibody was labeled with fluorescein-NHS ester. Absorbance measurements confirmed that there was an average of 4 fluoresceins per antibody. Cells known to express the antigen for BC8 did not stain with the antibody (i.e., the mean fluorescence intensity measured for cells incubated with the labeled BC8 by flow cytometry was similar to the isotype control). A second antibody, LR6, that recognizes the same antigen was similarly labeled to give about 4 fluoresceins per antibody. The same cells were efficiently stained by the labeled LR6 antibody.

Suggest why the labeled BC8 did not stain the cells even though LR6 did and describe what you could do to generate a labeled BC8 antibody that might stain the cells.

Answer: Most likely BC8 has one or more lysine residues that when fluorescein-labeled interfere with antigen binding. Thus, the BC8 labeled with fluorescein-NHS ester is unable to bind and the cells do not stain. One thing that could be tried is to label the BC8 with fluorescein-NHS ester for a shorter time and/or with less label to reduce the number of labels per protein. An alternative would be to partly reduce the antibody with something like TCEP and then label it with a cysteine-reactive fluorescein reagent.