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Hi
Your questions are about the chemical properties of amino acids, and the ways in which
these properties facilitate enzyme function (Qu. 1 and 2) and protein purification (Que 3).
1)
Because the alpha-carboxyl and alpha-amino groups are involved in forming the
peptide bond (the backbone of proteins), the functionality of amino acids within
proteins is dependent upon the chemical nature of the side chain groups. The
active site of an enzyme is a relatively small three-dimensional structure within
the enzyme, to which substrates bind by multiple weak interactions (and
sometimes covalent bonds). Polar residues are often found in active sites, to
create a specific electrostatic environment within the cleft that forms the
active site. Apparent pKa values for side chains of amino acids within proteins
are often different from those for free amino acids.
You have been given the pKa values for a glutamic acid residue and a histidine residue
within a dipeptidase. You are asked to determine the optimum pH for the enzyme based
on the criterion that both of these residues must be charged for substrate binding to the
active site. There are many examples of enzymes whose activity depends upon the
acid/base properties of two key amino acid side chains within the active site, as in this
example.
“Sensitivity to pH usually reflects an alteration in the ionization state of one or more
residues involved in catalysis or occasionally in substrate binding. Plotting reaction
velocity vs pH most often yields bell-shaped curve, where the inflection points
approximate the pKa values of ionizable groups in the active site. At the pH optimum
(midways between both pKa values) the greatest number of enzyme molecules are in
the active form.” (From
http://www.langara.bc.ca/biology/mario/Biol2315notes/biol2315chap7.html)
The pKa value tells you about the acid-base properties of a side chain: deprotonation of
acidic groups occurs at this pH. This makes sense if we look at the dissociation equation for
the carboxylic acid portion of glutamic acid’s side chain: deprotonation leads to a negative
charge on this group.
Since this deprotonation occurs at pH = pKa = 3.3, we must be at this pH or above to have
a charge on the glutamic acid side chain.
Similarly, in histidine we see a charge at pH values below the pKa 6.7, because
deprotonation neutralizes a charge on the imidazole ring of the side chain at this pH:
Since we know that the optimum pH for this type of situation is halfway between the
pKas for the two ionizable groups, we can easily calculate it as the average of the two
values given.
2(a).
The solution to this question involves the Henderson-Hasselbalch (H-H)
equation (for a derivation of this equation see this website
http://chemed.chem.purdue.edu/genchem/topicreview/bp/1biochem/amino2.ht
ml)
The H-H equation has a few useful forms, for your questions it is appropriate to use the
following:
pH = pKa + log10 ([A-]/[HA])
pH = pKa + log10 ([A-]/[HA])  rearrange to isolate log10 ([A-]/[HA])
pH – pKa = log10 ([A-]/[HA])  this form of the equation will allow you to plug in the
known values, which are pH =5 and pKa (aspartic acid) = 3.9
The left term of the equation becomes 1.1
1.1 = log10 ([A-]/[HA]) This can be solved by taking the antilog of both sides:
antilog 1.1 = ([A-]/[HA]), and [A-]/[HA] is the ratio of –COO- to –COOH groups present
for the aspartic acid side chains at pH 5
= 12.59.
so, if we round up, the ratio is 13:1, and for roughly every 14 side chains, 13 of them are
of the form –COO-. This makes sense because the pKa for this group is 3.9, which
means that deprotonation occurs at pH = 3.9, and therefore most of the molecules should
be deprotonated at pH = 5.
This means that 13/14 = x/100 and x = 92.86% are in the ionized form.
Similarly for glutamic acid:
0.7 = log10 ([A-]/[HA]) This can be solved by taking the antilog of both sides:
antilog 0.7 = ([A-]/[HA]), and [A-]/[HA] is the ratio of –COO- to –COOH groups present
for the glutamic acid side chains at pH 5
= 5.01
the ratio here is 5:1, and for roughly every 6 side chains, 5 of them are of the form –
COO-. Again, this makes sense because the pKa for this group is 4.3, therefore high
proportion of the molecules should be deprotonated at pH = 5.
5/6 = x/100 and x = 83.33% are in the ionized form.
But, for part (b), it is asked how then could the glutamic acid residue exist in the enzyme
in primarily its un-ionized form at the optimum pH of 5, as predicted by the major
proposed catalytic mechanism of lysozyme? One theory suggests that dissociation of the
acidic proton in this side chain is supressed by the very hydrophobic local environment of
this residue within the active site, giving this residue a much higher “apparent pKa” than
that predicted by the table you were using (from
http://www.food.rdg.ac.uk/online/fs916/lect7/lect7.htm.) In contrast, asp52 is in a polar
environment.
3).
This question asks about the purification of tripeptides using ion-exchange
chromatography. Since it is given in the question that you are using CM-cellulose, you
know you are performing cation-exchange chromatography. This means the charge on
the resin packing the inside of the column (stationary phase)is negative and will bind to
positively charged peptides. There are some good overviews of this technique here:
http://www.resonancepub.com/chromtutorial.htm
http://matcmadison.edu/biotech/resources/proteins/labManual/chapter_4/section4_4.htm
http://www.proteinchemist.com/tutorial/iec.html
Peptides are separated on the basis of their overall (net) charge when using this method.
CM-cellulose has dissociable –OH groups with a given pKa of 4.90. You know then
that all of the –OH groups have been dissociated to -O- at the 3 pH values used in the
column separations (because they are all at least 1 pH unit above the pKa of the column).
What you need to do is find the net charge on each of the peptides at all three pH values.
The peptides with the most electropositive charge will have the highest affinity for the
column material ie. the most negative ones will elute from the column first.
I will demonstrate the calculation for the first peptide at pH 6, so you will be able to see
how it is done and complete this question. It helps to draw the peptide if you can,
otherwise, have a diagram of each amino acid handy and imagine it in your mind…
When examining the net charge on a tripeptide, the important ionizable groups are: the
amino terminal (peptide formulas are always drawn so that this is at the left of the
molecule), the carboxyl terminal (at the right end), and any ionizable side chains present
in the amino acids comprising the peptide.
The net charge can be calculated for the first tripeptide, -Tyr-Arg-Ser-, at pH 6:
The ionizable groups I can identify on this peptide are: the α-amino goup of tyrosine
(pKa 9.1), the R-group of tyrosine (pKa 10.1), the R group of arginine (pKa 12.5), the Rgroup of Serine (pKa –13) and the α –carboxyl group of Serine (pKa 2.2). First, you
need to know the relative proportions of the protonated and unprotonated forms of each
ionizable group at the given pH. This can be calculated as shown in question 2 using the
Henderson-Hasselbalch equation, for the α-amino goup of tyrosine for example:
pH = pKa + log [A-]/[HA]
= 6 = 9.1 + log [A-]/[HA]
6 – 9.1 = log [A-]/[HA]
antilog (-3.1) = [A-]/[HA] = 0.00079 a very small number, so essentially all (100%) of
the groups are in the HA form. Because we are talking about the α-amino group, this
means it will be protonated and have a charge of +1. This is expected, because I am
looking at a pH of 6, which is well below the pKa for the α-amino group of tyrosine.
Similarly, the side chain of tyrosine has a high pKa, so we can be confident that all of
these groups will be protonated. Because the side chain of tyrosine has one -OH , this
means there will be no charge on this group. The side chain of arginine (see below, from
an excellent protein website http://www.cem.msu.edu/~reusch/VirtTxtJml/proteins.htm)
likewise has a high pKa and will be protonated at pH 6
At pH 6, the arginine side chain within this peptide carries a +1 charge. Serine has a very
acidic side chain and α-carboxyl group (low pKa), so they will both be deprotonated at
pH 6. For the carboxyl terminal this means a –1 charge, and for the side group of serine
it also means a –1 for a total charge on the serine residue of –2. To calculate the net
charge on the peptide at pH 6 we only need to add up all of the charges listed:
(+1) + (+1) + (-2) = 0 ie. there is no net charge and this peptide is zwitterionic at pH 6.
Note that this means the peptide will not interact with the negatively charged CM
cellulose and will be eluted (washed out) of the column at this pH.
The net charge on the other peptides at each pH can be calculated this way, making sure
you keep in mind the ratios of the acid to the base for each ionizable group in the peptide
(calculated from Henderson-Hasselbalch as above). You need to multiply the percent of
the charged form by the charge contributed by each ionizable group to get a fractional
charge when you are not certain it is 100%. A good rule of thumb is to use H-H to
calculate the acid/base ratio when the pH you are looking at is within 2 pH units of the
pKa of the ionizable group in question.
Once you have the net charge on each of the three peptides at a particular pH, you can
compare their affinity for the column material. The most highly positive peptides will be
the most tightly bound to the column and will elute last. A highly negative peptide would
be likely to elute before a neutral one. It is helpful when doig this question to have a
diagram of the amino acids so you can see what their side chains look like and figure out
what charge they would have at a given pH. Just remember, the pKa is the pH where a
dissociable hydrogen will come off; keep this in mind and it becomes easier to see where
the charges come in at the different pH values. If you can find a titration curve for a
specific amino acid, this gives you all the information you need to calculate and check the
ratios of acid to base and see what the charges on the groups look like.
I hope this is helpful, good luck with your studies.