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1. PROBLEM SET 1 (QUESTIONS 5 & 6)
TA: Hassan Allahrakha Badal Rana
hxa111230@utdallas.edu bxr112130@utdallas.edu

2. 5A)
The pentapeptide Val-Arg-Gln-Trp-Cys has N-terminal and C-terminal pKas
of 9.5 and 4.5 respectively. The side chain Pka’s are Arg 12.5 and Cys
8.0.
a. Write this peptide’s sequence using single letter abbreviations.
Example: Ala-Leu-Tyr-His
ALYH

3. b. Draw its predominant chemical structure at pH 7.4 including
charges.
-Properties of amino acids in proteins and peptides are determined by the R
groups but also by the charges carried by the titratable (R) groups, which will
ultimately affect the protein structure.
- Important to know which groups on peptides and proteins will be protonated
at a certain pH.
General rules:
No pKa : group uncharged
pH<pKa : Group predominantly protonated, H+ on
pH >pKa : Group predominantly deprotonated, H+ off
pH=pKa : Equal amounts of protonated and deprotonated species.

4. Val-Arg-Gln-Trp-Cys
N terminal pKa ≈ 9.5 > pH,
+1 charge
Valine (Val): No pKa,
Uncharged
Arginine (Arg): pKa 12.5 >
pH, +1 charge
Glutamine (Gln) : cont. with
other residues and then draw
the structure with
predominant charges. Draw
with the Fischer projection as
below.

5. c. What fraction of the peptide will exist in the predominant form at pH 7.4?
When the pKa differs from the pH by 2 units, the group can be considered to be
fully protonated (pKa < pH – 2) or fully deprotonated (pKa > pH + 2.
When the pKa differs from the pH by < 2 units it will be partially protonated.
In 5c. This will be true for Cys
In this case, use the H-H equation to calculate the fraction protonated.
푝퐻 = 푝퐾푎 + log
퐶푦푠−
퐶푦푠0
Then use the calculated ratio to determine the % protonated.
Example, if the ratio = 0.4/1, the % 퐶푦푠0 =
퐶푦푠0
( 퐶푦푠0 + 퐶푦푠−
푥 100 =
0.4
1 + 0.4
푥 100 = 29%
퐶푦푠− = 푐ℎ푎푟푔푒푑 퐶푦푠 푅 − 푔푟표푢푝
퐶푦푠0 = 푢푛푐ℎ푎푟푔푒푑 퐶푦푠 푅 − 푔푟표푢푝

6. d. Calculate the pI of the peptide.
• pI is the average of the pKa values representing the protonation and
deprotonation of the neutral form of the peptide. It is the pH value at which
the sum of their positive and negative electrical charges is 0.
• Figure out the pH range where your peptide has a net charge of 0. This
occurs
between 2 pKas.
• To find the pI, average these two pKa values that bracket the range.

7. Shortcut to finding pI
-Order pKas from smallest to highest
-Choose random pHs between the pKas and see what the charges on the R-groups
are
-Find in what range your pI will fall when the charges add up to neutral ,
then take the average of the two pKas it falls between.
An example is shown for Gly on the next page.
Apply this process to the al-Arg-Gln-Trp-Cys pentapeptide in the question

9. e. If this pentapeptide is reacted with NEM, what will be the pI of the
product?
-Results in addition of the –SH across the C=C double bond of the NEM,
resulting in a side chain with no chargeable group.
-Thus, in calculating the pI, ignore pKa of Cys
-Calculate pI using same method as step d.

10. f. To separate a mixture of the peptide and its NEM reaction product by ion
exchange chromatography at pH 7.4, would you choose a CM- or
DEAE-matrix? Explain why.
CM is a cation exchanger while DEAE is an anion exchanger.
-See what overall charge the peptide has at 7.4 for both the
unreacted peptide and NEM-peptide.
-Depending on the charges, you would use CM if the peptides
are positively charged and DEAE if they are negatively charged.

11. g. Which peptide would elute first? Explain why.
The greater the charge, the stronger it will bind to the matrix,
which will cause it to elute later.

12. h. If the peptide and its NEM reaction product were separated by ESI TOF
MS, which would reach the detector first? Explain why.
• ESI – Electrospray Ionization charges the NEM and unreacted
peptides by placing them in a pH 2 solution. Therefore determine
the , resulting in same charge for both.
• TOF MS- separates by mass to charge (m/z)
• Determine the charge on each peptide at pH 2
• NEM will have a greater mass than unreacted peptide.
• Consider the m/z ratios of the peptides.
• The lower the m/z ratio, the faster it arrives to the detector. See the
G&G text for a discussion of why.

13. i. If a solution of the pentapeptide, pH 6, has an absorbance of 0.5 at 280
nm in a standard 1-cm. width spectrophotometer cuvette, what is the
molar concentration of the peptide? What is its concentration in
mg/ml?
-Trp and Tyr are the only residues that absorb at 280 nm. According to
figure 4.10 in the textbook. The molar absorptivity 휺 is 5,000 푴−ퟏ풄풎−ퟏ.
Equation: A= 휺Cl
A=Absorbance, 휺= molar absorptivity, C= Concentration, l= path length
After you find concentration of the peptide, add the total molecular weight
of the amino acid residues using figure 4.18 in the lecture slide +the
molecular weight of 1 water for the H and OH not removed at the N- and C-termini
Grams peptide/liter = mg/ml = molar [peptide] x molecular weight (Mr)

14. Question 6
Assume you are given a mixture of proteins that you analyze by standard 2-
D electrophoresis, with the following results for isoelectric points and
apparent molecular weights: protein A (Mr 27,400; pI 7.21), protein B (Mr
41,350; pI 6.63), and protein C (Mr 102,200; pI 5.26).
2-D electrophoresis consists of isoelectric focusing followed by SDS-PAGE
http://highered.mheducation.com/sites/9834092339/student_view0/chapter2
0/electrophoresis.html- VIDEO FOR SDS PAGE electrophoresis

15. Isoelectric focusing separates proteins based on their pI’s. The
electrophoresis is carried out in a gel containing a pH gradient.
SDS PAGE separates on running gel based on molecular weight. The
smaller the molecule, the further it migrates on the running gel, as it is
able to move more more easily through the pores of the gel.

17. a. In the isoelectric focusing step, in what order did the protein bands,
starting with the anode and going to the cathode? Explain why.
The + anode will be in the acid tank at the low pH end of the isoelectric focusing
gel
- Molecules with a net - charge (where pH > pI) will move toward the anode
The - cathode will be in the base tank at the high pH end of the isoelectric
focusing gel
-Molecules with a net + charge (where pH < pI) will move toward the cathode
-Proteins will gain or loose protons as they move through the pH gradient finally
stopping and banding at their pI in a pH gradient.
The higher the pI, the closer to cathode and the lower the pI, the closer to
anode.

18. b. If the mixture is subjected to DEAE (pKa 11.5)- matrix chromotography,
what pH range should you use for the buffer and in what order will the
proteins emerge from the column? Explain why.
DEAE is an anion exchanger.
Anion chromatography principle: More negative the protein, the stronger it binds
to the (+) DEAE-matrix and the later it elutes.

19. b. cont.)
What pH?
For the DEAE to carry a + charge , the pH must be = , <, >? its pKa.
For a protein to carry a net negative charge, the pH must be = , <, >? its pI.
Eluation order?
The lower the pI below the pH, the greater the negative charge of the
protein. The greater the negative charge, the stronger the binding to the
DEAE-matrix, which means it elutes later.

20. c. Upon gel filtration of the mixture, protein A emerges from a G2000
Sephadex column at a Ve of 100 ml, protein B at 170 ml, and protein C at
105 ml. What can you conclude about the structure of protein A?
Explain.
• Gel filtration/size exclusion chromatography: Technique used for
protein purification that separates based on SIZE and SHAPE of
protein. It does not denature a protein. As a protein solution is passed
through the column of gel beads, molecules distribute based on ability
to enter the pores of the beads. If a molecule is too large, it is excluded
from the gel beads and unable to penetrate them and emerges from the
column sooner than the smaller molecules. The smaller molecules can
penetrate deep into the beads and thus encounters more of the column
buffer, which results in slower migration through the column and later
elution.

22. c (cont.):
From the elution pattern for the three proteins you can predict their relative
molecular weights, Mr’s. If you compare this to the Mr’s determined by
SDS-PAGE as given in the question, there is an anomaly for one of the
proteins. How can this be rationalized?
Remember : SDS-PAGE is denaturing and would separate subunits of
multimers; whereas, Gel filtration keeps the proteins in their native
structure.