PROBLEM SET 1 (QUESTIONS 5 & 6)

TA: Hassan Allahrakha Badal Ranahxa111230@utdallas.edu bxr112130@utdallas.edu

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.

ALYH

Example: Ala-Leu-Tyr-His

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 unchargedpH<pKa : Group predominantly protonated, H+ onpH >pKa : Group predominantly deprotonated, H+ offpH=pKa : Equal amounts of protonated and deprotonated species.

Val-Arg-Gln-Trp-Cys

N terminal pKa ≈ 9.5 > pH,+1 chargeValine (Val): No pKa,UnchargedArginine (Arg): pKa 12.5 > pH, +1 chargeGlutamine (Gln) : cont. with other residues and then draw the structure with predominant charges. Draw with the Fischer projection as below.

𝑝𝐻= 𝑝𝐾𝑎+ logሾ𝐶𝑦𝑠−ሿሾ𝐶𝑦𝑠0ሿ

Then use the calculated ratio to determine the % protonated. Example, if the ratio = 0.4/1, the % ሾ𝐶𝑦𝑠0ሿ =

ሾ𝐶𝑦𝑠0ሿ(ሾ𝐶𝑦𝑠0 +ሾ𝐶𝑦𝑠−ሿሻ𝑥 100 = 0.41+ 0.4𝑥 100 = 29%

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.

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.

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

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.

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.

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.

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.

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 absorptivityis 5,000

Equation: A= Cl

A=Absorbance, = 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)

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/chapter20/electrophoresis.html- VIDEO FOR SDS PAGE electrophoresis

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.

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.

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.

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.

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.

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.