Protein Structure and Analysis Exercise

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Question 1

Briefly describe why hydrophobic interactions are the primary driving force involved in protein folding. How do they counter act the thermodynamically unfavorable effect of a polypeptide chain being folded into a more ordered conformational?

When the protein folds into well defined structure, the majority of the hydrophobic side chains cluster together within the core of the protein. The rapid collapse of polypeptide chain, therefore largely reducing the configurational space to explore. After collapse, the side chains then repack more efficiently as they search a limited conformational space in the molten globule state. The removal of the hydrophobic side chain and the sequestration from the solvent generates sufficient energy to maintain the folded structure of the protein. The hydrophobic interaction is also a major stabilizing force contributing to the thermodynamic stability of the folded state (Szila´gyi et al., 2007).

Question 2

Briefly describe why polypeptides chains can fold into only a few different types of secondary structure (i.e. alpha-helices, beta sheets, beta turns, and loops).

As proteins fold, they test a variety of conformations before reaching their final form, which is unique and compact. Folded proteins are stabilized by thousands of non-covalent bonds between amino acids. In addition, chemical forces between a protein and its immediate environment contribute to protein shape and stability. A protein with an unpredictably variable structure and biochemical activity is unlikely to help the survival of a cell that contains it. Such proteins would therefore have been eliminated by natural selection through the enormously long trial-and-error process that underlies biological evolution. Proteins are so precisely built that the change of even a few atoms in one amino acid can sometimes disrupt the structure of the whole molecule so severely that all function is lost (Albert et al., 2002).

α-lactalbumin (14 200) Subunit L (19000)

Subunit H (19000) Tripsinogen (24000)

Egg albumin (45000) Bovine serum albumin (65000)

Ferritin (500000)

Question 3

You are instructed to run a gel-exclusion chromatography Sephadex G-50 column on the mixture containing the components below. All three components are colored, so you are able to observe the separation. The components are dissolved in water, and the column is eluted with water.

Blue dextran, a blue dye (MW = 2,000,000)

Cytochrome c, a red protein (MW = 12,400)

Flavin adenine dinucleotide, a yellow coenzyme (MW = 830)

Predict your result and explain your prediction.

Gel exclusion chromatography seperates molecules according to their size and shape. The seperations, with some exceptions, correlates with their molecular weight. Smaller particles spent more time with the pore of the beads and hence, move more slowly, than larger molecules. Higher molecular weight particles are eluted faster than the smaller particles (Priciples of gel filtration, n.d.). Thus, blue dextran will be eluted first, follow by cytochrome c and flavin adenine dinucleotide consequently. The largest size of blue dextran makes it unable to fit into the pores of the beads. Hence, it only flow through the beads without any interactions and elute fast. The same applies to flavin adenine dinucleotides, which has the smallest size among the others. It is retarded within the pores of the beads and hence, elute last. Cytochrome c has an intermediate size between blue dextran and flavin adenine dinucleotide. The elution time is in between that of blue dextran and flavin adenine dinucleotide.

Question 4

You have just completed an experiment using SDS-PAGE with the following proteins of know molecular weight. Draw the predicted result from the gel.





Subunit H


Subunit L






Egg albumin


Bovine serum albumin


SDS (sodium dodecyl sulfate) denatures the protein and the molecules are now negatively charge. When they are put into an electric field, they move towards the positive pole with separation by different sizes. Small molecules can move through the polyacrylamide gel faster than big molecules (SDS-PAGE (PolyAcrylamide Gel Electrophoresis, n.d.). Hence, the predicted result is α-lactalbumin, subunit L, subunit H, tripsinogen, egg albumin, bovine serum albumin and ferritin.

Question 5

You have successfully purified a protein to homogeneity. From gel permeation chromatography studies, you determine that the protein has a molecular weight for 235,000 daltons. When you run the protein on a SDS PAGE gel you detect the presence of three protein subunits. The α subunit has a molecular weight of 50,000 daltons, the ß subunit has a molecular weight of 15,000 daltons and the γ band has a molecular weight of 5,000 daltons. From this data determine the number of α, ß and γ subunits present in the native protein. Explain your results.

Total molecular weight of the native protein =235,000 Daltons.

There is at least one subunit for each α, β, and γ, then

the total molecular weight of α, β, and γ =50,000 Da + 15,000 Da + 5,000 Da

= 70,000 Da.

The molecular weight of remaining subunits =235,000 Da – 70,000 Da

= 165,000 Da.

Another three α subunits and one β subunits made up the remaining 165,000 Da. Hence, there are four α subunits, two β subunits and one γ subunit in the native protein.

Question 6

You are sequencing a polypeptide chain. You digest the polypeptide with trypsin to generate a mixture of six peptide fragments. You sequence the six peptides and come up the following sequences:


From the sequence of the five peptides, you suspect that the trypsin you used to digest the original polypeptide was contaminated with another enzyme. What enzyme is contaminating your trypsin enzyme stock? Explain how you came to this conclusion.

Trypsin only cleaves the peptide bonds after (on the C-terminal side of) the basic amino acids lysine and arginine. As a result, every peptide fragment generated by trypsin will have Arg (R) and Lys (K) at their C-terminus (Digestive enzymes, specificity and pH: Theory, n.d.).


M-G-P-A-D-L-I-R + H-V-E-S-K + D-T-C-K-M-R + A-A-E-Y-H-H-F-H-G

In this case, two more residue fragments appeared with different cleavage neither on the carbonyl sides of R or K.We suspect that the trypsin used to digest the original polypeptide was contaminated with another enzyme, which is the chymotrypsin. Chymotrypsin hydrolyzes the peptide bonds that follow large hydrophobic residues, e.g. Phenylalanine (F), Tyrosine(Y), Tryptophan(P) or Leucine(L) at their C-terminus. In summary, trypsin enzyme stock was contaminated by chymotrypsin enzyme because chymotrypsin and trypsin are serine proteases that have similar structure.

Trypsin and chymotrypsin are structurally very similar, although they recognise different substrates, both with extraordinary catalytic efficiency (McDowall, n.d.).

M-G-P-A-D-L-I-R- H-V-E-S-K-D-T-C-K-M-R-A-A-E-Y-H-H-F-H-G

M-G-P-A-D-L-I-R-H-V-E-S-K-D-T-C-K-M-R-A-A-E-Y + H-H-F + H-G

Hence, there are extra two fragments H-H-F and H-G from the chymotrypsin cleavage other than the trypsin cleavage.

Question 7

The initial rate for an enzyme catalyzed reaction has been determined at a number of substrate concentrations. Data are as follows: (20 marks)

[S] (µmoles/L)

V (µmoles/L/min)













  1. Use a Lineweaver-Burk plot (1/[S] vs 1/V) to determine Vmax and km. and label clearly the value and graph plotted.

b. Draw a line that would reflect what would happen if a competitive inhibitor were added. What is happening to the km and Vmax? Understand why these changes are occurring.

km increases while Vmax remain constant. In competitive inhibition, the inhibitor can only bind to free enzyme molecules which have not bound to any substrate (Moran et al., 2012). Therefore, the competitive inhibitor competes with the substrate for active sites. This cause reduces in the availability of free enzymes for the substrate binding and thus the concentration of substrate increases. km is the concentration of the substrate when the reaction velocity is half of the maximum velocity. Therefore, km increases when the concentration of substrate increases.


c. Draw a line that would reflect what would happen if an uncompetitive inhibitor were added. What is happening to the km and Vmax ? Understand why these changes are occurring.

Uncompetitive inhibitor does not have similar structure with the substrate. I may bind the free enzyme or enzyme substrate complex that is exposed to the inhibitor binding site. Although it binds away from the active site, the binding will cause structural distortion of the active and allosteric sites of the complexed enzyme and thus inactivates the catalysis. This results in a decrease in both km and Vmax (Sharma, 2012).

d. Draw a line that would reflect what would happen if a non-competitive inhibitor were added. What is happening to the km and Vmax? Understand why these changes are occurring.

Non-competitive inhibitor does not have structural similarity to the substrate. However, it can bind with both the free enzyme and the enzyme-substrate complex. Thus, its binding manner is not mutually exclusive with the substrate. Consequently, the presence of a non-competitive inhibitor has no influence on the ability of the substrate to bind with an enzyme and vice versa. However, although its binding is away from the active site, it will alter the conformation of the enzyme and reduces its catalytic activity because of the changes in the nature of the catalytic groups at the active site. As a result, km remains unchanged and Vmax decreases.

Question 8

The rate of an enzymatic reaction is measured with three similar but different substrates. From the experiments, the km and kcat values were determined for each substrate.





km (mM)




kcat (s-1)




kcat/ km (mM-1s-1)




Based on this data, answer the following questions:

  1. Which substrate binds tightest to the enzyme? Explain your answer.

Dibenzofuran binds the tightest to the enzyme. When the substrates are tightly bound to enzyme, k-1 and k2 decrease. From the formula km = (k-1+ k2) / k1, km is directly proportional to k-1 and k2. As the substrate tightly bound to enzyme give small value of k-1 and k2, which results in small value of km.

(b) Which substrate is the best substrate for the enzyme? Explain your answer. Biphenyl is the best substrate for the enzyme. The value of kcat/ km is used to measure the catalytic efficiency of an enzyme (Enzyme Kinetics, n.d.). High value of kcat and small value of km raised the value of kcat/ km. The value of kcat is known as the turnover number as it describes the number of reaction a molecule of enzyme can catalyze per second under optimal condition. As the value of kcat increases, the value of kcat/ km increase as well. Since the value of kcat/ km for biphenyl is highest among these three, biphenyl is the best substrate for the enzyme.

Question 9

Explain why very tight binding of substrate to an enzyme is not desirable for enzyme catalysis, whereas tight binding of the transition state is desirable.

A very tight binding of substrate to an enzyme formed a very stable enzyme-substrate complex. This results in very low energy and hard to reach the transition state. Therefore, the reaction is slow down due to the effectively increase of activation energy (KE0026 Biochemistry Exercises, n.d.).

A tight but not very tight binding of the transition state is necessary because it prevent the substrate ‘escape’ from the enzyme during the transition state. Besides, an adequate tightness can stabilize and lower down the substrate-enzyme complex. Therefore, the rate of reaction increased due to the effectively decreased in activation energy (Bke2 Biochemistry Exercises, n.d.).

Question 10

a) From the information in the table, calculate the specific activity (units/mg protein), yield (%) and purification fold of the enzyme solution after each purification step.


Total Protein (mg)

Activity (units)

Specific Activity (units/mg protein)

Yield (%)

Purification fold

Crude homogenate






(NH4)2SO4 precipitation






Acid precipitation






Ion-exchange chromatography






Affinity chromatography






Gel-filtration chromatography






Specific activity = Activity / Total Protein

Purification fold = (Final specific activity / Initial specific activity)

Yield = Final total activity unit / Initial total activity unit 100%

b) Which of the purification procedures used for this enzyme is most effective – gives the greatest increase in purity?

Affinity chromatography is the most effective purification procedure, it acquired greatest purity fold as well as the yield.

c) Which of the purification procedures is least effective? Is there any indication from the table that the enzyme is now pure? What else could be done to estimate the purity of the final preparation?

Gel-filtration is the least effective because its specific activity remained constant similarly to affinity chromatography.

There is no indication of pure purification of enzyme. The constant specific activity of gel-filtration remains constant which indicate that the remaining contaminants cannot be resolved on a size basis or the enzyme is close to homogeneity.

SDS-PAGE, 2D gel-electrophoresis and capillary isoelectric focusing (cIEF) can be use for estimation of purity.

d) By how much has the enzyme been purified overall? Assuming that the enzyme is now pure, what percentage of the total protein in the original crude homogenate does the enzyme represent? What assumptions must you make in order to perform this calculation? Hint: what are the recoveries of activity after each purification step?

The overall purification:

Final specific activity / Initial specific activity

= (15,000 units/mg)/(200 units/mg)

= 75-fold

Assume 100% of the enzyme's activity in the starting material had been recovered, then, the enzyme would, on the basis of these data, represent

of this material.

Overall recovery of enzyme activity: 675,000 units/4000,000 units = 16.9%.

The actual amount of enzyme in the starting material: (1.33)(100/16.9)% = 7.9%

Assume pure protein= 45 mg; Total activity= 675,000 units; Recovery yield= 16.9% of its activity

Theoretical activity: (100/16.9)(675,000)= 3,994,083 units

Theoretical specific activity: 3,994,083 units/45 mg= 88,757 units/mg

In short, we can only estimate that the abundance of the enzyme protein is between 45/20,000 = 0.225% and 7.9%. 0.225% is undoubtedly too low; 7.9% is undoubtedly too high. The former calculation assumes no loss of specific activity while the latter assumes no loss of enzyme protein. The truth is likely to be somewhere in between; namely losses of both specific activity and enzyme protein.


Albert. B., Johnson, A., Lewis, J., Raff, M., Roberts, K., & Walter, P. (2002). Biology of the cell (4th ed.). Retrieved from

Bke2 Biochemistry Exercises. (n.d.). Retrieved from

Digestive enzymes, specificity and pH: Theory. (n.d.). Retrieved from

Enzyme Kinetics. (n.d.). Retrieved from

KE0026 Biochemistry Exercises. (n.d.). Retrieved from

McDowall, J. (n.d.). Trypsin and Chymotrypsin. Retrieved from

Moran, L. A., Horton, H. R., Scrimgeour, K. G., & Perry, M. D. (2012). Principles of Biochemistry. New York, N. Y.: Pearson Education Inc. Principles of gel filtration. (n.d.). Retrieved from

SDS-PAGE (PolyAcrylamide Gel Electrophoresis). (n.d.). Retrieved from

Sharma, R. (2012). Enzyme Inhibition: Mechanisms and Scope, Enzyme Inhibition and Bioapplicatoins. Rijeka, Croatia: Intech Europe.

Szila´gyi, A., Kardos, J., Osva´th, S., Barna, L., & Za´vodszky, P. (2007). Protein Folding. Retrieved from