The following data describe the catalysis of cleavage of peptide bonds in small

Question

The following data describe the catalysis of cleavage of peptide bonds in small peptides by the enzyme elastase. the arrow indicates the peptide bond cleaved in each case. If a mixture of these three substrates was presented to elastase with the concentration of each peptide equal to 0.5 mM, which would be digested more rapidly? Which would be digested more slowly? (Assume the enzyme is present in excess.) Explain your reasoning. On the basis of these data, suggest what features of amino acid sequence dictate the specificity of proteolytic cleavage by elastase.

Explanation / Answer

Ans. Let the maximum velocity of the enzyme for hydrolyzing the most natural substrate (hypothetical) under physiological conditions = Vmax

Michaelis- Menten equation-         

                                                V0 =

Where, V0 = Initial velocity, depends on [S]

            [S] = substrate concentration, decrease enzyme catalyze it into product.

            Vmax = Maximum enzyme velocity conditions.

            Km = MM constant

#1. Using MM equation for substrate 1-

            Vo = Vmax [S] / (Km + [S])

            Or, Vo = (Vmax x 0.5 mM) (4.0 mM + 0.5 mM) = 0.111

So, the maximum velocity for substrate 1 under given conditions, Vmax1 = 0.111 Vmax

#2. Using MM equation for substrate 2-

            Vo = (Vmax x 0.5 mM) (1.5 mM + 0.5 mM) = 0.25 Vmax

Thus, Vmax2 = 0.25 Vmax

#3. Using MM equation for substrate 2-

            Vo = (Vmax x 0.5 mM) (0.64 mM + 0.5 mM) = 0.439 Vmax

Thus, Vmax3 = 0.439 Vmax

Step 2: using Kcat = Vmax/Eo            ; where, Eo = amount of enzyme in reaction nixture.

#1A. For substrate 1.

            26s-1 = Vmax1/ Eo = 0.111 Vmax/ Eo

            Or, Vmax = (26s-1 x Eo ) / 0.111 = 234.23Eo s-1

#2A. For substrate 2.

            37s-1 = 0.25 Vmax/ Eo

            Or, Vmax = (37s-1 x Eo ) / 0.25 = 148.0 Eo s-1

#3A. For substrate 3.

            18s-1 = 0.439 Vmax/ Eo

            Or, Vmax = (18s-1 x Eo ) / 0.439 = 41.00 Eo s-1

Assuming rate of velocity is directly proportion to enzyme concentration, the order of Vamx for substrates (from highest to lowest) is-

            S1 (234.23Eo s-1) > S2 (148.0 Eo s-1) > S3 (41.00 Eo s-1)

Therefore,

Ans. a. substrate 1 will be degraded most rapidly; and S3 degraded slowest.

Ans. b. Glycine is least non-polar among the variable residues G, A and F. Thus, elastase specifically hydrolyzes non-polar amino acids with small side chain (the lease being most effective).

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