Catalysis of the cleavage of peptide bonds in small peptides is described in the

Question

Catalysis of the cleavage of peptide bonds in small peptides is described in the following table. The arrow indicates the peptide bond cleaved in each case If the enzyme were added to a mixture of these peptides, with the concentration of each peptide being the same, which peptide would be digested most slowly, and which most rapidly? Substrate peptide MDSAIA MDSALF MDSAIG MDSALV 1.5 0.4 4.0 0.5 30 28 24 18 Most slowly: MDSAF; most rapidly: MDSAG Most slowly: MDSAA; most rapidly: MDSAV Most slowly: MDSAG; most rapidly: MDSAF They will all be digested at the same rate. Most slowly: MDSAV, most rapidly: MDSAA

Explanation / Answer

Ans. When determining the relative efficiency of the same enzyme towards different substrate, the term “catalytic efficiency” is preferred.

Catalytic efficiency = Kcat / Km

Greater is the value of catalytic efficiency for a substrate, the more efficient is the enzyme towards catalysis of that substrate.

#I. Catalytic efficiency for MDSA/A = 30 s-1 / 1.5 mM = 20.0 mM-1 s-1

#II. Catalytic efficiency for MDSA/F = 28 s-1 / 0.4 mM = 70.0 mM-1 s-1

#III. Catalytic efficiency for MDSA/G = 24 s-1 / 4.0 mM = 6.0 mM-1 s-1

#IV. Catalytic efficiency for MDSA/V = 18 s-1 / 0.5 mM = 36.0 mM-1 s-1

Result: Peptide MDSA/F with highest catalytic efficiency will be digested most rapidly. Peptide MDSA/G with lowest catalytic efficiency will be digest most slowly.

# So, correct option is- C. Most slowly: MDSAG           , Most rapidly: MDSAF

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