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Conformational instability of the N‐ and C‐terminal lobes of porcine pepsin in neutral and alkaline solutions

1993; Wiley; Volume: 2; Issue: 9 Linguagem: Inglês

10.1002/pro.5560020903

1469-896X

Xinli Lin, Jeffrey A. Loy, Fredy Sussman, Jordan Tang,

Enzyme Production and Characterization

Abstract Pepsin contains, in a single chain, two conformationally homologous lobes that are thought to have been evolutionarily derived by gene duplication and fusion. We have demonstrated that the individual recombinant lobes are capable of independent folding and reconstitution into a two‐chain pepsin or a two‐chain pepsinogen (Lin, X., et al., 1992, J. Biol. Chem. 267 , 17257–17263). Pepsin spontaneously inactivates in neutral or alkaline solutions. We have shown in this study that the enzymic activity of the alkaline‐inactivated pepsin was regenerated by the addition of the recombinant N‐terminal lobe but not by the C‐terminal lobe. These results indicate that alkaline inactivation of pepsin is due to a selective denaturation of its N‐terminal lobe. A complex between recombinant N‐terminal lobe of pepsinogen and alkaline‐denatured pepsin has been isolated. This complex is structurally similar to a two‐chain pepsinogen, but it contains an extension of a denatured pepsin N‐terminal lobe. Acidification of the complex is accompanied by a cleavage in the pro region and proteolysis of the denatured N‐terminal lobe. The structural components that are responsible for the alkaline instability of the N‐terminal lobe are likely to be carboxyl groups with abnormally high p K a values. The electrostatic potentials of 23 net carboxyl groups in the N‐terminal domain (as compared to 19 in the C‐terminal domain) of pepsin were calculated based on the energetics of interacting charges in the tertiary structure of the domain. The groups most probably causing the alkaline denaturation are Asp 11 , Asp 159 , Glu 4 , Glu 13 , and Asp 118 . Especially, the partially buried Asp 11 , which interacts with Asp 159 , could cause one of these two groups to have an abnormally high p K a and the other an abnormally low p K a value. Thus, the ionization of Asp 11 at a high pH may place two negatively charged residues in close vicinity. This unfavorable situation may be the trigger for the denaturation of the N‐terminal lobe of pepsin.