Task 5 Biochem

The intrinsic pathway is activated when the blood is exposed to anionic surfaces and this occurs when the blood vessel ruptures.

Extrinsic pathway triggered upon exposure of TF which is not normally present in the blood.

INTRINSIC: XII to XIIa which activates XI to XIa which activates IX to IXa which activates X to Xa.

EXTRINSIC:

Final common pathway: Prothrombin is activated to Thrombin by Xa, and Thrombin then activates Fibrinogen to Fibrin which is crosslinked into a fibrin clot by factor XIIIa

They are a class of enzymes that catalyze the hydrolysis of peptide bonds - serine comes from the fact that the serine residue is the one that catalyzes the reaction. The -OH on the serine is not a strong nucleophile on its own, it becomes an alkoxideion from the shift in electron density that occurs: Asp 102 carboxyl group pulls the H away from the Nh of His57 which then pulls on the H of the OH of Serine 192. In essence, the His57 is positioned in the best way to deprotonate the Serine 192 through the positioning by Aspartate.

Cleaves peptide bonds on the C terminal side of large hydrophobic residues: the scissile bond after the aromatic amino acid, specifically on the side of the peptide chain where the carboxyl (-COOH) group of the aromatic residue resides.

Step 1 is the nucleophilic attack from the alkoxide ion which allows the substrate to bind to Serine which dissociates from the triad. The recognition occurs at the N terminal. This displaces the double bond between the C and the O and the nitrogenn closest to the serine accepts the proton.

Step 2 is the formation of the tetrahedral intermediate where the O- is stabilized by an oxyanion hole - however, this intermediate is very unstable and quickly breaks down.

Step 3 occurs when electrons forming the C-N bond of the protein are displaced to form a bond with H and this forms the acyl enzyme (serine + acyl group of the protein). This is facilitated by the positive charge on the histidine, which helps trnasfer the H to the amino group. They remain hydrogen bonded.

Step 4 The amine component of the protein is freed - the acylation of the enzyme; aka the first step is done.

Step 5 Water will take the place of amine in the following reactions resulting in a new tetrahedral intermediate stabilized by an oxyanion hole (hydrogen bonds from other residues close by). The N from imidazole ring deprotonates water and the remaining OH displaces the C=O double bond, forming the intermediate.

Step 6 The tetrahedral intermediate collapses as the bond linking serine to the acyl is displaced, again facilitated by the positive charge on the imidazole ring which donates the H back to serine.

Step 7: Release of the carboxylic acid and the enzyme is back in its active form.