While you were on safari, your friend got bit on the leg by a posionous snake and nearly died from excessive hemolysis. Being a true scientist, you captured the snake and analyzed its venom to understand why it caused hemolysis. You found thwo enzymes, a protease (breaks down proteins), and a phospholipase. He isolated each of these enzymes and treated red blood cells with them. He found that only phospholipase caused red blood cell lysis. He also found that, after treatment of red blood cells with the phospolipase, there was excess amounts of free phosphophorylcholine (Choline phosphate group) and free diacylglycerol (glycerol with two hydrocarbon tails). 1.) What is the substrate for phospholipase and where does it cleave

Phospholipase is an enzyme, which is classified into phospholipase A2 and phospholipase A1.

In this case we are talking about phospholipase A2, that is why the sn-2 ester bond of glycerophospholipids will be taken as a substrate, giving fatty acids as a product, such as arachidonic acid.

Explanation:

The glycerophospholipids used by this enzyme are those found in cell membranes, these pathways are usually associated with inflammation pathways for the formation of arachidonic acid that functions as an immunological mediator.

On the other hand, it must be remembered that membrane phospholipids are amphipathic, that is, with hydrophilic and hydrophobic molecules and are responsible for balancing the internal homeostasis of the intracellular environment.