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Review Question - QID 100096

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QID 100096 (Type "100096" in App Search)
A 55-year-old woman presents with acute onset abdominal pain radiating to her back, nausea, and vomiting. CT scan suggests a diagnosis of acute pancreatitis. The pathogenesis of acute pancreatitis relates to inappropriate activation of trypsinogen to trypsin. Which of the following activates trypsin in normal digestion?

Amylase

4%

14/354

Lipase

5%

19/354

Cholecystokinin

7%

25/354

Enterokinase

80%

283/354

Secretin

3%

9/354

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Enterokinase (also called enteropeptidase) is an enzyme produced by duodenal epithelial cells that cleaves inactive pancreatic zymogen trypsinogen into its active form, trypsin. Trypsin in its active form is involved in breakdown of dietary amino acids and cleavage of other dietary proteases.

There are two forms of amylase, pancreatic amylase and salivary amylase. Salivary amylase breaks down starch to disaccharides. Pancreatic amylase breaks down starch to oligosaccharides and disaccharides. Lipase is a secreted pancreatic enzyme that breaks down fats.

Incorrect Answers:
Answer 1: Amylase degrades complex carbohydrates into mono-, di-, and oligosaccharides.
Answer 2: Lipase is necessary for triglyceride breakdown into monoglycerides and fatty acids.
Answer 3: Cholecystokinin stimulates bile secretion by the gallbladder in response to a fatty meal.
Answer 5: Secretin is released by the S cells of the duodenum, and regulates the pH of duodenal contents. It functions to increase pancreatic bicarbonate secretion and decrease gastric acid secretion. By decreasing gastric acid secretion and increasing bicarbonate production, it neutralize the gastric acid in the duodenum and hence allows the pancreatic enzymes to function (if not, the acidic environment can cause the pancreatic enzymes to degrade).

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