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2. Data interpretation | ![]() |
These questions have (and still be) used in examinations. You should be able to answer them in about 10 minutes. You should not have to write more than a few lines for an component of a question. |
question 1. | Top |
question 2. | Top |
NOTE: The question is marked out of 10 in total. Numbers in ( ) give the marks available for that element of the question. A. Describe the graph - Note: Cimetidine is a Histamine type 2 (H2) receptor blocker (1) B. What does line A show (0.5) C. What does line B show (0.5) D. Explain the rise in the concentration of secretin rise in A. (3) E. Where is the site of action of cimetidine (organ and cell type) (2) F. Explain the action of cimetidine on plasma [secretin] in B (2) G. What would be the likely consequences of taking cimetidine on gastric acid secretion. (1) |
question 3. | Top |
You will find this question quite taxing - you will
need to think and you may need to consult a text book or two.
If you can solve it you're doing well.
Note: The problem makes a big assumption. It
assumes that there is only one diffusional barrier between the
blood plasma and the gastric lumen. It assumes also that there
is only 1 cell (2 membranes) separating the blood and gastric
lumen. In reality there are 2 cells - the vascular endothelium,
the basement membrane and then the parietal cell.
Our subject is a healthy human being with a blood
plasma pH of 7.4. The intracellular [H+] of the parietal
cells is 10-10 M and the [H+] is 10-4
M in the gastric juice (a 1 million-fold difference in concentration).
The active transport of H+ against a 1 million-fold
concentration gradient require 85 kJ.mol-1 of H+.
This concentration of H+ from the blood
to the gastric lumen will obviously require energy, and glucose
is assumed to deliver that energy. One gram of glucose will produce
15.5 kJ of energy when oxidised (the molecular weight of glucose
is 180 g.mol-1). Oxidation needs oxygen!!!
It is assumed that 1 litre of oxygen (standard temperature
and pressure, dry; STPD) is equivalent to 20 kJ (for a person
burning a mixed diet). During a histmine test (remember histamine
will stimulate acid secretion) the total mass of perietal
cells secrete 35 mmol H+ h-1.
equilibrium potential (@ 37 o C) = 60
x log ([H]out/[H]in) |
question 4. | Top |
The human gut is about 6.5 m long with about 75% being the small intestine. The surface area available for absorption is increased by throwing the mucosa into folds, having villi (and crypts in the ileum), and by the micovilli brush border on the epithelial cells. Let’s assume the average diameter of the gut lumen is 3 cm and that the area of the lumen is increased x200 by the folds and villi. Let’s also assume that this area is covered in epithelial cells which for our purposes are 10 microns in diameter and square in cross-section (to pack together neatly). Each epithelial cell is replaced every 72 hours (3 days). Calculate:
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Copyright © 1998 University of Bristol. All rights reserved.
Author: Phil Langton |
Last modified: 11 Jun 1999 07:20 Authored in CALnet |