Cardiovascular System Page 2

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6. Given the results for questions 4 and 5 above,
a) How do you explain the discrepancy between the large increase in Dave’s cardiac output and
the relatively small change in his MAP? (Hint: MAP = CO X TPR) . Does Dave’s total peripheral
resistance increase or decrease during exercise?
Since C.O. increases by a large factor and MAP changes relatively little, this must mean that
total peripheral resistance decreases during exercise.
b) What specific physiological mechanism causes this change in peripheral resistance during
exercise?
Resistance decreases due to vasodilation of arterioles in the skeletal muscles during
exercise. This vasodilation results from: (1) local metabolic effects on vascular smooth
muscle (autoregulation); and (2) autonomic (sympathetic) effects on the blood vessels, via
binding of epinephrine and NE to beta-2 receptors in muscle.
Bonus Problem (Recall: concentration (C) = amount of solute (S) ; for dilutions: C
V
= C
V
.)
1
1
2
2
volume of solution (V)
7. Lisa has a plasma volume of 3 liters and a plasma glucose concentration of 50 mg/dL.
a) How many grams of glucose are contained in Lisa’s plasma?
S = C x V
(50 mg/dL) x (10 dL/L) x (3 L) = 1,500 mg = 1.5 g
b) Suppose a doctor determines that Lisa needs 3 more grams of glucose in her plasma.
What volume of a 5 g/dL (5%) glucose solution would be needed to deliver 3 grams of glucose?
Rearranging the equation to solve for volume:
V = S / C
(3 g) / (5 g/dL)
= 0.6 dL = 60 mL
c) Suppose Lisa receives an intravenous (IV) infusion of 50 mL of 5% glucose solution (5 g/dL).
How many grams of glucose has she received in this IV infusion?
S = C x V
(5 g/dL) x (50 mL) x (1 dL/100 mL)
= 2.5 g
d) Suppose you have 1 mL of a 20 g/dL glucose solution and you dilute it by adding 4 mL of
distilled water. What is the final volume and final concentration of your dilution?
Final volume (V
) = 1 mL + 4 mL = 5 mL
2
Rearrange the dilution equation to solve for the final concentration, C
= C
V
/ V
2
1
1
2
= (20 g/dL) x (1 mL) / (5 mL) = 4 g/dL

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