Session 4: Group Work Flashcards
What causes the high hydrostatic pressure (Pgc) in the specialised glomerular capillary bed?
The diameter of the afferent arteriole is larger than that of the efferent arteriole.
Name the two forces which oppose the movement of plasma across the leak glomerular capillary wall.
Oncotic pressure from the capillaries Hydrostatic pressure from the lumen of the tubules.
What would happen to GFR if the tubular hydrostatic (Pbc) pressure were to increase?
GFR would decrease due to lower net force into the tubules.
What would happen to GFR if the glomerular oncotic pressure (Pigc) would decrease?
GFR would increase because a higher net force into the tubules.
What does the term myogenic response mean?
A smooth muscular response to a stimulus.
What would happen to Pgc if the smooth muscle in the wall of the AA relaxed but EA tone remained the unchanged?
Pgc would increase and so would GFR
What would happen to Pgc if the smooth muscle in the wall of the AA and EA both relaxed?
It would stay around the same. Up to a point where it would decrease after that.
What would happen to the GFR if there is an increase in renal blood flow?
It would increase
What would the AA wall sense if there was an increase in renal blood flow?
More pressure via baroreceptors
To maintain GFR to its normal ranged what would happen to the AAs in response to an increase in renal blood flow?
The smooth muscle in the wall of the AA would constrict.
What would happen to the GFR if there is a decrease in renal blood flow?
It would decrease
To maintain GFR to its normal ranged what would happen to the AAs in response to a decrease in renal blood flow?
The AA would dilate.
What are the important characteristics of substances that an be used to measure GFR?
Endogenous and constant
Not secreted
Fully filtered
Not reabsorbed
The inulin clearance is a representation of the true GFR.
Plasma creatinine is 0.01 mg/ml.
GFR is 125 ml/min
This means that GFR x plasma creatinine gives the rate of filtration.
125 ml/min x 0.01 mg/ml = 1.25 mg/min
First GFR needs to be calculated by using the inulin measurements we have been given.
(10 x 1) / 0.0833 = GFR
GFR = 120
120 x 0.02 = 2.4
2.4 is the rate of filtration
The urine creatinine is 2.6 mg/ml
2.6-2.4=0.2
Rate of secretion is 0.2 mg/min
