What is the clearance of a substance when its concentration in the plasma is 10 mg/dL, its concentration in urine is 100 mg/dL, and urine flow is 2 mL/min?
Ans. c (20 ml/min) (Ref. Ganong physiology, 21st ed., 806)MEASURING GFR# The glomerular filtration rate (GFR) can be measured in intact experimental animals and humans by measuring the excretion and plasma level of a substance that is freely filtered through the glomeruli and neither secreted nor reabsorbed by the tubules.# The amount of such a substance in the urine per unit of time must have been provided by filtering exactly the number of milliliters of plasma that contained this amount.# Therefore, if the substance is designated by the letter X, the GFR is equal to the concentration of X in urine (UX) times the urine flow per unit of time (V.) divided by the arterial plasma level of X (PX), or UXV/PX. This value is called the clearance of X (CX).In our MCQ, the clearance will be Cx=100 x 2/ 10 = 20 ml/min.# PX is, of course, the same in all parts of the arterial circulation, and if X is not metabolized to any extent in the tissues, the level of X in peripheral venous plasma can be substituted for the arterial plasma level.SUBSTANCES USED TO MEASURE GFR# In addition to the requirement that it be freely filtered and neither reabsorbed nor secreted in the tubules, a substance suitable for measuring the GFR should be nontoxic and not metabolized by the body.# Inulin, a polymer of fructose with a molecular weight of 5200 that is found in dahlia tubers, meets these criteria in humans and most animals and is extensively used to measure GFR.UIN = 35 mg/mLV = 0.9 mL/minPIN = 0.25 mg/mLCIN = UINV----PIN=35 x 0.9------0.25CIN = 126 mL/min# Endogenous creatinine clearance is easy to measure and is a worthwhile index of renal function.# Normal GFR- The GFR in an average-sized normal man is approximately 125 mL/min.- Its magnitude correlates fairly well with surface area, but values in women are 10% lower than those in men even after correction for surface area.- A rate of 125 mL/min is 7.5 L/h, or 180 L/d, whereas the normal urine volume is about 1 L/d. Thus, 99% or more of the filtrate is normally reabsorbed.- At the rate of 125 mL/min, the kidneys filter in 1 day an amount of fluid equal to 4 times the total body water, 15 times the ECF volume, and 60 times the plasma volume.# Control of GFR- The factors governing filtration across the glomerular capillaries are the same as those governing filtration across all other capillaries, i.e.,# The size of the capillary bed,# The permeability of the capillaries, and# The hydrostatic and osmotic pressure gradients across the capillary wall.# For each nephron:- GFR = Kf [(PGC-PT)-(nGC-nT)# Kf, the glomerular ultrafiltration coefficient, is the product of the glomerular capillary wall hydraulic conductivity (ie, its permeability) and the effective filtration surface area.# PGC is the mean hydrostatic pressure in the glomerular capillaries,# PT the mean hydrostatic pressure in the tubule,# pGC osmotic pressure of the plasma in the glomerular capillaries, and# pT the osmotic pressure of the filtrate in the tubule.
2 mL/min
10 mL/min
20 mL/min
200 mL/min
Kidneys and Body Fluids