BIOL 2420 Lecture Notes - Lecture 8: Renal Function, Inulin, Creatinine
26 Apr 2018
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Unit 7 – Lecture 8
Clearance Is a Noninvasive Way to Measure GFR
- Clearance – the rate at which that solute disappears from the body by excretion or by
metabolism
o The general equation for clearance is:
Clearance of X =
o Clearance is mL plasma cleared of X per minute
o The units for clearance are mL plasma and time
o Substance X does not appear anywhere in the clearance units
- Any solute that is cleared only by renal excretion, clearance is expressed as the volume of
plasma passing through the kidneys that has been of that solute in a given period of time
o This is such an indirect way to think of excretion (how much blood has been cleared of X
rather than how much X has been excreted)
o Clearance is often a difficult concept to grasp!!
- Example: inulin
o Polysaccharide isolated from the tuberous roots of a variety of plants
o Inulin injected into the plasma filters freely into the nephron
o As it passes through the kidney tubule, inulin is neither reabsorbed nor secreted
o 100% of inulin that filters into the tubule is excreted
o if you inject inulin so the plasma concentration is 4 inulin molecules per 100 mL plasma
▪ is GFR is 100 mL plasma filtered per minute
▪ then filtration rate of inulin is:
filtered load of X = [X]plasma X GFR
= (4 inulin/100 mL plasma) X 100 mL plasma filtered/min
= 4 inulin/ min filtered
• see image on next page
- as filtered inulin and filtered plasma pass along the nephron, all the plasma is reabsorbed, but
all the inulin remains in the tubule
o reabsorbed plasma contains no inulin, so it has been totally cleared of inulin
▪ inulin clearance is 100 mL of plasma cleared/min
▪ excretion rate of inulin is 4 inulin molecules excreted per minute
- this information can be used to calculate the glomerular filtration rate
o for any substance that is freely filtered but neither reabsorbed nor secreted, its
clearance is equal to GFR
Mathematically proving inulin clearance is equal to GFR
1) filtered load of X = [X]plasma X GFR
2) filtered load of inulin = excretion rate of inulin
o 100% of the inulin that filters into the tubule is excreted
3) excretion rate of inulin = [inulin]plasma X GFR
o can be rearranged to give 4)
4) GFR =
o This is identical to the equation for clearance equation for inulin
5) Clearance of X =
find more resources at oneclass.com
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6) For inulin: inulin clearance =
7) GFR = inulin clearance
o This is because the two sides of equations 4 and 6 are identical!
- This shows that GFR can be measured in a living human by only blood and urine samples
- Inulin is not practical for routine clinical applications because it does not occur naturally in the
body and must be administered by continuous intravenous infusion
o Inulin is therefore limited to research
o No substance that occurs naturally in the human body is handled by the kidney exactly
the way inulin is handled
- In clinical settings, physicians use creatinine to estimate GFR
o Creatinine: a breakdown product of phosphocreatine, an energy-storage compound
found primarily in muscles
▪ Constantly produced by the body and need not be administered
▪ Normally the production and breakdown rates of phosphocreatine are relatively
constant, and the plasma concentration of creatinine does not vary by much
▪ Although always present in the plasma and ease of measure, it is not the perfect
molecule for estimating GFR because a small amount is secreted into the urine
• The amount secreted is small enough that in most people, creatinine
clearance is routinely used to estimate GFR
Clearance Helps Us Determine Renal Handling
- Oe we kow a perso’s GFR – we can determine how the kidney handles any solute by
easurig the solute’s plasa oetratio ad its eretio rate
- If we assume that the solute is freely filtered at the glomerulus, we know that:
Filtered load of X = [X]plasma X GFR
- By comparing the filtered load of the solute with tis excretion rate, we can tell how the nephron
handled that substance
o Example
▪ Less of the substance appears in the urine than was filtered
• net reabsorption occurred (excreted = filtered-reabsorbed)
▪ more of the substance appears in the urine than was filtered
• net secretion of the substance into the lumen (excreted = filtered +
secreted)
▪ same amounts of substance filtered and excreted
• substance is handled like inulin – not reabsorbed or secreted
- example
o suppose glucose is present in the plasma at 100 mg glucose /dL plasma
▪ GFR is calculated from creatinine clearance to be 125 mL plasma/min
▪ Filtered load of glucose = (100 mg glucose/ 100 mL plasma) X 125 mL
plasma/min
= 125 mg glucose/min
▪ No gluose i this perso’s urie
• Glucose excretion is zero
• Glucose was filtered at a rate of 125 mg/min but excreted at a rate of 0
mg/min
o It has to have been totally reabsorbed
- Clearance values can be used to determine how the nephron handles a filtered solute
find more resources at oneclass.com
find more resources at oneclass.com
Document Summary
Clearance is a noninvasive way to measure gfr. Clearance the rate at which that solute disappears from the body by excretion or by metabolism: the general equation for clearance is: Clearance of x = (cid:3032)(cid:3045)(cid:3032)(cid:3047)(cid:3042)(cid:3041) (cid:3045)(cid:3047)(cid:3032) (cid:3042)(cid:3033) (cid:3288)(cid:3171)i(cid:3172) [](cid:3287)(cid:3288)(cid:3288)(cid:3288)(cid:3043)(cid:3039)(cid:3046)(cid:3040: clearance is ml plasma cleared of x per minute, the units for clearance are ml plasma and time, substance x does not appear anywhere in the clearance units. Example: inulin: polysaccharide isolated from the tuberous roots of a variety of plants. = (4 inulin/100 ml plasma) x 100 ml plasma filtered/min. [(cid:3041)(cid:3048)(cid:3039)(cid:3041)](cid:3287)(cid:3288: clearance of x = (cid:3032)(cid:3045)(cid:3032)(cid:3047)(cid:3042)(cid:3041) (cid:3045)(cid:3047)(cid:3032) (cid:3042)(cid:3033) (cid:3288)(cid:3171)i(cid:3172) [](cid:3287)(cid:3288)(cid:3288)(cid:3288)(cid:3043)(cid:3039)(cid:3046)(cid:3040: this is identical to the equation for clearance equation for inulin, for inulin: inulin clearance = (cid:3032)(cid:3045)(cid:3032)(cid:3047)(cid:3042)(cid:3041) (cid:3045)(cid:3047)(cid:3032) (cid:3042)(cid:3033) (cid:4666)(cid:3040)(cid:3034)/(cid:3040)(cid:3041)(cid:4667) [](cid:3287)(cid:3288)(cid:4666)(cid:3040)(cid:3034)/(cid:3040) (cid:3043)(cid:3039)(cid:3046)(cid:3040)(cid:4667: gfr = inulin clearance, this is because the two sides of equations 4 and 6 are identical! This shows that gfr can be measured in a living human by only blood and urine samples.
