Renal Blood Flow
In 70 kg man, renal blood flow is 1000mVmin = 400 m8/ 100 gm:22% of cardiac output.
Table of Contents
oxygen consumption is twice that of the brain, which is mainly related to active sodium reabsorption by Na+ K+ pump and H+ secretion by H+ ATPase pump. Oxygen consumption depends on the Glomerular filtration rate and sodium filtered.
Renal blood flow
Blood is supplied to the kidney by the renal artery.
Distribution of Renal Blood Flow
Peculiarities of Renal Circulation
Inner Medulla is perfused by 1% blood caused by high resistance to flow due to.
- Long length of vasa recta (40mm).
- Increase in blood viscosity.
- Decrease in hydrostatic pressure.
It help in concentration of urine.
Renal circulation is very unusual in that it has 2 set of arterioles (afferent and efferent), and two set of capillaries (Glomerular and peritubular). So it becomes a Portal circulation.
Hydrostatic pressure in glomerulus is very high (60mmhg). It can be regulated by changing the diameter of the afferent and efferent diameter.
Regulation of renal blood flow: – It is regulated by following mechanism.
- Auto regulation
- Myogenic
- Tubuloglomerular feedback
- Sympathetic
- Hormonal and autacoids
- Diet
- Protein meal
- Sugars intake
Auto Regulation
Auto regulation is the regulation of constant blood flow despite changes in arterial blood pressure. Renal blood flow is auto-regulated over wide range of bp (70 – 180 mmHg).
Myogenic Autoregulation: The ability of individual blood vessel to resist stretching during increase blood pressure is known as myogenic autoregulation. Stretching causes calcium influx from extracellular fluid to iniracellular fluid, leading to increased vascular resistance and decrease blood flow.
Tubuloglomerular feedback: It is a feedback mechanism that controls constant blood flow, by sensing sodium chloride concentration at macula densa and accordingly changes the afferent and efferent arteriolar diameter. It ensures constant delivery of sodium chloride at the distal convoluted tubule. It has two components.
- Afferent arteriolar loop
- Efferent arteriolar loop
Sympathetic system:
Mild to moderate Sympathetic stimulation does not affect renal blood flow. But strong stimulation of the renal sympathetic nerve as occurs during severe hemorrhage, brain ischemia and severe exercise, releases epinephrine and norepinephrine. They constrict afferent and efferent arterioles thus decreasing blood flow and GFR.
Hormones and Autacoids:
Diet:
Measurement of Renal Plasma and Blood Flow
Para amino hippuric acid (PAH) clearance method
Fick’s Principle
Para amino hippuric acid (PAII) clearance test.
- PAH is an amide derivative-of the amino acid glycine and para aminobenzoic acid’
- It is not naturally found in humans; so need to be infused intravenously for diagnostic use.
- It is freely filtered, and secreted in urine.
- Its extraction ration is 0.9 i.e. 90% of PAH is removed in single circulation, venous concentration is nearly zero. So it is used to measure renal Plasma flow.
Fick’s principle
It states that amount of substance excreted by kidney per unit time is equal to its arteriovenous difference across the kidney times the blood flow.
Amount of substance excreted by kidney = arteriovenous difference across kidney (A-V) X Renal Blood flow (RBF).
RBF = Amount of substance excreted by kidney / (A-V)
As kidney filters plasma, we can replace RBF by Renal Plasma Flow (RPF)
So RPF : substance excreted by kidney / (A-V).
We know that substance excreted by kidney: Urinary concentration of a substance (Us) X Urinary Flow rate (V)’
RPF = Us X V / (A(PAH) -V(pAH))
PAH is used to measure RPF. Its extraction ratio is 0.9, i.e. 90% PAH is removed from the kidney in a single pass from the kidney so we can consider V(PAH) = 0. If we consider V(PAH) = 0, we denote RPF as Effective Renal Plasma Flow (ERPF).
Also, Kidney arterial plasma concentration of PAH is equal to the Peripheral venous plasma concentration of PAH i.e. we can replace A(PAH) in the equation by P(PHA).
Thus the ERPF : Us X V/ P (PAH)
egP(PAH): 0.0lmg/ml, Us = 7mglml, V = 0.9m1/min
ERPF : (7mg/ml X 0.9m/min) /0.01mg/ml = 630 ml/min
So, after doing cross multiplication, RPF : (ERPF X l)/0.9 = 630/0.9 = 730 ml/min.
As blood consist of plasma and Haematocrit, we can write following equation.
Blood: Plasma (P) + Haematocrit (H)
If we consider blood as I then we can write (1 : P+H) ——- {1}
We can rewrite equation as P : 1-H —- {2}
Considering equation 1 and 2
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