Module 4 Section 6 (Plasma Clearance and Urine Excretion) Flashcards
Describe what is meant by plasma clearance.
A healthy person filters 125 mL/min of plasma, 124 mL/min of which are reabsorbed, which leaves the final amount of urine produced to be 1 ml/min or 1.5 L/day.
During this reduction in volume there is a conc of waste products and other substances that are to be excreted in the urine.
- When we compare the plasma in the renal arteries and veins, many substances have been eliminated in the urine.
- These substances have been “cleaned” or cleared from the plasma.
- This is the concept of plasma clearance which can be defined for any substance as the volume of plasma cleared of that substance by the kidneys per minute.
• It’s the volume of plasma, not amount
• Plasma clearance expresses the effectiveness of the kidneys to remove a substance from the internal fluids.
It is calculated by this equation: Clearance rate (mL/min) = urine concentration (quantity/mL) x urine flow rate (mL/min) / plasma concentration (quantity/mL)
Describe what is meant by the vertical osmotic gradient and why it is important.
The ability to concentrate urine occurs because there is a vertical osmotic gradient in the interstitial fluid of the medulla.
The normal osmolarity of ECF fluids in the body = 300 mOsm/L.
- In the cortex, the ISF osmolarity = 300 mOsm/L but in the medulla, as you move from the cortex to the renal pelvis, there is a gradual gradient as the ISF osmolarity goes from 300 mOsm/L -> 1200 mOsm/L.
- This gradient allows the kidneys to produce urine with an osmolarity range of 100 mOsm/L to 1200 mOsm/L, depending on the hydration state of the body.
Describe the role of the loop of Henle and why its structure is important for its function.
The loops of Henle establish the vertical osmotic gradient, the vasa recta preserve the gradient, and the collecting ducts use the gradient, along with vasopressin, to produce urine of varying concentrations. Collectively, this is known as the medullary countercurrent system.
Describe countercurrent exchange.
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Describe the micturition reflex.
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How is plasma clearance calculated?
It is calculated by this equation:
Clearance rate (mL/min) = urine concentration (quantity/mL) x urine flow rate (mL/min) / plasma concentration (quantity/mL)
The plasma clearance rate varies for different substances, depending on how the kidneys handle each substance. These variations in plasma clearance can be classified into 3 types. What are they?
1) Substances that are filtered, not reabsorbed:
- Insulin is often used as an example to calculate plasma clearance
- It’s freely filtered, so its plasma clearance is used to estimate GFR.
- Since all glomerular filtrate is cleared of inulin, the volume of plasma cleared per min of inulin = the volume of plasma filtered per min.
- Clearance rate for inulin = (30 mg/mL urine x 1.25 mL urine/min) / 0.3 mg/mL plasma = 125 mL/min
- It is difficult to use inulin because it must be continually injected to maintain a steady plasma concentration.
• Therefore, creatine is generally used clinically to determine GFR. As an end product of muscle metabolism, it’s plasma conc is relatively steady and not reabsorbed with only minor secretion.
2) Substances that are filtered and reabsorbed:
- For substances that are reabsorbed, the plasma clearance has to be < than the GFR
Ex: glucose is typically fully reabsorbed so none of the plasma is cleared of glucose. It’s plasma clearance = 0.
- Ex: urea is partially reabsorbed (50%), only about half of the filtered plasma is cleared of urea. It’s plasma clearance is 62.5 ml/min.
3) Substances that are filtered and secreted, not reabsorbed
- For this type of substance, the plasma clearance will be > than the GFR.
- Ex: hydrogen ion tubular secretion. If we take the amount of secreted H+ to be the amount of H+ in 25 mL of plasma, and a GFR of 125 ml/min, we can calculate the plasma clearance of H+ to be 150 ml/min.
Using what you have learned about the molecules involved in determining plasma clearance, determine whether the molecules below are secreted and/or reabsorbed.
1) Hydrogen Ions
2) Glucose
3) Urea
4) Inulin
1) Hydrogen Ions = secreted
2) Glucose = reabsorbed
3) Urea = reabsorbed
4) Inulin = neither
True or false: the fundamental principle of concentrating urine is osmosis.
True
The ECF osmolarity depends on the relative amount of water compared to solute. The same principle applies to tubular fluid
Several unique anatomical arrangements allow the kidneys to use this vertical osmotic gradient to alter the conc of the urine.
What are the structural differences of these 2 different types of nephrons (cortical and juxtamedullary), specifically with reference to the Loop of Henle?
1) Cortical: The loop of Henle only dips slightly into the medulla.
2) Juxtamedullary: The loop of Henle dips all the way down to the renal pelvis. The vasa recta of these nephrons also goes all the way to the renal pelvis.
- Flow in the loop of Henle and the vasa recta goes in opposite directions in what is called countercurrent flow.
In both types of nephrons, the descending collecting ducts that go all the way to the renal pelvis. These anatomical arrangements, coupled with the permeability and transport properties of the different sections of the tubule, are what allow the kidneys to make urine of different concentrations.
We will follow the flow of filtrate through a juxtaglomerular (long loop) nephron to see how this structure establishes the vertical osmotic gradient.
What are the steps involved in establishing this gradient?
1) As soon as the fluid leaves Bowman’s capsule and enters the proximal tubule, there is a strong drive for osmotic reabsorption of water secondary to the active reabsorption of Na (water follows Na).
2) By the end of the proximal tubule, due to Na reabsorption, 65% of the filtrate volume has been reabsorbed. Osmolarity of the tubular fluid at this point = 300 mOsm/L, or isotonic, to other bodily fluids.
3) In the loop of Henle, an additional 15% of the filtered water will be reabsorbed during the establishment and maintenance of the vertical osmotic gradient.
- The descending and ascending limbs of the loop of Henle are distinct in their function.
• Ascending limbs are impermeable to water, but reabsorb N. In this case, water does not follow Na.
• The descending limbs, however, are highly permeable to water, but do not reabsorb Na.
Because the descending and ascending limbs are in close proximity, important interactions occur between them in order to establish the vertical osmotic gradient. Why does this process occur?
This process occurs because filtrate is constantly flowing.
What is the initial step of fluid movement through the loop of Henle and the 6 steps thereafter?
1) Fluid from the proximal tubule entering the descending loop of Henle is 300 mOsm/L and the interstitial space is also 300 mOsm/L.
- The descending limb allows both Na and water to pass, but since it is isotonic to the interstitial space, there is no net movement.
2) Due to the close proximity of the ascending limb, which actively reabsorbs Na but not water, Na moves into the interstitial space.
- Na can move until the ISF is 200 mOsm/L more concentrated than the ascending limb entering the distal tubule. So from the POV of the ascending limb, the tubular fluid is 200 mOsm/L and the ISF is 400 mOsm/L.
- B/c both Na and water can move across the descending limb wall, the osmolarity of the initial part of the descending limb equilibrates (remains isotonic to) the ISF so the tubular fluid in the descending limb is also now 400 mOsm/L.
3) As new fluid moves into the descending loop of Henle, the fluid all shifts forward so we now have 300 mOsm/L entering the descending limb, pushing the 400 mOsm/L fluid deeper into the medulla.
- As the 400 mOsm/L fluid moves around to the ascending limb, Na is reabsorbed until an osmotic difference of 210 mOsm/L is again established.
- Note, however, that we no longer have a 200 mOsm/l difference between descending and ascending limb fluids.
4) While the ascending limb is still transporting Na out, water continues to passively leave the descending limb until the 200 mOsm/L difference b/w the descending and ascending limbs is established at each horizontal level.
- Notice how the conc of the tubular fluid in the descending limb is gradually increasing to remain isotonic to the ISF, yet the fluid in the ascending limb is gradually decreasing to maintain the 200 mOsm/L difference.
5) As fresh 300 mOsm/L fluid enters the descending loop, all the fluid moves forward disrupting the conc gradient at all vertical levels until it again equilibrates.
6) Again, fresh filtrate enters, the osmolarity of the ISF incr, and the osmolarity of the ascending loop fluid decr to maintain the 200 mOsm/L difference.
7) Eventually, equilibrium is achieved such that even with 300 mOsm/L filtrate entering the descending limb, there is a vertical osmotic gradient that results in the tubular fluid being 1200 mOsm/L as it enters the ascending limb, and the tubular fluid is 100 mOsm/L as it enters the distal tubule.
- The maximum osmolarity is 4x > than the normal osmolarity of body fluids, while the osmolarity leaving the ascending limbs is 1/3 of normal osmolarity.
Order the steps of countercurrent multiplication in the loop of Henle:
- 200 mOsm/L gradient is first established between the interstitial fluid and the ascending limb
- Fluid flows forward several frames again
- 200 mOsm/L gradient is established once again at each horizontal level
- Ascending and descending limbs reestablish the 200 mOsm/L gradient
- Fluid flows forward several frames
- Vertical osmotic gradient is established and maintained in an ongoing fashion
1) 200 mOsm/L gradient is first established between the interstitial fluid and the ascending limb
2) Fluid flows forward several frames
3) Ascending and descending limbs reestablish the 200 mOsm/L gradient
4) Fluid flows forward several frames again
5) 200 mOsm/L gradient is established once again at each horizontal level
6) Vertical osmotic gradient is established and maintained in an ongoing fashion
If you consider only what happens to the tubular fluid as it flows through the loop of Henle, the whole process seems to be pointless. Why is that?
The isotonic fluid that enters the loop becomes progressively more concentrated as it flows down the descending limb, only to become progressively more dilute as it flows up the ascending limb.
