Module 4 Section 6

Question 1

plasma clearance

Answer 1

• a healthy person filters 125ml/min of plasma and 124ml/min are reabsorbed, leaving final amount of urine produced to be 1 ml/min or 1.5 L/day
• during this reduction in volume there is a concentration of waste products and other substances that are to be excreted in the urine
• when we compare the plasma in the renal arteries to the renal veins, many substances have been eliminated in the urine
• these substances have been “cleaned” or cleared from the plasma
  plasma clearance is the volume of plasma cleared of that substance by the kidneys per minute and expresses the effectiveness of the kidneys to remove a substance from the internal fluids

Question 2

types of plasma clearance

Answer 2

1. substances that are filtered, not reabsorbed
2. substances that are filtered and reabsorbed
3. substances that are filtered and secreted, not reabsorbed

Question 3

filtered, not reabsorbed

Answer 3

• it is freely filtered, but neither reabsorbed or secreted, so its plasma clarance is used to estimate glomerular filtration rate
• since all glomerular filtrate is cleared of inulin, the volume of plasma cleared per minute of inulin equals the volume of plasma filtered per minute
• it is difficult to use inulin because it must be continually injected to maintain a steady plasma concentration
• the endogenous substance creatine is generally used clinically to determine glomerular filtration rate
• as an end product of muscle metabolism, its plasma concentration is relatively steady and not reabsorbed with only minor secretion

Question 4

clearance rate for inulin

Answer 4

= (30 mg/ml urine x 1.25 ml urine/min) / 0.3 mg/ml plasma
= 125 ml/min

Question 5

filtered and reabsorbed

Answer 5

• for substances that are reabsorbed, the plasma clearance has to be less than the glomerular filtration rate
• for glucose, it is normally totally reabsorbed so none of the plasma is cleared of glucose. plasma clearance is 0
• for urea, which is partially reabsorbed only about half of the filtered plasma is cleared of urea. plasma clearance is 62.5 ml/min

Question 6

filtered, secreted, not reabsorbed

Answer 6

• the plasma clearance will be greater than the glomerular filtration rate
• if we take the amount of secreted H+ to be the amount of H+ in 25ml of plasma, and the glomerular filtration rate of 125 ml/min, we can calculate the plasma clearance of H+ to be 150 ml/min

Question 7

kidneys and urine concentration

Answer 7

• the fundamental principle of concnetrating uring is osmosis
• the ECF osmolarity depends on the relative amount of water compared to solute
• the same principle applies to tubular fluid
• given that water moves by osmosis, you would expect a concentrated uring to osmotically draw water from the surrounding tissues
• the ability to concentrate urine occurs because there is a vertical osmotic gradient in the interstitial fluid of the medulla

Question 8

vertical osmotic gradient

Answer 8

• the normal osmolarity of ECF fluids in the body are 300 mOsm/L
• in the cortex, the interstitial fluid osmolarity is 300 mOsm/L but in the medulla, as you move from the cortex to the renal pelvis, there is a gradual gradient as the interstitial fluid osmolarity goes from 300 mOsm/L to 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

Question 9

structural differences between the cortical nephron and juxtamedullary nephron

Answer 9

• cortical: the loop of henle only dips slightly into the medulla
• juxtamedullary: loop of henle dips all the way down to the renal pelvis. this 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

Question 10

medullary vertical osmotic gradient - step 1

Answer 10

as soon as the fluid leaves bowmans capsule and enters the proximal tubule, there is a stong drive for osmotic reabsorption of water secondary to the active reabsorption of Na+

(water follows Na+)

Question 11

medullary vertical osmotic gradient - step 2

Answer 11

• by the end of the proximal tubule, due to Na+ reabsorption, 65% of the filtrate volume has been reabsorbed
• the osmolarity of the tubular fluid at this point is 300 mOsm/L, or isotonic, to other bodily fluids

Question 12

medullary vertical osmotic gradient - step 3

Answer 12

• 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 asecnding limbs of the loop of henle are distinct in their function
• ascendng limbs are impermeable to water, but reabsorb Na+
• i this case, water does not follow Na+
• the descending limbs, are highly permeable to water, but do not reabsorb Na+

Question 13

mechanism of countercurrent multiplication

Answer 13

• because the descending and ascending limbs are in close proximity, important interactions occur between them in order to establish the vertical osmotic gradient
• this process occurs because filtrate is constantly flowing

Question 14

fluid movement - step 1

Answer 14

• 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

Question 15

fluid movement - step 2

Answer 15

• 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 interstitial fluid is 200 mOsm/L more concentrated than the ascending limb entering the distal tubule
• from the perspective of the ascending limb, the tubular fluid is 200 mOsm/L and the interstitial fluid is 400 mOsm/L
• because both Na+ and water can move across the descending limb wall, the osmolarity of the initial part of the descending limb remains isotonic to the interstitial fluid so the tubular fluid in the descending limb is also now 400 mOsm/L

Question 16

fluid movement - step 3

Answer 16

• 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
• no longer have a 200 mOsm/L difference between descending and ascending limb fluids

Question 17

fluid movement - step 4

Answer 17

• while the ascending limb is still transporting Na+ out, water continues to passively leave the descending limb until the 200 mOsm/L difference between the descneding and asending limbs is established at each horizontal level
• the concentration of the tubular fluid in the descneindg limb is gradually increasing to remain isotonic to the interstitial fluid, yet the fluid in the ascending limb is gradually decreasing to maintain the 200 mOsm/L difference

Question 18

fluid movement - step 5

Answer 18

• as fresh 300 mOsm/L fluid enters the descending loop, all the fluid moves forward disrupting the concnetration gradient at all vertical levels until it again equilibrates

Question 19

fluid movement - step 6

Answer 19

• again, fresh filtrate enters, the osmolarity of the interstitial fluid increases, and the osmolarity of the ascending loop fluid decreases to maintain the 200 mOsm/L difference

Question 20

fluid movement - step 7

Answer 20

• eventually, equilibriu is acheived such that even with 300 mOsm/L filtrate entering the descneidng limb, there is a vertcial 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 4 times greater than the normal osmolarity of body fluids, while the osmolarity leaving the ascending limbs is one-third or normal osmolarity
• once the incremental medullaty gradient is established, it remains constant due to the continuous flow of fluid and solute transport

Question 21

benefits of countercurrent multiplication

Answer 21

• 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

benefit 1
- it establishes a vertical osmotic gradient in the medullary interstitial fluid
- this gradient allows the collecting ducts to both form more concentrated and more dilute urine than normal bodily fluid
benefit 2
- it allows for the overall volume of urine to be significantly reduced, which also allows the body to conserve both salt and water

Question 22

vasopressin-controlled water reabsorption - step 1

Answer 22

• vasopresin, or antidiuretic hormone, is a hormone released from the posterior pituitary gland
• it is released in response to a water deficit, when the ECF is hypertonic
• its release is inhibited when the ECF is hypotonic

Question 23

vasopressin-controlled water reabsorption - step 2

Answer 23

• once released into the circulation, it travels to the kidneys where it acts on distal tubular cells to increase the number of aquaporin molecules in the luminal membrane
• this increases the amount of water reabsorbed into the epithelial cells

Question 24

vasopressin-controlled water reabsorption - step 3

Answer 24

• once inside the epithelial cells, water passively moves into the interstitial fluid and plasma
• vasopressin has no actions on the proximal tubule or the loop of henle where 80% of water is reabsorbed
• therefore, it can only increase water reabsorption in the distal and collecting tubules

Question 25

regulation of water reabsorption

Answer 25

- tubular fluid entering the distal tubule is around 100 mOsm/L, yet the interstitial fluid of the renal cortex is 300 mOsm/L and gets even higher approaching 1200 mOsm/L around the collecting tubules as they plunge toward the renal pelvis - these gradients mean that water wants to leave the tubular fluid due to osmosis, but it can only do so in the presence of vasopressin

Question 26

deficit of water

Answer 26

- when someone is very dehydrated, the release of vasopressin will increase the number of aquaporin channels in the distal and collecting tubules - under the maximum influence of vasopressin, the osmolarity of the tubular fluid at the end of the collecting ducts can be up to 1200 mOsm/L, isotonic to the interstitial fluid - this is the max concentration of urine that can be achieved by the body - urine production can be produced to as little as 0.3 ml/min

Question 27

excess of water

Answer 27

- the body fluid osmolarity is below 300 mOsm/L - the tubular fluid entering the distal tubule is still 100 mOsm/L - when the body fluids are so hypotonic that vasopressin secretion is completely supressed, this prevents the insertion of aquaporins in the luminal membrane of the distal and collecting tubules so no water is reabsorbed - urine with an osmolarity of 100mOsm/L can be produced with a volume of up to 25ml/min

Question 28

countercurrent exchange within the vasa recta

Answer 28

- the vasa recta, the blood supply to the renal medulla, supports the countercurent multiplier mechanism, due to important characteristics 1. the vasa recta is closely associated with the descneding and ascending loop of henle, in part due to the hairpin shape that allows it to dive down into the medulla 2. vasa recta is highly permeable to NaCl and H2O, and the blood flow through the vasa recta is opposite, to fluid flow through the loop of henle

Question 29

5 steps of countercurrent exchange

Answer 29

1. as efferent arteriolar blood leaves the renal cortex, its osmolarity is 300 mOsm/L, isotonic to the interstitial fluid 2. as the descending loop moves toward. the renal pelvis, the plasma remains isotonic to the surrounding interstitial fluid reabsorbing Na+ and water leaving 3. at the bottom of the loop, the plasma osmolarity is 1200 mOsm/L 4. as the blood flows up the ascending limb, the opposite occurs with water being reabsorbed and Na+ leaving, to keep the plasma isotonic with the different levels of the medulla 5. as the vasa recta re-enters the cortex its osmolarity is back to 300 mOsm/L, again isotonic to the interstitial fluid

Question 30

difference between countercurrent exchange and countercurrent multiplication of the loop of henle

Answer 30

countercurrent exchnage - the process of passive solute and H2O exchange between the two limbs of the vasa recta and the interstitial fluid countercurrent multiplication - flow does not estbalish the current gradient - blood enters the medulla for nutrient exchnage, yet the hypertonic gradient of the medulla is preserved

Question 31

water reabsorption

Answer 31

- in the tubular segments permeable to water, soluble reabsorption always leads to water reabsorption due to osmosis - the opposite is also true, in that solute excretion is always accompanied by water excretion, again due to osmosis - when there is excess, un-reabsorbed solute in the tubular fluid, it exerts and osmotic influence to retain excessive water in the tubule - this is the phenomenon of osmotic duiresis, which increases urinary excretion - there are 2 types of diuresis, osmotic diuresis and water diuresis

Question 32

osmotic diuresis

Answer 32

- the increased excretion of both water and excess un-reabsorbed solute - this is seen in diabetics with glucose levels high enough that not all the filtered glucose is reabsorbed - the excess glucose in the tubules attracts water and increases urine production - this is why a symptom of diabetes is excess urine production

Question 33

water diuresis

Answer 33

- the increased excretion of water when there is little or no chance in the excretion of solutes - this is what occurs following alcohol consumption as vasopressin secretion is suppressed - this causes the kidneys to excrete a dilute urine, generally in volume greater than the alcohol consumed, which explains how you can become dehydrated by drinking alcohol

Question 34

urine storage

Answer 34

- once formed in the kidneys, urine is transmitted through the ureters by peristaltic contractions to the bladder - urine does not normally flow backwards towards the kidneys, but it is possible if enough pressure is generated

Question 35

the bladder

Answer 35

- composed of smooth muscle with a specialized epithelial lining, and is capable of expanding to increase storage capacity - it is highly innervated by the parasympathetic nervous system, the stimulation of which causes bladder contraction - the exit through the urethra is guarded by the internal urethral sphincter and the external urethral sphincter

Question 36

internal urethra sphincter

Answer 36

- under involuntary control - although it is really part of the bladder wall and not a true sphincter, when the bladder is relaxed, the internal urethral sphincter closes the outlet to the urethra

Question 37

external urethral sphincter

Answer 37

- encircles the urethra and is supported by the pelvic diaphragm - this sphincter is kept closed by a constant, tonic firing of motor neurons - because it is comprised of skeltal muscle, it is under voluntary control in that it can be deliberately tightened to prevent urination, even when the bladder contracts

Question 38

micturition (urination) reflex

Answer 38

the process of the bladdr emptying, and is govered by 2 mechanisms

Question 39

micturition relfex

Answer 39

- the adult bladder normally holds between 250 and 400 ml before the internal pressure on the bladder wall initiates the micturition relfex - the stretch activates afferent fibres to the spinla cord where interneurons acitvate the parasympathetic system to stmulate bladder contraction and relaxation of the external sphincter - there is no mechanism to open the internal sphincter, it does so as the bladder changes shape during contraction - with both sphincters open, urine is expelled - this relfex is evident in infants; as soon as their bladders fill enough to activate, urine is released

Question 40

voluntary control

Answer 40

- the micturition relfex can be overridden by voluntary control - by learning the perception of bladder filling prior to the acitvation of the relfex, voluntary excitatory signals from the cerebral cortex can override the micturition relfecx - this can onyl continue for so long, however, as urine production is constant and eventually the pressure-activation of the reflex is stronger tan voluntary control and the bladder uncontrollably empties