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Physiology > Renal Physiology (Day 2) > Flashcards

Renal Physiology (Day 2) Flashcards

1
Q

Clearance

A
  • Volume of plasma from which a given substance is cleared by the kidneys per unit time
  • Every substance in blood has own distinct clearance value; units expressed as volume of plasma per unit time (ex. mL/min or L/day)
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2
Q

The concept of clearance can be used to calculate GFR, assuming one uses a substance that is: (three things)

A
  1. Freely filterable at the glomerulus (so that its concentration in glomerular filtrate is the same as in plasma
  2. Not secreted by the tubules
  3. Not reabsorbed by the tubules

–> Such a substance is the polysaccharide, inulin

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3
Q

Use of clearance of inulin to calculate GFR:

A

If mass excreted/time = mass filtered/time then, GFP = Conc(inulin) = (U(inulin))(V)/P(inulin)

  1. Infuse inulin into a subject such that it’s plasma concentration is 4 mg/mL
  2. Collect urine for 2 hours, collected 0.2 L
  3. Measure urine inulin concentration: 360 mg/mL
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4
Q

Clearance equation

A

clearance of a substance x = (mass of x in urine)(urine vol/time) / plasma conc. of x

Cx = (Ux)(V) / Px

–> Sample: C(urea) = (18.2mg/mL)(1mL/min) / .26mg/mL = 70mL/min

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5
Q

If inulin were secreted, would the calculated GFR be higher or lower than the true GFR?

A

?

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6
Q

Glucose is freely filtered at the glomerulus, but it is also totally reabsorbed int he proximal tubules and returned to the plasma. What is the clearance of glucose?

A

?

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7
Q

Reabsorption

A

Recall: GFR = 180 L / day.
Total body water = 40L.
–> Thus almost all of that 180 L (99%) must be returned to the circulation via reabsorption in tubular epithelium –filtration is nonselective (except for protein); reabsorption is highly selective

-low hydrostatic pressure in peritubular capillaries

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8
Q

What is reabsorbed?

A
  1. water
  2. electrolytes (ions): Na+, K+, Cl-, HCO3-, H+, Ca2+, PO43-
  3. small organic molecules: glucose, amino acids, etc.
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9
Q

What is the general pathway for reabsorption?

A

Tubule –> Epithelial Cells –> Interstitium –> Peritubular Capillaries

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10
Q

Tubular Reabsorption limitations

A

Tubular reabsorption of some substances cannot be physiologically controlled.

Ex. glucose filtered = 180 g/day, glucose excreted = 0 g/day

  • tubular capacity for reabsorption of glucose –> GFR, so reabsorption of glucose is always maximal
  • therefore, reabsorption of glucose is not adjusted/altered
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11
Q

Reabsorption of H2O and Na

A

In contrast, reabsorption of H2O and Na+ can be altered under normal conditions

Example: ingested water will be excreted into urine within a few hours
-Therefore, there is a control mechanism which acts to maintain plasma water within fairly narrow limits

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12
Q

Reabsorption: Epithelial Transport

A

a.k.a. trancellular transport

substances cross apical and basolateral membranes of the tubule epithelial cells

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13
Q

Reabsorption: Paracellular Pathway

A

Substances pass through the cell-cell junction between two adjacent cells

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14
Q

Reabsorption: Na Cotransport

A

-primarily glucose, but amino acids, other organic metabolites and some ions such as phosphate are also absorbed by Na+-dependent cotransport.

  1. Na moving down it’s electrochemical gradient uses the SGLT protein to pull glucose into the cell against it’s concentration gradient
  2. Glucose diffuses out the basolateral side of the cell using the GLUT protein
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15
Q

Reabsorption of Glucose

A

Glucose/Na+ cotransporters have a transport maximum (Tm).

  • If there is too much glucose in the filtrate, it will not be completely reabsorbed because all the carriers are in use (saturated).
  • Extra glucose spills over into the urine = glycosuria and is a sign of diabetes mellitus.
  • Extra glucose in the blood also results in decreased water reabsorption and possible dehydration.
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16
Q

Summary of Tubular Transport

A
  1. WATER transport entirely via PASSIVE diffusion
  2. As progress through tubular system, volume of tubular fluid DECREASES due to reabsorption of water
  3. Avid reabsorption of substances of nutritional significance (glucose, amino acids, etc.) takes place via secondary active transport in proximal tubules
  4. Reabsorption of metabolic wastes (urea, creatinine) POOR or nonexistant
  5. Reabsorption of electrolytes (sodium, potassium) by tubules is under PHYSIOLOGICAL CONTROL and can be altered under different conditions
17
Q

How do kidney’s control osmolarity?

A

Kidneys control osmolality (solute concentration) of body fluids by excreting either a concentrated (i.e. greater proportion of solutes) or dilute (i.e. greater proportion of water) urine.

–> The major determinant of plasma osmolality is Na. Therefore osm(plasma) is a function of Na(plasma), and when Na(plasma increases) so does osm(plasma).

18
Q

Collecting Duct and ADH

A
  1. Last stop in urine formation
  2. Impermeable to NaCl but permeable to water
    - Also influenced by hypertonicity of interstitial space – water will leave via osmosis if able to
    - Permeability to water depends on the number of aquaporin channels in the cells of the collecting duct
    - Availability of aquaporins determined by ADH (antidiuretic hormone)
  3. ADH binds to receptors on collecting duct cells –> cAMP –> Protein kinase –> vesicles with aquaporin channels fuse to plasma membrane. (Water channels are removed without ADH).
  4. ADH is produced by neurons in the hypothalamus but stored and released from the posterior pituitary gland—release stimulated by ↑ in plasma osmolality
19
Q

Overall mechanism for excreting concentrated urine

A
  1. increased {osm}plasma
  2. osmoreceptors
    (nerve signal)
  3. posterior pituitary
    (ADH vasopressin release)
  4. medullary collecting duct
    (insertion of aquaporin into epithelium)
  5. increased water reabsorption
  6. concentrated urine (conserves water)
20
Q

Overall mechanism for excreting dilute urine

A
  1. decreased {osm}plasma
  2. osmoreceptors
    (NO nerve signal)
  3. posterior pituitary
    (NO ADH release)
  4. medullary collecting duct
    (NO insertion of water pores into epithelium)
  5. decreased water reabsorption
  6. dilute urine (excretes water)
21
Q

Summary of effects of ADH on urinary excretion

A

increased ADH –> water reabsorbed –> concentrated urine

decreased ADH –> water excreted –> dilute urine

  • Excretion of concentrated urine is a major determinant of an organism’s ability to survive in a terrestrial environment
  • Human kidney can produce maximal urinary concentration of 1200-1400mOsm/kg, four to five times higher than plasma osmolality (300 mOsm/kg)
  • Urinary concentration takes place as fluid moves through collecting ducts
  • The medullary interstitial fluid around the collecting ducts is hyperosmotic
  • Therefore, when ADH increase, water diffuses out of the ducts
22
Q

Countercurrent multiplier and exchange system

A
  • sets up and maintains high medullary interstitial osmolality
  • present in 30-40% of nephrons whose loops are long, extending into the inner medulla. Glomeruli for these nephrons are in the juxtamedullary area and mid-cortex
23
Q

What is a countercurrent system?

A
  • fluid flowing in opposeite directions in adjacent tubes

- anatomy of tubules and blood vessels in medulla fit this definition

24
Q

What are the key elements in countercurrent multiplication & exchange mechanism for producing concentrated urine?

A

Descending Limb: passive absorption of water

Ascending Limb: active absorption of Na, no reabsorption of water

Medullary Collecting Duct: water permeability dependent on ADH

25
Q

Countercurrent Multiplication steps

A
  1. input
  2. pump
  3. equilibrate
  4. shift
  5. pump
  6. equilibrate
    (repeat 4-6)
  7. steady state
26
Q

Countercurrent Multiplication

A
  • Most important element is active pumping of sodium ions from ascending limb of Henle’s loop into medullary interstitium
  • Movement of water in response to increasing interstitial osmolality is passive; at any given level of the loop, acts to equilibrate osmolalities between interstitium and descending limb
  • New fluid is constantly entering; as it moves down the loop, its osmolality increases
27
Q

What is the final result of countercurrent multiplication?

A

FINAL: although the gradient across the loop is never more than 200 mOsm/kg, a large gradient (900-1100 mOsm/kg) exists from top to bottom

–>Thus, in the presence of ADH, water in the collecting duct readily diffuses into the hyperosmotic interstitium.

28
Q

Vasa Recta

A

A system of PASSIVE exchange capillaries closely associated with Henle’s Loop

  1. Solutes diffuse in, water comes out - as blood descends, it encounters the high medullary interstitial solute concentration
  2. As blood leaves, it’s {osm} only slightly higher than when it entered and only small amounts of solutes carried away from medulla
29
Q

Vasa Recta: Descending Limb

A

solutes (Na) diffuse in, water diffuses out, blood becomes hyperosmotic

30
Q

Vasa Recta: Ascending Limb

A

solutes diffuse out, water diffuses in, blood approaches isosmotic

Physiology (29 decks)

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