Renal System Flashcards

(81 cards)

1
Q

Describe Osmolarity.

A

It is the concentration of osmotically active particles present in a solution.
Units are osmol/l or mosmol/l (the latter is used for body fluids as these are weak salt solutions).

Can be calculate if:
1) The molar concentration of the solution is known.
2) The number of osmotically active particles present is known.

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2
Q

Compare osmolarity and osmolality with relation to body fluids.

A

For weak solutions (such as body fluids) osmolarity and osmolality are interchangeable.
Osmolality has units of osmol/kg water.
Osmolarity has units of osmol/l.

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3
Q

Describe tonicity.

A

Is the effect a solution has on cell volume, it also takes into consideration the ability of a solute to cross the cell membrane.

A solution can be:
- hypertonic = a decrease in cell volume as less water and higher salt concentration causes cell shrinkage.
- isotonic = no change in cell volume.
- hypotonic = an increase in cell volume as more water and lower salt concentration causes cell lysis (bursting).

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4
Q

Briefly describe TBW.

A

TBW -> total body water.
Males = ~60% of body weight
Females = ~50% of body weight (as they have higher body fate %)

Exists as two main compartments; intracellular fluid and extracellular fluid.

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5
Q

Describe the two compartments of TBW.

A

Intracellular fluid - 67% of TBW.

Extracellular fluid - 33% of TBW.
Includes:
Plasma (~20% of ECF)
Interstitial fluid (~80% of ECF)
Lymph and transcellular fluid (however, both male up such a low amount of ECF that they are negligible)

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6
Q

Describe the use of tracers in TBW.

A

We use ‘tracers’ to measure these by obtaining the ‘distribution volume’ of a tracer.

TBW = ECF + ICF
(therefore ICF=TBW-ECF)

TBW tracer - 3H2O
ECF tracer - Inulin
Plasma tracer - labelled Albumin

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7
Q

Describe the dilution principle.

A

V (litres) = Dose (D) / sample concentration (c)

Summary:
1) Add a known quantity of tracer X (Qx; mol or mg) to the body.
2) Measure the equilibration volume of X in the body ([X]).
3) Distribution volume (litres) = Qx (mol) / [X] (mol/litre)

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8
Q

Give an example of a water balance table.

A

Input (ml/day)
Fluid intake 1200
Food intake 1000
Metabolism 300

Output (ml/day)
Insensible loss:
Skin 350
Lungs 350
Sensible loss:
Sweat 100
Faeces 200
Urine 1500

Input=Output therefore homeostasis.

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9
Q

Give an example table of the effects of environment temperature on water loss.

A

Normal Temp
Skin 350
Lungs 350
Sweat 100
Faeces 200
Urine 1500
TOTAL = 2500

High Temp
Skin 350
Lungs 250
Sweat 1400
Faeces 200
Urine 1200
TOTAL = 3400

Prolonged Exercise
Skin 350
Lungs 650
Sweat 5000
Faeces 200
Urine 300
TOTAL = 6700

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10
Q

Explain what maintains water balance

A

Decreased excretion of water by the kidneys is insufficient to maintain water balance.
Water balance is maintained by increased water ingestion.

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11
Q

Give the overall ionic composition of ECF and ICF.

A

(values may differ between tissues)

ICF (mM)
Na+ 10
K+ 140
Cl- 7
HCO3- 10

ECF (mM)
Na+ 140
K+ 4.5
Cl- 11.5
HCO3- 28

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12
Q

Why can cells maintain different environments intracellularly to extracellularly?

A

The cell membrane and membrane transport mechanisms enable cells to maintain internal environments that differ in composition compared to their external surroundings.

Main ions in ECF are Na+, Cl-, HCO3-
Main ions in ICF are K+, Mg2+, -vely charged proteins

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13
Q

Why is the regulation of fluid balance and electrolyte balance intertwined?

A

Changes in solute concentrations lead to immediate changes in water distribution, therefore fluid balance and electrolyte balance are tightly intertwined.

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14
Q

Describe fluid shift.

A

Fluid shift; movement of water between the ICF and ECF in response to an osmotic gradient.

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15
Q

Explain challenges to fluid homeostasis.

A

Gain or loss of water = change in fluid osmolarity, similar changes in ICF and ECF volumes.

Gain or loss of NaCl = change in fluid osmolarity.
- Na+ ‘excluded’ from ICF
- Osmotic water movements

-> These two factors combine to produce opposite changes in ICF and ECF volumes:
ECF NaCl gain; ECF increase, ICF decreases.
ECF NaCl loss; ECF decreases, ICF increases

Gain or loss of isotonic fluid (0.9% of NaCl solution) = no change in fluid osmolarity. Change in ECF volume only.

-> Kidneys alter composition and volume of ECF. Regulation of ECF volume is vital for long term regulation of blood pressure.

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16
Q

Explain why electrolyte balance is important.

A

Total electrolyte concentrations can directly affect water balance (via changes in osmolarity).

The concentrations of individual electrolytes can affect cell function.
Na+ and K+ are particularly important:
- they are major contributors to the osmotic concentrations of the ECF and ICF (respectively).
- they directly affect the functioning of all cells.

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17
Q

Describe sodium balance.

A

More than 90% of the osmotic concentrations of the ECF results from the presence of sodium ions.
The amount of sodium in the ECF represents a balance between two factors (input and output).
Na+ is mainly present in the ECF therefore it is a major determinant of ECF volume.
Therefore it is vital to regulate Na+.

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18
Q

Describe Potassium balance.

A

Minor fluctuations in plasma [K+] can have detrimental consequences.
K+ plays a role in establishing membrane potential.
More than 95% of body K+ is intracellular; small leakages or increased cellular uptake may severely affect concentration of plasma K+ leading to:
- muscle weakness, in severe cases paralysis
- cardiac irregularities, worst case scenario cardiac arrest

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19
Q

Summarise the salt balance table (g/day).

A

Input
Fluids and Food 10.5

Output
Sweat and Faces 0.5
Urine 10

Input=Output therefore homeostasis.

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20
Q

Describe the impact of salt balance.

A

Salt imbalance is manifested as changes in extracellular fluid volume.
Regulation of ECF volume is important for long term regulation of blood pressure.

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21
Q

List Kidney functions.

A

1) Water balance
2) Salt balance
3) Maintenance of plasma volume
4) Maintenance of plasma osmolarity
5) Acid-base balance (regulate the concentrations of bicarbonate [HCO3-] and H+ ions)
6) Excretion of metabolic waste products
7) Excretion of exogenous/foreign compounds
8) Secretion of Renin (control of arterial blood pressure)
9) Secretion of erythropoietin (to stimulate red blood cell production)
10) Conversion of vitamin D into active form

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22
Q

State the primary function of the kidney.

A

Its primary function is to regulate the volume, composition and osmolarity of the body fluids.
The kidney’s role is the controlled excretion of substances.

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23
Q

Describe the composition of the urinary system.

A

The urinary system consists of the kidneys, that produce urine, and the structures that store and carry the urine from the kidneys to the outside for elimination from the body.

Each kidney receives 20-25% of the cardiac output.

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24
Q

Describe the nephron.

A

The nephron is the functional unit of the kidney - the smallest structural component of a system/organ/tissue capable or performing that systems/organs/tissues primary function.
Each kidney is composed of ~1 million nephrons.
It is the loop of Henle of each nephron that gives rise to the striated appearance of the kidney medulla cross section.

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25
List the components of a nephron - specifically noting those unique to cortical or juxtamedullary nephrons.
Renal artery Afferent arteriole Juxtamedullary apparatus Bowman's capsule Glomerulus Efferent arteriole Proximal tubule Loop of Henle Distal tubule Peritubular capillaries (cortical nephron) Vasa recta (juxtamedullary nephron) Renal vein Collecting duct
26
Compare cortical and juxtamedullary nephrons.
Juxtamedullary ~20% of nephrons, cortical ~80% of nephrons. Juxtamedullary nephrons has a much larger loop of Henle than cortical nephrons. Juxtamedullary nephrons are responsible producing a concentrated urine. Cortical nephrons have a network of peritubular capillaries surrounding it where as juxtamedullary nephrons just have one single capillary structure, the vasa recta.
27
Describe the juxtamedullary apparatus.
The juxtamedullary apparatus is the first step of glomerular filtration. The efferent arteriole is narrower than the afferent arteriole to ensure constant blood pressure therefore blood flows smoothly. Within the juxtamedullary apparatus there are two types of specialised cells: - macula densa cells - granular cells
28
Describe macula densa cells.
Found on the wall of the distal tubule that sits beside the juxtamedullary apparatus. Macula densa cells are salt sensitive and are able to release chemical messengers that influence the smooth muscle lining the wall of the afferent arteriole. Through relaxation or contraction of arterial smooth muscle it can regulate to blood flow into glomerular capillaries.
29
Describe Granular cells.
Granular cells are found between the macula densa cells of the distal tubule and the afferent arteriole. They secrete renin to initiate the renin angiotensin aldosterone system.
30
List what the kidneys incorporate and renal processes.
Filtration system Rich blood supply Mechanisms of tubular fluid modification; reabsorption and secretion. Renal process - glomerular filtration - tubular reabsorption - tubular secretion
31
Explain the renal tubule.
The renal tubule is a 'conveyor belt' ; substances are added or removed as urinary filtrate moves from proximal to distal ends. 80% of the plasma that enters the glomerulus is not filtered and leaves through the efferent arteriole. 20% of the plasma that enters the glomerulus is filtered, forming tubular fluid. It is only referred to as urine when it doesn't undergo any further modifications.
32
State the rate of excretion equation.
Filtration + secretion = reabsorption + excretion Rate of excretion = rate of filtration + rate of secretion - rate of reabsorption
33
Explain the movements of substances within the kidney when described in terms of concentration x flow. - rate of filtration - rate of excretion - rate of reabsorption - rate of secretion
1) rate of filtration of a substance For a freely filterable substance X: rate of filtration of X = [X]plasma x GFR where GFR is the glomerular filtration rate. Freely filterable = can pass through the molecular 'sieve' of the glomerulus. for a healthy adult GFR = 125ml/min. 2) rate of excretion of a substance rate of excretion of X = [X]urine x Vu where Vu is urine flow rate. This is highly dependent on hydration status but it is usually 1ml/min. Can be as low as 0.3ml and as high as 20-25ml. 3) rate of reabsorption of a substance if rate of filtration > rate of excretion, net reabsorption of that substance has occurred. rate of reabsorption of X = rate filtration of X - rate of secretion of X 4) rate of secretion of a substance if rate of filtration < rate of excretion, net secretion of that substance has occurred. rate of secretion of X = rate of excretion of X - rate of filtration of X Rate of filtration (1) and rate of excretion (2) are relatively easy to measure. Rate of reabsorption (3) and secretion (4) reflect tubular modification of filtrate; obtained as the difference between filtration and excretion.
34
Describe in detail the glomerular epithelial membrane and barriers to filtration.
Glomerular membrane is very 'leaky' as the gaps between endothelial cells are 100 times greater than other epithelial membranes in the body. Filtration barriers 1) Glomerular capillary endothelium (barrier to red blood cells). 2) Basement membrane/basal lamina (plasma protein barrier). 3) Slit processes of podocytes/glomerular epithelium (plasma protein barrier). Fluid filtered from the glomerulus into the Bowman's capsule must pass through the three layers that make up the glomerular membrane. Therefore glomerular filtrate should only contain water, small ions, glucose, amino acids and metabolic waste products.
35
Summarise the net filtration pressure that drives glomerular filtration.
Glomerular capillary blood pressure (BPgc) - drives inwards - 55mmHg Bowman's capsule hydrostatic pressure (HPbc) - drives outwards - 15mmHg Capillary oncotic pressure (COPgc) - drives outwards - 30mmHg Bowman's capsule oncotic pressure (COPbc) - drives inwards - 0mmHg
36
Describe net filtration pressure for glomerular filtration.
This pressure is entirely passive. Pressure remains constant in this capillary unlike others due to the narrower efferent arteriole. Net filtration pressure is the sum of inward pressures take away the sum of outward pressures. (55+0) - (15+30) = 10mmHg Value for Bowman's capsule oncotic pressure is 0 as there shouldn't be any plasma proteins in the Bowman's capsule as they shouldn't be able to get through the 3 filtration barriers. Oncotic pressure is pressure from plasma proteins.
37
Describe the determinants of glomerular filtration rate. (GFR)
The rate at which protein-free plasma is filtered from the glomeruli into the Bowman's capsule per unit time. GFR = Kf x net filtration pressure where Kf is the filtration coefficient (how 'holey' the glomerular membrane is). Normal GFR = 125ml/min Glomerular capillary fluid (blood) pressure is the major determinant of GFR.
38
Describe regulation of renal blood flow and glomerular filtration rate.
1) Extrinsic regulation of GFR -> sympathetic control via baroreceptor reflex 2) Autoregulation of GFR (intrinsic) -> myogenic mechanism -> tubuloglomerular feedback mechanism 2 is what keeps GFR constant.
39
Summarise the effect of vasoconstriction on GFR.
Vasoconstriction -> decreased glomerular capillary blood pressure -> decreased net filtration pressure -> decreased GFR.
40
Summarise the effect of vasodilation on GFR.
Vasodilation -> increased glomerular capillary blood pressure -> increased net filtration rate -> increased GFR.
41
Describe the pathway of GFR control by alterations in arterial blood pressure.
This is extrinsic control, mediated by increased sympathetic activity. 1) Fall in blood volume (such as a haemorrhage) 2) Decreased arterial blood pressure 3) Detected by aortic and carotid-sinus baroreceptors 4) Increased sympathetic activity 5) Generalised arteriolar vasoconstriction 6) Decreased glomerular capillary blood pressure 7) Decreased GFR 8) Decreased urine volume 8 helps compensate 1. Changes in systemic arterial blood pressure do not necessarily result in changes in GFR. Autoregulation prevents short term changes in systemic arterial pressure from affecting GFR.
42
Describe autoregulation pathways in the kidneys.
Intrinsic to kidneys; needs no external output. Tubuloglomerular feedback -> involves the juxtamedullary apparatus (although the mechanism remains uncertain) -> if GFR rises, more NaCl flows through the tubule leading to constriction of efferent arterioles. Myogenic -> if vascular smooth muscle is stretched (i.e. arterial pressure is increased) it contracts thus constricting the arteriole.
43
Summarise the effects of changing pressures on GFR.
Increasing HPbc (such as a kidney stone) = decreased GFR. Increased COPgc (such as diarrhoea) [as dehydration is loss of fluid therefore more proteins] = decreased GFR Decreased COPgc (such as severe burns) = increased GFR Decreased Kf (a change in surface area available for filtration) = decreased GFR
44
Describe plasma clearance.
A measure of how effectively the kidneys can 'clean' the blood of a substance. Equals the volume of plasma completely cleared of a particular substance per minute. Each substance that is handled by the kidney will have its own specific plasma clearance value.
45
State the equation to find clearance of substance X.
Clearance of substance X = [X]urine x Vu / [X]plasma Units [X]urine = mg/ml Vu = ml/min [X]plasma = mg/ml clearance = ml/min
46
Explain inulin clearance.
Inulin NOT insulin. - freely filtered at glomerulus - neither reabsorbed or secreted - not metabolised by kidney - not toxic - easily measured in urine and blood Therefore measurement of inulin clearance can be used clinically to determine GFR. Inulin enters the urine via filtration rate. Amount of inulin filtered per unit time = Amount of inulin excreted per unit time. [Inulin]plasma x GFR = [Inulin]urine x Vu GFR = [inulin]urine x Vu / [inulin]plasma Creatinine clearance works the same.
47
State and explain the expected clearance values for various substances in the kidney.
Glucose is not filtered and is not secreted, clearance = 0. For urea only a portion of the plasma is cleared therefore GFR > clearance. For H+ all of the filtered plasma is cleared of H+, and the peritubular plasma from which H+ is secreted is also cleared therefore clearance > GFR. Inulin = 125ml/min (equal to GFR) Creatinine is much the same. PAH = 650ml/min
48
Describe when it is tubular reabsorption or secretion.
If clearance < GFR then substance is reabsorbed. If clearance = GFR then substance is neither reabsorbed or secreted. If clearance > GFR then substance is secreted into tubule.
49
Describe the use of PAH to calculate renal plasma flow (RPF).
PAH - para-amino hippuric acid. It is an exogenous organic anion used clinically to measure renal plasma flow. Is usually = 650ml/min. PAH is: 1) freely filtered at glomerulus 2) secreted into tubule (not reabsorbed) 3) completely cleared from the plasma i.e. all the PAH in the plasma that escapes filtration is secreted from the peritubular capillaries.
50
Define renal plasma flow.
RPF is the volume of blood plasma that passes through the kidneys per unit time.
51
Define renal blood flow.
RBF is the total volume of blood that passes through the kidneys per unit time.
52
Describe clearance markers.
Any substance used as a clearance marker should have the following properties: - non-toxic - inert (i.e. not metabolised) - easy to measure A GFR marker should be filtered freely (not secreted or reabsorbed). A RPF marker should be filtered and completely secreted.
53
Describe how to find the filtration fraction.
Filtration fraction is the fraction of plasma flowing through the glomeruli that is filtered into tubules. Filtration fraction = GFR / RPF (renal plasma flow) 125ml/min / 650ml/min = 0.19 = 20% Therefore ~20% of the plasma that enters the glomeruli is filtered. The remaining 80% moves on to the peritubular capillaries. GFR is 60x plasma volume.
54
State how to find RBF.
RBF = RPF x (1/1-Hct) = 650 x 1.85 = ~1200 ml/min Cardiac output is 5litres/min therefore the kidneys receive ~24% of the CO. Hct is the haematocrit, all of the cells in a blood sample.
55
Why do we need reabsorption?
Plasma is filtered ~65 times per day. The kidneys reabsorb: - 99% of fluid - 99% of salt - 100% of glucose - 100% of amino acids - 50% of urea Reabsorption is specific while filtration is not.
56
List what is reabsorbed in the proximal tubule (PT).
Sugars Amino acids Phosphate Sulphate Lactate
57
List what is secreted in the proximal tubule (PT).
H+ Hippurates Neurotransmitters Bile pigments Uric acid Drugs Toxins
58
Describe transcellular transport.
This is when substances pass through cells of the epithelial membrane.
59
Describe paracellular transport.
This is when substances pass between cells of the epithelial membrane but do not enter them.
60
Describe primary active transport.
Energy is directly required to operate the carrier and move the substrate against its concentration gradient.
61
Describe secondary active transport.
The carrier molecule is transported coupled to the concentration gradient of an ion (usually Na+).
62
Describe facilitated diffusion.
Passive carrier-mediated transport of a substance down its concentration gradient.
63
Describe the reabsorption of sodium in the PT.
Na+ flows through Na+ channels on the apical membrane into tubular cells. Na+ is actively transported out (primary active transport) of the cell via Na+-K+ ATPase carrier. Therefore K+ is transported into the cell. Na+ crosses through interstitial fluid and diffuses into the peritubular capillary.
64
What drives Na+ reabsorption in the PT?
Iso-osmotic fluid reabsorption across 'leaky' proximal tubule epithelium due to: - standing osmotic gradient - oncotic pressure gradient
65
Describe the reabsorption of Cl- in the PT.
The transcellular movement of Na+ sets up an electrochemical gradient. Na+ can enter the cell via co-transport with glucose and amino acids (may be secondary active transport), through standard Na+ channels or via counter-transport with H+. The electrochemical gradient established by Na+ allows the paracellular transport of Cl-, water follows salt. Passive water absorption down NaCl osmotic gradient, oncotic drag of peritubular plasma.
66
Describe glucose reabsorption in the PT.
Glucose is co-transported with Na+ into the cell. Na+ is counter-transported with K+ to leave the cell. Glucose leaves through facilitated diffusion. Water moves across via paracellular transport and glucose, Na+ and water are reabsorbed in the peritubular capillary. Normally, 100% of glucose in the filtrate is reabsorbed in the proximal tubule.
67
Define transport maximum (Tm).
The maximum rate at which the kidney's renal tubules can reabsorb glucose.
68
Define renal threshold.
The renal threshold for glucose is the blood glucose level at which glucose starts appearing in the urine.
69
______ of all salt and water are reabsorbed in the PT.
~67%
70
What is the function of the loop of Henle?
Generates a cortico-medullary solute concentration gradient. This enables the formation of hypertonic urine.
71
Describe fluid flow in the loop of Henle.
Opposing flow in the the two limbs of the loop of Henle is termed countercurrent flow. The entire loop functions as a countercurrent multiplier. Together the loop and vasa recta establish a hyper-osmotic medullary interstitial fluid.
72
Describe the permeability of the ascending limb (AL) of the loop of Henle.
Along the entire length of the ascending limb Na+ and Cl- are being reabsorbed. In the thick upper AL this is achieved by active transport, in the thin lower AL this is passive. The ascending limb is impermeable to water therefore little or no water follows salt reabsorption.
73
Describe the permeability of the descending limb (DL) of the loop of Henle.
This segment does not reabsorb NaCl and is highly permeable to water.
74
Describe the Na+ reabsorption in the loop of Henle.
Na+ enters the cell via the Na+-2Cl--K+ triple co-transporter. Na+ leaves the cell via the counter-transport with K+. This region is impermeable to water so water stays in the filtrate.
75
Describe the triple co-transporter in the thick ascending limb of the loop of Henle.
K+ recycling means that NaCl is reabsorbed into the interstitial fluid. "Loop diuretics" block the triple co-transporter. Triple co-transporter pumps solute from the thick ascending limb of the loop of Henle. 1) Solute removed from lumen of ascending limb (water cannot follow). 2) Tubular fluid is diluted and osmolarity of interstitial fluid is raised. 3) Interstitial solute cannot enter the descending limb. 4) Water leaves the descending limb by osmosis. 5) Fluid in the descending limb is concentrated.
76
The _____ maintain the cortical-medullary concentration gradient (osmolarity).
Nephrons
77
The __________ contributes approximately half of the medullary osmolarity.
Urea cycle
78
What is the purpose of countercurrent multiplication and why do we have it?
To concentrate the medullary interstitial fluid. To enable the kidney to produce urine of different volume and concentration according to the amount of circulating antidiuretic hormone (ADH).
79
Describe the countercurrent exchanger.
Vasa recta runs alongside the long loop of Henle of juxtamedullary nephrons. Capillary blood equilibrates with interstitial fluid across the 'leaky' endothelium. Blood osmolarity rises as it dips down into the medulla (i.e. water loss, solute gained). Blood osmolarity falls as it rises back up into the cortex (i.e. water gained, solute lost).
80
Vasa recta acts as a countercurrent exchanger, together with the loop of Henle tis forms a countercurrent system. Essential blood flow through the medulla tends to wash away NaCl and urea. How do we minimise this problem?
1) Vasa recta capillaries follow hairpin loops. 2) Vasa recta capillaries are freely permeable to NaCl and water. 3) Blood flow to vasa recta is low (fewer juxtamedullary nephrons).
81
Describe passive exchange.
Passive exchange across the endothelium preserves medullary gradient-blood equilibrates at each layer. This ensure that the solute is not washed away.