Session 3

Question 1

What are the two types of nephrons?

Answer 1

All glomeruli are found in the cortex but there are 2 types of nephrons with the same filtration:

• Cortical nephrons (peritubular capillaries arranged haphazardly)
• Juxtamedullary capillaries (lie next to medullary border, vasa recta is arranged in parallel lines with flow of blood in opposite direction to the flow inside the tubules).

NB: Juxtamedullary nephrons also have a longer loop of Henle

Question 2

How does blood enter the kidney? What is the result of filtering?

Answer 2

Blood enters the kidney via the renal artery.

• Renal artery => Segmental arteries => Interlobar arteries => Arcuate arteries => Interlobular arteries => eventually millions of afferent arterioles.
• Each arteriole delivers blood to an individual kidney nephron
• The diameter of the afferent arteriole is slightly greater than the diameter of the efferent arteriole.
• The hydrostatic pressure of the blood inside the glomerulus is increased due to the difference in diameter of the incoming and out-going arterioles
• The increased hydrostatic pressure helps to force most of the water; most/all of the salts; most/all of the glucose; most/all of the urea of the blood out of the glomerular capillaries.

Result of filtering: the end product of filtration is identical to plasma without the large proteins and cells

Question 3

What becomes the ultra-filtrate? How fast is GFR?

Answer 3

• The water and solutes that have been forced out of the glomerular capillaries, pass into the Bowman’s space and become the glomerular filtrate, or the ultra-filtrate.
• GFR is 125ml/min
• ~20% of the plasma delivered to the glomerulus is filtered at any one time. 80% of the blood exits via the efferent arterioles.

Question 4

Describe the Filtration Barrier

Answer 4

• Capillary endothelium

Water, salts, glucose
Filtrate moves between cells

• Basement membrane

Acellular gelatinous layer of collagen/glycoproteins
Permeable to small proteins
Glycoproteins – their negative charge repel protein movement

• Podocyte layer
• Pseudopodia interdigitate to form filtration slits (fenestrations) so only small molecules can get through

Question 5

Why aren’t blood proteins and RBCs filtered?

Answer 5

The components are filtered in preference to other components of blood based on their size. Blood cells and plasma proteins are not filtered through the glomerular capillaries because they are by comparison larger in physical size.

The size limit for filtration is molecular weight 5,200 or an effective molecular radius of 1.48nm.

The basement membrane and podocytes glycocalyx have negatively charged glycoproteins, which repel protein movement

Question 6

What forces are involved in filtration? What is the net filtration pressure?

Answer 6

Plasma filtration is only due to three physical forces

• Hydrostatic pressure in the capillary (can be regulated) (P-GC)
• Hydrostatic pressure in Bowman’s capsule (P-BC)
• Osmotic pressure difference between the capillary and tubular lumen (osmotic pressure in this instance as well as including the osmotic pressure exerted by the solutes includes the oncotic pressure exerted by proteins)

REMEMBER: hydrostatic pressure pushes water away and osmotic pressure draws, attracts water in!

Net filtration pressure = 10mmHg

• Hydrostatic pressure in plasma favours filtration (50mmHg)
• Hydrostatic pressure in tubule opposes filtration (15mmHg)
• Osmotic pressure in glomerulus (includes oncotic pressure of proteins!) opposes filtration (25mmHg).
• 50-15-25 = 10mmHg

Question 7

What is the effect of charge? (On the filtration barrier)

Answer 7

• Neutral molecule: the bigger it is, the less likely to get through
• Anions: negative charge repels, more difficult to get through
• Cations: their positive charge allows slightly bigger molecules to get through
• In many disease processes, the negative charge on the filtration barrier is lost so that proteins are more readily filtered. This condition is called proteinuria

Question 8

Describe Tubular Reabsorption

Answer 8

• ~99% is reabsorbed back into the blood as it passes through the renal tubules. This recovery process is called tubular reabsorption and occurs via 3 mechanisms: osmosis, diffusion and active transport
• It is called reabsorption and not absorption as these substances have already been reabsorbed once (particularly in the intestines).
• Reabsorption in the PCT is isosmotic and driven by sodium uptake
• Other ions accompany sodium to maintain electroneutrality e.g. Cl- and HCO3- (Bicarbonate)
• Bulk Transport
• Solutes move from tubular lumen -> interstitium -> capillaries (blood).
• Reabsorption can either be transcellular or paracellular (around cells through tight junctions)

Question 9

Describe the tubular reabsorption of Na+

Answer 9

• Na+ is pumped out of tubular cells across the basolateral membrane by 3Na-2K-ATPase, lowering the luminal concentration of Na+.
• Na+ moves across the apical (luminal) membrane down its concentration gradient.
• This movement of Na+ utilises a membrane transporter or channel on the apical membrane.
• Water moves down the osmotic gradient created by reabsorption of Na+.

Question 10

Describe Tubular Secretion

Answer 10

[*] involves substances being added to the glomerular filtrate in the nephron tubule.

• Secretion provides a second route, other than glomerular filtration, for solutes to enter the tubular fluid.
• This is useful as only 20% of plasma is filtered each time the blood passes through the kidney.
• This removes excessive quantities of dissolved substances from the body such as H+ ions and maintains the blood at a normal healthy pH (typically in the range pH7.38-pH7.42).
• The substances are secreted into the glomerular filtrate (for removal from the body) include: K+ ions, H+ ions, NH4+ ions, creatinine, urea, some hormones and some drugs e.g. penicillin.
• Tubular secretion occurs from epithelial cells that line the renal tubules and collecting ducts into the glomerular filtrate.

Question 11

Describe the Model for Organic Cation (OC+) Secretion in the PCT

Answer 11

1. Entry by passive carrier

Mediated diffusion across the basolateral membrane down favourable concentration and electrical gradients by the 3Na-2K-ATPase pump

2. Secretion into the lumen

H+-OC+ exchanger that is driven by the H+ gradient created by the Na+-H+ Antiporter

• Several organic cation transporters are analogous to organic anion transporters. Cations compete with each other for transport – this adds a Tm limitations.*
• • Cations enter on the basolateral side by one of several OC (uniporters) and leave apical membrane (into glomerular filtrate in lumen) via antiporters which exchange H+ for organic cations.*

Question 12

Describe the isosmotic reabsorption of the PCT

Answer 12

• The majority of the volume of the glomerular filtrate solution is reabsorbed into the PCT. This includes some water and all of the glucose (except in diabetes).
• Reabsorption is an energy demanding process and most of the energy consumed by the kidneys is used in the reabsorption of sodium ions.
• The reabsorption of other things in the filtrate is coupled to the active reabsorption of sodium ions. Other ions accompany sodium to maintain electroneutrality e.g. Cl- and HCO3- (Bicarbonate)
• Following the movement of solutes, water is also reabsorbed by osmosis. About 70% of the glomerular filtrate-volume is reabsorbed in this way. As this part of the reabsorption process is not controlled by the proximal tubule itself, it is sometimes called obligatory water reabsorption.

By the end of PCT, the following have been reabsorbed:

• 100% filtered nutrients
• 80-90% of filtered HCO3-
• 67% filtered NA+
• 65% filtered water
• 50-65% filtered Cl-
• 65% filtered K+

Question 13

Describe the role of active transport and co-transport in tubular reabsorption and secretion

Answer 13

• Different segments of the tubule have different types of Na+ transporters and channels in the apical membrane.
• This allows Na+ to be the driving force for reabsorption, using the concentration gradient set up by 3Na-2K-ATPase (active transport)

Proximal tubule

• Na-H Antiporter
• Na-Glucose Symporter (SGLUT)
  These transporters rely on the Na+ gradient created by the sodium pump

Loop of Henle

• Na-K-2Cl Symporter

Early Distal Tubule

• Na-Cl Symporter

Late Distal Tubule and Collecting Duct

• ENaC (Epithelial Na-Cl)
• ROMK can also appear on either membrane depending on where we are in the tubule.*

Question 14

Describe the basic transport mechanisms for diffusion, facilitated diffusion (passive transport), primary active transport and secondary actie transport

Answer 14

Passive Diffusion

[Description: pingpong.tiff] Dependent on permeability and concentration gradient, with the rate of passive transport increasing linearly with increasing concentration gradient

Facilitated Diffusion
The permeability of the membrane for a substance is increased by the incorporation of a specific protein in the bilayer. Models for facilitated diffusion include carrier molecules (ping-pong) and protein channels.

Active Transport
Active transport allows the transport of ions or molecules against an unfavourable concentration and or/electrical gradient, requiring energy from the hydrolysis of ATP.
Whether or not energy is required is determined by the free energy change of the transported species. The free energy change is determined by the concentration gradient for the transported species and by the electrical potential across the membrane bilayer when the transported species is charged.
Some cells spend up to 30 – 50% of their ATP on active transport.

Secondary Active Transport
No direct coupling of ATP
Transporter protein couples the movement of an ion (typically Na+ or H+) down its celectrochemical gradient to the uphill movement of another molecule or ion against a concentration/electrochemical gradient
Thus energy stored in the electrochemical gradient of an ion is used to drive the transport of another solute against a concentration or electrochemical gradient.

Question 15

How does Sodium drive the reabsorption of other substances?

Answer 15

• The concentration of sodium ions in the glomerular filtrate solution is high. Sodium ions move from the glomerular filtrate in the lumen of the tubule, into the cells of PCT (through the interstitium and then into the blood)
• Na+ travels down its concentration gradient set up by 3Na-2K-ATPase from the tubule lumen into the interstitium. In the case of many Na+ ions, this occurs with the help of symporters (co-transporters) on the apical membrane.
• These symporters simultaneously facilitate passage through the PCT apical membranes of both Na+ and ions and other substances/solutes.
• Other such substances that are reabsorbed with sodium ions in this way include glucose, amino acids, water-soluble vitamins (B-complex and C0, lactate, acetate, ketone and Krebs cycle intermediates.
• These then move on through cells via diffusion and/or other transport processes.
• The operation of these symporters located on the apical membrane is dependent on the action of the 3Na+-2K+-ATPase transporter, located on the basolateral membrane.
• To summarise the above, the solutes are selectively moved from the glomerular filtrate to the plasma by active transport.

Question 16

Describe how Glucose is reabsorbed in the PCT, and what happens if there’s too much

Answer 16

• Glucose is reabsorbed in the PCT using the Na-Glucose Symporter SGLUT. This moves glucose against its concentration gradient into the tubule cells (from the glomerular filtrate). Glucose then moves out of the tubule cell on the basolateral side by facilitated diffusion
• 100% of glucose is normally reabsorbed but the system has a maximum capacity, or Transport Maximum ™.
• If the plasma concentration exceeds Tm, the rest spills over into the urine.
• If this happens, water follows into the urine, causing frequent urination (polyuria).
• The renal threshold for glucose is 200mg/100ml. Renal threshold is the plasma concentration of a substance at which T(m) of a substance is reached and the substance starts spilling into the urine.
• Transport Maximum for glucose in males = 375mg/min and 300mg/min in females, depending on age.
• If a patient had a plasma glucose concentration of 400mg/100ml, filtered load will be 500mg/min (400mg/100ml x 125ml/min).
• 500mg/min – 375mg/min = 125mg/min glucose will be lost to the urine.

Question 17

What is the concept of clearance?

Answer 17

• Clearance is the theoretical basis for the calculation of GFR: for any substance that enters the kidney in the renal artery, an equal amount must leave in the urine and the renal venous blood. THIS CAN ONLY HOLD FOR SUBSTANCES THAT ARE NOT METABOLISED OR SYNTHESISED.
• Clearance is only concerned with the amount of a Substance X that ends up in the urine.
• The definition of clearance: the volume of plasma from which Substance X can be completely cleared to the urine per unit time

Question 18

How do you calculate clearance and give an example

Answer 18

So clearance of Substance X is therefore calculated as (Amount in urine x Urine flow rate) / (Arterial plasma concentration)

• The excretion rate for substance X would be calculated = 100mg/ml x 1ml/min = 100mg/min
• If this substance (X) was present in the plasma at a concentration of 1 mg/ml then its clearance rate would be = (100mg/min) / (1mg/ml) = 100ml/min
• So 100ml of plasma would be completely cleared of substance X per minute. The concept of clearance is important because it allows you to calculate GFR and this, clinically, is used as a measure of kidney function particularly change over time (e.g. taking a GFR two weeks after the first sample and the result was decreased, could say patient had decreased kidney function) – relative measurement.

Question 19

How are GFR and Clearance related? Give an example

Answer 19

E.g. 125mls/min of plasma filtered containing “P”: clearance = volume of plasma cleared of substance per min therefore clearance = 125ml/min

If a compound behaves like “P’ and it is completely cleared from the plasma that is filtered, then the rate it appears in urine = GFR
Inulin is the Gold standard measurement but in order to use it you need to infuse inulin into the patient at a direct rate.

Not everything is like Inulin

Clearance of PAH is therefore equal to renal plasma flow

Question 20

What is eGFR and how do you calculate it?

Answer 20

• eGFR: estimated Glomerular Filtration Rate
• eConcentration of Creatinine = [(140-age) x (Mass –kg) x Constant)] / Serum [Creatinine (μmol/L)]
• Creatinine isn’t as good as Inulin – overestimates GFR before a tiny proportion is secreted but it’s considered good enough
• Constant for men is 1.23
• Constant for women is 1.04

Question 21

What is meant by Filtered Load? Give an example

Answer 21

(amount of stuff that goes through/ml) e.g. Glucose

• Normal plasma concentration for glucose is 100mg/100ml (or 1mg/ml)
• Freely filtered in Bowman’s capsule so ultrafiltrate has same concentration as plasma
• GFR = 125ml/min so filtered load will be (1mg/ml) x (125mg/min).
• If plasma concentration increased to 200mg/100ml (or 2mg/ml)
• Filtered load = (2mg/ml) x (125ml/min) = 250mg/ml

Question 22

What is Renal Blood Flow? What is Renal Plasma Flow? And what is the volume of fluid in urine?

Answer 22

Renal Blood Flow (RBF) is ~1.1l/min
Renal Plasma Flow:

• Haemaocrit (Ht or HCT or erythrocyte volume fraction (EVF)) is the volume percentage (%) of red blood cells in blood
• Normally ~0.45
• Therefore RPF = 0.55 x 1.1L/min = 605ml/min of plasma
• Flow of plasma in the kidney: every min 605ml of plasma goes to the kidney –renal plasma flow.
• 80% (~480ml) initially enters the peritubular capillaries (contains cells and proteins not filtered)
• 20% (125ml filtered = GFR)
• Fluid in urine is anything filtered and not reabsorbed + anything not filtered but secreted
• REMEMBER: composition of filtrate in the Bowman’s capsule is identical to plasma (- large proteins)

Question 23

What is GFR and what is its clinical significance?

Answer 23

• GFR: the volume of plasma from which any substance (X) is completely removed by the kidney in a given amount of time (usually 1 minute)
• E.g. clearance for urea in a normal healthy kidney is about 65ml/min i.e. the kidney removes all the urea from 65mls of plasma per min
• GFR is a measure of the filtration process of all the nephrons (kidney’s ability to filter a substance) and it is a measure of overall kidney function. A fall in GFR generally means kidney disease is progressing whilst a recovery in GFR indicates kidney function recovery. Knowing the GFR of a patient is clinically very important and helps to determine the severity and course of kidney disease.

Question 24

How do you calculate GFR? What are the normal GFRs for men and women?

Answer 24

• To measure GFR a substance X, which must be freely filtered across the glomerulus is required. This substance (X) must not be reabsorbed, secreted or metabolised by the cells of the nephron – it must pass directly into the urine.
• The amount of X excreted in the urine/min is equal to the amount filtered by the kidney/min.
• Creatinine, a by-product of skeletal muscle metabolism can be used to measure GFR even though it slightly overestimates GFR as the proximal tubule does secrete creatinine (but only a little). It is quick, cheap and easy to do so normally used.

Calculating GFR of creatinine: (amount of creatinine present in urine x flow rate) / arterial plasma concentration of creatinine

• Normal GFR for males = 115-125 ml/min
• Normal GFR for females = 90 -100ml/min

Question 25

What is the Filtration Fraction and how do you calculate it?

Answer 25

• Proportion of a substance that is actually filtered
• 605ml of plasma (RPF) enters the glomerulus and 20% is filtered
• Therefore 125ml is filtered through into Bowman’s space and 480ml passes into peritubular capillaries (efferent arterioles)
• To calculate FF: Fraction of GFR/RPF is 125/605 = ~20%

Question 26

What is Fractional Excretion? How do you calculate it?

Answer 26

• Low molecular weight solutes are readily filtered across the glomerular barrier. The filtered load of a solute depends, therefore, on both GFR and the concentration of the solute in the plasma.
• As the glomerular filtrate passes down the nephron tubule, its composition is modified by both reabsorption and secretion by tubular epithelial cells.
• The amount of solute appearing in the final urine is expressed as a fraction of the filtered load is termed the Fractional Excretion of this solute.
• FE is in effect the proportion of filtered solute that remained un-reabsorbed by the nephron.

FE is usually expressed as a percentage and can be calculated as follows:

• Amount of Substance Z = [(plasma concentration of substance Z mmol/L) x (urinary concentration of creatinine mmol/L) x (urinary volume ml/min)] / (plasma concentration of creatinine)
• Creatinine is here used to calculate GFR
• Amount of substance Z secreted = {Urinary concentration of Substance Z] x Urinary Volume = B
• Fractional excretion of Z (FEZ) (%) = (B x 100) / A
• Therefore, it follows that Fractional tubular reabsorption is that percentage of filtered solute that is reabsorbed and is simply 100 - FE

Question 27

Describe Autoregulation

Answer 27

• The kidney responds to changes in plasma volume and continuously regulates blood flow within the renal tissue. It is responsible for maintaining intra-renal blood flow over a wide range of systemic perfusion pressures.
• Selective vasoconstriction or dilatation must occur to maintain renal blood flow at a constant rate.
• These controls enabled a constant glomerular filtration rate to be maintained across a wide range of systemic blood pressures.
• Renal autoregulation is believed to be mediated by the combined and interacting contributions of two mechanisms: a faster myogenic and a slower tubuloglomerular (TG) feedback system.
• THE PURPOSE OF THE AUTOREGULATION OF RENAL BLOOD FLOW IS TO MAINTAIN GFR within normal limits
• Autoregulation is able to maintain GFR when blood pressure is within physiological limits (80-180mmHg)

Question 28

Describe the Myogenic Response and Tubuloglomerular Feedback System

Answer 28

• Myogenic response: smooth muscle in the afferent arterioles detect and respond to transmural pressure (affected by changes in arterial blood pressure) elevation with constriction (contraction) and to pressure reduction with dilation
• TG Feedback: Changes in tubular flow rate as a result of changes in GFR change the amount of NaCl that reaches the distal tubule. If there is an increase in arteriole pressure, macula densa cells in the distal convoluted tubule (near the glomerulus) sense the increased flow rate due to the increased NaCl concentration in lumen. They stimulate juxtaglomerular apparatus to release chemicals e.g. adenosine which acts a rapid response vasoconstrictor (NB: unique response in renal system) – to decrease GFR. If there is a fall in arteriole pressure, MD cells in DCT sense the decreased flow rate due to decreased NaCl concentration in lumen and stimulates juxtaglomerular apparatus to release chemicals such as prostaglandins which promote vasodilatation – to increase GFR so that GFR is unchanged by the change in blood pressure.
• NB: The relationship between GFR and distal salt concentration only applies in acute changes

Question 29

What is Aminoaciduria? What are the two broad types?

Answer 29

• Renal aminoaciduria is mainly confined to the dibasic acids (has 2 H+ ions) and is due to a genetically determined lack of the specific transport protein(s). For some reason cysteine is an abnormally insoluble amino acid, especially in acidic urine, and cystinuria may be associated with stone formation.
• General Overflow Aminoaciduria: characterised by raised plasma (and hence filtrate) levels of most or all acids; the renal absorptive processes become overloaded. This is typically due to inadequate deamination in the liver. It is also seen during early pregnancy, probably due to the greatly increased GFR at this time.
• Specific Overflow Aminoaciduria: normally confined to a single amino acid and reflect a genetically determined inability of the liver to metabolise the relevant acid. An important example is phenylketonuria in which the ability to metabolise phenylalanine is lacking (lack of phenylalanine hydroxylase). Phenylketonuria leads to severely retarded mental development, and all babies are routinely tested for this at birth (when, if the results are positive, treatment is possible).