Kidney 1

Question 1

Where does the tubule pass between due to the loop of henle folding back in on itself?
2) What is the function of this close contact between the tubule and arterioles?

Answer 1

the tubule passes directly between the afferent and efferent arterioles.
2) allows the arterioles to respond directly to the rate of low in the tubule

Question 2

What groups of cells mediate communication between the arteriole and tubule? + function

Answer 2

1) macular densa : epithelial cell in ascending limb of loop of Henle adj to arterioles that are modified to form a plaque
+detect rate of flow of fluid in distal tubule and release paracrines effecting afferent arteriole diameter.
2) Granular cells: a cluster of specialised smooth muscle cells in the afferent arteriole
+ secrete an enzyme called renin- an enzyme involved in control of vascular resistance & salt and water balance.

Question 3

Describe the tubuloglomerular feedback:

starting from Glomular filtrate rate increase

Answer 3

1. therefore flow through tubule increases
2. flow past macula densa increase
3. paracrine from macula densa to afferent arteriole
4. afferent arteriole constricts
5. resistance in afferent arteriole increase
6. hydrostatic pressure in glomerulus descreases
7. glomular filtrate rate falls

Question 4

1) What is the rate at which a substance is filtered from the glomerulus linked to?
2) How do we work out glomular filtrate rate from th excretion rate?
3) What is the equation to calculate the amount of solute excreted?

Answer 4

1) the rate at which it is excreted from the bladder. However the amount excreted is also affected by the amount of solute which is reabsorbed into the peritubular capillaries, and whether or not the solute is secreted by the peritubular capillaries
2) we need to measure the excretion of a solute which is not reabsorbed from or secreted into the kidney tubule.
3) amount filtered- amount reabsorbed+ amount secreted (from peritubular capillary to tubule)

Question 5

What is the equation to calculate the amount of solute excreted?

2) What molecule administered intravenously can be used to calculate GFR?
3) What variable from the equation must be removed
4) What is the new fomula for calculating GFR looking at inulin concentration?

Answer 5

amount filtered- amount reabsorbed+ amount secreted (from peritubular capillary to tubule)
2) inulin, a polymer a fructose and is not produced by the body
3) amount reabsorbed AND amount secreted (from peritubular capillary to tubule)
4) GFR= (concn inulin in urine *rate of urine production (ml/min).)
/
concn inulin in plasma

Question 6

Requirements of a substance to measure GFR:

2) Give 2 examples:
3) Which is worse?
4) why?

Answer 6

1)Must be freely filtered at glomerulus
Must not be reabsorbed nor secreted into nephron
Must not be subject to metabolism or produced by kidney
Must not alter GFR
2) inulin and Creatinine
3) causes GFR being overestimated by 10% as secreted by the proximal tubule from the peritubular capillaries
BUT
colorimetric test for creatine tends to overestimate it’s true concentration also by 10%, so the 2 errors cancel

Question 7

What within the body produces creatinine?

2) what happens if GFR falls due to renal disease to the amount of creatinine excreted??

Answer 7

1) produced at a constant rate by metabolism in skeletal mm.
2) there will be an initial drop in excretion of creatinine, but very quickly plasma concentration of creatinine will rise so that excretion of creatinine keeps up with production by the muscles.

Question 8

plasma inulin : 0.3mg/ml

urinary inulin: 30mg/ml

urine production: 1.25ml/min

What is her glomerular filtration rate?

Answer 8

1.25 * 30
/
0.3
=125ml/min

Question 9

Why do we measure GFR?

Answer 9

A change in GFR is often the first sign of kidney disease
GFR may alter in response to other pathologic conditions
GFR may alter in response to drugs
Knowledge of GFR allows us to analyse the way that the kidney handles any solute, by comparing the clearance rate of the solute to the GFR

Question 10

1) What is clearance rate?
2) What is the formula ?
3) What are its units?

Answer 10

the rate a solute disappears from the bidy by either excretion or by metabolism
2) excretion rate of X (mg/min)
/
[X] plasma (mg/mL plasma)
3) mL/min

Question 11

If there is net reabsorbtion of 4of a substance when there is 8 initially in tubule what will clearance of substance be compared to GFR?

Answer 11

it will be half of GFR
e.g. GFR=100mL/min
clearance rate = 50mL/min

Question 12

If there is net secretion of 2 of a substance when there is 4 initially in tubule already what will clearance of substance be compared to GFR?
e.g.

Answer 12

it would be 1.5 times greater then GFR
e.g. GFR=100mL/min
clearance rate = 150mL/min

Question 13

When will GFR equal clearance rate of a substance?

Answer 13

when there is no net reabsorption or secretion

Question 14

What formulas do you need to calculate clearance rate?

Answer 14

1. clearance rate = ecretion rate/ plasma concentration
2. excretion rate= urine concentration* urine vol/min
3. plasma concentration=filtration rate / GFR

Question 15

What are the units of :

1) excretion rate
2) clearance rate

Answer 15

1) mg/min

2) mL/min

Question 16

1) How do we calculate the filtered load of solute?
2) What does this value and excretion rate allow us to see when compared?
3) What other values allow us to figure this out?

Answer 16

1) concentration in plasma * GFR
2) if net reabsorption or secretion occurred
3) the comparison of inulin/ creatinine clearance rate with molecule if interest.

Question 17

If clearance rate of X is greater than inulin, What does that mean?

Answer 17

X has been secreted into tubule (netly)

Question 18

If clearance rate of X is less than inulin, What does that mean?

Answer 18

X has been reabsorbed (netly)

Question 19

What is starling’s hypothesis?

2) How does it apply to filtration in glomerulus?

Answer 19

the fluid movement due to filtration across the wall of a capillary is dependent on the balance between the hydrostatic pressure gradient and the oncotic pressure gradient across the capillary.
2) The hydrostatic pressure in the glomerular capillaries is greater than hydrostatic pressure in bowman’s capsule and the colloid osmotic pressure gradient.. =net flow out into bowmans

Question 20

Autoregulation of GFR

1) local or long distance?
2) What does it maintain?
3) What does it counteract?
4) What are the 2 mechanisms of autoregulation?

Answer 20

1) local
2) renal blood flow and glomerular filtration rate
3) BP fluctuations
4) Myogenic response and Tubuloglomerular feedback

Question 21

Describe the Myogenic response:

from systemic BP increases

Answer 21

increase in renal blood flow, afferent arteriole stretched, reflec contraction of vascular smooth muscle ,
increases in vascular tone and resistance therefore flow returns to normal.

Question 22

What makes up the glomerular barrier? from top to bottom where to is touching blood

Answer 22

1.Fenestrated endothelium
2. Collagen basement
membrane
3. Podocyte Filtration slits

Question 23

What is the pore size of the following in the glomerular barrier:

1) Fenestrated endothelium
2) Podocyte Filtration slits

Answer 23

1) 70nm diameter
2) 4-14nm in diameter
3) restricts negatively charged macromolecules and molecules weighing more than 5.5 kilodaltons

Question 24

What occurs in the following locations:

1. renal corpuscle
2. proximal tubule
3. Loop of Henle
4. Distal tubule & Collecting duct

Answer 24

1. production of filtrate
2. 70% H2O & solutes reabsorbed
3. urinary concentration
4. Control of Na+& H2O balance

Question 25

How would you calculate the net absorbtion or net secretion if you had the following values:

1) V: volume of urine produced a day e.g. 2L/day , therefore 1.39mL/min
2) [X]p: X’s plasma concentration e.g. 3mg/ml
3) [X]u: X’s urine concentration e.g. 20mg/mL
4) GFR e.g. 18.5mL/min

Answer 25

Filtered glucose = Px X GFR = 3mg/ml x 18.5ml/min = 55.5mg/min
Excreted glucose = Ux X V = 20mg/ml x 1.39ml/min = 27.8mg/min
Thus % glucose reabsorbed = [55.5-27.8]/55.5 = 50%

Question 26

Why are glucose levels high in the urine of diabetic patients?

Answer 26

Deficiency in insulin signalling leads to increased plasma concentration of glucose
Capacity for glucose reabsorption is exceeded – all transporters are saturated
Excess glucose is excreted in urine

Question 27

Why do diabetics produce large volumes of urine?

Answer 27

Increased concentration of glucose in tubule fluid
Water follows by osmosis (ie less H2O is reabsorbed in nephron)
Urine volume increases
(Dehydration triggers thirst response.)