renal - lecture 2 Flashcards

1
Q

compare plasma and filtrate

A

plasma = water and low molecular weight substances freely moves across filtration barrier to filtrate

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2
Q

what does plasma have that filtrate does not

A

cells =rbcs, wbcs = hematuria
proteins like albumins, globulins - if see albumin in pee = proteinuria= bad
protein bound substances = 1/2 calcium ion (since half bound to protein), fatty acids

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3
Q

name and describe forces involved in filtration

A

major force is pressure

pbs from bowmans space
= glomerular cap bp
no oncotic since no proteins
pcg and pigc from capillary
= fluid pressure in bowmans space, osmotic force due to protein in plasma = oncotic pressure
net = pgc-pbs-pigc = 16

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4
Q

what is gfr

A

the volume of fluid filtered from the glomeruli into Bowman’s space per unit time

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5
Q

what is gfr regulated by

A

net filtration pressure
membrane permeability
surface area available for filtration
last 2 = usually affected during disease, not usually regulated

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6
Q

what is normal gfr

A

70kg person = 180l/day (125ml/min)
plasma vol of person = 3.5l
180/3.5 = 51 = plasma filtered 51 times a day at glomeruli

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7
Q

describe regulation of gfr - decreased gfr

A

reduce net filtration pressure by constricting flow in efferent
and dilate efferent arteriole = outflow

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8
Q

describe regulation of gfr - increased gfr

A

constrict efferent arteriole = pressure increased
and dilate afferent arteriole = more blood flow

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9
Q

what is filtered load

A

total amount of any freely filtered substance = for individual substance
filtered load = gfr x plasma concentration of substance

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10
Q

give ex of filtered load

A

of glucose
180l/day x 1g/l = 180g/day

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11
Q

compare filtered load and amount excreted in urine

A

filtered load > amount excreted in urine = net reabsorption
filtered load < amount excreted in urine = net secretion

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12
Q

describe reabsorption - 2 ways

A

paracellular = through tight junctions connecting epithelial cells cuboidal
transcellular = through cells
basolateral = face blood side
luminal membrane = face tubular lumen

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13
Q

describe tubular reabsorption

A

water = 99% reabs
sodium = 99.5%
Glucose = 100%
urea = 44% (waste product )
net = 86.1, balance of secretion and reabsorption
k+ secreted too

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14
Q

describe 4 important facts about tubular resabsorption

A

filtered loads are enormous = generally greater than amounts of substance in body
reasb of waste products generally incomplete = urea
reasb of most useful plasma components is relatively complete = water, inorganic ions, organic nutrients
reabs of some substance not regulated = glucose, aas while others = highly regulated = water, inorganic ions

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15
Q

name 2 mechanisms for reabsorption

A

diffusion and mediated transport

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16
Q

describe diffusion - reabsorption

A

often across the tight junctions connecting the tubular epithelial cells = paracellular

17
Q

describe diffusion example - reabsorption

A

urea reabs in proximal tubule
urea freely filtered at glomerulus
in proximal tubule = water reabsorption occurs
urea concentration in tubular fluid becomes higehr
urea diffused into interstitial fluid and eventually peritubular caps

18
Q

describe mediated transport - gen - reabsorption

A

occurs across tubular cells = transcellular epithelial transport
requires participation of transport proteins in plasma membrane of tubular cells = usually coupled to the reabs of sodium

19
Q

describe mediated transport - specifics - reabsorption

A

nak atpase = pumps out sodium and in potassium = intracellular sodium goes down
Glucose and aas = transport proteins eventually has to get back to cap
sodium cotransporter - symporter same direction, aa and sodium too

20
Q

what is tm

A

when membrane transport proteins become saturated= tubule cannot reabsorb substance any more = limit = transport max = tm

21
Q

describe tm for ppl with diabetes mellitus

A

Uncontrolled diabetes mellitus = plasma concentration of glucose can be v high and filtered load of glucose exceeds capacity of the tubules to reasb glucose = tm exceeded
= glucose appears in urine glucosuria= dipstick test = first sign of diabetes mellitus

22
Q

describe tubular secretion and how its mediated

A

moves substances from peritubular capillaries into tubular lumen = opp of reabs
mediated by diffusion and transcellular mediated transport

23
Q

describe secretion - tubular secretion

A

most important substances secreted by tubules = hydrogen and potassium ions
tubular secretion usually coupled to reabs of sodium

24
Q

describe division of labor in tubules - gen

A

in order to excrete waste products adequately = gfr must be very large
filtered vol of water and filtered loads of all nonwaste plasma solutes are also v large

25
describe division of labor in tubules - proximal tubule
reabsorbs majority of this filtered water and solutes major site of secretion for various solutes EXCEPT potassium
26
describe division of labor in tubules - henles loop
reabsorbs large quantities of major ions, not much water mainly electrolytes
27
describe division of labor in tubules - dct/cd
volume of water and masses of solutes reaching here are small = fine tuning here determines final amounts excreted in urine by adjusting rates of reabs and in some cases secretions most homeostasis controls here
28
what is clearance
volume of plasma from which that substance is completely removed - cleared by kidneys per unit of time
29
what is clearance formula
clearance of s (Cs) = mass of 2 excreted per unit time/plasma concentration of s (ps) mass of 2 excreted per unit time = urine concentration of s US x urine volume per time V Cs = UsV/Ps
30
describe inulin clearance
polysaccharide that is given intravenously freely filtered at glomerulus but not reabs, secreted or metabolized by tubule so clearance of inulin - Cin = to vol of plasma originally filtered = GFR so Cin = GFR - most accurate marker of gfr
31
describe how to calculate gfr from inulin
urine vol = 2.4l/day inulin conc in urine = 300mg/l amount of inulin excreted in urine = 2.4l/day x 300mg/l = 720mg/day cin = 720mg/day divided by 4mg/l = 180l/day * 4mg/l = concentration of inulin in plasma gfr = cin = 180l/day not convenient
32
describe creatinine clearance
creatinine = waste product produced by muscle, endogenous filtered freely at glomerulus not reabs secreted at tubule but amount v small not metabolized by tubule thus, creatinine clearance used as clinical marker for gfr
33
describe creatinine clearance calculation
v=urine vol, ucr = urine conc of creatinine, pcr = plasma conc of creatinine creatinine clearance = UcrV/Pcr ~ equal to gfr ex = urine vol = 2l per day ucr = 9.6mmol/l pcr = 0.3mmol/l = 9.6 x 2/ 03 = 64l per day = person lost 2/3 gfr - normal = 180
34
compare clearance vs gfr
Clearance = for substance gfr = uniqe to person clearance of substance > gfr = secreted at tubule clearance of substance < gfr = reabsorbed at tubule
35
compare clearance vs gfr for specific substances
Clearance > gfr = secreted = like PAH, marker of renal plasma flow, clearance equal to renal plasma flow Clearance < gfr = reabsorption = water, sodium, glucose