Equations of Renal Function

Question 1

Glomerular Filtration Rate (GFR)

Answer 1

amount of plasma filtered per unit time

usually 100-125 mL/min

Question 2

Filtration Fraction

Answer 2

GFR/RPF

RPF=Renal Plasma Flow

usually 15-20% of entering plasma is filtered

Question 3

Renal Blood Flow(RBF) Formula

Answer 3

RPF/(1-HCT)

HCT=hematocrit

Question 4

Clearance

Answer 4

explains the ability of the kidney to excrete a substance, based on its plasma concentration

dimensions of volume and time in mL/min

Question 5

Clearance Formula

Answer 5

C=(UxV)/Px

mL/min

Question 6

Excretion rate

Answer 6

UxV

mg/min

Question 7

Ux

Answer 7

urinary concentration

mg/mL

Question 8

V

Answer 8

urine flow rate

mL/min

Question 9

Px

Answer 9

plasma concentration

mg/mL

Question 10

Inulin

Answer 10

freely filtered(does not alter GFR), not absorbed, not secreted, not metabolized, clearance of inulin measures GFR, must be infused

amt filtered per minute = amount excreted per minute

Question 11

Inulin formula

Answer 11

GFR = Clearance

GFR=(UinV)/Pin

Question 12

Creatinine

Answer 12

used to measure GFR, does not need to be infused, easily filtered with some secretion,

breakdown product of muscle metabolism

Question 13

Another clearance forumla

Answer 13

(GFRPx-reabsorbtion or +secretion)/Px

Question 14

GFR formula

Answer 14

(Urine conc. of inulin or creat)x(urine volume/collection time)/Px

Question 15

Tm

Answer 15

tubular transport maximum of glucose

375 mg/min

Question 16

Glucose Level less than 375 mg/min(Tm)

Answer 16

glucose is reabsorbed, clearance is 0

Question 17

Glucose level at or above 375 mg/min(Tm)

Answer 17

excess glucose appears in urine

Question 18

Splay

Answer 18

threshold point is not absolute for all nephrons, rounding of curve

Question 19

Glucose Clearance Formula

Answer 19

((GFRxPx)-Tm)/Px

Tm = 375 mg/min

if value obtained is 0 or negative then renal clearance is 0

Question 20

PAH

Answer 20

clearance>GFR, more PAH appears in urine than is filtered due to Secretion,

Tm for secretion is 80 mg/min,

Question 21

PAH concentration greater than 80mg/min(Tm)

Answer 21

some PAH remains in the peritubular capillaries

Question 22

PAH concentration lower than 80mg/min(Tm)

Answer 22

all PAH is cleared

Question 23

PAH clearance formula

Answer 23

((GFRxPx)+Tm)/Px

Tm = 80 mg/min

Question 24

Sodium Glucose Transporter 2 inhibitors (SGLT2)

Answer 24

blocks renal PCT reabsorbtion of glucose, type 2 diabetes treatment, lowers Blood Sugar, lowers hemaglobin A1C

pee away filtered glucose instread of reabsorbing it.