Equations, Brief Explanations, & Normal Values

Lab Exercise 20: Notes and Problems

Equations, Brief Explanations, & Normal Values:

*Cardiac Output (CO in L/min)

CO = Stroke Volume (SV) x Heart Rate (HR)
CO ≈ 5 L/min (Normal value used in lab)

*Renal Blood Flow (RBF in ml/min)

The amount of blood that travels toward the kidneys through the renal arteries.
RBF = ¼(CO) ∴ RBF = ¼(5 L/min) = 1.25 L/min = 1250 ml/min
RBF ≈ 1200 ml/min(Normal value used in lab)

*Effective Renal Blood Flow (ERBF in ml/min)

Only about 90% of the RBF actually reaches the nephrons at the glomerulus, therefore; this number indicates the estimated, or effective, blood flow that actually reaches the nephrons.
ERBF = 0.9(RBF) ∴ ERBF = 0.9(1200ml/min) = 1080 ml/min
ERBF ≈ 1080ml/min(Normal value used in lab)

*Renal Plasma Flow (RPF in ml/min)

Since it is the plasma that is actually filtered in the kidney, it is important to know the amount of plasma that actually reaches the kidneys which is represented by this number.
RPF = RBF(1 – HCT)
Normal HCT = .45 (men) ∴ RPF = 1200 ml/min(1-.45) = 660 ml/min
RPF ≈ 660ml/min(Normal value used in lab)

*Effective Renal Plasma Flow (ERPF in ml/min)

As in the blood flow, only about 90% of the plasma that flows through the renal artery actually reaches the nephrons. This figure represents the estimated amount of plasma that reaches the glomerulus.
ERPF = 0.9(RPF) ∴ ERPF = 0.9(660 ml/min) = 594 ml/min

ERPF ≈ 594 ml/min(Normal value used in lab)

*Glomerular Filtration Rate (GFR in ml/min)

This figure refers to the amount of plasma that is actually filtered into Bowman’s capsule. About 20% of the plasma that reaches the glomerulus is filtered but the lab manual figures this number by reducing the amount per day to amount per minute.
About 180 L of plasma is filtered daily.
GFR = 180 L/day = .125 L/min = 125 ml/min

1440 min/day

GFR ≈ 125 ml/min(Normal value used in lab)

*Renal Clearance Value (RCVs in ml/min)

Clearance refers to the amount of a specific substance that is removed from the plasma and excreted in urine in a time frame.
RCVs = Us x R

Us = Urine concentration of substance (s), R = Urine flow rate, & Ps = Plasma concentration of substance (s)
Note: if subject in question is inulin, the RCVinulin will equal the GRF. (See definition of inulin)

*Excretion (Es in mg/min)

This will determine the amount of a substance that is excreted in the urine in a specific time frame considering filtration, secretion, & reabsorption.
Excretions = filtrations + secretions - reabsorptions
Filtration Load or Tubular Loads = (GFR) x (Plasma Concentrations)
Secretions = Excretions – Filtrations + Reabsorptions
Excretions = (Urine Flow Rate)(Urine Concentrations)  This can also be used for excretion

Definitions & Relevance:

Inulin: This is a substance that is foreign to the body. It is freely removed through filtration but is not secreted or reabsorbed.It is important because the renal clearance rate (RCV) will equal the GFR. If a substance (not inulin) is reabsorbed, it will have a value less than the GFR, however; if a substance is secreted and not reabsorbed, it will have a higher value.

PAH (para amino hippurate): This is a weak acid that is filtered and secreted and is not reabsorbed, therefore; would result in a greater value then the GFR. When a PAH is used for testing the figures can be used to represent ERPF. The ERPF can be determined by RCV of the PAH.

Creatinine: This is a substance that is filtered normally and has a very slight secretion rate as well. Because of these characteristics, it acts somewhat like Inulin enabling it to be a, somewhat, useful tool for estimating GFR. Slightly over-estimates GFR.

Lab Problems:

Question 1:

Given:
Plasma inulin=0.14 mg/ml
Urine Inulin=20 mg/ml
Urine Flow Rate=9 ml/ 10 min = 0.9 ml/min
Solve to find…
GFR

*1st: Remember the renal clearance value (RCVinulin) of inulin is equal to the GFR (what we are

trying to find). So if you find the RCVinulin, you will essentially solve the problem.

*2nd: Find the equation for the RCV on the first page and copy it.

RCVs = Us x R

*3rd: Fill in the equation with the values given

RCVinulin = 20 mg/ml x 0.9 ml/min

0.14 mg/ml

*4th: Solve for the unknown variable. In this case the unknown is on the right side, therefore; you

simply have to do the math on the right side with a calculator.

RCVinulin ≈ 128.6 ml/min

*5th: Since the RCV is the same as the GFR then we can assume that the GFR, for the patient in

question is about 128.6 ml/minthe answer for the first question

***Other questions:

Is the GFR normal? To answer this, simply look at the normal values for the GFR on the first page.

The normal value for GFR is 125 ml/min, therefore; we can say that this patient DOES indeed have a normal GFR.

Does the GRF indicate any possible Goldblatt Renal Hypertension? Because the patients GRF is

Normal, it DOES NOT indicate Goldblatt Renal HTN

Question 2:

Given:
Tm (reabsorption) for glucose 320 mg/min
Plasma glucose is 450 mg %
The patient is the same as question 1
Solve to find…
How much glucose will spill into urine

*1st: Remember that glucose is not secreted, it is only filtered out in the glomerulus. Also

remember that mg % is the same as saying mg/100ml. Start by converting the 450 mg %. To do this, simply divide 450 mg by 100 ml and you end up with 4.5 mg/ml.

*2nd: Now think what the question is asking. It is looking for the amount that will spill, or be

excreted into the urine. Look at the equation for excretion.

Excretions = filtrations + secretions - reabsorptions

We know that the secretion is going to be 0 because glucose is not secreted, only filtered. The reabsorption rate was given as 320 mg/min. Knowing that, rewrite the equation putting in what we know.

Excretionglucose = filtrationglucose – 320 mg/min

I removed the secretion rate value because it was 0 and did not need to be represented.

*3rd: With the given information, we can find the filtration rate for the glucose. The equation is on

the first page.

Filtrations = (GFR) x (Plasma Concentrations)

The GFR can be found in the answer to the question 1 and the glucose plasma

concentration was the figure that we converted in step 1 (4.5 mg/min). Now fill in the equation for filtration rate using the known information that we have and solve.

Filtrations = (128.6 ml/min) x (4.5 mg/ml) = 578.7 mg/min

*4th: Now plug the new found value for the filtration rate into the equation from the 2nd step and

solve for the excretion rate.

Excretionglucose = 578.7 mg/min – 320 mg/min = 258.7 mg/min

*5th: The excretion rate of the glucose is 258.7 mg/min. This is indicating that the patient is

spilling glucose into the urine, which is abnormal. The most common cause for this is that patients blood sugar is too high, probably from diabetes.

Question 3:

Given:

Plasma inulin is 1.0 mg/ml
Urine inulin is 41 mg/ml
Urine flow rate is 2.0 ml/min
Tm (secretion) for PAH is 80 mg/min
Use 90% bladder clearance for PAH
Plasma PAH is 0.1 mg/ml
Urine PAH is 20 mg/ml
Hematocrit (HCT) is about 40%
MCV is 93 cubic microns

Solve to find…
A—The renal clearance value (RCV) for PAH
B—The secretion rate for PAH
C—The total whole blood flow through the renal system per minute

Part A:

*1st: For this part of the question we are looking for the renal clearance value for PAH.

To start the equation to find this is given on the first page. Copy the equation down below.

RCVPAH = UPAH x R

PPAH

*2nd: Fill in the equation above for Renal Clearance with the information given in the original

problem. To find out what each variable should be replaced with see the first page where the equation was originally given.

RCVPAH = (20 mg/ml) x (2.0 ml/min)

(0.1 mg/ml)

*3rd: To find the RCV, simply use a calculator and solve the right side of the equation. The answer

to this is the solution to part A of this problem.

RCVPAH = 400 ml/min

Part B:

*1st: In this part of the problem we are looking to find the secretion rate of the PAH. Refer to the

first page and find the equation for the secretion rate (Hint: it was formed by rearranging the excretion rate equation). Write this equation from below.

Secretions = Excretions – Filtrations + Reabsorptions

*2nd: Now this gets a little trickier because we must solve other equations to find the variable to fit

into the equation above. Lets start with excretion: the equation for it is on the first page. Since we only have given information to fill in urine flow rate and urine concentration you must keep this in mind when selecting equations to use, as there are two.

Excretions = (Urinary flow rate)(urine concentrations)

*3rd: Solve this equation to find excretion by filling in the variables with the given information.

Excretions = (2.0 ml/min)(20 mg/ml) = 40 mg/min

(The ml’s cancel each other out, making the final unit mg/min)

*4th: Now that we have the excretion we must find the filtered amount. Find the equation for the

filtered amount on the first page and copy it here.

FiltrationPAH = (GFR) x (Plasma ConcentrationPAH)

*5th: Again, it gets a little more complicated here because the GFR is not given. Remember from

the definition of inulin that the clearance rate of inulin is equal to the GFR. So we must again find the inulin RCV to find the GFR of the patient. Copy that equation from page one and fill in the variables with the given information.

RCVs = Us x R

RCVs = (41 mg/ml) x (2.0 ml/min) = 82 ml/min

1.0 mg/ml

*6th: Now that we have the GFR we can plug that into the equation for filtration with the given

plasma concentration of the PAH.

FiltrationPAH = (GFR) x (Plasma ConcentrationPAH)

FiltrationPAH = (82 ml/min) x (0.1 mg/ml) = 8.2 mg/min

*7th: Referring back to our original equation (solving for secretion), we now have two of the three

components needed: excretion and filtration. After all that hard work we finally get a break: if you look at the definition for PAH, there is no reabsorption making that value 0. We have all the components to the original equation that we need!!! Copy it down here and fill in all the variables and solve it.

Secretions = Excretions – Filtrations + Reabsorptions

Secretions = (40 mg/min) – (8.2 mg/min) + 0 = 31.8 mg/min

*8th: Write the answer to that on your paper and dance around the room because you just did the

most complicated math problem that you will hopefully ever have to do!!!

Part C:

*1st: The total whole blood flow refers to the total amount of blood that flows through the right

and left renal arteries. Basically we are going to start with the ERPF and work backward. If you look back at the definition of PAH you will see that the RCV of PAH is also the ERPF for the patient. We said earlier that 90% of the blood flow and plasma that reaches the glomerulus. Since we know the ERPF (400 ml/min  from Part A) we can start there and work backwards.

ERPF ≈ 0.9(RPF) ∴ RPF ≈ ERPF/0.9

RPF ≈ (400 ml/min) / (0.9) ≈ 444.44 ml/min

Part C (cont.):

*2nd: Now think about what we’re looking for. The RPF is the plasma flow and to get this figure

from the RBF (what we’re looking for), we took the RBF and subtracted the hematocrit (HCT) to find the plasma. Look back to the equation on the first page showing the conversion from RBF to RPF and then reverse it… (The HCT was given)

RPF ≈ RBF (1 – HCT) ∴ RBF ≈ RPF/(1 - HCT)

RBF ≈ (444.44 ml/min) / (1 - .4) ≈ 740.73 ml/min

*3rd: Once again you can get down and boogie because we have finished another difficult problem.

Additional Renal Problems

Renal Calculations Worksheet 2 (extra practice)

(TL glu) or Filtered Load = GFR X [S]p glucose

Filtered Load = (125 ml/min) X (4.5 mg/ml)

= 562.5 mg/min

Tm for glucose = 320 mg/min

Excretionglucose = 562.5 – 320 = 242.5 mg/min

a.) GFR = RCVinulin = 115 ml/min (given, trick question)

b.) (TL) or Filtered Load = GFR X [S]p

Filtered Load = (115 ml/min) X (2.0 mg/ml)

= 230 mg/min

c.)Excretion = Ru X [S]u

= (2.0 ml/min) X (190 mg/ml)

= 380 mg/min

d.)Secretion = Excretion – Filtration + Reabsorption

= (380 mg/min) – (230 mg/min) + 0

= 150 mg/min

RCV = [S]u X Ru.

[S]p

= (40 mg/ml) X (1.0 ml/min)

(0.4 mg/ml)

= 100 ml/min

a.) RCVPAH = [S]PAH X Ru

[S]PAH

= (110 mg/ml) X (1.5 ml/min)

(0.2 mg/ml)

= 825 ml/min

RCVPAH = ERPF

RPF ≈ ERPF

0.9

RPF ≈ 825 ml/min

0.9

RPF ≈ 916.67 ml/min

b.)RBF ≈ RPF

(1-HCT)

≈ 916.67 ml/min

1 – 0.42

≈ 1580.47 ml/min

a.) GFR = RCVinulin = [S]u X Ru.

[S]p

= (160 mg/ml) X (0.75 ml/ml

(2 mg/ml) = 60 ml/min

b.) RCVPAH = [S]PAH X Ru

[S]PAH

= (78 mg/ml) X (0.75 ml/min)

(0.2 mg/ml)

= 292.5 ml/min

RCVPAH = ERPF

RPF ≈ ERPF

0.9

RPF ≈ 292.5 ml/min

0.9

RPF ≈ 325 ml/min

RBF ≈ RPF

(1-HCT)

≈ 325 ml/min

1 – 0.43

≈ 570.18 ml/min