Renal Clearance answer key PDF

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Renal Physiol Physiology ogy Problems Equati Equations, ons, Brief Explanation Explanations, s, Normal Values, and Answer Key *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/mi ml/min n (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 glomeruli; therefore, this number indicates the estimated, or effective, blood flow that actually reaches the nephrons per minute.



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 = 0.45, or 45% (men) ∴ RPF = 1200 ml/min(1-0.45) = 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 glomeruli per minute. 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 glomeruli is filtered.



ml/min. About 180 L of plasma is filtered daily.



GFR =

180 L/day

This amount is often expressed either as L/day or

= .125 L/min = 125 ml/min

1440 min/day 

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

*Renal Clearance Value (RCVs,or Cs, in ml/min) 

Clearance refers to the amount of a specific substance “S” that is removed from the plasma and excreted in urine in a time frame.



RCVs = Us x V



Us = Urine concentration of substance (s), V = Urine flow rate, & Ps = Plasma concentration of substance (s)



Note: if subject in question is inulin, the RCV inulin will equal the GRF. (See definition of inulin)

Ps

*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, and reabsorption.

 

Excretions = filtrations + secretions - reabsorptions Filtration Load or Tubular Loads = (GFR) x (Plasma Concentrations)



Secretions = Excretions – Filtrations + Reabsorptions



Excretions = (Urine Flow Rate)(Urine Concentrations)  alternative fformula ormula used to calculate e excretion xcretion

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 value (RCV inulin) will equal 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 for renal clearance. PAH (para-aminohippuric acid): This is a weak acid that is filtered and secreted and is not reabsorbed; therefore, its clearance is greater than GFR. When PAH is used for testing, its RCV (RCV PAH ) can be used to represent ERPF. 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 overestimates GFR.

Lab Problems: Question 1a:  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 V Ps *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 ml/min /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 (i.e. not below 125ml/min).

Does the GRF indicate any possible Goldblatt Renal Hypertension? Because the patients GRF is normal, it DOES NO NOT T indicate Goldblatt Renal Hypertension.

Question 1b: 



Given:  Tm (reabsorption) for glucose 320 mg/min 

Plasma glucose is 450 mg %



The patient is the same as question 1a

Solve to find… 

How much glucose will spill into urine (i.e. excretion)

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

reabsorptio io ion ns Excretions = filtrations + secretions - reabsorpt 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.

Excretiongluco glucose se = filtrationglucose – 320 mg/min The secretion rate value was removed from the equation 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 1a 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.

Filtrationgluc glucose ose = (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 2 nd step and solve for the excretion rate.

Excretiongluco glucose se = 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 patient’s blood sugar is too high, probably from diabetes.

Question 2: 



Given:  Plasma inulin is 1.0 mg/ml (Pi)



Plasma PAH is 0.1 mg/ml (PPAH )



Urine inulin is 41 mg/ml (Ui)



Urine PAH is 20 mg/ml (UPAH)



Urine flow rate is 2.0 ml/min (V)



Hematocrit (HCT) is about 40%



Tm (secretion) for PAH is 80 mg/min



MCV is 93 cubic microns



Use 90% bladder clearance for PAH

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 V 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. Let’s 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.

ExcretionPAH = (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 V Ps

ml/min l/min = GFR ml/min) n) = 82 m RCVinulin = (41 mg/ml) x (2.0 ml/mi 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 clearance 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 *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)

4.

(TL glu) or Filtered Load = GFR X P glucose

RCVPAH = ERPF

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

RPF ≈ ERPF 0.9

= 562.5 mg/min 5.

6.

Tm for glucose = 320 mg/min

RPF ≈ 825 ml/min 0.9

Excretion glucose = 562.5 – 320 = 242.5 mg/min

RPF ≈ 916.67 ml/min

a.) GFR = RCVinulin = 115 ml/min (given, no need to calculate)

b.) RBF ≈ RPF (1-HCT) ≈ 916.67 ml/min 1 – 0.42

b.) (TL) or Filtered Load = GFR X PS Filtered Load = (115 ml/min) X (2.0 mg/ml)

≈ 1580.47 ml/min

= 230 mg/min 9.

a.) GFR = RCVinulin = Ui X V. Pi

c.) Excretion = V X US = (160 mg/ml) X (0.75 ml/ml) (2 mg/ml)

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

= 60 ml/min

= 380 mg/min d.) Secretion = Excretion – Filtration + Reabsorption = (380 mg/min) mg/min) + 0

–

b.) RCVPAH = UPAH X V PPAH = (78 mg/ml) X (0.75 ml/min) (0.2 mg/ml)

(230

= 292.5 ml/min = 150 mg/min 7.

RCV = US X V Ps = (40 mg/ml) X (1.0 ml/min) (0.4 mg/ml) = 100 ml/min

RCVPAH = ERPF RPF ≈ ERPF 0.9 RPF ≈ 292.5 ml/min 0.9 RPF ≈ 325 ml/min

8.

a.) RCVPAH = UPAH X V PPAH = (110 mg/ml) X (1.5 ml/min) (0.2 mg/ml) = 825 ml/min

RBF ≈ RPF (1-HCT) ≈ 325 ml/min 1 – 0.43 ≈ 570.18 ml/min...