Vet Homework #1, Digoxin

17 September 1995

ONE-COMPARTMENT MODEL CALCULATIONS:

You are presented with a 2.5 year old neutered male dog name Star that has moderate congestive heart failure and you decide to administer digoxin. Although you are also going to look up an accepted beginning dose for oral administration of tablets in a reference, just for the fun of it you decide to see what it might take to produce a plasma concentration of 2 ng/mL by intravenous infusion. The only clearance data you can find is 3.94 mL/min/kg [Craigmill, Sundlof, & Riviere, 1994, p477]. (Note: For purposes of this exercise, we are going to assume there was a misprint in the reference and that the CL is "per min" and "per h" as is implied on p. 34. The new Cl would be 236 ml/h/kg. This assumption is reasonable since Ravis, Pedersoli, and Turco [1987] found a CL after a single IV dose of digoxin of 207 ml/h/kg.) Star weighs 39#. Given these data what infusion rate would be required to produce the desired concentration?

ANS: What is the calculated initial infusion rate of digoxin?

Star = 39# = 17.7 kg

Convert CL to L/h: (3.94 mL/min/kg ) * (1 L/1000 mL) * (60 min/h) = 0.236 L/h/kg

Whole body clearance for Star is: 0.236 L/h/kg * 17.7 kg = 4.18 L/h

Convert mcg/mL to mcg/L: (0.002 mcg/mL) x (1000 mL/L) = 2.0 mcg/L

Infusion rate to produce 2.0 mcg/L is:

2.0 mcg/L * 4.18 L/h * = 8.36 mcg/h

How does the infusion rate you calculated compare to the normally recommended oral dosage regimen? A typical beginning regimen for oral administration is: 0.011 mg/kg q12h.

ANS: How does the calculated infusion rate compare to normally recommended PO dosage regimen?

8.36 mcg/h * 12 h = 100.3 mcg / 12h;

Convert to mg/12h: 100 mcg/12 * 1 mg/1000mcg = 0.100mg/12h infusion for the 17.7 kg dog.

A typically recommended dose regimen of digoxin for a dog is 0.011 mg/kg q12h.

For Star, this would be 0.195 mg PO q12h (0.011 mg/kg * 17.7 kg = 0.195 mg / dog q12h.).

Thus, the oral dose given q12h is computed to be 1.95 times greater than the amount given by intravenous infusion over the same period of time.

What reasons can you give for the difference in total dose required to maintain a therapeutic concentration of digoxin for the 12h.?

CL/F determined from oral dosage of digoxin was found to be 357 mL/h/kg [Ravis, Pedersoli, & Turco, 1987]. The F is necessary because the approached used could not separate the failure to absorb drug from the process of eliminating it from the central circulation. A calculation using this value to calculate a dose rate for digoxin given q12h would yield:

ANS: Star's whole body clearance/F: 357 mL/h/kg * 17.7 kg = 6,319 mL/h; Because the route by which we will give the drug is PO and the F in the CL/F accounts for this, we will drop the F term in the following calculations.

Dose Rate: 6,319 mL/h * 0.002 mcg/mL = 12.64 mcg / h

Dose/12h: 12.64 mcg/h * 12h = 152 mcg q12h

Note that this is actually reasonably close to a typically recommended dose.

What factors could account for the difference in estimated dose rate to produce a steady state concentration of 0.002 mcg/mL by intravenous infusion or an "average" concentration of 0.002 mcg/mL via oral administration? [No answer given here.]

Relationship between CL, Vd, Ke, and Half-life of elimination:

From orally determined CL/F and Vd/F [Ravis, Pedersoli, & Turco, 1987]

If Vd/F determined from the "oral administration data" is 16 L/kg and CL/F = 357 mL/h/kg, what would you estimate the Ke and, hence, elimination half-life of digoxin to be?

ANS:

Convert CL/F to L/h/kg: 357 mL/h/kg * 1 L/1000 mL = 0.357 L/h/kg

Vd/F = 16 L/kg; CL/F = 0.357 L/h/kg;

Ke = (0.357 L/h/kg) / (16 L/kg) = 0.0223 / h

Half-life = 0.693 / 0.00223 = 31.1 h

Note that this is very close to typically published values of half-life for digoxin, e.g., 31 h

From IV determined CL and Vd [Craigmill, Sundloff, & Riviere, 1994]

Vd(area)= 9.46 L/kg: This value for Vd was obtained in the same experiments that led to the CL used above. These were determined from IV administered drug. Using this Vd estimate and the "corrected" CL of 3.94 mL/kg/min, what would be the Ke and half-life?

ANS:

Convert CL to L/kg/h:

3.94 mL/kg/min * 1 L/1000 mL * 60 min /h = 0.236 L/kg/h

Ke = (0.236 L/kg/h) / (9.46 L/kg) = 0.0249 /h

Half-life = 0.693 / 0.0249/h = 27.8 h half-life

Assume that one began an infusion of digoxin into Star at a dose rate of 8.36 mcg/h, the rate estimated from "population" pharmacokinetic estimates. After an "incredibly long" (and one might say unrealistic!) infusion period of 6.5 days (155h), the concentration was found to be 0.001 mcg/mL rather than the desired concentration of 0.002 mcg/mL.

What new Dose Rate would you recommend to achieve the desired concentration? __________ mcg/h dose rate

ANS:

New Dose Rate = 8.36 mcg/h * [(0.002 mcg/mL) / (0.001 mcg/mL)] = 16.72 mcg/h

Assume you had predicted that it would be 6.5 days before the concentration would reach a steady state and decided that this was an intolerably long time before it would be near the steady state concentration. What dose of drug would you have given in a single injection (with digoxin you WOULD NOT ACTUALLY DO THIS!!!) to produce the desired concentration much more rapidly? That is, what would be a reasonable Loading Dose? (Use the initial population values for this estimate, not the "corrected" one since you would not have known it at the time you had to make a decision.)

ANS: mg/ml = mg / ml; mg = ml * mg/ml

Vd(area) = 9.46 L/kg (Craigmill, Sundloff, & Riviere, 1994) (Used because was computed from IV study)

Desired concentration = 0.002 mcg/mL; Convert to Liters: 2 mcg/L

The patient, Star, has a body mass of 17.7 kg

IV Loading Dose = 2 mcg/L * 9.46 L/kg * 17.7 kg/Star = 335 mcg digoxin / Star (the patient)

After giving this dose, one would begin the infusion.

Assume that you were going to give a loading dose to Star PO. What dose you give to obtain an "initial" concentration of 0.002 mcg/mL?

ANS:

Vd(area)/F = 16 L/kg

Desired Css(ave) = 0.002 mcg/mL or 2 mcg/L

PO Loading dose = 16 L/kg * 2 mcg/L * 17.7 kg/Star = 566 mcg/Star

Note: This assumes that absorption is rapid relative to the elimination rate. Is it likely that the loading dose will actually produce an initial concentration of 2 mcg/L when given orally?

Accumulation:

Assume that you are going to give Star, a 17.7 kg dog, a 0.011 mg/kg dose of digoxin IV (Note: You should NOT DO THIS for reasons that will be apparent later in your studies!). Further assume that you measure the concentration at 1 hour and again at 12 hours after giving the dose. Using a "population" estimate for elimination half-life of 31 h, what concentrations would you predict at the two times

ANS:

Elimination Half-life = 31h; (0.0224/h = Ke)

Vd(area, IV) = 9.46 L/kg

Dose = 0.011 mg/kg given IV

Predicted concentration at:

Note: We ignored the mass (kg) because both dose and Vd(area) were in the same units.

What "peak" concentration would you expect after more than 7 days of giving this dose q12h?

ANS:

Elmination Half-life = 31h; (0.0224/h = Ke)

Vd(area,IV) =9.46 L/kg

Dose = 0.011 mg/Kg given IV

Peak at Steady State (Css(max)) = F * D/Vd * 1 / [1-e(-Ke * T)]

= (1 * (0.011mg/kg) / (9.46 L/kg) ) * 1 / [1 - e(-0.0224/h * 12h)]

= 0.00116 mg/L * 4.243 = 0.0049 mg/L or 4.9 mcg/L

Trough at Steady State (Css(min)) = Css(max) * e(-Ke * T)

= 4.9 mcg/L * e(-0.0224/h * 12h) = 3.7 mcg/L at 12 h

Note that the normally recommended therapeutic window for digoxin is 0.9 to 3.0 mcg/L. Both the peak and trough at steady state exceed this value! Yet, the dose used was the one normally recommended! What is the difference? Remember that the usual route for this dose rate is PO! We gave it IV. Therefore, the F was 1.0 versus 0.71 [Craigmill, Sundloff, and Riviere 1994, p 478] as it would be for oral administration.

Using an F of 0.71, the respective values would be:

Css(max) = 4.9 mcg/L * 0.71 = 3.5 mcg/L

Css(max) = 3.7 mcg/L * 0.71 = 2.63 mcg/L

Note that this still predicts a potentially toxic concentration for the peak. However, because Ka for orally administered drug would be very slow compared to infinity for IV administered drug, the oscillation would be much lower, Therefore, the peaks would be lower and the trough slightly higher than the values we obtained in this estimate. The regimen would probably be ok. In fact, it is amazing that these theoretically derived values are as close as they are to what we use practically! Of course, they should be!

TWO COMPARTMENT MODEL:

Given the following values, compute the concentration of drug in the plasma of an animal at the listed times. [Baggot, 1977]

Drug = Oxytetracycline. given to a dog at a dose of 5 mg/kg IV

Alpha = 0.0125/min; (0.75/h)

A = 2.85 mg/L

Beta = 0.0020 /min, (0.12/h)

B = 1.75 mg/L

What is the half-life for --

alpha phase? ______h

beta phase? ______h

How long will be required for the "alpha" phase to be 97% complete? __________h

How long will be required for the "beta" phase to be 97% complete? __________h

Make a table for the values as follows:

TIME Conc of "A" term Conc of "B" term Sum of A & B A as Fraction of Sum

-------- ---------------------- --------------------- ------------------- -------------------------

0.5 h ________ _________ _______ __________

1.0 h ________ _________ _______ __________

2.0 h ________ _________ _______ __________

4.0 h ________ _________ _______ __________

8.0 h ________ _________ _______ __________

16.0 h ________ _________ _______ __________

How long does the "A" term continue to be a significant factor in determining the concentration?

ANS:

What is the half-life for --

alpha phase? 0.924h

beta phase? 5.775h

How long will be required for the "alpha" phase to be 97% complete? ___4.62__h

How long will be required for the "beta" phase to be 97% complete? ___28.9__h

Make a table for the values as follows:


  Time             A Term    B Term  Combined   alpha as frcn of  
 after                                               total        
  Dose                                                            

(hours)            (mg/L)    (mg/L)   (mg/L)       (Fraction)     

     0.5              1.959    1.648     3.607              0.543 

       1              1.346    1.552     2.898              0.464 

                                                                  

       2              0.636    1.377     2.013              0.316 

       4              0.142    1.083     1.225              0.116 

                                                                  

       8              0.007     0.67     0.677               0.01 

      16                 0.    0.257     0.257                 0. 

Note that the values for the A term and the fraction are     
NOT ACTUALLY  zero. They are very small, in the range of x   
* 10E-5!                                                     
                                                             



References:

Ravis, R. W.M. Pedersoli, and J.D. Turco. 1987. Pharmacokinetics and interactions of digoxin with phenobarbital in dogs. Am J Vet Res 48(8):1244-1249.

Craigmill, A.L., S.F. Sundlof, and J.E. Riviere. 1994. Handbook of Comparative Pharmacokinetics and Residues of Veterinary Therapeutic Drugs. CRC Press, Inc., Boca Raton, FL. pp 665.

Baggot, J.D., 1977. Principles of Drug Disposition in Domestic Animals, W.B. Saunders Co., Philadelphia,

 


index lecture menu

Send suggestions / questions
Last modified: 8/18/96 glc