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# Calculations and Pharmacokinetics

Calculations and Pharmacokinetics. Dr. J. Domenech CALCULATIONS AND PHARMACOKINETICS. Objectives. To emphasize the importance of calculations To review important calculation concepts related to Patient Parameters Measurements Concentrations Electrolyte Solutions

## Calculations and Pharmacokinetics

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### Presentation Transcript

1. Calculations and Pharmacokinetics Dr. J. Domenech CALCULATIONS AND PHARMACOKINETICS

2. Objectives • To emphasize the importance of calculations • To review important calculation concepts related to • Patient Parameters • Measurements • Concentrations • Electrolyte Solutions • Clinical Laboratory Tests • IV infusions flow rates • Parenteral Nutrition • To practice using the reviewed calculation concepts • To review important pharmacokinetic concepts

3. WHY SHOULD YOU CARE? • You can’t escape it • You need it to pass the NAPLEX • It can be applied to any pharmacy-related specialty • It may save your patient’s money • It may save your patient’s life (and your license)

4. Important Calculation Concepts PATIENT PARAMETERS

5. BSA and IBW • Body Surface Area (BSA), m2 • BSA = Weight (kg)0.425 x Height (cm)0.725 x 0.007184 • BSA = • Ideal Body Weight, kg • Males: 50 +( 2.3 x inches greater 60) • Females: 45.5 + (2.3 x inches greater than 60)

6. BODY MASS INDEX (BMI) • Measure of body fat to help categorize patients and assess risk of morbidity for certain diseases • BMI = weight(kg) ÷ [height (m)2]

7. Creatinine Clearance (CrCl) • Formula used to estimate renal function • Not useful when renal function is fluctuating rapidly • Used in the dosing of several medications • Tamiflu, Antibiotics, Low-molecular weight heparins Multiply by 0.85 if female

8. EXAMPLE Mack “Big Mack” Donald is a 55 year old man recently admitted to your service after choking on a Zebra Cake. He is 6’2” and 980 pounds. Calculate his BSA and BMI.

9. EXAMPLE Body Surface Area Body Mass Index Height in meters = 1.88 m Answer: 126 How do we classify Big Mack’s BMI? Underweight Normal Overweight Obese • Height in centimeters: 6’2” = 74 inches • 74 inches x 2.54 cm/inch = 187.96 cm • Weight in kilograms: 980 lbs • 980 lbs x 1 kg/2.2lbs = 445.45 kg • Answer: 4.82

10. EXAMPLE Mack had a DVT during admission. The doctors wish to start him on Enoxaparin (Lovenox), a renally dosed medication. His SCr is 2.3. Calculate his IBW and Creatinine Clearance

11. EXAMPLE Ideal Body Weight Creatinine Clearance Use IBW of 82.2kg Answer: 42.19 mL/min Dose of Lovenox is 50% if CrCl is less than 30 mL/min • Inches over 60 • 74 – 60 = 14 inches • IBW = 50kg + 2.3 (14) • Answer: 82.2 kg

12. Important Calculation Concepts MEASUREMENTS

13. The Basics Of Measurement • Irrelevant information is often given • Fundamental information is often left out • How many grams are in a kilogram? • How many milliliters are in a fluid ounce? • Start each problem by assessing what you know and what you want to know • Double check!

14. EXAMPLE • What information is irrelevant? • What fundamental information do you need to know to complete this problem? • What do you know? • What do you want to know? A cough syrup contains 10mg of dextromethorphan per 5 mL. Your pharmacy has 6 bottles left in stock. Each bottle contains 120mL of the syrup. How many grams of the drug are in one bottle?

15. EXAMPLE A cough syrup contains 10mg of dextromethorphan per 5 mL. Your pharmacy has 6 bottles left in stock. Each bottle contains 120mL of the syrup. How many grams of the drug are in one bottle?

16. Specific Gravity Definition Equations Grams = mL x SG mL = Grams ÷ SG SG = Grams ÷ mL Do not confuse with density or concentration • A ratio of the weight of any substance in relation to the weight of an equal volume of water • Water is used because 1g of water is equal to 1mL • If SG > 1 the substance is heavier than water • If SG < 1 the substance is lighter

17. EXAMPLE There is 10 grams of glycerin in a 500mL solution. 300 mL of glycerin weighs 165 grams. What is the specific gravity of glycerin? • Specific Gravity = grams / mL • 10 g / 500mL ? • 165 g/ 300mL ? • Think of SG as a conversion factor • 10 grams of glycerin does not EQUAL 500 mL of glycerin • 165 grams of Glycerin EQUALS 300 mL of glycerin

18. EXAMPLE There is 10 grams of glycerin in a 500mL solution. 300 mL of glycerin weighs 165 grams. What is the specific gravity of glycerin?

19. EXAMPLE What is the weight, in grams of a 2 fluid ounces of a liquid with a specific gravity of 1.118? • Grams = mL x SG mL SG

20. DOUBLE CHECK • DOES THE ANSWER MAKE SENSE? • If the SG is 1.118, the substance is • Heavier than water? • Lighter than water? • 2 fluid ounces of water (60mL) = 60 grams • The answer is 67.08 grams which is > 60 grams • Any answer below 60 would be WRONG

21. Important Calculation Concepts CONCENTRATIONS

22. PERCENTAGE PREPARATIONS

23. EXAMPLEWeight in Volume • 5% Dextrose by definition means 5 grams in 100mL How many grams of dextrose are required to prepare 4000 mL of a 5% solution?

24. ALLIGATION • Alligation is a method of solving problems that involves the mixing of solutions or substances with different percentage strengths • Alligation alternate is a method to calculate the number of parts of two or more components of different strengths mixed to prepare a desired strength.

25. EXAMPLEALLIGATION What is the percentage of zinc oxide in an ointment prepared by mixing 200 grams of a 10% ointment, 50 grams of a 20% ointment, and 100 grams of a 5% ointment 35 grams ÷ 350 grams = 0.10 x 100% = 10%

26. EXAMPLEALLIGATION ALTERNATE A pharmacist needs to prepare 50 mL of 3% hydrogen peroxide solution. He has 30% and 1.5% solutions in stock. How many mL of each should he use? • A – C = Y • C – B = X • X and Y are proportions of A and B (respectively) needed for the entire preparation

27. EXAMPLEALLIGATION ALTERNATE A pharmacist needs to prepare 50 mL of 3% hydrogen peroxide solution. He has 30% and 1.5% solutions in stock. How many mL of each should he use? • 1.5 parts of 30% • 27 parts of 1.5% • Total Parts: 28.5

28. EXAMPLEALLIGATION ALTERNATE • We need 1.5 of all 28.5 parts to contain 30% Hydrogen peroxide • 1.5/28.5 = x/50mL • X = 2.63mL of 30% • We need 27 parts of all 28.5 parts to contain 3% Hydrogen peroxide • 27/28.5 = x/50mL • X = 47.37mL of 1.5% A pharmacist needs to prepare 50 mL of 3% hydrogen peroxide solution. He has 30% and 1.5% solutions in stock. How many mL of each should he use?

29. Important Calculation Concepts ELECTROLYTE SOLUTIONS

30. MILLIEQUIVALENTS • Unit of measure related to the total number of ionic charges in a solution • Measures the chemical activity of an electrolyte relative to 1mg of hydrogen • 1 mEq represents the milligrams equal to its equivalent weight, taking into account the valency

31. EXAMPLE • Molecular weight of Sodium = 23 • Molecular weight of Chloride = 35.5 • Molecular weight of Sodium chloride = 58.5 • Valency = 1 (Na+, Cl+) What is the concentration, in milligrams per milliliter, of a solution containing 2mEq of NaCl per mL? 117 mg

32. MOLARITY How many milligrams would 3 mmol of monobasic sodium phosphate (MW = 138) weigh? • Millimoles • 1 mol = molecular weight in grams, therefore… • mmol = molecular weight in milligrams • Molarity is the number of mmols in a solution

33. OSMOLARITY • mOsmoles represent the number of particles in a solution when the substance dissociates • NaCl = 2 mOsmol • CaCl2 = 3 mOsmol • Anhydrous dextrose = 1 mOsmol • Osmolality is the milliosmoles of solute in a solution

34. Important Calculation Concepts CLINICAL LABORATORY TESTS

35. CALCIUM-ALBUMIN • Almost 50% of calcium is bound to plasma proteins • If levels of proteins are low, then the serum calcium may be inaccurate • If albumin is low, calcium will appear to be low, when it levels are actually within normal limits • Correct levels when Albumin is less than 4 g/dL • Corrected Ca2+ = Ca2+ + 0.8 [4 – albumin(g/dL)]

36. SODIUM-GLUCOSE • In patients with hyperglycemia, glucose does not enter the cell, causing a shift of fluid from intracellular to extracellular • The shift of fluid dilutes the concentration of sodium in the extracellular fluid • This type of hyponatremia (translational) does not need to be treated, instead control of glucose levels is indicated • Corrected Na+ = Na+ + 0.016(Serum Glucose – 100)

37. Important Calculation Concepts INFUSION FLOW RATES

38. EXAMPLE A physician orders 2 liters of D5W to be administered over 8 hours. The IV administration set in your pharmacy delivers 10 drops/mL. How many drops per minute should the patient be set to receive?

39. Important Calculation Concepts TOTAL PARENTERAL NUTRITION

40. TOTAL PARENTERAL NUTRUTION • Provides a patient with all nutritional requirements • Composition: • Fluids • Carbohydrates (Dextrose) • Protein (Amino acids) • Fats • Vitamins, minerals, trace elements • Electrolytes

41. TPN HIGHLIGHTS • Fluid Requirements: • Generally 30 – 40 mL/kg/day • Calculated: 1500 mL + 20mL (kg over 20) • Protein Requirements • Ambulatory: 0.8 – 1 g/kg/day • Hospitalized: 1.2 – 2 g/kg/day • Non-Protein Requirements (Total Energy Expenditure) • Male: 66.47 + 13.75(kg) + 5.0(cm) – 6.76 (yrs) • Female: 655.1 + 9.6(kg) + 1.85(cm) – 4.68(yrs) • Multiply by 1.2 if confined to a bed, and 1.3 if out of bed

42. TPN HIGHLIGHTS • Each TPN component provides a defined amount of calories • COMMIT THESE TO MEMORY!!

43. Important Calculation Concepts PHARMACOKINETICS

44. PHARMACOKINETCS • Pharmacokinetics describes what happens to a drug or substance inside of the body

45. FIRST-ORDER KINETICS • The amount of drug given is proportional to the change in concentration • The change in drug concentration with respect to time will create a rate constant (k) • C = C0e-kt

46. HALF-LIFE • The half-life (t1/2) is the time required for the concentration of the drug to decrease by one-half • t1/2 = 0.693/k (k is the rate constant) A patient is receiving an antibiotic for the treatment of a respiratory infection. The initial concentration of the drug was 17.9mg/L. The drug has an elimination half-life of 2 hours. How much of the drug is present after 8 hours?

47. EXAMPLE • C = C0e-kt SOLVE FOR “C” • t1/2 = 0.693/k (k is the rate constant) • 2h = 0.693/k • k = 0.347h-1 • C = 17.90e-0.347(8) = 1.11mg/L A patient is receiving an antibiotic for the treatment of a respiratory infection. The initial concentration of the drug was 17.9mg/L. The drug has an elimination half-life of 2 hours. How much of the drug is present after 8 hours?

48. QUESTIONS?

49. THANK YOU! • Joy A. Awoniyi • Email your questions, comments or concerns to Joy.awoniyi@va.gov

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