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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

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## Calculations and Pharmacokinetics

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**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 • Clinical Laboratory Tests • IV infusions flow rates • Parenteral Nutrition • To practice using the reviewed calculation concepts • To review important pharmacokinetic concepts**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)**Important Calculation Concepts**PATIENT PARAMETERS**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)**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]**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**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.**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**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**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**Important Calculation Concepts**MEASUREMENTS**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!**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?**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?**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**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**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?**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**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**Important Calculation Concepts**CONCENTRATIONS**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?**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.**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%**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**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**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?**Important Calculation Concepts**ELECTROLYTE SOLUTIONS**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**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**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**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**Important Calculation Concepts**CLINICAL LABORATORY TESTS**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)]**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)**Important Calculation Concepts**INFUSION FLOW RATES**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?**Important Calculation Concepts**TOTAL PARENTERAL NUTRITION**TOTAL PARENTERAL NUTRUTION**• Provides a patient with all nutritional requirements • Composition: • Fluids • Carbohydrates (Dextrose) • Protein (Amino acids) • Fats • Vitamins, minerals, trace elements • Electrolytes**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**TPN HIGHLIGHTS**• Each TPN component provides a defined amount of calories • COMMIT THESE TO MEMORY!!**Important Calculation Concepts**PHARMACOKINETICS**PHARMACOKINETCS**• Pharmacokinetics describes what happens to a drug or substance inside of the body**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**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?**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?**THANK YOU!**• Joy A. Awoniyi • Email your questions, comments or concerns to Joy.awoniyi@va.gov

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