Basic Pharmacy Calculations and Pharmacokinetes

CALCULATIONS AND PHARMACOKINETICS

Joy A. Awoniyi, PharmD.PGY1 Pharmacy Practice Resident

Miami VA Healthcare System

For The FAMU COPPS Class of 2013

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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

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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)

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PATIENT PARAMETERS

Important Calculation Concepts

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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)

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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]

BMI Classification

< 18.5 Underweight

18.5 to <25 Normal

25 to <30 Overweight

>30 Obese

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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

CrCl (mL/min) Interpretation

60 to 90 Mild renal impairment

30 to < 60 Moderate Renal impairment

15 to <30 Severe Renal impairment

< 15 Kidney failureMultiply by 0.85 if female

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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.

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EXAMPLE

Body Surface Area

• 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

Body Mass Index

• Height in meters = 1.88 m

• Answer: 126

• How do we classify Big Mack’s BMI?• Underweight

• Normal

• Overweight

• Obese

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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

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EXAMPLE

Ideal Body Weight

• Inches over 60• 74 – 60 = 14 inches

• IBW = 50kg + 2.3 (14)

• Answer: 82.2 kg

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

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MEASUREMENTSImportant Calculation Concepts

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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!

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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?

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EXAMPLE

10mg/5mL

120mL/ 1 bottle

1 gram/10

00mg

0.24 grams/ 1 bottle

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?

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Specific Gravity

Definition

• 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

Equations

• Grams = mL x SG

• mL = Grams ÷ SG

• SG = Grams ÷ mL• Do not confuse with density or

concentration

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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

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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?

165 grams

300 mL 0.55

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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

2 fl oz30mL/ 1 fl oz

1.118 67.08 grams

SGmL

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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

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CONCENTRATIONSImportant Calculation Concepts

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PERCENTAGE PREPARATIONS

Weight in volume• Amount of grams in 100mL of solution

Volume in volume• Amount of mL in 100mL of solution

Weight in weight• Amount of grams in 100 grams of substance

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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?

4000mL5

grams/ 100mL

200 grams

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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.

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EXAMPLEALLIGATION

0.10 x 200 g = 20 grams

0.20 x 50 g = 10 grams

0.05 x 100 g = 5 grams

350 g 35 grams

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%

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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?

Percent Available

Percent Desired

Proportions required

A X

C

B Y

• A – C = Y

• C – B = X

• X and Y are proportions of A and B (respectively) needed for the entire preparation

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Percent Available

Percent Desired

Proportions required

30% 1.5

3%

1.5% 27

• 1.5 parts of 30%

• 27 parts of 1.5%

• Total Parts: 28.5

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• 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%




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ELECTROLYTE SOLUTIONS


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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

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Ion Abbr. Valence Atomic Weight

Milliequivalent weight

Aluminum Al3+ 3 27 9

Magnesium Mg2+ 2 24 12

Sodium Na+ 1 23 23

Bicarbonate HCO32- 2 60 ???

MILLIEQUIVALENTS

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• 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?

EXAMPLE

117 mg

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MOLARITY

• Millimoles• 1 mol = molecular weight in grams, therefore…

• mmol = molecular weight in milligrams

• Molarity is the number of mmols in a solution

How many milligrams would 3 mmol of monobasic sodium phosphate (MW = 138) weigh?

3 mmol 138 mg/ 1 mmol 414 mg

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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

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CLINICAL LABORATORY TESTS


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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)]

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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)

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INFUSION FLOW RATES


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EXAMPLE

2 Liters/ 8hours

1000mL/ 1 Liter

1 hour/ 60 min

10 drops/ 1mL

42 drops/ min

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?

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TOTAL PARENTERAL NUTRITIONImportant Calculation Concepts

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TOTAL PARENTERAL NUTRUTION

• Provides a patient with all nutritional requirements

• Composition:• Fluids

• Carbohydrates (Dextrose)

• Protein (Amino acids)

• Fats

• Vitamins, minerals, trace elements

• Electrolytes

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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

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TPN HIGHLIGHTS

Component Units

Dextrose 3.4 kcal per gram

Amino Acids 4 kcal per gram

Lipids 9 kcal per gram

10% lipid emulsion 1.1 kcal per mL

20% lipid emulsion 2 kcal per mL

• Each TPN component provides a defined amount of calories

• COMMIT THESE TO MEMORY!!

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PHARMACOKINETICSImportant Calculation Concepts

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PHARMACOKINETCS

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

Absorption Distribution

Metabolism Excretion

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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

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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?

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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?

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QUESTIONS?

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THANK YOU!

• Joy A. Awoniyi

• Email your questions, comments or concerns to [email protected]

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Basic Pharmacy Calculations and Pharmacokinetes