2006 Pharmaceutics

7/29/2019 2006 Pharmaceutics

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DRUG STABILITY KINETICS

The basic equations wh ich describe the loss of drug from a dosage form on a she lf also apply to the

loss of drug from the body. The ones given are the only ones you are likely to need to know.

ZERO ORDER C = Co - kt k = zero order rate constant S lope = -k

Concentration vs. Time plot is linear. (Both Y and X axis areregular spacing; e.g., not logarithmic.)

The R ATE of change is constant.

The RATE = -dA = k (concentration or amount/time)

dt

The same AM OU NT of drug is lost per un it of tim e. In zero order,

the half life increases as the concentration increases.

FIRST ORDER C = (Co) (e )-kt

Concentration vs. Time plot is curved when using regular(e.g., linear) axes.

The R ATE of change is concentration dependent.

The RATE = -dA = kA

dt

Therefore the same FRACTION (or percent) o f drug is lost per un it

of time; independent of concentration.

For first order kinetics, the Concentration vs. Time plot is linear

when logarithmic Y axis is used and one plots the log of the

concentration versus time.

k = first order rate constant for eliminat ion (as fraction per unit of

time; e.g., k = per unit of time)

k = Clearance Clearance = Volume cleared

Vd Time

Vd = "Apparent" Volume of Distribution Log C = Log Co kt Slope of l ine is k

2.303 2.303

If you plot the natural log of Concentration versus Time, then Ln C = Ln Co kt. The slope of the line

is then determined by k alone.

HALF LIFE: Time required for elimination of one-half of a drug (by either zero or first order

degradation). Half life is often used in dosing situations to measure drug elimination from the body.

1/2 oZero order t = C = mg / ml = minutes

2K mg / ml / min

1/2First order t = 0.693 = 1 = minutes

k 1 / min


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SHELF LIFE: The expiration date is based on the time required for the drug product to degrade below

its USP established potency level. In the example below, this is assumed to be the time at which 90

percent of the drug remains (actual USP requirements range from 80% to 97% ).

90 90First order t = C = 90 = e t = 0.105-kt

oC 100 k

o 90 oZero order C - C = Kt t = 0.1 C

o oC - 0.9C = Kt k

TABLE 1. Stability of Drugs in Pharmaceutical Formulations

A. Factors Which Decrease Drug Stability

Heat (temperature) know storage guidelines

Light (especially u/v) light resistant containers

Moisture film coat tablets and do not store in bathroom (heat and moisture)

pH solution buffers

Excip ient incompatib ility modify formulation

B. Oxidation can be Reduced by

21. Small amounts of oxygen (O ) - autoxidation

2. Nitrogen, Carbon Dioxide used to replace air in container

3. Temperature 0-5 degrees (1/2 rate)

4. Trace amounts of heavy metals (chromium, copper, iron) catalyze oxidation and cause destruction

of: penicillin, epinephr ine, phenylephrine, lincomycin and procaine

5. Chelating agents - EDTA (ethylene diamine tetraacetic acid), dihydroethylglycine, citric, tartaric,

gluconic, and saccharic acid

6. Antioxidants

For aqueous solutions: Sodium sulfite (hi pH range), Sodium Metabisulfite (low pH range),

Sodium Bisulfite (intermediate pH range), Acetone Metabisulfite, Ascorbic Acid, Thioglycerol,

Sodium Thiosulfate, Cysteine Hydrochloride

For non-aqueous solutions: Ascorbyl Palmitate, Hydroquinone, Propyl Gallate,

Nordihydroguaiaretic, Butylated Hydroxytoluene, Butylated Hydroxyanisole and Alpha Tocopherol

(vitamin E)

C. Hydrolysis Rate can be Reduced by

Temperature (0 - 5 degrees C. ) Humidity < 40%

Dessicant in packaging Multi-layered tablets

Film coating Buffers for solutions

Suspensions for liquids

Absorbent diluent in tablet or capsule (Kaolin, Mg Oxide, Ca carbonate)

EXAMPLES

1. At room temperature the rate constant for loss of an antibiotic solution is k = 0.0357 days .-1

What is the shelf life at room temperature?A. 0.0357 days ANS: Units are reciprocal time; so First order

B. 3 days

90C. 14 days t = 0.105 = 0.105

D. 3 hours K 0.0357

E. cannot determine order

90t = 3 days at room temperature


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2. If refrigerated, the antibiotic solution has a rate constant for loss of k = .0075 days . What is the-1

new shelf life?

90A. 0.0357 days ANS: t = 0.105 = 14 days

B. 3 days 0.0075

C. 14 days

D. 3 hours

E. cannot determine order

3. After reconstitution, an antibiotic suspension of 250mg per 5 ml has a stability rate constant of

1 mg/ml/day. What is the half life of this drug in suspension?

A. 500 days

B. 250 days ANS: K = concentration / time; loss from

C. 100 days a suspension is a zero order process.

D. 50 days

50 oE. 25 days t = C = 50 mg / ml = 25 days

2K 2 (1 mg / ml / day)

Questions 4 to 8. The stability of a reconstituted antib iotic solution was evaluated for stability at

different temperatures and in different pH solutions. The data from these experiments is summarized

in the graph to the right. Use this information to answer questions 4 through 8.

4. From the graph it can be concluded that

I. The rate constant for the antibiotic's

degradation is the same for all solutions

II. Stability of the solutions is not affected

by temperature changes

III. Stability of the solutions increases

toward neutral pH

A. I only

B. III only I. No, if true only one line

C . I and II only II. No, h igher temperature

D. I I and I II only means increased rate of loss

E. I, II, and III only III. True, more stable at pH 7

than at pH 10

5. Based on the data shown, the marketed product should

I. contain a buffer to control the pH at neutral

II. be refrigerated after reconstitution

III. be reconstituted with 8.4% Sodium Bicarbonate Solution

A. I only

B. III only I. Yes

C. I and II only II. Yes

D. II and III only III. No, pH of 8 .4% Na Bicarb So lution would be alkaline (actually

E . I, II, and III on ly about 8.5) and would increase rate of decomposition

6. The data presented also indicate that

I. the only formulation suitable for use after 8 hours is the pH 7, 4 C productII. if the reconstituted solution is refrigerated, pH control is not required

III. if pH 7 is maintained, then refrigeration is not required for prolonged use

A. I only

B. III only I. True, only one above 90% at 8 hours

C. I and II only II. False, pH control required regardless of temperature

D. II and III only III. False, refrigeration required even at pH 4

E. I, II, and III only


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7. Which of the following statements are accurate based on the graph?

I. An increase in temperature slows the degradation process

II. For the conditions studied the best stability is at pH 7 and 4 C0

III. This study indicates that stability is a function of pH

A. I only

B. I II only I. No, it increases the degradation rate (universal truth)

C. I and II only II. Yes, line stays above 90% for longest period of time

D. II and III only III. Yes, it is a function of pH (and also of temperature)


8. Which of these statements may be concluded from the graph?

I. The rate of loss of drug is independent of the concentra tion present.

II. For the conditions studied the least stable solution is at pH 10 and 4 C.o

III. The stability of the cephalosporin is only a function of pH.

A. I only

B. III only I. True, because this is first order degradation

C. I and II only II. False, the least stable is pH 10 and 25 C.o

D. II and III only III. False, it is a function of pH, but also of temperature


pH and BUFFER CALCULATIONS

A. A buffer system consists of two components:

(1) a weak acid or weak base and (2) its corresponding salt form.

B. An infinitely dilute solution contains


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Basic Facts about logarithms and the numbers they represent

Log 1 = 0.0 pH = -Log [H ]+

Log 2 = 0.3 pH = 14 when [H ] = 1 x 10+ -14

Log 10 = 1.0 pH = 7 when [H ] = 1 x 10+ -7

Log 100 = 2.0 pH = 2 when [H ] = 1 x 10+ -2

w w wK = 1 x 10 pK = 14 pK = pH + pOH-14

SALT TYPES

1. A strong acid (e.g., HCl) + strong base (e.g., NaOH) forms a salt (e.g., NaCl) that are strong

electrolytes -- such salts are always 100% ionized in solution.

2. A weak acid (e.g., phenobarbital or penicillin G) + strong base (e.g., NaOH or KOH) will form salts

(e.g., sodium phenobarbital or K Penicillin G) that are weak electrolytes -- the degree of ionization

is a function of pH and pKa.

nonionized pH


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2 44. One hundred mLs (100 mls) of a buffer consisting of 0.01M Na HPO (pka 4.6) and 0.01M

2 4NaH PO is prepared. What is the pH of this buffer solution?

A. 2.6 ANSWER is C According to buffer guideline (C), if the ratio

2 4 2 4B. 6.6 of base (Na HPO ) to acid (NaH PO ) is 1,

C. 4.6 then the pH of the buffer is equal to the pKa

D. 9.2

E. 2.3

5. A drug is to be buffered in solution to a pH of 4.76. An acetate buffer is selected. If 0.05M Na

Acetate (pKa 4.76) is used, what strength (in moles) of Acetic Acid will be required?

A. 4.76 ANSWER is B Buffer gu ideline (C ), as rev iewed in Quest ion 5, indicates

B. 0.05 that the concentrations of Na Acetate and Acetic Acid

C. 0.10 should be the same.

D. 2.76

E. 1.00

6. What is the ratio of ionized to nonionized species present in 100 mLs of 0.1M Clindinium Br

(pKa = 6.5) at pH 6.2?

A. 1 to 2 ANSWER is D Salt of a weak base and strong acid

B . 0.3 to 1 pH = pKa + log (Nonionized/Ionized)

C. 10 to 1 6.2 = 6.5 + log (Nonionized/Ionized)

D. 2 to 1 - 0.3 = log (Nonionized/Ionized)

E. 1 to 10 10 = [Nonionized]- 0.3

[Ionized]

Note that the question is the reverse of answer. 1 = [Nonionized] = 1

10 [Ionized] 20.3

7. Which of the following solutions has the highest buffer capacity?

A . 0.1M HOAc + 0.1M NaC l ANSWER is B Buffer gu ideline (E ) says that if the ratios

B. 0.1M HOAc + 0.1M NaOAc of two buffers are the same the system

C. 0.01M HOAc + 0.01M NaOAc with the highest molar concentrations

D. (a) and (c) are equal will have the greatest buffer capacity.

E. (b) and (c) are equal NOTE: System (a) is not a buffer system.

8. One hundred mLs (100 mLs) of 0.4M NaOH is added to 300 mLs of 0.26M Acetic Acid (pKa 4.76).

What is the pH of the final solution?

A. 9 .52 ANSWER is D F irst , account for d ilution that occurs by adding the two

B. 3.84 solutions together (Steps 1 & 2). Then write out the chemical

C. 2.38 reaction which would occur (Step 3). Note that the reaction

D. 4 .76 forms 0.1M of Na Acetate and leaves 0.1M of Acetic Acid.

E. 7.04 Thus, the ratio between salt and base is 1 to 1; so pH = pKa.

Step 1. (X Molar) (400 mLs) = (0.4 Molar) (100 mLs) X = 0.1M NaOH

Step 2. (X Molar) (400 mLs) = (0.26 Molar) (300 mLs) X = 0.2M Acetic Acid

2Step 3. Acetic Acid + NaOH = Na Acetate + H O + Acetic Acid

0.2 molar + 0.1M = 0.1M + 0.1M + 0.1M

pH = pKa + log [Salt] pH = 4.76 + log 0.1M So, pH = 4.76

[Acid] 0.1M


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TABLE 2: Classification of Emulsifying Agents

Type Examples Mechanism

Synthetic ANIONIC: Coherent, flexible film formed by surface active agents.

(surface Soaps These agents marked ly lower in ter facia l tensions and th is

active agents) Potassium laurate contributes to emulsion stability. These are widely used,

agents) Triethanolam ine especially the nonionic type. Depending on the particular

Stearate agent(s) chosen, can prepare O/W or W/O emulsions.

Monomolecular Calcium oleate Prepare oil-in-water emulsions; oil-in-water emulsions are

Film Sulfates easier to make than are water-in-oil emulsions.

Sodium lauryl sulfate

Sulfonates

Dioctyl sodium sulfosuccinate (DOSS, D SS, Docusate)

CATIONIC:

Quarternary ammonium compounds

Cetyltrimethylammonium bromide, Benzalkonium Chloride

NONIONIC:

Polyoxyethylene fatty alcohol ethers (Tweens), Sorbitan fatty acid (Spans)

Natural Hydrophilic colloids Strong, rigid film form ed; m ostly by hydrocolloids

Multimolecular Oil-in-Water Emulsions colloids which produce O/W emulsions. Interfacial

Acacia, Gelatin tension is not reduced to any degree; stability is

Tragacanth due mainly to the strength of the interfacial film.

Finely D ivided Colloidal clays Film formed by solid particles that are small in size

Solids Bentonite compared to droplet of dispersed phase, The

Veegum particles must be wetted by both phases to some

extent in order to remain at the interface and form

Solid Particle Metallic hydroxides a stable film. Form either O/W or W/O emulsions,

Film Mg hydroxide depending on method of preparation.

BIOPHARMACEUTICSDefinitions

Bioavailability. The extent of absorption and the rate of absorption from a dosage form as reflected

by the time-concentration curve of the administrated drug in the systemic circulation. Three

parameters are used to establish bioavailability.

Area under the Curve (AUC)

maxC (maximum plasma concentration)

maxT (time at wh ich C max is reached)

max maxIf there is no statistically significant difference between AUC, T and C then two drug products

are considered to have equal bioavailability.

Absolute bioavailability is the bioavailability of a specific drug formulation given by any peripheral

route compared to IV administration of the same drug.

Relative bioavailability is the bioavailability of one drug formulation, of the same dosage form,

compared to another formulation, of the same dosage form, from a different manufacturer -or-

comparision of the bioavailability of different dosage forms; i.e. suspension versus tablet.

Bioequivalence. Chemical equivalents which, when administered to the same individuals in the

same dosage regimen, will result in comparable bioavailability.


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Chemical Equivalents. Drug products that contain the same amounts of the same therapeutically

active ingredients in the same dosage forms and that meet present compendial standards.

Deaggregation. The breaking up of granules or aggregates into fin e particles in aqueous fluid.

Disintegration. The breaking up of a tablet or capsule in to granules or aggregates in aqueous flu id.

Dissolution. The breaking down o f fine partic les into molecules or ions homogenously dispersed in

aqueous fluid.

Excipient. An inert substance used to give a preparation a suitable form or consistency.

Formulation. A complex mixture containing a selected chemical derivative of the drug compound,

in proper physical form, together with excipients, diluents, stabilizers, preservatives, or a variety of

other components.

General physicalCharacteristics Lipid Solubility; pKa (must be nonionized to be absorbed)

Intramuscular Dissolution; Compliance assured; Blood flow, increases if injection site massaged;

Advantage: larger volume than SubQ; depot with suspension

Oral Disintegration; Dissolution; Gastric Acid Stability; Gastric Fluid Volume;

Bioavailability Gastric Emptying Time; Meals; Gut Metabolism; First Pass Metabolism; Rate of GI

Passage; Concurrent Drug Administration

Pharmaceutical Equivalents. Drug products that contain the same amounts of the same

therapeutically active ingredients in the same dosage form and that meet standards established on

the basis of the best available technology.

Pharmacodynamics. The discipline that studies the biological and therapeutic effects of drugs.

Pharmacokinetics. The discipline that treats the rates of movement and biotransformation of a drug

and its metabolites within the body.

Rectal Disintegration; First Pass Metabolism; some vessels bypass liver;Dissolution - problem of availability if drug more oil soluble

Subcutaneous Dissolution (suspension = sustained release); Blood flow; Concurrent Drug Injection

(e.g., Lidocaine w ith Epinephrine)

Sublingual Disintegration ; Dissolution (Advantage: bypass liver; i.e. nitroglycerin, nifedipine)

Therapeutic Equivalents. Chemical equivalents which, when administered to the same individuals

in the same dosage regimen, will provide essentially the same efficacy and/or toxicity. In order for

a drug to act it must reach the site of action. Many factors stand in the way of achieving a

therapeutic response. Intravenous administration is not desirable as a route of chronic drug

administration. With other routes we encounter absorption variables which alter the availability of

drug to the systemic circulation. Factors influencing absorption by the various routes ofadministration are listed below:


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1. Use the data presented to answer the question below.

max maxt C T AUC(hr) (mcg/ml) (hr)

Formulation A 4.0 50 0.5 65

Formulation B 4.2 72 0.6 100

NS p < 0.05 NS p < 0.05

I. Formulation A and Formulation B are not bioequivalent.maxII. Formulation B has a greater T than Formulation A.

III. There is no difference between the AUC of Formulation A and Formulation B.

A. I only

B. III only Answer: A

C. I and II only I is true because of p < 0.05, so significant

D. II and III only II is false because NS (not significant)

E. I, II, and III only III is false because I is true

2. True statements about the graph to the left include

I. Drug A has a faster onset than Drug B.

maxI I. Drug B has a greater C than Drug C.

III. Drug B has a longer half life than Drug C.

A. I only

B. III only

C. I and II only

D. II and III only


ANS: C

maxI true due to shorter time to C

II true (you can see difference)

III false, B = 3 hrs, C = 4 hrs

maxDetermine by finding C for B is 70 at 5 hr, half of 70 is 35 at 8 hr so half-life is 3 hr

(8 hrs minus 5 hrs = 3 hrs); repeat the same process for C

A Little Review of Some Pharmacy MathWith the advent of an electronic calculator for use during the NAPLEX, there is an increased interest

in calculations problems. This review provides refreshment in some areas.

Moles and Millimoles

Gram M olecular Weight the Molecular Weight of a substance expressed in Grams

2Example: CaC1 = 111 Gms, (Ca = 40, C1 = 35.5) NaCl = 58.5 Gms (Na = 23, Cl = 35.5)

Gram A tomic Weight is defined as the number of grams numerically equal to the atomic weight.

Sulfur has an atomic weight of 32 atomic mass units, so 32 gms is 1 gram atom ic weight of sulfur.

A mole (M) of any substance is the gram-atomic or gram-molecular weight. The term millimole (mMol)

is used when the weight of the substance is expressed in milligrams. Thus, 58.5 milligrams of sodium

chloride will be one millimole of sodium chloride. Moles and millimoles are based solely on molecular

weight and the expression of that weight in grams (moles) or mill igrams (mill imoles).

A confounding factor that often occurs in determining the m olecular weight of drugs is water(s) of

hydration. Many inorganic molecules will take up water from the air and are in a state of permanent

hydration. Calcium chloride, for example, exists as either the dihydrate (two waters of hydration) or

the hexahydrate (six waters of hydration). The chemical formulas are usually written as shown below

(a period or asterisk is used to join the drug and the water):


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Calcium ch lor ide Calcium ch lor ide d ihydra te Calcium ch lor ide hexahydra te

2 2 2 2 2CaC1 CaC1 * 2 H O CaC1 * 6 H O

Molec Wt 111 Molec Wt 147 Molec Wt 249

4 41. What is the weight (in mg) of 3 mMol HP0 (Atomic weight for HP0 = 95.98)(-2) (-2)

4ANS: A millimole of HPO weighs 95.98 mg. Three times this amount would be 287.94 mg.

42. How many m Mol of P0 are present in 138 milligrams of sodium phosphate, monobasic,-3

2 4 2monohydrate? Molecular weight of NaH PO * H 0 = 138.

2 4 2Answer: One mMol of NaH PO *H 0 weighs 138 mg. Therefore, 138 mg of sodium phosphate,

monobasic contains one millimole of phosphate since there is only one phosphate radical per

molecule.

The use of mMol, rather than mEq, is suggested when calculations involving phosphorus (P) or

4phosphate (PO ) are required. The mEq of phosphate present cannot be determined unless the pH+3

2 4of the solution is known. The pH determines the relative proportions of monovalent (H PO ) and

4divalent (HPO ) phosphate present. The average valence of "phosphate" (between 1 and 2) in a

particular solution can be calculated, only if the pH is known. The use of mMol instead of mEq

eliminates the ambiguities present when the pH of the solution has not been specified.

Milliequivalents

The Combining Capacity of Electrolytes

An equivalent weight is the weight of an atom or radical expressed in grams, divided by the valence

(number of positive or negative charges, holding power, etc). The prefix "milli" means "one 1,000th

of." Therefore, a milliequivalent is 1/1000th of an equivalent.

Perhaps the most confusing element of m illiequivalent calculations involves recognizing that one mEq

of sodium chloride contains one mEq of sodium and one mEq of chloride. Similarly, two mE q of

calcium chloride contains two mEq of calcium and two mEq of chloride. Nothing seems to balance.

What does balance is the fact that the net charge on both sides of the equation is equal to zero. Forexample, NaCl ionizes to give Na (positive 1) and Cl (negative 1), usually written in the form of NaCl

= Na and Cl . On the NaCl side of the equation, the net charge is zero since the charge of the sodium+ -

and that of the chloride cancel each other. On the Na + Cl side of the equation, the charges also+ -

cancel out. Thus, the equation is in balance from the standpoint of milliequivalents (weights in

milligrams divided by number of charges).

Milliequivalent = Weight of Atom or Radical in milligrams

Weight (m Eq) Valence (num ber of charges)

1. Calculate the milliequivalent weight of Sodium (Na) given the atomic weights:

Na = 23, Cl = 35.5

Answer: milliequivalent weight = 23 milligrams = 23 milligrams per milliequivalent1

2. Calculate the milliequivalent weight (mEq. wt.) for Sodium and for Sodium Chloride.

Answer: Sodium mEq. Wt. 23 milligrams = 23 milligrams/mE q

1

NaCI mEq. Wt. = 23 mg + 35.5 mg = 58.5 mg/mEq

1


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3. Calculate the milliequivalent weight (mEq. wt.) for calcium. (Atomic weight = 40)

Answer: mEq. wt. = 40 mg = 20 mg/mEq

2

4. How many milliequivalents of calcium are there in 5 grams of anhydrous calcium chloride?

2Answer: CaC1 mEq. wt. = 40 mg + (2) * (35.5) mg = 55.5 milligrams/mEq.

2

5 grams = 5000 mg, so 5000 mg = 90.09 mEq

55.5 mg/mEq

Remember, earlier we said that calcium chloride usually exists in the hydrated form. See Question

Number 5 for how the waters of hydration change the answer to this problem.

2 25. How many milliequivalents of calcium are there in 5 grams of CaC1 * 2 H 0?

2 2Answer: CaCl * 2 H O mEq. wt. = 40 mg + (2) * (35.5) mg + (2) * (18) = 73.5 mg/mEq

2

5000 mg = 68.03 mEq

73.5 mg /mEq

Osmoles and Milliosmoles

Whenever a semipermeable membrane separates two solutions containing dissolved solids, as in

biologic systems and body compartments, fluid from one compartment moves across the membrane

to the other. Net movement of fluid is toward the compartment having a larger number of dissolved

particles. The force created by this movement of fluid (solvent) is termed osmotic pressure.

In osmotic relationships the important factor is the total number of particles present, both ions and

molecules dissolved in solution. The number of dissolved particles can be expressed as osmoles,

(Osm) or, usually in pharmacy, milliosmoles (m0sm). An osmole is the weight of a chemical substance

which, when dissolved in one liter of water, exerts an osmotic pressure equal to that exerted by a

gram-molecular weight of an unionized substance dissolved in one liter of water. One osmole contains

the same number of particles as one mole of molecules. (Avogadro's number of particles 6.023 x 10 ).23

Concentrations ofnonelectrolytes are usually expressed in millimoles. The concentration is the

same when expressed in milliosmoles. For example, one millimole of glucose (dextrose) is the same

as one milliosmole of glucose since glucose does not ionize. However, for electrolytes, the particles

formed by ionization must be considered in osmotic relationships. One m illimole of sodium chloride

ionizes into one millimole of sodium and one millimole of chloride, thus representing 2 m illiosmoles.

1. Dextrose 5% Injection, 1000 ml (MW = 198) (Dextrose as the monohydrate)

1000 ml x 0.05 = 50 gms/liter 50,000 milligrams = 252 millimoles per liter

250 grams = 50,000 mg 198 mg (MW of dextrose * H O)

252 mM ol per liter (x) 1 m0sm per mMol = 252 mOsm per liter (dextrose does not ionize)

2. 0.9% Sodium Choride Injection, 1000 ml (MW = 58.5) = 9 grams (= 9000 mg) NaCl per liter

9000 mg = 154 millimoles per liter (x) 2 mOsm per mMoI = 308 mOsm per liter

58.5 mg per mole

Blood plasma has an osmotic pressure of about 280 mOsm per liter, so both these solutions are

almost isotonic with plasma. Normal Saline is slightly hypertonic (308 mOsm/liter) and D-5-W is

slightly hypotonic (252 mOsm/liter). Both are considered to be clinically isotonic with blood.


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Parenteral Nutrition

Who Needs It? Any patient unable to orally consume the amount of nutrition needed for their

medical situation. Particularly, GI tract disease or surgery requiring bowel rest of >7 days, cannot

tolerate enteral feeding, unconscious, severe infections, newborn infants, hyperemesis gravidarium,

bulimia nervosa, burns, and other severe trauma.

Contraindications. If oral intake will resume in 75% of dailyneed), or there is a poor prognosis the patient does not warrant nutritional support

Choice of Peripheral or Central Route.

Peripheral: dextrose limited to 10%, phlebitis is an issue, days of therapy limited to 3 days at a single

IV site, compensate with IV fat

Central: invasive procedure, increased possibility of infection, best for long term, can use up to 25%

dextrose, reduces need for fat

Calculating the Basal Energy Expenditure (BEE) Harris Benedict Equations are used to calculate

specific energy requirements or do a round off of 20 to 25 kcal/kg/day

BEE for men 66.47 + 13.75 (x) weight in kg + 5 (x) height in cm 6.76 (x) age in years

(kcal per day)

BEE for women 655.1 + 9.56 (x) weight in kg + 1.85 (x) height in cm 4.68 (x) age in years

(kcal per day)

The Total Energy Expenditure (TEE) in kcal/day = BEE plus Stress Factor -or- Activity Factor

Stress or In jury Factor (% increase above BEE Activ ity Factors (% increase above BEE

Major surgery 10 to 20 percent Confined to bed 20 percent

Infection 20 percent Out of bed 30 percent

Fracture 20 to 40 percent Alternatively No stress 28 kcal/kg/day

Trauma 40 to 80 percent Mild stress 30 kcal/kg/day

Sepsis 80 percent Moderate stress 35 kcal/kg/day

Burns 80 to 100 percent Severe stress 40 kcal/kg/day

Energy Requirements: Adults 150 kcal per gram of nitrogen (range 130-200). Energy is best

supplied by one of the three choices below

1. Dextrose provides 3.4 kcal per gm, metabolic rate limit is 5 mg/kg/min, give 40-95% of non-

protein calories as dextrose; available as 50% and 70% for compounding.

2. Fat (soybean oil emulsion) provides 9 kcal per gm, available as 10% (550 kcal/500 m l) and 20%

(1000 kcal/500 ml); maximum is 60% of daily calories, minimum is 5% to supply essential fatty

acids (linoleic, arachidonic, linolenic).

3. Glycerin will provide 3.5 calories per gm and is popular in Europe (Procalamine is a product that

contains glycerin and amino acids for short term use, especially after surgery when tissue

rebuilding is required.)

Electrolyte Requirements (range per 24 hours)

Sodium 60 - 150 mEq Chloride 60 - 150 mEq Potassium 70 - 180 mEq Bicarbonate 25 - 160 mEq

Magnesium 10 - 30 mEq (as acetate or lactate salts)

Calcium 4 - 30 mEq Phosphate 7 - 10 mMol

Protein: is usually in the form of amino acids, provides 4 kcal per gm, give 1 to 1.25 (range 0.6 to

1.76 gms) per kg per day of a 50/50 mixture of essential and non-essential amino acids. Some

situations require the use of specific amino acid formulations. All amino acid formulations contain

some cations (Na, K) and anions (acetate, some phosphate).


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Renal failure only essential amino acids

(Nephramine, Aminosyn-RF, RenAmin)

Liver failure and trauma increase branched chain AA (isoleucine, leucine, valine)

(Hepatamine)

Stress and trauma increase branched chain AA (isoleucine, leucine, valine)

(FreAmine-HBC, Aminosyn-HBC, BranchAmine for stress or trauma)

Encephalopathy decrease aromatic AA (tryptophan, phenylalanine, tyrosine)

(Freamine-III is tyrosine free and lower in tryptophan and phenylalanine)

Water minimum daily requirement is 1200 mls of free water; usually give 1500 mls/M to assure2

adequate and dilute formulation to total volume of 2000-3000 m ls per day.

Vitamins several multi-vitamin mixtures are available, the most common is MVI-12 that contains

vitamins A, C (ascorbic acid), D (tachysterol), E (tocopherol), B-1 (thiamine), B-2 (riboflavin), B-3

(niacin/nicotinic acid or niacinamide/nicotinamde), B-5 (pantothenic acid), B-6 (pyridoxine),

B-12 (cyanocobalamin), biotin and folic acid. Vitamin K (phytonadione) is also routinely added.

Trace Elements (daily amounts)

Zinc, 2-6 mg Copper, 0.5-1 mg Chromium, 10-15 mcg

Manganese, 0.15-0.8 mg Selenium, 20-60 mcg

Molecular Weights and Chemical Formulas

Dextrose Monohydrate Injection, 70% Dextrose = 180, Water = 18

Amino Acid Injection, 10% No molecular weight

Na = 23, Cl = 35.5 NaCl Sodium Chloride Injection, 14.6% and 23.4%

3C = 12, H = 1, O = 16 CH COONa Sodium Acetate Injection, 16.4%

3NaHCO Sod Bicarbonate Injection, 8.4%

K = 39, Cl = 35.5 KCl Potassium Chloride Injection, 14.9%

3CH COOK Potassium Acetate Injection, 19.6%

4 2Mg = 24.3, S = 32 MgSO . 7 H O Magnesium Sulfate Injection, 50%

2 4 3 4 2 4P = 31 Na HPO / Na PO / NaH PO Sod Phosphate Injection, 3 mMol/ml

2 4 3 4 2 4K HPO / K PO / KH PO Pot Phosphate Injection, 3 mMol/ml

2 2Ca = 40 CaCl . 2 H O Calcium Chloride Injection, 10%

MVI-12 = vitamins A, C, D , E, B-1, 2, 3, 5, 6 , 12, biotin and folic acid in 10 m ls

Zn, Cu, Se, Cr and Mn 5 Trace Elements Injection, 5 mls

Vitamin K Injection Phytonadione Injection, 1 mg/0.5 ml

To make total final volume Sterile Water for In jection

Other Substances That Might be Added

1. Stomach Acid Reduction with H-2 Blocking Agent

Acid secretion reduction is indicated since no food into stomach can lead to ulcers

Zantac Injection, 25 mg/ml; Pepcid, 10 mg /m l or Tagamet, 300 mg/ml

Proton pump inhibitors often cannot be added since the parenteral forms are very unstable

2. Human Regular Insulin, 100 units/ml (Humalog/Novalog not approved for IV use)

Control of blood sugar is often required due to the high dextrose content

3. Iron only for very long term parenteral nutrition and best given separately


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4. Albumin raises serum protein levels quickly, but avoid routine use

5. Hydrocortisone can decrease phlebitis in peripherally administered solutions

6. Heparin non-therapeutic dose can decrease phlebitis

7. Carnitine for renal patients on prolonged parenteral nutrition (required for proper utilization

of long chain fatty acids)

8. Cysteine for neonates because the neonatal liver has not yet developed the ability to convert

methionine to cysteine

9. Drugs for therapy of diseases (e.g., antibiotics, anticancer drugs) are best administered

separately, rather than mixed directly into the parenteral nutrition formula, even though

they may be piggy-backed into the parenteral nutrition solution IV line

What follows is a formula for one 24-hour bag of parenteral nutrition solution. You should practice

doing the calculations required, using the product list provided, to determine the quantity of each

agent required to properly prepare this solution.

Total Parenteral Nutrition (TPN) Calculations

Prepare from the following parenteral products:

Dextrose Injection, 70% in 1000 ml bags 600 ml

Amino Acid In ject ion, 10% in 500 ml bottles 1200 ml

Sodium Chloride Injection, 2.5 mEq/ml 40 ml

Potassium Chloride Injection, 2 m Eq/m l 35 m l

Sodium Acetate Injection, 2 mEq/ml 20 ml

M agnesium Sulfate Injection, 8 mEq/2 m l 6 m l

Sodium Phosphate Injection, 3 m Mol/m l 6 m l

Potassium Acetate Injection, 2 mEq/ml 15 ml

Calcium Chloride In jection, 0 .465 mEq/ml 10 ml

Multivitamins Injection (MVI), 10 mls 10 ml

5 Trace Elements Injection, 5 mls 5 ml

Phytonadione Injection, 1 mg/0.5 ml 0.5 ml

Zantac Injection, 25 mg/ml 6 ml

Human Regular Insulin, 100 units/ml 0.5 ml

Ster ile Water for In ject ion in 1000 ml bags 446 ml

PHARMACOKINETICS

The chal lenge in treating the distribution of drugs and their elimination from the body is that we

cannot samp le all body tissues. Basically we are limited to blood, urine, saliva, and CSF.

Elimination of drugs fits zero order or first order equations (first order is most common).

k = elimination rate constant

1/2t = time required for a decline in concentration by one-half

C = concentration of drug

A = total amount of drug in the body

Parameters to enable us to estimate actual physiological conditions from those parameters which w e

can experimentally measure.

Ingredient Daily AmountDextrose 420 grams

Am ino Acids 120 grams

Sodium Chloride 100 mEq

Potassium Chloride 70 m Eq

Sodium Acetate 40 mEq

Magnesium Sulfate 24 m Eq

Sodium Phosphate 18 mMol

Potassium Acetate 30 mEq

Calcium Chloride 4.65 mEq

Multivitamins 10 mls

Trace Elements-5 5 mlsVitamin K-1 1 mg

Ranitidine 150 mg

Hum an Reg Insulin 50 units

S terile Water qs ad 2400 mls

Flow Rate 100 mls per hour


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Apparent Volume of Distribution. The theoretica l volume in which the total amount of drug in the

body would have to be dissolved if it were all present in the concentration measured in the blood.

d p poV = Dose = mg = 1 C = (C ) (e ) = Dose (e )-kt -kt

oC mg/L Vd

oC estimated from intercept log C vs. Time plot

Clearance. A measure of the volume of drug cleared per unit time. The total body clearance (TBC)

is similarly the sum of the individual clearance rates for each elimination pathway.

urine metab lungT B C = Cl + Cl + Cl + ...

Renal Clearance. The renal c learance rate provides a c lue as to the mechanism of elimination. The

glomerular filtration rate in a normal individual is approximately 125 ml/minute, therefore

Cl = 125 m l/min when drug is eliminated by glomerular filtration

Cl < 125 ml/min if glomerular filtration/tubular reabsorption

Cl > 125 ml/min if glomerular filtration/active secretion

The clearance rate of creatinine is approximately 125 m l/minute (same as glomerular filtration rate)

in a normal subject. It undergoes some excretion and reabsorption. Creatinine clearance is a good

estimate of renal function.

The creatin ine clearance can be estim ated by the Cockroft-Gault Equation.

CrCl (mls/min) = (140 - age) x weight (kg) Multiply by 0.85 for women

72 x serum Cr (mg/dl) Weight is ideal body weight

Ideal IBW in kilograms (males > 18 yrs) = 50 + (2.3 (x) height in inches over 5 feet)

Body

Weight IBW in kgms (females > 18 yrs) = 45.5 + (2.3 (x) height in inches over 5 feet)

Formulas

Children, 1-18 yrs and < 5 feet tall IBW in kg = (height in cm) (x) 1.652

1000

Males 1-18 yrs and > 5 ft tall IBW in kg = 39 + (2.27 (x) inches over 5 ft)

Females 1-18 yrs and > 5 ft tall IBW in kg = 42.2 + (2.27 (x) inches over 5 ft)

Elimination Rate Constant. Drugs are eliminated from the body via several pathways; i.e. kidney,

metabolism, lung metabolism, etc. The total elimination rate constant is the sum of the individual

total urine metab lungrate constants. K = k + k + k + ...

Important First Order Equations

o oA = A e C = A C = C e Vd = Dose-Kt -Kt

oVd C

1/2t = 0.693

u m u mK T B C = Cl + Cl K = k + k Cl = K Vd


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Pharmacokinetic Examples

1. After IV administration of 500 mg of Drug A the initial plasma concentration was 60 mcg/ml and

the k = 0.0024 hr . What is the volume of distribution and half life?-1

Vd = Dose = 500 mg = 500 x 10 mcg = 8333 ml = 8.3 liters3

oC 60 g/ml 60 g/ml

mg/L = mcg/ml =

g/ml1/2t = 0 .693 = 0 .693 = 288 hours ln 2 = 0.693

k 0.0024 hr-1

2. Drug B has a half-life of 24 hours. The drug is 70% eliminated by the kidney and

Vd = 20 liters. What is the renal clearance of this drug?

u m uK = 0.693 = 0.0288 hr-1 K = k + k k = 0.7 K = (.7) (.0288) = 0.0202 hr-1

t (24 hr)

renal renalCl = k Vd = (0.0202 hr ) (20 L) = 0.4 liters/hour-1

3. Drug C has a half life of 6 hours. The drug is 60% eliminated via kidney excretion.

A patient has lost 70% of her kidney function. What is the half life?

uK = 0.693 = 0.1155 k = (0.6) (0.1155) = 0.0693

1/2 m et k = K - k = 0.1155 - 0.0693 = 0.0462

u u mnew k = (0.3) (0.0693) = 0.0208 new K = new k + k = 0.0208 + 0.0462 = 0.067

1/2t = 0.693 = 10.3 hours in patient with 30% remaining kidney function

0.067

4. Drug B produces a peak concentration of 10 mcg/ml two hours after oral administration. This

drug is eliminated by first order kinetics and has a biological half-life of 4 hours. How long will

it take this drug to reach a drug level in the plasma of 1.25 mg/liter? mcg / ml = mg / L

10 mcg/ml at 2 hrs 2.5 mcg/ml at 10 hrs 12 hrs is wrong because drug level

5 mcg/ml at 6 hrs 1.25 mcg/ml at 14 hrs was 2 hrs after dose

5. Calculate a loading dose of phenytoin using the formula below:

Loading = (Vd/kg BW) * (kg of Body Weight) * (Cp desired) Given: Vd = 0.5 L/kg,

Dose F F = 1,

pC desired = 15 mcg/ml,

Loading = (0.5 L/kg) * (60 kg) * (15 mg/L) = 450 mg Weight = 132 lb

Dose 1

DRUG METABOLISM

Metabolism Biotransformation of a drug to

1. active products prodrug is administered and converted to the active form(s), or

2. inactive products usually done to make a drug less toxic or inactive. This generally makes the

drug more water soluble (hydrophilic or lipophobic) which makes it easier for the kidney to

excrete the metabolized form. Drugs that remain fat soluble (lipophilic or hydrophobic) are m ore

readily reabsorbed and are less likely to be excreted.

Drug metabolism occurs to some extent in the lung, kidney, blood plasma, brain, ovary, testes,

placenta and the adrenal glands. The primary sites for drug metabolism are the small intestine (ileum)

and the liver. Drugs are not generally excreted into the small intestine so any drug that will be

metabolized in the small intestine must be given orally. This is usually essential for prodrugs that will


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be converted to the active form. Examples of drugs that are converted to active forms, and remain in

the GI tract, would include sulfasalazine (Azulfidine) that releases 5-aminosalicylic acid or mesalamine

(Pentasa). Drugs that m ust undergo metabolism in the GI tract for conversion to an active form are

not likely to have any effect if given IV or by other parenteral routes.

Liver metabolism tends to be either high first pass (high intrinsic clearance) or low first pass (low

intrinsic clearance; most drugs). For drugs that undergo high rates of clearance, the liver will

metabolize as much drug as it receives and these are the drugs most affected by changes in liver

function and liver enzyme induction or inhibition. For these drugs the IV and oral dose are likely to

be quite different (propranolol, 1 mg IV vs 20 m g PO). Remember that drugs absorbed from the GI

tract will go through the liver on their way to the general circulation and these are the drugs that will

likely need increased oral doses in some patients. Conversely, patients with hepatic disease may need

to have doses of the same drugs decreased due to their inability to be metabolized.

Drugs that undergo low rates of liver metabolism (low intrinsic clearance) do not have a lot of drug

removed by the liver and so their oral and IV doses are likely to be about the same (a single dose of

cimetidine is 300 mg IV or PO). Similarly, these drugs are not very affected by induction or inhibition

of liver enzymes by other other drugs.

Sublingual administration of drugs is a standard method of avoiding first pass liver metabolism.

However, the same result occurs for drugs administered as nose drops and eye drops. This is the

reason that ophthalmic beta blockers can cause systemic problems (beta blockers normally undergo

significant first pass metabolism).

Drugs are metabolized in the liver by one of four primary mechanisms:

31. Oxidation oxygen is added to the structure as a lipid soluble CH becomes a water

2soluble CH OH

2. Reduction an example wou ld be addition of hydrogen to a lipophilic ketone =O to

become a h ydrophilic alcohol OH

3. Hydrolysis a molecule is broken into more than one part, such as when aspirin is

broken into salicylic and acetic acids

4. Conjugation this involves adding a water soluble group to the molecule and can be

attached to a water soluble site created by one of the first three reactions. Water soluble

groups comm only added include acetate, glycine, sulfate and glucuronate (most).

Pharmacokinetic Models

A. IV Bolus dose (single dose line A)

p po dFormulas C = C ( e ) Cl = K V-kt

p po p p oln C = - kt + ln C log C = - kt + log C

2.303

B. IV Infusion model (continuous dose line B)

Formulas

p o p ss o o oC = K (1 - e ) C = K = K = F S D K = Dose-kt

(k) (Vd) K Vd Cl T K Vd Time

F = fraction absorbed (bioavailab ility ) S = salt form percentage (e.g. , theophylline)

D = dose adm inistered T= dosing interval (in hours)

p ss aveC. Absorption model (single dose line C) Formula C = Dose

k Vd


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TRADE NAME DOSAGE FORMS

NAPLEX may ask you to iden tify a drugs dosing properties based on the dosage form o f the drug.

Some drug products have specific names that are used to identify specific activity of the dosage form

and these are listed below.

1. Coated Tablets, but not long acting Filmtab, Filmseals, Wyseals

2. Long Acting Tablets or Capsules (sustained release, not enteric coated) Gradumet, Endurets,

Duracaps, -DUR, Dospan, Sequels, EC pellets, -PM, -SR, -CD, -XL, Ten-Tab, Extentabs,

Extencaps, Timespan, Timecaps, Gyrocaps, Repetabs, -CR, Spansule, Tembids, Plateau Caps,

Del-Cap

3. Regular release tablets or capsules with fancy names Oralets, Perles, Pulvules, Gelseals, Gelcaps,

Inlay-Tabs, Bayer-TR Caplets, Caplet, Kapseal, Infatab

4. Enteric coated tablets or capsules (not long acting) Enseals, ENtabs

5. Sustained release liquid formulations Pennkinetic SR Microcapsule Suspension, T/SR

6. Injectable dosage forms in special packaging formats Abboject, Mix-o-Vial, Tubex, Carpuject,

Solu-, Depo-

7. Ophthalmic dosage forms Liquifilm, C-Cap, SOP, Ocumeter

8. Products for inhalation administration Rotacaps/Rotahaler, AeroBid, Respihaler; Turbinaire,

Medihaler

9. Packages of tablets for therapy use Dosepak, Enpak (Medrol)

Industrial Pharmacy and Dosage Forms

A sample of questions with answers following.

1. Which of the following IV drugs should be dispensed with light resistant covers?

A. metronidazoleB. co-trimoxazole

C. doxycycline

D. amphotericin B

E. all of the above

2. Propofol, an agent for induction and maintenance of anesthesia, is

A. a solution for inhalation

B. an emulsion for injection

C. a solution for injection

D. a gas for inhalation

E. a suspension for injection

3. Once in solution, which of the following agents should not be refrigerated?A. fluorouracil

B. mitomycin

C. cisplatin

D. ceftazidime

E. cyclophosphamide


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4. Which IV antiinfective is available already in solution and ready to use?

A. erythromycin

B. vancomycin

C. aztreonam

D. amphotericin B

E. gentamicin

5. Neut (Sodium Bicarbonate Injection) can be used to increase pH and reduce phlebitis associated

with the infusion of

I. aminophylline

II. metronidazole

III. erythromycin lactobionate

A. I only

B. III only

C. I and III only

D. II and III only

E. I, II and III only

6. What dose of Dilantin (phenytoin) Suspension, 125 mg per 5 ml, is equivalent to a Dilantin

(phenytoin sodium) Capsule dose of 300 mg? Assume equivalent bioavailability. Phenytoin is

15 12 2 2C H N O ; mol wts are C = 12, H = 1, N = 14, O = 16, Na = 23

A. 5 ml

B. 14 ml

C. 11 ml

D. 9 ml

E. 12 ml

7. Which is a sustained release nitroglycerin product?

A. Nitrostat

B. NitroBID

C. Nitrol

D. Tridil

E. Nitrolingual

8. Microencapsulation of drugs mayI. Provide release of drug over 12 hours

II. Increase surface area to minimize local irritation

III. Release drug in alkaline pH of small intestine

A. I only

B. III only

C. I and II only

D. II and III only


9. After reconstitution Suprax suspension should be stored

I. frozen

II. refrigerated

III. at room temperatureA. I only

B. III only

C. I and II only

D. II and III only



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Page 20

10. For which of the following is "dispense in the original container" related to drug stability

I. Cytotec

II. Accutane

III. Nitrostat

A. I only

B. III only

C. I and II only

D. II and III only


11. Which of the following are alcoholic or hydro-alcoholic preparations?

I. Syrup

II. Tincture

III. Elixir

A. I only

B. III only

C. I and II only

D. II and III only


12. Enteric coated tablets will NOT

A. Provide release of the drug over 12 hours.

B. Protect the drug from light and moisture.

C. Release drug in alkaline pH of small intestine.

D. Be affected by crushing for administration

E. Reduce irritation due to the drug in the stomach

13. Slow-K (Potassium Chloride) releases drug from a

A. microencapsulation

B. erosion core

C. porous inert carrier

D. multiple release rate granules

E. multiple layer tablet

14. Suspending agents primarily work byI. Increasing Viscosity

II. Decreasing caking

III. Neutralizing particle charges that form large flocculations

A. I only

B. III only

C. I and II only

D. II and III only


15. Fifty ml of Calcium hydroxide Solution and 50 ml of cottonseed oil are shaken. Which of

the following statements are true about the emulsion that is formed?

I. The emulsion could also be made with mineral oil

II. An oil in water emulsion is formedIII. The emulsifier is Calcium Oleate

A. I only

B. III only

C. I and II only

D. II and III only



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Page 21

16. What is a method used to reduce particle size?

A. levigation

B. fusion

C. geometric dilution

D. sublimation

E. incorporation

17. Which are major advantages of the hydrous absorption bases for sunscreen preparations?

I. The base will not wash off in water

II. Water soluble ingredients can be incorporated

III. The base is miscible with water on the skin

A. I only

B. III only

C. I and II only

D. II and III only


18. IV solutions having more than 5 mg/mL of theophylline in PVC bags may precipitate due to

A. drug degradation

B. formation of an insoluble complex

C. decreased solubility due to hydrogen bonding

D. decreased solubility due to a salting out effect

E. complexation between the theophylline and PVC plasticizers

19. A pharmacist in a hurry draws up two electrolyte solutions in the same syringe before making

a large volume I.V. Almost immediately, a white precipitate occurs in the syringe. The precipitate

is the product of a reaction between

A. calcium chloride and potassium chloride

B. potassium chloride and a multivitamin infusion product

C. calcium chloride and sodium phosphate

D. magnesium sulfate and potassium chloride

E. sodium acetate and potassium phosphate

20. If unprotected, UV light will catalyze the degradation of such that the product

must be d iscarded.I. norepinephrine

II. nitroprusside

III. nitroglycerin

A. I only

B. III only

C. I and II only

D. II and III only


21. The primary plastic polymer responsible for drug loss due to adsorption to IV solution containers

and administration set tubing is

A. PVP - polyvinylpyrrolidine

B. PSF - polysulfoneC. PVC - polvinylchloride

D. PC - polycarbonate

E. PS - polystyrene

22. Which of the following filter sizes is a "sterilizing" filter?

A. 0.22 micron filter D. 4.5 micron screen filter

B. 0.45 micron filter E. 5 micron filter

C. 2.2 micron filter


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Page 22

23. Which agent should NOT be administered with a 0.22 micron in-line filter IV set?

I. amphotericin B

II. alteplase

III. streptokinase

A. I only

B. III only

C. I and II only

D. II and III only


Answers to the Questions

1. E The light sensitivity of tetracyclines and metronidazole is well known. Some antineoplastic

agents require special light protection including carmustine (BCNU), decarbazine (DTIC),

daunorubicin, fluorouracil, methotrexate, mitomycin, and vinblastine. Any drug that is sold

in a brown ampule needs protection from light. Any drug with sulfur in its molecular

structure and can cause photosensitivity reactions and needs protection from light.

2. B Diprivan (propolol) is a colloidal soybean oil oil-in-water emulsion for intravenous injection.

The emulsif ier is egg lecithin. Bacterial contamination commonly occurs. All emulsions for

IV use are oil-in-water.

3. C Cisplatin may precipitate if refrigerated. Fluorouracil may precipitate during room

temperature storage but may be used after warming to dissolve. Cyclophosphamide is stable

after dilution in most solutions for 24 hours at room temperature and 6 days when

refrigerated. Mitomycin stability varies with diluent and final solution concentration.

4. E Gentamicin Injection, in both vials and ready-to-use piggyback containers, is already a

solution. All of the other drugs are provided in powders to be reconstituted prior to injection

or addition to IV piggyback containers. Amphotericin B in the fat emulsion is not a solution.

5. D Both metronidazole and erythromycin are highly acidic in solution. Neut a 4.2% sodium

bicarbonate solution can be used to neutra lize the pH. Flagyl ready-to-use bags are already

pH adjusted. Aminophylline is an alkaline solution and would not require Neut.

6. C Each Dilantin (Phenytoin Sodium) 100 mg Capsule contains 92 mg phenytoin. Dilantin

Suspensions and Infatabs are prompt release phenytoin free acid. The molecular weight ratio

between the free acid and its salt (0.92) can be used to adjust dose(s) when changing salt

forms. Divided doses would also be required. The 30 mg/5 ml and 125 mg/5 ml suspensions

are designed for children and are dosed by weight 5 mg/kg. NOTE: This is the same principal

that applies to converting between doses of theophylline and aminophylline.

Another form of the calculation is: MW phenytoin = 252, phenytoin Na = 275

Thus X mg of Phenytoin = 252 mg Phenytoin = 275 mg Phenytoin

300 mg of Phenytoin Na 275 mg Phenytoin Na

275 mg Phenytoin (div by) 125 mg Phenytoin per 5 ml (or 25 mg/ml) = 11 m l

7. B NitroBID is a controlled release capsule with a duration of action of approximately 8-12 hours.

Nitrostat is a short onset and duration sublingual tablet. Nitrol is the original brand name

for nitroglycerin ointment. Nitrolingual is a short onset and duration metered spray for

sublingual admin istration. Tridil is nitroglycerin injection.


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Page 23

8. C Microencapsulation can alter degree and extent of drug release in the intestine. However, no

presently marketed formulas act as a pH dependent enteric coating. Enteric coated products

are pH dependent release.

9. B Suprax is stable for 14 days at RT (15 - 30 C). More importantly refrigeration results in

increased viscosity of the suspending agents and the ability to pour an accurate dose may be

reduced. Cold storage includes freezer (- 20 to -10 C) and refrigerator (2 to 8 C). A cool place

is 8 to 15 C. Excessive heat is any temp above 40 C.

10. B The "dispense in the original tamper resistant container" cautions for Accutane and Cytotec

pertain to their severe teratogenic potential and manufacturer liability.

11. D Syrups are generally nonalcoholic. An exception is Phenergan Syrup with 7% alcohol.

Tinctures are by defin ition alcoholic di lutions of fluid extracts. Elixirs most frequently contain

alcohol. An exception, Tylenol with Codeine Liquid contains no alcohol.

12. A Enteric coating does not provide any sustained release effects, e.g. Ecotrin. Enteric coating

does release drug in the pH of the small intestine. As with any other film coating, enteric

coating with opacifying agents protects a drug from light and moisture and improves

palatability.

13. C Below are examples of various product formulations.

Microencapsulation Micro K

Porous Inert Carrie r/Wax Matr ix S low K , K lo tr ix , K lor-Con, Ferro-Grad

Multiple Layer Tab Peritrate SA, Tedral SA

Erosion Core Naldecon, Mestinon-Timespan, Extentabs

Mixed Release Rate Granules K-Dur, Theo-Dur

Coated or S low-Release Beads Spansu les (Various), S equels (Various)

14. A Suspending agents such as glycerin, propylene glycol, various polyethylene glycols increase

viscosity and decrease rate of particle settling. Flocculated suspensions settle faster but have

smaller surface area and do not form a solid cake so they readily resuspend. A colloid is the

most stable suspension (e.g., milk). If this is not possible using the optimum combination of

suspending and/or flocculating agents give the longest shelf life.

15. B Anionic soaps such as calcium oleate formed from fatty acids, e.g. oleic acid found in vegetable

oils and a salt such as Ca or Na will form W/O emulsions (see Table 1). Nonionic++ +

surfactants such as Tweens and Spans are very popular. In this problem m ineral oil does not

contain fatty acids and no emulsion would be formed unless a specific emulsifier were added

to the formula.

16. A Trituration, comminution, levigation , and pulverization by intervention are all methods of

particle size reduction. Incorpora tion and geometric dilution relate to mixing of ingredients.

Fusion is a process where waxes and petrolatum are melted and mixed to make ointments.

Sublimation is the heating of a powder until it vaporizes and then recrystalizes from the vapor

form. For further details consult Remington's Practice of Pharmacy.

17. E All properties given are advantages of W/O absorption bases. While still oily to the touchthere is prolonged sun protection compared to O/W water washable lotion bases which must

be reapplied frequently. NOTE: Water soluble drugs which cannot be incorporated into

oleaginous bases can be dissolved in water and incorporated readily into absorption bases.

(Remember this for method of preparation questions and for both wet or dry practical labs.)

18. D In electrolyte solutions eg. normal saline, the salt will compete for water association and

reduce the water available for solution of nonelectrolytes (e.g. aminophylline) resulting in

precipitation or "salting out" of the less soluble aminophylline.


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Page 24

419. C Reaction of a divalent cation (Ca ) and a divalent anion (HPO ) in high concentration forms+2 -2

4the insoluble precipitate CaHPO . This reaction is unlikely if the ions are well diluted before

mixing.

20. C Norepinephrine (Levophed) which has darkened on exposure to air and ultraviolet (UV) light

should not be used. It is very sensitive to alkaline pH and should not be mixed in saline alone.

In Dextrose 5% (pH 5.6) it is stable at room temperature (RT) for 24 hours. UV light causes

a color change of nitroprusside solutions from brown to blue due to a reduction of ferric to

ferrous ion and 20% drug loss within 4 hours. When pro tected it is stable 24 hours.

Nitroglycerin (TRIDIL) is stable at RT in clear glass bottles.

21. C Polyvinylchloride, a flexible plastic polymer readily adsorbs many drug molecules. However,

a more rigid polypropylene polymer bag shows no significant drug adsorption. Nitroglycerin

could be dispensed in this plastic IV container.

22. A The 0.22 micron filter pore diameter will prevent any bacterial contamination. It may not

remove some small viruses and pyrogens and requires aseptic manufacturing and handling

procedures.

23. E Many new large molecular weight protein drugs (e.g., Activase, or emulsions such as

Intralipid) may not pass through a 0.22 micron in-line filter. A 5 m icron filter can be used for

possible particulate contamination such as from am pule breakage ar stopper coring m issed

by visual inspection.

Some Statistical Information

1. Types of Statistics and Their Purpose

Descriptive (parametric or normal distribution) versus non-parametric (binomial distribution

describes the data through means, frequencies , modes, medians, variances or standard

deviation (spread), range and correlation/covariances. Example: How Y is affected by changes

in X.

Inferential the science of making decis ions while attempting to control or un derstand the

uncertainty (or variance); Decision making through the probability of being correct. Testing a

hypothesis about parameters.

2. Types of Measurements or Data

Nominal A number is used to designate a class or a category. An example would be

designating students by using 1 = Freshmen, 2 = Sophomores, 3 = Juniors and 4 = Seniors.

Ordinal In this case all measurements (numbers, values) are arranged in rank (numerical)

order from either highest to lowest or lowest to highest.

Interval Interval measurements are those that are based on a scale with equal units of

measurement such as degrees of temperature on a thermometer.

Ratio Ratio scales differ from interval scales in that they have an absolute zero (0). This is

seen when comparing the Kelvin scale of temperature and the Centigrade scale. The Kelvin scalehas an absolute zero (273 degrees Centigrade) where there is the complete absence of heat. The

Centigrade scale is a means of measuring heat and, theoretically, has no absolute limits in

either direction.

3. Distributions of Data

Normal Distribution Continuous distribution with means and variances of ratio or interval

data such as the Z-test, T-test, F-test, analysis of variance (ANOVA) or regression analysis.


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Binomial Distribution Discrete distribution (data takes the form of yes or no; live or dead;

improved or not improved, etc). Tests performed include the Chi Square test, contingency

tables, Fishers exactness test, simple probability, or frequency data. The data is either nominal

or ordinal data.

4. Hypothesis testing

It is easier to prove that something is NOT correct that it is to prove that something is ALWAYS

correct. E.g., the hypothesis that Joe is a liar vs Joe always tells the truth. To prove that Joe

is a liar, we only need one lie, but to prove Joe always tells the truth, we must study Joe for his

entire life. But, it is also difficult to prove negativity; the fact that something does not exist (Joe

never tells the truth).

The Null Hypothesis (Ho) is a derived statement tha t is opposite to the research hypothesis. An

example would be the statement that the height of students in Marys class is the same as the

height of students in Joes class (Marys = Joes). The Alternate Hypothesis (Ha) would be that

the height of the students is different for the two classes.

The Null Hypothesis (Ho) can be expressed as Ho: u1 = u2 (where u1 = the height of students

in Marys class and u2 = the height of students in Joes class).

The Alternate Hypothesis (Ha) can be expressed as Ha : u1u2 (where u1 = the height ofstudents in Marys class and u2 = the height of students in Joes class).

There are four possible outcomes for any hypothesis test based on whether the Null hypothesis

was accepted (correct or true) or rejected (not correct or false ) and whether or not the Null

hypothesis was true or false.

A Type I (alpha) error rejects Ho even

though the statement is actually true. This

error is controlled for by the statistical test.

A Type II (beta) error accepts Ho even

though the statement is actually false. This

error is NOT controlled for by the statisticaltest.

The POWER is the probability that the

statistical test will reject the Null

hypothesis when it is FALSE (1-beta).

5. Statistical Significance involves the likelihood (probability) of obtaining a given result by

chance (it may, or may not, be true). This statistical significance is commonly represented by

the letter p(for probability). The value ofpcan be between zero (0) and one (1).

Thepvalue. Think percentage when using the pvalue. This describes the likelihood that your

data is, or is not, valid. If you conver t the pvalue to a percent (p = 0.01 becomes 1 percent), this

means that there is only a 1% chance that your data is inaccurate or does not support theconclusion you have drawn. The smaller the pvalue (0.05 > 0.01 > 0.001), the more likely it is

that your conclusions are accurate and your hypothesis is correct. That is, your conclusion is

due to the data and is not simply an accidental finding. It is usually understood that a pvalue

must be 0.05, or below, for data to be statistically valid.

Ways of expressing statistical significance with a pvalue of 0.05 include:

a. the finding is significant at the 0.05 level;

b. the confidence level is 95% (I am 95% certain of my findings);

c. the Type I error rate is 0.05;

The Null Hypothesis (Ho) is:

TRUE FALSE

Accepted

Ho

Correct Type II error

(beta)

Rejected Ho Type I error(alpha)

Correct


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d. the alpha level is 0.05;

e. = 0.05;

f. there is a 1 in 20 chance of obtaining a false result;

g. the p value is 0.05; and

h. p = 0.05

Averages and related values

The Arithmetic Mean is the same as what we usually call the average of a set of values. The usual

method to calculate the arithmetic mean (average) is to add up all of the values and divide that total

by the number of values. This can be expressed in a formula as:

X = 3 X which is a shorter form of X = X1 + X2 + X3 + X4 + X5 * * *

N N

The Median is defined as the point in the distribution of data that has 50% of the data points on

each side. This is best done by arranging the various data values in numerical order and then

counting the data points. For the data points 1, 3, 5, 7, 9 the median is 5 because there two points

below and two points above five. This works when there is an odd number of data points.

If there are an even number of data points, then some calculation is required. For the data points 2,

4, 6, 8 here the median is again 5 because there are two points above and two points below. This

could be calculated by adding up the numbers and dividing by the number of observations: 20 div by

4 = 5.

Consider the more complicated series of data 14, 15, 16, 17, 17, 17, 18, 19, 20, 21 the presence of

the three 17's means we cannot chose either 16.5, 17, or 17.5 as our median. The three 17's are

presumed to be equally distributed between 16.5 and 17.5. Since we have three points (14, 15 and

16) below the 17's, we only need two m ore points (2 out of 3) to find our m edian. Therefore, we take

the 16.5 and add two-thirds (0.67) to determine a median of 17.2 for this case.

The Mode is the value that occurs most frequently in a set of data. For the set of data in the

paragraph above (14, 15, 16, 17, 17, 17, 18, 19, 20, 21), the mode is 17 because that value occurs

three times and no other value occurs more than once.

1 1Population S = (x X) Sample S = (x X)2 2 2 2

Variance (sigma) N Variance N-1

Population standard deviation = square root of population variance

Sample standard deviation = square root of sample variance

Degrees of freedom = means the freedom to vary. Suppose we have 44 data points. First, we compute

the mean, and then we take the deviation of each score from the mean and compute the standard

deviation. We had 44 degrees of freedom to begin with, but in the computations described, we used

up one degree of freedom. Thus, we now have 43 (N 1) degrees of freedom. Degrees of freedom will

be used in the t-test and Chi Square described below.

When the number of measurements (sample size) is very small (e.g., < 25), then the t-test statistic

is used to assess the difference between means. This is also caused the parametric test of means. Also

known as the t-ratioor as Students t, the t-test statistic is calculated from the formula:

Population mean (X) Sample Mean (Xo) with N-1 degrees of freedom.

Sample standard deviation the square root of the number of observations (%&N)


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The Chi Square (X ) test of significance. This is most useful when the data are in frequencies such2

as the number of responses in different categories. Chi Square is a non-parametric test of means. Two

illustrations can explain Chi Square.

First case, when one tosses a coin 100 times, you would expect to get 50 heads and 50 tails. But when

you actually tossed a coin 100 times, you got 40 heads and 60 tails . We make the null hypothesis that

the 40/60 split does not differ from the 50/50 split we would expect by chance.

The general formula for Ch i Square is X = 3 (Observed Expected)2 2

Expected

Thus, X = (40 50) + (60 50) = 4.002 2 2

50 50

Now one needs to use a Chi Square table and refer to the one degree of freedom line (a coin can only

come up heads or tails; hence, one degree of freedom). The table value for 4.00 lies between 0.05 and

0.02. Thus, we can be 5% confident that these results are different from those that would occur by

chance alone.

Second case, the null hypothesis (Ho) is that there is no relationship between the gender of a child

and the childs favorite TV commercial.

Data Table Group A B C Totals

Boys 30 (25.2) 29 (37.2) 16 (12.6) 75

Girls 12 (16.8) 33 (24.8) 5 (8.4) 50

Totals 42 62 21 125

The numbers in the parenthesis are calcu lated by us ing the totals in the columns and rows. The 25.2

in the Boys (A) group is calculated by taking the total number of boys (75) multiplied by the total

number in Group A (42) and then divided by the total number of boys and girls (125). Thus, 25.2 is

the expected value for this group.

X = (30 25.2) + (29 37.2) + ...... + ...... + ..... + ..... = 9.0972 2 2

25.2 37.2

The degrees o f freedom are calcu lated by the formula:

(No. of Rows 1) x (No. of Columns 1 ) = (2 1) x (3 1) = (1) x (2) = 2 degrees of freedom

Going to a Chi Square table; 9.097 with two degrees of freedom falls between 0.02 and 0.01. Thus,

at an alpha of 0.02, we would reject the null hypothesis that gender and favorite TV commercial are

independent. That is, they are not associated.

Sample from a Chi Square Table

Degrees of

Freedom

Probability

0.50

Probability

0.10

Probability

0.05

Probability

0.01

Probability

0.001

1 0.455 2.706 3.841 6.635 10.827

2 1.386 4.605 5.991 9.210 13.815

5 4.351 9.236 11.070 15.086 20.517

10 9.342 15.987 18.307 23.209 29.588

2006 Pharmaceutics

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