Download MIDW 125 Math Review and Equation Sheet

MIDW 125 Math Review and Equation Sheet 1. The Metric System Measures of weight are based on the gram (g): 1 kilogram = 1 kg = 1000 gram = 1000 g = 103g 1 milligram = 1 mg = 10‐3 g 1 microgram = 1 g = 10‐6 g 1 nanogram = 1 ng = 10‐9 g Measures of volume are based on the litre (L): 1 millilitre = 1 mL = 10‐3 L 1 microlitre = 1 L = 10‐6 L 1 nanolitre = 1 nL = 10‐9 L 2. Logarithms You will use logarithms (logs) for the Henderson‐Hasselbach equation and for solving some PK problems - uses a base of 10 (“a” is 10 in the examples below) - we assume “log” is the same as “log10” - you can also take the “antilog” of a number - be sure you can locate these buttons/functions on your calculator log a b = c  a c = b log (10) = 1  101 = 10 log (100) = 2  102 = 100 antilog (x) = 10x You will use natural logarithms (ln) for solving some PK problems - uses the base ‘e’ where e is an irrational constant equal to 2.718 - there are several (very complicated) ways to calculate ‘e’. For your purposes what is important is that you know the series of buttons to push on your calculator to get the 2.718 result. This involves taking the anti‐natural logarithm of 1 (the ‘inverse (inv)’ function may be useful). loge(x) = ln(x) ln(e) = 1  e1=e ln(1) = 0  e0=1 3. Weak acids / Weak bases Acid – a proton (hydrogen ion, H+) donor Base – a proton (hydrogen ion, H+) acceptor Most drugs are weak acids or weak bases. In an aqueous environment such as body fluids, blood and plasma, weak acids and bases exist in equilibrium (ie they fluctuate) between the protonated and unprotonated forms as shown below: Weak acid: HA ⇋ H+ + A‐ Weak base: B + H+ ⇋ BH+ Although HA and A‐ are both forms of the acid they behave differently. The uncharged (unionized) HA form is able to diffuse cross lipid membranes (ie be absorbed), whereas the charged (ionized) A‐ form cannot. Conversely, the charged (ionized) A‐ form is water soluble and is readily excreted in the urine whereas the uncharged (unionized) HA form is not. Similarly, although B and BH+ are both forms of the base they behave differently. The uncharged (unionized) B form is able to diffuse cross lipid membranes (ie be absorbed), whereas the charged (ionized) BH+ form cannot. Conversely, the charged (ionized) BH+ form is water soluble and is readily excreted in the urine whereas the uncharged (unionized) B form is not. The pH of the environment in which the acid or base (drug) is located determines the relative amount of each form that is present. - In an acidic environment (high H+ concentration) the equilibrium will shift to favour greater amounts of the protonated HA and BH+ forms. - In a basic environment (low H+ concentration) the equilibrium will shift to favour greater amounts of the unprotonated A‐ and B forms. 4. Henderson‐Hasselbach Equation For both acids and bases: pH = pKa + log [Unprotonated form] [Protonated form] ** Memory Aid: “UP” to remember “Unprotonated” over “Protonated” ** Note: Ka is the dissociation constant (inherent property of the drug – can look it up in a reference book) and pKa = ‐log Ka a) For weak acids: Dissociation equation: H‐H equation: HA ⇋ H+ + A‐ pH = pKa + log [A‐] [HA] [A‐] = ionized or excreted portion of drug [HA] = unionized or absorbed portion of drug To find the percentage of A‐ or HA: antilog (pH‐pKa) = [A‐] = x which is the same as [A‐] = _x_ [HA] [HA] 1 Since the only forms present are A‐ and HA: % HA = HA_ x 100 = _1_ x 100 A‐ + HA x + 1 % A‐ = __A‐__ x 100 = _x_ x 100 A‐ + HA x + 1 b) For weak bases: Dissociation equation: BH+ ⇋ B + H+ H‐H equation: pH = pKa + log [B]_ [BH+] [BH+] = ionized or excreted portion of drug [B] = unionized or absorbed portion of drug To find the percentage of B or BH+: antilog (pH‐pKa) = [B] = x which is the same as [B] = _x_ [BH+] [BH+] 1 Since the only forms present are B and BH+: % B = B__ x 100 = _x_ x 100 B + BH+ x + 1 % BH+ = _BH+ x 100 = _1_ x 100 B + BH+ x + 1 5. Pharmacokinetic Equations a) Body fluid compartments: Total body water (TBW) = 42 L Intracellular volume ~ 28 L Extracellular volume ~ 14 L Interstitial volume ~ 10 L Plasma volume ~ 4 L b) Volume of distribution: Remember that unless a drug specifically distributes to one of the physiological fluid compartments stated above, Vd is not necessarily a real, physiological entity. Rather, it is a proportionality constant that relates the concentration of drug measured in the plasma/blood to the amount of drug initially given. Vd = Dose / Concentration c) Concentration‐time equations: Exponential form: Ct = C0e‐kt Linear form: ln Ct = ‐kt + ln C0 The linear form gives a straight line graph where the y‐axis intercept = ln C0 (use this to solve for C0) and the slope = ‐k. d) Half‐life: t½ = 0.693/k Assumptions: it takes ~5 half‐lives for a drug to washout (be 97% eliminated) once administration is stopped, and ~5 half‐lives for drug to reach a steady state concentration (Css) if administered at repeat intervals approximating the half‐life. Derivation of the half‐life equation – for your interest only: lnCt = ‐kt + lnC0 ln(Ct/C0) = ‐kt multiply through by ‐1 ln(C0/Ct) = kt when t = t½ Ct = ½C0 or C0 = 2Ct ln(2Ct/Ct) = kt½ ln2 = kt½ t½ = ln2/k = 0.693/k