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By L. Sibur-Narad. Carnegie Institution of Washington. 2018.

A dose of drug is administered by rapid intravenous injection buy discount fincar 5mg online prostate cancer bone scan, and the concentrations shown in Table 6-1 result. The last four points form a straight line, (similar to Figure 6-5) so back-extrapolate a line that connects them to the y-axis. Then, for the first five points, extrapolated values can be estimated at each time (0. Subtracting the extrapolated values from the actual plasma concentrations yields a new set of residual concentration points, similar to those values shown in Table 6-2. Plot the residual concentrations (on the same semilog paper) versus time and draw a straight line connecting all of your new points (similar to Figure 6-7). Note that α must be greater than β, indicating that drug removal from plasma by distribution into tissues proceeds at a greater rate than does drug removal from plasma by eliminating organs (e. Plasma drug concentrations with a two-compartment model after an intravenous bolus dose. For a one-compartment model (Figure 6-8), we know that the plasma concentration (C) at any time (t) can be described by: -Kt Ct = C0e (See Equation 3-2. The equation is called a monoexponential equation because the line is described by one exponent. The two-compartment model (Figure 6-9) is the sum of two linear components, representing distribution and elimination (Figure 6-10), so we can determine drug concentration (C) at any time (t) by adding those two components. Therefore: -αt -βt Ct = Ae + Be This equation is called a biexponential equation because two exponents are incorporated. For the two-compartment model, different volume of distribution parameters exist: the central compartment volume (Vc), the volume by area (Varea, also known as Vβ), and the steady-state volume of distribution (Vss). As in the one-compartment model, a volume can be calculated by: For the two-compartment model, this volume would be equivalent to the volume of the central compartment (Vc). The Vc relates the amount of drug in the central compartment to the concentration in the central compartment. If another volume (Varea or Vβ) is determined from the area under the plasma concentration versus time curve and the terminal elimination rate constant (β), this volume is related as follows: This calculation is affected by changes in clearance (Cl). The Varea relates the amount of drug in the body to the concentration of drug in plasma in the post-absorption and post-distribution phase. Although it is not affected by changes in drug elimination or clearance, it is more difficult to calculate. One way to estimate Vss is to use the two-compartment microconstants: or it may be estimated by: using A, B, α, and β. Because different methods can be used to calculate the various volumes of distribution of a two- compartment model, you should always specify the method used. When reading a pharmacokinetic study, pay particular attention to the method for calculating the volume of distribution. Clinical Correlate Here is an example of one potential problem when dealing with drugs exhibiting biexponential elimination. Recall that A steeper slope equals a faster rate of elimination resulting in a shorter half-life. If a terminal half-life is being calculated for drugs such as vancomycin, you must be sure that the distribution phase is completed (approximately 3-4 hours after the dose) before drawing plasma levels. Plasma drug concentrations with a one-compartment model after an intravenous bolus dose (first-order elimination). Plasma drug concentrations with a two-compartment model after an intravenous bolus dose (first-order elimination). The plasma drug concentration versus time curve for a two- compartment model is represented by what type of curve? For a two-compartment model, which of the following is the term for the residual y-intercept for the terminal portion of the natural- log plasma-concentration versus time line? The equation describing elimination after an intravenous bolus dose of a drug characterized by a two-compartment model requires two exponential terms. A patient is given a 500-mg dose of drug by intravenous injection and the following plasma concentrations result.

A rate constant is a unit change per time expressed as reciprocal time units -1 (e buy fincar 5mg mastercard prostate cancer 3 monthly injection. We would expect the concentration to decrease from 120 mg/L to 60 mg/L, then to 30 mg/L, 15 mg/L, and finally, to 7. To find the T1/2 of 10 hours, find the interval of time necessary for the concentration to decrease from 100 to 50. When using the area method, it does not matter if the drug best fits any particular model. You may have neglected to divide the sum of 100 plus 50 by 2 before then multiplying by the width of 2. Remember, clearance has units of volume/time, so the units in the equation must result in volume/time. This area is estimated by dividing the drug concentration at 12 hours, -1 1 mg/L, by the elimination rate constant, 0. Two plasma concentrations are then determined, and the slope of the plasma concentration versus time curve is calculated. Drug X is given to two patients, and two plasma drug concentrations are then determined for each patient. Why is the half-life of most drugs the same at high and low plasma concentrations? The plasma concentration versus time curves for two different drugs are exactly parallel; however, one of the drugs has much higher plasma concentrations. Explain why a reduction of drug clearance by 50% would result in the same intensity of effect as doubling the dose. The plasma concentration at 9 hours after the dose estimated from a plot of the points on semilog graph paper is: A. This terminal area is calculated by taking the final concentration (at 8 hours) and dividing by K above. Be sure your x-scale for time is correct and that you extrapolated the concentration for 9 hours. First, estimate C0 by drawing a line back to time = 0 (t0) using the three plotted points. You may not have multiplied by 1/2 when calculating the area from 8 hours to infinity. Describe the principle of superposition and how it applies to multiple drug dosing. Calculate the estimated peak plasma concentration after multiple drug dosing (at steady state). Calculate the estimated trough plasma concentration after multiple drug dosing (at steady state). Using these models, we can obtain an elimination rate constant (K) and then calculate volume of distribution (V) and dosing interval (τ) based on this K value. Most clinical situations, however, require a therapeutic effect for time periods extending beyond the effect of one dose. The goal is to maintain a therapeutic effect by keeping the amount of drug in the body, as well as the concentration of drug in the plasma, within a fairly constant range (the therapeutic range). Although intermediate equations are used, only the final equation is important to remember. The first dose produces a plasma drug concentration versus time curve like the one in Figure 4-1. C0 is now referred to as Cmax, meaning maximum concentration, to group it with the other peak concentrations that occur with multiple dosing. If a second bolus dose is administered before the first dose is completely eliminated, the maximum concentration after the second dose (Cmax2) will be higher than that after the first dose (Cmax1) (Figure 4-2). The second part of the curve will be very similar to the first curve but will be higher (have a greater concentration) because some drug remains from the first dose when the second dose is administered.

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Variance-Ratio Test (or F-Test) A test that makes use of the ratio of the variances of two sets of results to determine if the standard deviations (s) are significantly different cheap 5mg fincar visa mens health 9. Its application may also be extended to compare precisely the results obtained either from two different laboratories or from two different analytical procedures. In both these instances, the physical characteristics are directly proportional to the concentration of the analyte under examination. In usual prac- tice, a number of solutions having known concentrations is prepared and the response of the instrument is subsequently measured for each standard solution. Finally, a standard curve or calibration curve is plotted between the observed response Vs concentration, which invariably gives rise to straight line. It has been noticed, that the experimental points rarely fall exactly upon a straight line by virtue of the indeterminate errors caused by the instrument readings. At this juncture, an analyst is confronted with the tedious problem to obtain the ‘best’ straight line for the standard curve based on the observed points so that the error in estimating the concentration of the unknown sample is brought down to the least possible extent. At this stage, instead of deciding to draw the line merely on an analyst’s judgement, statistics comes to the rescue by providing a mathematical relationship whereby the analyst not only may calculate the slope objectively but also can obtain the ‘best’ straight line. Presumably, the indeterminate errors caused by the instrument readings, y, are responsible for not allowing the ‘data points’ to fall exactly on the line. Therefore, the sum of the squares of the deviations obtained from the real instrument readings with respect to the correct values are minimized coinsiderably by adjusting adequately the values of the slope, m, and the intercept, b. Statistically, the slope (m) and intercept (b) of the straight line may be obtained by the help of the following equations : ∑ xy − (∑ x ∑ y) / n Slope : m = C 12 67. At this point, let us suppose that the ‘calibration curve’ is used to find out the concentration of the ‘unknown’. Assuming that three determinations have been carried out separately, thereby giving three y values of 5. Thus, two situations often arise, namely : (i) Number of replicates being small, and (ii) Number of replicates being large. Number of Replicates being Large In this instance, the analyst has the privilege of rejecting one value (i. They may be applied in a sequential manner as follows : (i) Calculate the mean ( x ) and average deviation (d ) of the ‘good’ results, (ii) Determine the deviation of the ‘suspected’ result from the mean of the ‘good’ results, (iii) In case, the deviation of the suspected result was found to be either 2. Rules Based on the Range The Q test, suggested by Dean and Dixon**** (1951) is statistically correct and valid, and it may be applied easily as stated below : (i) Calculate the range of the results, (ii) Determine the difference between the suspected result and its closest neighbour, (iii) Divide the difference obtained in (ii) above by the range from (i) to arrive at the rejection Quotient Q, (iv) Finally, consult a table of Q-values. In case, the computed value of Q is found to be greater than the value given in the table, the result in question can be rejected outright with 90% confidence that it was perhaps subject to some factor or the other which never affected the other results. Note : The Q-test administers excellent justification for the outright rejection of abnormally erroneous values ; however, it fails to eliminate the problem with less deviant suspicious values. Note : Youden* (1967) suggested that once the analytical error is reduced to 1/3rd of the sampling error, further reduction of the former is not required anymore. In order to have a meaningful ‘sampling plan’ the following points should be taken into consideration**, namely : (1) Number of samples to be taken : (2) Size of the sample, and (3) Should separate samples be analysed or should a sample made up of two or more increments (i. How many samples must be taken to give (at 95% confidence level) a sampling error of less than 0. Q Conclusion : From this test it has been established that at least 34 samples are required if the specifi- cations provided in the above cited example are to be fulfilled adequately. Sample Size : Another major problem associated with the sampling process is that of the sample size. In fact, the sample size withdrawn from a heterogeneous material is solely guided by two factors, namely : (a) Variation in particle size, and (b) Precision required in the results of the analysis. The sampling variance, V, is inversely proportional to the actual number of sampling increments (n) and may be expressed as : k V = n where, k = Constant entirely dependent on the size of the increment and variation within the bulk material. The following points with regard to sampling may be observed carefully : • A major source of error in sampling may be incorporated from the actual process of taking increments from the bulk material, • The accuracy of the sample is determined by its total size (based on Random Sampling Theory), and • The number of increments taken shall directly influence the sampling accuracy provided the bulk material comprises of varying particle sizes. What are the two major types of ‘errors’ invariably encountered in pharmaceutical analysis?

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