Multiple Short Term Infusion Homework # 5 PHA 5127
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1 Multipl Short rm Infusion Homwork # 5 PHA 527 A rug is aministr as a short trm infusion. h avrag pharmacokintic paramtrs for this rug ar: k 0.40 hr - V 28 L his rug follows a on-compartmnt boy mol. A 300 mg os of this rug is givn as a short trm infusion ovr 30 minuts. What is th infusion rat? What will b th plasma concntration at th n of th infusion? 2 How long will it tak for th plasma concntration to fall to 5.0 mg/l? 3 If anothr infusion is start 5.5 hours aftr th first infusion was stopp, what will th plasma concntration b just bfor th scon infusion? 4* What will th pak lvl b at th n of th scon infusion? 5* If th osing intrval is kpt at 6 hours, what will th trough lvl b bfor th thir infusion is start? 6 Calculat th pak an trough plasma lvls at stay-stat for this osing rgimn. 7 How long will it tak to rach stay-stat lvls? How many infusions ar rquir to rach stay-stat conitions? 8 o you think a loaing os woul b bnficial in this xampl? Suggst a loaing os. (Aum that th avrag stay-stat lvl is wll within th thraputic winow for this rug. 9* h sir pak an trough lvls at stay-stat ar 2 mg/l an 3 mg/l rspctivly. How oftn must th rug b infus to achiv ths lvls? How shoul th os b ajust for this nw rgimn? Aum that th infusion tim is kpt at 30 minuts (0.5 hr. *ifficult but o-abl. isktt #2/Problms/Homwork/Multipl short trm.oc
2 h infusion rat is simply th amount of rug (or volum of solution of givn concntration aministr pr unit tim, 300mg k 0 0mg 30min / min Aftr a singl infusion, th pak lvl is givn by: Cl k (max ( whr Cl k 300 mg (0.40hr x0.5hr [ (0.40hr (28L(0.5hr 9.7 mg/l V 2 Concntrations at any tim point t' aftr stopping th infusion may b calculat by multiplying (max by -k t', whr t' is th amount of tim that has laps sinc stopping th infusion: (t' (max -k t' his is poibl for a rug following a on-compartmnt boy mol with st orr limination. his is analogous to th quation w us for a singl i.v. bolus injction, (t 0 -k t Only hr w ar trating (max as th (0 Exampl: Calculat 30 minuts (0.5 hr aftr stopping th infusion t k t' ( ' (max whr t' 0. 5 hr (9.7mg / L 7.95 mg/l (0.4hr x0.5hr isktt #2/Problms/Homwork/Multipl short trm.oc 2
3 o trmin how much tim is rquir for th concntration to rop to 5.0 mg/l, w must solv th (t' xprion abov for t': (t' (max -k t' whr (t' 5.0 mg/l (max ( t' t ' tak th natural logarithm of ach sli, k ( t' ln (max t' t' is thn givn by t' k ( t' ln (max 5.0mg / L ln 0.4hr 9.7mg / L 60 min.66hr x 00 min hr 3 Hr w n to calculat th trough lvl just bfor th 2 n os. h osing intrval is apparntly 6 hours: isktt #2/Problms/Homwork/Multipl short trm.oc 3
4 I hav fin t' now as th tim btwn th stop of th first infusion an th start of th 2 n infusion. his may b xpr in trms of th infusion tim an th osing intrval : t' - ( hr 5.5 hr h quation for calculating (min is (min (max k (9.7 mg / L.08 mg/l ( (0.40hr x5.5hr 4 his calculation is mor involv than th prvious ons. W bgin anothr infusion which will incras th concntration. Howvr, w o not bgin with 0. At th start of th 2 n infusion, th plasma lvl is.08 mg/l. w must consir this concntration (which will continu to cras by st orr limination an a it to th incras in provi by th 2 n infusion: (max (min 2 + ( k V u to th complxity of ths quations, w oftn limit our iscuion to concntrations following on os or at stay-stat. (max `(min 2 + ( k V isktt #2/Problms/Homwork/Multipl short trm.oc 4
5 (.08mg / L (0.4hr x0.5hr 300mg + [ (0.40hr (28L(0.5hr (0.40 hr x0.5hr ( mg/l 0.59 mg/l 5 If th osing intrval 6 an th infusion tim 0.5 o not chang, ( - ( hours transpir btwn th pak lvl (at th n of th infusion an th trough (just bfor anothr infusion bgins. his is tru for th st os, th 2 n os, an th nth os. (min (max -k ( - n (min n (max -k ( - (min (max -k ( - h trough lvl aftr th 2 n os is thn 2 (min 2 (max -k ( - (0.59 mg/l -(0.40hr -x(6-0.5hr.7 mg/l 6 At stay-stat, th (max is (max Cl ( ( this is list as "Pak (multipl os on your quation sht. W know this is a multipl osing xprion at stay-stat sinc th accumulation factor quation. r ( appars in th isktt #2/Problms/Homwork/Multipl short trm.oc 5
6 Rcall: h accumulation factor tlls you how much th plasma concntration incras at stay-stat compar to th first os. Compar th two quations: (max ( vs Cl (max Cl ( ( h stay-stat max for this osing rgimn is thn 300mg [ (max x (0.40hr (28L(0.5hr [ 0.68 mg/l (0.40hr (0.40hr (0.5hr (6hr h trough at stay-stat is (min (max ( (0.68mg / L.8 mg/l (0.40hr x5.5hr 7 It taks roughly 5-7 half-livs to rach stay-stat for a givn osing rgimn. h half-lif for this rug is t ln k 0.40hr / 2 hus, to rach stay-stat lvls, 5 x.73 hours 8.65 hours.73 hours ar rquir. Sinc th osing intrval is 6 hours, 0 st os 6 2 n os * r os 2 o ar rquir to rach stay-stat. hat is, w ar at stay-stat lvls just aftr th 2 n infusion. isktt #2/Problms/Homwork/Multipl short trm.oc 6
7 8 Sinc lvls ar not that much iffrnt than thos aftr th st os (an sinc th stay-stat is rach aftr th 2 n os, a loaing os may not b n hr. Of cours, this woul pn on th thraputic winow. If w n th plasma concntration to b 0.68 mg/l aftr th st infusion, w woul n to incras th amount of rug aministr slightly. h short-trm infusion quation for singl osing is (max ( Cl (0.68mg / L(0.4hr (0.40hr [ x28l(0.5hr x0.5hr 330 mg his is solv for (which will b th loaing os an (max is rplac by (max: L (max Cl ( 9 Hr w want (max 2 mg/l an (min 3 mg/l. th fluctuation factor F is thn (max 2mg / L F 4 (min 3mg / L W can trmin th osing intrval using this fluctuation factor: F ln k + his quation is list unr "calculat rcommn osing intrval" on your quation sht. It is similar to th xprion us in i.v. bolus quations xcpt for short-trm infusions, th infusion tim must b a on. If is still 0.5 hr, ln hr 4 hours 0.40hr isktt #2/Problms/Homwork/Multipl short trm.oc 7
8 Having trmin th osing intrval, w can calculat th os n to rach ths staystat lvls. W can o this by solving th (max xprion for os : (max k V ( ( which rarrangs to giv (max k V ( ( (2mg / L(0.40hr (28L(0.5hr[ (0.40hr x0.5hr [ (0.40hr 296 mg hus, giving th sam os (296 mg 300 mg mor frquntly rai th concntrations slightly* an th fluctuation is cras**. x4hr *300 mg vry 4 hours 6 x mg vry 24hours vs th initial osing rgimn which was 300 mg vry 6 hours 4 x 300 mg 200 mg vry 24 hours. Obviously, giving mor rug uring th sam tim intrval will incras plasma lvls. **If th osing intrval is cras, thr is l tim btwn o for limination to tak plac. hus, (max an (min will b closr togthr with l fluctuation. isktt #2/Problms/Homwork/Multipl short trm.oc 8
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