Saturday, January 25, 2020

Preformulation Testing for Chemical Properties of Drug

Preformulation Testing for Chemical Properties of Drug PREFORMULATION STUDIES Preformulation testing involved investigation of physical and chemical properties of a drug substance alone and when combined with excipients. It was the first step in the rational development of dosage forms. These studies are categorised as under: 1. API characterization 2. Drug-Excipient Compatibility study API Characterization Organoleptic Evaluation These are preliminary characteristics of any substance which is useful in identification of specific material. Following physical properties of API were studied. a) Colour b) Odour Table no. : Characterization of API Test Observation Colour White Odour Odourless Particle size distribution Sieve analysis: The sieve analysis main concept is to know the different drug particles size in the sample. The standard sieves with larger pore size i.e., with less sieve number on the top position are placed one over the other and followed by sieves of decreasing pore size i.e., with larger sieve number towards the bottom. Procedure: Clean and dried sieves #40,#60,#80,#100,#120 and bottom meshes were collected Individual weight of each sieve was noted. These sieves were arranged in ascending order. Weighed quantity of API was placed in #40 meshes. Sieve shaker was set for 5 min at amplitude of 60. Remove the setup from sieve shaker after 5minutes. Each mesh was weighed individually and Calculate % retained in each size of mesh with following formula: % retained = Final weight – initial weight x 100 Total weight taken Table no. : Particle size distribution of API Sieve number Percentage of sample retained (%) Cumulative percentage of sample retained (%) 40 3.0 3.0 60 19.2 22.2 80 26.3 48.5 100 24.2 72.7 140 8.1 80.8 200 19.2 100.0 pH-Solubility Profile: The solubility studies for the drug were carried out using the orbital shaker. Solubility of the drug across different buffers was studied. The pH ranged from 1.2 to 6.8 (1.2 , 4.5, 6.8, and water). All the buffers were prepared according to USP 34 NF 29, 2011. Excess drug was added to 100 ml of water in stoppered conical flasks and were agitated continuously in a orbital shaker for 24 hrs at 200 rpm and room temperature (25Â ° C), till saturation was observed. Then, the samples were filtered using 0.45 Â µ Nylon (47 mm) syringe filters. Now the filtered samples were analyzed using UV spectrophotometer Table No.6.3 below describes the procedure of buffers preparation. Table 6.5 preparation of buffers Buffer Procedure PH 1.2 buffer 8.5 ml of Conc. HCL was added to 1000 mL volumetric flask. Then it was diluted and made up to volume with water PH 4.5 phosphate buffer 13.61 gm of KH2PO4 was added to 1000 mlvolumetric flask. Then it was made up to volumewith water PH 6.8 phosphate buffer 250 mL of 0.2 M monobasic potassium phosphate solution was taken in a 1000 mL volumetric flask. Then 112 mL 0.2M Sodium hydroxide solution was added to it and water was added to make up to the volume Table7.3 Solubility of API in buffers of different pHs PH Solubility(mg/ml) 1.2 42.36 4.5 44.96 6.8 0.80 Water 0.674 Fig 7.1 pH Solubility curve of API Drug excipient compatibility study There is always possibility of Drug excipient interaction in any formulation due to their intimate contact. It is also necessary to determine any possible interaction between excipients used in the formulation. This will also indicate success of stability studies. Preliminary studies: Method: Physical observation Condition: 40 Â ±2 o C and 75 Â ± 5% RH Procedure: Drug was mixed with excipients in 1: 1 ratios as indicated in the Table 6.6 These mixtures were kept in a 5ml glass vials and packed properly. In dry close method glass vials are closed with rubber stoppers These vials are exposed to 25ËÅ ¡C /60 % RH 40ËÅ ¡C /75 % RH. Blend (1gm) was prepared and filled in vials. Observations for physical appearance were made at the end 4 weeks. S.No EXCIPIENTS DRUG: EXCIPIENT RATIO 1 Polyethyleneoxide 1:1 2 HPMCK100M 1:1 3 MCC 1:1 4 Cellulose acetate 1:1 5 Sodiumchloride 1:1 6 Citric acid 1:1 7 Sodium lauryl sulphate 1:1 8 Magnesium stearate 1:1 9 Talc 1:1 Table 7.4 Results of Drug-Excipient compatibility at 25oC/60% RH S.no: Excipent Colour change Lumps Caking O C O C O C 1 Polyethylene oxide X X X X X X 2 HPMC X X X X X X 3 MCC X X X X X X 4 Cellulose acetate X X X X X X 5 Sodium chloride X X Lumps observed X X X 6 Citric acid X X Lumps observed X X X 7 Sodium lauryl sulphate X X Lumps observed X X X 7 Magnesium stearate X X X X X X 8 Talc X X X X X X Note: x – indicates no change, O- open condition ,C- close condition Table 7.5 Results of Drug-Excipient compatibility at 40oC / 75% RH S.no: Excipient Colour change Lumps Caking O C O C O C 1 Poly ethylene oxide X X X X X X 2 HPMC X X X X X X 3 MCC X X X X X X 4 Cellulose acetate X X X X X X 5 Sodium chloride X X X X Caking observed X 6 Citric acid X X X X Caking Observed X 7 Sodium lauryl sulphate X X X X Caking observed X 8 Magnesium stearate X X X X X X 9 Talc X X X X X X FTIR Study FTIR study: FTIR studies were carried out for pure drug alone and blend of drug excepients. The FTIR spectroscopy (BRUKER Optics FTIR spectrophotometer) is employed as analytical tool to check the drug-excepients interaction, using the KBr disc method. The FTIR spectra were scanned and recorded between 400 and 4000 cm-1 Method: Nearly to a fine alkali halide (example KBr) powder of 200 to 250 mg 0.1 to 1.0 % sample is mixed well. Later it is pulverized and in a pellet-forming die it is placed. Around an 8 tons force under a vacuum of several mm Hg is applied to form transparent pellets. FTIR spectroscopy of pure drug of Famotidine S.no Type of vibration Actual frequency (cm-1) Observed frequency (cm-1) Confirmation Table no. : Interpretation of FTIR spectra of pure famotidine FTIR spectroscopy of drug and excipient blends Table 7.6 Peaks of FTIR study Peaks ( cm –1) Functional groups 3506.13 –OH 3377.41-3400.95 –NH2 3238.03 -NH 1445.38- 1639.22 C=N 689.10 -606.6 C-S 1320.81 S(=O) 2 asymmetric stretching 1147.17 S(=O) 2 symmetric stretching Table no. : Interpretation of FTIR spectra of pure famotidine S.no Type of vibration Actual frequency (cm-1) Observed frequency (cm-1) Confirmation ANALYTICAL METHOD ESTIMATION OF FAMOTIDINE: A solution of Famotidine was prepared in 0.1 N HCl and Phosphate buffer pH 4.5and 6.8 UV spectrum was taken using Perkin Elmer UV/Vis double beam spectrophotometer.The UV maxima of Famotidine was found to be 265 nm in both 0.1N HCl pH 4.5. In pH 6.8 it was found to be 268 nm Preparation of standard curve of famotidine in 0.1N HCL pH 4.5 phosphate buffer: 100 mg Famotidine each was dissolved in 0.1 N HCl and pH 4.5 buffer and volume is made up to 100 with respective buffer. 10 mL of stock solution (1mg/ml) was further diluted upto 100 ml with respective buffer to obtained solution of 100 Â µg/mL.Now from stock 2 further dilutions were done with respective buffer to obtain solutions of 2, 5, 10, 15, 20 and 25 Â µg/ml Absorbance of each solution was measured at 265 nm using Perkin Elmer UV/Vis double beam Spectrophotometer. Preparation of standard curve in ph 6.8 phosphate buffer: 10 mg Famotidine each was dissolved in pH 6.8 phosphate buffer and volume is made up to 100 ml to obtain solution of 100 Â µg/ml. Now from this stock solution further dilutions were done with PH 6.8 to obtain solutions of 10 , 20 , 30 and 40 Â µg/ml Absorbance of each solution was measured at 268 nm using Perkin Elmer UV/Vis double beam Spectrophotometer. The experiment was performed in triplicate and based on average absorbance; the equation for the best line was generated. The results of standard curve prepared in pH 1.2, 4.5 6.8 were shown below Table 7.7 Standard curve of API in PH 1.2 , 4.5 6.8 buffers Concentration Absorbance in pH 1.2 Absorbance in pH 4.5 Absorbance in pH 6.8 2 0.085 0.082 4 0.141 0.148 5 0.189 0.186 10 0.333 0.341 0.251 15 0.510 0.497 20 0.701 0.651 0.467 25 0.852 0.806 30 0.746 40 0.989 FIG 7.4 standard curve at PH 1.2 Buffer Fig 7.5 standard curve at PH 4.5 Buffer Fig 7.6 standard curve at PH 6.8 Buffer Calculation of initial dose and maintenance dose for the design of elementary osmotic pump of famotidine for 12 hours: There are no sustained release formulations for famotidine in the market, hence the total dose (DT) consisting of initial (DI) and maintenance doses (DM) for formulating the famotidine sustained release was calculated as per Robinson and Eriksen equation with a zero order release principle36 . In this profile the rate of delivery is independent of the amount of drug remaining in the dosage form and constant over time as shown by the Eq. 6.1 Drug availability rate k0 = Rate in = Rate out Eq. 6.1 Where, k0 is the zero order rate constant for drug release (amount per time). DI is required to give initial rapid release of drug so as to attain the minimum therapeutic level immediately after dosing. Inital dose (DI) = CSSAVG Vd Eq. 6.2 F Where, C ssavg is the average steady state plasma level, V d is the volume of distribution and F is the fraction of dose absorbed. k0 = DIKel Eq. 6.3 Where, Kel is overall first order drug elimination rate constant (per hour). Hence k 0 should be equal to the elimination rate constant so as to maintain the steady state condition. In general the total dose required (D T) is the sum of maintenance dose (DM) and the initial dose (DI) DT = D I + D M Eq. 6.4 In practice, D M (mg) is released over a period of time and is equal to the product of H (the number of hours for which sustained action is desired after initial dose) and the zero order rate constant, k0 (mg/hr). Therefore the Eq. 6.4 can be expressed as DT = D I + k0H Eq. 6.5 Ideally the maintenance dose (DM) is released after DI has produced a minimum therapeutic blood level of the drug. However due to the limits of formulations, drug release even starts from DM also from the beginning i.e. at t=0, thus increasing the initial drug level in the blood. Hence it is necessary to reduce the initial dose of the drug to account for the excess release for drug from DM by using a correction factor, k0tp. This correction factor is the amount of drug provided by DM during the period from t=0 to the time of the peak drug level, tp. The corrected initial dose (DI*) becomes DI-(k0tp). Then the total dose is DT = DI* + k0H = (D I k0tp) + k0H Eq. 6.6 Pharmacokinetic parameters of famotidine: Elimination half life (t1/2) of famotidine is 3 hrs (average of 2.5 to 3.5 hrs), the time to reach peak plasma (t p) is 3 hrs and Vd = 80.5 L and F = 0.4 54,55 . From the literature of the PEPCID (innovator product of famotidine in USA) label and pharmacological review information 49,, it was found that the plasma levels after multiple doses are similar to those after single doses indicating the C max is similar to Cssavg , therefore Cmax of 0.07 mg/L was taken as C ssavg . Calculation of D I and DM: The initial dose (DI), corrected initial dose (DI*), maintenance dose (DM) and total dose (DT) were calculated according to calculations described above. Calculation of elimination rate constant: Elimination rate constant (K el ) = 0.693/t 1/2 = 0.693/3 = 0.231 hr -1 Calculation of initial dose: Inital dose (DI) = CSSAVG Vd Eq. 6.2 F = (0.07 X 80.5)/0.4 = 14.0875 mg Calculation of desired input rate (k 0): Desired input rate from maintenance dose (k 0) = DIKel = 14.0875 X 0.231 = 3.25 mg/hr Calculation of maintenance dose: Maintenance dose (DM) = k0H (Since, H = the number of hours for which sustained action is desired after initial dose = (12-1) = 11 hrs) = 3.25 X 11 = 35.796 mg Calculation of corrected initial dose DI*: DI* = DI – (k0tp) = 14.087 – (3.25 X 3 ) = 4.93 mg Calculation of total dose: Total dose (D T) = D I* + D M = 4.93 + 35.796 = 40.726 mg From the above calculations the total dose obtained for sustained release of famotidine for 12 hrs is 40.726 mg. The total dose was rounded off to 40mg for the convenience. Initially the dosage form should release the total initial dose (i.e. 4.93 mg ~ 5.0 of drug, means 11% of total 50 mg dose) in the first 1 hr followed by maintenance dose (i.e. 40-5=35 mg of drug) for up to 12 hrs there after at a release rate of 3.25 mg/hr (i.e. 8.125% of total 40 mg dose). Based on these assumptions the theoretical release profile was predicted and shown Table 6.7 Predicted theoretical release profile Time (hrs) % CDD 1 11.4 2 19.425 3 27.45 4 35.475 5 43.5 6 51.525 7 59.55 8 67.575 10 83.625 12 99.675 JNTUA-OTRI, Ananthapuramu 1

Friday, January 17, 2020

High School vs. College Essay

I think a good education is an important part of one’s life. To achieve a good education, one should attend both High School and College. The transition from High School to College is a step that a student will either adjust to or struggle with. Although, some people think High School has a lot in common with college, I find they have a few differences. There are also certain similarities as well, by which, one won’t feel as if College is a new world. The more prepared a person is to face the differences and similarities, the more successful they might be. High School and College are both educational grounds for a student to grow with knowledge. A student graduates from High School and again from College with a degree. Both places are full of experiences and filled with numerous memories. The government runs them. They both play an important role in making a person into a collected individual and a member of a society. High School students know that there are differences between High School and College, but sometimes what they think is not how it is. To begin with there are many ways in which the attitudes of the teachers in High School differ from the attitudes of the teachers in College. In High School, my teachers seemed to be stricter and have more rules for the students to follow. There was an everyday time schedule for each student to go by. Students go through drama in High School which some cannot get out of. Attendance is very important in High School as well as in College. Many teachers enforce it while others do not. I have noticed that it is the student’s responsibility to come to class. They believe that the students should be mature enough to make their own decision on whether to attend class or not and leave it to them to make that decision. When a student graduates from High School, a sense of maturity comes in them. They start realizing that everything in High School was materialistic, and College is practical. College is different than High School just by the personal freedoms, the classroom and the social life. In College, no one would be concerned about the basic everyday drama that would surround a student in High School. College prepares a student to face the real world, and how to handle it. It separates the mature people from the immature people. However, a person who wants to attend College has to pay to further her education. If a student doesn’t take College seriously and apply herself, she knows she wasted her hard earned money, or her parents. So, since students must pay to get into College, she works and studies harder than she did in High School. Therefore, she will study those required courses and finish her education with a degree and start a career. I don’t think I would ever want to go back to High School. I love College and all the freedom that comes with it. All there is in College is education. Now I am learning to be a better person and to improve and to learn different study habits. High School is only the first step into growing up and preparing you for College, whereas, College is preparing you for your career.

Thursday, January 9, 2020

Saint John Paul II ( Jpii ) Essay - 1760 Words

The human call to strive for holiness is intrinsic as we are ‘sanctified by God’s divine grace’ that makes us sharers in God’s life through the life, death and resurrection of Christ. This amazing grace makes us holy adopted children of God, the temple of the Holy Spirit and gives a right to eternal life. But as humanity is graced with the gift of God’s love, so too through ‘freewill’ human beings fall into sin. God’s ‘actual grace’ which is extrinsic, enlightens us and strengthens us to do good over evil. Saint John Paul II(JPII) lived a life which reflected how his understanding of God’s grace enabled him to live in a time of oppression and hardship in his childhood, to follow God’s call as priest, bishop, pope and saint. In this essay, I will reflect JPII’s understanding of grace and how he lived this out in his Christian practice. JPII’s pontificate was at a time when the people of God were still trying to understand the teachings of the Second Vatican Council. Through his teachings JPII empowers God’s children to understand what the true meaning of their human existence is about, to enter communion and be united to God whose self-giving love has given his own son to share in the suffering of humanity. This is like the triune relationship of self-giving love. Humans are created in God’s image and likeness from the time of creation, ‘although formed from the dust of the earth it is a manifestation of God in the world as a sign of his presence and glory.’ JPIIShow MoreRelatedA Brief Biography of Pope John Paul II1856 Words   |  8 Pages John Paul II On May 18, 1920, in a small Polish town just outside of Wadowice, a child was born to Karol Wojtyla (1879-1941, and Emilia Kaczorosks (1884-1929). His name was Karol Jozef Wojtyla. Little did his parents know that one day their child was destined not only to become a priest and a bishop, but the 264th pope of the Roman Catholic Church, and only the second non-Italian pope. Emilia, a schoolteacher, died in childbirth. Wojtyla was nine years old and the youngest of three children

Wednesday, January 1, 2020

The Advantages And Disadvantages Of The American Civil War

The Advantages and disadvantages of both sides during the civil war The American Civil War started in 1861 and ended in 1865. The two sides of the war are the Union (North) and Confederacy (South). The South wanted to keep slavery to maintain their economy and they worried that the North would end slavery. The South then seceded, starting the conflict. Around 620,000 people died in the war. Both sides of the war had advantages and disadvantages in the Civil War. Population in the South was small because it was based around farming communities, putting them at a disadvantage because they had fewer people to make up their army. On the other hand, they had great generals and a lot of morale. Meanwhile, the North’s large†¦show more content†¦Three weeks before the war ended the Confederacy eventually added slaves to their army in exchange for their freedom, but were not able to go into action. Their slaves did not want to fight to preserve slavery. Roughly 186,000 African Americans fought in the Civil War overall (Boyer 449). The transportation of products, supplies, Natural resources, and people in the North and South was a big factor in who won the war. The attached 1861 railroad map by James Lloyd shows how dense the railroads in the North were compared to the South. The North relied more on the railroad to transport their manufactured goods. The railroads also helped the union by sending ammunition and food quickly, but The there was a fear of being sabotaged. In the map, we can also see the South railroads were very scattered. The South didn t have much of a need to have as many railroads because they were farming communities. The Union also destroyed what little railroads the South had (Boyer 454). The Confederacy then could not transport supplies nor people to other parts of the South (Beringer 310). 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