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Chemistry Lab Report Example Essay Sample

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Chemistry Lab Report Example Essay Sample

Requirements:

  • Heating Block
  • Cyclo Mixer
  • Micropipettes
  • Depyrogenated Micro tips
  • Depyrogenated 10 x 75 mm Borosil glass reaction tubes
  • Depyrogenated 13 x 100 mm Borosil glass tubes
  • Lysate
  • Control Standard Endotoxins (CSE) E.coli (10 Ng/vial)
  • LAL Reagent Water (LRW)

Depyrogenation:

All the glass wares are depyrogenated in a hot-air oven at 250o C for 30 minutes.
The method used for detection:

Gel Clot Method.

Control Standard Endotoxin(CSE) stock solution:

  •  The CSE solution is prepared using Endotoxins reference standard.
  •  Preserve the reconstituted concentrate in the refrigerator or 2-8oC cooling cabinet.

Reconstitution of CSE:

  • Reconstitute entire content of CSE vial using 5 ml of LAL reagent water.
  • Mix for 30 minutes in a cyclo-mixer.
  • Initial dilution is of 20 EU/ml. i.e.

RSE/CSE Ratio of geometric mean end points: 10 EU/ng
10 EU / ng * 10 ng / vial = 100 EU/ vial
100 EU/ vial/ 5 ml/vial = 20 EU/ ml (CSE Reconstituted Potency)
(EU/ ml = IU/ ml)

  • Make serial dilutions with the help of this dilution.
  • Mix all the contents well for at least three mins.
  • Mix the dilution for at least 30 secs before proceeding to the next step.
  • CSE dilution of different concentration shall be prepared before performing the test.
  • Prepare the dilution as per the given table.

Preparation of Limulus amoebocyte lysate (lal) working solution:

  • Use a LAL reagent of conformed sensitivity.
  • Remove the LAL reagent vial from the refrigerator and place it to acclimatize with the room temperature.
  • Reconstitute the lyophilized LAL with LRW as indicated on the vial test.
  • Swirl gently to dissolve, avoiding liquid contact with stopper.
  •  Add 0.1 ml of reconstituted lysate into each tube.
  • Add LAL reagent first in the negative control, then samples, positive product control and followed by positive controls.

Testing the sensitivity of the LAL reagent:

A std. solution is made of the following concentrations equal to 2λ, λ, λ/2, λ/4 diluting std. endotoxins stock solution (20 EU/ml) + LRW.

2λ = 0.25 EU/ml
Λ = 0.125 EU/ml
λ/2 = 0.06 EU/ml
λ/4 = 0.03 EU/ml

  • Mix a volume (100 µ ltr.) of the lysate solution (0.125 EU/ml) and a volume (100 µ ltr.) of one std sol in different tubes of 10 x 75 mm.
  • Incubate the reaction mixture at constant temperature and time (usually at 37oC for one hour).
  • It is crucial to forming a firm gel; then the only result will be recorded as positive. For this, take 1 test tube from the incubator and completely invert that at an angle of 180o.

Preparation of Test Solution:

Dissolve the active materials using LRW up to maximum level (M V D) at which the endotoxins limit can be determined. The Endotoxins limit active substance administered parenterally is expressed as IU/ml or IU/mg. Etc. in monographs.

MVD (maximum valid dilution) = Endotoxins limit x Concentration of test solution
Lysate Sensitivity (X)

  • M VD depends on the LAL sensitivity and drug concentration.
  • Calculate the M VD of product with the help of above mention formula.
  • Sample dilution prepared at MVD/2 and sample tested at MVD (add 50 µ ltr. LRW in the final reaction tube to prepared MVD from M V 1)/2).
  • Prepared two replicates tube for test Positive Product Control, Negative control, and Sample.

Test (PPC) = 50µl Diluted sample MVD/2 + 50 µl CSE (0.125 EU/ml) + 100 µl lysate (O. 125 EU/mI. sensitivity)

Sample = 50 Diluted sample MVD/2 + 50 µl LRW + 100 µl Lysate

Negative control (blank) = 100 µl LRW + 100 µl Lysate.

Incubate all the tubes at 370C +- IOC for 60 minutes +2 minutes in the Multiblock heater.

Interpretation:

  • After incubation of 1 hour, check the result by carefully inverted each reaction tube at 1800.
    A stable clot remains intact when the container is inverted; the result is recorded as positive, if the test solution runs out of the tube, the result is recorded as negative.
  • The test is proved to be correct if the effect is negative for both cases: Positive in positive control & Negative in the negative control (Blank).

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