The Microsomal Isolation and Effects of Xenobiotics Upon Endoplasmic Reticulum from Rat Liver Cells | BIOL 515, Papers of Cell Biology

Download The Microsomal Isolation and Effects of Xenobiotics Upon Endoplasmic Reticulum from Rat Liver Cells | BIOL 515 and more Papers Cell Biology in PDF only on Docsity!

The Microsomal Isolation And The Effects

Of Xenobiotics Upon the Endoplasmic Reticulum

From Rat Liver Cells

Jim Zubricky Thursday, 25 March 2004

• BIOL 515: Th 9:30-1:

this paper, Murataliev and Feyereisen found that house fly P450 reductase “..can rapidly oxidize only one molecule of NADPH, wherease the rate of oxidation of a second molecule of NADPH is too slow to account for the observed rates of catalysis. This demonstrates that house fly P450 reductase does not require a priming reaction with NADPH for catalysis” (4). Finally, it was noted by Bertilsson, Berkenstam and Blomquist in 2001 that promoters during the perinatal stage of human development “… have coevolved to functionally conserve P450 gene induction in response to xenobiotics through CAR and PXR (two nuclear receptors that play a key role in cytochrome P gene induction). Thus, nuclear receptors for xenobiotics may not only play a role to provide a survival advantage during adulthood, but also to protect the embryo against endrogenous and exogenous toxins” (5).

Methods

The methods used to perform this experiment are as follows: A. Microsome Isolation a. Remove rat liver, wash with 0.9% ice cold saline solution and blot. b. Weigh out 5.0 grams of liver. c. Mince liver in mitochondrial medium. d. Transfer to homogenizer with teflon pestle. Add enough mitochondrial medium such that there is 10 mL per gram tissue. e. Homogenize at 4oC (on ice). f. Place homogenate in polycarbonate tubes. Balance tubes. g. Spin tubes at 420g for 15mins. h. Remove supernatant and place in clean polycarbonate tubes. Discard pellet. i. Balance tubes. j. Spin tubes at 12,000g for 30 mins. k. Carefully remove supernatant and place in beaker. Keep at 4oC. Discard pellet. l. Take supernatant from Step j and place in appropriate tubes. m. Spin samples at 100,000g for 45 mins. Discard supernatant. n. Suspend microsomal pellet in 2.0 mL of 0.01M Tris-HCL buffer (pH 7.5). o. Use microsomal fraction for enzyme assays. p. Determine the protein content of the microsomal fraction. B. Enzyme Assay – NADPH: Cytochrome C Reducatase a. Prepare appropriate dilutions of microsomal fraction using 0.01M Tris-HCl buffer (pH 7.5). b. Prepare two curvettes as given below:

Refernce curvette Sample curvette 0.25 mL buffer^1 0.25 mL buffer 0.10 mL cytochrome c^2 0.10 mL cytochrome c 0.02 mL enzyme^3 0.02 mL enzyme 0.61 mL water 0.61 mL water

0.02 mL water 0.02 mL NADPH^4 c. Prepare sample cuvette and start reaction with the addition of NADPH. d. Read activity in a dual-beam spectrophotometer set at 550 nm. e. Express enzyme activity as μmoles cytochrome c reduced/min/mg protein.

Results

A. Control Rat After excising the liver from the untreated rat, we carefully cut out and recorded the mass of a piece of that liver. The mass of our team’s piece of liver was 4.5715 g. After homogenizing the rat liver with the mitochrondrial medium, we obtained 45 mL of homogenate. After following the directions given to us to prepare the microsomal fraction, we performed a Bradford assay to determine the amount of protein in solution. We recorded the following data from the spectrophotometer: Dilution of Microsomal Fraction Absorbance at 595 nm 1:25 1. 1:50 1. 1:100 0. 1:200 0. 1:500 0. Graphing the absorbances of the standard solution gives us the following graph: (^1) 0.4 M Tris-HCl buffer (pH 7.5) (^2) cytochrome c, 3mg/mL (^3) microsomal preparation (^4) NADPH, 0.012 M Amount of BSA (μg/mL) Absorbance at 595 nm 20 0. 40 0. 60 0. 80 0. 100 1.

Discussion

A. Control Rat The first task was to perform a Bradford assay upon the microsomal fraction. Using the equation of the trendline from the graph on page 5, we should be able to use the obtained absorbances for our samples and calculate the amount of protein (μg/mL) by using the fomula: x = (y – 0.1382) / 0.0092, where x = concentration of protein (in μg/mL) and y is the absorbance: If the individual amount of proteins are multiplied by its’ dilution factor, then we can take an average of the amount of protein in each sample: Dilution s Absorbanc e Amount of Protein (μg/mL) Dilution Factor Amount of Protein (μg/mL) Amount of Protein (mg/mL) 1:25 1.539 152.261 25 3806.53 3. 1:50 1.068 101.065 50 5053.25 5. 1:100 0.618 52.1522 100 5215.22 5. 1:200 0.371 25.3043 200 5060.86 5. 1:500 0.150 1.28261 500 641.305 0. Average ^ 3296. μg/mL

mg/mL Looking at the printout from the dual-beam spectrophotometer, we can see that the starting absorbance value is 0.11 at time = 0 mins. If we draw a line tangent to the curve made by the spectrophotometer, we can see that the ending absorbance value is 0.34, and the ending time is 3 minutes 12 seconds, or 3.20 mins. First, we can calculate the Δabsorbance: Δabs = abs 2 – abs 1 = 0.34 – 0. =0. The molar extinction coefficient for this compound is 18.5; to obtain the concentration, we apply Beer’s Law: Δabs = ε b c where b = pathlength of light (1 cm) C = Δabs / ε C = 0.012432 mmoles. Dilutions Absorbance Amount of Protein (μg/mL) 1:25 1.539 152. 1:50 1.068 101. 1:100 0.618 52. 1:200 0.371 25. 1:500 0.150 1.

Taking the concentration and dividing by the time of the reaction gives: 0.012432 mmole / 3.2 mins = 0.003885 mmoles/min. Dividing this value by the amount of enzyme used (0.02 mL) gives: 0.003885 mmoles/min / 0.02 mL enzyme = 0.19425 mmol/min/mL. Multiplying this value by the dilution factor used for the spectrophotometer gives: 0.19425 mmol/min/mL x 5 = 0.97125 mmol/min/mL. Finally, dividing by the average volume of protein (mg/mL) gives: 0.97125 mmol/min/mL / 3.29619 mg/mL = 0.294658 mmol/min/mg. Converting mmoles to μmoles, we get: 0.294658 mmol/min/mg x 1000 μmoles/mmole = 294.658 μmol/min/mg. B. Treated Rat By following the same steps lined out in the previous section, we can calculate the amount of μmoles of cytochrome c being produced per minute per mg protein. Firstly, we need to convert the absorbances to μg/mL protein (using the same equation from the trendline): Dilutions Absorbance at 595 nm Amount of Protein (μg/ mL) Dilution Factor Amount of Protein (μg/ mL) Amount of Protein (mg/mL) 1:1 0.670 57.8043 1 57.8043 0. 1:5 0.408 29.3261 5 146.63 0. 1:10 0.240 11.0652 10 110.652 0. Average value  (^) 105.029 μg/ mL

mg/mL From the printout from the spectrophotometer, we can see that the: Δ absorbance = abs 2 – abs 1 = 0.455 – 0. = 0. and that the time needed to complete the reaction was 1.06 minutes. Following the calculations above: 0.335 / 18.5 = 0.018108 mmoles 0.018180 mmoles / 1.06 mins = 0.017083 mmoles/min 0.017083 mmole/min / 0.02 mL enzyme = 0.854156 mmol/min/mL