R1. (***Starting this questions)For each assay prepare: a) a table containing the dilution fraction and the absorbance. Table 2: Lipids assay Brillant Membrane Phosphate blue R Dilution Water suspension Sample pH 11, 1.5g/L fraction (mL) 0.2M (mL) (L) (L) 1) 0 1 0.95 0 50 Control A559 0.998 2) 1/250 .004 1 0.942 8 50 0.953 3) 1/125 0.008 1 0.934 16 50 0.828 4) 1/40 .025 1 0.900 50 50 0.521 5) 1/25 .04 1 0.870 80 50 0.392 Table 3: Protein assay Phosphate Membrane Water pH 11, suspension (mL) 0.2M (mL) (mL) Sample Dilution fraction 1) Blank 0 1 1 0 0 2) 1/200 0.005 1 0.99 0.01 0.052 3) 1/50 0.02 1 0.96 0.04 0.180 4) 1/20 .05 1 0.90 0.1 0.438 A280 b) a plot of the absorbance at 559 nm (lipid assay) and at 280 nm (protein assay) against the dilution fraction, including a point for the blank (reagent alone, no sample). In the protein assay plot, your control sample should be at zero. For the lipid assay plot, the linear region of the plot is normally found within the three or four most dilute samples. These slopes represent the decrease/increase in absorbance corresponding to a dilution factor of one, that is, the undiluted sample (x=1 in the linear regression equation y = mx). So with the tables you gave you basically just make 2 graphs (1 for lipid and 1 for protein): 1.2 1 559 Absorption 0.8 diluted fraction 0.6 559 absorption 0.4 Linear (diluted fraction) y = -19.827x + 1.0119 R² = 0.9891 0.2 0 0 0.01 0.02 0.03 0.04 0.05 280 Absorption 0.5 y = 8.6897x + 0.0046 R² = 0.9996 0.45 0.4 0.35 0.3 280 Absorption 0.25 0.2 Linear (280 Absorption) 0.15 0.1 0.05 0 0 0.02 0.04 0.06 Where x is undiluted fraction and y is absorption. We constructed the linear portion of the 559 absorption using the 3 most dilute samples. R6. Assuming an absorptivity (or extinction coefficient) at 280 nm of = 2.0 Lg1cm-1 for membrane proteins, calculate the protein concentration of your membrane preparation from the increase in absorption corresponding to your undiluted sample (x=1 in the linear regression equation y = mx). Show your calculation. Here we look at the equation from the graph above and if we set x=1 we will find that the undiluted sample has an absorption of 8.6897 + 0.0046 = 8.6943. We plug this into Beer’s Law (A=Ecl) with 8.6943 = (2.0 Lg-1cm-1) (concentration) (1cm) and then we solve for concentration, which = 4.34715 mg/mL or about 4.35 mg/mL. R7. Assuming that 1.0 g/mL of membrane lipids causes a decrease in Brilliant Blue R absorption at 559 nm of 0.011, calculate the concentration of lipids in the mitochondria-enriched fraction from decrease in absorption corresponding to your undiluted sample (x=1 in the linear regression equation y = mx). Show your calculation. This one’s a bit trickier, but basically you need to solve for the decrease in absorption, and then convert that back to concentration. Looking at our 559 graph we found that if we set x = 1 then y = -19.827 + 1.0119 = -18.8151. Above it was stated that a decrease of 0.011 corresponds to 1.0 g/mL, so if we divide: 18.8151/0.011 = 1710.46364 or about 1710.5 g/mL or about 1.7 mg/mL. R8. Estimate from your results the mass ratio lipids/protein (g/g) in your membrane preparation. Now you just divide the lipid concentration by the protein concentration or vice versa and solve to get the mass ratio. In this case I divided protein by lipid and then said it’s (1.7mg/mL):(4.35mg/mL) = 1lipid:2.55protein ratio, which makes sense because the mitochondrial fraction should have a considerable amount of protein. R9. From your results, estimate the yield for your mitochondria preparation in mg of mitochondrial proteins per gram of liver tissue (mg/g).