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).