Purification and Characterization of
Yeast Alcohol Dehydrogenase from Baker’s Yeast
By Jie Liu
Date: 2012.10.9
Propose:
To purify the Yeast Alcohol Dehydrogenase from the Baker’s Yeast.
Materials and Methods:
Part I

Take about 100 mL of yeast lysate in the beaker. Add about 50 mL of
room temperature ethanol (which is 33%), and stir at room temperature
for 15 min.

Put the solution in a ice bath until the it cools down to ~4 oC. when the
solution is at temperature, transfer it to a 250 mL centrifuge bottle,
adjust it with other guys and measure the centrifuge at 5713 rpm
(actually we run in 5700 rpm because the machine cannot set at 5713)
for 10 min.

Get the supernatant from the centrifugation and pout it in a
approximately beaker and set it in ice bath, add 70 mL of ice-cold
ethanol and stir the solution on ice for 10 min at 4 .

Transfer the solution we stirred into a 250 mL centrifuge bottle and
adjust the mass with other classmates, measure the centrifugation at
5000 g for 10 min at ~ 4 oC

Discard the supernatant as non-halogenated organic waste in the
centrifuge bottle.

Dissolve the precipitate from the centrifugation we got in 25 mL of cold
50 mM potassium phosphate buffer which PH=7. Add 7.5g ammonium
sulfate ((NH4)2SO4) and stir at room temperature for 45 min.

Transfer the solution to into a 250 mL centrifuge bottle (actually we use
the 50 mL centrifuge bottle because we don’t have such many sample)
and measure the centrifugation at 10000 g with adjusted mass for 10
min at 4 oC.

Pour the supernatant into 2 or 3 50 mL centrifuge bottle (actually need
only one of them), frozen them to solid with liquid nitrogen, then seal it
with a plastic bag and mark it then put it in the fridge.
Part II

Perform an activity assay on the sample from last week on the
GENESYS spectrophotometer. –How to do the assay: make up the
solution use 2780 μL of TRIS 20 mM pH= 8.5, 100 μL of 100% ethanol
(17M) and 80 μL of 40mM of NAD+. This is blank, make up another
identical solution and ass 40 μL of sample into the second solution and
test the absorbance of the solution run at 340 nm for 2 min.

Transfer the sample to an approximately beaker and add 2.5 g of solid
ammonium sulfate. Stir at room temperature for 30 min.

Transfer the slurry to a 50 mL centrifuge tube and centrifuge at 5000 g
for 10 min at 4 oC with adjusted mass.

Carefully pour the supernatant from the centrifugation we got into a
beaker and turn the centrifuge tube upside down on a kimwipe to let it
drop-dry for a few min. Perform a activity assay on a 40 μL aliquot of
the supernatant.

Redissolve the precipitate in a minimum volume of pH=7, 50 mM
potassium phosphate buffer. Add buffer in 100 μL aliquot until the
material is dissolved.

Then, use the graduated cylinder dilutes the sample to 10 mL with
ammonium sulfate buffer. Let it stand for 1 hour ( actually we did 45
min because time is late) at room temperature.

Transfer the slurry to a centrifuge tube and do centrifugation at 5000 g
for 10 min at 4 oC.

Discard the supernatant and dissolve the crystal in 1 mL of assay
buffer.

Filter the resulting solution through a syringe fitted with a micropore
filter.

Measure the absorbance of the protein we get at 280 nm with the blank
assay buffer and get A= 0.037.

Remove a 40 μL from the sample and perform an activity assay (we
skipped this activity assay because our adviser told us there is no
enough concentration of the protein to do the assay).
Calculation
The extinction of the alcohol dehydrogenase is 43810 M-1 cm-1 at 280 nm,
and because the protein is oligomeric state, so the extinction is four times
than the single sub unit. The alcohol dehydrogenase we used is homooligomeric because it consists of several identical subunits. According to the
𝐴
Beers law, A= abc, then, Conc = 𝑎 = 0.037/(43810*4) =0.2 μM which is barely
there, we cannot say this is concentration of the protein because this is too
small, and, the machine has its limit of detection we don’t know. The molar
weight of the protein is 149.5 kDa(refer to http://rrurl.cn/d0BHf6 ), if we don’t
consider the limit of detection of the instruction so the total weight is C*V*m =
0.2 X 10-6 X 1 X 10-3 X149.5 X 103 =3 X 10-5g, so there are 3 X 10-5g alcohol
dehydrogenase be purified from the 100 mL sample of yeast lysate.
I skipped the third activity assay because I don’t have enough protein there to
do the assay, so I cannot calculate how much ethanol per μg of YADH will
convert to acetaldehyde per min. But, if I have enough amount of protein and
already did the assay, we can calculate it.
The overall reaction is
Ethanol + NAD+  Acetaldehyde + NADH + H+
Set the wavelength at 340 nm and continue the reading 2 min, from the print
out of the machine, we can see that the absorption goes up versus time. And
we can cut 1 min length at the time-axis and read the absorbance A1 and A2
at two terminals as time goes from t1 to t2. Then use the absorption law, the
same as above, the extinction coefficient of NADH is 6220 M-1cm-1.we
calculate the concentration C1 and C2 of NADH. The changed concentration
of NADH indicates the changed concentration of acetaldehyde. Means how
many NADH produced, then same amount of ethanol convert to acetaldehyde.
N= ΔC*V , V is 1 mL. Then, divided by the mass of the alcohol
dehydrogenase, we can get the rate of the enzyme in per μg per min.
Discussion:
About the part I, we’ve answered the question in the write up.
About part II, we did two activity assays (we should do three, but we skipped
the third one because all of us don’t have enough protein), the first one just to
check whether the enzyme is live or not. The second assay we want to see it
close to zero because we want all the protein be precipitate and non of them
remain in the solution, but, in my print out, the line waves, even it’s not too
much, but it still have some protein. The third one would indicate how many
ethanol be converted to acetaldehyde by the alcohol dehydrogenase with
NAD+ at the room temperature (calculated on above).