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Kendall Maier
Title: Purification of Lactate Dehydrogenase (LDH) Using Affinity Chromatography
I. Objectives
1) Be able to tell how affinity chromatography is used to purify LDH and isolate a protein of
interest
2) Be able to identify what determines protein binding and elution of proteins from an affinity
column
3) Be able to use Cibacron Blue Affinity Column to isolate LDH from chicken muscle
4) Be able to measure LDH enzyme activity of the 12 fractions using the spectrophotometer
5) Be able to use enzyme activity data to see which fractions contain the protein of interest
II. Procedure
1) Placed waste container under Cibacron blue column
2) Using the P1000, pipetted 5mL or 5000uL of 0.01 M Tris HCl (pH=8.6) into the Cibacron blue
column
3) Made sure solution does not go below the gel band
*This will equilibrate the column
4) Labeled 12 microcentrifuge tubes 1-12 (these are the fractions)
5) Using 4ml of chicken muscle extract (containing LDH), pipetted 3.8ml (3800uL) into the
column and placed tube 1 underneath the column to collect 4ml
6) Placed tube 2 under the column and pipetted 1mL 0.01M Tris HCl into the column, tube 2
collected 1mL solution
7) Repeated step 5 for tubes (fractions) 3-6
8) For fraction 7, pipetted 2ml (2000uL) 0.01M Tris.HCl /0.2M NaCl into the column to wash it
-collected contents in fraction 7 by placing the tube under the column
9) For fraction 8, pipetted 2ml (2000uL) 0.01M Tris.HCl/ 0.5M NaCl into the column to wash it
-collected contents in fraction 8 by placing tube under column
10) For fraction 9, pipetted 1ml (1000uL) 0.01 M Tris.HCl/1.0 M NaCl into the column to wash it
– collected contents in fraction 9 by placing tube under column
11) Repeated step 10 for fraction 10
12) For fraction 11, pipetted 1ml (1000uL) 0.01M Tris HCl (pH=8.6) into the column to wash it
-collected contents in fraction 11by placing tube under column
13)Repeated step 12 for fraction 12
14) Prepared a 14mL stock solution in a new tube (7ml Lactate, 4.67 mL NAD, and 2.33mL
Bicarbonate)
15) Pipetted 1ml (1000uL) of the stock solution into the cuvette
16) Inserted cuvette into spectrophotometer and blanked it at 340nm
17) Pipetted 20 μl solution to be tested for LDH activity, and mixed by inverting cuvette
18) Timed the reaction for 30 seconds
-only added aliquots of fraction when ready to use spectrophotometer
19) Inserted cuvette in spectrophotometer and recorded absorbance right when reaction
starts(initial time) and after 30 seconds of the reaction (final time)
– If reading is more than 2, diluted sample 1/10 and measure activity again
20) Repeated 16-19 for every fraction (1-12) along with crude extract(USE 5uL for crude extract)
21) Labeled highest enzyme activity with initials ,date, and section to freeze, and left over crude
extract
22) Recorded absorbances in excel with start absorbance and absorbance after 30 second
reaction
23) Calculated umol NADH/min and umol NADH/min/mL
III. Results
umol NADH=absorbance *160
umol NADH/min=umol NADH*2
umol NADH/min/ml for crude= umolNADH/min*200
umolNADH/min/ml for fractions= umolNADH/min*50
IV. Discussion and Conclusion.
In this Affinity Chromatography lab, a sample of chicken muscle was used to detect LDH activity in the
sample. LDH stands for lactate dehydrogenase enzyme and it is a reversible reaction that converts
lactate to pyruvate and vice versa. The LDH in the chicken muscle sample was eluted from the affinity
column by
VI. Conclusion Questions(In discussion and conclusion section)
How is LDH eluted from the affinity column?
If you see enzyme activity in the first fraction (which flowed through the column when the crude extract
was added), how would you explain this?
Which of your fractions contain high enzyme activity level? Based on the solutions used for each wash,
why do you think the amount of LDH peaks in these fractions?
Fractions 8,9 and 11 contained high enzyme activity level with fraction 11 containing the highest.
Why do we multiply by 2 to get the μmol NADH/min value?
We multiply by 2 to get the umol NADH/min value because the reaction final absorbance was read at 30
seconds. One minute is 60 seconds so 30 seconds *2=1 minute.
What factor did you multiply by to find μmol/min/ml and why?
For the crude extract I multiplied by 200
For the fractions I multiplied by 50

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