Question

6. The pplied to a gel filtration column and the elution volumes determined. Generate a standard curve from this dat equation. following protein standards (Table 1) were a a. Show R2 and line Table 1. Protein mixture, Elution Volume, (mL) 20uL protein injected Thyroglobulin IgA IgG Bovine Serum Albumin Horse radish peroxidase Molecular Weight (Daltons) 670000 300000 158000 68000 40000 18.26 21.25 24.00 26.30 28.37 a. Using the line equation for your standard curve, calculate at what elution volume (mL) you would expect LDH to come off.of the column. b. If the column was re-equilibrated with buffer that possessed 4 M urea - what elution volume (mL) you would expect LDH to come off of the column.
Can you please show all calculations and any other thing you used to get the answer? Step by step.

Answer 1

Steps for generating the standard curve:

1. Open an Excel sheet and fill in the data as given in the table

2. Next, plot the molecular weight vs. elution volume as a X-Y scatter graph

3. Click on the graph area and using "Trendline" option under "Chart tools" insert a trendline that fits the points well; also click the "Display equation on chart" and "Display R-value" boxes to have the equation and R² value displayed on the graph. Note that I have used a logarithmic trendline. Your graph should look something like this:

Please note that I have formatted the X and Y axes to display only the range of values of interest to us; you can do this too using Chart Tools. Additionally, you may also choose to display the axis names or to use a different marker color for each protein.

So, now we have the line equation for our standard curve.

Please note that LDH is a tetrameric protein of molecular weight ~ 140000 Da and 4 M urea denatures it to its monomeric form, each monomer weighing ~ 35000 Da.

We substitute these values of x in the line equation to get our expected elution volume (y).

a. Expected elution volume for native LDH = -3.543ln(140000) + 65.959 = 23.97 mL

​b. Expected elution volume for denatured LDH = -3.543ln(35000) + 65.959 = 28.89 mL

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