But I thought I didn't need math anymore...
I harvested my samples yesterday to run all sorts of tests on them, which first includes determining the amount of protein in each tube. OK, assay complete. Oh drats, I used an old standard curve instead of my new one (which I also ran in the same test). So now I have all of these raw numbers in an excel spreadsheet and need to plug them into the new curve or run the assay again. I looked up the parabolic formula up online:
y= ax^2 + bx + c
however, I haven't been able to find any websites that allow you to solve for X (unless y = 0, which it isn't in this case). Frustrated, I went to run my samples again but the amount of buffer I made was short, meaning I had to start again, again. It was late in the day, I was frustrated and started to cry in the bathroom.
So I need your help friendly blog readers (read: Dr Arthur Lo, mathlete extraordinaire). Anybody that can solve this equation or point me to a website that can wins the prize.
Here are the details (or I can send you the spreadsheet if you prefer).
The quadratic equation of y= ax^2 + bx + c
produced A=-0.0149; B=0.2768; C=-0.0085; R-square=0.9992
(Check out that beautiful correlation!)
Where y is the raw number (indicating absorbance). Turn the following raw numbers into the appropriate calculated concentrations (x).
Raw (y); Calculated Concentration (x)
0.329; ?
0.372; ?
0.398; ?
Please help. Pretty pretty please. Don't make me do that assay for a third time.
posted by Lisa @ 8:54 AM
4 Comments:
Um, Lisa, why is your standard curve a quadratic?
For Beer's Law (absorbance is proportional to concentration) to be useful and vaguely accurate for quantifying protein (or anything else), you really need to be in the linear portion of the curve.
Come to think of it, the non-linear portion of an absorbance curve is logarithmic anyway. So now I'm really confused as to why you are using a quadratic fit.
What is your R-squared if you just do a linear regression of your standards?
BTW...if there is a good reason for you to be using a quadratic fit, then any of the old graphing calculators (TI-85, e.g.) have a functon where you can input coefficients and it will spit out the answers. I bet someone in your lab or near it has one lying around.
Our beloved Arthur... here's his response in an email to me:
-------------------------
"Not too clear on what you are trying to do. But,
if you wanted to solve for "x" and the "y" is not
zero, why don't you subtract the value of "y" from
both sides of the equation?
So:
0.329 = -0.0149 x^2 + 0.2768 x - 0.0085
becomes
0.329 - 0.329 = -0.0149 x^2 + 0.2768 x - 0.0085 - 0.329
simplifying
0 = -0.0149 x^2 + 0.2768 x - 0.3205
and x = 1.24 , 17.34
Is that the gist of it?"
--------------------
Hard equation, yada yada yada, answer. Wait, you yada yada'd over the hardest part! After some phone calls to California on the boss' dime, he leads me through the problem but basically tells me the same thing Brian did:
---------------------
"If the control raw absorbance is:
0.184
0.249
0.314
0.366
0.5
0.605
0.118
0.16
and the corresponding calculated concentration is:
0.261840177
0.506793204
0.75877241
0.965842534
1.524837025
1.992696152
0.019735339
0.173058558
Then the formula you want to use to is:
calculated conc. = 4.03802 * (raw absorbance) - 0.4844
Try that out and tell me how it works.
If the data is wrong, then my fit is off
and I'll fix it for you.
Why did you try to fit the data as a parabola in
the first place? A straight line works well."
--------------------------
Sweet. So both Brian and Arthur were right meaning after many wasted hours of looking up parabolic equations, I now have my answers and don't have to do that assay again.
In the process, I learned two things:
1. If it looks like a line, smells like a line and quacks like a line, don't use a parabolic equation.
2. The past 24 hours would best be summed up as the beginning of a joke told at a cocktail party... "How many PhD's does it take to find the slope of a line".
Doh.
Many thanks Arthur and Brian. You will both receive lovely parting gifts.
"How many PhD's does it take to find the slope of a line".LOL! Too true.
Glad you've got that sorted out.
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