Calculating Protein Concentration w/ Bradford Assay

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Discussion Overview

The discussion revolves around calculating protein concentration using the Bradford Assay, specifically focusing on a homework problem involving dilution and absorbance measurements. Participants are analyzing the calculations and potential errors in determining the concentration of protein A based on a standard curve and given absorbance values.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant presents their calculation method using the standard curve equation and absorbance value to determine the concentration of protein A, questioning the correctness of their result.
  • Another participant confirms the calculation steps appear correct but expresses skepticism about the reliability of textbook answer keys, suggesting they may contain errors.
  • A third participant notes that the original question may have omitted important information, which complicates providing accurate feedback.
  • A fourth participant suggests that gathering all relevant questions before seeking help could improve clarity in future discussions.

Areas of Agreement / Disagreement

There is no clear consensus on the correctness of the calculations, and participants express differing views on the completeness of the information provided in the original question.

Contextual Notes

Participants indicate that the problem may lack sufficient context or details, which could affect the accuracy of the responses and calculations discussed.

tvtokyo
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Homework Statement


Hi, I have one more question about the Bradford Assay. This one involves dilution.

Given 1ml of protein A, I am to determine its protein concentration (in mg/ml) using Bradford Assay (with known BSA protein standards).
Also, Under a 5 times dilution, the average absorbance is 0.497.

Homework Equations


The standard curve I obtained is:
upload_2014-12-26_10-35-0.png


The Attempt at a Solution


From the standard curve,
y = 66.717x
With absorbance = 0.497
0.497 = 66.717x
x = 0.00744 ug/ul
Since there is a dilution factor of 5
New x = 0.00744 * 5 = 0.0372 ug/ul = 0.0372 mg/ml ?? (is this correct?)

However, the correct answer is 0.372 mg/ml . What did I did wrongly??
 
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tvtokyo said:
x = 0.00744 ug/ul
Looks good.
tvtokyo said:
New x = 0.00744 * 5 = 0.0372 ug/ul = 0.0372 mg/ml ??
Looks good.

Answer keys in textbooks are "edited/proofed" by "slave labor," aka grad students and post-docs. Often, "half" changes are made after proofing --- problem statements change, and answers don't.
 
In your other question you clearly ignored part of the information given, hard to comment on this one not seeing the whole problem.
 
Is it another part of the same problem? Might be best in future to get all the questions together before asking for help.
 

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