Bradford Assay

  • Thread starter tvtokyo
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  • #1
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Homework Statement



View attachment 76905
View attachment 76906

Homework Equations

3. The Attempt at a Solution [/B]
Hi I have a question on Braford Assay:
I am given this question:
I did the caluclation as follows:
y = 66.717x
Since given absorbance = 0.562
0.562 = 66.717x
x = 0.00842ug/ul
But the answers given is 0.168mg/ml
I do not know what I did wrongly? Can anyone please help? Thanks!
My answer and the correct answer differ by a factor of 20? Why is that so?
 
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Answers and Replies

  • #2
Borek
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My answer and the correct answer differ by a factor of 20?
0.562 is not absorbance on the original solution.
 
  • #3
epenguin
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Yes, I was trying to answer this question but when I press your links this time I am told I don't have right to access the page. But I remember when I saw it before there was something about an 0.05 ml sample - which is 1/20 of an ml. Could these occurrences of 1/20 be in any easy connected?
 
  • #4
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I still dont get it though. Sorry. I will post the image again, Sorry for the trouble pal!
Anyway, even it is a factor of 20. My answer has a unit of ug/ul and the correct answer is in mg/ml so technically it is not a factor of 20??

upload_2014-12-26_10-21-24.png

upload_2014-12-26_10-21-47.png
 
  • #5
Borek
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Please reread the last paragraph of the attached text, it clearly tells you why the difference is 20.

50+750+200=1000.
 
  • #6
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I am suppose to find the protein concentration of the liver sample what do i do with the 1000ul ?
Is this correct?
Since x = 0.00842ug/ul and total volume of my assay --> 1000 ul
Then mass = 0.00842 * 1000 = 8.42 ug
Concentration of protein = 8.42ug/50ul (of protein in liver)
= 0.1684 ug/ul = 0.1684 mg/ml
 
  • #7
Borek
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Correct.

50 μl of your sample were diluted 20 times to 1000μl.
 
  • #8
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Thank you!
 

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