Need help with # 1, 3 and 4 Consider: Your Bradford Assay acquired using a 50x d
ID: 260953 • Letter: N
Question
Need help with # 1, 3 and 4
Consider: Your Bradford Assay acquired using a 50x dilution of your CFE sample gives a concentration of 0.37 mg/ml. What is the concentration of the undiluted CFE sample? 1. Refer to your lab notebook to find the volume of CFE you prepared for your purification. Based on this value, what is the total protein content of your CFE sample? 2. Consider: Your Bradford Assay acquired using a 5x dilution of your D2 elution fraction gives a concentration of 0.19 mg/ml. What is the concentration of the undiluted D2 fraction? 3. Assume this purified fraction contains only LDH, with a molecular weight of 36,500Da. What is the micro-molar (uM) concentration of LDH in this fraction? 4. 5. Which 3 amino acids possess intrinsic UV absorption properties that can be used to quantitate purified protein concentrations?Explanation / Answer
As you ask 1, 3 and 4 are the simple dilution problem and Molarity problem. You should know how to quantitatively determine the protein concentration via any method. For this you have to make a standard curve of the known protein such as BSA or correctly quantified and highly purified your protein. From that protein, you mix your protein different concentration with reagent, Bradford, BCA etc, and generate a linear reading graph and from the slope of that linear protein and readings of unknown protein you can determine the concentration of that protein.
Take the problem 1.
This problem is related with dilution. In this original solution is dilute 50 x or 50 time and that diluted protein is giving the reading that falls in the range of standard curve. From that protein amount was calculated. But as it was 50 fold diluted hence the original sample is 50 fold higher concentrated. Therefore we multiply with the concentration with 50
50 * 0.37 mg/ml= 18.5 mg/ml is the original concentration.
Problem 3,
Similarly it is 5 fold diluted then concentration is
=5 * 0.19 mg/ml=0.95 mg/ml
Problem 4 calculateMolarity
Formula is
M= (wt is gram * 1000)/ MW * volume ( in ml)
Her 0.95 mg/ml is present in 1ml
Volume is 1ml and wt is 0.95 mg
M= (0.95 * 10-3 * 1000) / 36500 * 1
M= 0.95/ 36500
M= 2.602 *10-5
M= 26.02* 10-6
M= 26.02 uM