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The specific activity of an enzyme in a crude extract of E.coli is 9.5 units/mg of protein. The specific activity increased to 68 units/mg of protein upon ion-exchange chromatography. The fold purification is ___.
Correct answer is '7.20'. Can you explain this answer?
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The specific activity of an enzyme in a crude extract of E.coli is 9.5...
Explanation:
To calculate the fold purification, we need to compare the specific activity of the enzyme before and after the ion-exchange chromatography.
Step 1: Calculate the fold increase in specific activity
The fold increase in specific activity is calculated by dividing the specific activity after purification by the specific activity before purification.
Fold Increase = Specific Activity after Purification / Specific Activity before Purification
In this case, the specific activity before purification is 9.5 units/mg of protein and the specific activity after purification is 68 units/mg of protein.
Fold Increase = 68 units/mg / 9.5 units/mg = 7.16
Step 2: Calculate the fold purification
The fold purification is the ratio of the final specific activity to the initial specific activity.
Fold Purification = (Specific Activity after Purification / Specific Activity before Purification) - 1
In this case, the fold purification is:
Fold Purification = (68 units/mg / 9.5 units/mg) - 1 = 7.16 - 1 = 6.16
Therefore, the fold purification is 6.16.
Step 3: Round off the fold purification to two decimal places
The correct answer is given as 7.20, which means the fold purification should be rounded off to two decimal places.
Rounding off 6.16 to two decimal places gives 6.16. Since the next digit after the second decimal place is less than 5, we keep the second decimal place as it is.
Therefore, the fold purification is 6.16.
Conclusion:
The fold purification of the enzyme after ion-exchange chromatography is 6.16.
To calculate the fold purification, we need to compare the specific activity of the enzyme before and after the ion-exchange chromatography.
Step 1: Calculate the fold increase in specific activity
The fold increase in specific activity is calculated by dividing the specific activity after purification by the specific activity before purification.
Fold Increase = Specific Activity after Purification / Specific Activity before Purification
In this case, the specific activity before purification is 9.5 units/mg of protein and the specific activity after purification is 68 units/mg of protein.
Fold Increase = 68 units/mg / 9.5 units/mg = 7.16
Step 2: Calculate the fold purification
The fold purification is the ratio of the final specific activity to the initial specific activity.
Fold Purification = (Specific Activity after Purification / Specific Activity before Purification) - 1
In this case, the fold purification is:
Fold Purification = (68 units/mg / 9.5 units/mg) - 1 = 7.16 - 1 = 6.16
Therefore, the fold purification is 6.16.
Step 3: Round off the fold purification to two decimal places
The correct answer is given as 7.20, which means the fold purification should be rounded off to two decimal places.
Rounding off 6.16 to two decimal places gives 6.16. Since the next digit after the second decimal place is less than 5, we keep the second decimal place as it is.
Therefore, the fold purification is 6.16.
Conclusion:
The fold purification of the enzyme after ion-exchange chromatography is 6.16.
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The specific activity of an enzyme in a crude extract of E.coli is 9.5 units/mg of protein. The specific activity increased to 68 units/mg of protein upon ion-exchange chromatography. The fold purification is ___.Correct answer is '7.20'. Can you explain this answer?
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