5. chemist discovers and purifies a new protein, generating the data below Proce

Question

5. chemist discovers and purifies a new protein, generating the data below Procedure Total protein Activity Specific Fold increase nmol/min activity n specific (mg) 1. Crude extract20,000 6,000,000 (homogenate) 4,500 1,670,000 salt) 3,500 600 100 80 1,450,000 4. Ion-exchange chromatography 810,000 325,000 270,000 5. Affinity chromatography 6. Size exclusion chromatography a) b) c) From the information given in the table above, calculate the specific activity of the enzyme solution after each procedure. Give the units of specific activity Calculate the fold increases by dividing specific activity at each step by specific activity in the homogenate. Which of the procedures used for this enzyme is most effective (i.e. Gives the greatest fold increase from one step to the next)? d) Which of the procedures is the least effective? e) Is the enzyme pure after step 6? Explain your reasoning f How can the purity of the enzyme be assessed? g) If you repeat this experiment what step would you eliminate? Please explain your reasoning.

Explanation / Answer

a) Specific activity of an enzyme refers to enzyme activity per unit protein. In the present case, this will be expressed as nmol/ min/ mg protein or nmol/min/ mg. As the enzyme purity increases, it's specific activity will increase.

Crude extract

Enzyme activity= 6000000, protein =20000

Sp. Activity= 6000000/ 20000= 300 nmol/ min/ mg

Since this is the original specific activity, so it's fold purification will be 1.

Salt ppt., enzyme activity=1670000, protein=4500

Sp. Activity= 1670000/4500= 371.11nmol/min/mg

Ppt pH, enzyme activity= 1450000, protein =3500

Sp. Activity= 1450000/3500=414.29 nmol/ min/ mg

Ion exchange chromatography,

Enzyme activity=810,000, protein= 600

Sp. Activity=810000/600= 1350

Affinity chromatography

Enzyme activity=325,000, protein= 100

Sp. Activity= 325000/ 100= 3250 nmol/ min/ mg

Size exclusion chromatography,

Enzyme activity= 270000, protein= 80

Sp. Activity= 270000/80= 3375 nmol/ min/ mg

B) fold increase = sp. activity of a particular stage/ sp. activity of homogenate.

Homogenate ( crude extract)= 300/300= 1

Salt ppt,

Fold increase= 371.11/ 350= 1.06

pH ppt,

Fold increase= 414.29/ 300= 1.38

Ion exchange chromatography,

Fold incease= 1350/300=4.5

Affinity chromatography,

Fold increase= 3250/300= 10.83

Size exclusion chromatography

Fold increase= 3375/ 300=11.25

c) The step of affinity chromatography is the most effective because it results into the maximum increase in fold purification (10.83 from 4.5) in a single jump.

d) The least effective step is salt precipitation, because it increases the purity to a very less extent (1.06 from 1).

e) There is a high probability that the enzyme is pure. However, it will depend on the type of ligand used for the affinity chromatography. If the ligand is highly specific for the enzyme, it is possible that the enzyme is pure. But if the ligand has less specificity for the enzyme, it may bind to unwanted proteins also resulting into a mixture of proteins ( so the enzyme will not be pure). Hence, one cannot be absolutely sure that the enzyme is pure and further investigations are required to establish the enzyme purity.

f) The enzyme purity can be assessed by subjecting the preparation from the last step to Native PAGE. Native PAGE will separate the proteins on the basis of their size and charge. If the preparation contains only the enzyme (i.e. it is pure), a single band of protein will be obtained. On the other hand if the enzyme preparation is impure, it will contain more than one protein and hence multiple bands will be obtained.

g) On repeating the experiment, the step of salt precipitation may be eliminated. This is because, it shows the least enhancement of purity (fold purification of 1.06) and maximum loss of enzyme activity ( i.e. lowest yield). For a step to be effective in enzyme purification, it should result into maximum fold increase in purity with minimum loss of yield.

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