Question

You perform an assay on the activity of the JAK tyrosine kinase using a 40 microliter aliquot of a lysate prepared from cells labeled for six minutes with radioactive cysteine. the total volume of the lysate is 5 milliliters (ml). In the aliquot, you measure the amount of all proteins and determine the presence of 13 micrograms of protein in 75 microliters of lysate. which of the following is the best calculation of the specific activity of the kinase in the 5 ml lysate? Your answer has the units of micromole/min/microgram protein. a. 6.6 b. 3.4 c. 17264 d. 136 e. 68

Answer 1

Answer:

Based on the information provided:

Total volume of lysate = 5ml

Volume of aliquot of lysate (labeled) prepared from cells and used for assay = 40ul

Amount of all protein in lysate = 13 ug / 75ul lysate

Amount of total protein in 5ml lysate = (13*5000)/75 = 866.66 ug protein / 5ml lysate

Specific activity of kinase = Total activity of kinase in 5ml lysate / total protein

Specific activity of kinase = Total acitivity of kinase in 5ml lysate (micromole/min) / 866.66 (ug)

*Substitute the value of total kinase activity (missing information) in the above eqution to get the specific activity.