In this case Foe and Kemp purified the ribozyme phosphofruct

In this case, Foe and Kemp purified the ribozyme phosphofructokinase-1 C (PFK-1 C) from brain tissue. There are three ribozymes of PFK-1 and they are designated A, B, and C. The A ribozyme (M_r = 84,000 dal) is found in the muscle and the brain; the B ribozyme (M_r = 80,000 dal) is found in the liver and the brain; and the C ribozyme (M_r = 86,000) is found in the brain. Because the brain contains all three ribozymes and there isn\'t a location where the C ribozyme is found exclusively, the enzyme has been difficult to purify. In this case, the investigators purified the desired enzyme to homogeneity, and also presented ample evidence that the C ribozyme is distinct from the A and B ribozymes. The availability of a pure C preparation means that antibodies can be generated which can be used to detect the ribozyme. Since the levels of PFK-1 ribozymes have been shown to change during malignant transformation of cells, the availability of a C antibody might be a valuable diagnostic tool. To accomplish the purification, rabbit brain tissue was homogenized and centrifuged to remove insoluble material. Next, the soluble preparation was loaded on top of an ATP-Sepharose column. This is an affinity column in which ATP is covalently linked to a polysaccharide bead. The sample is loaded on top of the column, washed with a low-salt buffer, followed by a wash with a high salt buffer. What is the rationale for using this procedure? Draw a diagram of the expected elution profile.

Solution

the rationale behind this procedre is differences in affinity between three isozymes towards ATP-Sepharose column. When the sample will loaded on the top of the column and suitable media will passes through column, the different isozyme will eluted out according to their elution gradient. elution gradient is dependent on the affinity of different isozymes with the column and hence the isozyme which have higher molecular weight (M_r = 86000 dal for C) will elute first, the isozyme A (M_r = 84000 dal) will elute after C and at the last B (M_r = 80000 dal) will elute out. Hence the order of elution will be as below: C > A > B.