Test 1 Q41 (SDS PAGE)

[Neplina94]

If you remember this question could you please assist me. It shows two SDS page analyses from muscle extracts visualized with either 14C or 32 P decay. The question states: Addition of Laforin to the reaction mixtures used to generate the data in figure 2 will result in?

It says the right answer is: a decrease in the intensity band representing WT without glucosidase when visualized with 32 P decay.

Im generally confused. Could someone explain this experiment to me? Im assuming the bands would decrease because laforin would cleave the molecule. So then would that result in multiple bands? And also, why only for the 32P and not the C14 band?

Any help appreciated!

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[aldol16]

Somewhere in the passage it should tell you that it's a phosphatase right? Adding a phosphatase to the mixture will cleave off the radioactive phosphate label and thus make it less intense. It wouldn't do anything to the carbon label.

 

[Neplina94]

Somewhere in the passage it should tell you that it's a phosphatase right? Adding a phosphatase to the mixture will cleave off the radioactive phosphate label and thus make it less intense. It wouldn't do anything to the carbon label.

Thank you! I finally get it, also, since it was also added to the carbon slide, can I conclude that cleavage occurred too? (its just that it wouldn't be visible because we're using Phosphate visualization).

Also, general question about SDS page and autoradiography. The intensity band would be decreased, I understand this. However, wouldn't there be multiple intensity bands because you've cleaved the molecule into bits? I figured the entire molecule was labeled with 32P (or is that not how this works?)

 

[aldol16]

Thank you! I finally get it, also, since it was also added to the carbon slide, can I conclude that cleavage occurred too? (its just that it wouldn't be visible because we're using Phosphate visualization).

Look into how carbon and phosphate visualization works.

Also, general question about SDS page and autoradiography. The intensity band would be decreased, I understand this. However, wouldn't there be multiple intensity bands because you've cleaved the molecule into bits? I figured the entire molecule was labeled with 32P (or is that not how this works?)

See above sentence. Look into how it works and it should be clear. Remember a phosphatase doesn't break up a molecule. A phosphatase just removes phosphate groups from the macromolecule.