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Apr 2 2013 08:29pm
You are studying a receptor tyrosine kinase and want to investigate the nature of the
autophosphorylation of this protein. To do this, you construct three forms of the receptor in an
expression vector: 1) a normal form with an active kinase domain and one tyrosine that is
phosphorylated upon activation, 2) a “kinase-dead” mutant that is larger than wildtype and carries an
inactivating mutation in the kinase domain, and 3) a truncated form that has an active kinase domain
but is missing the tyrosine residue (see Figure below). You express these forms in a cell line that
lacks expression of this receptor (that is, there is no endogenous receptor at the plasma membrane in
these cells). To test whether cis or trans phosphorylation occurs, you express the forms singly and in
pairs (see lanes on gel below). You then treat these cells with ligand and radioactive ATP (32P on the
γ-P) and then immunoprecipitate the receptor proteins from a cell extract. You analyze expression of
the receptors by immunoblotting with an antibody that recognizes all three forms of the receptor and
you detect phosphorylation by autoradiography. Your expression results are shown in the first blot
shown to the right in the diagram in the link.

I have been working on this question for a long time now and I have no idea how to figure out how you would go about figuring out what it would look like on a protein gel assay. Any help would be really appreciated. In the link, it is question 11, you just have to scroll down a bit. Many thanks to whoever looks at this.

Link (Question 11):

http://biomed.emory.edu/PROGRAM_SITES/BCDB/documents/quals/2008QE.pdf
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Apr 2 2013 08:45pm
When is it due by?
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Apr 3 2013 05:55am
First off, I'm assuming this is denaturing gel (link SDS-PAGE) and it migrates from top to bottom (loading lanes at the top). On that electrophoresis gel, you see that the largest #2 migrates the slowest by having its corresponding band higher than the others; the smallest #3 migrates fastest. Also, notice that the three forms are so different in M.W. that they can be separated on that gel.

The expression of these three receptor constructs in a cell line lacking this protein is important so you know the signal you see shouldn't be influenced by endogenous cell activities. At least that's the theory.. In other words, the results you obtain should be sufficiently specific so you don't need to worry about the host cell and its many proteins to screw the experiments.

The radioactive 32P-(gamma)-ATP (gamma means it the most terminal of the three phosphate groups that is the labeled one). Also, radioactivity is used because it does not influence the chemistry of this molecule of ATP.

The immunoprecipitation is to extract that protein from the mixture of whatever other proteins of the host cell. More important the immunoprecipitation isolates the receptor from unreacted 32P-(gamma)-ATP that stays in solution. Doing so reduces the background.

Immunoblotting has three components to it: separate the proteins on gel such as SDS-PAGE; transfer and affix the spread out proteins to a matching nitrocellulose membrane (chemically reactive paper); treat the membrane with a solution of antibodies that bind specifically to the receptor. Such antibodies are chemically modified with a labeling component such as peroxidase or whatever else.

The 32P is detected by autoradiography. The radioactive 32P naturally emits radiations that shows up on a photographic film. A band of 32P-labeled protein on a gel generates a corresponding band on the autoradiograph; the unlabeled similar band does not. Notice that only the constructs #1 and #2 have a tyrosine residue..

A kinase protein is an enzyme that chemically adds a covalently linked phosphate group to a target residue that is often part of specific residue sequence. For example, a Tyrosine kinase phosphorylates only a specific tyrosine residue and this often is done with sequence-specificity, i.e. the other tyrosines of that protein target may not be modified.


A ) Trans autophosphorylation means that a kinase enzyme molecule can only modify a different enzyme molecule, not exactly itself. In other words, if a kinase is self-inhibited by a part of its own sequence that wraps around and block itself, it won't activate itself. But another enzyme molecule that itself is active may phosphorylate the inactive one, even though they are the same in all other aspects. This is common in nature where a triggering factor is dimerization for example.

Form 1: Because the receptor activates other enzyme molecules, not itself, none gets activated. Therefore there is no phophorylation occurring.
Form 2: Dead kinase, so no phosphorylation occurring.
Form 3: Active kinase, but not tyrosine to phosphorylate. So no signal on the autoradiograph.
Forms 1+2: The form 1 is inactive, so it can't phosphorylate either 1 nor 2. Form 2 is still inactive.
Forms 1+3: The active form 3 phoysphorylates the form 1 that now show up on the autoradiograph. Form 3 does not have tyrosine to phosphorylate and thus is not detected.
Forms 2+3: The active form 3 phosphorylates the inactive form 2 that shows up on the autoradiograph. Form 3 is still a ghost not showing up because it does not have a tyrosine residue.




B ) A cis autophosphorylation means the kinase activates itself so it can be 'active' and phosphorylates whatever other proteins it can. Although it is not specified, the ligand is a triggering factor to signal to the kinase to do something such initiating its autophosphorylation in a cis mechanism.

Form 1: It self activates itself which leads to auto-phosphorylation and thus a band.
Form 2: Still inactive.. no band.
Form 3: Still no tyrosine to phosphorylate, so no band.
Forms 1+2: Form 1 auto-activates itself, thus showing a band for 1. In addition, that activated kinase now has the ability to phosphorylate other kinases, such as form 2! SO form 2 will show up as well on the autoradiograph.
Form 1+3: Only form 1 shows again as above. Form 3 still lacks a tyrosine and thus doesn't show up.
Forms 2+3: The active form 3 phosphorylates whatever it can find, such as the inactive form 2 that will show up on the autoradiograph. Form 3 is still without a tyrosine and thus will not show up.




C ) Yes the results would be different is 32P-(alpha)-ATP was used intend of the gamma variant. Three phosphates of an ATP molecule are names as alpha that is linked to the ribose-adenine; beta that is the middle one; gamma that is the third and terminal phosphate in the chain of three phosphates. Kinases enzymatically transfer that third, terminal phosphate group only to a receiving substrate like a protein tyrosine residue. Also, that gamma phosphate is the most reactive of the three phosphates of an ATP.


D ) Google that. It's easy.


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