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Radioactivity Questions
2005 Exam
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Measuring Glucose uptake using the radio-labelled 2-deoxyglucose method
You wish to measure the rate of glucose uptake into muscle cells in response to insulin. You have read about a method that uses an analog of glucose called 2‑deoxyglucose (2DG). 2DG differs from glucose in that it has an hydrogen instead of a hydoxyl at the C-2 position. Cells take up 2DG and convert it into 2‑deoxyglucose 6-phosphate (2DGP). The 2DGP cannot proceed down glycolysis and is ‘trapped’ in the cells after uptake. This is because 2DGP cannot be made into fructose 6-phosphate. In the method, cells are incubated in the presence of normal glucose containing a TRACER amount of radioactively labeled 2DG. After a particular time, the incubation is stopped so as to determine the amount of radioactivity that has become trapped in the cells.
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80. Principle of the Method
For the 2DG method to work properly, which of the following must be TRUE? It is important that hexokinase works faster on 2DG that it does on glucose The presence of trapped 2DGP should not affect the normal metabolism of glucose Muscle glucose transporters should behave differently towards 2DG than they do towards glucose 2DGP should be able to desphosphorylate back to 2DG 2DGP should be able to move freely in and out of the muscle cells For the method to work properly, yeast cells should not treat 2DG and differently from glucose (with respect to uptake and trapping). Also, the 2DG should be in tracer amounts so that it’s metabolism does not affect normal cellular metabolism. It must stay trapped inside the cells. So A and C are out because HK and GLUTs should treat 2DG the same as glucose. D and E are out because the 2DGP must not be allowed to get out of the cells once taken up.
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The Stock! You purchase 50 µCi of [U14-C] 2-deoxyglucose.
It arrives as a aqueous solution contained in a 3 mm thick glass vial. The label on the vial is as shown in the picture. 50 uCi 0.2 uCi/ul –so 1 uCi in 5 ul So 250 ul in pot. 0.2 uCi/ul – 200 nCi/ul 1 uCi is 2.2 million dpm 1 nCi is 2,200 dpm So 200 nCi is 440,000 dpm 200 Ci/mol 200 uCi/umol 50 uCi/250 nmol 250 nmol in 250 ul = 1 nmol/ul = 1 umol/ml = 1 mM
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81. Which is CORRECT? To avoid exposure to beta-particles, the glass vial should be encased in a lead container from now on Beta-particles will not penetrate the glass and gloved hands Drinking the entire contents will exceed your Annual Limit of Intake (ALI) for radiation Spilling the contents of this vial on your hands will require hospitalization You should maintain a distance of at least one metre away from this vial Because C14 beta-particles are pathetic (as you would know from reading the Radioactivity safety guide), A and E are wrong and B is right. We know the yearly limit of intake for 14C from the stimulus material (it is 40 MBq which is a touch over 1 mCi) it is easy to see that the 50 uCi in the pot would not be ‘over the limit’ (however, you would not, of course drink it!!!!). Hopefully you don’t think that D is correct.
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82-84 Simple dpm, moles & vol What is the volume of 2DG solution in the vial? How many dpm are contained in 1 µl of the 2DG solution? What is the concentration of 2DG in the vial? Vial contains 50 uCi Conc of radioactivity is 0.2 uCi/ul - so there’s 1 uCi in 5 ul So there must be 250 ul in vial. Conc of radioactivity is 0.2 uCi/ul = 200 nCi/ul 1 nCi is 2,200 dpm So 200 nCi is 440,000 dpm Specific activity = 200 Ci/mol = 200 uCi/umol So our 50 uCi must represent 250 nmol So there’s 250 nmol in 250 ul = 1 nmol/ul = 1 umol/ml = 1 mM
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85-87 Radioactive Solutions
We need to make up 1 ml of 50 mM glucose with tracer at 500 dpm/nmol. You dissolve 9 mg of glucose in 1 ml water to make a 50 mM solution. How many dpm SHOULD be taken from the 2DG vial to make this glucose solution the desired specific activity? What is the APPROXIMATE ratio of 2DG molecules to glucose molecules in the above 2DG/glucose mixture. After YOUR ASSISTANT makes up the 2DG/glucose mixture, you count 10 and 20 µl aliquots and find that these contain 25,123 and 52,715 dpm respectively. What is the ACTUAL specific activity of the mixture? 50 mM = 50 mmol/L = 50 umol/ml So we need 50 umol at a specific acitvity of 500 dpm/nmol (which is 500,000 dpm/umol) So we need 25 million dpm from the 2DG vial The Sp act of the 2DG in the vial is 200 uCi/umol = 200 nCi/nmol Which, since there are 2,200 dpm in a nmol, is equivalent to 200 x 2,200 = 440,000 dpm/nmol So the 25 million dpm that we took from the vial contains So there’s 25 million/440,000 = 57 nmol of 2DG… This is in comparison to 50,000 nmol of glu… So the ratio is about 1:1000 If 10 ul 25,000…. Then the radioactive concentration is 2,500 dpm/ul So, becaue the solution is 50 mM (ie, 50 nmol/ul) this makes the sp act = 50 dpm/nmol
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0.5 ml cells of suspension at 20,000 cells/ml
DISCARD 2DG/glucose spin incubate Final 1 ml 1 mM 2DG/glu COUNT
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DG Uptake APPROXIMATELY how much 2DG/glucose is trapped in the cells in Tube #4? What is the approximate rate of glucose uptake in Tube #4 (in nmol/min/10,000 cells)? In Tube #4, what percentage of the total dpm put into the tube has been taken up by the cells? 22000 dpm in tube – equivalent to 22,000/50 = 440 nmol This was done by 10,000 cells in 10 min – so the rate is 44 nmol per min per 10,000 cells How many dpm was put into the tube? We used 0.1 ml of the stock… which we know contained 2,500 dpm/ ul So we added 250,000 to the tube… So we only took up 7300 dpm out of 250,000 added…. – about 3%
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Tube 3 Scenarios Performing the centrifugation step when the clock showed 21 minutes Doubling the specific activity of the 2DG/glucose mixture Counting the cell pellet in the scintillation counter for twice as long Doubling the volume of all additions (ie, 2 ml incubation volume, 1 ml cells, 0.2 ml 2DG/glucose mixture and 0.8 ml buffer) Adding insulin insead of buffer X (ie, making it identical to Tube #5) dpm in cell pellet nmol 2DG/glucose in cell pellet rate of 2DG/glucose uptake into pellet (nmol/min) rate of 2DG/glucose uptake into pellet (nmol/min/10,000 cells) Tube 3 is a 10 min incubation (started at 1 and ending at 11)…. Stopping at at 21 min will make it a 20 min incubation. So there will be more uptake of 2DG into the cells and amount of 2DG in the pellet will rise… but the RATE won’t change (well, assuming it doesn’t run out… I guess you could make an argument that it will DECREASE!) Doubling the sp act simply increases the radioactivity but not the moles Counting for longer just increases the accuracy of the measurement Doubling EVERYTHING will increase the amount of 2DG in the cells and that includes the measured rate but it doesn’t increase the rate when expressed per cell. We know that insulin increases the rate… more 2DG uptake per cell… everything goes up.
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Consider Tube 3 Property A B C D E dpm in cell pellet INCREASE SAME
nmol 2DG/glucose in cell pellet rate of 2DG/glucose uptake into pellet (nmol/min) rate of 2DG/glucose uptake into pellet (nmol/min/10,000 cells)
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