2004 THEORY of PRACTICAL PAPER BCHM2072/2972 2004 THEORY of PRACTICAL PAPER MCQs 44 - 60

The following information refers to questions 44 to 59 You are investigating the regulation of lipogenesis in adipocytes. The pathway can be measured by following the incorporation of radioactivity from 14C-labelled glucose into fat. You have found this method in an old research paper. “Aliquots (either 50 μl or 100 μl) of a culture of adipocytes (100 mg cellular protein ml-1) were added to flasks containing cell culture medium (final volume = 4 ml). The culture medium contained 5 mM [U-14C] glucose (final specific activity 200 000 dpm umole-1) and/or other additions such as insulin. The flasks were incubated at 37oC in a shaking water bath. At 5 min intervals, 100 μl aliquots of the culture medium were removed and added to test-tubes containing 2 ml of chloroform/methanol (3:1 v/v ratio)*. After mixing and centrifugation, the lower chloroform layer was pipetted off and placed in a scintillation vial. The chloroform was then evaporated, leaving a residue of lipid in the vial. This fat was dissolved in 5 ml of scintillation fluid and the vials counted for 14C.” *This mixture smashes up the adipocytes. The released lipid dissolves in the chloroform whereas glucose stays in the aqueous phase. Centrifugation of the chloroform/methanol/culture medium mixture forces the aqueous and chloroform phases to separate. 1 nCi is equivalent to 2,220 dpm 1 Bq is equivalent to 1 dps Your task is to set up the experiment described above.

“Aliquots (either 50 μl or 100 μl) of a culture of adipocytes (100 mg cellular protein ml-1) were added to flasks containing cell culture medium (final volume = 4 ml). The culture medium contained 5 mM [U-14C] glucose (final specific activity 200 000 dpm umole-1) and/or other additions such as insulin. The flasks were incubated at 37oC in a shaking water bath. At 5 min intervals, 100 μl aliquots of the culture medium were removed and added to test-tubes containing 2 ml of chloroform/methanol (3:1 v/v ratio)*. After mixing and centrifugation, the lower chloroform layer was pipetted off and placed in a scintillation vial. The chloroform was then evaporated, leaving a residue of lipid in the vial. This fat was dissolved in 5 ml of scintillation fluid and the vials counted for 14C.” 100 ul or 50 ul Every 5 mins 100 ul All the fat in this is extracted and counted Cells 100 mg/ml 4 ml final volume 5 mM glucose 200 dpm/nmol

44. What is the specific activity of the 14C- glucose You have purchased a solution of 14C-glucose from a commercial supplier to set up these assays. The 14C- glucose has been supplied as a 20 mM solution. A 20 μl aliquot was counted and gave 400 000 dpm. What is the specific activity of this commercial preparation? 1 000 dpm per nmole B. 400 000 dpm per μmole C. 1 000 dpm per μmole D. 20 000 dpm per μmole E. 1 mCi per mmole 20 ul gives 400 000 dpm So 1 ul gives 20 000 dpm The [glucose] is 20 mM Which is 20 nmol per ul One ul contains 20 nmol and 20 000 dpm So sp act is 1000 dpm/nmol

45. How much 14C- glucose per flask How much commercial 14C- glucose would you need to add to each flask to set up the experiment described on the previous page? 500 ul B. 50 ul C. 100 ul D. 200 ul E. 400 ul final volume = 4 ml Culture medium contained 5 mM [U-14C] glucose Final specific activity 200 000 dpm umole-1 5 mM = 5 umol/ml So there’s 20 umol in the 4 ml culture medium Final specific activity is 200 000 dpm/umol So 20 umol needs 4 million dpm Activity of the stock glucose is 20 000 dpm per ul (from stem of last question) 4 million dpm will come from 200 ul

46. How much unlabelled glucose per flask The final specific activity of 14C-glucose in each flask is 200 000 dpm μmole-1. If the commercial preparation above is used in the assay set up, what volume of 100 mM glucose (unlabelled) would you need to add to each flask? none B. 200 μl C. 160 μl D. 50 μl E. 40 μl final volume = 4 ml Culture medium contained 5 mM [U-14C] glucose Final specific activity 200 000 dpm umole-1 5 mM = 5 umol/ml So there’s 20 umol in the 4 ml culture medium Stock glucose is 100 mM = 100 umol/ml So perhaps we need 1/5th of a ml (200 ul)… But the radioactive glucose stock contained SOME glucose…we put in 200 ul of a 20 mM solution = 4 umol So we really only need 160 ul

47. How many dpm in 10μl If a 10 μl aliquot of the final cell/culture medium mixture was counted how many dpm would it contain? 100 000 B. 1 000 C. 10 000 D. 20 000 E. 40 000 5 mM [U-14C] glucose final specific activity 200 000 dpm umole-1) 5 mM = 5 umol/ml = 5 nmol/ul So 10 ul contains 50 nmol Specific activity is 200 dpm/nmol So 50 nmol will give 10,000 dpm

Below are the results from the experiment. Remember: The Specific Activity of the 14C-glucose in the flasks is 200 000 dpm μmole-1. Final assay volume = 4 ml. dpm in chloroform layer from 100 ml samples of cell/glucose mix Time (min) Flask 1 50 ml cells Flask 2 100 ml cells 41 39 5 6 035 12 030 10 12 042 24 050 15 18 045 25 735 20 26 052 25 26 149 26 156

48. How much glucose was incorporated in flask 1 How many nmoles of glucose have been incorporated into fat in the 5 min sample taken from Flask 1? 0.03 B. 50 C. 30 D. 150 E. 100 Five min sample gives about 6,000 dpm This represents glucose that’s been made into lipid 200 dpm would represent 1 nmol glucose 6,000 dpm represents 30 nmol glucose

49. How much glucose was incorporated in flask 2 How many μmoles of glucose have been incorporated in 5 mins in Flask 2? Consider the total contents of the flask. 2.4 B. 60 C. 0.06 D. 600 E. 2400 Five min sample gives about 12,000 dpm Which represents lipid. These 12,000 dpm came from 100 ul of the flask’s contents Extracting the lipid from the entire flask would have given 12,000 x 4/0.1 = 480,000 dpm Specific activity of the glucose is 200,000 dpm/umol So 480,000 dpm = 2.4 umol glucose

50. What is the initial rate of lipogenesis What is the initial rate of lipogenesis (in μmol glucose per min per g cell protein) in Flask 2? (2 marks) A. 4.8 B. 12 C. 24 D. 480 E. 48 From the last question, the entire flask of 4 mls has incorporated 2.4 umol glucose into lipid in 5 mins So the rate is 2.4/5 = 0.48 umol/min Flask 2 contained 100 ul cells And the cells were 100 mg of protein/ml So Flask 2 contained 10 mg protein So 10 mg protein gives a rate of 0.48 umol/min So 1 mg protein gives 0.048 umol/min So 1 g protein gives 48 umol/min

51. How much glucose was used in 25 min How much glucose (μmoles) was used up by 25 min in both flasks? 5.2 B. 1305 C. 1.31 D. 130.5 E. 5220 At the end of the incubations, 100 ul samples from the flasks contained 26,100 dpm as lipid All 4 mls of the flasks would have contained 26,100 x 4/0.1 = 1,044,000 dpm as lipid 200,000 dpm is equivalent to 1 umol So 1,044,000 dpm is equivalent to 5.22 umol

52. What is the [glucose] after 25 min What is the concentration of glucose in mM at the end of the 25 min incubations? 20 B. 1.3 C. 3.7 D. 5.2 E. 14.8 The flasks start with 20 umol glucose 5.22 umol was used during the incubation (from previous Q) 14.78 umol left…. In 4 ml that’s 3.7 mM

53. How many dpm in flask 1 after 15 min if not extracted Consider the 15 min sample from Flask 1. There were 18 045 dpm in the chloroform extract. If this sample had not been extracted in chloroform, but had been counted without lipid/aqueous separation, how many dpm would you have counted? A. 200 000 B. 18 045 C. negligible D. 100 000 E. 1 000 This would just be like counting a 100 ul sample of the flask Remember we have 5 mM glucose at 200,000 dpm/umol 5 mM = 5 umol/ml So 100 ul contains 0.5 umol Which would be 100,000 dpm

54. 18 045dpm The 15 min sample from Flask 1 contained 18 045dpm. This could also be expressed as: A. 20 050 cpm B. 1 082 700 dps C. 300.75 Bq D. 8.12 μCi E. 3.96 mCi Dpm are > cpm. Conversion factor varies. Multiplied by 60 18 045 dpm = 18,045/60 = 300.75 dps = 300.75 Bq Remember 2,200 dpm = 1 nCi So 18,045 dpm would be 8.2 nCi

55. How to reduce specific activity by factor of 2 Which of the following changes in the assay procedure would REDUCE the specific activity of the 14C- glucose in the incubation medium by a factor of 2? The procedure in each case is otherwise unchanged from that described earlier. A. Diluting the flask contents with an equal volume of culture medium (ie final incubation volume = 8 ml) B. Adding half the volume of commercial 14C glucose to the assay (final incubation volume still adjusted to 4 ml) C. Adding half the volume of unlabelled glucose to the flask (again final incubation volume adjusted to 4 ml) D. Adding half the volume of both the commercial 14C glucose and unlabelled glucose to the flask (final incubation volume = 4 ml) E. Only taking 50 μl instead of 100 μl aliquots to the test tubes containing chloroform/methanol Looking for something that changes the ratio of labelled glucose to unlabelled glucose Ratio stays the same Ratio changes and in right direction! Ratio changes but in favour of the labelled! Ratio stays the same Doesn’t affect the ratio

Use the options below to answer questions 56 to 59. An increase is considered significant if it is >10% Consider the sample taken at 10 min in Flask 1. Which of the options (A-E) in the table above would be observed if:

56. Incubation time reduced to 5 min The incubation time was reduced from 10 to 5 mins? A lower incubation time means less glucose has gone to fat when the sample is taken But the rate at which it happened hasn’t changed

57. Aliquot of cells reduced to 25 μl If the aliquot of cells added was reduced from 50 μl to 25 μl? A smaller number of cells will make the reaction go slower So lower counts in the vial and lower rate when expressed per flask But the rate when expressed per cell (or, in this case, per mg protein) won’t change

58. [adipocytes] reduced by ½ If the adipocytes culture concentration was reduced from 100 mg/ml to 50 mg/ml? A smaller number of cells will be put in the flask This makes the reaction go slower So lower counts in the vial and lower rate when expressed per flask But the rate when expressed per cell (or, in this case, per mg protein) won’t change

59. 14C glucose has twice the specific activity If the 14C glucose solution had twice the specific activity? Every fat molecule that’s made will contain twice as many counts But the rate at which glucose goes to fat won’t change

60. Scanning check As a check on the scanning procedure enter an answer of A on the answer sheet to question 60. An anti-climax!!!