NMTCB PET CERTIFICATION REVIEW
QUESTIONS AND ANSWERS (GRADED
A)
Rb-82 Genertaor Alert Level - ANSWER-14L of eleute volume reached, or Sr-82 level
0.002uCi/mCi of Rb-82, or Sr-85 level 0.02uCi/mCi of Rb-82
Actions for Rb82 Generator Alert Level - ANSWER-Test after every 750 mL has passed
through generator
Stop using Rb-82 generator if - ANSWER-17L of eleute volume reached, or 42 days
post calibration date, or Sr82 levenis 0.01uCi/mCi of Rb82, or Sr85 level reaches
0.1uCi/mCi of Rb82
Cyclotron produced isotopes and half-life - ANSWER-C11, 20.3 min; N13 9.97 min; O15
2.03 min; F18, 109.7 min
Advantages of negative ion cyclotrons - ANSWER-1) less nuclear activation of cyl
potion cavity and surroundings, 2) can self-focus beam of particles, 3) can irradiate 2
targets at once
Any product with a half-life greater than ____ minutes must pass quality control testing
before released for human use - ANSWER-20 minutes
According to USP, F18-FDG must have a labeling efficiency of greater than ___ -
ANSWER-90%
What quality control tests for F18-FDG cannot be performed before patient use -
ANSWER-Sterility
The initial precursor molecule for producing FDG is - ANSWER-mannose triflate
Due to high radioactivity, the production of PET Rx's must be done in lead-shield hot
cells with automated systems called - ANSWER-synthesis modules
Isotopes produces by negative ion cyclotrons - ANSWER-C, N, O = metabolites
Reaction for chemical labeling of F18-FDG - ANSWER-Nucleophilic substitution
Chemical reaction of FDG - ANSWER-Electron rich group of atoms change place on
molecule with a positively charged leaving group
,Precursor molecule for FDG - ANSWER-Mannose triflate
Chem formula of FDG - ANSWER-2-deoxy-2-[18F]fluoro-D-glucose. (Fludeoxyglucose
F-18)
Intermediate step in FDG synthesis - ANSWER-Alkaline hydrolysis of 4 acetyl groups by
replacing with OH
Size filter used after purification - ANSWER-0.22 um
FDA ref for PET pharma production - ANSWER-General chapter 823
FDG QA appearance - ANSWER-Colorless liquid
FDG QA radiochemical identity - ANSWER-Thin-layer chromatography Rf value to
reference standard
FDG QARadionuclide identity - ANSWER-Half-life 105 and 115 minutes
FDG QA bacterial endotoxins test - ANSWER-LAL (Limulus amebocyte lysate) <175
USP endotoxins units per max volume injection; endotoxins< 17.5 EU
FDG QC pH - ANSWER-Between 4.5 to 7.5
FDG QA Radiochemical purity - ANSWER-Thin-layered chromatography radioactive Rf
should be not less than 90% of total radioactivity of chromatography plate
FDG QC Radionuclideic purity - ANSWER-should be >99.5% of radionuclidie specified.
tested using multi-channel analyzer with peaks at 0.511 and 1.022 MeV
FDG QC chemical purity - ANSWER-testing depends upon synthesis method used to
create FDG
Kryptofix 222 - ANSWER-FDG QC chemical purity test - thin-layer paper
chromatography spot test. 50 ug/mL and FDG solution must demonstrate a less intense
response than standard
2-Chloro-2-deoxy-D-glucose - ANSWER-FDG QC chemical purity test used if hydrolysis
step was performed with hydrochloric acid. Use high-performance liquid
chromatography using an electrochemical detector. Mus not have more than 1 mg
CLDG: 2-deoxy-2-D-glucose
FDG QC Residual solvents - ANSWER-solvents can be in FDG from cleaning or
synthesis solutions. solutions may contain acetonitrle, ether anddehydrated alcohol. no
more 0.04% acetonitrle, and less than 0.5% of ether or ethanol
, Production of C-11 Palmitate - ANSWER-Produce 11CO2 in cyclotron. React with
pentadecylmagnesium bromide to form 11C-Palmitate and purified.
Procedure for imaging with C-11 Palmitate - ANSWER-NPO 6 hrs. Place patient on
scanner and injected and 5 min uptake then scan.
F-18 FDG method of localization - ANSWER-Glucose analog transported across cell
membrane. Cells metabolized FDG my phosphorylation by hexokinase to make FDG-6-
phosphate and stays in cells.
Normal biodistribution of FDG - ANSWER-Brain, myocardium, moderate in
nasopharyngeal region,mild in muscle, liver, spleen and testes (suv appox 2). FDG is
glomerulus filtered so in bladder, ureter and kidneys
FDG QC Sterility - ANSWER-not performed before dosing due to need of 14 days to
perform. Regulations require final produce to be placed in soy broth mixture with a
thioglycollate growth media for 30 hrs. look for observable bacteria
FDG QC Filter membrane integerity - ANSWER-final sterility filter used at end of
manufacturing process. Filter must remain intact as pressure increased to 50 psi
(bubble point test)
FDG synthesis room requirements - ANSWER-room: cGMP, ISO class 5 clean room air
hood, 21 CFR 212 for manufacturing of IV drugs
which radiouclides are possible radioactive contaminates in eluent of Rb82 generators -
ANSWER-Sr82 and Dr85
negative ion PET cyclotrons accelerate which partiles - ANSWER-negatively charged
hydrogen atoms
chemical method used to label F18 to glucose - ANSWER-nucleophilic substitution
The least amount of energy needed by bombarding particles used for nuclear reactions
occurring within a cycltron - ANSWER-8 Mev
Most common irradiation process of producing F18 with an 109.7 min half-life -
ANSWER-O15(p,n)F18
1 mCi = X MBq - ANSWER-37
Webster Ped dose - ANSWER-Dose(adult) x (Age+1)/(Age+7)
Clark Ped Dose - ANSWER-Dose(Adult) x (child weight/150 pds)
QUESTIONS AND ANSWERS (GRADED
A)
Rb-82 Genertaor Alert Level - ANSWER-14L of eleute volume reached, or Sr-82 level
0.002uCi/mCi of Rb-82, or Sr-85 level 0.02uCi/mCi of Rb-82
Actions for Rb82 Generator Alert Level - ANSWER-Test after every 750 mL has passed
through generator
Stop using Rb-82 generator if - ANSWER-17L of eleute volume reached, or 42 days
post calibration date, or Sr82 levenis 0.01uCi/mCi of Rb82, or Sr85 level reaches
0.1uCi/mCi of Rb82
Cyclotron produced isotopes and half-life - ANSWER-C11, 20.3 min; N13 9.97 min; O15
2.03 min; F18, 109.7 min
Advantages of negative ion cyclotrons - ANSWER-1) less nuclear activation of cyl
potion cavity and surroundings, 2) can self-focus beam of particles, 3) can irradiate 2
targets at once
Any product with a half-life greater than ____ minutes must pass quality control testing
before released for human use - ANSWER-20 minutes
According to USP, F18-FDG must have a labeling efficiency of greater than ___ -
ANSWER-90%
What quality control tests for F18-FDG cannot be performed before patient use -
ANSWER-Sterility
The initial precursor molecule for producing FDG is - ANSWER-mannose triflate
Due to high radioactivity, the production of PET Rx's must be done in lead-shield hot
cells with automated systems called - ANSWER-synthesis modules
Isotopes produces by negative ion cyclotrons - ANSWER-C, N, O = metabolites
Reaction for chemical labeling of F18-FDG - ANSWER-Nucleophilic substitution
Chemical reaction of FDG - ANSWER-Electron rich group of atoms change place on
molecule with a positively charged leaving group
,Precursor molecule for FDG - ANSWER-Mannose triflate
Chem formula of FDG - ANSWER-2-deoxy-2-[18F]fluoro-D-glucose. (Fludeoxyglucose
F-18)
Intermediate step in FDG synthesis - ANSWER-Alkaline hydrolysis of 4 acetyl groups by
replacing with OH
Size filter used after purification - ANSWER-0.22 um
FDA ref for PET pharma production - ANSWER-General chapter 823
FDG QA appearance - ANSWER-Colorless liquid
FDG QA radiochemical identity - ANSWER-Thin-layer chromatography Rf value to
reference standard
FDG QARadionuclide identity - ANSWER-Half-life 105 and 115 minutes
FDG QA bacterial endotoxins test - ANSWER-LAL (Limulus amebocyte lysate) <175
USP endotoxins units per max volume injection; endotoxins< 17.5 EU
FDG QC pH - ANSWER-Between 4.5 to 7.5
FDG QA Radiochemical purity - ANSWER-Thin-layered chromatography radioactive Rf
should be not less than 90% of total radioactivity of chromatography plate
FDG QC Radionuclideic purity - ANSWER-should be >99.5% of radionuclidie specified.
tested using multi-channel analyzer with peaks at 0.511 and 1.022 MeV
FDG QC chemical purity - ANSWER-testing depends upon synthesis method used to
create FDG
Kryptofix 222 - ANSWER-FDG QC chemical purity test - thin-layer paper
chromatography spot test. 50 ug/mL and FDG solution must demonstrate a less intense
response than standard
2-Chloro-2-deoxy-D-glucose - ANSWER-FDG QC chemical purity test used if hydrolysis
step was performed with hydrochloric acid. Use high-performance liquid
chromatography using an electrochemical detector. Mus not have more than 1 mg
CLDG: 2-deoxy-2-D-glucose
FDG QC Residual solvents - ANSWER-solvents can be in FDG from cleaning or
synthesis solutions. solutions may contain acetonitrle, ether anddehydrated alcohol. no
more 0.04% acetonitrle, and less than 0.5% of ether or ethanol
, Production of C-11 Palmitate - ANSWER-Produce 11CO2 in cyclotron. React with
pentadecylmagnesium bromide to form 11C-Palmitate and purified.
Procedure for imaging with C-11 Palmitate - ANSWER-NPO 6 hrs. Place patient on
scanner and injected and 5 min uptake then scan.
F-18 FDG method of localization - ANSWER-Glucose analog transported across cell
membrane. Cells metabolized FDG my phosphorylation by hexokinase to make FDG-6-
phosphate and stays in cells.
Normal biodistribution of FDG - ANSWER-Brain, myocardium, moderate in
nasopharyngeal region,mild in muscle, liver, spleen and testes (suv appox 2). FDG is
glomerulus filtered so in bladder, ureter and kidneys
FDG QC Sterility - ANSWER-not performed before dosing due to need of 14 days to
perform. Regulations require final produce to be placed in soy broth mixture with a
thioglycollate growth media for 30 hrs. look for observable bacteria
FDG QC Filter membrane integerity - ANSWER-final sterility filter used at end of
manufacturing process. Filter must remain intact as pressure increased to 50 psi
(bubble point test)
FDG synthesis room requirements - ANSWER-room: cGMP, ISO class 5 clean room air
hood, 21 CFR 212 for manufacturing of IV drugs
which radiouclides are possible radioactive contaminates in eluent of Rb82 generators -
ANSWER-Sr82 and Dr85
negative ion PET cyclotrons accelerate which partiles - ANSWER-negatively charged
hydrogen atoms
chemical method used to label F18 to glucose - ANSWER-nucleophilic substitution
The least amount of energy needed by bombarding particles used for nuclear reactions
occurring within a cycltron - ANSWER-8 Mev
Most common irradiation process of producing F18 with an 109.7 min half-life -
ANSWER-O15(p,n)F18
1 mCi = X MBq - ANSWER-37
Webster Ped dose - ANSWER-Dose(adult) x (Age+1)/(Age+7)
Clark Ped Dose - ANSWER-Dose(Adult) x (child weight/150 pds)
