Exam 1 Flashcards

Question

Supersaturated potassium iodide solution may be administered to a patient for therapy with which of the following radiopharmaceuticals? a. Sm153 Lexidronam b. Sr89 Chloride c. I131 Sodium Iodide d. I131 Tositumomab

Answer 1

d. To minimize thyroid uptake of unbound iodine that may be present in the radiopharmaceutical, supersaturated potassium iodide is administered to a patient receiving I131 Tositumomab for treatment of non-Hodgkin's lymphoma. The blocking agent is administered one day before and for the next 14 days during therapy.

Answer 2

c. call for help

Answer 3

a. a total parenteral nutrition mixture because tracer may form a complex with components of the total parenteral nutrition mixture, because the chemical form of the tracer is now altered, its distribution in the body will be changed

Answer 4

b. In111 Pentetate Liquid Phase : Tc99m Sulfur Colloid Solid Phase When Liquid and gastric emptying times are to be determined simultaneously, the solid phase labeled with Tc99m Sulfur Colloid, and the liquid phase is mixed with In111 Pentetate

Answer 5

d. hanging it from the imaging table so that it is lower than the level of the bladder for proper drainage, he urine collection bag must be lower than the urinary bladder. if it is lower, the urine may reflux back into the bladder, which may result in a urinary tract infection

Answer 6

d. Tc99m pentetate Radionuclide cystography is used to assess vesicoureteral reflux. This study maybe performed by the indirect method. In the indirect method, a renal agent is administered intravenously. When the tracer has cleared from the kidneys into the bladder, imaging of the ureters and kidneys is performed during and after the patient voids.

Answer 7

b. 0.159 mL convert 175lbs to kg by dividing 175 by 2.2 which equals 79.54kg figure out how much CCK is needed: 79.54kgs x 0.02ug/kg = 1.59 ug total needed for the patient figure out how many mLs are needed: 1.59ug x 1mL/10ug = 0.159mL needed for the patient

Answer 8

d. oral the term parenteral means "other than through the intestine"

Answer 9

d. hemolytic anemia the red cell survival test is used to study the lifespan of red blood cells in patients with suspected hemolytic anemia. these patients will have a cell life span shorter than normal

Answer 10

a. locating the order for the test in the patients' medical record the referring physician initiates orders for diagnostic testing and enters those orders in the patients' medical record. it is the technologists responsibility to conform the order for a nuclear medicine test for each patient

Answer 11

d. flat field a probe system equipped with a flat field collimator should be used to obtain organ counts, this will allow adequate statistics in reasonable time periods and exclude counts coming from areas outside of the region of interest

Answer 12

d. infiltration Depending on the body weight and gender, total blood volume normally ranges from about 3-6L. If a plasma volume has been determined to be 15L, there most likely some technical error that caused this result. The most plausible explanation is that the tracer was infiltrated. When this occurs, it appears as if the amount of radioactivity thought to be injected has been diluted in a large volume, but what has actually happened is that not all the tracer was added to the circulation. Then, when a plasma sample in which the tracer has been diluted in a smaller volume. Or consider the formula used to calculate a plasma volume: Plasma Volume (mL) = volume injected x cpm in mL std x dilution all divided by cpm in mL plasma The lower the counts in the plasma sample. the smaller the number in the denominator of the plasma volume formula, hence, the larger the calculated plasma volume

Answer 13

d. unit of blood released for clinical use biohazard warning labels are attached to any container of regulated materials. These include refrigerators, freezers, and anything else used to store, transfer, or ship blood or infectious materials. Labels are not required on containers of blood that have been released for clinical use like blood transfusion

Answer 14

d. corrected plasmacrit the formula for calculating the total blood volume (TBV) based on the plasma volume uses the corrected plasmacrit in the denominator. Correction is needed to account for trapped plasma and to adjust the venous hematocrit to an average whole-body hematocrit. the formula is: TBV = plasma volume / 1 - (HCT x 0.97 x 0.91)

Answer 15

c. falsely decreased When residual radioactivity is present in the circulation, it appears as if the amount of radioactivity injected has been diluted in a smaller volume than it actually has. Then when a plasma sample is counted, the counts are relatively high compared with a sample in which the tracer has been diluted in a large volume. The larger the counts in the plasma sample, the larger the number in the denominator of the plasma volume formula, hence, the smaller the calculated plasma volume.

Answer 16

c. 25-50 mg

Answer 17

d. head to upper femurs because detection of both the primary tumor and any metastases is the goal of the examination, if indicated the extremities may be included

Answer 18

a. right antecubital area you want to inject at a site distant from the breast or lymph nodes, which may be involved as well as on the contralateral side of the body

Answer 19

b. 35% % activity in ROI = counts in ROI/ (total counts in all ROIs) and the answer is multiplied by 100 so... = 175,363 counts/501,034 counts = 0.35 x 100 = 35%

Answer 20

b. 91% Elution does not remove all Tc99m present on the generator column. Elution efficiency is expressed as the percentage of the Tc99m activity on the column that is eluted: elution efficiency = Tc99m activity eluted / Tc99m activity on column and the answer is multiplied by 100 In this case: elution efficiency = 342mCi / 375mCi = 0.912 x 100 = 91.2 %

Answer 21

b. 6.76uCi ``` 1Ci = 37GBq 1mCi = 37MBq 1uCi = 37kBq ``` to convert kilobecquerels to microcuries: 250kBq x 1uCi/37kBq = 6.75uCi

Answer 22

c. 0600 Tuesday maximum Tc99m activity obtained after the last elution is about 24hrs

Answer 23

d. only after the first elution

Answer 24

c. the eluate may be administered to patients @ 0600 the Mo99 contamination in the eluate is below the maximum allowable limit of 0.15 uCi Mo99 per mCi of Tc99m: 10uCi Mo99/416mCi Tc99m = 0.024uCi Mo99/mCi Tc99m To determine the Mo99 concentration in the eluate at 1700, decay correct both Mo99 and Tc99m activities obtained @ 0600 for 11hr (0600 to 1700 = 11hr) Mo99: 10uCi x 0.891 = 8.9 uCi Tc99m: 416mCi x 0.282 = 117 mCi At 1700 then, Mo99 concentration is equal to: 8.9uCi Mo99/ 117mCi Tc99m = 0.08 uCi Mo99/mCi Tc99m

Answer 25

b. Aluminum is a chemical impurity that is measured using a colorimetric test consisting of a special indicator paper and a reference solution of known aluminum ion concentration. No color change represents an aluminum ion concentration below 10ug/mL of eluate, the USP limit. Therefore, this eluate may be administered to patients and used to prepare other Tc99m labeled radiopharmaceuticals. Tc99m does contain radionuclide impurities such as Mo99, but these are assayed in a different manner.

Answer 26

d. Meckel's diverticulum localization You use Tc99m pertechnetate to localize Meckel's diverticulum

Answer 27

c. In111 oxine and I123 sodium iodide I131 sodium iodide is the agent used for treatment of hyperthyroidism. I123 sodium iodide is the agent of choice for thyroid uptake and may be used for thyroid imaging as well. In111 pentetate is administered for cisternography.

Answer 28

b. patients will receive more MAA particles per milliliter of Tc99m MAA The volume added to an MAA kit governs the number of particles in each milliliter or radiopharmaceautical and, therefore, the number of particles a patient will receive. For example, if a vial of MAA contains 2 million particles and 2 mL Tc99m pertechnetate are added to the vial, there will be approximately 1 million particles per milliliter. If only 1 mL is added, there will be twice as many particles in each milliliter. Therefore, a patient would receive twice as many particles in a unit dose from the second preparation.`

Answer 29

d. both the in vivo method and the modified in vivo method In both the in vivo and modified in vivo techniques, reconstituted "cold" pyrophosphate, containing stannous ions that permit Tc99m pertechnetate to permeate red cell membranes, is administered intravenously to the patient. In the in vitro method, the entire labeling process takes place outside the patient. A blood sample is collected, to which stannous ions are added. After an incubation period Tc99m pertechnetate is added to the red cells after they have been isolated from plasma. After another incubation period, the red cells are washed to remove unbound tracer, reconstituted, and injected back into the patient.

Answer 30

c. mertiatide the contents in the mertiatide reaction vial are light sensitive and must be protected from light

Answer 31

a. 0 The Rf value is the distance traveled by a given radiochemical compound compared with the solvent front: Rf = distance from origin to radiochemical component / distance from origin to solvent front In this example, the radiochemical component is question is the radiochemical impurity. Because the impurity remained at the origin, the distance from the origin to the impurity is zero. Thus, the value for the radiochemical impurity is: Rf = 0 cm / 8.5 cm = 0

Answer 32

d. 1.1mL The elapsed time between calibration and unit dose administration is 4hr. From the decay factor table. the decay factor for 4hr is 0.631. The initial concentration is decay corrected to determine the concentration at 1100: C(t) = C (o) x DF = 11.8mCi/mL x 0.631 = 7.4mCi/mL @ 1100

Answer 33

c. 1.0mL The elapsed time between calibration and unit dosage administered is 46hr. From a decay factor table, the decay factor for Tl201 at 46hr is approximately 0.646. The initial concentration is decay corrected for 46hr to determine the concentration at 1000, February 16. C (t) = C (o) x DF = 55.5Mbq/mL x 0.646 = 35.8Mbq/mL at 1000 on Feb 16

Answer 34

b. 1.35mL The elapsed time between calibration and dosage calculation is 52hr. In this example, the radiopharmaceutical is being used before the calibration time. Therfore, the initial concentration on June 29 is the unkown, and the known concentration is divided by the decay factor. The concentration on July 1 is 1.8mCi/mL (10mCi/5.5mL): C (o) x 0.609 = 1.8mCi/mL => C (o) = 1.8mCi/mL / 0.609 = 2.96mCi/mL then, determine the required volume volume = activity required / C (t) = 4mCi / 2.96mCi/mL = 1.35

Answer 35

c. Tc99m pertechnetate The three methods of labeling red blood cells with Tc99m all use Tc99m pertechnetate. Tc99m pertechnetate enters the red blood cell and binds to hemoglobin. Part of the red blood cell labeling process involves administration of a reconstituted "cold" pyrophosphate kit to provide the stannous compound necessary for labeling to occur. Tc99m albumin is a blood-pool agent; that is, it remains is the circulation for a period of time after intravenous injection. It does not bind to the red blood cells. Tc99m exametazime is used to label white blood cells

Answer 36

b. Tc99m exametazime Tc99m exametazime is used to label white blood cells. Tc99m pertechnetate is used to label red blood cells. Tc99m bicisate is used to demonstrate regional brain perfusion. Tc99m sestamibi is used for myocardial imaging as well as other applications.

Answer 37

c. methylene blue stabilizer

Answer 38

d. I, II, III, or IV The NRC states that the administration dosage must fall within the prescribed dosage range or may not differ from the prescribed dosage more than 20%. If the prescribed dosage is 4mCi, the measured dosage must be between 3.2 and 4.8mCi.

Answer 39

a. patient blood, radiochromium, ascorbic acid Radiochromium should not be added to the ACD solution before the patient's blood because ACD may change the valence of the radiochromium before labeling can take place. Ascorbic acid is added at the end of the labeling procedure to stop the tagging before the blood is returned to the patient.

Answer 40

c. 650,000 The concentration at 1000 is: 148MBq/0.75mL = 197MBq/mL At 1600, after correcting for decay, the concentration is: 197MBq/mL x 0.5 = 98.5MBq/mL Therefore, the volume of unit dosage at 1600 is: 148MBq/98.5MBq/mL = 1.5mL The number of particles in 1mL remains constant over time and is: 325,000 particles/0.75mL = 433,333 particles Therefore, the 1600 dosage contains twice as many particles as the 1000 dosage, because the 1600 dosage contains twice the volume to obtain the same unit dosage of 148MBq: 433,333 particles/mL x 1.5mL = 650,000 particles

Answer 41

d. 0.27mL Determine the concentration: concentration = total activity/total volume = 4.5mCi/1.2mL = 3.75mCi/mL Determine the volume that will contain the required activity: Volume = activity required/concentration = 3.5mCi/3.75mCi/mL = 0.93mL Determine the volume that will need to be removed from the syringe: Total volume - volume required = volume removed 1.2mL - 0.93mL = 0.27mL

Answer 42

c. 4.4mL Determine the concentration: 350mCi/7mL = 50mCi/mL Determine the volume of Tc99m pertechnetate needed to obtain the required activity: 30mCi / 50mCi/mL = 0.6mL because the activity must be contained in a specific volume 5mL, sterile, preservation-free saline must be added to make up the required volume: 5mL - 0.6mL = 4.4mL

Answer 43

c. 10min In this case, the centrifuge cannot provide necessary relative centrifuge force (RCF) identified in the protocol. However, by adjusting the time, the same degree of centrifugation may be administered. The relationship is: T1G1 = T2G2 where T1 and G1 are the minimum time centrifugation and RCF, respectively, and T2 and G2 are the alternative time and RCF, respectively. Solving for the alternative time: T2 = T1G1 / G2 = (5min) (5000g) / 2500g = 10min

Answer 44

b. "Caution: Radioactive Materials" The NRC requires that restricted areas, such as imaging rooms, be posted with specific signs depending on the radiation level present. A "Caution: Radioactive Materials" sign should be posted in areas where certain quantities of radioactivity materials are used or stored. These quantities are those exceeding 10 times the quantities specified in the Appendix C to to CFR Part 20. The quantities for commonly used radiopharmaceuticals that will results in required posting include an excess of 10mCi of Tc99m, Ga67, Xe133, or Tl201; an excess of 1mCi of I123 or In111; more than 10uCi of I131.

Answer 45

a. 56mR/hr @ surface and 3.5mR/hr @ 1m The label types and exposure rate limits for packages containing radioactive materials are: ``` Cat 1 (white) less than or equal to 0.5 @ surface and no detectable radiation @ 1m Cat 2 (yellow) less than or equal to 50 @ surface and less than or equal to 1 @ 1m Cat 3 (yellow) less than or equal to 200 @ surface and less than or equal to 10 @ 1m ```

Answer 46

b. 4.2mR/hr Use the inverse square law = (I1) (D1) squared = (I2) (D2) squared For this example it is: (30mR/hr) (15cm) squared = (I2) (40cm) squared Isolate the unkown and solve: I2 = (30mR/hr) (15cm) squared / (40cm) squared = 4.2mR/hr

Answer 47

Each half-value layer (HVL) will decrease the activity to one-half of the previous activity. 0 HVL = 12mR/hr 1 HVL = 6mR/hr 2 HVL = 3mR/hr 3 HVL = 1.5mR/hr Thus 3 HVLs are required to decrease the exposure rate to less than 2mR/hr. Hence, of lead required 3 times the HVL, or 3 x 0.3cm = 0.9cm.

Answer 48

d. The situation describes a medical event requiring notification of NRC. A licensee must notify the NRC of medical events, which are defined as an administration of licensed material that results in an effective equivalent dose greater than 5rem, a dose to an organ or tissue greater than 50rem, or shallow dose exposure to the skin greater than 50rem because of the following: The total dosage delivered differs from the prescribed dosage by 20% or more The total dosage falls outside the prescribed range The wrong radioactive drug was administered by the wrong route The radioactive dosage was administered to the wrong person When a medical event (misadministration) occurs, NRC requires the licensee to notify the NRC, the referring physician, and the involved individual, unless the referring physician determines, on the basis of medical judgement, that such notification would be harmful to the patient.

Answer 49

b. 150mSv According to the NRC, the annual occupational dose limit to the eye is 15rem or 150mSv

Answer 50

b. Return the vial to storage According to NRC regulations, decay in storage of radioactive materials requires that the materials remain in storage until the radioactivity is indistinguishable from background radiation using a survey meter on the most sensitive scale and without any shielding in place.

Answer 51

a. on a reasonable basis NRC regulations require survey of areas where radiopharmaceuticals that require a written directive are prepared or administered. This would include therapeutic radiopharmaceuticals and dosages of I131 greater than 30uCi. As of October 2002, the NRC no longer requires daily surveys and weekly wipe tests. Instead, it allows facilities to establish s survey and wipe test schedule. The schedule must be reasonable, based on the need to determine radiation levels, concentrations or quantities of radioactive materials, the potential radiological hazards.

Answer 52

b. put on disposable gloves because it has not been determined that the package is free of contamination, gloves should be put on before the pakcage or its contents are handled

Answer 53

a. no more contamination can be removed from the area decontamination of a radioactive spill should continue until the residual radioactivity is below the action level set by the licensee or until no further contamination can be removed by cleaning the area. If residual radioactivity is present after thorough decontamination, the area should be covered as a reminder for personnel to limit their time in that area

Answer 54

c. likely to exceed 10% of the annual maximum allowable occupational exposure the NRC requires monitoring of occupationally exposed individuals if they are likely to exceed 10% of the annual allowable limit for external exposure or internal uptake

Answer 55

c. 0.5rem According to the NRC, an individual who has received a therapeutic dosage of radiopharmaceutical can be released if no other individual is likely to receive a total effective dose equivalent exceeding 0.5rem (5 mSv) as a result of contact with the person who received the radiopharmaceutical.

Answer 56

c. 6 half-lives ``` 0 HVLs = 3mR/hr 1 HVLs = 1.5mR/hr 2 HVLs = 0.75mR/hr 3 HVLs = 0.37mR/hr 4 HVLs = 0.19mR/hr 5 HVLs = 0.09mR/hr 6 HVLs = 0.05mR/hr ```

Answer 57

b. 3 years

Answer 58

c. tungsten Tungsten is recommended because it will absorb more radiation than the same thickness of lead

Answer 59

d. The exposure from the X-ray examination must not be included in the occupational exposure record Only radiation exposure directly related to the practice of nuclear medicine as an occupation is recorded in the individual's dosimetry record. Personal diagnostic and therapeutic exposures are not included

Answer 60

c. use a correction factor to determine true activities The NRC requires that dose calibrators be tested for proper functioning according to nationally recognized standards. The standard of practice dictates that the measured activity should be within plus or minus 10% of the expected activity when the instrument is tested for linearity. If the instrument exceeds this level of error, a correction factor should be calculated and applied to all measurements in the affected activity range.

Answer 61

b. constancy Using a long-lived source, a reading is obtained for each of the commonly used preset buttons on the dose calibrator. Although this does not provide an accurate (true) reading, it shows whether the instrument is performing with precision (reproducibility). The reading should be withing plus or minus 10% of the expected decay-corrected reading.

Answer 62

d. recalibrate the operating voltage Voltage may drift slightly from one day to the next, causing the gamma ray spectrum to shift to the right or left, out of the preset window. From the data, it appears that this is what may have happened. Before using the instrument for clinical studies or sending it for repair, the technologist should recalibrate the operating voltage by observing the counting rate at the most recent voltage setting and voltage settings just above and below that setting. If the counting rate increases when counts are obtained off the original voltage, the voltage should be adjusted until a maximum counting rate is reached. The voltage at which a maximum counting rate is obtained is the new operating voltage. If there is significant voltage fluctuation from day to day, the instrument should be serviced.

Answer 63

b. quarterly

Answer 64

b. Cs137 The reference source used for monitoring dose calibrator constancy must be long-lived to permit comparison of measured and predicted activities over an extended period. Co57 half-life = 270 days and Cs137 half-life = 30years.

Answer 65

b. the lower the cutoff frequency, the nosier the image The cutoff frequency of a filter determines which frequencies are passed unchanged and which are altered or suppressed. Lowering the cutoff frequency eliminates more of the high-frequency signal that gives the image its sharp detail. As more of the high frequencies are removed, the image becomes smoother or less detailed, eventually decreasing resolution.

Answer 66

c. framing rate Temporal resolution is the ability to demonstrate changes in radiopharmaceutical distribution over time. Therefore, this is a factor in dynamic and multiple-gated acquisitions, both of which acquire data in sequential frames. Teh framing rate is the acquisition time per frame. If the framing rate is too long, changes in radiopharmaceutical distribution can be missed. If the framing rate is too short, too few counts will be acquired in each frame, and the resolution of the images may be affected.

Answer 67

b. gated-equilibrium cardiac function study Temporal smoothing is used in dynamic studies to produce a gradual variation in counts and smoother motion from one frame to the next. Therefore, it is useful in gated studies, but it is not useful in static imaging, such as thyroid or whole-body images or nongated SPECT.

Answer 68

b. 2.73 x 2.73 mm Pixel size is determined by the matrix dimensions and the size of the field of veiw: 350/128pixels = 2.73mm Because pixels are square, the dimensions of each pixel in this 128 x 128 matrix are 2.73 x 2.73.

Answer 69

c. well counter When Gieger-muller counter readings no longer decrease as the cleanup progresses, a wipe test, which is measured in a well counter, is performed. If the wipe test shows activity does not exceed background levels, then all removable contamination has been eliminated. If nonremovable activity remains, shielding of the spill area may be necessary.

Answer 70

a. 64 x64 byte A blood flow study of the feet would be acquired at a frame rate of 2-4 sec/frame, so few counts would be obtained in each frame. Therefore, a small matrix must be used: 64 x 64. Because there will be few counts per frame, byte mode will be adequate, as it can store up to 255 counts per pixel. in word mode, about 65,000 counts can be acquired for each pixel; however, twice as much computer storage space is needed. Because this level of activity will not be present, it is wasteful - but not harmful - to use word mode.

Answer 71

b. 3-5 million For a small-field-of-view camera, 3 million counts should be acquired; for a large field of view, 5 million counts should be acquired. More counts are required as the crystal becomes larger to ensure of noise and assures an adequate count density so that uniformity can be accurately evaluated.

Answer 72

b. the gland will appear smaller

Answer 73

a. 0.84 When the geometric variation varies from the expected by more than 10%, a correction factor must be calculated and applied whenever that configuration and activity range are measured. To determine the correction factor equation is used: correction factor = expected activity/actual activity for this example: 212uCi/253uCi = 084