ST:1 Flashcards

Question

Supersaturated potassium iodide solution may be administered to the patient for therapy with which of the following radiopharmaceuticals? (a)[153Sm]lexidronam (b) [89Sr]chloride (c) [131I]sodium iodide (d) [131I]tositumomab (e) [90Y]chloride

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 131I tositumomab for treatment of non-Hodgkin's lymphoma. The blocking agent is administered 1 day before and for the next 14 days during therapy.

Answer 2

(d) American Heart Association standards and guidelines indicate that once verification of cardiac arrest has occurred, the rescuer should call for help first and then initiate CPR.

Answer 3

(b) 111In-pentetreotide should not be administered through an intravenous line containing total parenteral nutrition (TPN) mixtures, because the tracer may form a complex with components of the TPN mixture. Because the chemical form of the tracer is now altered, its distribution in the body will change.

Answer 4

(a) When liquid and gastric emptying times are to be determined simultaneously, the solid phase is labeled with 99mTC-sulfur colloid and teh liquid phase is mixed with 111In-pentetate.

Answer 5

(e) For proper drainage, the urine collection bag must always be lower than the urinary bladder. If it is not lower, the urine may reflux back into the bladder, which may result in a urinary tract infection.

Answer 6

(c) Radionuclide cystography is used to assess vesicoureteral reflux. This study may be performed by the indirect of the direct 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 after the patient voids.

Answer 7

(e) There are 2.2 lbs per kg. Therefore: How many kg = 175 lbs x 1 kg/2.2 lbs = 79.54 kg How many micrograms of CCK needed = 79.54 kg x 0.02 ug/1 kg - 1.59 ug total needed for patient How many mL needed = 1.59 ug needed x 1 mL/10 ug = 0.159 mL volume needed for patient.

Answer 8

(a) The term parenteral means "other than through the intestine."

Answer 9

(a) 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 lifespan shorter than normal.

Answer 10

(c) The referring physician initiates orders for diagnostic testing and enters those orders in the patient's medical record. It is the technologist's responsibility to confirm the orders for a nuclear medicine test for each patient.

Answer 11

(d) 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 the region interest.

Answer 12

(e) Depending on body weight and gender, a patient's total blood volume normally ranges from about 3 to 6 L. If a plasma volume has been determined to be 15 L, some technical error probably 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 occurred is that not all the tracer was added to the circulation. When a plasma sample is counted, the counts are relatively low compare with a sample in which the tracer has been diluted in a smaller volume. Or, consider the formula used to calculate a plasma volume: PV (mL) = volume injected x cpm in 1 mL std x dilution / (cpm in 1 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) Biohazard warning labels are attached to any container of regulated materials. These include refrigerators, freezers, and anything 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, such as blood transfusion.

Answer 14

(d) 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

(a) 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. When a plasma sample is counted, the counts are relatively high compared with a sample in which the tracer has been diluted in large volume. Or, consider the formula used to calculate a plasma volume: PV (mL) = volume x cpm in 1 mL std x dilution / (cpm in 1 mL plasma)

Answer 16

(c) 25-50 mg tablet, usually given orally 1 hr before hypertension renal study.

Answer 17

(d) Imaging with 111In-pentetreotide routinely includes the 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 also be included

Answer 18

(c) If the suspected disease is in the left breast, it would be best to use the contralateral side of the body for injection at a site distant from the breast or lymph nodes that may also be involved.

Answer 19

(b) % activity in ROI = (counts in ROI / (total counts in all ROIs)) x 100 % activity right lung = (175,362 cts / (175,362 cts + 325,672 cts)) x 100 = (175,362 cts / 501,034 cts) x 100 = 35%

Answer 20

(b) Elution does not remove all 99mTc present on the generator column. Elution efficiency is expressed as the percentage of 99mTc activity on the column that is eluted: elution efficiency = (99mTc activity eluted / 99mTc activity on column) x 100 In this case: elution efficiency = (342 mCi / 375 mCi) x 100 = 91%

Answer 21

(c) Conversions between the CGS and SI systems can be accomplished easily if you remember the three following relationships: 1 Ci = 37 GBq 1 mCi = 37 MBq 1 uCi = 37 kBq Then to convert kilobecquerels to microcuries: 350 kBq x 1 uCi / 37 kBq = 6.75 uCi

Answer 22

(a) Maximum 99mTc activity is obtained about 24 hours after the last elution of 99mTc

Answer 23

(c) According to the NRC, 99Mo breakthrough must be measured in the first elution from a 99Mo/99mTc generator

Answer 24

(e) At 0600, the 99Mo contamination in the eluate is below the maximum allowable limit of 0.15 uCi 99Mo per mCi 99mTc: 10 uCi 99Mo / 416 mCi 99mTc = 0.024 uCi 99Mo/mCi 99mTc To determine the 99Mo concentration in the eluate at 1700, decay correct both the 99Mo and 99mTc activities obtained at 0600 for 11 hours (0600 to 1700 = 11 hr): 99Mo: 10 uCi x 0.891 = 8.9 uCi (0.891 is the 99Mo decay factor for 11 hours) 99mTc: 416 mCi x 0.282 = 117 mCi (0.282 is the 99mTc decay factor for 11 hours) At 1700, then, the 99Mo concentration is equal to: 8.9 uCi 99Mo / 117 mCi 99mTc = 0.08 uCi 99Mo/mCi 99mTc This level is below the maximum allowable limite of 99Mo contamination set by the NRC. Therefore, the eluate may be administered to patients at 1700.

Answer 25

(c) 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 10 ug/mL of eluate, the USP limit. Therefore, this eluate may be administered to patients and used to prepare other 99mTc-labeled radiopharmaceuticals. 99mTc does contain radionuclidic impurities such as 99Mo, but these impurities are assayed in a different manner.

Answer 26

(d) Gastric emptying and gastroesophageal reflux studies and gastrointestinal bleeding localization all may be performed with 99mTc-sulfur colloid. 99mTc-pertechnetate is used to localized a Meckel's diverticulum.

Answer 27

(a) 131I-sodium iodide is the agent used for treatment of hyperthyroidism. 123I-sodium iodide is the agent of choice for thyroid uptake and may be used for thyroid imaging as well. 111In-pentetate is administered for cisternography.

Answer 28

(a) The volume added to an MAA kit governs the numbers of particles in each milliliter of radiopharmaceutical and, therefore, the number of particles a patient will receive. For example, if a vial of MAA contains 2 million particles and 2 mL 99mTc-pertechnetate are added to the vial, there will be approximately 1 million MAA particles in each 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

(b) In both the in vivo and modified in vivo techniques, reconstituted "cold" pyrophosphate, containing stannous ion that permits 99mTc-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, 99mTc-pertechnetate is added to the red cells after they have been isolated from the plasma. After another incubation period, the red cells are washed to remove unbound tracer, reconstituted, and injected back into the patient.

Answer 30

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

Answer 31

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

Answer 32

(c) According to the USP, most 99mTc-labeled radiopharmaceuticals should have a radiochemical purity of at least 90%

Answer 33

(a) The elapsed time between calibration and unit dosage administration is 4 hours (0700-1100). From a decay fact-table from 99mTc, the decay factor for 4 hours is 0.631. The initial concentration is decay-corrected to determine the concentration at 1100: C(t) = C(0) x DF =11.8 mCi/mL x 0.631 = 7.4 mCi/mL at 1100 Then the volume to be administered is calculated based on the prescribed patient dosage and the concentration at the time of administration: volume = patient dosage / C(t) = 8 mCi / 7.4 mCi/mL = 1.1 mL

Answer 34

(a) The elapesed time between calibration an unit dosage administration is 46 hours (1200 on February 14 - 1000 on February 16). From a decay factor table for 201Tl, the decay factor for 46 hours is approximately 0.646. The initial concentration is decay-corrected for 46 hours to determine the concentration at 1000 on February 16: C(t) = C(0) x DF = 55.5 MBq/ml x 0.646 = 35.8 MBq/mL at 1000 on 2/16 Then the volume to be administered is calculated based on the prescribed patient dosage and the concentration at the time of administration: volume = patient dosage / C(t) = 37 MBq / 35.8 MBq/mL = 1.0mL

Answer 35

(b) The elapsed time between calibration and dosage calculation is 52 hours (0800 on June 29 - 1200 on July 1). In this example, the radiopharmaceutical is being used before the calibration time. Therefore, the initial concentration on June 29 is the unknown, and the known concentration is divided by the decay factor. The concentration on July 1 is 1.8 mCi/mL (10.0 mCi / 5.5 mL): C(0) x 0.609 = 1.8 mCi/mL C(0) = 1.8 mCi/mL / 0.609 = 2.96 mCi/mL Then, determine the required volume: volume = activity required / C(t) = 4.0 mCi / 2.96 mCi/ml = 1.35 mL

Answer 36

(b) The three methods of labeling red blood cells with 99mTc all use 99mTc-pertechnetate. 99mTc-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. 99mTc-albumin is a blood-pool agent; that is, it remains in the circulation for a period of time after intravenous injection. It does not bind to the red blood cells. 99mTc-exametazime is used to label white blood cells.

Answer 37

(e) 99mTc-exametazime is used to label white blood cells. 99mTc-pertechnetate is used to label red blood cells. 99mTc-bicisate is used to demonstrate regional brain perfusion. 99mTc sestamibi is used to myocardial imaging as well as other applications.

Answer 38

(d) When 99mTc-exametazime is used to label white blood cells, methylene blue stabilized is omitted from its preparation.

Answer 39

(e) The NRC states that the administered dosage must fall within the prescribed dosage range or may not differ from the prescribed dosage by more than 20%. If the prescribed dosage is 4 mCi, the measured dosage must be between 3.2 and 4.8 mCi (+-20% of 4 mCi).

Answer 40

(d) 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 41

(e) The concentration at 1000 is: 148 MBq / 0.75 mL = 197 MBq/mL At 1600, after correcting for decay, the concentration is: 197 MBq/mL x 0.5 = 98.5 MBq/mL Therefore, the volume of a unit dosage at 1600 is: 148 MBq / 98.5 MBq/mL = 1.5 mL The number of particles in 1 mL remains constant over time and is: 325,000 particles / 0.75 mL = 433,333 particles/mL 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 148 MBq: 433,333 particles/mL x 1.5 mL = 650,000 particles

Answer 42

(c) Determine the concentration: concentration = total activity / total volume = 4.5 mCi / 1.2 mL = 3.75 mCi/mL Determine the volume that will contain the required activity: volume = activity required / concentration = 3.5 mCi / 3.75 mCi/mL = 0.93 mL Determine the volume that will have to be removed from the syringe: Total volume - volume required = volume to be removed 1.2 mL - 0.93 mL = 0.27 mL

Answer 43

(b) Determine the concentration of the eluate: concentration = total activity / total volume = 350 mCi / 7 mL = 50 mCi/mL Determine the volume of 99mTc-pertechnetate needed to obtain the required activity: volume = activity required / concentration = 30 mCi / 50 mCi/mL = 0.6 mL Because the activity must be contained in a specific volume (5 mL), sterile, preservative-free saline must be added to make up the required volume: Reconstituting volume = 99mTc-pertechnetate volume + saline volume Saline volume = reconstituting volume - 99mTc-pertechnetate volume Saline volume = 5 mL - 0.6 mL = 4.4 mL

Answer 44

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

Answer 45

(d) 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 radioactive materials are used or stored. There quantities are those exceeding 10 times the quantities specified in Appendix C to 10 CFR Part 20. The quantities for commonly used radionuclides that will result in required posting include: as excess of 10 mCi of 99mTc, 67Ga, 133Xe, or 201 Tl; an excess of 1 mCi of 123I or 111In; more than 10 uCi of 131I.

Answer 46

(b) (AND a) The label types and exposure rate limits for packages containing radio active materials are: ``` Label category - At surface (mR/h) - At 1 m Category I (white) - <= 0.5 - No detectable radiation Category II (yellow) - <= 50 - <= 1.0 Category III (yellow) - <= 200 - <= 10 ```

Answer 47

``` (a) Use the inverse square law: (I1)(D2)^2 = (I2)(D2)^2 where I1 = intensity at original distance (D1) and I2 = intensity at new distance (D2). In this example: (30 mR/hr)(15 cm)^2 = (I2)(40 cm)^2 Isolate the unknown and solve: I2 = (30 mR/hr)(15cm)^2 / (40cm)^2 I2 = 4.2 mR/hr ```

Answer 48

(d) Each half-value layer (HVL) will decrease the activity to one-half of the previous activity. No. of HVLs - Exposure rate (mR/h) 0 - 12 1 - 6 2 - 3 3 - 1.5 Thus three HVLs are required to decrease the exposure rate to less than 2 mR/h. Hence, the thickness of lead required is three times the HVL, or 3 x 0.3 cm = 0.9 cm.

Answer 49

(e) A licensee must notify the NRC of medical events, which are defined as an administration of a licensed material that results in an effective equivalent dose (EDE) greater than 5 Rem, a dose to an organ or tissue that is greater than 50 Rem, or shallow dose exposure (SDE) to the skin greater than 50 Rem because of any 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. The radioactive drug was administered by the wrong route. The radioactive dosage was administered to the wrong person. When a medical event (misadministration) occurs, the 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 judgment, that such notification would be harmful to the patient.

Answer 50

(a) According to the NRC, the annual occupational dose limit to the eye is 15 Rem or 150 mSv.

Answer 51

(c) According to NRC regulations, decay in storage of radioactive materials requires that the materials remain in storage until the radioactivity is indistinguishable form background radiation, using a survey meter on the most-sensitive scale and without any shielding in place.

Answer 52

(a) NRC regulations required a survey of areas where radiopharmaceuticals that require a written directive are prepared or administered. This would include therapeutic radiopharmaceuticals and dosages of 131I greater than 30 uCi. As of October 2002, the NRC no longer requires daily surveys and weekly wipe tests. Instead, it allows facilities to establish a survey and wipe test schedule. The schedule must be reasonable, based on the need to determine radiation levels, concentrations or quantities of radioactive materials, and potential radiological hazards.

Answer 53

(b) The measured exposure rate is the product of the number indicated by the needle on the meter and the scale. In this example, the reading on the meter is 1.7 on a scale of 0.1. Thus, 1.7 mR/h x 0.1 = 0.17 mR/h

Answer 54

(c) Because it has not been determined that the package is free of contamination, gloves should be put on before the package or its contents are handled.

Answer 55

(e) 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 56

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

Answer 57

(a) 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 (TEDE) exceeding 0.5 Rem (5mSv) as a result of contact with the person who received the radiopharmaceutical.

Answer 58

``` (b) For each half-life that passes, the activity decreases to half of the previous activity. No. of half-lives - Exposure rate (mR/h) 0 - 3 1 - 1.5 2 - 0.75 3 - 0.37 4 - 0.19 5 - 0.09 6 - 0.05 After 6 half-lives, the original activity has decreased to background activity. ```

Answer 59

(e)According to the NRC, the results of all surveys must be retained for three years.

Answer 60

(b) Tungsten is recommended, because it will absorb more radiation than the same thickness of lead.

Answer 61

(b) 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 62

(b) The NRC requires that dose calibrators be tested for proper function according to nationally recognized standards. The standard of practice dictates that the measured activity should be within +- 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 63

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

Answer 64

(e) 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 voltages settings just above and below that setting. If the counting rate increases when counts are obtained off the original operating 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 65

(b) Dose calibrator linearity should be performed quarterly.

Answer 66

(d) The round photopenic area corresponds to failure of a single photomultiplier tube. Damage to the crystal would appear more irregular. Because this is an intrinsic flood, the collimator is not in place. However, collimator or damage would also have an irregular appearance. Failure of the x,y localization hardware or software would result in a misshapen flood image.

Answer 67

(a) 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. 57Co (half-life = 270days) and 137Cs (half-life - 30 years) are sufficiently long-lived to meet this criterion.

Answer 68

(e) 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 details. As more of the high frequencies are removed, the images becomes smoother or less detailed, eventually decreasing resolution. (a and b?)

Answer 69

(c) 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. The framing rate is the acquisition time per frame (for example, 20 sec/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 70

(a)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 71

(a) Pixel size is determined by the matrix dimensions and the size of the field of view. pixel size (mm) = diameter of field of view (mm) / number of pixels In this example: 350 mm / 128 pixels = 2.73 mm Because pixels are square, the dimensions of each pixel in this 128 x 128 matrix are 2.73 x 2.73 mm.

Answer 72

(b)When Geiger 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 73

(b) A blood-flow study of the feet would be acquired at a framing 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, since 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 74

(c) For a small-field-of-view camera, 3 million counts should be acquired; for a large field of view, 5 million counts should be obtained. More counts are required as the crystal becomes larger to ensure a minimum of 10,000 counts per pixel. This minimizes the effect of noise and assures an adequate count density so uniformity can be accurately evaluated.

Answer 75

(c) As pinhole collimator is moved form a source, the image of the source becomes smaller. As it is moved closer, the image becomes larger.

Answer 76

(d) 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, the following equation is used: correction factor = expected activity / actual activity For this example: 212 uCi / 253 uCi = 0.84

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ST:1 Flashcards

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