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IMPERIAL COLLEGE LONDON

MSc Examinations 2011

Biomedical Engineering

BE9-MNMED Nuclear Medicine

Wednesday, 11 May 2011 10.00 am - 12.00 noon

Students have 2 hours to complete the exam.

Answer THREE (3) Questions

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Imperial College MSc 2011

Question 1

1(a) Prior to administration of I-131 for thyrotoxicosis, what aspects ofthe patients lifestyle should be taken into consideration whendeciding upon radiation protection advice? (30 marks)1. Pregnancy/lactation/nursing status2. Possibility of interfering medications (e.g. thyroid hormone, antithyroiddrugs, iodinecontaining medications)3. Prior iodinated contrast

4. Ingestion of iodine-rich foods5. Pertinent laboratory data, including results of thyroid function tests6. Results of prior thyroid imaging tests7. Results of prior thyroid uptake8. Recently administered radionuclides

1(b) Explain the principles of SIRTEX microspheres therapy. Includein your answer details of the isotope, its relevant properties, theadministration method, dose localisation technique, and the radiationprotection issues for staff and the patient at discharge. (40 marks)Treatment of liver metastases from e.g. bowl cancerBeads coated with Y-90 injected into hepatic artery become lodged in fine

capillary network of liver mets.Patient specific dose (~2GBq) based on body surface area and liver-lungshunt scan calculation2 day procedure for recovery after angiographyPerspex shielding jig used for RP during administration

Although pure emitter, dose from Bremstrahlung radiationImaging of Bremstrahlung to confirm dose localisedSimple RP advice given at dischargeY-90 ideal therapeutic properties

Mean range tissue = 2.5mm (max 11mm).

Half life = 64 hrs (effective treatment time of 92 hrs; 94% of radiationdelivered in 11 days).

20-60 microns in diameter (10-6m)

SIR-Spheres microspheres consists of biocompatible microspheresdesigned to be between 20-60 m (microns) in diameter, containing

yttrium-90, a high-energy pure beta emitting isotope with noprimary gamma emission. The upper size limit of the microspheres allowsdelivery to the tumors via the hepatic artery. The lower size limit prevents

the microspheres passing from the arterial circulation, through the tumor

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vasculature and into the venous circulation. The microspheres remaintrapped within the vasculature of the tumors and deliver a radiation dose tothe surrounding tissue. The microspheres do not degrade and remainpermanently implanted. They are not retrievable unless the tumor is

resected at a later stage. The microspheres are biocompatible but havedemonstrated a mild dermal sensitivity in an animal model. This has notbeen demonstrated in humans. The microspheres are supplied for singlepatient use with an activity of 3GBq 10% at the calibration time and date.SIR-Spheres microspheres are suspended in pyrogen free water forinjection to a total of ~5ml per 3GBq. This allows the activity required forimplantation into individual patients to be measured as a volume. Thedevice is supplied with a decay graph to allow for estimation of theremaining activity of the product on arrival. This should be separatelyverified. The device forms a suspension of microspheres in the water forinjection. Each device is moist heat sterilized and single useonly.SIR-Spheres microspheres are 3GBq (10%) of activity as a singledose device from which the individual patient dose is calculated and drawn.The activity of the microspheres, rather than their weight or volume,determines the number of microspheres delivered to any individual patient.The total radiation required by a patient is dependent on the extent of tumortissue and is at the discretion

1(c) A patient is assessed as having the following height, weight, liver-lungshunt and percentage tumour involvement:(30 marks)

Height = 1.7mWeight = 64KgLiver-lung shunt = 6%Tumour involvement = 30%

(i) Using the equation for body surface area given below, calculate the totaldose to be delivered.

(BSA = 0.20247 x height(m)0.725x weight(kg)0.425

)

BSA=0.20247*(1.7)^0.725*(64)^0.425=1.7421

Activity to administer (A)in GBq = (BSA 0.2) + ( % tumourinvolvement/100)(1.7421-0.2)+0.06=1.6021GBq

Most patients will receive between 1.3GBq -2.5GBq

(ii) What would the dose be adjusted to if the patient had a liver-lungshunt of 13%?

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Reduce 20%

(iii) At which level of liver-lung shunt would the therapy not beadministered and why?

Because a lot of Y90 will go to other place in our body. Not only injuryour body but also can give the most dose to the tumor

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Question 2

2(a) Describe why a radionuclide generator is useful in Nuclear Medicine.(25 marks) Cheap because: The parent radionuclide is only manipulated onceper system Only transported once per system Only way of using radionuclides with anextremely short half-life e.g. Kr-81m (13 seconds) Can produce radionuclides that decay by

isomeric transition Tc-99m, Kr-81m Produces radionuclides with a high specificactivity

2(b) Describe with the aid of a schematic diagram a Tc-99m/Mo-99radionuclide generator. Discuss how each component contributes to theelution process.

(30 marks)

2(c) Outline how the eluate from a Tc-99m/Mo-99 radionuclide generator isused to prepare commonly used radiopharmaceuticals.

(20 marks)

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2(d) Outline a test that is performed to measure the amount of Mo-99 in thefinal product.(25 marks)

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Question 3

3(a) Discuss relative contrast in planar and tomographic imaging.(20 marks)

3(b)Describe how data is acquired for SPECT and list the variousacquisition parameters that have to be set. (20 marks)

Matrix size resolution Number of projections Resolution, study duration Time per projection SNR, study duration 180 or 360 orbit

SNR, contrast, study duration Elliptical or circular

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resolution Collimator Resolution, SNR, contrast

3(c) From a SPECT acquisition a series of count profiles is created.Back-projection is the simplest way to reconstruct the data. Describe thismethod (using sketches to illustrate your answers) and list its limitations.

(20 marks)

a backprojected image is very blurry. A single point in the true image isreconstructed as a circular region that decreases in intensity away from thecenter. In more formal terms, the point spread function of backprojection iscircularly symmetric, and decreases as the reciprocal of its radius

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3(d)Explain how filtering can be used to overcome these limitations.

(20 marks)

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3(e) Explain what you understand by attenuation effect and describe twomethods used to correct for attenuation effect in SPECT.

(20 marks)The consequence of attenuation is that there are less photons detectedfrom deeper structures in the body than from tissues near the body surface.

Reconstruction of a uniform source will demonstrate that counts are lowtowards the centre. For example, in reconstruction of brain scans there willbe lower counts from the deeper structures in the brain (basal ganglia). Inthe case of the brain or lower abdomen, where attenuation can beconsidered to be constant, then a relatively straightforward means ofcorrection can be applied. This method (Chang [1]) is supplied with mostcommercial systems. The method is only an approximate method, whichsimply calculates the average attenuation for photons travelling from eachpoint in the body at different angles (see Figure 3). The method involvesmultiplication by a correction factor at each point, with typically slightover-correction, even when using an effective attenuation coefficient.Operators must ensure that the correct body boundary is selected for the

part of the body being corrected, since the algorithm will assume a constantm within that boundary. For example, the required boundary is the edge ofthe head, not the edge of the brain, and this boundary is different in eachtransverse slice through the head.

It should be noted that the attenuation correction factors in SPECT are notsmall. The average attenuation factor in the head using Technetium-99m isof the order of 2.5, increasing to a factor of roughly 5 in the abdomen.

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Non-uniform attenuation

The problem with attenuation correction is that use of a constantattenuation coefficient is inappropriate if the attenuation is non-uniform inthe body. This is particularly the case for the chest where attenuation inlungs is approximately 1/3 of the attenuation in other tissues, since the lungis full of air. Figure 4 shows an area of increased activity, such as the heart,and an area of decreased attenuation, such as lung. If there is activitythroughout the body tissues then there will be increased counts detected bythe gamma camera not only due to the increased activity in the heart, butalso due to the increased transmission of photons through the area ofdecreased attenuation. This will be true of all projections. The gammacamera cannot distinguish between the possible reasons for the increasedcounts; so during reconstruction there will be increased counts backprojected resulting in the appearance of increased counts in the area of lowattenuation. This is an artefact, which the Chang algorithm cannot correct.The Chang algorithm simply multiplies each point in the image by somecorrection factor. The only way this can be corrected is to actually measurethe attenuation and then correct for the exact measured values. Thisnormally involves using iterative reconstruction [see 2]. Readers shouldrefer to the useful reviews by King et al for further detail on attenuationcorrection [3,4].

Figure 4a: The gamma camera cannot distinguish increased countsoriginating from increased activity or decreased attenuation. As a resultareas of low attenuation are reconstructed with artificially increased counts.Similarly nearby areas with high attenuation will have reduced counts.

Figure 4b: Example of whole body positron emission tomography (PET)study which clearly demonstrates artificial increased counts in lung anddecreased counts in heart when no attenuation correction is performed.This is rectified after attenuation correction.

Attenuation measurementMost manufacturers now offer, as optional equipment, a means to measureattenuation. Usually they offer a method that permits measurement ofattenuation simultaneous with the emission measurement so that thepatient is in the identical position for both measurements. To measureattenuation requires a transmission scan, essentially identical to a CTmeasurement. This involves shining gamma rays (rather than x-rays)through the body and measuring the fraction of the gammas stopped by thebody. Referring to Figure 5, if a source of radiation is placed opposite thedetector with no intervening tissue, a certain countrate will be detected.

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If the radiation has to pass through the body the number of detected countswill be reduced according to the equation:

C = C0 e-(.x)

This time the countrate detected through the patient (C) is reduced due toattenuation through the complete thickness of the body, passing throughsmall sections with possibly different values. C0 is the countrate with nopatient present. .x represents the sum of the attenuation coefficienttimes distance for the different tissues; normally referred to as theattenuation path. If we take the log of both sides of the above equation andrearrange the equation then we see that this provides a similar form to theemission case except that we have a sum of unknown attenuationcoefficients instead of a sum of unknown counts:

log (C0 / C) = .x (emission projections represent just Scounts)

Reconstructing the transmission projections using the normal filtered backprojection provides a reconstructed map of attenuation coefficients(essentially the same as the Hounsfield numbers reconstructed in CT).

There are now a number of methods employed by manufacturers to directlymeasure attenuation, usually simultaneous with the emission measurement.Briefly these are:

a) Flood source or multiple line sources [5,6]These transmit photons through the patient irradiating the complete crystalat any time. A major problem with these systems is the scatter from theemission source in the transmission energy window; assuming emissionenergy is higher than the energy used for transmission. This is referred toas down-scatter.

b) Fixed line source with fanbeam collimator [7]This still irradiates the complete detector so down-scatter is still a problem.Also the geometry results in a high chance that the recorded data will betruncated; i.e. there will be missing data on the sides of the study due to the

magnification of the fanbeam.

c) Scanning line or point source [8]These are used in conjunction with a scanning electronic window, whichspatially isolates transmission counts from the emission counts. This needshigher activity than the fixed line source with fanbeam unless a scanningpoint source is used in conjunction with a half fanbeam collimator [9].

d) Combined SPECT / CT system [10]

An expensive solution, only recently made commercially available, is tohave a full x-ray CT system mounted on the SPECT gantry so that an

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actual CT scan can be performed just prior to the SPECT study.

In nuclear cardiology attenuation correction is particularly important since

attenuation can give rise to a posterior artefact which is hard to distinguishfrom an actual perfusion defect. Only recently the Society of NuclearMedicine and the American Society of Nuclear Cardiology have released aposition statement regarding the need for attenuation correction inmyocardial SPECT [11]. Based on the published literature on attenuationcorrection they conclude that attenuation correction is justified (see Figure7). However they note the need to pay particular attention to the quality ofattenuation measurement

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Question 4

4(a) Sketch a simple model of an ionisation chamber, including the mainfeatures and the effect of incident radiation. Name a use of an ionisationchamber in a nuclear medicine department.

(25 marks)

An ionization chamber is an instrument constructed to measure chargefrom the number of ions within a medium (which we will consider to begaseous, but can also be solid or liquid). It usually consists of a gas filledenclosure between two conducting electrodes (the anode and cathode).The electrodes may be in the form of parallel plates (Parallel PlateIonization Chambers: PPIC), or coaxial cylinders to form a convenientportable detector; in some cases one of the electrodes may be the wall ofthe vessel itself.When gas between the electrodes is ionized by any means, such as by

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alpha particles, beta particles, X-rays, or other radioactive emission, theions and dissociated electrons move to the electrodes of the oppositepolarity, thus creating an ionization current which may be measured by agalvanometer or electrometer. Each ion essentially deposits or removes a

small electric charge to or from an electrode, such that the accumulatedcharge is proportional to the number of like-charged ions. A voltagepotential that can have a wide range from a few volts to many kilovolts,depending on the application, can be applied between the electrodes. Theapplied voltage allows the device to work continuously by mopping upelectrons and preventing the device from becoming saturated. The currentthat originates is called a bias current, and prevents the device fromreaching a point where no more ions can be collected.

4(b) Draw a graph showing the regions of operation of a gas-filled detectorsat different voltages. Explain what happens at very low voltage.

(15 marks)

count GM counter

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4(c) Name the good and bad properties of NaI(Tl) as a scintillation materialused for nuclear medicine applications.

(15 marks)

4(d) What properties of a well counter need to be considered in radioactivesample counting? (15 marks)

4(e) Describe with the aid of a diagram what is meant by energy resolution.

(15 marks)

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4(f) A sample of Tc-99m is counted using a multi channel analyser. Thepeak is located in channel 248 with peak counts of 2408. The widths of thepeak at 30%, 50% and 80% of the maximum peak height are 42, 28 and 16channels respectively. The channel to keV calibration factor is 1.75.Determine the energy resolution for Tc-99m. (15 marks)

Peak=248*1.7550%=28*1.75

FWHM=28*1.75/(248*1.75)*100%=11.29

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Question 5

5(a) How would you determine the NEMA intrinsic flood field uniformityparameter for a gamma camera? Give another method for measuring floodfield uniformity. What are the advantages and disadvantages of eachmethod? Distinguish between central field of view and the useful field ofview. (35 marks)

central Field of View (CFOV): the area defined by scaling all lineardimensions of the Useful Field of View (UFOV) by a factor of 75%Useful Field of View (UFOV): the area of the detector that is used forimaging gamma rays and x-rays. It is defined by a dimensioned figuresupplied by the manufacturer

The intrinsic uniformity of the system shall be measured for the CFOV andUFOV. The measured values shall meet or exceed the specification.The intrinsic uniformity is the response of the system without a collimator toa uniform flux of radiation rom a point source. Two different uniformityparameters shall be determined: integral uniformity and ifferential uniformity.Integral uniformity is a measure of the maximum pixel count deviation in theCFOV r UFOV. Differential uniformity is a measure of the maximumdeviation over a limited range designed to approximate the size of aphotomultiplier tube.

2.3.1 Test Conditions

The radionuclide used to measure intrinsic uniformity is to be c-99m. Anyother radionuclide used shallbe reported separate. The count rate shall not exceed 20,000 counts per

second through a symmetric 15percent photopeak window. The status of uniformity corrections used shallbe stated with the results. If other radionuclides are tested , the energywindow settings recommended by the manufacturer shall be used.

2.3.2 Test EquipmentThe test equipment required for this measurement consists of a sourceholder, a lead mask for the detector, and a computer or multi-channel

analyzer. The source holder shall consist of a lead shield to prevent backand side scatter but be open at the front so that it does not restrict thegamma flux from the source to the detector. The source holder is shown inFigure 2-1. The lead mask for the detector is a leadaperture of at least the dimensions of UFOV.

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2.3.3 Measurement ProcedureThe detector shall be masked using a lead mask described above. Thesource in the source holder shall be placed on the central axis of thedetector. The distance from the detector to the source shall be at leastfive times the largest dimension of the UFOV. The flood field image shall bestored in a matrix size whichproduces pixel sizes with a linear dimension of 6.4 mm :!:: 30%.he pixelsshall be square.A minimum of 10,000 counts shall be collected in the center pixel oftheimage.2.3.4 Calculations and AnalysisPrior to performing the uniformity calculations, the pixels for inclusion shallbe determined as describedbelow. First, any pixels at the edge of UFOV containing less than 75% ofthe mean counts per pixel in the CFOVshall be set to zero.Second, those pixels which now have at least one of their four directlyabutted neighbors containing zerocounts, will be also set to zero. The remaining non-zero pixels are thepixels to be included in the analysis

for the UFOV. This step shall be performed only once. Any pixel that has atleast 50% of its area inside the CFOV shall be included within the CFOV

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analysis.

2.3.4.1 Data PreparationThe flood field image, after removing the edge pixe shall be smoothed

once by convolution with a 9-point filter function of the following weightings:121242121The weighting factor for a pixel outside the analyzed area in the 9-pointfilter function shall be zero. The smoothed value shall be normalized bydividing by the sum of non-zero weighting factors.

2.3.4.2 Integral UniformityFor pixels within each area (CFOV and UFOV), the maximum and theminimum values are to be found from the smoothed data. The differencebetween the maximum and the minimum is divided by the sum of these twovalues and multiplied by 100.

2.3.4.3 Differential UniformityFor pixels within each area (FOV and UFOV) the largest differencebetween any two pixels within a set of 5 contiguous pixels in a row orcolumn shall be calculated. The calculation shall be done for the X and theY directions independently and the maximum change expressed as apercentage using the following:

The filtered data are treated as a number of rows (X slices) and columns (Yslices). Each slice is

processed by starting at the beginning pixel for the respective field of view.A set of five contiguous pixelsis examined to find the maximum and minimum pixels.he differentialuniformity is calculated using thesevalues. he next set of five pixels is analyzed by stepping forward one

pixel and again determining thepercent uniformity. This is repeated until the outermost pixel is reached.The maximum differential uniformity is found in the slice. This process isthen repeated for all of the slices.

5(b) Explain the term intrinsic spatial resolution of a gamma camera.

Describe, using a diagram, how you would determine the intrinsic spatialresolution.

(20marks)intrinsic: a term used to describe performance characteristics of ascitillation camera that exclude external variables which affect thesespecifications, for example collimators or display devices.

Intrinsic spatial resolution measurements shall meet or exceed thespecification.2.1.1 Test ConditionsThe radionuclide employed for this test shall be Tc-99m.A source hold which shields the source from walls ,

ceilings and personnel without restricting the gamma

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flux from the source to the camera (as shown in Figure 2-1 , shall beemployed. One or more copper plates, as shown in Figure 2.1 , may beused to adjust the count rate. The energy window for Tc-99m shall be 15percent centered on the photopeak.he count rate shall not exceed 20,000

cps through the energy window. If other radionuclides are used, the energywindow should be set according to the manufacturer's recommendations.2.1.2 Test Equipmentthe test pattern shall consist of a lead mask in closest possible proximity tothe crystal , covering the entire UFOV with 1 millimeter wide parallel slits.Adjacent slit centers shall be 30 millimeters from each other (seeFigure 2-2 which shows the geometry for rectangular field of view). Thethickness of the mask shall be 3 millimeters for Tc-99m.2.1.3 Measurement ProcedureThe lead mask with the parallel slits shall be positioned on the cameradetector with one of the slits centered perpendicular to the axis ofmeasurement. he radionuclide shall be a point source centered at leastfive times the largest linear dimension of the UFOV above the lead maskwith the parallel slits. The digital resolution perpendicular to the slits isrecommended to be less than or equal to 0.1 FWHM. The digitalresolution parallel to the slits shall correspond to a channel width less thanor equal to 30 millimeters (i.e. a profile up to 30 mm wide is used). At least1,000 counts shall be collected in the peak channel of each linespread function measurement. Full Width at Half Maximum (FWHM) andFull Width at Tenth Maximum (FWTM) of the line-spread function shall bemeasured

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5(c) Explain the term overall (extrinsic) spatial resolution of a gammacamera. Describe, using a diagram, how you would determine the overallspatial resolution for a particular collimator at a particular distance from thecollimator face. What is the significance of making these measurements inair and scatter material? (35 marks)

2.6 SPECT RECONSTRUCTED SPATIALRESOLUTION WITHOUT SCATTERThe reconstructed spatial resolution of the system shall be measured atthree specified points in air. FWHM values of resolution in the X, Y and Zdirection for these three points shall be reported. the measured values shall

meet or exceed thespecification2.6.1 Test ConditionsThree c-99m or Co-57 point sources shall be used for thesemeasurements. The count rate shall not exceed 20,000 cps through asymmetric 15 percent energy window(s). No table shall come between thesources and the detector in any view.2.6.2 Test EquipmentThree thin-walled glass capillary tubes or equivalent shall be used to formand position the point sources as defined in Figure 2-5. These three pointsources shall be made as spherically symmetric as possible

with a maximum dimension no larger than 2 mm

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2.6.3 Measurement ProcedurePosition the plane of the three point sources parallel to the plane of thetable, with the central point source positioned on the axis of rotation andcentered in the field of view within :!:5 mm. The radius of rotation for thecircular orbit shall be 150:!: 5 mm. The data shall be collected andreconstructed in a matrix with an effective pixel size of less than or equal to2.5 mm. At least 20,000 total counts must be acquired in each of at least120 different projection angle images over 3600 utilizing step and shootmode. One 1305 mm thick transverse slice, centered on the central point

source, shall be reconstructed from raw projection data using the filtered

back projection technique with a ramp filter. in addition , otherreconstruction techniques are used, they shall be specified. This transverseslice will contain an image of the three point sources.Likewise, one 180 :!: 5 mm thick sagittal slice, centered on the central pointsource, shall be reconstructed. This slice will also contain an image of thethree point sources. Likewise, one 30 :!: 5 mm thick coronal slice shall bereconstructed , with the slice thickness centered on the plane of the threepoint sources.2.6.4.1 Analysis of Point ImagesEach of these nine point source images contained in the three slices shallbe analyzed individually within asquare region of interest (ROI) centered on the maximum count pixelassociated with this point. The

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dimension of the square ROI must be at least four times the anticipatedFWHM to be analyzed. Each point image shall be integrated in the Ydirection to determine the X direction point spread function andintegrated in the X direction to determine the Y direction point spread

function.2.6.4.2 Full Width at Half Maximum (FWHM) CalculationsThe FWHM in X and Y for each of the above nine point images shall becalculated according to themethod described in Section 2.1.4. Record these measured values inworksheet 2-1 for the central point and on worksheet 2-2 for the peripheralpoint

5(d) Using a diagram, define the Full Width Half Maximum (FWHM).(10 marks)

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full width at half maximum (FWHM): the measure of the spread of a point orline spread function measured between locations 50 percent down on eachside from the peak amplitude