Tutorial Answers

Question 1

Write a short note on (a) DICOM

Answer 1

(a) DICOM
• DICOM is an international standard for handling, storing, printing and transmitting
information in medical imaging.
• It includes a file format definition and a network communication protocol
• DICOM covers most image formats for all of the medicine
• A single DICOM file contains a header and image data

Question 2

Write a short note on (b) PACS

Answer 2

(b) PACS
• PACS is an integrated computer system for storage, transfer, and display of
radiological images
• PACS consists of;
o A digital archive to store medical images
o Display workstations to permit physicians to view the images
o A computer network to transfer images and related information between the
imaging devices and the archive, and between the archive and the display
workstation

Question 3

Magnification results in degradation of spatial resolution. How the magnification and spatial
resolution changes with (a) focal spot size (b) Source to Object Distance (c) Object to Image
Distance

Answer 3

(a) Smaller the focal spot, sharper the image
(b) Increases SOD decreases magnification (better spatial resolution)
(c) Smaller the OID, magnification decreases (better spatial resolution)

Question 4

Contrast resolution is affected by noises in radiography imaging. Name three types of noises in an
X-ray image?

Answer 4

• Quantum or mottle noise (fluctuation in x-ray, photons, electrons etc),
• Electronic noise (exists in all electronic circuits)
• Anatomical noise (unwanted anatomy)

Question 5

Briefly explain;

a) Fourier Transform (FT

Answer 5

• Fourier transform decomposes a function of time ( e.g. a signal or an image) into the
sum of a number of sine waves
• FT transforms the spatial domain signal to frequency domain
• FT widely used in medical imaging
• Inverse Fourier Transform, which converts frequency domain to spatial domain is
used in MRI imaging

Question 6

Briefly explain; (b) The relationship between Signal to Noise Ratio(SNR) and the number of photons (N)

Answer 6

SNR = square root of N

i.e. to double the SNR, the number of photons (hence dose) must be increased by a
factor of 4.
( SNR = Signal/Noise.
Higher SNR means, higher signal and lower noise, therefore a better image
SNR = √𝑁 where N is the number of photons.
Higher exposure → Higher N → Higher SNR →better image → but higher dose )

Question 7

Briefly explain; Detective Quantum Efficiency (DQE)

Answer 7

• The detective quantum efficiency (DQE) is a measure of the combined effects of the
signal and noise performance of an imaging system
• DQE describes how effectively an x-ray imaging system can produce an image with a
high signal-to-noise ratio (SNR)
• In radiology, DQE is a good measure of the radiation dose efficiency of a detector
(NB: DQE measures the SNR and MTF at various spatial frequencies. High DQE values indicate that less
radiation is needed to achieve a good image quality. The ideal detector would have a DQE of 1, meaning that
all the radiation energy is absorbed and converted into image information)

Question 8

Briefly explain; Bone densitometry scan (DEXA)

Answer 8

refer lecture notes

Question 9

Write a short note on Nyquist frequency?

Answer 9

• Nyquist frequency is the highest frequency that can be accurately measured on the
imaging system
• If Δis the centre to centre spacing between each detector elements, the Nyquist frequency
is given by; FN=1/ 2Δ
• If the input frequency incident on the detector is higher than FN, the true frequency will
not be recorded, aliasing effect occurs
• If the input frequency exceeds FN, the measured frequency will be lower than FN by the
same amount by which the input exceeds FN
• e.g. A certain imaging system with FN = 10 is used.
If input frequency = 8 , output frequency = 8 (input freq. less than FN)
input frequency = 9 , output frequency = 9 (input freq. less than FN)
input frequency = 11 , output frequency = 9 (input freq. is FN+1, so output will be FN -1 )
input frequency = 12 , output frequency = 8 (input freq. is FN+2, so output will be FN -2 )and so on…)

Question 10

For a digital imaging system, the distance between each detector element is 20µm. Estimate
the highest frequency (in cycles/mm), the imaging system can accurately show?

Answer 10

Nyquist frequency; FN=1/ 2Δ
Δ = 20µm = 0.020mm
FN=1/ (2*0.020) = 25 cycles/mm

Question 11

(a)Why pulse sequences are employed in MR imaging? Name three basic pulse sequences
commonly used in MR imaging?

Answer 11

MRI signal depends on T1 and T2 decay constants and proton density. These parameters are
fundamental properties of tissues. By emphasising the differences between T1 and T2 relaxation
time constants and proton density, the contrast of the MR image can be changed. This is done by
using a pulse sequence, where the nature and timing of the RF signal that generates the transverse
magnetization are changed. Pulse sequences dramatically impact the appearance of the image.
Three basic pulse sequence used in MRI:
i) Spin Echo (900
inversion pulse + 1800
refocusing pulse)
ii) Gradient Echo (generally less than 600
inversion pulse + 1800
refocusing using gradient reversal)
iii) Inversion Recovery (1800
inversion pulse + 900
inversion pulse + 1800
refocusing pulse)

Question 12

b) The following figure represents a spin echo pulse sequence. Indicate TE and TR?

Answer 12

see diagram

Question 13

Describe the Spin Echo pulse sequence? Why T2* doesn’t occur in case of SE sequence?

Answer 13

see diagram
 Spin Echo pulse sequence consists of a 900 RF pulse to excite the transverse
magnetization (Mxy) followed by a 1800 RF pulse to refocus the spins to generate an
echo signal
– 2 –
 First, a 900 RF pulse is applied
o The 900 RF pulse converts Mz into Mxy (i.e. longitudinal to transverse
magnetization)
o Soon after 900 RF pulse, the transverse magnetization decays due to loss of
coherence of protons, mainly due to T2* (extrinsic inhomogeneities)
o This generates an FID signal
 Second, a 1800 RF pulse is applied at TE/2
o The 1800 RF pulse applied at TE/2 inverts the spin system hence results in the
cancellation of the extrinsic inhomogeneities and associated dephasing effects.
i.e dephasing effects due to T2*
o As a result, recovery of transverse magnetization occurs, which generates an
echo signal at time TE, in opposite direction to Mxy
(The 1800
pulse will not refocus protons which lose coherence due to T2 relaxation. This is because T2 effect is due
to intrinsic inhomogeneity which is a random process. On the other hand T2* effect is due to external factors, which
is constant over time, will be reversed due to the introduction of 1800
pulse)

Question 14

Explain in detail T1, PD and T2 weighted image acquisition for Spin Echo pulse sequence? (The
following figure represents a typical T1 and T2 relaxation pattern for different types of tissue.
Use this graph to answer this question. Also, mention the appearance of at least Fat and CSF in
each case)

Answer 14

By changing the pulse sequence parameters TR and TE, the contrast dependence can be weighted
towards T1, proton density or T2 characteristics of the tissue

See diagram

T1 weighting
 A T1 weighted SE sequence produces contrast mainly based on the T1 characteristic of
tissue with de-emphasis of T2 and proton density contribution to the signal
 Achieved by using a short TR to maximise the difference in longitudinal magnetization
recovery and short T2 to minimise T2 decay
 T1 weighted image produces good contrast between soft tissue types (because different
tissues have different T1 values)
 Fat with short T1 has a large signal because of the greater recovery of Mz
 CSF with long T1 has a low signal
 i.e. Fat appear white and CSF appear dark
PD weighting
 Proton density contrast weighting relies mainly on differences in the number of
magnetized protons per unit volume of tissue
 Achieved by reducing the contribution of T1 recovery and T2 decay
 T1 differences are reduced by selecting a long TR
 T2 differences of the tissue are reduced by selecting a short TE
 Signal strength (contrast) depends on the number of protons
 CSF appear bright and Fat appear dark
T2 weighting
 T2 weighted image demonstrates the good contrast between normal tissue and pathology
 Reduce T1 differences in tissue with long TR
 Emphasize T2 difference with long TE
 CSF produces a maximum signal (appear white) while fat appear dark [opposite to T1
weighted image]
[NB: Some additional information
 T1 effects are connected to TR
 T2 effects are connected to TE
 Long TR minimises T1 effects, since all tissues have time to fully recover between
excitations
 Short TE minimise T2 effects, since there is little time for T2 decay differences to appear]

Question 15

Gradient Echo technique uses a low initial flip angle (<60 degrees) and a magnetic gradient to
induce an echo signal. What are the advantage and disadvantage of Gradient Echo pulse
sequence?

Answer 15

Advantages:
 GE technique has great versatility- A variety of contrasts can be produced while imaging
rapidly
 Deposition of RF energy in the patient is lower since the 1800 RF pulses are not used
(less heating of patient tissues)
 3D or volume imaging can be accomplished
Disadvantages:
 Static inhomogeneity of the magnet and inhomogeneity caused by the magnetic
susceptibility of patient tissue are not corrected by GE (echo in the same direction as
FID). i.e. T2* is not cancelled.

Question 16

Inversion Recovery pulse sequence uses 1800
-900
-1800
. What is the advantage of using Inversion
Recovery pulse sequence?

Answer 16

 Inversion recovery emphasize T1 relaxation time by using an initial 1800
excitation pulse
 IR pulse sequence creates a heavily T1 weighted image
 IR pulse sequence is useful for the suppression of selected tissues
 Disadvantages: Long scan time, also more RF energy deposition within the patient (both
due to the additional pulse)

Question 17

Explain MRI localization?

Answer 17

 Gradient magnetic fields(1-50mT/m) are used for signal localization in MRI
 This gradient fields superimposed on the main magnetic filed
 i.e. the total magnetic field at any point is the result of the main magnetic field and
gradient magnetic field
 As a result, the proton precessional frequencies vary slightly at different points
 A selective narrow band of RF pulse excites protons from a specific location where the
RF frequency matches the precessional frequency.
Three gradient coils are used
 Slice Selection Gradient (SSG) is applied along the z-axis
 Frequency Encode Gradient(FEG) is applied in the x-axis
 Phase Encode Gradient (PEG) is applied in the y-axis

Question 18

Briefly explain k-space image acquisition and image reconstruction in MRI?

Answer 18

MR data are initially stored in the “k space” matrix
 “k -space” matrix is a frequency domain repository
 The spatial frequency signals acquired during the evolution and decay of echo is stored in
“k space”
– 2 –
 Data are deposited in the k space matrix determined by FEG (x-axis gradient) and PEG
(y-axis gradient)
 A process known as “inverse two dimensional Fourier Transform” converts data into a
visible image. i.e. convert frequency domain to space domain
 The final image is scaled and adjusted to represent the proton density, T1 and T2
characteristic of the tissue using a grayscale range, where each pixel represent a voxel

Question 19

What are the major safety and biohazard concerns in MRI imaging

Answer 19

Refer lecture notes

Question 20

Explain why protons act like tiny magnets?

Answer 20

Magnetic fields are created by electric currents, ie, by moving electrically charged particles. A
proton may be thought of as a positively charged spinning sphere. The spinning motion of the
positive electric charge creates the magnetic field of the proton.

Question 21

Which of the following atoms have a net spin and why? 15P
31
, 6C
12
, 6C
13
, 8O
16 and 2He4
?

Answer 21

15P
31 and 6C
13. Only atoms with an odd number of protons or neutrons or both have net spin.

Question 22

Give two reasons, why 1H is best suited for MR imaging?

Answer 22

(a) 1H is very abundant in biological tissues. The body contains mostly fat and water, both of
which contain hydrogen
(b) 1H has a large magnetic moment and therefore provides a strong MRI signal compared to
other nuclei. This is because the nucleus of 1H contains only one proton and no neutron,
therefore there is no neutron present to cancel out (or shield) the proton spin value.

Question 23

(a) Explain the process of energy level splitting of protons in a strong magnetic field and the
formation of longitudinal magnetization (Mz)?

Answer 23

 In a strong external magnetic field (>1 Tesla), protons will either line up its magnetic
moment µ parallel or antiparallel to the external magnetic field
 The parallel state has lower energy than the antiparallel state
 This splitting in the spin energy level of a nucleus, when placed in an external B field, is
known as the Zeeman effect
 Slightly more than half of the protons will be in lower energy state (i.e. aligned with B0)
 For a 1 Tesla field, the relative excess is ~3 per million
 Hence, a net magnetisation (Mz) is generated in the direction of the external magnetic
field

Question 24

(b)Why it is better to have a higher strength magnet for MRI imaging (e.g. 3T vs 0.5T )?

Answer 24

The stronger the external magnetic field, the larger the excess nuclei align parallel to the
external magnetic field, hence stronger the net magnetization Mz. This provides a
stronger signal hence better image quality and contrast

Question 25

What is meant by the precessional motion of protons and Larmor frequency? What is the Larmor
frequency of hydrogen protons in 1 Tesla magnetic field?

Answer 25

When placed in an external magnetic field, other than spinning its own axis, protons also undergo
a precessional motion around the external magnetic field. The frequency of this precessional
motional is known as Larmour frequency.
Larmor frequency; f = Gyromagnetic ratio x External magnetic field
Gyromagnetic ratio is unique to each element
The Larmor frequency of hydrogen proton in 1Tesla field is 42.6MHz
(i.e. to flip the protons in a 1T field, we use a radio frequency pulse of frequency 42.6MHz,
in 2 T field RF pulse of frequency, 2 x 42.6 = 85.2MHz and so on)

Question 26

Explain the T2 relaxation (spin-spin)process and FID with the help of a simple diagram? What is
T2* relaxation?

Answer 26

T2 Relaxation
(for the detailed figure, refer to lecture slides)
 With the 900 RF pulse, a maximum net magnetization Mxy is generated in the transverse
direction
 The transverse magnetic field Mxy induces a signal in the receiver coil due to Faraday’s
induction
 This damped sinusoidal signal is known as Free Induction Decay (FID)
 After the RF signal applied, Mxy decays due to loss of coherence
 Loss of phase coherence is due to intrinsic micromagnetic inhomogeneities in the sample
 Due to the loss of phase coherence, the transverse magnetization Mxy decays
 T2 relaxation time is defined as the time for the transverse magnetization to reduce to
37% of its initial value

T2* relaxation:
 Extrinsic magnetic inhomogeneities (imperfect main magnetic field, the presence
of magnetic materials etc) also causes loss of phase coherence
 This process is known as T2* decay.
 T2* decay time is much shorter than T2 decay
 T2* decay doesn’t provide useful information about tissues in most cases (except
in some cases)
 Moreover, it will mask the signal due to T2 decay

refer figure

Question 27

T2 relaxation depends on molecular size and motion. Explain

Answer 27

 T2 relaxation is due to the loss of phase coherence as a result of intrinsic micro magnetic
inhomogeneities, whereby individual protons precess at different frequencies arising
from the slight changes in local magnetic field strengths
 T2 values depend on the molecular structure of the sample
 Amorphous structures (CSF, edemous tissue, water) contain mobile molecules with fast
and rapid motion. These tissues do not support intrinsic magnetic field inhomogeneities,
hence exhibit long T2
 Large macromolecules (bone) are non-moving and have a large intrinsic field, hence very
short T2
 T2 relaxation time doesn’t depend on field strength (very small dependence)

Question 28

Describe T1 relaxation (spin-lattice) with the help of a simple diagram?

Answer 28

 Longitudinal magnetization Mz begins to recover simultaneously with transverse (Mxy)
decay
 T1 relaxation is due to the energy dissipation of protons into the surrounding lattice
 T1 relaxation time is defined as the time needed for the recovery of 63% of Mz after 900
pulse
 T1 is also known as spin-lattice relaxation time
 T1 value is affected by the external magnetic field: higher B0 longer T1
refer figure

Question 29

Why T1 relaxation is different for different tissues? (use the following graph to answer this
question) refer figure

Answer 29

 T1 relaxation time depends on the rate of energy dissipation into the surrounding
molecular lattice
 Molecules are tumbling through space with certain frequencies
 Energy transfer is most efficient when these tumbling frequencies(vibrational
frequencies) are comparable to Larmour frequency
 Small molecule (water, CSF) have a broad distribution of motional frequencies with poor
matching with Larmour frequencies, hence shows long T1 values
 Medium-sized molecules (proteins, fatty tissues) have a narrow distribution of tumbling
frequencies and good matching with Larmour frequencies. Hence,short T1 values
 Large-sized molecules tumble too slowly and almost no matching with Larmour
frequency, hence very long T1

Question 30

What are the main advantages and disadvantages of Brachytherapy, compared to external
beam therapy?

Answer 30

Advantages of brachytherapy
 Improved localisation of dose
 Better sparing of overlying tissues
 Sharp dose fall-off outside the target volume
Disadvantages of brachytherapy
 Only for well localised tumours
 Only for relatively small tumours (depending)
 Very labour intensive

Question 31

Briefly explain the classification in brachytherapy based on (a) treatment duration and (b)
dose rate.

Answer 31

(a) Based on treatment duration, Brachytherapy is classified into temporary & permanent
Temporary implant
 The dose is delivered over a relatively short time compared to the half-life of the
source
 Sources are removed once the prescribed dose has been delivered
Permanent Implant
 Sources are implanted and remain there until the patient dies
 Dose is delivered over the full lifetime of the sources (i.e they are usually non
radioactive when the patient dies)
(b) Based on dose rate, Brachytherapy is classified into LDR, MDR & HDR
 Low Dose Rate, LDR - (0.4 – 2 Gy/h)
 Medium Dose Rate, MDR- (2 – 12 Gy/h)
 High Dose Rate, HDR- (>12 Gy/h) – treatment time is significantly reduced and
often minutes. Much superior in terms of patient comfort

Question 32

Briefly explain hot loading, manual after loading and remote after loading

Answer 32

Hot loading
 The applicator/catheter comes with the radioactive sources already inside
 This is then implanted into the patient at the desired site
 Common for LDR treatments of the prostate
Manual afterloading
 One or many applicator/catheter is first placed into the patient at the treatment site,
sources are loaded later

 This can be done by hand (manual afterloading) – not generally done anymore due to
radioprotection issues
Remote afterloading
 Applicators/catheter is first placed into the patient at the treatment site
 Sources are loaded later using a machine (aka remote afterloading)
 Due to safety concerns, this is the preferred option as the source can be controlled
from a shielded room away from the patient

Question 33

Write a short description of a remote afterloading HDR unit and explain how the desired dose
distribution is achieved within the patient

Answer 33

Nucletron micro Selectron-HDR Unit (one of the most common HDR unit)
 Single Ir-192 source (10Ci or 370 GBq, few mm long) contained in tungsten shielded
safe in the head of the machine.
 Source capsule is laser welded to a long drive cable, which is connected to a
computer-controlled stepper motor.
 This can position the source to sub-mm accuracy along a transfer tube and catheter
combination
 The Ir-192 source moves step by step through the catheter, controlled by the
computerised motor
 The dwell times at each location determine the dose distribution within the patient

Question 34

What decay type would you prefer for the following applications; (a) Intracavitary(b)
intravascular and why?

Answer 34

(a) The treatment volume is several cms in size, hence gamma radiation (most commonly
done using Ir-192 HDR) is preferred
(b) Need to deliver a uniform dose to the arterial wall (few mm ), hence beta source is
preferred. (Sr/Y-90, P-32 etc)
(note: For most cases in Brachytherapy, the treatment volume is several centimetres in size, hence
a gamma source is needed to penetrate the whole target. A beta source is only used for
intravascular and ophthalmic applications because the treatment volume is few mm, hence less
penetrating radiation preferred.)

Question 35

Write a short note on various types (at least five) of Brachytherapy applications? (Include
common sites and at least one isotope used in each case)

Answer 35

Intracavitary:
 Used for gynaecological cancers (most common), rectum etc.
 Use specialised applicators such as the Fletcher-Suit applicator, which consists of a
long tube (tandem) which extends up into the uterus and lateral ‘ovoids’ which allow
the dose to the cervix to be boosted while protecting the surface of the vagina from
high doses
 Generally, Ir-192 HDR technique is used
Interstitial
 Used for Prostate(most common), Breast & Head and Neck
 Sources are directly implanted into the tumour volume
 This might be via an applicator/catheter or by directly placing the source in the tissue
 For low-grade prostate I-125 permanent implant is used
 For high-grade prostate cancer, Ir-192 HDR is used
Intraluminal:
 Various lumen in the body can also be treated with brachytherapy including
Oesophagus, Bile duct, Lung
 A long catheter is inserted via the respective orifice to the treatment site
 The source is then driven to the required positions to achieve the desired dose
distribution
 Usually done in an HDR setting using Ir-192
Eye plaque:
 used to treat ocular melanomas and retinoblastomas
 Ru-106 is commonly used. Ru-106 decays via beta with rapid dose fall-off, hence
providing good tumour control and preserving a degree of visual acuity
 I-125 seeds are also used
Intravascular:
 Primarily used to combat restenosis of the coronary arteries
 A high energy beta source is preferred (or intermediate energy gamma source)
 Commonly used isotopes; Sr/Y90 source pellet, radioactive stents impregnated with
P-32

Question 36

List three factors that influence the choice of radiation detector for a given application.

Answer 36

 Radiation type
 Energy
 Desired quantity to be measured
 Dose-rate or exposure-rate of radiation
 Useful range of the detector
 Precision

Question 37

(a) . Briefly describe (i) gas-filled detectors,(ii) semiconductors (iii) luminescence detectors
(iv) film detectors

Answer 37

Gas-filled detectors: A volume of (usually inert) gas is housed between two electrodes.
Ionisation radiation generates charged particles within the gas. These are accelerated toward
the electrodes and produce an electric current, which is then measured (e.g. by an
electrometer).
Semiconductor detectors: A reverse bias is applied to a p-n junction (diode), producing a
depletion region between the n¬-type and ¬p¬-type regions of the crystal. The n¬-type and
¬p¬-type regions now serve as electrodes on either side of a region containing no free
electrons/holes. Ionisation radiation produces free electrons/holes in the depletion region
which travel toward the electrodes, producing an electric current in a similar fashion to gasfilled detectors. This current is then measured using an electrometer.
Luminescence detectors: Luminescence detectors are crystals that emit light when exposed
to ionising radiation. This light is then converted to electric current using a PMT or similar.
The current is then measured using an electrometer.
Film detectors: 2D detectors that change colour (darken) when exposed to ionising radiation.
Optical transmission through the film is a measure of the dose.

Question 38

(b).List one example of detector type for each of the four categories.

Answer 38

Gas-filled: Any one of the following:
•Ionisation chamber
•Proportional counter
•Geiger-Müller counter
 Semiconductor: Any one of the following:
•Silicon diode
•Germanium detector
•MOSFET
 Luminescence: Any one of the following:
•Inorganic scintillator
•Organic scintillator
•Thermoluminescence detector
•Optically stimulated luminescence detector
Film: Any one of the following:
•Radiographic film
•Radiochromic film

Question 39

(a).What are the three types of gas-filled detectors? What is the major difference between
them?

Answer 39

Ionisation chambers, proportional counters, and Geiger-Müller counters.
The major difference is the voltage applied between the electrodes.

Question 40

(b). Briefly describe the three regions of operation (the ‘useful’ regions of the voltage-
response curve) of gas-filled detectors

Answer 40

Ionisation chamber region: Voltage is relatively low → ionised particles drift peacefully
toward the electrodes
Proportional region: Voltage somewhat higher → ionised particles energetic enough to
create local avalanches, amplifying the signal.
Geiger-Müller region: Highest useful voltage → ionised particles so energetic that a
single ionisation creates a cascade of avalanches through the gas volume.

Question 41

List one advantage and one disadvantage of semiconductor detectors.

Answer 41

Advantages: High density → improved sensitivity/resolution OR energy discrimination
capability
Disadvantages: Expensive (depending on application) OR temperature dependence.

Question 42

.Briefly explain the following types of luminescence detectors. For each detector, briefly
describe the luminescence process?(a) Thermoluminescence detectors (b) Optically stimulated
luminescence detectors

Answer 42

Thermoluminescence detectors: Ionising radiation excites electrons bound to the valence
band up to the conduction band, where they roam freely. Some of the electrons that
subsequently de-excite are trapped within impurities introduced into the TLD crystal.
These trapped electrons can later be stimulated to de-excite, resulting in light emissions
that can be measured and converted to dose. TLDs are stimulated by heating the crystal
in a light-tight enclosure in front of a PMT. The emitted light is proportional to the dose
absorbed by the TLD. Lithium fluoride (LiF) most common base material
Optically stimulated luminescence detectors. Ionising radiation excites electrons bound
to the valence band up to the conduction band, where they roam freely. Some of the
electrons that subsequently de-excite are trapped within impurities introduced into the
crystal. These trapped electrons can later be stimulated to de-excite, resulting in light
emissions that can be measured and converted to dose. OSLDs are stimulated by
scanning the crystal with a laser beam and capturing the different coloured light using a
PMT. The emitted light is proportional to the dose absorbed by the TLD. Aluminium
oxide (Al2O3) is the most common material.

Question 43

. a. Briefly describe the two real-time detection modes: pulsed and current?

Answer 43

Detectors operating in pulsed mode process each interaction individually.
Detectors operating in current mode collect the signals of multiple interactions together
into a single, averaged signal.

Question 44

b. Which detectors can only be operated in one or the other of these modes?

Answer 44

GM counters only operate in pulsed mode → each interaction saturates the detector for a
brief period, producing a long dead time.
Ion chambers only operate in current mode → insufficient sensitivity for individual
interaction detection (no amplification).

Question 45

Which mode is used for spectroscopy and why?

Answer 45

Pulsed-mode is used for spectroscopy since individual particle energy information is
preserved.

Question 46

What are the principal applications of electron beam therapy?

Answer 46

a) treatment of skin and lip cancer
b) chest wall irradiation for breast cancer
c) administering boost dose to nodes
d) treatment of head and neck cancer
e) total skin electron beam therapy for Mycosis Fungoides

Question 47

Briefly explain the major components of LINAC and describe electron therapy mode?

Answer 47

a) Electron gun- Generates pulses of electrons and injected into the accelerator tube
b) Magnetron/Klystron- Pulsed microwaves produced in the Magnetron or Klystron is
injected into the accelerator tube.
c) Modulator- Synchronises electron gun and Magnetron/Klystron pulses
d) Accelerator structure – Consists of a copper tube with it’s interior divided by copper
discs. This section is evacuated to a high vacuum. As the electrons are injected into the
accelerator structure with an initial energy of about 50 keV, the electron interacts with
the electromagnetic field of microwave and gain energy in the range of MeV
e) Bending magnet- To bent (900 or 2700
) high energy electron between the accelerator
tube and the target.
Electron Mode: The narrow electron beam(~3mm) exits the window of the accelerator tube
and strike on a scattering foil to spread the beam as well as to get uniform electron fluence
across the treatment field. Scattering foil consists of a thin metallic foil, usually made up of
lead. Ion chamber monitors the dose and dose-rate delivered and electron applicator
collimates the electron beam close to the surface

Question 48

Explain the major electron interactions with matter? Define stopping power?

Answer 48

Inelastic collisions (results in energy loss)
Inelastic collisions with atomic electrons – ionization, excitation etc.
Inelastic collisions with atomic nuclei – Bremsstrahlung
Elastic collision (no considerable energy loss, mainly results in scattering)
Elastic collisions with atomic nuclei – nuclear Coulomb scattering
Elastic collisions with atomic electrons – electron-electron scattering
Stopping power (S/ρ)tot of a material for charged particles is defined as the total energy lost by
the particle in traversing through a material of density ρ,
(S/ρ)tot = (S/ρ)col + (S/ρ)rad
where(S/ρ)col and (S/ρ)rad apply to collisional (energy loss due to ionisation and excitation)and
radiation losses(due to Bremsstrahlung) respectively
NB: Stopping power is often expressed in MeV.cm2
/g. Stopping power tables for electron,proton and other
particles can be found https://www.nist.gov/pml/stopping-power-range-tables-electrons-protons-and-heliumions .

Question 49

Discuss the main differences between the photon and electron depth dose curve with the help of a
diagram?

Answer 49

See diagram
Photons:
Shows skin sparing effect (surface dose ~50%)
Maximum dose at a depth,dmax.
dmax increases with energy (i.e skin sparing effect increases with energy)
Dose fall off → after dmax approximately exponential dose fall off
Electrons:
Shows moderate skin sparing effect compared to photons (surface dose in the range of 80-85%).
Maximum dose at a depth dmax.
dmax decreases with energy(i.e skin sparing effect decreases with energy)
Also, unlike photons dmax depends on field size and other factors
Dose fall off→ rapid dose fall off dose after dmax. But at higher energy the rapid fall of becomes
less steep.

Question 50

Why do electron beams have a virtual source point?

Answer 50

Unlike x-ray , an electron beam does not emanate from a physical source in the accelerator head
An electron beam appears to diverge from a point called the virtual source. The virtual source may be
defined as an intersection point of the back projections along the most probable directions of
electron motion at the patient surface

Question 51

Explain why >20MeV electrons are not very useful in external beam therapy?

Answer 51

For electrons, skin sparing effect decreases with energy.
Moreover, the effect of rapid dose falls- off decreases with higher energy. Therefore, more than
20MeV electron is not very useful

Question 52

What is Bragg’s peak? Sketch a diagram comparing the depth dose curve of a 6MV photon and
250MeV proton beam?

Answer 52

refer diagram
Proton or carbon beam deposit most of its energy at a certain depth(near the end of its range),
known as Bragg’s peak. The position of Bragg’s peak depends on the initial energy of the ion.

Question 53

(a) What are the advantages and disadvantages of proton therapy?
(b) What is the added advantage of using carbon ions?

Answer 53

(a) Advantages and disadvantages of proton therapy
• Due to the Bragg’s peak effect, we can deliver a higher dose to a tumour while sparing
the normal tissue and critical organs
• The integral dose is less. i..e total dose delivered (total amount of tissue irradiated) is
much less compared to conventional x-ray treatment → reduced secondary cancer
induction probability
• Sparing of the normal tissues, critical organ, and reduction in integral dose is important,
especially in the case of paediatric tumours and complex head and neck cancers
• Disadvantage- higher cost is the major issue at present. Others include range
uncertainties, RBE issue etc.
(b) Due to the higher charge and mass, carbon ion has a higher LET and higher RBE and
hence produces more ionization tracks and more DNA damage compared to proton beam.
Therefore, carbon ion is more effective in treating radioresistant tumours.

Question 54

Background counts can significantly degrade resolution. List 4 sources of added background
radiation?

Answer 54

a. Scattered radiation
b. Septal penetration in collimator
c. Inadequately shielded sources in vicinity of camera
d. Overlying tissue background

Question 55

Disregarding the consideration of cost, suggest a way to increase the sensitivity and therefore the
detectability of a scintillation camera?

Answer 55

Use of dual or triple headed camera to give 2 or 3 times increase in sensitivity

Question 56

Why is the detectability for tumor detection in the brain similar in nuclear medicine and
computed tomography when the resolution for computed tomography is an order of magnitude
higher than that in nuclear medicine?

Answer 56

Because the contrast is much higher in Nuclear Medicine (with high target/background ratio of
radiopharmaceutical uptake). e.g. PET imaging of metastasis

Question 57

Very high count-rate studies can result in images with reduced contrast. Why? Give an example
of a method used to reduce these dead-time effects.

Answer 57

Very high count rate can give mispositioning of events and this increases the noise and degrades
contrast.
One solution: mask out, using lead, any areas of high counts that are not useful for diagnostic.
E.g. face activity in brain imaging

Question 58

Sketch an ROC curve showing the ROC for (i) a near perfect test, and (ii) one equivalent to
guessing. Where would the curve of a typical diagnostic test lie?

Answer 58

refer image

Question 59

Consider figure 10.38 from Bushberg (given below)
What effect does moving the decision threshold to the left (towards the normal population
peak) have on (i) sensitivity and (ii) specificity of a test
b. Illustrate the direction of movement on the ROC curve of the test.

Answer 59

Moving threshold to left

(i) Increases TPF, i.e increases sensitivity
(ii) Increasing FPF i.e. decreasing specificity which = 1-FPF

refer diagrams

Question 60

Explain FWHM with the help of a simple diagram?

Answer 60

Refer lecture notes – Full Width Half Maximum (FWHM)
is a useful index of spatial resolution. Smaller the FWHM, better the spatial resolution, hence
better image. In case of gamma camera, FWHM is very useful in expressing relationship of
resolution to collimator parameters (e.g. hole size and length etc).

Question 61

Define MTF?

Answer 61

The Modulation Transfer Function (MTF) is a plot of Modulation (M) Vs Spatial frequency of
the image, where M = image contrast/object contrast obtained (using Fourier Analysis) at each
frequency. MTF gives most complete characterisation of resolution of imaging device.
Both FWHM and MTF are two useful indices of spatial resolution
(NB: Fourier Transform and MTF will be covered in Week 11)

Question 62

What are the two major factors affecting the intrinsic spatial resolution of a gamma camera?

Answer 62

(a) Multiple scattering of photons within detector

(b) Statistical fluctuations in distribution of light photons among PM tubes

Question 63

With the help of a simple diagram show how the FWHM curve (and hence spatial resolution)
varies (a) as a point source is moved away from a parallel hole collimator (b) decreasing the
hole length (c) increasing the hole diameter of the collimator

Answer 63

refer diagram
Effect of moving the source away. Similarly, draw pictures for other two cases
Spatial Resolution R2 of a parallel hole collimator
where, d-dia. of the holes, L-length of the collimator holes, F- distance from source to collimator
face c-thickness of the crystal
(a) As a point source is moved away, FWHM increases (worse image)
(b) Decreasing hole length-increases FWHM
(c) Increasing hole diameter increases FWHM
In other words, long narrow holes provide best spatial resolution (lowest FWHM). Also in
nuclear medicine studies patient should be placed as close to the collimator as possible to
provide the best spatial resolution

Question 64

The MTF of three scintillation cameras at given spatial frequency are 0.8, 0.5, and 0.7
respectively. Which of these scintillation cameras has the best spatial resolution at this spatial
frequency?

Answer 64

First camera (MTF =0.8) - higher the value of MTF, more accurately it reproduces the object.

Question 65

Determine the overall spatial resolution of a scintillation camera if the spatial resolution of the
collimator is 10mm and the intrinsic spatial resolution is (a) 10mm (b) 5mm and (c) 1mm

Answer 65

Rs
 R1  R
 (a)
Rs 10 10 200 14.1 mm
2 2
   
 (b)
Rs 5 10 125 11.1 mm
2 2
   
 (c)
Rs 1 10 101 10 mm

Question 66

Which of the two collimators both designed for the same spatial resolution, but one designed for
use with low energy γ –rays and the other designed for high energy γ –rays, will have higher
sensitivity?

Answer 66

Low energy collimator has higher sensitivity as it requires thinner septa between holes to stop γ –
ray penetrating it(septal penetration). i.e. more holes of higher sensitivity.

Question 67

Why is about 1inch of crystal at the edge not useable for imaging in scintillation camera?

Answer 67

Edge packing is seen around the edge of an image as a bright ring and results in non-uniformity
of the image. This result from the fact that more light photons are reflected near the edge of the
detector to the PM tube. Normally a 2” wide lead ring is attached around the edge of the
collimator to mask this effect. In modern cameras, electronic masking is employed.

Question 68

The point source to detector distance (a) must be how many times greater than the largest length
of the detector when measuring the intrinsic uniformity of the scintillation camera? Why?

Answer 68

To minimise the geometric effect (reduced counts on the edges due to longer path length and
inverse square law effect on count rate), make ‘a’ as large as possible – ie. minimize difference in
path lengths between outer edge (b) and centre(a).

Question 69

In addition to the expected loss of counts, what problem does the dead time of a scintillation
camera cause that does not occur in other counting situations?

Answer 69

“Pulse pile-up” : mispositioning of counts at high count rate
Higher count rate increases the probability of two events being accepted in dead time period and
taken as one count (if each is not full energy – as would occur if both had undergone Compton
scatters). Position would be somewhere between two primary event locations

Question 70

Define exposure? What are the limitations of exposure

Answer 70

The exposure (X) is the total amount of electric charge (of one sign) produced by ionizing
electromagnetic radiation per mass of air
 X = Q/m
Limitation of exposure- Exposure applies only to photons (not charged particles) and in air (not
any other medium). A conversion factor is needed to convert Exposure to Absorbed dose.

Question 71

Describe KERMA? How KERMA is related to absorbed dose, in diagnostic radiology energy
range?

Answer 71

KERMA is defined as the sum of the kinetic energies transferred to charged particles liberated by
indirectly ionizing radiation per unit mass
K = KEtras/m
In diagnostic radiology range, KERMA ~ Absorbed Dose
(In diagnostic energy range (30-150kV), the kinetic energy transferred to secondary electrons is low and
the secondary electrons will travel only a very small distance. i.e. all its energy will be deposited locally
which is same as absorbed dose. However, in radiotherapy energy range(MV range), the KE energy
acquired by secondary electrons is significantly higher and they may deposit its energy away from the
point of interaction. Moreover, the high energy electrons may produce Bremsstrahlung radiation which
may deposit energy even further away. Therefore in radiotherapy energy range, KERMA ≠ Absorbed Dose)

Question 72

Explain equivalent dose and effective dose?

Answer 72

Equivalent Dose:
 Biological damage depends on the type of radiation
 High LET radiation produces dense ionisation tracks and more complex damages, hence
more biological damage
 Radiation Weighting factor (wR) is used to account for this effect
 Equivalent dose = Absorbed Dose x Radiation Weighting Factor
H= D x wR
 Unit of Equivalent dose is Sv
Effective Dose
 Biological tissues vary in sensitivity to the effects of radiation
 Therefore, biological damage also depends on the type of biological tissue
 Tissue Weighting factor (wT) is used to account for this effect
 Effective dose = Equivalent Dose x Tissue Weighting Factor
E= H x wT
 Unit of Effective dose is Sv

Question 73

What is the dose limit for occupational worker and the general public?

Answer 73

For occupational worker – 20mSv per year, averaged over a period of 5 consecutive calendar
years, with no single year exceeding 50mSv
For the general public- 1mSv /yr

Question 74

Explain three main duties of the Radiation Safety Officer?

Answer 74

 To assist the specific employer in complying with the requirements of the Act and
Regulation
 To advice, the specific employer on all aspects of radiation safety applicable to the
activities carried out by specific employer
 To perform the duties imposed upon a radiation safety officer by these regulations

Question 75

Briefly describe (a) Workload (b) Use factor (c) Occupancy factor

Answer 75

(a) Workload (W)- is the time integral of the x-ray tube current in milli ampere minute over a
period of 1 week (mA- min/wk). i.e the total amount of x-ray produced in a week. More
workload, more shielding required
(b) Use factor (U) – indicates the fraction of time during which the radiation under
consideration is directed at a particular direction/barrier. U ranges between 0 and 1 for
primary, depending upon the primary is directed at that particular wall. U=1 for secondary
barrier because scattered radiation is always present in all directions. Higher use factor,
more shielding required
(c) Occupancy factor (T)- indicates the fraction of time a particular place is occupied by staff,
patients or public. e.g. work area, offices staff room (T=1), corridors, employee lounge
(T=1/5), Toilets, unattended waiting room (T=1/20) etc. When the occupancy factor is
more, we need more shielding for that area.

Question 76

What is the shielding material used for beta and neutrons? Explain why high z material such as
Pb is not used for shielding beta rays and neutrons?

Answer 76

For beta rays and alpha – low z material such as Perspex is used
For neutrons- hydrogen-rich materials are used because neutron will be absorbed by them.
Commonly used materials are water, paraffin wax and concrete
Beta rays, which are high energy electrons, will produce bremsstrahlung radiation when
interacting with high z material.
Neutrons are chargeless particle and therefore will not be absorbed by high z materials.

Question 77

(a) Write a short note on Y-90 targeted therapy for hepatic tumours?
(b)Explain the radiation protection issues related to Y-90 targeted therapy [pure β emitter, Eav ~
933kV, T1/2 = 2.7days, range in water~ 4.33mm]

Answer 77

(a) Y-90 targeted therapy for hepatic tumour
 Hepatocellular carcinoma and metastatic hepatic tumours are difficult to treat with
external radiation therapy or surgery
 These tumours are often treated with Y-90 labelled microspheres
 Through a small incision near the groin, a catheter is inserted and guided through the
artery into the liver
 Y-90 microspheres are delivered directly to liver tumour via hepatic artery (blood
supply to tumour site is provided by hepatic artery)
 A higher dose is delivered to the tumour, while exposure to the remaining healthy
liver tissue is minimised (as the range of beta particles are couple of mm in tissue)
(b) Radiation protection issue associated with Y-90 therapy
 Y-90 is a pure beta emitter
 A low Z material (e.g. Perspex, acrylic) should be used as primary shielding material
 However, a small fraction of β energy will be converted to bremsstrahlung x-ray (
due to very high energy of beta rays,933kV)
 A thin lead shield outside of primary shielding may be used to reduce bremsstrahlung
 Administration of Y-90 is generally performed in interventional radiology suites
 Composite shielding (e.g.Perspex + Pb) is recommended for vial, syringe and
transport container.
 Shielded manual injector is often used for increased distance and for finger shielding

Question 78

Determine the thickness of concrete wall required to protect a controlled area, which is 4m from
the target of a 150kVp diagnostic unit having a weekly workload of 102.4mA-min. The wall has a
use factor of 1 and the occupancy factor of the area beyond the wall is 0.5? (HVL of concrete at
150kVp =2.24 cm)[Transmission Factor = Pd2
/WUT]

Answer 78

Transmission factor = Pd2
/WUT
P= 0.1 mGy/wk (goal for controlled area), d=4m, W=102.4mA-min/wk , U=1, T=0.5
Transmission factor= 0.1x42
/102.4x1x0.5 = 0.03125
Thickness of concrete required to get a transmission factor of 0.03125 = 5HVL=5x2.24=11.2cm
[Number of HVL required to get a transmission factor of 0.03125(i.e. 3.125%) can be estimated as follows;
(a) with 1HVL, transmission factor is 0.5 (i.e.50%), 2 HVL0.25, 3 HVL-0.125, 4 HVL0.0625,
5HVL0.03125
(b) Or you may use the formula 1/2n
=0.03125, where n is the number of HVL required ]

Question 79

Explain the challenges in mammography imaging?

Answer 79

Mammography is used to detect subtle changes in breast pathology with fine details and minimal
glandular dose. Most signs of breast pathology are either in the form of soft tissue masses, which
are not very different from the surrounding tissues or in the form of microcalcifications (early
marker). Visualisation of these structures is very challenging and requires special imaging
equipment to provide images with high contrast and high resolution. Dose to the patient must be
minimal as mammography examination is often done for screening purposes.

Question 80

What are the different targets used in film-screen mammography and digital mammography
and explain why?

Answer 80

refer diagram
Screen film mammography- Mo (17.5keV and 19.6keV) and Rh (20.2keV and 22.7 keV) are the
most commonly used target material. Characteristic x-ray production in the optimal range is the
major reason for using Mo and Rh targets. The numbers of X-ray in the optimal energy range for
breast imaging (15-25 keV) is significantly increased by characteristic x-ray emission ( as shown
in the fig below).
Digital mammography – W target is preferred in digital mammography due to the following
properties (a) increased x-ray production efficiency due to its higher atomic number (b) improved
heat loading, due to its higher melting point.
Digital detectors have a large dynamic range (i.e. produce acceptable image quality even at low
exposure). Moreover, through post-processing contrast of the image can be enhanced. Therefore
characteristic radiation is not important in digital mammography as it is with screen film

Question 81

Explain the function of filters used in mammography and how it is different from filters used
in general radiography. Explain k-edge absorption?

Answer 81

Added filtration is used in mammography to improve energy distribution. Filtration selectively
removes the lowest and highest energy x-rays from the spectrum and allow transmission of
continuous and characteristic x-ray energy in the optimal range. (high energy photons often
undergo Compton Effect, which produces scattered radiation and hence reduces contrast). Filters
used in mammography are based on the principle of k-edge absorption. This is accomplished by
using elements with k absorption edge energies between 20 and 27keV (Mo, Rh, Ag etc.). In
general radiography, filters are used to remove the low energy part (soft x-rays) only to reduce
the patient dose (low energy photons will be abrobed by the patient and will not contribute to
image formation).
Common target/filter used in mammography
Screen film:
Mo/Mo- used for thin breast
Mo/Rh- used for thicker and denser breast
Rh/Rh-used for thickest and densest breast
Digital Mammo:
W/Rh, W/Ag
k edge absorption is the sharp increase in photon absorption by an element when the incoming photon
energy is just above the k-shell binding energy. This is because the probability of PE effect is very high
when the incoming photon energy is just above the binding energy. e.g. Mo has a K shell binding energy of
20keV. Therefore, Mo will absorb most of the photons with energies slightly more than 20keV through
photo electric effect. (For more details on k edge absorption refer Bushberg Chapter 3)

Question 82

Sketch a diagram showing a typical Mo/Mo spectrum and Mo/Rh spectrum, operating at 30kVp
(Mo binding energies K=20keV, L=2.5keV, M=0.4keV
Rh binding energies K=23.2keV, L= 3keV, M=0.5keV)

Answer 82

refer diagram
(Note: You should be able to draw the spectrum for any target/filter combination. The only information you
need is the binding energies. Characteristic peaks and cut-off energies can be calculated from binding
energies. Plot the spectrum for a hypothetical target/ filter combination A/B operating at 30kVp, where the
binding energies are A (K=20,L=3,M=1) and B(K=25,L=2,M=0.5)
For A/B combination, the characteristic peaks are positioned at 17kVp&19kVp, cut-off is at 25kVp
(note: when you draw the spectrum make sure that the shape of the spectrum is reasonably ok. )

Question 83

Why breast orientation is important in mammography?

Answer 83

• Due to Heel effect → high intensity near cathode side and low intensity near the anode
• Breast positioning is important for better uniformity of transmitted x-rays
• Cathode over the chest wall and anode over the anterior portion (nipple)
• This orientation also decreases the equipment bulk near patients head

Question 84

In a mammography unit, often the x-ray tube is physically tilted. Why?

Answer 84

To achieve adequate field coverage (24x40) on the anterior side of the field, x-ray tube must be
physically tilted (although most of the x-ray tube uses anode tilt, this is not enough as the spot size
is small in mammography)
[NB: the requirement of the special target (Mo/Rh) in conventional mammography and tube tilt are the consequences
of small focal spot used in mammography.
Small focal spot→small mA→ less number of photons→needs characteristic x-ray to enhance the number of photons
Small focal spot → small field coverage → use tube tilt to increase field coverage. Also, tube tilt allows the central axis
(half beam geometry is used in mammography) of the x-ray beam to run perpendicular to the plane of the detector at
the chest wall. ]

Question 85

With the help of a simple diagram, explain why half-beam geometry is employed in
mammography

Answer 85

To avoid exposure to patients torsos and to maximise the amount of breast tissue near the
chest wall, all dedicated mammography system utilises a’ half- field’ geometry, achieved by
collimation
refer diagram

Question 86

Give three reasons why breast compression is essential in mammography?

Answer 86

(a)Breast compression reduces overlapping anatomy (b) decreases tissue thickness (c) reduces
inadvertent motion of the breast. This results in less scattered x-rays, less geometric blurring of
anatomical structures and lowers radiation dose to the breast tissue

Question 87

Describe magnification mammography? What are the advantages and disadvantages?

Answer 87

• Geometric magnification is used to improve resolution, typically for better visualisation
of microcalcification. Typically 1.5x to 2x magnification is used
• Position the breast closer to the focal spot, mag=SID/SOD
• Select a small focal spot (0.1mm) to reduces geometric blurring
• Remove anti-scatter grid → air gap reduces scattered radiation
Advantages
• Improves system resolution by the magnification factor, better visualisation of
microcalcification
• Reduction in noise
• Reduction in scattered radiation
Disadvantage
• Long exposure time, ~4s (due to small focal spot) – patient motion and blur

Question 88

Describe Digital Breast Tomosynthesis?

Answer 88

• In conventional mammography, anatomical structures are superimposed on the pathology,
which obscures the visualization and detection of cancer or other abnormalities
• Digital Breast Tomosynthesis (DBT) reduces the superimposition
• DBT is a modification of a digital mammography unit to enable the acquisition of 3D
volume of thin section data
• Acquire images at different angles during single motion of x-ray tube and reconstructed
using reconstruction algorithm
How it is done:
• X-ray tube moves in an arc around the breast, typically up to -500
to +500
• Acquire multiple low dose images at several angular positions, up to ~50 images
• With high-speed digital readout TFT, this can be completed in ~20s
Images are processed with limited angle reconstruction algorithm (shift &add method,
filtered back projection, maximum likelihood algorithm etc.) to obtain a tomogram at a given
depth in the breast

Question 89

What is Average Breast Glandular Dose (AGD)? State MQSA recommended dose regulation for
mammography examination?

Answer 89

Mammography is widely used for screening. Therefore dose optimisation in mammography is
very important. Average Glandular Dose is the most commonly used dose index in mammography
The Average Glandular Dose,
Dg=DgNx XESAK
XESAK – entrance skin Air Kerma in mGy
DgN – Air Kerma to average glandular dose conversion factor
MQSA (US) regulation – the average glandular dose for a compressed breast thickness of
4.2cm and a breast composition of 50% glandular and 50% adipose tissue should be
maximum of 3mGy per image

Question 90

Briefly describe three Acute Radiation Syndromes?

Answer 90

Depending on an acute total body dose above ~1Gy, the response is described as a specific
radiation syndrome known as Acute Radiation Syndrome.
Bone Marrow Syndrome
Occurs at doses of around 2-10 Gy,
Characterised by nausea
Caused by bone marrow depletion, which leads to a drop in platelets and WBC counts
Gastro Intestinal (GI) Syndrome
Occurs at doses above ~10 Gy
Nausea, vomiting, and prolonged diarrhea
Caused by loss of the intestinal villi, which absorbs nutrient and fluid from the bowel (villi also
form a barrier between the blood and bacteria in the gut lumen)
Above 10 Gy- death in a couple of weeks
Central Nervous System (CNS) syndrome
Occurs at doses of around 60-100 Gy
Severe nausea and vomiting (within minutes) followed by disorientation, loss of coordination of
muscular movement, respiratory distress, diarrhea, convulsive seizures, coma
Caused by brain edema or dysfunction of critical cells in the CNS
Death in 2-3 days

Question 91

Radiation damage to mammalian cells can be classified into three categories, based on repair.
Explain?

Answer 91

1. Lethal Damage- irreversible, irreparable and leads to cell death
2. Sublethal Damage (SLD)- can be repaired in hours (unless additional sublethal damage
   added and eventually leads to lethal damage)
3. Potentially Lethal Damage- can be manipulated by repair

Question 92

Describe the Linear Quadratic Model and sketch a graph showing a typical cell survival curve?

Answer 92

*A cell survival curve describes the relationship between the surviving fraction of cells and dose
*Linear Quadratic model is most often used to describe the cell survival curve
*Linear Quadratic model consists of two components: linear term & quadratic term
*Linear term is proportional to dose, quadratic term proportional to dose squared
Mathematically, the fraction of surviving cells (SF), following a single dose of radiation D is
observed to follow the relationship:
SF(D)=exp(-αD-βD2
)
α – a constant describing the initial slope of the cell survival curve
β – a constant describing the quadratic component of cell killing
refer diagram
In the linear region, cell killing is proportional to D and in the quadratic region, it is proportional
to D2
. This is because, in the quadratic region, biological damages produced are more complex in
nature and difficult to repair compared to the linear region.

Question 93

(a) Illustrate, typical cell survival curves demonstrating the effects of (a) X- rays (low LET) and
α-particles (high LET)exposure
(b) Define RBE

Answer 93

refer diagram
(b) RBE is used to compare the biological effectiveness of different types of radiation.
RBE= Dose from standard radiation (250kVp) to produce a given biological effect/Dose from
a test radiation to produce the same biological effect
(From the above graph, to produce 90% cell kill, around 1Gy and 6Gy are required, for high LET alpha
radiation and low LET X-rays, respectively. i.e. the RBE of high LET alpha is around 6)

Question 94

(a) Oxygen is a radio sensitizer. Explain?
(b) The following graph represents typical cell survival curves for hypoxic and oxic
conditions. Estimate the oxygen enhancement ratio (OER) for 90% cell kill

Answer 94

refer diagram
(a) When oxygen is present, cells are more sensitive to radiation because (i) Damaging effects of
free radicals are enhanced by the presence of oxygen (ii) Oxygen reduces the probability of
free radical recombination to form harmless chemical species and inhibit repair of damages
caused by free radicals
(b) For 90% cell killing (10% cell survival) @ hypoxic condition, dose required ~ 6 Gy
For 90% cell killing @ oxic condition, the dose required ~ 2 Gy
Therefore, OER = 6/2 =3
i.e. almost 3times more dose is required to produce 90% cell kill in hypoxic condition

Question 95

Radiotherapy is often delivered in fractions. Explain why (4R’s of radiotherapy)?

Answer 95

There is no significant intrinsic difference in the radiosensitivity of normal cells and tumour
cells and may produce nearly the same amount of damage if we deliver the total dose at once. By
giving the total dose in small fraction allows producing maximum damage to the tumour cells
and minimal damage to the normal cells by exploiting some of the differences in the “living
conditions” of tumour and normal cells, often known as 4R’s of radiotherapy.
Repair
*Mammalian cells can repair radiation damage between dose fractionation
*Fractionation allows normal tissue to repair all repairable radiation damage prior
to giving next fraction of radiation (Most of the repair occurs around 2-6 hrs after radiation)
Redistribution (Reassortment)
*Cells have different radiation sensitivities in different parts of the cell cycle
*Fractionation allows redistribution of tumour cells to sensitive part of cell cycle
*Some of the tumour cells are in radioresistant phase during irradiation (G1, S)→ less cell kill
*Cells move to radiosensitive phases (G2, M) for next fraction
Reoxygenatation
*Tumour cells are fast proliferating, hence higher demand for oxygen and very hypoxic
(especially tumour cells away from capillaries)
*Therefore, tumour cells are radioresistant
*Reduction in hypoxic cells leads to reoxygenation of tumour cells between fractions
and opening of compressed blood vessels
– 2 –
*Making tumour cells more radiosensitive to subsequent doses of radiation
Repopulation
*Fractionation allows normal tissue repopulation
*This is an important mechanism to reduce acute side effects from the irradiation of
skin, mucosa etc.

Question 96

Explain early and late effects of radiation on normal tissues?

Answer 96

Early (Acute) effects
• Manifests themselves soon after radiation (within few days/weeks of radiation)
• Occurs mainly in tissues with rapid cell turnover
• e.g. Skin, gastrointestinal epithelium, mucosa, bone marrow
• Damage may be repaired because of the rapid proliferation of stem cells
• Completely reversible
Late(Chronic) effects
• Appears after a delay of months or years
• Occurs mainly in slowly proliferating tissues
• e.g. Lung, spinal cord, liver, CNS bone, cartilage
• Late damage is never completely repaired

Question 97

What are radiosensitizers and radioprotectors?

Answer 97

Radioprotectors
*Chemical agents that reduce normal cell response to radiation
*Generally influence the indirect effect of radiation by scavenging the production of free
radicals
*Helps to reduce normal tissue complications
Radiosensitizers
*Chemical agents that enhance the tumour cell response to radiation
*Generally promoting both direct and indirect effects of radiation

Question 98

(a) Estimate the BED for late responding normal tissues for the following treatment protocol?
Total Dose=60Gy, number of fractions=30, Dose/fr =2Gy, α/β = 3,
(b) If the dose per fraction is increased to 3Gy, how many fractions are required to deliver the
same biologically effective dose to normal tissue?

Answer 98

see formula image
(a) = 302(1+ 2/3) =100 Gy
(b) d= 3Gy , α/β=3, n=?
We want to produce the same biological effect, i.e. BED =100
100 = n3 (1+ 3/3) = n6
n =100/6 =16.7 fractions
i.e. total dose = 16.73Gy= 50Gy required to produce the same effect
i.e. 60Gy/30 fraction = 50Gy/16.7fr - produces the same biological effect on late
responding normal tissue
[Please note that BED formulae given here is a very basic one and can be applied for limited
situations. More comprehensive equations are available to account for several influencing factors.
Radiotherapy students are encouraged to refer radiobiology text books for further information]