Tutorial Answers Flashcards

(98 cards)

1
Q

Write a short note on (a) DICOM

A

(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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Write a short note on (b) PACS

A

(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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

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

A

(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)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

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

A
  • Quantum or mottle noise (fluctuation in x-ray, photons, electrons etc),
  • Electronic noise (exists in all electronic circuits)
  • Anatomical noise (unwanted anatomy)
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

Briefly explain;

a) Fourier Transform (FT

A

• 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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

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

A

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 )

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

Briefly explain; Detective Quantum Efficiency (DQE)

A

• 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)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Briefly explain; Bone densitometry scan (DEXA)

A

refer lecture notes

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Write a short note on Nyquist frequency?

A

• 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…)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

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?

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

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

A

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)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

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

A

see diagram

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

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

A

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)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

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)

A

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]

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

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?

A

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.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

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

A

 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)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

Explain MRI localization?

A

 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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

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

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

What are the major safety and biohazard concerns in MRI imaging

A

Refer lecture notes

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

Explain why protons act like tiny magnets?

A

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.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q
Which of the following atoms have a net spin and why? 15P
31
, 6C
12
, 6C
13
, 8O
16 and 2He4
?
A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

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

A

(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.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

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

A

 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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

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

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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?
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)
26
Explain the T2 relaxation (spin-spin)process and FID with the help of a simple diagram? What is T2* relaxation?
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
27
T2 relaxation depends on molecular size and motion. Explain
 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)
28
Describe T1 relaxation (spin-lattice) with the help of a simple diagram?
 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
29
Why T1 relaxation is different for different tissues? (use the following graph to answer this question) refer figure
 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
30
What are the main advantages and disadvantages of Brachytherapy, compared to external beam therapy?
``` 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 ```
31
Briefly explain the classification in brachytherapy based on (a) treatment duration and (b) dose rate.
(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
32
Briefly explain hot loading, manual after loading and remote after loading
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
33
Write a short description of a remote afterloading HDR unit and explain how the desired dose distribution is achieved within the patient
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
34
What decay type would you prefer for the following applications; (a) Intracavitary(b) intravascular and why?
(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.)
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)
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
36
List three factors that influence the choice of radiation detector for a given application.
```  Radiation type  Energy  Desired quantity to be measured  Dose-rate or exposure-rate of radiation  Useful range of the detector  Precision ```
37
(a) . Briefly describe (i) gas-filled detectors,(ii) semiconductors (iii) luminescence detectors (iv) film detectors
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.
38
(b).List one example of detector type for each of the four categories.
``` 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 ```
39
(a).What are the three types of gas-filled detectors? What is the major difference between them?
Ionisation chambers, proportional counters, and Geiger-Müller counters. The major difference is the voltage applied between the electrodes.
40
(b). Briefly describe the three regions of operation (the ‘useful’ regions of the voltage- response curve) of gas-filled detectors
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.
41
List one advantage and one disadvantage of semiconductor detectors.
Advantages: High density → improved sensitivity/resolution OR energy discrimination capability Disadvantages: Expensive (depending on application) OR temperature dependence.
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
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.
43
. a. Briefly describe the two real-time detection modes: pulsed and current?
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.
44
b. Which detectors can only be operated in one or the other of these modes?
``` 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). ```
45
Which mode is used for spectroscopy and why?
Pulsed-mode is used for spectroscopy since individual particle energy information is preserved.
46
What are the principal applications of electron beam therapy?
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
47
Briefly explain the major components of LINAC and describe electron therapy mode?
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
48
Explain the major electron interactions with matter? Define stopping power?
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 .
49
Discuss the main differences between the photon and electron depth dose curve with the help of a diagram?
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.
50
Why do electron beams have a virtual source point?
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
51
Explain why >20MeV electrons are not very useful in external beam therapy?
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
52
What is Bragg’s peak? Sketch a diagram comparing the depth dose curve of a 6MV photon and 250MeV proton beam?
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.
53
(a) What are the advantages and disadvantages of proton therapy? (b) What is the added advantage of using carbon ions?
(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.
54
Background counts can significantly degrade resolution. List 4 sources of added background radiation?
a. Scattered radiation b. Septal penetration in collimator c. Inadequately shielded sources in vicinity of camera d. Overlying tissue background
55
Disregarding the consideration of cost, suggest a way to increase the sensitivity and therefore the detectability of a scintillation camera?
Use of dual or triple headed camera to give 2 or 3 times increase in sensitivity
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?
Because the contrast is much higher in Nuclear Medicine (with high target/background ratio of radiopharmaceutical uptake). e.g. PET imaging of metastasis
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.
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
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?
refer image
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.
Moving threshold to left (i) Increases TPF, i.e increases sensitivity (ii) Increasing FPF i.e. decreasing specificity which = 1-FPF refer diagrams
60
Explain FWHM with the help of a simple diagram?
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).
61
Define MTF?
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)
62
What are the two major factors affecting the intrinsic spatial resolution of a gamma camera?
(a) Multiple scattering of photons within detector | (b) Statistical fluctuations in distribution of light photons among PM tubes
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 ```
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
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?
First camera (MTF =0.8) - higher the value of MTF, more accurately it reproduces the object.
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
``` 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 ```
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?
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.
67
Why is about 1inch of crystal at the edge not useable for imaging in scintillation camera?
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.
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?
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).
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?
“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
70
Define exposure? What are the limitations of exposure
``` 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. ```
71
Describe KERMA? How KERMA is related to absorbed dose, in diagnostic radiology energy range?
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)
72
Explain equivalent dose and effective dose?
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
73
What is the dose limit for occupational worker and the general public?
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
74
Explain three main duties of the Radiation Safety Officer?
 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
75
Briefly describe (a) Workload (b) Use factor (c) Occupancy factor
(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.
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?
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.
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]
(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
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]
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 ]
79
Explain the challenges in mammography imaging?
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.
80
What are the different targets used in film-screen mammography and digital mammography and explain why?
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
81
Explain the function of filters used in mammography and how it is different from filters used in general radiography. Explain k-edge absorption?
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)
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)
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. )
83
Why breast orientation is important in mammography?
* 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
84
In a mammography unit, often the x-ray tube is physically tilted. Why?
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. ]
85
With the help of a simple diagram, explain why half-beam geometry is employed in mammography
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
86
Give three reasons why breast compression is essential in mammography?
(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
87
Describe magnification mammography? What are the advantages and disadvantages?
• 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
88
Describe Digital Breast Tomosynthesis?
• 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
89
What is Average Breast Glandular Dose (AGD)? State MQSA recommended dose regulation for mammography examination?
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
90
Briefly describe three Acute Radiation Syndromes?
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
91
Radiation damage to mammalian cells can be classified into three categories, based on repair. Explain?
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
92
Describe the Linear Quadratic Model and sketch a graph showing a typical cell survival curve?
*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.
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
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)
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
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
95
Radiotherapy is often delivered in fractions. Explain why (4R’s of radiotherapy)?
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.
96
Explain early and late effects of radiation on normal tissues?
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
97
What are radiosensitizers and radioprotectors?
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
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?
see formula image (a) = 30*2(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 = n*3 (1+ 3/3) = n*6 n =100/6 =16.7 fractions i.e. total dose = 16.7*3Gy= 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]