Long Qs

Question 1

Aggressive periodontal disease is rare and rapid.

a) Give features of its localised and generalised forms
b) Define the term “pathogenicity”
c) Discuss the circumstances in which an opportunistic pathogen may cause infection

Answer 1

a) 
Localised:
- Functional abnormalities of neutrophiles
- Sparse levels of plaque 
- Aggregatibacter actinomycetemcomitans serotype b
-Leukotoxin production 
Enzymes; modulate host defences
- JP2 Clone- overproduces leukotoxins 

Generalised:
- more diverse microflora

b) Capacity of a microbe to cause damage in a host

c) - Damage the epithelium
- Transfer to other site
- Suppression of immune system
- Presence of foreign body
- Disruption of normal microbiota
- Impairment of host defences by exogenous pathogen

Question 2

The Periodontal Ligament has three main components; Cellular, Nerves and Vasculature and
Extracellular Matrix (ECM).

a) Summarise the hierarchy of Type 1 Collagen structure

Answer 2

• Pro- collagen (intracellular)
• Tropocollagen (Extracellular)
• Microfibrils (Extracellular)
• Fibrils (Extracellular)
• Fibres (Extracellular)
  Fibre Bundles (Extracellular)

Question 3

b) Unlike in other connective tissues there is no conversion to non-reducible cross-links with age in
the PDL.

Give some possible reasons for this.

Answer 3

• Due to CHO on HYL
• Due to rapid turnover of collagen in PDL
• Other suitable response

Question 4

Compare collagen formation and breakdown in the PDL to the gingiva, skin and alveolar bone.

Answer 4

Collagen formation in the PDL is:
- 2 x Subdermal (beneath skin) gingiva.
- 4 x Skin.
- 6 x Alveolar bone.
Collagen breakdown in the PDL is:
- 5 x Gingiva.
- 15 x Skin.
- 6 x Alveolar bone.

I.e. the PDL forms and breaks down collagen at a much higher rate than other tissues.

Question 5

X-rays are commonly used in dentistry, as X-ray imaging is useful in the diagnosis of common
oral problems, such as cavities.

a. Describe the following electron interactions.

Answer 5

Photoelectric -

• Here the x-ray photon interacts with an inner shell electron and deposits all its energy.
• The electron is ejected and the vacancy left is filled by an outer shell electron.
• It is the photoelectric effect which is responsible for contrast; it does however come at the expense of patient dose.

Compton -

• This is where the x-ray photon interacts with an outer shell electron which is ejected with some absorption of energy.
• The remainder is scattered and another electron is captured.
• The Compton Effect doesn’t contribute to contrast because the outer shell electron energy isn’t
  proportional to the atomic number cubed.
• There is also incomplete absorption so the patient dose is less.
• The scatter however can degrade the image.

Question 6

Discuss the three principles of radiation protection.

Answer 6

Principle
- description

1. Justification
   - Any exposure should bring about a positive net benefit
2. Optimisation
   - Technique should be optimised to keep dose as low as possible. Quality should be assured so you don’t have to repeat the film.
3. Limitation
   - NO dose limit for patients, only for staff and the public

Question 7

Discuss the four main cells involved in the homeostatic process of alveolar bone production and
resorption and briefly explain their function.

Answer 7

1. Osteoclast – This is a bone resorbing cell which is often large and multi-nucleated. The cells have a
   brush border to increase surface area at the site of bone resorption and normally lie is resorption pits called Howship’s Lacunae.
2. Osteoblast – Responsible for bone formation. Normally found in a single cell layer lining the bone surface. It is an active protein producing cell.
3. Fibroblast – This cell is responsible for forming and maintaining the extra cellular matrix (collagenous) of the ligament and lamina propria.
4. Macrophage – For defence.

Question 8

Describe the process of osteoclast formation.

Answer 8

1. The matrix is highly vascularised, circulating monocytes are recruited,
2. Under the control of growth factor RANKL (produced by adjacent osteoblasts) they fuse and form osteoclasts.
3. RANKL may be free or bound to the osteoblast cell membrane, and binds to a specific receptor on the monocytes membrane.
4. Modulation in RANKL production is influenced by the applied force exerted on the bone

Question 9

Summarise the effects of IL-1.

Answer 9

• Acts on osteoblasts to up regulate RANKL production.
• Acts on fibroblasts to up regulate IL-1 production.
• Acts on fibroblasts to up regulate IL-6 production which potentiates RANKL.
• Acts on fibroblasts to induce collagenase production.
• Act on osteoblasts to up regulate cathepsin K production thereby increasing effectiveness of individual cells in degrading bone matrix.

Loss of loading on the bone can result in further bone loss.

Question 10

The supportive therapy routine generally comes in three stages. Describe each stage in detail.

Answer 10

Examination and Evaluation

• This firstly looks at the patient plaque control – Plaque distribution and obtaining a plaque free score.
• Plaque retention sites should be managed.
• Gingival indices.
• Pocket depths.
• Bleeding on Probing.
• Suppuration.
• Furcation.
• Mobility.
• Recession.
• Attachment levels.
• Take radiographs if necessary for justification and optimization (only taken when clinically justified).

Supportive Therapy

• Information and motivation.
• Instruction in methods for plaque control.
• Scaling and polishing.
• Demonstrations in patients own mouth.
• Encouragement no criticism, vary the lecture!
• Healing will result in a change in gingival morphology.
• Post-surgery, good OH is critical to success.
• New restorations can be placed.
• Smoking cessation - Current status, did they reduce smoking or quit in the initial phase of therapy?
  Have they maintained reduction/quitting? Has there been progression from the initial phase?
• Repeated subgingival scaling may leave grooves and furrows on the root surfaces which act as PRFs.
  Avoidance of need for repeated instrumentation is preferable

Treatment of Recurrent Diseases

• May be generalized or localised!

Question 11

A digital image is essentially made up of lots of pixels (picture elements) all with their own
value corresponding to a certain shade.

a. Discuss the categories which digital imaging can be divided into and describe briefly how a CCD
works.

Answer 11

• Indirect – Flat-bed scanner, CCD-video camera. Converting an analogue image into a digital image. Often degrades the image in some way.
• Semi-direct – Relatively cheap, uses a photostimulable phosphor plate which stores the light energy before emitting it. This is then computed.
• Direct – CCD/CMOS based, most have a wire to a computer. They are also known as solid state.
• A CCD is lots of individual pixel element which are sensitive to light or x-rays.
• Each is given a charge which is read out.
• First it goes along the horizontal axis then it piles up at the vertical axis.
• A computer algorithm computes the information to produce the image.
• A damaged pixel element can affect the entire image.
• VERY EXPENSIVE

Question 12

A digital image is essentially made up of lots of pixels (picture elements) all with their own
value corresponding to a certain shade.

b. Describe the main advantages and disadvantages of direct digital radiology.

Answer 12

Advantages of direct digital radiology:

• Instant imaging.
• Low patient doses (90% dose reduction).
• Image processing/ enhancement/ archiving.

Disadvantages of direct digital radiology:

• Requires connecting wire.
• Physically large detectors compared to film.
• Low doses but narrow exposure latitude.
• Small detector area.

Question 13

A digital image is essentially made up of lots of pixels (picture elements) all with their own
value corresponding to a certain shade.

c. Quality is task dependant and relies on the ability of the receptor to produce the desired
information. Give four requirements of an image in order to make it good quality.

Answer 13

• Proper visual characteristics.
• Geometric accuracy.
• Anatomical accuracy.
• Absence of artifacts.
• Adequate coverage.

Question 14

Human immunodeficiency virus (HIV) is a lentivirus (a member of the retrovirus family) that causes acquired
immunodeficiency syndrome (AIDS). HIV is present as both free virus particles and a virus within infected
immune cells. 

Describe the transition from the Acute and Asymptomatic stages of HIV pathogenesis leading to
AIDS.

Answer 14

• In the early stages of HIV infection, the virus enters the T helper cells as we’ve already described (it is also able to colonize macrophages but T Helper cell infection is the primary route and cause of the
  eventual disastrous consequences ).
• Once inside the T helper cells, the virus replicates. At this stage, the levels of virus (referred to as the
  “viral load”) dramatically increase while the numbers of T helper cells decrease. This is because when
  the Cytotoxic T lymphocytes (CTLs – online session 2) detect the presence of HIV antigens in Class I MHC complexes on the infected T Helper cells’ membranes, they become activated and kill the T helper cells as they would any other cells which are infected by a virus.
• The CTLs themselves will increase in numbers, due to clonal expansion (see online session 2).
• T helper cells are also killed by the virus itself.
• B-cells will also be activated by meeting up with HIV-antigen and will begin producing immunoglobulins
  to neutralise free virus. They themselves will also increase in numbers, due to clonal expansion (see
  online session 1).
• The combined approach of CTLs killing infected cells and B-cells producing immunoglobulins begins
  to have its effect and at the end of this initial (or “acute”) phase, the T cells are replaced by the body
  and their numbers start to rise whereas the viral load falls.
• In the acute phase, only about a third of individuals infected by the virus will show symptoms and these
  are very non-specific, resembling those you might expect from a cold or flu, such as swollen glands
  and fever.
• The infection, if untreated, now moves in to the second stage.
• This is the “asymptomatic stage”, where the individual will experience few, if any, unusual symptoms.
• During the asymptomatic stage, the immune system continues to fight against the virus but without
  treatment, this will ultimately be a losing battle.
• At the start of the asymptomatic stage (end of the acute stage), viral levels are low but these slowly increase with time. T Helper cell numbers gradually decline, due to the effects of the virus and because the infected helper cells are continually being killed by cytotoxic T lymphocytes.
• At the end of the asymptomatic stage, the immune system collapses and full strength AIDS takes hold.

Question 15

What does the term BPE refer to? Also state the internationally recognised term.

Answer 15

• Basic Periodontal Examination

- Internationally recognised term = CPITN or community periodontal index of treatment needs

Question 16

Give the full title of the probe used to undertake the BPE procedure.

Answer 16

WHO 621 probe

Question 17

Describe the features of the normal C-type probe. (BPE)

Answer 17

1) Ball ended, thus no trauma to sulcus

2) Markings at 0.5, 3.5, 5.5, 8.5, 11.5mm

Question 18

Describe the divisions of the mouth whilst using the screen. (BPE)

Answer 18

1) Into sextants

2) 17 – 14 
13 – 23 
24 – 27
47 – 44 
43 – 33 
34 – 37

Question 19

During the screen a code is allotted to each sextant, define the 6 codes and give brief descriptions of
the clinical interpretations. (BPE)

Answer 19

Code and Description:

Code 4 Deep pockets of 5.5mm or more (black band not visible)

Code 3 Shallow pockets 3.5 to 5.5mm (band partly visible)

Code 2 Supraging / subging calculus, overhanging restns, no PPD’s > 3.5mm

Code 1 Bleeding on probing

Code 0 No disease detectable using WHO 621 probe

Code * Furcation, or, Recession plus PPD of 7mm or more

Question 20

The Periodontal Ligament has three main components; Cellular, Nerves and Vasculature and Extracellular
Matrix (ECM).

Describe the structure of Type I Collagen.

Answer 20

Type one collagen is made up of three alpha chains wound into a triple helix, they aggregate together to
form fibres.

Question 21

The Periodontal Ligament has three main components; Cellular, Nerves and Vasculature and Extracellular
Matrix (ECM).

Outline the full process of Type I Collagen biosynthesis.

Answer 21

INTRACELLULAR:
- Firstly individual pro alpha chains are synthesised at the RER.
- Then proline and lysine are hydroxylated – HYDROXYLASES VIT C DEPENDANT.
- Glycosylation of HYL and ASN.
- Alignment of pro alpha chains – S-S bonds for to stabilise the helix.
- Golgi – More GLYCOLYLATION
EXTRACELLULAR:
- SECRETION.
- Cleavage of terminal pro-peptides – PRO-COLLAGEN PEPTIDASES.
- This forms TROPOGOLLAGEN.
- AGGREGATION – ¾ staggered array.
- Formation of intermolecular cross links – LYSYL OXIDASE.
- FIBRIL – further aggregation to form…
- FIBRES

Question 22

The Periodontal Ligament has three main components; Cellular, Nerves and Vasculature and Extracellular
Matrix (ECM).

What would happen to Collagen in the absence of Vitamin C?

Answer 22

In the absence of Vitamin C the hydroxyproline will not be formed as the enzymes required are dependent on
vitamin C. This leads to vascular fragility and connective tissue pathology – SCURVY.

Question 23

Describe the three principles of radiation protection.

Answer 23

• Justification – Any exposure should bring about a positive net benefit.
• Optimisation – Technique should be optimised to keep dose as low as possible. Quality should be assured
  so you don’t have to repeat the film.
• Limitation – NO dose limit for patients, only for staff and the public (people in the waiting room are classed as the general public so they have a dose limit).

Question 24

Give the ideal positions of the film, object and x-ray source needed to achieve the best image. You may
use diagrams to help you.

Answer 24

• Film as close to the object as possible.
• Film parallel to object.
• Film beyond tooth apexes.
• X-ray source as far from object as possible.
• X-ray beam perpendicular to film.

Question 25

Which technique is most commonly used to perform periapical radiographs? Describe the technique.

Answer 25

Paralleling technique.

• The film is held parallel to teeth by a holder with a bite block for the patient.
• Only in the lower molars, where the film can slot into the floor of the mouth, can the film be close to the
  teeth.
• A film size 2 is normally used for posteriorly (horizontally) and sizes 0-1 are used anteriorly (vertically).
• The x-ray beam should always be centered on the film holder and cotton wool is sometimes used
  between the bite block and opposing teeth in order to get the correct positioning.
• It is also important to move the lips down over the teeth in order to avoid lip shadow on the radiograph.

Question 26

Describe a disadvantage of the Paralleling technique for performing periapical radiographs.

Answer 26

Uncomfortable for some patients / Careful and accurate placement is required / Film holders are expensive and they have to be sterilized / cannot use a short cone / it takes longer to perform a whole mouth survey / Smaller films used anteriorly

Question 27

List what is recorded in a patient’s periodontal history.

Answer 27

• Complains of/Reasons for attendance (why referred if necessary). C/O
• History of presenting complaint. HPC
• Past dental history. PDH
• Social history.
• Medical history.

Question 28

When taking a patient medical history there are certain conditions which should be given greater
consideration. Give examples and explain why.

Answer 28

• ANGINA – Drugs for angina are gylceryl trinitrate which increases the risk of heart attack.
• TYPE 1 DIABETES – Increased risk of hypoglycaemia.

Question 29

Give examples of notes which would be made for the following sections in a patient periodontal status:
Probing depth, oral hygiene, recession, mobility.

Answer 29

• Probing depth – Note teeth with > 6mm.
• Oral hygiene – Good, fair, poor. CALCULUS
• Recession – Which teeth?
• Mobility – Which teeth?

Must record all radiographic assessments as well, following this you must make a diagnosis which could be
any number of conditions (not just periodontally related).

Question 30

NAD’ indicates what on a patient’s periodontal status?

Answer 30

NAD – Nothing Abnormal Diagnosed

Question 31

Describe and explain the purposes of good communication between patient and dentist in Dentistry.

Answer 31

Purposes of communication:

• Reassurance
• Alleviate distress
• Make decisions
• Give and gather information
• Form and maintain relationships
• Convey feelings
• Persuade
• Solve problems

Why is good communication important?

• To make accurate and appropriate diagnosis.
• Detect emotional distress in patients and respond accordingly.
• Increase patient satisfaction with care provided.
• Increase compliance with advice and instructions.

Putting the patient at ease:

• Comfortable setting.
• Being told where to sit.
• Handshake greeting.
• Introduce yourself and explain procedure.
• Easy first question.
• Appear interested.

Question 32

Describe the advantages and disadvantages of asking open questions to the patient.

Answer 32

Advantages:

• Open – More info can be obtained, patient feels more involved and can express concerns and
  anxieties.
• Closed – Quick and precise.

Disadvantages:

• Open – Can take too long, hard to control and not all info may be relevant.
• Closed – Info is restricted by questions, the interview is controlled by the interviewer and the patient
  has little opportunity to express concerns and feelings.

Question 33

Radiation protection

Upon which four main details does patient dose optimisation depend on?

Answer 33

• Film + processing.
• X-ray equipment.
• Technique.
• Quality assurance.

Question 34

Radiation protection

What is a classified person and what are their requirements?

Answer 34

Those workers who are likely to receive doses greater than:
- 6 mSv effective dose.
- 30% of any relevant limit on equivalent dose.
- Classification is unlikely to be necessary for dental x-ray work.
Requirements
- Personal dose assessment.
- Medical surveillance.
- Dose records kept for 50 years.
- Dose summary sent annually to HSE.

Question 35

Describe and explain the benefits and disadvantages of direct and indirect radiography.

Answer 35

A conventional radiograph is an x-ray film which is exposed to light or x-rays. Light sensitive salts in the emulsion record the image. The image becomes visible to the naked eye following processing.
A digital image is also produced by a passage of x-rays through the patient but the receptor is different. It may be a phosphate plate or a CCD/CMOS which stores the image on a computer (it can then be viewed on the monitor or printed).

Disadvantages of film:

• Long processing time.
• Variable image quality.
• Larger radiation dose.
• Space taken up with storage.

Advantages of direct digital radiology:

• Instant imaging.
• Low patient doses (90% dose reduction).
• Image processing/ enhancement/ archiving.

Disadvantages of direct digital radiology:

• Requires connecting wire.
• Physically large detectors compared to film.
• Low doses but narrow exposure latitude.
• Small detector area.

Disadvantages of digital cephalometry:

• High cost initially.
• Unknown life expectancy of sensor – phosphor plates get damaged quickly.
• Special training required for use.
• Must be close to computer – wiring.

Advantages of digital:

• Consistent image quality, no processing errors.
• Immediate image viewing.
• Elimination of dark room costs.
• Elimination of dark room mess.
• Improved detection of lesions.

Question 36

Discuss the structure and function of immunoglobulin IgG.

Answer 36

• All antibodies have the same basic Y-shaped structure comprised of four polypeptide chains, two heavy chains (50-70k) and two light chains (25k).
• One end acts as the recognition and binding site for antigens (amino terminal) and the other acts to instigate the effector function (biological activities – carboxyl terminal).
• The chains are joined by covalent inter-chain disulphide bonds.
• The humoral response is mainly where immunoglobulins recognise invading pathogens as foreign particles. They bind to them and bring about their eliminations.
• An antigen is any molecule that can specifically bind to an immunoglobulin generating an immune response. Hence antigens are sometimes called antibody generators.
• Each antibody is made up of globular domains. At the amino terminal the light and heavy chains have a variable domain. Other than these, all other domains are fairly constant. The light chains tend to only have 2 domains whereas the heavy chains can have up to 6 domains.
• The flexible region is located between CH1 (constant domain 1) and CH2 (constant domain 2) of the heavy chain. This permits maximum interactions to occur between immunoglobulins and their antigens.
• Light chains have two Isoforms, k and ʎ, but never a mixture of both.
• IgG has 4 H chain domains. It is monomeric with 2-3% CHO. IgG is a major immunoglobulin in serum and is produced in large amounts in response to an antigenic challenge.
• It activates compliment, binds to macrophages and granulocytes, crosses placenta in man: Passive immunity of new-born.
• IgG has 4 subclasses.

Question 37

Describe and explain the Bremsstrahlung and Characteristic Radiation theories of x-ray production.

Answer 37

To understand the mechanisms first we need to understand the atom.

Nucleus with neutrons and protons (+ve)

Surrounded by orbiting electrons (-ve)

Orbits fill from inside out, 2 in innermost, 8 on next out, 8 next

Bremsstrahlung is produced in an electron-nucleus interaction.

Whenever an electron changes direction or speed (sharply) it “screeches” x-rays. The electron loses energy in this process.

There are many possible paths for the incoming electron around the nucleus and this results in x-rays with a range of energies, as the incoming electron can lose any amount up to its full energy.

The incoming electron has high energy and slows down as it is diverted from its original path. The energy lost by the electron has to transfer somewhere else and this is emitted as x-ray photons.

Use of Bremsstrahlung curve – notation – it is a function of the energy of incoming electrons – the maximum photon energy is the total energy of the incoming electron.

Characteristic X-rays

Electron-electron interaction

An incoming high-energy electron collides with an orbital electron in an inner shell. Both electrons are ejected from the atom leaving a vacancy in an inner electron shell.

An electron from an outer shell loses energy (as an x-ray photon) and fills the gap in the inner shell.

Likewise, the vacancy left by this is filled by another electron falling into it by losing energy as an x-ray photo, and so on. The x-rays emitted by the electrons falling from one shell to another are of fixed energies, which are characteristic of the difference between the shells

The characteristic spectrum has two peaks at 59keV and 69keV characteristic of tungsten used as the target atom in dental x-ray tubes.

Question 38

Outline the functions of PG and GAG in connective tissue.

Answer 38

• PG’s control fibrillogenesis via their ability to bind to collagen fibrils.
• High GAG content is associated with unimodal distribution of small fibrils (50nm).
• DS (Decorin) is usually associated with large diameter fibrils (200nm).
• CS rich PG’s are usually associated with small fibrils.
• As connective tissues mature, total GAG content decreases and the proportions of CS:DS shift.
• CS predominates in foetal tissue; DS predominates in mature tissues.

PDL has small fibrils within ground substance, which is relatively CS rich (DS still predominates in healthy tissue).

Control of Mineralisation:

• GAGs can inhibit deposition of HAP crystals in soft connective tissues
• This may be due to their high anionic charge density binding mineral ions and/or their binding to collagen preventing nucleation on the fibrils (NB. GAGs can also promote mineralisation in other tissues)
• Removal of GAG from PDL leads to deposition of hydroxyapatite crystals in and around collagen fibrils

Generation of the Eruptive Force

• GAGs may be responsible, at least in part, for generation of the forces required for tooth eruption.
• This is related to their large hydrodynamic volume due to their high anionic charge density
• High concentrations of GAG are associated with increased eruption rates in rodents.
• Low concentrations of GAG are associated with reduced eruption rates in rodents.

Structure:

PDL contains mainly two small leucine rich proteoglycans, DECORIN and BIGLYCAN. Decorin is the major proteoglycan in the PDL. Decorin is dermatan sulphate rich.
Proteoglycans in the ground substance don’t exist individually. The GAG and core protein first bond covalently to each other. This forms a proteoglycan monomer. The proteoglycan then bonds with a link protein and a hyaluronan monomer non-covalently to form a proteoglycan aggregate.

A proteoglycan is a protein core with a glucosylaminoglycan (GAG). They are carbohydrate heteropolymers (copolymer) with repeated disaccharide monomers.

Two examples: Hexosamine (modified) and Uronic acid (modified). They are highly negatively charged, hydrophilic and have a high degree of hydration.

Question 39

Describe the association between gram-positive and gram-negative bacterial virulence factors on cell walls.

Answer 39

Virulence factors in different parts of the cell:

• Capsule – K-Antigens, adhesion, resistance to phagocytosis and complement.
• Fimbriae – Adhesion, gene transfer and receptors.
• Extra-cellular products – Nutrition, host damage, interaction with host cells.
• Cell wall – Gram positive cytokine induction, gram negative cytokine induction inflammation.
• Flagellum – Motility and H-Antigens.

GRAM POSITIVE:

• Thick peptidoglycan layer:
  o Induction of cytokines
• Teichoic/lipoteichoic acids:
  o Adhesion
  o Induction of cytokines
  o Resistance to host defence peptides (e.g. Defensins).

- Cell wall anchored proteins:
o Adhesion
o Evasion of host defences
o Induction of cytokines
o Host tissue destruction

GRAM NEGATIVE

• Thin peptidoglycan layer:
  o Induction of cytokines

- Lipopolysaccharides:
o O-Antigens
o Induction of cytokines and inflammatory response
o Resistance to phagocytosis
o Complement and other host responses
o Adhesion
o Initiation of mediation of bone resorption
o Killing macrophages

Question 40

List and describe the main functions of information explanation to patients.

Answer 40

• Provide information
• Simplify complexities
• Correct mistaken beliefs
• Give advice
• Aid patient compliance
• Highlight the important elements of any procedure
• Offer reassurance and reduce uncertainty
• Justify one’s actions and recommendations
• Increase patient satisfaction
• Ensure patient understanding
• Be realistic in the objectives you set
• Advice and instructions should be given early in the session, the most important information should be given first
• Use short words and sentences
• Avoid jargon plus don’t overload your patient (seven chucks is retention limit).
• Give information in a structured way – in a set of ‘chunks’
• Use visual aids, leaflets, posters
• Check that your patient understands your message and allow them to ask for clarification
• Be friendly, not officious – sufficient to establish rapport

DEMONSTRATE – I hear I forget, I see I remember, I do I understand.

Get patient to repeat advice – 30% increase in retention.

Patient recall is increased by signposting, summarizing, repetition, clarity and using diagrams.

Always leave time for the explanation to end properly.

Essential Features:

• Summarise what the patient has told you
• Ask them to check the accuracy of what you have said
• Ask them to repeat in their own words what advice or information they have been given
• Enquire if they would like to add anything
• End by thanking the patient.

Introducing yourself to a patient:

• Call your patient from the waiting room
• Say who you are
• Escort your patient onto the clinic
• Show your patient where to sit
• Position yourself in front of your patient
• Explain what you are going to do today
• Say who your tutor is
• Small talk

Question 41

Describe the procedural outline of the following periodontal indices:

Bleeding on Probing

Answer 41

Bleeding on probing is recorded following gentle insertion of the probe to the base of the pocket for measurement of pocket depths at 6 sites around each tooth (mesio-buccal, mid- buccal, disto-buccal, mesio-palatal, mid-palatal, disto-palatal).

Bleeding from the base of the pocket at the above sites is recorded on the periodontal chart. Ideally, record with a red pen. If using a four point measuring system, the deepest lingual/palatal site is noted at the correct location.

Question 42

Describe the procedural outline of the following periodontal indices:

Plaque Free Score

Answer 42

Disclosing the teeth

Apply a little petroleum jelly to the patient’s lips with a cotton wool roll. Give the patient a plaque-disclosing tablet and ask them to chew it and then swish it around the mouth for 30 seconds. Then ask the patient to rinse once. The plaque will be coloured and the patient is given a hand held mirror and shown the presence of the plaque.

The presence of plaque at the gingival margin is recorded for all teeth at 4 sites: mesio- buccal, buccal, disto-buccal, lingual/palatal on the Plaque Free and Marginal Bleeding Free Chart. The number of plaque free sites is expressed as a percentage of the total number of sites in the mouth to give a plaque free score. This allows the patient to get a higher score as the mouth becomes cleaner, indicating improved plaque control and tooth cleanliness.

Question 43

Describe the procedural outline of the following periodontal indices:

Marginal Bleeding

Answer 43

This is assessed by running a blunt periodontal probe such as the Hu Friedy PCP10 or Williams probe around the gingival margin and noting marginal gingival bleeding up to 20 seconds later. Bleeding indicates marginal gingival inflammation and therefore gingivitis of the gingival units. It is easier to record this index before disclosing the plaque

Question 44

Describe the procedural outline of the following periodontal indices:

Furcation Involvement

Answer 44

Lower molars should be probed from the buccal and lingual aspect to check for furcation involvement between mesial and distal roots. For upper molars, the trifurcation needs probing in three locations: buccally between the mesio-buccal and disto-buccal roots; mesially between the mesio-buccal and palatal roots; distally between the disto-buccal and palatal roots. The horizontal classification is routinely used. Special furcation probes facilitate measurement, but if not available, a graduated periodontal probe can be used.

Horizontal classification

. F1 = horizontal furcation measurement less than 3mm
. F2 = horizontal furcation measurement greater than 3mm
. F3 = horizontal through and through involvement.