# MEQ 2003b.doc by sacardozo

VIEWS: 71 PAGES: 9

• pg 1
```									MEQ Paper 2003b
Question 1
Local mechanisms are most important in regulating coronary vascular resistance. Left ventricular coronary blood flow takes place during systole because this is the point where back flow of blood through the aortic sinuses takes place. In coronary heart disease, the regulation of blood flow is impaired by the narrowing of the lumen of the coronary artery so therefore normal regulation does not have such a great effect. If blood flow through the right coronary artery is impaired then the patient will be susceptible to irregular heartbeats. If blood flow through the left coronary artery is impaired, then the left ventricle’s ability to pump adequate amounts of blood through the systemic circulation will be diminished due to inadequate oxygen supply leading to heart failure. The possible outcome of heart failure is death.

Question 2

Mr Elks’ BMI is 87kg/(1.67m)2 = 31.2kg/m 2 (1dp) I would classify him as being obese as his BMI is greater than 30kg/m 2. Physical Activity Level (PAL) Total Energy Expenditure (TEE)/BMR = (12MJ/day)/(8.4MJ/day) = 1.4 (1dp) 12MJ/day – 10MJ/day = 2MJ/day = 14MJ/week = 14000kJ/week 0.8 x 37kJ/g = 29.6kJ from triacylglycerides 0.05 x 17kJ/g = 0.85kJ from protein = 30.45kJ/g adipose tissue (14000kJ/week)/(30.45kJ/g) =459.8g/week (1dp) PAR x BMR = (3 x 1/12days) x 8.4MJ/day = 2.1MJ/day Additional energy expenditure/energy in adipose tissue = (14700kJ/week)/(30.45kJ/g) = 482.8g/week (1dp)

Reduction in energy intake

1g adipose tissue contains Weight loss/week Additional energy expenditure from walking Adipose tissue lost as result of walking

Question 3

In diagram 1, area B is usually supplied by the left coronary artery. In diagram 2, area J is usually supplied by the right coronary artery. Left coronary artery = E Branch of the right coronary artery to the sinuatrial node = G Posterior interventricular artery = I Circumflex artery = D You should palpate the apex beat of a normal heart in the 5 th intercostals space. You would auscultate the aortic valve in the 2nd right intercostals space.

Question 4

To examine Mr Elks’ chest, I would:  Introduce myself  Explain what I am going to do  Get permission  Wash my hands  Get patient into the right position – lying with chest at 45° angle  Ask patient to expose the chest  Inspect the patient for asymmetry or deformity of the chest, scars, movement of the chest  Count the respiratory rate  Feel for the apex beat  Check chest expansion  Percuss the chest comparing side to side and listening for changes in resonance  Auscultate all areas of the chest listening for vesicular or bronchial breath sounds, wheeze, crackles or rubs  Test for vocal resonance  Tell Mr Burns that I have finished and tell him that he can get dressed  Thank the patient To measure Mr Elks’ peak expiratory flow rate:  Introduce myself  Explain what I am going to do  Get permission

       

Wash hands Place a clean mouthpiece into the peak flow meter Ensure that the pointer is at zero Ask Mr Elks to stand up Tell Mr Elks to take a large breath in through his mouth and then to make a tight seal around the mouthpiece and breath out as quickly and forcefully as possible Ask Mr Elks reposition the pointer and repeat procedure a further two times Take the highest reading of the three as the patient’s peak expiratory flow rate Thank the patient

Question 5

Patient Patient Patient Patient Patient Patient Patient Patient Patient Patient Patient Patient Patient

B is normal C is normal A has an obstructive disease B has an obstructive disease C has an obstructive disease A has a restrictive disease C has a restrictive disease A might have bronchitis A might have fibrosis of the lungs B might have fibrosis of the lungs B might have asthma C might have emphysema C might have fibrosis of the lungs

False True False True False True False False True False True False False

I would expect patient B to derive benefit from inhalation of a bronchodilator drug such as salbutamol.

Question 6

Factors that may influence young peoples smoking behaviour:  Peer pressure  Family  Role models i.e. celebrities  Urge to rebel  Advertising Health promotion strategies that can be implemented to reduce rates of smoking among adolescents:  Education about risks – school workshops to teach the consequences and effects of smoking on individuals and those around them  Advertising campaigns – to promote healthy living in general with no smoking as a theme  Free advice and support to those who wish to quit – for example offering free nicotine replacement therapy Agencies outside the NHS that have an important role to play in reducing adolescent smoking are:  Schools

 

Question 7
Double blind means that neither the patient nor the doctor is aware of which treatment the patient is receiving. Therefore, this reduces bias as if the doctor or patient is aware which treatment is being prescribed, they may exaggerate the results. An intention to treat analysis specifies how to handle noncompliant patients in a randomized control trial and requires that patients be analysed in the groups they were randomized into, regardless of whether they complied with the treatment they were given. It is important so that the baseline characteristics of the control and intervention group stay similar so that comparison can take place. The number needed to treat is the number intervention to prevent one outcome. % people on nicotine patches that gave up % people on placebo that gave up of patients needing to receive the

(130/260) x 100 = 50% (60/240) x 100 = 25% Absolute risk reduction (ARR) 50% - 25% = 25% Number needed to treat (NNT) 100/ARR = 100/25 =4 For this trial four patients would need to be treated with nicotine patches to prevent one person continuing to smoke.

Question 8
Very low density lipoprotein (VLDL) is formed in the liver and contains newly synthesised triacylglycerol, cholesterol, cholesteryl esters, phospholipids and lipids from chylomicron remnants and transports them to peripheral tissues which have lipoprotein lipase, phospholipase and cholesterol esterase. Low density lipoprotein (LDL) is formed from the loss of lipids from very low density lipoprotein (VLDL) and carries cholesterol to the peripheral tissues. High density lipoprotein (HDL) takes up surplus cholesterol from peripheral tissues and returns it to the liver for catabolism. Cholesterol is synthesised in the liver from acetoacetyl CoA and acetyl CoA so as long as there is adequate amounts of these substrates, cholesterol synthesis can occur. The administration of an HMG CoA reductase inhibitor would lead to a reduction in plasma LDL concentration due to the synthesis of cholesterol in the liver being inhibited leading to a reduction in cholesterol concentration in the liver so promoting LDL receptor synthesis and the clearance of LDL from the plasma.

Question 9 Question 10 Question 11
The gastric cells that produce hydrochloric acid are parietal cells.

Parietal cells are found in the fundus, body and antrum of the stomach situated in gastric glands. Three functions of gastric acid:  Helps in the breakdown of connective tissue and muscle fibres of ingested meat  Activation of pepsinogen  Provision of the optimal conditions for the activity of pepsins

In the parietal cell, hydrogen ions are derived from the dissociation of water with the cell and are then actively pumped into the gastric lumen against a concentration gradient in exchange for potassium ions. Hydroxyl ions are also produced as a result of the dissociation of water and combine with carbonic acid to form bicarbonate, a reaction that is catalysed by carbonic anhydrase, which leaves the cell at the basolateral membrane in exchange for chloride ions. This secretion of bicarbonate ions into the blood means that the venous blood of the stomach is more alkaline than the arterial blood which is known as the alkaline tide. The chloride ions brought into the parietal cell then leave for the gastric lumen via co-transport with potassium or through the chloride channel on the canalicular membrane.

Question 12
Efferent fibres whose cell bodies lie within the dorsal motor nuclei of the vagus nerve stimulate the release of gastric acid by releasing acetylcholine from the postganglionc fibres at the myenteric plexus stimulating the output of the gastric glands. Also, vagal stimulation stimulates the release of gastrin which reaches the gastric glands through the bloodstream and stimulates them to secrete gastric acid. Furthermore, both acetylcholine and gastrin stimulate histamine activity which acts on parietal cells via H2 receptors to stimulate hydrogen ion secretion. Habitual aspirin ingestion can cause peptic ulceration by interfering with the production of certain prostaglandins, which especially those of the E series, increase the thickness of the mucus gel layer, stimulate the production of bicarbonate and cause vasodilation of the microvasculature thus improving the supply of nutrients to any damaged areas of mucosa. Processes which reduce gastric acid secretion are:  Inhibition of gastrin secretion when the pH of gastric contents reaches pH2-3 mediated by the release of somatostatin  Secretion of secretin by the duodenal mucosa in response to acid inhibits the release of gastrin as well as reducing the parietal cells’ sensitivity to gastrin The gastric mucosa is normally protected from attack by gastric acid by the mucosal barrier which is made up of tight junctions between the cells of the mucosal

epithelium to prevent gastric juice from leaking to the underlying tissue. There is mucus secreted by the surface epithelial cells and neck cells of the gastric gland which is alkaline due to its contents of bicarbonate and potassium ions and also there are prostaglandins which especially those of the E series, increase the thickness of the mucus gel layer, stimulate the production of bicarbonate and cause vasodilation of the microvasculature thus improving the supply of nutrients to any damaged areas of mucosa. Reasons why patients with peptic or duodenal ulceration are usually anaemic are: Increased blood loss due to the ulcer bleeding Damage to epithelium inhibits absorption of digestion products including those required for red blood cell formation, for example iron and vitamin B12.

Question 13
An irreversible enzyme inhibitor is a substance that causes chemical modification of the enzyme protein. It does this by binding covalently to the enzyme and cannot be removed. Many are substrate analogues that undergo part of the reaction then form a transition state covalent intermediate that does not break down to yield product. The effect of the inhibitor is prolonged and only diminishes gradually as the enzyme protein is catabolised and replaced. To test if an inhibitor is reversible or irreversible, place together enzyme and inhibitor in a solution within a semi-permeable membrane that will allow the inhibitor to diffuse through it. After a period of time, withdraw a sample and add substrate to the sample. If the inhibitor was reversible, there should be enzyme activity whereas if the inhibitor was irreversible, there would not be enzyme activity. Therefore, with a reversible inhibitor, there will be product formation, but with an irreversible inhibitor, there will be significantly reduced/no product formation. This happens because an irreversible enzyme permanently inactivates the enzyme, while a reversible inhibitor does not. Advantages to a patient taking an irreversible inhibitor instead of a reversible inhibitor would be that as the enzyme would be inactivated permanently, the dose of inhibitor would be less frequent than the dose required of a reversible inhibitor, depending on how fast new enzyme protein is synthesised. The possible disadvantages to using an irreversible inhibitor as a drug would be difficultly in adjusting the dose of an irreversible inhibitor to match the patient’s needs due to the long duration of action. The patient has to take regular doses of an irreversible enzyme inhibitor due to the fact that there is continuous protein synthesis which includes synthesis of the target enzyme so the patient needs to take regular doses to ensure continued inhibition.

Question 14 Question 15
Hypertonicity is having a higher osmotic pressure than a comparison solution. Plasma osmolality: 280-290mosmol/kg Urine osmolality: 100-1250mosmol/kg Regions of the nephron involved in the production of hypertonic urine are the proximal tubule, descending loop of Henle and distal tubule. Tubular processes in the nephron involved in the production of hypertonic urine include:

   

Absorption of solutes and electrolytes in the proximal tubule allowing water to be reabsorbed via osmosis Water movement from descending limb of loop of Henlé to interstitial fluid Absorption of solutes and electrolytes in the distal tubule allowing water to be reabsorbed via osmosis Absorption of water in the collecting duct dependent on the number of water channels due to the action of ADH

Approximate normal water content of a 70kg male is 56% which is 39.2kg. Plasma Intracellular fluid Extracellular fluid 7% 67% 24%

You could determine plasma volume by injecting Peter with a known quantity of Evans Blue, a solution that does not leave the blood vessels to any significant extent for 2-3 hours. After 20 minutes, take a blood sample and determine the concentration of Evans Blue in the blood sample then divide the concentration found by the initial amount injected.

Question 16

Anti-diuretic hormone (ADH) is important in minimising Peter’s excretion of water. It is secreted by the posterior pituitary gland in response to increased plasma osmolality and is regulated by osmoreceptors located in the hypothalamus. An increase of plasma osmolality of 3mosmol/kg from the normal 280-290mosmol/kg is enough to stimulate ADH secretion. ADH increases the permeability of the last third of the distal tubule and the whole of the collecting duct to water resulting in movement of water down its osmotic gradient into the tubular cells and interstitial fluid and plasma. This movement of water is independent of solute uptake and so results in an increase of urine osmolality and decrease in plasma osmolality. Fast acting hormones work by:  Interacting with cell surface receptors which behave as allosteric proteins in that specific binding of hormone results in a conformational change in the receptor which initiates subsequent signalling events  Signalling mechanisms include G-protein coupled systems and receptor tyrosine kinase mediated systems  G-protein coupled systems work by the G-protein dissociating after interaction with a hormone which then activates or inhibits an effector  Receptor tyrosine kinase mediated systems work by the activation of protein tyrosine kinase activity on binding of the hormone leading to phosphorylation of cytosolic proteins  Protein phosphorylation and de-phosphorylation of the signalling pathways  Amplification through protein phosphorylation cascades

Question 17

Question 18

Myoglobin and haemoglobin are both oxygen-binding proteins but myoglobin has a much higher affinity for oxygen than haemoglobin. Each myoglobin molecule contains one haem prosthetic group whereas haemoglobin contains four haem prosthetic groups. Also, when oxygen binds to the first subunit of deoxyhemoglobin it increases the affinity of the remaining subunits for oxygen and as additional oxygen is bound to the second and third subunits oxygen binding is further strengthened explaining the sigmoid shape of the haemoglobin curve. This is unlike the dissociation curve for myoglobin which is a hyperbolic shape due to there being only one oxygen molecule biding to one molecule of myoglobin so no influence on the affinity of oxygen binding to myoglobin. A decrease in plasma pH can be translated as an increase in plasma hydrogen ion concentration which decreases affinity of haemoglobin for oxygen thereby shifting the oxygen dissociation curve to the right.

Question 19

mRNA: RNA, synthesized from a DNA template during transcription, that mediates the transfer of genetic information from the cell nucleus to ribosomes in the cytoplasm, where it serves as a template for protein synthesis. tRNA: short-chain RNA molecules present in the cell (in at least 20 varieties, each variety capable of combining with a specific amino acid) that attach the correct amino acid to the protein chain that is being synthesized at the ribosome of the cell (according to directions coded in the mRNA). DNA replication: the process of copying a double-stranded DNA strand, prior to cell division (in eukaryotes, during the S phase of mitosis and meiosis) resulting in two identical double strands. Transcription: the process by which mRNA is synthesized from a DNA template resulting in the transfer of genetic information from the DNA molecule to mRNA. Translation: the process by which mRNA, tRNA, and ribosomes effect the production of a protein molecule from amino acids, the specificity of synthesis being controlled by the base sequences of the mRNA.

Ribosome: an organelle in the cytoplasm of a living cell which attach to mRNA and move down it one codon at a time and stop until tRNA brings the required amino acid. Polysome: a number of ribosomes attached to the same mRNA chain so translating and forming the same protein. Differences between bacterial and human protein synthesis include the size of the ribosome subunits so the antibiotic only recognises bacterial ribosomes and not human ribosomes. This can cause inhibit the formation of an initiation complex with mRNA or cause misreading of the mRNA message, leading to the production of nonsense peptides. Furthermore, binding to the bacterial ribosome can inhibit certain enzymes required for protein synthesis to take place

Question 20

Primary antibody deficiency is consistent with the clinical and laboratory findings. In a healthy baby, the B-lymphocytes which are deficient in this baby are made in the bone marrow. The baby had IgG molecules in her body but not IgM molecules because IgG molecules are transported in colostrum if the baby had been breast fed. Antibodies have four polypeptide units which consist of two light chains and two heavy chains. The light chains can either by κ or λ but not both and there are five variants of the heavy chain in correspondence with the five groups of antibodies (IgG, IgM, IgE, IgA, IgD). The N terminal of the heavy chain and all of the light chain is known as the Fab region and makes up the antigen binding site which contains the hypervariable region made up of minigenes. The C terminal of the heavy chains makes up the Fc region that among other things, binds with complement and macrophages.

Forces     

which contribute to the binding of a specific antibody and its antigen include: The relative concentration of antigen and antibody Complementary shapes Hydrophobic interactions Van der Waal forces Hydrogen bonding

```
To top