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					MEQ Paper 2004b
Question 1
Creatinine clearance ([creatinine in urine] x urine flow rate)/[creatinine in plasma] = (4mmol/L x 2ml/min)/(0.5mmol/L) = 16ml/min

The difference between inulin and creatinine clearance is that creatinine is secreted to a small amount into the tubular fluid. The metabolic source of creatinine is skeletal muscle. Creatinine excretion is relatively constant from day to day in a normal subject due to the fact that it is proportional to muscle mass which stays relatively constant. Urinary creatinine measurement can be used as a measurement of glomerular filtration rate Mr Patel was unable to produce hypertonic urine due to his chronic renal failure where due to the proportion of damaged nephrons, the kidney is not able to reabsorb water as is normal. Water absorption in the nephron occurs in the proximal tubule, descending limb of the loop of Henle and the distal tubule and is linked to the movement of osmotically active substances such as glucose and electrolytes, for example sodium and chloride ions. These substances are moved from tubule to the interstitial fluid. The result of the increase in the osmolality of the interstitial fluid compared to the tubular lumen is water moving from tubule to interstitial fluid to balance osmolality both through the cells (transcellular) but also through tight junctions (paracellular).

Question 2

Fibrous capsule Major calyx Renal pelvis Renal column Cortex

Renal pyramid Renal papilla Renal artery Renal vein Ureter

Bowman’s capsule Glomerulus Afferent arteriole Renal artery Renal vein Proximal limb Descending thin limb of loop of Henle

Collecting duct Thick ascending limb of loop of Henle Thin ascending limb of loop of Henle

Question 3

Mr CF’s BMI is 90kg/(1.65m)2 = 33.1kg/m 2 (1dp) I would classify her as being obese as her BMI is greater than 30kg/m 2. Basal Metabolic Rate (BMR) 13.75L/hour x 20kJ/L = 275kJ/hour = 6600kJ/day = 6.6MJ/day Total Energy Expenditure (TEE)/BMR = (8MJ/day)/(6.6MJ/day) = 1.2 (1dp) 0.8 x 37kJ/g = 29.6kJ from triacylglycerides 0.05 x 17kJ/g = 0.85kJ from protein = 30.45kJ/g adipose tissue = 30450kJ/kg adipose tissue 8MJ/day – 6MJ/day = 2MJ/day = 14MJ/week = 14000kJ/week (14000kJ/week)/(30.45kJ/g)WRONG =459.8g/week (1dp) PAR x BMR = (2 x 3/24days) x 6.6MJ/day******wrong = 1.65MJ/day

Physical Activity Level (PAL)

Adipose tissue contains

Reduction in energy intake

Weight loss/week Additional energy expenditure from walking

Adipose tissue lost as result of walking

Additional energy expenditure/energy in adipose tissue = (11550kJ/week)/(30.45kJ/g) = 379.3g/week (1dp)

Question 4
The dark-brown urine is due to the presence of increased amounts of bilirubin in the urine which are normally excreted in the faeces while the pale, greasy looking stool is due to incomplete fat digestion causing excess fat to be excreted in the faeces and also the absence of bilirubin in the faeces. Two functions of the gallbladder are to store bile and to concentrate bile. The major differences in the composition of hepatic and gallbladder bile are that the bile salts present in the gallbladder may be concentrated as much as 20-fold and that there is a lower concentration of bicarbonate and chloride ions, while increased concentrations of sodium ions, cholesterol and lecithin. Hepatocytes secrete hepatic bile which is brought into the hepatocytes by cotransport with sodium. It then passes along bile ducts via the bile canaliculus where the ductal epithelial cells modify the primary secretion by secreting a watery bicarbonate-rich fluid. The rate at which this occurs depends on the rate at which bile salts are returned via the enterohepatic circulation. Three functions of bile are:  A route for cholesterol excretion from the body  A vehicle for the elimination of bile pigments and other waste products such as less polar molecules of high molecular weight  The formation of a stable emulsion from large fat globules so that fat can be digested by pancreatic lipases The enterohepatic circulation is the pathway by which bile salts are reabsorbed into the portal circulation by active transport in the distal ileum which is more energy efficient then synthesising new bile. Bile acids may be recycled up to 20 times before finally being excreted and the importance of this in GI function is that the amount of bile salt synthesis is reduced.

Question 5

Gallbladder Hepatic duct Portal vein Bare area Inferior vena cava

Quadrate lobe Ligamentum teres Fissure for ligamentum teres Hepatic artery Caudate lobe

Common hepatic duct Cystic duct Free margin of lesser omentum Fundus of gallbladder

Left hepatic duct Left portal vein Hepatic artery Common bile duct Pancreatic duct Duodenal papilla

Question 6

Mr AC is suffering from angina which is due to a narrowing of his coronary arteries so that oxygen demand of the cardiac myocytes cannot be met in times of exertion.

Question 7

The overall pattern of acid-base disturbance is metabolic acidosis characterised by a large fall in plasma bicarbonate concentration. The role of the kidneys in acid-base balance is to maintain plasma bicarbonate levels. The role of the respiratory system in acid-base balance is to regulate arterial PCO 2 so the low arterial PCO2 of the patient is due to the response of the respiratory system to the low plasma bicarbonate. In order to maintain a normal pH of 7.4, if the plasma bicarbonate drops, so must the arterial PCO 2 to compensate. A way of reducing arterial PCO2 is to excrete more via the lungs and more specifically through an increased respiration rate. Potassium levels are elevated due to there being low plasma bicarbonate so less is exerted in the urine therefore in order to retain the electroneutrality of the urine, less potassium is secreted into the tubular fluid so there is an increase in the plasma. The low bicarbonate is maybe due to the fact that Mr Brown has just suffered a myocardial infarction which could cause tissue hypoxia and a release of lactic acid into the circulation. in K+ due to large amounts in cell, cell die K+ released????

Question 8

Ketone bodies comprise acetone, acetoacetate and β-hydroxybutyrate and are produced in the liver normally in fasting to provide metabolic fuel to other tissues in the body including muscles. Ketone bodies are produced in large amounts during fasting STARVING in a normal individual. The substrate used for the synthesis of ketone bodies is acetyl CoA. The liver synthesises ketone bodies. Three tissues that use ketone bodies are heart muscle, the brain and skeletal muscle. One tissue that can never use ketone bodies as a metabolic fuel are red blood cells. Two ways in which Marlene’s ketotic state might be easily detected would be the ‘fruity’ smell of his breath due to acetone and also the detection of ketone bodies in the urine. The blood concentration of ketone bodies is high due to Marlene’s diabetes. Even though she is hyperglycaemic, she is unable to utilise the glucose due to a lack of insulin so in response the body reacts as if in the fasting state causing gluconeogenesis and releasing non esterified fatty acids which in the liver are synthesised to ketone bodies and subsequently released into the bloodstream increasing plasma levels of ketone bodies. Marlene has glucose in her urine due to the high plasma concentration. This means more glucose enters the nephron as part of the glomerular filtrate and the glucose transporters in the proximal tubule even working at their maximum rate are unable to reabsorb all the glucose in the filtrate so unlike in a normal situation, glucose remains in the filtrate after the proximal tubule and is so excreted in the urine.

Question 9

Randomisation is where subjects chosen for a clinical trial are assigned to one or other of the alternative treatment groups to be compared which is normally achieved by using random number tables or a computer selection program. Therefore, the patient has the same probability as any other patient of being assigned to a particular group in order to make the treatment groups as alike in every respect but the treatment they receive. Thus any differences in outcome can be attributed to the treatment and not a characteristic of the groups, although sometimes the groups can differ in their ‘baseline characteristics’. The number needed to treat is the number intervention to prevent one outcome. Simvastatin event rate Control group event rate Absolute risk reduction (ARR) of patients needing to receive the

9% 13% 13% - 9% = 9% Number needed to treat (NNT) 100/ARR = 100/4 = 25% For this trial 25 patients would need to be treated with simvastatin to prevent one major coronary event. 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.

Question 10

Over the last 30 years, more and more people can expect to live in into retirement and this number as a proportion of the total population is expected to rise. Older people can also expect to lead a similar life to those experienced by other people in the population which is known as the squaring of the rectangle of survival, even though there is a connection between old age and physical disability and acute health problems. Furthermore, it is more likely now than 30 years ago that those who are retired and so considered old have money to spend due to pensioners’ income growing at a faster rate than other sectors of the population although this varies between different groups of pensioners. For example, couples will receive twice as much as single pensioners while the rises in occupational pensions have been much greater than those of state pensions. Furthermore, there are differing theories to account for the social position of older people:  Disengagement theory – older people ‘disengaged’ from society so younger people could take over  Structured dependency theory – the dependency of older people is structured by society so decisions made the government have a dramatic effect on how older people live their lives  Third ageism – the proportion of time spent in retirement is increasing all the time which provides possibilities for undertaking self-enriching activities that there was not enough time for when occupied with working or raising children

Question 11

In order to take a medical history from Ms SD, I would:  Explain who I was  The nature of the interview  Get her permission  Ask if she has any queries and that it is ok for her to stop the interview at any point  Begin the information gathering by asking her open questions to gain a general oversight into her medical history  Ask her to clarify any points that she raised  Ask her closed questions to gather any specific evidence that has not already been gathered  Close the interview by telling her I had finished asking all that I wanted to ask  Thanking for her time To examine Ms SD’s respiratory system, I would:  Introduce myself  Explain what I am going to do  Get permission  Ask her to clarify any points of the examination that she is unclear about  Ask patient to expose the chest which may be a source of misunderstanding as she may not know how much of the chest area to expose which could be solved by clarifying that her chest area needs to exposed but that she can leave any undergarments on  When asking her to breathe in and out, I could resolve any misunderstanding by demonstrating the procedure myself  Tell Ms SD that I have finished and tell her that she can get dressed  Ask Ms SD if she has any further questions  Thank the patient

Question 12

Five independent factors that would decrease the abilities of the lung to transport oxygen from the alveoli to the venous blood:  Ventilation/perfusion inequality  Decrease in diffusing capacity of the alveoli  Decrease in surface area available for gas exchange  Difference in partial pressure of oxygen between the alveoli and lung capillaries  Heart failure  FiO2  Thickness of the barrier between alveoli and capillary In a standing subject, pulmonary blood flow is greatest at the bottom of the lung. During inspiration the inspired air goes preferentially to alveoli at the bottom of the lungs. This is due to the fact that alveoli at the apex of the lung have smaller compliance than those at the base. Due to these regional variations in ventilation and blood flow the Ventilation/Perfusion ratio is normally greatest at the top of the lung. If the blood flow to a group of alveoli was blocked by a small clot of blood, the transfer factor (diffusing capacity) for the lung will be decreased. These alveoli will have a higher than normal Ventilation/Perfusion ratio. The lung has intrinsic mechanisms to help optimise V/Q relations. Blood flow to alveoli which are underventilated will decrease and this is due to the low alveolar PO 2 acting on the local pulmonary arterioles to produce vasoconstiction. Ventilation to alveoli which are overventilated will decrease due to the low alveolar PCO acting on 2 the local bronchial smooth muscle to constrict the airways.

Question 13

Health promotion is the improvement of health through environmental and behavioural changes brought about by specific interventions by health education. This covers prevention, health education and public health policy and is primarily concerned with the attempt to reduce risk factors associated with disease to bring about a reduction in the social and economic burden of disease. This consists of primary, secondary and tertiary prevention. Health education is one of the means by which health promotion can be achieved to provide advice in how to prevent or live with disease states. A GP’s role in health education may be:  Through the advice given to patients regarding specific diseases and how to minimise chances of being diagnosed with the disease, for example bowel cancer  Through the advice given to patients regarding their health in general to minimise of morbidity, for example advising people to ea five portions of fruit and vegetables a day  Relaying public policy laid down by the government to patients and their families  Giving talks to various groups of individuals about health in general or more specific topics  Holding workshops whereby patients can ask questions relating to their health and healthcare Four other primary healthcare workers involved in health promotion:  Nurses: educating about various topics related to the patients that they visit, for example wound care management  Pharmacists: giving advice to patients regarding how to take their medications

 

Dentists: giving advice as to how to maintain healthy teeth Opticians: through advice on how to maintain vision

Question 14
Agglutination is the sticking together of target cells by antibodies through the antibodies binding to epitopes on the target cells. Coagulation is the formation of a clot which is due to the clotting cascade in response to the puncture of a blood vessel for example, forming an insoluble mesh. When a blood vessel is cut, platelets adhere to the site of puncture releasing clotting factors and attracting nearby platelets to the site of injury. The clotting factors initiate the clotting cascade which results in the formation of thrombin leading to fibrinogen becoming fibrin, enmeshing the platelet aggregates and forming a stable plug which is a thrombus. Blood normally remains fluid by the release of coagulation factor inhibitors such as tissue factor pathway inhibitor (TFPI) and antithrombin.

Question 15

A cause of the diarrhoea and vomiting may be food poisoning due to the Salmonella bacterium.

Question 16

Three classes of anti-retroviral therapy to treat AIDS are:  Nucleoside reverse transcriptase inhibitors  Non-nucleoside transcriptase inhibitors  Protease inhibitors An example of an anti-retroviral drug is Zidovudine. Six recognised risk groups associated with HIV infection:  Male homosexuals  Injecting drug users  Haemophiliacs  Infants born to infect mothers  Promiscuous heterosexual couples  Healthcare workers Other     consequences experienced by the patient due to HIV infection: Diarrhoea Oral ulceration Tuberculosis Pneumonia

The major class of lymphocyte that declines during HIV infection are CD4 lymphocytes due to HIV gaining entry into them through the CD4 molecule present on the cell surface and then replicating inside them before emerging from them, killing the CD4 lymphocyte. Resistance of HIV to anti-retroviral drugs is monitored in the laboratory by viral genotyping of plasma RNA determining the genomic basis of resistance.

Question 17

Mrs AP may be vitamin B12 deficient as it is primarily found in meat, eggs and dairy products which are non-vegetarian. Vitamin B12 is absorbed in the ileum when bound to intrinsic factor which is secreted by the parietal cells of the gastric mucosa. The resulting complex is recognised by a

receptor protein in the brush-border membrane of ileal cells to which the complex binds before slowly entering the cell and eventually the blood where it is bound to transcobalamn II. Mrs AP may be iron deficient as it is found mostly in meat which is not eaten by her. Iron can only be absorbed bound to haem or free in the ferrous form (Fe 2+). The low pH of the stomach lumen solubilises the iron salts from the Fe3+ form to the Fe2+ form which can also be carried out by vitamin C by acting as a reducing agent. In the upper intestine, iron is taken up by a carrier mediated process with a hydrogen ion into the enterocyte where the iron is then either stored in the cell as ferritin or transferred to the bloodstream where it is bound to trans-ferrin via another transport protein. The absorption of dietary iron is limited due to the negative effects of iron overload which is toxic but can take place in the genetic disease haemochromatosis. Women are more likely to suffer from iron deficiency than men due to blood loss from menstruation. Iron tablets may be prescribed with vitamin C supplements because it promotes iron absorption and bioavailability by reducing Fe3+ to Fe2+, the form in which it can be absorbed. Microcytic means that the smaller than is usual Hypochromic means that the cells are not as pigmented as is usual It is more likely that Mrs AP is iron deficient.

Question 18
Arterial blood pressure = cardiac output x total peripheral resistance Arterial blood pressure Cardiac output Total peripheral resistance 130mmHg 5L/min 26mmHg/L/min

The normal heart beat is initiated by cells of the sinu-atrial node. Sympathetic effect: Transmitter: sympathetic preganglionic nerves from spinal segments T1-T6 Hormone: noradrenaline Receptor: β-adrenergic Noradrenaline increases the rate of discharge from the cells of the SA node by increasing the permeability of the pacemaker cells to sodium and chloride ions increasing the slope of the pacemaker potential so that the cells reach threshold more quickly and the interval between successive actions potentials is reduced. Parasympathetic effect: Transmitter: Vagus nerve Hormone: acetylcholine Receptor: muscarinic cholinergic Acetylcholine increases the permeability of the nodal cell to potassium, hyperpolarising the membrane potential and decreasing the slope of the pacemaker potential. Therefore, there is an increase in the time taken for the pacemaker potential to reach threshold and the interval between successive action potentials is longer and the heart rate falls.

Increased inotropy – sympathetic stimulation Decreased inotropy – following myocardial infarction

Question 19

Autosomal refers to any of the chromosomes except the sex chromosomes. Recessive refers to the pattern of inheritance where a condition only appears if an individual has two copies of a mutant gene. A carrier of an autosomal recessive genetic disease is an individual who has one copy of the normal gene and one copy of an the autosomal gene so does not display symptoms of that disease. The probability that Marcel’s sister is a carrier is 0.5 C c C CC Cc c Cc cc The probability that Marcel’s mother is a carrier is 1 The probability that Marcel’s father is a carrier is 1 A hydrophobic α-helix is a protein which is does not combine with water where the peptide backbone of the protein adopts a spiral form with hydrogen bonds being formed between peptide bonds which are near to each other in the primary structure. Four hydrophobic amino acids are: Valine Alanine Leucine Isoleucine Three hydrophilic amino acids are: Aspartate Arginine Asparagine

Question 20

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. There is an increase in the plasma concentration of lactate due to the exercising muscle not receiving adequate amounts of oxygen so resorting to anaerobic metabolism to provide energy of which lactic acid is a product. 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.

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