Body fluid compartments in a 70kg male:
Total volume=42L (60% body weight)
Intracellular fluid compartment (ICF) =28L
Extracellular fluid compartment (ECF) = 14L

ECF comprises:
Intravascular fluid (plasma) = 3L
Extravascular fluid = 11L

Extravascular fluids comprises:
Interstitial fluid = 10.5L
Transcellular fluid = 0.5L

Bilirubin is the tetrapyrrole and a catabolic product of heme. 70-90% of bilirubin is end product of haemoglobin degradation in the liver.

Bilirubin circulates in the blood in 2 forms; unconjugated and conjugated bilirubin.

Unconjugated bilirubin is insoluble in water. It travels through the bloodstream to the liver, where it changes from insoluble into a soluble form (i.e.; unconjugated into conjugated form).

This conjugated bilirubin travels from the liver into the small intestine and the gut bacteria convert bilirubin into urobilinogen and then into urobilin (not urobilin to urobilinogen). A very small amount passes into the kidneys and is excreted in urine.

Glycolysis occurs in the cytoplasm of the cell. It converts 1 glucose molecule (6-carbon) to pyruvate (two 3-carbon molecules) and produces 4 ATP molecules and 2NADH but uses 2 ATP in the process with an overall net energy production of 2 ATP.

Pyruvate is then oxidised to acetyl coenzyme A (generating 2 NADH per pyruvate molecule). This takes place in the mitochondria and then enters the Krebs cycle (citric acid cycle). It produces 2 ATP, 8 NADH and 2 FADH2 per glucose molecule.

Electron transport phosphorylation takes place in the mitochondria. The aim of this process is to break down NADH and FADH2 and also to pump H+ into the outer compartment of the mitochondria. It produces 32 ATP with an overall net production of 36ATP.

In anaerobic respiration which occurs in the cytoplasm, pyruvate is reduced to NAD producing 2 ATP.

The patient in the case has a fluid volume requirement of 30 mL/kg/day. His basic electrolyte requirement per day is:

Sodium at 2 mmol/kg/day x 110 = 220 mmol/day
Potassium at 1 mmol/kg/day x 110 = 110 mmol/day

His energy requirement per day is:

35 kcal/kg/day x 110 kg = 3850 kcal/day

One gram of glucose in fluid can provide approximately 4 kilocalories.

The following are the electrolyte components of the different intravenous fluids:

Fluid Na (mmol/L) K (mmol/L)
0.9% Normal saline (NSS) 154 0
0.45% NSS + 5% dextrose 77 0
0.18% NSS + 4% dextrose 30 0
Hartmann’s 131 5
5% dextrose 0 0

1000 mL of 5% dextrose has 50 g of glucose

Option B is inadequate for his sodium and caloric requirements (30 mmol of Na+ and 560 kcal). It is adequate for his K+ requirement (120 mmol of K+).

Option C is in excess of his Na+ requirement (462 mmol of Na+). Moreover, it does not provide any K+ replacement.

Option D is inadequate for his caloric requirement (600 kcal) and K+ requirement (60 mmol of K+). Moreover it does not provide any Na+ replacement.

Option E is in excess of his Na+ requirement (393 mmol of Na+), and is inadequate for his potassium requirement (15 mmol of K+)

Option A has adequate amounts for his Na+ (231 mmol of Na+) and K+ (120 mmol of K+) requirements. It is inadequate for his caloric requirement (600 kcal).

In any alveolar unit, the V/Q ratio affects alveolar oxygen (PAO2) and carbon dioxide tension (PACO2).

The partial pressure of alveolar carbon dioxide (PACO2) is plotted against the partial pressure of alveolar oxygen in a Ventilation-Perfusion (V/Q) ratio graph (PAO2). Given a set of model assumptions, the curve represents all of the possible values for PACO2 and PAO2 that an individual alveolus could have.

In the case of an infinity V/Q ratio (ventilation but no perfusion or dead space), the PACO2 of the alveolus will equal zero, while the PAO2 will approach that of external air (150mmmHg). At the apex of the lung, the V/Q ratio is 3.3, compared to 0.67 at the base.

PACO2 and PAO2 approach the partial pressures for these gases in the venous blood when the V/Q ratio is zero (no ventilation but perfusion). At the base of the lung, the V/Q ratio is 0.67, whereas at the apex, it is 3.3.

PAO2 at the apex is typically 132mmHg, and PACO2 is typically 28mmHg.

The average PAO2 at the base is 89 mmHg, while the average PACO2 is 42 mmHg.

Osmolarity refers to the osmotic pressure generated by the dissolved solute molecules in 1 L of solvent. Measurements of osmolarity are temperature dependent because the volume of the solvent varies with temperature. The higher the osmolarity of a solution, the more it attracts water from an opposite compartment.

Osmolarity can be computed using the following formulas:

Osmolarity = Concentration x number of dissociable particles; OR
Plasma osmolarity (Posm) = 2([Na+]) + (glucose in mmol/L) + (BUN in mmol/L)

Posm = 2 (144) + 3.5 + 9.5 = 301 mOsm/L

Suppose there is electrical neutrality, the formula will double the cation activity to account for the anions.

Plasma osmolarity (Posm) = 2([Na+] + [K+]) + (glucose in mmol/L) + (BUN in mmol/L)

Posm = 2 (144 + 6) + 3.5 + 9.5 = 313 mOsm/L

Total body water (TBW) is about 50% to 70% in adults depending on how much fat is present. ECF is relatively contracted in an obese person.

The simple rule is 60-40-20. (60% of weight = total body water, 40% of body weight is ICF and 20% is ECF)

For this woman, the total body water is 36 litres (0.6 × 60). ECF is 12 litres (1/3 of TBW) and 24 litres (2/3 of TBW) is intracellular fluid .

Sodium concentration is approximately 135-145 mmol/L in the ECF.

The ECF is made up of both intravascular and extravascular fluid and plasma proteins is found in both.

The derived unit of energy, work or amount of heat is joule (J). It is defined as the amount of energy expended if a force of one newton (N) is applied through a distance of one metre (N·m)

J = 1 kg·m/s2·m = 1 kg·m2/s2 or 1 kg·m2·s-2

Kinetic energy (KE) = ½ MV2

An object with a mass of 1500 kg moving at 30 m/s correlates to 675 kJ:

KE = ½ (1500) × (30)2 = 750 × 900 = 675 kJ

Total energy released when 1 kg fat is metabolised to CO2 and water is 37 MJ. 1 g fat produces 37 kJ/g, therefore 1 kg fat produces 37,000 × 1000 = 37 MJ.

Raising the temperature of 1 kg water from 0°C to 100°C correlates to 420 kJ. The amount of energy needed to change the temperature of 1 kg of the substance by 1°C is the specific heat capacity. We have 1 kg water therefore:

4,200 J × 100 = 420,000 J = 420 kJ

In order to calculate the energy involved in raising a 100 kg mass to a height of 1 km against gravity, we need to calculate the potential energy (PE) of the mass:

PE = mass × height attained × acceleration due to gravity
PE = 100 kg × 1000 m × 10 m/s2 = 1 MJ

The heat generated when a direct current of 10 amps flows through a heating element for 10 seconds when the potential difference across the element is 1000 volts can be calculated by applying Joule’s law of heating:

Work done (WD) = V (potential difference) × I (current) × t (time)
WD = 10 × 10 × 1000 = 100 kJ

Sodium concentration for different fluids
3% N saline 513 mmol/L
5% N saline 856 mmol/L
0.9% N saline 154 mmol/L
Hartmann’s solution 131 mmol/L
0.45% N saline with 5% glucose 77 mmol/L

This means that:

500 mL 5% N saline contains 428 mmol of sodium
1000 mL 3% N saline contains 513 mmol of sodium
3500 mL 0.9% N saline contains 539 mmol of sodium
4000 mL Hartmann’s contains 524 mmol of sodium
6000 mL 0.45% N saline with 5% glucose contains 462 mmol of sodium.

The respiratory quotient is the ratio of CO2 produced to O2 consumed while food is being metabolized:

RQ = CO2 eliminated/O2 consumed

Most energy sources are food containing carbon, hydrogen and oxygen. Examples include fat, carbohydrates, protein, and ethanol. The normal range of respiratory coefficients for organisms in metabolic balance usually ranges from 1.0-0.7.

Granulated sugar is a refined carbohydrate with no significant fat, protein or ethanol content.

The RQ for carbohydrates is = 1.0

The RQ for the rest of the compounds are:

Fats RQ = 0.7
The chemical composition of fats differs from that of carbohydrates in that fats contain considerably fewer oxygen atoms in proportion to atoms of carbon and hydrogen.

Protein RQ = 0.8
Due to the complexity of various ways in which different amino acids can be metabolized, no single RQ can be assigned to the oxidation of protein in the diet; however, 0.8 is a frequently utilized estimate.