How can these results be reconciled


(Instructors’ name)



Lecture 1

How can these results be reconciled?

Which of the following explanations do you regard as being most likely and why?

The mode of action of secretin from American dogs is different from that in English dogs.

Normally secretin evokes a pancreatic secretion rich in enzymes and in bicarbonate, but the procedure used by the English workers altered its characteristics to such an extent that its enzyme stimulating properties were inhibited

The secretin preparation SI in fact contained 2 hormones: secretin and another (now known as CCK-PZ). CCK-PZ stimulates the secretion of enzymes hence the secretion resulting from the SI injection contains enzymes and bicarbonate. M contains only one hormone, secretin, responsible for the production of a secretion rich in bicarbonate but poor in enzymes.

M contains two hormones. One of them, secretin, stimulates bicarbonate secretion and another unnamed hormone inhibits the release of enzymes from the pancreas. SI contains only one hormone, secretin, which stimulates the release of both bicarbonate and enzymes from the pancreas.

The most likely explanation of the difference in results is the second which postulates that the procedure used by the English researchers altered the characteristics of the hormone so much to the extent that it inhibited its triggering capacities to lead to production of little enzymes. Usually, secretin triggers the production of pancreatic juices rich in bicarbonate and enzyme (Bayliss and Starling 325-53). The fact the English researchers could not extract a hormone with the same abilities only means that the resulting enzyme had diminished activity as a result of some alteration during extraction (Hacki 609- 32).

Lecture 2

How do you think we can measure the resting potential of a neuron?

In a number of cells, the membrane potential is never constant. As it follows, there is never any resting potential and it is usually a theory. However, other cells containing little functions in membrane transport that change less frequently have a resting potential that one can measure by putting or inserting an electrode inside the cell. One can also measure the potential optically using dyes that alter their colours according to the potential of the membrane.

Draw a simple diagram to illustrate your explanation.

(Caldwell and Keynes 12-13).

What is the typical value of the resting potential of a neuron (give units)?

The resting potential of a neuron has a typical value of between -70 millivolts to -80 millivolts (Wright 139- 142).

The charged particles that are responsible for the RMP are called ions:

The positive ion is called the potassium and sodium ions

The negative ion is called the p-, chloride ions

Generation of the Resting Membrane Potential

Effect of K+ on the RMP




Concentration gradient for K+




Label the concentration gradient for movement of K+ across the membrane

What happens to the P-? Label this on the diagram.

A potential will now exist across the membrane: label this on the diagram.

Label the electrical gradient for movement of K+ across the membrane.

Effect of Na+ on the RMP



Concentration gradient for Na+

Electrical gradient




_ +


Label the concentration gradient for movement of Na+ across the membrane

Excess Cl- will be left outside the cell; label the potential difference that now exists across the plasma membrane

Label the electrical gradient for movement of Na+ across the membrane

Calculating the equilibrium potential

Calculate the equilibrium potential for K+ and Na+

EK = Potassium ion== 26.6 In 0.03

= -40.43

ENa = Sodium ion =

26.6 In 156/13

=26.6 In 12

=26.6 x 1.08


The real situation: the effect of K+ and Na+ combined:






Label the concentration gradient for K+ (K+ attempts to establish EK)

Label the concentration gradient for Na+ (Na+ attempts to establish ENa

Why is the resulting resting membrane potential of –70mV much closer to EK than ENa?

Because Ek is more and tending to drag the resting membrane potential towards itself (Hartline and Colman 25- 35). Even though most membranes are not that permeable to these ions, the resting membrane is usually more permeable to potassium ions than sodium ions (Wright 139- 142).

At the RMP neither K+ nor Na+ is at equilibrium, therefore they will continue to diffuse down their concentration gradients. After a period of time, the RMP would be dissipated.

How is this prevented in the cell?

The membranes solve this by having an active transport mechanism that actively transports potassium to the inside and sodium ions to the outside. Hence the tendency of these ions to diffuse down the gradient is counterbalanced by the sodium, potassium ion pump, transporting the ions against their concentration gradients (Huxley 479- 95).

The Chloride ion (Cl-) is the principal extracellular anion. It has an ECl of –70mV, which is the same as the RMP

How is Cl- distributed in the cell?

Chloride ions are usually more concentrated outside the cells or extracellularly. Just like sodium ions. The higher concentration outside cells of sodium ions is usually counterbalanced by the high concentration of chloride ions outside the cell (Zoidi and Dermietzel 137-42).

The concentrations of ions in the ECF compartments of a skeletal muscle fibre are given below:

Ion Concentration (mM/L)


Na+ 156 13

Cl- 130 9.5

Ca2+ 1 0.1

K+ 5 160

Calculate the equilibrium potentials for each ion:

Sodium ion =

26.6 In 156/13

=26.6 In 12

=26.6 x 1.08


Chloride ion=26.6 In 130/9.5

26.6 In 13.7


Calcium ion= 26.6 In 10


Potassium ion== 26.6 In 0.03

= -40.43

Lecture 3

What are the primary determinants of serum osmolality?

Sodium salts like bicarbonates and chlorides, urea and glucose.

If serum glucose increases, what will happen to serum osmolality?

It increase

If serum osmolality increases what will happen to the ICF?

It shifts its fluids towards ECF

As fluid shifts out of the ICF and into the ECF, what will happen to the serum?

It becomes more concentrated

If water is lost in excess of electrolytes, what will happen to the serum osmolality?

It increases

What is the estimated serum osmolality for the above example?


g. What is the calculated osmolality for the above example?


h. How do these figures compare to the laboratory value? Why do you think they are slightly different?

They are higher than the laboratory value. This is because in a laboratory setting the loss of water can be controlled (Estad 1085- 6).

Example Two.

If a substance such as mannitol is added to the ECF, what will happen to its osmolality?

It increases

If very little enters the ICF, what will happen initially to the osmolality of the ICF?

It will increase

There is now a concentration difference between the two compartments. Remember that the two compartments are separated by a semi-permeable membrane that allows the movement of water. In which direction will the water move?

Towards the ECF compartment

Can you think of a clinical situation where these properties of mannitol could be life-saving?

In cases where lungs become filled with water

Lecture 4

Question 1: A man with normal lungs and a right to left shunt is found at catheterization to have an O2 concentration in his arterial and mixed venous blood of 18 and 14ml O2/100ml blood. If oxygen concentration of blood leaving the pulmonary capillaries is calculated to be 20mlO2/100ml blood, how large is the shunt?

Question 2. A patient whose arterial PO2 and PCO2 values are both 40mmHg when breathing air PO2 150mmHg, has a marked ventilation perfusion inequality but no shunt. On measuring his CO2 produced at rest it was found to be 240ml/min and his O2 consumption was found to be 300ml/min. Taking the correction factor for the alveolar gas equation to be 2mmHg, calculate the degree of his ventilation perfusion inequality.

Lecture 5

Question 1. For the control period, mark on the figure on page 3:-

the baseline

the amplitude of contractions

the frequency of the contractions

Question 2. What properties of the muscle do these measurements correspond to?

Work as syncytium, larger sheets, and have bridges with low resistance.

Question 3. Calibrate the tension of the preparation and complete the table below.

Calibration of tension: 1mm = g

Normal Saline Control ACh

(50ng/ml) Control NAd

(50ng/ml) Control

Resting Tone (g) 0 0.8 0 0.4 0

Amplitude (g) 0 0.9 0 0.5 0


(per minute) 0 1 1 1 0


(50ng/ml) Control ACh

(50ng/ml) ACh

(500ng/ml) Control NAd


Resting Tone (g) 0 0.5 0.7 0 0.2

Amplitude (g) 0 0.7 0.8 0 0.2


(per minute) 0 1 2 0 1

Question 4. Do ACh and NAd affect all three measurements i.e. baseline, amplitude and frequency of contractions? If so, do they affect them in the same way?


Question 5. Using the information about Atropine, explain the results and deduce the possible sites of action of ACh and NAd in the preparation.

Question 6. What conclusion can you draw about the origin of the spontaneous activity and the way in which it is modified in the jejunum?

Lecture 7

If the plasma concentration of a substance X is 200mg/100ml and the GFR is 125ml/min, what is the filtered load of the substance?



2 x 125


Practice Calculation 2.

A good example of a substance like A is a polysaccharide called insulin. Suppose you want to measure GFR in a human subject. A reasonable procedure would be to infuse the subject intravenously with a solution of insulin that is enough to give you a plasma concentration of 3.9mg/litre. Over a two hour period, you manage to collect 0.3L of urine from your subject and after measuring the concentration of insulin in that urine sample, you find that you have a urine concentration of 351mg/litre. What is your subject’s GFR? What are its units?

3.9 x0.3 = 351 x GFR

1.17= 351 GFR

= 0.003

Practice Example 3.

If the urine concentration of a substance is 7.5mg/ml of urine, its plasma concentration is 0.2mg/ml of plasma, and the urine flow rate is 2ml/min, what is the clearance rate of the substance?

`7.5/2= 3.75



Practice Example 4.

A patient is found to have a urine creatinine concentration of 196mg/ml: plasma concentration that equals 1.4mg/ml: and a urine flow that equals 1ml/min. what is the clearance of creatinine?



Complete the following Table:

You can find the value for Tm from the glucose titration curve above. It corresponds to the rate at which glucose reabsorption starts to plateau. Read off the value from the Movement of glucose axis. (Clue: the value is 300, 350, 375 or 400 mg/min? Which is it?)

Plasma Concentration (mg/100ml) GFR

(ml/min) Filtered Load

(mg/min) Tm

(mg/min) Amount


(mg/min) Amount



80 125 80 80 375 0

100 125 100 100 375 0

200 125 200 200 370 0

300 125 300 300 340 20

400 125 400 400 375 50

500 125 500 500 375 200

It is 375

Practise Example 5.

In 1 hour, 54 ml of urine are collected from a test subject. The concentration of PAH in the plasma is 0.02mg/ml and in the urine it is 14mg/ml. Haematocrit value is 43% (o.45 as a fraction). What is the effective renal blood flow?



= 98.18

Lecture 8

Assess the validity of each of the five hypotheses (I-V) by commenting on the information in the figure and tables. Deal with each of the hypotheses in turn and state clearly why it is refuted or supported by the data provided…. Either partially or fully. Try to think of other experiments that could be done to test each hypothesis.

Hypothesis I: The buccal cavity itself does not play a major role in digestion and absorption

This hypothesis is supported by the experiments and it is true because its units of hydrolysis are low, its absorption is low and because it is mostly made up of complex muscles that do not have any absorption abilities.

Hypothesis II: The oesophagus does not play a major role in the digestion and absorption. The major function is propulsion of food into the stomach

This is also true because its absorption ability is low and because it is mainly composed of circular and longitudinal muscles on its walls used for pushing food downwards.

Hypothesis III: the main role of the stomach is digestion. No absorption of the products of digestion occurs in the stomach.

This is true because it mostly has muscle fibres as its main components and goblet and parietal cells for digestion.

Hypothesis IV: The most important region for the absorption of products of digestion is the intestine, and no enzymes are produced here

This is true because absorption is made possible by villi, internal folding and epithelial cells for absorption. The little enzymes are as a result of less goblet cell.

Hypothesis V: The colon plays no major role in digestion or absorption; its main function is removal of water and storage of faeces.

This is true because it has thin walls that cannot allow for absorption or digestion because of lack of muscles.

Work cited

 Bayliss, W. and Starling, H. “The mechanism of pancreatic secretion”. J. Physiol. 28 (1902): 325–353. Print.

Caldwell, C. and Keynes, D. “The utilization of phosphate bond energy for sodium extrusion from giant axons”. J. Physiol. (London) 137.1 (1957): 12–13P. Print.

Erstad, L. “Osmolality and osmolarity: narrowing the terminology gap.” Pharmacotherapy 23. 9 (2003): 1085–6. Print.

Garcia, A. et al. “The Oxygen Concentrations Delivered by Different Oxygen Therapy Systems”. Chest Meeting 128.4 (2009): 389S–390S. Print.

Häcki, H. “Secretin”. Clin Gastroenterol 9.3 (1980): 609–32. Print.

Hartline, K. and Colman, R. “Rapid conduction and the evolution of giant axons and myelinated fibers”. Curr. Biol. 17.1 (2007): R29–R35. Print.

 Huxley, A. and Stämpfli, R. “Direct determination of membrane resting potential and action potential in single myelinated nerve fibers”. Journal of Physiology 112. 3-4 (1949): 476–95. Print.

Patarinski, D. “Indications and contraindications for oxygen therapy of respiratoryinsufficiency”. Vŭtreshni bolesti 15.4 (1976): 44–50. Print.

Wright, H. Advances in Physiology Education, 28. 1-4 (2004): 139-142. Print.

Zoidl, G. and Dermietzel, R. “On the search for the electrical synapse: a glimpse at the future”. Cell Tissue Res. 310. 2(2002): 137–42. Print.