Cell physiology

The word physiology is derived from physiologicos

It means the discourse of natural knowledge

Def

It is the science, which deals with normal functions of tissues and organs of living organisms

Branches of physiology

1. Viral physiology
2. Bacterial physiology
3. Plant physiology
4. Animal physiology
5. Human physiology

Organization of Human body

Cell----------- tissue ------ Organs ---- Organ systems---- ----human body

System

1. CNS
2. ANS
3. CVS
4. Respiratory system
5. Digestive system
6. Urinary system
7. Endocrine glands
8. Circulatory system - Blood
9. Reproductive system
10. Autacoids

Basic characteristics of Human  or living organisms

1. Irritability  or excitability
2. Conductivity
3. Contractility
4. Absorption
5. Digestion
6. Secretion
7. Excretion
8. Growth
9. Reproduction

Body fluid comportments

There are two fluid comportments

1. Extracellular fluids

               ECF is divided into 

1. Plasma
2. Interstitial fluid

2. Intracellular fluids

The ECT & ICF remain separate and do not mix up by a very thin membrane  - cell membrane

ECF contains nutrients , electrolytes, oxygen and all other substances essential for survival of cells

Homeostasis

It means maintenance static or constant conditions in the internal environment

The internal environment in the human body is the ECT, in which all the cell of the body live, it constantly moves or interchanges throughout the body 

It includes blood and fluid present in between the cells

ECF contains nutrients, electrolytes, oxygen, and all other substances essential for survival of cells

Factors involved in homeostasis

1. Maintenance of PH
2. Maintenance of temperature
3. Maintenance of body water
4. Maintenance of electrolyte balance
5. Supply of oxygen, nutrients and hormones
6. Maintenance of normal blood volume
7. Maintenance of normal arterial blood pressure
8. Removal of waste products from body
9. Maintenance constant osmotic pressure
10. Coagulation of blood

Control mechanism of homeostasis

1. Sensor-  Which detects or finds out the change in  the normal value and function
2. Comparator- which compares the changed value or function with normal
3. Effector system- which returns the changed value or function to normal
4. Variable- Which brings down the changed value to normal value

Ex

1. Blood sugar  - Normal vale 80-120 mg/ml

1. Sensor- It senses the raised blood sugar due to excess intake of sugar from normal –  

    80-120 mg/80-120mg/100mg

2. Comparator- It compares the increased blood sugar level with normal

3. Effector system- There is increased secretion of insulin form pancreas, which brings

    the raised blood sugar to normal value

4. Variable- The variable value – raised blood sugar is brought to normal and constancy

    of internal environment is maintained

2. Body temperature- Normal value 37 degree C when a person is exposed to

    hypothermia leads to low body temperature

1. Sensor- Detects the fall of body temperature
2. Comparator- Compares reduced temperature with normal
3. Effector system – There is increased secretion of thyroid hormone T3 &T4 , which raises the body temperature by caloriginic action to normal value.

Mechanism of action of Homeostatic control system

All control systems involved in homeostatic control system operate through following feedback mechanism

1.      – ve feedback mechanism – If the activity of particular system is increased regulatory mechanism will soon reduce it

2.      + VE feed back mechanism – Coagulation of blood

Various systems take part in homeostasis

1. Respiratory system – Control PH of blood

2. Skin, respiratory , digestive , renal and CNS – Involve in body temperature

3. Liver- It changes chemical composition of many absorbed substances to more useful

               form

4. Endocrine system – By secretion of various hormones such as insulin, ADH, cortisol,

    T3 & T , Aldosterone keep homeostasis normal

Cell

It is the structural and functional unit of all living beings

Cell membrane

It is a protective thin sheet  enveloping the cell body

Its thickness ranges from 75A- 110A

Composition of cell membrane

It contains of

1. Proteins                    -  55%

2. Lipids                       -  40%

3.Carbohydrates           -  5%

Structure of cell membrane

It is a 3 layered structure

1. Lipid layer
2. Protein layer
3. Basement layer

1. Protein layer

1. Integral protein – Provides integrity to cell membrane

2. Channel protein- Provides channels for diffusion o water- soluble substances

    E.g – glucose, amino acids

3. Carrier protein- Helps in transport of substances across cell membrane (facilitated  

    diffusion)

4. Receptor proteins - They are receptors for hormones and neurotransmitters and they

    form receptor protein hormone complex

5. Enzyme protein- Molecules form enzymes

   Antigen protein- some proteins act as antigens and produce antibodies

2. Lipid layer

Lipid layer forms a semi permeable membrane, it allows fat – soluble substances such as oxygen, CO₂, and alcohol to pass through the cell membrane

Phospholipids permit lipid-soluble materials to easily enter or leave the cell by diffusion through the cell membrane.

3. Carbohydrate layer

These molecules are attached to either protein or lipid layer- glycol- protein & glycol – lipid

Carbohydrate layer forms a covering on the cell membrane called glycocalyx

Carbohydrate molecules are negatively charged ions to keep substances outside the membrane . They form tight junctions

 Smooth endoplasmic reticulum

As sacroplasmic reticulum they play important role in skeletal and cardiac muscles

Rough endoplasmic reticulum

They contain ribosome’s and they are the sites for protein synthesis

E.g – Enzymes and hormones

Golgi apparatus

They are packaging department of the cells

They produce secretion granules, which store hormones and enzymes

Mitochondria

They are energy generating cells

Lysosomes

They act as digestive system of the cells-  bacteria, worm & exogen substances

Centrioles or centrosomes

They are concerned with movements of chromosomes during cell division

Nucleus

It is also the site of RNA synthesis. They are 3 types RNA- m, r,t RNAs

Nuclei

It contains some protein and RNA. Nuclei are sites for synthesis of ribosome’s and transfer them to cytoplasm where ribosome’s synthesize proteins

Ways of transport across cell membrane

 

Passive transport

Substances are transported from region of higher concentration to lower concentration . It is called Down Hill movement

It is 2 types

1. Simple diffusion
2. Facilitated diffusion

A. Simple diffusion

Diffusion through the lipid layer

Lipid soluble substances are transported through by simple diffusion across the lipid layer

E. g – O2, CO2 and alcohol

Diffusion through protein layer – Water – soluble substances diffuse through the protein layer

Eg- Glucose and electrolytes

B. Facilitated diffusion

Larger molecules of water- soluble substances cannot diffuse through protein channels such substances diffuse with the help of carrier protein, hence this type of diffusion is called facilitated or carrier mediated diffusion

E.g

Glucose and amino acids are transported by facilitated diffusion

These molecules bind with receptor protein lining the channels

 2. Active transport

The movements of substances are against the chemical or electrical or electro chemical gradient is called active transport

Substances are transported from region of lower concentration to higher concentration . It is called UP  Hill movement

It requires expenditure of energy which is liberated by breakdown of ATP into ADP and inorganic phosphate. It is faster than passive transport

Types active transport

A. Primary active transport

They directly use energy obtained from hydrolysis of ATP

They consists

1. Na- K Pump

2. Ca – pump

3. K- H  pump

1. Na – k pump

It is an electrogenic pump present in all the cells of the body

Whenever a nerve cell or nerve fiber or muscle fiber is stimulated, the Na channels open and the Na ions enter inside the cell

Simultaneously K 9ions leave the cell through K channels and come outside

It leads to depolarization of cell membrane which causes origin of action potential

The depolarization does mot and should not continue and it is followed by reploraization to obtain second stimulus

So for the repolarizaiton , Na ions must go outside the cell and K ions enter the cell, It happens with the help of Na – K pump

Na – K pump extrudes out 3 Na ions for every 2 K ions which return inside of cell membrane, resulting in a net removal of + ve charged Na ions so that inside of the cell membrane again becomes more –ve than outside

Na – K pump derives energy by breakdown of ATP into ADP + Pi with the help of ATP ase enzyme which is present in cell membrane

B. Secondary active transport

When Na is transported by carrier protein, another substance is also transported by the same protein simultaneously either in the same direction of Na or in opposite direction

This type of transport of substance along Na ion by means of carrier protein is called secondary active transport

It is of two types

A. Na – Co- transport

Along with Na ion another substance is carried by carrier protein in same direction

Substance carried by Na co-transport are glucose, amino acids, CL, I, Fe , urte ions are transported

B. Na  Counter transport

The substances are transported across cell membrane in exchange of Na ions

1. Na- Ca counter transport

2. Na- H counter transport

3. Na- Mg counter transport

4. Na- K counter transport

3. Endocytosis

The process involved in endocytosis is as follows

A.    Pinocytosis

B.     Phagocytosis

A. Pinocytosis

Large molecules are transported into the cell by cell drinking molecules dissolve in fluids bind to the outer surface of the cell membrane

Cell membrane evaginates around the droplet and they get engulfed by membrane and are converted into vacuoles

They come into contact with lysozomes and get ruptured and released inside the cell

B. Phagocytosis

The large particles are  engulfed into the cell, it is called cell eating larger bacteria are antigen when enter the body, phagocytic cell sends cytoplasmic pseudopodium around the bacteria. The particles are engulfed inside the cell and are digested

Circulatory system

Heart and blood vessels together form CVS . Blood vessels include arteries, veins and their derivatives

Artery        -- Artery is a vessel which caries oxygenated blood

Vein          -- Vein is a vessel which carries deoxygenated blood

Capillary   -- Capillary is the channel which connects the arterial system and venous

                         system together

 

Blood vessels

Histologically, a typical blood vessel shows three different areas in it

Tunica intima   

It is the innermost layer of the blood vessel lined by simple squamous epithelium or endothelium

Tunica media

It is the middle layer of a blood vessel seen just below the tunica intima

Made up of either smooth muscles or elastic fibers

Tunica adventitia

It is an outermost layer of a vessel which is made up of dense irregular connective tissue  

S No	Artery	Vein
1	Carries oxygenated blood	Carries de- oxygenated blood
2	Tunica adventitia is smaller	Tunica adventitia is larger
3	Tunica media is larger	Tunica medial is smaller
4	Lumen  is almost circular	Tunica adventitia is larger

 

Function

Endothelial layer of tunica intima is anti- thrombotic and nutritive in function

Elastic fiber of the tunica medial  expand the arterial wall thus helps in maintaining normal arterial pressure and blood flow

Connective tissue layer of the tunica adventitia prevents undue stretching and rupture of the artery

Buffer Solutions

Introduction

            Any solution that contains both a weak acid HA and its conjugate base A– in significant amounts is a buffer solution.  A buffer is a solution that will tend to maintain its pH when small amounts of either acid or base are added to it.  Buffer solutions can be made to maintain almost any pH, depending on the acid-base pair used.  The pH of a buffer solution is determined by the Ka of the acid and by the ratio of concentrations of HA and A–.  This can be calculated by rearranging the expression for the Ka of the conjugate acid of the buffer:

                                      

The rearranged equation shows that the H3O+ ion concentration of the buffer solution can be found by multiplying the Ka of the acid by the ratio of the molar concentrations of the two components.  To solve directly for the pH of the buffer, the equation can be put into logarithmic form. If the above equation is rearranged and the negative log of both sides is taken, a new form of the equation known as the Henderson-Hasselbalch equation results.

                                     

           Henderson-Hasselbalch equation


A buffer solution can maintain its approximate pH when an acid or a base is added to it because it can react with both acids and bases. If a strong acid (H3O+) is added, the basic component of the buffer (A–) can react with it, and if a strong base (OH–) is added, the acidic component of the buffer (HA) will react with it.

            H3O+ (aq)  +  A– (aq)    ¾®   HA (aq)  +  H2O (l)

            OH– (aq)    +  HA (aq)   ¾®  H2O (l)   +   A– (aq)

In this way, any strong acid or strong base that is added to the buffer solution is converted into a weak acid or weak base.  The ratio of weak acid to weak base changes, which causes the pH to change slightly, but not drastically.

            Buffer solutions are most effective when both components, the conjugate acid and the conjugate base, are present in reasonably large concentrations.  If this is the case, the buffer is said to have a high buffer capacity.  Also, a buffer is most effective when there are approximately equal concentrations of the two buffer components (a ratio of [X–]/[HX] close to 1/1) because in this case the solution will guard against large pH changes equally well whether acid is added or base is added.

In order for the solution to be considered a buffer, this ratio must be between a 1/10 and a 10/1 ratio of [X-]/[HX].  This restriction means that the [H3O+] in the buffer will be within a factor of 10 of the Ka value of the conjugate acid. It also means that the pH of the buffer will be within one pH unit of the value of the pKa of the conjugate acid.  For example, suppose you wanted to make a buffer containing phosphoric acid (H3PO4) and dihydrogen phosphate (H2PO4–).  Since the Ka of H3PO4 is 7.5 ´ 10-3 and the pKa is 2.12, as shown in the table below, the pH of this buffer would have to be between 1.12 and 3.12.  If you wanted to make a buffer with a pH outside of this range, you would have to choose a different conjugate acid-base pair.

Buffer Solutions That Can Be Made With H3PO4 and H2PO4–

Ka	pKa
7.5 ´ 10–3	2.12	1/10	7.5 ´ 10–2	1.12
		1/1	7.5 ´ 10–3	2.12
		10/1	7.5 ´ 10–4	3.12

            In this experiment, you will be assigned a pH value, and you will prepare a buffer solution having that pH. There will be two sets of acid-base pairs available in the lab. These are:

1.      Acetic acid (CH3CO2H, Ka = 1.8 ´ 10–5) and sodium acetate (NaC2H3O2).

2.      Ammonium chloride (NH4Cl, Ka for NH4+ = 5.6 ´ 10–10) and ammonia (NH3).

The buffer will be prepared by choosing the appropriate acid-base pair, calculating the molar ratio of acid to base that will produce the assigned pH, and then mixing the calculated amounts of the two compounds with enough deionized water to make 200. mL of buffer solution.

            A solution with approximately the same pH as the buffer solution will be prepared by diluting a solution of a strong acid or a strong base.  This solution will not be a buffer solution, as can be shown by comparing its buffering ability to that of the buffer solution.  Finally, a known amount of strong acid or base will be added to the buffer solution.  Before adding this acid or base, the pH change that the addition should cause will be calculated.  The observed pH change will be compared to the calculated value.

Experimental Procedure

SAFETY PRECAUTIONS:  Wear your SAFETY GOGGLES.  Use the concentrated acetic acid and ammonia solutions in the FUME HOOD.  If any acid or base solution splashes on your skin, wash it off immediately with copious amounts of running water.

WASTE DISPOSAL:  All waste from this experiment should be poured down the drain, followed by plenty of running water.

Part 1 - Preparing the Buffer Solution

Prelab Calculations

1.      Obtain a pH assignment from your instructor.

2.      Decide which of the two available buffer systems you should use.

3.      Determine the ratio of [A–]/[HA] needed in the buffer.

4.      Which component will be more concentrated, A– or HA?  Choose a value for the concentration of that component anywhere in the range 0.2 – 0.6 M. Calculate what the concentration of the other component should be.

5.      Calculate the volume (in mL) of 17.5 M CH3CO2H or 14.8 M NH3 needed to make 200 mL of your buffer.  Calculate the mass (in g) of solid NaCH3CO2·3H2O or NH4Cl needed to make 200 mL of your buffer.

Procedure

            Measure out the amounts of the buffer components needed, add them to about 100 mL of deionized water in a 250-mL Erlenmeyer flask.  Swirl until all the solids are dissolved.  Pour the mixture into a large graduated cylinder, and dilute it to a total volume of 200 mL.  Pour it back and forth between the graduated cylinder and your flask so that it is thoroughly mixed.  Store the buffer solution in the flask.

Set up a pH meter at your desk and calibrate it according to the instructions.  Remember that the electrode of the pH meter is FRAGILE!  Also, remember to rinse the electrode off with deionized water when transferring it between different solutions.

            Pour about 50 mL of your buffer solution into a small beaker and measure the pH. The pH of the solution should be close to the expected value but might not be exactly the same.  A minor discrepancy will not affect the results of the remainder of the experiment.

            There are several possible reasons for the pH meter reading being different from the expected pH.  One is that some of the volatile component of your buffer may have escaped from the solution into the air.  Another is the fact that the Ka values for acids vary with temperature.  The values usually reported are accurate for 25°C.

Part 2- Preparing a Solution with the Same pH by Dilution of a Strong Acid or Base

The pH of a typical acetic acid-acetate buffer solution is between 4 and 5.  A solution with a pH in this range can easily be prepared by diluting a solution of strong acid, such as HCl, until the hydronium ion concentration is between 10–4 and 10–5 M.  Similarly, a typical ammonium ion-ammonia buffer solution has a pH between 9 and 10.  A dilute solution of NaOH with [OH–] between 10–4 and 10–5 M will also be in this pH range.  These solutions, however will not act as buffer solutions.  In this section of the experiment, you will prepare a solution with approximately the same pH as your buffer solution by diluting a more concentrated solution of strong acid or base.

Prelab Calculations

If your buffer is acidic, calculate the volume (in mL) of 0.0010 M HCl needed to prepare 100 mL of a solution that has the same pH as your buffer.

If your buffer is basic, calculate the volume (in mL) of 0.0010 M NaOH needed to prepare 100 mL of a solution that has the same pH as your buffer.

Procedure

Prepare your unbuffered solution, by diluting the calculated amount of 0.0010 M HCl or 0.0010 M NaOH solution to a final volume of 100 mL.  Measure its pH.  Again, the reading on the pH meter may not display the pH expected for the solution.  The pH of solutions as dilute as this can be affected by small amounts of impurities, including CO2 dissolved in the water.  The pH can be adjusted if desired, but the comparison of this solution to the buffer solution will work even if the pH values of the two are not the same.

Part 3 - Comparing the Buffering Abilities of the two Solutions

Procedure

            To test the buffering abilities of the solutions, 0.1 M HCl or 0.1 M NaOH will be added.  If the solutions are acidic, add 0.1 M NaOH, and if they are basic, add 0.1 M HCl.

            Place the pH electrode in the 50 mL of buffer solution tested earlier.  Re-measure the pH, then add 5 drops of 0.1 M NaOH or 0.1 M HCl solution (see above).  Carefully swirl the solution, making sure not to bump the electrode, and record the new pH.  Repeat the procedure with another 5 drops of NaOH or HCl.  How much did the pH change upon the addition of these amounts of HCl or NaOH?

            Rinse off the electrode with deionized water, place it in the nonbuffered solution, and measure the pH.  Follow the same procedure as you used for the buffer solution.  Add 5 drops of drops of HCl or NaOH, and record the new pH.  Add 5 more drops, and record the pH.  It is important to follow the same procedure so that you can more accurately compare the two solutions.  How much did the pH change for the unbuffered solution?  How do the pH changes compare for the buffered and unbuffered solutions?

Part 4 - Changing the pH of a Buffer Solution

            The pH of a buffer solution does change when large amounts of strong acid or strong base are added to it.  Addition of strong acid uses up the conjugate base of the buffer, and addition of strong base uses up the conjugate acid.  The pH of the buffer changes because the ratio of conjugate acid to conjugate base has been changed.

            In this part of the experiment, enough 1.0 M HCl will be added to 100. mL of the buffer solution to react with half of the conjugate base in the solution.  The observed change in pH will be compared to the pH change expected on the basis of the prelab calculation.

Prelab Calculations

1.      How many moles of base (NH3 or C2H3O2–) are in 100 mL of the buffer solution you prepared?  What is half of this amount?  This is the number of moles of HCl that you will add to the solution.

2.      Calculate the volume (in mL) of 1.0 M HCl needed to supply this number of moles.  That is the amount needed to react with half of the conjugate base in 100 mL of your buffer solution.

3.      What should the pH be after this amount of HCl is added?  What is the expected change in pH of the buffer?

Procedure

Add this volume of 1.0 M HCl to 100 mL of  the buffer solution, and measure the pH.  Record the change in pH.  Compare the expected (calculated) pH change to the actual pH change of the buffer.

Sample Calculations

Part 1 – Preparing a Buffer Solution

Sample calculations will be done for an assigned pH of 2.50.

            A pH of 2.50 corresponds to a [H3O+] = 10–2.50 = 3.162 ´ 10–3 M.  Phosporic acid (H3PO4) and sodium dihydrogen phosphate (NaH2PO4) could be used to make this buffer, because Ka for H3PO4 (Ka = 7.5 ´ 10–3) is close to the desired [H3O+].

            The  ratio should be  =  =  =

            H2PO4– will be the most concentrated component of this buffer system.  We will arbitrarily choose its concentration to be 0.50 M.  This means that the concentration of H3PO4 should be 0.50 M ¸ 2.372 = 0.2108 M

            To make 200 mL of the buffer solution, we will need to measure out a certain volume of concentrated phosphoric acid, which is available as a 14.7 M aqueous solution.  And we will need to weigh out a certain mass of sodium dihydrogen phosphate, which is available as the hydrated solid, NaH2PO4·2H2O (the waters of hydration are included in the molar mass).

            To make 200 mL of a 0.2108 M phosphoric acid solution,

                        200 mL ´  = 2.87 mL of concentrated phosphoric acid needed.

            To make 200 mL of a 0.50 M sodium dihydrogen phosphate solution,

                        0.200 L ´  ´  = 15.60 g of NaH2PO4·2H2O needed.

            So, to prepare the buffer solution, we will mix 2.87 mL of concentrated H3PO4 and 15.60 grams of NaH2PO4·2H2O with enough water to make 200 mL of solution.

Part 2- Preparing a Solution with the Same pH by Dilution of a Strong Acid or Base

            To make 100 mL of a solution with pH = 2.50 ([H3O+] = 3.162 ´ 10–3 M), we will dilute a 0.10 M HCl solution.  (Note, in your experiment, you will be diluting an 0.0010 M HCl solution or an 0.0010 M NaOH solution.)

            100 mL ´  = 3.16 mL of 0.10 M HCl needed

            So, to prepare the unbuffered solution, we will mix 3.16 mL of 0.10 M HCl with enough water to make 100 mL of solution.

Part 4 - Changing the pH of a Buffer Solution

            We will take 100 mL of the buffer solution, and add enough 1.0 M HCl to react with half of the moles of H2PO4–, the conjugate base.

                        0.100 L ´  = 0.050 moles of H2PO4– are present in 100 mL

                        0.025 moles of H2PO4– will be react with 0.025 moles of HCl.

                        0.025 mol HCl ´  = 0.025 L or 25 mL of 1.0 M HCl will be added.

            So, to change the pH of our buffer solution, we will take 100 mL of our original buffer solution, and add 25 mL of 1.0 M HCl to it.

            All of the added HCl will react with the H2PO4–.  Some H2PO4– will be used up, and more H3PO4 will be formed.

                                          H3O+ (aq)  +  H2PO4– (aq)  ¾®  H3PO4 (aq)  +  H2O (l)

            Before reaction        0.025 mol         0.050 mol               0.021 mol              lots

                                         –0.025 mol       –0.025 mol            +0.025 mol

            After reaction          0                      0.025 mol               0.046 mol

             =  = 1.38 ´ 10–2 M

            So, after the addition of HCl, the new pH will be –log (2.01 ´ 10–2) = 1.86.  The expected change in pH will be 1.86 – 2.50 = –0.64 pH units.

Additional Questions (for the finished lab report)

1.      For each of these desired pH values, choose a weak acid-conjugate base pair from your textbook that could be used to prepare a buffer solution with that pH.  Calculate the desired [A–]/[HA] ratio in each case.

(a) pH = 6.00                  (b) pH = 8.00               (c) pH = 11.00

2.      500 mL of a buffer solution contains 0.050 mol NaHSO3 and 0.031 mol Na2SO3.

(a) What is the pH of the solution?

(b) Write the net ionic equation for the reaction that occurs when NaOH is added to this buffer.

(c) Calculate the new pH after 10. mL of 1.0 M NaOH is added to the buffer solution.

         (d) Calculate the new pH after 10. mL of 1.0 M NaOH is added to 500. mL of pure water.

(e) Explain why the pH of the water changed so much as compared to the pH of the buffer.

FILE LINKS


buffer solutions 1

buffer solutions 2

Cardiovascular drugs

Digitalis

Cardiac glycosides include digoxin, digitoxin and ouabin. These are derived from plant Foxglove (Digitalis purpurea). Cardiac glycosides composed of steroid nucleus linked to a lactone ring and a series of sugars.

Clinical uses: treatment of congestive heart failure and management of supraventricular rhythm disturbances.

Pharmacokinetics: digoxin is taken orally, well absorbed but in some patients (10%) the presence of some enteric bacteria may cause degradation of the drug and the use of broad spectrum antibiotics may cause sudden increase in serum digoxin level causing toxicity. Cardiac glycosides are widely distributed to tissues including CNS. Digoxin is excreted unchanged by the kidney; the dose should be adjusted in patients with renal impairment. Digitoxin is metabolized in the liver and excreted in bile.

Digoxin has narrow therapeutic index; the toxic dose is 2 ngm/ml which is near to the therapeutic dose which is 1.1 ngm/ml.

Cardiac effect

1-    Mechanical effect

Through inhibition of Na­⁺/K⁺ ATPase enzyme (membrane bound enzyme). This enzyme keep K⁺ inside the cell and Na­⁺ outside the cell; so when this enzyme is inhibited K­⁺ transport back into the cell is blocked and its concentration in the extracellular fluid increases, at the same time Na ions will enter the cell and this will promote or facilitate the entry of Ca ⁺² which are essential for the contraction of actin and myosine.

2-     Electrical effect  ( direct effect and autonomic effect)

a-    Direct effect

It causes brief prolongation of the action potential followed by a period of shortening especially the plateau phase. The decrease in the action potential duration is probably the result of increase potassium conductance that is caused by increase intracellular ca ions. All these effects can be observed at therapeutic concentration in the absence of over toxicity.

Shortening of the action potential contributes to the shortening of atrial and ventricular refractoriness.

At higher concentration resting membrane potential is redused (made less negative) as aresult of inhibition of sodium pump and reduced intracellular K⁺, this lead to appearance of ascillatory depolarizing afterpotential which followed normally evoked action potentials. The afterpotentials also known as delayed afterdepolarizations are associated with overloading of the intracellular Ca⁺² store.

·        When below threshold, these afterpotentials may interfere with normal conduction.

·        When afterpotential reach threshold it elicits an action potential (premature depolarization or ectopic beat).

·        If afterpotentials in the purkinje conducting system regularly reach threshold bigeminy will be recorded on the electrocardiogram (ECG).

NSR: an inverted T wave and depressed ST segment are present.

PVB: is a manifestations of depolarization evoked by delayed oscillatory afterpotential.

With further intoxication, each afterpotential evoked action potential will itself elicit a suprathreshold afterpotential, lead to tachycardia and fibrillation.

b-    Autonomic action (indirect)

Indirect action include sympathetic and parasympathetic.

At low dose parasympathomimetic effects predominate, lead to decrease heart rate, decrease conduction velocity, prolongation of refractory period so digoxin used in the treatment of supraventricular arrhythmia.

At toxic level sympathetic out flow is increased  lead to increase heart rate and contraction.

Manifestation of digitalis toxicity

1-     Gastrointestinal effect: anorexia, nausea and vomiting. They are the earliest signs of toxicity.

2-    Visual effect: blurred vision, loss of visual acuity and yellow-green halos.

3-    CNS effect: headache, fatigue and confusion.

4-    Cardiac effect: sever dysrhythmia moving from decreased or blocked atrioventricular nodal conduction, paroxysmal supraventricular tachycardia to the conversion of atrial flutter to atrial fibrillation, premature ventricular depolarization, ventricular fibrillation and finally complete heart block.

5-    Endocrinological effect: gynecomastia due to antiandrogenic effect of the drug.

Factors predisposing to digitalis toxicity

1-    Electrolyte disturbances: hypokalemia, hypomagnesemia and hypercalcemia predispose digitalis toxicity.

2-     Hypothyroidism, hypoxia, renal failure and myocarditis are predisposing factors to digitalis toxicity.

3-     Drugs: quinidine can cause digitalis intixicationboth by displacing digitalis from plasma protein binding sites and by competing with digitalis for renal excretion.

Verapamil also diplace digitalis from plasma protein binding site and can increase digoxin levels by 50-75%, this may require a reduction in the dose of digoxin.

Potassium depleting diuretics, corticosterois and a variety of drugs can also increase toxicity.

4-     Use of antibiotic lead to kill microorganisms like Eubacterium lentum so lead to toxicity.

Treatment of digitalis toxicity

1-     Removing of ingested drug by vomiting, gastric lavage, use of adsorbant agents eg. Activated charcoal, cholestyramine and colestepol.

2-     Maintenance of a normal potassium concentration: hypokalemia is more common after chronic digitalis toxicity, while massive acute overdoses often causes hyperkalemia. Potassium compete with digoxin on Na⁺/K⁺-ATPase pump so decrease K⁺ level and increase digoxin activity and toxicity.

Hyperkalemia may require treatment with insulin, dextrose, bicarbonate and sodium polystyrene sulfonate. In case of hypokalemia continuous potassium replacement may be sufficient. Potassium administration may correct arrhythmias restoring intracellular concentration.

3-     Reversal of arrhythmias: for atrial and ventricular arrhythmias that do not respond to potassium therapy the treatment of choice includes phenytoin and lidocaine. Phenytoin increase AV nodal conduction and directly reverse the toxic action of digitalis at AV node without interfering with its inotropic action. While quinidine and procainamide are not used because they slow AV nodal conduction.

If digitalis has produced AV block, the vagolytic action of atropine may increase heart rate and AV conduction. Catecholamines are contraindicated for the treatment of bradyarrhythmias because they increase the risk of precipitating more serious ectopic arrhythmias. Β-blockers such as propranolol are useful to suppress  supraventricular and ventricular arrhythmias induced by digitalis toxicity.

4-     Increase removal of unabsorbed drug: the use of dieresis or hemodialysis have not been successful because of the large volume of distribution for digitalis. Hemodialysis may be equired to control hyperkalemia.

5-     Use of specific antidote digoxin immune Fab: these are antibody fragments prepared by conjugation of digoxin to human or bovine serum albumin. This is then used to immunize sheep, which produce antibodies. Their sera are obtained and purified yielding the drug.

The fragments are less immunogenic and can be eliminated by glomerular filtration.

Adverse effects to digoxin immune Fab are minimal including