CARDIAC PHYSIOLOGY AND ECG INTERPRETATION (First of Two Parts)

I.     Basic Concept

A.      Electrical Activity of the Heart

       -  Each contraction of the heart is preceded by excitation waves ofelectrical activity that originate in the sinoatrial (SA) node.

ü  SA node

·         is a small, flattened, ellipsoid strip of specialized cardiac muscle about 3 millimeters wide, 15 millimeters long, and 1 millimeter thick.

·         located near the junction of the superior vena cava and RA.

·         located in the superior posterolateral wall of the right

atrium immediately below and slightly lateral to the opening of the superior vena cava.

·         The fibers of this node have almost no contractile muscle filaments

and are each only 3 to 5 micrometers in diameter, in contrast to a diameter of 10 to 15 micrometers for the surrounding atrial muscle fibers.

·         However, the sinus nodal fibers connect directly with the atrial muscle fibers, so that any action potential that begins in the sinus node spreads immediately into the atrial muscle wall.

·         Consist of 2 cell types

1.       pacemaker of P cells – ovoid in shaped thought to perform the actual pacemaker function.

2.       Transitional or T cells – elongated lie between P cells and typical atrial muscle cells.

·         has an intrinsic firing rate of 90-120 beats/min being higher in young patients.

               - The waves of electrical activity spread through the atria and reach the atrioventricular (AV) node.

ü  AV node

·      located in the posterior wall of the right atrium immediately behind the tricuspid valve.

·      located beneath the endocardium

·      right side of the interatrial septum

·      only pathway for excitation to proceed from atria to ventricles.

       Note that the SA node tracing shows no steady resting potential, as does the ventricular muscle tracing.

       The SA node’s spontaneous depolarization and repolarization provides a unique and miraculous automatic pacemaker stimulus that activates the atria and the AV node, which conducts the activation current down the bundle branches to activate the ventricular muscle mass.

ü  Bundle of His (Fig. 1)
·   divides into R and L bundle branches
·   lie beneath the endocardium on the two respective sides of the ventricular septum.
·   Each branch spreads downward toward the apex of the ventricle, progressively dividing into smaller branches. These branches in turn course sidewise around each ventricular chamber and back toward the base of the heart.
·   The ends of the Purkinje fibers penetrate about one third the way into the muscle mass and finally become continuous with the cardiac muscle fibers.

Fig. 1. Sinus node, and the Purkinje system of the heart, showing also the AV node,

          atrial internodal pathways, and ventrical bundle branches.

                                     

                                     

  Cardiac cells outside the SA node normally do not exhibit spontaneous depolarization; thus they must be activated.


Here then is the conduction pathways of the heart:
SA node-->AV node --> Right and Left Bundle Branches --> Purkinje Fibers --> Ventricle


        Physiologic delay at AV node
·     To allow for ventricular filling. Without this normal delay at AV node and if there is an immediate ejection, air embolus may happen (Martin).
     ·      The AV node provides a necessary physiologic delay of the electrical currents, which allows the atria to fill the ventricles with blood before ventricular systole.
     ·     (Mechanism) The reason why there is a physiologic delay at AV node is because the fibers of AV node are extremely small. Thus, conduction velocity is slow. Also, the Purkinje fibers conduct very slowly (Martin)
     ·     The transient halt and slowing of conduction through the specialized AV
           node fibers play an important protective role in patients with atrial flutterand atrial fi brillation. In these common conditions, a rapid atrial rate of approximately 300 to 600 beats/min reaches the AV node; this AV “tollgate” reduces the electrical traffi c that reaches the superhighway that traverses the ventricles to approximately 120 to 180 beats/min, and serious life-threatening events are prevented.


       Heart block (AV block): Cardiac dromotropic incompetence; conduction interference
        ·   There is discrepancy in P-R interval; prolonged P-R interval.
        ·   Normal P-R interval is 0.12 to 0.20 sec.
            a.   First degree heart block
                 ·     involves Purkinje fibers and ventricles
                 ·     can still perform  exercise (light aerobic exercise)

            b.   Second degree heart block (Mobitz type 1)
                 ·      involves AV Bundle of His, right and left bundle branches, Purkinje Fibers, and Ventricles.
                 ·       P on P; absence of QRS complex:
                       Ø  Wenckebach phenomenon (named after the Dutch Internist Karl Wenckebach)      describes a disturbance in conduction where the atrioventricular node conducts each successive impulse earlier and earlier. Eventually, an impulse arrives when the node is not able to conduct it. The following impulse will then be conducted normally, causing the cycle to begin again.
                       Ø  The QRS complex and the P wave are normal, but not all P waves are followed by a QRS complex. The PR interval gets longer until an impulse is not conducted.
     Ø  The ECG trace below shows sinus rhythm at a rate of 67 bpm (arrowed), with a lengthening PR interval (shaded) preceding a nonconducted P wave.

(Fig.2)  Electrocardiogram: Second Degree Block — Mobitz Type I (Wenckebach

                                 

                     ·      exercise is contraindicated

             c.  Third degree heart block (Mobitz type II)

                   ·      complete heartblock; from SA node to ventricles are affected

                   ·       More on P on P phenomenon than Mobitz type I

                   ·       artificial cardiac pacemaker needed

       Pharmacology: Effects of meds on heart block

         ·      Heart block may lead to cardiac arrest. Both HR and BP decrease.

         ·       Meds should be administered to increase both the HR and BP.

         ·       Pharmacologic effect: Abrupt; 

 § Cardiac terminology:

    a) Abrupt:  rapidly increasing

    b) Shunted-Blunted: gradually decreasing

    c) Plateau: steady state

                 §  Abrupt increase in HR and BP

                    ü  Anticholinergic drug: sympathomimetic and parasympatholytic

                         Ø  Atropine SO4: antidote to cholinergic crisis; it blocks the acethylcholine release in the parasympathetic.

       Artificial cardiac pacemaker (for third degree heart block)

        ·     Poorly insulated pace maker (in the past): very sensitive; when a pt is grounded, it reverts to default settings. Ultrasound should be applied 6 inches away from the pacemaker. Use of electrical modalities is relatively contraindicated.

        ·     Well-insulated pacemaker (2011): nothing to worry about. ES, US, and other electrical modalities, except MRI, are not contraindicated. These electrical modalities can be used as long as they are more than 2 inches from the apparatus.

       Cardiac Action Potential

        -    Our heart is excitable tissue.

        -     Normal resting membrane potential of the heart (RMP): -60 to -80 mV

        -      resting membrane potential is in the polarized state.

        -      In a resting cardiac muscle cell, molecules dissociate into positively charged ions on the outer surface and negatively charged ions on the inner surface of the cell membrane; the cell is in an electrically balanced or polarized resting state.

        -      Cardiac action potential has 5 phases.

        -        Phase 4 is an rmp phase which is brought about by sodium-potassium pump and by the presence of the negatively charged protein which is imperpeable. Thus, rmp phase is in the negative state. 

        -        Phase 0 is a rapid depolarization. At this point, the membrane is less negative brought about by the sodium (Na+) influx. Sodium then gets inside the membrane making it less negative until it reaches +30mV, a peak potential, which then undergoes an overshoot.

        -       This peak potential immediately goes down losing its positivity which is now the initial repolarization phase (phase 1). Thus, phase 1 is the initial repolarization.

        -       And since the membrane loses positivity, it then loses positive ion because potassium (K+) goes outside (potassium efflux). (Sodium displaces potassium, hence potassium efflux occurs.)

        -       Once potassium is outside the membrane, calcium then goes inside since both potassium and calcium are positively charged ions (cat ion) They cannot stay together. There is repulsion.

       -         The influx of Calcium then creates a steady state which is the plateau phase. (Nerve action potential doesn’t have a plateau phase.)

       -        This plateau phase (phase 2) is only momentary because Calcium has only a scanty amount of concentration, and therefore it is easily used up while potassium efflux continues

       -        As the potassium continues to go outside the membrane, repolarization (phase 3, the real repolarization) occurs

       -       Next depolarization does not happen if sodium (and calcium) remain inside while potassium stays outside the membrane. They have to be brought back to where they came from. Thus, sodium-potassium happens again. It pumps 3 Na+ from the intracellular to the extracellular fluid compartment. It also pumps 2 K+ from extracellular to the intracellular fluid compartment. So the net diffusion is 1, that is, 1 Na+. If you have five cycles of pump, then you have 15 Na+ pumped to the outside and 10 K+ to the inside. The net diffusion is 5 Na+ outside, thus the outer surface of the cell membrane is positively charged. The inner surface is negative. Hence, the rmp is always negative.

   fig. 3. Action potential

                                   

Fig. 4. Sodium infl ux, potassium eff ux, the action potential, and the electrocardiogram.(From Khan, M. Gabriel: On Call Cardiology, 3rd ed., Philadelphia, 2006, WB Saunders, Elsevier

             Science.

        - When the cardiac impulse passes through the heart, electrical current also spreads from the heart into the adjacent tissues surrounding the heart. A small portion of the current spreads all the way to  the surface of the body. If electrodes are placed on the skin on opposite sides of the heart, electrical  potentials generated by the current can be recorded; the recording is known as an electrocardiogram.

 Fig. 5(From Guyto and Hall). A normal electrocardiogram for two beats of the heart.

                                  
12 Leads to record ECG
a. Limb leads
    1) Unipolar = aVleads: aVR, aVL, aVF

         
      
    2) Bipolar  = I(right/left UE), II(right UE/left LE), III(left UE/right LE)\

        


 b. Chest Leads:
     V1 = red (4^th interscostal space right from the sternum)          
     V2 = yellow (4^th intercostals space left from the sternum)                         
     V3 = grey (between V2 and V4)
     V4 = brown (apex of the heart; 5^th intercostals space left mid-clavicular line)                                   
     V5 = black (5^th intercostals space left anterior axillary line)
     V6 = purple (5^th intercostals space left mid axillary line) – hidden when shoulder is adducted.

   

  Leads represent different parts of the heart
  aVR = superior wall of the heart
  I, aVL, V5, V6 = lateral wall of the heart
  II, III, aVF= inferior wall of the heart
  V1 & V2 = septal regions of the heart; also send reflection to the posterior wall of the heart.
  V1, V2, V3, V4 = anterior wall of the heart


 Leads changes may indicate wall infarct
WALL INFARCTS                                         LEADS AFFECTED
Lateral                                                               I, aVL, V5, V6
Inferior                                                              II, III, aVF
Posterior                                                           V1, V2
Anterior                                                             V1, V2, V3, V4 


       Posterior wall infarct: a tall T wave, a tall R wave, and a persistent ST depression.
       Components of an ECG waveform
         ·    ECG strip is like a graphing paper
         ·    traces of evoked potential could be seen.
         ·    There are two axes:
              §     X axis = horizontal axis
              §      Y axis = vertical axis

•  Each of the strip (graphing-like paper) has a small square

·      The area of this small square is 1mm x 1mm. or one square mm.

·      The horizontal line of this square is the wave duration which is 0.04 sec.

·      The vertical line of this square is the wave amplitude which is 0.1 mV.

·      So, each small square has 0.1 mV and 0.04 sec.

·      A small is too small to read on an ECG strip. A block makes reading an ECG easier.

·      A block is composed of small squares.

The dimension of a block:

·      Its area is 25 square mm; thus, it is a 5x5 mm.

·      It comprises of one isoelectric line.

·      It is easy to recognize the block because it is highlighted.

·      The wave duration of one isoelectric line (horizontal) is 0.20 sec.

·      It is 0.20 sec because the wave duration in each one small square is 0.04 sec. And since the block has a measurement of 5mm horizontally, 0.04 sec then is multiplied by 5. The product is 0.20 sec.

·      The wave amplitude of one isoelectric line is 0.5mV.

·      It is 0.5 mV because the vertical length of a small square is 0.1 mV. And since a block has a vertical length of 5mm, then 0.1 mV is multiplied by 5 which is equal to 0.5 mV.

·      Therefore, one isoelectric line is 0.5 mV @ 0.20 sec

       Examples:

       v  If you are reading a 1 sec strip, how many small squares are there?    Answer: 25 small small squares. Or 5 (big) blocks because five 0.20 sec is equal to 1 sec.

       It looks like this (elementary math J):

               0.20 (1 block)

               0.20 (1 block)

               0.20 (1 block)

               0.20 (1block)

               0.20 (1 block)

             ----------------------

      1.00 sec

Thus, there are 5 blocks in 1 sec. strip.

v  How many big blocks if you are reading a 5 sec strip?

Answer: 25 big blocks or 125 small squares

v  How many big blocks if you are reading a 1 min. strip?

Answer: 300 big blocks

1 min. is equal to 60 sec. And when you multiply 60 sec by 5, the product is 300.

v  How many big blocks in a 45 sec strip?

Answer: 225 big blocks

45 sec multiplied by 5 is equal to 225.

Always remember that 1 sec = 5 block.

v  How many big blocks in a 30 sec strip.

Answer: 150 big blocks

v  How many blocks in a 15 sec strip?

Answer: 75 blocks.

v  This then is the sequence:

0 à 150 à 225 à 300

                                      (15s)    (30s)     (1min)

v  How many small squares in 3 mV?

Answer: 30 small squares or 6 big blocks

Each block has a wave amplitude of 0.5 mV. 1 mV then is equal to 2 blocks. So, if you have 3 mV, you have to multiply 2 blocks by 3 mV which is equal to 6 blocks.

                                       To get the number of small squares, multiply 6 blocks by 5. The answer is 30 small squares.

v  If you have 13 small squares, how many millivolts?

Answer: 1.3 mV

There is 1 mV in 2 big blocks or 10 small squares. And there are 3 remaining squares since what is given in the problem above is 13 small squares. So, 3 small squares are equal to 0.3 mV, since each small square has a wave amplitude of 0.1 mV.

Therefore, the answer is 1.3 mV. Simply put, 1 mV plus 0.3 mV is equal to 1.3 mV.

v  How many millivolts in 15 small squares?

Answer: 1.5 mV

15 small squares or 3 big blocks.

3 big blocks are multiplied by 0.5 mV. The product is 1.5 mV.

v  How many big blocks in 2.3 mV?

Answer: 4 blocks  and 3 small squares

Again as explained above, each 1mV is equal to 2 blocks. So, if you have 2 mV, then you have 4 big blocks.

The remaining 0.3 mV corresponds to 3 small squares, since each small square is 0.1 mV.

v  How many big blocks in wave amplitude of 0.9 mV?

Answer: 1 big block and 4 small squares

0.5 mV = 1 block

0.4 mV = 4 small squares

v  If you have 0.65 mV, how many big blocks?

Answer: 1 block and 1 ½ small squares

A wave amplitude of 0.50 mV is confined to 1 block. The remaining 0.15 mV includes 1 small square and a half. Hence, 1 block and 11/2 small squares.

0.15 mV contains 1 small square and a half, since each small square has a wave amplitude of 0.10 mV and the remaining 0.05 is half of 0.10 mV. So the answer is 1 small square and a half.

v  Given: 0.20 mV @ 0.16 sec.

How many big blocks and small squares?

Answer: 0.20 = 2 small squares; 0.16 sec = 4 small squares

There are two small squares in 0.20 mV, because each small square has 0.1 mV (or 0.10 mV). Since the give wave amplitude above is 0.20 mV, the answer is 2 small squares.

There are 4 small squares in 0.16 sec, because each small square has a wave duration of 0.04 sec.

0.16 sec divided by 0.04 sec is equal to 4 small squares. Or you may use multiplication to make it easier for a manual computation. So, 4 x 0.04 sec is equal to 0.16 sec.

v  Given:

a)       Wave amplitude: 6mm

b)       Wave duration: 31/2 mm

Answer:

a)       0.6 mV

Remember, the vertical length of each block is 5mm which is equal to 0.5 mV. And each small square has a vertical length of 0.1 mV. So, add 0.5 mV to 0.1 mV, then you have a wave amplitude of 0.6 mV.

b)       0.14 sec.

Again, the horizontal length of each small square is 1 mm with a wave duration of 0.04 sec. So, if you have horizontal length of 3mm, then you have to multiply 0.04 sec by 3mm. The product is 0.12 sec. The remaining ½ mm is 0.02, which is half of 0.04 sec.

It looks like this:

     0.12 sec (3mm)

+ 0.02 sec (1/2mm)

------------------------

    0.14 sec

v  Given:

a)       12 small squares going up (vertical direction)

b)       2 big blocks and 21/2 small squares going horizontally

Answer:

a)       1.2 mV

b)       0.50 sec

It is 1.2 mV that is confined to 12 small squares going up, because each block has wave amplitude of 0.5 mV and each small square has wave amplitude of 0.1 mV. So, if you have 12 small squares, then you have two big blocks and two small squares. There is 1 mV in two big blocks and 0.2 mV in two small squares. Therefore, the answer is 1.2 mV.

The wave duration is 0.50sec which corresponds to two big blocks and 21/2 small squares in the horizontal line, because each big block corresponds to 0.20 sec. And since, there are two big blocks, the wave duration then is 0.40 sec, that is, 0.20 plus .20.

The 21/2 remaining small squares correspond to 0.10 sec. It is so because each small square has wave duration of 0.04 sec. So, two small squares have a wave duration of 0.08 sec. And half of the small square is 0.02. Therefore, the answer is 0.50 sec.

Simply put:

 0.40 sec (two big blocks) + 0.08 (two small squares) + 0.02 (half square) = 0.50 sec

References

Gabriel Khan, M. Rapid ECG Interpretation, 3^rd Ed. Totowa, New Jersey: Human Press, 2008.

Guyton, Arthur C., and John E. Hall. Text Book of Medical Physiology, 11^th Ed. Philadelphia, Pennsylvania: Elsevier Inc., 2006.

Martin, Gerard L. “Cardiac Physiology and ECG Interpretation.” Class lecture, SLRC, Sampaloc, Manila, May 6, 2011. 

Pablo-Santos, Ramona Luisa, PT-OT Reviewer, Manila, Philippines: Ramona Luisa P. Santos, M.D., 1995.

Paz, Jaime C.  and Michelle P. West. Acute Care Handbook for Physical Therapist, 3^rd Ed. Boston: Butterworth-Heinemann, 2009.

_________. “Rhythms and Arrythmias.” Cardionetics. http://www.cardionetics.com/cardiology/second-degree-block.php (accessed June 30, 2011).

                                                                                                                             R.O.L.