Measuring Intracellular [Ca2+]

The following fundamentals are common to all techniques that use absorption or fluorescence.


Is an example of a first order process. Light in a homogeneous medium is absorbed in proportion to the existing amount (in other words, photons are absorbed independently from one another).

This leads to an important equation and concept.

Water does not absorb light. Some dissolved molecules do (dyes). These absorb proportionally to their concentration C.

Light in the figure travels from left to right. The difference between the intensity at position x + dx and at position x is represented by dI. In the case of absorption it is negative.

dI = -I C dx (l). (1)

This is the fundamental differential equation of absorption. It contains many pieces of information.

First, there is only one spatial coordinate, x, meaning that the interesting things occur only along the path of propagation. Perpendicularly to it the process just reproduces itself. The intensity of light in this case must be understood as the intensity in a region of fixed area.

The change in intensity is negative (absorption, dispersion, scattering, all can be grouped together).

The change is proportional to I, (number of incident photons). This is the essence of a first order process.

The change is proportional to the number of absorbing molecules ( C dx).

The proportionality constant , is a function of the wavelength (absorbers have color).

Another way of putting eqn. 1 is dI/dx = -C I.

We know that the function I that satisfies this differential equation is

I = Constant e-C()x = I0 e-C()x (Lambert and Beer's law). (2)

Because absorption leads to an exponential decay, the measure of absorption is defined logarithmically.

Absorption - log (I/I0) = C ( ) x log10e (3)

() log10e is called the extinction coefficient, (), hence Absorption = C x

ABSORBANCE = absorption / x

hence, () = absorbance/C

therefore, () is also called the molar absorbance

Recipe, to find the extinction coef. of a dye, make a 1M solution, put it in a 1cm cuvette and get the absorption graph in Fred's spectrograph.

(That would not work. Why ? Hint, the extinction coef of ApIII at 550 nm is 2.55 104 M-1 cm-1 ).


In an ion sensitive dye the extinction coefficient of the dye, D(), and that of the dye:ion complex, DI(), are different.

In a cell of diameter p, with a cytoplasmic dye that is free of ion, the absorption will be

A (0) = [D]T 0.7 p D() +Ai (0) (4)

where [D]T is dye concentration and the 0.7 is needed because only 0.7 of the volume of the cell is free aqueous cytoplasm. Ai is the intrinsic absorption, not related to the dye. The (0) argument indicates that the cell is at rest, time 0.

When the cell is stimulated, things change, Ca is released and increases in the cytoplasm, other concentrations may change.. Part of the dye, if it is Ca-sensitive, binds and converts to CaD. The absorption becomes

A(t) = 0.7 p [CaD] CaD() + 0.7 p {[D]T-[CaD]}D( ) + Ai(t) (5)

Subtracting (5)-(4)

A = A(t) - A(0) = 0.7p {CaD() - D()}[CaD] + {Ai(t) - Ai(0)} (6)

The fist term in eqn 6 is the Ca:Dye signal "S", the second is the "intrinsic signal".

The intrinsic signal can be studied in the absence of any dyes, and one can subtract it, obtaining the pure Ca:Dye signal

S= A-Ai = 0.7p {CaD() - D()}[CaD] = 0.7p () [CaD] (7)

After this correction one is left with the Ca:dye signal, and in this and most cases, the important concept is that it is proportional to [CaD]. (In more general cases the change in ion:dye signal will be proportional to the change in [ID]).

Obviously, from (6), in order to calculate [CaD] it is necessary to know the fiber diameter (vertical) and the change in of the dye upon binding Ca.(which is found in calibrations).

From [CaD] it is usually easy to derive the [Ca2+]. If the dye responds rapidly relative to the speed of the change in [Ca2+] one may assume that the Ca:dye reaction is always in equilibrium, or


Here one substitutes [D]T-[CaD] for [D] and after some work:


All that is needed to go from a light measurement to a [Ca2+] measurement (or other ion) is equation (9), the "chemical" equation, and equation (6), the "optical" equation.

One can substitute [CaD] from (7) into (9), to obtain [Ca2+] in terms of the signal. (10)

note that [D]T 0.7p is the maximum possible signal (11)

This equation is not limited to absorption dyes, it is valid for fluorescence also, and more generally appears as (12)

Note how the concentration first grows linearly with the signal, and as saturation is reached, grows beyond all bounds.

For this method to be used, one needs to know [D]T (or Smax). "Ratioing" dyes allow the determination of dye concentration by measuring absorbance at a different wavelength (isosbestic), where = 0, then using eqn (4).

An application.

Simultaneous recording of [Ca2+] and [H+].

The recording of [H+] in the presence of high [EGTA] is a way to record Ca release (Chandler and coworkers). Unlike our methods, it records cummulated release, not release rate. Two H+ are displaced from EGTA for every Ca2+, so that [H+] is proportional to total increase in Ca:EGTA, which in turn is approximately equal to total released Ca when [EGTA] is very high. Some advantages of this method over ours will be seen with the records.

The Ca sensitive dye is Antipyrylazo III, and the H-sensitive dye is Phenol Red (PR). Both are absorption dyes. In the figure are graphs of absorption vs. at different [H+]. For PR is very high at 560 nm (in cells, this peak shifts to ~575 nm. Its KD = 20 nM. ApIII is almost insensitive to H+. In blue is its absorption in the presence of saturating Ca. ApIII has an isosbestic point at 590 nm and a peak of at 720 nm. Hence one can obtain a pure Ca signal, without interference from PR at 720 nm and a not completely pure H signal at 575 nm, with some interference from ApIII.

The interference is two-fold: at rest, ApIII absorbs at 575 nm, hence it will reduce the light and the H signal. During Ca release, the absorbance of ApIII will increase, hence add its own signal to the change in absorbance due to PR. These interferences can be corrected for.

This arrangement of filters is used to do the measurements

Ignore DM1 and the light source H2, used for other measurements.

A halogen bulb, H1, of wide spectrum, generates the light, that is focused on a slit on the fiber. The various filters will be described.

The records are the raw intensity changes at three wavelengths, during a 100 ms pulse to -30 mV. To be used quantitatively, the records first have to be converted to absorption. For this we need to record the baseline intensity, not shown in the figure. (It was ~7 V at 720 nm, 1 V at 575 and 3 V at 850 nm).

At what wavelength(s) is there an increase in absorption?

Which signal is biggest?

Because the concentration of ApIII at this time was very low, the interference with the signal at 575 nm is very small.

In addition to these records, one must do other measures of absorption at rest, to calculate the concentration of the two dyes, which is constantly increasing during the experiment.

The next figure shows the change in absorption at 575 during a set of pulses designed to test the effect of depletion on Ca release. A brief reference pulse, followed by a 900 ms pulse to +50 mV, and a test pulse, same as the reference.

The following figure shows Ca analysis. In red is Ca release flux calculated from the Ca transients (not shown, calculated from the Ca:dye signal at 720 nm). In black is

the time derivative of the H:dye absorbance at 575 (meaning that I subtracted the intrinsic signal and the Ca:dye signal component.) While the two signals are somewhat similar, there are important differences.

The change in dA(575)/dt accompanying the test pulse decays much more after depletion than the release record.

We probably won't have time to discuss this, but the release calculation in the presence of 40EGTA becomes very simple, and the result very similar to the Ca transient itself, which in turn is very similar to the raw intensity signal. This can be shown in fig. 3 of previous page, where the signal at 720 nm is already very similar to a release flux record. Hence the release flux is probably correct.

The conclusion is that the PR signal is not proportional to release. IT SATURATES.

This would be expected because the pKA of phenol red is 7.7, which is equivalent to saying that the KD is 20nM. This implies that at pH 7, [H+]=100 nM, the fraction of PR bound to H, which is equal to

[H+]/([H+]+KD) (from eq. 8), is 83%.

I therefore decided not to use the absorbance at 575 as a measure of release, but the [H+], calculated from the A(575) in the same way as we calculate [Ca2+] from A(720), using eqn (12).

The figure shows in black release flux records and in red the derivatives of [H+](t), for a pulse pattern as in the previous figure, but for increasing values of depleting pulse duration. It can be seen that, once the [H+] is used, the agreement with release flux is much better.

Finally we can do some physiology. Work is in progress.