MTSH discussion

Understanding and Using the MTSH survival equation

At the end of last week's lecture I discussed the multitarget, single-hit model for radiation-induced damage to cells. The discussion was a bit hurried and left some significant points out, and I want you to understand these points when you take the midterm, so I'd like to run through this material in a slightly different format--namely, this document.

Derivation: I won't produce the derivation of the MTSH survival-curve equation here, other than to say that it follows logically from the set of assumptions in the Lea target theory. Alpen lays out those assumptions clearly, and I encourage you to think carefully about them. The gist is this: in order for the damage to a cell to become serious enough to lead to clonogenic death, there have to be multiple hits onto the DNA. If we count probabilities of interaction with DNA in terms of the ratio of the volume occupied by DNA in a single cell to the total volume irradiated, the MTSH equations arise.

The equation: The survival curve equation can be written
S / S₀ = 1 - (1 - exp(-kD))ⁿ
or
S / S₀ = 1 - (1 - exp(-D/D₀))ⁿ
or
ln(S / S₀ ) = ln(1 - (1 - exp(-D/D₀))ⁿ)
where
S = mumber of clonogenically viable cells after exposure to dose D
S₀ = number of clonogenically viable cells in the absence of exposure
k = inactivation constant, discussed below
D₀ = inactivation dose, discussed below; equals 1/k
n = experimentally determined number of targets per cell

So the measured values are dose (the independent variable) and the survival fraction, S / S₀ (the depenedent variable). The parameters we need to determine from experiment are the inactivation constant kor its reciprocal, D₀; and n, the nominal number of targets per cell.

The graph: It is conventional to plot ln(S/S0) against dose. S = S0 at D = 0, so a plot of ln(S/S0) against dose begins at (D=0, ln(S/S0) = 0), since ln(1) = 0. The curve will become linear with a negative slopeat high dose, for reasons we will explain below. In between the origin and the linear region, it will curve. The curve begins with a slope of 0 at D = 0, for reasons I will outline below. Thus the curve looks like this:
MTSH curve

Single-hit case: The equation simplifies substantially in the case of single-hit kinetics, for which n = 1. In that instance 1 - (1 - exp(-D/D₀))ⁿ = 1 - (1 - exp(-D/D₀)) = exp(-D/D₀)
which is the same equation as we already discussed for the single-hit Lea model.
The log-survival is very simple: ln(S/S₀) = ln(exp(-D/D₀)) = -D/D₀.
The derivative is simple too:
d(S/S0)/dD = (-1/D0)exp(-D/D0)
and
d(ln(S/S0)/dD = d(ln(exp(-D/D₀) / dD = d(-D/D₀)/dD = -1/D₀So the slope of log(S/S₀) versus D is is constant and nonzero at all doses.

Slope of the curve: As usual, we want to know the slope of the survival curve. We'll determine the slope of S/S₀ versus dose and ln(S/S₀) versus dose. For n > 1 (the only case we're interested in other than the one we just covered),

d(S/S₀) / dD = d(1 - (1 - exp(-D/D₀))ⁿ)/dD = -n(1 - exp(-D/D₀))^n-1(-1/D0)exp(-D/D₀)
d(S/S₀) / dD = (n/D₀)exp(-D/D₀)(1 - exp(-D/D₀))^n-1
To compute the slope of ln(S/S₀) versus D, we remember that
d ln(u) / dx = (1/|u|) du/dx
So
d(ln(S/S₀))/dD = [1/(S/S₀)](n/D₀)exp(-D/D₀)(1 - exp(-D/D₀))^n-1

Low-dose limit: As we already discussed, S = S₀ at D = 0. We can show easily that d(S/S₀)/dD = 0 and d (ln(S/S₀))/dD = 0 at D = 0 for n > 1:
d(S/S₀)/dD = d(1 - (1 - exp(D/D₀))ⁿ) / dD = (n/D₀)exp(-D/D₀)(1 - exp(-D/D₀))^n-1Therefore at D = 0, d(S/S0)/dD = (n/D₀)exp(-0/D₀)(1 - exp(-0/D₀))^n-1 = (n/D₀)(1)(1 - 1)^n-1 = 0
Furthermore, at D = 0,
d(ln(S/S₀))/dD = [1/(S/S₀)](n/D₀)exp(-0/D₀)(1 - exp(-0/D₀))^n-1 = [1](n/D₀)(1)(1 - 1)^n-1 = 0,
since we already know that S/S₀ = 1 at D = 0. Note that this analysis requires n > 1, because otherwise the derivative we just discussed isn't correct.
Thus the slope of the survival curve is flat at very low dose. This fact has subjected the MTSH model to substantial criticism from many analysts, who believe that survival should have a nonzero dependence on dose even at very low dose. There are both scientific and political reasons to dislike a zero slope at low dose, but (in my opinion) they are insufficiently potent to dismiss the MTSH model on that grounds. We will soon discuss an alternative model of cell survival known as the linear-quadratic model for which the slope at low dose is nonzero.

High-dose limit: Here we consider the case where D >> D₀, so that D/D₀ is large and exp(-D/D₀) is small. Let u = exp(-D/D₀), so that u << 1. We can write
S/S₀ = 1 - (1 - u)ⁿ
But since u << 1, we can expand (1 - u)ⁿ as a binomial expansion in u:
(1 - u)ⁿ ~ 1 - un + u²n(n-1)/2 - . . .
where the ellipsis indicates terms involving powers of u higher than 2. Therefore
S / S0 ~ 1 - ( 1 - un + u²n(n-1) /2 - . . . ) ~ un,
where we have eliminated even the term in u². Thus for large enough values of D that the terms that are quadratic in u are negligible,
S / S₀ = nu = nexp(-D/D₀)
and
ln(S/S₀) = ln(nexp(-D/D₀)) = ln n + ln(exp(-D/D₀)) = ln n - D / D₀
This brings us to an important result: the plot of ln(S/S₀) versus dose is linear at high dose, with slope -1/D₀, and the y intercept of the resulting line is ln n.

This provides us with the best way of determining whether a real experiment exhibits MTSH characteristics. If we plot ln(S/S₀) against dose and find that the curve has a small slope at low dose, and a constant negative slope at high dose, then the curve suggests that MTSH kinetics might be operating. The slope of that curve will be -1 / D₀, and the y intercept extrapolated back will be ln n, from which we can easily measure n.

Quasi-threshold dose: A final parameter should be defined: the quasi-threshold dose. This is defined as the dose that is observed by extrapolating the linear portion of the MTSH log-survival curve back to the point where it crosses the ln(S/S₀) = 0 point. It is easy to determine what that dose value is, since the linear portion of the curve is defined by ln(S/S₀) = ln n - D / D₀
so the point on that line where ln(S/S₀) = 0 is 0 = ln n - D / D₀, D = D₀ ln n. We call this value D_q, and we see that D_q = D₀ ln n. Examination of the curve given above shows that the slope of the curve is almost zero all the way from D = 0 up through D_q; accordingly we tend to think of the D_q value as the one at which the dose starts to really have a significant effect on survival. That is why we describe it as the quasi-threshold dose. It is not a true threshold, since some cell killing occurs even at D < D_q, but there isn't much. The larger the value of n, the flatter the curve is at low dose. There is no threshold at all at n = 1 (since D_q = 0 in that case), but for larger values like n = 5, the curve really does look nearly flat until we get to D = D_q.