(1943; Genetics 28:491-511)

Dean W. Gabriel

When a human acquires immunity to disease, he starts out in a state in which he is susceptible, but winds up resistant. Is the resistant state heretible? A bacterial population may be used to model the situation. The starting bacterial culture is susceptible, and yet it winds up resistant, after most are killed off. The question is, does exposure to the disease

Luria & Delbruck used *E. coli* bacteria and a virus (phage)
that lyses the bacteria. The observation was that if you start with a pure
(from single colony) culture, add the phage, it will clear the culture.
After a few hours or days, the culture is back. Nothing was added. Where
did the resistant bacteria come from? It wasn't an introduction, so it
was either mutation or some kind of resistance response (say, similar to
the way we might resist or die from, the plague).

Hypothesis 1: (Mutation hypothesis): There is a finite and small probability
(a) that a mutation from susceptible to resistant will occur at a constant
rate.

Hypothesis 2: (Physiological resistance hypothesis): there is a finite
and small probability that each cell in the culture might survive the phage
attack, and all subsequent progeny from survivors will be resistant.

Hypothesis 1 corollaries:

---will generate resistant clones, and their prevalence will be determined by how early in culture growth the mutation event occurred.

---the proportion of resistant bacteria will increase over time, prior to selection.

--if you study a large number of small cultures, the proportion of resistant
bacteria in each culture will vary or fluctuate wildly, depending on how
early the mutation occurred.

Hypothesis 2 corollaries:

--no clones. Some bacteria will survive and reproduce resistant daughter cells. Selection begins only after the addition of phage, not before.

--the proportion of potential survivors (resistant) bacteria will remain constant.

--if you study a large number of small cultures, the proportion of resistant
bacteria will remain constant; size of the culture is irrelevant.

The experiment:

First, to demonstrate good sampling techniques, a large culture was
used. A carefully measured sample was drawn, plated on an agar plate loaded
with phage, and the total # of survivors was counted. A dilution series
was also performed to get the total number of bacteria per ml of culture.

Typically:

Experiment 1 | Experiment 2 | Experiment 3 | |

Sample 1 | 14 | 46 | 4 |

Sample 2 | 15 | 56 | 2 |

Sample 3 | 13 | 52 | 2 |

Sample 4 | 21 | 48 | 1 |

mean | 16.7 | 51.4 | 3.3 |

variance | 15 | 27 | 3.8 |

Note that in each experiment, the sampling error is low. But between
experiments, there is variation, indicating some support for the mutation
theory.

They then grew very small cultures of E. coli, and sampled and plated
as before. The results were typically:

Exp 1 | Exp 3 | Exp 4 | |

Culture 1 | 10 | 30 | 1 |

Culture 2 | 18 | 10 | 0 |

Culture 3 | 225 | 40 | 0 |

Culture 4 | 10 | 45 | 7 |

Culture 5 | 14 | 153 | 303 |

mean | 26.7 | 62 | 26.2 |

variance | 1,217 | 3,498 | 2,178 |

With a large number of small cultures, the total number of resistant bacteria fluctuates wildly from culture to culture, within a given experiment (same medium, same day, same phase of the moon, etc.) Note the variance. This provides good evidence for the mutation hypothesis. A possible objection: maybe the probability of survival by the second hypothesis is highly variable in small cultures. This is possible, but remained a poor possibility in light of the authors' demonstration that despite the wild fluctuations in numbers among cultures, the mutation rate remained constant.

(The calculations provided below are for your benefit, and will not be required for exams: )

The calculation is based on two things: the generation time--the
time for doubling (a common calculation in bacteriology), and on a determination
of the mutation rate (a difficult calculation that is not ordinary):.

First, the doubling time:

N_{t}= 2^{n} (N_{o}), where N_{t} is
total number (measured), N_{o} is starting number (measured) ,
and n is # of bacterial generations (calculated).

Then, ln N_{t} = n(ln 2) + ln N_{o}

n =__ ln N _{t} - ln N_{o}__

ln 2

If a = mutation rate or chance of mutation per individual per generation,
then

a = __m__

n

where m = number of mutations (not number of
mutants [which could be measured directly]!)

The problem is how to correlate the total number
of __mutants__ with the number of __mutations__. The total number
of mutants depends on the growth rate of cells carrying "old" mutations
("mutants") plus cells with the new mutations. That is, dm (increase in
number of mutants) = M_{t}dt (total number of mutants at time t)
+ a N_{t }dt (new mutants added).

But we cannot determine exactly how far back
old mutations occurred, and their subsequent clonal replication began;
that is, we cannot directly measure the number of mutations, m.

We can, however, apply a probability function
to estimate the average number of mutations per milliliter. This is because
there are two possible "states", mutant or wild type, resistant or susceptible.
Let P = the probability of the mutant state; and Q = the probability of
wild type. Since the number of mutations is exceedingly small, we cannot
apply a binomial distribution here. Instead we use a Poisson distribution,
where

P_{o} = e^{-y}, P_{1}
= e^{-y} y, P_{2} = __e__^{-y}__ y__^{2}
etc.

Y!

in the above, y represents the average number
of events per unit of space or time. In our case, it represents m, the
average number of mutations per ml.

If we look at experiment 4, above, for example, we see that

P_{o} = 2/5. Therefore, 2/5 = e^{-m}.
It is then a simple matter to solve for m.

In the small culture experiments, the mutation rates were:

Exp. 1 | 1.8 X 10^{-8} |

Exp 2 | 1.4 X 10^{-8} |

Exp 3 | 4.1 X 10^{-8} |

Exp. 4 | 2.1 X 10^{-8} |

Exp.5 | 1.1 X 10^{-8} |