The Luria-Delbrück Experiment

Introduction

Today’s question is: do mutations happen randomly, or are they caused by some selective pressure? In other words, are mutations a Darwinian event where they happen by chance and then natural selection selects those that are favorable to pass on to the offspring, or are mutations Lamarckian where they happen because they help a species survive (like a giraffe constantly stretching its neck to reach the leaves at the top of the tree, thereby making its neck longer, and then passing that acquired trait to its offspring). To determine which of these two hypotheses is correct, we need an experimental test.

Let’s examine one famous experiment. To make things simple, consider a specific case. Assume we start with just one individual, who is not a mutant. Furthermore, let each parent have two offspring, and only analyze three generations. For the first two generations there is no selective pressure, and only in the third generation the selective pressure is present. To make the analysis really simple, assume the probability of a mutation, p, is very small.

The most common case is shown in the figure below. Blue circles represent the individuals in each generation, starting in the first generation with just one. Locations where lines branch represent births. (Wait, you say, each child should have two parents, not one! Okay, we are making a simple model. Assume an individual reproduces asexually by splitting into two. We should talk about “splittings” and not “births.”) The green dashed line represents when the selective pressure begins. So our picture shows one great-grandparent, two grandparents, four parents, and eight children. A mutation is indicated by changing a blue circle to red. 

Because p

 

Lamarckian Evolution

In the case when mutations are caused by some selective pressure (Lamarckian), you can get a more interesting situation like shown below. No one above the dashed line undergoes a mutation because there was no selective pressure then. A child below the dashed line in the bottom row might have a mutation. There are eight children, so the probability of one of the eight having a mutation is 8p. The probability of two offspring having mutations will go as p², but since we are assuming p is small the odds of having multiple mutant offspring will be negligible. We’ll ignore those cases.  

Let’s calculate some statistics for this case. Let n be the number of mutant offspring in the last generation (below the dashed line). To find the average value, or mean, of n over several experiments, which we’ll call , you sum up all the possible cases, each multiplied by its probability. In general, we could have n = 0, 1, 2, …, 8, each with probability p₀, p₁, …, p₈, so is 

= p₀ (0) + p₁ (1) + p₂ (2) + … + p₈ (8).

But in this case p₂, p₃, …, p₈ are all negligibly small, so we have only the first two terms in the sum to worry about.

For each individual, the odds of not mutating is (1 – p). In the last generation below the dashed line there are 8 offspring, so the probability of none of them having a mutation, p₀, is (1 – 8p). The probability for one mutation (p₁) is 8p because there are 8 offspring, each with probability p of mutating. So

= (1 – 8p) (0) + 8p (1) = 8p .

We will also be interested in the variation of results between different trials. For this, we need n²>

n²> = (1 – 8p) (0)² + 8p (1)² = 8p .

The variance is the mean of the square of the variation from the mean. In Appendix G of Intermediate Physics for Medicine and Biology, Russ Hobbie and I call the variance σ² and we prove that σ² = n²> – n>². In our case

σ² = n²> – n>² = 8p – (8p)² .

But remember, p p. Therefore, the mean and variance are the same. You may have seen a probability distribution with this property before. Appendix J of IPMB states that the Poisson distribution has the same mean and variance. Basically, the Lamarckian case is a Poisson process. 

 

Darwinian Evolution

Now consider the case when mutations occur randomly (Darwinian). You still can get all the results shown earlier in the Lamarckian case, but you get others too because mutations can happen all the time, not just when the selective pressure is operating. Suppose one of the parents (just above the dashed line) mutates. Their mutation gets passed to both offspring. The odds of mutating back (changing from red to blue) are very small (p

You could also have one of the two grandparents give rise to four mutant offspring below the dashed line, as shown below.

Let’s do our statistics again. As before, the vast majority of the cases have no mutations. There are now 14 cases, each of which could have the mutation in one of the offspring. All the cases are shown below.

The probability of having no mutations ever (the bottom right case) is (1 – 14p). The probability of one of the offspring having a mutation is 8p (the eight cases in the top row). The probability of any one of the parents having a mutation is p and there are 4 parents, so the probability of a mutation among the parents is 4p, and each would give rise to two mutants below the dashed line (the four cases on the left in the bottom row). Finally, one of the two grandparents could mutate (the fifth and sixth cases in the bottom row), each with probability p. If a grandparent mutates it results in 4 mutants below the dashed line. So, the mean number of mutants in the final generation is

n> = (1 – 14p) (0) + 8p (1) + 4p (2) + 2p (4) = 24p .

The odds of a mutant appearing in the final generation is three times higher in the Darwinian case than in the Lamarckian case. What about the variance?

n²>  = (1 – 14p) (0)² + 8p (1)² + 4p (2)² + 2p (4)² = 56p .

The variance is

σ² = n²> – n>²  = 56p – 24²p² = 56p

(remember, terms in p² are negligible). Now the variance (56p) is over twice the mean (24p). It is not a Poisson process. It’s something else. There is much more variation in the number of mutants because of mutations happening early in the family tree that pass the mutation to all of the subsequent offspring. 

Conclusion

In an experiment, p may not be easy to determine. You need to know how many individuals you start with (in our example, one) and how many generations you examine (in our example, three), as well as how many mutants you end up with. But you can easily compare the variance to the mean; just take their ratio (variance/mean). If they are the same, you suspect a Lamarckian Poisson process. If the variance is significantly more than the mean, you suspect Darwinian selection.  In our example, variance/mean = 2.3.

There are some limitations. The probability is not always very small, so you might need to extend this analysis to cases where you have more than one mutation occurring. Also, in many experiments you will want to let the number of generations be much larger than three. There is also the possibility of a mutant mutating back to its original state. Finally, during sexual reproduction you have the in-laws to worry about, and you could have more than two offspring. So, to be quantitative you have some more work to do. But even in the more general case, the qualitative conclusion remains the same: Darwinian evolution results in a larger variance in the number of mutants than does Lamarckian evolution.

I suspect you now are saying “this is an interesting result; has anyone done this experiment?” The answer is yes! Salvador Luria and Max Delbrück did the experiment using E. coli bacteria (so the asexual splitting of generations is appropriate). The selective pressure applied at the end was resistance to a bacteriophage (a virus that infects bacteria). Their result: there was a lot more variation than you would expect from a Poisson process. Evolution is Darwinian, not Lamarckian. Mutations happen all the time, regardless of if there is some evolutionary pressure present.

The Luria-Delbrück experiment, described by Doug Koshland of UC Berkeley. 

https://www.youtube.com/watch?v=slfLeKqE3Bg

Source: http://hobbieroth.blogspot.com/2024/12/the-luria-delbruck-experiment.html