Statistically, why is the number of mutated genes in eggs treated with chemical mutagenesis one?

Question

Excerpted from the Guide to Research Techniques in Neuroscience ^[1]:

In chemical mutagenesis, a scientist applies a mutagenizing chemical, such as ethyl methane sulfonate (EMS) or N-ethyl-N-nitrosourea (ENU), to thousands of eggs/larvae, which statistically creates lines of ani- mals with mutations in a single gene in the genome.

My hypothesis is that the discrete probability distribution of the number of mutated genes can be approximated by a normal distribution with mean=1, thus it is said statistically the number of mutated genes is one. But how can the scientists know which organism or cells contains only one mutated gene, or for that matter one mutated gene that is caused only by chemical mutagenesis?

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^[1] Carter, Matt, and Jennifer C. Shieh. Guide to Research Techniques in Neuroscience. Academic Press, 2015. ISBN: 978-0-12-374849-2

Answer 1

The expected number of mutation would not follow normal distribution (ND) as you speculated because a ND would have negative values too as it ranges from $(-\infty,\infty)$, as Remi has also pointed out. In many cases a ND can be approximated because the variance would be so less that you can practically find no negative values. However, it is a wrong assumption to consider all biological statistics as ND by default. Please see this post. Basically, ND approximations can be made on multiple measurements that follow the central limit theorem. Now, in your case of a ND centered at 1, I'll tell you how that is impossible. You can have 2 and more than 2 mutations but you cannot have less than 0 mutations. The distribution is not symmetric and is skewed towards right. You also cannot have fractional mutations and therefore the distribution has to be discrete and not continuous.

Mutagenesis itself can be modelled as a Poisson process. For more information on this please see this post. So the probability of $k$ mutations in a time window, $t$, can be represented by the following equation:

$$P(n=k)=\frac{(\lambda t)^ke^{-\lambda t}}{k!}$$

Where $\lambda$ is the rate of mutagenesis.

The mean number of mutations (and also the variance for a Poisson distribution) would be $\lambda t$.

Now, the rate of mutagenesis would be proportional to the concentration of the mutagen. I speculate that the relationship between them would follow a saturation kinetics (like Michaelis-Menten) but at lower concentrations it would be essentially linear.

Moreover, the rate of mutagenesis would also depend on the cell type.

Therefore, standardization is required and labs working in this area for a long time develop standardized protocols (with specific mutagen concentrations and treatment times) which would let them avoid multiple mutations. Even then, they have to screen a lot of individuals to filter out the non-mutants.

Basically, there is no magic number.

Answer 2

With the forward genetic screen approach you would treat the organism with your mutagen, and cross them to produce mutants. With forward genetics, you're using a phenotype to map out a gene. You isolate the mutants based on phenotype, and perform a complementation test to determine if the mutation was on the same gene or different genes. This was classically done with Drosophila.

By the classical genetics approach, a researcher would then locate (map) the gene on its chromosome by crossbreeding with individuals that carry other unusual traits and collecting statistics on how frequently the two traits are inherited together. Classical geneticists would have used phenotypic traits to map the new mutant alleles. Eventually the hope is that such screens would reach a large enough scale that most or all newly generated mutations would represent a second hit of a locus, essentially saturating the genome with mutations. This type of saturation mutagenesis within classical experiments was used to define sets of genes that were a bare minimum for the appearance of specific phenotypes. However, such initial screens were either incomplete as they were missing redundant loci and epigenetic effects, and such screens were difficult to undertake for certain phenotypes that lack directly measurable phenotypes. Additionally a classical genetics approach takes significantly longer.

Source: Wikipedia

The reason that I posted this answer is due to the context of chemical mutagenesis, provided on 335 of your source:

Chemical mutagenesis The use of chemical agents, such as EMS or ENU, to mutagenize hundreds or thousands of eggs/larvae for the purpose of performing a forward genetic screen.

Answer 3

What I think the text is saying

The text you quote is a little unclear (note that it is not a peer-reviewed article given the language used). I would think based on the quotation that the treatment is chosen so that the expected number of mutations is 1.

But how can the scientists know which organism or cells contains only one mutated gene?

You assume they are screening through the lines to find out those who contain a single mutation. This is not what the quoted text is claiming. The quoted text only states that the treatment create mutations at a rate of 1 per genome.

If they would have to screen through the lines to select those that have had one single mutation somewhere, then sequencing would be the only solution.

Note on Stats

The number of mutated genes after treatment is probably Poisson distributed, not normally distributed. A normal distribution is a continuous probability distribution anyway and is bounded between $-\infty$ and $+\infty$, while a negative number of mutations makes no sense.

Note on the design of the post

You should link to you source directly. It will be easier for those willing to answer