Null model for relative abundance

Let us consider a simple scenario of two co-cultures of species A and B growing together in two separate demes, each containing a single nutrient (labeled 1 and 2). The fractions of A and B in nutrient/deme one are f_A,1 = n_A,1 /(n_A,1 + n_B,1) and f_B,1 = n_B,1/(n_A,1 + n_B,1), respectively, and similarly, the fractions of A and B in nutrient/deme two are f_A,2 = n_A,2/(n_A,2 + n_B,2) and f_B,2 = n_B,2 /(n_A,2 + n_B,2) (where n is the total number of cells of species A or B). If we consider the two-deme system as a whole (i.e. if we pool together the amount of species in each nutrient/deme), the fractions of A and B in the mixture are given by: f_A,12 = (n_A,1 + n_A,2)/(n_A,1 + n_B,1 + n_A,2 + n_B,2) and f_B,12 = (n_B,1 + n_B,2)/(n_A,1 + n_B,1 + n_A,2 + n_B,2).

We can define n_t,1 = n_A,1 + n_B,1 and n_t,2 = n_A,2 + n_B,2 as the total number of cells in the nutrient demes 1 and 2, respectively. We can thus write f_A,12 = (n_A,1 + n_A,2)/(n_t,1 + n_t,2). Defining w₁ = n_t,1/(n_t,1 + n_t,2) and w₂ = n_t,2/(n_t,1 + n_t,2), it is straightforward to show that: f_A,12 = w₁ f_A,1 + w₂ f_A,2. By the same reasoning, we find that f_B,12 = w₁ f_B,1 + w₂ f_B,2.