Effects of three heavy metals on the bacteria growth kinetics: a bivariate model for toxicological assessment

Abstract

The effects of three heavy metals (Co, Ni and Cd) on the growth kinetics of five bacterial strains with different characteristics (Pseudomonas sp., Phaeobacter sp. strain 27-4, Listonella anguillarum, Carnobacterium piscicola and Leuconostoc mesenteroides subsp. lysis) were studied in a batch system. A bivariate model, function of time and dose, is proposed to describe simultaneously all the kinetic profiles obtained by incubating a microorganism at increasing concentrations of individual metals. This model combines the logistic equation for describing growth, with a modification of the cumulative Weibull’s function for describing the dose-dependent variations of growth parameters. The comprehensive model thus obtained—which minimizes the effects of the experimental error—was statistically significant in all the studied cases, and it raises doubts about toxicological evaluations that are based on a single growth parameter, especially if it is not obtained from a kinetic equation. In lactic acid bacteria cultures (C. piscicola and L. mesenteroides), Cd induced remarkable differences in yield and time course of characteristic metabolites. A global parameter is defined (ED_50,τ: dose of toxic chemical that reduces the biomass of a culture by 50% compared to that produced by the control at the time corresponding to its semi maximum biomass) that allows comparing toxic effects on growth kinetics using a single value.

Appendices

Appendix 1. Re-parameterizations of the logistic equation

The explicit form of the logistic Eq. 2 can be written as follows:

$$ X = \frac{{{X_m}}}{{1 + \exp \left( {c - {\mu_m}t} \right)}}\,;\,c = \ln \left( {\frac{{{X_m}}}{{{X_0}}} - 1} \right) $$

(8)

For determining the maximum growth rate, it is necessary: (1) to obtain the abscissa (τ) of the inflection point by isolating it from the expression that results by equating the second derivative of the function to zero and (2) to insert the value τ in the first derivative of the function. The results are:

$$ \tau = \frac{c}{{{\mu_m}}}\,{;}\,{v_m} = \frac{{{X_m}{\mu_m}}}{4} $$

(9)

For determining the lag phase, it must be kept in mind that the ordinate of τ is K/2. Thus, the equation of the tangent at the inflection point and its intersection (τ) with the abscissa axis are:

$$ X = \frac{{{X_m}}}{2} + {v_m}\left( {t - \tau } \right)\,;\,\lambda = \frac{{c - 2}}{{{\mu_m}}} $$

(10)

Thus, the re-parametrized logistic equation, with explicit v _m and τ, requires isolating μ _m and c in 9 and 10, respectively, and to insert the corresponding values into 8:

$$ X = \frac{{{X_m}}}{{1 + \exp \left[ {2 + \frac{{4{v_m}}}{{{X_m}}}\,\left( {\lambda - t} \right)} \right]}} $$

(11)

Moreover, by inserting X = X _m/2 in 8, we obtain c = μ _mτ, where τ (abscissa of the inflection point) is the time needed to reach the semi-maximum biomass. By replacing c by μ _mτ in 8, we obtain another re-parameterized form:

$$ X = \frac{{{X_m}}}{{1 + \exp \left[ {{\mu_m}\left( {\tau - t} \right)} \right]}} $$

(12)

Or, in general, to make explicit the time τ _q necessary to achieve a proportion q of the maximum biomass:

$$ X = \frac{{{X_m}}}{{1 + \exp \left[ {\ln \left( {\frac{1}{q} - 1} \right) + {\mu_m}\left( {{\tau_q} - t} \right)} \right]}} $$

(13)

Appendix 2. Calculation of DE_50,τ

Once the solution of the examined system is obtained by means of a model 6, the ED_50,τ, or dose that reduces the biomass to 50% of that produced by the control in time τ, can easily be calculated as follows:

• 1. Fit the kinetic data of the control to the growth equation in the parametric form L5 to obtain the semi-maximum biomass (X _m,0/2) and the time needed to reach it (τ ₀). For another proportion q of the maximum biomass, use the form 13.

• 2. Set an arbitrary initial value (^IED_50,τ, see next point 5).

• 3. Calculate the value of biomass (X) that results from applying model 6 by assigning the values ^IED_50,τ and τ ₀ to the variables D and t.

• 4. Calculate the absolute value of the difference $H = \left| {X - \frac{{{X_{{{\rm m}, 0}}}}}{4}\,} \right|$

• 5. Calculate, using the Solver macro in Microsoft Excel, the value of ED_50,τ that minimizes H. For ensuring that the algorithm finds the absolute minimum, it is advisable to start with an ^IED_50,τ value associated with a reasonably small value of H.