A glimpse into yeast growth kinetic models

Eureka, science post! Some math, model building and biology: I would like to start talking about yeast growth kinetic models today. In general, growth kinetics describes how different conditions (substrate, oxygen amount, metabolism products, inoculation rate, growth inhibitors etc) influence the growth of a microorganism in a time dependent manner. At the end one can construct mathematical models to describe the growth behaviour. In most cases one begins with experiments with fixed conditions and the growth of the organism is observed over time. A next experiment can be conducted with the change of one condition and the growth over time is observed once again. And so forth.

All these kinetic models can be very powerful tools. Not only can one improve for example the efficiency of yeast propagation but also investigate the effect of different parameters such as the inoculation rate of yeast, the substrate concentration, oxygen- or ethanol concentration on yeast growth. One application for growth kinetics could be to calculate yeast propagation like done in other yeast growth calculators for homebrewers. 

This post is a general introduction about yeast kinetics and I would like to show you how one can get from experimental data to a simple growth kinetic model. Future posts will go into more detail and I would like to give a general introduction first.

Exponential growth of microorganism

One of the most basic models in mathematical biology is the exponential growth equation for microorganisms. This equation can be used to calculate the amount of microorganisms or biomass (X) after a certain amount of time (t). Biomass can be the physical mass of the microorganisms, the cell concentration or any other measurement related to specify how many microorganisms there are (like optical density). I will stick to the physical mass of microorganisms as biomass in this post. In the exponential growth equation model no inhibition or substrate limitation is included. Substrate by the way is a term for any food source for the microorganisms such as glucose, maltose etc.

X_t = X_{t0} \cdot e^{\mu_{max} \cdot t - t_0}

  • Xt is the biomass concentration at the time point t [g L-1]
  • X0 is the biomass concentration at the time point zero [g L-1]
  • µmax is the maximum specific growth rate [h-1]
  • t Time [h]
  • t0 Time where the experiment begins. Normally zero [h]

Lets make an example. Lets say you have a 1 L yeast starter (with indefinite amount of substrate) and add 10 g of yeast (X0) at the beginning (t0). You might ask yourself how many yeast cells you have after waiting for 90 min. The only thing you might not know is the specific growth rate (µmax). These coefficients can be looked up and for yeast µmax is somewhere around 0.5 h-1. My solution for this question would be 21.2 g L-1. So roughly double the amount.

You can even calculate the doubling time (tD) after changing the previous equation (doubling time specifies the time needed to double the initial amount of microorganisms):

ln(\frac{X_t}{X_{t0}}) = \mu_{max} \cdot t

By the way, the equation above will be important later on to get µmax. Moving on, at the time of doubling (tD), Xt equals 2 times Xt0:

ln(\frac{2 X_{t0}}{X_{t0}}) = \mu_{max} \cdot t_D

ln(2) = \mu_{max} \cdot t_D

t_D = \frac{ln(2)}{\mu_{max}}

Common doubling times for Saccharomyces cerevisiae are around 90 min [1]. This gives you a µmax of roughly 0.5 h-1.

In the exponential growth equation model substrate limitations are not included and therefore makes it not that useful to fully describe the growth behaviour if you want to investigate the effect of the substrate concentration itself. Let go to the next model.

Monod equation for biomass

A very widely used model to describe the growth of microorganisms is the Monod model. The whole model is based on the Michaelis-Menten equation widely used in enzyme kinetics. I don’t want to go into further details here how one came up with these equations. In the end, the Monod model is again a simplified version of reality (like every model is). The first equation for the Monod model is:

\mu = \mu_{max} \frac{S}{K_S + S}

  • µ is known as the specific growth rate [h-1]
  • µmax is the maximum specific growth rate [h-1]
  • S is the substrate concentration [g L-1]
  • KS is the substrate saturation constant [g L-1]

We already know µmax, S is just the substrate concentration. For example the amount of dry malt extract you use for your yeast starter. Or glucose or whatever you are interested in. Just keep in mind that µmax depends on the used substrate. The definition of µ is:

\mu = \frac{dX}{dt} \cdot \frac{1}{X}

  • µ is known as the specific growth rate [h-1]
  • X is the biomass concentration [g L-1]
  • dX is the change of the biomass concentration [g L-1]
  • dt is the change of time when dX happens [h]

and depends on the change of biomass within an indefinite amount of time (dt). You could therefore write the Monod equation like:

\frac{dX}{dt}= \mu_{max} \frac{S}{K_S + S} \cdot X

and you already have your first differential equation for your model to describe the change of biomass in a time dependent manner.

This leaves KS. This is similar to the Michaelis constant Km in the Michaelis-Menten equation. In the case of the Michaelis-Menten equation, Km describes the substrate concentration at which the enzyme reaction is equal to half of the maximum speed. In the case of Monod, KS describes the substrate concentration (therefore S as index) where you have half of µmax. This is all so far for the first part of Monod.

Monod equation for substrate

In the previous section, I introduced the basic Monod equation which can be used to describe the change of biomass over time. Next we like to include the change of substrate. In case of yeast one might be interested to investigate the behaviour of dry malt extract. For that we have to introduce another coefficient:

Y_{X/S} = \frac{\frac{dX}{dt}} {\frac{dS}{dt}}

  • YX/S Biomass on substrate yield coefficient [gX gS-1]
  • dX is the change of the biomass concentration [g L-1]
  • dS is the change of the substrate concentration [g L-1]
  • dt is the change of time when dX and dS happen [h]

YX/S simply defines how much biomass you can get from substrate (thus the index X and S). For example how much biomass you get from one gram of substrate. Looking at the previous equations, you can already see how to get a differential equation for the substrate:

\frac{dS}{dt} = - \frac{\mu_{max} \frac{S}{K_S + S} \cdot X}{Y_{X/S}}

Per definition, the equation is multiplied by minus one because the slope of dS over dt is negative (substrate is metabolized and therefore only decreases).

This leaves us to look at product kinetics. In case of yeast, one might be interested to describe the production of ethanol as a product. This is very similar to the substrate shown above. In this case you need another yield factor:

Y_{P/X} = \frac{\frac{dP}{dt}}{\frac{dX}{dt}}

  • YP/X Product on biomass yield coefficient [gP gX-1]
  • dP is the change of the product concentration [g L-1]
  • dX is the change of the biomass concentration [g L-1]
  • dt is the change of time when dX and dP happen [h]

In this case the product depends on the biomass and not on the substrate. Once again one can write down the differential equation for the product formation:

\frac{dP}{dt} = \mu_{max}\frac{S}{K_S + S} \cdot X \cdot Y_{P/X}

Are you still reading? Yes? You must either be very interested and/or be a math geek like me… I would like to stop with the introduction here and discuss other things such as inhibition etc in a future post. I now would like to share some information how you could/can get the coefficients for the equations above from empirical data.

Coefficient determination from experimental data

All the data below is from a S. cerevisiae batch cultivation I cultivated during my undergraduate studies. I don’t want to get much into detail here but some information to understand the values you get afterwards. The batch cultivation was done in a 16 L reactor (4.2 gal) under sterile conditions and glucose as the only carbon source (substrate). All the values below therefore are for glucose only.

During the cultivation (8 h), every 30 min the optical density (OD) and glucose concentration was measured and every 60 min the dry mass was determined. The optical density is another measurement method to get an idea about the yeast concentration. The glucose was measured enzymatically and the dry mass was determined by filtration, drying of the filter paper and finally determining the mass of yeast on the filter paper.

µmax determination based on dry mass

As previously mentioned, the following equation can be used to get the µmax value by plotting the logarithmic ratio of the biomass against the time (t). This should give you a linear function with the slope µmax as shown in Fig 1:

ln(\frac{X_t}{X_{t0}}) = \mu_{max} \cdot t


Fig 1: Log dry mass ratio against time. µmax determined from slope of linear fit function

One can easily see that the amount of yeast grew steadily up to the time point of six hours where the yeast concentration stayed the same (Fig 1). After six-hour the whole glucose was already metabolized (not shown) and no further yeast growth could be observed (Fig 1). The linear fit function therefore only makes sense between time point zero and six hours. The slope of the linear fit function was 0.43 h-1. Please remember, this µmax value is for the dry mass and is close to known values [2].

Lets quickly calculate the doubling time for this µmax. This give you a doubling time of roughly 97 min. Not that far away from the 90 min stated in source [1].

µmax determination based on optical density (OD)

The same can be done for the OD (Fig 2).


Fig 2: Log OD ratio against time. µmax determined from slope of linear fit function

Once again after five to six hours, the yeast growth came to a stop (Fig 2). This time the slope of the linear fit function was 0.53 h-1. A nice example that not every µmax is the same. One has to be very careful how the specific µmax was determined. Either based on the optical density, dry mass weight or even the cell count.

Biomass on substrate yield coefficient (YX/S)

To determine the yield coefficient, one can plot the measured yeast dry mass or OD against the measured glucose values and use the linear slope to determine the yield factors (Fig 3).


Fig 3: Determine the biomass to glucose yield factor

The yield factor based on the dry mass was -0.5 g g-1, and -0.1 g g-1 for the OD. Known yield factors for yeast based on glucose and dry mass are between -0.3 and – 0.5 g g-1 [3].

Building growth models

Now its time to put all the equations and values into a model. The only coefficient not known so far is KS. This value is hard to determine and the easiest way to get this value is by iterative approaches. You simply use the model to get KS. I use Berkley Madonna for this purpose. What you have to do is input all the different differential equations, the measured values and run an iterative algorithm to let the model function approximate the measured values. If you do this the right way you might get graphs like below (Fig 4).


Fig 4: Yeast kinetic model 1, explanation in text below

In this case the substrate concentration (S) and the yeast concentration (X) are plotted against time. In addition, the black dots correspond to the measured glucose concentrations. I included the values I used for the different coefficients in the graph as well. Unfortunately, one cannot fit the measured glucose concentration curve with the numerical values of the determined coefficients. I therefore had to use slightly different values for the yield coefficient (in this case 0.3 g g-1) and 0.51 h-1 for µmax. The differences between the numerical values of the parameters might be due to measuring inaccuracies for the glucose concentration and/or yeast concentration. In the end, one can determine KS to be around 0.031 g L-1.

You now can use the model to investigate how the different conditions affect the growth of the yeast. For instance different substrate conditions or inoculation rates. All this can be useful to understand the behaviour of yeast growth under different conditions. However, one has to keep in mind that all this is based on a simplified model and it does not have to represent reality. Model building is sometimes hard work because iteration processes might get stuck and lead to wrong results (such as negative substrate concentrations).


Growth kinetics models can be used to describe the growth of microorganisms. Because biological systems can hardly be approximated by simple fit functions, more sophisticated methods need to be applied. Such as the models described in this post.


This was just a basic introduction about growth kinetics and models. Future posts will go into more details covering additions to the basic Monod model. The next post concerning growth kinetics will be about Brettanomyces growth and is a nice organism to introduce inhibition effects.

I would like to do some small-scale experiments with yeast propagation and determine the individual model-parameters in the future. The resulting models therefore might be useful to approximate yeast starters under different conditions. Just be patient, I am currently really busy with my real scientific work. Thanks for reading and comment if you like. Cheers!


[1] : http://dbb.urmc.rochester.edu/labs/sherman_f/yeast/4.html, “An Introduction to the Genetics and Molecular Biology of the Yeast Saccharomyces cerevisiae” (2012)
[2] : http://www.atcc.org (2012)
[3] : B. Sonnleitner, Lecture slides Bioprozesstechnik 1, ZHAW Wädenswil, 2010


8 thoughts on “A glimpse into yeast growth kinetic models

  1. Hi Sam, great post. I was wondering, does the fact that yeast cells have a specific life-span influence your model on exponential growth? Saccharomyces is different from lets say bacteria in this regard, I have wondered about what effect this has on the culture for a while, but never had the time to look into this. You seem to be the guy to ask.

    • Hi Jaapie,
      thank you for reading and commenting. I assume you have thought about the capacity of Saccharomyces to bud to a certain point followed by cell death or senescence, right? I actually have no idea if this limited life-span has any (observable) effect on the kinetics. I have the feeling that this effect is minor and doesn’t influence the kinetics significantly. However, this is pure speculation. In my opinion, the older cells are only a small part in the whole population and therefore should not have a huge impact. Especially in the log-phase.

      By the way, your recent isolate screening post is very cool. I would like to screen my isolated yeasts as well but haven’t found useful small-sized fermenters so far.
      Cheers, Sam

  2. Mistake:

    Per definition, the equation is multiplied by minus one because the slope of dX over dt is negative (substrate is metabolized and therefore only decreases).

    You certainly meant the slope of dS over dt because in your examination dX stands for the biomass and not the substrate mass.

  3. Hi there… I just need to confirm about the equation on “Exponential growth of microorganism” (its the first equation in this article). I found in some other references that the specific growth rate in the equation is not “maximum specific growth rate” (umax) but “specific growth rate” (u), and to calculate “maximum specific growth rate” we have to use Monod Equation by plotting 1/s vs 1/u (check: 1. Stanbury et al. 1995. Principles of Fermentation Technology; 2. Tempest DW (1978) Dynamics of microbial growth. In: Norris, JR, Richmond MH (eds) Essays in microbiology. John Wiley, New York, pp 1 -32; 3. Widdel 2010: https://www.mpi-bremen.de/Binaries/Binary13037/Wachstumsversuch.pdf). Or do you have a reasoning for that equation?

    • Hi furanosa,
      sorry for the late reply. The reason for choosing umax instead of u is the following. In case of choosing a very high substrate concentration [S] >> KS, the growth occurs at umax. Meaning, the batch cultivation is unlimited and the Monod term (S/S+KS) is close to one. Which gives us u = umax. This is a trick to determine umax using ln(Xt/X0) = umax*(t-t0) in case of using a batch cultivation with S >> KS. I am sorry I did not mention in the post that the exponential equation we are talking about here is only valid if S >> KS is true. Thanks for pointing that out. Hope this explains the reason for the equation. In general (no specific choosing of S), umax should be replaced by u in the exponential equation as you rightly mention.

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