Mathematical Proof of using a Starter Culture (Heavy Math Stuff)

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toastermm

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I’ve decided to look into the debate of using a starter or not. I, personally, do not usually use a starter very often, and have not noticed a difference. (Not to my eyes at least). But just because we can’t see it, doesn’t mean it’s not helping. Seeing as I love mathematics, I decided to let some famous dead mathematicians do some work for me. (I’m staring at you Liebniz and Newton).

The overall hypothesis is that the initial innoculation (starter size) has an impact on doubling time. And in fact, it does, as shown below.

Here’s what the plan is, if we can mathematically describe what is going on in a fermenter (according to the yeast population), maybe some interesting results will come about. The hardest part of this is translating a real system into mathematics, and then translating back to reality. But in between, since math follows a logic system (1+1=2 and so on…) we can manipulate the math anyway we please as long as the logic rules are not broken.

Maybe this is overkill, and you can just ask any seasoned brewer (not me… yet), and they will tell you how a starter is better or not. But I like my facts with a sweet side dish of math.


********************WARNING*********************
Plenty of math ahead. Not just adding and subtracting gravity readings, I'm talking calculus. So hold on to your pants, strap yourself in, and no refunds on the ride! WHEEE!
************************************************


We are going to start with the simplest idea possible. Yeast in a fermenter grow exponentially until saturation. Saturation is an iffy word and we need to clarify that. The basic idea is that the closer the yeast get to saturation, the slower and slower they grow. Now some naysayers will scream, “no wait!, my yeast grow faster and faster the longer they are in the fermenter!”. That is not the case- because if that did happen, the yeast would keep growing, blast out of the fermenter, and exponentially take over the world. Yikes. So all I’m saying is the yeast ‘start-off’ exponentially growing, then ‘slow-down’ until saturation.

I can now say that the growth of yeast is proportional to how much yeast there is. Think of it as this: if there is 1 yeast cell, after one division there will be 2. If there are 10 yeast cells, after one division for all cells (1 generation), there will be 20 cells. But it’s not that simple, yeast cells do not divide synchronously, and we have to factor in how long it takes them to divide. For those scienctists out there, this means that ‘continuous models’ are much better than ‘discrete models’. All right, so let’s write out what this proportionality means.

(The change in population over time) is proportional to (the amount of yeast already existing).

This ‘change in population over time’ is known in calculus as the ‘derivative of the population’. (Equivalently, think of velocity as being the derivative or change in position over some time.)

(dP/dt) is proportional to (P).

Proportionality basically means, “multiple of”. So we throw in a constant to get the following. Also note that ‘is’ means ‘equality’ in math.

(dP/dt) = r * P.

“HOLD ON, what is r?” Good question. This ‘r’ value basically tells us how strong the proportionality is. We will call this ‘r’ value the ‘growth rate’. Just a definition. Well, this is almost where we want to be. If we ask a computer to graph this differential equation, we end up with exponential growth… at all times. If that was the case in reality, we would be covered in bacteria, and they would take over the universe.


Hmm… problem. Well we can safely say that they grow exponentially in the beginning, but the closer they come to this ‘saturation’, they slow down. This is where an idea called the ‘logistic equation’ comes along. Most people attribute this equation to the famous population geneticist, “Alfred Lotka” in the 1930’s, but it was actually derived 100 years earlier by a French Math guy, “Pierre Verhulst”.


Enough dabbling in history. What is this ‘logistic’? It is the idea that the population is self-limiting. The more population you have, the slower they grow. This population amount is measured relative to some maximum population that can exist. We will call this maximum population, K. So if the population of yeast ever does reach K, it will not grow at all. Drum roll please…

dP/dt = (logistic equation) *(r)* (P).

or

dP/dt = (1-P/K)*(r)*(P)

What? Well, it’s just an equation with two KEY properties, if P=K, then the logistic equation part is equal to zero, and if P is really close to zero, the logistic equation is really close to 1.


YAY! We have our equation. The most obvious question to ask (if you can follow my horrible explanation so far) is to ask how does the initial innoculum come into play?


The above equations are called “differential equations”. In order to solve these, you need a point on the graph, i.e., our initial innoculum. Here’s where the detials get messy, but as they say, you can’t have a good time until you get your feet wet.


******************Calculus-Haters skip the next section******************

We have

dP/dt=rP(1-P/K)

Here we will use the technique of solving called ‘separating of variables’. We will bring our P’s and t’s to separate sides and integrate.

(dP)/(P*(1-P/K)) = r*dt

We can’t quite integrate the left side yet, unless you can sporadically guess the solution, so we will use partial fractions to get,

(dP/P) + ((1/P)(dP)/(1-P/K)) = r *dt

Now we can integrate both sides directly (remembering our integrating constants),

ln(P/(1-P/K)) = r*t + C

Solving for P,

P = (C*exp(r*t)) / (1+ (C/K)*exp(r * t))

Solving for our integrating constant, we note that at t=0, P = (P_0), where (P_0) is our initial innoculum size.

C = (K*P_0)/(K-P_0)

And finally:

P = (K*P_0*exp(r*t)) / (K+P_0*(exp(r*t)-1))

Please note that is is really hard to express an equation in a word document without some sort of equation editor.
***************************************************************


We are 95% done. All we have to know now is the doubling time. We have our final equation,

P = (K*P_0*exp(r*t)) / (K+P_0*(exp(r*t)-1))

Where
P = Population size at a time t.
K = Saturation Limit.
P_0 = Initial population or Innoculum size. (Starter Size)
r = growth rate.
t = time.

To find the doubling time, set P = 2*P_0. I’m going to pull a fast one on you, and use the famous “It can be shown that”.

Doubling Time = DT = (1/r) * ln(2) + ln(K-P_0) - ln (K-2*P_0)

Ok the final punch line is this:

If I want to decrease my doubling time (DT), then I have to increase my starting population (starter size, P_0). If you can't see this, just convince yourself that the last term is an increasing function of P_0.

And there you have it. Either go through all the equations above, and verify my work, or go ask a brewer.
 
Yep.

I was kinda bored today, and thought this would be a good preface to making a program that could determine your yeast-doubling rate from some simple measurements.

I might create this program if people are interested.
 
Somebody posted a graph/chart before. That's what I want to see.

DTExample.jpg


This is me playing around with fake data and fitting the population curve.

Those blue lines that intersect the curve show where it reaches 90% saturation. This was because I wanted to know when it would be good to pitch the starter.
 
Yep.

I was kinda bored today, and thought this would be a good preface to making a program that could determine your yeast-doubling rate from some simple measurements.

I might create this program if people are interested.

+1. Highly interested in that if you are willing to make it happen. :mug:

-Tripod
 
i would have thought that somewere in there you would have used some form of f(x)= e^x as it concerns to yeast growth?

You are quite right. It was hidden in the details,

The formula you mention is the solution to the following equation I first considered (with r = 1):
(dP/dt) = r * P

So yes, you are right. Also, the final solution of the system has a form of exp(blah), which is the same thing as e^(blah), just notational differences.

:mug:
 
I randomly found this thread searching on google for "logistic growth homebrew"

I have just recently started a batch with propagated yeast.

About the only thing I remember from differential equations in college is the phrase "logistic growth model", which is how organisms will grow exponentially in their little closed world until they run out of food.

i'm interested in re-learning some of this stuff and applying it to home brewing.
 
I actually understood that (for the most part)and I had the same nun teach me math for eight years until high school. She'd be proud!
 
HEY!! I became a Member. No more Junior Member for this guy. Where's Gnome? I'm sure he'd have something witty to say. Sorry. Done now.
 
I like Beer.

I like Math.

I don't like Beer and Math. Makes my head go




Good explanation but, what benifit would the program provide? Would this solve for a cell count over time?

Meaning, could I tell the program that I started with an "x"mL sample and have a target cell count of "y" then it would return a value in time plus the volume of "food" required to acheive the desired count?

Yes? No? :confused:
 
I like that there is someone else who applies math to brewing. And I like the way you explain things. I oftentimes read only through the explanations to see where it is coming from and where it is going w/o really having to do the heavy math.

But I would have expected a different outcome since you applied a brake to the growth by introducing the "logistic equation" which basically slows down the rate of growth r based on how close the population is to the max. So one would expect that the closer you already are to your max population, i.e. larger initial population size, the slower the growth will be and the longer the doubling time should be. I don't have the time now to check the equations, but two graphs drawn for two different values of P_0 should be enough to show that.

In practical fermentation it has been shown that larger pitching sizes lead to less relative yeast growth.

Kai
 
WOW i'm gladd i the explanation finished when it did! I was about two more lines away from shooting myself in the face. Awsome work but damn... I'm not even done with school yet and i feel like i need to ask for a refund.
 
I like the mathematical equation behind this.

However, for practical purposes, we also need to factor in the yeast strain, the temperature during the reproductive period, and the gravity of the wort, and what particular sugars compose this gravity.

But, like we did on all our engineering exams in college, we state our assumptions for a "controlled system." Thanks again for posting this.
 
Guy, this math is interesting, but is nothing more than a model, because the "real" explanation is in biochemistry and in the yeast metabolism.
Listen to me, I'm a chemical engineer, so I love math, I use a lot of it, so I don't wanna blame you or minimize your work in any way, but I wanna add some more.
First of all I wanna remember that yeast is an unicellular fungi, aerobic but capable of growing in anaerobic condition.
His source of energy and carbon are carbohydrates, usually the simple ones, so you give it starch its metabolism cannot do nothing, otherwise if you give it maltose or glucose this saccharides can be easily metabolised.
But it is not so easy, because to use a specif disaccharides, as maltose, the yeast need some specific enzymes, that is different from the one that is used to hydrolize glucose, so if you make a starter based on glucose, when you pitch it to the wort the yeast is "conditioned" to eat glucose and need to spend some time, called "the lag", to learn how eat maltose.
But this is not enough, as starting from a singular moleculas of glucose, in aerobic condition the yeast can produce 16 times the energy that can produce in anaerobic condition, so his metabolism is less efficient in condition of lack of oxigen.
So your starter should be aereated in the right way in order to obtain a good cell proliferation.
Pitching a healthy yeast population in the full exponential growth, already conditioned to metabolize the same nutrients that will find in the worth are key factor in obtaining a good fermentation and reducing off flavours.
Also pH and osmotic pressure of the wort can influence on the metabolic function of the yeast, influencing the population dynamics.
 
So, when I have to take Calculus 1, 2, 3 and Differential Equations in the near future, I can come here for help??
 
I like Beer.
Good explanation but, what benifit would the program provide? Would this solve for a cell count over time?

Meaning, could I tell the program that I started with an "x"mL sample and have a target cell count of "y" then it would return a value in time plus the volume of "food" required to acheive the desired count?

Yes? No? :confused:

Good question. Most of the reasons for me doing this include boredom at work. But this program could do a few things:

1. Estimate the doubling rate of strains based on "simple measurements".

2. Given a doubling time, determine the time needed to let a starter sit before it is ready.​

I'm not sure #1 is particularly useful, but it could be related to the flocculation of the strain. Whereas, #2 could be useful to some brewers.

What are these "simple measurements"? Good question, they have to be related to how much yeast is in there. Normally people go two routes:

1. Take something called an "optical density"- this is from an expensive piece of machinery called a spectrophotometer. yikes.

2. OR the only other way that I can think of at the moment is to culture small samples onto plates (petri dishes).
So to take these measurements one would need access to petri dishes (relatively inexpensive), and agarose (jello-type substance, can be expensive.) Then they would combine a starter substance with agarose, heat it up to dissolve the agarose, and pour it into plates, let them set until hard and then they could use the cell counts on these plates to give an estimate to concentration in the starter.

Lots of work, I know, and probably not worth it.


BUT if you knew your yeast's doubling time, the program could estimate the length of time needed for the starter. Would that be useful? Maybe.

So the simple answer is- yes it could
 
But I would have expected a different outcome since you applied a brake to the growth by introducing the "logistic equation" which basically slows down the rate of growth r based on how close the population is to the max. So one would expect that the closer you already are to your max population, i.e. larger initial population size, the slower the growth will be and the longer the doubling time should be. I don't have the time now to check the equations, but two graphs drawn for two different values of P_0 should be enough to show that.

You are quite right. But there is a distinction to be made between the "growth rate", r, and the doubling time. The time to double is assumed to be an "inherent" feature in these populations. BUT the growth rate is different, it is best explained by saying the following,

"The growth rate is an expression of how the doubling time and resource limitation are coupled."

That was a very touchy-feely explanation. But there are two cases to illustrate this:

Case 1: No resource limitation. Here the growth rate and doubling time are the same thing. This is nothing changing the growth rate.​

Case 2: Zero resources. Here the gwoth rate and doubling time are at the opposite ends of the spectrum. If a yeast cell some how gets the resources to double, it will double in X hours. That doesn't change. But because of the limited resources, it has to sit around and wait until it can find more.​
 
First of all I wanna remember that yeast is an unicellular fungi, aerobic but capable of growing in anaerobic condition.
His source of energy and carbon are carbohydrates, usually the simple ones, so you give it starch its metabolism cannot do nothing, otherwise if you give it maltose or glucose this saccharides can be easily metabolised.
But it is not so easy, because to use a specif disaccharides, as maltose, the yeast need some specific enzymes, that is different from the one that is used to hydrolize glucose, so if you make a starter based on glucose, when you pitch it to the wort the yeast is "conditioned" to eat glucose and need to spend some time, called "the lag", to learn how eat maltose.
But this is not enough, as starting from a singular moleculas of glucose, in aerobic condition the yeast can produce 16 times the energy that can produce in anaerobic condition, so his metabolism is less efficient in condition of lack of oxigen.
So your starter should be aereated in the right way in order to obtain a good cell proliferation.
Pitching a healthy yeast population in the full exponential growth, already conditioned to metabolize the same nutrients that will find in the worth are key factor in obtaining a good fermentation and reducing off flavours.
Also pH and osmotic pressure of the wort can influence on the metabolic function of the yeast, influencing the population dynamics.

Yes yes yes and yes. I personally have no experience with this. Most of the time, using math means simplification of systems.

But to further back up your point to others, it is obvious when we realize that the inherent doubling time will change if the yeasties are fed different food. That much makes perfect sense.

To compensate for this, we have to make the system much more complex. We move from one ODE equation to a system of multiple coupled ODE's that make analytic solutions cry in fear and throw themselves out the nearest window, drying their pee-stained pants out in the cold air.

But I do not want to do this, nor could I. My background is in math and biology. One day though, version 15 of this proposed program could take into account the chemical and physical properties of this system. I guess I should start making friends in the chem department. (By bringing them beer! :mug:)
 
Yes, I know that structurated model would require a lot of pain, but I think that there are a lot of basic biochemical issue that can be understood and that can give some insight on the yeast life without too much math.
 
If I assume these 2 cases:

(1) P_0 = 10; K=100; r=0.25
(2) P_0 = 40; K=100; r=0.25

Then I find that (1) reaches 2*P_0 = 20 after about 3 time units while (2) reaches its 2*P_0 = 80 after about 7 time units. If P_0 >= 50, it will never double.

Am I missing something in the assumptions that were made?

Kai
 
If I assume these 2 cases:

(1) P_0 = 10; K=100; r=0.25
(2) P_0 = 40; K=100; r=0.25

Then I find that (1) reaches 2*P_0 = 20 after about 3 time units while (2) reaches its 2*P_0 = 80 after about 7 time units. If P_0 >= 50, it will never double.

Am I missing something in the assumptions that were made?

Nope, you're right. Maybe the term "doubling time" is misleading. By this I am talking about what would happen if the yeasties had infinite food available. But of course, they don't.

You've pointed out a nuance that I did not want to go into. I should have pointed out that it makes no sense to calculate a "doubling time" if the growth rate changes. I'm doing this calculation because it is a good representation to how fast the yeast grow.

So continuing with the calculus:
Since the yeasties are halfway or more to saturation we can no longer talk about a "doubling time", but we can generalize this equation. Let's talk about a ratio of P to P_0. We were calling this ratio 2 because we wanted doubling time. Now we have to have this ratio limited. We end up with the following equation with a generic ratio:

(1/r)*ln(P/P_0) + ln(K-P_0) - ln(K-P)

Thanks for bringing this oversight to my attention... I goofed on that one.:drunk:

But the point that I want to make, is that to decrease the doubling time, we have to increase the starter size.
 
But the point that I want to make, is that to decrease the doubling time, we have to increase the starter size.


But even that I’m willing to contest. In case of unlimited resources, the yeast will grow exponentially which means that the amount of time it takes to double the initial population is independent from the initial population size. Basically the idea of exponential growth.

But does it really matter what the doubling time is. If we try to mathematically prove the benefit of a starter, shouldn’t we look one step further and evaluate the time it takes to ferment a particular volume of wort? There I can clearly see that a large starting population size means faster fermentation.

Kai
 
This is getting heavy but I don't understand any of it. I guess I will just stick to counting yeast and make plenty of them. :)
 
But even that I’m willing to contest. In case of unlimited resources, the yeast will grow exponentially which means that the amount of time it takes to double the initial population is independent from the initial population size. Basically the idea of exponential growth.

Yes, doubling time is doubling time as long as there are unlimited resources. I don't know for brewing yeast, but for lab yeast, doubling time is ~ 1hr, E.coli is 30 min. I doesn't matter how many cells you start from.

I think one of the keys to good brewing is to have the yeast do less reproducing (aerobic) in your wort, and more fermenting (anaerobic). This is the benefit of a larger starter.

Now this makes my wonder then, if during reproductive growth, do the yeast produce more potential "off-flavors" (assuming optimal temp.). Presumably then these are comsumed to some extent during actual fermentation. Perhaps the longer it takes to begin fermentation, the more of the off flavors are produced and the less able the yeasts are to clean them up
 
Now this makes my wonder then, if during reproductive growth, do the yeast produce more potential "off-flavors" (assuming optimal temp)


In case of ester production, this is where home brewers and brewing scientists disagree. The literature reports that lower pitching rates lead to lower esters b/c the yeast will be able to grow for longer which consumes the ester precursor Acyl-CoA. Homebrewers report the opposite.

I wonder if there is another affect at work. Home brewers may not have adequate aeration to sustain a good growth of a smaller pitch but that is compensated when they pitch more yeast.

I still have to make an experiment to see for myself how pitching rate affects esters.

Kai

 
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