About Biochemical Engineering
Biochemical engineering questions in a recruitment paper cover the whole breadth of the subject - the design of the bioreactor, the modes of operation, growth and product kinetics, yield coefficients, oxygen transfer, mixing and heat transfer, sterilisation, scale-up, and the downstream processing that recovers the product - and most of them turn on a relation that has to be applied to the numbers given. This topic covers that breadth on one page, with the arithmetic that goes with it.
What you need to understand
- A bioreactor has to do four things: hold the organism, supply it with nutrients, supply it with oxygen, and remove heat. Everything else in the design follows from those four.
- Batch culture has nothing added and nothing taken out except gas, so the substrate falls and the product rises through the run. Productivity is limited by the downtime between batches.
- Fed-batch adds the substrate during the run, which keeps its concentration low. That avoids both substrate inhibition and the overflow metabolism that comes from too much sugar at once.
- Continuous culture, or a chemostat, feeds and removes at a constant rate. At steady state the organism grows at exactly the dilution rate, so the dilution rate is the control on growth.
- The dilution rate is the feed rate divided by the working volume. Setting it above the maximum specific growth rate washes the culture out entirely, which is called washout.
- Growth follows the Monod relation: the specific growth rate is the maximum rate times the substrate concentration divided by the saturation constant plus the substrate. At the saturation constant the organism grows at half its maximum.
- A yield coefficient says how much biomass is made per unit of substrate. The rest of the substrate goes to carbon dioxide, product and heat, so the biomass plus the product plus the carbon dioxide cannot exceed what went in.
- Volumetric productivity is the product made in each litre in each hour, and it decides how large a vessel has to be. Raising the titre without raising the rate does not raise the productivity.
- Oxygen dissolves poorly in water, so the transfer rate sets the limit on how dense a culture can be. The rate is the transfer coefficient times the driving force: the further the broth is below saturation, the faster oxygen goes in.
- The transfer coefficient rises with stirring and with the air flow, but stirring harder costs power and can damage shear sensitive cells, which is the central trade-off in reactor design.
- Mixing has to be good enough that the nutrients, the oxygen and the pH are the same everywhere in the vessel. A large vessel mixes worse than a small one, and that is the heart of scale-up.
- Heat is released by growth, roughly in proportion to the oxygen taken up, and it has to be removed through a jacket or a coil. A vessel that is too large for its cooling surface simply cannot be run any harder.
- Sterilisation is by steam, usually at 121 degrees Celsius for fifteen minutes, and the medium is sterilised in place while the vessel is empty of organisms.
- Downstream processing recovers the product: cell separation by centrifugation or filtration, then concentration, then purification by chromatography. It is usually the larger part of the cost of a biological process.
- A mass balance is the first check on any process calculation: what goes in either comes out or accumulates. A balance that does not balance means something has been left out.
How to work through these questions
- Name the mode of operation first - batch, fed-batch or continuous - because it decides which relations apply. A dilution rate belongs only to a continuous process.
- For a rate question, write down which quantity is per what. A specific rate is per unit of biomass, a volumetric rate is per unit of volume, and mixing the two up is the commonest error in the subject.
- For a yield, check that the answer is smaller than the substrate it came from. A yield above one gram per gram would mean mass appearing from nowhere.
- For an oxygen transfer calculation, subtract the broth concentration from the saturation value before multiplying. Multiplying by the saturation alone gives the rate at which oxygen would enter a broth with no oxygen in it at all.
- For a mass balance, add up everything that leaves and compare it with what went in before working out what accumulated. The comparison is the check.
- Check the scale of every answer: a dilution rate is a fraction of an hour, a productivity is a small number of grams per litre per hour, and a transfer rate is of the order of the oxygen demand of the culture.
Mistakes that cost marks
- Leaving the saturation constant out of the denominator of the Monod relation, which makes the growth rate exceed its own maximum.
- Inverting the dilution rate, dividing the volume by the feed rate instead of the feed rate by the volume.
- Adding the two outflows to the feed in a mass balance instead of subtracting them, which hides the fact that material is accumulating.
- Multiplying the transfer coefficient by the saturation concentration rather than by the driving force, which overstates the oxygen entering the broth.
- Forgetting that a yield coefficient is a fraction of the substrate, not a multiple of it.
- Dividing the product by the volume alone, or by the time alone, when the productivity needs both.
- Running a chemostat above the maximum growth rate and expecting steady state. The culture washes out, and no amount of waiting brings it back.
- Confusing the working volume with the total volume of the vessel: the headspace above the broth takes no part in the dilution rate, and it is the working volume that belongs in the calculation.
Worked example
A chemostat with a working volume of 20 litres is fed at 5 litres per hour. What is the dilution rate? An organism with a maximum specific growth rate of 0.4 per hour has a saturation constant of 5 gram per litre and is growing on 5 gram per litre: what is its specific growth rate?
- For the chemostat, the dilution rate is the feed rate divided by the working volume: 5 / 20 = 0.25 per hour.
- Check what that means: a dilution rate of 0.25 per hour replaces a quarter of the volume every hour, so the whole vessel turns over in four hours. That is a sensible figure for a continuous culture.
- For the growth, the substrate concentration equals the saturation constant, and whenever the two are equal the Monod relation gives exactly half the maximum rate.
- The rate is therefore 0.4 / 2 = 0.2 per hour. Working it out in full: 0.4 x 5 / (5 + 5) = 2 / 10 = 0.2 per hour, the same answer.
- That also shows the two answers together nicely: if the dilution rate were set at 0.2 per hour the culture would hold a substrate concentration of exactly 5 gram per litre, because at steady state the growth rate has to equal the dilution rate.
Answer: Dilution rate 0.25 per hour; specific growth rate 0.2 per hour
Practice questions with answers
A few Biochemical Engineering questions with the full solution shown, so you can see
how the method is applied before you attempt the timed set.
Question 1
An organism has a maximum specific growth rate of 0.3 per hour and a saturation constant of 10 gram per litre. Growing on 10 gram per litre of substrate, what is its specific growth rate, per hour?
Answer: Option D — with explanation
The Monod relation is the growth version of enzyme kinetics: the specific growth rate is the maximum rate times the substrate concentration, divided by the saturation constant plus the substrate concentration, mu = mu max S / (Ks + S).
That is 0.3 x 10 / (10 + 10) = 3.00 / 20 = 0.15 per hour.
At the saturation constant the organism grows at exactly half its maximum rate, and when the substrate is well above the constant the rate approaches the maximum. That is why a fermenter is run with the substrate in excess when the aim is the fastest possible growth.
Common mistakes
- 1 is not the answer: 1
- 0.07 is not the answer: 0.07
- 0.30 is not the answer: 0.30
Question 2
A chemostat with a working volume of 100 litres is fed at 10 litres per hour. What is the dilution rate, per hour?
Answer: Option B — with explanation
The dilution rate is the rate at which the medium is replaced, which is the feed rate divided by the working volume of the vessel: D = F / V.
That is 10 / 100 = 0.10 per hour - which also means the whole volume is replaced once every hour divided by that figure.
In a chemostat at steady state the organism grows at exactly the dilution rate, because it is being washed out at one rate and growing at the other. That is what makes the dilution rate the control: set it above the maximum growth rate and the culture is washed out altogether, which is called washout.
Common mistakes
- 100 is not the answer: 100
- 10 is not the answer: 10
- 1000 is not the answer: 1000
Question 3
A culture consumes 50 gram of substrate with a yield coefficient of 0.25 gram of biomass per gram of substrate. What mass of biomass does it produce, in gram?
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A
0.25
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B
200
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C
12.50
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D
50
Answer: Option C — with explanation
A yield coefficient says how much biomass is made from each unit of substrate used, so the biomass produced is the yield multiplied by the substrate consumed. It is a mass balance for the carbon: the rest of the substrate goes into carbon dioxide, product and heat.
That is 0.25 x 50 = 12.50 gram of biomass.
The coefficient is not a constant of nature. It depends on the organism, on the substrate, and on how much of the substrate is being diverted into product rather than into cells - which is why a process making a product usually has a lower biomass yield than one growing cells.
Common mistakes
- 0.25 is not the answer: 0.25
- 50 is not the answer: 50
- 200 is not the answer: 200
Question 4
A fermenter holding 25 litres makes 250 gram of product in 25 hours. What is the volumetric productivity, in gram per litre per hour?
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A
10
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B
0.40
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C
0.20
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D
0.80
Answer: Option B — with explanation
Volumetric productivity is how much product is made in each litre of broth in each hour, so it is the total product divided by the volume and by the time.
That is 250 / (25 x 25) = 250 / 625 = 0.40 gram per litre per hour.
It is the figure that decides how big a fermenter has to be for a given output, and it is why a process is judged on the productivity of the vessel rather than on the total it produces. A longer run makes more product but does not raise the productivity unless the rate itself improves.
Common mistakes
- 0.20 is not the answer: 0.20
- 10 is not the answer: 10
- 0.80 is not the answer: 0.80
Question 5
A fermenter has an oxygen transfer coefficient of 300 per hour. The saturation concentration of oxygen is 0.3 millimol per litre and the broth is held at 0.05 millimol per litre. What is the oxygen transfer rate, in millimol per litre per hour?
Answer: Option A — with explanation
Oxygen dissolves poorly in water, so a fermenter has to keep transferring it in from the gas as fast as the organism takes it out. The transfer rate is the transfer coefficient times the driving force, which is how far the broth is below saturation: OTR = kLa (C* - C).
The driving force is 0.3 - 0.05 = 0.25 millimol per litre, so the rate is 300 x 0.25 = 75 millimol per litre per hour.
The whole of oxygen transfer engineering is keeping that number above the organism's demand: raising the coefficient by stirring harder and sparging more air, or raising the driving force by enriching the air with oxygen or by running the broth cooler.
Common mistakes
- 15 is not the answer: 15
- 90 is not the answer: 90
- 105 is not the answer: 105
Frequently asked questions
Why does the growth rate equal the dilution rate in a chemostat?
Because the vessel is in steady state, so the amount of biomass has to stay the same. The organism is being washed out at the dilution rate and growing at the growth rate, and the only way for the two to cancel is for them to be equal. That is what makes the dilution rate the control lever.
What happens if the dilution rate is set above the maximum growth rate?
The organism cannot grow fast enough to replace what is being washed out, so the biomass falls steadily until the culture is gone. It is called washout, and it is irreversible without starting the culture again, because there is nothing left to grow.
Why is fed-batch often preferred to batch?
Because adding the substrate during the run keeps its concentration low, which avoids substrate inhibition and avoids the wasteful overflow metabolism that comes from a large amount of sugar at once. It also allows a much higher final titre than a batch can reach.
Why is oxygen transfer usually the limiting factor?
Because oxygen hardly dissolves in water, so the broth can hold only a few parts per million while the organism can consume them in seconds. The rate at which oxygen can be put back in therefore sets the ceiling on how dense and how fast the culture can be.
Why is downstream processing often the more expensive half?
Because a fermenter handles a dilute broth in one vessel, while recovery has to separate a small amount of product from a large volume of water and then purify it to a specification. Each step loses some product, so the yield and the cost both suffer.
What is the difference between the working volume and the total volume?
The working volume is the liquid in the vessel, and the rest is headspace left for foam and for the air. The dilution rate, the productivity and the heat release all refer to the working volume, so using the total volume makes every one of them wrong.
