CHEN 5751, Spring 2007

Stoichiometry of Growth and Production and Medium Design

Prof. Wei-Shou Hu

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Stoichiometry and Energetics ....................................................................................................... 1

Diversity in Stoichiometry........................................................................................................ 3

Material balance on biochemical reactions................................................................................... 7

Stoichiometry for cell growth....................................................................................................... 8

Biomass Formula ...................................................................................................................... 8

Energetic of Biomass Synthesis.................................................................................................. 10

Conversion and yield coefficient ................................................................................................ 13

THEORETICAL YIELD OF A PRODUCT .............................................................................. 18

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STOICHIOMETRY AND ENERGETICS

Chemical reactions can be written as a single or a system of stoichiometric equations.

0=∑ dicibiaii ONHCα In the equation the stoichiometric coefficient (α) has opposite sign for reactants

and products. By convention the products are given positive stoichiometric coefficients, while

reactants negative.

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The heat of reaction );( orHΔ is positive if the reaction is endothermic; or negative if it’s

exothermic. From a reaction point of view the formulation and solution of traditional chemical

reactions and biochemical reactions are virtually identical. Nevertheless, some characteristics of

biological reactions are worth mentioning. In most cases we will only consider C, H, O, N, and P in

biochemical reactions, as other elements participate only in a small fraction of all biochemical

reactions and contribute only a very small quantity to the biomass.

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Furthermore, most biological reactions (almost all) occur in aqueous solution. In treatment of

biological systems biochemical reactions rely on enzymes to reduce the activation energy, thus the

Gibb’s free energy change is of more interest to us, but not the heat of reaction. The reactions can only

occur if the corresponding enzymes are present. A large number of reactions are coupled. Many

require cosubstrates or cofactors, which are largely “recycled” or “regenerated” by another reaction or

reactions. Those reactions are thus “coupled”.

Diversity in Stoichiometry

Since the biological reactions are catalyzed by enzymes in the organism, the starting point in

setting up the stoichiometric equation is to consult the biochemical pathways. In addition to various

textbook, a number of websites also provide valuable data; some are more comprehensive and covers

important pathways of general interest, including the KEGG (http://www.genome.ad.jp/kegg), Brenda,

and ExPASy (http://www.expasy.ch or

http://www.expasy.com/prodinfo_fst.htm?/techserv/metmap.htm).

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In the course of evolution, most key metabolic pathways involved in energy metabolism,

biosynthesis of building blocks, (amino acids, fatty acids, nucleic acids) are very well conserved,

meaning that the basic reaction mechanisms and the enzymes are very similar. The pathways are used

from the most primitive organisms to mammals. For example, for organisms utilizing glucose to derive

energy aerobically, they all convert each mole of glucose to six moles of CO2 and six moles of H2O,

using basically the same set of reactions. However, amongst this seemingly very similar reaction

network, there are also many important differences between different species.

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Consider glucose metabolism, both yeast Sacchryomyces cerevisiae and bacterium Xymomonas

mobilis can convert glucose to ethanol, but they do have some subtle differences in their conversion

pathway. In Sacchanomyces the typical glycolysis pathway (Embden-Meyerhoff-Parnas (EMP)

pathway) is used, whereas in Xymomonas mobilis, the Entner Doudoroff pathway is used.

C6H12O6→2C2H5OH + 2CO2 ∆Go = -56.5 kcal/mole

ADP + Pi → ATP + H2O ∆Go = 7.3 kcal/mole

In Sacharomyces cerevisiae the net reaction is:

C6H12 O6 + 2 ADP + 2Pi →2 C2H5OH + 2CO2 + 2ATP + 2H2O

∆Go = -41.9 kcal/mole

Macroscopically one mole of glucose is converted to 2 moles each of ethanol and CO2. ATP is

used in intracellular chemical work and recycled back to ADP. Whereas in Xymomonas mobilis, the

net reaction is:

C6H12O6 + ADP + Pi →2 C2H5OH + 2CO2 + ATP + 2H2O

∆Go = 49.2 kcal/mole

It is clear that the pathways that the two microorganisms use have different energetic

efficiencies even though they both produce two moles each of ethanol and carbon dioxide from one

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mole of glucose. Different pathways with different efficiencies are not only seen in different

microorganism, but may also be used under different environmental or physiological conditions in the

same organism. For example, there are two pathways for ammonia assimilation into organic nitrogen

in E. coli. When the ammonia concentration in the environment is high, E. coli assimilates ammonia by

reductive amination of α – ketoglutarate. The reaction is catalyzed by NADP-dependent L – glutamate

dehydrogenase

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3 2 22 2

2 2

0COOH COOHC H N CH

NH NADPH NADP H OCH CHCH CH

COOH COOH

+

= ⋅++ + → + +

The product, L-glutamate, can then be the amino group donor to transfer the amino group to

other α-keto acid to form other amino acids.

When the ammonia concentration is low, E. coli employs an energetically more expensive

pathway with a higher affinity for ammonia (a lower km) to assimilate ammonia. The enzyme

catalyzing the reaction is glutamine synthetase.

The amino group in glutamine can be transferred to α position of α-ketoglutarate to form

glutamate as catalyzed by glutamate synthase.

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3 22

22

2

i

COOHCOOHH NCHH NCH

ATP NH ADP PCHCHCHCH

CONHCOOH

+ + → + +

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Glutamine + α-ketaglutarate + NADPH2+→2 glutamate + NADP+

The second molecule of glutamate is recycled to be used in the slutamine synthetase catalyzed

reaction. The net reaction is

3 2 iketoglutarate ATP NH NADPH glutamate NADP ADP Pα + +− + + + → + + +

In the above two examples the same chemical conversions, from glucose to ethanol and from α-

ketoglutarate and ammonia to glutamate, are accomplished. But different biochemical pathways are

used with different energetic efficiency.

MATERIAL BALANCE ON BIOCHEMICAL REACTIONS

The universal principle that materials cannot be created or destroyed (unless there is a nuclear

reaction) holds in biochemical systems. In any closed system the total mole of every element, C, N, O,

H, P, etc, is constant over time. As a first step in analyzing a biochemical reaction system, the system

and its boundary and the surroundings must be defined. Often a physical entity, such as a cell or a

bioreactor, is defined as a system, and material and energy balance is performed on the system.

Material balance can be performed on either a compound (a chemical species) or an element. For a

chemical species ί, the material balance is the rate of accumulation of i = rate of inputs of i – rate of

outputs of i + net rate of production consumption of i from all reaction. For a given element, such as C,

N, O, it can shift from one compound to another due to chemical reaction, but the total amount is

conserved. The balance is the rate of accumulation = rate of inputs – rate of outputs

In many cases the system on which the material balance is to be performed is “abstract”,

without a real physical entity. In other cases the material balance is performed on a physical system,

such as a bisector, a cell or a population of cells. For example, one can balance all glycolysis reactions

in the cell. In this case the system is a set of reactions catalysed by the glycolytic enzymes, not a

physical entity. The inputs are glucose and other compounds needed to carry out the reactions,

including NAD+, ADP, Pi, and the outputs are pyruvate, NADH2+, ATP.

Here one can also see a difference between balancing on a reaction system and on a cell with

respect to the same pathways. Take fermentation of glucose by yeast, Saccharomyces cerevisiae, as an

example. Balance on the cell using glucose as feed (input) and ethanol and CO2 as the products

(output). If the yeast is resting (not growing) and all the glucose consumed goes to ehanol, then the

balance on the cell as a system is glucose → 2 ethanol + 2CO2. On the other hand, if one balances on

the reaction system or implicitly balancing on intracellular glycolysis and fermentation reactions, then

the balance equation is:

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glucose + 2ADP + 2Pi → 2 ethanol + 2CO2 + 2ATP+2H2O

ADP, Pi and ATP do not cross cell cytoplasmic membrane. Macroscopically glucose is the only input

into the cell, and 2CO2 and ethanol are the only outputs. One can view the cell as an overall system,

the intracellular ethanol production as a subsystem. There must be another subsystem intracellularly

that convert ATP to ADP and Pi, so that the sum of the two subsystems equals the overall system.

That subsystem is the many intracellular reactions or “work” which consumes ATP. For example the

mechanical work and osmotic work both require ATP. Thus, two subsystems, one generating ATP the

other consumes ATP, are coupled.

STOICHIOMETRY FOR CELL GROWTH

From the stoichiometric and energetic point of view, cells degrade and convert the nutrients

(i.e. carbon and energy sources for chemotrophs) to derive chemical energy and produce biomass and

product. The chemical energy is used to assimilate NH3 (in microbes and plants), to make other

building blocks (fatty acids, amino acids, nucleic acids, etc.) and to make polymeric components of

biomass (proteins, lipids, nucleic acids, etc.) and to grow.

The cost of making polymers (protein, nucleic acids) from building blocks is rather high. The

incorporation of each mole of nucleotide into DNA or RNA requires 2 moles of ATP. The synthesis of

each peptide bond or the addition of each amino acid, through translation costs at leat 4 ATP. The

energetic cost of synthesizing a unit amount of biopolymers which constitute the bulk of biomass thus

can be estimated, at least in principle.

In addition to assembling the materials to make cell mass, cells also maintain a membrane

potential (pH and small electric gradient) and an osmotic pressure across membrane, they also take up

or excrete some compounds against gradient, and then all require energy. Thus, a large portion of

nutrients taken up by cells is used for generation of energy, to sustain their functions.

Biomass Formula

A chemical formula can be written for any compound or a chemical species with a defined

composition, such as glucose, glutamic acids, a protein like human insulin, a molecule of DNA. For a

mixture of molecules with known composition, such as a mixture of proteins, one can derive a

weighted average molecular formula. For example for a mixture of proteins consisted of n species Cαi

Hβi Nγi Oδi, each with a mole fraction ai., the average molecular weight can be expressed as

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MWAve = ∑∑

i

ii

aMWa

The average molecular formula Cα’i Hβ’i Nγ’i Oδ’i , will have their coefficient as

iα’=

i

ii

a

a

∑

∑ α

and likewise for other coefficients. However, when the molar composition of the protein

mixture is uncertain, as in the case of the cellular proteins, this approach is not applicable. One has to

resort to various methods of estimation. In the same vein, if the molar composition of the cell,

including all the biopolymers and building blocks, is all known, by summing up the equations for all

those components, one will be able to write a “formula” of the cell. In practice the precise molar

formula cannot be easily written by summing up all components in the cell and developing a weighted

average formula. Rather, a cell cormula can be written, but largely by doing an elemental analysis

(usually only on key component elements, C, N, O, H, maybe also P, S).

With a formula for biomass, one can develop a stoichiometric equation for converting nutrients

to cells. Considering the case cells take up organic nutrients, and use oxygen and ammonium

additionally.

The stoichiometric equation for “synthesizing” the cell, from nutrients, and those for

transforming building blocks into polymer and other components of cells.

i∑ 2 3 ' ' ' ' 2 2i i i iC H N O aO bNH cC H N O dCO eH Oα β γ δ α β γ δ+ + → + +

Where subscript i indicates different nutrients. The formula of biomass is basically the elemental

composition of cells.

Although the stoichiometric equation for making each major biopolymetric material (protein,

DNA, etc) can be written, the sum of those equations does not give rise to the biomass stoichiometric

equation. In addition to the cost of making materials there are other costs associated with

“maintenance” or for taking up nutrients, excreting metabolites. There are many other “costs”, for

example in DNA replication, there is extensive “proof reading” and editing to remove mismatched

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nucleotides, i.e. the nucleotides that are mistakenly incorporated into replicating DNA and not a

correct match, are removed. The extent of such repair and the energetic cost of such repair is still

uncertain. The energetic cost of maintaining cells homeostasis (osmotic pressure, pH gradiant etc.) is

also difficult to estimate.

ENERGETIC OF BIOMASS SYNTHESIS

The energy for biosynthesis of biomass is derived from catabolism of nutrients. In most living

organism, carbohydrate metabolism through glycolysis and tricarboxylic acid (TCA) cycle contributes

most to energy generation. Upon complete oxidation of glucose to CO2 and H2O via TCA cycle, the

energy efficiency is calculated from the stoichiometric equation

C6H12O6 + 6O2 + (30~33)ADP +36 Pi →6CO2 + 6H2O + (30~33)ATP H2O

The stoichiometric coefficient of ATP varies with organisms. It depends on the efficiency of ATP

sythase.

If one considers biomass as an assembly of all the other constituents, one can write the

equation”

""""'

δγβα ONHCa (average protein formula) + """""

δγβα ONHCa

(average DNA formula) + … (RNA) + …→biomass

The energetics of assembling the constituents into cells cannot be easily estimated. With the

uncertainty about the energetic efficiencies as described above, it is very difficult to obtain a biomass

equation with a high degree of confidence.

Another difficulty in developing a stoichiometric equation for biomass arises from “turn-over”

of cellular materials. Both protein and mRNA “turns over” in the cells. In other words, after being

synthesized they serve their function for a period of time, and then get degraded. Some components

become degradation products and are excreted or catabolized, others become monomers like amino

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acids and recycled to build cellular biopolymers. The balance in the intracellular concentration of

protein P is described as

α=dtdP -βP-μP

where α is the average synthetic rate, β is the rate constant for degradation (turn over), and μ is the rate

of cell expansion which causes P, the intracellular concentration of proteins, to decrease due to the

expansion of cell volume. The degradation rate of protein and mRNA and varies widely. The formula

for the synthesis of biopolymers can reasonably accurately incorporate energetic cost (i.e. specify the

number of moles of ATP used to synthesize a mole of biopolymer), or can specify the energetic cost

associated with α. However, knowing steady state cellular composition P is not sufficient to determine

α unless β and μ are also determined. In practice, even though μ can be determined accurately, β is

largely undetermined. In other words, cells with the same P and the same μ may have very different α

and β even both have the same dP/dt.

Other factors that contribute to the wide range of diversity in the biomass formation equation,

even among microorganisms of similar composition, is the energetic components which are not

directly related to “cell composition”, or to the synthesis of biomass. Such factors include the

energetic cost of transport and maintenance of membrane potential and osmotic pressure. Nutrient

uptake and product excretion often require energy. Some nutrients are transported across cell

membrane by passive or facilitated diffusion for which there is not a net energy cost. For many others,

the transport is coupled to the concentration gradient of Na+/K+ or H+ across the membrane. In other

cases the transport is directed coupled to the hydrolysis of ATP. The maintenance of the concentration

gradient of a chemical species requires energy (otherwise, no matter how small the permeability the

cellular membrane is, eventually the two sides of the membrane will reach equilibrium). Transport of

molecules utilizing H+ or Na+/K+ gradient is thus energy dependent indirectly. In many cases, the cost

of those transport process are not clearly known. In Eukaryote, organelles provides

compartmentalization in all the cell. Electric or pH potential exists across those membranes separating

the cytosol and lumen of organells. Reaction intermediates are transported across these organelles.

Maintaining these gradients and transport across orgenell is also energy dependent.

The stoichiometric equation for biomass can be written to include many atomic species,

especially P, S. However, usually it includes C, H, N, and O. For each elemental species, an

elemental balance equation can be written. In the case that a biomass equation considers only C, H, N,

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and O, four conservation equations can be derived for C, H, N, O. In the stoichiometric equations for

biomass using NH2, O2 and a carbon source five to six chemical species are involved in the reaction.

The four elemental balance equations are certainly not sufficient to determine the five stoichiometric

coefficients. The equation is thus undetermined. One has to rely on experimentally measured values to

obtain the stoichiometric coefficients. Unlike simple stoichiometric chemical reaction equations like

CH3 CO COOH → CH3 CHOH COOH, for which a unique set of stoichiometric coefficient exists, a

wide range of the combination of stiochiometric coefficients exist for biomass equation. There is thus

no single equation for all biomass forms. There is a wide diversity on how organisms use the same raw

materials to build cells.

The biomass equation presented in this chapter describes only the conversion of substrate into

biomass. Implicitly we considered only the organic matters in the cells only. In addition to their

organic constituent, cells contain a large amount of water. In fact, the water content of cells is typically

from 75-85%. In plant cells the fraction can be even higher under some conditions. The stoichiometric

equation described above considers only the dry biomass, not wet biomass. In general, the symbol, x,

will be used to denote dry biomass.

Like writing an average molecular formula for a mixture of compounds, there are many

different ways to present the biomass formula. Conventionally one can set one atomic coefficient

(usually carbon) to be 1.0 or set the formula weight to be equal to a convenient number such as 100.

Biomass is made up of more than the four elements of C,H,N,O. Other major constituents include P, S,

Mg, Fe, and other minerals. Those are often referred to as ashes, since their mass is determined after

treatment of cells in a high temperatures (~450oC) furnace to complete oxidize the combustible portion

of the cells. Therefore with a formula Cα, Hβ, Nr, Oб, the formula weight accounts for only the portion

excluding the ashes.

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It should be obvious that since the biomass composition is dependent on the cultivation

condition and on the growth stages, both the biomass formula and the values of the stoichiometric

coefficients for the biomass equation change accordingly. The biomass composition varies with other

physical, chemical variables in the environment. When a complex medium is used for cultivation or

when additional products (in addition to CO2 and H2O )are produced, other terms are added to the

equation accordingly.

Biomass formation equation is basically a summary of the material balance (i.e. stoichiometric

relation) of nutrient uptake and the outputs of cells and excreted products. Experimentally, the equation

is merely an approximation, since a precise measurement of biomass formula is very difficult to obtain.

Furthermore, the equation always entails production of H2O, CO2 and consumption of O2. For CO2 and

O2, experimental measurements are at least available, although accurate determination is not easy. For

H2O, the measurement is extremely difficult since cells grow in aqueous environment. Therefore, the

utility of the biomass formula is largely in the analysis of material’s flow using material balances

(metabolic flux analysis) for which the formation of biomass must be accounted for in the overall

balance.

CONVERSION AND YIELD COEFFICIENT

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The stiochiometric equation of biomass can be seen as a conversion of new materials (or input)

into cells and other products (outputs). A yield coefficient is often used to describe the conversion

efficiency. This term yield is loosely defined and used in different units. Virtually any pair of output

and input gives a yield coefficient. The most commonly used ones are shown in Table III.

Units Example

Yp/s Yield of product based on substrate Kg product/kg substrate, mole product/mole substrate

Yethanol/glucose

Yx/s Yield of biomass based on substrate

Kg biomass/kg substrate Ybiomass/glucose

Yx/ATP Yield of biomass based on ATP (i.e., ATP generated from substrate consumed)

Kg bimass/mole ATP Ybiomass/glucose

Yx/o Yield of biomass mased on oxygen Kg biomass/kg oxygen, kg biomass/mole oxygen

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The basic form of yield coefficient is

SXY sx Δ

Δ=/

In the case that multiple nutrients and products are invovlved, for example, in mammalian cell

culture, a stoichiometric ratio (βίј) for a pair of nutrients/products is often used. For example, in

mammalian cell culture, a large fraction of glucose consumed by cells is converted to lactate. Under

the condition that glucose concentration is controlled at a very low level, much less lactate is produced.

The stoichiometric ratio of lactate to glucose is often used to characterize the metabolism of glucose.

Yield coefficient and stoichiometric ratio can be applied to a wide variety of substrates, with a variety

of units. Referring to the biomass formation equation, on a mass basis it is basically the product of the

stoichiometric coefficient of the biomass and the formula weight of biomass divided by the product of

the stoichiometric coefficient and the formula weight of the substrate. The biomass composition and

the biochemical pathways used for cell growth may change under different conditions. As a result, the

biomass stoichiometric equation, and the yield, may also change. Thus, yield can be an instantaneous

value obtained at a particular time point; alternatively, it can be an average value over a long period of

cultivation.

dsdxY sx =/

2

1

2

1

/

t

t

t

t

x

xx s s

s

dxY

ds=∫∫

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Depending on the nature, a nutrient (or commonly called substrate), may be either completely

conserved in biomass, or only partially converted to biomass. In the latter case, the rest is excreted as

converted product. For example, carbon source usually serves as energy source (except in

photosynthetic organism), thus, part of it is converted to biomass, and the rest is excreted as CO2 and

other products. The fraction of carbon atom conserved in biomass is always less than 1.0. For Na, K,

Fe, basically all that is taken up by cells is retained in biomass for nearly all chemoorganotrophs. The

yield coefficient based on atom basis is 1.0 mole cellular atom/mole uptake atom. For nitrogen source,

the yield is variable depending on the organism and the growth conditions. If there is no excreted

nitrogen containing compound, then the yield is basically 1.0 mole cellular N/mole uptake N.

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THEORETICAL YIELD OF A PRODUCT

Knowing the theoretical maximum efficiency that a raw material can be converted to a product

is important in process design and innovation. The return on the investment of improving the

production yield is higher if the yield is substantially below the theoretical maximum than if the yield

is already approaching the maximum. To estimate such a maximum one needs to consider both

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material and energetic aspects. Largely one needs to consult the biochemical pathways of the

organism. There are different issues to be addressed for different types of compounds.

a. Products that are converted directly from the raw materials in an energetically favorable

fashion. In other words, the producing cell confers a biochemical pathway to convert the raw

materials to the product with a net generation of chemical energy. For example, for ethanol

production by yeast, one mole glucose can be converted to two moles of ethanol, and generates

two moles of ATP. The reaction in thus energetically feasible, and the theoretical maximum is

(2x48/180=) 0.51 Kg ethanol/Kg glucose. The true yield in process is lower because some

glucose will be diverted to make biomass and for the cells other energetic needs. The basic

way of calculating the theoretical yield of such products is to tabulate the pathway and examine

its energetic conversion. The second example, production of glutamic acid from glucose and

ammonia in corynebacterium glutamican encompasses the following pathways.

Glucose + 2ADP + 2Pi→2 Pyruvate + 2 ATP

Pyruvate + NAD→acetylCoA + CO2+NADH

Pyruvate + CO2 + ATP + H2O→oxaliacetate +ADP + Pi

Acetyl CoA + oxaliacetate + NAD→ α-ketoglutarate + CO2 + NADH

α-ketoglutarate + NADPH + NH3 → glutamate +NADP + H2O

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Net reaction is:

Glucose + NH3 + ADP + Pi+ NADPH + 2NAD → glutamate +CO2 + ATP + 2NADH + NADP

Overall, it is energetically feasible since a net production of 1 ATP and 1 NADH. So, the

theoretical maximum conversion based on carbon is 5/6 (mole carbon in glutamate/mole carbon

in glucose), or on a mass basis (143/180=) 0.79

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b. Products which require energetic contribution of energy derived from the catabolism of

raw materials.Some products from biosynthetic pathways require the catabolism of

substrate to make the reaction energetically favorable. In this case, from the

biochemical pathways one derives the biosynthetic reaction equation:

icPcADPproductNOHCcATPsubstrateNOHbCsubstrateNOHaC ++→++ )()2()1( """"'''' δγβαδγβαδγβα

The energy is supplied by catabolism of the substrate, assuming catabolism is primarily derived

from substrate:

ATPdCOcPdADPdObNOHC i'

2'''

2' +→+++δγβα

The maximum yield is the combination of the two reactions that gives a net production of ATP

equivalent is zero. For first appropriation, one can assume one mole NADH is equivalent to three

moles of ATP.

a. Metabolic engineering of the pathways. In the past two decades our ability to alter the genetic

makeup and their metabolic pathway has been greatly enhanced. This has enabled one to

construct new pathways by introducing enzymes which are not normally present in the

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producing organism. Many of such efforts are, of course, aimed at producing new products.

Many others are aimed at alternating the carbon or energy efficiency of the process. The

estimation of the maximum theoretical yield of a product from a substrate is now not just based

on the pathways in the producing organism, but on pathways, or the combination of segments

of pathways, available in the living system. One may even take one step further by combining

with protein engineering to create new enzymes to facilitate improved carbon flow or energy

efficiency for product formation. Therefore, the question to be posed is the thermodynamic

efficiency for the conversion of substrates to product. Take ethanol fermentation as an example,

2526126 22 COOHHCOHC +→ ∆G=

2526126 22222 COATPOHHCPADPOHC i ++→++ ∆G=

Much of the chemical energy of glucose (%) is preserved as ATP when converting to ethanol.

The carbon efficiency is 66%. The ∆G of the reaction is sufficient to ensure the reaction occurs

spontaneously. But how much ∆G is sufficient to drive a reaction forward? One also needs to keep in

mind that eventually the energetic cost of producing a compound may also entail the uptake of nutrient

and the excretion of the product.

STOICHIOMETRY 1. A microorganism is capable of utilizing glucose, methanol and hexadecane. Its

average cell composition on a weight basis was analyzed to be 47% carbon, 6.5% hydrogen, 31% oxygen, 10% nitrogen and the rest is ash. During the cultivation it converts the substrate, O2 and ammonia into cells, carbon dioxide and water. A batch culture was carried out to estimate the cell yield based on substrate and oxygen (kg cells/kg substrate or oxygen). Air was supplied continuously and the exhaust gas was vented from the top of the bioreactor. A mass spectrometer was available for gas composition analysis. The water vapor was removed from both inlet and outlet gases before analysis. The inlet gas was composed of 21% oxygen and 79% nitrogen. The composition of the off-gas on a volumetric basis is listed below.

Substrate % nitrogen %carbon dioxide %oxygen glucose 78.8 10.2 11.0 hexadecane 85.6 9.0 5.4 methanol 85.3 8.5 6.1

Write down the stoichiometric reaction equation for each substrate. What are the estimated cell yields based on substrate and oxygen for the three substrates? What are the R.Q.s? (R.Q. is respiratory quotient = mole carbon dioxide produced/mole of oxygen consumed.)

2. Calculate the theoretical maximum yield of L-lysine production based on glucose

using glucose and NH3 as the raw materials. 3. A yeast can be grown in glucose and hexane. A student performed a series of

experiments to determine the yield coefficients. Unfortunately he forgot to label the results properly and have only the numerical values for biomass yield based on glucose, hexane and ammonia (as the nitrogen sole nitrogen source) listed below. The ammonia yield is the same for both carbon sources. Identify which one is most likely to be for glucose, hexane and ammonia respectively compound and explain. 45% of the biomass is elemental carbon, 7.0% is nitrogen.

i. 0.5 g cell/g ii. 0.8 g cell/g

iii. 11.8 g cell/g 4. A batch culture of E. coli was initiated with a medium containing 10 g/l glucose, 2 g/l

ammonia, and 0.1 g/l of cells. At the end of culture, 0.1 g/l of glucose remained in the medium. A total of 5.1 g/l of E. coli were recovered. The elemental composition of the cells was determined to be 53% (by mass) C, 7.3% H, 12.0% N and 19.0% O. The balance was ashes. Calculate the yield (by mass) of biomass based on glucose and nitrogen. Write down the stoichiometric equation for biomass formation from glucose and ammonia. Assume all the carbon not incorporated into biomass was used for energy generation and got completely oxidized to carbon dioxide, and assume

each mole of glucose produces 38 mole of ATP when completely oxidized to carbon dioxide. Calculate the biomass yield coefficient based on ATP.

5. The biosynthetic pathway of phenylalamine in E. coli is available in a standard

biochemistry textbook. Estimate the theoretical maximum yield on phenylalamine based on glucose when E. coli is grown in glucose and ammonia.

6. The experimental yield of yeast Candida grown on glucose, methanol and hexadeane

are 0.51, 0.41, and 0.84 (g dry weight/g substrate), respectively. The nitrogen source is NH3. The biomass composition is determined to be 47.3% (by mass). 6.7% H, 10.0% N, and 28.8% O. The balance is ashes. Estimate the yield coefficient of biomass based on oxygen (Y x/o) is kg dry biomass/kg oxygen. Can you generalize the magnitude of yield coefficients for biomass based on substrate and oxygen with respect to the degree of the oxidative state of the substrate?

7. The biomass formation equation for a microorganism using glucose and NH3 as carbon and

nitrogen sources are shown below.

a. Calculate the yield coefficient for biomass base on glucose and NH3 (kg cells/kg substrate).

6 12 6 3 2 1.7 0.181 0.475 2 2 2C H O + 0.705NH + 1.867O 3.899 CH N O + 0.088O + 2.103CO + 3.75H O→

b. In a one liter batch culture that has a cell concentration 20 g/l and a doubling time

of 1 hour. How much glucose (in grams) has to be fed to the culture in the next hour in order to sustain the growth rate? Assume the glucose can be added in a very concentrated solution so that the culture volume is not affected by glucose addition.

8. A batch culture of E. coli was initiated with a medium containing 10 g/l glucose, 2 g/l

ammonia, and 0.1 g/l of cells. At the end of culture, 0.1 g/l of glucose remained in the medium. A total of 5.1 g/l of E. coli were recovered. The elemental composition of the cells was determined to be 53% (by mass) C, 7.3% H, 12.0% N and 19.0% O. The balance was ashes. Calculate the yield (by mass) of biomass based on glucose and nitrogen. Write down the stoichiometric equation for biomass formation from glucose and ammonia. Assume all the carbon not incorporated into biomass was used for energy generation and got completely oxidized to carbon dioxide, and assume each mole of glucose produces 38 mole of ATP when completely oxidized to carbon dioxide. Calculate the biomass yield coefficient based on ATP.

9. The biosynthetic pathway of phenylalamine in E. coli is available in a standard biochemistry

textboos and on KEGG, BRENDA website. First sketch the pathway (you can copy and paste) starting from glucose. Estimate the theoretical maximum yield on phenylalamine based on glucose when E. coli is grown in glucose and ammonia.

10. The yeast, Saccharomyces cerevisae, and bacterium, Zymomonas mobilis, can both

ferment glucose to produce ethanol. The ethanol fermentation process in both

microorganisms is growth associated, i.e. the production of ethanol is an energy generating process accompanying cell growth. Saccharomyces uses EMP which produces 2 mol ATP/mol glucose converted to ethanol. Zymomonas uses Entero-Doudoroff pathway which produces only 1 mol ATP/mol glucose converted to ethanol. The biomass yield based on ATP can be assumed to be 100 mmole ATP/g biomass for both. The cellular composition (% of dry biomass) for both microorganisms is listed below. In a medium containing 80 g/l of glucose, 3 g /l of NH3, 3 g/l of KH2PO4, and with plenty of other minerals, how much ethanol (in g/l) can be produced by each microorganism respectively? You must justify that the medium actually contains sufficient elemental sources for producing the biomass and ethanol.(Note: atomic weight K = 39, P = 31) Saccharomyces Zymomonas

C 50 45N 8 9O 30 31H 7 11P 1 1.5 11. A genetically engineered bacterium is used to produce fine cellulose. It grows on

glucose and converts glucose to cellulose. The reaction to synthesize cellulose is nC6H12O6 + 2(n-1)ATP C6H11O6(C6H10O5)n-2C6H11O5 + 2(n-1)ADP + 2(n-1)Pi +(n-1)H2O

The average chain length, n, of the cellulose molecules is 100. Assume one mole of glucose can be completely oxidized to generate 36 moles of ATP. Calculate the theoretical maximum yield (accounting for the energy requirement for its synthesis) of cellulose (g cellulose/g glucose). This bacterium is grown in a medium containing 40 g/l of glucose. Cells grow at a maximum rate until glucose is exhausted and cell concentration reaches 5.0 g/l. The initial cell concentration (innoculum) is 0.05 g/l. At that point the gene for cellulose synthesis is "turned on" (or induced) and the growth immediately stops. The final cellulose concentration is 21 g/l. Assume that there is no maintenance energy requirement, calculate the yield coefficient of biomass (g cells/g glucose).

12. Cells excised from plants, especially from embryos in the seeds, can be cultured in

bioreactors. After being propagated to increase biomass, the cells can be induced to form embryos by controlling the concentration of growth regulators in culture. The embryos are then encapsulated and become artificial (or synthetic) seeds. These synthetic seeds can give rise to virtually identical clones. Cells of Douglas fir are grown in culture. The medium contains 200 kg glucose and 25 kg NH3 per m3.

Initially, 0.05 m3 of cells were inoculated into 0.95 m3 of medium to initiate the culture. After two weeks when glucose was depleted, 0.35 m3 of biomass was recovered. At that moment, the NH3 concentration was 0.001 kg/m3. The specific density of the wet biomass is 1,020 kg/m3 and the water content of the biomass is 90%. Calculate the yield of the biomass based on glucose and NH3 (express the biomass in dry weight).

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