Subject Code: CHE-513
Course Title: Biochemical Engineering
Relative
Weightage
CWS 20-35
Credits: 4
MTE 20-30
ETE 40-50
Lectures: 3/week
Tutorial : 1/week
2
Course Content
S. No.
Content
1.
Introduction
2.
Microbiology
3.
Biochemistry
4.
Kinetics of Enzyme Catalysed Reactions
5.
Microbial Fermentation Kinetics
6.
Sterilization
7.
Aeration and Agitation
8.
Scale-up of Bioreactors
9.
Aerobic and Anaerobic Fermentations
10.
Downstream Processing
3
Suggested books
S. No.
Authors / Name of Book / Publisher
Year of
Publication
1.
Bailey J.E. and Olis D.F., “Biochemical Engineering
Fundamentals”,2nd Ed., McGraw-Hill.
1987
2.
Doble M. and Gummadi S.N., “Biochemical Engineering”,
Prentice Hall.
2007
3.
Schuler M.L. and Kargi F., “Bioprocess Engineering”, 2nd
Ed., Prentice Hall.
2002
4
Chapter 1
S. No.
Content
1
Introduction
1.1
History of Biochemical Engineering
1.1.1
Early Development of Biochemical Engineering
1.1.2
Renewed and Expanded Efforts on Biomass Conversion
1.2
Biochemical Engineering Fundamentals
1.3
Role of biochemical engineering in the biochemical product synthesis
1.3.1
Biochemical Engineer’s Responsibilities
1.3.2
Further Advances in Biochemical Engineering
1.4
Bioprocess Economics
1.4.1
Process Economics
1.4.2
Bioproduct Regulation
1.4.3
General Fermentation Process Economics
5
1 Introduction
 Biochemical engineering, also known as bioprocess engineering, is a field
of study with roots stemming from chemical engineering and biological
engineering
 Processing of biological materials and processing using biological agents
such as cells, enzymes, or antibodies are the central domains of biochemical
engineering.
 Success in biochemical engineering requires integrated knowledge of
governing biological properties and principles, and of chemical engineering
methodology and strategy.
 Work at the forefront captures the latest, best information and technology
from both areas and accomplishes new syntheses for bioprocess design,
operation, analysis, and optimization. Reaching this objective clearly requires
years of careful study and practice.
6
Biotechnology and biochemical engineering
Biotechnology is "the integration of natural sciences and engineering sciences in
order to achieve the application of organisms, cells, parts thereof and molecular
analogues for products and services"
Biotechnology is technology based on biology - biotechnology harnesses
cellular and biomolecular processes to develop technologies and products that help
improve our lives and the health of our planet.
Biochemical engineering, mainly deals with the design, construction, and
advancement of unit processes that involve biological organisms or organic
molecules and has various applications in areas of interest such as biofuels,
food, pharmaceuticals, biotechnology, and water treatment processes.[1][2]
The role of a biochemical engineer is to take findings developed by biologists and
chemists in a laboratory and translate that to a large-scale manufacturing process.
7
Continued…
Cells
Enzyme
Antibody
8
• Antibody, also called immunoglobulin, a protective protein produced
by the immune system in response to the presence of a foreign substance,
called an antigen. Antibodies recognize and latch onto antigens in order to
remove them from the body. A wide range of substances are regarded by
the body as antigens, including disease-causing organisms and toxic
materials such as insect venom.
• When an alien substance enters the body, the immune system is able to
recognize it as foreign because molecules on the surface of the antigen
differ from those found in the body. To eliminate the invader, the immune
system calls on a number of mechanisms, including one of the most
important—antibody production.
• Antibodies are produced by specialized white blood cells called
B lymphocytes (or B cells). When an antigen binds to the B-cell surface, it
stimulates the B cell to divide and mature into a group of identical cells
called a clone.
• The mature B cells, called plasma cells, secrete millions of antibodies into
the bloodstream and lymphatic system.
Vaccine production
Vaccine production has several stages. Process of vaccine manufacture has the
following steps:
oInactivation – This involves making of the antigen preparation
oPurification – The isolated antigen is purified
oFormulation – The purified antigen is combined with adjuvants, stabilizers and
preservatives to form the final vaccine preparation.
The initial production involves generation of the antigen from the microbe. For this the
virus or microbe is grown either on primary cells such as chicken eggs (e.g. in
influenza) or on cell lines or cultured human cells (e.g. Hepatitis A).
Bacteria against which the vaccines are developed may be grown in bioreactors (e.g.
Haemophilus influenzae type b).
The antigen may also be a toxin or toxoid from the organism (e.g. Diphtheria or
tetanus) or it may be part of the microorganism as well.
Proteins or parts from the organism can be generated in yeast, bacteria, or cell
cultures.
Bacteria or viruses may be weakened using chemicals or heat to make the vaccine
(e.g. polio vaccine).
After the antigen is generated, it is isolated from the cells used to
generate it.
For weakened or attenuated viruses no further purification may
be required.
Recombinant proteins need many operations involving
ultrafiltration and column chromatography for purification before
they are ready for administration
11
1.1 History of Biochemical Engineering
 The phrase, “biochemical engineering”, first appeared in the late 1940s
and early 1950s. That was the time shortly after aerobic submerged
culture was launched as a way of increasing the production capacity of
penicillin, used to cure battle wounds of World War II (Shuler and Kargi
1992).
 Commercial scale biological processing of biomass materials is an
activity as old as human civilization.
 In more recent years, utilization of lignocellulosic biomass has become
an actively pursued subject, because of the concerns of future exhaustion
of non-renewable fossil fuels.
 Since the energy crisis caused by the oil embargo by the OPEC countries
in 1974, there have been expanded efforts in research and development
aimed at improved conversion efficiency of lignocellulosics as an
alternative material resource. The future payoff from successful
utilization of lignocellulosics will be enormous.
12
1.1.1 Early Development of Biochemical
Engineering
 Following the commercialization of penicillin, a large number of
antibiotics were also discovered from extensive screening programs by
many pharmaceutical companies worldwide in the 1950s and 1960s.
 There was a strong demand for better designs of aeration systems and
deeper understanding of the process of oxygen transfer in biological
systems so that many new drugs could be manufactured efficiently.
 Oxygen transfer became in those years a popular subject for biochemical
engineers to engage in.
 In those years, one of the most important references in oxygen transfer is
the comprehensive review by Professor Robert Finn (1954).
13
1.1.2 Renewed and Expanded Efforts on
Biomass Conversion
 Renewable biomass and coal were looked upon for possible replacement
of fuels and chemicals from petroleum.
 At the time, George Tsao was on assignment at the U.S. National Science
Foundation, on leave from Iowa State University, managing several
funding programs as a part of the RANN (Research Applied to National
Needs) initiatives.
 Recognizing the need for biomass research, George Tsao invited
Professor Charles R.Wilke of the University of California at Berkeley to
conduct a conference on the use of cellulose as a potential alternative
resource of fuels and chemicals.
 It took place in 1974 and the proceedings of that conference were later
published as a special volume of Biotechnology and Bioengineering
(Wilke 1975).
14
Continued…
 That conference served an important function in stimulating renewed
interest in the conversion of biomass into fuels and chemicals. In 1978,
the first of a series of conferences on Biotechnology for Fuels and
Chemicals was organized by researchers of the Oak Ridge National
Laboratory, Oak Ridge, Tennessee, under the leadership of Dr. Charles D.
Scott. Later, researchers of the National Renewable Energy Laboratory,
Golden, Colorado also joined the effort and this series of conferences has
been held annually ever since.
15
1.2 Biochemical Engineering Fundamentals
 Biology is amazing at transforming one
chemical to another, Biochemical Engineering
is the harnessing of this process to make
products on an industrial scale.
 One of the earliest examples was brewing
where
humankind took grain and water and
used yeast to change it into beer.
 Over the last hundred years the challenge has
been translating discoveries into processes,
such as taking an exciting discovery like
antibiotics and finding a way to manufacture
them on a large scale. Without a viable way of
making antibiotics quickly, safely and cheaply
they would never have saved as many lives as
they have.
16
Continued…
 Biochemical engineering involves the use of a catalyst, either an enzyme
or whole cell, which is either freely suspended in an aqueous medium or
“immobilized” in a gel or attached to a solid surface.
 In contrast to an enzyme, whose activity decreases over time, cells are
autocatalytic (dx/dt = μx, where x = cell concentration and μ = growth
rate, t–1) and can grow exponentially with excess nutrients.
17
1.3 Role of Biochemical Engineering in the
Biochemical Product Synthesis
 Biochemical engineering is concerned with conducting biological
processes on an industrial scale. This area links biological sciences
with chemical engineering. The role of biochemical engineers has
become more important in recent years due to the dramatic
developments of biotechnology.
 We need to design an effective bioreactor to cultivate the cells in the
most optimum conditions. Therefore, biochemical engineering is one
of the major areas in biotechnology important to its
commercialization.
 Biochemical engineers research and develop processes to use
microbial organisms and enzyme systems for manufacturing
chemical products. It is necessary to have advanced knowledge of
biochemistry, microbiology and chemical engineering to work in this
field. By understanding the exact duties of biochemical engineers.
18
Continued…
Figure 1.1: To illustrate the role of a biochemical engineer, let's look at a typical
biological process (bioprocess) involving microbial cells as shown.
19
Continued…
To carry out a bioprocess on a large scale, biochemical engineers need to
work together with biological scientists:
1. To obtain the best biological catalyst (microorganism, animal cell,
plant cell, or enzyme) for a desired process.
2. To create the best possible environment for the catalyst to perform by
designing the bioreactor and operating it in the most efficient way.
3. To separate the desired products from the reaction mixture in the most
economical way.
20
Continued…
 Biochemical engineers follow strict safety protocols since they work with
potentially dangerous chemicals, materials and equipment. They may
assist with creating safety regulations to ensure product manufacturing
safety. Some of the places that employ biochemical engineers are:
o Hospitals
o Aerospace companies
o Research centres
o Chemical companies
o Universities
o Petrochemical engineering
companies
o Textile companies
o Medical instrumentation
companies
o Food companies
o Mining companies
o Paper companies
o Oil and natural gas companies
o Plastics companies
o Pharmaceutical companies
21
1.3.1 Biochemical Engineer’s Responsibilities
The work responsibilities of biochemical engineers can vary by the industry
they work in and their experience level and skills. Some of the typical duties
include the following:
 Researching interactions of biological materials in different environments
and ways to inhibit or promote their growth.
 Predicting material interaction results by using computer simulation and
statistical models and replicating the results in the laboratory.
 Studying cells, proteins, enzymes and other biological materials and
using them to build compounds for public consumption.
 Preparing, overseeing, directing and recording laboratory experiments
and writing technical reports on research findings.
 Training staff and providing technical support in the use of biomedical
equipment and evaluating its safety and effectiveness.
22
Continued…
 Working in collaboration with biologists, chemists, research staff and
manufacturing personnel to develop new products and technologies.
 Designing equipment with computer-aided design (CAD) software,
developing production processes and preparing budgets.
 Checking the accuracy of information about products and product
development processes and preparing operating manuals for mass
production.
 Publishing research papers, applying for patents, making research
recommendations and presenting research findings.
 Maintaining a research database and staying updated on the latest
industry research and scientific and technological developments.
23
1.3.2 Further Advances in Biochemical
Engineering
 The required engineering expertise in manufacture of ethanol and bulk
chemicals is somewhat different from that in producing high priced
health products. For low value products, the required efficiency is
important in determining the overall process economics.
 For pharmaceuticals, assurance of the product purity and safety is of top
priority. The price of a new drug can always be properly adjusted to give
manufacturers the desired profit. For products like ethanol, process
efficiency requirements stimulated a great deal of engineering research
and development work over the years.
 In order to improve reactor efficiency, multiple stage, fed batch reaction
systems with internal cell recycle has been developed for ethanol
production in the industry. For reducing energy cost spent in dehydration
of ethanol to produce fuel grade products, the technique of adsorption by
corn grits was invented (Ladisch 1984).
24
Continued…
 Biological agents almost always suffer from product feedback inhibition.
The SSF (simultaneous saccharification and degradation) concept avoids
the feedback inhibition of cellulase activities.
 For ethanol, a number of techniques were proposed and tested, including,
for instance, membrane separation of the ethanol product (Matsumuro
1986), evaporation of ethanol from processing beer by gas stripping
(Dale 1985, Zhang 1992), pervaporation (Mori 1990), solvent extraction
(Bruce 1992), and others.
25
Fed-batch culture is, in the broadest sense, defined as an operational technique in
biotechnological processes where one or more nutrients (substrates) are fed
(supplied) to the bioreactor during cultivation and in which the product(s) remain in
the bioreactor until the end of the run.[1]
An alternative description of the method is that of a culture in which "a base medium
supports initial cell culture and a feed medium is added to prevent nutrient
depletion".[2]
It is also a type of semi-batch culture.
In some cases, all the nutrients are fed into the bioreactor. The advantage of the fedbatch culture is that one can control concentration of fed-substrate in the culture
liquid at arbitrarily desired levels (in many cases, at low levels).
Generally speaking, fed-batch culture is superior to conventional batch culture when
controlling concentrations of a nutrient (or nutrients) affects the yield or productivity
of the desired metabolite.
26
1.4 Bioprocess Economics
 We examine the continuing role which economics plays bioprocess
research, development, and commercialization.
 Subsequently, characteristic features of particular fermentation processes
are discussed, including fine chemicals, bulk chemicals, and single cell
protein in the examples, the relative cost importance of substrate
feedstocks, equipment, utilities, and bioreactor vs. recovery sections will
be examined, since identification of major cost areas frequently pinpoints
process economic weaknesses and thereby indicates whether an
engineering process improvement and/or strain development would be
most logical to pursue in process optimization.
 Ethanol from biomass processes include not only fermentation and
recovery sections but may also requires substantial pre-treatment process
to hydrolyze and solubilize the biomass components.
27
Continued…
 Moreover, different countries (including the United States) have
developed various forms of tax subsidies or credits for ethanol plants:
such device have a clear impact on process economics.
 Biological waste treatment provides the main process example for which
substrate conversion (rather than biomass production or product
formation) is the operating goal; here, biomass (cell sludge) disposal is
responsible for a major fraction of plant operating costs.
28
1.4.1 Process Economics
 For a project which eventually achieves commercialization, these stages
include inception of process or product idea, a preliminary evaluation of
economics and markets, the development of any additional data needed
for final detailed design.
 Inception may arise from any source: a sales department suggestion,
emergence of a competing product, a customer request, an offshoot of a
current process, or a new idea from research and development.
 If the preliminary evaluations are promising, the development of data
necessary for final design normally follows. This development includes
both market and cost refinement.
29
Continued…
Table 1.1 Stages in plant design project
1. Inception
2. Preliminary evaluation of economics and market
3. Development of data necessary for final design
4. Final economic analysis
5. Detailed engineering design
6. Procurement
7. Construction
8. Start-up and trial runs
9. Production
30
1.4.2 Bioproduct Regulation
 Involvement of government in bioproduct regulation stems from
legislated agency responsibility for worker health and safety, consumer
safety, environmental statutes applicable to biotechnology
 Specific regulations concerning newer aspects of biotechnology
(recombinant DNA in particular).
 The three most important U.S. agencies in regulation are the Food and
Drug Administration (FDA), the U.S. Department of Agriculture
(USDA), and the Environmental Protection Agency (EPA).
31
Continued…
Table 1.2 U.S. Agencies involved in biotechnology regulation
1. FDA
(a) Office of New Drug Evaluation (all new drugs require animal and human testing; any rDNA
drug products are all "new").
(b) Office of Biologics: clinical trials required, followed by licensing of both product and
producer.
(c) National Center for Devices and Radiologic Health: regulates medical devices and medical
diagnostics (including monoclonal antibody diagnostics, except those related to blood bank
operation). Monoclonal antibody (MAb) systems currently must demonstrate performance
equivalence to prior product, rather than safety and efficacy.
2. USDA
Primary regulatory authority over animal biologics, with FDA secondary. MAb and rDNA open
grey areas of agency responsibility uncertainty (eg, USDA regulates interstate marketing, but
FDA may regulate intrastate conditions.)
3. EPA
Authority "to identify and evaluate potential hazards and then to regulate the production, use,
distribution, and disposal of these substances, including chemicals, herbicides, and pesticides."
4. OSHA (Occupational Safety and Health Administration.)
Responsible for worker health and safety regulation. As of 1984, no bioprocess standards issued,
nor regulatory stance toward biotechnology declared.
32
Food Safety and Standards Authority of India (FSSAI) : It is a statutory body
established under the Ministry of Health & Family Welfare, Government of India. The
FSSAI has been established under the Food Safety and Standards Act, 2006, which
is a consolidating statue related to food safety and regulation in India.
The Central Drugs Standard Control Organisation(CDSCO): It is the
National Regulatory Authority (NRA) of India under Directorate General of
Health Services, Ministry of Health & Family Welfare, Government of India.
CPCB
1.4.3 General Fermentation Process
Economics
 While considerable differences clearly exist between different
fermentations, the plant design is most easily discussed from a common
flow sheet, given in Fig. 1.2. This simple diagram indicates that the
process is largely divisible into fermentation (or bioreaction) section, and
a product recovery section.
34
Continued…
Figure 1.2: Generalized process flow sheet.
35
Thanks
Chapter 2
S. No.
Content
S. No.
Content
2
Microbiology
2.3.6
Difference Between Plant cell and Animal cell
2.1
Cell theory
2.4
RDNA Technology
2.1.1
Importance of the Cell Theory
2.4.1
Enzymes for Manipulating DNA
2.2
Structure of Microbial Cells
2.4.2
Vectors for Escherichia coli
2.2.1
Procaryotic Cells
2.4.3
Characterization of Cloned DNAs
2.2.2
Eucaryotic Cells
2.4.4
Expression of Eucaryotic Proteins in E. coli
2.2.3
Cell Fractionation
2.4.5
Genetic Engineering Using Other Host
Organisms
2.3
Cell Types
2.4.6
Concluding Remarks
2.3.1
Bacteria
2.5
Genetically Engineered Microbes
2.3.2
Yeasts
2.5.1
Virus and Phages: Lysogeny and Transduction
2.3.3
Molds
2.5.2
Commercial Applications of Microbial
Genetics and Mutant Populations
2.3.4
Algae and Protozoa
2.5.3
Utilization of Auxotrophic Mutants
2.3.5
Animal and Plant Cells
2.5.4
Mutants with Altered Regulatory Systems
2
2. Microbiology
 Microbiology is the study of living organisms too small to be seen clearly
by the naked eye. As a rough rule of thumb, most microorganisms have a
diameter of 0.1 mm or less.
 Many conclusions have been reached in numerous experimental studies
involving methods adapted from the physical sciences.
 Since this approach has proved so fruitful, the applicability of the
principles of chemistry and physics to biological systems is now a
widespread working hypothesis within the life sciences.
3
2.1 Cell Theory
 A development critical to the understanding of living systems started in
1838, when Schleiden and Schwann first proposed the cell theory.
 This theory stated that all living systems are composed of cells and their
products.
 This notion of a common denominator permits an important
decomposition in the analysis of living systems: first the component
parts, the cells, can be studied, and then this knowledge is used to try to
understand the complete organism.
 The value of this decomposition rests on the fact that cells from a wide
variety of organisms share many common features in their structure and
function.
 All cells are not alike. Muscle cells are clearly different from those found
in the eye or brain.
4
Continued…
 Equally, there are many different types of single-celled organisms.
 These in turn can be classified in terms of the two major types of cellular
organization.
 The cell theory definition states that cells are the building blocks of life.
Cells both make up all living things and run the processes needed for life.
Hair, skin, organs, etc. are all made up of cells.
 In fact, each person is estimated to be made up of nearly 40 trillion cells!
Each part of a cell has a different function, and cells are responsible for
taking in nutrients, turning nutrients into energy, removing waste, and
more. Basically, everything what body does, it does because cells are
directing the action.
5
2.1.1 Importance of the Cell Theory
 Knowing that all living things are made up of cells allows us to
understand how organisms are created, grow, and die. That
information helps us understand how new life is created, why organisms
take the form they do, how cancer spreads, how diseases can be
managed, and more.
 Cells even help us understand fundamental issues such as life and death:
an organism whose cells are living is considered alive, while one whose
cells are dead is considered dead
 Before the cell theory existed, people had a very different view of
biology. Many believed in spontaneous generation, the idea that living
organisms can arise from non-living matter. An example of this would be
a piece of rotten meat creating flies because flies often appear around
rotten meat.
6
Continued…
 Additionally, before cells and the cell theory were known, it wasn’t
understood that humans, as well as all other living organisms, were made
up of billions and trillions of tiny building blocks that controlled all our
biological processes.
7
2.2 Structure of Microbial Cells
Cells
Procaryotic Cells
Eucaryotic Cells
8
2.2.1 Procaryotic Cells
 Procaryotic cells, or procaryotes, do not contain a membrane-enclosed
nucleus. Procaryotes are relatively small and simple cells.
 They usually exist alone, not associated with other cells. The typical
dimension of these cells, which may be spherical, rodlike, or spiral, is
from 0.5 to 3 um.
 In order to gain a qualitative feel for such dimensions, it is instructive to
compare the relative sizes of cells with other components of the universe.
9
Angstrom
units, Å
1036
Distance seen by
200-in telescope
Figure 2.1:
Characteristic
dimensions of the
universe. The biological
world encompasses
abroad spectrum of sizes
−109 light years
Angstrom
units, Å
Our galaxy
1030
Cosmic world
1012
−1 light year
Length of man
Radius of an amino
Solar system
109
1020
Radius of giant algal cell
Earth
106 - 100 μm radius of giant amoeba
1010
20 μm
Biological world
1 μm
Radius
of most
cells
Eucaryotes
Procaryotes
Resolution of light microscope
0
10
Hydrogen atoms
103 - Radius of smallest bacterium
Molecular world
Unit membrane thickness
Resolution of electron microscope
Atomic nucleus
10−6
Atomic world
100
Radius of an amino acid
Continued…
 Microorganisms of this type grow rapidly and are widespread in the bio
sphere. Some, for example, can double in size, mass, and number in 20
min.
 Typically, procaryotes are biochemically versatile; i.e, they often can
accept a wide variety of nutrients and further are capable of selecting the
best nutrient from among several available in their environment.
 The rapid growth and biochemical versatility of procaryotes make them
obvious choices for biological research and biochemical processing.
 In Fig. 2.2 the basic features of a procaryotic cell are illustrated.
11
Continued…
Cell Membrane
70 A
Cell wall
200 A
Ribosome
Nuclear region
Figure 2.2: Electron micrograph of a procaryote, (Bacillus subtilis)
12
Continued…
 The cell is surrounded by a rigid wall, approximately 200 A thick.
 This wall lends structural strength to the cell to preserve its integrity in a
wide variety of external surroundings.
 Immediately inside this wall is the cell membrane, which typically has a
thickness of about 70 A. It is sometimes called a plasma membrane.
 These membranes play a critical role: they largely determine which
chemical species can be transferred between the cell and its environment
as well as the net rate of such transfer.
 Within the cell is a large, ill-defined region called the nuclear zone,
which is the dominant control center for cell operation.
13
Continued…
 The grainy dark spots apparent in the cell interior are the ribosomes, the
sites of important biochemical reactions.
 The cytoplasm is the fluid material occupying the remainder of the cell.
 Not evident here but visible in some photographs are clear, bubblelike
regions called storage granules.
14
2.2.2 Eucaryotic Cells
 Eucaryotic cells, or eucaryotes, make up the other major class of cell
types. Eucaryotes may be defined most concisely as cells which possess a
membrane enclosed nucleus.
 As a rule these cells are significantly larger than procaryotes. All cells of
higher organisms belong to this family.
 The internal structure of eucaryotes is considerably more complex than
that of procaryotic cells, as can be seen in Figure 2.3. The internal region
is divided into a number of distinct compartments.
 They have special structures and functions for conducting the activities
of the cell. The cell is surrounded by a plasma, or unit, membrane similar
to that found in procaryotes. On the exterior surface of this membrane
may be a cell coat, or wall. The nature of the outer covering depends on
the particular cell.
15
Continued…
Figure 2.3: A typical eucaryotic cell
16
Continued…
 Important to the internal specialization of eucaryotic cells is the presence
of unit membranes within the cell.
 A complex, convoluted membrane system, called the endoplasmic
reticulum, leads from the cell membrane into the cell. The nucleus here is
surrounded by a porous membrane.
 Ribosomes, biochemical reaction sites seen before in procaryotes, are
embedded in the surface of much of the endo plasmic reticulum.
 A major function of the nucleus is to control the catalytic activity at the
ribosomes.
 The nucleus is one of several interior regions surrounded by unit
membranes. These specialized membrane-enclosed domains are known
collectively as organ ells.
17
Continued…
 The mitochondria are organelles with an extremely specialized and
organized internal structure.
 They are found in all eucaryotic cells which utilize oxygen in the
process of energy generation.
 Chloroplasts and mitochondria are the sites of many other
important biochemical reactions in addition to their role in energy
production.
 The Golgi complex, lysosomes, and vacuoles are the remaining
organelles illustrated in Figure. 2.3.
 In general, they serve to isolate chemical reactions or certain
chemical compounds from the cytoplasm.
18
2.2.3 Cell Fractionation
 A major problem in analyzing the characteristics of a particular organelle
from a given type of cell is obtaining a sufficient quantity of the
organelle for subsequent biochemical analysis. Typically this requires
that a large number of organelles be isolated from a large number of
cells, or a cell population.
 Let us follow a common procedure for this purpose: First a cell
suspension is homogenized in a special solution using either a rotating
pestle within a tube or ultrasonic sound.
 Here an attempt is made to break the cells apart without significantly
disturbing or disrupting the organelles within. Fractionation of the
resulting suspension, which now ideally contains a variety of isolated
intact organelles, is the next step.
 As process engineers, we know that any separation process is based upon
exploitation of differences in the physical and/or chemical properties of
the components to be isolated.
19
Continued…
 The standard centrifugation techniques for fractionating cell organelles
rely upon physical characteristics: size, shape, and density. A rudimentary
analysis of centrifugation is presented in the following example.
Example 2.1: Analysis of particle motion in a centrifuge. Suppose that a
spherical particle of radius r and density 𝜌𝑝 , is placed in a centrifuge tube
containing fluid medium of density 𝜌𝑓 , and viscosity 𝜇𝑐 . If this tube is then
placed in a centrifuge and spun at angular velocity 𝜔 (see Fig. 1E1.1), we
may calculate the particle motion employing the following force balance
(what approximations have been invoked here?):
Solution:
Drag Force on particle = bouncy force
4𝜋𝑟 3
6𝜋𝜇𝑐 𝑟𝜇𝑟 =
𝐺(𝜌𝑝 − 𝜌𝑓 )
3
2.1
20
Continued…
Figure 2.1.1: When a centrifuge is spun at high speed, particles suspended in the
centrifuge tubes move away from the centrifuge axis. Since the rate of movement of
these particles depends on their size, shape, and density, particles differing in these
properties can be separated in a centrifuge.
21
Continued…
where u, is the particle velocity in the r direction
𝑑𝑟
𝑢𝑟 =
𝑑𝑡
2.2
and G is the acceleration due to centrifugal forces
𝐺 = 𝜔2 𝑟
2.3
Stokes' law has been used in Eq. (2.1) to express the drag force since particle
velocities (and therefore particle Reynolds numbers) are usually very low in this
situation. The usual gravitational force term does not appear in Eq. (2.1) because the
direction in Fig. 2.1.1 is normal to the gravity force. Rotation of the centrifuge at
high speed, however, produces an acceleration G usually many times larger than the
acceleration of gravity g; for example, G = 600-600,000 g. Integration of this
expression gives the time required for movement of the particle from position 𝑟1 to
𝑟2 :
22
Continued…
9
𝜇𝑐
𝑟2
𝑡=
𝑙𝑛
2 𝜔 2 𝑅2 (𝜌𝑝 − 𝜌𝑓 ) 𝑟1
Spheres with different sizes and/or densities will take different times to
traverse the same distance in the centrifuge tube. This is the basis for the
method of differential centrifugation.
Since the larger particles such as nuclei and unbroken cells sediment more
rapidly, they can be collected as a precipitate by spinning the suspension for
a limited time at relatively low velocities.
The supernatant suspension is then subjected to additional centrifugation at
higher rotor speeds for a short time, and another precipitate containing
mitochondria is isolated. By continuing this procedure a series of cell
fractions can be obtained.
The overall process is illustrated schematically in Fig. 2.1.2. More
sophisticated centrifugation methods employ liquid media with density
gradients along the centrifuge tube.
23
Continued…
These techniques are also applicable for continued subdivision and
fractionation of smaller cell constituents such as particular types of
macromolecules. RPM to RCF conversion
Centrifuge rotor speed is often expressed as
Relative centrifugal field (RCF)
RCF in units of gravity (x g)
G = (1.118 x 10-5) R S2
where R is the rotor
radius (cm) and S is
the speed (RPM):
Speeds range from
0-7,500 RPM for low-speed
centrifuges, all the way to
20,000 RPM or higher
Figure 2.1.2: The steps in a typical differential centrifugation separation of
cell constituents. Smaller components are isolated as the process proceeds.
24
2.3 Cell Types
 Taxonomy is the art of biological classification. The basic unit in this
classification scheme is the species which is characterized by a high
degree of similarity in physical and biochemical properties and
significant differences from the properties of related organisms.
Procaryotes
Bacteria
Table: 2.1
Classification of
microorganisms
belonging to the
kingdom of protists
Eucaryotes
PROTIST
KINGDOM
Blue-green
algae
Fungi
Molds
Algae
Protozoa
Yeasts
25
2.3.1 Bacteria
 Bacteria are typically unicellular, but they may exist in three basic
morphological forms (Fig. 2.4). Most can not utilize light energy, are
capable of motion and reproduce by division into two daughter cells
(binary fusion).
 There are many subdivisions of bacteria: some of the general groups of
bacteria and some of their distinguishing characteristics are given in
Table 2.2.
Figure 2.4: The three forms of bacteria
26
Table 2.2:
Dominant
Morphologic
al form
Some
nutritional
habits
Common
habitat
Most members
require 𝑶𝟐
(aerobic)
Pho
tosy
nth
etic
Form
s
endos
pores
Gram
reactio
n
Acetic acid
bacteria
Rods
Often consume
alcohol
Decaying plants
Yes
No
No
Negative
Bacillus
Rods
Versatile
Soil
Yes
No
Yes
Positive
Clostridium
Irregular form
Many varieties
Soil
Most species
cannot tolerate 𝑶𝟐
No
Yes
Positive
Corynebacte
rium
Rods
Simple
requirements
Soil, human
body
No, but use 𝑶𝟐 if
present
No
Yes
Positive
Enteric or
coliform
bacteria
Rods or cocci
Simple organic
compounds
Naturally in
intestine of
higher animals
No, but can use
𝑶𝟐 if present
No
No
Negative
Lactic acid
Rods
Species acidtolerant
Plants
No
No
No
Positive
Pseudomona
s
Rods
Some extremely
versatile
Soil, water
Yes
No
No
Negative
Rhizobium
Rods
Fixes nitrogen
Soil, Plants
Yes
No
No
Negative
Rhodospirill
ium
Rods, spirals
Can fix 𝑵𝟐 or
produce 𝑯𝟐
Specialized
aqueous
environment
No
Yes
No
Negative
Zymomonas
Rods
Ferments glucose
to ethanol
Soil
No, can tolerate
𝑶𝟐
No
No
Negative
Group
27
Continued…
 The aerobic Bacillus species are extremely widespread and adaptable.
 Several Clostridium species, which normally function under anaerobic
conditions, die in the presence of oxygen in the vegetative state but form
spores unaffected by oxygen.
 Some bacteria whose vegetative forms are rapidly killed at 45°C can
form spores which survive boiling in water for several hours.
 Therefore, when we are attempting to kill microorganisms by heating
(heat sterilization) the spore forming capability of bacteria demands use
of higher temperatures, typically achieved by boiling under pressure in
an autoclave to give T > 120 °C.
28
2.3.2 Yeasts
 Yeasts form one of the important subgroups of fungi.
 Fungi, like bacteria, are widespread in nature although they usually live
in the soil and in regions of lower relative humidity than bacteria.
 They are unable to extract energy from sunlight and usually are freeliving.
 Although most fungi have a relatively complex morphology, yeasts are
distinguished by their usual existence as single, small cells from 5 to 30
μm long and from 1 to 5 μm wide.
 The various paths of reproduction of yeasts are asexual (budding and
fission) as shown in Fig. 2.7, and sexual.
 In budding, a small offspring cell begins to grow on the side of the
original cell; physical separation of mature offspring from the parent may
not be immediate.
29
Continued…
Example 2.2: Yeast cells grow in the exponential phase. The cell mass
concentrations are given in Table 4.4. Calculate the specific growth rate,
𝜇𝑚𝑎𝑥 .
Time (h)
0
5
10
15
20
25
Cell concentration, 𝐶𝑥
0.5
1.3
4.0
10.2
24,0
52.5
Kg dry cell 𝑚−3 medium.
Solution:
A plot of a 𝐶𝑥 /𝐶𝑥𝑜 against cultivation time (Figure 4.2) gives a straight
line, the slope of which can be used to evaluate 𝜇𝑚𝑎𝑥 .
𝜇𝑚𝑎𝑥 = 0.20 ℎ−1
30
Continued…
Figure E2.2: Determination of a specific growth rate in the exponential growth phase.
31
2.3.3 Molds
 Molds are higher fungi with a vegetative structure called a mycelium. As
illustrated in Fig. 2.6, the mycelium is a highly branched system of tubes.
 Within these enclosing tubes is a mobile mass of cytoplasm containing
many nuclei.
 The mycelium may consist of more than one cell of related types. The
long, thin filaments of cells within the mycelium are called hyphae.
 In some cases the mycelium may be very dense. Molds, like yeasts, do
not contain chlorophyll, and they are generally non motile.
 Reproduction, which may be sexual or asexual, is typically accomplished
by means of spores.
32
Continued…
Figure 2.6: The mycelial structure of molds
33
Continued…
 The most important classes of molds industrially are Aspergillus and
Penicillium (Fig. 2.7). Major useful products of these organisms include
antibiotics ,organic acids, and biological catalysts.
Penicillium
Figure 2.7: The hyphae of Aspergillus and Penicillium
Aspergillus
34
2.3.4 Algae and Protozoa
 These relatively large eucaryotes have sophisticated and highly organized
structures.
 For example, Euglena has flagella for locomotion, lacks a rigid wall, and
has an eyespot sensitive to light,
 Considerable commercial interest in algae is concentrated on their
possible exploitation as foodstuffs and food supplements.
 In Japan, several processes for algae food cultivation are in operation
today. Also important in Asia is use of seaweed in the human diet.
 Just as algae may be viewed as primitive plants, protozoa, which cannot
exploit sunlight's energy, are in a sense primitive animals.
 Although protozoa are not now employed for industrial manufacture of
either cells or products, their activities are significant among the
microorganisms which participate in biological waste treatment.
35
2.3.5 Animal and Plant Cells
 Many vaccines and other useful biochemicals are produced by growth of
animal cells in process reactors; i.e., by cell propagation outside of the
whole animal.
 Improvements in cultivation techniques for these tissue-derived cells and
emerging methods for genetic manipulation of animal and plant cells
offer great potential for expanded commercial utilization of these higher
cells.
 When a piece of animal tissue, perhaps after disruption to break the cells
apart, is placed in appropriate nutrient liquid, many cell types, such as
blood cells, die within a few days, weeks, or months. Other cells multiply
and are called primary cell lines.
36
Continued…
 It is also possible to grow certain plant cells in culture, either as a callus
or as aggregated cells in suspension.
Cell line designation
Animal
Tissue
HeLa (CCL 2)
Human
Cervix carcinoma
HLM
Human
Fetal liver
FS-4
Human
Foreskin fibroblast
MK2
Monkey
Kidney
CHO (CCL 61)
Chinese hamster
Ovary
L-M (CCL 2)
Mouse
Connective tissue
Table: 2.3 A
sampling of
standard
animal cell
lines and their
origins
Since plants produce many commercially important compounds including
perfumes, dyes, medicinals, and opiates, there is significant potential for
future applications of plant cell culture.
Cultured plant cells have several potential agricultural applications including
whole plant regeneration.
37
2.3.6 Difference Between Plant cell and
Animal cell
 Unicellular organisms are believed to be one of the earliest forms of life
on earth. Eventually, more complex multicellular organisms evolved
from these unicellular life forms over the aeons. Multicellular organisms
have specialized cells with complicated cell organelles, which unicellular
organisms typically lack.
 In an ecosystem, plants have the role of producers while animals have
taken the role of consumers. Hence, their daily activities and functions
vary, so do their cell structure. Cell structure and organelles vary in
plants and animals, and they are primarily classified based on their
function. The difference in their cell composition is the reason behind the
difference between plants and animals, their structure and functions.
 Each cell organelle has a particular function to perform. Some of the cell
organelles are present in both plant cell and the animal cell, while others
are unique to just one. Most of the earth’s higher organisms are
eukaryotes, including all plant and animals. Hence, these cells share
some similarities typically associated with eukaryotes.
38
Continued…
 For example, all eukaryotic cells consist of a nucleus, plasma membrane,
cytoplasm, peroxisomes, mitochondria, ribosomes and other cell
organelles.
Plant cell
Animal cell
Figure 2.8: Plant cell vs Animal cell
39
2.4 RDNA Technology
 Recombinant DNA technology provides carefully controlled and novel
genetic alterations in certain organisms.
 This approach has made possible manufacture of valuable animal
products (proteins) in simple bacteria.
 These examples indicate the central importance of genetics to the
biochemical engineer.
 This subject occupies a large fraction of the present efforts in both
university and industrial biological research.
 The practical importance of designing genetic modifications (or, in other
situations, of avoiding such changes) emphasizes the need for close
cooperation between engineers, biologists, and biochemists in
biochemical process design and evaluation.
40
Continued…
 A brief schematic of a method for cloning a segment of DNA shown in
Fig. 2.9. Here cloning means obtaining a colony of genetically identical
cells which contain the DNA segment of interest.
 This segment, called "foreign DNA" in Fig. 2.9, is obtained by cutting a
larger piece of DNA into fragments using specific endonuclease enzymes
or by enzymatic or chemical synthesis.
 The foreign DNA is joined in the test tube (in vitro) to a vector or vehicle
which will carry the foreign DNA as a passenger into a bacterial cell.
 In this example the vector is a bacterial plasmid which has been
manipulated to facilitate the cloning process. The recombinant DNA
molecule formed by the foreign DNA-plasmid construct is then
introduced into a bacterial cell by transformation.
41
Continued…
Figure 2.9: Schematic illustration of the major steps in cloning a
foreign DNA segment.
42
2.4.1 Enzymes for Manipulating DNA
One key to recent breakthroughs in genetic engineering is the discovery and
the subsequent commercial supply of several enzymes which can be used to
cut, alter, and join DNA molecules in the test tube.
Prominent among these crucial enzymes are the restriction endonucleases
which recognize and cut specific nucleotide sequences within the DNA
molecule. For example the restriction enzyme called EcoRI recognizes the
double-stranded six-nucleotide sequence
∙ ∙ ∙ −G − A − A − T − T − C − ∙ ∙ ∙
∙ ∙ ∙ −C − T − T − A − A − G − ∙ ∙ ∙
and cleaves each strand between the G and A residues as marked above by
arrows. The two ends may then separate
∙ ∙ ∙ −G
∙ ∙ ∙ −C − T − T − A − A
A−A−T−T−C−∙∙∙
G−∙∙∙
43
Continued…
 Notice that in this case each end contains a short single-stranded
segment. These are called "sticky ends" or "cohesive termini" since the
two ends are complementary single-stranded nucleotide sequences which
will naturally tend to base pair and to join, by hydrogen bonding, to each
other.
 Whether or not these hydrogen bonded sequences tend to stay together or
separate can be adjusted by manipulating the temperature or the pH of
the solution.
 More than 100 different restriction enzymes have now been identified.
Some of the more commonly used ones, their recognition sites, and the
points of DNA backbone hydrolysis are listed in Table 2.4. We should
observe here that not all restriction enzymes leave sticky ends.
44
Continued…
Table 2.4: Recognition sites and rage points (arrows) for some on endonuclease enzymes
Enzymes
Target site
BumHΙ
G G A T C C
C C T A G G
BglΙΙ
A G A T C T
T C T A G A
EcoRΙ
G A A T T C
C T T A A G
HaeΙΙΙ
G G C C
C C G G
HindΙΙΙ
A A G C T T
T T C G A A
HpaΙ
G T T A A C
C A A T T G
PstΙ
C T G C A G
G A C G T C
45
Continued…
 Suppose the sticky ends obtained by earlier EcoRI hydrolysis of different
DNA molecules are allowed to anneal-to come together and base pair.
This configuration is not covalently closed: the phosphodiester bonds
(indicated by dashes) are missing between the G and A residues in both
strands. DNA ligase catalyzes the condensation of a 3’− hydroxyl group
with a 5'−phosphate group to add these missing links. Annealed fragments
∙ ∙ ∙ −G
A−A−T−T−C−∙∙∙
∙ ∙ ∙ −C − T − T − A − A
G−∙∙∙
are covalently closed by DNA ligase
∙ ∙ ∙ −G − A − A − T − T − C − ∙ ∙ ∙
∙ ∙ ∙ −C − T − T − A − A − G − ∙ ∙ ∙
46
2.4.2 Vectors for Escherichia coli
 Vectors are DNA molecules which provide propagation of a DNA
fragment in a growing cell population. A useful vector for cloning should
have the following properties:
1. Ability to replicate in the host cell
2. Ability to accommodate foreign DNA of various sizes without damage
to replication functions
3. Easy insertion in host cell after in vitro DNA manipulations
4. Contains a selection marker to facilitate rapid, positive selection of cells
which contain the vector
5. Contains only one target site for one or more different restriction
endonucleases
Two different classes of vectors-plasmids and bacteriophages have been
developed for cloning DNA in E. coli. We will concentrate here on plasmid
vectors.
47
Continued…
 Selection markers are also important. By including, for example, a gene
for antibiotic resistance in the plasmid, we can rapidly and positively
select for cells containing plasmids.
 This is done by growing the cells on medium containing a level of the
antibiotic which kills plasmid-free cells but which allows growth of cells
containing plasmids and hence the antibiotic resistance gene.
 Another common strategy for selection is to use a mutant host lacking an
enzyme required for growth in a particular medium and to include on the
recombinant plasmid the normal gene for that enzyme.
 Plasmid vectors for cloning in E. coli have been highly developed in
recent years. Among the more popular plasmids is pBR322 (Fig. 2.13)
which includes a number of unique restriction enzyme sites, genes for
tetracycline and for ampicillin resistance, and an origin of replication
which allows amplification of the plasmid.
48
Continued…
 Here "amplification" means a large increase in the ratio of plasmid to
chromosomal DNA. Amplification is possible using pBR322 and related
plasmids since the plasmid can replicate in the absence of protein
synthesis in the host cell but the chromosome cannot.
Figure 2.11: Map of the plasmid pBR322 showing genetic loci and
some of the unique restriction enzyme sites. The complete nucleotide
sequence of the 4362 base-pair plasmid is known.
49
2.4.3 Characterization of Cloned DNAs
 First, we can distinguish between clones which have plasmids with
inserts and those which contain only reclosed plasmids by applying
insertional inactivation during the cloning procedure. Suppose, for
example, that DNA fragments were inserted at the PstΙ restriction
enzyme site in plasmid pBR322.
 Since inserts this position interrupt the gene for ampicillin resistance,
cells containing recombinant plasmids will be ampicillin sensitive and
tetracycline resistant. On the other hand, cells with insert-free plasmids
will be resistant to both ampicillin and tetracycline.
 Screening for specific nucleotide sequences in the clone may be
accomplished by several different procedures which are all based on
hybridization of denatured (ie, single-stranded) DNA from the clone with
a nucleic acid probe, typically labelled with 3 2𝑃 to make it radioactive.
50
Continued…
 In the in situ colony hybridization method, a replica plate of the clones
is grown on nitrocellulose paper laid over the Petri plate with nutrient
medium. Cells are then lysed by treatment with lysozyme, and NaOH is
used to denature the DNA. The filter paper is then removed, contacted
with the probe, washed, and exposed to x-ray film. Spots appear on the
film where probe hybridization to complementary DNA has occurred,
identifying interesting clones for further study.
A probe is a nucleic acid which has been labeled i.e., chemically modified in
some way which allows it and hence anything it hybridizes to, to be
detected. There are three major types of probe: Oligonucleotide
probes, DNA probes and. cRNA probes (riboprobes)
51
In the Southern blotting method, the DNA in the gel is
denatured with alkali, and the gel is covered with cellulose
nitrate filter paper. Buffer solution drawn from moist filter
paper below this sandwich to dry filter paper above it carries
DNA the gel to cellulose nitrate where binds. The bound,
denatured DNA then probe and film before to identify
complementary that the probe.
52
Continued…
 In the Maxam-Gilbert method, the double-stranded DNA is first
labelled radioactively at one end of each strand. The DNA is then
denatured, and a preparation of one of the two types of strands is divided
into four aliquots, each which is treated with different chemicals. Each of
the four chemical treatments destroys selectively one or two bases and
cleaves the DNA at those points. It is critical to the method that this
destruction be incomplete-ideally the chemical treatment to cleave each
strand only once. The fragments resulting from these four parallel
treatments are then run on a long polyacrylamide g Gel band positions
are visualized by exposure to x-ray film.
53
Continued…
Figure
2.12:
MaxamGilbert DNA sequencing:
application of four different
reagents
which
attack
different bases gives a
mixture of fragments with
relative
lengths
corresponding
to
the
destroyed
bases.
The
pattern of bands on the four
lane gel permit direct
reading
of
the
corresponding nucleotide
sequence running, from top
to bottom, toward the
labelled end of the starting
oligonucleotide.
54
Continued…
 The Sanger DNA sequencing method, There are three main steps to
Sanger sequencing.
1. DNA Sequence For Chain Termination PCR. The DNA sequence of
interest is used as a template for a special type of PCR called chaintermination PCR.
2. Size Separation by Gel Electrophoresis.
3. Gel Analysis & Determination of DNA Sequence.
55
2.4.4 Expression of Eucaryotic Proteins in
E. coli
 Procaryotes such as E. coli do not possess the biochemical machinery to
do RNA splicing Accordingly, a eucaryotic gene placed directly in E. coli
(with appropriate bacterial expression controls to be discussed next) will
not be properly expressed.
 There are two quite different strategies for accomplishing this. The first
is chemical synthesis of the desired nucleotide sequence. While details of
chemical synthesis of genes are best left for the references, we should
observe that DNA synthesis chemistry as of this writing does not permit
direct synthesis of the entire gene. Instead, overlapping oligonucleotides
are designed which will tend to self-assemble by base pairing into a
complete double stranded coding sequence.
56
Continued…
 The other way to obtain a bacterial gene for a eucaryotic protein is to
work backward from the mature, spliced eucaryotic messenger RNA. As
shown schematically in Fig. 2.15, the mRNA for the desired protein
provides a template from which the enzyme reverse transcriptase can
synthesize a complementary DNA strand.
 After separating the mRNA and single DNA strand, the complementary
DNA strand is added using DNA polymerase, and the enzyme S1
nuclease then serves to clip off a small single-stranded loop at one end.
57
Continued…
Figure 2.13: In eucaryotic
gene expression, the primary
transcript mRNA is spliced
to obtain the mature mRNA
which is translated. An
abbreviated hypothetical
example is shown here.
58
2.4.5 Genetic Engineering Using Other Host
Organisms
There is great interest in developing cloning and expression systems for
organisms other than E. coli. In all cases it is important to keep in mind the
following potential barriers to expression of a foreign gene:
1. Host cell nuclease attack on foreign DNA or RNA
2. Vector replication malfunction
3. Poor activity of promoter or transcription terminator
4. Incomplete mRNA splicing
5. Inefficient translation
6. Protease attack.
Also, we need an efficient procedure for introducing the vector into host
cells. with high yields.
59
Continued…
 Methods exist for molecular coloning and gene expression in several
bacteria. Perhaps the most extensively studied is Bacillus subtilis, a gram
positive nonpathogenic, nonparasitic, industrial microorganism used
before the advent of genetic engineering to manufacture several enzymes
and polypeptide antibiotics B. subtilis secretes some of its proteins into
the surrounding medium.
 Several plasmids and bacteriophages are available for cloning in B.
subtilis Transformation, transduction, and protoplast fusion methods can
be used to introduce foreign DNA.
In transformation, a bacterium takes up a piece of DNA floating in its
environment. In transduction, DNA is accidentally moved from one bacterium
to another by a virus. In conjugation, DNA is transferred between bacteria through a
tube between cells.
Somatic fusion, also called protoplast fusion, is a type of genetic modification in
plants by which two distinct species of plants are fused together to form a new
hybrid plant with the characteristics of both, a somatic hybrid
60
Continued…
Several mammalian proteins including insulin and interferons have been
successfully expressed in B. subtilis, but, as of this writing export of foreign
proteins from B. subtilis remains problematic.
 Cloning technology is also available for several Pseudomonas and
Streptomyces strains. A combination of genetic engineering and
mutation/selection methods has generated Pseudomonas species with
new metabolic capabilities-for example, the ability to grow on ordinarily
toxic chlorinated hydrocarbons as their sole carbon source.
 The last examples introduce a new theme-metabolic engineering-which
de serves further comment. With genetic engineering we can not only
produce proteins useful in their own right, but we can also add to a living
cell more of some particular enzymes, regulatory proteins, permeases, or
any other protein.
 Further, we may be able to give to the cell completely new enzymatic,
regulatory, or transport activity which we would never expect to find in
nature or through use of random mutagenesis.
61
Continued…
 Because of all the available methodology and its rapid growth, E. coli re
mains the organism of choice for cloning and characterizing DNAs.
 Thus, when the final objective is to propagate the vector and express its
genes in another organism, it is convenient to construct and to utilize a
shuttle vector which can be propagated both in E. coli and in the other
organism of choice.
 Clearly, a shuttle vector must contain two origins of replication-one for
each host organism. Shown in Fig. 2.17 is the shuttle vector developed to
express immune interferon (INF-y) in monkey cells.
62
Continued…
Figure 2.17: Diagram of the shuttle vector for E. coli and monkey cells
which gives expression and secretion of immune interferon in monkey cells
63
2.4.6 Concluding Remarks
 Given the potential significance of recombinant DNA technology in
future activities in biotechnology, further study of the underlying
molecular biology and methods is essential to the biochemical engineer.
 Gene expression levels, intracellular protease activities, genetic stability,
and other factors which determine the productivity of recombinant
populations all depend on the nucleotide sequence of the vector, on the
genetic properties of the host cell, on medium composition, and on
bioreactor configuration and operating conditions.
 Combined efforts of biochemical engineers and molecular biologists are
essential to understanding causes and effects and interactions in
genetically engineered organisms.
64
2.5 Genetically Engineered Microbes
 The use of genetically engineered microorganisms (GEMs, often referred
to as Genetically Modified Microorganisms or GMMs) has become
widespread in the production of food processing aids and other food
ingredients.
 GEMs are advancing food production by increasing efficiency, reducing
waste and resource requirements, and ultimately enabling beneficial
innovations such as the cost-effective fortification of food with essential
nutrients, vitamins, and amino acids, and delivery of tailored enzymes to
achieve unique food processing capabilities.
 In addition to mutation, a variety of processes change the genetic
material within a cell. These have both positive and negative implications
from the standpoint of industrial biotechnology.
65
Continued…
 From the former perspective, any method for altering a cell's DNA
content provides an opportunity for development of more productive
strains. On the other hand, an unplanned modification of a desired
species' DNA can cause expensive failures of industrial bioprocesses.
 Much of what follows pertains only to bacteria, E. coli serving as the
subject of most research on these topics
66
Thank you
Chapter 3
S. No.
Content
S. No.
Content
3
Biochemistry
3.3.1
Building Blocks, an Energy Carrier, and
Coenzymes
3.1
Lipids
3.3.2
Biological Information Storage
3.1.1
Properties of Lipids
3.4
Amino Acids into Proteins
3.1.2
Lipid Structure
3.4.1
Amino Acid Building Blocks and Polypeptides
3.1.3
Classification of Lipids
3.4.2
Protein Structure
3.1.4
Fatty Acids and Related Lipids
3.4.3
Primary Structure
3.1.5
Fat soluble Vitamins, steroids and
Other Lipids
3.4.4
Three-Dimensional Conformation: Secondary
and Tertiary Structure
3.2
Sugars and Polysaccharides
3.4.5
Quaternary Structure and Biological
Regulation
3.2.1
D-Glucose and Other
Monosaccharides
3.5
Hybrid Biochemicals
3.2.2
Disaccharides to Polysaccharides
3.5.1
Cell Envelopes: Peptidoglycan and
Lipopolysaccharides
3.2.3
Cellulose
3.5.2
Antibodies and Other Glycoproteins
3.3
From Nucleotides to RNA and DNA
2
3. Biochemistry
 The four main classes of polymeric cell compounds are the fats and
lipids, the polysaccharides (cellulose, starch, etc.), the informationencoded polydeoxyribonucleic and polyribonucleic acids (DNA, RNA),
and proteins.
 The various biological polymers may be usefully regarded as being either
repetitive or nonrepetitive in structure.
 Repetitive biological polymers contain one kind of monomeric subunit.
 The major function of repetitive polymers in the cell is to provide
structures with the desired mechanical strength, chemical inertness, and
permeability.
 Repetitive polymers also provide a means of nutrient storage.
 In the latter function, for example, a 1 M glucose solution can be stored
as the polymer glycogen, a cell polysaccharide reserve, with a concurrent
reduction in molarity by a factor of 10,000 or greater.
3
Continued…
 Nonrepetitive polymers may contain from several up to 20 different
monomer species.
 The predominant elements (hydrogen, oxygen, nitrogen, and carbon)
form chemical bonds by completing their outer shells with one, two,
three, and four electrons, respectively.
 They are the lightest elements in the periodic table with such properties,
and (except for hydrogen) they can form multiple bonds as well.
 The variety of chemicals which can be assembled from these four
elements include, if we add a little phosphorus and sulfur, the four major
biopolymer classes.
 An addition to variety, the biochemical compounds assembled from these
elements are quite stable, reacting only slowly with each other, water,
and other cellular compounds.
4
Continued…
 Chemical reactions involving such compounds are catalyzed by
biological catalysts: proteins which are called enzymes (recall that a
catalyst is a substance which allows an increase in a reaction rate without
itself undergoing a permanent change).
Element
% of dry weight
Element
% of dry weight
Carbon
50
Sodium
1
Oxygen
20
Calcium
0.5
Nitrogen
14
Magnesium
0.5
Hydrogen
8
Chlorine
0.5
Phosphorus
3
Iron
0.2
Sulfur
1
All other
≈ 0.3
Potassium
1
Table 3.1: The elemental composition of E.coli (Escherichia coli)
5
3.1 Lipids
 By definition, lipids are biological compounds which are soluble in
nonpolar solvents such as benzene, chloroform, and ether, and are
practically insoluble in water.
 Consequently, these molecules are diverse in their chemical structure and
their biological function.
 Fats, which simply serve as polymeric biological fuel storage, are lipids,
as are several important mediators of biological activity.
 Lipids also constitute portions of more complex molecules such as
lipoproteins and liposaccharides, which again appear predominantly in
biological membranes of cells and the external walls of some viruses.
 The hydrocarbon chain is nearly insoluble in water, but the carboxyl
group is very hydrophilic.
6
Continued…
Figure 3.1: Lipids
7
3.1.1 Properties of Lipids
Lipids are a family of organic compounds, composed of fats and oils. These
molecules yield high energy and are responsible for different functions
within the human body. Listed below are some important characteristics
of Lipids.
1. Lipids are oily or greasy nonpolar molecules, stored in the adipose
tissue of the body.
2. Lipids are a heterogeneous group of compounds, mainly composed of
hydrocarbon chains.
3. Lipids are energy-rich organic molecules, which provide energy for
different life processes.
4. Lipids are a class of compounds characterised by their solubility in
nonpolar solvents and insolubility in water.
5. Lipids are significant in biological systems as they form a mechanical
barrier dividing a cell from the external environment known as the
cell membrane.
8
3.1.2 Lipid Structure
Lipids are the polymers of fatty acids that contain a long, non-polar
hydrocarbon chain with a small polar region containing oxygen. The lipid
structure is explained in the diagram below:
Figure 3.2: Lipid Structure – Saturated and Unsaturated Fatty Acids
9
3.1.3 Classification of Lipids
Lipids can be classified into two main classes:
1. Non-saponifiable lipids
2. Saponifiable lipids
Non-saponifiable Lipids
A nonsaponifiable lipid cannot be disintegrated into smaller molecules
through hydrolysis. Nonsaponifiable lipids include cholesterol,
prostaglandins, etc.
Saponifiable Lipids
A saponifiable lipid comprises one or more ester groups, enabling it to
undergo hydrolysis in the presence of a base, acid, or enzymes, including
waxes, triglycerides, sphingolipids and phospholipids.
Phospholipids and sphingolipids are two classes of lipids mainly found in the cell membrane
of all living organisms.
10
3.1.4 Fatty Acids and Related Lipids
• The main difference between phospholipids and sphingolipids is that phospholipids consist of
a glycerol backbone whereas sphingolipids consist of a sphingosine backbone. (Sphingosine (2amino-4-trans-octadecene-1,3-diol) is an 18-carbon amino alcohol with an
unsaturated hydrocarbon chain,)
• Furthermore, phospholipids are the major component of lipids in the cell membrane while
sphingolipids are the second major component of lipids in the cell membrane.
 Saturated fatty acids are relatively simple lipids with the general formula:
𝑂
𝐶𝐻3 − 𝐶𝐻2
𝑛
−𝐶
𝑂𝐻
 The hydrocarbon chain is constructed from identical two-carbon
monomers, so that fatty acids may be regarded as noninformational
biopolymers with a terminal carboxylic group.
 The value of n is typically between 12 and 20 (even numbers) in biological
systems.
11
Continued…
 Unsaturated fatty acids are formed upon replacement of a saturated
(−𝐶 − 𝐶−) bond by a double bond (−𝐶 = 𝐶−).
 For example, oleic acid is the unsaturated counterpart of stearic acid (n =
16).
𝐶𝐻3 − (𝐶𝐻₂)16 −𝐶𝑂𝑂𝐻
Stearic acid
𝐶𝐻3 − (𝐶𝐻2 )7 −𝐻𝐶 = 𝐶𝐻 − (𝐶𝐻2 )7 − 𝐶𝑂𝑂𝐻
Oleic acid
12
Continued…
Lipid monolayer at an air-water interface
Lipid micelle in water
Figure 3.3: Some stable configurations of fatty acids in water
 These hydrophilic-hydrophobic lipid molecules possess very small
solubilities: elevation of the solution concentration above the
monomolecular solubility limit results in the condensation of excess
solutes into larger ordered structures called micelles (Fig. 3.3).
13
Continued…
 Important as reservoirs of fuel, fats are esters formed by condensation of
fatty acids with glycerol.
𝑂
𝐶𝐻₂𝑂𝐻
𝐻𝑂 − 𝑂𝐶 𝐶𝐻2
𝑛1
− 𝐶𝐻3
𝐶𝐻𝑂𝐻
+
𝐻𝑂 − 𝑂𝐶 𝐶𝐻2
𝑛2
− 𝐶𝐻3
𝑛3
− 𝐶𝐻3
+
−3𝐻2 𝑂
𝐶𝐻2 𝑂 − 𝐶 − 𝐶𝐻2
𝑂
𝑛1
− 𝐶𝐻3
𝐶𝐻2 𝑂 − 𝐶 − 𝐶𝐻2
𝑂
𝑛2
− 𝐶𝐻3
𝐶𝐻2 𝑂 − 𝐶 − 𝐶𝐻2
𝑛3
− 𝐶𝐻3
+
𝐶𝐻₂𝑂𝐻
Glycerol
𝐻𝑂 − 𝑂𝐶 𝐶𝐻2
Fatty acids
A Fat
14
3.1.5 Fat soluble Vitamins, steroids and
Other Lipids
 A vitamin is an organic substance which is required in trace amounts for
normal cell function.
 The vitamins which cannot be synthesized internally by an organism are
termed essential vitamins: in their absence in the external medium, the
cell cannot survive.
 The water-soluble vitamins such as vitamin C (ascorbic acid) are not
lipids by definition.
 However, vitamins A, E, K, and D are water-insoluble and dissolve in
organic solvents. Consequently these vitamins are classified as lipids.
 Steroids are a class of lipid biochemicals with the general structure
shown in Fig.3.4(a).
 Of these, a subgroup (hormones) constitutes some of the extremely
potent controllers of biological reaction rates: hormones may be effective
at levels of 10−8 M in human tissue.
15
Continued…
Steroids
 Our bodies possess chemical messengers known as hormones, which are
basically organic compounds synthesized in glands and transported by
the bloodstream to various tissues in order to trigger or hinder the desired
process.
 Steroids are a kind of hormone that is typically recognized by their
tetracyclic skeleton, composed of three fused six-membered and one
five-membered ring, as seen above.
 The four rings are assigned as A, B, C & D as observed in the shade blue,
while the numbers in red indicate the carbons.
 Related sterol compounds have been shown to alter cell plasma
membrane permeabilities.
16
Continued…
a)
General steroid base: perhydroxycyclopentano
phenathrene
𝐶𝐻3
b)
O
C
𝐶𝐻3
Progesterone
𝐶𝐻3
𝐶𝐻3
O
C
𝐶𝐻3
c)
HC
O
(𝐶𝐻2 )3 −𝐶𝐻(𝐶𝐻3 )2
O
OH
𝐶𝐻3
Cholesterol
𝐶𝐻3
𝐶𝐻3
𝐶𝐻3
Cortisone
O
HO
progesterone can be converted into cortisone in a two step
process (microbial, then chemical) (Fig. 3.4b).
Figure 3.4: Some examples of steroid structure
17
Examples of Lipids
Waxes
 Waxes are “esters” (an organic compound made by replacing the
hydrogen with acid by an alkyl or another organic group) formed from
long-alcohols and long-chain carboxylic acids.
 Waxes are found almost everywhere. The fruits and leaves of many
plants possess waxy coatings, that can safeguard them from small
predators and dehydration.
 Fur of a few animals and the feathers of birds possess the same coatings
serving as water repellents.
 Carnauba wax is known for its water resistance and toughness
(significant for car wax).
18
Continued…
Cholesterol
 Cholesterol is a wax-like substance, found only in animal source
foods. Triglycerides, LDL, HDL, VLDL are different types of cholesterol
found in the blood cells.
 Cholesterol is an important lipid found in the cell membrane. It is a sterol,
which means that cholesterol is a combination of steroid and alcohol. In the
human body, cholesterol is synthesized in the liver.
 These compounds are biosynthesized by all living cells and are essential for
the structural component of the cell membrane.
 In the cell membrane, the steroid ring structure of cholesterol provides a
rigid hydrophobic structure that helps boost the rigidity of the cell
membrane. Without cholesterol, the cell membrane would be too fluid.
 It is an important component of cell membranes and is also the basis for the
synthesis of other steroids, including the hormones estradiol and
testosterone, as well as other steroids such as cortisone and vitamin D.
19
3.2 Sugars and Polysaccharides
 The carbohydrates are organic compounds with the general formula
(CH₂O)n , where 𝑛 ≥ 3.
 These compounds are found in all animal, plant, and microbial cells; the
higher-molecular-weight polymers serve both structural and storage
functions.
 The formula (CH₂O)n is sufficiently accurate to be useful in calculating
overall elemental balances and energy release in cellular reactions.
 When photosynthesis occurs, carbon dioxide is converted to simple
sugars 𝐶3 , to 𝐶9 in reactions driven by the sunlight.
 These sugars are then polymerized into forms suitable for structure
(cellulose) or sugar storage (starches).
 By these processes, radiant solar energy is stored in chemical form for
subsequent utilization.
20
3.2.1 D-Glucose and Other Monosaccharides
 Monosaccharides, or simple sugars, are the smallest carbohydrates.
Containing from 3 to 9 carbon atoms, monosaccharides serve as the
monomeric blocks for noninformational biopolymers with molecular
weights ranging into the millions.
 D(+)-Glucose, the optical isomer which rotates polarized light in the +
direction, is a polyhydroxyalcohol derivative like other simple sugars
(Table 3.2).
 Although D-glucose is by far the most common monosaccharide, other
simple sugars are also found in living organisms (Table 3.2).
 These common sugars are all either aldehyde or ketone derivatives.
 Thus, glucose is an aldohexose; the notation D refer ring to a particular
optical isomer occurring almost exclusively in living systems.
21
Continued…
Table 3.2:
Aldose (aldehyde derivatives; prefix aldo-)
𝐶𝐻𝑂
Triose (three-carbon)
𝐶𝐻2 𝑂𝐻
𝐶=𝑂
𝐻𝐶𝐻𝑂
𝐶𝐻2 𝑂𝐻
𝐶𝐻2 𝑂𝐻
D-Glyceraldehyde
Dihydroxyacetone
𝐶𝐻2 𝑂𝐻
𝐶=𝑂
𝐻𝐶𝑂𝐻
𝐶𝐻𝑂
𝐻𝐶𝑂𝐻
Pentose (five-carbon)
𝐻𝐶𝑂𝐻
𝐻𝐶𝑂𝐻
𝐻𝐶𝑂𝐻
𝐶𝐻2 𝑂𝐻
𝐶𝐻2 𝑂𝐻
D-Ribose
Hexose (six-carbon)
𝐶𝐻𝑂
𝐻𝐶𝑂𝐻
𝐻𝑂𝐶𝐻
𝐻𝐶𝑂𝐻
𝐻𝐶𝑂𝐻
𝐶𝐻2 𝑂𝐻
D-Glucose
𝐶𝐻𝑂
𝐻𝑂𝐶𝐻
𝐻𝑂𝐶𝐻
𝐻𝐶𝑂𝐻
𝐻𝐶𝑂𝐻
𝐶𝐻2 𝑂𝐻
Ketone
(ketone derivatives; prefix
keto)
D-Ribulose
𝐶𝐻𝑂
𝐻𝐶𝑂𝐻
𝐻𝑂𝐶𝐻
𝐻𝑂𝐶𝐻
𝐻𝐶𝑂𝐻
𝐶𝐻2 𝑂𝐻
D-Mannose D-Galactose
𝐶𝐻2 𝑂𝐻
𝐶=𝑂
𝐻𝑂𝐶𝐻
𝐻𝐶𝑂𝐻
𝐻𝐶𝑂𝐻
𝐶𝐻2 𝑂𝐻
D-Fructose
22
Continued…
 Five-membered sugars D-ribose and deoxyribose are major components
of the nucleic acid monomer of DNA and RNA and other biochemicals.
5
𝐶𝐻2 𝑂𝐻
𝐶𝐻2 𝑂𝐻
O
O
1
4
3
OH
2
OH
OH
D- Ribose
OH
OH
Deoxyribose
23
3.2.2 Disaccharides to Polysaccharides
 Because the ringed form of many simple sugar predominates in solution,
they do not exhibit the characteristic reactions of aldehydes or ketones.
 In the D-gluco-pyranose ring above, the −𝑂𝐻 attached to position 1 is
relatively reactive.
 As shown below, this hydroxyl group, here attached in the 𝛼 − 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛,
can condense with an − 𝑂𝐻 on the 4 carbon of another sugar to eliminate
a water molecule and form an 𝛼 − 1,4 − 𝑔𝑙𝑦𝑐𝑜𝑠𝑖𝑑𝑖𝑐 𝑏𝑜𝑛𝑑 ∶
𝐶𝐻2 𝑂𝐻
O
HO
𝐶𝐻2 𝑂𝐻
O
OH
+
OH
OH
OH
α-D-Glucose
HO
𝐶𝐻2 𝑂𝐻
O
OH
Condensation
OH
OH
α-D-Glucose
⇀
↽
Hydrolysis
HO
O
𝐶𝐻2 𝑂𝐻
O
OH
OH
OH
OH
+ 𝐻2 𝑂
α-1,4-glycosidic bond
α-Maltose
24
Continued…
 The condensation product of two monosaccharides is a disaccharide.
 In addition to maltose, which is formed from two D-glucose molecules,
the following disaccharides are relatively common:
𝐶𝐻2 𝑂𝐻
HO
O
OH
O
OH
β-D-Galactose
𝐶𝐻2 𝑂𝐻
O OH
OH
OH
β-D-Glucose
Lactose
 Among all disaccharides, sucrose is the easiest to hydrolyze: the resulting
mixture of glucose and fructose monosaccharides is called invert sugar.
25
3.2.3 Cellulose
 Cellulose, a major structural component of all plant cells from algae to
trees, is the most abundant organic compound on earth.
 Cotton and wood are two common examples of materials rich in
cellulose.
 Each cellulose molecule is a long. unbranched chain of D-glucose
subunits with a molecular weight ranging from 50,000 to over 1 million.
 One of the common products of enzymatic cellulose hydrolysis is
cellobiose, a dimer of two glucose units joined by a B-1,4-glycosidic
linkage.
 Most of the cellulose is organized into highly ordered crystalline regions,
in which cellulose chains or fibrils are so tightly packed that even water
molecules scarcely penetrate.
 Cellulose is, accordingly, water insoluble. Less ordered portions of the
assembly, called amorphous regions.
26
Continued…
C
O
C
O
O
O
C
O
O
β - 1,4 linkage
The glucose chain of cellulose
 The cellulose fibrils are clustered in microfibrils, which are often
depicted in cross section as in Fig. 3.5(a).
 Here the solid lines denote the planes of the glucose building blocks, and
the broken lines represent orientations of another important group of
polysaccharides called hemicellulose.
27
Continued…
Figure 3.5: Diagram of the structure of cellulose microfibrils
 Hemicellulose molecules are found in wood and other cellulosic
materials surrounding clusters of microfibrils, and these structures are in
turn encased in an extensively cross-linked coating of lignin.
 Because of this packaging, the term lignocellulose is often used to
describe these materials.
28
3.3 From Nucleotides to RNA and DNA
 The informational biopolymer DNA (deoxyribonucleic acid) contains all
the cell's hereditary information.
 When cells divide, each daughter cell receives at least one complete copy
of its parent's DNA, which allows offspring to resemble parent in form
and operation.
3.3.1 Building Blocks, an Energy Carrier, and Coenzymes
 Three components make up all nucleotides: (1) phosphoric acid, (2)
ribose or deoxyribose (five-carbon sugars), and (3) a nitrogenous base,
usually derived from either purine or pyrimidine.
H
C
N
N
C
HC
C
N
H
C
CH
N
H
Purine
N
CH
HC
CH
N
Pyrimidine
29
Figure 3.6: (a) General structure of ribonucleotides and deoxyribonucleotides. (b) The
five nitrogenous base found in DNA and RNA.
5
HO
O
H
H
(a)
Base
1’
H
Sugar
2’
H
DNA only
Purines
H
Pyrimidines
O
Thymine
H
H
N
H
O
H
H
H
RNA only
N
N
Guanine
H
N
H
N
H
H
N
N
H
OH
Ribonucleotides
(General structure)
H
N
N
N
O
H
OH
O
N
H
H
N
Adenine
N
H
N
N
H
H DNA and RNA
H
H
𝐻3 𝐶
O
Phosphoric
acid
Deoxyribonucleotides
(General Structure)
(b)
𝑂𝐻 − 𝑃 − 𝑂 − 𝐶𝐻2 O
Base
H
3’
OH
OH
Base
𝑂𝐻 − 𝑃 − 𝑂 − ’𝐶𝐻2 O
H
O
N
N
O
H
Uracil
Cytosine
H
30
Continued…
 These three components are joined in a similar fashion in two nucleotide
types (Fig. 3.6) which are distinguished by the five-carbon sugar
involved.
 Ribonucleic acid (RNA) is a polymer of ribose-containing nucleotides,
while deoxy ribose is the sugar component of nucleotides making up the
biopolymer deoxyribonucelic acid (DNA).
 Also shown in Fig. 3.6 are the five nitrogenous bases found in DNA and
RNA nucleotide components.
 Three of the bases, adenine (A). guanine (G), and cytosine (C), are
common to both types of nucleic acids.
 Thymine (T) is found only in DNA, while uracil (U), a closely related
pyrimidine base, is unique to RNA.
31
Continued…
Table 3.3: Nomenclature of nucleosides and nucleotides. As indicated in the last row,
the prefix deoxy- is used to identify those molecules containing deoxyri-bose.
Base
Nucleoside
Nucleotide
Adenine
Adenosine
Adenylate (AMP)
Cytosine
Cytidine
Cytidylate (CMP)
Guanine
Guanosine
Guanylate (GMP)
Uracil
Uridine
Uridylate (UMP)
Thymine
Deoxythymidine
Deoxythymidylate (dTMP)
 Both series of nucleotides are strong acids because of their phosphoric
acid groups.
 Of particular biological significance is the nucleoside adenosine, made
with ribose and adenine.
32
Continued…
 Sketched in Fig. 3.7, along with its important derivatives, is adenosine
monophosphate (AMP).
 One or two additional phosphoric acid groups can condense with AMP to
yield ADP (adenosine diphosphate) and ATP, respectively.
 The phosphodiester bonds connecting the phosphate groups have
especially useful free energies of hydrolysis.
 For example, the reaction of ATP to yield ADP and phosphate is
accompanied by a standard Gibbs free-energy change of -7.3 kcal/mol at
37 °F and pH 7 [recall that pH = −log (molar 𝐻 + concentration)].
33
Continued…
2. Adenosine
OH
OH
diphosphate
(ADP)
Adenine
𝑂𝐻 − 𝑃 − 𝑂 − 𝑃 − 𝑂 − 𝐶𝐻2
OH
𝑂𝐻 − 𝑃 − 𝑂 − 𝐶𝐻2
O
1. Adenosine
monophosphate
(AMP)
O
H
H
OH
H
OH
H
O
O
H
H
OH
H
OH
H
OH
OH
OH
Adenine
𝑂𝐻 − 𝑃 − 𝑂 − 𝑃 − 𝑂 − 𝑃 − 𝑂 − 𝐶𝐻2 O
Figure 3.7: The
phosphate of
adenosine. AMP,
ADP and ATP are
important cellular
energy transfer
process, and cyclic
AMP serves a
regulatory Function
O
O
H
O
H
H
O
𝐶𝐻2
H
𝑃
OH
Adenine
O
H
H
H
O
Adenine
O
O
OH
H
OH
3. Adenosine
triphosphate
(ATP)
OH
4. Cyclic AMP
34
3.3.2 Biological Information Storage
 DNA and RNA As with the polysaccharides, the polynucleotides are
formed by condensation of its monomers.
 For both RNA and DNA, the nucleotides are connected between the 3'
and 5' carbons of successive sugar rings.
 Figure 3.8 illustrates a possible sequence in a trinucleotide. The DNA
molecules in cells are enormously larger: all the essential hereditary
information in procaryotes is contained in one DNA molecule with a
molecular weight of the order of 2 × 109 .
 In eucaryotes, the nucleus may contain several large DNA molecules.
 The negative charges on DNA are balanced by divalent ions
(procaryotes) or basic amino acids (eucaryotes).
 It is important to notice in Fig. 3.8 that the sequence of nucelotides has a
direction or polarity, with one free 5' -end and the other terminus
possessing a 3' -end not coupled to another nucleotide.
35
Continued…
𝑂𝐻𝐶𝐻2 O
𝐵𝑎𝑠𝑒1
5 ' End
𝐵𝑎𝑠𝑒1
O
OH HO
−𝑂
𝑂𝐻𝐶𝐻2 O
O
OH
𝑂𝐻𝐶𝐻2 O
P
P
−𝑂
O
𝑂𝐶𝐻2 O
𝐵𝑎𝑠𝑒2
𝐵𝑎𝑠𝑒2
O
OH HO
−𝑂
O
OH
𝑂𝐻𝐶𝐻2 O
P
P
OH
𝐵𝑎𝑠𝑒3
−𝑂
O
𝑂𝐶𝐻2
O
𝐵𝑎𝑠𝑒3
OH
3 ' End
Figure 3.8: (a) Condensation of several nucleotides to form a chain linked by
phosphodiester bonds
36
Continued…
 As indicated in Fig. 3.8(b), it is conventional to write the nucleotide
sequence in the 5' to 3' direction.
 More abbreviated notations are often used. Suppose in the example in
Fig. 3.8(b), which contains a phosphate group esterified to the 5'hydroxyl, that 𝐵𝑎𝑠𝑒1 is cytosine, 𝐵𝑎𝑠𝑒2 is adenine, and 𝐵𝑎𝑠𝑒3 is
thymine.
Schematic form
𝐵𝑎𝑠𝑒1
𝑃
5'
𝑃
𝐵𝑎𝑠𝑒2
𝐵𝑎𝑠𝑒3
𝑃
3'
Figure 3.8: (b) Another schematic for nucleotide chain
37
Continued…
 Then the same nucleotide sequence information is often written pCpApT,
or, even more briefly, CAT.
 As James Watson and Francis Crick deduced in 1953, the DNA molecule
consists of two polynucleotide chains coiled into a double helix (Fig.
3.9).
 The regular helical backbone of the molecule is composed of sugar and
phosphate units.
 In the interior of the double helix are the purine and pyrimidine bases.
 It is the sequence of the four different nitrogenous bases along the
polynucleotide chains which carries genetic information–namely the
instructions for ribonucleic acid and protein synthesis.
38
Continued…
Figure 3.9: Several views of the double-helical structure of DNA.
39
Continued…
 The function of DNA is to store instructions for synthesis of RNA
molecules of specific nucleotide sequence and length, some of which
then coordinate synthesis of a variety of specific proteins.
 A segment of DNA coding for an RNA molecule's sequence is called a
gene.
 There are three distinct classes of ribonucleic acid (RNAs) in all cells and
a fourth group found in some viruses.
 RNAs are comprised of ribonucleotides in which ribose rather than
deoxyribose is the sugar component.
 The RNA sequences are constructed using information contained in
segments of DNA.
 Again, base pairing is the central device in transmission of information
from DNA to RNA.
40
Continued…
DNA
(temple)
RNA
(being assembled)
T
A
A
U
C
G
G
C
Figure 3.10: A simplified diagram of DNA replication.
41
42
Continued…
 The various RNAs which participate in normal cell function serve the
purpose of reading and implementing the genetic instructions of DNA.
 Messenger RNA (mRNA) is complementary to a base sequence from a
gene in DNA.
 Each mRNA molecule carries a message from DNA to another part of the
cell's bio chemical apparatus.
 Since the length of these messages varies considerably, so do the size of
mRNAs: a chain of 10³-104 nucleotides in length is typical.
 RNAs are typically single-stranded.
 However, intermolecular and intramolecular base pairing of RNA
nucleotide sequences often play a role in the structure or function of the
RNA.
43
Continued…
Table 3.4: Properties of E.coli RNAs
Type
Sedimentation
coefficient
Mol wt.
No. of
nucleotide
residues
Percent of total
cell RNA
mRNA
6S-25S
25000-1000000
75-3000
~2
tRNA
~4S
23000-30000
35-90
16
rRNA
5S
~35000
~100
16S
~550000
~1500
23S
~1100000
~3100
}
82
44
Continued…
 The message from mRNA is read in the ribosome.
 Up to 65 percent of the ribosome is ribosomal RNA (rRNA), which in
turn can be separated in a centrifuge into several RNA species.
 Transfer RNA species (tRNAs), the smallest type, with only 70 to 95
nucleotide components, are found in the cell's cytoplasm and assist in the
translation of the genetic code at the ribosome,
 Table 3.4 characterizes the various RNAs in the. E. coli bacterium.
 The nucleotide sequences and thereby the structures and functions of the
various rRNAS and tRNAs found in the cell are, like mRNA sequences,
dictated by nucleotide sequences of the corresponding genes in the cell's
DNA.
 The end result of this intricate information transmitting and translating
system is a protein molecule.
 Proteins are the tangible biochemical expression of the information and
instructions carried by DNA.
45
3.4 Amino Acids into Proteins
 Proteins are the most abundant organic molecules within the cell:
typically 30 to 70 percent of the cell's dry weight is protein.
 All proteins contain the four most prevalent biological elements: carbon,
hydrogen, nitrogen, and oxygen. Average weight percentages of these
elements in proteins are C (50 percent), H (7 percent). O (23 percent),
and N (16 percent).
 In addition, sulfur (up to 3 percent) contributes to the three-dimensional
stabilization of almost all proteins by the formation of disulfide (S-S)
bonds between sulfur atoms at different locations along the polymer
chain.
 The two major types of protein configuration, fibrous and globular, are
shown in Fig. 3.11.
 A critical role of many proteins is catalysis.
46
Continued…
a) Fibrous
proteins
𝛼- Helix
Antiparallel
𝛽-pleated-sheet
Collagen
Triple helix
b) Globular
proteins
Single chain
(myoglobin)
Oligomeric protein
containing four chain
(hemoglobin)
Figure 3.11: The two types of protein structure with variations: (a) fibrous and (b) globular
47
Table 3.5: The diverse biological function of proteins
Protein
Occurrence or function
Enzymes (biological
catalysts):
Protein
Occurrence or function
Toxins:
Glucose isomerase
Isomerizes glucose to
fructose
Bacillus thuringiensis
Kills insects
Trypsin
Hydrolyzes some peptides
E. Coli ST toxin
Causes disease in pigs
Alcohol dehydrogenase
Oxidizes alcohols to
aldehydes
Clostridium botulinum toxin
Causes bacterial food
poisoning
Regulatory proteins:
Storage proteins:
Iac repressor
Controls RNA synthesis
Ovalbumin
Eggs-white protein
Catabolite activator protein
Catabolite repression of
RNA synthesis
Casein
Milk protein
Interferons
Induce virus resistance
Zein
Seed protein of corn
Insulin
Regulates glucose
Bovine growth protein
Stimulates growth and
lactation
Transport protein:
Contractile proteins:
Dynein
Cilia and flagella
Structural proteins:
Lactose permease
Transports lactose across cell
membrane
Collagen
Cartilage, tendons
Myoglobin
Transports 𝑂2 in muscle
Gylcoproteins
Cell walls and coats
Hemoglobin
Transports 𝑂2 in blood
Elastin
Ligaments
48
Continued…
 Protein catalysts called enzymes determine the rate of chemical reactions
occurring within the cell.
 Enzymes are found in a variety of cell locales: some are suspended in the
cytoplasm and thus dispersed throughout the cell interior.
 Some are localized by attachment to membranes or association with
larger molecular assemblies.
 Other membrane-associated proteins called permeases aid in transport of
specific nutrients into the cell.
 Proteins are isolated, purified, and characterized by several different
types of physical and chemical techniques.
 Protein separation methods are based on differences in molecular
properties, which in turn are partly due to the characteristics of the
constituent amino acids.
49
3.4.1 Amino Acid Building Blocks and
Polypeptides
 The monomeric building blocks of polypeptides and proteins are the aamino acids, whose general structure is
H
H₂N---C---COOH
H
 Thus, amino acids are differentiated according to the R group attached to
the carbon (the one closest to the -COOH group). Since there are
generally four different groups attached to this carbon (the exception
being glycine, for which R-H), it is asymmetrical.
 Many organic compounds of biological importance, sugars and amino
acids included, are optically active: they possess at least one asymmetric
carbon and therefore occur in two isomeric forms, as shown below for
the amino acids:
50
Continued…
H
H₂N---C---COOH
H
L form
H
COOH---C---NH₂
H
D form
 Solutions of a pure isomer will rotate plane polarized light either to the
right (dextro- or d-) or left (levo- or l-).
 This isomerism is extraordinarily important since viable organisms can
usually utilize only one form, lacking the isomerization catalysts needed
to interconvert the two.
 In Fig. 3.12 are the 20 amino acids commonly found in proteins.
 In addition to the carboxyl and amine groups possessed by all amino
acids (except proline), the R groups of some acids can ionize.
51
Continued…
H−
1
CH3 −
2
3
CH3
CH −
CH3
Glycine
(Gly)
Alanine
(Ala)
Valine
(Val)
4
Cyclic
C −CH2 −
CH3
CH −CH2 −
CH3
CH
N
H
Leucine
(Leu)
Tryptophan
(Trp)
CH3 − S − CH2 − CH2
−CH2 −
Methionine
(Met)
Hydrophobic
0
Phenylalanine
(Phe)
CH3 − CH2 − CH −
Isoleucine
Figure 3.12: The 20 amino acids found in proteins. (Ile)
CH3
52
Continued…
4
Cyclic
HO
HO − CH2 −
Serine
(Ser)
NH2
Asparagine
(Asn)
HO
HO − CH2 −
Threonine
(Thr)
Tyrosine
(Tyr)
NH2
C − CH2 − CH2 −
C − CH2 −
O
O
Glutamine Acid O
-O
(Glu)
C − CH2 −
O
HS − CH2 − Cysteine
(Cys)
Figure 3.12: (continued)
−CH2 −
Glutamine
(Glu)
Hydrophilic
3
Hydroxyl
2
Admide
1
+
H3 N − (CH2 )4 −
Lysine
O
C = C − CH2 −
(Lys)
HN NH
H2 N − C − NH − (CH2 )3 −
C
Arginine
O
H
C
(Arg)
C
Histinine
C C
Others
(His)
Proline
C
H O
N
(Pro)
H
(entire molecules)
C − CH2 − CH2 −
Acidic or Base
0
53
Continued…
 Polypeptides are short condensation chains of amino acids (Fig. 3.13).
 A little reflection suggests that as the length of the chain increases, the
physicochemical characteristics of the polymer will be increasingly
dominated by the R groups of the residues while the amino and carboxyl
groups on the ends will shrink in importance.
 Many hormones such as insulin, growth hormone, and somatostatin are
polypeptides.
 It has been found that not all 20 amino acids in Fig. 3.12 are found in all
proteins.
 Amino acids do not occur in equimolar amounts in any known protein.
For a given protein, however, the relative amounts of the various amino
acids are fixed.
54
Continued…
+
6-7 HN
4-5-
COO
4-5
-
7-8
+
H3 N
H
CH2
COO
O
C
C
N
H
H
H
O
H
CH2
C
N
N
H C
C
C
N
O
CH2
H
H
N
NH3
C
C
N
H
3-
COO
C
N H
C
C
O
CH2
H
O
CH3
CH2
CH2
CH2
+
H
H
CH2
C
CH2
CH2
O
H
CH2
C
NH
C
10 - 11
H2 N
12 - 13
+
NH2
Gly-Asp-Lys-Glu-Arg-His-Ala
Figure 3.13: A hypothetical polypeptide, showing the ionizing groups found among the amino
acids. Notice that the listing sequence for the amino acid residues starts at the N-terminal end.
Numbers near each ionizing group denote the corresponding pK value.
55
Continued…
 Amino acids are not the only constituents of all proteins.
 Many proteins contain other organic or even inorganic components.
 Hemoglobin, the oxygen-carrying molecule in red blood cells, is a
familiar example of a conjugated protein: it has four heme groups, which
are organometallic complexes containing iron.
 A related smaller molecule, myoglobin, has one heme group.
56
3.4.2 Protein Structure
 The structure of many proteins is conveniently described in three levels.
 A fourth structural level must be considered if the protein consists of more than a
single chain.
 As seen from Table 3.7 each level of structure is determined by different factors,
hence the great range of protein structures and functions.
Table 3.7: Protein structure
1. Primary
Amino acid sequence, joined through peptide bond
2. Secondary
Manner of extension of polymer chain, due largely to hydrogen bonding
between residues, not widely separated along chain
3. Tertiary
Folding, bending of polymer chain, induced by hydrogen, salt, and
covalent disulfide bonds as well as hydrophobic and hydrophilic
interactions
4. Quaternary
How Different polypeptide chains fit together; structure stabilized by
same forces as tertiary structure
57
3.4.3 Primary Structure
 The primary structure of a protein is its sequence of amino acid residues.
 The first protein structure completely determined was that of insulin
(Sanger and co-workers in 1955).
 It is now generally recognized that every protein has not only a definite
amino acid content but also a unique sequence in which the residues are
connected.
 By a variety of techniques, the protein is broken into different
polypeptide fragments.
 For example, enzymes are available which break bonds between specific
residues within the protein.
 Automated spinning cup analyzers can determine an amino acid sequence
up to 100 residues in length; about two hours are required per amino
acid. Recent gas phase protein sequencing technology has greatly
reduced the amount of protein sample required to obtain sequence
information.
58
Continued…
 The sequence of amino acid residues in many polypeptides and proteins
is now known.
 So far no repeating pattern of residues has been found, although the
available evidence suggests some patterns may exist in fibrous proteins.
 As an example, Fig. 3.14 gives the amino acid sequence of an enzyme
called lysozyme found in egg white.
 The folding of the polypeptide chain into a precise three-dimensional
structure often yields a complex, convoluted form as seen in Fig 3.14b
for the Iysozyme molecule.
 This elaborate geometrical structure is dictated primarily by the protein’s
amino acid sequence.
59
Continued…
(b)
The corresponding three-dimensional
molecular structure (tertiary structure)
of crystalline lysozyme
(a)
The amino acid sequence, or primary
structure, of eggs white lysozyme, an
enzyme.
Figure 3.14:
60
3.4.4 Three-Dimensional Conformation: Secondary
and Tertiary Structure
 Secondary structure of proteins refers to the structural configuration of
the amino acid residues which are close neighbors in the linear amino
acid sequence.
 We recall that the partial double-bond character of the peptide bond holds
its neighbors in a plane.
 Consequently, rotation is possible only about two of every three bonds
along the protein backbone (Fig. 3.15).
 This restricts the possible shapes which a short segment of the chain can
assume.
 There are two general forms of secondary structure: helices and sheets.
 The configuration believed to occur in hair, wool, and some other fibrous
proteins arises because of hydrogen bonding between atoms in closely
neighboring residues.
61
Continued…
C- Terminal end
N- Terminal end
Planar peptide
group
Figure 3.15: The planar nature of the peptide bond restricts rotation in the
peptide chain
62
Continued…
 If the protein chain is coiled, hydrogen bonding can occur between the CO group of one residue and the NH group of its neighbor four units down
the chain.
 Because of the relative weakness of individual hydrogen bonds, the ahelical structure is easily disturbed.
 Proteins can lose this structure in aqueous solution because of
competition of water molecules for hydrogen bonds.
 If wool is steamed and stretched, it assumes a different, pleated-sheet
structure.
 Achievement of the secondary structure of fibrous proteins is of
importance in the food industry.
 The complete three-dimensional configurations, or tertiary structures, of
a few proteins have been determined by painstaking and laborious
experiments.
63
Continued…
 After crystals of pure protein have been prepared, x-ray crystallography
is employed.
 Each atom is a scattering center for x-rays; the crystal diffraction pattern
contains information on structural details down to 2 A resolution.
 In order to obtain this detail, however, it was necessary in the case of
myoglobin (Fig. 3.16) to calculate 10,000 Fourier transforms, a task
which had to wait for the advent of the high-speed digital computer.
 Interactions between R groups widely separated along the protein chains
determine how the chain is folded or bent to form the compact
configuration typical of globular proteins.
 Several weak interactions, including ionic effects, hydrogen bonds, and
hydrophobic interactions between nonpolar R groups (Fig. 3.17). are
prime determinants of protein three-dimensional structure.
64
Continued…
Figure 3.16:
The three-dimensional structure
of myoglobin. Dots indicate the
α-carbons of the 121 amino acids
residues in this oxygen-carrying
protein. A single heme group,
with its central iron atom larger
and labeled, is shown in the
upper central portion of the
molecule.
65
Continued…
Figure 3.17: A summary of the bonds and interactions which stabilize protein
molecular structure.
66
Continued…
 Protein conformation is a major determinant of protein biological
activity.
 There is considerable evidence that in many cases a definite threedimensional shape is essential for the protein to work properly.
 This principle is embodied, for example, in the lock-and-key model for
enzyme specificity.
 This simple model suggests why the proteins which catalyzes
biochemical reactions are often highly specific.
 That is, only definite reactants, or substrates, are converted by a
particular enzyme.
 The lock-and-key model (Fig. 3.18) holds that the enzyme has a specific
site (the "lock") which is a geometrical complement of the substrate (the
"key") and that only substrates with the proper complementary shape can
bind to the enzyme so that catalysis occurs.
67
Continued…
Figure 3.18: Simplified diagram of the lock-and-key model for enzyme action.
Here the shape of the catalytically active site (the lock) is complementary to the
shape of the substrate key.
68
Continued…
Native form
Figure 3.19:
After a protein has been
denatured by exposure to an
adverse environment, it will
often return to the native,
biologically active confirmation
following restoration of suitable
conditions. This result indicates
that primary protein structure.
Native form
69
Continued…
 When protein is exposed to conditions sufficiently different from its
normal biological environment, a structural change, again called
denaturation, typically leaves the protein unable to serve its normal
function (Fig. 3.19).
 Relatively small changes in solution temperature or pH, for example,
may cause denaturation, which usually occurs without severing covalent
bonds.
 If a heated dilute solution of denatured protein is slowly cooled back to
its normal biological temperature, the reverse process or, renaturation,
with restoration of protein function, often occurs.
70
3.4.5 Quaternary Structure and Biological
Regulation
 Proteins may consist of more than one polypeptide chain, hemoglobin
being perhaps the best known example.
 How these chains fit together in the molecule is the quaternary structure.
 As Table 3.8 indicates, a great many proteins, especially enzymes
(usually indicated by the -ase suffix), are oligomeric and consequently
possess quaternary structure.
 The forces stabilizing quaternary structure are believed to be the same as
those which produce tertiary structure.
 While disulfide bonds are sometimes present, as in insulin, most of the
proteins in Table 3.8 are held together by relatively weak interactions.
 Many oligomeric proteins are known to be self-condensing; e.g,
separated hemoglobin a chains and B chains in solution rapidly reunite to
form intact hemoglobin molecules.
71
Table 3.8: Characteristics of several oligomeric proteins
Protein
Molecular weight
No. of chains
No. of −𝐒 − 𝐒 − Bonds
Insulin
5798
1+1
3
Ribonuclease
13683
1
4
Lysozyme
14400
1
5
Myoglobin
17000
1
0
Papain
20900
1
3
Trypsin
23800
1
6
Chymotrypsin
24500
3
5
Carboxypeptidase
34300
1
0
Hexokinase
45000
2
0
Taka-amylase
52000
1
4
Bovine serum albumin
66500
1
17
Yeast enolase
67000
2
0
Hemoglobin
68000
2+2
0
Liver alcohol dehydrogenase
78000
2
0
Alkaline phosphatase
80000
2
4
72
Hemerythrin
107000
8
0
Glyceraldehyde-phosphate dehydrogenase
140000
4
0
Lactic dehydrogenase
140000
4
0
ϒ-Globulin
140000
2+2
25
Yeast alcohol dehydrogenase
150000
4
0
Tryptophan synthetase
159000
2+2
Aldolase
160000
4
Phosphorylase b
185000
2
Threonine deaminase (Salmonella)
194000
4
Fumarase
200000
4
0
Tryptophanase
220000
8
4
Formyltetrahydrofolate synthetase
230000
4
Aspartate transcarbamylase
310000
4+4
0
Glutamic dehydrogenase
3160000
6
0
Fibrinogen
330000
2+2+2
Phosphorylase a
370000
4
Myosin
500000
2+3
Galactosidase
540000
4
Ribulose diphosphate carboxylase
557000
24
0
0
73
3.5 Hybrid Biochemicals
 Many biologically and commercially important biochemicals do not fit
clearly into any of the chemical classes described above.
 These diverse compounds are hybrids which contain various
combinations of lipid, sugar, and amino acid building blocks.
 As indicated in Table 3.9, hybrid biochemicals provide several different
significant functions.
74
Continued…
Table 3.9: Some examples of the nomenclature, subunit composition, and biological
function of hybrid biochemicals
Name
Building blocks
Location, function
Peptidoglycans
Crosslinked disaccharide and
peptides
Bacterial cell walls
Proteoglycans
95% carbohydrate, 5%
protein
Connective tissue
Glycoprotein
1-30% carbohydrate,
remainder protein
Diverse; includes antibodies,
eucaryotic cell outer
Surface components
Glycolipids
Lipids which contain from
one to seven sugar residues
Membrane components
Lipoprotein
1-50% protein, remainder
lipid in plasma lipoproteins
Membrane, blood plasma
components
Lipopolysaccharides
Lipid plus a highly variable
oligosaccharide chain
Outer portion of gramnegative bacterial envelop
75
3.5.1 Cell Envelopes: Peptidoglycan and
Lipopolysaccharides
 From a natural biological viewpoint, microorganisms, such as bacteria
that live in environments much more diverse and much less controlled
than cells in liver tissue in an animal, must possess greater physical
rigidity and resistance to bursting under large osmotic pressure.
Accordingly, bacteria have much different outer envelopes than animal
cells,
 Looking at cell envelopes from a biochemical process engineering
perspective, we find several points of interest. The properties of outer cell
surfaces determine the cells' tendency to adhere to each other or to walls
of reactors, pipes, and separators.
 Such phenomena have important influences in formulation of
immobilized cell catalysts, operation of continuous microbial reactors,
and cell separation from liquid.
76
Continued…
 As indicated in part by their opposite responses to gram staining, the
envelopes of gram-positive and gram-negative cells are very different
(Fig. 3.20).
 While the envelope of a gram-positive bacterium has a single membrane,
there are two membranes of this type in gram-negative cells. The outer
and cytoplasmic membranes in gram-negative bacteria enclose a region
called the periplasmic space or periplasm.
 Adjacent to the outer surface of the cytoplasmic membrane in both types
bacteria is a layer of peptidoglycan. The building blocks of peptidoglycan
area 1,4 linked disaccharide of N-acetylmuramic acid (NAM) and Nacetylgluco amine (NAG), a bridging pentapeptide comprised entirely of
glycyl residues, a tetrapeptide.
77
Continued…
Figure 3.20: Schematic diagrams of the cell envelope structure of gram-positive
and gram-negative bacteria (CM cytoplasmic membrane, PG peptidoglycan, PS
periplasmic space, OM outer mem brane).
78
Continued…
 Lysozyme, the enzyme is an effective antibacterial agent because it
catalyzes hydrolysis of the f-1,4 glycosidic linkage between the NAM
and NAG components of peptidoglycan.
 This causes breakdown and removal of peptidoglycan from the bacterium
which results in cell bursting or lysis in natural hypotonic solutions.
 Lipopolysaccharides (LPS) are important components of the outer
portion of the outer membrane of gram-negative bacteria.
 Some of these compounds are called endotoxins: they produce powerful
toxic effects in animals. LPS toxicity is one reason why an infection by
E. coli in the bloodstream is extremely dangerous.
79
Continued…
Outer membrane LPS molecules have three different regions:
(1) a lipid-A component which consists of six unsaturated fatty acid chains,
which extend into the membrane, joined to a diglucosamine head,
(2) a core oligosaccharide region containing ten sugars, some of them rare,
and
(3) an O side chain comprised many repeating tetrasaccharide building
blocks. The core region and O side chain are extended outward, away
from the cell.
 This is not yet the end of the story of envelope layers in bacteria. Beyond
the outer membrane of many species is a capsule or slime layer which
consists polysaccharide.
80
Continued…
Figure 3.21: (a) Structure of the building block for peptidoglycan; (b) schematic
illustration of highly cross-linked peptidoglycan in the cell wall of Staphlococcus
aureus, a gram-positive bacterium.
81
Continued…
 Turning now to cell envelopes of yeast cells, the cytoplasmic membrane
consists of lipids, protein, and polysaccharide containing mannose.
Moving outward from this membrane through the periplasmic space
toward the cell exterior, we find next a cell wall.
⋯O
𝐶𝐻2 𝑂𝐻
O
OH
𝐶𝐻3
𝐶𝑂
𝑁𝐻
O
OH
O
𝑁𝐻
𝐶𝑂
𝐶𝐻3
O⋯
𝐶𝐻2 𝑂𝐻
Chitin
 Animal cells, because they normally live in a well-controlled isotonic
environment, do not possess cell walls. In their plasma membranes, in
addition to phospholipids and proteins, we find 2 to 10 percent
carbohydrate
82
3.5.2 Antibodies and Other Glycoproteins
 Proteins with covalently coupled monosaccharides or short
oligosaccharide chains, the glycoproteins, are found in considerable
variety in and around eucaryotic cells.
 Several enzymes such as glucose oxidase produced by the mold
Aspergillus niger are glycoproteins. Collagen, the biological coaxial
cable mentioned earlier, consists of glycosylated protein.
 Antibodies, primary agents in immunological defenses of vertebrates, are
glycoproteins (Fig. 3.22).
 As part of the vertebrate immune system, B cells, one of two classes
lymphoid cells found in the body, differentiate in the presence of a
foreign substance, virus, or cell (the antigen) to form plasma cells which
secrete antibodies.
83
Continued…
Figure 3.22:
(a) Schematic diagram of the structure of an
immunoglobulin. C and V denote constant and
variable regions, respectively, while subscripts
identify heavy (H) and light (L) chains.
Disulfide bonds are shown, as is the attachment
of carbohydrate groups (CHO) to the heavy
chains.
(b) Diagram of the molecular
structure of immunoglobulin
G, or IgG, an antibody.
84
Continued…
 These antibodies bind specifically to the antigen and closely related
molecule serving to coagulate the invading species for subsequent
capture and removal, Antibodies are members of a class of proteins
called immunoglobulins have the common structural features indicated in
Fig. 3.22.
 The immunoglobulin molecule consists of two identical, longer, "heavy"
chains linked together disulfide bonds and noncovalent interactions to
two identical, shorter, "light" chains.
 The primary differentiation within a class of antibodies occurs at the
amino terminal ends of the chains, named the variable regions. The
antigen-binding site of the antibody is formed by the variable regions of
the light and heavy chains
85
Mechanism of bio-removal of metals
contd..
Mechanism of protein synthesis
Promoter
DNA
P
RNA
Ribosome
binding
sequence
Structural gene protein
coding sequence
Stop codon
Start codon
terminator
RB
Transcription
RB
Translation
Protein
86
Mechanism of bio-removal of metals
Operator gene
Promoter gene
P
O
R
Structural gene
S
Regulator gene
contd..
Inducer absent
RNA
polymerase
Repressor
Inducer present
O
P
S
R
Repressor
RNA
polymerase
Inducer
Inactive
Repressor
Transcription
mRNA
Translation
Enzyme
87
Mechanism of bio-removal of metals
Native biosorption
_
M+2
_
_-_
__
M+2
Metal binding proteins
M
M+2
M+2
Enzymatic
transformation
contd..
M
M
M-CH3
M (ox)
eMetal precipitation
M (red)
M(0)
M (red)
M (ox)
H2S
HPO42+M+2
MS
MHPO4
(M. Valls, V.D. Lorenzo, FEMS Microbiology Review. 26(2002), 327 – 338)
88
Mechanism of bioadsorption
65
2913.05
3439.00
3427.92
40
3734.41
3673.93
30
25
A
663.76
B
20
15
1700.20
1652.80
5
4000
3712.39
3649.22
0
-5
1399.71
10
3885.47
3837.20
%T
35
2328.72
45
1651.96
50
996.59
1653.11
55
473.55
418.64
60
3500
3000
2500
2000
Due to protein & lipid Harz et al.,
2005; Ellerbrock et al., 1999
1500
1000
500
Wavenumbers (cm-1)
FTIR spectra of GAC before (A) and after (B) SABA
(a
)
SEM of GAC before SABA (a) & after SABA (b)
(b
)
89
Thank you
4 Kinetics of Enzyme Catalysed Reactions
 A breakthrough in enzymology occurred in 1897 when Büchner first
extracted active enzymes from living cells.
 Büchner's work provided two major contributions: first, it showed that
catalysts at work in a living organism could also function completely
independently of any life process.
 The first successful isolation of a pure enzyme was achieved in 1926 by
Sumner.
 Studies of the isolated enzyme revealed it to be protein, a property now
well established for enzymes in general.
 There are six major classes of reactions which enzymes catalyst.
 These units form the basis for the Enzyme Commission (EC) system for
classifying and as signing index numbers to all enzymes (Table 4.1).
1
Table 4.1: Enzyme Commission classification system for enzymes (class names, Enzyme Commision
type numbers, and type of reactions catalyzed)
1.
Oxidoreductases
(oxidation-reduction reaction)
1.1 Acting on −CH − OH
3.5 Other C − N bonds
1.2 Acting on −C = O
3.6 Acid anhydrides
1.3 Acting on −CH = CH −
2.
3.4 Peptide bonds
4.
1.4 Acting on −CH − NH2
4.1 −C = C −
1.5 Acting on −CH − NH −
4.2 −C = O
1.6 Acting on NADH; NADPH
4.3 −C = N −
Transferases
(transfer of functional groups)
5.
2.1 One-carbon
2.2 Aldehydic or ketonic groups
3.
Lyases
(addition to double bonds)
Isomerases
(isomerization reaction)
5.1 Racemases
6.
Ligases
(formation of bonds with ATP cleavage)
2.3 Acyl groups
6.1 C − O
2.4 Glycosyl groups
6.2 C − S
2.7 Phosphate groups
6.3 C − N
2.8 S-containing groups
6.4 C − C
Hydrolases
(hydrolysis reaction)
3.1 Esters
3.2 Glycosidic bonds
Continued…
 Although the common enzyme nomenclature established by past use and
tradition is often employed instead of the "official" names, the EC system
provides a convenient tabulation and organization of the variety of
functions served by enzymes.
 A catalyst is a substance which increases the rate of a chemical reaction
without undergoing a permanent chemical change.
 While a catalyst influences the rate of a chemical reaction, it does not
affect reaction equilibrium (Fig. 4.1).
 Equilibrium concentrations can be calculated using only the
thermodynamic properties of the substrates (remember that substrate is
the biochemist's term for what we usually call a reactant) and products.
 Reaction kinetics, however, involve molecular dynamics and are
presently impossible to predict accurately without experimental data.
3
Continued…
Figure 4.1: By lowering the
activation
energy
for
reaction, a catalyst makes it
possible
for
substrate
molecules with smaller
internal energies to react.
4
https://www.chegg.com/homework-help/questions-and-answers/given-exothermic-reactionactivation-energy-forward-reaction-compare-activation-energy-rev-q66689538
5
Continued…
 For design and analysis of a reacting system, we must have a
mathematical formula which gives the reaction rate in terms of
composition, temperature, and pressure of the reaction mixture.
 The similarities between synthetic catalysts and enzymes go beyond the
use of a common technique for modeling reaction kinetics.
 The rate expressions eventually obtained for both types of catalysts are
very similar and sometimes of identical forms.
 This arises because, in both cases, it is known that the reacting molecules
form some sort of complex with the catalyst.
 This general phenomenon in catalysis accounts for such similarity in rate
expressions.
 Another distinguishing characteristic of enzymes is their frequent need
for cofactors.
6
Continued…
 A cofactor is a nonprotein compound which combines with an other wise
inactive protein (the apoenzyme) to give a catalytically active complex.
 While this complex is often called a holoenzyme by biochemists, we
shall often simply call it an enzyme.
 Two distinct varieties of cofactors exist. The simplest cofactors are metal
ions (Table 4.2). On the other hand, a complex organic molecule called a
coenzyme may serve as a cofactor.
 Enzymes with the same name but obtained from different organisms
often have different amino acid and hence different properties and
catalytic activities.
 It is often asserted that enzymes are more active, i.e., allow reactions to
go faster, than nonbiological catalysts.
7
Continued…
Table 4.2: Some enzymes containing or requiring metal ions as cofactors
Ca2+
𝛼-Amylase (swine pancreas)
Collagenase
Lipase
Micrococcal nuclease
Co2+
Glucose isomerase (Bacillus coagulans) (also requires Mg 2+ )
Cu2+ (Cu2+ )
Galactose oxidase
Tryosinase
Fe2+or Fe3+
Catalase
Cytochromes
Peroxidase
Mg 2+
Deoxyribonuclease (bovine pancreas)
Mn2+
Arginase
Na+
Plasma membrane ATPase (also requires K + and Mg 2+ )
Zn2+
Alcohol dehydrogenase
Alkaline phosphatase
Carboxypeptidase
8
Continued…
 A common measure of activity is the turnover number, which is the net
number of substrate molecules reacted per catalyst site per unit time. To
permit some comparison of relative activities, turnover numbers for
several reactions are listed in Table 4.3, which shows that at the ambient
temperatures where enzymes are most active they are able to catalyze
reactions faster than the majority of artificial catalysts.
 When the reaction temperature is increased, however, solid catalysts may
become as active as enzymes.
 Unfortunately, enzyme activity does not increase continuously as the
temperature is raised.
 Instead, the enzyme usually loses activity at quite a low temperature,
often only slightly above that at which it is typically found.
9
Continued…
Table 4.3: Some turnover numbers for enzyme- and solid-catalyzed reactions
𝑁 = molecules per site per second
Enzyme
Reaction
Range of reported
N values at 0 − 37 °𝐶
Ribonuclease
Transfer phosphate of a
polynucleotide
2 − 2 × 103
Trypsin
Hydrolysis of peptides
3 × 10−3 − 1 × 102
Papain
Hydrolysis of peptides
8 × 10−2 − 1 × 101
Bromelain
Hydrolysis of peptides
4 × 10−3 − 5 × 10−1
Carbonic anhydrase
Reversible hydration of
carbonyl compounds
8 × 10−1 − 6 × 105
Fumarate hydratase
L-Malate ⇌ fumarate + H2 O
1 × 103 (forward)
3 × 103 (reverse)
10
Continued…
Table 4.3: continue
Solid Catalyst
Reaction
N
Temp.,℃
SiO2 − Al2 O3
(𝑖𝑚𝑝𝑟𝑒𝑔𝑛𝑎𝑡𝑒𝑑)
Cumene cracking
3 × 10−8
2 × 104
25
420
Decationized zeolite
Cumene cracking
~ 103
~108
25
325
V2 O3
Cyclohexene dehydrogenation
7 × 10−11
102
25
350
Treated Cu2 Au
HCO2 H dehydrogenation
2 × 107
3 × 1010
25
327
AlCl2 − Al2 O3
n-Hexane isomerization
(liquid)
1 × 10−2
1.5 × 10−2
25
60
11
4.1 Simple Enzyme Kinetics With One and
Two Substrates
 Our objective in this section is to develop suitable mathematical
expressions for the rates of enzyme-catalyzed reactions.
 Naturally, the crucial test of a reaction rate equation is comparison with
experimentally determined rates.
Let us be quite clear on what we mean by a reaction rate. If the reaction is
S
P
the reaction rate u, in the quasi-steady state approximation, is defined by
𝑑𝑠 𝑑𝑝
𝑣=−
=
𝑑𝑡 𝑑𝑡
(4.1)
where lowercase letters denote molar concentration.
The dimensions of u are consequently moles per unit volume per unit time.
12
Continued…
 The reaction rate is an intensive quantity, defined at each point in the
reaction mixture.
 Consequently, the rate will vary with position in a mixture if
concentrations or other intensive state variables change from point to
point.
 At time zero, solutions of substrate and of an appropriate purified
enzyme are mixed in a well stirred, closed isothermal vessel containing a
buffer solution to control pH.
Since the reaction conditions including enzyme and substrate concentrations
are known best at time zero, the initial slope of the substrate or product
concentration vs. time curve is estimated from the data:
𝑑𝑝
𝑑𝑠
= −
= 𝑣
𝑑𝑡 𝑡 = 0 𝑑𝑡 𝑡 = 0
𝑡=0
where e = eo, s = 𝑠𝑜 , and p = 0
(4.2)
13
Dextrose produced, mg/ml
Continued…
Reaction time, min
Figure 4.2: Batch reactor data for hydrolysis of 30 % starch by glucoamylase
14
Continued…
 Figure 4.2 shows actual data for such an experiment.
 Notice that some product is present initially, indicating that the reported
zero reaction time is not the actual time when the reaction started.
 The problem of reproducible enzyme activity is an important one which
also has serious implications in design of reactors employing isolated
enzymes.
 Since we have already mentioned that proteins may denature when
removed from their native biological surroundings, it should not surprise
us that an isolated enzyme in a "strange" aqueous environment can
gradually lose its catalytic activity (Fig. 4.3).
 Although this gradual deactivation is known to occur for many enzymes,
the phenomenon is scarcely mentioned in many treatments of enzyme
kinetics
15
Relative activity, %
Continued…
Treated time, min
Figure 4.3: Loss of activity in solution. At 37°C, aspartase in
solution loses more than 5 percent of its natural activity within 1 h.
16
4.1.1 Michaelis-Menten Kinetics
 Now let us suppose that armed with a good set of experimental rate data
(Fig 4.4) we face the task of representing it mathematically.
 From information of the type shown in Fig. 4.4, the following qualitative
features often emerge:
1. The rate of reaction is first order in concentration of substrate at
relatively low values of concentration. [Recall that if 𝑣 = (𝑐𝑜𝑛𝑠𝑡)(𝑠 𝑛 ),
n is the order of the reaction rate.]
2. As the substrate concentration is continually increased, the reaction
order in substrate diminishes continuously from one to zero.
3. The rate of reaction is proportional to the total amount of enzyme
present.
 Henri observed this behavior, and in 1902 proposed the rate equation
𝑣𝑚𝑎𝑥 𝑠
𝑣=
𝐾𝑚 + 𝑠
𝑤ℎ𝑒𝑟𝑒 𝑣𝑚𝑎𝑥 = 𝛼𝑒𝑜
(4.3)
 which exhibits all three features listed above. Notice that 𝑣 = 𝑣𝑚𝑎𝑥 /2 when
s is equal to 𝐾𝑚 .
17
Continued…
(a)
(b)
Figure 4.4: Kinetic data for enzymecatalyzed reaction. (a) Enzyme concentration
is held constant when studying substrate
concentration dependence; (b) the converse
holds for investigation on the influence of
enzyme concentration.
18
Continued…
Although Henri provided atheoretical explanation using a hypothesized reaction mechanism, his
derivation and the similar one offered in 1913 by Michaelis and Menten are now recognized as
Michaelis and Menten equation (not rigorous in general, however, useful in the derivation of
more complicated kinetic models .
As a starting point, it is assumed that the enzyme E and substrate S combine
to form a complex ES, which then dissociates into product P and free (or
uncombined) enzyme E:
S+E
ES
𝑘1
⇀
↽
𝑘−1
𝑘2
ES
P+E
(4.4a)
(4.4b)
This mechanism includes the intermediate complex discussed above, as well
as regeneration of catalyst in its original form upon completion of the
reaction sequence.
While perhaps considerably oversimplified, Eqs. (4.4) are certainly
reasonable.
19
Continued…
Henri and Michaelis and Menten assumed that reaction (4.4a) is in
equilibrium which, in conjunction with the mass-action law for the kinetics
of molecular events, gives 𝑣 = − 𝑑𝑠 = k1se-k-1(es) = 0
Or (es) = k se/k
1
𝑑𝑡
𝑠𝑒
𝑘−1
=
= 𝐾𝑚 = dissociation constant
(𝑒𝑠)
𝑘1
-1
(4.5)
Here, s, e, and (es) denote the concentrations of S, E and ES respectively.
Decomposition of the complex to product and free enzyme is assumed
irreversible.
The rate limiting step is decomposition of (es)
𝑣=−
𝑑𝑠
= 𝑘2 (𝑒𝑠)
𝑑𝑡
(4.6)
Since all enzyme present is either free or complexed, we also have
𝑒 + (𝑒𝑠) = 𝑒𝑜
where e, is the total concentration of enzyme in the system.
e + k1se/k-1 = e0
Or e = eo/(1+ k1s/k-1)
(4.7)
𝑣 = 𝑘2 (𝑒𝑠)
𝑣 = 𝑘2 k1se/k−1
20
Continued…
𝑣 = (𝑘2 k1s/k−1 ) ∗eo/(1+ k1s/k−1)
𝑣𝑚𝑎𝑥 𝑠
𝑣=
𝐾𝑚 + 𝑠
𝑣 = 𝑘2 eo s/ (k−1/k1+ s)
𝑣𝑚𝑎𝑥 = 𝑘2 eo
𝐾𝑚 = k−1/k1
 This is known from the amount of enzyme initially charged into the
reactor. Equation (4.3) with 𝑣𝑚𝑎𝑥 equal to 𝑘2 𝑒0 can now be obtained by
eliminating (es) and e from the three previous equations.
 We should note here that a reaction described by Eq. (4.3) is commonly
referred to as having Michaelis-Menten kinetics.
For reaction in a well-mixed closed vessel we can write the following mass
balances for substrate and the ES complex: (Briggs Haldane approach)
𝑣=−
𝑑𝑠
= 𝑘1 𝑠𝑒 − 𝑘−1 (𝑒𝑠)
𝑑𝑡
𝑎𝑛𝑑
𝑑(𝑒𝑠)
= 𝑘1 𝑠𝑒 − (𝑘−1 + 𝑘2 )(𝑒𝑠)
𝑑𝑡
Using Eq. (4.7) in the previous two equations gives a closed set: two
simultaneous ordinary differential equations in two unknowns, s and (es).
The appropriate initial conditions are, of course,
21
Continued…
𝑠(0) = 𝑠𝑜
(𝑒𝑠)(0) = 0
These equations cannot be solved analytically but they can be readily
integrated on a computer to find the concentrations of S, E, ES, and P as
functions of time. In calculated results illustrated in Fig. 4.5, it is evident
that to a good approximation
𝑑(𝑒𝑠)
=0
𝑑𝑡
(4.8)
Proceeding from Eq. (4.8). as Briggs and Haldane did, it is possible to use
Eq. (4.7) to eliminate e and (es) from the problem, leaving
𝑑𝑠
𝑣=−
=
𝑑𝑡
𝑘2 𝑒0 𝑠
𝑘
𝑘−1 + 2 + 𝑠
𝑘1
(4.9)
22
Continued…
Consequently, the Michaelis-Menten from results, where Eq.(4.9) shows
now
(4.10)
𝑣𝑚𝑎𝑥 = 𝑘2 𝑒𝑜
𝐾𝑚 =
𝑘−1 + 𝑘2
𝑘1
(4.11)
Notice that 𝐾𝑚 no longer can be interpreted physically as a dissociation
constant.
With the Michaelis-Menten rate expression at hand, the time course of
the reaction can now be determined analytically by integrating
𝑑𝑠 −𝑣𝑚𝑎𝑥 𝑠
=
𝑑𝑡
𝐾𝑚 + 𝑠
𝑤𝑖𝑡ℎ 𝑠(0) = 𝑠𝑜
(4.12)
to obtain
𝑠0
𝑣𝑚𝑎𝑥 𝑡 = 𝑠0 − 𝑠 + 𝑘𝑚 ln
𝑠
(4.13)
23
Continued…
Figure 4.5: The time course of the reaction
24
𝑠/𝑠0
Continued…
𝑇𝑖𝑚𝑒, 𝑡 × 102 𝑠
Figure 4.6: Computed time course of batch hydrolysis of acetyl L-phenylalanine either by
chymotrypsin. Considerable discrepancies between the exact solution and the quasi-steadystate solution arise when eo/so = (𝛼) is not sufficiently small.
25
Continued…
Figure 4.7: These data on esterase activity show deviations from
Michalis-Menten kinetics at large values of initial enzyme content.
26
Continued…
Naturally this equation is most easily exploited indirectly by computing the
reaction time t required to reach various substrate concentrations, rather than
vice versa.
 As Fig. 4.6 reveals, the deviation is significant when the total enzyme
concentration approaches 𝑠0 .
 We show the dependence of esterase activity on enzyme concentration in
Fig. 4.7.
 The linear dependence predicted by the Michaelis-Menten model is
followed initially but does not hold for large enzyme concentrations.
Remember that the slope at the linear portion is equal to 𝑘2 𝑠/(𝐾𝑚 + 𝑠).
27
4.1.2 Evaluation of Parameters in the
Michaelis-Menten Equation
 The Michaelis-Menten equation in its original form [Eq. (4.3)] is not well
suited for estimation of the kinetic parameters 𝑈𝑚𝑎𝑥 and Km. As Fig. 4.4
shows, it is quite difficult to estimate 𝑣𝑚𝑎𝑥 accurately from a plot of 𝑣
vs. s. By rearranging Eq. (4.3) we can derive the following options for
data plotting and graphical parameter evaluation:
1
1
𝐾𝑚 1
=
+
𝑣 𝑣𝑚𝑎𝑥 𝑣𝑚𝑎𝑥 𝑠
𝑠
𝐾𝑚
1
=
+
𝑠
𝑣 𝑣𝑚𝑎𝑥 𝑣𝑚𝑎𝑥
𝑣
𝑣 = 𝑣𝑚𝑎𝑥 − 𝐾𝑚
𝑠
(4.14)
(4.15)
(4.16)
 Each equation suggests an appropriate linear plot.
 In evaluation of the kinetic parameters of the model using such plots,
however, several points should be noted.
 Plotting Eq. (4.14) as 1/v vs.1/𝑠 (known as a Lineweaver-Burk plot)
cleanly separates dependent and independent variables (Fig. 4.8a).
28
Continued…
(a)
(b)
Figure 4.8: (a) Lineweaver-Burk plot of experimental data for pepsin. (b) Eadie-Hofstee
plot of data for hydrolysis of methyl hydrocinnamte catalyzed by chymotrypsin
29
Continued…
 The third method uses the Eadie-Hofstee plot: 𝑣 is graphed against 𝑣/𝑠
(Fig. 4.8b).
 Both coordinates contain the measured variable v, which is subject to the
largest errors.
 These considerations suggest the following strategy for evaluating 𝑣𝑚𝑎𝑥
and 𝐾𝑚 : determine 𝑣𝑚𝑎𝑥 from a plot of Eq. (4.14) (find intercept
accurately) or Eq. (4.15) (find slope accurately).
 Then return to a graph of v vs. s and find 𝑠1/2, the substrate concentration
where v is equal to 𝑣𝑚𝑎𝑥 /2.
 Recalling the comment following Eq. (4.3), we see that Km is equal in
magnitude and dimension to 𝑠1/2.
 It is important to have a feel for the magnitudes of these kinetic
parameters Table 4.4 shows the range of values encountered for several
different enzymes. While 𝑘2 varies enormously, note that 𝐾𝑚 has typical
values of 2 − 10 × 10−3 𝑀.
30
Table 4.4: Kinetic parameters for several enzymes
Enzyme
Substrate
Temp,
℃
pH
𝒌𝟐 , 𝒔−𝟏
1/𝑲𝒎 ,
𝑴−𝟏
Pepsin
Carbobenzoxy-L-glutamyl-L-tyrosine
ethyl ester
31.6
4.0
0.00108
530
Carbobenzoxy-L-glutamyl-L-tyrosine
31.6
4.0
0.00141
560
Benzoyl-L-argininamide
25.5
7.8
27.0
480
Chymotrypsinogen
19.6
7.5
2900
<770
Sturin
24.5
7.5
13100
400
Benzoyl-L-arginine ester
25.0
8.0
26.7
12500
Methyl hydrocinnamate
25.0
7.8
0.026
256
Methyl dl-𝛼-chloro-𝛽phenylpropionate
25.0
7.8
0.135
83.3
Methyl d-𝛽-phenyllactate
25.0
7.8
0.139
28.6
Methyl-l-𝛽-phenyllactate
25.0
7.8
1.38
100
Benzoyl-L-phenylalanine methyl ester
25.0
7.8
51.0
217
Acetyl-L-tryptophan ethyl ester
25.0
7.8
30.7
588
Benzoyl-L-o-nitrotyrosine ethyl ester
25.0
7.8
3.27
31.2
Benzoyl-L-tyrosine ethyl ester
25.0
7.8
78.0
90.9
tyrosine
Chymotrypsin
Table 4.4: continued
Benzoyl-L-phenylalanine ethyl
ester
25.0
7.8
37.4
167
Benzoyl-L-methionine ethyl
ester
25.0
7.8
0.77
1250
Benzoyl-L-tyrosinamide
25.0
7.8
0.625
23.8
Acetyl-L-tyrosinamide
25.0
7.8
0.279
12.3
Carbobenzoxyglycyl-L-
25.0
7.5
89
196
tryptophan
25.0
8.2
94
164
Carbobenzoxyglycyl-Lphenylalanine
25.0
7.5
181
154
Carbobenzoxyglycyl-L leucine
25.0
7.5
10.6
37
Adenosine
triphosphatase
ATP
25.0
7.0
104
79000
Urease
Urea
20.8
7.1
20000
250
20.8
8.0
30800
256
Carboxypeptida
se
4.1.3 Kinetics for Reversible Reactions, TwoSubstrate Reactions, and Cofactor Activation
 In the simplest model for reversible enzyme kinetics, we begin with the
model reaction sequence
𝑘1
S+E
ES
(4.17a)
⇀
↽
ES ⇀
↽
𝑘−1
𝑘2
𝑘−2
P+E
(4.17b)
which is the same as the Henri/Michaelis-Menten sequence in Eq. (4.4)
except that now formation of ES complex from product combination with
the free enzyme E is included in Eq. (4.17b).
Writing the material balances on S, ES, and P in a closed, well-mixed vessel
as before, taking advantage of the mass balance on different enzyme forms
in Eq. (4.7), and invoking the quasi-steady-state approximation Eq. (4.8), we
readily obtain
𝑣𝑝
(𝑣𝑠 /𝐾𝑠 )𝑠 − 𝐾 𝑝
𝑑𝑠 𝑑𝑝
𝑝
𝑣=−
=
=
𝑠
𝑑𝑡 𝑑𝑡
1 + 𝐾 + 𝑝/𝐾𝑝
𝑠
(4.18)
33
Continued…
𝑒 + (𝑒𝑠) = 𝑒𝑜
(4.7)
𝑑(𝑒𝑠)
=0
𝑑𝑡
(4.8)
Here, 𝑣𝑠 and 𝐾𝑠 are identical to 𝑣𝑚𝑎𝑥 and 𝐾𝑚 in Eqs. (4.10) and (4.11),
respectively, and
𝑘−1 𝑣𝑠
𝑣𝑝 =
𝑘2
𝑣𝑠
𝐾𝑠
𝑣𝑚𝑎𝑥 = 𝑘2 𝑒𝑜
𝑘−1 + 𝑘2
𝐾𝑚 =
𝑘1
𝑘1 𝐾𝑠
𝐾𝑝 =
𝑘−2
(4.19)
(4.10)
(4.11)
34
Michaelis-Menten approach:
Reversible Reactions
Alternatively
v= rp
CE= e
Cs= s
CES= es
CEo = eo
Cp = p
Assuming the first reversible reaction is in equilibrium gives
By substitution
And
Briggs-Haldane approach:
Reversible Reactions
Assume that the change of the complex concentration with time, dCES/dt, is
negligible. Then,
We
have
Inserting the above equation in
By substitution
We get
If the first step of the reaction, the complex formation
step, is much faster than the second, the product
formation step, and k2 will be much larger than k3 and k4.
Therefore,
And the final expression
is simplified to Michaelis
Menten Equation
Two-Substrate Reactions
In many two-substrate reactions it appears that a ternary complex
may be formed with both substrates attached to the enzyme. One
possible sequence in this situation is
Dissociation equilibrium
constant
E + 𝑆1
E + 𝑆2
E𝑆1 + 𝑆2
E𝑆2 + 𝑆1
E𝑆1 𝑆2
⇀
↽
⇀
↽
⇀
↽
⇀
↽
𝑘
E𝑆1
𝐾1
E𝑆2
𝐾2
E𝑆1 𝑆2
𝐾12
E𝑆1 𝑆2
𝐾21
𝑃+𝐸
(4.20)
eo = e + es1 + es2 + es1s2
37
Continued…
so that
(4.21)
𝑣 = 𝑘(𝑒𝑠1 𝑠2 )
Assuming equilibria in the first four reactions in (4.20) leads to
𝑣=
𝑘𝑒0
1
𝐾21 𝐾12 2 (𝐾2 𝐾21 + 𝐾1 𝐾12 )
1+ 𝑠 + 𝑠 +
𝑠 𝑠
1
2
(4.22)
1 2
Equation (4.22) can be shortened slightly by noting that equilibrium requires
The ratio of the affinity of
substrate molecules are
same
𝐾1 𝐾12 = 𝐾2 𝐾21
(4.23)
38
Continued…
Rearrangement of (4.22) results in the now familiar form
∗ 𝑠
𝑣𝑚𝑎𝑥
1
𝑣= ∗
𝐾1 + 𝑠1
(4.24)
where
∗ =
𝑣𝑚𝑎𝑥
𝑘𝑒0 𝑠2
𝑠2 + 𝐾12
𝐾21 𝑠2 + 𝐾1 𝐾12
∗
𝐾1 =
𝑠2 + 𝐾12
(4.25)
(4.26)
From the previous three equations it is apparent that if 𝑠2 , is held constant
and is varied, the reaction will follow Michaelis-Menten kinetics.
However, Eqs. (4.25) and (4.26) show that the apparent maximum rate and
Michaelis constant are functions of 𝑠2 , concentration.
39
Continued…
 Assuming that the above sequence and the rate expression for a twosubstrate reaction are correct, we can verify the original MichaelisMenten form (4.3) when one substrate is in great excess.
∗
∗
 In this case, 𝑣𝑚𝑎𝑥 becomes = 𝑘𝑒0 , while 𝐾1 approaches the constant
value 𝐾21 . Therefore, the two-substrate reaction can be treated as though
𝑠1 were the only substrate when 𝑠2 ≫ 𝐾12 .
Lesser amount substrate has more control on rate of multi substrate
biochemical reaction
40
Cofactor Activation
For two substrate reactions
𝑘𝑒0
𝑣=
1
𝐾21 𝐾12 2 (𝐾2 𝐾21 + 𝐾1 𝐾12 )
1+
+
+
𝑠1
𝑠2
𝑠1 𝑠2
(4.27)
If s1 = substrate concentration, s and s2 = cofactor concentration, c
Assuming that substrate complexes only with the apoenzyme-coenzyme complex, the final
result is
𝑘𝑒0 𝑐𝑠
𝑣=
𝑐𝑠 + 𝐾𝑠 (𝑐 + 𝐾𝑐 )
If substrate concentration s is considered fixed, this rate expression shows a MichaelisMenten dependence on cofactor concentration c.
41
Continued…
Figure 4.9: Schematic illustration of a plausible mechanism for
enzyme requiring a co-factor.
42
Continued…
 Thus, if there is little cofactor present (𝑐 ≪ 𝐾), the reaction velocity is
first order in c.
 On the other hand, for 𝑐 ≫ 𝐾𝑐 ,, the single-substrate equation is
recovered, and the rate is independent of cofactor concentration.
43
4.2 Determination of Elementary Step Rate
Constants
 we explored convenient plotting techniques for determining the
parameters 𝑣𝑚𝑎𝑥 and 𝐾𝑚 which appear in the Michaelis-Menten rate
equation.
 As the Briggs-Haldane derivation reveals, however, these parameters
depend on the elementary step rate constants 𝑘1 , 𝑘−1 and 𝑘2 .
 Examination of the specific relationships in Eqs. (4.10) and (4.11) shows
that knowledge of 𝐾𝑚 and 𝑣𝑚𝑎𝑥 is not sufficient to determine the
elementary rate constants.
 An additional, independent relationship between the elementary step rate
constants and experimental data is required.
 These more fundamental kinetic parameters are of interest for at least
two reasons:
1. They provide a better picture of what occurs during the process of
enzyme catalysis.
44
Continued…
2. We saw earlier that the simplification leading to the Michaelis-Menten
equation is rigorously justified only when the total enzyme
concentration is relatively small. In cases where this is not true, a
careful treatment of enzyme reaction kinetics would involve integration
of the mass balances for s, p, and (es).
 In the pre-steady-state kinetics method, attention is focused on the short
period (~1 𝑠 or less) just after the reaction is initiated when the quasisteady-state approximation does not apply.
 The stopped-flow method perfected by Britton Chance and colleagues is
the central experimental component of this method.
 After rapid mixing of substrate and enzyme solutions in an absorption
cell to initiate the reaction, product composition changes on a time scale
down to milliseconds are monitored by a spectrophotometer.
45
Continued…
 Relaxation methods, developed and widely applied by Eigen and coworkers, can explore reactions which occur on a time scale of
nanoseconds.
 Although there are several variations of the method, all are based on the
same principles: an external perturbation of some reaction condition such
as temperature, pressure, or electric field density is applied to a reaction
mixture at equilibrium (or steady state).
46
4.2.1 Relaxation Kinetics
 Step change relaxation
Let us consider for the moment only the equilibrium between substrate,
enzyme and enzyme-substrate complex:
E+S
⇀
↽
𝑘1
𝑘−1
ES
For this reaction, we know that
𝑠 + (𝑒𝑠) = 𝑠0
and
𝑒 + (𝑒𝑠) = 𝑒0
so that in a well-mixed batch reactor, the mass balance for s
𝑑𝑠
= 𝑘−1 𝑒𝑠 − 𝑘1 𝑠𝑒
𝑑𝑡
(4.28)
47
Continued…
Figure 4.10: In a relaxation experiment, a small step change in reaction
conditions, e.g., in temperature, cause a transient approach to a new steady state.
48
Continued…
becomes
𝑑𝑠
= 𝑘−1 𝑠0 − 𝑠 − 𝑘1 𝑠(𝑒0 + 𝑠 − 𝑠0 ) ≡ 𝑓(𝑠)
𝑑𝑡
(4.29)
If we let 𝑠∗ denote the s concentration at equilibrium after the step change in
reaction conditions, s can be determined
by solving
∗
𝑓(𝑠∗) = 0
(4.30)
where 𝑓(𝑠) is the right-hand side of Eq, (4.29) and is equal to 𝑑𝑠/𝑑𝑡. In any
relaxation experiment s will be close to s ∗, and so we can approximate 𝑓(𝑠)
by the first terms in its Taylor expansion
zero because of Eq, (4.30)
𝑑𝑓 𝑠
∗
𝑓 𝑠 ≈𝑓 𝑠 +
𝑑𝑠
∗
𝑠 − 𝑠∗ + terms of order 𝑠 − 𝑠∗
2
and higher (4.31)
Letting x be the deviation from the equilibrium concentration
𝜒 = 𝑠 − 𝑠∗
(4.32)
49
Continued…
and remembering that s∗ is time-invariant, we can combine Eqs. (4.29) to
(4.32) to obtain the linearized mass balance [squared and higher-order terms
neglected in (4.31)]
𝑑𝜒
= −[𝑘−1 −𝑘1 (𝑒0 − 𝑠0 + 2𝑠∗)]𝜒
𝑑𝑡
(4.33)
If the system is initially at the old equilibrium corresponding to conditions
before the step perturbation
(4.34)
𝜒(0) = ∆𝜒0
and consequently
where
𝜒(𝑡) = ∆𝜒0 𝑒 −𝑡/𝜏
1
= 𝑘−1 + 𝑘1 𝑒0 − 𝑠0 + 2𝑠
𝜏
∗ + 𝑘 (𝑒 ∗ + 𝑠 ∗ )
= 𝑘−1
1
(4.35)
(4.36)
In Eq. (4.36), 𝑒 ∗ represents the concentration of uncomplexed enzyme at the
final equilibrium state.
50
Continued…
From Eq. 4.35, it is clear that 𝜏 can be determined from the slope of a
semilog plot of 𝜒(𝑡)/Δ𝜒0 .
Alternatively, 𝜏 is the time required for the deviation to decay 𝜒(𝑡) to 37
percent of its initial value. After 𝜏, 𝜒 ∗ , and 𝑒 ∗ (or 𝑒0 , and 𝑠0 ) have been
measured, Eq. (4.36) provides a relationship between 𝑘1 and 𝑘−1 which is
independent of the equilibrium equation. Consequently, both 𝑘1 , and 𝑘−2
can be calculated.
51
Substrate activation and inhibition
Substrate activation
At low substrate concentration the bonding of one substrate molecule enhances the binding of
the next one (dv/ds increases), it is assumed that the enzyme has multiple binding sites for
substrate and that the first substrate molecule bound to the enzyme alters the enzyme’s
structure so that the remaining sites have a stronger affinity for the substrate.
Substrate activation
Sometimes when a large amount of substrate is present the enzyme catalyzed reaction is
diminished by the excess substrate. s is larger that the s ( v = vmax).
52
53
54
4.3 Modulation and Regulation of Enzymatic
Activity
 Chemical species other than the substrate can combine with enzymes to
alter or modulate their catalytic activity.
 Such substances, called modulators or effectors, may be normal
constituents of the cell.
 In other cases they enter from the cell's environment or act on isolated
enzymes.
 Although most of our attention in this section will be concentrated on
inhibition, where the modulator decreases activity, cases of enzyme
activation by effectors are also known.
 The combination of an enzyme with an effector is itself a chemical
reaction and may therefore be fully reversible, partially reversible, or
essentially irreversible.
55
Continued…
 Known examples of irreversible inhibitors include poisons such as
cyanide ions, which deactivate xanthine oxidase, and a group of
chemicals collectively termed nerve gases, which deactivate
cholinesterases (enzymes which are an integral part of nerve transmission
and thus of motor ability).
 Several different combinations and variations on the two basic types of
reversible inhibitors are known.
 Some of these are listed in Table 4.6.
56
Continued…
Table 4.6: Partial classification of reversible inhibitors
Type
Description
Result
Ia
Fully competitive
Inhibitor adsorbs at substrate binding site
Increase in apparent
value of 𝐾𝑚
b
Partially
competitive
Inhibitor and substrate combine with
different groups; inhibitor binding affects
substrate binding
Increase in apparent
value of 𝐾𝑚
IIa
Noncompetitive
Inhibitor binding does not affect ES
affinity, but ternary
EIS (enzyme-inhibitor-substrate)
complex does not decompose into
products
No change in 𝐾𝑚 ,
Decrease of 𝑣𝑚𝑎𝑥
b
Noncompetitive
Same as IIa except that EIS decomposes
into product at a finite rate different from
that of ES
No change in 𝐾𝑚 ,
Decrease of 𝑣𝑚𝑎𝑥
III
Mixed inhibitor
Affects both 𝐾𝑚 ,
and 𝑣𝑚𝑎𝑥
57
Reversible modulation of enzyme activity is one control mechanism employed by
the cell to achieve efficient use of nutrients.
For present purposes, we shall employ control of a single sequence of reactions as
an example.
Figure 4.11 shows a five-step sequence for the biosynthesis of the amino acid Lisoleucine.
Regulation of this sequence is achieved by feedback inhibition: the final product,
L-isoleucine, inhibits the activity of the first enzyme in the path.
58
Continued…
Figure 4.11: In this feedback-inhibition system, the activity of enzyme 𝐸1 is reduced
by the presence of the end product, L-isoleucine.
59
4.3.1 The Mechanisms of Reversible Enzyme
Modulation
 Many known competitive inhibitors, called substrate analogs, bear close
relationships to the normal substrates.
 Thus it is thought that these inhibitors have the key to fit into the enzyme
active site, or lock, but that the key is not right so the lock does not work;
i.e., no chemical reaction results.
 For example, consider the inhibition of succinic acid dehydrogenation by
malonic acid:
COOH
COOH
CH2
CH2
Catalyzed by
Succinic dehydrogenase
Inhibited competitively
By malonic acid
COOH
Succinic acid
CH
CH
COOH
Fumaric acid
COOH
CH2
COOH
Malonic
acid
60
Continued…
 The malonic acid can complex with succinic dehydrogenase, but it does
not react.
 The action of one of the sulfadrugs, sulfanilamide, is due to its effect as a
competitive inhibitor.
 Sulfanilamide is very similar in structure to p-aminobenzoic acid, an
important vitamin for many bacteria.
 By inhibiting the enzyme which causes p-aminobenzoic acid to react to
give folic acid, the sulfadrug can block the biochemical machinery of the
bacterium and kill it.
HO
O
C
NH2
p-Aminobenzoic acid
H2 N
O
O
C
NH2
Sulfanilamide
61
Continued…
 Another mechanism, called allosteric control, yields behavior typical of
non competitive inhibition and is thought to be the dominant mechanism
for non competitive inhibition and activation.
 Allosteric control may either inhibit (reduce) or activate (increase) the
catalytic ability of the enzyme.
 One schematic view of this process is depicted in Fig. 4.12.
 Experimental studies on one oligomeric allosteric enzyme, aspartyl
transcarbamoylase (ATCase), have provided the most dramatic evidence
to date in support of the allostery theory of enzyme activity control.
 The enzyme was physically separated into two subunits.
 It was found that the larger subunit, which is catalytically active, is not
influenced by CTP (cytidine triphosphate), an inhibitor of the intact
ATCase enzyme.
 The smaller subunit, on the other hand, has no catalytic activity but does
bind CTP.
62
Continued…
" Symmetry principle"
excludes a mixed T-R
state
Figure 4.12: In the symmetry model for allosteric control of enzyme activity, binding of an
activator A and substrate S gives a catalytically active R enzyme configuration, while binding of
an inhibitor changes all subunits in the oligomeric protein molecule to an inactive T state.
63
4.3.2 Analysis of Reversible Modulator
Effects on Enzyme Kinetics
 A major contribution of the Michaelis-Menten approach to enzyme
kinetics is accounting quantitatively for the influence of modulators.
 To begin our analysis, we shall assume that the following sequence
reasonably approximates the interactions of a (totally) competitive
inhibitor and substrate with enzyme.
Here E and S have the usual significance, and I denotes the inhibitor.
Also, EI is an enzyme inhibitor complex.
Dissociation
Reaction steps at equilibrium:
contant
E+S
+
I
ES
E +P
E+S
E+I
⇀
↽
⇀
↽
ES
𝐾𝑠
(4.37)
EI
𝐾𝑖
(4.38)
EI
64
Continued…
Slow step:
ES
𝑘
E+P
(4.39)
In this case binding of substrate and inhibitor to enzyme are mutually
exclusive.
Making use of the equilibrium relationships indicated in Eq. (4.36) and a
total mass balance on enzyme
𝑒 + 𝑒𝑠 + 𝑒𝑖 = 𝑒0
(4.40)
we can write the reaction rate in terms of total enzyme concentration e, and
free substrate and inhibitor concentrations (s and i, respectively):
𝑘𝑒0 𝑠
𝑣=
𝑠 + 𝐾𝑠 (1 + 𝑖/𝐾𝑖 )
𝑡𝑜𝑡𝑎𝑙𝑙𝑦 𝑐𝑜𝑚𝑝𝑒𝑡𝑖𝑡𝑖𝑣𝑒 𝑖𝑛ℎ𝑖𝑏𝑖𝑡𝑖𝑜𝑛 (4.41)
65
Continued…
Comparing this with the original Michaelis-Menten form shows that u, is
𝑎𝑝𝑝
unaffected, but the apparent Michaelis constant 𝐾m
, here
𝑖
𝑎𝑝𝑝
(4.42)
𝐾m = 𝐾𝑠 (1 + )
𝐾𝑖
is increased by the presence of the competitive inhibitor. Stated in different
terms, the rate reduction caused by a competitive inhibitor can be
completely offset by increasing the substrate concentration sufficiently; the
maximum possible reaction velocity is not affected by the competitive
inhibitor.
We can obtain a useful model for (totally) noncompetitive inhibition
by adding the following equilibrium steps to the reaction sequence :
EI + S
ES + I
⇀
↽
⇀
↽
EIS
Dissociation
𝐾 contant
EIS
𝐾𝑖
𝑠
(4.43)
(4.44)
66
Continued…
In this situation, inhibitor and substrate can simultaneously bind to the
enzyme, forming the ternary complex designated EIS.
We assume in this simplest non competitive inhibition model that binding of
either inhibitor or substrate does not influence the affinity of either species
to complex with the enzyme.
Also, it is assumed here that the EIS complex does not react to give product
P.
Using the now familiar procedure of writing the complex concentrations in
terms of 𝑒0 , s, and i, we obtain the following rate expression:
𝑘𝑒0 𝑠/(1 + 𝑖/𝐾𝑖 )
𝑣=
𝑠 + 𝐾𝑠
𝑡𝑜𝑡𝑎𝑙𝑙𝑦 𝑛𝑜𝑛𝑐𝑜𝑚𝑝𝑒𝑡𝑖𝑡𝑖𝑣𝑒 𝑖𝑛ℎ𝑖𝑏𝑖𝑡𝑖𝑜𝑛
(4.45)
67
Continued…
Now the Michaelis constant is unaffected, but the maximum reaction
velocity which can be obtained is reduced to
𝑘𝑒
app
(4.46)
𝑣𝑚𝑎𝑥 =
1 + 𝑖/𝐾𝑖
 In the presence of a noncompetitive inhibitor, no amount of substrate
addition to the reaction mixture can provide the maximum reaction rate
which is possible without the inhibitor.
 As an example, examine the data in Fig. 4.13 for starch hydrolysis.
 Comparing the data plots for different inhibitors with the no inhibitor
line, we see that a-dextrin is a competitive inhibitor, while
noncompetitive inhibition is caused by both maltose and limit dextrin.
 Already mentioned in Table 4.6 are the possibilities of other inhibitor
effects on enzyme kinetics.
68
Continued…
Figure 4.13: Competitive (a-dextrin) a
noncompetitive (maltose, limit dextrin)
inhibition of 𝛼-amylase.
69
Continued…
In Eqs. (4.36) and (4.37) the following reactions which are assumed to be in
Dissociation
equilibrium:
contant
(4.47)
EI + S
EIS
𝐾𝑖𝑠
⇀
↽
⇀
↽
ES + I
and the slow step:
k
EIS
EIS
EI + P
𝐾𝑠𝑖
(4.48)
(Note that it can be shown 𝐾𝑖𝑠 = 𝐾𝑠 𝐾𝑠𝑖 /𝐾𝑖 .)
In another noncompetitive inhibition situation (case IIb), the
noncompetitive inhibition model above is altered to take into account
formation of product from the EIS complex, although with a smaller rate
than ES complex reacts:
EIS
𝑘𝑖
EI + P
(4.49)
70
Continued…
Table 4.7: Intercepts of reciprocal (Lineweaver-Burk) plots in presence of inhibitor
Type
Ia
Purely competitive
Ib
Partially competitive
IIa
Purely noncompetitive
IIb
Partially noncompetitive
Ib and IIa
Mixed
1
𝑣𝑚𝑎𝑥
1
𝑣𝑚𝑎𝑥
1 + 𝑖/𝐾𝑖
𝑣𝑚𝑎𝑥
1 + 𝑖/𝐾𝑖
𝑣𝑚𝑎𝑥 + 𝑘 𝑖 𝑒0 /𝐾𝑖
1 + 𝑖𝐾𝑠 /𝐾𝑖 𝐾𝑠
𝑣𝑚𝑎𝑥
1
𝐾𝑠 (1 + 𝑖/𝐾𝑖 )
1 + 𝑖𝐾𝑠 /𝐾𝑖 𝐾𝑖𝑠
𝐾𝑠 (1 + 𝑖/𝐾𝑖 )
1
𝐾𝑠
1
𝐾𝑠
1 + 𝑖𝐾𝑠 /𝐾𝑖 𝐾𝑖𝑠
𝐾𝑠 (1 + 𝑖/𝐾𝑖 )
71
Continued…
 One mixed inhibitor model which combines pure noncompetitive with
partially competitive inhibition is obtained by deleting reaction (4.47)
from the partial competition model above.
 The results for all of these cases may be written in Michaelis-Menten
app
form
𝑣𝑚𝑎𝑥 𝑠
𝑣=
(4.50)
app
𝑠 + 𝐾𝑚
 with apparent parameters PP and KPP that depend on inhibitor
concentration and on various equilibrium and reaction rate parameters
(see Table 4.7).
 Figure 4.14 shows schematically how various types of inhibition
influence different forms of kinetic data plots.
72
Figure 4.14: Effects of various
types of inhibition as reflected in
different rate plots. Subscripts i
here denote the presence of
inhibitor.
cept
4.4 Other Influences on Enzyme Activity
 We have already explored how different chemical compounds which bind
with enzymes can influence the rate of enzyme-catalyzed reactions.
 Many other factors can influence the catalytic activity of enzymes,
presumably by affecting the enzyme's structural or chemical state.
Included among these factors are:
1. pH
2. Temperature
3. Fluid forces (hydrodynamic forces, hydrostatic pressure and interfacial
tension)
4. Chemical agents (such as alcohol, urea and hydrogen peroxide)
5. Irradiation (light, sound, ionizing radiation)
 Sometimes the change in catalytic activity caused by a shift in pH, for
example, is reversed by returning to original reaction conditions.
74
4.4.1 The Effect of pH on Enzyme Kinetics in
Solution
 These biochemical units possess basic, neutral, or acidic groups.
 For the appropriate acid or base catalysis to be possible the ionizable
groups in the active site must often each possess a particular charge; i.e.,
the catalytically active enzyme exists in only one particular ionization
state.
 Thus, the catalytically active enzyme may be a large or small fraction of
the total enzyme present, depending upon the pH.
 Figure 4.15 illustrates the influence of pH on several enzymes.
 In several cases shown there, catalytic activity of the enzyme passes
through a maximum (at the optimum pH) as pH is increased.
 We can obtain a useful form for representing such pH effects on enzyme
kinetics using the following simple model of the active site ionization
state:
−H +
−H +
(4.51)
−
2−
E
E
E
+
+
⇀ ⇀
↽
↽
+H
K1
+H
K2
75
Continued…
Figure 4.15: Enzyme activity is strongly pH dependent, and the range
of maximum activity differs greatly depending on the enzyme involved.
76
Continued…
In these acid-base reactions, E − denotes the active enzyme form while E and
E 2− are inactive forms obtained by protonation and deprotonation of the
active site of E, respectively. 𝐾1 and 𝐾2 are equilibrium constants for the
indicated reactions. Further ionizations away from the E − state of the
enzyme are not considered since the first ionization is assumed here to
completely eliminate enzyme activity.
After writing the equilibrium relations for the two ionizations
ℎ+ 𝑒 −
= 𝐾1
𝑒
ℎ+ 𝑒 2−
= 𝐾2
𝑒−
(4.52)
we can determine the fraction of total enzyme present which is active. When
we let 𝑒0 denote the total enzyme concentration,
𝑒0 = 𝑒 + 𝑒 − + 𝑒 2−
(4.53)
77
Continued…
 The active fraction 𝑦 − is 𝑒 − / 𝑒0 , and is given by
𝑦− =
1
ℎ+ 𝐾2
1+
+
𝐾1 ℎ+
(4.54)
 This is one of the Michaelis pH functions. The other two, y and 𝑦 2− ,
give the fraction of enzyme in the acid and base forms, respectively.
 The 𝑦 − function depends upon pH in a manner quite similar to the
enzyme activity-pH data shown in Fig. 4.15. There is a single maximum
of 𝑦 − with respect to pH which occurs at the value
+
ℎ𝑜𝑝𝑡𝑖𝑚𝑢𝑚
=
𝐾1 𝐾2
or
1
(pH)optimum = (𝑝𝐾1 − 𝑝𝐾2 )
2
(4.55)
 The y function declines smoothly and symmetrically as pH is varied
away from the optimum pH.
78
Continued…
Consequently, the influence on the maximum reaction velocity max is
obtained by replacing the total enzyme concentration e, with the total active
form concentration 𝑒0 𝑦 − :
𝑣𝑚𝑎𝑥 = 𝑘𝑒0 𝑦 − =
𝑘𝑒0
ℎ+ 𝐾2
1+𝐾 + +
ℎ
1
(4.56)
Using this relationship, enzyme activity data may be used to evaluate the
parameters K, and K₂. The pH of maximal activity is related to 𝐾1 , and K₂
by Eq. (4.54).
79
4.4.2 Enzyme Reaction Rates and
Temperature
In any study of chemical kinetics, a recurring theme is the Arrhenius form
relating temperature to a reaction-rate constant
𝑘=
Where
𝐸𝑎
−𝑅𝑇
𝐴𝑒
(4.57)
Ea = activaton energy
R = gas−law constant
A = frequency factor
T = absolute temperature
In an Arrhenius plot, log 𝑘 is graphed against 𝐼/𝑇 to give a straight line with
−𝐸
a slope of 𝑎 [assuming, of course, that Eq. (4.57) holds].
𝑅
The Arrhenius dependence on temperature is indeed satisfied for the rate
constants of many enzyme-catalyzed reactions, as exemplified by the data in
Fig. 4.16.
80
Continued…
Figure 4.16: Arrhenius plot for an enzyme catalyzed reaction
(myosin catalyzed hydrolysis of ATP)
81
Continued…
 It should be noted, however, that the temperature range of Fig. 4.16 is
quite limited.
 No temperatures significantly larger than the usual biological range were
considered.
 What would happen if we attempted to push the enzyme farther and try
for higher rates via higher temperatures?
 The result would in most cases be disastrous, as the example of Fig. 4.17
illustrates.
 Thermal deactivation of enzymes may be reversible, irreversible, or a
com bination of the two.
 A simple model of reversible thermal deactivation often suffices to
represent T-activity data for enzyme kinetics over a wide range of
temperatures.
82
Continued…
Figure 4.17: Arrhenius rate dependence breaks down at high temperatures, above
which enzyme deactivation predominates (H₂O₂ decomposition catalyzed by catalase).
83
Continued…
 In this approach we assume that the enzyme exists in inactive (1) and
active (a) forms in equilibrium:
𝐸𝑎
𝐸𝑖
with equilibrium constant
⇀
↽
𝑒𝑖
−∆𝐺𝑑
−∆𝐻𝑑
∆𝑆𝑑
= 𝐾𝑑 = exp
= exp(
)exp(
)
𝑒𝑎
𝑅𝑇
𝑅𝑇
𝑅𝑇
(4.58)
In Eq. (4.58) ∆Gd , ∆Hd , and ∆Sd denote the free energy, enthalpy, and
entropy of deactivation, respectively.
For such ∆Hd values, the enzyme deactivates almost totally over a range of
30 Celsius degrees.
Since all enzyme present is either in active or inactive forms,
𝑒𝑎 + 𝑒𝑖 = 𝑒0
(4.59)
84
Continued…
We may obtain, using Eq. (4.58)
𝑒𝑎 =
𝑒0
1 + 𝐾𝑑
(4.60)
According to transition state theory, the rate for reactions (4.4) at large
substrate concentration can be written as
𝑣𝑚𝑎𝑥 = 𝑒𝑎 . 𝑘(𝑇)
where
𝑘𝐵 𝑇 ∆𝑆 ∗ − 𝐸𝑎
𝑘(𝑇) = 𝛼(
)𝑒 𝑅 𝑒 𝑅𝑇
ℎ
(4.61)
(4.62)
𝑘𝐵 and h are Boltzmann's and Planck's constants, respectively, and 𝛼 is a
proportionality constant. Combining Eqs. (4.58) and (4.60) through (4.52),
we obtain,
𝐸
− 𝑎
𝛽𝑇𝑒 𝑅𝑇
𝑣𝑚𝑎𝑥 =
1+
∆𝑆𝑑
𝑒 𝑅
−∆𝐻𝑑
𝑒 𝑅
(4.63)
85
Continued…
where the overall proportionality factor 𝛽 includes 𝛼 , 𝐾𝐵 , h, 𝑒0 , and
exp(∆𝑆/𝑅).
Equation (4.63) is the desired relationship for representing behavior
like that shown in Fig. 4.17.
The solid curve drawn through those data points was in fact calculated from
Eq. (4.63) after estimating the parameters 𝛽 , E, ∆𝑆𝑑 , and ∆𝐻𝑑 , from the
data. The slope at large 1/T values is approximately −𝐸/𝑅 (the error is
equal to 𝑇 (𝐾), which is usually not significant).
The slope of the other straight line obtained at higher temperatures is
approximately equal to (∆𝐻𝑑 − 𝐸)/𝑅.
∆𝑆𝑑 , may be estimated after noting that, at the temperature T where log mas
is maximized,
𝐾𝑑 (𝑇𝑚𝑎𝑥 ) =
𝐸 + 𝑅𝑇𝑚𝑎𝑥
∆𝐻𝑑 − 𝐸 − 𝑅𝑇𝑚𝑎𝑥
(4.64)
86
Continued…
(This is obtained simply by setting 𝑑(log 𝑣)/𝑑𝑇 equal to zero). Since 𝑇𝑚𝑎𝑥
is known from the measurements and E and ∆𝐻𝑑 have already been
estimated, we can evaluate the right-hand side of Eq. (6.63). Now, knowing
𝐾𝑑 (𝑇𝑚𝑎𝑥 ) ,we return to Eq. (4.49) and calculate ∆𝑆𝑑 .
87
Continued…
Example 4.1: Rates of hydrolysis of 𝑝-nitrophenyl-𝛽-𝐷-glucopyranoside by
𝛽 -glucosidase, an irreversible unimolecular reaction, were measured at
several concentrations of the substrate. The initial reaction rates were
obtained as given in Table E4.1. Determine the kinetic parameters of this
enzyme reaction.
Table E4.1: Hydrolysis rate of 𝑝-nitrophenyl-𝛽-𝐷-glucopyranoside by 𝛽-glucosidase
Substrate concentration (gmol 𝑚−3 )
Reaction rate (gmol 𝑚−3 𝑠 −1 )
5.00
4.84 × 10−4
2.50
3.88 × 10−4
1.67
3.18 × 10−4
1.25
2.66 × 10−4
1.00
2.14 × 10−4
88
Continued…
Solution: The Lineweaver-Burk plot of the experimental data is shown in
Figure E4.1. The straight line was obtained by the method of least squares.
From the figure, the values of 𝐾𝑚 and 𝑉𝑚𝑎𝑥 were determined as
2.4 𝑔𝑚𝑜𝑙 𝑚−3 and 7.7 × 10 − 4𝑔𝑚𝑜𝑙 𝑚−3 𝑠 −1 , respectively.
Figure E4.1: Linewever-Burk plot for hydrolysis of substrate by 𝛽-glucosidase.
89
Continued…
Example 4.2: A substrate 𝐿-benzoyl arginine 𝑝-nitroanilide hydrochloride
was hydrolyzed by trypsin, with inhibitor concentrations of 0, 0.3, and 0.6
𝑚𝑚𝑜𝑙𝑙 −1 . The hydrolysis rates obtained are listed in Table E4.2. Determine
the inhibition mechanism and the kinetic parameters (𝐾𝑚 , 𝑉𝑚𝑎𝑥 and 𝐾𝐼 ) of
this enzyme reaction.
Table E4.2: Hydrolysis rate of L-benzoyl arginine p-nitroanilide hydrochloride by
trypsin with and without an inhibitor.
Inhibitor concentration (mmoll−1 )
Substrate concentration
(mmoll−1)
0
0.3
0.6
0.1
0.79
0.57
0.45
0.15
1.11
0.84
0.66
0.2
1.45
1.06
0.86
0.3
2.00
1.52
1.22
Hydrolysis rate, 𝜇mol 𝑙 −1 𝑠 −1 .
90
Continued…
Solution: The results given in Table E4.2 are plotted as shown in Figure
E4.2. This Lineweaver-Burk plot shows that the inhibition mechanism is
competitive inhibition. From the line for the data without the inhibitor, 𝐾𝑚 ,
and 𝑉𝑚𝑎𝑥 are obtained as 0.98 gmol𝑚−3 and 9.1 mmol 𝑚−3 𝑠 −1 respectively.
From the slopes of the lines, 𝐾𝐼 is evaluated as 0.6 gmol𝑚−3 .
Figure E4.2: Lineweaver-Burk
plot of hydrolysis reaction by
trypsin.
91
Immobilization of enzyme
92
93
94
95
96
97
98
99
100
101
Thank you
5. Microbial Fermentation Kinetics
 Cell kinetics deals with the rate of cell growth and how it is affected by
various chemical and physical conditions.
 During the course of growth, the heterogeneous mixture of young and old
cells is continuously changing and adapting itself in the media
environment which is also continuously changing in physical and
chemical conditions.
 As a result, accurate mathematical modeling of growth kinetics is
impossible to achieve.
 Even with such a realistic model, this approach is usually useless because
the model may contain many parameters which are impossible to
determine.
 Therefore, we must make assumptions to be able to arrive at simple
models which are
useful for fermenter design and performance
predictions.
1
Continued…
 Various models can be developed based on the assumptions concerning
cell 'components and population as shown in Table 5.1 (Tsuchiya et al.,
1966).
 The simplest model is the unstructured, distributed model which is based
on the following two assumptions:
1. Cells can be represented by a single component, such as cell mass, cell
number, or the concentration of protein, DNA, or RNA. This is true for
balanced growth, since a doubling of cell mass for balanced growth is
accompanied by a doubling of all other measurable properties of the cell
population.
2. The population of cellular mass is distributed uniformly throughout the
culture. The cell suspension can be regarded as a homogeneous solution.
The heterogeneous nature of cells can be ignored. The cell concentration
can be expressed as dry weight per unit volume.
Balanced-growth means "every extensive property of the growing system increases
by the same factor over a time interval".
2
Continued…
Table 5.1: Various Models for Cell Kinetics
Population
Cell Components
Unstructured
Structured
Distributed
Cells are represented by a
single component, which is
uniformly distributed
throughout the culture
Multiple cell components,
uniformly distributed
throughout the culture
interact with each others
Segregated
Cells are represented by a
single component, but they
form a heterogeneous
mixture
Cells are composed of
multiple components and
form a heterogeneous
mixture
3
Continued…
 In this text, the following nomenclature is adopted:
𝐶𝑋 Cell concentration, dry cell weight per unit volume
𝐶𝑁 Cell number density, number of cells per unit volume
𝜌 Cell density, wet cell weight per unit volume of cell mass
Accordingly, the growth rate can be defined in several different
ways, such as:
𝑑𝐶𝑋 /𝑑𝑡 Change of dry cell concentration with time
𝑟𝑋 Growth rate of cells on a dry weight basis
𝑑𝐶𝑁 /𝑑𝑡 Change of cell number density with time
rn Growth rate of cells on a number basis
𝛿 Division rate of cells on a number basis, 𝑑 log 2 𝐶𝑁 /𝑑𝑡.
 It appears that 𝑑𝐶𝑋 /𝑑𝑡 and 𝑟𝑋 are always the same, but this is not true.
4
Continued…
 The former is the change of the cell concentration in a fermenter, which
may include the effect of the input and output flow rates, cell recycling,
and other operating conditions of a fermenter.
 Sometimes, the growth rate can be confused with the division rate, which
is defined as the rate of cell division per unit time. If all of the cells in a
vessel at time 𝑡 = 0(𝐶𝑁 = 𝐶𝑁0 ) have divided once after a certain period
of time, the cell population will have increased to 𝐶𝑁0 × 2. If cells are
divided n times after the time t, the total number cells will be
𝐶𝑁 = 𝐶𝑁0 × 2𝑁
(5.1)
 and the average division rate is
𝛿=
𝑛
𝑡
(5.2)
 Since 𝑁 = log 2 𝐶𝑁 − log 2 𝐶𝑁0 according Eq. (5.1), the average division
rate is
5
Continued…
1
𝛿 = (𝑁 = log 2 𝐶𝑁 − log 2 𝐶𝑁0 )
𝑡
(5.3)
𝑑 log 2 𝐶𝑁
𝛿=
𝑑𝑡
(5.4)
and the division rate at time t is
 Therefore, the growth rate defined as the change of cell number with
time is the slope of the 𝐶𝑁 versus t curve, while the division rate is the
slope of the log 2 𝐶𝑁 versus curve.
 As explained later, the division rate is constant during the exponential
growth period, while the growth rate is not.
 Therefore, these two terms should not be confused with each other.
6
5.1 Growth Cycle for Batch Cultivation
 If you inoculate unicellular microorganisms into a fresh sterilized
medium and measure the cell number density with respect to time and
plot it, you may find that there are six phases of growth and death, as
shown in Figure 5.1. They are:
1. Lag phase: A period of time when the change of cell number is zero.
2. Accelerated growth phase: The cell number starts to increase and the
division rate increases to reach a maximum.
3. Exponential growth phase: The cell number increases exponentially as
the cells start to divide. The growth rate is increasing during this phase,
but the division rate which is proportional to 𝑑 𝑙𝑜𝑔 𝐶𝑁1 /𝑑𝑡, is constant at
its maximum value, as illustrated in Figure 5.1.
4. Decelerated growth phase: After the growth rate reaches a maximum, it
is followed by the deceleration of both growth rate and the division rate.
5. Stationary phase: The cell population will reach a maximum value and
will not increase any further.
7
Continued…
6. Death phase: After nutrients available for the cells are depleted, cells
will start to die, and the number of viable cells will decrease.
8
5.1.1 Lag phase
 The lag phase (or initial stationary, or latent) is an initial period of
cultivation during which the change of cell number is zero or negligible.
 Even though the cell number does not increase, the cells may grow in
size during this period.
 The length of this lag period depends on many factors such as the type
and age of the microorganisms, the size of the inoculum, and culture
conditions.
 The lag usually occurs because the cells must adjust to the new medium
before growth can begin.
 If microorganisms are inoculated from a medium with a low nutrient
concentration to a medium with a high concentration, the length of the
lag period is usually long.
 This is because the cells must produce the enzymes necessary for the
metabolization of the available nutrients.
9
Continued…
Figure 5.1: Typical growth
curve of unicellular
organisms: (A) lag phase; (B)
accelerated growth phase;
(C) exponential growth
phase; (D) decelerated
growth phase; (E) stationary
phase; (F) death phase.
10
Continued…
 If a small amount of cells are inoculated into a large volume, they will
have a long lag phase.
 If they are moved from a high to a low nutrient concentration, there is
usually no lag phase.
 Another important factor affecting the length of the lag phase is the size
of the inoculum.
 For large-scale operation of the cell culture, it is our objective to make
this lag phase as short as possible.
 Therefore, to inoculate a large fermenter, we need to have a series of
progressively larger seed tanks to minimize the effect of the lag phase.
 At the end of the lag phase, when growth begins, the division rate
increases gradually and reaches a maximum value in the exponential
growth period, as shown by the rising inflection at B in Figure 5.1.
 This transitional period is commonly called the accelerated growth phase
and is often included as a part of the lag phase.
11
5.1.2 Exponential growth Phase
 A bacterial culture undergoing balanced growth mimics a first-order
autocatalytic chemical reaction (Carberry, 1976; Levenspiel, 1972).
Therefore, the rate of the cell population increase at any particular time
is proportional to the number density of bacteria present at that time:
𝑟𝑁 =
𝐶𝑁
= 𝜇𝐶𝑁
𝑑𝑡
(5.5)
where the constant is known as the specific growth rate [hr −1 ].
 The specific growth rate should not be confused with the growth rate,
 The growth rate is the change of the cell number density with time, while
the specific growth rate is,
𝜇=
1 𝐶𝑁 𝑑 ln 𝐶𝑁
=
𝐶𝑁 𝑑𝑡
𝑑𝑡
(5.6)
12
Continued…
which is the change of the natural log of the cell number density
with time. Comparing Eq. (5.4) and Eq. (5.6) shows that
𝑑 ln 𝐶𝑁
𝑑 log 𝐶𝑁
𝜇=
= l𝑛 2(
) = 𝛿 ln 2
𝑑𝑡
𝑑𝑡
(5.7)
Therefore, the specific growth rate 𝜇 is equal to 𝐼𝑛2 times of the
division rate, 𝛿 .
If 𝜇 is constant with time during the exponential growth period, Eq.
(5.5) can be integrated from time 𝑡1 to t as
𝐶𝑁
𝑡
𝑑𝐶𝑁
න
= න 𝜇𝑑𝑡
𝐶
𝐶𝑁0 𝑁
𝑡0
(5.8)
To give
𝐶𝑁 = 𝐶𝑁0 exp[𝜇(𝑡 − 𝑡0 )]
(5.9)
13
Continued…
where 𝐶𝑁0 is the cell number concentration at 𝑡1 when the exponential
growth starts. Eq. (5.9) shows the increase of the number of cells
exponentially with respect to time.
The time required to double the population, called the doubling time (𝑡𝑑 ) ,
can be estimated from Eq. (5.9) by setting 𝐶𝑁 = 2𝐶𝑁0 and 𝑡1 = 0 and solving
for t:
𝑡𝑑 =
𝑙𝑛2 1
=
𝜇
𝛿
(5.10)
 The doubling time is inversely proportional to the specific growth rate
and is equal to the reciprocal of the division rate.
14
5.1.3 Factors Affecting the Specific Growth
rate
 Substrate Concentration: One of the most widely employed
expressions for the effect of substrate concentration on 𝜇 is the Monod
equation, which is an empirical expression based on the form of equation
normally associated with enzyme kinetics or gas adsorption:
𝜇=
𝜇𝑚𝑎𝑥 𝐶𝑆
𝐾𝑆 + 𝐶𝑆
(5.11)
where 𝐶𝑆 is the concentration of the limiting substrate in the medium and
𝐾𝑆 is a system coefficient.
The value of 𝐾𝑆 is equal to the concentration of nutrient when the specific
growth rate is half of its maximum value (μmax ).
While the Monod equation is an oversimplification of the complicated
mechanism of cell growth, it often adequately describes fermentation
kinetics when the concentrations of those components which inhibit the cell
growth are low.
15
Continued…
According to the Monod equation, further increase in the nutrient
concentration after 𝜇 reaches μmax does not affect the specific growth rate,
as shown in Figure 5.2.
However, it has been observed that the specific growth rate decreases as the
substrate concentration is increased beyond a certain level.
Figure 5.2: Dependence of the specific
growth rate on the concentration of the
growth limiting nutrient:, 𝜇𝑚𝑎𝑥 = 0.935
ℎ𝑟 −1 ; 𝐾𝑠 = 0.22 x 1(T −4 mol/L).
16
Continued…
Several other models have been proposed to improve the Monod equation.
They are:
𝜇𝑚𝑎𝑥 𝐶𝑆
𝐾11 + 𝐶𝑆 + (𝐶12 𝐶𝑆 )2
(5.12)
𝜇 = 𝜇𝑚𝑎𝑥 [1 − exp(−𝐶𝑆 /𝐾𝑆 )]
(5.13)
𝜇=
𝜇=
𝜇=
𝜇𝑚𝑎𝑥
1 + (𝐾𝑆 𝐶𝑆 −𝜆 )
𝜇𝑚𝑎𝑥 𝐶𝑆
𝛽𝐶𝑁 + 𝐶𝑆
(5.14)
(5.15)
If several substances can limit the growth of a microorganism, the following
model can be employed
𝐶1
𝐶2
𝜇 = 𝜇𝑚𝑎𝑥
…
𝐾1 + 𝐶1 𝐾2 + 𝐶2
(5.16)
17
Continued…
If the limiting nutrient is the energy source for the culture, a certain
amount of substrate can be used for purposes other than growth. Some
models include a term, e, which accounts for the maintenance of cells as,
𝜇𝑚𝑎𝑥 𝐶𝑆
𝜇=
− 𝑘𝑒
𝐾𝑆 + 𝐶𝑆
(5.17)
When 𝐶𝑆 is so low that the first term of the right-hand side of Eq. (5.17)
is less than or equal to 𝑘𝑒 , the specific growth rate is equal to zero.
These alternative models give a better fit for the growth of certain
microorganisms, but their algebraic solutions are more difficult than for the
Monod equation.
 Product Concentration: As cells grow, they produce metabolic byproducts which can accumulate in the medium.
The growth of microorganisms is usually inhibited by these products,
whose effect can be added to the Monod equation as follows:
18
Continued…
𝜇 = 𝜇𝑚𝑎𝑥
𝐶𝑆
𝐾𝑆 + 𝐶𝑆
𝐶𝑃
𝐾𝑃 + 𝐶𝑃
𝜇 = 𝜇𝑚𝑎𝑥
𝐶𝑆
𝐾𝑆 + 𝐶𝑆
𝐶𝑃
1−
𝐶𝑃𝑚
(5.18)
or
𝑛
(5.19)
The preceding equations both describe the product inhibition fairly
well. The term is the maximum product concentration above which cells
cannot grow due to product inhibition.
 Other Conditions: The specific growth rate of microorganisms is also
affected by medium pH, temperature, and oxygen supply. The optimum
pH and temperature differ from one microorganism to another.
19
5.1.4 Stationary and Death Phase
 The growth of microbial populations is normally limited either by the
exhaustion of available nutrients or by the accumulation of toxic products
of metabolism.
 As a consequence, the rate of growth declines and growth eventually
stops. At this point a culture is said to be in the stationary phase.
 The transition between the exponential phase and the stationary phase
involves a period of unbalanced growth during which the various cellular
components are synthesized at unequal rates.
 The stationary phase is usually followed by a death phase in which the
organisms in the population die.
 Death occurs either because of the depletion of the cellular reserves of
energy, or the accumulation of toxic products.
 Like growth, death is an exponential function.
 In some cases, the organisms not only die but also disintegrate, a process
called lysis.
20
5.2 Kinetics of Balanced Growth
In our beginning attempts to model and understand cell population kinetics,
we shall use unstructured models in which only cell mass or number
concentrations will be employed to characterize the biophase. The net rate
of cell mass growth, 𝑟𝑥 , is often written as 𝜇x where x is the cell mass per
unit culture volume and 𝜇, which has the units of reciprocal time, is the
specific growth rate of the cells. Using this representation in a steady-state
CSTR material balance for cell mass gives
𝐷𝑥𝑓 = 𝐷 − 𝜇 𝑥
(5.20)
Often the liquid feed stream to a continuous culture consists only of sterile
nutrient, so that xf = 0. In this case Eq. (5.20) reveals that a nonzero cell
population can be maintained only when
𝐷=𝜇
(5.21)
21
Continued…
i.e., when the culture has adjusted so that its specific growth rate is equal to
the dilution rate. When Eq. (5.21) is satisfied, it appears that Eq. (5.20) does
not determine x when the feed is sterile. Experiments with continuous
culture of Bacillus linens have confirmed the indeterminate nature of the
population level. After a steady continuous operation was achieved at the 6h point (Fig. 5.3), two subsequent interruptions of the culture were imposed.
Figure 5.3: As Eq. (5.21)
indicates, the population
concentration in continuous
culture is indeterminate so
long as μ is constant (B.
linens at 26°C with
D = 0.417 h 1).
22
Kinetics of Balanced Growth
23
5.2.1 Monod Growth Kinetics
 The behavior displayed in Fig. 5.3 can occur only when the specific
growth rate is independent of x and s. (Here s denotes the mass
concentration of limiting substrate). The indeterminate nature of the
system disappears if a particular nutrient is growth-rate limiting.
 We distinguish two types of media according to their makeup. A
synthetic medium is one in which chemical composition is well defined.
As Table 5.2 shows, such media can be constructed by supplementing a
mineral base with the necessary carbon, nitrogen, and energy sources as
well as any necessary vitamins.
 In addition to providing necessary ions for proper cell function, the
mineral base of the medium also contains buffering compounds to reduce
large pH fluctuations during growth. Complex media contain materials of
undefined composition.
24
Continued…
Table 5.2: Some examples of synthetic and complex media
Additional ingredients
Common
ingredients
(Mineral base)
Medium 1
Medium 2
Medium 3
Medium 4
Water, 1 L
NH4 Cl, 1 g
Glucose, 5 g
NH4 Cl, 1 g
Glucose, 5 g
NH4 Cl, 1 g
Nicotinic acid, 0.1 mg
Glucose, 5 g
Yeast extract,
5g
K 2 HPO4 , 1 g
MgSO4 . 7H2 O,
200 mg
CaCl2 , 10 mg
Trace elements
(Mn, Mo, Cu, Co,
Zn) as inorganic
salts, 0.02-0.5 mg
of each
25
Continued…
 The general goal in making a medium is to support good growth and/or
high rates of product synthesis. Contrary to intuitive expectation, this
does not necessarily mean that all nutrients should be supplied in great
excess. For one thing, excessive concentration of a nutrient can inhibit or
even poison cell growth.
 If the concentration of one essential medium constituent is varied while
the concentrations of all other medium components are kept constant, the
growth rate typically changes in a hyperbolic fashion, as Fig. 5.4 shows.
A function relationship between the specific growth rate u and an
essential compound, concentration was proposed by Monod in 1942.
 Of the same form as the Langmuir adsorption isotherm (1918) and the
standard rate equation for enzyme-catalyzed reactions with a single
substrate (Henri in 1902 and Michae and Menten in 1913), the Monod
equation states that
26
Continued…
𝜇=
𝜇𝑚𝑎𝑥 𝑠
𝐾𝑠 + 𝑠
(5.22)
 Here is the maximum growth rate achievable when 𝑠 ≫ 𝐾𝑠 , and the
concentrations of all other essential nutrients are unchanged.
The particular form of Eq. (5.22) is appealing; its simplicity urges several
warnings upon the user. First, the value of K s is often rather small. Thus s
can be ≫ 𝐾𝑠 , rather easily, and the term s/(K s + s) may be regarded simply
as an adequate description for calculating the deviation of 𝜇 from 𝜇max as
the concentration s becomes smaller.
The relation also suggests that the specific growth rate is finite (𝜇 ≠ 0) for
any finite concentration of the rate-limiting component. Generally, this
implied behavior is not well tested for 𝑠 ≪ 𝐾𝑠 .
27
Continued…
Figure 5.4: Dependence of E. coli specific growth rate on the
concentration of the growth-limiting nutrient: (a) glucose medium and
(b) tryptophan medium (for a tryptophan-requiring mutant).
28
Continued…
When population growth rate is related to limiting nutrient equation by a
relationship such as the one proposed by Monod, definite connections
emerge among reactor operating conditions and microbial kinetic and
stoichiometric parameters.
𝑌𝑥/𝑠
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑒𝑙𝑙𝑠 𝑓𝑜𝑟𝑚𝑒𝑑
=
𝑚𝑎𝑠𝑠 𝑓 𝑠𝑢𝑏𝑡𝑟𝑎𝑡𝑒 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑
(5.23)
The steady-state mass balance on substrate is then
1
𝐷 𝑠𝑓 − 𝑠 −
𝜇𝑥 = 0
𝑌𝑥/𝑠
(5.24)
or, taking 𝜇 from Eq. (5.22),
𝐷 𝑠𝑓 − 𝑠 −
𝜇𝑚𝑎𝑥 𝑠𝑥
=0
𝑌𝑥/𝑠 (𝑠 + 𝐾𝑠 )
(5.25)
29
Continued…
The corresponding cell mass balance is
𝜇𝑚𝑎𝑥 𝑠
− 𝐷 𝑥 + 𝐷𝑥𝑓 = 0
𝐾𝑠 + 𝑠
(5.26)
Equations (5.25) and (5.26) are often called the Monod chemostat model
For the common case of sterile feed (x, 0), these equations can readily be
solved for x and s to yield
𝑥𝑠𝑡𝑒𝑟𝑖𝑙𝑒 𝑓𝑒𝑒𝑑 = 𝑌𝑥/𝑠
𝑠𝑓 −
𝐷𝐾𝑠
𝜇𝑚𝑎𝑥 − 𝐷
(5.27)
and
𝑠𝑠𝑡𝑒𝑟𝑖𝑙𝑒 𝑓𝑒𝑒𝑑 =
𝐷𝐾𝑠
𝜇𝑚𝑎𝑥 − 𝐷
(5.28)
30
Continued…
Equations (5.27) and (5.28) contain the explicit steady-state dependence of
and 𝑠 on flow rate (𝐷 = 𝐹/𝑉). For very slow flows at a given volume, 𝐷 →
0; thus 𝑠 tends to zero. Since nearly all feed substrate is consumed by the
cells, the resulting effluent cell-mass concentration is 𝑥 = 𝑠𝑓 𝑌𝑥/𝑠 .
The dilution rate 𝐷 has just surpassed the maximum possible growth rate,
and the only steady-state solution is 𝑥 = 0. This condition of loss of all
cells at steady state, termed wash out, occurs for 𝐷 greater than 𝐷𝑚𝑎𝑥 ,
where, if 𝑥 = 0 in Eq. (5.27)
𝐷𝑚𝑎𝑥
𝜇𝑚𝑎𝑥 𝑠𝑓
=
𝐾𝑠 + 𝑠𝑓
(5.29)
The character of these solutions for substrate and cell-mass concentration
is shown in Fig. 5.5 for the following parameter values:
𝜇𝑚𝑎𝑥 = 1.0ℎ−1
𝑌𝑥/𝑠 = 0.5
𝐾𝑠 = 0.2 𝑔/𝐿
𝑠𝑓 = 10 g/L
31
Continued…
Figure 5.5: Dependence of effluent substrate concentrations, cell concentration 𝑥, and
cell production rate 𝑥𝐷 on continuous culture dilution rate 𝐷 as computed from the
Monod chemostat model (𝜇𝑚𝑎𝑥 = 1 ℎ−1, 𝐾𝑠 = 0.2 𝑔/𝑙, 𝑌𝑥/𝑠 = 0.5, 𝑠𝑓 = 10𝑔/𝑙).
32
33
Continued…
Notice, that near washout the reactor is very sensitive to variations in 𝐷; a
small change in 𝐷 gives a relatively large shift in 𝑥 and/or s.
The rate of cell production per unit reactor volume is 𝐷𝑥. This quantity is
illustrated in Fig. 5.5, and there is a sharp maximum. We can compute the
maximal cell output rate by solving
𝑑(𝐷𝑥)
=0
𝑑𝐷
(5.30)
where Eq. (5.27) is used to write x as a function of D. Solution of Eq. (7.17)
gives
𝐷𝑚𝑎𝑥−𝑜𝑢𝑡𝑝𝑢𝑡 = 𝜇𝑚𝑎𝑥
𝐾𝑠
1−
𝐾𝑠 + 𝑠𝑓
(5.31)
If s,> K,, as often is the case, the value of 𝐷𝑚𝑎𝑥−𝑜𝑢𝑡𝑝𝑢𝑡 approaches 𝜇𝑚𝑎𝑥 and
consequently is near washout.
34
35
36
37
38
39
Continued…
This situation, evident in Fig. 5.5, may require us to forgo maximal biomass
production in order to avoid the region of large sensitivity.
When analyzing the production of end products from a continuous
fermentation, we introduce another yield coefficient 𝑌𝑝/𝑥 , defined by
𝑌𝑝/𝑥 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑓𝑜𝑟𝑚𝑒𝑑
𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑐𝑒𝑙𝑙 𝑚𝑎𝑠𝑠
(5.32)
Use of the product yield coefficient allows us to write the steady-state
product mass balance in the form
So that
𝐷 𝑝𝑓 − 𝑝 + 𝑌𝑝/𝑥 𝜇𝑥 = 0
(5.33)
𝑌𝑝/𝑥 𝜇𝑥
𝑝 = 𝑝𝑓 +
𝐷
(5.34)
40
41
5.2.2 Kinetic Implications of Endogeneous
and Maintenance Metabolism
 The data presented in Fig. 5.7 for A. aerogenes show a marked decline in
cell concentration as dilution rate becomes small. Similar behavior has
also been observed for the food yeast Torula utilis. This trend, which is
contrary to the Monod chemostat model, can be explained by including
the possibility of endogeneous metabolism in the model.
Thus, we might write
𝜇𝑚𝑎𝑥 𝑠𝑥
𝑟𝑥 =
− 𝐾𝑒 𝑥
𝑠 + 𝐾𝑠
(5.35)
to account for this effect. Notice that the additional term in Eq. (5.35) can
also be interpreted formally as a cell death rate.
Such a modification of the Monod model is also consistent with other
available data. For example, if the rate of respiration of an aerobic culture is
proportional to the rate of substrate utilization
42
Continued…
Figure 5.7: Cell concentration decreases as the dilution rate is decreased, a trend
contrary to the Monod chemostat model (continuous cultivation of A. aerogenes in
glycerol medium).
43
Continued…
𝑟𝑟𝑒𝑠𝑝
𝛽𝜇𝑚𝑎𝑥 𝑠𝑥
=
𝑠 + 𝐾𝑠
(5.36)
then from Eqs. (5.20), (5.35), and (5.36) it follows that the specific
respiration rate is
𝑟𝑟𝑒𝑠𝑝
= 𝛽(𝐷 + 𝑘𝑒 )
𝑥
(5.37)
Observed variations in the yield coefficient 𝑌 with 𝐷 also support a growth
rate of the form given in Eq. (5.35). If the rate of substrate disappearance is
−𝑟𝑠 =
1 𝜇𝑚𝑎𝑥 𝑠𝑥
′
𝑌𝑥/𝑠
𝑠 + 𝐾𝑠
(5.38)
44
Continuous culture of Aspergillus aerogenes
shows liner relationship between rrep and D
Respiration is defined as a metabolic process
wherein, the living cells of an organism obtains
energy (in the form of ATP) by taking in oxygen
and liberating carbon dioxide from the
oxidation of complex organic substances.
45
Continued…
Variation of yield coefficient with D
′
where 𝑌𝑥/𝑠
is the "true" coefficient, using Eqs. (5.35) and (5.38) and the
definition of 𝑌𝑥/𝑠 (remember 𝑌𝑥/𝑠 is an overall stoichiometric coefficient
equal to the total mass of cells produced divided by the total mass of
substrate consumed, so that with sterile feed −𝑟𝑠 = 𝐷𝑥/𝑌𝑥/𝑠 ) gives
𝑌𝑥/𝑠 =
′
𝐷𝑌𝑥/𝑠
𝐷 + 𝑘𝑒
(5.39)
Such dependence of Y on D has been verified experimentally for several
microorganisms.
Another possible scenario, is parallel consumption of substrate for
growth and for other energetic, or maintenance, requirements. In this case,
the rate of substrate utilization is
−𝑟𝑠 =
1
′ 𝜇𝑥 + 𝑚𝑥
𝑌𝑥/𝑠
(5.40)
46
Continued…
Figure 5.8: The specific respiration rate of the bacterium A. aerogenes in continuous
culture depend linearly on dilution rate as Eq. (7.24) indicates.
47
Continued…
where 𝑚 is the specific maintenance rate. Presuming now that 𝑟𝑥 is equal to
𝜇𝑥, it follows that
𝑌𝑥/𝑠 =
′
𝑌𝑥/𝑠
𝐷
′
𝐷 + 𝑚𝑌𝑥/𝑠
(5.41)
which is exactly the same functional relationship between 𝑌𝑥/𝑠 and 𝐷 as was
found for the endogeneous metabolism model [Eq. (5.41)].
48
49
50
5.2.3 Other Forms of Growth Kinetics
Other related forms of specific growth rate dependence have been proposed
which in particular instances give better fits to experimental data. For
example, Tessier, Moser, and Contois suggest the following models:
Tessier
Moser
𝜇 = 𝜇𝑚𝑎𝑥 (1 − 𝑒 −𝑠/𝐾𝑠 )
(5.42)
𝜇 = 𝜇𝑚𝑎𝑥 (1 + 𝐾𝑠 𝑠 −𝜆 )−1
(5.43)
Contois
𝜇 = 𝜇𝑚𝑎𝑥
𝑠
𝐵𝑥 + 𝑠
(5.44)
The first two examples render algebraic solution of the growth equations
much more difficult than the Monod form. The equation of Contois contains
an apparent Michaelis constant which is proportional to biomass
concentration x.
51
Continued…
This last term will therefore diminish the maximum growth rate as the
population density increases, eventually leading to 𝜇 ∝ 𝑥 −1 .
The specific growth rate may be inhibited by medium constituents such as
substrate or product. An example due to Andrews proposes that substrate
inhibition be treated by the form
𝜇 = 𝜇𝑚𝑎𝑥
𝑠
𝐾𝑖 + 𝑠 + 𝑠 2 /𝐾𝑝
(5.45)
Alcohol fermentation provides an example of product inhibition; the
anerobic glucose fermentation by yeast has been treated by Aiba, Shoda, and
Nagatani with specific-growth function of the type
𝜇 = 𝜇𝑚𝑎𝑥
𝐾𝑝
𝑠
𝐾𝑖 + 𝑠 𝐾𝑝 + 𝑝
(5.46)
52
Continued…
It is possible that two (or more) substrates may simultaneously be growth
limiting. While few data are available, a Monod dependence on each
limiting nutrient may be proposed, so that
𝜇 = 𝜇𝑚𝑎𝑥
𝑠1
𝑠2
⋯
𝐾𝑖 + 𝑠1 𝐾𝑝 + 𝑠2
(5.47)
In the absence of convincing data for this form, we may regard it simply as a
useful indicator that growth depends on several limiting nutrients.
53
5.2.4 Other Environmental Effects on
Growth Kinetics.
During balanced growth, only a single parameter 𝜇 [or the population
doubling time 𝑡ഥ𝑑 (= In 2/𝜇)] is required to characterize population growth
kinetics. For this reason, the magnitude of the specific growth rate 𝜇 is
widely used to describe the influence of the cells' environment on the cells'
performance.
Consider first the influence of temperature: the range of temperature
capable of supporting life a we know it lies between roughly -5 and 95°C.
Procaryotes can be classified according to the temperature interval in which
they grow.
As seen in Table 5.3, each class has an optimum temperature where
growth is maximal and upper and lower temperature bounds beyond which
the population cannot grow.
The data in Fig. 5.9 for growth of E. coli dramatically illustrate the strong
influence of temperature.
54
Continued…
Table 5.3 Classification of microorganisms in terms of growth-rate dependence on
temperature
Temperature, ℃
Group
Minimum
Optimum
Maximum
Thermophiles
40 to 45
55 to 75
60 to 80
Mesophiles
10 to 15
30 to 45
35 to 47
Psychrophiles:
Obligate
Facultative
-5 to 5
-5 to 5
15 to 18
25 to 30
19 to 22
30 to 35
55
Continued…
Figure 5.9: (a) Up to a point, the growth rate of E. coll increases with increasing
temperature, but the cells die if the temperature is too high. (b) An Arrhenius plot
of the data in (a).
56
Continued…
Notice in the Arrhenius plot that classical Arrhenius behavior appears at low
temperature whereas there is a rapid decrease in growth rate as the
temperature approaches the upper limit for survival of the bacterium. The
similarity of the temperature dependence for growth in Fig. 5.9 with the
enzyme-activity-temperature relationship depicted is unescapable.
Since protein configuration and activity are pH dependent, we expect
cellular transport processes, reactions, and hence growth rates to depend on
pH. Bacterial growth rates are usually maximum in the range of pH from 6.5
to 7.5. This is exemplified by the data for E. coli shown in Fig. 5.10. There
are exceptions, however, including acidophilic bacteria which grow at pH
2.0. Typically, a change in 1.5 to 2.0 pH units above or below the pH for
maximum growth rate results in negligible growth (Fig. 5.10).
57
Continued…
Figure 5.10: Effect of pH and temperature on the generation time of E. coli.
58
5.3 Deviation from Monod model
 At large dilution rates, continuous-culture behavior can deviate
significantly from the Monod chemostat model, as Fig. 5.11 shows. Not
only is the predicted cell concentration in error near washout, but the
population may be maintained at substantially larger dilution rates theory
indicates.
 Moreover, the yield coefficient decreases as D approaches the critical
maximum value.
 Among the probable explanations for these difficulties is the relatively
high substrate concentration which prevails at large dilution rates.
 Under these circumstances, the substrate may not limit growth, and the
members of the cell population may shift their metabolic patterns in
recognition of some other limiting factor in their environment.
59
Continued…
Figure 5.11: This experimental
data on continuous growth of
Aerobacter cloacae shows non
zero cell concentrations at
dilution rates exceeding the
calculated critical dilution rate
60
5.7 Transient Growth Kinetics
 During certain intervals in batch cultivation or during startup or
disturbances to continuous flow reactors, cell populations grow in a
transient state in which more complicated kinetic behavior may be
observed.
 Different types of kinetic models may be required to describe different
transient growth situations.
61
5.7.1 Growth-Cycle Phases for Batch
Cultivation
 In a typical batch process the number of living cells varies with time, as
shown in Fig 5.25.
 After a lag phase, where no increase in cell numbers is evident, a period
of rapid growth ensues, during which the cell numbers increase
exponentially with time.
 Although this stage of batch culture is often called the logarithmic phase,
we prefer the more descriptive term exponential growth, which we shall
use in the following discussion.
 Naturally in a closed vessel the cells cannot multiply indefinitely, and a
stationary phase follows the period of exponential growth.
 At this point the population achieves its maximum size.
 Eventually a decline in cell numbers occurs during the death phase.
 Here an exponential decrease in the number of living individuals is often
observed.
62
Continued…
Figure 5.25: Typical growth curve
for batch cell cultivation.
Figure 5.26: In a medium containing
initially equal amounts of glucose and
xylose, diauxic growth of E coli is
observed in batch culture.
63
Continued…
 Multiple lag phases may sometimes be observed when the medium
contains multiple carbon sources (Fig. 5.26).
 This phenomenon, known as diauxic growth, is caused by a shift in
metabolic patterns in the midst of growth.
 After one carbon substrate is exhausted, the cell must divert its energies
from growth to “retool” for the new carbon supply.
 Design to minimize culture and process times normally includes
minimization of the lag times associated with each new batch culture.
 From the previous general discussion and other relevant data, the
following generalizations can be drawn:
1. The inoculating culture should be as active as possible and the
inoculation carried out in the exponential-growth phase.
2. The culture medium used to grow the inoculum should be as close as
possible to the final full-scale fermentation composition.
64
Continued…
3. Use of the reasonably large inoculum (order of 5-10 percent of the new
medium volume) is recommended to avoid undue loss by diffusion of
required inter mediate or activators.
At the end of the lag phase the population of microorganisms is well
adjusted to its new environment.
The cells can then multiply rapidly, and cell mass, or the number of living
cells, doubles regularly with time. The equations
𝑑𝑥
= 𝜇𝑥
𝑑𝑡
𝑥 = 𝑥0
or
1 𝑑𝑥
=𝜇
𝑥 𝑑𝑡
(5.85a)
at
𝑡 = 𝑡𝑙𝑎𝑔
(5.85b)
with
65
Continued…
describe the increase in cell numbers during this period. Thus the rate of
increase in x is proportional to x. From the integrated form of Eqs. (5.87)
𝑥
ln = 𝜇(𝑡 − 𝑡𝑙𝑎𝑔 )
𝑥0
or
𝑥 = 𝑥0 𝑒 𝜇(𝑡−𝑡𝑙𝑎𝑔 )
𝑡 > 𝑡𝑙𝑎𝑔
(5.86)
we can readily deduce that time interval 𝑡𝑑 required to double the
population is given by
𝑡𝑑 =
ln 2
𝜇
(5.87)
As is the case for steady state growth in a CSTR, only a single parameter μ
(or 𝑡𝑑 ) is required to characterize the population during exponential batch
growth.
To a reasonable approximation, growth is balanced during this stage of
batch cultivation,
66
Continued…
 Deviations from exponential growth eventually arise when some
significant variable such as nutrient level or toxin concentration achieves
a value which can no longer support the maximum growth rate.
 Exhaustion of a particular critical nutrient may appear rather sharply at a
given time since the cells are rapidly increasing the total rate of nutrient
consumption in the exponential-growth phase.
To formulate a rough analysis of this event, we suppose that the rate of
nutrient A consumption is proportional to the mass concentration of living
cells until the stationary phase is reached:
𝑑𝑎
= −𝑘𝑎 𝑥
𝑑𝑡
(5.88)
Next we assume that exponential growth continues unabated until the
stationary phase is reached, and we take the time when exponential growth
begins as time zero. Then
𝜇𝑡
𝑥 = 𝑥0 𝑒
(5.88)
67
Continued…
where 𝑥0 is the mass concentration of living cells when exponential growth
starts.
If the concentration of A at time zero is 𝑎0 , we can determine from Eq (5.87)
and (5.88) that A is completely consumed when
𝑘𝑎
𝑎0 =
(𝑥 − 𝑥0 )
𝜇 𝑠
(5.89)
where 𝑥𝑠 , is the mass of the population when A is exhausted, and the
population enters the stationary phase. Consequently, 𝑥𝑠 is the maximum
population size achieved during the batch culture (review Fig. 5.25).
Rearranging Eq. (5.89), we find the maximum population in the case of
nutrient depletion to be given by
𝜇
𝑥𝑠 = 𝑥0 + 𝑎0
𝑘𝑎
(5.90)
68
Continued…
Figure 5.27: Dependence of
maximum batch population size on
the initial concentration of growthlmiting nutrient: (a) Pseudomomas
sp. in fructose medium, and A.
Aerogenes in (b) lactose and (c)
ammonium tartrate media..
(b)
(c)
Influence of concentration on total population of Aerobacter aerogenes
69
Continued…
Linear dependence of x, on initial nutrient level has been observed
experimentally in many cases (often xo is so small that it is imperceptible).
Figure 5.27 illustrates this behavior.
If a toxin accumulates which slows the rate of growth from exponential,
an equation of the form
𝑑𝑥
= 𝑘𝑥 [1 − 𝑓(𝑡𝑜𝑥𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛)]
𝑑𝑡
is found useful. In the particular case where toxin linearly decreases the
growth rate
1 𝑑𝑥
𝑥 𝑑𝑡
= 𝑘 [1 − 𝑏𝑐𝑡 ]
(5.91)
Where 𝐶𝑡 , is toxin concentration and b is a constant. A plausible assumption
is that the rate of toxin only on x and is proportional to it
𝑑𝐶𝑡
= 𝑞𝑥
𝑑𝑡
Or
𝑡
Ct= 𝑞 ‫׬‬0 𝑥𝑑𝑡
(5.92)
70
Continued…
𝑡
so that 𝐶𝑡 = 𝑞 ‫׬‬0 𝑥𝑑𝑡 (assuming 𝐶𝑡 = 0 at 𝑡 = 0) and the growth equation
𝑡
becomes
𝑑𝑥
𝑑𝑡
= 𝑘𝑥 1 − 𝑏𝑞 න 𝑥𝑑𝑡
(5.93)
0
The instantaneous value of the effective specific-growth rate 𝜇𝑒𝑓𝑓
𝜇𝑒𝑓𝑓
𝑡
1 𝑑𝑥
=
= 𝑘 1 − 𝑏𝑞 න 𝑥𝑑𝑡
𝑥 𝑑𝑡
0
(5.94)
diminishes more and more rapidly with time, i.e.,
𝑑𝜇𝑒𝑓𝑓
= −𝑘𝑏𝑞𝑥 < 0
𝑑𝑡
𝑑 2 𝜇𝑒𝑓𝑓
𝑑𝑥
= −𝑘𝑏𝑞
< 0 (Max.)
𝑑𝑡
𝑑𝑡
(5.95)
Growth halts when dx/dt = 0
1 𝑑𝑥
𝑥 𝑑𝑡
= 𝑘 [1 − 𝑏𝑐𝑡 ] Or Ct = 1/b
Thus, growth ceases only when 𝐶𝑡 , reaches a particular level 𝐶𝑡 , = 1/b.
71
72
5.7.2 Unstructured Batch Growth Models
In the simplest approach to modeling batch culture, we suppose that the rate
of increase in cell mass is a function of the cell mass only. Thus
𝑑𝑥
= 𝑓(𝑥)
𝑑𝑡
(5.97)
As we shall soon see, such a form does not require us to neglect changes
occurring in the medium during growth.
One of the simpler models belonging to the general form given in Eq. (5.97)
is Mathus’s law, which uses
𝑓 𝑥 = 𝜇𝑥
(5.98)
with 𝜇 as a constant.
Verlhurst in 1844 and Pearl and Reed in 1920 contributed to a theory which
included an inhibiting factor to population growth. Assuming that inhibition
𝑑𝑥
proportional to 𝑥 2 , they used
= 𝑘𝑥 − 𝑃𝑥2
P = k𝛽
𝑑𝑡
𝑑𝑥
= 𝑘𝑥(1 − 𝛽𝑥)
𝑑𝑡
(5.99)
𝑥 0 = 𝑥0
73
Continued…
Figure 5.28: The logistic curve (k > 0, 𝛽 > 0).
74
Continued…
This is a Riccati equation which can be easily integrated to give the logistic
curve
𝑘𝑎 𝑒 𝑘𝑡
𝑥=
1 − 𝛽𝑥0 (1 − 𝑒 𝑘𝑡 )
(5.100)
As illustrated schematically in Fig. 5.28, the logistic curve is sigmoidal and
lead to a stationary population of size 𝑥𝑠 = 1/𝛽.
One possible interpretation of the logistic curve can be formulated by
assuming that the production rate of a toxin is proportional to the population
growth rate
So that if
𝑑𝐶𝑡
𝑑𝑥
=𝛼
𝑑𝑡
𝑑𝑡
(5.101)
𝐶𝑡 (0) = 0
(5.102)
𝐶𝑡 = 𝛼(𝑥 − 𝑥0 )
(5.103)
75
76
Continued…
Another class of unstructured models which approach a stationary
population level can be formulated based on limiting nutrient exhaustion.
Assuming a constant overall yield factor 𝑌𝑥/𝑠 and 𝜇 = 𝜇(𝑠), such as Monod
kinetics, nutrient and cell material balances can be combined into a single
equation.
A drawback of the logistic equation is its failure to predict a phase of decline
after the stationary population has exhausted all available resources.
This feature sound in one model developed by Volterra early in this century.
In this model, integral term of the form
𝑡
න 𝐾 𝑡, 𝑟 𝑥 𝑟 𝑑𝑟
(5.104)
0
77
Continued…
is added to Eq. (5.98). If K is independent of t, a term such as (5.104) may
be viewed as representative of the influence of a component of the culture
whose concentration follows
𝑑𝐶
= [𝑘 𝑡 𝑥(𝑡)]
𝑑𝑡
𝑐 0 =0
(5.105)
Specifically, let us suppose that K is a constant equal to Ko and that the
history or memory term Eq. (5.105) is added to Eq. (5.99). Then the
population size is described by
𝑡
𝑑𝑥
= 𝑘𝑥 1 − 𝛽𝑥 + 𝐾0 න 𝑥 𝑟 𝑑𝑟
𝑑𝑡
0
𝑥 0 = 𝑥0
(5.106)
As Eq. (5.101) indicates, the sign of Ko is taken as negative for an inhibitor
and positive for a compound which promotes growth.
This problem can be solved numerically with results as given in Fig. 5.20.
So long as Ko is negative, the population size declines after passing through
the maximum.
78
Continued…
Figure 5.29: Response of the Vol-terra model,
which includes the population history.
79
5.4 Stirred-Tank Fermenter
 The bioreactor is frequently called a fermenter whether the
transformation is carried out by living cells or in vivo cellular
components (enzymes).
 However, in this text, we call the bioreactor employing living cells a
fermenter to distinguish it from the bioreactor which employs enzymes,
the enzyme reactor.
 For a large-scale operatior L, the stirred-tank fermenter (STF) is the most
widely used design in industrial fermentation.
 It can be employed for both aerobic or anaerobic fermentation of a wide
range of cells including microbial, animal, and plant cells.
 Figure 5.12 shows a diagram of a fermenter used for penicillin
production.
 The mixing intensity can be varied widely by choosing a suitable
impeller type and by varying agitating speeds.
80
81
Continued…
Figure 5.12: Cutaway diagram of a fermenter
used for penicillin production
 The mechanical agitation and aeration are effective for the suspension of
cells, oxygenation, mixing of the medium, and heat transfer.
 The STF can be also used for high viscosity media.
 It was one of the first largescale fermenters developed in the
pharmaceutical industries.
82
5.4.1 Batch or Plug-Flow Fermenter
 An ideal stirred fermenter is assumed to be well mixed so that the
contents are always uniform in composition.
 In a tubular-flow fermenter, nutrients and microorganisms enter one end
of a cylindrical tube and the cells grow while they pass through.
 Since the long tube and lack of stirring device prevents complete mixing
of the fluid, the properties of the flowing stream will vary in both
longitudinal and radial direction.
 However, the variation in the radial direction is small compared to that in
the longitudinal direction. The ideal tubular-flow fermenter without
radial variations is called a plug-flow fermenter (PFF).
 In reality, the PFF fermenter is hard to be found.
 However, the packed-bed fermenter and multi-staged fermenter can be
approximated as PFF.
83
Continued…
 Even though the steady-state PFF is operated in a continuous mode, the
cell concentration of an ideal batch fermenter after time t will be the
same as that of a steady-state PFF at the longitudinal location where the
residence time i is equal to t (Figure 5.13).
 Therefore, the following analysis applies for both the ideal batch
fermenter and steady-state PFF.
 If liquid medium is inoculated with a seed culture, the cell will start to
grow exponentially after the lag phase.
 The change of the cell concentration in a batch fermenter is equal to the
growth rate of cells in it:
𝑑𝐶𝑋
= 𝑟𝑋 = 𝜇𝐶𝑋
𝑑𝑡
(5.48)
84
Continued…
To derive the performance equation of a batch fermentation, we need to
integrate Eq. (5.20) to obtain:
𝐶𝑋
𝐶𝑋
𝑡
𝑑𝐶𝑋
𝑑𝐶𝑋
න
=න
= න 𝑑𝑡 = 𝑡 − 𝑡0
𝑟
𝜇𝐶
𝑋
𝐶𝑋0 𝑋
𝐶𝑋0
𝑡0
(5.49)
According to Eq. (5.49), the batch growth time t - 𝑡0 is the area under
the 1/ R X versus 𝐶𝑋 curve between 𝐶𝑋0 and 𝐶𝑋 as shown in Figure 5.14.
The solid line in Figure 5.14 was calculated with the Monod equation and
the shaded area is equal to t − 𝑡0 .
The batch growth time is rarely estimated by this graphical method since the
𝐶𝑋 versus t curve is a more straightforward way to determine it.
just note that the curve is U shaped, which is characteristic of auto catalytic
reactions:
𝑆+𝑋
𝑋+𝑋
85
Continued…
Figure 5.13: Schematic diagram of (a) batch stirred-tank fermenter and (b) plugflow fermenter
86
Continued…
The rate for an autocatalytic reaction is slow at the start because the
concentration of X is low.
It increases as cells multiply and reaches a maximum rate.
As the substrate is depleted and the toxic products accumulate, the rate
decreases to a low value.
If Monod kinetics adequately represents the growth rate during the
exponential period, we can substitute Eq.(5.11) into Eq.(5.49) to obtain,
𝐶𝑋
𝑡
(𝐾𝑆 + 𝐶𝑆 )𝑑𝐶𝑋
න
= න 𝑑𝑡
𝜇
𝐶
𝑚𝑎𝑥 𝑋
𝐶𝑋0
𝑡0
(5.50)
The growth yield (𝑌𝑋/𝑆 ) is defined as
𝑌𝑋/𝑆
∆𝐶𝑆
𝐶𝑋 − 𝐶𝑋0
=
=
−∆𝐶𝑆 −(𝐶𝑆 − 𝐶𝑆0 )
(5.51)
87
Continued…
Substitution of Eq.(5.51) into Eq.(5.50) and integration of the resultant
equation gives a relationship which shows how the cell concentration
changes with respect to time:
𝐾𝑆 𝑌𝑋
𝑡 − 𝑡0 𝜇𝑚𝑎𝑥
𝐾𝑆 𝑌𝑋
𝐶
𝐶𝑆0
𝑋
𝑆
𝑆
=
+ 1 ln
+
ln
𝐶𝑋0 + 𝐶𝑆0 𝑌𝑋
𝐶𝑋0 𝐶𝑋0 + 𝐶𝑆0 𝑌𝑋 𝐶𝑆
𝑆
(5.52)
𝑆
Figure 5.14: A graphical representation of the batch growth time 𝑡 − 𝑡1 (shaded
area).
88
Continued…
The Monod kinetic parameters, 𝜇𝑚𝑎𝑥 and 𝐾𝑆 , cannot be estimated with a
series of batch runs as easily as the Michaelis-Menten parameters for an
enzyme reaction.
In the Michaelis-Menten equation,
𝑑𝐶𝑝
𝑟𝑚𝑎𝑥 𝐶𝑆
=
𝑑𝑡
𝐾𝑀 + 𝐶𝑆
(5.53)
while in the Monod equation,
𝑑𝐶𝑋 𝜇𝑚𝑎𝑥 𝐶𝑆 𝐶𝑋
=
𝑑𝑡
𝐾𝑆 + 𝐶𝑆
(5.54)
 There is a 𝐶𝑋 term in the Monod equation which is not present in the
Michaelis-Menten equation.
89
5.4.2 Ideal Continuous Stirred-tank
Fermenter
 Microbial populations can be maintained in a state of exponential growth
over a long period of time by using a system of continuous culture.
 Figure 5.15 shows the block diagram for a continuous stirred tank
fermenter (CSTF).
 The growth chamber is connected to a reservoir of sterile medium.
 Once growth has been initiated, fresh medium is continuously supplied
from the reservoir.
Figure 5.15: Schematic diagram
of continuous stirred-tank
fermenter (CSTF)
90
Continued…
 Continuous culture systems can be operated as chemostat or as
turbidostat.
 In a chemostat the flow rate is set at a particular value and then rate of
growth of the culture adjusts to this flow rate.
 In a turbidostat the turbidity is set at a constant level by adjusting the
flow rate.
 It is easier to operate chemostat than turbidostat, because the former can
be done by setting the pump at a constant flow rate, whereas the latter
requires an optical sensing device and a controller.
 However, the turbidostat is recommended when continuous fermentation
needs to be carried out at high dilution rates near the washout point, since
it can prevent washout by regulating the flow rate in case the cell loss
through the output stream exceeds the cell growth in the fermenter.
91
Continued…
The material balance for the microorganisms in a CSTF (Figure 5.7) can be
written as
𝑑𝐶𝑋
𝐹𝐶𝑋𝑖 − 𝐹𝐶𝑋 + 𝑉𝑟𝑋 = 𝑉
𝑑𝑡
(5.55)
where is the rate of cell growth in the fermenter and represents the change of
cell concentration in the fermenter with time.
For the steady-state operation of a CSTF, the change of cell concentration
𝑑𝐶
with time is equal to zero 𝑑𝑡𝑋 = 0 since the microorganisms in the vessel
grow just fast enough to replace those lost through the outlet stream, and Eq.
(5.55) becomes
𝜏𝑚 =
𝑉 𝐶𝑋 − 𝐶𝑋𝑖
=
𝐹
𝑟𝑋
(5.56)
where D is known as dilution rate and is equal to the reciprocal of the
residence time (𝜏𝑚 ).
1 1
𝜏𝑚 =
𝜇
=
𝐷
(5.57)
92
Continued…
If the growth rate can be expressed by Monod equation, then
𝐷=𝜇=
1
𝜇𝑚𝑎𝑥 𝐶𝑆
=
𝜏𝑚 𝐾𝑆 + 𝐶𝑆
(5.58)
From Eq. (5.30), 𝐶𝑆 can be calculated with a known residence time and the
Monod kinetic parameters as:
𝐶𝑆 =
𝐾𝑆
𝜏𝑚 𝜇𝑚𝑎𝑥 − 1
(5.59)
Figure 5.16: A graphical
representation of the estimation
of residence time for the CSTF
93
Continued…
It should be noted, however, that the preceding expression is only valid
when 𝜏𝑚 𝜇𝑚 > 1. If 𝜏𝑚 𝜇𝑚 < 1, the growth rate of the cells is less than the rate
of cells leaving with the outlet stream.
Consequently, all of the cells in the fermenter will be washed out and
Eq.(5.59) is invalid.
If the growth yield (𝑌𝑋/𝑆 ) is constant, then
𝐶𝑋 = 𝑌𝑋/𝑆 (𝐶𝑆𝑖 − 𝐶𝑆 )
(5.60)
Substituting Eq.(5.31) into Eq.(5.32) yields the correlation 𝐶𝑆 for as
𝐶𝑋 = 𝑌𝑋/𝑆
𝐶𝑝 = 𝐶𝑝𝑖 + 𝑌𝑋/𝑆
𝐶𝑆𝑖 −
𝐾𝑆
𝜏𝑚 𝜇𝑚𝑎𝑥 − 1
𝐾𝑆
𝐶𝑆𝑖 −
𝜏𝑚 𝜇𝑚𝑎𝑥 − 1
(5.61)
(5.62)
Again, Eqs. (5.61) and (5.62) are valid only when 𝜏𝑚 𝜇𝑚𝑎𝑥 > 1.
94
5.5 Cell Recycling
 For the continuous operation of a PFF or CSTF, cells are discharged with
the outlet stream which limits the productivity of fermenters. The
productivity can be improved by recycling the cells from the outlet
stream to the fermenter.
Figure 5.17: Schematic diagram of PFF with cell recycling.
95
96
5.5.1 PFF with Cell Recycling
A PFF requires the initial presence of microorganisms in the inlet stream as
a batch fermenter requires initial inoculum.
The most economical way to provide cells in the inlet stream is to recycle
the part of the outlet stream back to the inlet with or without a cell
separation device.
Figure 5.8 shows the schematic diagram of a PFF with cell recycling.
Unlike the CSTF, the PFF does not require the cell separator in order to
recycle, though its presence increases the productivity of the fermenter
slightly as will be shown later.
The performance equation of the PFF with Monod kinetics can be written
as:
𝐶𝑋𝑓
𝐶𝑋𝑓
𝜏𝑝
𝑉
𝑑𝐶𝑋
𝐾𝑆 + 𝐶𝑆 𝑑𝐶𝑋
=
=න
=න
(1 + 𝑅)𝐹 1 + 𝑅
𝑟𝑋
𝜇𝑚𝑎𝑥 𝐶𝑆 𝐶𝑋
𝐶𝑋
𝐶𝑋
(5.63)
97
Continued…
where 𝜏𝑝 is the residence time based on the flow rate of the overall
system. The actual residence time in the fermenter is larger than 𝜏𝑝 due to
the increase of the flow rate by the recycling.
If the growth yield is constant,
𝐶𝑆 =
𝐶𝑆′
1
−
(𝐶𝑋 − 𝐶𝑋′ )
𝑌𝑋/𝑆
(5.64)
Substituting Eq.(5.63) into Eq.(5.36) for 𝐶𝑆 and integrating it will result:
𝐾𝑆 𝑌𝑋
′
𝐾𝑆 𝑌𝑋/𝑆
𝐶𝑋𝑓
𝜏𝑃 𝜇𝑚𝑎𝑥
𝐶
𝑆
𝑆
=
+ 1 ln ′ + ′
ln
1+𝑅
𝐶𝑋 + 𝐶𝑆 𝑌𝑋
𝐶𝑋 𝐶𝑋 + 𝐶𝑆 𝑌𝑋 𝐶𝑆𝑓
𝑆
(5.65)
𝑆
where 𝐶𝑋′ and 𝐶𝑆 can be estimated from the cell and substrate balance at
the mixing point of the inlet and the recycle stream as
98
Continued…
𝐶𝑋′ =
𝐶𝑋𝑖 + 𝑅𝐶𝑋𝑅
1+𝑅
(5.66)
𝐶𝑆′ =
𝐶𝑆𝑖 + 𝑅𝐶𝑆𝑅
1+𝑅
(5.66a)
The cell concentrations of the outlet stream, can be estimated from the
overall cell balance
𝐶𝑆𝑓 =
1
[𝐶 + 𝑌𝑋 (𝐶𝑆𝑖 − 𝐶𝑆𝑓 )]
𝛽 𝑋𝑖
𝑆
(5.67)
The cell concentrations of the recycle stream, can be estimated from the
cell balance over the filter as
𝐶𝑋𝑅 =
1+𝑅+𝛽
𝐶𝑋𝑓
𝑅
(5.68)
where is 𝛽 the bleeding rate which is defined as
𝛽=
𝐵
𝐹
(5.69)
99
Continued…
Figure 5.18 shows the effect of the recycle rate (R) on the residence time
of a PFF system with recycling.
 Note that is calculated based on the inlet flowrate that is the true
residence time for the fermenter system.
 The actual 𝜏 in the PFF is unimportant because it decreases with the
increase of the recycle rate.
 When β = 1, the bleeding rate is equal to the flow rate and the flow rate
of filtrate stream L is zero, therefore, the recycle stream is not filtered.
 The residence time will be infinite if R is zero and decreases sharply as R
is increased until it reaches the point the decrease is gradual.
 In this specific case, the optimum recycle ratio may be somewhere
around 0.2.
100
Continued…
Figure 5.18: The effect of the recycle rate (R) and the bleeding rate (β = B/F)
on the residence time (𝜏 = 𝑉/𝐹 hr) of a PFF system with recycling.
101
Continued…
Another curve in Figure 5.18 is for 𝛽 = B/F = 1.8. The required residence
time can be reduced by concentrating the recycle stream 25 to 40% when R
is between 0.2 and 1.0.
When 𝑅 ≤ 1.2, the part of curve is noted as a dotted line because it may
be difficult to reduce the recycle ratio below 0.2 when 𝛽 = 0.8.
For example, in order to maintain R = 0.1, The amount of 1.3F needs to
be recycled and concentrated to 0.1F which may be difficult depending on
the cell concentration of the outlet.
The outlet stream of the cell settler will be equal to 𝐹 = 𝐵 + 𝐿 and its
𝐵
concentration will be
𝐶𝑋𝑓 = 𝛽𝐶𝑠𝑓 .
𝐹
102
Continued…
Example 5.1: Draw dimensionless growth curves (𝑋 against 𝜇𝑚𝑎𝑥 𝑡 ) for
𝐾𝑠 𝑌𝑥𝑆
= 0 , 0.2, and 0.5, when 𝑋0 = 0.05.
𝐶𝑥0 +𝑌𝑥𝑆 𝐶𝑠0
Solution:
In Figure E5.1 the dimensionless cell concentrations X are plotted on semi
logarithmic coordinates against the dimensionless cultivation time for the
𝐾𝑠 𝑌𝑥𝑆
different values of
. With an increase in these values, the growth
𝐶𝑥0 +𝑌𝑥𝑆 𝐶𝑠0
rate decreases and reaches the decelerating phase at an earlier time. Other
models can be used for estimation of the cell growth in batch fermentors.
103
Continued…
Figure E5.1: Dimensionless growth curves in batch fermenter
104
5.5.2 CSTF with Cell Recycling
 The cellular productivity in a CSTF increases with an increase in the
dilution rate and reaches a maximum value.
 If the dilution rate is increased beyond the maximum point, the
productivity will be decreased abruptly, and the cells will start to be
washed out because the rate of cell generation is less than that of cell loss
from the outlet stream.
 Therefore, the productivity of the fermenter is limited due to the loss of
cells with the outlet stream.
 One way to improve the reactor productivity is to recycle the cell by
separating the cells from the product stream using a cross-flow filter unit
(Figure 5.19).
 The high cell concentration maintained using cell recycling will increase
the cellular productivity since the growth rate is proportional to the cell
concentration.
105
106
107
Continued…
Figure 5.19: Schematic diagram of CSTF with cell recycling.
108
Continued…
 However, there must be a limit in the increase of the cellular productivity
with increased cell concentration because in a high cell concentration
environment, the nutrient-transfer rate will be decreased due to
overcrowding and aggregation of cells.
 The maintenance of the extremely high cell concentration is also not
practical because the filter unit will fail more frequently at the higher cell
concentrations.
 If all cells are recycled back into the fermenter, the cell concentration
will increase continuously with time and a steady state will never be
reached.
 Therefore, to operate a CSTF with recycling in a steady-state mode, we
need to have a bleeding stream, as shown in Figure 5.19.
 The material balance for cells in the fermenter with a cell recycling unit
is For a steady-state CSTF with cell recycling and a sterile feed,
109
Continued…
𝐹𝐶𝑋𝑖 − 𝐵𝐶𝑋 + 𝑉𝜇𝐶𝑋 = 𝑉
𝑑𝐶𝑋
𝑑𝑡
(5.70)
Figure 5.20: The effect of bleeding ratio on the cellular productivity (βDCX )·
110
Continued…
𝛽𝐷 =
𝛽
=𝜇
𝜏𝑚
(5.71)
Now, 𝛽 D instead of D is equal to the specific growth rate. When 𝛽 = 1
cells are not recycled, therefore, D = 𝜇.
If the growth rate can be expressed by Monod kinetics, substitution of Eq.
(5.11) into Eq. (5.71) and rearrangement for 𝐶𝑆 yields
𝐶𝑆 =
𝛽𝐾𝑆
𝜏𝑚 𝜇𝑚𝑎𝑥 − 𝛽
=𝜇
(5.72)
which is valid when 𝜏𝑚 𝜇𝑚𝑎𝑥 > 𝛽 The cell concentration in the fermenter
can be calculated from the value of Cs as
𝑌𝑋
𝐶𝑋 =
𝑆
𝛽
(𝐶𝑆𝑖 − 𝐶𝑆 )
(5.73)
Figure 5.20 shows the effect of bleeding ratio on the cell productivity for the
Monod model. As is reduced from 1 (no recycling) to 0.5, the cell
productivity is doubled.
111
5.6 Multiple Fermenters Connected in Series
 A question arises frequently whether it may be more efficient to use
multiple fermenters connected in series instead of one large fermenter.
 Choosing the optimum fermenter system for maximum productivity
depends on the shape of the 1/𝑟𝑋 versus 𝐶𝑋 curve and the process
requirement, such as the final conversion.
 In the 1/𝑟𝑋 versus 𝐶𝑋 curve, if the final cell concentration is less than
𝐶𝑋,𝑜𝑝𝑡 one fermenter is better than two fermenters connected in series,
because two CSTFs connected in series require more residence time than
one CSTF does.
 However, if the final cell concentration is much larger than 𝐶𝑋,𝑜𝑝𝑡 the
best combination of two fermenters for a minimum total residence time is
a CSTF operated at 𝐶𝑋,𝑜𝑝𝑡 followed by a PFF, as shown in (Figure 5.21a).
 A CSTF operated at 𝐶𝑋,𝑜𝑝𝑡 followed by another CSTF connected in series
is also better than one CSTF (Figure 5.12b).
112
Continued…
(a)
(b)
Figure 5.21: Graphical illustration of the total residence time required
(shaded area) when two fermenters are connected in series: (a) CSTF and
PFF, and (b) two CSTFs.
113
5.6.1 CSTF and PFF in Series
 Figure 5.22 shows the schematic diagram of the two fermenters
connected in series, CSTF followed by PFF. The result of the material
balance for the first fermenter is the same as the single CSTF which we
have already developed. If the input stream is sterile (𝐶𝑋𝑖 = 0), the
concentrations of the substrate, cell and product can be calculated from
Eqs. (5.74),(5.75) and (5.76) as follows:
𝐶𝑆1 =
𝐾𝑆
(5.74)
𝜏𝑚 𝜇𝑚𝑎𝑥 − 1
𝐶𝑋1 = 𝑌𝑋/𝑆
𝐶𝑆𝑖 −
𝐶𝑝1 = 𝐶𝑝𝑖 + 𝑌𝑃/𝑆
𝐾𝑆
𝜏𝑚1 𝜇𝑚𝑎𝑥 − 1
𝐶𝑆𝑖 −
𝐾𝑆
𝜏𝑚𝑖 𝜇𝑚𝑎𝑥 − 1
(5.75)
(5.76)
114
Continued…
Figure 5.22: Schematic diagram of the two fermenters,
CSTF and PFF, connected in series.
For the second PFF, the residence time can be estimated by
𝐶𝑋2
𝜏𝑃 2 = න
𝐶𝑋1
𝐶𝑋2
𝑑𝐶𝑋
𝐾𝑆 + 𝐶𝑆 𝑑𝐶𝑋
=න
𝑟𝑋
𝜇𝑚𝑎𝑥 𝐶𝑆 𝐶𝑋
𝐶𝑋1
(5.77)
115
Continued…
Since growth yield can be expressed as
𝑌𝑋/𝑆 =
𝐶𝑋2 − 𝐶𝑋1
𝐶𝑋1 − 𝐶𝑋2
(5.78)
Integration of Eq. (5.77) after the substitution of Eq.(5.78) will result,
𝜏𝑃2 𝜇𝑚𝑎𝑥
𝐾𝑆 𝑌𝑋
𝐾𝑆 𝑌𝑋/𝑆
𝐶𝑋2
𝐶𝑆1
𝑆
=
+ 1 ln
+
ln
𝐶𝑋𝑖 + 𝐶𝑆𝑖 𝑌𝑋
𝐶𝑋1 𝐶𝑋1 + 𝐶𝑆1 𝑌𝑋 𝐶𝑆2
𝑆
(5.79)
𝑆
If the final cell concentration ( 𝐶𝑋2 ) is known, the final substrate
concentration (𝐶𝑆2 ) can be calculated from Eq. (5.78).
The residence time of the second fermenter can then be calculated using Eq.
(5.79). If the residence time of the second fermenter is known, Eqs. (5.78)
and (5.79) have to be solved simultaneously to estimate the cell and
substrate concentrations.
Another approach is to integrate Eq. (5.50) numerically until the given 𝜏𝑃2
value is reached.
116
5.6.2 Multiple CSTFs in Series
 If the final cell concentration is larger than 𝐶𝑋,𝑜𝑝𝑡 , the best combination
of two fermenters for a minimum total residence time is a CSTF operated
at 𝐶𝑋,𝑜𝑝𝑡 followed by a PFF, as explained already.
 However, the cultivation of microorganisms in the PFF is limited to
several experimental cases, such as the tubular loop batch fermenter
(Russell et al., 1974) and scraped tubular fermenter (Moo-Young et al.,
1979).
 Furthermore, the growth kinetics in a PFF can be significantly different
from that in a CSTF.
 Another more practical approach is to use multiple CSTFs in series, since
a CSTF operated at 𝐶𝑋,𝑜𝑝𝑡 followed another CSTF connected in series is
also better than one CSTF.
 Hill and Robinson (1989) derived an equation to predict the minimum
possible total residence time to achieve any desired substrate conversion.
117
Continued…
 They found that three optimally designed CSTFs connected in series
provided close to the minimum possible residence time for any desired
substrate conversion.
 Figure 5.23 shows the schematic diagram of the multiple CSTFs
connected in series.
Figure 5.23: Schematic diagram of the multiple CSTFs
connected in series.
118
Continued…
For the nth steady-state CSTF, the material balance for the microorganisms
can be written as
𝐹 𝐶𝑋𝑛−1 − 𝐶𝑋𝑛 + 𝑉𝑛 𝑅𝑋𝑛 = 0
𝑟𝑋𝑛 =
𝜇𝑚𝑎𝑥 𝐶𝑆𝑛 𝐶𝑋𝑛
𝐾𝑆 + 𝐶𝑆𝑛
(5.80)
(5.81)
Growth yield can be expressed as
𝑌𝑋/𝑆 =
𝐶𝑋𝑛 − 𝐶𝑋𝑛−1
𝐶𝑆𝑛−1 − 𝐶𝑆𝑛
(5.82)
By solving Eqs. (5.80), (5.81), and (5.82) simultaneously, we can calculate
either dilution rate with the known cell concentration, or vice versa.
From Eq. (5.80), the dilution rate of the first reactor when the inlet stream is
sterile is
𝐹 𝑟𝑋1
𝐷1 = =
𝑉1 𝐶𝑋1
(5.83)
119
Continued…
which can be represented by the slope of the straight line connecting the
origin and 𝐶𝑋𝑖 𝑆𝑋𝑖 in Figure 5.24. Similarly, for the second fermenter
𝑟𝑋2
𝐹
𝐷2 =
=
𝑉2 𝐶𝑋2 − 𝐶𝑋1
(5.84)
which is the slope of the line connecting (𝐶𝑋1 , = 0) and (𝐶𝑋2 , 𝑟𝑋2 ).
 Therefore, by knowing the dilution rate of each fermenter, you can
estimate the cell concentration of each fermenter, or vice versa.
Figure 5.24: Graphical solution for a twostage continuous fermentation
120
5.10 Thermal-Death Kinetics of Cells and
Spores
 The activity of cells, spores, and viruses in air or liquids can be
diminished by destruction (heat, radiation, or chemical or mechanical
means), removal (filtration or centrifugation), or inhibition (refrigeration,
dessication, water depletion, chemicals).
 Figure 5.38 illustrates experimental data for vegetative-cell and spore
death over a range of temperatures. In general, cells have a greater
susceptibility for destruction than spores, consistent with earlier
qualitative remarks concerning endospore formation as a survival
mechanism of some cells.
 The death of a particular cell is probably due to the thermal denaturation
of one or more kinds of essential proteins such as enzymes. The kinetics
of such cooperative transitions for these molecules may be rather
complex in time. Also, the rate of the molecular processes ultimately
leading to cell death will depend on environ ment composition, including
the solvent concentration itself.
121
Continued…
Figure 5.2938: Experimental data for
thermal death of E. coli (a) and
inactivation of vegetative spores of
Bacillus subtilis (b) and B.
sterothermophilus (c).
122
Continued…
The most reliable data come from measurements on systems approximating
those of the application as closely as possible.
To begin analysis of death-rate kinetics, note that the straight lines
through the data in the semilog plots in Fig. 5.38 are consistent with a firstorder decay for the viable population level n
𝑑𝑛
= −𝑘𝑑 𝑛
𝑑𝑡
(5.154)
𝑛(𝑡) = 𝑛0 𝑒 −𝑘𝑑 𝑡
(5.155)
which, for constant 𝑘𝑑 , yields
where n is the concentration of spores or vegetative cells at 𝑡 = 0. The slope
of the semilog plots is − 𝑘𝑑 ; a plot of ln𝑘𝑑 vs. reciprocal temperature
provides another straight line, so that the temperature dependence of 𝑘𝑑 ,
may be taken as that of the familiar Arrhenius form
123
Continued…
Table 5.9: Time required for coagulation of albumin as a
function of water content
Water content, %
Approximate coagulation 𝑇, ℃
50
56
25
76
15
96
5
149
0
165
124
Continued…
Figure 5.39: Thermal
destruction of S. aureus.
125
Continued…
𝑘𝑑 = 𝑘𝑑0 𝑒 −𝐸𝑑 /𝑅𝑇
(5.156)
The range of E values for many spores and vegetative cells lies between 50
and 100 𝑘𝑐𝑎𝑙/𝑚𝑜𝑙.
In a deterministic model, the fraction of the population remaining viable
(assuming no current growth under the applied lethal conditions) is seen
from the previous equation to be
𝑛
𝑉𝑖𝑎𝑏𝑙𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 =
= 𝑒 −𝑘𝑑 𝑡
𝑛0
(5.157)
If each organism's fate is independent of the others in that no reproduction
occurs, and if the lethal condition is uniformly applied, a stochastic
approach to sterilization shows that the probability that at any time the
remaining population contains N viable organisms is given by
126
Continued…
𝑃𝑁 (𝑡) =
𝑁0 !
(𝑒 −𝑘𝑑 𝑡 )(1 − 𝑒 −𝑘𝑡 )(𝑁0−𝑁)
𝑁0 − 𝑁! 𝑁!
(5.158)
with 𝑁0 the initial viable number of organisms in the fluid being sterilized.
The 𝑘𝑑 , which appears here is the same as the rate constant in Eq.
(5.158): in the stochastic model 𝑘𝑑 may be interpreted as the reciprocal of
the mean life span of the organism. As the number of organisms falls to a
low value, the assumptions of a uniform population characterizable by a
single 𝑘𝑑 value begins to fail.
Setting 𝑁 = 0 in Eq. (5.158) gives
𝐸𝑥𝑡𝑖𝑛𝑐𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 = 𝑃0 (𝑡) = 1 − (1 − 𝑒 −𝑘𝑑 𝑡 ))𝑁0
(5.159)
from which it follows that the probability of at least one organism's survival
is
(5.160)
1 − 𝑃0 (𝑡) = (1 − 𝑒 −𝑘𝑑 𝑡 )𝑁0
127
Continued…
For the usual situation of 𝑁0 ≫ 1, this becomes
1 − 𝑃0 (𝑡) = 1 − 𝑒 −𝑁𝑡
(5.161)
where
𝑁𝑡 = 𝑁0 𝑒 −𝑘𝑑 𝑡
(5.162)
We can interpret 1 − 𝑃0 physically as the fraction of sterilizations which are
expected to fail to produce a contaminant-free product.
A statistical analysis shows that the expected value of the population size is
𝐸[𝑁(𝑡)] = 𝑁0 (1 − {1 − exp 𝑘𝑑− 𝑡 𝑡 }𝑟 )𝑤
(5.163)
where 𝜔 = number of kinds of essential subcellular structures
i = number of units of each structure in each organism
𝑘𝑑− 𝑡 = mean specific death rate for each kind of structure
128
Continued…
Note that this expression expands into a series of decaying exponentials
and is not thus expected to be linear over the major fraction of a semilog
plot.
129
Thank you
6.1 Sterilization
 Sterilization of fermentation media or equipment can be accomplished by
destroying all living organisms by means of heat (moist or dry), chemical
agents, radiation (ultraviolet or X-rays), and mechanical means (some or
ultrasonic vibrations). Another approach is to remove the living
organisms by means of filtration or high-speed centrifugation.
 Heat is the most widely used means of sterilization, which can be
employed for both liquid medium and heatable solid objects. It can be
applied as dry or moist heat (steam). The moist heat is more effective
than the dry heat, because the intrinsic heat resistance of vegetative
bacterial cells is greatly increased in a completely dry state.
 As a result, the death rate is much lower for the dry cells than for moist
ones. The heat conduction in dry air is also less rapid than in steam.
Therefore, dry heat is used only for the sterilization of glassware or
heatable solid materials.
1
Continued…
 Chemical agents can be used to kill microorganisms as the result of their
oxidizing or alkylating abilities. However, they cannot be used for the
sterilization of medium because the residual chemical can inhibit the
fermentation organisms.
 Many cellular materials absorb ultraviolet light, leading to DNA damage
and consequently to cell death. Wavelengths around 265 nm have the
highest bactericidal efficiency.
 Sonic or ultrasonic waves of sufficient intensity can disrupt and kill cells.
This technique is usually employed in the disruption of cells for the
purpose of extracting intracellular constituents rather than as a
sterilization technique.
 Filtration is most effectively employed for the removal of
microorganisms from air or other gases.
2
6.2 Pasteurization
 Compared to sterilization, pasteurizing is a comparatively low order of
heat treatment, generally at a temperature below the boiling point of
water. The general objective of pasteurization is to extend product shelflife by inactivating all non-spore-forming pathogenic bacteria and the
majority of vegetative spoilage microorganisms, as well as inhibiting or
stopping microbial and enzyme activity.
 To be effective, pasteurization is frequently combined with another
means of preservation such as concentration, acidification, chemical
inhibition, etc.
 In pasteurizing, two types of processes can be used: slow and rapid. Slow
pasteurization uses pasteurization temperatures for several minutes; e.g.
typical temperature–time combinations are 63 to 65°C over 30 minutes
or 75°C over 8 to 10 minutes.
 Rapid, high or flash pasteurization uses pasteurization temperatures of
about 85 to 90°C or more for a time only in the order of seconds.
3
6.3 Differences between sterilization and
pasteurization
 In general terms, we can say that both pasteurization and sterilization are
conservation techniques that are based on the eradication of
microorganisms and enzymes by applying heat for a certain time to foods
placed in hermetic containers.
 Its main difference lies in the fact that sterilization seeks to eliminate all
microorganisms and spores, while in pasteurization, the most resistant
forms and some spores remain present.
 Therefore, sterilized foods are kept at room temperature for long periods
of time, while pasteurized foods, in contrast, require refrigerated
preservation to delay the proliferation of possible microorganisms or
spores present in the food.
4
Continued…
The characteristics of the food
 Each product has different characteristics that will determine the type of
sterilization technique to be used and the combinations of temperature
and time to try to preserve the maximum properties and minimize the
nutritional changes of the packaged food.
 One of these properties is the acidity level of the food (pH). Foods with a
higher pH need high temperatures, such as vegetables, meat or fish.
 Juices, for example, have an acidic pH, which conditions the growth of
different types of pathogens, especially sporulated microorganisms, the
most resistant to high temperatures. This means that milder temperatures
can be applied and therefore they can be pasteurized.
 On the other hand, the physical shape of the food also determines the
technique to be applied, since, for example, the spherical shape requires
more temperature.
5
Continued…
Food shelf life
• As we have already mentioned, in sterilization, the product can be kept in
perfect condition for more than four months and in pasteurization, from
two to three weeks and always kept cold.
6
6.4 Batch Sterilization
Let us begin by considering a well-mixed closed volume containing a cell or
spore suspension. The fluid is to be sterilized by heating, and then cooled to
a suitable temperature for subsequent processing. The concentration of
surviving organisms resulting from this process can readily be computed
−𝐸𝑑
𝑑𝑛
= −𝑘𝑑0 𝑒 𝑅𝑇(𝑡) 𝑛
𝑑𝑡
6.1
where we have explicitly included time variations in the fluid temperature.
Separating the variables in Eq. (6.1) and integrating, we find
𝑡𝑡 −𝐸𝑑
𝑛0
ln = න 𝑒 𝑅𝑇(𝑡) 𝑑𝑡
𝑛𝑓
0
6.2
where the f subscript denotes final conditions.
Common batch-sterilization designs include one or more of the following
heat sources:
7
Continued…
steam sparging (bubbling of live steam through the medium), electrical
heating, and heating or cooling with a two-fluid heat exchanger. Dein
doerfer and Humphrey have associated with each heating or cooling mode a
particular time-temperature profile; these functions are shown in Table 6.1.
The integral on the right-hand side of Eq. (6.2) can be evaluated by
segmentation into three intervals of heating, holding, and cooling. A total of
four integral forms arise: constant temperature and each of the three
transient modes (hyperbolic, linear, and exponential).
Constant temperature
6.3
𝑡𝑡
−𝐸𝑑
−𝐸𝑑
𝑛𝑓
ln = න 𝑘𝑑0 𝑒 𝑅𝑇 𝑑𝑡 = 𝑘𝑑0 𝑡𝑓 𝑒 𝑅𝑇
𝑛0
0
8
Continued…
Table 6.1: Temperature – time profile in batch sterilization
Type of heat transfer
Stream spring
Temperature-time profile
𝑎𝑡
𝑇 = 𝑇0 1.0 +
1 + 𝑏𝑡
Parameter
ℎ𝑠
𝑎 = 𝑀𝑇 𝜌𝐶
0
𝑃
𝑠
𝑏=𝑀
,
hyperbolic
Electrical heating
𝑇 = 𝑇0 1.0 + 𝑎𝑡
𝑎=
𝑞
𝑀𝑇0 𝜌𝐶𝑃
linear
Steam (heat exchanger)
𝑇 = 𝑇0 1.0 + 𝑏𝑒 −𝑎𝑡
𝑈𝐴
𝑎 = 𝑀𝑇 𝜌𝐶
0
𝑃
𝑏=
𝑇0 −𝑇𝑁
𝑇𝑁
exponential
Coolant (heat exchanger)
𝑇 = 𝑇0 1.0 + 𝑏𝑒 −𝑎𝑡
exponential
𝑤𝑐 ′
′
𝑎=
(1 − 𝑒 −𝑈𝐴/𝑤𝑐 )
𝑀𝑇0 𝜌𝐶𝑃
𝑏=
𝑇0 −𝑇𝑒0
𝑇𝑒0
9
Continued…
Hyperbolic
𝑛𝑓
ln
𝑛0
6.4
𝐸𝑑
1 + 𝑏𝑇𝑓
𝐸𝑑
𝑎
𝑅𝑇0 −(𝐸𝑑 /𝑅𝑇0)[𝑏/(𝑎+𝑏)]
=
𝑒
×
𝐸
−
2
(𝑎 + 𝑏)2
𝑅𝑇0 1 + 𝑎 + 𝑏 𝑡𝑓 𝑎 + 𝑏
𝑘𝑑0 𝑎
− 𝐸2
𝐸𝑑 𝑎
𝑅𝑇0 𝑎 + 𝑏
Linear increasing
𝑛𝑓 𝑘𝑑0 𝐸𝑑
ln =
𝑛0
𝑅𝑇0
Exponential
ln
𝐸𝑑
𝐸𝑑
𝑅𝑇0
𝐸2
− 𝐸2
1 + 𝑎𝑡𝑓
𝑅𝑇0
6.5
𝑛𝑓
𝑛0
𝐸𝑑
𝐸𝑑
𝑘𝑑0
𝑅𝑇𝐻
𝑅𝑇𝐻
=
𝐸1
− 𝐸1
𝑎
1+𝑏
1 + 𝑏𝑒 −𝑎𝑡𝑓
𝐸𝑑
𝑘𝑑0 −𝑅𝑇
𝐸𝑑 𝑏
𝐸𝑑 𝑏𝑒 −𝑎𝑡𝑓
𝐻 −𝐸
−
𝑒
+ 𝐸1
1
𝑎
𝑅𝑇0 1 + 𝑏
𝑅𝑇𝐻 1 + 𝑏𝑒 −𝑎𝑡𝑓
6.6
10
Continued…
Table E6.1 Temperature versus time relations in batch sterilization by various heating
Heating methods
Heating by direct steam sparging
Heating with a constant rate of heat flow, for
example, electric heating
Indirect heating by saturated steam
Temperature (T)-time (t) relations
Hms t
T = T0 +
cP (M + ms t)
qt
T = T0 +
cP M
T = TN + T0 − TN exp −
UA
t
McP
A = area for heat transfer; cP = specific heat capacity of medium; H=heat content of steam
relative to initial medium temperature; ms = mass flow rate of steam; M = Initial mass of
medium; q = rate of heat transfer; t = time: T = temperature; T0 = initial temperature of
medium; TN = temperature of heat source; and U = overall heat transfer coefficient.
11
Continued…
Example 6.1: Derive an equation for the temperature-time relationships of a
medium in a fermentor during indirect heating by saturated steam [Figure
E6.1(c)]. Use the nomenclature given in Table E6.1.
Figure E6.1: Modes of heat transfer for batch sterilization: (a) direct steam sparging,
b) constant rate heating by electric heater, (c) heating by condensing steam (isothermal
heat source), and (d) cooling by water (nonisothermal heat sink).
12
Continued…
Solution:
The rate of heat transfer q to the medium at 𝑇 ℃ from steam condensing at
TN ℃ is
𝑞 = 𝑈𝐴(TN − 𝑇)
Transferred heat raises the temperature of the medium (mass M and specific
heat cP ) at a rate dT/dt. Thus,
𝑀cP
𝑑𝑇
= 𝑈𝐴(TN − 𝑇)
𝑑𝑡
Integration and rearrangement give
T = TN + T0 − TN exp −
UA
t
McP
where T0 is the initial temperature of the medium.
13
6.5 Continuous Sterilization
Two basic types of continuous-sterilizer designs are shown schematically in
Fig 6.2. In Fig. 6.2(a), direct heating is provided by steam injection, with the
heated fluid then passing through a holding section before cooling by
expansion. The system in Fig. 6.2(b) features indirect heating using a plate
heat exchanger.
Two different design approaches are available if we assume that the
temperature in the continuous sterilizer is approximately uniform. This may
be valid at least in the holding sections. In the first method we apply the
dispersion model,
The resulting value of n at sterilizer effluent is given by
𝑛(𝐿)
=
𝑛0
4𝑦 exp[𝑃𝑒/2]
𝑃𝑒 𝑦
2 exp − 𝑃𝑒 𝑦
(1 + 𝑦)2 exp
−
1
−
𝑦
2
2
6.16
14
Continued…
Figure 6.2: Two different
continuous sterilizer designs:
(a) Direct steam injection
(b) Plate heat exchanger.
15
Continued…
With
4 Da
𝑦 = 1+
Pe
1/2
6.17
Where here the DamkÖhler number Da is defined by
𝑘𝑑 𝐿
𝐷𝑎 =
𝑢
6.18
For small deviations from plug flow (Pe−1 small), Eq. (6.16) reduces to the
simpler form
𝑛(𝐿)
𝐷𝑎 2
= exp −𝐷𝑎 +
𝑛0
𝑃𝑒
6.19
These solutions are conveniently displayed as a plot of remaining viable
fraction 𝑛(𝐿)/𝑛0 vs. the dimensionless group Da for various values of the
Peclet number (Fig. 6.3)
16
Continued…
Figure 6.3: Effect of axial
dispersion on organism
destruction in a continuous
sterilizer.
17
Continued…
We can compute the effluent surviving-organism concentration n. Rewriting
this equation in terms of organism concentration, we have
∞
𝑛 = න 𝑛𝑏 𝑡 𝜉 𝑡 𝑑𝑡
6.20
0
where 𝜉 𝑡 is the RTD of the continuous sterilizer. Recalling that 𝑛𝑏 𝑡 is
the organism concentration at time t in a batch sterilizer with 𝑛𝑏 𝑡 = 𝑛0 ,
∞
𝑛
= න 𝜉 𝑡 𝑒 −𝑘𝑑 𝑡 𝑑𝑡
𝑛0
0
6.21
In some instances evaluation of the right-hand side of Eq. (6.21) is
facilitated by noting that it is formally identical to the Laplace transform of
𝜉 with the usual Laplace transform parameter s replaced by 𝑘𝑑 .
18
6.5.1 Plate Heat Exchanger Type Sterilization
 Plate heat exchangers (figure 6.2b) are systems that separates two air
flows with solid elements (plates).
 The heat exchange occurs through the plates, which can be built with fins
to maximize the contact surface and increase the exchange efficiency.
 Depending on the flows direction the system will be parallel flows or
counter current flows.
 The plate heat exchanger efficiency is around 75%.
19
6.5.2 Direct Steam Injection Type Sterilization
• During the direct high-heating (figure 6.2a), steam is injected into the
product by means of a DSI unit (DSI = direct steam injector).
• In this way, the product is suddenly heated and flashed against a low
pressure after a very short holding period and in this way it is cooled
again just as quickly.
• This sterilization process shows significant advantages compared to an
indirect high-heating by means of heat exchangers.
• In case of indirect high heating the holding time at high temperatures is
much longer.
• The short retention time and the possibility to treat high viscous products
are the main advantages of this technology.
20
Continued…
Example 6.2: A medium is to be continuously sterilized at a flow rate of
2 m³ h−1 in a sterilizer by direct steam injection (Figure E6.2a). The
temperature of a holding section is maintained at 120°C, and the time for
heating and cooling can be neglected. The bacterial count of the entering
medium, 2 × 1012 m−3 , must be reduced to such an extent that only one
organism can survive during 30 days of continuous operation. The holding
section of the sterilizer is a tube, 0.15 m in internal diameter. The specific
death rate of bacterial spores in the medium is 121 h−1 at 120°C, the
medium density ρ = 950 kgm−3 ; the medium viscosity = 1 kgm−1 h−1 .
Calculate the required length of the holding section of the sterilizer
a) assuming ideal plug flow;
b) considering the effect of the axial dispersion.
21
Continued…
Figure E6.2.1: Continuous sterilizing systems. (a) Heat exchanger and
continuous steam injection and (b) heat exchanger and indirect heater.
22
Continued…
Solution:
(a)
12
2 × 10
𝑛0
𝑙𝑛
= ln
𝑛
𝑚
−3
𝑚3
ℎ
×2
× 24
× 30 𝑑𝑎𝑦
ℎ
𝑑𝑎𝑦
1
= 35.6
Average holding time:
𝜏ℎ𝑜𝑙𝑑
𝑛
𝑙𝑛 𝑛0
=
= 0.294 ℎ
𝑘𝑑
The average medium velocity in the holding section:
2
𝑣=
= 113 𝑚ℎ−1
2
(𝜋 × 0.075 )
The required length of the holding section:
𝐿 = 131 × 0.294 = 33.2 m
23
Continued…
Figure E6.2.2: Correlation for axial dispersion coefficient
in pipe flow, (Pe) −1 versus (Re)
24
Continued…
𝑛
Figure E6.2.3: Theoretical plots of 𝑛 as a function of (Pe) and (Da).
0
25
Continued…
(b) The Reynolds number of the medium flowing through the holding tube
is
0.15 × 113 × 950
𝑅𝑒 =
1
= 1.64 × 104
Figure E6.2.2 (Pe) is approximately 2.5 at Re = 1.64 × 104 , and
𝐸𝐷𝑧
(113 × 0.15)
=
= 6.8
2.5
Assuming a length of the holding section of 36 m, the Peclet number is
given as
And
𝑣𝐿
(113 × 36)
(𝑃𝑒 ) =
=
= 598
𝐸𝐷𝑧
6.8
𝑘𝑑 𝐿 (121 × 36)
(𝐷𝑎 ) =
=
= 38.5
𝑣
113
From Figure E6.2.3, the value of
𝑛
,
𝑛0
is about 3 × 10−16 and almost equal to
that required in this problem. Thus, the length of the holding section should
be 36 m.
26
Aeration and agitation
1
7.1 Gas-liquid mass transfer
 A sparingly soluble gas, usually Oxygen, is transferred from a source, say
a rising air bubble, into a liquid phase containing cells.
These arise from different combinations of the following resistances
1. Diffusion from bulk gas to the gas-liquid interface (Fig.
2. Movement through the gas-liquid interface.
3. Diffusion of the solute through the relatively unmixed liquid region
adjacent to the bubble into the well-mixed bulk liquid.
4. Transport of the solute through the bulk liquid to a second relatively
unmixed liquid region surrounding the cells
5. Transport through the second unmixed liquid region associated with the
cells.
6. Diffusive transport into the cellular floc, mycelia, or soil particle.
7. Transport across cell envelope and to intracellular reaction site.
2
Continued…
Figure 7.1: Schematic diagram of steps involved in transport of oxygen from a
gas bubble to inside a cell
3
7.1.2 Mass-Transfer Coefficient
The mass flux, the rate of mass transfer 𝑞𝑐 per unit area, is proportional to a
concentration difference. If a solute transfers from the gas to the liquid
phase, its mass flux from the gas phase to the interface 𝑁𝐺 is
𝑁𝐺 =
𝑞𝑐
= 𝑘𝐺 𝐶𝐺 − 𝐶𝐺𝑖
𝐴
7.5
where 𝐶𝐺 and 𝐶𝐺𝑖 , is the gas-side concentration at the bulk and the interface,
respectively,
Figure 7.2: Concentration profile near a gaas-liquid interface and an equilibrium curve
4
Continued…
Similarly, the liquid-side phase mass flux 𝑁𝐿 is
𝑁𝐿 =
𝑞𝐿
= 𝑘𝐿 𝐶𝐿𝑖 − 𝐶𝐿
𝐴
7.6
Since the amount of solute transferred from the gas phase to the interface
must equal that from the interface to the liquid phase,
𝑁𝐺 = 𝑁𝐿
7.7
Substitution of Eq. (7.5) and Eq. (7.6) into Eq. (7.7) gives
𝐶𝐺 − 𝐶𝐺𝑖
𝑘𝐿
=
𝐶𝐿𝑖 − 𝐶𝐿
𝑘𝐺
7.8
which is equal to the slope of the curve connecting the (𝐶𝐿 , 𝐶𝐺 ) and
(𝐶𝐿𝑖 , 𝐶𝐺𝑖 ), as shown in Figure 7.2.
5
Continued…
It is hard to determine the mass-transfer coefficient according to Eq. (7.5) or
Eq. (7.6) because we cannot measure the interfacial concentrations, 𝐶𝐿𝑖 or
𝐶𝐺𝑖 Therefore, it is convenient to define the overall mass-transfer coefficient
as follows:
𝑁𝐺 = 𝑁𝐿 = 𝑘𝐺 𝐶𝐺 − 𝐶𝐺∗ = 𝑘𝐿 𝐶𝐿∗ − 𝐶𝐿
7.9
Where,
𝐶𝐺∗ is the gas-side concentration
𝐶𝐿∗ is the liquid-side concentration.
Figure 7.3: The Equilibrium curve explaining
the meaning of 𝐶𝐺∗ and 𝐶𝐿∗
6
Oxygen absorption rate
The oxygen absorption rate per unit volume q/ v can be estimated by
To maximize the oxygen absorption rate, we have to maximize KL, a, CL*- CL.
However, the concentration difference is quite limited for us to control because the
value of CL* is limited by its very low maximum solubility. Therefore,
the main parameters of interest in design are the mass-transfer coefficient and the
interfacial area.
The maximum concentration of oxygen in water which is in equilibrium with air CL* at
atmospheric pressure can be determined according to the Henry's law,
7
7.2 Bubble Column
Akita and Yoshida (1973) correlated the volumetric mass-transfer
coefficient 𝑘𝐿𝑎 for the absorption of oxygen in various aqueous solutions in
H : vol. fraction of gas phase
bubble columns, as follows:
𝑘𝐿 𝑎 = 0.6𝐷𝐴𝐵 0.5 𝑉𝑐 −0.12
𝜎
𝜌𝑐
−0.62
𝐷𝑇0.17 𝑔0.93 𝐻1.1
Vc the
which applies to columns with less effective spargers.
In bubble columns, for 0 < 𝑉𝑠 < 0.15
al. (1980) correlated the 𝑘𝐿 𝑎 as
𝑚
𝑎𝑛𝑑
𝑠
7.10
kinematic
viscosity of the
continuous phase
100 < 𝑃𝑔 / 𝑣 < 1100 𝑊/𝑚3, Botton et
𝑃𝑔
𝑘𝐿 𝑎
= 𝑣
0.08
800
0.75
Vs is superficial gas velocity
V is the volume of the dispersion
7.11
Where 𝑃𝑔 is the gas power input, which can0.75
be calculated from
𝑃𝑔
𝑉𝑠
= 800
𝑣
0.1
7.12
8
7.2.1 Mechanically Agitated Vessel
For aerated mixing vessels in an aqueous solution, the mass-transfer
coefficient is proportional to the power consumption (Lopes De Figueiredo
and Calderbank, 1978) as
𝑃𝑚
𝑘𝐿 ∝
𝑣
0.33
7.13
The interfacial area for the aerated mixing vessel is a function of agitation
conditions. Therefore,
𝑃𝑚
𝑎∝
𝑣
0.4
𝑉𝑠 0.5
7.14
by combining the above two equations, 𝑘𝐿 𝑎 will be
𝑃𝑚
𝑘𝐿 𝑎 ∝
𝑣
0.77
𝑉𝑠 0.5
7.15
9
Continued…
Numerous studies for the correlations of 𝑘𝐿 𝑎 have been reported and their
results have the general form as
𝑘𝐿 𝑎 = 𝑏1
𝑃𝑚
𝑣
𝑏2
𝑉𝑠 𝑏3
7.16
where 𝑏1 , 𝑏2 and 𝑏3 vary considerably depending on the geometry of the
system, The values of 𝑏2 and 𝑏3 are generally between 0 to 1 and 0.43 to
0.95, respectively, as tabulated by Sideman et al. (1966).
Van't Riet (1979) reviewed the data obtained by various investigators
and correlated them as follows:
1. For "coalescing" air-water dispersion,
𝑃𝑚
𝑘𝐿 𝑎 = 0.026
𝑣
0.4
𝑉𝑠 0.54
7.17
10
Continued…
2. For "noncoalescing" air-electrolyte solution dispersions,
𝑃𝑚
𝑘𝐿 𝑎 = 0.002
𝑣
0.7
𝑉𝑠 0.2
7.18
both of which are applicable for the volume up to 2.6 𝑚3 ; for a wide variety
𝐷
of agitator types, sizes, and 𝐼 ratios; and 500 < 𝑃𝑚/𝑣 <10,000 𝑊/𝑚3 .
𝐷𝑇
These correlations are accurate within approximately 20 percent to 40
percent.
11
7.3 Design and Power requirement
The power consumption for mechanical agitation can be measured using
a torque table as shown in Figure 7.4.
The torque table is constructed by placing a thrust bearing between a
base and a circular plate, and the force required to prevent rotation of the
turntable during agitation F is measured. The power consumption P can be
calculated by the following formula
𝑃 = 2𝜋𝑟𝑁𝐹
7.19
Figure 7.4: A torque table to measure the power consumption for mechanical agitation
12
Continued…
where N is the agitation speed, and r is the distance from the axis to the
point of the force measurement.
Power consumption by agitation is a function of physical properties,
operating condition, and vessel and impeller geometry. Dimensional
analysis provides the following relationship:
𝑃
𝜌𝑁𝐷2 𝑁 2 𝐷𝐼 𝐷𝑇 𝐻
=𝑓
,
,
, ,…
𝜇
𝑔 𝐷𝐼 𝐷𝐼
𝜌𝑁 3 𝐷𝐼5
7.20
The dimensionless group in the left-hand side of Eq. (7.20) is known as
power number 𝑁𝑃 . The first term of the right-hand side of Eq. (7.20) is the
impeller Reynolds number 𝑁Re𝑖 and the second term is the Froude number
𝑁𝐹𝑟 . The gravity force affects the power consumption due to the formation
of the vortex in an agitating vessel. The vortex formation can be prevented
by installing baffles.
13
Continued…
For fully baffled geometrically similar systems, the effect of the Froude
Number on the power consumption is negligible and all the length ratios in
Eq. (7.20) are constant. Therefore, Eq. (7.20) is simplified to
𝑁𝑃 = 𝛼 𝑁𝑅𝑒
𝛽
7.21
Figure 7.5 shows Power number-Reynolds number correlation in an agitator
with four baffles (Rushton et al., 1950) for three different types of impellers.
The power number decreases with an increase of the Reynolds number
and reaches a constant value when the Reynolds number is larger than
10,000. At this point, the power number is independent of the Reynolds
number.
Therefore, Eq. (7.21) is simplified to,
𝑁𝑃 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑓𝑜𝑟
𝑁𝑅𝑒 > 10,000
7.22
14
Continued…
Figure 7.5: Power number-Reynolds number correlation in an agitator with four baffles.
15
Continued…
The power required by an impeller in a gas sparged system Pm is
usually less than the power required by the impeller operating at the
same speed in a gas-free liquids Pmo.
The 𝑃𝑚 for the flat-blade disk turbine can be calculated from 𝑃𝑚0 (Nagata,
1975), as follows:
𝑃𝑚
𝐷𝐼
log 10 =
= −192
𝑃𝑚0
𝐷𝑇
4.38
0.115
𝐷𝐼2 𝑁
𝑣
𝐷
1.96 𝐷 𝐼
2
𝑇
𝐷𝐼 𝑁
9
𝑄
𝑁𝐷𝐼3
7.23
16
Continued…
Example 7.2: Calculate the power requirements, with and without aeration,
of a 1.5m diameter stirred tank, containing water 1.5m deep, equipped with
a six-blade Rushton turbine that is 0.5m in diameter d, with blades 0.25 d
long and 0.2 d wide, operating at a of 180 rpm. Air is supplied from the tank
bottom at a rate of 0.6 m3 min−1 . Operation is at room temperature. Values
of water viscosity μ = 0.001 kg m−1 s −1 and water density ρ =
μ
1000 kgm−3 ; hence = v = 10−6 m2 s −1 can be used.
ρ
Solution:
(a) The power requirement without aeration can be obtained using Figure
E7.2.
𝑁
3
2
2
5
𝑅𝑒 = 𝑑
𝑣
= 0.5 ×
10−6
= 7.5 × 10
This is in the turbulent regime. Then, from Figure E7.2,
𝑁𝑃 = 6
P0 = 6ρN 3 d5 = 6 1000 33 0.5
5
= 5060 kgm2 s−3 = 5060 W
17
Continued…
Figure E7.2: Correlation between Reynolds number (Re) and Power number (Np).
Curve (a): six-flat blade turbine, four baffles, W=0.1 D;
curve (b): two-flat blade paddle, four baffles, W=0.1 D;
curve (c): three-blade marine propeller, four baffles, W=0.1 D.
18
Continued…
(b) Power requirement with aeration is estimated using Equation 7.51,
log
𝑃𝐺
1
= −192
𝑃0
3
4.38
0.52 ×
3
10−6
0.115
2
× 0.5 ×
3
9.8
1.96
3
0.01
3 × 0.53
= −0.119
𝑃𝐺
= 0.760
𝑃0
Here,
PG = 5060 × 0.760 = 3850 W
For comparison, calculation by eq.
𝑃𝐺
𝑄
= 0.10
𝑃0
𝑁𝑉
Gives
𝑃𝐺
= 0.676
𝑃0
−1/4
𝑁 2 𝑑4
−1/5
2
𝑔𝑏𝑉 3
PG = 3420 W
19
Continued…
Example 7.3: Estimate 𝑘𝐿 𝑎 for oxygen absorption into water in the sparged
stirred tank of Example 7.2. Operating conditions are the same as in
Example 7.2.
Solution:
Equation 𝑘𝐿 𝑎 𝑠
−1
= 0.026
𝑃𝐺 0.4 0.5
𝑈𝐺
𝑉
is used.
From Example 7.2
PG = 3850 W
volume of water = 2.65 𝑚3
𝑃𝐺 3850
𝑃𝐺
=
= 1453 𝑊𝑚−3 ,
𝑉
2.66
𝑉
0.4
= 18.40
Since the sectional area of the tank = 1.77 𝑚²,
0.6
𝑈𝐺 =
= 0.00565 𝑚𝑠 −1 , 𝑈𝐺
(1.77 × 60)
0.5 =
0.0752
20
Continued…
k L a = 0.026 1453
0.4
(0.00565)0.5 = 0.0360 s−1 = 130 h−1
Although this solution appears simple, it requires an estimation of PG which
was given in Example 7.2.
21
Continued…
Example 7.4: A stirred-tank reactor equipped with a standard Rushton
turbine of the following dimensions contains a liquid with density ρ =
1.0 g cm−3 and viscosity μ = 0.013 gcm−1 s −1 . The tank diameter D =
2.4 m, liquid depth HL = 2.4m, impeller diameter d = 0.8 m, and liquid
volume = 10.85 m³. Estimate the stirrer power required and the mixing
time, when the rotational stirrer speed N is 90 rpm, that is, 1.5 s −1 .
Solution:
The Reynolds number
𝜌𝑁𝑑 2
1
𝑅𝑒 =
= 1.5 × 802 ×
= 7.38 × 105
𝜇
0.013
From Figure E7.2 the power number NP = 6.
22
Continued…
Figure E7.4: Correlations for mixing times
(with standard Rushton turbine).
23
Continued…
The power required
P = 6ρN 3 d5 = 6 × 1000 × 1.53 × 0.8
5
= 6650 kgm2 s−3 = 6650 W
P = 6.65 KW
From Figure E7.4 values of Nt m , for the above Reynolds number should be
about 30. Then, t m = 30/1.5 = 20s.
24
7.4 Gas Hold Up
Gas hold-up is one of the most important parameters characterizing the
hydrodynamics in a fermenter.
Gas hold-up can be determined easily by measuring level of the aerated
liquid during operation 𝑍𝐹 and that of clear liquid 𝑍𝐿 .Thus, the average
fractional gas hold-up 𝐻 is given as
𝑍𝐹 − 𝑍𝐿
𝐻=
𝑍𝐹
7.24
Gas Sparging with No Mechanical Agitation:
In a two-phase system where the continuous phase remains in place,
the hold-up is related to superficial gas velocity s and bubble rise velocity t
(Sridhar and Potter, 1980):
𝐻=
𝑉𝑠
𝑉𝑠 + 𝑉𝑡
7.25
25
Continued…
Akita and Yoshida (1973) correlated the gas hold-up for the
absorption of oxygen in various aqueous solutions in bubble columns,
as follows:
𝐻
= 0.2
(1 − 𝐻)4
1
2
𝑔𝐷𝐶 𝜌𝑐 8
𝜎
1
3
𝑔𝐷𝐶 𝜌𝑐 12
𝑉𝑐2
𝑉𝑠
7.26
𝑔𝐷𝐶
Gas Sparging with Mechanical Agitation:
Calderbank (1958) correlated gas hold-up for the gas-liquid
dispersion agitated by a flat-blade disk turbine impeller as
𝑉𝑠 𝐻
𝐻=
𝑉𝑡
1/2
+ 2.16 × 10−4
𝑃𝑚
𝑣
0.4
𝜎 0.6
𝜌𝑐0.2
𝑉𝑠
𝑉𝑡
1/2
7.27
26
Continued…
where 2.16 × 10−4 has a unit (m) and 𝑉𝑡 = 0.265 m/s when the bubble size
is in the range of 2-5 mm diameter.
For high superficial gas velocities (𝑉𝑠 > 0.02 m/s), replace 𝑃𝑚 and
𝑉𝑡 of Eq. (7.10) with effective power input P, and 𝑉𝑡 + 𝑉𝑠 , respectively
(Miller, 1974).
27
8.1 Scale-up and Dimensionless numbers for
Scale-up
For the optimum design of a production-scale fermentation system
(prototype), we must translate the data on a small scale (model) to the large
scale. The fundamental requirement for scale-up is that the model and
prototype should be similar to each other.
Two kinds of conditions must be satisfied to insure similarity between
model and prototype. They are:
1. Geometric similarity of the physical boundaries: The model and the
prototype must be the same shape, and all linear dimensions of the model
must be related to the corresponding dimensions of the prototype by a
constant scale factor.
2. Dynamic similarity of the flow fields: The ratio of flow velocities of
corresponding fluid particles is the same in model and prototype as well as
the ratio of all forces acting on corresponding fluid particles. When
dynamic similarity of two flow fields with geometrically similar
boundaries is achieved, the flow fields exhibit geometrically similar flow
patterns.
1
Continued…
The first requirement is obvious and easy to accomplish, but the second is
difficult to understand and also to accomplish and needs explanation. For
example, if forces that may act on a fluid element in a fermenter during
agitation are the viscosity force 𝐹𝑉 , drag force on impeller 𝐹𝐷 , and gravity
force 𝐹𝐺 , each can be expressed with characteristic quantities associated
with the agitating system. According to Newton's equation of viscosity,
viscosity force is
𝑑𝑢
𝐹𝑉 = 𝜇
𝐴
𝑑𝑦
8.1
where 𝑑𝑢/𝑑𝑦 is velocity gradient and A is the area on which the viscosity
force acts. For the agitating system, the fluid dynamics involved are too
complex to calculate a wide range of velocity gradients present. However, it
can be assumed that the average velocity gradient is proportional to
agitation speed N and the area A is to 𝐷𝐼2 , which results.
2
Continued…
𝐹𝑉 ∝ 𝜇𝑁𝐷𝐼2
8.2
The drag force 𝐹𝐷 can be characterized in an agitating system as
𝐹𝐷 ∝
𝑃𝑚𝑜
𝐷𝐼 𝑁
8.3
Since gravity force 𝐹𝐺 is equal to mass m times gravity constant g,
𝐹𝐺 ∝ 𝜌𝐷𝐼2 𝑔
8.4
The summation of all forces is equal to the inertial force 𝐹𝐼 as,
න 𝐹 = 𝐹𝑉 + 𝐹𝐷 + 𝐹𝐺 = 𝐹𝐼 ∝ 𝜌𝐷𝐼4 𝑁 2
8.5
Then dynamic similarity between a model (m) and a prototype (p) is
achieved if
(𝐹𝑉 )𝑚 (𝐹𝐷 )𝑚 (𝐹𝐺 )𝑚 (𝐹𝐼 )𝑚
=
=
=
(𝐹𝑉 )𝑝
(𝐹𝐷 )𝑝
(𝐹𝐺 )𝑝
(𝐹𝐼 )𝑝
8.6
3
Continued…
or in dimensionless forms:
𝐹𝐼
𝐹𝑉
=
𝑝
𝐹𝐼
𝐹𝑉
𝑚
𝐹𝐼
𝐹𝐷
=
𝑝
𝐹𝐼
𝐹𝐷
𝑚
𝐹𝐼
𝐹𝐺
=
𝑝
𝐹𝐼
𝐹𝐺
8.7
𝑚
The ratio of inertial force to viscosity force is
𝐹𝐼 𝜌𝐷𝐼4 𝑁 2 𝜌𝐷𝐼2 𝑁
=
=
= 𝑁𝑅𝑒𝑖
𝐹𝑉
𝜇
𝜇𝑁𝐷𝐼2
8.8
which is the Reynolds number. Similarly,
𝐹𝐼
𝜌𝐷𝐼4 𝑁 2
𝜌𝑁 3 𝐷𝐼5
1
=
=
=
𝐹𝐷 𝑃𝑚𝑜 /𝐷𝐼 𝑁
𝑃𝑚𝑜
𝑁𝑃
8.9
𝐹𝐼 𝜌𝐷𝐼4 𝑁 2 𝐷𝐼 𝑁 2
=
=
= 𝑁𝐹𝑟
𝐹𝐺
𝑔
𝜌𝐷𝐼3 𝑔
8.10
4
Continued…
Dynamic similarity is achieved when the values of the nondimensional
parameters are the same at geometrically similar locations.
(𝑁𝑅𝑒𝑖 )𝑃 = (𝑁𝑅𝑒𝑖 )𝑚
(𝑁𝑃 )𝑃 = (𝑁𝑃 )𝑚
8.11
(𝑁𝐹𝑟 )𝑃 = (𝑁𝐹𝑟 )𝑚
Therefore, using dimensionless parameters for the correlation of data has
advantages not only for the consistency of units, but also for the scale-up
purposes.
However, it is difficult, if not impossible, to satisfy the dynamic
similarity when more than one dimensionless group is involved in a system,
which creates the needs of scale-up criteria.
5
Example 8.1:
The power consumption by an agitator in an unbaffled vessel can be
expressed as
𝑃
𝜌𝑁𝐷𝐼2 𝑁 2 𝐷𝐼
=𝑓
,
𝜇
𝑔
𝜌𝑁 3 𝐷𝐼5
8.12
Can you determine the power consumption and impeller speed of a 10,000gallon fermenter based on the findings of the optimum condition from a
geometrically similar one-gallon vessel? If you cannot, can you scale up by
using a different fluid system?
Solution:
Since
𝑉𝑝
𝑉𝑚
= 1,000, the scale ratio is,
(𝐷𝐼 )𝑝
= 10001/3 = 10
(𝐷𝐼 )𝑚
8.13
6
Continued…
To achieve dynamic similarity, the three dimensionless numbers for the
prototype and the model must be equal, as follows:
𝑃𝑚𝑜
𝜌𝑁 3 𝐷𝐼5
𝜌𝑁𝐷𝐼2
𝜇
𝑁 2 𝐷𝐼
𝑔
𝑝
𝑝
𝑝
𝑃𝑚𝑜
=
𝜌𝑁 3 𝐷𝐼5
𝜌𝑁𝐷𝐼2
=
𝜇
𝑁 2 𝐷𝐼
=
𝑔
8.14
𝑚
8.15
𝑚
8.16
𝑚
If you use the same fluid for the model and the prototype, 𝜌𝑝 = 𝜌𝑚 and
𝜇𝑝 = 𝜇𝑚 . Canceling out the same physical properties and substituting Eq.
(8.14) to Eq. (8.16) yields,
7
Continued…
(𝑃𝑚𝑜 )𝑝 = 105 (𝑃𝑚𝑜 )𝑚
𝑁𝑝
𝑁𝑚
3
8.17
The equality of the Reynolds number requires
𝑁𝑃 = 0.01𝑁𝑚
8.18
On the other hand, the equality of the Froude number requires
𝑁𝑃 =
1
10
𝑁𝑚
8.19
which is conflicting with the previous requirement for the equality of the
Reynolds number. Therefore, it is impossible to satisfy the requirement of
the dynamic similarity unless you use different fluid systems.
8
Continued…
If 𝜌𝑝 ≠ 𝜌𝑚 and 𝜇𝑝 ≠ 𝜇𝑚 , to satisfy Eqs. (8.15) and (8.16), the following
relationship must hold.
𝜇
𝜌
=
𝑚
1 𝜇
31.6 𝜌
8.20
𝑝
Therefore, if the kinematic viscosity of the prototype is similar to that of
water, the kinematic viscosity of the fluid, which needs to be employed for
the model, should be 1/31.6 of the kinematic viscosity of water. It is
impossible to find the fluid whose kinematic viscosity is that small. As a
conclusion, if all three dimensionless groups are important, it is impossible
to satisfy the dynamic similarity.
As an example, for a fully baffled vessel when 𝑁𝑅𝑒 > 10,000, the power
number is constant. For a geometrically similar vessel, the dynamic
similarity will be satisfied by,
9
Continued…
𝑃𝑚𝑜
𝜌𝑁 3 𝐷𝐼5
𝑝
𝑃𝑚𝑜
=
𝜌𝑁 3 𝐷𝐼5
8.21
𝑚
If the fluid employed for the prototype and the model remains the same, the
power consumption in the prototype is
(𝑃𝑚𝑜 )𝑝 = (𝑃𝑚𝑜 )𝑚
where 𝐷𝐼 𝑝 / 𝐷𝐼
𝑚
𝑁𝑝
𝑁𝑚
3
𝐷𝐼
𝐷𝐼
5
𝑚
𝑝
8.22
is equal to the scale ratio.
With a known scale ratio and known operating conditions of a model,
we are still unable to predict the operating conditions of a prototype because
there are two unknown variables, 𝑃𝑚𝑜 and N. Therefore, we need to have a
certain criteria which can be used as a basis.
10
8.2 Scaling of Mass-Transfer Equipment
 As discussed by Oldshue, the various quantities which may influence the
product 𝑘𝑙 𝑎 in an agitated industrial reactor do not scale in the same way
with reactor size or impeller rate.
1. The turbulent Reynolds number Re𝑡 , determines 𝑢𝑟𝑚𝑠 and thus bubble
mass transfer coefficients 𝑘𝑙
𝜌𝑢𝑟𝑚𝑠 𝐷 𝜌𝐷 𝐷 𝑃
Re𝑡 =
∝
𝜇
𝜇 𝜌𝑉
1/3
8.23
2. The impeller tip velocity 𝜋𝑁𝑖 𝐷𝑖 , determines the maximum shear rate 𝛾,ሶ
which in turn influences both maximum stable bubble or microbial floc
size and damage to viable cells.
3. The power input per unit volume 𝑃/𝑉 through Re𝑡 , determines masstransfer coefficients and particulate sizes. In laminar and transition
regimes of aerators
𝑃 ∝ 𝑁𝑖2 𝐷𝑖3
8.24
11
Continued…
Figure 8.1: Different relationships among air bubbles (B), n-hexadecane droplets
(O), and yeast cells (C), at different stages of batch culture of petrophilum (1 = log
phase, 2 = first half phase, 3 = second half of exponential phase, 4 = after nhexadecane exhaustion).
12
Continued…
For turbulent regimes, the power number is constant; thus
𝑃 ∝ 𝑁𝑖3 𝐷𝑖5
8.25
Then taking 𝑉reactor to scale with 𝐷𝑖3 gives
𝑃
𝑉
∝ 𝑁𝑖2
𝑃
𝑉
∝ 𝑁𝑖3 𝐷𝑖2
laminar, transition aeration
8.26
turbulent aeration
4. The power input during aeration is
Thus,
𝑃𝑎 = 𝑚′
𝑃𝑎
≈
𝑉
𝑃2
0.44
𝑁𝑖 𝐷𝑖3
𝑁𝑎0.56
0.45
3 0.44
𝑁
𝐷
𝑖 𝑖
𝑚′ 2.22
𝑉
𝑁𝑎0.56
𝑃2
8.27
0.45
8.28
which will determine the motor size needed during fermentation.
13
Continued…
5. If the vessel liquid is well mixed internally, a characteristic circulation
time exists. The liquid recirculation flow rate 𝐹𝑙 , through the impeller
region varies as a cross-sectional area 𝜋𝐷𝑖2 and tank average impeller
velocity varies as 𝑁𝑖 𝐷𝑖 . Thus
Fl
Ni D3i
∝
3 = Ni
V
Di
time−1
8.29
a quantity of importance since it is inversely proportional to the time that
fluid may spend away from the homogenizing influence of the impeller.\
Here by “basis for scale-up” we mean the quantity which, by choice of
operating conditions in the larger unit, will be maintained at the same value
as in the smaller scale unit.
For example, if we scale-up on the basis of constant power per unit volume,
for mechanical agitators in the turbulent regime this means
3 2
3 2
Ni1
Di1 = Ni2
Di2
8.30
14
Continued…
where 1 and 2 denote the values in the small and large scale vessel,
respectively.
As indicated in Fig. 8.2a this gives very similar yields of penicillin for
vessels from 5 liters to 200 gallons. However, from the curves in Fig. 8.2a,
we see that, at different P/V values, there are significant differences in yield
at different scales.
Another frequently applied basis for scale-up is constant volumetric
transfer coefficient k l a. Fig. 8.2b shows vitamin B12 yields from bacterial
fermentations at different scales versus corresponding values of (k l ap).
The total pressure is included here to correct for the greater driving force for
oxygen transfer at higher pressures which are encountered in large-scale
bioreactors.
15
Continued…
Figure 8.2: (a) Penicillin yields vs. power input in various sized vessels
(After E Gaden, Super Sanita, vol. 1, p. 61, 1961). (b) Vitamin B12 yield
(ug/g) vs. mass-transfer group (𝑘𝑙 𝑎𝑝)
16
Continued…
If we scale up on one basis, we must be aware that other mixing and
flow properties are different. This point is dramatically illustrated in an
example of Oldshue which considers scale-up from an 80 L to a 10,000 L
agitated bioreactor. Here, 𝐷𝑖 increases fivefold and V increases by a factor
of 125.
We must try to select as a scale-up basis the transport property most
critical to the performance of the bioprocess. If the bulk liquid composition
is uniform and the bubbles are uniformly dispersed throughout the vessel,
the mass-transfer rate is simply
𝑉k l 𝑎(𝑐 ∗ − 𝑐𝑙𝑖𝑞 )𝑒 = 𝑜𝑥𝑦𝑔𝑒𝑛 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒, 𝑚𝑜𝑙 𝑂2 /𝑠
8.31
where subscript e refers to gas exit compositions.
17
Continued…
When the bubbles rise in plug flow through the vessel but the impeller
still maintains perfect mixing in the liquid phase, 𝑐 ∗ varies with position.
Over a differential reactor height dz, the instantaneous loss of oxygen from
the bubble is
𝐻 𝐴 𝑑𝑧
which equals
𝑑𝑝𝑂2 1
𝑔𝑎𝑠 𝑣𝑜𝑙
𝑑𝑖𝑓𝑓𝑟𝑒𝑛𝑡𝑖𝑎𝑙
=
𝑑𝑡 𝑅𝑇 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 𝑣𝑜𝑙 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 𝑣𝑜𝑙𝑢𝑚𝑒
𝑐𝑜𝑛𝑐. 𝑟𝑎𝑡𝑒 𝑜𝑓
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑏𝑢𝑏𝑏𝑙𝑒
8.32
−k l 𝑎(𝑐 ∗ −𝑐𝑏 )𝐴𝑑𝑧
the mass transfer rate into the liquid. Since 𝑝𝑂2 = 𝑀𝑐𝑙∗ (M is Henry’s law
constant), we have
𝐻 𝑑𝑝𝑂2 𝐻𝑀 𝑑𝑐𝑙∗
=
= −𝑎k l (𝑐𝑙∗ − 𝑐𝑏 )
𝑅𝑇 𝑑𝑡
𝑅𝑇 𝑑𝑡
8.33
18
Continued…
For constant bubble-rise velocity 𝑢𝑏 , 𝑑𝑡 = 𝑑𝑧/𝑢𝑏 , and the 𝑧 variation 𝑐 ∗
seen to be
∗
(𝑐𝑙 − 𝑐𝑏 )𝑧
−𝑎k l 𝑅𝑇 𝑧
𝑙𝑛 ∗
=
(𝑐𝑙 − 𝑐𝑏 )𝑖𝑛𝑙𝑒𝑡
𝐻𝑀 𝑢𝑏
or
(𝑐𝑙∗
− 𝑐𝑏 )𝑧 =
(𝑐𝑙∗
−𝑎k l 𝑅𝑇 𝑧
− 𝑐𝑏 )𝑖𝑛𝑙𝑒𝑡 exp
𝐻𝑀 𝑢𝑏
8.34
8.35
The overall mass-transfer rate in the volume Ah is therefore
ℎ
න
0
𝑎k l 𝑐𝑙∗
− 𝑐𝑏
𝐻𝑀𝑢𝑏 ∗
−𝑎k l 𝑅𝑇 ℎ
𝑧 𝐴 𝑑𝑧 =
(𝑐𝑙 − 𝑐𝑏 )𝑖𝑛𝑙𝑒𝑡 𝐴 1 − exp
𝑅𝑇
𝐻𝑀 𝑢𝑏
8.36
19
8.3 Criteria for Scaling-Up Fermentors
Few steps are available for the so-called “scale-up” of fermentors,
starting from the glass apparatus used in fundamental research
investigations. Usually, chemical engineers are responsible for building and
testing pilot-scale fermentors (usually with capacities of 100 − 1000 𝑙), and
subsequently for designing industrial-scale fermentors based on data
acquired from the pilot-scale system.
One criterion for scaling-up this type of bioreactor is the power input per
unit liquid volume of geometrically similar vessels, which should be
proportional to 𝑁 3 𝐿2 for the turbulent range and to 𝑁 2 for the laminar
range, where 𝑁 is the rotational stirrer speed and L is the representative
length of the vessel.
In order to minimize any physical damage to the cells, the product of the
diameter 𝑑 and the rotational speed 𝑁 of the impeller - which should be
proportional to the tip speed of the impeller and hence to the shear rate at
the impeller tip - becomes an important criterion for scale-up.
20
Aerated stirred tanks, bubble columns, and airlifts are usually used for
aerobic fermentations. One criterion of scaling-up aerated stirred tank
fermentor is 𝑘𝐿 𝑎, approximate values of which can be estimated. For the
turbulent range, a general correlation for 𝑘𝐿 𝑎 in aerated stirred fermentors is
of the following type:
𝑘𝐿 𝑎 = 𝑐𝑈𝐺𝑚 (𝑁 3 𝐿2 )𝑛
8.37
where 𝑈𝐺 is the superficial gas velocity over the cross-sectional area of the
tank, 𝑁 is the rotational stirrer speed, and 𝐿 is the representative length of
geometrically similar stirred tanks.
For the turbulent regime (𝑁 3 𝐿2 ) should be proportional to the power
requirement per unit liquid volume.
21
Continued…
The main operating factor of a bubble column fermentor is the superficial
gas velocity 𝑈𝐺 over the column cross section, which should be kept equal
in scaling up.
𝑘𝐿 𝑎 in bubble columns < 60 cm in diameter will increase with the
column diameter 𝐷 to the power of 0.17. As this trend levels off with larger
columns, it is recommended that 𝑘𝐿 𝑎 values estimated for a 60 cm column
are used.
If heat transfer is a problem, then heat transfer coils within the column,
or even an external heat exchanger, may become necessary when operating a
large, industrial bubble column-type fermentor.
22
Continued…
Example 8.2: Two geometrically similar stirred tanks with flat-blade
turbine impellers of the following dimensions are to be operated at 30°C as
pilot-scale and production-scale aerobic fermentors.
Scale
Tank diameter and
liquid depth (m)
Impeller
diameter (m)
Liquid volume
(𝑚3 )
Pilot
0.6
0.24
0.17
Production
2.0
0.8
6.28
Satisfactory results were obtained with the pilot-scale fermentor at a
rotational impeller speed N of 1.5𝑠 −1 and air rate (30°𝐶) of 0.5 𝑚3 𝑚𝑖𝑛−1 .
The density and viscosity of the broth are 1050 𝑘𝑔 𝑚3 and 0.002 𝑃𝑎𝑠,
respectively. Data from the turbine impellers showed that 𝑘𝐿 𝑎 can be
correlated by Equation 8.37, with values 𝑚 = 𝑛 = 2/3.
23
Continued…
Using 𝑘𝐿 𝑎 as the scale-up criterion, estimate the impeller speed and the air
rate for the production-scale fermentor that will give results comparable
with the pilot-scale data.
𝑘𝐿 𝑎 = 𝑐𝑈𝐺𝑚 (𝑁 3 𝐿2 )𝑛
Solution:
In the pilot-scale fermentor: 𝑈𝐺 = 0.5/
𝜋
4
0.6
2
× 60 = 0.0295 𝑚𝑠 −1 .
At equal 𝑈𝐺 , the air rate to the production-scale fermentor should be
𝜋
0.0295 × 60 ×
× (2.0)2 = 5.56 𝑚3 𝑚𝑖𝑛−1
4
With the pilot fermentor 𝑁 3 𝐿2 = 1.53 × 0.622 = 1.215.
With this equal value of 𝑁 3 𝐿2 and 𝐿2 = 2.02 = 4 for the production
fermentor, the production fermentor should be operated at
𝑁3 =
1.215
= 0.304
4
𝑁 = 0.672 𝑠 −1
24
Continued…
Incidentally, the impeller tip speeds in the pilot and production fementors
are calculated as 1.13 and 1.69 𝑚𝑠 −1, respectively.
25
8.5 Design estimation of various Scale-up
Parameters
Most often, power consumption per unit volume 𝑃𝑚0 /𝑣 is employed as a
criterion for scale-up. In this case, to satisfy the equality of power numbers
of a model and a prototype,
𝑃𝑚0
𝐷𝐼3
𝑝
𝑃𝑚0
=
𝐷𝐼3
𝑚
𝑁𝑝
𝑁𝑚
3
𝐷𝐼
𝐷𝐼
2
𝑝
8.39
𝑚
Note that 𝑃𝑚0 /𝐷𝐼3 represents the power per volume because the liquid
volume is proportional to 𝐷𝐼3 for the geometrically similar vessels. For the
constant 𝑃𝑚0 /𝐷𝐼3 ,
𝑁𝑝
𝑁𝑚
3
=
𝐷𝐼
𝐷𝐼
2
𝑚
8.40
𝑝
26
Continued…
As a result, if we consider scale-up from a 20-gallon to a 2,500gallon
agitated vessel, the scale ratio is equal to 5, and the impeller speed of the
prototype will be,
𝑁𝑝 = 𝑁𝑚
𝐷𝐼
𝐷𝐼
2/3
𝑚
𝑝
𝑁𝑝 = 0.34 𝑁𝑚
8.41
which shows that the impeller speed in a prototype vessel is about one third
of that in a model.
For constant 𝑃𝑚0 /𝑣, the Reynolds number and the impeller tip speed
cannot be the same. For the scale ratio of 5,
(𝑁𝑅𝑒𝑖 )𝑝 = 8.5 (𝑁𝑅𝑒𝑖 )𝑚
8.42
(𝑁𝐷𝐼 )𝑝 = 1.7 (𝑁𝐷𝐼 )𝑚
8.43
27
Continued…
Table 8.1 shows the values of properties for a prototype (2,500 gallon) when
those for a model (20-gallon) are arbitrarily set as 1.0 (Oldshue, 1966).
The parameter values of the prototype depend on the criteria used for
the scale-up. The third column shows the parameter values of the prototype,
when 𝑃𝑚0 /𝑣 is set constant. The values in the third column seem to be more
reasonable than those in the fourth, fifth, and sixth columns, which are
calculated based on the constant value of 𝑄/𝑣 , 𝑁𝐷𝐼 and 𝑁𝐷𝐼 2 𝜌/𝜇 ,
respectively.
For example, when the Reynolds number is set constant for the two
scales, the 𝑃𝑚0 /𝑣 reduces to 0.16 percent of the model and actual power
consumption 𝑃𝑚0 also reduces to 20 percent of the model, which is totally
unreasonable.
28
Continued…
As a conclusion, there is no one scale-up rule that applies to many different
kinds of mixing operations. Theoretically we can scale up based on
geometrical and dynamic similarities, but it has been shown that it is
possible for only a few limited cases. However, some principles for the
scale-up are as follows (Oldshue, 1985):
1. It is important to identify which properties are important for the optimum
operation of a mixing system. This can be mass transfer, pumping
capacity, shear rate, or others. Once the important properties are
identified, the system can be scaled up so that those properties can be
maintained, which may result in the variation of the less important
variables including the geometrical similarity.
2. The major differences between a big tank and a small tank are that the
big tank has a longer blend time, a higher maximum impeller shear rate.
29
Continued…
3. For homogeneous chemical reactions, the power per volume can be used
as a scale-up criterion. As a rule of thumb, the intensity of agitation can
be classified based on the power input per 1,000 gallon as shown in Table
8.1.
Table 8.1: Properties of Agitator on scale up
30
Continued…
4. For the scale-up of the gas-liquid contactor, the volumetric mass-transfer
coefficient kLa can be used as a scale-up criterion. In general, the
volumetric mass-transfer coefficient is approximately correlated to the
power per volume. Therefore, constant power per volume can mean a
constant kLa.
5. Typical impeller-to-tank diameter ratio 𝐷𝐼 /𝑇 for fermenters is 0.33 to
0.44. By using a large impeller, adequate mixing can be provided at an
agitation speed which does not damage living organisms. Fermenters are
not usually operated for an optimum gas-liquid mass transfer because of
the shear sensitivity of cells.
Table 8.2: Criteria
of Agitation
Intensity
31
8.6 Power Estimation for Ungassed Liquids
There are well-established empirical correlations for stirrer power
requirements. Figure 7.8 is a log-log plot of the power number N, for
ungassed liquids versus the stirrer Reynolds number (Re). These
dimensionless numbers are defined as follows:
𝑁𝑃 =
𝑃
(𝜌𝑁 3 𝑑5 )
𝑁𝑑 2 𝜌
𝑅𝑒 =
𝜇
8.44
8.45
where 𝑃 is the power required (𝑀𝐿2 𝑇 −3 ), 𝑁 is the number of revolutions of
the impeller per unit time (𝑇 −1 ), d is the impeller diameter (𝐿), p is the
liquid density (𝑀𝐿−3 ), and 𝜇 is the liquid viscosity (𝑀𝐿−1 𝑇 −1 ). Since the
product 𝑁 disproportional to the peripheral speed (i.e., the tip speed of the
rotating impeller).
32
Continued…
Figure 8.6: Correlation between Reynolds number (Re) and Power number (𝑁𝑝 ).
Curve (a): six-flat blade turbine, four baffles, 𝑊𝑏 = 0.1 𝐷 ,
curve (b): two-flat blade paddle, four baffles, 𝑊𝑏 =0.1 D,
curve (c): three-blade marine propeller, four baffles, 𝑊𝑏 =0.1 D.
33
Continued…
In Figure 8.6, the curves a, b, and c correlate data for three types of
impellers, namely, the six-flat blade turbine, two-flat blade paddle, and
three-blade marine propeller, respectively. It should be noted that, for the
range of (Re) >104 , 𝑁𝑃 is independent of (Re). For this turbulent regime it
is clear from Equation 8.44 that
𝑃 = 𝑐1 𝜌𝑁 3 𝑑 5
8.46
where 𝑐1 , is a constant that varies with the impeller types. Thus, 𝑃 for a
given type of impeller varies in proportion to 𝑁 3 , 𝑑 5 and liquid density
𝜌,but is independent of liquid viscosity 𝜇.
For the ranges of Re below approximately 10, the plots are straight lines
with a slope of −1; that is, 𝑁𝑃 , is inversely proportional to (𝑅𝑒). Then, for
this laminar regime, we can obtain Equation 7.33 from Equations 8.44 and
8.45:
2 3
8.47
𝑃 = 𝑐2 𝜇𝑁 𝑑
34
Continued…
Thus, 𝑃 for the laminar regime varies in proportion to liquid viscosity 𝜇,
𝑁 2 , 𝑑 3 and to a constant 𝑐2 , which varies with impeller types, although 𝑃 is
independent of the liquid density 𝜌.
It is worth remembering that the power requirements of geometrically
similar stirred tanks are proportional to 𝑁 3 𝑑 5 in the turbulent regime and to
𝑁 2 𝑑 3 in the laminar regime.
In the case of scaling-up of geometrically similar stirred tanks, the equal
power input per unit liquid volume, which should be proportional to
𝑁 3 𝑑 2 for the turbulent regime and to 𝑁 2 for the laminar regime, is
sometimes used as a criterion for scale-up, because it leads to the same
values of 𝑘𝐿 𝑎 in a larger stirred tank with that in the smaller stirred tank.
35
8.7 Power Estimation for Gas-Sparged
Liquids
The ratio of the power requirement of gas-sparged (aerated) liquid in a
stirred tank,𝑃𝐺 , to the power requirement of ungassed liquid in the same
stirred tank, 𝑃0 , can be estimated using Equation 8.48.
This is an empirical, dimensionless equation based on data for six-flat
blade turbines, with a blade width that is one fifth of the impeller diameter
𝑑, while the liquid depth 𝐻𝐿 , is equal to the tank diameter. Although these
data were for tank diameters up to 0.6 𝑚, Equation 8.48 would apply to
larger tanks where the liquid depth-to-diameter ratio is typically in the
region of unity.
𝑃𝐺
𝑑
log
= −192
𝑃0
𝐷
4.38
𝑑2𝑁
𝑣
0.115
𝑑𝑁 2
𝑔
1.96(𝑑/𝐷)
𝑄
𝑁𝑑 3
8.48
where 𝑑 is the impeller diameter (L), 𝐷 is the tank diameter (L), 𝑁 is the
rotational speed of the impeller (𝑇 −1 ), g is the gravitational constant
(𝐿 𝑇 −2 ), Q is the gas rate (𝐿3 𝑇 −1 ), and 𝑣 is the kinematic viscosity of the
liquid (𝐿2 𝑇 −1 ).
36
Continued…
The dimensionless groups include (𝑑 2 𝑁/𝑣)= Reynolds number (Re);
(𝑑 𝑁 2 /𝑔) = Froude number (Fr);
(𝑄/𝑁 𝑑 3 ) = aeration number (Na),
which is proportional to the ratio of the superficial gas velocity with respect
to the tank cross section to the impeller tip speed.
The ratio 𝑃𝐺 /𝑃0 for flat-blade turbine impeller systems can also be
estimated by
𝑃𝐺
𝑄
log
= 0.10
𝑃0
𝑁𝑉
−1/4
𝑁 2 𝑑4
2
𝑔𝑏𝑉 3
−1/5
8.49
where V is the liquid volume (𝐿3 ) and 𝑏 is the impeller blade width (𝐿). All
other symbols are the same as in Equation 8.48.
37
Continued…
Example 8.3: After a batch fermentation, the system is dismantled and
approximately 75% of the cell mass is suspended in the liquid phase (2 l),
while 25% is attached to the reactor walls and internals in a thick film (ca.
0.3 cm). Work with radioactive tracers shows that 50% of the target product
(intracellular) is associated with each cell fraction. The productivity of this
reactor is 2 g product/l at the 2 l scale. What would be the productivity at
20,000 l scale if both reactors had a height-to-diameter ratio of 2 to 1?
Solution:
Both tanks are geometrically similar, so we can calculate the diameter and
the resulting surface area and volume in both tanks
1 2
1 2
1
𝑉 = 𝜋𝐷 . 𝐻 = 𝜋𝐷 . 2𝐷 = 𝜋𝐷3
4
4
2
𝑆 = 𝜋𝐷. 𝐻 = 𝜋𝐷 . 2𝐷 = 2𝜋𝐷2
38
Continued…
For the 2 𝑙 system
𝐷 = 10.8 𝑐𝑚
𝑆 = 738 𝑐𝑚2
𝑉 = 2000 𝑐𝑚3
For the 20,000 𝑙 systems,
𝐷 = 233 5𝑐𝑚
𝑆 = 342600𝑐𝑚2
𝑉 = 2 × 107 𝑐𝑚3
At the bench scale (2 𝑙) the amount of product made by surface-attached
cells is
g
1
2 l . 2 l. 2 = 2 g,
for 738 cm2 of surface area
39
Continued…
The amount of product formed at 20,000 𝑙 due to surface-attached growth is
342600 𝑐𝑚2
× 2𝑔 = 928 𝑔
738𝑐𝑚2
The overall yield in the 2 𝑙 system is
2
𝑔
×2𝑙 =4𝑔
𝑙
The overall yield in the 20,000 𝑙 system is
𝑔 1
928 𝑔 + 2 . . 20000 𝑙 = 20928 𝑔
𝑙 2
If no wall growth had been present, the 20,000 𝑙 tank would have yielded
40,000 𝑔. Thus, wall growth, if present, can seriously alter the productivity
of a large-scale system upon scale-up.
40
Continued…
Example 8.4: Consider the scale-up of a fermentation from a 10 𝑙 to
10,000 𝑙 vessel. The small fermenter has a height-to-diameter ratio of 3.
The impeller diameter is 30% of the tank diameter. Agitator speed is 500
rpm and three Rushton impellers are used. Determine the dimensions of the
large fermenter and agitator speed for:
a. Constant P/V
b. Constant impeller tip speed
c. Constant Reynolds number Assume geometric similarity
Solution:
Assume the vessel is a cylinder.
Thus,
𝑉=
𝜋
𝐷𝑡 2 𝐻
4
41
Continued…
but 𝐻 = 3 𝐷𝑡 , so
𝑉 = (𝜋/4) 3 𝐷𝑡 3 = 10,000 𝑐𝑚3
Solving for 𝐷𝑡 , 𝐻, and 𝐷𝑖 = 0.3 𝐷𝑡 gives:
𝐷𝑡 = 16.2 𝑐𝑚
𝐻 = 48.6 𝑐𝑚
𝐷𝑖 = 4.9 𝑐𝑚
The scale-up factor is the cube root of the ratio of tank volumes, or
10000 𝑙
10 𝑙
1/3
= 10
To maintain geometric similarity the larger vessel will have its dimensions
increased by a factor of 10. That is:
𝐷𝑡 = 1.62 𝑚
𝐻 = 4.86 𝑚
𝐷𝑖 = 0.49 𝑚
42
Continued…
a. For constant 𝑃/𝑉 to apply 𝑁 3 𝐷𝑡2 must be the same in both vessels. Let
subscript 1 refer to the small vessel and 2 to the large vessel.
𝑁2 = 𝑁1
𝐷𝑡1
𝐷𝑡2
2/3
1
= 500 𝑟𝑝𝑚
10
2/3
= 107𝑟𝑝𝑚
b. For constant tip speed to apply 𝑁𝐷𝑡 must be the same in both vessels.
𝑁2 = 𝑁1
𝐷𝑡1
1
= 500 𝑟𝑝𝑚
= 50 𝑟𝑝𝑚
𝐷𝑡2
10
c. For constant Re to apply 𝑁𝐷𝑡2 must be the same in both vessels.
𝑁2 = 𝑁1
𝐷𝑡1
𝐷𝑡2
2
1
= 500 𝑟𝑝𝑚
10
2
= 5 𝑟𝑝𝑚
Scale-up on the basis of constant 𝑃/𝑉 would be the most likely choice
unless the culture was unusually shear sensitive. Scale-up based on constant
Re would not be used.
43
9.1 Aerobic Fermentation
 Aerobic fermentation is a metabolic process by which cells metabolize
sugars via fermentation in the presence of oxygen and occurs through the
repression of normal respiratory metabolism.
 Aerobic fermentation occurs in the presence of oxygen. It usually occurs
at the beginning of the fermentation process. Aerobic fermentation is
usually a shorter and more intense process than anaerobic fermentation.
Aerobic fermentation is also essential for multiple industries, resulting in
human domestication of several yeast strains.
 Fermentation has several kinds of process steps and equipment to
consider such as
 Media prep (dealing with blending-tanks, agitators, or heat exchangers);
 Propagation (dealing with agitators and other vessels); production
fermenters (dealing with vessels or agitators, and internal & external heat
exchangers);
1
Continued…
 Separations (dealing with evaporators, centrifuges, distillation, or
membrane/ultra-filtration).
 Purification (dealing with crystallizers, heat exchangers, and ultrafiltration); and product packaging (dealing with bulk containers (totes),
packing lines, and other things.
2
9.1.1 Aerobic Fermenter's design
There are two types of materials which can be used in construction a
fermenter:
(i) Stainless steel; according to American iron and Steel Institute (AISI), if a
steel contains 4% chromium, it is called stainless. The long and
continuous use of stainless steel sometimes shows pitting it is also
important to consider the material used for aseptic seal.
(ii) Glass; It can be made between glass and glass and metal, or using metal
for joints between a vessel and detachable top or base plate (glass) On
pilot scale, any material to be used will have to be assessed in their
ability to withstand pressure, sterilization, corrosion and their potential
toxicity and cost.
3
Continued…
This is a typical stirred tank bioreactor's design which generally
contains,
1. A cooling system
2. Media inlet
3. Gas exhaust
4. Agitation system (stirrer & baffles)
5. Airation system (sparger)
6. Foam breaker.
7. pH & O2 concentration sensors.
4
Continued…
Figure 9.1: An industrial
Anerobic fermentor
(internal view)
5
9.2 Anaerobic Fermentation
 Anaerobic fermentation, which is common to all bacteria and eukaryotes,
is a metabolic process that converse carbohydrates (sugar) to organic
acids, gases or alcohols under anaerobic conditions.
 Therefore, microorganisms are used to produce nutraceuticals or
bacteriocins through accumulation of various primary metabolites, as
well as complex secondary metabolites.
 Metabolic engineering strategies are usually involved so that the
production of a certain metabolite can be emphasized or newly induced
through overexpression and/or disruption of relevant metabolic genes.
Anaerobic fermentation has a broad range of applications.
 It could be used for production of various industrial chemicals, such as
ethanol, butyl alcohol, lactic acid, acetic acid, hydrogen gas and various
nutraceutical or antimicrobial molecules with medical or health benefit.
The process is also able to degrade different types of biomass.
6
Continued…
• A variety of microorganisms are involved to produce many industrial
materials using several bacterial strains:
1. Saccharomyces
2. Lactobacillus
3. Escherichia coli
4. Clostridium
Each strain could be used alone, or in a combination with others. Most
industrial fermentation uses batch or fed-batch procedures. Continuous
fermentation can be more economical if challenges, particularly the
difficulty of maintaining sterility, is properly handled.
7
Continued…
Some examples of fermentation metabolites are listed below:
1. Lactic Acid
2. Glycerol
3. Methane
4. Fatty acids
5. Biogas production
6. Modified/unmodified peptides, proteins
8
9.3 Manufacture of Antibiotics
Use of Fermentor in Production of Antibiotics:
Antibiotics are generally produced in stainless steel fermentors (30,000200,000 1 medium volume) used in the batch or fed-batch mode. Agitation
is mostly by impellers, but air-lift system is also used. Water cooling is often
used to maintain the temperature between 24-26°C for most antibiotic
producers.
Generally, the fermentor is maintained at above atmospheric pressure;
this reduces contamination risk and enhances O2 supply in the medium.
Sterile air is supplied as per need, and for some processes.
The final stage fermentor is preferably used for antibiotic production
for the longest possible period. But the initial stages of fermentation are
designed for considerable microbial growth; typically these are carried out
in seed-stage fermentors of smaller size.
9
Continued…
Table 9.1: A list of selected antibodies produced commercially
10
Continued…
Medium used for Antibiotic Production:
Antibiotic production employs a variety of media, a different one for each
stage of operation (Table 9.2). A considerable research effort is directed at
developing seed-stage and production media to reduce costs and to enhance
yields. A typical production medium has about 10% (w/v) solids.
Biotransformation of Antibiotics:
Biotransformation is routinely used for commercial production of several
useful antibiotics.
BiotransPenicillin G
formation
and
Penicillin V (Deacylation)
6-Aminooenicillinic
Acid
Chemical
Semixythetic
Penicillins and
Cephalosporins
11
Continued…
Table 9.2: Example of some media used in the different stages of antibiotic production
Stage
Medium constituents
Storage of spores
Buffered salt, or glucose (10-20%), or glycerol (10-20%) solution
Growth and sporulation
Carbohydrate (0.5-2%), N-source (0.2-1%), chalk (0-1%), NaCl
(0-0.5%), phosphate (0-1%), various salt
Spore recovery from solid
surface
Water/buffer solution: usually a surfactant is used
Shaker-flask testing
Lactose (5-8%), CSL (5-8%), chalk (0-1.6%), ammonium sulphate
(0-0.5%), phenylacetic acid (added daily)
Seed stage fermentor
(growth medium)
Sugar (glucose, sucrose or lactose; 3-5%), Pharmamedia (2-5%),
and/or CSL (3-5%), chalk (0-1%), phosphate (0-1%), ammonium
sulphte (0-0.5 %) edibale oil (0-0.5%)
Final stage fermentor
Sugar (5-10%), CSL (5-9%) and/or Pharmamedia (3-5%), chalk (01%), ammonium sulphate (0-1%), phosphate (0-0.5%), edible oil
(0.3-0.8%),Phenoxyacetic acid (0.3-0.6%)
12
9.4 Manufacture of Ethanol
Ethanol is the alcohol found in beer, wine and spirits. It is also used as a fuel
for vehicles, either on its own or mixed with petrol. Ethanol is also used as a
solvent.
Ethanol is produced by fermentation and concentrated by fractional
distillation.
Fermentation is an anaerobic process:
Fermentation takes place in which the glucose is broken down to ethanol by
the action of enzymes in the yeast.
The equation for the reaction is:
glucose
𝐶6 𝐻12 𝑂6(𝑎𝑞)
yeast
yeast
ethanol + carbon dioxide
2𝐶2 𝐻5 𝑂𝐻(𝑎𝑞) + 2𝐶𝑂2(𝑎𝑞)
13
Continued…
The fermentation reaction requires the following conditions:
Temperature: The temperature must be between the range of 25°C and
50°C. Enzymes are affected a great deal by temperature.
Substrate (the glucose solution): Enzymes work best when there is a high
enough substrate concentration for the reaction they catalyse.
Absence of Oxygen: Air must be excluded from the vessel in which
fermentation is being carried out. Air contains a large proportion of bacteria
called Acetobacter. Acetobacter bacteria use atmospheric oxygen from air to
oxidise ethanol in the wine, producing a weak solution of ethanoic acid
(vinegar).
Yeast: The fermentation of the glucose solution to ethanol cannot take place
without the presence of yeast. Yeast contains the enzyme zymase which acts
as a catalyst for the reaction
14
9.5 Organic and Amino Acid Manufacture
Organic acid production examples illustrate the important effect of yield on
cost. A submerged fermentation process with high citric acid yields of 70 kg
acid from 100 kg sugar has fermentation capital costs which are 15-25
percent less than a surface fermentation, but requires 16-26 percent greater
operating costs.
A more modest yield of 40 g/L itaconic acid led to consideration of
recovery by evaporation, ion exchange, or electrodialysis based on a
continuous fermentation.
The economic aspects of amino acid production are now considered. Of the
twenty amino acids, only two (lysine and glutamic acid) were produced
primarily by fermentation in 1980 (in Japan), although more are potentially
producible (arginine, glutamine, histidine, leucine, ornithine, phenylalanine,
proline, thre onine, and valine)(Table 9.3).
15
Table 9.3: Operating characteristics of selected single-cell protein
processes
Process
Item
Algal
Spirulina
maxima
Bacterial
Methylophilus
methylotrophus
(methanol)
Yeast
Candida utilis
(ethanol)
Mold
Paecilomyces
varioti (sulfite
waste liquor)
Type of process
Batch or
semicontinuous
Continuous
Continuous or
batch
Continuous
Sterility
Nonseptic
Aseptic
Aseptic
Nomaseptic
Fermentor
Ponds
Airlift
Agitated
Agitated
Feedstock
utilization
Partially or fully
utilized
Fully utilized
Fully utilized
Partially utilized
Temperature, °C
Ambient
35 to 42
30 to 40
38 to 39
pH
9 to 11
6.0 to 7.0
4.6
4.5 to 4.7
Product recovery
Filtration
Agglomeration
and
centrifugation
Centrifugation
Filtration
16
9.7 Modes of Fermentor Operation
Batch operation of fermentors is much more common than continuous
operation, although theories for continuous operation are well established
(as will be indicated later in this section).
The reasons for this are as follows:
1. The operating conditions of batch fermentors can be more easily
adjusted to required specific conditions, as the fermentation proceeds.
2. The capacities of batch fermentors are usually large enough, especially
in cases where the products are expensive.
3. In the case of batch operation, damages by the so-called
“contamination”- that is, the entry of unwanted organisms into the
systems, with resultant spoilage of the products is limited to the
particular batch that was contaminated.
17
Continued…
In the fed-batch operation of fermentors (which is also commonly
practiced), the feed is added either continuously or intermittently to the
fermentor, without any product withdrawal, the aim being to avoid any
excessive fluctuations of oxygen demand, substrate consumption, and other
variable operating conditions.
18
Fermentation and energy production
Different energy sources from biomass / wastes through fermentation
 Fermentation is a metabolic process that converts sugar to acids, gases, or alcohol
Organic wastes /
• Ethanol
Syngas
biomass
Gasification
• Butanol Fermentation
Fermentation
Anaerobic digestion
• Ethanol
• Ethanol can be used as blend with gasoline (10 % by volume) (26.8 MJ/kg)
• Butanol can be used in pure form or blended with gasoline at higher
concentrations (16 % by volume), (32.5 MJ/kg)
Heating value of gasoline 42.9 MJ/kg
19
19
Fermentation and energy production
Ethanol fermentation
 It is a series of chemical reactions for converting sugars to ethanol in the presence
of suitable microbial strain, which feed on the sugars.
 Ethanol and carbon dioxide are produced as the sugar is consumed.
The simplified fermentation reaction for a 6-carbon sugar, glucose, is as follows:
Yeast is normally used for hexose
Source of sugar
Sugars may contain
Sugar crops
Starchy crops
Lignocellulosic biomass
6 carbons (i.e., Hexose )
5 carbons (i.e., Pentose)
20
20
Production of ethanol from three different feedstocks
Sugar crops/starchy
crops/lignocellulosic
biomass
Ethanol concentration in:
Preprocessing
Fermentation
Byproducts
Distillation
Fermented mash ~ 4-10%
After distillation ~ 95-96% Recovery section
After dehydration~ 99.5%
Byproducts
Dehydration
Ethanol
21
21
 The ethanol concentration in the fermented mash can attain a level of only 4%
to 10% depending on the operating conditions and the strain of yeast being
used.
 Ethanol is therefore recovered through distillation but only hydrous ethanol of
about 95-96% can be produced through steam distillation of the fermented
mash due to the formation of water-ethanol azeotrope.
 To make ethanol fully miscible with gasoline, it is necessary to further remove
the residual water to produce anhydrous ethanol with a concentration of at
least 99.5%. To attain this concentration, the hydrous ethanol has to undergo a
suitable dehydration process or operation.
22
22
Preprocessing steps for different sugar sources
Sugar crops
Sugar
Fermentation
Starchy crops
Starch
Hydrolysis to sugar
Lignocellulosic
biomass
Pretreatment
Cellulose
Fermentation
Hydrolysis to
sugar
Fermentation
Steps of pre-processing increases as :
Sugar crops < Starchy crops< Lignocellulosic biomass
23
23
wet process
Production of ethanol from starchy crops (corn)
Step 1 Cleaning and steeping of grains
water
Step 2 Grinding
Step 3 Washing and filtering
water
Step 4 Gluten separation
Heat and Enzyme
Yeast
Steam
Step 5 Cooking
Step 6 Fermentation
Step 7 Distillation and dehydration
Ethanol
Waste water
Germ
Fiber
Gluten
Starch to sugar conversion
CO2
Glucose, Emissions , wastewater
https://www.osha.gov/dts/osta/otm/otm_iv/otm_iv_5.html
24
24
Wet process
Steeping in dilute sulphuric acid and water helps disintegrating different parts of the corn
kernel
Grinding is used to liberate the intact germ from the kernel. Germ separation is done by a
series of hydrocyclones, High oil content of the germ imparts low density, which makes it tend
to float in water, enhanced additionally by starch, suspended in water. After a washing of the
germ, oil can be extracted (usually with hexane).
Fibre separation consists of four steps:
an initial screening of coarse fibre; a milling or grinding step to liberate and disperse all the
starch and gluten from the remaining fibre; another washing step and a dewatering and drying
step with the addition of corn steep liquor for the production and sale of gluten feed.
After the fibre removal, starch, protein and miscellaneous soluble components (e.g. salts,
minerals, and vitamins) are left.
alpha-amylase may be used for cooking
Starch is separated from the gluten (protein) in several steps and used for cooking
25
25
Dry process
Production of ethanol from starchy crops (corn)
Step 1 Cleaning
Shelled Corn
Step 2 Milling
Step 3 Liquefaction
water
Heat and Enzyme
Yeast
Steam
Step 4 Saccharification
Step 5 Fermentation
Step 6 Distillation and dehydration
Step 7 Coproduct processing
Animal feed and other coproducts
Out of 180 corn or starch
ethanol plants operating in the
U.S. 173 are dry milling
process , and only 11 are wet
milling process (EPA, 2010).
Starch to sugar conversion
CO2
Ethanol
In step 4, Enzymes like gluco-amylase
converts
liquefied
starch
to
fermentable sugars (dextrose)
26
26
Production of ethanol from lingo cellulosic biomass and wastes
Step 1 Lignocellulosic biomass / wastes
• Pre treatment processes may be
physical, chemical or biological
Step 2 Drying and grinding
Step 3 Slurry formation (10-15 % dry solid)
Step 4 Pre treatment of slurry
Step 5 Detoxification and hydrolysis
Step 6 Fermentation
Step 7 Product recovery
• Water is used for slurry formation
CO2
• Hydrolysis may be carried out by
acid or enzymes
• Fermenting
added
microorganisms
Ethanol
Solid residue
Waste water
are
Adapted
from
Margeot et al. 2009,
Chandel et al. 2007,
Hamelinck et al. 2005
27
27
Lignocellulosic biomass
Common lignocellulosic biomass
Agricultural
residues and agro
wastes
Leaves, stovers, straws, seed pods in field, processing
residues like bagasse, husks, cobs, seeds, roots etc.,
and solid cattle manure
MSW
Food wastes, papers, sorted refuse
Forest biomass and
wastes
Soft wood like pine, cedar etc., hard wood like poplar
oak etc, saw dust, wood chips, slashes, branches from
dead trees, forest thinning
Industrial wastes
Chemical pulp and waste water solid
Whole plant
Perennial grass, aquatic plants etc.,
28
28
Pre-treatment
(Adapted from (Mosier et al. 2005, Hsu et al. 1980).
• It is applied to remove the protecting shield of lignin–hemicellulose and make
the cellulose more readily available for enzymatic hydrolysis.
• It alters the macroscopic, sub–microscopic and microscopic structure of the
biomass to make enzyme easily accessible for hydrolyzing and converting the
carbohydrate polymers into fermentable sugars during subsequent hydrolysis step
Oligosaccharides
Lignin
 Altered cellulose
Cellulose
Hemicellulose from
hemicellulose Altered cellulose and simple sugar
from
hemicellulose
Pretreatment
both
are
produced during
Amorphous Crystalline
Lignin removal pretreatment
region
region
Simple sugar from hemicellulose
29
29
Pre-treatment
Energy efficiency of any pretreatment process can be determined from total sugar
yield and energy consumption during the process as shown in the following equation
(Zhu and Pan 2010):
 TSY / TEC
where, η = Pretreatment energy efficiency (kg/MJ);
TSY = Total soluble (monomeric) sugar yield (kg);
TEC = Total energy consumption during pretreatment (MJ)
30
30
Favourable characteristics of pretreatment processes
 Obtainment of cellulose with high digestibility
 Avoiding destruction of hemicelluloses and cellulose
 No or little degradation of sugars and sugar derived products
 No loss of sugar
 Production of minimum amount of toxic compounds.
 No requirement of biomass size reduction before pretreatment
31
31
Favourable characteristics of pretreatment processes
 Capable of operating in reasonable size and moderate cost reactors.
 No production of solid–waste residues
 Effectiveness at low moisture content.
 Recovery of high amounts of carbohydrates.
 Fermentation compatibility.
 High lignin recovery for conversion into valuable coproducts
 Minimum heat and power requirements
32
32
Different types of pre-treatment techniques
Pre-treatment Characteristics
method
Remarks
Physical
Increases accessible surface area and pore
volume
decreases the degrees of polymerization of
cellulose and its crystallinity
Cutting, grinding
Chemical
Increases delignification, decreases the degree
of polymerization and crystallinity of cellulose
associated with its swelling, and porosity
growth
Acid pretreatments solubilizes the
hemicellulosic fraction of the biomass and
makes the cellulose more accessible to
enzymes.
Commonly used acid and
base are H2SO4 and NaOH,
respectively.
Organic acids such as
fumaric or maleic acids are
also used as alternatives to
enhance cellulose
hydrolysis
33
33
Pre-treatment
method
Characteristics
Remarks
Physicochemical Exploits the use of conditions and
compounds that affect the physical
and chemical properties of biomass
Increases accessible surface area of
the biomass for enzyme accessibility,
decreases cellulose crystallinity,
removes hemicelluloses and lignin
Steam explosion, ammonia
fiber explosion (AFEX),
ammonia recycling
percolation (ARP), soaking
aqueous ammonia (SAA), wet
oxidation, CO2 explosion
Biological
White rot, brown rot and soft
rot fungi are used Brown rot
attacks cellulose while white
and soft rots attack both
cellulose and lignin
Alters the structure of lignin and
cellulose and separate from the
lignocellulosic matrix.
Slow rate
34
34
Detoxification
• Several unwanted by-products such as pentose and hexose sugars, sugar acids,
acetic acid, formic acid, levulinic acid, hydroxymethyl furfural (HMF) and furfural
may form due to partial breakdown of lignin and hemicellulose during
pretreatment.
• Some of these products act as inhibitors for both fermenting microorganisms and
cellulose degrading enzymes when their accumulation is sufficiently high.
• The inhibitors are categorized into three major groups based on the origin such as
aliphatic acids, furan derivatives and phenolic compounds.
• Biological, physical, and chemical methods are used to remove the inhibitors.
• Chemical detoxification may be carried out by using sodium dithionite, sodium
sulfite, and sodium borohydride
35
35
Hydrolysis
The major portions of cellulose and hemicellulose are glucan and xylan, respectively.
Digestibility of both glucan and xylan can be determined from the yield of their
respective monomers and disaccharides using the equations (Wyman et al. 2005).
Digestibility of glucan (%) 
Amount of glu cos e ( g )  1.053  Amount of cel lub iose ( g )
 100
1.111 Amount of glucan ( g )
Digestibility of xylan (%) 
Amount of xylose ( g )  1.064  Amount of xylobiose ( g )
 100
1.136  Amount of xylan ( g )
Where, 1.053, 1.111, 1.064, and 1.136 are the hydration factors of cellubiose,
glucan, xylobiose and xylan, respectively.
Zhu et al., 2008, Bioresource Technology 99(9): 3817-3828.
36
36
Total SugarConversion(%) 

Glu cos e yield  Xylose yield
Theoretical glu cos e yield  Theoretical xylose yield
(G   G 0) V  (X   X 0) V
W  Glucan content W  Xylan content

0.9
0.88
Where,
[G], is glucose concentration in hydrolysis liquid (mg/mL);
[G]0, is the initial glucose concentration, can be assumed as 0 (mg/mL);
[X], is xylose concentration in hydrolysis liquid (mg/mL);
[X]0, is initial xylose concentration, can be assumed as 0 (mg/mL);
V, is initial volume of biomass slurry (mL);
W, is the initial dry weight of biomass (mg);
0.9, is conversion factor of glucose to equivalent glucan;
0.88, is the conversion factor of xylose to equivalent xylan.
37
37
Fermentation
C 6 H 12O6 (Glu cos e)  2C 2 H 5OH  2CO2
3C5 H 10O5 ( Xylose, arabinose)  5C 2 H 5OH  5CO2
(6.7)
C12 H 22O11 (Cel lub iose)  H 2O  5C 2 H 5OH  5CO2
(6.8)
3C10 H 18O9 ( Xylobiose)  3H 2O  5C 2 H 5OH  5CO2
Xylose
Arabinose
(6.9)
Where, YEtOH is ethanol yield (g/kg),
Ethanol yield
C is ethanol concentration (g/L),
YEtOH 
CV
m
V is initial volume of liquid medium (L),
Glucose
and m is the mass of the substrate (kg).
38
38
Ethanol productivity
Amount of ethanol produced per unit of substrate utilized per unit of time. It is
typically determined when ethanol concentration is maximal.
QEtOH 
CV 1000YEtOH

mt
t
(6.11)
Where, QEtOH is the ethanol productivity (mg/kg h), and t is the time at which the
ethanol concentration produced on substrates is maximum (h).
The maximum theoretical ethanol yield :
Y max (%) 
Ethanol produced in reactor ( g )
100
Initial sugar in reactor ( g )  0.511
(6.12)
39
39
Microorganisms for fermentation of LCB
• Sugar for starch based ethanol production is mostly hexose sugar (glucose),
however, for LCBs mixer of pentose and hexose sugars are available.
• The most common and traditional microorganism for fermenting sugars into
ethanol is a yeast, Saccharomyces cerevisiae, However, the natural strains of this
yeast can ferment only six carbon sugars and hardly pentoses.
• Some yeast strains such as Pichia stipitis, Candida shehatae, and Candida
parapsilosis can ferment xylose but not efficient for hexose fermentation
• Bacterial strains used for ethanol fermentation are Zymomonas mobilis,
Escherichia coli and Klebsiella oxytoca, which have attracted a particular interest
due to their capability to ferment hexose sugars rapidly
• Genetic modification of bacteria and yeast has produced strains capable of co–
fermenting both pentoses and hexoses
40
40
Product recovery
Product
recovery
Thermal route
Ethanol
Low temperature route
Solid residue
Fermenter
Steam
stripper
Distillation
tower
Waste water
• Membrane processing
• Molecular sieve water adsorption
Vapour
phase drying
41
41
• The first process that was used in many of the earlier ethanol plants is the socalled azeotropic distillation or ternary distillation process (as opposed to a binary
or two component distillation process). It consists of introducing a third
component, benzene or cyclohexane, to the azeotropic solution which forms a
heterogeneous azeotropic mixture. When this mixture is distilled, anhydrous
ethanol is produced at the bottom of the distillation column.
 (B.P. of ethanol 78.4 oC , water 100 oC, mixture 78.1 oC (95.5 % ethanol)
 (B.P. of cyclohexane 80.7 oC, water 100 oC, mixture 69.8 oC (80 % cyclohexane)
• Another early method is called extractive distillation, which consists of adding a
ternary component that increases the relative volatility of ethanol. In this case,
anhydrous alcohol is produced and withdrawn at the top of the distillation column.
 ethylene-glycol
 Some other methods: Salt distillation; Pressure swing distillation; Pervaporation
42
42
Butanol production from lignocellulosic biomass
Process
• Butanol is a major co-product of ABE (acetone, butanol and ethanol) fermentation
as the typical proportions of acetone, butanol and ethanol are 3:6:1..
• Clostridium spp., particularly C. acetobutylicum or C. beijerinckii are capable of
converting broad-ranging carbon sources (e.g. glucose, galactose, cellobiose,
mannose, xylose and arabinose) to fuels and chemicals
• The ABE fermentation is biphasic involving acidogenesis and solventogenesis.
• Acidogenic phase involves the production of acids (e.g. acetic and butyric acid),
• Solventogenic phase is related to the production of solvents (e.g. acetone, butanol
and ethanol).
• The ABE-producing bacteria can utilize both starchy and lignocellulosic substrates
43
43
Process
• The butanol-producing clostridia can metabolize both pentose and hexose sugars
in the biomass hydrolysates.
• Butanol toxicity to the fermenting microorganisms is a major barrier for the
commercial bio production of the alcohol that limits its concentration in the
fermentation broth.
• Butanol-producing bacterial cultures can rarely tolerate more than 2% butanol.
• Application of molecular techniques to the solventogenic clostridia has resulted
in the development of hyper-butanol-producing strains such as C. beijerinckii
P260 and C. beijerinckii BA101.
• Process for product recovery is also different with respect to ethanol
44
44
Butanol recovery
• The partial solubility of butanol in water creates impediments in its downstream
processing and extraction from the fermentation broth.
• The high energy requirements subsequently add to the process expenditures making
the overall butanol production expensive.
• To make ABE fermentation profitable, many in situ product recovery systems have
been developed; a few of which include adsorption, liquid-liquid extraction,
pervaporation, gas stripping and supercritical fluid extraction.
• The research on these butanol recovery techniques is still at infancy stage, and none
of these technologies have been implemented at the commercial level.
45
45
Representation of butanol recovery by adsorption
Heating
Spent
adsorbent
Butanol lean solution
Regeneration column
• Activated carbon, zeolite and polymeric
resins, silicalite (790–810 g/L ),
adsorbent resins, amberlite, polyvinyl
pyridine, are thoroughly studied for
butanol recovery
Fermentor
Adsorption column
• Adsorption of butanol onto a packed
column from the dilute solution or
fermentation broth followed by
desorption by heat treatment or
displacer
Product
butanol
46
Representation of butanol recovery by pervaporation
• It is a membrane-based recovery technique that employs a non-porous or
molecularly porous membrane. One side of the membrane is brought into contact
with the fermentation broth while on the other side vacuum or gas purge is applied
to create permeate vapor stream. The components from the feed liquid diffuse
through the membrane and evaporate into the permeate vapor phase
• The use of pervaporation
Liquid state
for butanol extraction is
Retentate
Vapor state based on High selective
Permeate
permeation
and
Species A hydrophobicity of butanol
Feed
Species B through the hydrophobic
membrane
Membrane
Adapted from
Selective for Species A J. Chem. Technol. Biotechnol. 80:603-29(2005).
47
Representation of butanol recovery by supercritical fluid extraction
CO2 recycle
Liquid- gas
separator
Feed
• Supercritical fluids, having
both liquid-like and gas-like
properties, can dissolve organic
materials efficiently.
• Water
above
its
critical
temperature (Tc > 374°C) and
critical pressure (Pc > 22 MPa)
is called SCW.
Product
• Similarly, CO2 above its critical
temperature (Tc > 30.95°C) and
Raffinate
Extraction column
critical pressure (Pc > 7.38
MPa) is known as SCCO2.
Adapted from J. Supercrit. Fluids 15:245-52 (1999)
48
Representation of butanol recovery by gas stripping
Fermentor
Gas + product vapor
Gas
recycle
Condenser
Products
(Adapted from Ezeji, Chem. Rec. 4:305-14
• In principle, gas can be purged into the
fermentor through a rotating shaft and the
volatiles are condensed and recovered
from the condenser.
• As the butanol fermentation is anaerobic,
extreme care should be taken to avoid the
entry of O2 into the fermentation medium
through gas stripping.
• Beneficial in its simplicity, costeffectiveness, non-toxicity, no fouling or
clogging due to the presence of biomass
or cells.
49
Representation of butanol recovery by liquid-liquid extraction
• Acetone, butanol and ethanol are
separated by an extracting solvent and
Extract
Fermentor
further recovered through backextraction by another extracting
solvent or via distillation.
Extractor
Regenerator • Extractants explored for butanol
recovery
include
corn
oil,
polyoxyalkylene ethers, vegetable oil,
Product
hexanol,
aromatic
hydrocarbons,
Raffinate Solvent
ketones, alkanes, esters , dibutyl
phthalate, decanol and oleyl alcohol,
(Adapted from Vane, 2008)
Biofuels, Bioprod. Bioref. 2:553-88
crude palm oil and biodiesel
• It relies on the differences between the distribution coefficients of the solvents.
50
Representation of butanol recovery by distillation
• Distillation is a traditional method for butanol recovery from the aqueous
fermentation broth.
• Distillation requires high energy as butanol has a boiling point (117.7°C) greater
than that of water.
• A heterogeneous distillation, especially involving an azeotrope is used to separate
butanol from water mixture.
B.P. of n-butanol 117.8 oC , water 100 oC , mixture 92.4 oC (55 % butanol)
51
BIOGAS
Biogas is a combustible mixture of gases. It
consists mainly of methane(CH4) and carbon
dioxide (CO2) and is formed from the anaerobic
bacterial decomposition of organic compounds,
i.e. in absence of oxygen.
Major components of biogas
Gas
%
Methane
50-75%
Carbon
dioxide
25-50%
Nitrogen
0-10%
Hydrogen
0-1%
Hydrogen
sulphide
0-3%
Oxygen
0-1%
52
52
Steps of AD
Complex organic compounds(carbohydrates , fats, proteins)
Pretreatment
for (LCB)
Hydrolytic bacteria/ enzymes
Hydrolysis
Simple organic compounds(sugar, amino acids, peptides)
Fermentative bacteria
Acidogenesis
Short chain fatty acids
Acetogenesis
H2 producing acetogenic bacteria
Acetic acid /Acetate
Methanogensis
H2, CO2
CH4, CO2
Methane producing
bacteria
53
53
Quantitative description of biomass degradation
Organics Compounds
(76%)
Hydrolysis & Acidogenesis
(4%)
Short chain fatty Acids
(24%)
Acetogenesis &
Dehydrogenation
(52%)
Acetic Acid
Hydrogen
(28%)
Methanogenesis
(20%)
(72%)
Methane
54
54
Elementary description of AD
Source- http://www.fao.org/
•In case the chemical composition of the organic matter is known, the
stoichiometric equation according to McCarty (Pavlostathis et al., 1991):
CnHaObNc + (2n + c - b – 9sd/20 - ed/4) H20
de/8 CH4 + (n - c sd/5 - de/8) CO2
+ sd/20 C5H7O2N + (c - sd/20) NH4++ (c - sd/20) HCO3where: d = 4n + a - 2b – 3c,
s = fraction of waste converted into cells,
e = fraction of waste converted into methane for energy (s + e = 1),
CnHaObNc = empirical formula of waste being digested
C5H7O2N = empirical formula of bacterial dry mass (VSS)
55
55
Some factors affecting biogas yield
• Solid(dry matter) content in feed wastes
• C/N ratio
• Temperature
• pH of the solution
• Retention time
• Type of digester
• Microbes types
56
56
Groups of biogas microbes
Biogas microbes
Methane forming
Non methane forming
Hydrolysing and
fermenting
bacteria
Acetogenic bacteria
Acetoclasticacetate
degrading
bacteria
Hydrogenotrophic
-hydrogen utilising
bacteria
Hydrolytic
Hydrogen producing acetogens
Acidogenic
Homoacetogens
57
57
Role of microbes involved in the main 3 stages of biogas production
1st stage: Fermentative bacteria
Hydrolysis and fermentation of
organic substances
Cellulose
decomposing
bacteria
Protein
decomposing
bacteria
Fat
decomposing
bacteria
• Monomers
• Oligomers
• Saccharides
• Amino Acids
• Fatty acids
• Alcohols
Short chain organic acids,
aldehydes
H2 & CO2
58
58
2nd stage: Hydrogen producing acetogenic bacteria
Decomposes the substances
produced in 1st stage
Acetic acid, H2, CO2
3rd stage: Methane producing bacteriaConvert the substances
produced in 1st & 2nd
stage
CH4 & CO2
59
59
Hydrolysing and fermenting bacteria
• The first step in the fermentation of complex substrates is the hydrolysis of
polysaccharides to oligosaccharides and monosaccharides, of proteins to
peptides and amino acids, of triglycerides to fatty acids and glycerol, and of
nucleic acids to heterocyclic nitrogen compounds, ribose, and inorganic
phosphate
• Relative anaerobes of genera like Streptococcus and Enterobacterium, Bacillus,
Cellulomonas, Mycobacterium are some important hydrolytic bacteria.
• Hydrolytic enzymes like
cellulases, hemicellulases, amylases, lipases and proteases from these
bacteria depolymerizes carbohydrates, lipids, and proteins to soluble molecule.
60
60
Hydrolysing and fermenting bacteria
Complex compounds in feedstocks
Typical Hydrolysis products
Carbohydrates
Sugars, alcohols
Cellulose
Glucose, cellobiose
Proteins
Amino acids
Peptides
Fats, Fatty acids, glycerol
Lignin
Degraded very slowly
61
61
Hydrolysing and fermenting bacteria
• Obligate anaerobes including Bacteroides, Bifidobacterium, Butyrovibrio, Eubacterium, and
Lactobacillus act as acidogenic bacteria.
• Many enteric coliform bacteria, classically represented by the pathogen Escherichia coli in
the genus Escherichia and pathogens in the genera Salmonella and Shigella, are also some
common fermentative(acidogenic) bacteria.
Hydrolysis product
Conversion during acidogenesis
Sugar
Fatty Acids (succinate, acetate, proprionate, lactate,
formate), carbon dioxide, hydrogen
Alcohols
Fatty acids, CO2
Amino acids
ammonia, sulphides, CO2, H2
Fatty acids
Glycerol
Acetate , CO2
62
62
Acetogenic bacteria
 Hydrogen producing acetogens
• Proprionate
CH3CH2COOH+2H2OCH3COOH+CO2+3H2
Clostridia and Acetivibrio
are some example of
acetogenic bacteria
• Propanol
CH3CH2CH2OH+3H2OCH3COOH+CO2+5H2
• Butyrate
CH3CH2CH2COOH+4H2OCH3COOH+2CO2+6H2
 Homoacetogens
4H2+ 2CO2CH3COOH + 2H2O
63
63
Methanogenic bacteria
 Acetoclastic methanogens
CH3COOHCH4+CO2
 Hydrogenotrophic methanogens
• ~70 % of methane is produced by this route
• These are highly sensitive and having
slowest growth rate
• ~30 % of methane is produced by this
route
4H2+CO2CH4+2H2O
• These are less sensitive and having
higher growth rate
 Formic acid and methanol can also be
converted to methane by bacteria
4HCOOH
4CH3OH
CH4 + 3CO2 +2H2O
3CH4 + CO2 +H2O
64
64
Methanogenic bacteria
Methanogenic bacteria are unicellular, Gram-variable, strict anaerobes that do not
form endospores. Their morphology, structure, and biochemical makeup are quite
diverse. More than ten different genera have been described.
The methanogens have been divided into three groups based on the fingerprinting of
their 16S ribosomal RNA (rRNA) and the substrates used for growth and
methanogenesis
Group I contains the genera Methanobacterium and Methanobrevibacter;
Group II contains the genus Methanococcus; and
Group III contains several genera, including Methanomicrobium, Methanogenium,
Methanospirillum, and Methanosarcina
65
65
Pathways for fermentation
Embden–Meyerhof–Parnas (EMP pathway) is responsible for the production of
pyruvate from sugar as an intermediate product, which is further converted to
acetate, fatty acids, carbon dioxide, and hydrogen
•At low partial pressures of hydrogen, acetate is favoured.
• At higher partial pressures, propionate, butyrate, ethanol, and lactate are
favoured, generally in that order (McInerney and Bryant, 1981)
There is also a special mode of cleavage of intermediate pyruvic acid to formic
acid by enteric bacteria that is not found in other bacterial fermentations
66
66
Kinetics of methane production
• One of the steps might be rate-limiting among hydrolysis, acidogenesis,
acetogenesis and methanogenesis for high rate digestion.
• Normally the methanogenesis is the rate limiting step.
• If cellulose content is higher in feedstock, hydrolysis may be rate limiting step
•Presence of lignin decreases the rate of hydrolysis.
•Transfer of the gaseous products to the gas phase may also be the rate-limiting
step for some cases.
67
• Theoretical substrate conversion rates per unit reactor volume can be estimated
as:
R = So (µmaxθ -1) – Ks
θ(µmaxθ - 1)
where R is the substrate converted per liquid volume at hydraulic retention time θ,
So is the substrate concentration in the feed,
µmax is maximum specific growth rate
and Ks is saturation constant or substrate concentration at which the specific growth
rate is ½ µmax.
At 30 – 37oC , optimum conversion of glucose can be achieved at θ 's of 4 h and 4
days in the acid- and methane-phase reactors respectively.
68
For cellulosics, the corresponding θ 's are much higher, about 1 to 2 days for acidphase digestion and 5 to 8 days for methane-phase digestion
Kinetic
constant
Acidogenesis of
Glucose
Cellulose
Methanogenesis of acetate
µmax, day-1
Ks, g/L
7.2
0.4
0.49
4.2
1.7
36.8
Kinetic constants are: µmax, maximum specific growth rate; and Ks, saturation
constant or substrate concentration at which the specific growth rate is ½ µmax.
Source: Donald L. Klass, Biomass for Renewable Energy, Fuels, and Chemicals
69
Some important facts
• Methane fermentation at thermophilic temperatures increases the methane
production rate because of higher reaction rates.
• Reduced pressure also provides little or no benefit (Hashimoto, 1982).
• Low pH and short retention time reduces methane production.
•Rapid, continuous agitation of anaerobic digesters is not necessary, and in some
cases is even harmful (Stafford, 1982).
• Specific methane yield is directly proportional to biodegradable COD or VS loading
in the digester.
70
70
Anaerobic digester
Biogas
Effluent gas
Effluent substrate
Substrate flow
Fluid
zone
Sludge
zone
Mixing
zone
Mixing zone
Ground sludge
pipe
Ground injection
pipe
71
71
Mixing helps to reduce the natural stratification that occurs in a low-profile tank.
Mixing can be accomplished with a variety of gas mixers, mechanical mixers, and
draft tubes with mechanical mixers or simply recirculation pumps.
Completely mixed reactors can have fixed covers, floating covers or gas holding
covers.
Most municipal digesters have floating covers.
Floating covers are more expensive than fixed covers and standard diameters range
from 15 to 125 ft.
72
72
Types of anaerobic digesters
 Low rate (Conventional digesters)
• Intermittent mixing, sludge feeding and
sludge withdrawal
• Detention time = 30-60 days
• Feeding rate 0.5 to 1.5 kgVS/m3.day
 High-rate digesters
• Continuous mixing (homogenous)
• Continuous or intermittent sludge
feeding and sludge withdrawal
• Detention time = < 15 days
• Feeding rate 1.6 to 6.4 kgVS/m3.day
Most digesters are heated and operated in the mesophilic range and usually
made of concrete or steel
73
73
Digester operation
• Most digesters operate in the temperature range of 30-380 C to optimize time.
• The digester gas produced from the process may be used for the heating purposes.
• Optimum pH range is 7.0 - 7.2.
 Maintained by properly seeding with fresh added sludge and not excessively
withdrawing sludge (should not exceed 3-5 % of dry solids weight in digester).
 Temporary solution for acidification: lime addition.
• Heavy metals may inhibit digestion process. Must be eliminated at the source.
• The supernatant liquor is the water released during digestion.
 May have BOD as high as 2000 mg/l and SS concentration as high as 1000 mg/l.
 Fed back to the influent to primary clarifiers.
74
74
Two stage digester
In a single-phase digester optimum pH is 7
The acidogenic reactions favours at the pH range of 5.5 -5.9
Methanogenic reactions favour at pH 7.4 -7.9
Gas out
Gas out
Sludge in
Thus, two phase
reactor improves
the over all gas
yield
Scum collection
Mixed liquor
Mixing zone
and actively
digested
sludge
Supernatant
Sludge out
Sludge out
Supernatant
removal
Stabilized
sludge
75
75
Sludge volume
• The volume of a digesting sludge in a digester is a function of a volume of fresh
sludge added daily , the volume of digested sludge produced daily and required
digestion time.
• Empirical evidence (batch experiments) has shown that if supernatant is removed
from a batch of digesting sludge as it is produced, volume of remaining digested
sludge vs. digestion time is a parabolic function.
• For a parabolic function, average volume = initial volume – (2/3)(final volume –
initial volume) therefore,
Vavg = Average volume of digesting sludge (m3/day)
V1 = Volume of fresh sludge added daily (m3/day)
Vavg = V1 – (2/3)(V1 –V2)
V2 = Volume of digested sludge produced daily (m3/day)
76
76
•The volume of total sludge in the reactor is given by adding both the digesting and
the digested sludge.
Vs = Vavg. td + V2. ts
where
Vs = Total sludge volume (m3)
Vavg = average volume of digesting sludge (m3/day)
V2 = volume of digested sludge produced daily(m3/day)
td = Time required for digestion(days)
ts = Time provided for sludge storage(days)
•The sludge volume normally occupies the bottom half of the digester, and the
supernatant liquor occupies the upper half therefore the total digester volume, Vt
(m3):
Vt = 2Vs
where
Vt = total digester volume (m3)
77
77
Cross section of digester
Vcol
Vgst
Where
Vcol is gas collection volume
Vgst is gas storage volume
Vd is active digestion volume
Vd
Mixing zone
Vsl is sludge layer volume
Vsl
Total volume (V) = Vcol + Vgst + Vd + Vsl
78
78
• V1 is volume of top head
• V2 is Volume of bottom head
S1
• S1 is surface area of top head
h
V1
1
• S2 is surface area of bottom head
• V3 is the volume of cylindrical part
R1
V3
H
• R1 is the characteristic diameter of top head
• R2 is the characteristic diameter of bottom head
R2
Mixing zone
• H is the height of cylindrical portion
V2
h2 • h1 is the height of top head
S2
• h2 is the height of bottom head
D
R
• R is radius of the cylindrical part
Geometrical dimensions of the • D is the diameter of the cylindrical part
cylindrical
shaped
biogas All the above parameters can be related to “D”
differently on the basis of head types
digester body
Reactor dimension
79
79
Comparison between cross section and dimension
Vcol
Vgst
S1
V3
Vd
R1
H
R2
Mixing zone
Mixing zone
Vsl
h1
V1
H1
V2
S2
R
h2
D
80
80
Reactor dimension
For volume
Vcol ≤ 5 % of V
Vsl ≤ 15 % of V
Vgst + Vd = 0.8*V
For a typical case
Source
http://www.sswm.
info/sites/default/f
iles/reference_atta
chments/BRC%20n
y%20Design%20Bi
ogas%20Plant.pdf
Vgst = 0.5 *(Vgst + Vd + Vsl)*K
Or = 0.5*TS*M
Where K is gas production rate per m3 digester volume per day
and M is gas production rate per kg of TS per day.
For Bangladesh K is 0.4 m3/m3/d and M is 0.28 m3/kg/d
For geometrical dimension
D = 1.3078*V 1/3
V1 = 0.0827 D3
V2 = 0.05011 D3
V3 = 0.3142 D3
R1 = 0.725D
R2 = 1.0625D
h1 = D/5
h2 = D/8
S1 = 0.911 D2
S2 =0.8345 D2
81
81
Volume calculation of digester chamberDetermine the volume of the digester chamber and dimension of the chamber for
the biogas production from the cow dung of 10 cows having body weight of 200 kg
each. Assume the temperature is 30 oC , the solid content of the cow dung is 16 %,
HRT is 40 days. One cow produces 10 kg cow dung daily.
Solution:
Total discharge = 10 kg X 10 = 100 Kg/day
TS of fresh discharge = 100 kg X 0.16 = 16 Kg.
In 8% concentration of TS ( To make favourable condition )
8 Kg. Solid = 100 Kg. Influnt
1 Kg. Solid = 100 / 8 Kg influent
16 Kg Solid = 100 x 16/ 8 = 200 Kg. Influent.
Total influent required = 200 Kg.
82
82
Water to be added to make the discharge 8% concentration of TS
=200 Kg - 100 Kg. = 100 Kg.
Working volume of digester = Vgst + Vd
Vgst + Vd = Q.HRT
= 200 Kg/day X 40 days
= 8000 Kg. ( 1000 Kg = 1 m3 ) = 8 m3.
From geometrical assumptions:
Vgst + Vd = 0.80 V
Or V= 8/0.8 = 10 m3. (Putting value Vgst + Vd = 8 m3)
& D = 1.3078 V 1/3 = 2.8176 m ~ 2.82 m.
83
83
Again
V3 = 3.14*D2*H/4
Putting V3 = 0.3142D3 = 7.046 m3
3.14*D2*H/4 =0.3142D3
Or H = 4*0.3142D/3.14 = 1.129 m
h1 = 0.564
V1 =
m
1.86 m3
R1= 2.045
V3 =
h1=D/5 = 0.564 m
h2 = D/8 = 0.352 m
R1 = 0.725D= 2.045 m
R2 = 1.0625D= 2.996 m
V1 = 0.0827D3 =1.855 m3
V2= 0.05011 D3 = 1.124 m3
7.05 m3
m
R
Mixing zone 2= 2.996
H=
1.129 m
m
V2 = 1.12
h2=
D= 2.82 m
0.352 m
R
84
84
Vcol= 0.05*V= 0.5 m3
Vgst = 0.5 *(Vgst + Vd + Vsl)*K = 0.5 *(V- Vcol)*K= 0.5*(9.5)*0.4 = 1.9 m3
Alternatively,
Vgst = 50 % of daily gas yield = 0.5 * TS* gas production rate per kg TS/day
= 0.5*100*0.16*0.28 m3/kg TS= 2.24 m3
Vgst + Vcol = 0.5 + 2.24 = 2.74 m3
Further,
V1 = [(Vgst + Vcol ) - (3.14*D2H1/4)] = 2.74 – 6.24 H1
Or H1 =(2.74-1.855)/6.24 = 0.142 m
Vd = (V- Vgst -Vcol – Vsl)
3
Assuming Vsl = 15 % of V = 10*0.15 = 1.5 m
= 10-2.74-1.5= 5.76 m3
85
85
10.1 Introduction
 Certain foodstuffs and pharmaceutical materials are produced by
fermentations that include the culture of cells or microorganisms. When
the initial concentrations of these products are low, a subsequent
separation and purification by the so-called “downstream processing” is
required to obtain them in their final form.
 In many cases, a high level of purity and biological safety are essential
for these products.
 Very often, many separation steps (as shown schematically in Figure
10.1) are required to achieve the requirements of high purity and bio
logical safety during industrial bioprocessing.
 Thus, downstream processing may often account for a large proportion of
the production costs.
1
Continued…
Figure 10.1: Main steps in
downstream processing
2
Continued…
 Figure 10.2 shows, as an example, several steps in the production of
interferon a. In the first step, Escherichia coli (which actually produces
the interferon) is obtained genetically, and a strain with high productivity
is then selected.
 Next, the strain is cultivated first on a small scale (flask culture), and
then increasingly on larger scales, usually with approximately 10-fold
increases in culture volume at each step.
3
Continued…
Figure 10.2: The steps of
interferon a production.
4
10.2 Filtration
 Filtration separates particles by forcing the fluid through a filtering
medium on which solids are deposited. Filtration can be divided into
several categories depending on the filtering medium used, the range of
particle sizes removed, the pressure differences, and the principles of the
filtration, such as conventional filtration, microfiltration, ultrafiltration,
and reverse osmosis.
 A wide variety of filters are available for the cell recovery. There are
generally two major types of filters: pressure and vacuum filters.
 The two types of filters most· used for cell recovery are the filter press
and rotary drum filters.
Assuming the laminar flow across the filter, the rate of filtration (𝑑𝑉𝑓 /𝑑𝑡)
can be expressed as a function of pressure drop ∆𝑝 by the modified D'Arcy's
equation as (Belter et al., p. 22, 1988),
5
Continued…
1 𝑑𝑉𝑓
∆𝑝
=
𝐴 𝑑𝑡
𝜇(𝐿/𝐾)
10.1
Where,
A is the area of filtering surface,
K is D'Arcy's filter cake permeability
L is the thickness of the filter cake.
Eq. (10.1) states that the filtration rate is proportional to the pressure drop
and inversely proportional to the filtration resistance which is the sum of the
resistance L/K by the filter medium 𝑅𝑀 and that by the cake 𝑅𝐶 ,
𝐿
= 𝑅𝑀 + 𝑅𝐶
𝑘
10.2
6
Continued…
The value of 𝑅𝑀 is relatively small compared to 𝑅𝐶 , which is proportional to
the filtrate volume as
𝛼𝜌𝑐 𝑉𝑓
𝐿
≈ 𝑅𝐶 =
𝑘
𝐴
10.3
where 𝛼 is the specific cake resistance and 𝜌𝑐 is the mass of cake solids per
unit volume of filtrate.
which can be used to relate the pressure drop to time when the filtration rate
is constant.
Therefore, biological feeds may require pretreatments such as:
1. Heating to denature proteins
2. Addition of electrolytes to promote coagulation and flocculation
3. Addition of filter aids (diatomaceous earths or perlites) to increase the
porosity and to reduce the compressibility of cakes
7
10.3 Centrifugation
Centrifugation is an alternative method when the filtration is ineffective,
such as in the case of small particles. Centrifugation requires more
expensive equipment than filtration and typically cannot be scaled to the
same capacity as filtration equipment.
Two basic types of large-scale centrifuges are the tubular and the disk
centrifuge as shown schematically in Figure 10.6.
 The tubular centrifuge consists of a hollow cylindrical rotating element
in a stationary casing.
 The disk centrifuge is the type of centrifuge used most often for bioseparations. It has the advantage of continuous operation.
When a suspension is allowed to stand, the particles will settle slowly
under the influence of gravity due to the density difference between the
solid and surrounding fluid, a process known as sedimentation.
8
10.4 Chromatography
 Chromatography separates mixtures into components by passing a fluid
mixture through a bed of adsorbent material. We will be interested
primarily in elution chromatography.
 Typically a column is packed with adsorbent particles, which may be
solid, a porous solid, a gel, or a liquid phase immobilized in or on a solid.
 Solutes that interact weakly with the matrix pass out of the column
rapidly (see Fig. 11.27), while those that interact strongly with the matrix
exit slowly.
 These differential rates of migration separate the solutes into different
peaks that exit at a characteristic retention time (which will change if
operating conditions are altered).
9
Continued…
Figure 10.7: Concentrations in elution
chromatography. Three solutes, shown
schematically as circles, squares, and
triangles, are injected into one end of a
packed bed. When solvent flows through
the bed from left to right, the three solutes
move at different rates because of
different adsorption. They exit at different
times, and hence are separated.
10
Continued…
For the remainder of our discussion we focus on elution chromatography as
a batch process. Some important types of chromatographic methods are:
1. Adsorption chromatography (ADC) is based on the adsorption of solute
molecules onto solid particles, such as alumina and silica gel, by weak
van der Waals forces and steric interactions.
2. Liquid–liquid partition chromatography (LLC) is based on the
different partition coefficients (solubility) of solute molecules between
an adsorbed liquid phase and passing solution. Often the adsorbed liquid
is nonpolar.
3. Ion-exchange chromatography (IEC) is based on the adsorption of ions
(or electrically charged compounds) on ion-exchange resin particles by
electrostatic forces.
11
Continued…
4. Gel-filtration (molecular sieving) chromatography is based on the
penetration of solute molecules into small pores of packing particles on
the basis of molecular size and the shape of the solute molecules. It is
also known as size exclusion chromatography.
5. Affinity chromatography (AFC) is based on the specific chemical
interactions between solute molecules and ligands (a functional
molecule covalently linked to a support particle). Ligand–solute
interaction is very specific, such as enzyme– substrate interaction,
which may depend on covalent, ionic forces or hydrogen-bond
formation. Affinity binding may be molecular size and shape specific.
6. Hydrophobic chromatography (HC) is based on hydrophobic
interactions between solute molecules (e.g., proteins) and functional
groups (e.g., alkyl residues) on support particles. This method is a type
of reverse phase chromatography which requires that the stationary
phase is less polar than the mobile phase.
12
Continued…
7. High-pressure liquid chromatography (HPLC) is based on the general
principles of chromatography, the only difference being high liquid
pressure applied to packed column. Owing to high-pressure liquid (high
liquid flow rate) and dense column packing, HPLC provides fast and
high resolution of solute molecules.
13
10.4.1 Chromatography for Bioseparation
 Liquid column chromatography is the most commonly used method in
bio-separation. As shown in Figure 10.8, the adsorbent particles are
packed into a column as the stationary phase, and a fluid is continuously
supplied to the column as the mobile phase.
 For separation, a small amount of solution containing several solutes is
supplied to the column (Figure 10.8a).
 Each solute in the applied solution moves down the column with the
mobile phase liquid at a rate determined by the distribution coefficient
between the stationary and mobile phases (Figure 10.8b). and then
emerges from the column as a separated band (Figure 10.8c).
 Depending on the type of adsorbent packed as the stationary phase and
on the nature of the interaction between the solutes and the stationary
phase, several types of chromatography may be used for bio-separations,
as listed in Table 10.1.
14
Figure 10.8: Schematic diagram showing the chromatographic
separation of solutes with different distribution coefficients.
15
Continued…
Table 10.1 Liquid column chromatography used in bio-separation.
Liquid
chromatography
Mechanism of
partition
Elution method
Characteristics
Gel filtration
chromatography (GFC)
Size and shape of
molecule
Isocratic
Relatively long column for
separation
Separation under mild
condition High recovery
lon-exchange
chromatography (IEC)
Electrostatic
interaction
Gradient Stepwise Short column, wide
applicability
Separation depending
on elution conditions High
capacity
Hydrophobic interaction
chromatography (HIC)
Hydrophobic
interaction
Gradient Stepwise Adsorption at high salt
concentration
Affinity chromatography
(AFC)
Biospecific
interaction
Stepwise
High selectivity High
capacity
16
Continued…
 Chromatography is operated in several ways, depending on the strength
of the interaction (affinity) between the solutes and the stationary phase.
 In the case where the interaction is relatively weak as in gel
chromatography the solutes are eluted by a mobile phase of a constant
composition (isocraticelution).
 The performance of a chromatographic system is generally evaluated by
the time required for the elution of each solute (retention time) and the
width of the elution curve (peak width), which represents the
concentration profile of each solute in the effluent from a column.
Although several models are used for evaluation, the equilibrium, stage,
and rate models are discussed here.
17
10.4.2 General Theories on Chromatography
10.4.2.1 Equilibrium Model
In this model, the rate of migration of each solute along with the mobile
phase through the column is obtained on the assumptions of instantaneous
equilibrium of solute distribution between the mobile and the stationary
phases, with no axial mixing. The distribution coefficient K is assumed to be
independent of the concentration (linear isotherm), and is given by the
following equation:
𝐶𝑆 = 𝐾𝐶
10.8
where 𝐶𝑆 , (kmol per 𝑚3 -bed) and C (kmol 𝑚3 ) are the concentrations of a
solute in the stationary and the mobile phases, respectively. The fraction of
the total solute that exists in the mobile phase is given by
𝐶𝜀
1
𝑋𝑆 =
=
𝐶𝜀 + 𝐶𝑆 (1 − 𝜀) 1 + 𝐸𝐾
10.9
18
Continued…
where 𝜀 (−) is the interparticle void fraction and 𝐸 = (1 − 𝜀)/𝜀 . The
moving rate of the solute 𝑑𝑧/𝑑𝑡 (𝑚𝑠 −1 ) along with the mobile phase
through the column is proportional to 𝑋𝑆 , and the interstitial fluid velocity
𝑢 (𝑚𝑠 −1 ).
𝑑𝑧
𝑢
= 𝑋𝑆 𝑢 =
𝑑𝑡
1 + 𝐸𝐾
10.10
If the value of 𝐾 is independent of the concentration, then the time required
to elute the solute from a column of length Z (m) (retention time 𝑡𝑅 ) is given
as
𝑍(1 + 𝐸𝐾)
𝑡𝑅 =
10.11
𝑢
The volume of the mobile phase fluid that flows out of the column during
the time 𝑡𝑅 − that is, the elution volume 𝑉𝑅 (𝑚3 ) is given as
𝑉𝑅 = 𝑉0 + 𝐾(𝑉𝑡 − 𝑉0 )
10.12
= 𝑉0 + (1 + 𝐸𝐾)
19
Continued…
where 𝑉0 (𝑚³) is the interparticle void volume and 𝑉𝑡 (𝑚³) is the total bed
volume.
 Solutes flow out in the order of increasing distribution coefficients, and
thus can be separated. Although a sample is applied as a narrow band, the
effects of finite mass transfer rate and flow irregularities in practical
chromatography result in band broadening, which may make the
separation among eluted bands of solutes insufficient.
 These points must be taken into consideration when evaluating
separation among eluted bands and can be treated by one of the following
two models, namely, the stage model or the rate model.
20
10.4.2.2 Stage Model
 The performance of a chromatography column can be expressed by the
concept of the theoretical stage at which the equilibrium of solutes
distribution between the mobile and stationary phases is reached.
 For chromatography columns, the height of packing equivalent to one
equilibrium stage (i.e., the column height divided by the number of
theoretical stages N) is defined as the height equivalent to an equilibrium
stage, Hs (m).
 The shape of the elution curve for a pulse injection can be approximated
by the Gaussian error curve for 𝑁 > 100, which is almost the case for
column chromatography.
 The value of N can be calculated from the elution volume 𝑉𝑅 (𝑚3 ), and
the peak width W (𝑚3 ), which is obtained by extending tangents from
the sides of the elution curve to the baseline and is equal to four times the
standard deviation 𝜎𝑣 𝑚3 = (𝑉𝑅 2 /𝑁)1/2, as shown in Figure 10.5.
21
Continued…
Figure 10.9: Elution curve and parameters for evaluation of N.
22
Continued…
𝐻𝑠 =
𝑍
𝑍
=
𝑁
𝑉𝑅
𝜎𝑣
2
𝑍
=
16
𝑉𝑅
𝑊
2
10.13
With a larger N, the width of the elution curve becomes narrower at a given
𝑉𝑅 , and a better resolution will be attained. The dependence of 𝐻𝑠 on these
parameters can be obtained by the following rate model.
23
10.4.2.4 Resolution Between Two Elution
Curves
Since the concentration profile of a solute in the effluent from a
chromatography column can be approximated by the Gaussian error curve,
the peak width W (𝑚3 ) can be obtained by extending tangents at inflection
points of the elution curve to the base line and is given by
4𝑉𝑅
𝑊=
= 4𝜎𝑣
(𝑁)1/2
10.15
where 𝑉𝑅 (𝑚3 ) is the elution volume, 𝑁 is the number of theoretical plates,
and 𝜎𝑣 , is the standard deviation based on elution volume (𝑚3 ). The
separation of the two successive curves of solutes 1 and 2 can be evaluated
by the resolution 𝑅𝑠 , as defined by the following equation:
𝑅𝑠 =
𝑉𝑅,1 − 𝑉𝑅,2
(𝑊1 + 𝑊2 )/2
10.16
24
Continued…
Values of 𝑉𝑅,2 and 𝑉𝑅,1 given by Equation 10.12 are substituted into
Equation 10.16 and, if 𝑊1 is approximated as equal to 𝑊2 , then substitution
of Equation 10.15 into Equation 10.16 gives
𝐸(𝐾2 − 𝐾1 )𝑁 1/2
𝑅𝑠 =
4(1 + 𝐸𝐾1 )
10.17
The separation behaviors of two curves with change in the number of
theoretical plates are shown in Figure 10.10. For any given difference
between the values of 𝐾1 , and 𝐾2 , in Equation 10.17, larger values of 𝑁 lead
to a higher resolution, as shown in Figure 10.10. In order to attain good
separation in chromatography − that is, a good recovery and high purity of a
target − the value of the resolution between the target and a contaminant
must be above 1.2.
25
Continued…
Figure 10.10: Resolution of two elution curves at three theoretical plates,
interparticle void fraction 0.36.
26
Continued…
Example 10.2: A chromatography column of 10 mm i.d. and 100 mm
height was packed with particles for gel chromatography. The interparticle
void fraction was 0.20. A small amount of a protein solution was applied to
the column and elution performed in an isocratic manner with a mobile
phase at a flow rate of 0.5 𝑐𝑚³ 𝑚𝑖𝑛−1 . The distribution coefficient 𝐾 of a
protein was 0.7. An elution. curve of the Gaussian type was obtained, and
the peak width W was 1.30 𝑐𝑚³. Calculate the 𝐻𝑠 value of this column for
this protein sample.
Solution:
The elution volume of the protein is calculated by Equation 11.18 with the
values of 𝐾 and 𝐸 = (1 − 𝜀)/𝜀.
𝑉𝑅 = 1.57 1 + 4 × 0.7 = 5.97 𝑐𝑚3
Equation 10.13 gives the 𝐻𝑠 value as follows:
𝐻𝑠 =
10
= 0.03 𝑐𝑚
16(5.97/1.3)2
27
10.4.3 Factors Affecting the Performance of
Chromatography Columns
Liquid chromatography can be operated under mild conditions in terms of
pH, ionic strength, polarity of liquid, and temperature. The apparatus used is
simple in construction and easily scaled up. Moreover, many types of
interaction between the adsorbent (the stationary phase) and solutes to be
separated can be utilized, as shown in Table 10.1. Liquid chromatography
can be operated isocratically, step-wise, and with gradient changes in the
mobile phase composition.
In order to estimate resolution among peaks eluted from a
chromatography column, those factors that affect N must first be elucidated.
By definition, a low value of Hs will result in a large number of theoretical
plates for a given column length.
28
10.4.3.3 Sample Volume Injected
In industrial-scale chromatography, one very desirable aspect is the
ability to handle large amounts of samples, without any increase in Hs . For
sample volumes of up to a small percentage of the total column volume, Hs
remains constant, but: it then increases with the sample volume. It has been
reported that the effect of sample volume on Hs should rather be treated as
the effect of the sample injection time t 0 (s).
𝑍(𝜎𝑡2 + 𝑡02 /12)
𝐻𝑠 =
(𝑡𝑅 + 𝑡0 /2)2
10.20
where Z is the column height (m), 𝜎𝑡2 is the dispersion of elution curve at a
small sample volume (𝑠 2 ),and 𝑡𝑅 is the retention time (s).
The dispersion a, is obtained from the peak width (40 𝜎𝑡 ) of an elution
curve on the plot of solute concentrations versus time. From the relative
magnitudes of 𝜎𝑡2 and 𝜎𝑡2 /12 , a suitable injection time 𝑡0 without
significant loss of resolution can be determined.
29
Continued…
Example 10.3: In a gel chromatography column packed with particles of
average radius 22 𝜇m at an interstitial velocity of 1.2 𝑐𝑚 𝑚𝑖𝑛−1 of the
mobile phase, two peaks show poor separation characteristics, that is, 𝑅𝑆 =
0.85.
A. What velocity of the mobile phase should be applied to attain good
separation (𝑅𝑆 above 1.2)with the same column length?
B. What length of a column should be used to obtain 𝑅𝑆 = 1.2 at the same
velocity?
C. The linear correlation of ℎ = 𝐻𝑠/2𝑟0 versus v shown in Figure 10.8 can
be applied to this system, and the diffusivities of these solutes of 7 ×
10−7 𝑐𝑚2 𝑠 −1 can be used.
Solution:
Under the present conditions
(2𝑟0 𝑢) (2 × 0.0022 × 1.2)
𝑣=
=
= 125
𝐷
(60 × 7 × 10−7 )
30
Continued…
From Figure 10.8,
𝐻𝑠
ℎ=
= 9.0
(2 × 0.0022)
𝐻𝑠 = 0.040
A. R S increases in proportion to (Hs)−1/2 . Thus, the value of Hs must be
lower than 0.020 to obtain a resolution R S larger than 1.2. Again, from
Figure 10.8.
𝑣 = 46
B. Then
𝑢 = 0.0073 cm 𝑠 −1 = 0.44 𝑐𝑚 𝑚𝑖𝑛−1
B. At the same value of Hs, R S increases in proportion to 𝑁 1/2 , that is,
𝑍1/2 . To increase the value of R S from 0.85 to 1.2, the column should be
(1.2/0/85) 2 = 2.0 times longer.
31
Mechanism of adsorptive removal
 Adsorptive Filtration contd..
Step-1. Transport of adsorbate from bulk to the surface of adsorbent
Step-2. Diffusion of adsorbate into the interior part of adsorbent
Step-3. Adsorption of adsorbate by the following possible routes
OH
OH
OH
Fe
Fe
O
Physical adsorption
Fe
O
O
O
C
O
OH-
H2AsO4-1, HAsO4-2
Attachment through exchange of
hydroxyl ion
Chemisorption
-
-
O AsO3
Fe
+
+
O
O AsO3
O AsO3
O
Fe
O
Fe
O
C
O
H
Mondal et al., 2007, Ind. Eng. Chem. Res. 46(8)
32
 Adsorptive Filtration contd..
Factors affecting adsorption
• Temperature
• Concentration of pollutants/adsorbents
• Contact time of adsorbate and adsorbent
• Adsorbent dose
• pH of the solution
 Performance of the adsorption process largely depends on the surface chemistry of the
adsorbent and solution chemistry of the adsorbate/pollutant
 Zero charge potential of the adsorbent helps to understand the effect of pH on the
adsorbent process
33
33
 Adsorptive Filtration contd..
GAC as adsorbent and its surface modification
GAC has wide application in water treatment due to its low cost and its capacity to remove
various types of pollutants including color & odour etc.
At neutral pH the arsenic removal efficiency of GAC is very less due to its predominantly
negatively surface charge (In our experiment with 16 g/l adsorbent dose and 24 h of agitation
the arsenic concentration in the treated water could not be reduced from 188 ppb to 50 ppb)
δ+
 Positive charge density on the surface of
GAC can be increased at neutral pH by
impregnating metal ion having more than one
positive ions.
δ_
δ+
δ+
δ+
δ_
δ+
Ca+2
δ+
δ_
δ+
Ca+2
δ+
δ+
δ_
δ+
δ+
δ_
b. Surface charge with medium impregnation (NPC=6δ+)
δ+
2007, Ind. Eng. Chem. Res. 46(8)
δ_
a. Surface charge without impregnation (NPC=2δ+)
δ+
Ca+2
Schematic presentation of the increase in positive charge density
on adsorbent surface due to surface modification Mondal et al.,
δ+
δ+
δ_
Ca+2
δ+
δ+
δ+
δ_
δ+
Ca+2
δ+
δ_
c. Surface charge with optimum impregnation (NPC=8δ+)
34
 Adsorptive Filtration contd..
Performance evaluation of adsorbent
• Adsorption capacity (specific uptake)
• Adsorption kinetics
• Spent adsorbent regeneration
Specific uptake is determined through isotherm study . Important isotherm models are
Langmuir isotherm: It assumes a mono-layer adsorption on a uniform adsorbent surface with
energetically identical sorption sites.
𝑞 =𝑞 𝑏 c /(1+𝑏 c )
𝑒
𝑜 𝐿 e
𝐿 e
The linear form of the Langmuir isotherm equation is given by
𝐶𝑒
𝑞𝑒
=
1
𝑞𝑜 𝑏𝐿
+
𝐶𝑒
𝑞𝑜
Where Ce is the equilibrium concentration of the adsorbate (μg.L-1) in solution, qe is the
amount of adsorbate per unit mass of adsorbent (μg.g-1), qo and bL are Langmuir constants
related to adsorption capacity and rate of adsorption respectively.
35
35
 Adsorptive Filtration contd..
The feasibility of Langmuir isotherm can be described by a separation factor RL, which can be
determined by the following equation:
1
R L = 1+b
𝐿 Co
Where, Co is the initial concentration of arsenic or fluoride (μg.L-1) and bL is the Langmuir
constant (L.g-1).
The value of separation factor RL, indicates the isotherm’s shape and the nature of the
adsorption process, unfavorable for (RL > 1), linear for (RL = 1), favorable for (0 < RL < 1) and
irreversible (RL = 0).
36
36
 Adsorptive Filtration contd..
Freundlich adsorption isotherm describes equilibrium on heterogeneous surfaces and hence
does not assume mono layer capacity. The logarithmic form of the Freundlich isotherm
expression is given by the following equation:
1
lnqe = lnK F + 𝑛 lnCe
Where, KF and n are Freundlich constants which show the adsorption capacity and favorability
of adsorption respectively. The value of 1/n obtained from the slope of the above Eq. should
lie between 0 and 1 for favorable adsorption process
For multicomponent system modified competitive Langmuir model, extended Langmuir
model and extended Freundlich model are used
37
37
 Adsorptive Filtration contd..
Adsorption kinetics
In order to study the adsorption kinetics some important models used are
Pseudo first order, pseudo second order as well as intra particle diffusion models (Webber
and Morris model )
Pseudo first order rate equation or linearized Lagergren equation can be written as:
ln 𝑞𝑒 − 𝑞𝑡 = 𝑙𝑛𝑞𝑒 − 𝑘1 𝑡
Where, qe and qt are the amount of pollutants adsorbed at equilibrium (μg.g-1) and at time t
respectively, k1 is rate constant for pseudo first order adsorption (sec-1).
The linear form of Pseudo second order kinetic model can be represented as:
𝑡
𝑞𝑡
=𝑘
1
2
2 𝑞𝑒
+
𝑡
𝑞𝑒
Where, k2 is the rate constant for pseudo second order adsorption (g.µg-1.min-1).
38
38
 Adsorptive Filtration contd..
Webber and Morris model
According to this model the relationship between qt and t1/2 is written as
1
2
𝑞𝑡 = 𝐾𝑖𝑑 𝑡 + 𝐶
Where, Kid (μg.g-1.sec-1/2) is the rate constant for intra particle diffusion and C (μg.g-1) is a
constant that gives an idea about the thickness of the boundary layer, i.e. the larger the value of
C the greater is the boundary layer effect.
Useful to investigate the pore diffusion in adsorption (intra particle diffusion) and to
understand the implication of data for improved adsorbent and process design
39
39
 Adsorptive Filtration contd..
Spent adsorbent regeneration
Management of spent adsorbent, which is reach in pollutant concentration, is an important
issue for the success of this process
Adsorbents may be regenerated through different chemical treatment, however
management of residual material is an issue if the residual material is toxic in nature.
Solidification of the spent adsorbent
Landfilling/ hazardous waste management procedures
40
40
Adsorption types and reactors
 Adsorptive Filtration contd..
Powdered
adsorbent
Batch
liquid
desorbate
Feed
Slurry to
filtration
Adsorption
step
Fixed
beds
Desorbate
step
Wash water
Treated water
Batch operation
Continuous operation
41
 Adsorptive Filtration contd..
Fixed bed adsorption
Effluent
Stoichiometric front
LES & WES : length and weight of the bed
section upstream of the front
Spent
Adsorbent
Concentration front at some time t
Feed
Solute concentration
Assumptions:
LUB & WUB : length and
Internal and external mass transfer resistances are very small
Plug flow is achieved
weight of the bed section
Axial dispersion is negligible
downstream of the front
The adsorbent is initially free from adsorbate
Adsorption isotherm initially begins at the origin then local equilibrium between the fluid and
adsorbent is achieved instantaneously
Unused
adsorbent
(In equilibrium
with entering
fluid )
WES
WUB
LES
LUB
0
L
Distance through the bed
Stoichiometric
(equilibrium )
Concentration
front for ideal
Fixed bed
LB adsorption
42
 Adsorptive Filtration contd..
Fixed bed adsorption
Break point & breakthrough curve: After a period of time , called the stoichiometric time , the
stoichiometric wave front reaches the end of the bed , the concentration of the solute in the
fluid abruptly rises to the inlet value cF, no further adsorption is possible and the adsorption
step is terminated. This point is referred to as the break point and the stoichiometric wave
front becomes the ideal breakthrough curve.
Material balance of the adsorbate before
breakthrough occurs :
Solute in entering feed = adsorbate
QF cF tideal = qF SLideal/LB
Location of L as a function of time , is
obtained solely by material balance and
adsorption equilibrium considerations
Where QF is the volumetric flow rate of feed; cF is concentration of the solute in feed; tideal is
time for an ideal front to reach Lideal < LB ; qF is the loading per unit mass of adsorbent that
is in equilibrium with the feed concentration; S is the total mass of adsorbent in the bed; LB
is total bed length
43
43
 Adsorptive Filtration contd..
LUB = LB - LES
Lideal= LES = (QF cF tideal /qF S) LB
Fixed bed adsorption
WUB = S - WES
WES = S (LES/ LB)
In a real fixed bed adsorber , the above assumptions are not valid. Internal transport
resistance and, in some cases, external transport resistance are finite. Axial dispersion can
also be significant, particularly at low flow rates in shallow beds.
Unused bed
at t2
Breakthrough
curve
Solute wave
fronts in a
fixed bed
adsorbers with
mass transfer
effects
(a)Concentrati
Ls < L < Lf
on-distance
Mass transfer zone
profiles. (b)
(MTZ) where adsorption Breakthrough
takes place
curve
At t= t2
L < Ls
Bed is almost saturated
L > Lf
Bed is almost clean
44
 Adsorptive Filtration contd..
Fixed bed adsorption
Lf can be taken where c/ cF = 0.05 with Ls at c/cF =0.95
At tb the leading point of MTZ reaches the end of the bed. This is the breakthrough point
Breakthrough concentration is the maximum allowable solute concentration in the effluent
fluid.
Outlet to inlet solute concentration as a function of time, S shaped curve, is called
breakthrough curve
Prior to tb, the outlet solute concentration is less than some maximum permissible value, say
Cout/CF = 0.05
At tb this value is reached and the regeneration is initiated
If the adsorption is continued at t > tb , the outlet solute concentration would be observed to
rise rapidly, eventually approaching the inlet concentration as the outlet end of the bed
became saturated
Time to reach Cout/CF = 0.95 is designated as te
45
 Adsorptive Filtration contd..
Scale up of adsorption bed for constant pattern front
For the ideal case, a solute mass balance for a
cylindrical bed of diameter D gives
𝑐𝐹 𝑄𝐹 𝑡 = 𝑞𝐹 𝜌𝑏 𝜋𝐷2 ∗ (𝐿𝐸𝑆)Τ4
Where t is the time of breakthrough, from which LES
can be determined
Loading = qF*
LUB
Loading = 0
Determination of length of full scale adsorbent bed from break through curves
obtained in lab scale experiments
LES
Total bed length LB = LES + LUB
LES
For an ideal fixed bed adsorber, with MTZ = 0, LUB
LES
is not necessary
B LES
A
But if LB > LES, then LUB is the length of unused
bed
When MTZ is present then LUB is necessary and is
1
referred to as the equivalent length of unused bed
LES
Area A = Area B
0
tb
ts
te
t, time for adsorption step
Determination of bed length from laboratory measurements
46
 Adsorptive Filtration contd..
Scale up of adsorption bed for constant pattern front
LUB = Ideal wave front velocity *(ts – tb) = (Le/ts) *(ts – tb)
Le is the length of the experimental bed
𝑡𝑒
𝑡𝑠 = න
0
1−
𝑐
𝑑𝑡
𝑐𝐹
If the shape of the experimental breakthrough curve below area B is equal to the curve
above area A, then LUB = MTZ/2.
In absence of experimental breakthrough data , a conservative estimate of MTZ is 4 ft.
47
10.6 Membrane Separation
We can use the same filtration principle for the separation of small particles
down to small size of the molecular level by using polymeric membranes.
Depending upon the size range of the particles separated, membrane
separation processes can be classified into three categories: microfiltration,
ultrafiltration, and reverse osmosis, the major differences of which are
summarized in Table 10.1.
Table 10.2: Pressure driven membrane separation processes
Process size
Cutoff
Molecular
Wt. cutoff
Pressure
Drop (psi)
Material
Retained
Microfiltration
0.02 – 10 𝜇𝑚
-
10
Suspended
Material including
microorganisms
Ultrafiltration
10 - 200Å
3 – 300 k
10 – 100
Biologicals, colloids,
macromolecules
Reverse
Osmosis
1 - 10Å
< 300
100 - 800
All suspended and
dissolved material
48
Continued…
Microfiltration refers to the separation of suspended material such as
bacteria by using a membrane with pore sizes of 0.02 to 10 If the pore size
is further decreased so that the separation can be achieved at the molecular
level, it is called ultrafiltration. The typical molecular weight cutoff for
ultrafiltration is in the range of 300 to 300,000, which is in the size range of
about 10 to 200 Å.
Reverse osmosis is a process to separate virtually all suspended and
dissolved material from their solution by applying high pressure (100 to 800
psi) to reverse the osmotic flow of water across a semipermeable membrane.
The molecular weight cutoff is' less than 300 (1 to 10 Å).
The solvent flux across the membrane 𝐽 is proportional to the applied force,
which is equal to the applied pressure reduced by the osmotic pressure ∆Π
as (Belter et al., p. 255, 1988),
𝐽 = 𝐿𝑝 (Δ𝑝 − (𝜎∆Π))
10.37
49
Continued…
where 𝐿𝑝 is the membrane permeability and 𝜎 is the reflection coefficient.
If the solution is dilute the osmotic pressure is,
∆Π = 𝑅′ 𝑇𝐶 ∗
10.38
∗
where 𝐶 is the solute concentration at the membrane surface.
One of the major problems with the membrane separation technique
is concentration polarization, which is the accumulation of solute
molecules or particles on the membrane surface.
At steady state, it will be countered by the molecular diffusion of the solute
away from the membrane surface as
𝐶𝐽 = −𝐷
𝑑𝐶
𝑑𝑧
10.39
which can be solved with the boundary conditions,
𝐶 = 𝐶∗
𝑎𝑡 𝑧 = 0
𝐶 = 𝐶𝑏
𝑎𝑡 𝑧 = 𝛿
10.40
50
Continued…
to yield
𝐷 𝐶∗
𝐽 = 𝑙𝑛
𝛿 𝐶𝑏
10.41
where 𝛿 is the thickness of the laminar sublayer near the membrane surface
The concentration ratio, 𝐶 ∗ / 𝐶𝑏 is known as the polarization modulus.
51
10.7 Electrophoresis
When a mixture of solutes is placed in an electrical field, the positively
charged species are attracted to the anode and the negatively charged ones to
the cathode. The separation of charged species based on their specific
migration rates in an electrical field is termed electrophoresis.
It is one of the most effective methods of protein separation and
characterization. The chief advantages of this method are that it can be
performed under very mild conditions.
For globular proteins, the drag force can be approximated from Stokes law.
Therefore, the balance is
𝑞𝐸𝑓 = 3𝜋𝜇𝑑𝑝 𝑢𝐸
10.42
so that
𝑢𝐸 =
𝑞𝐸𝑓
2𝜋𝜇𝑑𝑝
10.43
52
Continued…
where the steady velocity of particle 𝑢𝐸 is known as electrophoretic
mobility, which is proportional to the electrical field strength and electric
charge but is inversely proportional to the viscosity of the liquid medium
and the particle size.
53