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CPE240 > Living systems > Flashcards

Living systems Flashcards

1
Q

what is biotechnology

A

the use of biological processes to create useful products

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2
Q

What is the process of creating a biotech product from raw materials?

A
  • Raw material  upstream processing  fermentation  downstream processing
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3
Q

examples of biotech products

A

wine, cheese, vinegar, insulin and citric acid

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4
Q

red biotech?

A

the pharmaceutical branch which uses bacteria to produce drugs e.g. disease preventation

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5
Q

White biotech?

A
  • White biotech is the aid of living organisms and enzymes e.g. biofuels to biodiseal
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6
Q

Green biotech?

A
  • Green biotech is the use of agriculture such as plants less suspectible to pests e.g. water purification
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7
Q

What is cell growth

A

essential element of fermentation for the growth of a organism

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8
Q

Growth is an orderly increase of…

A

cellular constituents

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9
Q

what does optimising the environment for growth require?

A

balancing the economic feasibility of maintaining a suitable chem environment

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10
Q

What does growth depend on?

A

the ability to form new protoplasm from the nutrients available

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11
Q

What does cell growth involve

A

An increase in the cell mass and number of ribosomes and cell division

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12
Q

Methods of measurement of cell mass

A
  • Direct physical measurement of weight after centrifugation
  • Direct chemical measurement of chemical component
  • Indirect measurement of chemical activity
  • Turbidity measurements to determine amount of light scattered by suspension of cells
  • Microscopic count
  • Viable cell count
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13
Q

Lag phase

A

after inoculation of cells the population remains unchanged so no cell division occurs but the metabolic activity may increase

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14
Q

expo phase

A

cells divide by binary fission at constant rate

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15
Q

stationary phase

A

exhaustion of available nutrients and space growth stops

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16
Q

death phase

A

population declines

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17
Q

What is a closed system

A

It is a batch culture where there is a lack of input and output of materials once batch medium has been inoculated, a discontinous process which environment and organisms are changing continually

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18
Q

What is an open system

A

It is a continous culture which is capable of obtaining steady state where there is balanced input growth substrate and ouput of organisms

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19
Q

equation for increase in cell mass

A

ln x = ln x0 + n ln 2

*n= t/td, ie no of generations = total time/doubling time

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20
Q

equation for increase in cell mass in terms of specific growth rate

A

ln x = ln x0 + µt
or 2=exp(ut) doubling
t=generation time

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21
Q

What is a growth limiting substrate

A

nutrient medium containing one limiting substrate at low concentrations

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22
Q

what limits cell synthesis

A

nutrient concentration in culture vesssel

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23
Q

Monod kinetics eq for specific growth rate

A

µ = µmax . s/(Ks + s)

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24
Q

Lineweaver burke plot
slope?
x intercept?
y intercept?

A

ks/umax
-1/ks
1/umax
plot is /u v 1/s

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25
Q

what is cell yield coefficent

A

amount of cell mass produced per unit amount of substrate consumed

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26
Q

Yield factor eq

A

Yfg= - deltaF/deltaG

f= moles produced
G= moles consumed
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27
Q

observed biomass yield

A

Y’xs=-DeltaX/deltaSG
X=biomass produced
SG= mass of substrate

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28
Q

what is a continuous culture

A

a continuous feed of influent solution containing substrate and nutrients and a drain of effluent solution containing cells and metabolites

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29
Q

what is a chemostat

A

allows control of rate of growth which can be used to optimise the production of products

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30
Q

What is a primary metabolite

A

produced at high flow/dilutuion rate stimulating cell growth

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31
Q

What is a secondary metabolite

A

produced at low flow/dilution maintaining cell numbers

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32
Q

What two parameters are controlled in a chemostat and why?

A

Dilution rate and influent substrate conc. By controlling the dilution rate we can control the specific growth rate. By controlling the substrate conc we can control number of cells produced or cell yield in chemostat

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33
Q

dx/dt=?

A

ux-Dx

u=d is steady state
u>D utilization of substrate exceeds supply so slow growth
U

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34
Q

critical dilution eq

A

Dc=Umax(s/(s+Ks))

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35
Q

Why do we have critical dilution

A

when a chemostat is operating at Dc , if the dilution rate is increased further, the growth rate will not be able to increase (since it is already at µmax ) to offset the increase in dilution rate.

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36
Q

what is maintence energy

A

the level of energy required to maintain a cell

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37
Q

purpose of a fermentor

A

control temp and ph
reduce evapouration
minimise the use of labour
maximise computer control

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38
Q

scales of operation for a fermentor/bioreactor

A

laboratory 20L
pilot plant 5000L
production 20,000L

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39
Q

Most important factor in fermentor design

A

height to diameter ratio

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40
Q

why do we implement impellers in bioreactors

A

to reduce bubble size hence increase surface area for oxygen transfer

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41
Q

What is genetic engineering

A

the direct manipulation of an organisms genes using biotechnology to change the genetic make up if cells

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42
Q

what is metabolic engineering

A

the direct improvement of cellular properites through direct modifications of biochem reactions

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43
Q

what are plasmids

A

self replicating pieces of DNA

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44
Q

what is gene cloning

A

genes are inserted into a plasmid then added to bacteria for replication

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45
Q

Steps of gene cloning

A
  1. Isolate the plasmid
  2. Isolate gene of interest
  3. Insert the gene into the plasmid
  4. Bacteria will take it through transformation
  5. Cells divide along with the plasmid and form a clone of cells
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46
Q

Why is bacteria grown in a culture

A

to produce copies of the isolated gene of interest

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47
Q

what is Metabolism

A

the summation of chemical reactions in a organism

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48
Q

what are autotrophs

A

organisms that use co2 as their sole carbon

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49
Q

what are chemotrophs

A

life forms that obtain their energy by injecting carbon

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50
Q

Phototrophs

A

i.e. plants via sun

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51
Q

Chemotrophs

A

oxidation of co2 and lipids

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52
Q

anabolism

A

construction path

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53
Q

Catabolism and how it works

A

degradation path, it is the breakdown of macromolecules into building blocks then the oxidation of building blocks to produce acetyl coA which is further oxidised to co2 and water

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54
Q

gycolysis under aerobic conditions

A

pyruvate further oxidises and enters the krebs cycle

55
Q

gycolysis under anaerobic conditions

A

pyruvate is converted into a reduced end product

lactate in muscle) = homolactic fermentation. In yeast = alcoholic fermentation yield ethanol and CO2

56
Q

As cells have limited quantities of NAD+ what should happen?

A

it must be recycled after reduction to NADH

57
Q

what happens to NAD+ under aerobic conditions

A

oxidised in mitochondria

58
Q

what happens to NAD+ under anaerobic conditions

A

NADH is replenished by reduction of pyruvate

59
Q

what is pathway flux

A

Is the rate at which input metabolites are processed to form output metabolites

60
Q

what are enzymes?

A

catalysts that preform a wide variety of biochemical reactions in the cell

61
Q

How do enzymes differ from chemical catalysts

A

they catalyse under mild conditions
have a high degree of specificity
enzyme activity is regulated in the cell

62
Q

oxidases

A

oxidation reduction

63
Q

what enzyme involves the transfer of function groups

A

Transferases

64
Q

what enzyme involves the hydrolysis of reactions

A

hydrolysis

65
Q

lyases

A

group elimination of double bonds

66
Q

isomerases

A

isomerisation

67
Q

what enzyme involves bond formation

A

ligases

68
Q

what is catalytic potential equal to

A

enzyme activity

69
Q

how do enzymes bind to substrates

A

through a series of non covalent bonds

70
Q

how is enzyme activity measured

A

substrate depletion, product production and can also be measured by coupling with an enzymatic reaction to others which transforms the product into a more detectable analyte

71
Q

enzyme catalysed reaction

A

E+s–> ES–> E+P

where [s] is high enough to convert E to ES so the second step is the rate limiting step

72
Q

reaction velocity eq

A 
vo=vmax[S]/Km+S
km=[S] when v0=vmax/2
enzymes with low km achieve max catalytic efficency 
Km varies with temp, ph and enzyme 
at high [S] vo approaches vmax
73
Q

eadie hofstee plot parameters

A

V vs V/[S]
slope= -km
y-vmax
x=vmax/km

74
Q

Detergent enzymes and their conditions

A

proteases, lipases and amylases are added to detergents where they catalyse the breakdown of chemical bonds on the addtion of water
Condtions –> ph 9-11, 60 degrees

75
Q

proteases

A

remove protein stains such as blood, grass and egg. Hydrolyse the protein and break them down into more soluble polypeptides

76
Q

Amylases

A

remove residue of starch foods such as mash potato

77
Q

cellulose

A

modify the structure of cellulose fibrils such as those found on cotton

78
Q

Lipases

A

remove fatty stains such as lipstick

79
Q

detergents should be produced using?

A

Bacillus

80
Q

what was the first cost effective detergent and what was it based on?

A

lipases based on the use of rDNA technology. it has made is possible to alter the dna sequence randomly or at specific sites of lipase enzyme in order to change the amino acid sequence

81
Q

what properties of lipase has protein engineering improved?

A

substrate specificity
thermostability
proteases stability

82
Q

enzyme production process

A

gene+plasmid–>recombinant plasmid –> bacterial transformation –> protein

83
Q

steps of enzyme production

A

screening - choosing appropriate micro organism
modification - improving microbial stain
lab scale - determine optimum conditions
pilot plant - large scale

84
Q

upstream processing

A

cell factory(bioreactor) sepration of cells (centurifgation) enzymes secreted from cells downstream processing

enzymes secreted in cells –> enzymes inside cells –> cell lysis –> downstream

85
Q

downstream processing

A

LL extraction chromatography purified enzyme  three by producs(solid, liquid formulation and the reaming enzyme is imbolized)

86
Q

biomass -> …..->biofeul

A

sugars

87
Q

examples of biomass

A

cereal crops and lignocellulose

proteins and carbs

88
Q

why is biomass not compatible with existing engines

A

it is a solid whereas conventional feuls are gases and has a low energy density and high moisture and oxygen content

89
Q

examples of biofeuls

A

gasoline, hydrogen gas and propane

90
Q

production of bioethanol

A

extraction of sugar cane –> saccharification of corn –> pre treatment

(grains)milling 
(water and enzymes)saccharification 
(yeast and sugar)fermentation 
(mash from fermentation)distillation 
can be dehydrated to bioethanol
91
Q

what is milling

disadvantages

A

mechanical crushing of cereal grains to release starch

time consuming, expensive and slow

92
Q

saccharification

A

heating and addition of water and enzymes for conversion into fermentable sugar

93
Q

fermentation

A

of the mash using teast converting bioethanol and c02

94
Q

saccharification of starch

A

starch is mixed with water at 60 degrees for 5-10mins
starch dissolves in water to form mash
liquefication of mash by adding amylase at 80 degrees for 2 hours to degrade starch
saccharification of glucoamylase added at 65 degrees for 30 mins to produce glucose

95
Q

common microbes in glycolysis fermentation

A

saccharmyces cerevisae and zymommonis mobillis

96
Q

fermentation process

A

1) in yeast fermentation glucose solution is obtained from starch saccharification and is cooled to 32 degrees
2) yeast culture is added and glucose is metabolised to ethanol and co2
3) some of the released energy and glucose are utilised by the yeast cells to support growth the rest of the energy becomes heat to fermentation

97
Q

normal inoculation ratio for fermentation

A

5-10% 6-8hrs

98
Q

how long does active fermentation last

A

12 hours then fermentation activity slows down due to less glucose being available

99
Q

what happens during slow fermentation

A

cells do not grow anymore biochem reactions are limited by substrate conc

100
Q

sources for bioethanol production

A

corn farm in us and brazil

60% is based on sugar cane conversion

101
Q

three constituents for lignocellulose

A

celluose, hemi-cellulose and lignin which enhances strength and compactness

102
Q

what is the structure in lignocellulose like

A

complex structure due to strong linkages between molecules

103
Q

advantages of using lignocellulose

A

increases available surface area
reduces particle size
pre treatment leads to solubising hemi-cellulose and increasing enzymatic hydrosybility of cellulose

104
Q

biological pre treatment

A 
fungi and bacteria 
slow 
no chemicals 
solubises micro-organisms 
brown white and soft rot fungi 
white and soft rot both attack cellulose and lignin but white is more effective for pre treatment
105
Q

Chemical pre treatment

A

dilute sulfric acid can acheive high reaction rates

but there is high costs and neutralisation is required

106
Q

Alkaline pre treatment

A

can disrupt lignin structure and decrease crystalinity

but there is less sugar degradation and naoh can be recovered

107
Q

why are advanced biofeuls good

A

can drop into existing infrastructure because they have similar properties to petroleum but requires large biorefineries to produce high yields

108
Q

what leads to corrosion

A

ethanol

109
Q

what is the main challenge in isobutanol production

A

is using native hosts to convert feedstocks into advanced bio fuels to overcome regulation of biofuel pathways to improve yields.

110
Q

what is isobutanol

A

a next generation product which builds on the foundation and provides additional solutions to various challenges not met by first generation products

111
Q

advantages of isobutanol

A
  • Isobutanol has more hydrocarbons then ethanol therefore has a better energy content and is less polar
  • Absorbs less water
  • Volatility
  • good Phase separation
  • Good platform molecule
  • Good blend properties
112
Q

what 3 pathways does the isobutanol pathway consist of

A
  • Glycolysis to provide pyruvate
  • Valine biosynthesis to metabolise pryruvate to KIV
  • Ehrilich pathway which is required for degradation of KIV to isobutanol
113
Q

why is a high activity of KDC essential is isobutanol production

A

it links valine metabolism and ehrilich pathway so is essential for high levels of isobutanol

114
Q

How can we increase isobutanol production

A

by over expression of genetic engineering by increasing KDC and ADH.

115
Q

Salmonella

A

distributed in domestic wild animals such as pigs and cattle
in humans it is generally contacted through consumption of contaminated food of animal orgin
symptoms - diarrhoea, fever and abdominal cramps

116
Q

Listeria Monocybogenes

A

found in soil and water and animal digestive tracts
food with long shelf life under refridgeration
symptoms- gastroentritis, meningitis
onset 9-48hrs
duration
2-10 days

117
Q

Campylobaceri-jeuni

A

distributed in warm blooded animals
symptoms - abdominal pain, diarrhoea and fever
onset 2-5 days
duration 2-10 days

118
Q

Esherichia coli

A 
found in cattle
food implicated with this are uncooked hamburger, dry cure salami, yogurt, cheese made from raw milk 
symptoms - bloody diarhea 
onset - 1-8 days 
duration 5-10days
119
Q

Staphyloccus aureus

A 
found on skin and nose 
foods at high risk of transmitting toxins are those that do not cook but handle such as sliced meat and sanwiches 
symptoms - vomitting and diarrhoea 
onset 6hrs
duration 24hrs
120
Q

Bacillus cereus

A

found in rice
causes vomitting and diarrhoea
onset 8-16hrs
duration is 24hrs

121
Q

Clostridium prefringens

A

commonly found in raw meat such as beef, poultry, gravy
symptoms - abdominal pain, stomach cramps followed by diarrhoea
onset 8-16hrs
duration 24hrs

122
Q

Clastidium

A

anerobic bacteria so can only grow in the absence of oxygen

onset 12-72 hours

123
Q

what should be considered for foods that carry food borne pathogens

A

paesturization such as milk and fruit juices

124
Q

standard sterilisation

what eq can describe it

A

steam sterilisation for reactors 121 degrees for 15mins at 103kpa
arrhenius eq

125
Q

Bacillus stearothermophilus

A

spores are commercially available for validation for steam sterilisation unit operations

126
Q

number of viable organisms eq

A

Nt/No=exp(-kt)

127
Q

sanatation

control variables

A

reduction of the oppourtunity for bacterial contamination by removing it from process equipment

Control variables are cleaning agent conc, cleaning solution temp, time and velocity
Need uniform flow over tank surface which is function of application rate, surface and viscocity.

128
Q

what does cooking do

A

Cooking kills viruses, bacteria, parasites and destroys toxins. Uncooked food contains viruses, live bacteria, parasites and toxins

129
Q

what happens when re-heating food

A

if they do not reach 70 degrees bacteria will survive and toxins remains.

130
Q

corrective action

A

reduce impact of an outbreak

131
Q

preventive action

A

stop it from happening again by finding where it went wrong

132
Q

method for cultivation

A
  1. Rinse the seed(useful source for contamination) and remove the damaged seed
  2. Soak for 8hrs
  3. Rinse
  4. Grow for 3 days at 15 degrees (ideal for salmonella growth)
  5. Package
  6. Distributed
133
Q

are metabolic pathways reversible or irreversible?

A

irreversible – highly exergonic (ΔG = -ve) so reactions go to completion. This provides the pathway with direction.

134
Q

Metabolite concentrations are a function of

A

Thermodynamics

Reaction kinetics

CPE240 (1 decks)

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