PRODUCTION OF BUTANOL FROM LOWER TERMITE PROTISTS AND

E. COLI FERMENTATION

Team Gold All On My Polypeptide Chain:

Derek Weidlein

Ricky Rodriguez

Michelle Empleo

Michael McClurg

Jerrell Ross

PURPOSE

There is an increase demand for alternative

biofuel.

Utilize otherwise useless plant material for energy

Produce a higher efficiency product

“In 2011, the United

States consumed about

134 billion gallons of

gasoline.”

ETHANOL AS A FUEL SOURCE

One of the cleanest

sources of energy

Established method of

production

Takes away from human

food sources

Low energy density

Absorbs moisture in the air

OUR IDEA: BUTANOL

Clean source of energy

Would mix well with gasoline

Not hygroscopic

Higher energy density than ethanol

High cellulose

feedstock

Cellulose breakdown

with termite hindgut

protists and bacteria

Extract simple sugars

PLAN OVERVIEW

Use genetically altered

E. Coli to ferment sugars

into alcohol product

Distill product to

isolate butanol for use as

biofuel

OUR FEEDSTOCK?

% Cellulose

50

95

63

70

78

71

76

73

Hardwood

Cotton

Flax

Hemp

Henequen

Jute

Ramie

Sisal

Useful or Not?

Useful

Not Useful

Potentially

Potentially

Not Useful In U.S.

Useful

Not Useful

Not Useful

BREAKING DOWN CELLULOSE

Endocellulase-breaks

crystal structure of complex

cellulose (breaks H-bonds)

Exocellulase-cleaves 1

glucose off of 3-ringed

cellulose

Cellobiase-splits 2-ringed

cellobiose into 2 glucose

molecules

Invasive species found in Southeast

Asia (China, Japan, and Taiwan), but

can also be found in the continental

United States, South Africa, and

Hawaii

Grouped in the Lower Termite class

which means it relies on a symbiotic

relationship with protists and

bacteria to complete its metabolic

functions (cellulose breakdown and

glucose fermentation)

TERMITES (COPTOTERMES FORMOSANUS)

DIGESTION OVERVIEW

Some cellulose activity in the salivary glands and midgut of termite, but most

activity is done in the hindgut by the protozoan fauna (full decomposition of

cellulose to glucose), which feed by phagocytosis.

Pseudotrichonympha grassi- largest protozoan that decomposes highly

polymerized cellulose

Holomastigotes hartmannii- breaks down low molecular weight

cellulose

Spirotrichonympha leidyi- smallest protozoan breaks down low

molecular weight cellulose

Trichomitopsis termopsidis- ferments the glucose into acetate, carbon

dioxide, and hydrogen

PROTISTS IN HINDGUT

IN MORE DETAIL

Cellulose-containing product would first be chipped and then

ground as small as possible with machinery

This sawdust would be mixed with water into a cellulose soup

This soup would sit in a large pool where it will be aerated to

maintain an oxygen-rich environment, and inoculated in many

separate locations with about one total gram of p. grassii, h.

hartmannii, and s. leidyi

This mixture would be maintained at a pH of about 6.5, and kept

slightly above room temperature

PROTIST ASSUMPTIONS

Characteristic Assumption

Unit mass 1 protist≈1 nanogram

Doubling time 2 hours

Daily consumption 3.685E-13kg cellulose per protist per day

All assumptions based on typical protists, since we could not find specific information on our protists

1 colony consumes .3685kg wood per day, there are 1,000,000 termites per colony, each termite weighs about 2mg, and “termites can be up to 50% digestive protists and bacteria”

CONSUMPTION RATE VS T IME (UNDER IDEAL CONDITIONS)

0 10 20 30 40 50 600

50

100

150

200

250

300

Rate of Consumption vs. Time

Time (hours)

Rate

of

Consu

mpti

on

(kg c

ell

ulo

se p

er

hour)

EXPECTED YIELD

In order to make about 10,000kg of glucose , we would need about

18.7 metric tons of hardwood feedstock. This translates to about 7

large trees, about 2/3 the capacity of one semi-truck

With a 1g inoculum and an assumed doubling time of about 2 hours,

the protists would be capable of consuming 6200kg of cellulose per day

after the first two days.

Total estimated time to consume cellulose without lag phase or death

phase is about 47.2 hours

1. Rotating drum filtration to

remove the solid non-

cellulose wood components

2. Precipitate acetate via a

sodium salt

3. Filter out the salt product

REMOVAL OF GLUCOSE FROM THE BATCH

Naturally converts glucose

to valine via a pyruvate and

2-Keto-isovalerate

intermediate

We know a lot about E.

Coli’s DNA, so we can add in

new genetic material to

utilize this 2-Keto-isovalerate

intermediate

E. COLI

PATHWAY

Glucose Pyruvate2-Keto-

isovalerate

Valine

Isobutyraldehyde

Isobutanol

2-keto-acid decarboxylase (KDC)

alcohol dehydrogenase (ADH)

ADDING NEW DNA

KDC gene comes from Lactococcus lactis

bacterium

ADH gene comes from Saccharomyces cerevisiae

bacterium

Cut new DNA and E. Coli plasmid with restriction

enzyme with selectible marker, creating new

recombinant DNA in a single bacterium when

treated with DNA ligase

BIOREACTOR CONDITIONS

A glucose-water solution would be continuously fed

into bioreactor inoculated with 1g of genetically

altered E. Coli

The bioreactor would be heated to maintain a

constant 37 degrees Celsius

The bioreactor will be aerated for maximum E. Coli

function

Valine-melts at 295°C

Isobutanol-107.89°C

Isobutyraldehyde-63°C

Ketoisovalerate-170.5°C

Pyruvate-165°C

Drum centrifugation will

remove any solids (“excess

glucose, cell metabolites,

biomass”)

Any valine will be removed

prior to distillation (solid)

The isobutanol can be

collected in the second fraction.

It has a significant boiling point

difference from other main

components

CENTRIFUGATION AND DISTILLATION

Boiling points of different

components:Procedure:

E. COLI ASSUMPTIONS

Characteristic Assumption

Unit Mass .3 picograms dry weight

Doubling Time 20 minutes

Consumption 20mmol glucose per g dry weight per hour

EXPECTED YIELD

Taking the 10,000kg of glucose as a basis for this

process:

Ideally, .41g of isobutanol would be formed per

gram of glucose

Assuming about 75% efficiency due to byproducts

and lost product during seperation, the 10,000kg of

glucose would yield about 1,015 gallons of

isobutanol. Mixed at 15% per gallon of gasoline, this

is 6,762 gallons of mixed fuel gasoline.

SCHEMATIC PART 1

Covered Cellulose

Fermenter

Covered Cellulose

Fermenter

Covered Cellulose

Fermenter

Wood Processin

g

Rotating

Drum

Boiler

Schematic Part 2

SCHEMATIC PART 2

Incoming Glucose Solution

Bioreactor

CentrifugeWast

e

Distillation

Isobutanol

Boom.

Water and

waste liquid

s

FEASIBILITY

Using only one cellulose fermenter would not make

a big difference

With multiple cellulose fermenters, the plan

becomes more and more reasonable. The E. Coli

functions faster than the protists, and could easily

handle an increase in feed glucose. With a significant

plant size, a plant could produce tens of thousands of

gallons of mixed biofuel per day.

THANKS FOR LISTENING

QUESTIONS?