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Unformatted text preview: CEE
266:
Environmental
Biotechnology
 Homework
#
4

 
 1. [5
points]

The
following
are
produced
by
algae
 a. biodiesel
 b. ethanol
 c. hydrogen
 d. all
of
the
above
 

 2. [5
points]

Which
one
is
the
correct
sequence
of
reductive
dechlorination
 A. PCE
–
DCE
–

TCE–

VC
–

ethene
 B. PCE
–

TCE
–

DCE
–
VC
–
CO2
 C. PCE
–

TCE
–

DCE
–
VC
–

ethene
 D. PCE
–

TCE
–

DCE
–

VC
–

ethane
 
 3. [5
points]

When
a
compound
is
concentrated
up
a
food
chain,
it
is
said
to
be

 A. biotransformation
 B. bioaccumulation
 C. bioconcentration
 D. biomagnification
 
 4. [5
points]
Secondary
metabolites
appear
during
which
of
the
following
stages?

 a. at,
near,
or
in
the
lag
phase

 b. at,
near,
or
in
the
stationary
phase

 c. at,
near,
or
in
the
exponential
phase

 d. all
of
the
above

 
 5. [5
points]

Which
is
the
better
engineered
bioremediation
technology
for
petroleum
 hydrocarbons
strictly
in
the
vadose
zone
 a. Bioventing
 b. Biosparging
 c. Bioslurping
 d. Water
circulation
with
glucose
added
as
electron
donor
 

 6. [5
points]

Which
is
the
better
engineered
bioremediation
technology
for
petroleum
 hydrocarbons
with
residual
in
the
saturated
zone
 a. Bioventing
 b. Biosparging
 c. Bioslurping
 d. Water
circulation
with
glucose
added
as
electron
donor
 
 7. [5
points]

Which
is
the
better
engineered
bioremediation
technology
for
TCE
as
DNAPL

 a. Bioventing
 b. Air
sparging
 c. Bioslurping
 d. Water
circulation
with
glucose
added
as
electron
donor
 
 8. [20
points]

Will
the
natural
attenuation
of
trichloroethylene
be
inhibited
by
the
presence
of
 toluene
in
(a)
aerobic
environment,
and
(b)
anaerobic
environment?


 Answer:
 Under
aerobic
conditions,
the
natural
attenuation
of
TCE
will
be
inhibited
in
the
 presence
of
toluene
because
toluene
is
more
easily
biodegraded
than
TCE,
and
will
consume
 oxygen
rapidly.

Many
bacteria
that
oxidize
TCE
also
oxidize
toluene
using
the
same
enzymes,
so
 these
two
compounds
will
experience
competitive
inhibition.


 
 
 Under
anaerobic
conditions,
toluene
can
serve
as
electron
donor
for
reductive
 dehalogenation
of
TCE.

Consequently,
toluene
will
promote
TCE
attenuation.

 

 9. [20
points]

How
much
oxygen
will
be
required
and
how
much
hydrogen
will
be
produced
if
 wastewater
containing
240
mg/L
acetate
is
treated
using
the
following
designs:
(a)
microbial
 fuel
cell
and
(b)
microbial
electrolysis
cell.


 e. Microbial
fuel
cell
 Anode:


 C2H4O2
+
2H2O
 

2CO2
+
8e‐
+
8H+

 Cathode:

 O2
+
4e‐
+
4H+

 
2H2O
 O2
required
=
(2
mol
O2/mol
acetate)
*
32
g/mol
*
(240
mg/L)
/
(60
g/mol)
=
256
mg/L
 Hydrogen
produced
=
0
 
 f. Microbial
electrolysis
cell
 Anode:


 C2H4O2
+
2H2O

 

2CO2
+
8e‐
+
8H+

 Cathode:

 8H+
+
8e‐


 
4H2
 Oxygen
required
=
0
 H2
produced
=
(4
mol
H2/mol
acetate)
*
2
g/mol
*
(240
mg/L)
/
(60
g/mol)
=
32
mg/L
 
 10. 
[25
points]
A
groundwater
plume
contains
18.4
mg/L
toluene.

Calculate
how
much
the
 following
would
be
necessary
for
full
mineralization
of
the
toluene
if
each
reaction
were
the
 only
reaction
responsible
for
accepting
electrons
from
toluene.

You
may
ignore
biomass
 growth.

Express
the
result
as
a
concentration
in
the
groundwater
(mg/L)
 
 
 
 
 
 Electron
donor
reaction:

C7H8
+
14H2O


7CO2
+
36e‐
+
36H+
 a. Aerobic
respiration
(mg
O2/L)
 O2
+
4e‐
+
4H+

2H2O
 9
moles
of
oxygen
needed
per
mole
of
toluene
 Concentration
of
O2
=
9
*
(32
g/mol)
*
(18.4
mg/L)/(92
g/mol)

=

57.6
mg/L
 b. Sulfate
reduction
to
sulfide
(mg
SO42‐/L)
 SO42‐
+
8H+
+
8e‐


S2‐
+
4H2O
 4.5
moles
of
sulfate
needed
per
mole
of
toluene
 Concentration
of
SO42‐
=
4.5
*
(96
g/mol)
*
(18.4
mg/L)/(92
g/mol)

=

86.4
mg/L
 c. Complete
nitrification+denitrification
(mg
NO3‐/L)
 NO3‐
+
5e‐
+
6H+


½N2
+
3H2O
 7.2
moles
of
nitrate
needed
per
mole
of
toluene
 Concentration
of
NO3‐
=
7.2
*
(62
g/mol)
*
(18.4
mg/L)/(92
g/mol)

=

89.3
mg/L
 d. Ferric
hydroxide
reduction
to
ferrous
(mg
Fe/L)
 Fe3+
+
e‐

Fe2+

 36
moles
of
ferric
ions
needed
per
mole
of
toluene
 Concentration
of
Fe
=
36
*
(56
g/mol)
*
(18.4
mg/L)/(92
g/mol)

=

403.2
mg/L
 
 

 
 e. Methanogenesis
(mg
CH4/L)
 CO2
+
8e‐
+
8H+

CH4
+
2H2O
 4.5
moles
of
carbon
dioxide
needed
per
mole
of
toluene
 Concentration
of
CH4
=
4.5*
(16
g/mol)
*
(18.4
mg/L)/(92
g/mol)

=

14.4
mg/L
 ...
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This note was uploaded on 02/02/2012 for the course CEE 266 taught by Professor Shailymahendra during the Fall '11 term at UCLA.

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