Homework1answers

Homework1answers -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CEE
266:
Environmental
Biotechnology
 Homework
#
1
 
 
 
 
 
 
 1. [10
points]

Which
statement
is
true?


 a. Ribose
is
the
six‐carbon
sugar
found
in
DNA.


 b. Ribose
is
the
six‐carbon
sugar
found
in
RNA.


 c. Ribose
is
the
five‐carbon
sugar
found
in
DNA.


 d. Ribose
is
the
five‐carbon
sugar
found
in
RNA.


 2. [10
points]

Denaturation
destroys
all
but
which
level
of
protein
structure?
 a. Primary
 b. Secondary
 c. Tertiary
 d. Quarternary
 3. [10
points]

Which
of
these
statements
is
true
for
Gram‐positive
cell
wall?


 a. Contains
two
membranes.


 b. Contains
a
thick
layer
of
peptidoglycan.


 c. Periplasmic
space
is
present.


 d. Is
more
complex
than
the
Gram‐negative
wall.


 4. [10
points]

A
nucleoside
is
like
a
nucleotide
except
that
the
nucleoside
does
not
include

 a. one
or
more
phosphates.


 b. the
sugar.


 c. the
nitrogen
base.


 d. the
phosphate
and
sugar.
 5. [10
points]

When
S
is
much
smaller
than
Ks,
the
reaction
is

 a. zeroeth
order
 b. first
order
 c. mixed
order
 d. steady
state
 6. [15
points]

Describe
the
structure
of
the
phospholipid
bilayer.

Include
molecular
 orientation
and
embedded
proteins.

Identify
the
hydrophobic
and
hydrophilic
regions.

 Answer:

Drawing
or
written
description
should
be
similar
to
figure
on
lecture
slide
#
42.

 Fatty
acids
point
inward
to
form
hydrophobic
environment;
hydrophilic
portions
remain
 exposed
to
external
environment
or
the
cytoplasm.

Transport
proteins
are
embedded
 across
the
membrane.


 
 
 
 
 
 
 7. [15
points]

Construct
balanced
redox
reactions
for
the
microbial
oxidation
of
ethanol
to
 acetic
acid
by
sulfate
reducing
bacteria.


 Answer:
 2(CH3CH2OH
+
H2O


 


 CH3COOH
+
4H+
+
4e‐)
 SO42‐
+
10H+
+
8e‐


 


 H2S
+
4H2O
 2CH3CH2OH
+
SO42‐
+
2H+

 

 2CH3COOH
+
H2S
+
2H2O
 8. [20
points]

The
velocities
of
an
enzyme‐catalyzed‐reaction
at
various
concentrations
are
 given
below:

Calculate
Vmax
and
Km.

(Hint:
okay
to
use
Lineweaver‐Burk
plots).


 Answer:

Vmax
=
1/0.2
=
5
mM/min
&
Km
=
0.1
x
5
=
0.5
mM
 
 S
(mM)
 V
(mM/min)
 
 
 0.1
 0.833
 0.3
 1.875
 1
 3.333
 3
 4.286
 10
 4.762
 30
 4.918
 100
 4.975
 
 
 
 ...
View Full Document

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

Ask a homework question - tutors are online