1. The name for an agent that causes
disease is a(n)
Ba
og
en
25%
Pa
th
25%
ct
er
ia
25%
An
t ig
en
y
25%
An
t ib
od
A. Antibody
B. Antigen
C. Bacteria
D. Pathogen
An antibody is a protein that is
produced to target a specific
foreign particle (cell surface,
virus etc.).
An antigen is a foreign particle
that the body produces
antibodies against. Not all
antigen are in fact diseasecausing.
Not all bacteria cause disease.
2. An example of a water-borne
pathogen is
25%
Tu
be
rc
ul
o
tit
i
sis
sC
25%
He
pa
le
ra
25%
Ch
o
al
ar
ia
Malaria
Cholera
Hepatitis C
Tuberculosis
M
A.
B.
C.
D.
25%
3. The largest water pollution source
by volume is
25% 25% 25% 25%
lan
St
ts
ee
lm
an
uf
ac
tu
rin
g
Ve
hi
c le
ex
ha
us
t
Co
a
lp
ow
er
p
ul
tu
re
Agriculture
Coal power plants
Steel manufacturing
Vehicle exhaust
Ag
r ic
A.
B.
C.
D.
Sediment from agriculture is the biggest pollutant by volume.
Clouds water (more turbidity) and reduces rate of
photosynthesis.
Conventional tillage (turning of soil) increases the erosion of
soil and sediment pollution.
Conservation tillage reduces the turning of soil and leaves
plant matter on top, decreasing the rate of erosion.
Other pollution from
agriculture includes:
• Pesticides
• Excess fertilizers
4. An example of an oxygen demanding waste is
25%
25%
s
et
al
ym
He
av
M
an
ur
e
25%
tic
en
t
25%
Pl
as
Pesticides
Plastic
Manure
Heavy metals
Se
di
m
A.
B.
C.
D.
Organic matter such as manure is decomposed
by aerobic organisms that use up oxygen in the
water.
A
5. For the above graph of DO (dissolved O2) in a stream,
which of the following is correct regarding point A?
C
an
d
C
B
an
d
Bo
th
A
th
Bo
lo
gi
ca
l
ni
c
(B
io
D
BO
h
Hi
g
ox
yg
..
er
m
at
t
te
r
at
m
or
ga
an
ic
Lo
w
or
g
h
Hi
g
A. High organic matter
B. Low organic matter
C. High BOD (Biological
oxygen demand)
D. Both A and C
E. Both B and C
20% 20% 20% 20% 20%
Sources of organic waste:
Combined sewage overflow (during high
amounts of rain)
Livestock
Untreated waste
BOD (Biological Oxygen Demand) is a specific measure of how much
oxygen is consumed due to the amount of organic matter. (Incubated
for 5 days at 22 degrees C to see change in DO)
A
B
6. Compared to the stream in graph A, the stream in graph B is
(assume the same scale on graphs)
A
w
C
B
an
d
C
d
an
A
tre
am
ar
m
er
s
tre
am
ng
s
ov
i
te
rm
ng
s
A
fa
s
er
m
ov
i
s lo
w
A slower moving stream
A faster moving stream
A warmer stream
A and C
B and C
A
A.
B.
C.
D.
E.
tre
am
20% 20% 20% 20% 20%
Turbulence in a faster moving stream can help
restore oxygen through mixing the water more
with surface air.
Slower streams also may be warmer, which
decreases the dissolved oxygen.
7. Which of the following can increase
the amount of oxygen-demanding
waste in water?
25%
25%
25%
rB
B
Ne
ith
er
A
no
an
d
A
th
Bo
De
te
rg
en
ts
iliz
er
s
Fertilizers
Detergents
Both A and B
Neither A nor B
Fe
rt
A.
B.
C.
D.
25%
Fertilizer contains contain nitrogen and phosphates.
Detergents may contain phosphates.
These elements are both often limiting factors in
aquatic ecosystems and can cause an overgrowth of
algae which creates an oxygen-demand when it
decomposes.
Key to low oxygen
8. A lake with a low amount of
nutrients is called _______ and these
lakes generally have ______ amounts
of oxygen.
25% 25% 25% 25%
igh
c;
h
c;
l
ph
i
ph
i
tro
igo
ol
ol
igo
tro
ph
i
eu
tro
er
er
igh
c;
h
lo
op
hi
c;
ow
er
we
r
eutrophic; lower
eutrophic; higher
oligotrophic; lower
oligotrophic; higher
eu
tr
A.
B.
C.
D.
9. The addition of an excess of
nutrients into a lake from a human
source is called
33% 33%
33%
rB
Ne
ith
er
A
no
tio
n
le
utr
op
hic
a
le
utr
op
h ic
at i
on
A. Cultural eutrophication
B. Natural eutrophication
C. Neither A nor B
Eutrophication= too much of a good thing (nutrients)
Cultural eutrophication – Anthropogenic (human-caused) increase in
nutrients from runoff of fertilizer, livestock waste, sewage(untreated
runoff, leaking septic systems etc.), detergents.
10. An area that is a dead zone due to
low oxygen levels is called
c
on
i
Hy
po
to
n
25%
Hy
pe
rt
25%
ic
25%
xic
25%
Hy
pe
ro
Hypoxic
Hyperoxic
Hypotonic
Hypertonic
Hy
po
xic
A.
B.
C.
D.
Hypo= Low
Hypoxic = Low oxygen levels
11. A toxic substance that can be
naturally occurring in groundwater
25%
25%
25%
CF
C’
s
nz
en
e
Be
Ar
se
ni
c
ni
d
e
Cyanide
Arsenic
Benzene
CFC’s
Cy
a
A.
B.
C.
D.
25%
12. Parking lots can be a major
contributor to water pollution because
they are
33%
33%
33%
rB
no
er
A
Ne
ith
pe
rm
Se
le
ct
ive
ly
Im
pe
rv
io
us
t
ea
bl
e
o
w
to
.. .
at
er
A. Impervious to water
B. Selectively
permeable to water
C. Neither A nor B
Impervious= not allowing something to pass through
13. In primary sewage treatment
...
in
se
ttl
sp
ec
ia
liz
ed
a
nd
ee
ns
a
sc
r
lp
gt
an
it h
w
fe
ct
ed
sin
di
is
as
te
w
ro
c..
.
33% 33%
.. .
33%
ch
em
ic a
A. waste is disinfected with
chlorine.
B. screens and a settling tank
are used to remove solids
C. specialized chemical
process are performed to
remove pollutants
14. The first step in the secondary
sewage treatment is to
di
o
Us
e
ge
. ..
la
ica
sp
ec
ifi
cc
et
or
in
ch
l
Us
e
33%
he
m
sin
fe
ct
.
to
te
r ia
ba
c
bi
c
ae
ro
33%
..
. ..
33%
Us
e
A. Use aerobic bacteria to
break down the organic
material
B. Use chlorine to disinfect
the potential pathogens
C. Use specific chemical
agents to remove harmful
water pollutants.
15. Harmful chemicals in the waste
water system are
A. always treated.
B. only rarely treated –
requires advanced
treatment for specific
pollutants.
50%
on
ly
ra
re
ly
al
tre
at
w
ed
ay
st
–r
eq
u
...
re
at
ed
.
50%
Study the diagram below to answer
the following questions
Note that the waste in the
digester could be used to
generate heating or electricity
16. Explain why the digester portion
of the sewage treatment can be used
for heating or electricity
50%
50%
es
te
rt
di
g
th
e
th
e
di
g
es
te
rt
an
k
an
k
is
is
tre
...
tre
at
..
A. the digester tank is
treated with aerobic
bacteria
B. the digester tank is
treated with anaerobic
bacteria
The liquid portion of the sewage can be aerated
and treated with aerobic bacteria.
The solid portion of the sewage must be treated
with anaerobic bacteria which generates
methane. Some wastewater treatment plants
harvest this methane for power generation.
17. What are different methods used
to deal with the solid waste
25%
25%
ov
e
n
Al
lo
ft
he
ab
tio
in
er
a
nd
fil
l
s
25%
La
ul
tu
re
25%
In
c
Agriculture
Landfills
Incineration
All of the above
Ag
r ic
A.
B.
C.
D.
g/
L
m
4
g/
L
m
3
g/
L
m
2
g/
L
m
1
m
8 mg/L
1 mg/L
2 mg/L
3 mg/L
4 mg/L
8
A.
B.
C.
D.
E.
g/
L
18. The amount of dissolved oxygen in a pond is
measured at 7 mg/L. Two BOD water sample bottles are
filled up with pond water. One is left in the light while
the other is covered. After 5 days at 20 °C, the covered
bottle has a DO concentration of 4 mg/L and the
uncovered bottle has a DO concentration of 8 mg/L. The
20% 20% 20% 20% 20%
BOD for this pond was
Biological Oxygen Demand (BOD) is a way of
comparing the amount of organic matter in an
aquatic ecosystem by measuring the
decrease in oxygen in a BOD sample bottle
after 5 days at 20 C in a sample that is not
exposed to light.
The dissolved oxygen level will drop if organic
matter is present for decomposers to break
down (and any living plants will rely on
respiration of their stored sugars).
The bottles starting DO (dissolved oxygen)
level was 7 mg/L and the ending DO level was
4 mg/L so the BOD was 3 mg/L.
19. What was the NPP for the pond being
tested? Remember that the initial DO was 7
mg/L and after 5 days the covered bottle’s DO
was 4 mg/L and the uncovered bottles DO was
8mg/L.
m
g/
L
20%
8
m
g/
L
20%
4
m
g/
L
20%
3
m
2
g/
L
m
20%
g/
L
20%
1 mg/L
2 mg/L
3 mg/L
4 mg/L
8 mg/L
1
A.
B.
C.
D.
E.
Net Primary Productivity is the
increase in the products of
photosynthesis (glucose or
oxygen).
While some of the glucose and
oxygen is used by immediately by
aerobic organisms and the plants
themselves, the increase in
glucose or oxygen is the NPP.
Since the initial DO was 7 mg/L
and the final DO in the
uncovered bottle was 8 mg/L the
NPP was 1 mg/L.
20. What was the GPP for the pond being
tested? Remember that the initial DO was 7
mg/L and after 5 days the covered bottle’s DO
was 4 mg/L and the uncovered bottles DO was
8mg/L
m
g/
L
20%
4
m
g/
L
20%
3
m
g/
L
20%
2
1
m
g/
L
20%
m
g/
20%
8 mg/L
1 mg/
2 mg/L
3 mg/L
4 mg/L
8
A.
B.
C.
D.
E.
The Gross Primary Productivity (GPP) is the overall
rate of photosynthesis. The GPP cannot be measured
directly since some of the products of photosynthesis
(oxygen or glucose) is used up immediately by other
organisms or the plants themselves.
NPP = GPP – R (R= rate of respiration which is seen in
BOD measurement)
GPP = NPP + R
Since the NPP (shown in uncovered bottle) is 1 mg/L
and the R (shown in the covered bottle is 3 mg/L then
the GPP = 4 mg/L)
1.D
2.B
3.A
4.C
5.D
6.B
7.C
8.D
9.A
10.A
11.B
12.A
13.B
14.A
15.B
16.B
17.D
18.B
19.A
20.E