1. The name for an agent that causes disease is a(n)

A. AntibodyB. AntigenC. BacteriaD. Pathogen

Antibody

Antigen

Bacteria

Pathogen

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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 disease-causing.

Not all bacteria cause disease.

2. An example of a water-borne pathogen is

A. MalariaB. CholeraC. Hepatitis CD. Tuberculosis

Malaria

Cholera

Hepatitis C

Tuberculosis

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3. The largest water pollution source by volume is

A. AgricultureB. Coal power plantsC. Steel manufacturingD. Vehicle exhaust

Agricu

lture

Coal power p

lants

Steel manufactu

ring

Vehicle exhaust

25% 25%25%25%

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

A. PesticidesB. PlasticC. ManureD. Heavy metals

Sediment

Plastic

Manure

Heavy metals

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Organic matter such as manure is decomposed by aerobic organisms that use up oxygen in the water.

5. For the above graph of DO (dissolved O2) in a stream, which of the following is correct regarding point A?

A. High organic matterB. Low organic matterC. High BOD (Biological

oxygen demand)D. Both A and CE. Both B and C

High organic

matter

Low organic

matter

High BOD (Biologica

l oxy

g..

Both A and C

Both B and C

20% 20% 20%20%20%

A

Sources of organic waste:Combined sewage overflow (during high amounts of rain)LivestockUntreated 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)

6. Compared to the stream in graph A, the stream in graph B is(assume the same scale on graphs)

A. A slower moving streamB. A faster moving streamC. A warmer streamD. A and CE. B and C

A slower m

oving stre

am

A faste

r moving st

ream

A warm

er stre

am

A and C

B and C

20% 20% 20%20%20%

A

B

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?

A. FertilizersB. DetergentsC. Both A and BD. Neither A nor B

Ferti

lizers

Detergents

Both A and B

Neither A

nor B

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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.

A. eutrophic; lowerB. eutrophic; higherC. oligotrophic; lowerD. oligotrophic; higher

eutrophic;

lower

eutrophic;

higher

oligotro

phic; lo

wer

oligotro

phic; high

er

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9. The addition of an excess of nutrients into a lake from a human source is called

A. Cultural eutrophicationB. Natural eutrophicationC. Neither A nor B

33% 33%33%

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

A. HypoxicB. HyperoxicC. HypotonicD. Hypertonic

Hypoxic

Hyperoxic

Hypotonic

Hypertonic

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Hypo= Low

Hypoxic = Low oxygen levels

11. A toxic substance that can be naturally occurring in groundwater

A. CyanideB. ArsenicC. BenzeneD. CFC’s

Cyanide

Arsenic

Benzene

CFC’s

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12. Parking lots can be a major contributor to water pollution because they are

A. Impervious to waterB. Selectively

permeable to waterC. Neither A nor B

Impervi

ous to w

ater

Selective

ly permeable to

...

Neither A

nor B

33% 33%33%

Impervious= not allowing something to pass through

13. In primary sewage treatment

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

waste is

disinfecte

d with

...

screens a

nd a settlin

g tan...

specia

lized ch

emical p

roc..

.

33% 33%33%

14. The first step in the secondary sewage treatment is to

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.

Use ae

robic

bacteria

to ...

Use ch

lorine to

disinfe

ct ...

Use sp

ecific c

hemical a

ge...

33% 33%33%

15. Harmful chemicals in the waste water system are

A. always treated. B. only rarely treated –

requires advanced treatment for specific pollutants.

always

treate

d.

only rare

ly tre

ated –requ...

50%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

A. the digester tank is treated with aerobic bacteria

B. the digester tank is treated with anaerobic bacteria

the digeste

r tank is

treat..

the digeste

r tank is

tre...

50%50%

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

A. AgricultureB. LandfillsC. IncinerationD. All of the above

Agricu

lture

Landfills

Incinera

tion

All of t

he above

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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 BOD for this pond was

A. 8 mg/LB. 1 mg/LC. 2 mg/LD. 3 mg/LE. 4 mg/L

8 mg/

L

1 mg/

L

2 mg/

L

3 mg/

L

4 mg/

L

20% 20% 20%20%20%

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.

A. 1 mg/LB. 2 mg/LC. 3 mg/LD. 4 mg/LE. 8 mg/L

1 mg/

L

2 mg/

L

3 mg/

L

4 mg/

L

8 mg/

L

20% 20% 20%20%20%

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

A. 8 mg/LB. 1 mg/C. 2 mg/LD. 3 mg/LE. 4 mg/L

8 mg/

L1 m

g/

2 mg/

L

3 mg/

L

4 mg/

L

20% 20% 20%20%20%

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.D2.B3.A4.C5.D6.B7.C8.D9.A10.A11.B

12.A13.B14.A15.B16.B17.D18.B19.A20.E