Category: Class 12 Biology Objective Questions

Biology Multiple Choice Questions on “Properties of Genetic Material (DNA versus RNA)”.

1. Which of the following criterion needn’t be fulfilled by the genetic material?
a) Occurrence of replication
b) Should be able to express itself in the form of “Bohr’s characters”
c) Provision of scope for the mutation that is required for evolution
d) It should be chemically and structurally stable
Answer: b
Clarification: A genetic material must be both chemically and structurally stable. It must be able to replicate. It should provide scope for the evolutionary changes. But it should also be able to express itself in the form of “Mendelian characters” and not “Bohr’s characters”. These Mendelian characters refer to the three laws that were proposed by Gregor Mendel.

2. Which of the following is not a characteristic of RNA?
a) It has ribose sugar molecules in the nucleotides
b) It is a single stranded molecule
c) It is not stable under alkaline conditions
d) All the 3 types of RNA are involved in protein synthesis
Answer: c
Clarification: The hydrogen atom which is attached to the hydroxyl group on the 2’- C atom of the RNA can be easily deprotonated. The presence of larger grooves in RNA is also a reason for it not being stable under alkaline conditions. But under the same condition, the DNA molecule will be stable for it has smaller grooves which cannot be easily attacked or penetrated.

3. Which of the following criterion cannot be fulfilled by protein?
a) Formation of polypeptide chains
b) Generation of its replica
c) Formation of alpha helix and beta sheets
d) Non covalent bonds are present between amino acids
Answer: b
Clarification: Proteins form polypeptide chains by the peptide bond formation between amino acids. Their different folding statures results in the formation of non-covalent bonds as well as the different forms of proteins. Despite these capabilities, proteins lack the ability to generate their replica.

4. Stability is not a property of genetic material.
a) True
b) False
Answer: b
Clarification: The genetic material of an organism should be stable and not change during the course of an organism or person’s lifetime. If any changes are to occur, then variations and problems will arise. It is also clearly inferred from the “transforming principle” of Griffith from his experiments revolving around the R and S strains of the bacterium that stability persists.

5. Stability of the genetic material can also be inferred from the “transforming principle”.
a) True
b) False
Answer: a
Clarification: Heat killed the bacterium but, it didn’t destroy all of its genetic material characteristics. This was proved by Griffith in his experiments concerning about the transformation principles. From this it can be inferred that the stability of the genetic material (either DNA or RNA) cannot be entirely killed. The stability of the genetic can be seen.

6. What will happen when the 2 complementary strands of the DNA are heated up and brought together?
a) They will repel each due to the charges formed
b) They will attract each other due to the charges formed
c) They will become non-complementary to each other
d) They will combine with each other under appropriate conditions
Answer: d
Clarification: From the Griffith’s experiment related to the transforming principle, we could understand that the genetic material still possessed some stability. Likewise, the genetic material DNA which is present in this case also has the capability to join the 2 complementary strands despite it being heated up under appropriate conditions.

7. Which of the following statements is correct regarding DNA and RNA?
a) DNA is highly reactive
b) RNA is not catalytic
c) RNA cannot be easily degraded
d) DNA is a better genetic material than RNA
Answer: d
Clarification: DNA is comparatively less reactive than RNA. The 2’-OH group which is present at the nucleotide of the RNA is the reactive group. This makes the RNA highly reactive in comparison to the DNA. RNA is also catalytic and liable. It cannot be easily degraded. Due to these reasons, the conclusion is that that DNA is the better genetic material than the RNA.

8. The presence of which base in the place of uracil makes the DNA more stable?

a) Adenine
b) Cytosine
c) Thymine
d) Guanine
Answer: c
Clarification: In the case of DNA, thymine is found in the place of uracil. Uracil is found to make a double hydrogen bond with Adenine in RNA. The resistance showed by thymine towards all the photochemical mutations is what makes the DNA more stable.

9. At what rate does the RNA mutate as compared to DNA?
a) Faster rate
b) Slower rate
c) Moderate rate
d) Depending on the medium
Answer: a
Clarification: Mutations can be seen in both DNA and RNA. But RNA mutates at a much faster pace in comparison to the DNA. This is due to the lack of stability that is seen in RNA. Likewise, the viruses which contain RNA as their genetic material also mutate at a faster rate. Their life span will stand short on the contrary.

10. RNA codes are used for translation into which of the following?
a) Genes
b) DNA
c) Proteins
d) Carbohydrate
Answer: c
Clarification: The three types of RNA are r-RNA, t-RNA and m-RNA. These 3 different types of RNA are responsible for the process of translation (translation of RNA into proteins). RNA codes for different amino acid sequences which finally when combined together forms the protein.

11. Which of the following is responsible for the transmission of genetic information?
a) DNA
b) RNA
c) Proteins
d) Mitochondrion
Answer: b
Clarification: The storage of the genetic information takes place in the DNA. It carries all the necessary information needed for heredity. The stability of the DNA is more too, this helps in carrying information. During the central dogma, DNA in transcripted to RNA after which RNA will be responsible for the transmission of the genetic material.

12. Which of the following is responsible for the storage of genetic material?
a) DNA
b) RNA
c) Proteins
d) Ribosome
Answer: a
Clarification: DNA is a stable genetic molecule in comparison to the RNA. So, the carriage of the genetic information will be carried out by the DNA. The transmission of the genetic material will be done by RNA as it is comparatively less stable.

Biology Assessment Questions and Answers on “Origin and Evolution of Man – 3”.

1. What is another name of human evolution?
a) Neogenesis
b) Anthropogenesis
c) Metagenesis
d) Fossilizes
Answer: b
Clarification: Human evolution or anthropogenesis is part of biological evolution which studies the emergence of Homo sapiens sapiens. They were the distinct species from other hominids, great apes and placental mammals.

2. Propliopithecus originated in _____ epoch.
a) Oligocene
b) Miocene
c) Pliocene
d) Azaacene
Answer: a
Clarification: Origin and evolution of Propliopithecus were in an Oligocene epoch so-called as Oligocene apes. They originated about 30-35 million years ago. Origin of man started from Propliopithecus and diverged to many other species.

3. Who was known as Miocene apes?
a) Proconsul
b) Agyptopithecus
c) Ramapithecus
d) Dryopithecus
Answer: b
Clarification: Origin and evolution of Agyptopithecus occurred in late Oligocene and Miocene epoch, so-called as Miocene apes. They were fossils that remained in Miocene rocks. They evolved after Propliopithecus.

4. Who discovered Dryopithecus and Kenyapithecus?
a) Leakey
b) Lewis
c) Joseph
d) Sam
Answer: a
Clarification: Dryopithecus and Kenyapithecus fossils were discovered by Leakey. Dryopithecus was discovered from Eastern Africa near Victoria lake in Kenya. Kenyapithecus was discovered from Kenya.

5. Which of the following is not a characteristic of Dryopithecus?
a) Semi erect posture
b) Evolution about 15-20 million years ago
c) Teeth larger and sharper
d) Meat eater
Answer: d
Clarification: They were not meat eaters whereas they were vegetarian by nature. They only ate fruits. The remaining options are true characters of Dryopithecus.

6. Ramapithecus and Shivapithecus were discovered from which place in India?
a) Himalayan hills
b) Shivalik hills
c) Nagaraj hills
d) Ganga hills
Answer: b
Clarification: Ramapithecus and Shivapithecus were discovered from Shivalik hills in India. They were discovered by Lewis. These species spent most of the time on land. They originated in Pliocene epoch.

7. Ramapithecus was more ape-like whereas Dryopithecus was more man-like.
a) True
b) False
Answer: b
Clarification: It was the opposite. Ramapithecus was more man-like whereas Dryopithecus was more ape-like. Ramapithecus was the earliest hominid fossil showed characters of man whereas Dryopithecus was the earliest fossil ape showed the characters of an ape.

8. Who discovered Australopithecus?
a) Alexander Fleming
b) Lewis Jackson
c) Prof. Raymond dart
d) Gregor Mendel
Answer: c
Clarification: Professor Raymond dart discovered a fossil skull of a 5-6-year-old baby from the old Pliocene rocks. It was located in the Tuang region in South Africa. He first named it Tuang baby and he renamed it, A. Africanus.

9. Australopithecus was also known as the ______
a) first ape man
b) first human
c) human baby
d) humanoid
Answer: a
Clarification: Australopithecus was known as the first ape-man. They had both man and ape-like characters. They had the same dental formula and had a thick growth of hair on the body.

10. Which of the following is the correct statement regarding Australopithecus?
a) They lived in West African grasslands
b) They lived about 5-6 million years ago
c) They essentially ate fruits
d) They hunted with guns
Answer: c
Clarification: They lived about 3-2 million years ago in East African grasslands. They essentially ate fruits and hunted with stones. They had a less cranial capacity of 600 c.c.

11. Which fossil of the ape was known as the connecting link between apes and man?
a) Ramapithecus
b) Dryopithecus
c) Shivapithecus
d) Australopithecus
Answer: d
Clarification: Australopithecus was known as the connecting link between apes and man. They had complete erect posture and showed bipedal locomotion. It was the first man to stand erect.

Biology Assessment Questions,

Biology Multiple Choice Questions on “Human Health and Disease – Humoral Immunity-1”.

1. What is the full form of AMI?
a) Antibody-Mediated Immunity
b) Antigen Mediated Immunity
c) Automated Immunity
d) Acquired Mediated Immunity
Answer: a
Clarification: The full form of AMI is Antibody-Mediated Immunity. Antibodies are immunoglobulins which are produced in response to antigenic stimulation.

2. which of the following immunities is also called as Antibody-Mediated Immunity?
a) Acquired Immunity
b) Cell-Mediated Immunity
c) Humoral Immunity
d) Innate Immunity
Answer: c
Clarification: Humoral Immunity is also called as Antibody-mediated Immunity. Our immune system comprises of the antibodies present in the body humour-blood, lymph. Antibodies protect our body from pathogens that happen to enter blood and lymph.

3. Number of antibodies produced per day during an infection can be _______
a) 2 trillion
b) 20 trillion
c) 200 trillion
d) 2000 trillion
Answer: b
Clarification: Number of antibodies produced per day during an infection can be 20 trillion. Antibodies are immunoglobulins which are produced in response to antigenic stimulation.

4. Which structure is depicted by the following figure?

a) Antigen
b) Neutrophil
c) Antibody molecule
d) Basophil
Answer: c
Clarification: The above given structure is of an antibody molecule. Each antibody molecule has four peptide chains, two small called as light chains and two longer called heavy chains. Both chains are linked by disulphide bonds.

5. An antibody is represented by which of the following formula?
a) H2L2
b) C1C2
c) A1A2
d) H1L2
Answer: a
Clarification: An antibody is represented by H2L2 formula. Each antibody molecule has four peptide chains, two long chains called heavy or H chains and two short chains called light or L chains.

6. Antigenic determinants are ___________
a) Large and complex molecules
b) Proteins or carbohydrates
c) Specific products of pathogen
d) Recognisable sites over antigens
Answer: d
Clarification: Antigenic determinants are recognisable sites over antigens. Antigens are generally large molecules. The majority of them are made up of Proteins and Polysaccharides found on the cells walls of bacteria and other cells or the coat of viruses.

7. Which of the following set of antibodies are responsible for providing Natural Passive Immunity to the foetus?
a) IgD and IgE
b) IgM and IgA
c) IgA and IgE
d) IgG and IgA
Answer: d
Clarification: IgG and IgA antibodies are responsible for conferring Natural Passive Immunity to the foetus. IgG crosses the placenta during pregnancy and IgA antibodies are present in the yellowish fluid colostrum secreted by mother during the initial days of lactation.

8. Which structure of the antibody is represented by the following figure?

a) IgA
b) IgM
c) IgD
d) IgG
Answer: b
Clarification: The given structure represents the structure of IgM. It is present in pentameric form and has 10 paratopes (Antigen Binding Sites).

9. Which of the following is the largest antibody?
a) IgA
b) IgG
c) IgM
d) IgE
Answer: c
Clarification: IgM is the largest antibody amongst all because it has 10 paratopes or antigen binding sites. IgG antibody is the smallest amongst all.

10. Which of the following statements is not correct regarding IgM antibody?
a) Responsible for the initial activation of B-cells
b) It is an effective agglutinator of antigen
c) It makes up 7-10% of our total antibodies
d) It is the last antibody to be released during the primary response
Answer: d
Clarification: IgM is the first antibody which is released during an infection site. IgM is also responsible for the initial activation of B-cells, macrophages and the complement system. It also makes up 7-10% of our total antibodies.

11. Which of the following is not the function of an antibody?
a) Lysis
b) Neutralisation
c) Assimilation
d) Precipitation
Answer: c
Clarification: Assimilation is not the function of an antibody. The functions of antibodies include Lysis, Neutralisation, Opsonisation, Precipitation and Agglutination.

12. Which of the following antibodies shows Opsonisation?
a) IgG
b) IgE
c) IgD
d) IgA
Answer: a
Clarification: IgG shows Opsonisation property. It is a property of Adherence where non-palatable antigen becomes palatable which helps in the endocytosis of pathogen by the macrophages.

Biology MCQs for MBBS Entrance Exams on “Human Health and Disease – Drugs and Alcohol Abuse-4”.

1. Cocaine is obtained from __________
a) Erythroxylum coca
b) Papaver somniferum
c) Datura
d) Atropa belladonna
Answer: a
Clarification: Cocaine is a local anaesthetic, vasoconstrictor and powerful stimulant which is obtained from the dried leaves of Coca plant, Erythroxylum coca which is native to South America. Its leaves can be chewed, eaten, sniffed in powdered form or taken as a drink.

2. Which of the following symptoms does not occur by the intake of small quantities of cocaine?
a) Increases hunger
b) Increases Mental alertness
c) Relaxes fatigued muscles
d) Reduces hunger
Answer: d
Clarification: In small quantities, as present in the leaves, cocaine relaxes fatigued muscles, increases mental alertness and physical strength but it reduces hunger. Its purified content was once used as a local anaesthetic, especially in dentistry.

3. Which of the following is a powdered form of cocaine?
a) Crack
b) Speedball
c) Coke
d) Betelnut
Answer: c
Clarification: Powdered cocaine is coke. The drug is a bitter, white, crystalline powder which is often adulterated with glucose or lactose and local anaesthetics. It is inhaled or injected.

4. Cocaine suppresses the brain activity and produces a feeling of calmness, relaxation and drowsiness.
a) True
b) False
Answer: b
Clarification: Cocaine has a potent stimulating action on the central nervous system producing a sense of euphoria and increased energy. Cocaine gives a feeling of pleasure followed by hallucinations.

5. Which of the following is not a physiological symptom associated with the intake of cocaine?
a) Increase in heartbeat
b) Increase in blood pressure
c) A decrease in body temperature
d) Increase in body temperature
Answer: c
Clarification: Cocaine increases heartbeat, blood pressure and body temperature. In high dose, the rise in temperature or blood pressure may be lethal. The drug alters membrane transport, prevents uptake of catecholamines at adrenergic nerve endings and interferes with the transport of dopamine.

6. Which of the following coca alkaloids is present in an endosperm?
a) Crack
b) Speedball
c) Coke
d) Betelnut
Answer: d
Clarification: The betelnut is the seed of Areca catechu. The endosperm or kernel contains a number of alkaloids, the major being arecoline. It is mildly stimulant and vermifuge.

7. Cocaine is a non- toxic drug and enhances the physiological functions of our body.
a) False
b) True
Answer: a
Clarification: Cocaine is a toxic drug and deteriorates the health of a person who is addicted to it. Cocaine is toxic to the brain, produces severe headache, convulsions, loss of appetite and causes insomnia. Cardiac arrhythmia and respiratory depression are also common.

8. Which of the following statement is false regarding hallucinogens?
a) These drugs are also known as psychedelics
b) They produce a true sensory stimulus
c) They cause Intensified sensory perceptions
d) They are also responsible for Perceptual Distortion
Answer: b
Clarification: Hallucinogens are formulations which changes one’s perceptions, thoughts and feelings without any true sensory stimulus. They are also called psychedelics. These drugs are characterised by intensified sensory perception, perceptual distortion, hallucination and intense feeling of either despair or euphoria.

9. Which of the following plant is depicted in the diagram given below?

a) Datura
b) Atropa belladonna
c) Opium poppy
d) Cannabis sativa
Answer: a
Clarification: The given diagram shows the flowering branch of the Datura plant. It is a well-known plant with hallucinogenic properties. The drugs obtained from this plant changes one’s thoughts and feelings without actually sending any true sensory stimulus.

10. What is the full form of LSD?
a) Lanthanide stimulated drug
b) Lysergic acid diethylamide
c) Lanthanide steric drug
d) Lysergic drug
Answer: b
Clarification: LSD stands for Lysergic acid diethylamide. LSD leads to CNS damage, non-coordination of body parts, chromosomal aberrations, foetal abnormalities and psychosis. It may lead to unconsciousness and even death.

11. LSD is extracted from which of the following organism?
a) A fungus
b) A plant
c) An animal
d) A bacterium
Answer: a
Clarification: LSD is a crystalline amidated alkaloid derived from ergot, an extract got from the fruiting body of fungus Claviceps purpurea. LSD was found to be psychedelic in 1947. There is marked hallucinations, ecstasy and emotional outburst.

12. Which of the following is not a physiological effect observed by the intake of LSD?
a) High blood pressure
b) Rapid heartbeat
c) Pupillary dilation
d) Reduced heart rate
Answer: d
Clarification: Psychological and physiological effects are found even in a small dose of only 20 μg taken orally. The drug brings about rapid heartbeat, high blood pressure, high body temperature, pupillary dilation, vomiting, tremors. There are mood changes, visual illusions, conflicting often bizarre perceptual changes.

To practice Biology MCQs for MBBS Entrance Exams,

Biology Multiple Choice Questions for Schools on “Food Production Strategies – Plant Breeding-2”.

1. ___________ is the race of a species which is superior to all other existing varieties in one or more traits.
a) Improved variety
b) Variety
c) Character
d) Genes
Answer: a
Clarification: Improved variety is the race of a species which is superior to all other existing varieties in one or more traits. A variety is the race of organisms of a species having the same genotype which differs from other groups of individuals of the same species in one or more traits. A trait or character is any morphological, anatomical, behavioural and biochemical feature.

2. Which of the following processes is not related to hybridisation?
a) Emasculation
b) Selection of parents
c) Bagging
d) Crossing or artificial pollination
Answer: b
Clarification: Hybridisation is the crossing of two or more types of plants for bringing their traits together in the progeny. The various steps involved in hybridisation are Emasculation, Bagging, Crossing or Artificial Pollination, the Desired combination of variations, backcrossing.

3. Which of the following statements is true for Emasculation?
a) Removal of stigma and anther from a plant
b) Removal of male parts from the future female plant
c) Removal of female parts from the future male plant
d) Removal of all parts of the plant
Answer: b
Clarification: Emasculation is the removal of male parts from the future female plants in the young state so as to avoid chances of contamination from their pollen. In the case of unisexual plants, the male plants are not allowed to grow near the female plants.

4. The selection process in hybridisation is crucial to the success of the breeding objective and requires careful scientific evaluation of the progeny.
a) True
b) False
Answer: a
Clarification: Selection and testing of superior recombinants consist of selecting, among the progeny of the hybrids, those plants that have the desired character combination. The selection process is crucial to the success of the breeding objective and requires careful scientific evaluation of the progeny. This step yields plants that are superior to both of the parents.

5. Which of the following is not a method of evaluation of new crops?
a) Irrigation
b) Fertiliser application
c) Harvesting them
d) Growing them in research fields
Answer: c
Clarification: The newly selected lines are evaluated for their yield and other agronomic traits of quality, disease resistance, etc. This evaluation is done by growing these in the research fields and recording their performance under ideal fertiliser application, irrigation, and other crop management practices.

6. Which of the following scientists developed the semi-dwarf wheat variety?
a) Herbert Boyer
b) Stanley Cohen
c) M.S. Swaminathan
d) Norman E. Borlaug
Answer: d
Clarification: Nobel laureate Norman E. Borlaug, at International Centre for Wheat and Maize Improvement in Mexico, developed semi-dwarf wheat varieties.

7. HYV stands for ___________
a) High Yak Variety
b) Heat Yak Variety
c) High Yielding Varieties
d) Heat Yielding Varieties
Answer: c
Clarification: HYV stands for High Yielding Varieties. A green revolution occurred in India with the introduction of HYVs of wheat and rice from outside. Continually, new HVYs are bring added so that there is a continuous increase in crop output.

8. During which period, wheat production increased from 11 million tonnes to 75 million tonnes?
a) 1960-2000
b) 1980-1990
c) 1942-1980
d) 1920-1950
Answer: a
Clarification: During the period 1960-2000, wheat production increased from 11 million tonnes to 75 million tonnes while rice production went up from 35 million tonnes to 89.5 million tonnes. This was due to the development of semi-dwarf varieties of wheat and rice.

9. Which of the following is not a High Yielding Variety of wheat?
a) Jaya
b) Ratna
c) Sonalika
d) Jowar
Answer: d
Clarification: In 1963, several HYVs of wheat such as Sonalika and Kalyan Sona, which were disease resistant were introduced all over the wheat-growing belt of India. Later better-yielding semi-dwarf varieties Jaya and Ratna were developed in India.

10. Which variety of sugarcane was originally grown in North India?
a) Saccharum barberi
b) Saccharum officinarum
c) Saccharum spontaneum
d) Kalyan sona
Answer: a
Clarification: Saccharum barberi was originally grown in North India, but had poor sugar content and yield. Tropical canes grown in South India Saccharum officinarum had thicker stems and higher sugar content but did not grow well in South India.

11. Semi-dwarf varieties were derived from IR-8 and Taichung Native-1.
a) True
b) False
Answer: a
Clarification: Semi-dwarf varieties were derived from IR-8, (developed at International Rich Research Institute, Philippines) and Taichung Native-1 (from Taiwan). The derivatives were introduced in 1966.

12. Which of the following is not a variety of Millet’s?
a) Hybrid maize
b) Jowar
c) Bajra
d) Jaya
Answer: d
Clarification: Hybrid maize, jowar and bajra have been successfully developed in India. Hybrid breeding has led to the development of several high yielding varieties resistant to water stress.

To practice Biology Multiple Choice Questions for Schools,

Biology Multiple Choice Questions on “Microbes in Sewage Treatment”.

1. What is the major component of waste-water?
a) Pure water
b) Human excreta
c) Sand
d) Clay
Answer: b
Clarification: A major component of this waste-water is human excreta. This municipal waste-water is also called as sewage. It contains a large amount of organic matter and microbes. Many of which are pathogenic.

2. What is the full form of STPs?
a) Sexually transmitted problems
b) Sewage treatment plants
c) Serum tissue plasminogen
d) Sebaceous tissue plasminogen
Answer: b
Clarification: STP stands for sewage treatment plants. Before the disposal of waste-water into the natural water bodies, sewage is treated in STPs to make it less polluting.

3. In how many stages the treatment of waste-water in STPs is carried out?
a) One
b) Two
c) Three
d) Four
Answer: b
Clarification: The treatment of waste-water in STPs is carried out in two stages namely primary treatment and secondary treatment. Treatment of waste-water is done by the heterotrophic microbes naturally present in the sewage.

4. Primary treatment involves filtration and distillation.
a) False
b) True
Answer: a
Clarification: Primary treatment involves the basic removal of particles- large and small-from the sewage through filtration and sedimentation. These particles are removed further in various stages.

5. Which of the following is not correct regarding the primary treatment of waste-water?
a) Initially floating debris is removed by distillation
b) Grit and pebbles are removed by sedimentation
c) Solids that settle form the primary sludge
d) Supernatant forms the effluent
Answer: a
Clarification: In primary treatment, initially floating debris is removed by sequential filtration. Then grit and small pebbles are removed by sedimentation. All solids that settle form the primary sludge while the supernatant forms the effluent.

6. The effluent from the primary settling tank is taken for secondary treatment.
a) True
b) False
Answer: a
Clarification: The effluent from the primary settling tank is taken for secondary treatment. The primary effluent is then passed into large aeration tanks where it is constantly agitated mechanically and the air is pumped into it.

7. What are flocs?
a) Masses of fungi
b) Masses of algae
c) Masses of animals
d) Masses of bacteria
Answer: d
Clarification: The constant agitation of primary effluent allows vigorous growth of useful aerobic microbes into flocs which are the masses of bacteria associated with fungal filaments to form mesh-like structures.

8. What is the full form of BOD?
a) Biochemical oxygen demand
b) Bionatal oxygen demand
c) Biological disease
d) Biological oxygen deficiency
Answer: a
Clarification: BOD refers to the Biochemical oxygen demand. It is the amount of oxygen that would be consumed if all the organic matter in one litre of water were oxidised by bacteria.

9. Which of the following statements is not true regarding BOD?
a) Sewage water is treated until the BOD is reduced.
b) BOD is a measure of organic matter present in the water
c) Greater the BOD, lesser is the polluting potential of water
d) BOD measures the rate of uptake of oxygen
Answer: c
Clarification: The sewage water is treated until the BOD is reduced. The BOD test measures the rate of uptake of oxygen by micro-organisms in a sample of water and thus, BOD is indirectly a measure of the organic matter present in the water. The greater the BOD of waste-water more is its polluting potential.

10. What is activated sludge?
a) Distilled bacteria
b) Fermented bacteria
c) Sediment bacterial flocs
d) Fungi
Answer: c
Clarification: Activated sludge are the sediment bacterial flocs which were present in the effluent. When the biochemical oxygen demand of sewage water is reduced significantly, the effluent is then passed into a settling tank where the bacterial flocs are allowed to sediment.

11. Where is the major part of sludge pumped in secondary treatment?
a) Aeration tank
b) Anaerobic sludge digesters
c) Rivers
d) Drains
Answer: b
Clarification: In the secondary treatment, a small part of the activated sludge is pumped back into the aeration tank to serve as inoculum. The remaining major part of the sludge is pumped into large anaerobic sludge digesters.

12. How many gallons of waste-water is treated before being released into the natural water-bodies?
a) Millions
b) Billions
c) Trillions
d) Quadrillion
Answer: a
Clarification: Millions of gallons of waste water is treated in various sewage plants where microbes play a major role in decomposing the waste. This methodology has been practised for a century, in almost all parts of the world.

13. What is the full form of GAP?
a) Ganglion active potential
b) Ganga active plan
c) Ganglion action potential
d) Ganga Action Plan
Answer: d
Clarification: The Ministry of Environment and Forests has initiated Ganga Action Plan (GAP) and Yamuna Action Plan (YAP) to save these major rivers of our country from pollution.

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