2nd PUC Biology Previous Year Question Paper March 2014

Students can Download 2nd PUC Biology Previous Year Question Paper March 2014, Karnataka 2nd PUC Biology Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Previous Year Question Paper March 2014

Time: 3 hrs 15 min
Max. Marks: 70

General Instructions

  • This question paper consists of four parts A, B, C and D. Part – D consists of two sections. Section – I and Section – II.
  • All the parts are compulsory.
  • Draw diagrams wherever necessary. Unlabelled diagrams or illustrations do not attract any marks.

Part – A

Answer the following questions in one word or one sentence each. (10 × 1 = 10)

Question 1.
What is ovulation?
Answer:
The process of release of ovum from the ovary into the pelvic cavity is called ovulation.

Question 2.
Name the cells that secrete androgens.
Answer:
Cells of Leydig.

Question 3.
What is the gene pool?
Answer:
Gene pool refers to the sum total of different kinds of genes pooled by all the members of a population.

2nd PUC Biology Previous Year Question Paper March 2014

Question 4.
Name the causative agent of elephantiasis.
Answer:
Wuchereria bancrofti.

Question 5.
What is an allergy?
Answer:
Allergy can be defined as the exaggerated or hypersensitive reaction of the immune system to certain antigens present in the environment.

Question 6.
Plant cells are totipotent. Why?
Answer:
The ability of a plant cell to give rise to an entire plant on the medium is called totipotency.

Question 7.
Mention the bacteria responsible for the large holes seen in ‘Swiss Cheese’.
Answer:
Propionibacterium sharmanii.

Question 8.
Define the polymerase chain reaction.
Answer:
It is a reaction in which multiple copies of the gene(or DNA) of interest are synthesised in-vitro using two sets of primers and the enzyme DNA polymerase.

2nd PUC Biology Previous Year Question Paper March 2014

Question 9.
What is global warming?
Answer:
Increase in the level of greenhouse gases has led to considerable heating of earth known as global warming.

Question 10.
Define Mortality.
Answer:
The number of deaths in the population during a given period.

Part – B

Answer any five of the following questions in 3 to 5 sentences each: Wherever applicable: (5 × 2 = 10)

Question 11.
Distinguish between homogametes and heterogametes.
Answer:

Homogametes Heterogametes
Gametes formed are similar in appearance. Gametes formed are dissimilar in appearance.
Also known an Isogametes. Also known as Anisogametes.

Question 12.
What is the test cross? Mention its significance.
Answer:
Crossing the F1 individual with a recessive parent is known as a test cross. It is used to test whether the F1 of any given plant is homozygous or heterozygous for a particular character and also to test the validity of Mendel’s laws of heredity.

Question 13.
Write a note on selection and testing of superior recombinants in plant breeding.
Answer:
Selection and testing of superior recombinants:

  • First, the individuals with the desired combination of characters have to be selected from among the progeny of hybrids.
  • Such hybrids are superior to both of the parents (hybrid vigourlheterosis).
  • They arc self-pollinated for several generations tilt they reach a state of homozygosity so that there will be no segregation of characters in the progeny.

2nd PUC Biology Previous Year Question Paper March 2014

Question 14.
What are homologous organs? Mention any two examples.
Answer:
Homology or homologous organs is/are the result of divergent evolutions, i.e., the evolutionary process where the same structure develops along with different directions due to adaptations for different needs.

Question 15.
Differentiate between active immunity and passive immunity.
Answer:
Active immunity:
The immunity provided by antibodies produced by B-cells in their encounter with antigens is called active immunity. It may be natural or artificial.

Naturally acquired active immunity results from any bacterial or viral infection. During infection, antibodies are produced against the pathogen. If the individual contacts with the same pathogen again, the disease will not affect the person as the antibodies are already present in the body.

Artificially active immunity is introduced into the body through a process called vaccination or immunization. In this process, dead or attenuated (living but extremely weakened) pathogens the introduced into the body in the form of the vaccine. This induces immune response and antibodies are produced in the body.

Passive immunity:
The immunity provided to an individual by the introduction of borrowed antibodies obtained from an immunized animal is called passive immunity. Passive immunity may be natural or artificial.

Naturally acquired passive immunity results through the natural transfer of antibodies. The transfer of antibodies from the mother’s body into the foetus through the placenta is a typical natural passive immunity. The baby is protected for several months after birth.

Artificially acquired passive immunity results by transfer of antibodies produced in the body of one individual into other through injections. The donated antibodies provide protection for short period. The serum contaiñing antibodies (globulins) is used for treating snake bites, rabies and tetanus etc.

2nd PUC Biology Previous Year Question Paper March 2014

Question 16.
Draw a neat labelled diagram of striped tank bioreactor.
Answer:
2nd PUC Biology Previous Year Question Paper March 2014 Q16

Question 17.
Introduction of some alien species causes biodiversity loss. Justify the statement with an example.
Answer:
Intentional or chance introduction of exotic species into new territories or countries by humans adversely affects the native species.
For eg: Introduction of Nile perch into lake victoria, led to the extinction of more than 200 species of cichlid fish in the lake.

Question 18.
What are sacred groves? Mention any two examples.
Answer:
The tracts of forest conserved and protected based on the religious faith ground.
Examples:

  • Khasi and Jaintia Hills in Meghalaya.
  • Aravalli hills of Rajasthan.

Part – C

Answer any five of the following questions in about 40 to 80 words each wherever applicable: (5 × 3 = 15)

Question 19.
(a) What are hermaphrodites? Mention one example.
Answer:
The animals in which both the sex organs are present in the same body are called hermaphrodites.
e.g.: Earthworm

(b) Offsprings of asexual reproduction are called clones. Why?
Answer:
The offsprings formed by asexual reproduction are genetically identical to each other, hence are called clones.

Question 20.
Briefly explain the structure of pollen grains.
Answer:
Structure of pollen grain: Pollen grain is the haploid unicellular, uninucleated spherical shaped body bounded by outer thick, spiny or reticulate callose wall called exine and inner thin, smooth cellulose walL called intine. In the exine, thin areas called germ pores are present. Intine encloses peripheral cytoplasm and central nucleus.

Pollen grain represents the first stage of male gametophyte. Development by pollen grain takes place when it is still present in the anther locule. During the development, it divides unequally into a large vegetative cell and a small generative cell. The generative cell then divides and give rise to two non-motile male gametes. This takes place either in the pollen grain or in the pollen tube after pollination.

2nd PUC Biology Previous Year Question Paper March 2014

  1. Pollen grains of many species cause severe allergies and chronic respiratory disorders like asthma, bronchitis etc. e.g. parthenium (carrot grass)
  2. Pollen grains are rich in nutrients. Thus pollen tablets are used as nutrient supplements.

Question 21.
Mention the causes and effects of phenylketonuria.
Answer:

  • It is caused by a recessive mutant allele on chromosome 12.
  • The affected individuals lack an enzyme that catalyses the conversion of the amino acid phenylalanine into tyrosine.
  • Consequently, phenylalanine is metabolised into phenylpyruvate and other derivatives.
  • Accumulation of these chemicals in the brain results in mental retardation.
  • These are also excreted in the urine as they are not absorbed by the kidney.

Question 22.
List the period, brain capacity and probable food of the Homo erectus stage in the human evolution.
Answer:

  • Their fossils were found in Java (Java man) in 1981.
  • They probably lived about 1.5 million years ago.
  • They had a brain capacity of about 900cc.
  • They probably ate meat.

Question 23.
Mention the different steps of the process of recombinant DNA technology.
Answer:

  • Isolation of genetic material (DNA)
  • Cutting of DNA at specific locations
  • Gel electrophoresis.
  • Amplification of gene of interest using PCR
  • Insertion of recombinant DNA into the host
  • Obtaining the foreign gene product.
  • Downstream processing.

Question 24.
(a) Draw an ideal pyramid of energy and mention the Units.
Answer:
2nd PUC Biology Previous Year Question Paper March 2014 Q24

(b) What is ecological succession?
Answer:
The occurrence of a relatively definite sequence of same resulting in the establishment of stable or climax community is known as ecological or biotic succession.

2nd PUC Biology Previous Year Question Paper March 2014

Question 25.
(a) Distinguish between primary productivity and secondary productivity of an ecosystem.
Answer:
A constant input of solar energy is the basic requirement of any ecosystem to function and sustain.
Primary productivity is expressed in terms of gm-2 yr-1 or k Cal m-2 y-1.
Primary productivity has two aspects:

  • Gross Primary Productivity (GPP) is the rate of production of organic matter during photosynthesis.
  • Net Primary Productivity (NPP) is the amount of energy left in the producers after utilisation of some energy for respiration, i.e., GPP – R = NPP. Tt is the amount of energy available in the producers, for the consumption of herbivores.

Primary productivity depends on

  • the plant species in a particular area,
  • availability of nutrients,
  • photosynthetic capacity of plants and
  • a number of environmental factors.

The annual net primary productivity oÍthe whole biosphere is approximately 170 billion tons (dry weight) of organic matter.

Of this, the productivity of oceans is only 55 billion tons, while the rest is from the land.
Secondary productivity is defined as the rate of assimilation and the formation of new organic matter by consumers.

(b) Define food web.
Answer:
Food Web:
Food chains remain interconnected at various levels. The web-like interlocking system of several food chains in a community is called food web, This maintains ecological homeostasis. It allows an organism to feed on two or more organisms of the lower trophic level and provides for the survival of organisms.

Question 26.
Briefly explain any three prevention and control measures of drug/alcohol abuse.
Answer:
Some of the measures useful for the prevention and control of alcohol and drug abuse are:
(a) Avoid undue peer pressure:
Every child has his/her own choice and personality which should be respected and nurtured. A child should not be pushed unduly to perform beyond his/her threshold limits, be it studies, sports or other activities.

(b) Education and counselling:
Educating and’ counselling him/her to face problems and stresses and to accept disappointments and failures as a part of life. It would also be worthwhile to channelise the child’s energy into healthy pursuits like sports, reading, music, yoga and other extracurricular activities.

(c) Seeking help from parents and peers:
Help from parents and peers should be sought immediately so that they can guide appropriately. Help may even be sought from close and trusted friends.

Part – D

Section – I

Answer any four of the following questions in about 200 to 250 words each, wherever applicable: (4 × 5 = 20)

Question 27.
Describe the outbreeding device that prevents autogamy.
Answer:
Contrivances for cross-pollination: [Out breeding devices]
The following contrivances ensure cross-pollination:
1. Dicliny: It is the condition where one of the two sexes is absent in the flower and flower becomes unisexual male or female (diclinous). Such diclinous flowers may be borne either on the same plant or on the two different plants. In such cases, cross-pollination is the rule. e.g.. maize, Cucurbita.

2. Herkogamy: (Herkos = barrier) It is the condition where the style of the gynoecium extend far beyond the anthers or stamens may face outward or pollens may aggregate into pollinia. In such cases, self-pollination is impossible. e.g. gloriosa, calotropis.

2nd PUC Biology Previous Year Question Paper March 2014

3. Dichogamy: It is the condition where androecium and gynoecium in a flower mature at different times. In such a case, self-pollination is ineffective, however, it may take place at a later stage if cross-pollination fails. Dichogamy may be of two types:

  • Protandry: Anthers mature earlier than the carpels. e.g. sunflower, cotton.
  • Protogyny: Carpels mature earlier than anthers. e.g. Michela, ficus.

4. Self sterility: It is the condition where pollen grains fail to germinate on the stigma of the same flower. In such cases, self-pollination is ineffective and cross-pollination is a must. e.g. tobacco, potato.

5. Heterostyly: It ¡s the condition where the flowers on the same plant have styles of different sizes, where one has short style and long stamens, while another has long style and short stamens. In such cases self-pollination is impossible and cross-pollination is a must. e.g. oxalis, primula.

Question 28.
Draw a neat labelled diagram of the human sperm.
Answer:
2nd PUC Biology Previous Year Question Paper March 2014 Q28

Question 29.
(a) What is the medical termination of pregnancy? Mention the safe period for medical termination of pregnancy.
Answer:
Medical Termination of Pregnancy (MTP): Termination of pregnancy is called abortion. Medical termination of undesired pregnancy is a method of birth control. However, termination of advanced pregnancies is dangerous. This should be done within 2 – 2½ months of pregnancy.

(b) Write three simple principles to prevent sexually transmitted diseases.
Answer:

  • Multiple sexual contacts should be avoided and a condom must be used as a protection against the disease.
  • Never inject drugs, share razors and toothbrushes.
  • Use thoroughly sterilised equipment for injection, acupuncture, tattooing and piercing of ear and nose.

Question 30.
Write the schematic representation of the dihybrid cross and mention the result.
Answer:
Dihybrid Cross:
In a dihybrid cross, Mendel selected two pairs of contrasting characters. He selected the shape and colour of the seed as two characters, for his cross. He recognised round and wrinkled shape of the seeds as one pair of contrasting characters and yellow and green colour as another pair of contrasting characters.

He crossed a plant having Yellow and Round seeds with a plant having Green and Wrinkled seeds and got F1 generation. In the F2 he got all the plants producing Round and Yellow seeds. Thus he found that yellow and round traits were dominant over green and wrinkled, respectively.

Later, he self-crossed the F1 hybrids and raised F2 generation. In this generation, he got four types of plants producing Yellow-Round, seed Yellow-Wrinkled seed, Green Round seed, and Green Wrinkled seed. Thus he got two new characters-Yellow-Wrinkled seed and Green Round seed, in addition to the parental characters.

By using appropriate letters to denote dominant and recessive genes, the dihybrid cross is represented as shown below:
2nd PUC Biology Previous Year Question Paper March 2014 Q30
Thus in the F2 generation, he got Yellow Round, Yellow wrinkled, Green round and Green wrinkled in the ration of 9 : 3 : 3 : 1, respectively.
Number of pairs of alleles
involved = two pairs
Base Number = 16
Phenotypicratio = 9 : 3 : 3 : 1
Genotypic ratio = 4 : 2 : 2 : 2 : 1 : 1 : 1 : 1.
Phenotypic classes = Four (Yellow Round, Yellow Wrinkled, Green Round, Green Wrinkled)
Genotypic classes = 9
Based on the results of a dihybrid cross, he came to the conclusion that the four genes for two pairs of contrasting characters are separated independently in the F1 hybrids because the genes were present on different chromosomes and cómbined in the F2 generation to produce four characters.

Based on the results of the dihybrid cross, he formulated the second law of heredity namely, the Law of Independent Assortment. It states that “When genes or alleles for two or more separate pairs of contrasting characters are brought together in a hybrid, they separate independently of each other”.

2nd PUC Biology Previous Year Question Paper March 2014

Question 31.
List the salient features of the human genome project.
Answer:

  1. The human genome contains 3164.7 million nucleotide bases.
  2. The average gene consists of 3000 bases.
  3. The largest known human gene being dystrophin at 2.4 million bases.
  4. The total number of the gene is estimated at 30,000.
  5. 99.9 per cent of nucleotide base sequences are the same in all the people.
  6. The function of 50% of the genes discovered is still unknown.
  7. Less than 2 per cent of the genome, codes for proteins.
  8. Repeated sequences make up a very large portion of the human genome.
  9. Chromosome I has the most genes (2968) and Y has the fewest (231).
  10. It is identified about 1.4 million locations where single-base DNA differences (SNPs – single nucleotide polymorphism) occur in humans.

Question 32.
Mention the steps involved in DNA fingerprinting.
Answer:
2nd PUC Biology Previous Year Question Paper March 2014 Q32
This is a technique which helps in the establishment of the identity of a person by detecting and comparing spécific nucleotide sequences in the DNA of individuals. This technique was developed by Alec Jeffreys and his associates.

Steps:
a. Collection of sample: Sample like blood, semen, hair, saliva or any other tissue even in the dried state are collected.

b. Extraction of DNA: The material collected is subjected to high-speed centrifugation which separates the DNA from the cell.

c. Fragmenting DNA: The DNA in the sample is treated with restriction endonucleases to obtain smaller restriction fragments.

d. Separating DNA fragments: The smaller fragments of DNA are made to run on an agarose gel plate by electrophoresis technique in which depending on molecular size and shape, the nucleotides produce distinct bands in the gel.

e. Extracting DNA from gel: The double-stranded DNA fragment in the gel it denatured into single strands by exposing the gel into an alkaline solution (NaOH). Then by Southern blotting technique, these single-stranded DNA fragments are transferred onto a nitrocellulose filter or a nylon membrane.

f. Tagging radioactive probes: A probe is radioactively labelled synthetic DNA. The nitrocellulose filter is now incubated with these radioactive probes. The probes act as specific nucleotide detectors and bind or hybridise with the specific nucleotides present in DNA fragments on the filter.

g. Auto radiography: An x-ray film is placed on the filter to obtain autoradiographic prints. The film is then processed to observe the visible hybridised pattern of bands. This is the genetic fingerprint or DNA fingerprint.

2nd PUC Biology Previous Year Question Paper March 2014

h. Comparison: Comparison of two genetic fingerprints with great care can provide information about the degree of similarities or differences.

Section – II

Answer any three of the following about 200 to 250 words each wherever applicable: (3 × 5 = 15)

Question 33.
(a) Discuss the MOET technique of animal breeding.
Answer:
Multiple Ovulation Embryo Transfer (MOET):
It is a method to improve herds.
The steps ¡n the method are as follows:

  • A cow is administered hormones (like FSI-I) to induce follicular maturation and superovulation, i.e., production of 6-8 ova in one cycle.
  • The cow is mated with the selected bull or artificially inseminated.
  • The fertilised eggs at 8-32-celled stages are recovered and transferred to surrogate mothers.

This technology has been used for cattle, rabbits, mares, etc.
High milk-yielding breeds of females and high-quality meat-yielding bulls have been bred successfully to increase the herd size in a short time.

(b) Differentiate between inbreeding and outbreeding.
Answer:
Inbreeding:
The breeding strategy includes the identification of superior males and females of the same breed and mating them in pairs.
The progeny of such matings are evaluated and superior males and females are identified for further mating.
Inbreeding increases homozygosity and thus inbreeding is necessary for evolving pure line in any animal.
Inbreeding exposes the harmful recessive alleles, which become eliminated by selection.
Inbreeding also helps in the accumulation of superior genes and the elimination of less desirable genes.
But, continued inbreeding causes inbreeding depression that reduces vigour, fertility and even productivity.
Under such a situation, the selected animals of the breeding population are mated with unrelated superior animals of the same breed to restore fertility and yield.

Outbreeding:
Outbreeding refers to the breeding of unrelated animals either of the same breed having no common ancestors for 4-6 generations (outcrossing) or of different breeds (cross-breeding) or even different species (interspecific).

Outbreeding is of the following types:
(i) Outcrossing:
Outcrossing is the practice of mating of animals of the same breed, but that have no common ancestors on either side of their pedigree upto 4-6 generations.
The offspring of outcrossing is called an outcross.
A single outcross helps to overcome inbreeding depression.
It is the best breeding method for animals that are below average in productivity and growth rate.

(ii) Cross-breeding:
It is a method of outbreeding in which superior males of one breed are mated with the superior females of another breed of the same species.
This helps in combining the desirable qualities of the two different breeds into the hybrid progeny.
The hybrid progeny may be directly used for commercial production or they may be subjected to some form of inbreeding and selection, to develop new stable breeds.
One example of cross-breeding is Hisardale, a new breed of sheep developed by crossing Bikaneri ewes and Marino rams.

2nd PUC Biology Previous Year Question Paper March 2014

(iii) Interspecific hybridisation:
It is a method of outbreeding in which male and female animals of two different species are crossed to combine the desirable features of both the parents into one.
e.g. Mule is produced by a cross between a male donkey and a female horse.

Question 34.
Briefly describe the role, of biological control of pests and diseases.
Answer:
Microbes As Biocontrol Agents:
Biocontrol refers to the use of biological methods for controlling plant diseases and pests.
These methods rely on natural predation rather than on the introduced chemicals.

An organic gardener works to create a system where the insects (that are pests) are not eradicated but are kept at manage.able levels by a complex system of checks and balances within the ecosystem.
The beneficial predatory and parasitic insects which depend on these insect pests are able to survive; for example, the beetle is useful to get rid aphids and dragonflies control mosquitoes.
Bacillus thuringiensis is a bacterium whose spores are toxic to certain insect larvae and kill them but is not harmful to other insects.

The toxin-producing genes of this bacterium are transferred (genetic engineering) into crop plants, which become resistant to insect pests; Bt cotton is an example.
Another biological contrõl being developed for treatment/control of plant diseases in the fungus Trichoderma, which is free-living in the soil and root ecosystems and is effective against several plant pathogens.

Question 35.
“One of the applications of biotechnology is the production of insect-resistant crop plants.” Justify the statement with reference to Bt-cotton.
Answer:
Bt cotton:

  1. Bacillus thuringiensis produces crystal proteins called Cry proteins which are toxic to larvae of insects like tobacco budworm, armyworm, beetles and mosquitoes.
  2. Cry proteins exist as inactive protoxins and get converted into active toxin when ingested by the insect, as the alkaline pH of the gut solubilises the crystals.
  3. The activated toxin binds to the surface of epithelial cells of midgut and creates pores.
  4. This causes swelling and lysis of cells leading to the death of the insect (larva).
  5. The genes (cry genes) encoding this protein are isolated from the bacterium and incorporated into several crop plants like cotton, tomato, corn, rice, soybean, etc.
  6. The proteins encoded by the following cry genes control the corresponding pests.
    • Cry I AC and Cry II Ab control cotton bollworms.
    • Cry I Ab controls corn borer.
    • Cry III Ab controls Colorado potato beetle.
    • Cry III Bb controls corn rootworm.

2nd PUC Biology Previous Year Question Paper March 2014

Question 36.
(a) State “Gauss competitive exclusion principle”.
Answer:
Competition: Competition is defined as ‘an interaction that occurs between two or more organisms when the resources necessary for them are limited and adversely affect them.

Gause’s Competition Exclusion Principle, states that two closely related species competing for the same resources cannot co-exist independently and competitively and the inferior one will get eliminated eventually.
e.g. When two species of Paramecium Le, Paramecium caudal and Paramecium aurelia are kept together in one habitat for a long time, it was observed that one species got eliminated.

The Abingdon tortoise ¡n Galapagos Islands became extinct within a decade after goats were introduced on the island, apparently due to the greater browsing efficiency of the goats. The larger and competitively superior barnacle Balanus dominates the intertidal area and excludes the smaller barnacle Chathamalus from that zone. in general, herbivores and plants appear to be more adversely affected by competition than carnivores.

(b) Mutualism is a method of population interaction. Explain with examples.
Answer:
Mutualisna or symbiosis in which both the species are mutually benefitted but the association is obligatory and they cannot live separately under normal conditions.
e.g., mycorrhizae with the association of fungus (e.g., Boletus) and roots of higher plants. The fungi help the plant in the absorption of water and essential nutrients from the soil while the plant, in turn, provides the fungi with carbohydrates.

Lichens represent an intimate mutualistic relationship between a fungus and photosynthesising algae or cyanobacteria. Root nodules of legume plants, Casuarina. Alnus and leaf nodules of Ardisia have symbiotic association of nitrogen fixing bacteria – Rhizobium. The plants provide food and shelter to the bacteria and the bacteria fix free atmospheric nitrogen to the plant.

The most interesting example of mutualism is found in plant-animal relationships. Plants take the help of animals for pollinating their flowers and dispersing their seeds. Animals get rewards or fees in the form of pollen and nectar (for pollinators) and juicy and nutritious fruits (for seed dispersals).

In many species of fig trees, there is a tight one-to-one relationship with the pollinator species of wasp. The female wasp uses the fruit not only as an oviposition (egg-laying) site but uses the developing seeds within the fruit for nourishing its larvae. The wasp pollinates the fig inflorescence while searching for suitable egg-laying sites. In return for the favour of pollination, the fig offers the wasp some of its developing seeds as food for the developing wasp larvae.

The Mediterranean orchid ophrys employs sexual deceit to get pollination done by a species of bee. One petal of its flower bears a un carry resemblance to the female of the bee in size, colour and markings. The male bee is attracted to what it perceives as a female, ‘pseudo copulates’ with the flower and during that process is dusted with pollen from the flower. When this same bee pseudoscope lates with another flower, it transfers pollen to it and
thus pollinates the flower.

2nd PUC Biology Previous Year Question Paper March 2014

Question 37.
Define pollution. With reference to water pollution, explain biochemical oxygen demand, algal bloom, biomagnification and eutrophication.
Answer:
As Odum defines, pollution is the undesirable change in physical, chemical and biological characteristics of our environment, adversely affecting human health and life of domestic animals.
Pollution is also defined as the unfavourable alteration or contamination of our environment, largely as a result of human activities.

Sewage is a biodegradable pollutant concentrated with human faeces, animal excreta, food residues, domestic wastes, detergents and several dissolved chemicals (sulphates phosphates, alkalies, nitrates, acids). Sewage also contains pathogenic bacteria, eggs, cysts and spores of parasites. Sewage is discarded from homes, hotels, and choultries and hostels being dumped constantly into nearby water resources like lakes, ponds, rivers and oceans causing water pollution.

Domestic sewage contains nutrients like nitrogen and phosphorus whiêh favours the excessive growth of algae (free-floating) called an algal bloom.

The harmful effects:

  • Deterioration of water quality and imbalance of the water ecosystem.
  • Fish mortality rate increases.
  • Some bloom-forming algae are extremely toxic to human beings and animals.

The magnitude of Biodiversity:

  • The predicted number of total species on this earth varies from 5 to 50 million and average at 14 million, but only 1.7 million have been described till today and the distribution is highly uneven.
  • About seven per cent of the world’s total land area is home to half of the world’s species, with the tropics alone accounting for 5 million.
  • About 61 per cent of the known species are insects, but, only 4650 species of mammals are known to us.
  • A large number of plant species (2,70,000) and vertebrates are known.
  • About 40,000 species of algae and 72,000 species of fungi are known.
  • About 2,50,000 species of angiosperms are known but only 750 species of Gymnosperms are familiar to us.
  • Information about bacteria, viruses, protists and archaea is just fragmentary.

However, new species are being discovered faster than ever before due to the efforts of projects like Global Biodiversity.

Leave a Comment

error: Content is protected !!