Maharashtra Class 12th Biology 2023 : Important Long Answer Questions With Answers

Maharashtra Class 12th Biology 2023 : Important Long Answer Questions With Answers

Maharashtra Board Biology Class 12th Practice Questions are available to download here. In which Long Type Questions are given with fully detailed solutions.

Maharashtra Board 12th Class Practice Question Papers are very helpful to get an overview of the real exam question papers. The available practice question papers can be very useful for the students to prepare for the exam.

Chapter 1 : Reproduction in Lower and Higher Plants

Q.1. Explain the stages involved in the maturation of microspore into male gametophyte.

OR

Describe the development of male gametophyte before pollination in angiosperms.

Ans. (1) Microspore or pollen grain is first cell of male gametophyte.

(2) The protoplast of pollen grain divides mitotically to form two unequal cells – a small thin walled generative cell and a large naked vegetative or tube cell.

(3) The generative cell possesses thin cytoplasm and a nucleus. It separates and floats in the cytoplasm of vegetative cell.

(4) The vegetative cell, possesses thick cytoplasm, irregular shaped nucleus and the reserved food.

(5) In majority of the angiosperms, the pollen grains are liberated at two-celled stage after the dehiscence of the anther.

(6) The generative cell of the pollen grain divides by mitosis to form two male non-motile gametes.

Q.2 Describe outbreeding devices which encourage cross pollination.

Ans. (1) Unisexuality, dichogamy, prepotency, self sterility, heteromorphy and herkogamy are the outbreeding devices.

(2) Unisexuality : The plants bear either male or female flowers. Due to unisexual nature, self-pollination is avoided. Plants are either dioecious, e.g. Papaya or monoecious, e.g. maize.

(3) Heteromorphy : In same plants different types of flowers are produced. In these flowers, stigmas and anthers are situated at different levels. There is heterostyly and heteroanthy. This prevents self-pollination e.g. Primrose.

(4) Herkogamy : In bisexual flowers we may come across mechanical device to prevent self-pollination. Natural physical barrier avoids contact of pollens with stigma. e.g. Calotropis where pollinia are situated below the stigma.

(5) Prepotency : In prepotency pollens of other flower germinate rapidly rather than from the same flower. E.g. Apples.

(6) Self sterility or Self incompatibility : It is agenetic mechanism in which pollen germination is inhibited on the stigma of the same flower. E.g. Tobacco, Thea.

Chapter 2 : Reproduction in Lower and Higher Animals

Q.1 (a) Sketch and label female reproductive system of human and answer the following questions :

Ans.

Fig. Human Female Reproductive System

(b) Name the hormones that regulate ovarian function.

Ans. FSH and LH.

(c) What are the three layers of uterus?

Ans. Perimetrium, Myometrium, Endometrium.

(d) Enlist the parts of external genitalia.

Ans. Labia majora, labia minora, mons veneris, clitoris and vestibule.

Q.2. Explain the histological structure of testis. :

Ans : Histological structure of testis :

(1) The external covering of testis is a fibrous connective tissue called tunica albuginea.

(2) Then there is an incomplete peritoneal covering called tunica vaginalis.

(3) Interior to this there is a covering called tunica vascularis formed by capillaries.

(4) The testis is composed of many seminiferous tubules that are lined by cuboidal germinal epithelial cells.

(5) In the seminiferous tubules various stages of developing sperms are seen as spermatogenesis takes place here. These stages are spermatogonia, primary and secondary spermatocytes, spermatids and sperms.

(6) Large, pyramidal subtentacular cells, nurse cells or Sertoli cells are present between germinal epithelium. Sperm bundles remain attached to Sertoli cells with their heads.

(7) Seminiferous tubules form sperms whereas Sertoli cells provide nourishment to the sperms till maturation.

(8) In between the seminiferous tubules there are interstitial cells of Leydig which are endocrine in nature. They secrete testosterone.

Chapter 3 : Inheritance and Variation

Q.1 What is dihybrid cross ? Explain with suitable example and checker board method.

Ans. (1) A cross which involves two pairs of alleles is called a dihybrid cross. A phenotypic ratio of 9 : 3 : 3 : 1 obtained in the F₂ generation of a dihybrid cross is called a dihybrid ratio.

(2) Thus for example, when we cross a true breeding pea plant bearing round and yellow seeds with a true breeding pea plant bearing wrinkled and green seeds we get pea plants bearing round and yellow seeds in the F₁ generation.

(3) When F₁ plants are selfed, we get a ratio of 9 : 3 : 3 : 1 in the F2 generation, where 9 plants bear yellow round seeds, 3 plants bear yellow wrinkled seeds, 3 plants bear green round seeds and 1 plant bears green wrinkled seeds.

(4) Parents (P₁) : RRYY × rryy

Gametes of P₁ RY and ry

F1 generation : RrYy (Yellow round)

On selfing F₁ : RrYy  × RrYy

Gametes of F₁ : RY, Ry, rY, ry

P₂ generation :

Round Yellow : 9 Round green : 3 Wrinkled yellow : 3 Wrinkled green : 1

Phenotypic ratio : 9 : 3 : 3 : 1

Genotypic ratio : 1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1.

Q.2 Explain with suitable diagram how test cross is used to find out genotype of dominant plant.

Ans. Test cross is used to find out the exact genotype by crossing the F₁ individual with homozygous recessive one.

E.g. To find out the genotype of unknown violet flower obtained in

F₁ generation, one can conduct two crosses as follows :

I. Unknown flower considering as RR (homozygous dominant)

RR × rr (Homozygous recessive)

In such cross, all the flowers will be violet.

II. Unknown flower considering as Rr (heterozygous)

Rr × rr (Homozygous recessive)

In such cross half the flowers will be violet and half will be white.

Chapter 4 : Molecular Basis of Inheritance

Q.1 (1) Describe Griffith’s transformation experiment.

OR

In the light of Griffith’s experiment, explain the action of two strains of streptococcus pneumoniae and give his conclusion.

Ans. (1) In 1928, Frederick Griffith, carried out experiments with bacterium Streptococcus pneumoniae (which causes pneumonia in humans and other mammals).

(2) Griffith used two strains of Streptococcus pneumoniae :

(a) S-type (Virulent, smooth, pathogenic and encapsulated).

(b) R-type (Non-virulent, rough, non-pathogenic and non-capsulated).

(3) Experiments carried out by F. Griffith :

(a) Mice were injected with R-strain bacteria and they survived (no pneumonia).
(b) Mice injected with S-strain bacteria developed pneumonia and died.
(c) When heat-killed S-strain bacteria were injected in mice, the mice survived.
(d) On injecting a mixture of heat-killed S-bacteria and live R-bacteria, the mice died.

(4) Griffith obtained live S-strain bacteria from the blood of the dead mice.

(5) He concluded that the live R-strain bacteria must have picked up something (which he called transforming principle) from the heat killed S bacterium, and got changed into S-type. Transforming principle allowed R-type to synthesize capsule and it became virulent.

(6) Thus, Griffith first demonstrated genetic transformations.

Q.2 Describe Avery, McCarty and MacLeod’s experiments.

Ans. (1) In 1944, Oswald T. Avery, Colin M. MacLeod and Maclyn McCarty proved that the DNA is a genetic material (transforming principle).

(2) They purified DNA, RNA, proteins (enzymes) and other materials from heat killed S-strain cells and mixed them with cells of R-strain bacteria separately to confirm which one could transform living R cells into S cells.

(3) Only DNA was able to transform harmless R-strain into virulent S-strain.

(4) They also demonstrated that proteases, RNases did not affect transformation. Thus it was proved that the transforming substance was neither a protein nor-RNA.

(5) When DNA was digested with DNase, there was no transformation.

(6) These experiments proved that the transformation of Live R-strain bacteria into S-strain type was because of DNA of bacteria of S-strain.

(7) Thus, they proved that the DNA is transforming principle.

Chapter 5 : Origin and Evolution of Life

Q.1 Explain modern Synthetic Theory of Evolution in brief.

Ans. (1) Modern synthetic theory of evolution is the result of modification of Darwinism and theory of mutations by taking into consideration studies of genetics, ecology, anatomy, geography and palaeontology.

(2) Five key factors of modern synthetic theory are gene mutations, mutations in the chromosome structure and number, genetic recombinations, natural selection and reproductive isolation. All these finally contribute in the evolution of new species or process of speciation.

(3) Population or Mendelian population is the small group of ‘interbreeding populations’. For every Mendelian population, there is a gene pool which is constituted by total number of genotypes in it. The genotype of an organism in a population is constant, but the gene pool constantly undergoes change due to different factors such as mutations, recombination, gene flow, genetic drift, etc.

(4) Every gene has two alleles. The proportion of a particular allele in the gene pool, to the total number of alleles at a given locus, is called gene frequency. Thus, any change in the gene frequency in the gene pool affects population.

(5) The five main factors are broadly divided into three main concepts as follows :

(i) Genetic variations caused due to various aspects of mutation, recombination and migration. Such variations cause change in the gene frequency. Gene mutations or point mutation change the phenotype of the organism, leading to variation. Recombination is caused due to crossing over in which new genetic combinations are produced. Sexual reproduction due to fertilization of gametes also cause recombinations. All these lead to variations. Gene flow is movement of genes into or out of the population, either due to migrations or dispersal of gametes. Gene flow, therefore, change the gene frequencies of the population. Genetic drift is a random change which occurs by pure chance. It occurs in small populations but change the gene frequency. Chromosomal aberrations are structural or morphological changes in the chromosomes causing rearrangement of the sequence of genes.

(ii) Natural selection is said to be the main driving force in evolution. It brings about evolutionary changes by selecting favourable gene combinations by differential reproduction of genes. This brings about changes in gene frequency from one generation to next generation.

(iii) Isolation means the separation of the population of a particular species into smaller units which prevents interbreeding between them. This over a long time period leads to speciation or formation of new species.

Q.2 What are different types of chromosomal aberrations?

Ans. Chromosomal aberrations :

(1) The structural, morphological change, which take place in chromosome due to rearrangement, is called chromosomal aberrations.

(2) The aberrations change the sequence of the genes. This causes variations.

(3) Chromosomal aberrations are mainly of following four types :

(i) Deletion : Loss of genes from chromosome.

(ii) Duplication : Genes are repeated or doubled in number on chromosome.

(iii) Inversion : A particular segment of chromosome is broken and gets reattached to the same chromosome in an inverted position due to 180° twist. There is no loss or gain of gene complement of the chromosome.

(iv) Translocation : Transfer or transposition of a part of chromosome or a set of genes to a non-homologous chromosome is called translocation. It is effected naturally by the transposons present in the cell.

Chapter 7 : Plant Growth and Mineral Nutrition

Q.1 What are plant growth regulators? Give the characteristics of plant growth regulators.

Ans. 1. Plant growth regulators : Plant growth regulators or phytohormones or plant hormones as they are also called are organic compounds synthesised by the plants which promote, inhibit or control the growth and other physiological processes.

2. Characteristics of plant growth regulators :

(1) Plant growth regulators are organic compounds other than nutrients.

(2) They are synthesised at the apices of root, stem and leaves from where they are transported to other parts of plants where they produce their effects.

(3) They are required in minute quantities.

(4) A single plant growth regulator can control or regulate the various aspects of growth.

Q.2 Write an account of auxins as growth regulators

Ans. (1) Auxins are plant growth regulators produced naturally by plants.

(2) They are weak organic acids capable of promoting cell elongation during the growth of stem and root.

(3) Auxins are synthesized in shoot and root apices besides young leaf primordia.

(4) Auxins may be natural or synthetic.

(5) Naturally occurring auxins are indole-3-acetic acid (IAA) and its derivatives.

(6) NAA, 2, 4-D and 2,4, 5-T are synthetic auxins.

(7) Auxins in higher concentration promote the growth of stem.

(8) Auxins play an important role in initiation and promotion of cell division.

(9) Auxins help in the formation of adventitious roots from cuttings when applied in lower concentration.

(10) Auxins play an important role in apical dominance.

(11) Auxin prevents abscission by preventing the action of hydrolytic enzymes in abscission layer.

(12) Auxins are used to produce parthenocarpic fruits in plants like orange, apple, tomato and grapes.

Chapter 8 : Respiration and Circulation

Q.1. Observe the table and fill the appropriate blocks and rewrite.

Ans.

Q.2 Answer the following questions :

(a) Describe the structure of lymphocytes and mention its types.

Ans. (1) Lymphocytes are smallest WBC which have large spherical nucleus. They are about 25 to 30% of total WBCs.

(2) Lymphocytes are of two types : B-lymphocytes and T-lymphocytes. This division is according to their function.

(3) B-lymphocytes mature in bone marrow and form antibodies. They are responsible for humoral immunity.

(4) T-lymphocytes mature in thymus and is responsible for cell-mediated immunity. Four different types of T-lymphocytes are :

Helper – T cells, Killer – T cells, Memory – T cells and Suppressor – T cells.

(b) Name the disorder caused due to abnormal and uncontrolled increase in number of WBCs.

Ans. Leukemia.

(c) State the functions of neutrophils.

Ans. Neutrophils perform phagocytosis by amoeboid movements. They destroy pathogens by such phagocytosis.

Chapter 9 : Control and Coordination

Q.1 Describe the structure of T.S. of spinal cord.

Ans. (1) Externally there are three meninges covering spinal cord : Dura mater, arachnoid membrane and pia mater.

(2) Dorsoventrally there are two fissures, the shallow dorsal or posterior fissure and the deeper ventral or anterior fissure.

(3) From dorsal fissure a dorsal septum extends inside.

(4) Neurocoel or central canal is situated in the centre of spinal cord.

(5) The central canal is filled with cerebro-spinal fluid and is lined by cuboidal epithelial cells called layer of ependyma.

(6) There is inner grey and outer white matter in the spinal cord. This grey matter is in the shape of ‘H’ with two dorsolateral horns and two ventro-lateral horns.

(7) Dorsal horns form dorsal roots and ventral horns form ventral roots.

(8) White matter is divided into three columns, viz., the dorsal funiculi, ventral funiculi and lateral funiculi on either side.

(9) Ascending and descending tracts of nerve fibres arise from dorsal and ventral roots of the spinal cord. Ascending tracts are sensory while descending tracts are motor in nature.

Q.2 Answer the following questions :

(a) Name the nerve fibres internally connecting the cerebral hemispheres.

Ans. Corpus callosum.

(b) Name the sulci which divide each cerebral hemisphere into 4 lobes.

Ans. (i) Central sulcus : Between frontal and parietal lobe.

(ii) Parieto-occipital sulcus : Between parietal and occipital lobes.

(iii) Sylvian sulcus : Demarcating temporal lobe from the parietal and frontal lobes.

(c) Describe the various functional areas found in the different lobes of cerebral hemispheres.

Ans. Functional areas of cerebrum :

(1) In frontal lobes, there is motor area which control voluntary motor activities, such as muscular movements.

(2) There is association area for coordination between sensation and movements.

(3) Broca’s area which is motor speech area which translates thoughts into speech is present in cerebrum.

(4) In frontal lobe – there are areas for emotions, intelligence, will power, memory and personality.

(5) Parietal lobes have centre for somasthetic sensations of pain, pressure, temperature and taste.

(6) Temporal lobes contain centres for olfactory and auditory sensations, speech and emotions.

(7) Occipital lobes have visual areas for sense of vision.

Chapter 10 : Human Health and Diseases

Q.1 Describe the different types of immunity.

Ans. There are two basic types of immunity, viz. innate immunity and acquired immunity.

(A) Innate immunity :

(1) Innate immunity is natural, inborn immunity, which helps the body to fight against the invasion of microorganisms.

(2) Innate immunity is non-specific because it does not depend on previous exposure to foreign substances.

(3) Innate immunity mechanisms consist of various types of barriers such as anatomical barriers, physiological barriers, phagocytic barriers and inflammatory barriers. They prevent entry of foreign agents into the body.

(B) Acquired immunity :

(1) The immunity that an individual acquires during his life is called acquired immunity or adaptive immunity or specific immunity. It helps the body to adapt by fighting against specific antigens hence it is called adaptive immunity. Since it is produced specifically against an antigen, it is called specific immunity.

(2) Acquired immunity takes long time for its activation.

(3) This type of immunity is seen only in vertebrates.

(4) Due to acquired immunity, the body is able to defend against any invading foreign agent.

Q.2 What are the various public health measures, which you would suggest as safeguard against infectious diseases?

Ans. Infectious diseases spread through pathogens, therefore, it is an important duty of each person to decrease the risk of infecting our own self or others. This can be achieved by

(1) Washing hands often, especially whenever, we are in contact with food and water. Before and after preparing food, before eating and after using the toilet, hand wash is a must.

(2) Vaccinations : Immunization helps us to protect against contracting many diseases. Therefore, timely vaccination should be taken. Especially at the time of epidemic, one must keep distance from infected area or get vaccinated.

(3) One must be at home if there are signs and symptoms of an infection. By going out, we may infect other healthy persons.

(4) Proper diet and exercise should be followed to improve one’s own immunity.

(5) Hygiene should be utmost in the kitchen and dining area. One must take care while eating uncovered and leftover food.

(6) Bathroom and toilet should be cleaned daily as there can be a high concentration of bacteria or other infectious agents in these areas.

(7) One should have responsible sexual behaviour to avoid sexually transmitted diseases.

(8) Personal items such as toothbrush, comb, towel, undergarments or razor blade should never be shared.

(9) Travelling should be avoided because we may infect other passengers during travel. Moreover, our illness can be aggravated. Some special immunizations are needed during certain travels, such as anti-cholera vaccine while going to Pandharpur during Ashadhi.

Chapter 11 : Enhancement of Food Production

Q.1 (1) Which dairy products are prepared using microorganisms ? How ?

Ans. (1) Dairy products prepared using microorganisms are curds, yoghurt, buttermilk and cheese.

(2) The starter or inoculum used in preparation of dairy products contains millions of lactic acid bacteria (LAB).

(3) Curd is prepared by inoculating milk with Lactobacillus acidophilus. It ferments lactose sugar of milk into lactic acid. Lactic acid causes coagulation and partial digestion of milk protein casein. Thus, milk is changed into curd. It also checks growth of disease causing microbes.

(4) Yoghurt is produced by curdling milk with the help of Streptococcus thermophilus and Lactobacillus bulgaricus.

(5) Buttermilk is the acidulated liquid left after churning of butter from curd, is called buttermilk.

(6) During the preparation of cheese, the milk is coagulated with LAB. The curd formed is filtered and whey is separated. The solid mass is then ripened with growth of mould that develops flavour in it. Characteristic texture, flavour and taste of cheese are developed by different specific microbes.  The ‘Roquefort cheese is ripened by blue-green mold Penicillium roquefortii. Camembert cheese is ripened by blue-green mold P. camembertii. The large holes in Swiss cheese are developed due to production of a large amount of CO2 by a bacterium known as Propionibacterium shermanii.

Q.2 Explain the process of sewage water treatment before it can be discharged into natural bodies. Why is this treatment essential?

Ans. Sewage treatment includes following steps :

(1) Preliminary treatment :

(a) Screening : The larger suspended or floating objects are filtered and removed in screening chambers by passing the sewage through screens or net in the chambers.

(b) Grit Chamber : Filtered sewage is passed into series of grit chambers which contain large stones (pebbles) and brickballast. Coarse particles which settle down by gravity are removed.

(2) Primary treatment (physical treatment) :

(a) The sewage water is pumped into the primary sedimentation tank where 50–70% of the suspended solid or organic matter get sedimented and about 30–40% (in number) of coliform organisms are removed.

(b) The organic matter which is settled down is called primary sludge.

(c) Primary sludge is removed by mechanically operated devices.

(d) Dissolved organic matter and microorganisms in the supernatant (effluent) are then removed by the secondary treatment

Chapter 12 : Biotechnology

Q.1 (1) Enlist different types of restriction enzymes commonly used in r-DNA technology ? Write on their role.

Ans. (1) Different types of restriction enzymes commonly used in r-DNA technology are Alu I, Bam HI, Eco RI, Hind II, Hind III, Pst I, Sal I, Taq I, Mbo II, Hpa I, Bgl I, Not I, Kpn I, etc.

(2) They are the molecular scissors which recognize and cut the phosphodiester back bone of DNA on both strands, at highly specific sequences.

(3) The sites recognized by them are called recognition sequences or recognition sites.

(4) Different restriction enzymes found in different organisms recognize different nucleotide sequences and therefore cut DNA at different sites.

(5) Restriction cutting may result in DNA fragments with blunt ends or cohesive or sticky ends or staggered ends (having short, single stranded projections).

(6) Restriction endonucleases like Bam HI and EcoRI produce fragments with sticky ends.

(7) Restriction endonucleases like Alu I, Hind III produce fragments with blunt ends.

(8) Type I restriction endonucleases function simultaneously as endonuclease and methylase e.g. EcoK.

(9) Type II restriction endonucleases have separate cleaving and methylation activities. They are more stable and are used in r-DNA technology e.g. EcoRI, BgII. They cut DNA at specific sites within the palindrome.

(10) Type III restriction endonucleases cut DNA at specific nonpalindromic sequences e.g. Hpa I, Mbo II.

(11) In bacterial cells, REs destroy various viral DNAs that might enter the cell, thus restricting the potential growth of the virus.

Chapter 13 : Organisms and Population

Q.1 What are the different ways in which organisms adapt to the changes in the environment ?

Ans. To survive and propagate further in any environment, organisms show one of the four possible ways, viz. regulate, conform, migrate and suspend.

(1) Regulate : In this method, organisms maintain homeostasis by physiological and behavioural changes. Due to homeostatic regulation, they can perform thermoregulation orosmoregulation. E.g. All birds and mammals show constant body temperature and osmotic concentration irrespective of external temperature.

(2) Conform : Most of the animals and plants are unable to maintain a constant internal environment. Their body parameters change according to outside environment. E.g. Poikilothermic animals cannot maintain body temperature but they are simple conformers. In few aquatic animals, the osmotic concentration of the body fluids changes according to surrounding osmotic concentration. Few conformers can regulate the parameters in limited range.

(3) Migrate : When organism is unable to cope up with surrounding temperatures, they migrate temporarily from such stressful habitat to a more favourable habitat. After the stressful period is over, they return back. Birds show long-distance migrations during severe winter.

(4) Suspend : Suspending the life activities for particular period is one of the methods to cope up with stressful conditions.Seeds of plants remain dormant over unfavourable period and once favourable conditions are resumed they start growing. This state is called dormancy during which metabolic activities are suspended.  Hibernation and aestivation seen in some animals is also for escaping severe winter or summer respectively. E.g. Polar bear shows hibernation while snails and fish show aestivation. These are also suspension measures.

Q.2 Explain the ‘Competitive Exclusion Principle’ given by Gause.

Ans. Gause’s ‘Competitive Exclusion Principle’ :

(1) This principle states that two closely related species competing for the same resources cannot co-exist indefinitely and the competitively inferior one will be eliminated eventually.

(2) The Gause’s principle may be true if resources are limiting, but not otherwise. More recent studies do not support such gross generalisations about competition. The species during competition also show resource partitioning.

(3) In resource partitioning, the species facing competition might evolve mechanisms that promote co-existence rather than exclusion. If two species compete for the same resource, they could avoid competition by choosing, for instance, different times for feeding or different foraging patterns. E.g. Five closely related species of warblers living on the same tree were able to avoid competition and co-exist due to behavioural differences in their foraging activities. If there are two competing species and one is comparatively superior than the other, then the inferior one remains restricted to smaller geographical area. If this superior species is removed then only the inferior species expands its range.

Chapter 14 : Ecosystems and Energy Flow

Explain the following terms with reference to ecological succession.

(1) Seral stages.

Ans. The developmental stages of the ecological succession are known as seral stages.

(2) Pioneers.

Ans. The organisms belonging to first seral stage in the ecological succession are known as pioneers.

(3) Hydrosere.

Ans. Hydrosere or hydrarch succession is a type of ecological succession which is determined by the amount of water available during succession. Hydrosere occurs when there is abundant water available in the area where organisms reside.

(4) Xerosere.

Ans. Xerosere or xerarch succession is a type of ecological succession which is determined by the amount of water available during succession. Xerosere occurs when there is very little water available in the area where organisms reside. Such succession is observed in desert regions.

Chapter 15 : Biodiversity, Conservation and Environmental Issues

Q.1 Species richness goes on decreasing as we move from equator to pole. Explain.

Ans. (1) In tropical regions, there are lesser climatic changes throughout the year and availability of plenty of sunlight.

(2) Moreover, in tropical areas there are lesser disturbances like periodic glaciations as compared to those seen in the polar regions.

(3) In tropical regions, there is a stability over millions of years which favoured speciation and hence there is more species richness.

(4) Also in tropical regions, there are lesser migrations which reduce gene flow between geographically isolated regions. This too favoured speciation.

(5) There is more availability of intense sunlight, warmer temperatures and higher annual rainfall in tropics. These factors have brought higher species richness in tropics.

(6) Constant climatic conditions and abundance of resources in tropical regions provide more food preferences for animal species. E.g. Fruits are available throughout the year in rain forests, therefore, variety of frugivorous animals are seen here, as compared to the temperate regions. In short, species richness or diversity for plants and animals decreases as we move away from equator to the poles.

Q.2 Give importance of conservation in terms of utilitarian reasons.

Ans. The conservation of biodiversity can be done in utilitarian way or for ethical reasons. Utilitarian reasons are further classified into narrowly utilitarian and broadly utilitarian reasons :

1. Narrowly utilitarian reasons :

(1) Humans always reap material benefits from biodiversity in the form of resources for basic needs such as food, clothes, shelter.

(2) Industrial products like resins, tannins, perfume base, etc. are also obtained through biodiversity resources.

(3) For making ornaments or artefacts for aesthetic purpose, again biodiversity is sacrificed.

(4) Many medicines are also obtained through biodiversity resources which shares 25% of global medicine market.

(5) Around 25000 species are used for traditional medicines by tribal population worldwide.

(6) Bioprospecting which is a systematic search for development of new sources of chemical compounds, genes, microorganisms, macroorganisms and other valuable products from nature which is of economically important species is also due to biodiversity.

2. Broadly utilitarian reasons :

(1) Production of oxygen done by all green plants helps human beings to thrive. Amazon forest alone gives 25% of the oxygen to the entire world.

(2) Insects carry out pollination and seed dispersal.

(3) If insects do not carry out pollination and seed dispersal, man would go hungry without crops and fruits.

(4) Biodiversity also is useful in recreation of human beings.

3. Taking all these aspects in consideration, conservation of biodiversity becomes essential. Therefore, to protect and conserve our rich biodiversity on the planet, we have to remember all the utilitarian reasons.

Maharashtra Board Class 12 Study Material

Maharashtra Board Class 12 Study Material
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Model Paper	Revision Notes
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