CBSE Class 12 Biology Revision Notes Chapter 12: Ecosystem

An ecosystem is a functional unit in which living organisms interact with one another and with their physical environment. Its major processes include productivity, decomposition and the transfer of energy through different trophic levels.

Ecosystems differ greatly in size. A small pond, a grassland, a forest and a large sea can each function as an ecosystem. Together, all ecosystems form the biosphere.

These CBSE Class 12 Biology Revision Notes Chapter 12 follow the current 2026–27 chapter sequence. They explain ecosystem components, productivity, decomposition, food chains, energy transfer and ecological pyramids through structured notes and comparison tables.

Key Takeaways

  • 170 billion tonnes: Approximate annual net primary productivity of the biosphere.
  • 2%–10%: Proportion of photosynthetically active radiation captured by plants.
  • 10%: Energy transferred from one trophic level to the next.
  • Always upright: The pyramid of energy cannot be inverted.

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Class 12 Biology revision infographic explaining the energy pyramid, trophic levels and food web in an ecosystem.

Ecosystem Class 12 Notes Overview

The Ecosystem Class 12 Notes explain how biotic and abiotic components combine to form a functional ecological unit.

The chapter also traces how solar energy enters an ecosystem, passes through organisms and is gradually lost as heat.

Chapter Area Main Focus
Structure and function Biotic and abiotic components
Productivity Formation of biomass
Decomposition Breakdown of dead organic matter
Energy flow Food chains, trophic levels and energy transfer
Ecological pyramids Number, biomass and energy relationships

Access Class 12 Biology Chapter 12 Ecosystem Notes in 30 Minutes

Use this sequence for rapid revision:

  1. Learn the definition and types of ecosystems.
  2. Revise biotic and abiotic components.
  3. Understand species composition and stratification.
  4. Learn the structure of a pond ecosystem.
  5. Compare GPP, NPP and secondary productivity.
  6. Memorise the five steps of decomposition.
  7. Compare grazing and detritus food chains.
  8. Learn trophic levels and standing crop.
  9. Revise the 10 percent law.
  10. Compare the three ecological pyramids and their limitations.

Ecosystem Structure and Function Class 12 Notes

An ecosystem is a structural and functional unit of nature.

Within an ecosystem:

  • Living organisms interact with one another.
  • Organisms interact with physical conditions.
  • Energy moves through food relationships.
  • Dead matter is decomposed.
  • Nutrients become available for reuse.

Types of Ecosystems

Ecosystems may be natural or human-made.

Ecosystem Type Examples
Terrestrial ecosystem Forest, grassland, desert
Aquatic ecosystem Pond, lake, wetland, river, estuary
Human-made ecosystem Crop field, aquarium

The biosphere is considered a global ecosystem because it includes all local ecosystems on Earth.

Components of an Ecosystem

The components of an ecosystem are broadly divided into biotic and abiotic components.

Abiotic Components

Abiotic components are the non-living physical and chemical parts of an ecosystem.

They include:

  • Water
  • Air
  • Soil
  • Sunlight
  • Temperature
  • Minerals
  • Organic substances
  • Inorganic nutrients
  • Climatic conditions

Biotic Components

Biotic components are the living organisms in an ecosystem.

They include:

  • Producers
  • Consumers
  • Decomposers

Producers

Producers are autotrophic organisms that manufacture organic food from inorganic substances.

Examples include:

  • Green plants
  • Algae
  • Phytoplankton
  • Photosynthetic bacteria

Consumers

Consumers are heterotrophic organisms that obtain food from producers or other consumers.

They include:

  • Herbivores
  • Carnivores
  • Omnivores

Decomposers

Decomposers break down dead organic matter into simpler inorganic substances.

The major decomposers are:

  • Bacteria
  • Fungi

Species Composition and Stratification

The interaction of biotic and abiotic components produces a characteristic physical structure in each ecosystem.

Species Composition

Species composition refers to the identification and number of plant and animal species present in an ecosystem.

Different ecosystems have different species compositions because their environmental conditions differ.

Stratification

Stratification is the vertical distribution of species at different levels in an ecosystem.

In a forest:

  • Trees occupy the upper layer.
  • Shrubs occupy the middle layer.
  • Herbs and grasses occupy the lower layer.

Stratification allows species to use available light, space and other resources at different levels.

Four Major Functions of an Ecosystem

An ecosystem functions as a unit through four major processes:

  1. Productivity
  2. Decomposition
  3. Energy flow
  4. Nutrient cycling

The current chapter studies productivity, decomposition and energy flow in detail before explaining ecological pyramids.

Pond Ecosystem Structure and Function

A pond ecosystem is a simple and largely self-sustaining aquatic ecosystem.

It contains all the major structural components and ecological functions found in larger ecosystems.

Abiotic Components of a Pond

The abiotic components include:

  • Water
  • Dissolved organic substances
  • Dissolved inorganic substances
  • Nutrient-rich soil at the bottom
  • Solar energy
  • Temperature
  • Day length
  • Climatic conditions

These factors regulate the functioning of the pond.

Producers in a Pond Ecosystem

The autotrophic components include:

  • Phytoplankton
  • Algae
  • Floating plants
  • Submerged plants
  • Marginal plants

They convert inorganic substances into organic matter using solar energy.

Consumers in a Pond Ecosystem

Consumers include:

  • Zooplankton
  • Free-swimming organisms
  • Bottom-dwelling animals

They obtain food directly or indirectly from producers.

Decomposers in a Pond Ecosystem

Decomposers are especially abundant near the bottom of the pond.

They include:

  • Fungi
  • Bacteria
  • Flagellates

They break down dead organic matter and release nutrients for reuse by producers.

Functions Performed in a Pond

A pond ecosystem performs the following functions:

  • Autotrophs convert inorganic matter into organic food.
  • Consumers feed on producers and other consumers.
  • Decomposers break down dead material.
  • Minerals return to the environment.
  • Energy moves towards higher trophic levels.
  • Energy is gradually lost as heat.

Productivity in Ecosystem Class 12 Notes

A constant input of solar energy is necessary for an ecosystem to function.

Productivity in ecosystem refers to the rate at which biomass or organic matter is produced.

Productivity is generally expressed as:

  • g m⁻² year⁻¹
  • kcal m⁻² year⁻¹

Primary Productivity in an Ecosystem

Primary productivity is the rate at which producers form biomass through photosynthesis.

Primary production refers to the amount of organic matter produced per unit area during a given period.

It may be expressed as:

  • Weight per unit area
  • Energy per unit area

Primary productivity depends on:

  • Plant species
  • Availability of nutrients
  • Environmental conditions
  • Photosynthetic capacity
  • Temperature
  • Water availability
  • Light availability

Gross Primary Productivity and Net Primary Productivity

Primary productivity is divided into gross primary productivity and net primary productivity.

Gross Primary Productivity

Gross primary productivity, or GPP, is the total rate of organic matter production through photosynthesis.

A portion of this organic matter is used by plants during respiration.

Net Primary Productivity

Net primary productivity, or NPP, is the biomass left after subtracting respiratory losses from GPP.

The equation is:

GPP − R = NPP

Where:

GPP = Gross primary productivity
R = Respiration losses
NPP = Net primary productivity

NPP represents the biomass available to:

  • Herbivores
  • Other consumers
  • Decomposers

GPP and NPP Compared

Feature Gross Primary Productivity Net Primary Productivity
Meaning Total organic matter produced Organic matter remaining after respiration
Includes respiratory use Yes No
Equation Total photosynthetic production GPP − R
Ecological significance Total energy fixed Biomass available to heterotrophs

Secondary Productivity

Secondary productivity is the rate at which consumers form new organic matter.

Consumers cannot manufacture food from inorganic material. They convert the food they consume into new body tissue.

Secondary productivity therefore depends on the organic matter available from producers or other consumers.

Productivity of the Biosphere

The annual net primary productivity of the entire biosphere is approximately 170 billion tonnes of dry organic matter.

Oceans cover about 70% of Earth’s surface but contribute only about 55 billion tonnes.

The remaining net primary productivity occurs on land.

The productivity of oceans is relatively low in many regions because nutrient availability limits phytoplankton growth.

Decomposition in Ecosystem Class 12 Notes

Decomposition in ecosystem is the breakdown of complex organic matter into simpler inorganic substances.

The products include:

  • Carbon dioxide
  • Water
  • Inorganic nutrients

Decomposers play the main role in this process.

What Is Detritus?

Detritus is the raw material for decomposition.

It includes:

  • Fallen leaves
  • Bark
  • Flowers
  • Dead plant parts
  • Dead animal remains
  • Faecal matter

Steps of Decomposition

The major steps of decomposition are:

  1. Fragmentation
  2. Leaching
  3. Catabolism
  4. Humification
  5. Mineralisation

These processes may operate simultaneously on detritus.

Fragmentation and Leaching

Fragmentation and leaching begin the breakdown and movement of material through the soil.

Fragmentation

Fragmentation is the breakdown of detritus into smaller particles.

Detritivores such as earthworms carry out fragmentation.

Smaller particles provide a larger surface area for microbial action.

Leaching

During leaching, water-soluble inorganic nutrients move downward through the soil.

These nutrients may precipitate as unavailable salts in deeper soil layers.

Catabolism in Decomposition

Catabolism is the microbial breakdown of detritus into simpler inorganic substances.

Bacteria and fungi secrete enzymes that degrade complex organic compounds.

This process releases substances that can eventually re-enter nutrient cycles.

Humification and Mineralisation

Humification and mineralisation occur during the decomposition of organic matter in soil.

Humification

Humification leads to the formation of humus.

Humus is:

  • Dark in colour
  • Amorphous
  • Colloidal
  • Resistant to microbial action
  • Decomposed very slowly
  • A reservoir of nutrients

Mineralisation

Mineralisation is the microbial breakdown of humus and other organic material.

It releases inorganic nutrients into the soil.

These nutrients become available for absorption by plants.

Humification and Mineralisation Compared

Feature Humification Mineralisation
Main result Formation of humus Release of inorganic nutrients
Rate Very slow Depends on microbial activity
Ecological role Stores nutrients Makes nutrients available to plants
Product Dark, resistant organic material Inorganic mineral substances

Factors Affecting Decomposition

The rate of decomposition depends on:

  • Chemical composition of detritus
  • Temperature
  • Soil moisture
  • Oxygen availability
  • Microbial activity

Chemical Composition of Detritus

Decomposition is slower when detritus is rich in:

  • Lignin
  • Chitin

Decomposition is faster when detritus is rich in:

  • Nitrogen
  • Sugars
  • Water-soluble substances

Climatic Conditions

Warm and moist conditions favour decomposition because they support microbial activity.

Decomposition slows under:

  • Low temperature
  • Dry conditions
  • Anaerobic conditions

Anaerobiosis inhibits decomposition and causes organic matter to accumulate.

Energy Flow in Ecosystem Class 12 Notes

The energy flow in ecosystem begins with solar energy.

Except for deep-sea hydrothermal ecosystems, the Sun is the main source of energy for ecosystems.

Less than 50% of incident solar radiation is photosynthetically active radiation, or PAR.

Plants capture only about 2% to 10% of PAR.

This small proportion sustains most life on Earth.

Unidirectional Flow of Energy

Energy moves in one direction:

Sun → Producers → Consumers → Decomposers

Energy does not return to the Sun or move back to lower trophic levels.

At every transfer:

  • Some energy supports biological activity.
  • Some energy is lost as heat.
  • Less energy reaches the next trophic level.

Ecosystems therefore require a continuous input of solar energy.

Producers and Consumers in Energy Flow

Green plants are producers because they trap solar energy and form organic matter.

Terrestrial Producers

Major terrestrial producers include:

  • Herbaceous plants
  • Trees
  • Shrubs

Aquatic Producers

Major aquatic producers include:

  • Phytoplankton
  • Algae
  • Aquatic plants

Types of Consumers

Consumer Type Food Source Trophic Position
Primary consumer Producers Second trophic level
Secondary consumer Primary consumers Third trophic level
Tertiary consumer Secondary consumers Higher trophic level

Herbivores are primary consumers.

Carnivores feeding on herbivores are secondary consumers.

Carnivores feeding on secondary consumers may be called tertiary consumers.

Grazing Food Chain in an Ecosystem

A grazing food chain begins with a living green plant.

A simple example is:

Grass → Goat → Human

Here:

  • Grass is the producer.
  • Goat is the primary consumer.
  • Human is the secondary consumer.

In aquatic ecosystems, the grazing food chain is often the main pathway of energy transfer.

Detritus Food Chain in an Ecosystem

A detritus food chain begins with dead organic matter.

Its main organisms are decomposers such as:

  • Bacteria
  • Fungi

Decomposers secrete digestive enzymes onto detritus.

They convert dead and waste matter into simple inorganic substances and absorb the products.

In terrestrial ecosystems, a large proportion of energy passes through the detritus food chain.

Grazing and Detritus Food Chains Compared

Feature Grazing Food Chain Detritus Food Chain
Starting point Living producers Dead organic matter
First consumers Herbivores Decomposers and detritivores
Major organisms Plants and animals Bacteria, fungi and detritivores
Major role Direct transfer from producers Recycling of dead biomass
More prominent in Aquatic ecosystems Terrestrial ecosystems

Food Chains and Food Webs

Food chains in natural ecosystems are interconnected.

A detritus food chain may connect with a grazing food chain because:

  • Detritivores may be eaten by other animals.
  • Omnivores may feed through more than one pathway.
  • Some consumers use several food sources.

These connections form a food web.

A food web increases ecosystem stability by providing organisms with alternative food sources.

Trophic Levels in an Ecosystem

A trophic level is the functional position occupied by an organism in a food chain.

The position depends on its source of nutrition.

Trophic Level Organisms
First trophic level Producers
Second trophic level Herbivores
Third trophic level Primary carnivores
Fourth trophic level Secondary carnivores

A species may occupy more than one trophic level.

For example, a sparrow is:

  • A primary consumer when it eats seeds
  • A secondary consumer when it eats insects

Trophic level represents a feeding position, rather than a fixed species category.

Standing Crop in an Ecosystem

The standing crop is the amount of living material present at a trophic level at a particular time.

It may be measured as:

  • Biomass
  • Number of organisms per unit area

Biomass may be expressed as:

  • Fresh weight
  • Dry weight

Dry weight gives a more accurate estimate because water content varies between organisms.

10 Percent Law of Energy Transfer

The 10 percent law states that only about 10% of the energy available at one trophic level is transferred to the next.

For example:

Trophic Level Available Energy
Producers 10,000 units
Primary consumers 1,000 units
Secondary consumers 100 units
Tertiary consumers 10 units

Most energy is lost through:

  • Respiration
  • Movement
  • Metabolism
  • Heat
  • Waste material

This energy loss limits the number of trophic levels in a grazing food chain.

Ecological Pyramids Class 12 Notes

Ecological pyramids represent relationships between organisms at different trophic levels.

The base represents producers, while the apex represents top-level consumers.

Ecological pyramids may represent:

  • Number
  • Biomass
  • Energy

Pyramid of Numbers

A pyramid of numbers represents the number of organisms present at each trophic level.

Upright Pyramid of Numbers

An upright pyramid occurs when producers are more numerous than consumers.

A grassland ecosystem may contain:

  • Millions of grasses
  • Fewer herbivores
  • Still fewer carnivores

Inverted Pyramid of Numbers

An inverted pyramid may occur in a tree ecosystem.

One large tree may support:

  • Many insects
  • Several small birds
  • Fewer larger birds

The number of producers is lower than the number of primary consumers.

Pyramid of Biomass

A pyramid of biomass represents the mass of living material at each trophic level.

Upright Pyramid of Biomass

The biomass pyramid is usually upright in terrestrial ecosystems.

Producers have the greatest biomass, followed by herbivores and carnivores.

Inverted Pyramid of Biomass

The pyramid of biomass may be inverted in aquatic ecosystems.

The standing biomass of phytoplankton may be lower than the biomass of zooplankton and fishes.

Phytoplankton reproduce and are consumed rapidly, allowing a small standing crop to support a larger consumer biomass.

Pyramid of Energy

A pyramid of energy represents the amount of energy available at each trophic level over a given period.

The energy pyramid is always upright.

Energy decreases at every successive trophic level because some energy is lost as heat during each transfer.

A higher trophic level can never contain more usable energy than the level below it.

Ecological Pyramids Compared

Pyramid Represents Can Be Inverted?
Pyramid of numbers Number of organisms Yes
Pyramid of biomass Mass of living material Yes
Pyramid of energy Energy available No

Limitations of Ecological Pyramids

Ecological pyramids simplify complex feeding relationships.

Their major limitations are:

  • They assume a simple food chain.
  • They do not represent food webs.
  • They do not account properly for species occupying more than one trophic level.
  • They do not give decomposers a separate position.
  • They may oversimplify the role of omnivores.
  • They cannot represent all seasonal changes in an ecosystem.

Ecosystem Class 12 Quick Revision Tables

The following tables summarise the key concepts from the Class 12 Biology Chapter 12 Notes.

Ecosystem Components

Component Examples Main Role
Abiotic Water, light, soil, minerals Physical and chemical environment
Producers Plants, algae, phytoplankton Form organic matter
Consumers Herbivores, carnivores Transfer energy
Decomposers Bacteria, fungi Break down dead matter

Productivity Terms

Term Meaning
Primary production Biomass formed by producers
Productivity Rate of biomass formation
GPP Total photosynthetic production
NPP GPP remaining after respiration
Secondary productivity New organic matter formed by consumers

Steps of Decomposition

Step Main Process
Fragmentation Detritus broken into smaller particles
Leaching Soluble nutrients move into deeper soil
Catabolism Microbial enzymes break down detritus
Humification Humus is formed
Mineralisation Inorganic nutrients are released

Food Chain Terms

Term Meaning
Grazing food chain Begins with living plants
Detritus food chain Begins with dead organic matter
Food web Interconnected food chains
Trophic level Feeding position in a food chain
Standing crop Living material at a trophic level
10 percent law About 10% energy reaches the next level

Important Ecosystem Figures

Figure Meaning
Less than 50% Solar radiation that is PAR
2%–10% PAR captured by plants
170 billion tonnes Annual biosphere NPP
55 billion tonnes Approximate ocean NPP
70% Earth’s surface covered by oceans
10% Energy transferred to the next trophic level

Useful Links for Class 12 Biology

Section Useful Links
Syllabus CBSE Class 12 Biology Syllabus
Revision Notes CBSE Class 12 Biology Revision Notes
Biology Notes CBSE Class 12 Biology Revision Notes Chapter 1
NCERT Solutions NCERT Solutions for Class 12 Biology
Sample Papers CBSE Sample Papers for Class 12 Biology
Important Questions Important Questions Class 12 Biology
NCERT Books NCERT Books for Class 12 Biology
Class 12 Support CBSE Class 12 Syllabus

Q.1 Give three advantages of using genetically engineered human insulin?

Ans

Advantages of using genetically engineered human insulin are:

1. It causes very less or no allergic reactions.

2. It is produced in larger quantity.

3. Strict vegetarians can use it because no animals are involved in its production.

Q.2 Why is Bt-toxin considered more ecofriendly than the conventional chemical pesticides?

Ans

Unlike chemical insecticides which are generalised and nonspecific in their effects, Bt-toxin is very selective in nature and kills only the target pest species. Non-target species remain unharmed, thus the environmental balance is maintained.

Q.3 In Gujarat some strains of pink bollworm have shown resistance against Bt-cotton. What is the reason behind this resistance development?

Ans

The resistance was developed due to mutation in a gene encoding a Cadherin receptor which binds “Cry” protein.

Q.4 What is ‘Dicer’?

Ans

Dicer is an endoribonuclease that cleaves double-stranded RNA (dsRNA) into short double-stranded RNA fragments called small interfering RNA (siRNA) which are 20-25 nucleotides long.

Q.5 What is ‘Transposon’?

Ans

Transposons are sequences of DNA that can move around to different positions within the genome of a single cell, by the process called transposition. Transposons were also once called jumping genes.

Q.6 Why RNAi is considered as an important technique of Biotechnology?

Ans

RNAi is considered as an important technique of Biotechnology because:

1. This technique can be used to trace the function of a gene.

2. It can be used to discover new drugs which can suppress or enhance the expression of a gene or set of genes.

Q.7 During the process of downstream processing, how are the two chains (A and B) linked together to form activated enzyme.

Ans

Downstream processing of biosynthetic insulin includes the extraction and combination of its two separate chains through disulphide linkage. Disulphide bonds are generated through thiol-disulphide exchange reactions between the “cystein” moieties of the two peptides.

Q.8 Write a short note on genetically engineered insulin and its production by Eli Lilly.

Ans

– Insulin is an important life saving drug for diabetic patients.
– Insulin consists of two short polypeptide chains: chain A and chain B, that are linked together by disulphide bridges.
– In mammals, including humans, insulin is synthesised as a prohormone which contains an extra stretch called the C peptide.
– This C peptide is not present in the mature insulin and is removed during its maturation.

– The main challenge for production of insulin using rDNA techniques was getting insulin assembled into a mature form.

– In 1983, Eli Lilly, an American company prepared two DNA sequences corresponding to A and B, chains of human insulin and introduced them in plasmids of E. coli to produce insulin chains.

– Chains A and B were produced separately, extracted and combined by creating disulfide bonds to form human insulin.

Q.9 Why does insecticidal “Cry proteins” not affect mammals?

Ans

Cry proteins are absorbed into epithelial cells of insect midgut through Cadherin receptors expressed on their surface. Mammals do not have such receptors hence they are not affected by “Cry proteins”.

Q.10 What are the critical research areas in biotechnology?

Ans

Three critical research areas of biotechnology are:
(i) Providing the best catalyst in the form of improved organism usually a microbe or pure enzyme.
(ii) Creating optimal conditions through engineering for a catalyst to act.
(iii) Downstream processing technologies to purify the protein/organic compound.

Q.11 What are the key applications of biotechnology?

Ans

The key applications of biotechnology include therapeutics, diagnostics, genetically modified crops for agriculture, processed food, bioremediation, waste treatment, and energy production.

Q.12 How can biotechnology benefit agriculture?

Ans

Biotechnology increases the food production by

(i) organic agriculture; and

(ii) genetically engineered crop-based agriculture.

Q.13 Expand and explain GMO.

Ans

Plants, bacteria, fungi and animals whose genes have been altered deliberately for human welfare and development are called Genetically Modified Organisms (GMO).

Q.14 How are genetic modifications beneficial in plants?

Ans

Genetic modifications in plants are beneficial in many ways:
(i) make crops more tolerant to biotic stresses (cold, drought, salt, heat).
(ii) reduced reliance on chemical pesticides (pest-resistant crops).
(iii) help to reduce post harvest losses.
(iv) increase efficiency of mineral usage by plants (this prevents early exhaustion of fertility of soil).
(v) enhanced nutritional value of food, e.g., Vitamin ‘A’ enriched rice.

Q.15 What is Bt toxin?

Ans

Bt toxin is produced by a bacterium called Bacillus thuringiensis. Bt toxin gene has been cloned from the bacteria and has been incorporated in crop plants to provide resistance to their parasitic insects without the need of insecticides.

Q.16 Write a short note on Bt Cotton.

Ans

Bt Cotton is considered as a genetically modified strains of cotton plant which produces proteins that kill certain insects such as lepidopterans (tobacco budworm, armyworm), coleopterans (beetles) and dipterans (flies, mosquitoes).

Q.17 Explain the mode of action of the Bt toxin. Why does it not kill the Bacillus?

Ans

B. thuringiensis forms protein crystals during the spore forming phase of their growth. This crystal contains toxic insecticidal protein.These protein molecules exist as inactive protoxins but when an insect ingest the inactive toxin, it is converted into an active form of toxin due to the alkaline pH of the insect gut through solubilisation of the crystals. The activated toxin binds to the surface of midgut epithelial cells and create pores that cause cell swelling and lysis, eventually causing insect death.

Q.18 What does the gene named ‘cry’ code for?

Ans

The gene ‘cry’ code for the crystalline proteinaceous toxin called Bt-toxin in Bacillus thuringiensis.

Q.19 What does RNAi stands for? Explain RNAi technology using an example.

Ans

RNAi stands for RNA interference.
RNAi takes place in all eukaryotic organisms as a method of cellular defense. This method involves silencing of a specific mRNA due to a complementary dsRNA molecule that binds to and prevents translation of the mRNA (silencing). The source of this complementary RNA could be from an infection by viruses having RNA genomes or mobile genetic elements (transposons) that replicate via an RNA intermediate.
Example : Using Agrobacterium vectors, nematode-specific genes were introduced into the host plant. The introduction of DNA was such that it produced both senseand anti-sense RNA in the host cells. These two RNA’s being complementary toeach other formed a double stranded (dsRNA) that initiated RNAi and thus, silenced the specific mRNAof the nematode. The consequence was that the parasitecould not survive in a transgenic host expressing specific interfering RNA. Thetransgenic plant, therefore, got itself protected from the parasite.

Q.20 What are the various applications of biotechnology in medicine?

Ans

Biotechnology has widespread applications in the field of medicine:
Many pharmaceutical products have been produced by using recombinant DNA technology. Monoclonal antibodies, vaccines, antibiotics, genetically engineered insulin, steroids are produced with the help of biotechnology.

Q.21 A person is born with a genetic disease. Can he be cured? Explain how?

Ans

Person can be cured with the help of Gene Therapy.
– In Gene therapy, genes are inserted into a person’s cells and tissues to treat a disease.
– Correction of a genetic defect involves delivery of a normal gene into the individual or embryo to take over the function of and compensate for the non- functional gene.

Q.22 What is Gene Therapy?

Ans

Gene therapy is a collection of methods which allows correction of a genetic defect through various techniques of recombinant DNA technology.

Q.23 Expand ADA. What are the therapies available for it? What role can gene therapy play in curing ADA?

Ans

– ADA stands for Adenosine deaminase deficiency.
– The therapies available for curing ADA is Gene therapy, Enzyme replacement therapy and Bone marrow transplantation.

Role of gene therapy in curing ADA:

– The first clinical gene therapy was given in 1990 to a 4-year old girl with Adenosine deaminase (ADA) deficiency.
– The lymphocytes from the blood of the patient are grown in a culture outside the body.
– A functional ADA cDNA (using a retroviral vector) is then introduced into these lymphocytes, which are subsequently returned to the patient.

– As these cells are not immortal, the patient requires periodic infusion of such genetically engineered lymphocytes.

– If the gene isolate from marrow cells producing ADA is introduced into cells at early embryonic stages, it could be a permanent cure.

Q.24 How does biotechnology serve the purpose of early detection of diseases?

Ans

Recombinant DNA technology, Polymerase Chain Reaction (PCR) and Enzyme Linked Immuno-Sorbent Assay (ELISA) are some of the techniques of biotechnology that serve the purpose of early disease detection.

Q.25 Expand and describe ELISA.

Ans

-ELISA stands for Enzyme-Linked Immuno Sorbent Assay.
– ELISA, is a biochemical technique used mainly in immunology to detect the presence of an antibody or an antigen in a sample. The ELISA has been used as a diagnostic tool in medicine and plant pathology, as well as a quality control check in various industries.

Q.26 How does PCR help in diagnosis of diseases?

Ans

– PCR stands for Polymerase Chain Reaction. It is used for amplification of DNA.
– It is a powerful techqnique to identify many genetic disorders like gene mutation in cancer etc.
– A very low concentration of a bacteria or virus ( before the development of symptoms ) can be detected by amplification of their nucleic acid by PCR.

Q.27 What are transgenic animals? What are their applications?

Ans

Animals that have had their DNA manipulated to possess and express an extra (foreign) gene are known as transgenic animals.

Applications of transgenic animals:

  • Transgenic animals can be specifically designed to allow the study of genes regulation, and their affect on the normal functions of the body and its development, e.g., study of complex factors involved in growth such as insulin-like growth factor.
  • Many transgenic animals are designed to show genes contribution in the development of disease.
  • Transgenic animals also produce useful biological products that can be created by the introduction of the portion of DNA (or genes) which codes for a particular product such as human protein (α-1-antitrypsin) used to treat emphysema.
  • Transgenic mice are being developed for use in testing the safety of vaccines before they are used on humans.
  • Transgenic animals are made that carry genes which make them more sensitive to toxic substances than non-transgenic animals which help in testing the toxicity of drugs in shorter period of time.

Q.28 Expand GEAC. What does it do?

Ans

– GEAC stands for Genetic Engineering Approval Committee.
– It is an Indian government organization which will make decisions regarding the validity of GM research and the safety of introducing GM-organisms for public services.

Q.29 Write a short note on the conflict over patent of Basmati rice.

Ans

In 1997, an American company got patent rights on Basmati rice through the US Patent and Trademark Office. This allowed the company to sell a ‘new’ variety of Basmati, in US and abroad. This ‘new’ variety of Basmati had actually been derived from Indian farmer’s varieties. Indian Basmati was crossed with semi- dwarf varieties and claimed as an invention or a novelty. The patent extends to functional equivalents, implying that other people selling basmati rice could be restricted by the patent.

Q.30 What is Biopiracy?

Ans

The use of bio-resources by multinational companies and other organisations with no proper authorisation from the countries and people concerned without compensatory payment, is termed as Biopiracy.

Q.31 Which living system is popularly used for the production of recombinant or modified proteins?

Ans

Bacterial cells are popularly used for the expression of recombinant or modified proteins.

Q.32 Why are bacterial based systems used for producing genetically engineered insulin?

Ans

Bacterial based systems are used for producing genetically engineered insulin because:

1. They have small Generation time.

2. Recombinant protein (i.e., insulin) can be easily extracted.

3. Human body shows very less or no resistance against recombinant proteins produced in bacterial systems.

Q.33 Write a short note on patent issues related to Basmati rice.

Ans

Basmati rice, sought-after for its fragrant taste, was developed by Indian farmers over hundreds of years, but the Texan company RiceTec obtained a patent on September 2, 1997 for a cross-breed with American long-grain rice. This patent was granted by US Patent and Trademarks Office (USPTO). June 2000, Indian government has filed 50,000 pages of scientific evidence to the US Patents and Trademarks Office, insisting that most high quality basmati varieties already possess these characteristics. Claims filed by Ricetec numbered 1 to 3; 5 to 7; 10, 12 to 14 and 18 to 20 of the Basmati Patent have been rejected by the USPTO.

Q.34 Observe the flow chart of a clinical therapy given to a four year old patient for an enzyme.

Try to answer the following questions:

  1. Which clinical therapy is being outlined in the given flowchart?
  2. Identify the disease which can be cured by this treatment.
  3. Does the given treatment completely cure the disease?

Ans

  1. Gene therapy is being outlined in the given flowchart.
  2. Adenosine deaminase (ADA) deficiency disease can be cured by the given therapeutic treatment.
  3. No, the given treatment does not cure the disease completely as genetically engineered lymphocytes are not eternal. So, the patient requires periodic infusion of cells.

The disease can be cured permanently cured if the gene isolated from bone marrow cells producing ADA is introduced into cells at early embryonic stages.

Q.35 Bt crops are genetically modified crops, which are produced by the insertion of bacterial genes into plant cell.

  1. Is it possible to insert bacterial gene into a plant DNA?
  2. Suggest an appropriate answer how is it possible?

Ans

  1. Yes, it is possible to insert bacterial gene into a plant cell by a technique known as genetic engineering.
  2. Genetic engineering makes it possible to transfer cry gene of Bacillus thuringiensis into plant DNA by Agrobacterium mediated transformation. The steps involved are:
  1. Isolation of bacterial DNA
  2. Generation of rDNA by incorporating the target DNA from bacteria
  3. Transfer of rDNA into the host cell
  4. Selection of transformed host cell to produce a large amount of insert

Q.36 A technique ‘X’ has been used in producing transgenic plants resistant to nematodes.

In this technique, nematode specific genes are introduced into the host plant in such a manner that it produces both sense and antisense RNA.

Based on the given information, answer the following questions:

  1. Name the vector used in the above mentioned technique.
  2. How is RNAi initiated?
  3. How does RNAi strategy work?

Ans

  1. Agrobacterium is a vector through which nematode specific genes are introduced into a host plant.
  2. Sense and antisense RNA, being complementary to each other, form double stranded RNA (dsRNA) that initiates RNAi and thus; silences a specific mRNA of a nematode.
  3. RNAi prevents translation of a specific mRNA (silencing) due to a complementary dsRNA molecule.

Q.37 Read the statements from ‘A’ to ‘F’ and fill the correct word to complete them.

  1. The technique by which genetic material of an organism is manipulated for yielding useful products is ______.
  2. Production of insulin includes a technique in which human DNA is inserted into ____ which then grows and produces a synthetic version of human insulin.
  3. The technology which uses biological processes to produced desired product for human welfare is known as ______.
  4. The ____ genes code for Bt cotton.
  5. The _____ is a technique in which multiple copies of nucleotide sequence are synthesised under in-vitro conditions using complementary primers and DNA polymerase enzyme.

Ans

  1. Genetic engineering
  2. E.coli
  3. Biotechnology
  4. Cry
  5. Polymerase chain reaction (PCR)

Q.38 Read the statement given below and answer the following questions:

‘X’ is used to cut DNA at specific loci.

  1. What is ‘X’ in the given statement?
  2. How can we use ‘X’ to detect differences in DNA of different people?

Ans

  1. ‘X’ is a restriction enzyme in the given statement. Restriction enzymes are used to cut DNA at specific loci.
  2. We can use restriction enzymes to detect differences in DNA of different people as given below:
  1. Restriction enzymes are sensitive to changes in DNA sequences.
  2. They recognise only a particular site.
  3. Change of a single base at a restriction site can cause it to be “lost”, i.e., no longer recognised by the enzyme.
  4. It results in varying lengths of DNA fragments in different people having differences in DNA.

Q.39 The given below steps (a to e) are involved in a technique that helps to produce genetically modified products. Fill up the blanks with appropriate and correct answer.

Ans

a) Plasmid

b) Desired gene of interest from source organism

c) Plasmid + gene of interest (modified plasmid)

d) Insertion of modified plasmid into bacterial cell for multiplication and division

e) Bacterial cell

Q.40 The following is a Southern blot of EcoRI-digested DNA of wheat plants from two different inbred lineages, K and Z.

Autoradiogram (I) shows the bands that was developed after probing the blot with 32P-labelled cDNA 1, while autoradiogram (II) shows the same blot after it was exposed to probe and re-probed with 32P-labelled cDNA 2.

  1. Which bands are expected to be visible in the autoradiogram of a similarly probed Southern blot prepared by using EcoRI-digested DNA from F1 hybrid plants yielded by crossing the two inbred lineages?
  2. What can be inferred about the genes depicted by band b1 on blot I of two inbred lineages, K and Z?
  3. Suppose F1 progeny were crossed with plants possessing bands b1, b4, and m3. DDNA was isolated from several individual progenies and digested with EcoRI. Then, the resulting DNA fragments were separated by gel electrophoresis, transferred to a nylon membrane and hybridised with radioactive cDNA 1 and cDNA 2 probes.

The following table summarises the bands present in autoradiograms obtained by using DNA from individual progeny:

Does this data provide evidence for RFLPs?

Ans

  1. b1, b2, b3, b4, m1, m2 and m3 bands are expected to be visible in the autoradiogram of a similarly probed Southern blot prepared by using EcoRI-digested DNA from F1 hybrid plants yielded by crossing the two inbred lineages
  2. Band b1 represents a locus whose DNA is homologous to cDNA 1. Since the marker is not polymorphic in the parents, it cannot be mapped in this cross.
  3. cDNA 1 probe detects one RFLP locus with alleles that are visualised as band b4 and b2/b3. cDNA 2 detects another RFLP locus with alleles that are visualised as band m3 and m1/m2. The two loci are linked with a recombination frequency of 20%.

Q.41 If you have to clone an unidentified human gene, which procedure of biotechnology can help you to place this gene on the cytological map of human genome without carrying out pedigree analysis? Also, design the steps that can be followed for this purpose.

Ans

If a clone of gene is available, fluorescent in situ hybridisation (FISH) can be opted to determine the human chromosome that carries the gene and localise the gene on the chromosome.

Steps involved in this method are:

  1. Couple the single-stranded copies of the clone to a fluorescent probe.
  2. Hybridise it to a denatured DNA in chromosomes.
  3. Remove the free probe by washing.
  4. Locate the fluorescent probe by using a fluorescence microscope

Q.42 As per the recent discovery, CF gene and its product can help in the treatment of cystic fibrosis by the method of somatic-cell gene therapy.

Being a biology student, try to enlist the possible obstacles that scientists may face while carrying out somatic-cell gene therapy to treat this genetic disorder?

Ans

The possible obstacles that scientists may face while carrying out somatic-cell gene therapy to treat this genetic disorder are:

  1. Size of CF gene: Size of CF gene is approximately 250 kb, which can be a matter of concern as it is extremely large to fit in the standard gene transfer vectors.
  2. Insertion of transgene into enough cells: Another major obstacle can be insertion of transgene into enough of the target cells of the patient in order to reduce the symptoms.
  3. Development of expression vector: One more challenge can be the development of an expression vector containing the gene that can result in long-term expression of the introduced gene into the transgenic cells.

Q.43 Currently, a strategy of anti-sense RNA has been used to block the expression of certain genes in both prokaryotic and eukaryotic organisms.

  1. Is this strategy applicable for complete blockage of expression of a gene? Elaborate.
  2. Apply this strategy on vine ripened tomatoes and explicate how can it be helpful to increase their shelf life?
  3. Can this strategy prevent fatal genetic disorders? Apply your knowledge of this strategy to justify your answer.

Ans

  1. Yes, this strategy is applicable for complete blockage of expression of a gene. It can be achieved:
  1. either by reintroducing multiple copies of ‘anti-sense gene’ into host cells or,
  2. by using a very strong promoter to allow the transcription of the inverted coding sequence.
  1. Strategy of ‘anti-sense RNA’ reduces the expression of polygalacturonase gene in tomatoes by 10% of its normal level due to which, the firmness of vine ripened tomatoes enhances and shelf life increases.
  2. Yes, this strategy can prevent fatal genetic disorders.

When the genetic sequence of a particular gene is known to be a cause of a particular disease, it is possible to synthesise a strand of nucleic acid that can bind to the mRNA produced by that gene and inactivate it. This is because mRNA has to be single stranded in order to carry out translation.

Q.44 Scientists have used specially designed DNA chips to find the SNPs in 73 protein coding genes in 70 organisms.

They scanned approximately 185 kb of genomic sequence comprising comprising 82 kb of coding, 31 kb of non-coding and 72 kb of untranslated sequences. They identified 858 possible SNPs, out of which 350 were within the coding sequences; which are demarcated as cSNPs.

Out of the total cSNPs, 200 would change the sequence of amino acid in one of the proteins.

Analyse the data and answer the following questions:

  1. Calculate the frequency of SNPs in the sample.
  2. Is there equal distribution of SNPs in coding and non-coding sequences? Carry out the required calculations to support your answer.
  3. In humans, approximately 40,000 ESTs (expressed Sequence Tags) have been identified and the estimated number has to reach 75,000. Based on the given statistical data, predict the number of SNPs in:
  1. human genes,
  2. non-coding regions and
  3. coding regions that do not affect the structure of proteins.

Ans

  1. In 185,000 bp, there are 858 possible SNPs.

So, average number DNA sequence per SNP = 185,000/858 ≈215 bp of DNA sequence between two SNPs

  1. In coding sequences, percentage of SNPs= (350/ 858) X 100 ≈ 41%

In non-coding sequences, percentage of SNPs= [(858 -350)/ 858] X 100 ≈ 59%

From the calculations, it can be observed that smaller percentage of SNPs lie in the coding sequences, which signifies that there is lesser variation in these sequences.

  1. If there are 75,000 genes, we can expect about 8.58 X 105 SNPs in a human genome.

So, (858 SNPs/ 73 genes) X 75,0000 genes ≈ 881506. 84

  1. No. of SNPs in non-coding regions = [(858 -350)/ 858] X 100 ≈ 59%
  2. No. of SNPs in coding regions = (350/ 858) X 100 ≈ 41%.

Approx. [(350-200)/ 858] X 100 ≈ 17% would not affect the structure of protein because they do not change the sequence of amino acid in a protein.

Q.45 Read the case study on patent of an anti-cancer drug carefully.

A British pharmaceutical manufacturer is leading the production of an anti-cancer drug, which is exported to major developed countries for the last 20 years, and recently to developing countries, particularly in Southeast Asia.

The active ingredient of the drug was patented by him, however the original patent expired 3 years ago.

A new improved process for making the drug was patented 10 years ago, and this patent is still effective in various countries like Singapore, Malaysia and Indonesia.

Two years ago, the manufacturer observed that a generic manufacturer based in Vietnam was making and exporting the anti-cancer drug to Malaysia, which are being sold for half the price of his own drug. This posed a serious adverse effect on the sales.

  1. Apply your knowledge of intellectual property rights and advice the British manufacturer so that he can take possible actions against a generic manufacturer based in Vietnam.
  2. What could be the learning outcome for the British manufacturer from this experience?

Ans

  1. The possible advices are as follows:
  1. We can inform the British manufacturer that patents are the national rights, and therefore possible action can be taken only in the countries where the patent existed. In the given case, the manufacturer had not registered any patent in Southeast Asia.
  2. We can tell him that a patent for a product and process is infringed if a third party makes, imports, sells, offers to sell, stores or uses the product without the permission of the patent owner.
  3. We can advise him to contact a local expert to determine the exact process being used by the generic manufacturer before engaging in litigation.
  1. The learning outcome for the British manufacturer from this experience are:
  1. Now, he will submit his patent applications in potential future markets in Southeast Asia.
  2. He would have realised that the cost of litigation can be high and may be even more than the cost of lost business, so making a settlement is comparatively more cost-effective.

Q.46 A linear double stranded DNA has three restriction sites, which can either be BamHI or HaeIII site.

The DNA strand is digested completely with the help of both the ezymes. The products are purified and allowed to undergo an end-filling reaction using the Klenow fragment and [Ɑ-32P]-dCTP.

Then, the product of end filling reaction are purified and seperated by electrophoresis, stained with EtBr and subjected to autoradiography.

In the sample lane, the number below each band represent their main signal intensity in arbitary units.

Based on the information given, design an appropriate restriction map of the DNA, which is 8 kbp long?

Ans

Q.47 Identify the correct statements in each of the following:

  1. In the development of insect-resistant plant,
  1. It is easy to grow and maintain multiple copy integration events rather than maintaining single copy events of insect resistant plants.
  2. The level of expression of transgene does not influence the efficiency of the transgenic plant developed.
  3. A transgenic plant developed using multiple genes may be more effective in developing and maintaining resistance.
  4. The targeted insects cannot develop resistance against transgenic plants using a single gene.
  1. In naturally occurring Agrobacterium strains,
  1. all the virulent genes are constitutively expressed.
  2. Agrobacterium-induced galls do not require bacterial perseverance for their growth.
  3. The T-DNA region of Agrobacterium does not contain functional genes.
  4. Agrobacterium-induced galls require the application of plant hormones for their growth.
  1. In a binary plasmid vector, containing the expression cassette of a selectable marker gene, orientation of a cloned passenger gene cassette can be checked by PCR by using
  1. selectable marker gene-specific primers.
  2. vector-specific primers.
  3. passenger gene-specific primers.
  4. combination of passenger gene-specific and vector-specific primers.

Ans

  1. (iii) A transgenic plant developed using multiple genes may be more effective in developing and maintaining resistance.
  2. (ii) Agrobacterium-induced galls do not require bacterial perseverance for their growth.
  3. (iv) a combination of passenger gene-specific and vector-specific primers

Q.48 RNA transcripts are synthesised in the nucleus of eukaryotes and are then transported to the cytoplasm. Being a student of biotechnology, frame an experiment to demonstrate the same.

Ans

We can perform a simple pulse-and pulse/chase-labeling experiment which can demonstrate that RNA is synthesised in the nucleus and is then transported to the cytoplasm.

This experiment involves:

  1. pulse labelling eukaryotic culture cells by growing them in 3H-uridine for a few minutes and localising the incorporated radioactivity by autoradiography,
  2. repeating the experiment by adding huge amount of non-radioactive uridine to the medium in which the cells are growing after the labelling period and allowing the cells to grow in the non-radioactive medium for about an hour and,
  3. localising the incorporated radioactivity by the method of autoradiography.

Q.49 A scientist isolated an entire RNA from the nuclei of human cells growing in culture. He mixed this RNA with a purified denatured DNA fragment, that possessed a large intron of a house-keeping gene, and incubated the mixture for half day under renaturation conditions.

Read the experimental summary carefully and answer the following questions:

  1. During incubation period, is there any probability of formation of DNA-RNA duplex? Support your answer with a reason.
  2. If you have to perform the same experiment by taking total cytoplasmic RNA from the same cells. Is there any probability of formation of RNA–DNA duplex in this experiment? Support your answer with a reason

Ans

  1. In an experiment performed by a scientist, there is probability of formation of DNA-RNA duplex. Some RNAs will hybridise because all of them have not been completely processed and hence; will still contain the intron sequence.
  2. If cytoplasmic RNA is used in the same experiment, there is no probability of formation of DNA-RNA duplex because all of the mRNA in the preparation have been processed as processing occurs in the nucleus of a cell.

Q.50 A biotechnologist has designed a synthetic gene which codes for an enzyme that degrades the herbicide glyphosate. If another biotechnologist wishes to introduce this gene into Arabidopsis plant and test for the resistance to glyphosate, how would he produce a transgenic plant extracting the synthetic gene by Agrobacterium tumefaciens mediated transformation?

Ans

The possible steps involved would be:

  1. First of all, he would construct a chimeric gene containing a synthetic gene fused to a plant promoter and a plant transcription termination and poly-adenylation signal.
  2. The, he would insert this chimeric gene into the T-DNA of a Ti plasmid containing a dominant selectable marker gene, which provides resistance to kanamycin and introduce into the cells of Agrobacterium tumefaciens by the method of transformation.
  3. After that he would co-cultivate tissue explants from Arabidopsis plant and cells of Agrobacterium tumefaciens, and extract the recombinant Ti plasmid.
  4. Then, he would select the plant cells carrying T-DNAs inserted into their chromosomes on medium containing kanamycin.
  5. Ultimately, he would regenerate the transgenic plants from the transformed cells and test them for their resistance to glyphosate.

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FAQs (Frequently Asked Questions)

Plants use part of the organic matter produced during photosynthesis for respiration. NPP is the biomass remaining after these respiratory losses are subtracted from GPP.

Low temperature reduces microbial activity. Waterlogging creates anaerobic conditions, which inhibit the oxygen-dependent processes involved in decomposition.

Phytoplankton have a small standing biomass but reproduce quickly. Their rapid turnover supports a larger standing biomass of zooplankton and fishes.

Energy is lost as heat during every transfer between trophic levels. Therefore, each higher trophic level receives less energy than the level below it.

Ecological pyramids assume a simple linear food chain. Natural ecosystems contain interconnected food chains, omnivores and species occupying more than one trophic level.