Chapter 15, Biodiversity, Conservation and Environmental Issues, hsc, biology, maharashtra board, 12th std, balbharathi solution,

Chapter 15: Biodiversity, Conservation and Environmental Issues

Choose the correct option

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1.Observe the graph and select the correct option.

OPTIONS

• Line A represents, S = CA²
• Line B represents, log C = log A + Z log S
• Line A represents, S = CA^Z
• Line B represents, log S= log Z + C log A

2.Select the odd one out on the basis of Ex-situ conservation.

OPTIONS

• Zoological park
• Tissue culture
• Sacred groves
• Cryopreservation

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3.Which of the following factors will favour species diversity?

OPTIONS

• Invasive species
• Glaciation
• Forest canopy
• co extinction

4.The term “terror of Bengal’ is used for ___________.

OPTIONS

• algal bloom
• water hyacinth
• increased BOD
• eutrophication

5.CFC are air polluting agents which are produced by ___________.

OPTIONS

• Diesel trucks
• Jet planes
• Rice fields
• Industries

Very short answer type question

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1.Give two examples of biodegradable materials released from the sugar industry.

SOLUTION

Bagasse (dry pulpy residue left after extraction of juice from sugarcane), molasses (the liquid left after the first extraction of sugar) and press mud (organic waste) are biodegradable materials released from the sugar industry.

2.Name any 2 modern techniques of protection of endangered species.

SOLUTION

i. Tissue culture

ii. In vitro fertilization

3.Where was the ozone hole discovered?

SOLUTION

The ozone hole was discovered over the Antarctic region, wherein a depletion of the ozone layer has resulted in the formation of a large area of the thinned ozone layer, commonly called the ozone hole.

4.Give one example of natural pollutants.

SOLUTION

Natural pollutants: Dust (fine particles from sand), fog, mist. fungi, bacteria, moulds, algae, viruses, etc.

5.What do you understand by the EW category of a living being?

SOLUTION

Extinct in the Wild (EW): A category containing those species whose members survive only in captivity

Short answer type question

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1.Dandiya raas is not allowed after 10.00 pm. Why?

SOLUTION

i. During dandiya raas, the use of amplifiers or loudspeakers create lots of noise.

ii. Noise pollution may have many ill effects on human health:

a. Noise causes psychological and physiological changes in human beings.

b. Exposure to extremely high sound level (150 decibels or more) like that generated during a jet plane or rocket take off, may damage eardrums and cause permanent hearing loss.

c. Noise also can cause sleeplessness, increased heartbeat, altered breathing pattern, and psychological stress.

d. Noise may negatively interfere with a child’s learning and behaviour pattern.

iii. There is a need for creating awareness about noise pollution caused during festivals and processions in our society. Thus the Govt. of India has rules and regulations against firecrackers and loudspeakers.

iv. The Supreme court of India banned loudspeakers at public gatherings after 10:00 pm.

v. Also, playing loudspeakers or having public gatherings like dandiya raas after 10 p.m. violates the supreme court orders. Hence, dandiya raas is not allowed after 10.00 pm.

2.Tropical regions exhibit species richness as compared to polar regions. Justify.

SOLUTION

i. Factors like overall stability of tropical regions for millions of years, lesser climatic changes throughout the year, and availability of plenty of sunlight have favoured speciation.

ii. Tropical areas have less often experienced drastic disturbances like periodic glaciations observed at poles. Such stability over millions of years might have favoured speciation.

iii. Lesser migrations in tropics might have reduced gene flow between geographically isolated regions and favoured speciation.

iv. Scientists also have considered the availability of more intense sunlight, warmer temperatures and higher annual rainfall in tropics, as factors responsible for the bountifulness of these regions.

v. Some animals enjoy food preferences under climatic conditions and abundance of resources. e.g. Fruits being available throughout the year in rain forests, a variety of frugivorous organisms is obviously more as compared to the temperate regions.

vi. In short, species richness or diversity for plants and animals decreases as we move away from the equator to the poles. It is maximum in tropical rain forests.

e.g. Amazon rain forest (40,000 plants, 1300 birds, 427 mammals, 3000 species).

Hence, tropical regions exhibit species richness as compared to polar regions.

3.How does genetic diversity affect the sustenance of a species?

SOLUTION

Genetic diversity includes variation within a population and diversity between populations that are associated with adaptation to local conditions.

Genetic variations (e.g. allelic genes) lead to individual differences within species. Such variations eventually lead to evolution. They also improve the chances of continuation of species in the changing environmental conditions or allow the best adapted to survive.

The greater the genetic diversity, the better would be the sustenance of a species.

e.g.

a. Existence of subspecies or races

b. There are about 1000 varieties of mangoes and 50,000 varieties of rice or wheat in India.

c. A medicinal plant Rauwolfia vomitoria which secretes active component reserpine is found in different Himalayan ranges. This plant shows variations in terms of potency and concentration of the active chemical, from location to location.

4.The greenhouse effect is boon or bane? Give your opinion.

SOLUTION

i. Greenhouse effect is responsible for the heating of the earth’s surface and atmosphere. Without the greenhouse effect, the average temperature of Earth would have been -18°C rather than the current average of 15°C. Hence, the greenhouse effect can be considered a boon for keeping Earth warm.

ii. However, increasing concentrations of greenhouse gases result in global warming which causes unfavourable climatic changes i.e., melting of polar ice caps. Hence, the greenhouse effect is also a bane. Thus, it can be said that the greenhouse effect is a boon only until the emission of greenhouse gases is kept under control.

5.How does CO cause giddiness and exhaustion?

SOLUTION

Carbon monoxide is a poisonous gas which binds with haemoglobin of the blood more readily than oxygen to form carboxyhaemoglobin. The presence of CO, therefore, reduces the amount of haemoglobin available in the blood for the transport of oxygen to the body cells. The harmful effects of inhaling increased amount of CO include giddiness, exhaustion, weak eyesight, headache, nervousness and cardiovascular disorders.

6.Name two types of particulate pollutants found in the air. Add a note on ill effects of the same on human health.

SOLUTION

i. Two types of particulate pollutants:

a. Natural pollutants: Dust (fine particles from sand), fog, mist. fungi, bacteria, moulds, algae, viruses, etc.

b. Manmade pollutants: Smoke, smog, pesticides, heavy metals, radioactive elements, etc.

ii. Adverse effects of particulate pollutants:

a. Particulates of about 1.0 µm in size enter lungs easily and those greater than 5 µm get lodged in nasal passage causing irritation in the respiratory tract. Viable particulate matters such as fungi, bacteria, moulds, and algae cause various air-borne diseases.

b. Heavy metal – mercury (Hg) particulate causes heaviness, headache, fatigue and nervousness along with a number of other problems. Prolonged exposure may cause CNS (Central Nervous System) breakdown. Accumulation of heavy metal – lead (Pb) in human tissue may disrupt normal functioning of RBC (Red Blood Corpuscles), which leads to anaemia. It also damages organs like liver, kidneys, intestines and also affects the CNS.

c. Women exposed to fine particulate matter (having a diameter less than 10 µm) give birth to children with small heads and bodies. These children also suffer from learning disability and have an increased risk of cancer. Polynuclear hydrocarbon coated particulates cause irreversible damage to DNA of the growing foetus.

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Long answer type question

1.Montreal protocol is an essential step. Why is it so?

SOLUTION

i. Recognising the harmful effects of ozone depletion, an international treaty, known as the Montreal Protocol was signed at Montreal (Canada) in 1987 to control the emission of ozone-depleting substances.

ii. Later, many more efforts have been made and protocols have laid down definite roadmaps separately for developing and developed countries for reducing the emission of CFCs and other ozone-depleting chemicals.

iii. Montreal Protocol primarily focuses on chloro or Bromo derivates of hydrocarbons which are the main reasons for the depletion of the ozone layer.

iv. Montreal Protocol has provided a mechanism to reduce and phase-out the global production and consumption of substances that deplete the ozone layer.

v. Montreal Protocol has helped in successfully reducing the global production, consumption, and emissions of substances that deplete the ozone layer.

vi. Encouraging evidence for recovery of stratospheric ozone has been found. If the Montreal Protocol was not brought in, ozone depletion likely would be much greater than observed today. Hence, the Montreal Protocol is an essential step.

2.Name any 2 personalities who have contributed to control deforestation in our country. Elaborate on the importance of their work.

SOLUTION

Saalumarada Thimmakka, an Indian environmentalist from the state of Karnataka, and Moirangthem Loiya from Manipur have contributed to control deforestation in our country.

i. Saalumarada Thimmakka, an Indian environmentalist from the state of Karnataka noted for her work in planting and tending to 385 banyan trees along a 4 km stretch of highway between Hulikal and Kudur. She has also planted nearly 8000 other trees. Her work has been honoured with the National Citizens Award of India. She was also conferred with Padma Shri in 2019.

ii. Moirangthem Loiya from Manipur dedicated 17 years of his life to restore the Punshilok forest. He left his job and took over the task of bringing back the lost glory of 300 acres of forest land. He planted a variety of trees like, bamboo, oak, Ficus, teak, jackfruit, and Magnolia. Today the forest has over 250 varieties of plants including 25 varieties of bamboo. It is now selected as home by the great diversity of animals too.

3.How BS emission standards changed over time? Why is it essential?

SOLUTION

i. According to the new fuel policy, the norms are set to reduce sulphur and aromatic content of petrol and diesel. Another provision is the up-gradation of engines. For this, Bharat stage emission standards (BS) are set. These standards are equivalent to Euro norms and have evolved on similar lines as Bharat Stage II (BS-II) to BS-VI from 2001 to 2017.

ii. It is essential to change BS emission standards in order to limit the release of air pollutants from the internal combustion engine.

iii. In 2001, Bharat stage II emission norms were set for CNG and LPG vehicles. As per Bharat Stage II, the emission of sulphur should be controlled at 50 ppm in diesel and 150 ppm in petrol. Aromatic hydrocarbons should be just 42% in concerned fuel.

iv. The aim was to reduce sulphur emission to 50 ppm in petrol and diesel along with aromatic hydrocarbons to 35%. Hence, the Government of India directly adapted BS-VI in the year 2018, skipping BS V. These efforts decreased the levels of CO₂and SO₂ in Delhi.

v. BS emission standards in cities of India:

Vehicle	Norms	Cities of Implementation
4 wheelers	Bharat Stage II	All metro cities
4 wheelers	Bharat Stage III	Throughout the country since October 2010
4 wheelers	Bharat Stage IV	13 megacities (Delhi and NCR, Mumbai, Kolkata, Chennai, Bengaluru, Surat, Kanpur, Agra, Lucknow, Solapur) since April 2010.
2 wheelers	Bharat Stage III	Throughout the country since October 2010
3 wheelers	Bharat Stage III	Throughout the country since October 2010

4.During large public gatherings like Pandharpur vari mobile toilets are deployed by the government. Explain how this organic waste is disposed of.

SOLUTION

i. During large public gatherings like Pandharpur vari mobile toilets are deployed by the Government. These mobile toilets are an example of ecological sanitation.

ii. This is a practical, efficient, and cost-effective solution for human waste disposal.

iii. In order to conserve water and prevent the creation of sewage, ecological sanitation (ecosan) is a sustainable system for handling human excreta using dry composting toilets.

iv. Organic waste disposal:

Ecological sanitation (Ecosan) is an approach to sanitation provision which safely reuses excreta in agriculture. This reduces the need for chemical fertilizers.

Ecosan toilet is a closed system that does not need water. It is an alternative to leach pit toilets in a place where water is scarce or where there is a risk of groundwater contamination. It is based on the principle of recovery and recycling of nutrients from excreta to create a valuable resource for agriculture. When the pit of an ecosan toilet fills up, it is closed and sealed.

After about 8-9 months, the faeces are completely composted to organic manure.

5.How Indian culture and traditions helped in bio-diversity conservation?

SOLUTION

i. Indian culture and traditions are always connected with nature, and rituals are laid down to protect biodiversity.

ii. In many cultures, stretches of forests were set aside and protected in the name of Almighty, which is called sacred groves.

iii. Such sacred groves are found in Khasi and Jaintia hills in Meghalaya, Western ghat regions of Maharashtra and Karnataka, Aravalli hills of Rajasthan and Bastar, and Chanda and Sarguja areas of Madhya Pradesh.

iv. Sacred groves serve the only chance of survival for some endangered varieties of animal and plant species. Tribals do not allow to cut even a single branch of a tree from a sacred grove.

6.Give the importance of conservation in terms of utilitarian reasons.

SOLUTION

The reasons for the conservation of biodiversity can be classified into three categories:

i. Narrowly utilitarian reasons:

a. Since ancient times, humans are reaping material benefits from biodiversity.

b. This includes, deriving resources for basic needs such as food, clothes, shelter, or industrial products like resins, tannins, perfume base, etc. or aesthetic use like ornaments and artifacts.

c. The medicinal use of plants and animals is another major factor. It shares 25% of the global medicine market. Around 25000 species are put to use by tribals worldwide as traditional medicines. Several are yet to be explored for their potential as medicinal plants.

d. Nowadays, bioprospecting of economically important species is carried out. Bioprospecting is a systematic search for the development of new sources of chemical compounds, genes, micro-organisms, macro-organisms, and other valuable products from nature.

ii. Broadly utilitarian reasons:

a. Animals play a crucial role in pollination and seed dispersal.

b. Amazon forest is estimated to produce 20% of the total oxygen of Earth’s atmosphere. We need to consider the recreational use of biodiversity.

c. Devastating fires in the amazon rainforest were reported in August 2019. Such fires are mainly caused in Brazil and are more manmade than natural. The slash and burn policy of locals to reclaim forestland has caused a towering 906000 hectares of forest devastation, only in the year 2019.

12.BIOLOGY FULL CHAPTER COMPLETED

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Chapter 14, Ecosystems and Energy Flow, hsc, biology, maharashtra board, 12th std, balbharathi solution,

Chapter 14: Ecosystems and Energy Flow

Multiple choice question

1.Which one of the following has the largest population in a food chain?

OPTIONS

• Producers
• Primary consumers
• Secondary consumers
• Decomposers

2.Secondary consumers are __________.

OPTIONS

• Herbivores
• Producers
• Carnivores
• Autotrophs

3.The second trophic level in a lake is ________________.

OPTIONS

• Phytoplankton
• Zooplankton
• Benthos
• Fishes

4.What is the % of photosynthetically active radiation in the incident solar radiation?

OPTIONS

• 100%
• 50 %
• 1-5%
• 2-10%

5.Give the term used to express a community in its final stage of succession?

OPTIONS

• End community
• Final community
• Climax community
• Dark community

6.After landslide which of the following type of succession occurs?

OPTIONS

• Primary
• Secondary
• Tertiary
• Climax

7.Which of the following is most often a limiting factor of the primary productivity in any ecosystem.

OPTIONS

• Carbon
• Nitrogen
• Phosphorus
• Sulphur

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Very short answer question

1.Give an example of an ecosystem that shows an inverted pyramid of numbers.

SOLUTION

A tree ecosystem is an example of an inverted pyramid of numbers.

2.Give an example of an ecosystem that shows an inverted pyramid of biomass.

SOLUTION

The oceanic ecosystem is an example of an inverted pyramid of biomass.

3.Which mineral acts as a limiting factor for productivity in an aquatic ecosystem.

SOLUTION

Phosphorus.

4.Name the reservoir and sink of carbon in the carbon cycle.

SOLUTION

i. Reservoir of carbon in the carbon cycle in the atmosphere and ocean.

ii. Carbon which is present in the rock and fossil fuels like oil, coal and natural gas are the sink of the carbon cycle.

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Short answer question

1.Distinguish between the upright and inverted pyramid of biomass.

SOLUTION

	Upright Pyramid of Biomass	Inverted Pyramid of Biomass
i.	It is the type of ecological pyramid where the producers have maximum biomass and occupy a broad base and the consumers decrease in terms of biomass.	It is the type of ecological pyramid where the producers have less biomass and form a narrow base, while the consumers are more in terms of biomass.
E.g.	Upright pyramid of biomass in the grassland ecosystem.	The inverted pyramid of biomass in the oceanic ecosystem.

2.Distinguish between Food chain and food web

SOLUTION

	Food Chain	Food Web
i.	The food chain is a definite sequence of interaction between producers, consumers, and decomposers (saprophytes).	The Food web is a network of food chains that are interconnected at various levels forming an intricate web instead of a linear chain.
ii.	If any of the intermediate organisms are removed from the chain it affects the whole food chain.	In the food web, there is more than one alternative of food to most of the organisms; hence the removal of an organism does not affect the food web directly.

Long answer question

1.Define ecological pyramids.

SOLUTION

An ecological pyramid is a graphical representation of various environmental parameters such as the number of individuals present at each trophic level, the amount of energy, or the biomass present at each trophic level. Ecological pyramids represent producers at the base, while the apex represents the top-level consumers present in the ecosystem.

2.Describe with examples of yramids of number, and biomass.

SOLUTION 1

There are three types of pyramids:

1. Pyramid of numbers
2. Pyramid of energy
3. Pyramid of biomass

• Pyramid of numbers:
  It is a graphical representation of the number of individuals present at each trophic level in a food chain of an ecosystem. The pyramid of numbers can be upright or inverted depending on the number of producers. For example, in a grassland ecosystem, the pyramid of numbers is upright. In this type of a food chain, the number of producers (plants) is followed by the number of herbivores (mice), which in turn is followed by the number of secondary consumers (snakes) and tertiary carnivores (eagles). Hence, the number of individuals at the producer level will be the maximum, while the number of individuals present at top carnivores will be least.

On the other hand, in a parasitic food chain, the pyramid of numbers is inverted. In this type of a food chain, a single tree (producer) provides food to several fruit eating birds, which in turn support several insect species.

• Pyramid of biomass
  A pyramid of biomass is a graphical representation of the total amount of living matter present at each trophic level of an ecosystem. It can be upright or inverted. It is upright in grasslands and forest ecosystems as the amount of biomass present at the producer level is higher than at the top carnivore level. The pyramid of biomass is inverted in a pond ecosystem as the biomass of fishes far exceeds the biomass of zooplankton (upon which they feed).

SOLUTION 2

1. Pyramid of number:
   The number of individuals at each trophic level is shown in the pyramid. The pyramid of number (for example of a grassland) is upright. In this, there is a decrease in the number of organisms starting from primary producers (plants) to top consumers (carnivores).
2. Pyramid of biomass :
   The pyramid of biomass is a graphic representation of the amount of biomass per unit area sequence wise in rising trophic levels with producers at the base and top carnivores at the apex. Pyramids of the biomass of a tree or. grassland ecosystem is upright and the pyramid of a pond ecosystem is inverted.

SOLUTION 3

• Pyramid of numbers:

1. The relative number of individuals per unit area at different trophic levels constitutes the pyramid of number.
2. In a most well-balanced ecosystem, the pyramid of number is upright i.e. producers are more in number than herbivores, and herbivores are more in number than carnivores.

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Upright pyramid of number

c. However pyramid of numbers can also be inverted in some cases.

For e.g. a tree ecosystem represents an inverted pyramid of number. If we plot the number of insects on a single tree, smaller birds feeding on insects, and parasites on those birds, we get an inverted pyramid.

• Pyramid of biomass:
  a. The amount of biomass per unit area at different trophic levels constitutes the pyramid of biomass.
  
  b. In a most well-balanced ecosystem, the pyramid of biomass is upright i.e. biomass of producers is more than that of herbivores, and biomass of herbivores is more than that of carnivores.

Upright pyramid of biomass

c. However pyramid of biomass can also be inverted in some cases. For e.g. Oceanic ecosystem shows an inverted pyramid of biomass. In this case, the biomass of phytoplanktons (producer) is less than that of zooplanktons and fishes.

3.What is primary productivity?

SOLUTION

Primary productivity:

Primary productivity is the rate of generation of biomass in an ecosystem that is expressed in units of mass per unit surface (or volume) per unit time i.e. g/m²/day.

The mass unit may relate to dry matter or to the mass of carbon generated.

4.Give a brief description of the factors that affect primary productivity.

SOLUTION

Factors that affect primary productivity are as follows:

a. It depends on the plant species inhabiting a particular area.

b. It depends upon environmental factors such as light, temperature, water, precipitation, etc.

c. It depends upon the availability of nutrients.

d. It also depends upon the photosynthetic capacity of plants. The greater the photosynthetic activity, the higher will be the primary productivity.

5.Define decomposition.

SOLUTION

i. Decomposition is the process of breakdown of complex organic matter into inorganic substances like carbon dioxide, water, and nutrients by the decomposers.

ii. Raw materials for decomposition are dead remains of plants and animals, fecal matter, detritus.

iii. This process requires oxygen. Temperature and soil moisture are important factors that indirectly help soil microbes for decomposition.

iv. Warm and the moist environment favours decomposition whereas, low temperature and anaerobic conditions inhibit the process.

6.describe the processes and products of decomposition.

SOLUTION

The steps of decomposition are fragmentation, leaching, catabolism, humification, and mineralization.

a. Fragmentation: Detritivores like earthworm breakdown detritus into smaller fragments or particles.

b. Leaching: In this process, water-soluble inorganic nutrients percolate into the soil horizon and get precipitated as unavailable salts.

c. Catabolism: The bacterial and fungal enzymes degrade detritus into simpler inorganic substances. All of the above steps occur simultaneously.

d. Humification: It leads to the accumulation of particularly decomposed, a dark coloured, amorphous, colloidal organic substance called humus. Humus serves as a reservoir of nutrients. It is resistant to microbial action and undergoes decomposition at an extremely slow rate. Humus changes the soil texture and increases the capacity of water holding in the soil.

e. Mineralization: Some microorganisms degrade humus and release inorganic nutrients by the process of mineralization.

Decomposition cycle

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7.Write important features of a sedimentary cycle in an ecosystem.

SOLUTION 1

Sedimentary cycles have their reservoirs in the Earth’s crust or rocks. Nutrient elements are found in the sediments of the Earth. Elements such as sulphur, phosphorus, potassium, and calcium have sedimentary cycles.

Sedimentary cycles are very slow. They take a long time to complete their circulation and are considered as less perfect cycles. This is because during recycling, nutrient elements may get locked in the reservoir pool, thereby taking a very long time to come out and continue circulation. Thus, it usually goes out of circulation for a long time.

SOLUTION 2

Features of a sedimentary cycle in an ecosystem are as follows:

a. Earth’s crust is the main reservoir of phosphorus and other minerals, such as calcium and potassium that undergo sedimentary cycles.

b. The rate of release of minerals that take part in the sedimentary cycle is regulated by various environmental factors temperature, moisture, and nature of the soil.

c. Sedimentary cycles are slower than the gaseous cycles therefore they take more time to complete.

d. Sedimentary cycles are considered as less perfect cycles as, during recycling, nutrient elements may get locked in the reservoir pool, thereby taking a very long time to come out and continue circulation

8.Describe the carbon cycle and add a note on the impact of human activities on the carbon cycle.

SOLUTION

i. Reservoir of carbon:

a. All life forms on earth are carbon-based because carbon is the main component of all the organic compounds of protoplasm. It constitutes 49% of the dry weight of organisms.

b. 71% of carbon is found dissolved in oceans. The oceanic reservoir regulates the amount of carbon dioxide in the atmosphere.

c. Carbon present in the rock and fossil fuels like oil, coal, and natural gas has been away from the rest of the carbon cycle for a long time. These long term storage places are known as the sink.

d. The element carbon is a part of seawater, the atmosphere, rocks such as limestone and coal, soils, as well as all living things.

ii. Cyclic pathway of carbon:

a. Carbon as CO₂ moves from the atmosphere to plants during the process of photosynthesis to produce food.

b. Carbon moves from plants to animals, through food chains.

c. At the time of exhalation, the CO₂ gas is released into the atmosphere. Thus, carbon moves from living things to the atmosphere.

d. Decomposers also contribute substantially to CO₂ in the atmosphere, by their processing of waste materials and dead organic matter of land and oceans.

e. When fossil fuels burn to power factories, power plants, motor vehicles, most of the carbon quickly enters the atmosphere as carbon dioxide gas.

f. Most of the remainder is dissolved in seawater and deposited as calcium or magnesium carbonate compounds that make up shells of marine animals.

g. The additional sources for releasing CO₂ in the atmosphere are the burning of wood, forest fire and combustion of organic matter, fossil fuel, and volcanic activity.

h. The ocean absorbs some carbon in the form of CO₂ from the atmosphere. This carbon gets dissolved in the ocean water. Some amount of the carbon which is fixed is lost to sediments and removed from circulation.

iii. The impact of human activities on the carbon cycle

a. Carbon cycle is significantly influenced by human activities.

b. Rapid deforestation and the massive burning of fossil fuel for energy and transport have significantly increased the rate of release of carbon dioxide into the atmosphere.

Carbon cycle

COMPLETED

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Chapter 13: Organisms and Populations, hsc, biology, maharashtra board, 12th std, balbharathi solution,

Chapter 13: Organisms and Populations

Multiple choice question

1.Which factor of an ecosystem includes plants, animals, and microorganisms?

OPTIONS

• Biotic factor
• Abiotic factor
• Direct factor
• Indirect factor

2.An assemblage of individuals of different species living in the same habitat and having functional interactions is _______________.

OPTIONS

• Biotic community
• Ecological niche
• Population
• Ecosystem

3.Association between sea anemone and Hermit crab in gastropod shell is that of _______________.

OPTIONS

• Mutualism
• Commensalism
• Parasitism
• Amensalism

4.Select the statement which explains the best parasitism.

OPTIONS

• One species is benefited.
• Both the species are benefited.
• One species is benefited, other is not affected.
• One species is benefited, other is harmed.

5.Growth of bacteria in a newly innoculated agar plate shows ____________.

OPTIONS

• exponential growth
• logistic growth
• Verhulst-Pearl logistic growth
• zero growth

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Define the following term and give one example for each:

Commensalism

SOLUTION 1

Commensalism:Commensalism is an interaction between two species in which one species gets benefited while the other remains unaffected. An orchid growing on the branches of a mango tree and barnacles attached to the body of whales are examples of commensalisms.

SOLUTION 2

Commensalism:

Commensalism is the interaction in which one species gets benefited and the other is neither harmed nor benefited.

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2. Parasitism

SOLUTION 1

Parasitism is a kind of relationship between two species in which one species which is termed as parasite, derives its food from the other species which is termed as host. Parasitism also involves shelter, in addition to food obtained by a parasite. Parasites may be ectoparasites or endoparasites. Ectoparasites live on the surface of their host while endoparasites live inside the body of the host.

Examples of Parasitism

(i) Cuscuta growing on shoe flower plant: Cucuta grows on the stem of shoe and derive nutrition from the plant.

(ii) Head lice is an ectoparasite and suck human blood

(iii) Ascaris, Taenia, Plasmodium causing diseases in humans

(iv) Koel laying its eggs in crow’s nest is an example of Brood parasitism. Birds lay egg in the nest of its host and host incubate it.

SOLUTION 2

Parasitism:

Parasitism is the interaction in which only one species (parasite) is benefited and the interaction is detrimental to other species (host).

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Camouflage

SOLUTION

Camouflage:

Camouflage is the cryptic coloration or patterns adopted by prey species to blend with the surroundings or background so as to escape their predators.

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Mutualism

SOLUTION

Mutualism:

E.g. Lichens represent an intimate, mutualistic relationship between a fungus and photosynthetic algae or cyanobacteria.

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3.Give one example for Interspecific competition

SOLUTION

Interspecific competition:

E.g. Competition between leopards and lion, resident fish competing with migratory birds Flamingos for common food i.e. zooplankton

4.Name the type of association: Clownfish and sea anemone

SOLUTION

Clown fish and sea anemone: Commensalism

5.Name the type of association: Crow feeding the hatchling of Koel

SOLUTION

Crow feeding the hatchling of Koel: Brood parasitism

6.Name the type of association: Humming birds and host flowering plants

SOLUTION

Humming birds and host flowering plants: Mutualism

7.What is the ecological process behind the biological control method of managing with pest insects?

SOLUTION

The ecological process behind the biological control method of managing with pest insects is Predation. Predators regulate the population of prey in a habitat, thus helping in the management of pest insects.

Short answer question

1.How is the dormancy of seeds different from hibernation in animals?

SOLUTION

i. Seed dormancy is the inability of viable seeds to germinate even under suitable environmental conditions, whereas hibernation in animals is a state of reduced activities to escape cold winter conditions.

ii. During seed dormancy, growth and development of an embryo are arrested temporarily, whereas in hibernation animals enter a state of inactivity by slowing their metabolism.

2.If a marine fish is placed in a freshwater aquarium, will it be able to survive? Give reason.

SOLUTION

i. If a marine fish is placed in a freshwater aquarium, fish would not be able to survive because marine fishes are adapted to high salt concentrations of the marine environment.

ii. Marine fishes have more osmotic concentration (more salt concentration) than marine water which prevents marine water to enter into the body.

iii. When marine fish is placed in a freshwater aquarium, water enters into the body of marine fish due to osmosis, as freshwater creates a hypotonic environment outside the fish’s body.

iv. Entry of water into the body causes its body to swell leading to the death of the marine fish.

3.Name important defense mechanisms in plants against herbivores.

SOLUTION

Defense mechanisms in plants against herbivores can be morphological like thorns (in Acacia, Cactus) or chemicals like poisonous cardiac glycosides (produced by Calotropis), secondary metabolites (for e.g.nicotine, caffeine, quinine, strychnine, opium, etc.)

4.An orchid plant is growing on the branch of the mango tree. How do you describe this interaction between the orchid and the mango tree?

SOLUTION 1

An orchid growing on the branch of a mango tree is an epiphyte. Epiphytes are plants growing on other plants which however, do not derive nutrition from them. Therefore, the relationship between a mango tree and an orchid is an example of commensalisms, where one species gets benefited while the other remains unaffected. In the above interaction, the orchid is benefited as it gets support while the mango tree remains unaffected.

SOLUTION 2

i. An orchid plant is growing on the branch of a mango tree represents Commensalism.

ii. In this interaction, one species gets benefited and the other is neither harmed nor benefited.

iii. Orchid is an epiphytic plant. While growing on mango tree it gets support but does not derive any nutrition from the mango tree.

iv. Thus, the orchid plant is benefited, while the mango tree is neither benefited nor harmed.

5.Distinguish between the following:

Hibernation and Aestivation

SOLUTION

	Hibernation	Aestivation
i.	It is also called as winter sleep.	It is also called as summer sleep.
ii.	It is a state of reduced activities in some organisms to escape cold winter conditions.	It is a state of reduced activities in some organisms to escape desiccation due to heat in summer.
iii.	Animals rest in warm places.	Animals rest in cool, shady, and moist places.
iv.	It is shown by bears inhabiting cold regions.	It is shown by some fishes and snails.

6.Distinguish between the following:

Ectotherms and Endotherms

SOLUTION

	Ectotherms	Endotherms
i.	These are cold-blooded animals.	These are warm-blooded animals.
ii.	Ectotherms do not possess the ability to generate sufficient heat to keep them warm, thus their body temperature varies with surroundings	Endotherms do possess the ability to generate heat and keep them warm, thus they can maintain constant body temperature.
iii.	They are also known as poikilothermic.	They are also known as homeothermic.
iv.	They are affected by changes in environmental temperature	They remain unaffected by changes in environmental temperature.
	E.g. Most of the fishes, amphibians, reptiles	E.g. Birds, mammals

7.Distinguish between the following:

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Parasitism and Mutualism

SOLUTION

	Parasitism	Mutualism
i.	In parasitism, only one species (parasite) is benefited and the interaction is detrimental to other species (host).	In mutualism, both species are benefited.
ii.	The parasite needs a host, but the host does not need the parasite.	Both species need the presence of each other.
E.g.	Cuscuta, a parasitic plant commonly found growing on hedge plants.	Lichen represents the mutualistic relationship between a fungus and photosynthetic algae or cyanobacteria.

8.Write a short note on Adaptations of desert animals

SOLUTION

Adaptations of animals for desert habitats:

1. Desert animal-like Kangaroo rat inhabiting the Arizona deserts has the potential to concentrate its urine to conserve water. This animal never drinks water in its life.

2. Snakes and desert lizards bask in the sun early in the morning and burrow themselves in the sand in the afternoons to escape the heat of the day, to prevent water loss.

3. Camels can store fat in the hump which can be metabolised for energy. A camel can survive for many days without water. Long eyelashes, ears lined with hair, and slit-like nostrils help to keep out sand.

9.Write a short note on the Adaptations of plants to water scarcity.

SOLUTION 1

Adaptations of plants to water scarcity:

Plants found in deserts are well adapted to cope with water scarcity and scorching heat of the desert. Plants have an extensive root system to tap underground water. They bear thick cuticles and sunken stomata on the surface of their leaves to reduce transpiration. In Opuntia, the leaves are modified into spines and the process of photosynthesis is carried out by green stems. Desert plants have special pathways to synthesize food, called CAM (C₄ pathway). It enables their stomata to remain closed during the day to reduce water loss by transpiration.

SOLUTION 2

Adaptations of plants for desert habitats:

1. Many desert plants have a thick cuticle on their leaf surfaces and have their stomata in deep pits to minimize loss of water through transpiration.

2. They show a special photosynthetic pathway (CAM – Crassulacean Acid Metabolism) that enables their stomata to remain closed during the daytime.

3. Some desert plants like Opuntia have their leaves reduced (modified) to spines and the photosynthetic function is taken over by the flattened stems.

10.Write a short note on Behavioural adaptations in animals

SOLUTION 1

Behavioural adaptations in animals

Certain organisms are affected by temperature variations. These organisms undergo adaptations such as hibernation, aestivation, migration, etc. to escape environmental stress to suit their natural habitat. These adaptations in the behaviour of an organism are called behavioural adaptations. For example, ectothermal animals and certain endotherms exhibit behavioral adaptations. Ectotherms are cold blooded animals such as fish, amphibians, reptiles, etc. Their temperature varies with their surroundings. For example, the desert lizard basks in the sun during early hours when the temperature is quite low. However, as the temperature begins to rise, the lizard burrows itself inside the sand to escape the scorching sun. Similar burrowing strategies are exhibited by other desert animals. Certain endotherms (warm-blooded animals) such as birds and mammals escape cold and hot weather conditions by hibernating during winters and aestivating during summers. They hide themselves in shelters such as caves, burrows, etc. to protect against temperature variations.

SOLUTION 2

Behavioural adaptations in animals.

a. To cope up with extreme variations in their environment, some organisms respond through behaviourally (like migration, hibernation, and aestivation).

b. For e.g. Desert lizards manage to keep their body temperature fairly constant by behavioural adaptations.

c. They bask in the sun and absorb heat when their body temperature drops below the comfort zone. But when the ambient temperature starts increasing, they move into the shade.

d. Some species burrow into the sand to hide and escape from the heat.

11.Define Population.

SOLUTION

Population: Organisms of the same kind inhabiting a geographical area constitute the population.

OR

Individuals live in groups in a well-defined geographical area, share or compete for similar resources, potentially interbreed and thus form a population.

OR

The population is defined as a group of individuals of a species occupying a definite geographic area at a given time.

12.Define Community.

SOLUTION

Community: Several populations of different species in a particular area constitute a community that interacts with one another in several ways.

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Long answer question

1.With the help of a suitable diagram describe the logistic population growth curve.

SOLUTION

Logistic growth curve of population

i. Resources like food and space are not always unlimited. They may be plenty in the beginning; but as the population density increases, competition for those resources starts, resulting in a slowdown in the rate at which the original population was growing. This results in a logistic or sigmoid growth curve.

ii. Competition between individuals for limited resources will weed out the ‘weaker’ ones. Only the ‘fittest’ individuals will survive and reproduce.

iii. A given habitat has enough resources to support a maximum possible number, beyond which no further growth is possible. This limit can be called nature’s carrying capacity (K) for that species in that habitat.

iv. A population growing in a habitat with limited resources shows initially a lag phase, followed by phases of acceleration and deceleration and finally an asymptote when the population density reaches the carrying capacity.

v. A plot of population density (N) in relation to time (t) results in a sigmoid curve. This type of population growth is called VerhulstPearl Logistic Growth.

vi. Since resources for the growth of most animal populations, are finite and become limiting sooner or later, the logistic growth model is considered a more realistic one.

2.Enlist and explain the important characteristics of a population. www.asterclasses.com

SOLUTION

i. The important characteristics of a population are population size, population density, natality, mortality, sex ratio, immigration, emigration, age pyramids, expanding population, population growth forms, and biotic potential.

ii. Some important characteristics of the population are:

a. Population density: Population density tells us the number of individuals presents per unit space, in a given time.

OR

The density of a population is the total number of individuals in that population present per unit area at a specific time.

b. Natality: Natality is the birth rate of a population.

c. Mortality: Mortality is the death rate of a population.

d. Age distribution and Age pyramids:

1. A population consists of individuals of different ages. The entire population is divided into three age groups – pre-reproductive (0-14 years), reproductive (age 15-44 years), post-reproductive (45-85+years)The relative proportion of individuals of various age groups in the population is referred to as the age structure of the population.

2. If the age distribution (percent individuals of a given age or age group) is plotted for the population, the resulting structure is called as age pyramid.

e. Sex Ratio:

Sex ratio is the ratio of the number of individuals of one sex to that of the other sex.

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COMPLETED

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Chapter 12, Biotechnology, hsc, biology, maharashtra board, 12th std, balbharathi solution,

Chapter 12: Biotechnology

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Choose the correct option

1.The bacterium which causes a plant disease called crown gall is _______________.

OPTIONS

• Helicobacter pylori
• Agrobacterium tumifaciens
• Thermophilus aquaticus
• Bacillus thuringienesis

2.The enzyme nuclease hydrolyses _____________ of polynucleotide chain of DNA.

OPTIONS

• hydrogen bonds
• phosphodiester bonds
• glycosidic bonds
• peptide bonds

3.In vitro amplification of DNA or RNA segment is known as ___________.

OPTIONS

• chromatography
• Southern blotting
• polymerase chain reaction
• gel electrophoresis

4.Which of the following is the correct recognition sequence of restriction enzyme hind III?

OPTIONS

• ^5′—A-A-G-C-T-T—^{3′
  3′}—T-T-C-G-A-A—^5′
• ^5′—G-A-A-T-T-C—^{3′
  3′}—C-T-T-A-A-G—^5′
• ^5′—C-G-A-T-T-C—^{3′
  3′}—G-C-T-A-A-G—^5′
• ^5′—G-G-C-C—^{3′
  3′}—C-C-G-G—^5′

5.Recombinant protein ……………… is used to dissolve blood clots present in the body.

OPTIONS

• insulin
• tissue plasminogen activator
• relaxin
• erythropoietin

6.The recognition sequence of restriction enzymes is generally __________ nucleotide long.

OPTIONS

• 2 to 4
• 4 to 8
• 8 to 10
• 14 to 18

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Very short answer type question

1.Name the vector which is used in the production of human insulin through recombinant DNA technology.

SOLUTION

pBR322 vector is used in the production of human insulin through recombinant DNA technology.

2.Which cells from Langerhans of the pancreas do produce a peptide hormone insulin?

SOLUTION

Insulin is a peptide hormone produced by β-cells of islets of Langerhans of the pancreas.

3.Give the role of Ca⁺⁺ ions in the transfer of the recombinant vector into the bacterial host cells.

SOLUTION

Ca⁺⁺ ions assist the transfer of the recombinant vector into a bacterial host cell.

4.Expand the following acronym which is used in the field of protechnology.

YAC

SOLUTION

YAC: Yeast Artificial Chromosome

4.1.Expand the following acronym which is used in the field of protechnology.

RE

SOLUTION

RE: Restriction Enzyme

4.2.Expand the following acronym which is used in the field of protechnology.

dNTP

SOLUTION

dNTP: Deoxyribonucleoside triphosphate

4.3.Expand the following acronym which is used in the field of protechnology.

PCR

SOLUTION

PCR: Polymerase Chain Reaction

4.4.Expand the following acronym which is used in the field of protechnology.

GMO

SOLUTION

GMO: Genetically Modified Organism

4.5.Expand the following acronym which is used in the field of protechnology.

MAC

SOLUTION

MAC: Mammalian Artificial Chromosome

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Very short answer type question

Fill in the blanks and complete the chart.

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GMO	Purpose
i. Bt cotton	____________
ii. _____________	Delay the softening of tomato during ripening.
iii. Golden rice	____________
iv. Holstein cow	____________

SOLUTION

GMO	Purpose
i. Bt cotton	Insect pest resistance
ii. Flavr savr tomato	Delay the softening of tomato during ripening.
iii. Golden rice	High vitamin A content
iv. Holstein cow	High milk yield capacity

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Short answer type question

1.Explain the properties of a good or ideal cloning vector for rDNA technology.

SOLUTION

Following characteristic properties, a cloning vector must possess in order to be used in rDNA technology:

i. A good vector should have the ability of independent replication so that as the vector replicates (through ori gene) and a large number of copies of the DNA insert will be formed.

ii. The vector should be able to easily introduce into host cells.

iii. A vector should have marker genes for antibiotic resistance.

iv. A vector must contain a unique cleavage site in one of the marker genes for the restriction enzyme.

v. It should have at least suitable control elements like a promoter, operator, ribosomal binding sites, etc.

vi. The plasmids obtained naturally do not possess all the characteristics. Hence, they are constructed by inserting a gene for antibiotic resistance.

e.g. pBR322, pBR320, pACYC177 are the constructed plasmids. pBR322 is mostly used in rDNA technology in plants.

2.A PCR machine can raise the temperature up to 100°C but after that, it is not able to lower the temperature below 70°C automatically. Which step of PCR will be hampered first in this faulty machine? Explain why?

SOLUTION

i. If a PCR machine fails to lower the temperature below 70°C then the annealing step (in which primers attach to the respective ends of the DNA template) would be affected.

ii. The annealing step of PCR requires temperature ranging between 40°C to 60°C therefore if a PCR machine fails to lower the temperature below 70°C, primers will not attach to the DNA templates.

3.In the process of rDNA technology, if two separate restriction enzymes are used to cut vector and donor DNA then which problem will arise in the formation of rDNA or chimeric DNA? Explain.

SOLUTION

i. Restriction enzymes cut the DNA at a specific recognition site and produce either sticky ends or blunt ends.

ii. For e.g., if we use the restriction enzyme EcoR I to cut vector DNA then it will produce vector DNA having sticky ends and if we use the Hind II restriction enzyme to cut donor DNA then it will produce donor DNA with blunt ends.

iii. A vector DNA to get ligated to the desired gene, they both must possess similar kind of ends i.e. both should have either blunt ends or sticky ends.

iv. If two separate restriction enzymes are used to cut vector and donor DNA then they will fail to form complementary base pairing and chimeric DNA will not be formed.

Short answer type question

Match and write the pairs.

Recombinant protein	It’s used in or for
i. platelet-derived growth factor	a. Anemia
ii. α-antitrypsin	b. cystic fibrosis
iii. Relaxin	c. Haemophilia A
iv. Erythropoietin	d. Diabetes
v. Factor VIII	e. Emphysema
vi. DNase	f. Parturition
	g. Atherosclerosis

SOLUTION

Recombinant protein	It’s used in or for
i. platelet-derived growth factor	g. Atherosclerosis
ii. α-antitrypsin	e. Emphysema
iii. Relaxin	f. Parturition
iv. Erythropoietin	a. Anemia
v. Factor VIII	c. Haemophilia A
vi. DNase	b. cystic fibrosis

Long answer type question

Define and explain the term

1.Biopiracy

SOLUTION

Biopiracy is defined as ‘theft of various natural products and then selling them by getting a patent without giving any benefits or compensation back to the host country’.

i. For proper and lawful working of biopatent, the nation should be rich in bio-diversity, people residing there should have traditional knowledge and the nation should also have sufficient financial resources.

ii. However, it is generally observed that industrialized nations are rich in financial resources and technology but lack bio-diversity, whereas developing countries are rich in biodiversity and traditional knowledge but are short of financial resources and advanced technology. These situations lead to biopiracy.

iii. Industrialized nations have always been enjoying immense profits by patenting the indigenous biomedical knowledge and bioresources of third world communities without paying any compensation to the indigenous group who originally developed such knowledge.

Example:

i. Texmati case: A strain of Basmati rice was patented by Texas-based company Rice Tec Inc with trade name Texmati. This patenting was illegal and unethical as Basmati is a long-grained, aromatic variety of rice indigenous to the Indian subcontinent. India fought a long legal battle after which the patent was canceled.

ii. Turmeric: Since ancient times, Indians have been using Haldi (Turmeric powder) as an antiseptic for healing wounds for killing pests and medicinal purposes. However, American companies have patented Turmeric and many medicinal plants of India. After a long legal battle, most of the patents have been revoked.

iii. Neem (Azadirachta indica): The patenting of the fungicidal properties of Neem was an example of biopiracy. The USDA and an American MNC W.R. Grace in the early 90s sought a patent from the European Patent Office (EPO) on the “method for controlling on plants by the aid of hydrophobic extracted neem oil.”

Define and explain the termwww.asterclasses.com

2.Biopatent

SOLUTION

i. Biopatent is a biological patent awarded for strains of microorganisms, cell lines, genetically modified strains, DNA sequences, biotechnological processes, product processes, products, and product applications.

ii. Biopatents are awarded to recognize real innovative contributions made by the inventor to the cause of human welfare.

iii. The awards are given to inculcate encouragement and values in developing scientific culture and in emphasizing the role of biology in shaping human society.

iv. Indian patent allows ‘process patent’ and not the ‘product patent’. Biopatent allows the patent holder to exclude others from making, using, selling, or importing a protected invention for a limited period of time.

v. The duration of biopatents is five years from the date of the grant or seven years from the date of filing the patent application, whichever is less.

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3.Bioethics

SOLUTION

i. Bioethics helps to study moral vision, decisions, and policies of human behaviour in relation to biological phenomena or events.

ii. Ethics deals with ‘Life’ e.g. in vitro fertilization, sperm bank, gene therapy, cloning, gene manipulations, euthanasia, death, maintaining those who are in a comatose state, prenatal genetic selection, etc.

iii. The era of biotechnology has brought a wide spectrum on new topics like cloning, transgenic, gene therapy, eugenics, rDNA technology, etc.

iv. The use of all these has drawn a wide range of reactions in the society.

v. Ethical aspects pertaining to the use of biotechnology seems to be more controversial and frightening.

vi. These concerns are broadly summarized below: Use of animals causes great sufferings to them; violation of integration of species caused due to transgenesis; transfer of human genes into animals and vice versa; indiscriminate use of biotechnology poses risk to the environment, health, and biodiversity.

vii. The introduction of Genetically Modified Organisms (GMOs) has led to a wider debate on bioethical concerns affecting social, economic, and environmental spheres.

viii. These include the effects on non-target organisms, insect resistance crops, gene flow, and the loss of diversity as well as the issue of interfering with nature.

ix. Ethics in biotechnology also includes the general subject of what should and should not be done in using recombinant DNA techniques.

2.Explain the steps in process of rDNA technology with suitable diagrams.

SOLUTION

The steps involved in gene cloning are as follows:

i. Isolation of DNA (gene) from the donor organism:

a. The desired gene to be cloned is obtained from the source organism (donor).

b. Initially, the cells of the donor organism are sheared with the blender and treated with a suitable detergent.

c. Genetic material from the donor is isolated and purified using several techniques.

d. Isolated DNA can be spooled on to a glass rod.

ii. Cutting of the desired gene:

a. Isolated purified DNA is then cleaved by using restriction enzymes i.e. restriction endonucleases.

b. These enzymes cleave DNA at restriction sites and break the DNA into fragments.

c. There are several types of restriction endonucleases.

d. Cleaved DNA fragments have cohesive, sticky, staggered ends or blunt ends.

e. From cleaved DNA fragments, a fragment containing the desired gene is isolated and selected for cloning. This is now called foreign DNA or passenger DNA.

f. The desired gene can also be obtained directly from genomic library or cDNA library.

iii. Insertion of a desired foreign gene into a cloning vector (vehicle DNA):

a. The foreign DNA or passenger DNA is now inserted into a cloning vector or vehicle DNA.

b. The most commonly used cloning vectors are plasmids of bacteria and bacteriophage viruses like lambda phage and M13.

c. The most commonly used plasmid is pBR322.

d. Plasmids are isolated from the vector organisms i.e. bacterium.

e. By using the same restriction enzyme (which is used in the isolation of the desired gene from the donor), plasmid i.e. vector DNA is cleaved.

f. Now by using enzyme DNA ligase, foreign DNA is inserted/ integrated into the vector DNA.

g. The combination of vector DNA and foreign DNA is now called Recombinant DNA or Chimeric DNA and the technology is referred to as rDNA technology.

iv. Transfer of rDNA into suitable competent host or cloning organism:

a. Finally, the recombinant DNA is transferred for expression into a competent host cell which is usually a bacterium.

b. The host cell takes up naked rDNA by process of ‘transformation’ and incorporates into its own chromosomal DNA which finally expresses the trait controlled by passenger DNA.

c. The transfer of rDNA into a bacterial cell is assisted by divalent Ca⁺⁺ .

d. The cloning organisms used in plant biotechnology are E. coli and Agrobacterium tumefaciens.

e. The host/ competent cell which has taken up rDNA is now called a transformed cell.

f. Foreign DNA can also be transferred directly into the naked cell or protoplast of the competent host cell, without using vector.

g. This is done by using techniques like electroporation, microinjection, lipofection, shotgun, ultra-sonification, biolistic method, etc. But in plant biotechnology, the transformation is through Ti plasmids of A. tumefaciens.

v. Selection of the transformed host cell:

a. The transformation process generates a mixed population of transformed (recombinant) and non-transformed (non-recombinant) host cells.

b. For the isolation of recombinant cells from non-recombinant cells, the marker gene of the plasmid vector is employed.

c. For example, the pBR322 plasmid vector contains different marker genes (Ampicillin resistant gene and Tetracycline resistant gene).

d. When the PstI restriction enzyme is used, it knocks out Ampicillin resistant gene from the plasmid, so that the recombinant cell becomes sensitive to Ampicillin.

vi. Multiplication of transformed host cell:

a. Once transformed, host cells are separated by the screening process.

b. In this step, the transformed host cells are introduced into fresh culture media.

c. At this stage, the host cells multiply along with the replication of the recombinant DNA carried by them.

vii. Expression of the gene to obtain the desired product:

a. The next step involves the production of desired products like alcohol, enzymes, antibiotics, etc.

b. Finally, the desired product is separated and purified through downstream processing using a suitable bioreactor.

Outline of the process of recombinant DNA technology

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3.Explain the gene therapy.

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SOLUTION

Gene therapy is the treatment of disease by replacing, altering, or supplementing a gene that is absent or abnormal and whose absence or abnormality is responsible for the disease.

The following are the different ways through which gene therapy is being used for the treatment of a disease or disorder:

i. Replace missing or defective genes;

ii. Deliver genes that speed up the destruction of cancer cells;

iii. Supply genes that cause cancer cells to revert back to normal cells;

iv. Deliver bacterial or viral genes as a form of vaccination;

v. Deliver DNA to antigen expression and generation of immune response;

vi. Supply of gene for impairing viral replication;

vii. Provide genes that promote or impede the growth of new tissue; and

viii. Deliver genes that stimulate the healing of damaged tissue. www.asterclasses.com

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4.Give two types of gene therapy.

SOLUTION

There are two forms of gene therapy based on the types of cells in which genes are delivered:

i. Germline gene therapy:

a. In this method, healthy genes can be introduced into germ cells like sperms, eggs, early embryos.

b. It allows the transmission of the modified genetic information to the next generation.

c. Though it is highly effective in counteracting genetic disorders, it is not encouraged for application in human beings due to a variety of technical and ethical reasons.

ii. Somatic cell gene therapy:

a. In this type the gene is introduced only in somatic cells like bone marrow cells, hepatic cells, fibroblasts endothelium, and pulmonary epithelial cells, central nervous system, endocrine cells, and smooth muscle cells of blood vessel walls.

b. Modification of somatic cells only affects the person being treated and the modified chromosomes cannot be passed on the future generations.

c. Somatic cell gene therapy is the only feasible option and the clinical trials have already employed for the treatment of acquired disorders such as cancer and rheumatoid arthritis and blood disorders including SCID, Gaucher’s disease, familial hypercholesterolemia, haemophilia, phenylketonuria, cystic fibrosis, sickle cell anemia, Duchenne muscular dystrophy, emphysema, thalassemia, etc.

5.How are the transgenic mice used in cancer research?

SOLUTION

Transgenic mice:

a. Transgenic mice that have been modified using a particular oncogene (cancer-causing gene) and thus developed a certain type of cancer, is useful to answer questions concerning the relationship between oncogenes and cancer development.

b. Theoretically, such animals can also be used for research into cancer treatment and prevention of malignancy.

c. In the laboratory, one such a transgenic mouse model for the investigation of breast cancer was developed. The oncogenes Myc and ras were analyzed to find out if they lead to breast cancer in mice transformed with these genes.

6.Give the steps in PCR or polymerase chain reaction with suitable diagrams.

SOLUTION

Polymerase Chain Reaction (PCR) is the process of in vitro amplification of the gene of interest using a PCR machine.

i. PCR can generate a billion copies of the desired segment of DNA or RNA, with high accuracy and specificity, in a few hours.

ii. The process of PCR is completely automated and involves automatic thermal cycles for denaturation and renaturation of double-stranded DNA.

iii. The device required for PCR is called a thermal cycler.

iv. Requirements for polymerase chain reaction:

a. DNA containing the desired segment to be amplified

b. several molecules of four deoxyribonuclueoside triphosphates (dNTPs)

c. excess of two primer molecules

d. heat-stable DNA polymerase and

e. appropriate quantities of Mg⁺⁺ ions.

Mechanism of PCR:

At the start of PCR, all the requirements are mixed together in ‘eppendorf tube’ and the following operations are performed sequentially:

Step i: Denaturation

The reaction mixture is heated to a temperature (90–98oC) to separate two strands of desired DNA. This is called denaturation.

Step ii: Annealing

The mixture is allowed to cool (40–60oC) that permits the pairing of the primer to the complementary sequences in DNA. This step is called annealing.

Step iii: Primer extension / Polymerization

The temperature (70–75°C) allows thermostable Taq DNA polymerase to use single-stranded DNA as a template and adds nucleotides. This is called primer extension. It takes around two minutes duration.

v. One cycle takes around 3 to 4 minutes.

vi. To begin the second cycle, DNA is again heated to convert double-stranded DNA into single strands.

vii. In an automatic thermal cycler, the above three steps are automatically repeated 20-30 times. Thus, at the end of ‘n’ cycles, 2n copies of DNA segments are produced.

viii. The machine performs the entire operations automatically and precisely.

DNA replication through a polymerase chain reaction.

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7.What is a vaccine?

SOLUTION

A vaccine is a biological preparation that provides active acquired immunity against a certain disease.

8.Give advantages of oral vaccines or edible vaccines.

SOLUTION

The benefit of such vaccines is the comfort of administration, low cost, and ease of storage.

9.Enlist different types of restriction enzymes commonly used in rDNA technology? Write about their role.

SOLUTION

There are three types of restriction enzyme:

i. Type I – These enzymes function simultaneously as endonuclease and methylase e.g. EcoK.

ii. Type II – These enzymes have separate activities for cleaving and methylation; they are more stable and are used in rDNA technology

e.g. EcoRI, BglII; these enzymes cut DNA at specific sites within the palindrome. There are thousands of type II restriction enzymes that are recognized/ discovered.

iii. Type III – These enzymes cut DNA at specific non-palindromic sequences e.g. HpaI, MboII.

Role of restriction enzymes:

Restriction enzymes either cut straight across the DNA in the region of palindrome to give blunt ends or cuts producing short, single-stranded projections at each end of DNA to produce, cohesive or sticky ends or staggered ends.

10.Enlist and write in brief about the different biological tools required in rDNA technology.

SOLUTION

The different biological tools required in rDNA technology:

i. Instruments: PCR, Agarose Gel Electrophoresis, SDS-PAGE 

Polymerase Chain Reaction (PCR) is the process of in vitro amplification of the gene of interest using a PCR machine.

i. PCR can generate a billion copies of the desired segment of DNA or RNA, with high accuracy and specificity, in a few hours.

ii. The process of PCR is completely automated and involves automatic thermal cycles for denaturation and renaturation of double-stranded DNA.

iii. The device required for PCR is called a thermal cycler.

iv. Requirements for polymerase chain reaction:

a. DNA containing the desired segment to be amplified

b. several molecules of four deoxyribonuclueoside triphosphates (dNTPs)

c. excess of two primer molecules

d. heat-stable DNA polymerase and

e. appropriate quantities of Mg⁺⁺ ions.

DNA replication through a polymerase chain reaction.

ii. Biological tools: Enzymes, Cloning Vectors, Competent host

Different enzymes used in rDNA technology are as follows: Lysozymes, Nucleases such as exonucleases, endonucleases, restriction endonucleases, DNA ligases, DNA polymerases, alkaline phosphatases, reverse transcriptase, etc.

i. Enzymes that cut the phosphodiester bonds of polynucleotide chains are called nuclease.

ii. These are of two types- exonuclease and endonuclease.

iii. Exonucleases cut nucleotides from the ends of DNA strands whereas endonuclease cut DNA from within.

iv. The phosphodiester backbone at highly specific sites on both strands of the duplex is cut by these enzymes called restriction endonucleases or simply restriction enzymes.

v. The restriction enzymes are thus the molecular scissors that are used to recognize and cut DNA at specific sequences.

vi. The sites recognized by them, are called recognition sequences or recognition sites.

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

The following characteristic properties a cloning vector must possess in order to be used in rDNA technology:

i. A good vector should have the ability of independent replication so that as the vector replicates (through ori gene) and a large number of copies of the DNA insert will be formed.

ii. The vector should be able to easily introduce into host cells.

iii. A vector should have marker genes for antibiotic resistance.

iv. A vector must contain a unique cleavage site in one of the marker genes for the restriction enzyme.

v. It should have at least suitable control elements like a promoter, operator, ribosomal binding sites, etc.

vi. The plasmids obtained naturally do not possess all the characteristics. Hence, they are constructed by inserting a gene for antibiotic resistance.

e.g. pBR322, pBR320, pACYC177 are the constructed plasmids. pBR322 is mostly used in rDNA technology in plants.

Competent hosts (cloning organisms) used are usually bacteria like Bacillus haemophilus, Helicobacter pylori, and E. coli. Mostly E. coli is used for the transformation with recombinant DNA.

COMPLETED

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Chapter 11, Enhancement of Food Production, hsc, biology, maharashtra board, 12th std, balbharathi solution,

Chapter 11: Enhancement of Food Production

Multiple Choice Question

1.Antibiotic Chloromycetin is obtained from ____________.

OPTIONS

• Streptomyces erythreus
• Penicillium chrysogenum
• Streptomyces venezuelae
• Streptomyces griseus

2.Removal of large pieces of floating debris, oily substances, etc. during sewage treatment is called _______________.

OPTIONS

• primary treatment
• secondary treatment
• final treatment
• amplification

3.Which one of the following is a free-living bacterial biofertilizer?

OPTIONS

• Azotobacter
• Rhizobium
• Nostoc
• Bacillus thuringiensis

4.Most commonly used substrate for industrial production of beer is ___________.

OPTIONS

• barley
• wheat
• corn
• sugarcane molasses

5.Ethanol is commercially produced through a particular species of _____________.

OPTIONS

• Aspergillus
• Saccharomyces
• Clostridium
• Trichoderma

6.One of the free-living anaerobic nitrogen-fixer is _________.

OPTIONS

• Azotobacter
• Beijerinckia
• Rhodospirillum
• Rhizobium

7.Microorganisms also help in production of food like ______________.

OPTIONS

• bread
• alcoholic beverages
• vegetables
• pulses

8.MOET technique is used for _____________.

OPTIONS

• production of hybrids
• inbreeding
• outbreeding
• outcrossing

9.Mule is the outcome of __________________.

OPTIONS

• inbreeding
• artificial insemination
• interspecific hybridization
• outbreeding

Very Short Answer Question

1.What does make idlies puffy?

SOLUTION

The bubbles of CO₂ trapped in dough during fermentation make idlies puffy.

2.Name any two bacterial biofertilizers.

SOLUTION

Biofertilizers:

a. Rhizobium:

1. Rhizobia are rod-shaped, motile, aerobic, gram-negative, non-spore-forming, nitrogen-fixing bacteria containing Nod genes and Nif genes.

2. They form symbiotic associations with the roots of leguminous plants.

3. They bring about nodule formation on the roots and multiply inside the nodule.

4. They fix atmospheric nitrogen into organic forms, which can be used by plants as nutrients.

5. For e.g. R. leguminosarum is specific to pea and R. phaseoli to beans.

b. Azotobacter:

1. It is the important and well known free-living, nitrogen-fixing, aerobic, non-photosynthetic, non-nodule forming bacterium, intimately associated with roots of grasses and certain plants.

2. It is used as a bio-fertilizer for all non-leguminous plants especially rice, cotton, vegetables, etc.

c. Azospirillum:

1. It is a free-living, an aerobic nitrogen-fixing bacterium associated with roots of corn, wheat, and jowar.

2. It fixes the considerable quantity of nitrogen (20-40kg N/ha) in non–leguminous plants such as cereals, millets, cotton, oilseed, etc.

3.What is the microbial source of vitamin B₁₂?

SOLUTION

The microbial source of Vitamin B₁₂ is – Pseudomonas denitrificans.

4.What is the microbial source of enzyme Invertase?

SOLUTION

The microbial source of enzyme Invertase is – Saccharomyces cerevisiae

5.Milk starts to coagulate when Lactic Acid Bacteria (LAB) is added to warm milk as a starter. Mention any two other benefits of LAB.

SOLUTION

Production of dairy products:

i. Curd: Indian curd is prepared by inoculating milk with Lactobacillus acidophilus. It also checks the growth of disease-causing microbes.

ii. Yogurt (yogurt): It is produced by curdling milk with the help of Streptococcus thermophilus and Lactobacillus bulgaricus.

iii. Butter Milk: The acidulated liquid left after churning of butter from curd, is called buttermilk.

iv. Cheese:

a. The milk is coagulated with lactic acid bacteria and the curd formed is filtered to separate whey.

b. The solid mass is then ripened with the growth of mould that develops flavor in it.

c. Different varieties of cheese are known by their characteristic texture, flavor, and taste which are developed by different specific microbes.

d. The ‘Roquefort and Camembert cheese’ are ripened by Bluegreen molds Penicillium roqueforti and P. camemberti respectively.

e. The large holes in Swiss cheese are developed due to the production of a large amount of CO₂ by a bacterium known as Propionibacterium shermanii.

6.Name the enzyme produced by Streptococcus bacterium. 

SOLUTION

Streptokinase:

Streptokinase enzyme (TPA) is produced by the bacterium Streptococcus spp.

7.Explain importance in medical sciences.

SOLUTION

It has a fibrinolytic effect. Hence, it is used as a ‘clot buster’ for clearing blood clots in the blood vessels of patients, which may cause a heart attack.

8.What is the breed?

SOLUTION

A group of animals related by descent and similar in most characters like general appearance, features, size, configuration, etc., are said to belong to a breed.

9.Define estuary.

SOLUTION

Estuaries are places where rivers meet the sea and may be defined as areas where saltwater is measurably diluted with freshwater.

10.What is shellac?

SOLUTION

Shellac is a pure form of lac obtained by washing and filtering.

Short Answer Question

1.Many microbes are used at home during the preparation of food items. Comment on such useful ones with examples.

SOLUTION

i. Many household preparations involve the use of microbes.

e.g. idli, dhokla, jalebi, dosa, etc.

ii. Microbes such as Leuconostoc and Streptococcus help in the fermentation of batter of idli and dosa.

iii. Microbes like Lactobacilli help in the preparation of jalebi and nan.

iv. Milk is fermented to make curd, yogurt, and cheese using lactobacilli.

v. Saccharomyces cerevisiae (yeast) is commonly used for making bread.

2.What is biogas?

SOLUTION

Biogas is a non-conventional and renewable source of energy and is obtained by microbial fermentation.

3.Write in brief about the production process.

SOLUTION

Process of biogas production:

Anaerobic digestion involves three processes:

a. Hydrolysis or solubilization:

1. In the initial stage, the raw material (cattle dung) is mixed with water in equal proportion to make a slurry which is then fed into the digester.

2. Here anaerobic hydrolytic bacteria (e.g. Clostridium, Pseudomonas) hydrolyse carbohydrates into simple sugars, proteins into amino acids, and lipids into fatty acids.

b. Acidogenesis: In this stage, facultatively anaerobic, acidogenic bacteria and obligate anaerobic organisms, convert simple organic material into acids like formic acid, acetic acid, H₂, and CO₂.

c. Methanogenesis: This is the last stage in which anaerobic Methanogenic bacteria like Methanobacterium, Methanococcus convert acetate, H₂, and CO₂ into Methane, CO₂ and H₂O, and other products.

1. 

12mol CHX3COOH(Acetic acid)⟶12CHX4Methane+12COX2

2. 

4mol H.COOH(Formic acid)⟶CHX4+3COX2+2HX2O

3. 

COX2+4HX2⟶CHX4+2HX2O

4.Write a note on biocontrol agents.

SOLUTION

i. The agents which are employed for biological control are called biocontrol agents.

ii. Microbes like bacteria, fungi, viruses, and protozoans act as biocontrol agents. They act in three ways, either they cause the disease to the pest or compete or kill them.

Example:

i. Bacillus thuringiensis

a. It is used to get rid of the butterfly, caterpillars where dried spores of Bacillus thuringiensis are mixed with water and sprayed onto vulnerable plants such as Brassica and fruit trees.

b. These spores are then eaten by insect larvae.

c. In the gut of the larvae, the toxin (cry protein) is released and the larvae get killed eventually.

ii. Trichoderma

a. Trichoderma species are free-living fungi found in the root ecosystem (rhizosphere).

b. These are effective as biocontrol agents of several soil-borne fungal plant pathogens.

c. The fungus produces substances like viridin, gliotoxin, gliovirin, etc. that inhibit the other soil-borne pathogens attacking root, rhizomes, etc. causing rot disease.

5.Name any two enzymes and antibiotics with their microbial source.

SOLUTION

No.	Enzyme	Microbial Source
i.	Invertase	Saccharomyces cerevisiae
ii.	Pectinase	Sclerotiana libertine, Aspergillus niger
iii.	Lipase	Candida lipolytica
iv.	Cellulase	Trichoderma koningi

No.	Antibiotic	Microbial source
i.	Chloromycetin	Streptomyces venezuelae
ii.	Erythromycin	Streptomyces erythreus
iii.	Penicillin	Penicillium chrysogenum
iv.	Streptomycin	Streptomyces griseus
v.	Griseofulvin	Penicillium griseofulvum
vi.	Bacitracin	Bacillus licheniformis
vii.	Oxytetracycline/ Terramycin	Streptomyces aureofaciens

6.Write the principles of farm management.

SOLUTION

i. Farm management begins with the selection of high yielding breeds, food requirements, the supply of adequate nutritional sources, and cleanliness of the environment, and maintenance of health.

ii. Management of farm animals includes veterinary supervision, vaccination, high yielding crossbreed development, production and preservation of products, distribution, and marketing.

7.Give the economic importance of fishery.

SOLUTION

Economic importance of fishes:

i. Fishes are caught, processed, raised, and marketed under fisheries. It provides good job opportunities and self-employment.

ii. Culturing of fishes on a large scale in ponds, lakes, and reservoirs boost the productivity and economy of the nation.

iii. Fishes are a source of nutritious food as they are rich in proteins, vitamins (A, D, and K), carbohydrates, fats, and minerals.

iv. They also yield a number of by-products that hold commercial value.

v. The by-products obtained from fishes include fish oil, fish meal, fertilizers, fish guano, fish glue, and isinglass, which are widely used in paints, soaps, oils, and medicines.

vi. Prawns and lobsters have a market value all over the world.

8.Enlist the species of honey bee mentioning their specific uses.

SOLUTION

Apiculture or beekeeping deals with an artificial rearing of honey bees to obtain bee products like honey, wax, pollen, bee venom, propolis (bee glue), and royal jelly as well as pollinating agents for crop plants.

The four species of honey bees commonly found in India are Apis dorsata (rock bee or wild bee), Apis florea (little bee), Apis mellifera (European bee), and Apis indica (Indian bee).

For beekeeping, Apis mellifera and Apis indica are the suitable species and are known as domesticated species.

9.What are A, B, C, and D in the table given below :

Types of microbe	Name	Commercial product
Fungus	A	Penicillin
Bacterium	Acetobacter aceti	B
C	Aspergillus niger	Citric acid
Yeast	D	Ethanol

SOLUTION

Types of microbe	Name	Commercial product
Fungus	Penicillium chrysogenum	Penicillin
Bacterium	Acetobacter aceti	Acetic acid (vinegar)
Fungus	Aspergillus niger	Citric acid
Yeast	Saccharomyces cerevisiae	Ethanol

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

Long Answer Question

Explain the process of sewage water treatment before it can be discharged into natural water bodies.

SOLUTION 1

The sewage in sewage treatment plants is carried out in two ways:

(a) Primary treatment: It involves the physical removal of particles from sewage through filtration and sedimentation. Floating debris is removed by filtration and grit is removed by sedimentation. Thus, all solids which settle form the primary sludge and the supernatant forms the effluent.

Secondary treatment: The effluent from primary treatment is passed to aeration tanks where the air is pumped into it. This allows the growth of useful aerobic microbes into flocs (masses of bacteria associated with fungal filaments) and microbes consume the major part of the organic matter in the effluent. This reduces the BOD (biological oxygen demand) of the effluent. The effluent is then passed into the settling tank where bacterial flocs are allowed to sediment. This sediment is called activated sludge. The small portion of this activated sludge is again passed to the aeration tank to serve as inocula. The remaining major part of this sludge is pumped into large anaerobic sludge digesters. Here, anaerobic bacteria digest bacteria and fungi in the sludge. During this digestion, bacteria produce a mixture of gases such as methane, H₂S, and CO₂. This treatment is essential as the sewage or municipal waste discharged into rivers, streams and other water bodies contains human excreta, organic wastes, and several pathogenic microbes.

SOLUTION 2

Before wastewater is made available for human use, it has to be treated properly, so as to remove organic matter, inorganic salts, and pathogens as well.

The sewage treatment process includes four basic steps as follows:

i. Preliminary Treatment: The preliminary treatment includes Screening and Grit Chamber.

a. Screening:

1. Sewage and wastewater contain plenty of suspended, floating materials, coarse and solid particles along with dissolved substances.

2. The suspended objects are filtered and removed. This is done in screening chambers.

3. The sewage is passed through screens or nets in the chambers. Larger suspended or floating objects are held back in the screening chambers.

4. These have to be removed before the biological treatment.

b. Grit Chamber:

1. After the screening, the filtered sewage is then passed into a series of grit chambers that contain large stones (pebbles) and brick ballast.

2. Coarse particles settle down by gravity.

3. Thus, the passage of filtered sewage removes much of the coarse particulate matter.

ii. Primary treatment (physical treatment):

a. After the preliminary treatment, the sewage water is pumped into the primary sedimentation tank.

b. The sedimentation of suspended solids or organic matter occurs in this tank.

c. About 50-70% of the solids settle down. There is a reduction of about 30-40% (in number) of coliform organisms.

d. The organic matter which is settled down is called primary sludge which is removed by mechanically operated devices.

e. The supernatant (effluent) in the primary sedimentation tank still contains a large amount of dissolved organic matter and micro-organisms which can then be removed by the secondary treatment.

iii. Secondary treatment (biological treatment):

a. The primary effluent is passed into large aeration tanks. Here it is constantly agitated mechanically and the air is pumped into it.

b. Aerobic bacteria grow vigorously and form flocs.

c. Flocs are the masses of bacteria held together by slime and fungal hyphae to form mesh-like masses.

d. These aerobic microbes consume the major part of the organic matter present in the effluent, as they grow.

e. Due to this BOD (Biochemical Oxygen Demand) of the effluent is significantly reduced.

iv. Tertiary treatment:

a. Once the BOD of wastewater is reduced, it is passed into a settling tank.

b. Here the bacterial flocs are allowed to sediment which is now called activated sludge.

c. A small part of this is passed back into the aeration tank and the major part is pumped into large tanks called anaerobic sludge digesters.

d. In these tanks, anaerobic bacteria grow and digest the bacteria and fungi in the sludge.

e. During this anaerobic digestion, gases such as methane, hydrogen sulphide, CO₂, etc. are produced.

f. Effluents from these plants (digester) after chlorination are released in natural water bodies like rivers and streams.

g. Chlorination kills pathogenic bacteria. h. Digested sludge is then disposed of.

2.Why is this sewage water treatment essential?

SOLUTION

Importance of sewage water treatment:

1. The sewage water contains pathogenic microorganisms like bacteria, viruses, protozoa, and parasitic worms, which can spread a variety of diseases.
2. Also, it contains harmful chemicals that may cause toxicity if mixed with natural resources of water. Therefore, sewage water treatment is essential before it can be discharged into natural bodies.

3.Write a note on lac culture.

SOLUTION

i. Lac is produced by an insect Tachardia lacca, which is quite small in size and colonial inhabit.

ii. Resin like substance is produced by the dermal glands of female lac insect.

iii. The insect feeds on succulent twigs of certain plants like ber, peepal, palas, Kusum, babool, etc. and secretes pink coloured resin, that hardens on coming in contact with air forming lac.

iv. Lac is a complex substance having a large amount of resin together with sugar, water, minerals, and alkaline substances.

v. Natural lac is always contaminated. vi. Shellac is a pure form of lac obtained by washing and filtering.

vii. Products of lac play a vital role in the economy of the farmers.

viii. Lac is used in bangles, toys, woodwork, inks, mirrors, etc. ix. Production of lac requires an artificial inoculation of plants which gives a better and regular supply of good quality and quantity of lac.

4.Describe various methods of fish preservation.

SOLUTION

After catching the fishes, fish spoilage is prevented by different preservation methods like:

i. Chilling with ice: It is a method of refrigeration. Due to the lowering of temperature, putrefaction is prevented and the quality of fish is maintained.

ii. Freezing: Freezing of fishes helps to retain the natural appearance of fishes.

iii. Freeze drying: Fishes are frozen and dried.

iv. Smoke drying: This removes the moisture and prevents the growth of bacteria.

v. Sun-drying: Fishes are dried in sun to remove moisture.

vi. Salting: By adding salt, dehydration takes place by osmosis.

vii. Canning: Fishes are preserved in cans with salt or other artificial preservatives.

5.Give an account of poultry diseases.

SOLUTION

Following are the different types of poultry diseases:

i. Viral diseases like Ranikhet, Bronchitis, Avian influenza (bird flu), etc.

ii. Bacterial diseases mainly include Pullorum, Cholera, Typhoid, TB, CRD (chronic respiratory disease), Enteritis, etc.

iii. Fungal diseases are Aspergillosis, Favus, and Thrush.

iv. Parasitic diseases include lice infection, roundworm, caecal worm infections, etc.

v. Protozoan diseases e.g. Coccidiosis is a protozoan disease.

6.Give an account of mutation breeding with examples.

SOLUTION

The mutation is a sudden heritable change in the genotype. In mutation breeding, desirable mutations are induced in crop plants with the use of physical or chemical mutagens.

i. Natural (physical) mutagens are:

High temperature, high concentration of CO2, X rays, UV rays.

ii. Chemical mutagens are:

Nitrous acid, EMS (Ethyl- Methyl- Sulphonate), Mustard gas, Colchicine, etc.

iii. Seedlings or seeds are irradiated by CO-60, exposed to UV bulbs, X-ray machines, etc.

iv. Mutagens cause gene mutations and chromosomal aberrations.

v. The treated seedlings are then screened for resistance to diseases/ pests, high yield, etc.

e.g. Jagannath variety of rice, NP 836 variety of wheat (rust-resistant), Indore-2 variety of cotton (resistant to bollworm), Regina-II variety of cabbage (resistant to bacterial rot), etc.

7.Describe briefly various steps of plant breeding methods.

SOLUTION

Hybridization involves the following steps:

i. Collection of variability:

a. Wild species and relatives of the cultivated species having desired traits are collected and preserved.

b. The entire collection having all the diverse alleles (i.e. variations) for all genes in a given crop is called germplasm collection.

c. Variations are useful in the selection. Germplasm conservation can be done in the following ways.

a. In situ conservation: It can be done with the help of forests and natural reserves.

b. Ex-situ conservation: It is done through botanical gardens, seed banks, etc.

ii. Evaluation and selection of parents:

a. The collected germplasm is evaluated (screened) to identify plants with desirable characters.

b. The selected parents must be healthy, vigorous, and should show desirable but complementary features.

c. The selected parents are selfed for three to four generations to make them pure or homozygous.

d. It is made sure that only pure lines are selected, multiplied, and used in the hybridization.

iii. Hybridization:

a. The variety showing maximum desirable features is selected as a female (recurrent) parent and the other one as male parent (donor) which lacks good characters found in a recurrent parent.

b. The pollen grains from anthers of male parents are collected and then artificially dusted over stigmas of emasculated flowers of the female parents.

c. Pollination is followed by seed and fruit formation in due course.

d. The seed thus obtained represents the hybrid generation.

e. The hybrid F₁ progeny is selected and evaluated for the desired combinations of characters.

iv. Selection and testing of superior recombinants:

a. The F₁ hybrid plants showing superiority over both the parents and having high hybrid vigour are selected.

b. Such hybrids are then selfed for a few generations to make them homozygous for the said desirable characters till there is a state of uniformity so that the characters will not segregate further.

v. Testing, release, and commercialization of new cultivars:

a. The newly selected lines are evaluated for productivity and other features like disease resistance, pest resistance, quality, etc.

b. Initially, these plants are grown under controlled conditions of water, fertilizers, etc. and their performance is recorded.

c. The selected lines are then grown for three generations at least in the natural field, in different agroclimatic zones.

d. Finally, variety is released as a new variety for use by the farmers.

COMPLETED

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Chapter 10: Human Health and Diseases, hsc, biology, maharashtra board, 12th std, balbharathi solution,

Chapter 10: Human Health and Diseases

Multiple Choice Question

1.Which of the following is NOT caused by unsterilized needles?

OPTIONS

• Elephantiasis
• AIDS
• Malaria
• Hepatitis B

2.Opium derivative is ____________.

OPTIONS

• Codeine
• Caffeine
• Heroin
• Psilocybin

3.The stimulant present in tea is ___________.

OPTIONS

• tannin
• cocaine
• caffeine
• crack

4.Which of the following is caused by smoking?

OPTIONS

• Liver cirrhosis
• Pulmonary tuberculosis
• Emphysema
• Malaria

5.An antibody is ___________.

OPTIONS

• the molecule that binds specifically an antigen
• WBC which invades bacteria
• secretion of mammalian RBC
• the cellular component of blood

6.The antiviral proteins released by a virus-infected cell are called ___________.

OPTIONS

• histamines
• interferons
• pyrogens
• allergens

7.Both B-cells and T-cells are derived from __________.

OPTIONS

• lymph nodes
• thymus glands
• liver
• stem cells in bone marrow

8.Which of the following diseases can be contracted by droplet infection?

OPTIONS

• Malaria
• Chicken pox
• Pneumonia
• Rabies

9.A confirmatory test used for detecting HIV infection is _________.

OPTIONS

• ELISA
• Western blot
• Widal test
• Eastern blot

10.Elephantiasis is caused by ____________.

OPTIONS

• W. bancrofti
• P. vivax
• Bedbug
• Elephant

11.Innate immunity is provided by __________.

OPTIONS

• phagocytes
• antibody
• T- Lymphocytes
• B- Lymphocytes

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Very Short Answer Question

1.What is the source of cocaine?

SOLUTION

Cocaine is an alkaloid obtained from the coca plant – Erythroxylum coca.

2.Name one disease caused by smoking?

SOLUTION

Cancer of mouth, lips, and lungs

3.Which cells stimulate B-cells to form antibodies?

SOLUTION

Helper T cells stimulate B cells to form antibodies.

4.What does the abbreviation AIDS stand for?

SOLUTION

The abbreviation of AIDS stands for Acquired Immuno Deficiency Syndrome.

5.Name the causative agent of typhoid fever?

SOLUTION

Salmonella typhi

6.What is the Rh factor?

SOLUTION

i. Rh is the most complex of the blood group system.

ii. Rh – factor is an antigenic protein present on the surface of the red blood cells in human beings.

iii. It was first discovered by Landsteiner and Wiener (1940), on the surface of RBCs of Rhesus monkey, so it is called the Rh factor/ Rhesus factor (also called D antigen).

iv. The person having Rh factor (D antigen) are called Rh-positive (Rh +ve) and those lacking D antigens are called Rh negative (Rh -ve).

7.What is schizont?

SOLUTION

Schizont is a cell formed from a trophozoite during the asexual stage of the life cycle of Plasmodium.

8.Name the addicting component found in tobacco.

SOLUTION

Nicotine is the addicting component found in tobacco.

9.Name the pathogen causing Malaria.

SOLUTION

Plasmodium is the pathogen that causes malaria.

10.Name the vector of Filariasis.

SOLUTION

Wuchereria bancrofti is transmitted to the human body by female Culex mosquito.

11.Give the name of the causative agent of ringworm.

SOLUTION

Dermatophytes of the genera Trichophyton and Microsporum are the most common causative agents of ringworm.

12.Define health.

SOLUTION

Health is defined as the state of complete physical, mental, and social well-being and not merely the absence of disease or infirmity.

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Short Answer Question

1.What are acquired diseases?

SOLUTION

Acquired diseases develop after birth. These diseases can be subdivided into the following categories:

a. Communicable or infectious diseases: The diseases which are transmitted from an infected person to another healthy person either directly or indirectly are known as Communicable or Infectious diseases. Communicable diseases are caused by pathogens like viruses, bacteria, fungi, helminth worms, etc.

e.g. Malaria, typhoid, ascariasis, etc.

b. Non-Communicable or Non-infectious diseases: The diseases that cannot be transmitted from an infected person to another healthy one either directly or indirectly are known as Non- Communicable or Non – Infectious diseases.

e.g. Cancer, deficiency diseases, etc.

Differentiate between 

2.Antigen and Antibody.

SOLUTION

Sr. No.	Antigen	Antibody
i.	Any foreign material which brings about an immune response in the body is called an antigen	Glycoproteins formed in the body to fight against antigen is called an antibody.
ii.	It triggers the formation of antibodies.	It is produced against antigens.
iii.	It may be a free molecule or component of the microbial cell surface.	It is extruded from the surface of plasma cells.
iv.	The structure of the antigen is variable.	The antibody is Y-shaped.

3.Name the infective stage of Plasmodium. Give any two symptoms of malaria.

SOLUTION

i. For humans, sporozoites are the infective stage of malaria.

ii. Signs and symptoms of malaria: Symptoms of malaria begin to appear about 7 to 15 days after the bite of an infective mosquito.

a. Initial symptoms are fever, headache, and chills.

b. The classical symptom of malaria is cyclic occurrence of high fever followed by sweating and sudden shivering. Such an entire episode lasts for four to six hours and recurs every two days or three days.

c. Vomiting and convulsions.

d. Arthralgia ( joint pain), anemia due to rupturing of RBCs.

e. Haemoglobinuria, hepatomegaly (liver enlargement).

f. Retinal damage (eye).

g. Cerebral malaria (brain infection)

4.Explain the mode of infection and cause of elephantiasis.

SOLUTION

The life cycle of Wuchereria bancrofti:

i. Wuchereria bancrofti has a complicated life cycle that consists of five stages.

ii. After mating of male and female worms, the female gives birth to thousands of live microfilariae.

iii. These microfilariae are taken up by vector insect (intermediate host), as a blood meal.

iv. In the intermediate host, microfilariae moult and develop into 3 rd stage (infective) larvae, in the intermediate host.

v. When the vector bites a healthy person, infectious larvae are injected in the dermis of the skin.

vi. After about one year, the larvae moult through two more stages maturing into adult worms.

Mode of transmission:

Wuchereria bancrofti is transmitted to the human body by female Culex mosquito. The larvae escape the mosquito bodies and arrive on the human skin.

They penetrate the skin, undergo two moultings before they become adults, and settle in the lymphatic system. The incubation period can be as long as 8-16 months.

5.Why is smoking a bad habit?

SOLUTION

i. Smoking tobacco is the greatest risk factor for cancer mortality.

ii. Tobacco smoke has more than 7000 chemicals and at least 250 are known to be harmful and more than 50 of them are known to cause cancer.

iii. Tobacco smoking causes many types of cancers, including cancers of the lung, oesophagus, larynx (voice box), mouth, throat, kidney, bladder, pancreas, stomach, and cervix. Hence, for all these reasons smoking is considered a bad habit.

6.What do the abbreviations AIIMS denote?

SOLUTION

AMIS: Antibody-mediated immune system

6.1What do the abbreviations CMIS denote?

SOLUTION

CMIS: Cell-mediated immune system

7.What is a carcinogen? 

SOLUTION

Factors that are known to cause cancer are called carcinogens.

7.1Name one chemical carcinogen with its target tissue.

SOLUTION

Carcinogen	Organ affected
Soot	Skin, lungs
Coal tar (3, 4 benzpyrene)	Skin, lungs
Cigarette smoke (N-nitrosodimethylamine)	Lungs
Cadmium oxide	Prostate gland
Aflatoxin (a metabolite of Aspergillus flavus, a mould)	Liver
2 – naphthylamine and 4 – aminobiphenyl	Urinary bladder
Mustard gas	Lungs
Nickel and chromium	Lungs
Asbestos	Lungs
Diethylstilbestrol (DES)	Vagina
Vinyl chloride (VC)	Liver

8.Distinguish between active immunity and passive immunity.

SOLUTION

No.	Active Immunity	Passive Immunity
a.	When resistance is developed by individuals as a result of an antigenic stimulus it is called active immunity.	When ready-made antibodies are directly given to protect the body against foreign agents, immunity is called ‘Passive immunity’.
b.	The types of active immunity are natural acquired active immunity and artificial acquired active immunity.	The types of passive immunity are natural acquired passive immunity and artificially acquired passive immunity.
c.	It has no side effects.	It may cause a reaction.
d.	It provides relief only after a long period.	It provides immediate relief.
e.	It is long-lasting immunity.	It is short-lived immunity.
e.g.	Polio vaccine, BCG vaccine, etc.	Rabies vaccine, maternal antibodies, etc.

Short Answer Question

1.Differentiate between B-cells and T-cells.

SOLUTION

No.	B-cells	T-cells
i.	B-cells mature in bone marrow	T-cells mature in the thymus gland.
ii.	They are involved in humoral immunity	They are involved in cell-mediated immunity.
iii.	B cells originate and mature in the bone marrow.	T cells originate in the bone marrow and mature in the thymus.
iv.	They occupy 20% of total lymphocytes	They occupy 80% of lymphocytes
v.	B-cells produce specific plasma cells that produce antibodies.	T-cells produce clone T-cells and can kill the infected cell.

2.What are the symptoms of malaria?

SOLUTION

Signs and symptoms of malaria:

Symptoms of malaria begin to appear about 7 to 15 days after the bite of an infective mosquito.

a. Initial symptoms are fever, headache, and chills.

b. The classical symptom of malaria is the cyclic occurrence of high fever followed by sweating and sudden shivering. Such an entire episode lasts for four to six hours and recurs every two days or three days.

c. Vomiting and convulsions.

d. Arthralgia ( joint pain), anemia due to rupturing of RBCs.

e. Haemoglobinuria, hepatomegaly (liver enlargement).

f. Retinal damage (eye).

g. Cerebral malaria (brain infection).

2.1How does malaria spread?

SOLUTION

Mode of transmission:

The malaria parasite life cycle involves two hosts. Plasmodium is transmitted from one person to another through an insect vector- female Anopheles mosquito.

When an infected female Anopheles sucks the human blood, it may transfer sporozoites to human circulation. Sporozoites reproduce asexually through the fission (schizogony) in the liver cells or erythrocytes.

The cells formed are now called merozoites. The cells formed within erythrocytes function as gametocytes (gamogony).

Besides, it forms gametocytes within erythrocytes (gamogony). Gametocytes if taken up by female Anopheles, fertilization occurs in its gut.

A diploid zygote is formed which transforms into an oocyst. Oocyst forms a large number of haploid sporozoites through meiosis (sporogony). Sporozoites migrate to salivary glands and are ready to infect the new human hosts.

Stages in the life cycle of Plasmodium in mosquito and human

3.Write a short note on AIDS.

SOLUTION

Modes of transmission of AIDS:

i. Unsafe sexual contact: Including oral, vaginal, and anal sex.

ii. Blood: Through blood transfusions or needle sharing.

iii. From mother to child (Transplacental): A pregnant woman can transmit the virus to her fetus through their shared blood circulation, or a nursing mother can transmit it to her baby from her breast milk.

iv. Other methods of spreading the virus are rare and include accidental needle injury, artificial insemination with infected donated semen, and transplantation with infected organs.

v. AIDS virus has been found in urine, tears, saliva, breast milk, and vaginal secretions but it seems not to be transmitted by these fluids unless it gets into injuries.

The clinical manifestations (symptoms) of AIDS have been classified into four broad categories.

i. Initial infection with the virus and formation of antibodies, usually lasting for 2-8 weeks after the initial infection.

ii. Asymptomatic carrier state in which no signs of disease, are seen. The incubation period ranges from 6 months to 10 years.

iii. AIDS-related complex (ARC) with one or more of the following clinical signs; recurrent fever for longer than one month, fatigue, unexplained diarrhoea, night sweats, shortness of breath, loss of more than 10 per cent body weight, etc.

iv. AIDS is the end stage of HIV infection. It is characterised by life-threatening opportunistic infections (like pneumonia, tuberculosis, Kaposi’s sarcoma, etc.).

Preventive measures:

AIDS has no cure; hence prevention is the best choice. The following steps help in preventing AIDS:

i. People, particularly those in a high – risk group, should be educated about HIV transmission.

ii. Disposable needles and syringes should be used and disposed of properly and immediately.

iii. Unsafe sexual habits should be changed.

iv. High-risk groups should refrain from donating blood.

v. Toothbrushes, razors, other articles that can become contaminated with blood should not be shared.

vi. Before receiving blood, ensure that it has been screened for not containing HIV infections.

vii. Routine screening must be done for:

a. Blood donors.

b. Organ donors (kidney, liver, lung, cornea).

c. Donors of semen and growth hormone.

d. Patients undergoing haemodialysis and females in the high-risk group who are pregnant or contemplating pregnancy.

i. At first, a test ELISA (Enzyme-Linked Immunosorbent Assay) is used to detect HIV antibodies.

ii. The second confirmatory test is usually a Western Blot. It is used to eliminate any false-positive results. It is a highly specific test and it is based on detecting specific antibody to viral core protein and envelope glycoprotein.

Although AIDS has no cure, certain medicines called Antiretroviral drugs can help in reducing the viral load and prolong the life of HIV patient. e.g. TDF (Tenofovir), EFV(Efavirenz), Lamivudine (3TC), etc.

4.Give the symptoms of cancer.

SOLUTION

Symptoms of cancer are as follows:

i. Cancer symptoms are quite varied and depend on where the cancer is located, where it has spread, and the size of the tumour.

ii. Some cancers can be felt or seen through the skin – a lump on the breast or testicle can be an indicator of cancer in those locations.

iii. Melanoma (skin cancer) is often noted by a change in a wart or mole on the skin.

iv. Some oral cancers present as white patches inside the mouth or white spots on the tongue.

v. Metastasis of cancer can produce additional symptoms in the newly affected area, like swollen or enlarged lymph nodes.

vi. Vertigo, headache, seizures may be experienced if cancer spreads to the brain.

vii. In addition, the liver may become enlarged and cause jaundice.

viii. The bones become painful, brittle, and break easily. ix. Coughing and shortness of breath may occur if the lungs are affected due to cancer.

5.Write a note on antigens on blood cells.

SOLUTION

There are several known antigens on the surface of human red blood cells. These antigens give rise to different blood groups.

ABO Blood Groups:

The A, B, and O blood groups were discovered by Karl Landsteiner in 1900.

He found two antigens or agglutinogens on the surface of human red blood cells and named them as antigen A and antigen B. He also noticed the corresponding antibodies or agglutinins in the serum called ‘a’ and ‘b’.

Blood group	Genotype	An antigen on the Surface of RBC	Antibody in Serum	Can donate blood to	Can receive blood from
A	IA IA orIA IO	A	Antibody b	A, AB	A, O
B	IB IB orIB IO	B	Antibody a	B, AB	B, O
AB (universal acceptor)	IA IB	A and B	Nil	AB	A, B, AB, O
O (universal donor)	IO IO	Nil	Both Antibody a and Antibody b	A, B, AB, O	O

In ABO system, the blood groups are determined by the presence or absence of antigen A and antigen B. The blood group of a person is classified into four groups A, B, AB, and O.

Blood group A: Individuals, with blood group ‘A’, has the antigen A on the surface of their red blood cells (RBCs) and antibody ‘b’ in their plasma.

Blood group B: Individuals with blood group ‘B’ have the antigen B on the surface of their RBCs and antibody ‘a’ in their plasma.

Blood group AB: Individuals with blood group ‘AB’ have both antigens A and B on the surface of their RBCs and no antibodies in their plasma.

Blood group O: Individuals with blood group ‘O’ lack both antigens A and B on the surface of their RBCs and show the presence of both ‘a’ and ‘b’ antibodies in their plasma.

Rh factor: Rh – factor is an antigenic protein present on the surface of the red blood cells in human beings.

6.Write a note on the antigens-antibody complex.

SOLUTION

i. Each antibody is specific for a particular antigen.

ii. Combining sites of antigen, called antigenic determinants (epitopes) react with the corresponding antigen-binding sites of antibodies called paratopes.

iii. The antigen-binding sites (paratopes) are located on the variable regions of the antibody.

iv. Small variations in the variable regions make each antibody highly specific for a particular antigen.

v. The variable region enables the antibody to recognize the specific antigen and bind to a specific antigen in a lock and key manner forming an antigen-antibody complex.

Formation of antigen-antibody complex

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

SOLUTION

Some public health measures that must be taken to safeguard the population against infectious diseases are as follows:

i. Improved access to vaccinations and contraceptives (to prevent STDs)

ii. Facilitation of screening, counseling, and education of those at risk of infection.

iii. Support to access to treatment

iv. Following good hygiene practices like hand washing

v. Infection control standard, contact, droplet, and airborne precautions

vi. Procedures for decontamination of persons and disinfection of equipment and environment, if needed.

vii. Quarantine of contacts (if necessary)

viii. Prophylaxis of exposed individuals

ix. Control of the vectors of transmission of infection

x. Spreading awareness about the route of transmission of infected diseases

xi. Practice good food – safety techniques

xii. Precautionary measures while traveling to areas of known epidemics.

xiii. Controlling spread by rodents like rats

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8.How does the transmission of the following diseases take place?

Amoebiasis

SOLUTION

i. Causative organism: Entamoeba histolytica (protozoan).

ii. Mode of transmission:

a. Amoebiasis is usually transmitted by the faeco-oral route.

b. It can be transmitted indirectly through contact with dirty hands or objects.

c. It is transmitted through contaminated food and water. The infection spreads through the ingestion of the cyst form of the parasite.

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8.1How does the transmission of the following diseases take place?

Malaria

SOLUTION

Mode of transmission :

The malaria parasite life cycle involves two hosts.

Plasmodium is transmitted from one person to another through an insect vector- female Anopheles mosquito.

When an infected female Anopheles sucks the human blood, it may transfer sporozoites to human circulation. Sporozoites reproduce asexually through the fission (schizogony) in the liver cells or erythrocytes.

The cells formed are now called merozoites. The cells formed within erythrocytes function as gametocytes (gamogony). Besides, it forms gametocytes within erythrocytes (gamogony). Gametocytes if taken up by female Anopheles, fertilization occurs in its gut.

A diploid zygote is formed which transforms into an oocyst.

Oocyst forms a large number of haploid sporozoites through meiosis (sporogony).

Sporozoites migrate to salivary glands and are ready to infect the new human hosts.

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8.2How does the transmission of the following diseases take place?

Ascariasis

SOLUTION

Life cycle of Ascaris lumbricoides

The life cycle of Ascaris involves only one host, i.e. human.

i. Ascaris is present in the gastrointestinal tract and faecal matter. Adult worms live in the lumen of the small intestine.

ii. Ingestion of infective eggs takes place from contaminated vegetables and water is the primary route of infection.

iii. A female worm may produce approximately 2, 00,000 eggs per day, which are passed with the feces.

iv. Unfertilized eggs may be ingested but are not infective.

v. Fertile eggs develop into embryos and become infective after 18 days to several weeks depending on the environmental conditions (Optimum conditions: moist, warm, shaded soil).

vi. After swallowing the infective eggs, the larvae hatch and are carried via. circulation to lymphatics and the lungs.

vii. The larvae mature further in the lungs (10 to 14 days), penetrate the alveolar walls, ascend the respiratory tract to the throat, and are swallowed. Upon reaching the small intestine, they develop into adult worms.

viii. Two-three months are required from the ingestion of the infective eggs to oviposition by the adult female. Adult worms can live for 1 to 2 years.

ix. The eggs appear in stools after 60-70 days. In larval ascariasis, symptoms occur 4-16 days after infection.

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Mode of transmission

a. Food and drinks contaminated with the eggs of these worms is the main mode of transmission.

b. Eggs hatch inside the intestine of the new host.

c. The larvae pass through various organs and settle as adults in the digestive system.

8.3How does the transmission of the following diseases take place?

Pneumonia

SOLUTION

i. Pneumonia mostly spreads by direct person to person contact.

ii. It can also spread via droplets released by an infected person or even by using shared clothes and utensils.

9.What measure would you take to prevent water-borne diseases?

SOLUTION

Waterborne diseases can be prevented in the following ways:

i. Drink only boiled water

ii. If not boiled, water should be chlorinated and filtered.

iii. Practice good personal hygiene

iv. Take food safety precautions like proper washing and cooking of food.

v. Use disposable glass and plates wherever possible while eating food outside

vi. Take vaccinations against preventable diseases like typhoid, hepatitis A, polio, etc.

10.Write a short note on typhoid.

SOLUTION

• Salmonella typhi is the agent of typhoid fever.
• Signs and symptoms of typhoid are as follows:

1. Prolonged fever is as high as 1040 F.
2. General nausea, fatigue and headache.
3. Abdominal pain, constipation or diarrhoea.
4. Rose-coloured rash on the skin.
5. Whitecoat on tongue, cough.
6. Anorexia (loss of appetite).
7. vii. If not treated on time, the patient may experience breathlessness, irregular heartbeats and haemorrhage.

• Modes of transmission of typhoid are as follows:

1. It is a food and water-borne disease.
2. Insects like housefly and cockroaches feeding on faecal matter may transfer the bacteria to the food material.
3. Poor hygiene habits and poor sanitation conditions are responsible for the spread of typhoid.

• Typhoid fever can be confirmed by the Widal Test.
• Antibiotics like Chloromycetin is a helpful treatment. Treatment of typhoid involves surgical removal of the gall bladder in severe cases.
• The two vaccines for the prevention of typhoid are:

1. Live, oral Ty21a vaccine (sold as Vivotif Berna)
2. Injectable Typhoid polysaccharide vaccine (sold as Typhim Vi by Sanofi Pasteur and Typherix by GlaxoSmithKline).

1.Match the following

Column I	Column II
a. AIDS	i. Antibody production
b. Lysozyme	ii. Activation of B-cells
c. B-cells	iii. Immunoglobulin
d. T-helper cells	iv. Tears
e. Antibody	v. Immunodeficiency

SOLUTION

Column I	Column II
a. AIDS	v. Immunodeficiency
b. Lysozyme	iv. Tears
c. B-cells	i. Antibody production
d. T-helper cells	ii. Activation of B-cells
e. Antibody	iii. Immunoglobulin

Long Answer Question

1.Describe the structure of an antibody.

SOLUTION

Structure of antibody

Antibodies are glycoproteins which are highly specific to specific antigens. 

Antibodies are also known as Immunoglobulins (Igs), produced in response to antigenic stimulation. 

Antibodies are produced by plasma cells which in turn are formed by B–lymphocytes. 

The mature plasma cells produce antibodies at an extremely rapid rate i.e. about 2000 molecules per second.

Structure of antibody:

i. The antibody is a ‘Y’ shaped molecule.

ii. Each immunoglobulin molecule is made up of four polypeptide chains.

iii. There are two heavy or H-chains and two light or L-chains.

iv. The four polypeptide chains are held together by disulfide bonds (-s s-) to form a ‘Y’ shaped structure.

v. The region holding together arms and stem of the antibody is termed as the hinge.

vi. Each chain of the antibody includes two distinct regions, the variable region, and the constant region.

vii. Variable regions constitute the antigen-binding site (paratope).

viii. This part of the antibody recognizes and binds to the specific antigen to form an antigen-antibody complex.

ix. Since most antibodies carry two antigen-binding sites, they are said to be bivalent.

2.Write a note on Vaccination.

SOLUTION 1

i. Administration of vaccine (i.e. inactivated pathogen or antigenic protection of particular pathogen) to protect against a particular pathogen, is called vaccination.

ii. Normally, the body’s immune system helps to protect against pathogens that cause infection.

iii. However, some pathogens can overwhelm the immune system. This results in serious illness.

iv. The pathogens most likely to cause illness, are the ones the body doesn’t recognize and which go undetected by the immune system.

v. Vaccination is a method in which the immune system is exposed to a safe antigenic preparation that helps it recognize and eliminate a pathogenic organism when it is encountered.

vi. Vaccination is an important form of primary prevention, that can protect people from getting sick.

vii. Vaccination has helped control deadly diseases like measles, polio, tetanus, and whooping cough.

viii. The more people that are vaccinated, the more protected the individuals and the society are against diseases by a phenomenon known as herd immunity.

SOLUTION 2

i. Vaccination teaches the body to recognize new pathogens causing diseases.

ii. It stimulates the body to make antibodies against antigens of pathogens.

iii. It also primes immune cells to remember the types of antigens, allowing a faster response to the pathogen in future encounters.

iv. Vaccines work by exposing a person to a safe version of a pathogen.

v. The different forms of vaccines are as follows:

a. A protein or sugar from the pathogen.

b. A dead or inactivated form of a pathogen

c. A toxoid containing toxin produced by a pathogen.

d. A weakened (attenuated) pathogen. When the body responds to the vaccine, it builds an adaptive immune response. This helps to equip the body to fight off an actual infection.

3.What is cancer?

SOLUTION

An uncontrolled growth and multiplication of cells resulting in a harmful tumour are called cancer.

3.1.Differentiate between a benign tumor and a malignant tumor.

SOLUTION

No.	Benign cancer	Malignant cancer
i.	It grows slowly	The growth rate of this tumour is rapid
ii.	The mortality rate is extremely low	The mortality rate is comparatively high
iii.	These cells remain restricted to the site of origin and do not spread to other parts of the body.	These cells spread from one organ to another via blood or lymph by metastasis
iv.	They may become malignant	They form secondary tumours
v.	They are non-malignant/ noncancerous.	They are malignant/ cancerous.

3.2.Name the main five types of cancer.

SOLUTION

The five types of cancer are:

i. Carcinoma

ii. Sarcoma

iii. Lymphoma

iv. Leukemia

v. Adenocarcinomas

4.Describe the different types of immunity.

SOLUTION

There are two types of immunity i.e. Innate or Inborn (inherited) immunity and Acquired or Adaptive immunity.

i. Innate immunity or Inborn Immunity:

Innate immunity is the resistance to infections that an individual possesses genetically.

It is the natural (inborn) defense system of the body. 

It is not affected by prior contact with microorganisms or immunization. 

It is non-specific when it indicates a degree of resistance to infection in general. 

Innate immunity comprises the various types of barriers that prevent the entry of foreign agents into the body.

a. Epithelial surface: The intact skin and mucous membranes (secrete mucously) covering the body, protects against invasion by a microorganism(s). 

The healthy skin possesses bactericidal activity due to the presence of high concentrations of salt in drying sweat. Sebaceous secretions and a long chain of fatty acids have bactericidal and fungicidal properties. 

The mucosa of the respiratory tract has several innate mechanisms of defense.

The nose prevents entry of microorganisms to a large extent, the inhaled particles being arrested through the hair at or near the nasal orifices. 

Those foreign particles that pass the nasal orifices are held by the mucus lining the epithelium and are swept back to the pharynx where they tend to swallow or coughed out. 

The cough reflex is an important defense mechanism of the respiratory tract. The mouth is constantly bathed in saliva which has an inhibitory effect on microorganisms. The acidity of gastric secretions in the stomach destroys micro-organisms.

The flushing action of urine eliminates bacteria from the urethra. 

Spermine and zinc present in semen are antibacterial.

b. Antimicrobial substances in blood and tissues:

The complement system contains more than 30 serum proteins, circulating in the blood in an inactive state. The presence of microbial pathogens activates the ‘Complement cascade’ to eliminate pathogens. The interferons are a class of cytokines (soluble proteins) released by viral cells infected with viruses and certain white blood cells to stimulate other cells to protect themselves from viral infection.

c. Cellular factors in innate immunity:

Natural defense against the invasion of blood and tissues by microorganisms and other foreign particles is mediated to a large extent by phagocytic cells which ingest and destroy them. Phagocytic cells are grouped as macrophages and macrophages. These cells remove foreign particles that enter the body. A class of lymphocytes called Natural killer (NK) cells is important in non-specific defense against viral infections and tumors.

d. Fever:

An increase in body temperature following the infection is a natural defense mechanism. It helps to accelerate the physiological processes to destroy the invading pathogens. Fever stimulates the production of interferons and helps in recovery from viral infections.

e. Acute-phase proteins (APPs):

Infection on injury leads to a sudden increase in the concentration of certain plasma proteins, collectively called acute-phase proteins. These include C Reactive Protein (CRP), Mannose-binding protein, Alpha-1-acid glycoprotein, Serum Amyloid P, etc. APPs are believed to enhance host resistance, prevent tissue injury, and promote repair of inflammatory lesions.

ii. Acquired immunity:

The resistance that an individual acquires during life is known as Acquired immunity or Adaptive or Specific immunity. It involves the formation of antibodies in the body, which neutralize the antigens.

Acquired immunity is of two types Active and Passive.

a. Active immunity:

It is the resistance developed by individuals as a result of an antigenic stimulus (exposure to antigen).

It also is known as “Adaptive immunity”. Active immunity may be natural or artificial.

1. Natural Acquired Active immunity: Immunity acquired due to infection is called natural active immunity. It is developed after the entry of pathogens in the body. It is long-lasting immunity. e.g. Person who has recovered from an attack of measles develops naturally acquired active immunity to measles, for a lifetime.

2. Artificial Acquired Active immunity: It is the resistance induced by vaccines. The vaccine is introduced into the body to stimulate the formation of antibodies by the immune system. e.g. Polio vaccine, BCG vaccine, etc. such immunity may be temporary or permanent.

b. Passive immunity: Passive immunity is acquired when ready-made antibodies are received by the body cells. i.e. Body cells do not take any active part in the production of immunity. Passive immunity can be acquired either naturally or artificially.

1. Natural Acquired Passive immunity: Before birth maternal antibodies are transferred from mother to foetus through the placenta. After birth, antibodies are transferred from mother to infant through colostrum (first milk of mother) and continue throughout the period of breastfeeding. The antibodies received by the baby from the mother remain in the body for a short time. Therefore, natural acquired passive immunity is short-lived.

2. Artificially Acquired Passive immunity: This immunity is developed by injecting previously prepared antibodies using serum from humans or animals. e.g. Antibodies obtained from hyper immunised horses are injected to humans against rabies pathogens. It is short-lived.

5.Describe the ill –effects of alcoholism on health.

SOLUTION

Effects of Drug/ Alcohol Abuse:

i. Behavioural changes: The immediate adverse effects of drugs and alcohol abuse are manifested in the form of reckless behaviour, vandalism and violence.

ii. Coma/ Death: Excessive doses of drugs may lead to coma and death due to respiratory failure, heart failure or cerebral haemorrhage. A combination of drugs or their intake along with alcohol generally results in overdose and even deaths.

iii. Social changes: The most common warning signs of drug and alcohol abuse among youth include a drop in academic performance, unexplained absence from school/college, lack of interest in personal hygiene, withdrawal, isolation, depression, fatigue, aggressive and rebellious behaviour, deteriorating relationships with family and friends, loss of interest in hobbies, change in sleeping and eating habits, fluctuations in weight, appetite, etc.

iv. Crimes: If an abuser is unable to get money to buy drugs/ alcohol he/she may turn to crime. At times, a drug/alcohol addict becomes the cause of mental and financial distress to his/her entire family and friends.

v. Adverse health effects/ Diseases: Those who take drugs intravenously (direct injection into the vein using a needle and syringe) are likely to acquire serious infections like HIV and hepatitis B. Use of alcohol during adolescence may also have longterm effects like loss balance, liver cirrhosis, pancreatitis. It could lead to heavy drinking in adulthood. Chronic use of drugs and alcohol damages nervous system and liver (cirrhosis). Use of drugs and alcohol during pregnancy adversely affects the foetus.

vi. Misuse in Sports: Another misuse of drugs is that certain sportspersons use drugs to enhance performance. They (mis)use narcotic analgesics, anabolic steroids, diuretics and certain hormones to increase muscle strength and bulk and to promote aggressiveness and an overall improvement in their performance.

vii. Side-effects of the use of anabolic steroids in females: Masculinization (features like males), increased aggressiveness, mood swings, depression, abnormal menstrual cycles, excessive hair growth on the face and body, enlargement of the clitoris, deepening of the voice.

viii. Side-effects of the use of anabolic steroids in males: Acne, increased aggressiveness, mood swings, depression, and reduction of the size of the testicles, decreased sperm production, kidney and liver dysfunction, breast enlargement, premature baldness, enlargement of the prostate gland. These effects may be permanent with prolonged use.

6.In your view, what motivates youngsters to take to alcohol or drugs and how can this be avoided?

SOLUTION

i. Addiction is a psychological attachment to certain effects such as euphoria and a temporary feeling of well being associated with drugs and alcohol.

ii. These feelings drive people to take them even when these are not needed, or even when their use becomes self-destructive.

iii. With repeated use of drugs, the tolerance level of the receptors present in our body increases.

iv. Consequently, the receptors respond only to higher doses of drugs or alcohol leading to greater intake and addiction.

Prevention:

i. Habits such as smoking, taking drugs, or alcohol are more likely to be taken up at a young age, more during adolescence.

ii. It is best to identify the situations that push an adolescent towards the use of drugs or alcohol and to take remedial measures well in time. In this regard, the parents and the teachers have a special responsibility.

7.Do you think that friends can influence one to take alcohol/drugs? If yes, how may one protect himself/herself from such an influence?

SOLUTION

Yes, friends can influence one to take alcohol/ drugs.

One can protect oneself from such influence by the following ways:

i. Counseling: Seek counseling to deal with mental illness or traumas experienced. One can also attend support groups, read self – help books, and seek advice from a therapist. There are many specific national helpline numbers that can also offer help.

ii. Support: Share your fears and problems with a reliable person like a teacher or a parent who can support you and help you. This will make it easier to avoid stress and the desire to adopt an addiction.

iii. Choose friends carefully: Peer pressure is one of the main causes of addictions amongst teenagers. Hence, it is best to avoid people who you know are addicted. Carefully choose your friend circle and who you decide to confide in.

iv. Avoid starting young: Those who start drinking or using drugs at an early age (even in small amounts) are most likely to develop an addiction later in life.

v. Be aware of the consequences of addictions: Stay aware of the effects of drug and alcohol addiction on your social, physical, and mental health and prevent adverse impact on yourself by refraining from it.

vi. Develop strong bonds: Social connections can help protect you against the risk of substance use.

vii. Participate in anti-drug, tobacco, and alcohol programs: Early interventions can help a person. If you have adopted a bad habit, or feel you are on the path of addiction, admit the problem and seek timely help to prevent it from turning into an addiction.

COMPLETED

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Chapter 9: Control and Co-ordination, hsc, biology, maharashtra board, 12th std, balbharathi solution,

Chapter 9: Control and Co-ordination

Multiple choice question.

1.The nervous system of mammals uses both electrical and chemical means to send signals via neurons. Which part of the neuron receives impulse?

OPTIONS

• Axon
• Dendron
• Nodes of Ranvier
• Neurilemma

2.___________ is a neurotransmitter.

OPTIONS

• ADH
• Acetyl CoA
• Acetyl choline
• Inositol

3.The supporting cells that produce myelin sheath in the PNS are _________.

OPTIONS

• Oligodendrocytes
• Satellite cells
• Astrocytes
• Schwann cells

4.A collection of neuron cell bodies located outside the CNS is called _________.

OPTIONS

• Tract
• Nucleus
• Nerve
• Ganglionl

5.Receptors for protein hormones are located ________.

OPTIONS

• in cytoplasm
• on cell surface
• in nucleus
• on Golgi complex

6.If the parathyroid gland of man is removed, the specific result will be ___________.

OPTIONS

• onset of aging
• disturbance of Ca⁺⁺
• onset of myxoedema
• elevation of blood pressure

7.Hormone thyroxine, adrenaline, and nonadrenaline are formed from _____________.

OPTIONS

• Glycine
• Arginine
• Ornithine
• Tyrosine

8.Pheromones are chemical messengers produced by animals and released outside the body. The odour of these substance affects _____________.

OPTIONS

• skin colour
• excretion
• digestion
• behaviour

9.Which one of the following is a set of discrete endocrine glands?

OPTIONS

• Salivary, thyroid, adrenal, ovary
• Adrenal, testis, ovary, liver
• Pituitary, thyroid, adrenal, thymus
• pituitary, pancreas, adrenal, thymus

10.After ovulation, the Graafian follicle changes into ___________.

OPTIONS

• Corpus luteum
• Corpus albicans
• Corpus spongiosum
• Corpus callosum

11.Which one of the following pairs correctly matches a hormone with a disease resulting from its deficiency?

OPTIONS

• Parathyroid hormone – Diabetes insipidus
• Leutinising hormone – Diabetes mellitus
• Insulin – Hyperglycemia
• Thyroxine – Tetany

12.___________ is in direct contact of brain in human.

OPTIONS

• Cranium
• Duramater
• Arachnoid
• Piamater

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Very very short answer question.

1.What is the function of the red nucleus?

SOLUTION

Near the centre of the midbrain is a mass of grey matter scattered within the white matter. It is called the red nucleus. It plays an important role in controlling posture and muscle tone, modifying some motor activities, and motor coordination.

2.What is the importance of Corpora quadrigemina?

SOLUTION

The corpora quadrigemina are four rounded elevations on the dorsal surface of the midbrain. The two superior colliculi are involved in visual reflexes and the two inferior colliculi are relay centres for auditory reflexes that operate when it is necessary to move the head to hear sounds more distinctly.

3.What does the cerebellum of the brain control?

SOLUTION

Cerebellum: It is the second-largest part of the brain and consists of two lateral hemispheres and a central vermis. It is composed of white matter with a thin layer of grey matter, the cortex. The white matter intermixes with the grey matter and shows a tree-like pattern called arborvitae. The surface of the cerebellum shows convolutions (gyri and sulci) a number of nuclei lie deep within each lateral or cerebellar hemisphere. Over 30 million neurons lie in the cortex. Three pairs of myelinated nerve bundles called cerebullar penduncles to connect the cerebellum to the other parts of the CNS.

Functions: It is an important centre that maintains the equilibrium of body, posture, balancing orientation, moderation of voluntary movements, maintenance of muscle tone. It is a regulatory center for neuromuscular activities and controls rapid activities like walking, running, speaking, etc. All activities of the cerebellum are involuntary (though may involve learning in early stages).

4.Name the three ossicles of the middle ear.

SOLUTION

Middle ear: It consists of a chain of three ear ossicles called Malleus (hammer), Incus (anvil), and Stapes (stirrup-the smallest bone). On receiving the vibrations from the tympanic membrane, the ear ossicles amplify the vibrations and transfer these to the cochlea.

A short Eustachian tube connects the middle ear to the pharynx. It equalizes air pressure on both sides of the eardrum.

5.Name the hormone which is an anti-abortion hormone.

SOLUTION

Progesterone

6.Name an organ that acts as a temporary endocrine gland.

SOLUTION

Placenta

7.Name the type of binding to DNA and alter gene expression.

SOLUTION

Steroids

8.What is the cause of abnormal elongation of long bones of arms and legs and of the lower jaw?

SOLUTION

Excessive secretion of Growth Hormone causes abnormal elongation of long bones of arms and legs and of the lower jaw.

9.Name the hormone secreted by the pineal gland.

SOLUTION

Melatonin

10.Which endocrine gland plays an important, role in improving immunity?

SOLUTION

Thymus gland

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1.Match the organism with the type of nervous system found in them.

1. Neurons	a. Earthworm
2. Ladder-type	b. Hydra
3. Ganglion	c. Flatworm
4. Nerve net	d. Human

SOLUTION

1. Neurons	d. Human
2. Ladder-type	c. Flatworm
3. Ganglion	a. Earthworm
4. Nerve net	b. Hydra

^{Very short answer question.}

1.Describe the endocrine role of islets of Langerhans.

SOLUTION

Islets of Langerhans are endocrine cells of the pancreas. They are four types of cells in Islets of Langerhans which have an endocrine role i.e. they secrete hormones.

i. Alpha cells (α cells): They constitute 20% of Islets of Langerhans. They secrete hormone glucagon. Glucagon stimulates glycogenolysis (the breakdown of glycogen) in the liver which causes hyperglycemia.

ii. Beta cells (β cells): They constitute 70% of Islets of Langerhans. They secrete insulin which stimulates glycogenesis (formation of glycogen) in the liver and muscles. Insulin causes hypoglycemia by increasing the uptake of glucose by cells.

iii. Delta cells (δ cells): They constitute 10% of Islets of Langerhans. These cells secrete somatostatin which inhibits the secretion of insulin and glucagon. It also lowers gastric secretions, motility, and absorption in the digestive tract. Somatostatin inhibits the release of growth hormone.

iv. PP cells or F cells: These cells secrete pancreatic polypeptide (PP) and inhibit the release of pancreatic juice.

2.Mention the function of testosterone?

SOLUTION

The group of hormones secreted by testis is androgens such as testosterone.

Functions of androgens:

i. It is also responsible for the appearance of secondary sexual characters such as facial and pubic hair, deepening of the voice, broadening of shoulders, male aggressiveness, etc.

ii. It involves in development of testis.

3.Give symptoms of the disease caused by hyposecretion of ADH.

SOLUTION

This condition causes excessive micturition or polyuria, polydipsia (increased thirst), etc.

Short answer question

1.Rakesh got hurt on his head when he fell down from his motorbike. Which inner membranes must have protected his brain? What other roles do they have to play?

SOLUTION

i. When Rakesh fell down from his motorbike the inner membranes called meninges protected his brain from injury.

ii. These meninges form a protective covering around the brain and spinal cord. They act as shock absorbers.

2.Give a reason – Injury to medulla oblongata may prove fatal.

SOLUTION

i. The medulla oblongata is a part of the brain stem.

ii. It controls involuntary vital functions like heartbeat, respiration, vasomotor activities, and peristalsis.

iii. It also controls non-vital reflex activities like coughing, sneezing, swallowing, vomiting, yawning, etc.

iv. Thus, damage or injury to medulla oblongata may disrupt these vital functions. Therefore, injury to medulla oblongata may prove fatal.

3.Distinguish between the sympathetic and parasympathetic nervous system on the basis of the effect they have on: a. Heartbeat b. Urinary Bladder

SOLUTION

i. Heart: Sympathetic nervous system accelerates the heartbeat whereas the parasympathetic nervous system decelerates the heartbeat.

ii. Urinary bladder: Sympathetic nervous system inhibits bladder contraction whereas the parasympathetic nervous system stimulates bladder contraction.

4.While holding a teacup Mr. Kothari’s hands rattle. Which disorder he may be suffering from and what is the reason for this?

SOLUTION

i. Mr. Kothari may be suffering from Graves’ disease.

ii. It is caused due to increased levels of thyroid hormone or hyperthyroidism.

5.List the properties of the nerve fibres.

SOLUTION

i. Excitability/Irritability: Nerve fibres have polarized membranes, thus they have the ability to perceive stimulus and enter into a state of activity.

ii. Conductivity: It is the ability of the nerve to transmit impulses along the whole length of the axon.

iii. Stimulus: It is any detectable, physical, chemical, electrical change in the external or internal environment which brings about excitation in a nerve/muscle/organ/organism. A stimulus must have a minimum intensity called threshold stimulus, in order to be effective. The subliminal (weak) stimulus will have no effect while the supraliminal (strong) stimulus will produce the same degree of impulse as the threshold stimulus.

iv. Summation effect: A single subliminal stimulus will have no effect but when many such weak stimuli are given again and again they may produce an impulse due to summation of effects.

v. All or none law: The nerve will either conduct the impulse along its entire length or will not conduct the impulse at all. This occurs in the case of a subliminal or weak stimulus.

vi. Refractory period: It is the time interval (about a millisecond) during which a nerve fails to respond to a second stimulus even if it is strong.

vii. Synaptic delay: The impulse takes about 0.3 to 0.5 milliseconds to cross a synapse. It is required for the release of neurotransmitters from the axon terminal and excitation in the dendron of the next neuron.

viii. Synaptic fatigue: The transmission of nerve impulses across the synapse stops temporarily due to the depletion of the neurotransmitter.

ix. Velocity: The rate of transmission of impulse is higher in long and thick nerves. It is higher in homeotherms than in poikilotherms. The velocity of transmission is higher in voluntary fibres (100 – 120 m/s in man) as compared to autonomic or involuntary nerves (10-20 m/s). In medullated nerve fibre, the velocity of transmission is higher as an impulse has to jump from one node of Ranvier to the next.

6.How does the tongue detect the sensation of taste?

SOLUTION

The tongue detects the sensation of taste due to gustatoreceptors.

7.State the site of production and function of Secretin, Gastrin, and Cholecystokinin.

SOLUTION

i. Site of production: Secretin, gastrin, and cholecystokinin are secreted in the gastrointestinal tract

ii. Functions:

Secretin: It is responsible for the secretion of pancreatic juice from the pancreas and bile from the liver.

Gastrin: It stimulates gastric glands to produce gastric juice.

Cholecystokin in CCK/ Pancreozymin PZ: It stimulates the pancreas to release enzymes and also stimulates the gall bladder to release bile.

8.An adult patient suffers from low heart rate, low metabolic rate, and low body temperature. He also lacks alertness, intelligence, and initiative. What can be this disease? What can be its cause and care?

SOLUTION

i. The patient may be suffering from myxoedema.

ii. It is caused due to the deficiency of thyroid hormones (hypothyroidism) in adults.

iii. Care: Patients should take prescribed medications regularly and eat food rich in iodine.

9.Where is the pituitary gland located? Enlist the hormones secreted by the anterior pituitary.

SOLUTION

The pituitary gland located just below the hypothalamus.

Hormones secreted by anterior pituitary:

i. Somatotropic Hormone (STH) / Somatotropin / Growth Hormone (GH): The secretion of GH is high till puberty later its secretion becomes low. However, it is continuously secreted throughout life for repair and replacement of body tissue or cells.

Functions:

a. It stimulates the growth of the body and the development of all tissues.

b. It accelerates protein synthesis and cell division.

c. It stimulates the release of growth hormone.

ii. Thyroid Stimulating Hormone (TSH) / Thyrotropin:

Function:

It stimulates the thyroid gland to produce the hormone thyroxine.

iii. Adrenocorticotropic Hormone (ACTH) / Adrenocorticotropin:

Functions:

a. It stimulates the adrenal cortex to produce its hormones.

b. It maintains the functioning of the adrenal cortex.

iv. Prolactin / Luteotropin/ Mammotropin : The secretion of this hormone is regulated by PIF (Prolactin inhibiting factor) of the hypothalamus.

Functions:

a. Activates the growth of mammary glands during pregnancy (mammotropin).

b. Stimulates milk production and secretion of milk (lactogenic) by the mammary gland after childbirth.

v. Gonadotropins:

a. Follicle Stimulating Hormone (FSH): In males, it stimulates the development of seminiferous tubules. In females, it stimulates the growth of ovarian follicles.

b. Luteinizing hormone (LH): LH induces the ruptured follicles to develop into corpus luteum and for the production of progesterone FSH and LH are responsible for the stimulation of ovaries to produce estrogen.

c. ICSH: In males, it stimulates the testes to produce the androgen called testosterone. Testosterone is responsible for the development of secondary sexual characters.

10.Explain how the adrenal medulla and sympathetic nervous system function as a closely integrated system.

SOLUTION

i. The adrenal medulla is the inner region of the adrenal gland. It is the modified sympathetic ganglion of the autonomic nervous system (ANS).

ii. The chromaffin cells of the adrenal medulla secrete hormones rather than releasing a neurotransmitter. These cells are innervated by sympathetic pre-ganglionic neurons of the autonomous nervous system (ANS).

iii. The autonomous nervous exerts direct control over the chromaffin cells, thus the hormones – adrenaline and nor adrenaline can be released quickly into the blood.

iv. The impulses from the hypothalamus stimulate sympathetic pre-ganglionic neurons which in turn stimulate the chromaffin cells to secrete adrenaline to nor-adrenaline.

v. The fight-or-flight response is initiated by nerve impulses from the hypothalamus to the sympathetic nervous system, including the adrenal medulla. This response rapidly increases circulation, promotes ATP production, and decreases non-essential activities.

Thus, the adrenal medulla and sympathetic nervous system function in a closely integrated manner.

11.Name the secretion of alpha, beta, and delta cells of islets of langerhans. Explain their role.

SOLUTION

Islets of Langerhans are endocrine cells of the pancreas.

They are four types of cells in Islets of Langerhans which have an endocrine role i.e. they secrete hormones.

i. Alpha cells (α cells): They constitute 20% of Islets of Langerhans. They secrete hormone glucagon. Glucagon stimulates glycogenolysis (the breakdown of glycogen) in the liver which causes hyperglycemia.

ii. Beta cells (β cells): They constitute 70% of Islets of Langerhans. They secrete insulin which stimulates glycogenesis (formation of glycogen) in the liver and muscles. Insulin causes hypoglycemia by increasing the uptake of glucose by cells.

iii. Delta cells (δ cells): They constitute 10% of Islets of Langerhans. These cells secrete somatostatin which inhibits the secretion of insulin and glucagon. It also lowers gastric secretions, motility, and absorption in the digestive tract. Somatostatin inhibits the release of growth hormone.

iv. PP cells or F cells: These cells secrete pancreatic polypeptide (PP) and inhibit the release of pancreatic juice.

12.Which are the 2 types of goitre? What are its causes?

SOLUTION

Hypersecretion of thyroid hormones: It is caused by an increase in the levels of thyroid hormones. This increases metabolic rate, sensitivity, sweating, flushing, rapid respiration, bulging of eyeballs, and affects various physiological activities. Graves’ disease: Hyperthyroidism in adults results in this disorder. It is characterised by protruding eyeballs, increased BMR, and weight loss. Increased BMR produces a range of effects like increased heartbeat, increased B.P., higher body temperature, nervousness, irritability, and tremor of fingers.

Simple goitre: It is iodine deficiency goitre. Iodine is required for the synthesis of thyroid hormone and if there is a deficiency of iodine in the diet, it causes enlargement of the thyroid gland leading to simple goitre. This disease is common in hilly areas. Addition of iodine to table salt prevents this disease. The size of the thyroid gland is increased but the total output of thyroxine is decreased.

13.Name the ovarian hormone and give their functions.

SOLUTION

Ovaries secrete the following hormones:

i. Progesterone: It is secreted by the corpus luteum of the ovary after ovulation. It is essential for the thickening of the uterine endometrium, thus preparing the uterus for implantation of the fertilized ovum. It is responsible for the development of mammary glands during pregnancy. It inhibits uterine contractions during pregnancy.

ii. Oestrogen: It is secreted by developing follicles. Estradiol is the main oestrogen. It is responsible for the development of secondary sexual characters in females.

iii. Relaxin: It is secreted by the corpus luteum of the ovary at the end of the gestation period. It relaxes the cervix of the pregnant female and the ligaments of the pelvic girdle to ease out the birth process.

iv. Inhibin: It is secreted by the corpus luteum. Inhibin inhibits FSH and GnRH production.

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Answer the following.

1.Complete the table.

Location	Cell Type	Function
PNS	___________	Produce myelin sheath
PNS	Satellite cells	_____________
___________	Oligodendrocytes	Form myelin sheath around central axon
CNS	___________	Phagocytose pathogens
CNS	___________	Form the epithelial lining of brain cavities and central canal.

SOLUTION

Location	Cell Type	Function
PNS	Schwann cells	Produce myelin sheath
PNS	Satellite cells	Support the function of neurons
CNS	Oligodendrocytes	Form myelin sheath around the central axon
CNS	Microglia or brain macrophages	Phagocytose pathogens
CNS	Ependymal cells	Form the epithelial lining of brain cavities and central canal.

Long answer question.

1.Explain the process of conduction of nerve impulses up to the development of action potential.

SOLUTION

Polarisation and Depolarisation along a nerve

i. The nerve impulse is a wave of bioelectrical or electrochemical disturbances passing along a neuron.

ii. Neurons have a charged cellular membrane with a voltage that is different on the outer and inner side of the membrane. The plasma membrane separates the outer and inner solutions of different chemical compounds but having approximately the same total number of ions.

iii. The external tissue fluid has both Na⁺ and K⁺ but there is a predominance of Na⁺ and Cl^–, while K⁺ is predominant within the fibre or in the intracellular fluid. This condition of a resting nerve is also called a polarised state.

iv. The polarized state of a neuron is established by maintaining an excess of Na⁺ on the outer side. On the inside, there is an excess of K⁺ along with large negatively charged protein molecules and nucleic acid.

v. Some amount of Na⁺ and K⁺ always leaks across the membrane. The Na⁺/K⁺ pump actively maintains the ionic gradient across the resting membrane. The sodium pump or Na-K allows the entry of K⁺ inside the membrane and exit of Na⁺.

vi. The difference in the distribution of Na⁺ and K⁺ on the two sides of the membrane produces a potential difference of – 50 to –100 millivolts (average is – 70 millivolts).

vii. The potential difference seen in a resting nerve is thus called resting potential (–70 millivolts) and it is mainly due to differential permeability of the resting membrane, which is much more permeable to K⁺ than to Na⁺. This results in slightly more K⁺ diffusing out than Na⁺ moving inside and causing a slight difference in polarity.

viii. Also, ions like negatively charged proteins and nucleic acids inside the cell make the overall charge negative on the inside and positive charge on the outside. The nerve membrane not only has leakage channels but also has many gated channels for Na⁺ /K⁺. These are also called voltage-gated channels. These channels enable the neuron to change its membrane potential to active potential in response to stimuli. The Na⁺ /K⁺ gated channels are separate so the transport of both these ions is separately done. However, during resting potential, both these gates are closed and the membrane resting potential is maintained.

ix. The resting potential of the membrane is maintained unless the stimulus reaches the neuron. Any change or disturbance to the membrane will cause Na⁺ to enter into the membrane and lower the potential difference (lesser than –70 millivolts). Thus, the membrane becomes more permeable to Na⁺

x. During resting potential, both gates are closed, and resting potential is maintained. However, during depolarization, the Na⁺ gates open and the K⁺ gates remain closed. This causes Na⁺ to rush into the axon and bring about depolarisation (opposite of polarity).

xi. The Extra Cellular Fluid (ECF) becomes electronegative with respect to the inner membrane which becomes electropositive. The value of action potential reaches +30 millivolts to +60 millivolts. This triggers depolarization in the next part while it itself starts going to repolarisation.

2.Draw the neat labelled diagram of the human ear.

SOLUTION

Structure of the human ear

2.1.Draw the neat labelled diagram of the Sectional view of the human eye.

SOLUTION

A sectional view of the human eye

2.2.Draw the neat labelled diagram of L. S. of the human brain.

SOLUTION

Sagital section of the brain

2.3.Draw the neat labelled diagram of Multipolar Neuron.

SOLUTION

3.Answer the question after observing the diagram given below.

What do the synaptic vesicles contain?

SOLUTION

Synaptic vesicles contain neurotransmitter molecules.

3.1Answer the question after observing the diagram given below.

What process is used to release the neurotransmitter?

SOLUTION

A neurotransmitter is released by the process of exocytosis.

3.3Answer the question after observing the diagram given below.

What should be the reason for the next impulse to be conducted?

SOLUTION

a. When a neuron receives an impulse, it passes it to the next neuron. The impulse travels along the axon of the pre-synaptic neuron to the axon terminal.

b. Pre-synaptic neurons or axons have synaptic knobs at their ends or terminals. These synaptic knobs have synaptic vesicles that contain neurotransmitter molecules.

c. When the impulse reaches a synaptic knob, Ca⁺ channels open and Ca⁺ ions diffuse inward from the extracellular fluid.

d. This causes the release of neurotransmitters that bind to the receptors of the post synaptic cell.

e. The neurotransmitter is destroyed by the enzyme cholinesterase. A new impulse/ next impulse is generated and conducted to the synaptic gap.

3.4Answer the question after observing the diagram given below.

Will the impulse be carried by the postsynaptic membrane carried even if one pre-synaptic neuron is there?

SOLUTION

a. A pre synaptic neuron when receives an impulse, releases a neurotransmitter into the synaptic cleft which is required to cross the gap between the axon terminal and the next neuron.

b. For further transmission of impulse, a pre-synaptic neuron is required that initiates the release of neurotransmitters that facilitate the movement of impulses across synapses.

3.5.Answer the question after observing the diagram given below.

Can you name the channel responsible for their transmission?

SOLUTION

The ligand-gated ion channel is responsible for the transmission of the impulse.

4.Explain the Reflex Pathway with the help of a neat labelled diagram.

SOLUTION

i. The reflex pathway comprises at least one afferent neuron (receptor) and one efferent (effector or excitor) neuron appropriately arranged in a series.

ii. The afferent neuron receives a signal from a sensory organ and transmits the impulse via a dorsal nerve root into the CNS (at the level of the spinal cord).

iii. The efferent neuron then carries signals from CNS to the effector. The stimulus and response thus form a reflex arc as shown below in the knee jerk reflex.

5.Krishna was going to school and on the way, he saw a major bus accident. His heartbeat increased and hands and feet become cold. Name the part of the nervous system that had a role to play in this reaction.

SOLUTION

The sympathetic nervous system controls body activities during fight, fright, or flight situations. It activates the release of the hormones adrenaline and nor-adrenaline due to which the heartbeat increased, hands and feet become cold.

6.What will be the effect of thyroid gland atrophy on the human body?

SOLUTION

Thyroid gland atrophy causes hypothyroidism.

For effects of thyroid gland atrophy on the human body:

Hyposecretion of thyroid hormone: It is caused by a deficiency of thyroid hormones or removal of the thyroid gland (Thyroidectomy).

a. Cretinism: It is caused due to deficiency of thyroid hormones in infants. A cretin (individual suffering from cretinism) has reduced BMR and low oxidation. They are short-statured because the skeleton fails to grow. They are mentally retarded, show stunted growth and delayed puberty. They show dry skin, thick tongue, prolonged neonatal jaundice, lethargy and constipation. This can be treated by early administration of thyroid hormones.

b. Myxoedema: It is caused due to the deficiency of thyroid hormones in adults. It is characterised by a peculiar thickening and puffiness of skin and subcutaneous tissue particularly of the face and extremities. The patient lacks alertness, intelligence. The patient suffers from slow heart rate, low B.P., low body temperature (feels cold) and stunted sexual development.

c. Simple goitre: It is iodine deficiency goitre. Iodine is required for the synthesis of thyroid hormone and if there is a deficiency of iodine in the diet, it causes enlargement of the thyroid gland leading to simple goitre. This disease is common in hilly areas. Addition of iodine to table salt prevents this disease. The size of the thyroid gland is increased but the total output of thyroxine is decreased.

7.Write the names of hormones and the glands secreting them for the regulation of the following functions.

a. Growth of thyroid and secretion of thyroxine.

b. Helps in relaxing pubic ligaments to facilitate the easy birth of young ones.

c. Stimulate intestinal glands to secrete intestinal juice.

d. Controls calcium level in the blood

e. Control tubular absorption of water in kidneys.

f. Urinary elimination of water.

g. Sodium and potassium ion metabolism.

h. Basal Metabolic rate.

i. Uterine contraction.

j. Heartbeat and blood pressure.

k. Secretion of growth hormone.

l. Maturation of Graafian follicle.

SOLUTION

Functions	Hormones	Glands
Growth of thyroid and secretion of thyroxine	TSH (Thyroid-stimulating hormone)	Pituitary
Helps in relaxing pubic ligaments to facilitate the easy birth of young ones.	Relaxin	Ovaries
Stimulate intestinal glands to secrete intestinal juice	Cholecystokinin	Gastrointestinal tract
Controls calcium level in the blood	Calcitonin, Parathormone	Thyroid, parathyroid respectively
Controls tubular absorption of water in kidneys	ADH/ Vasopressin	Hypothalamus
Urinary elimination of water	Antidiuretic hormone	Hypothalamus
Sodium and potassium ion metabolism	Aldosterone (mineralocorticoid)	Adrenal cortex
Basal Metabolic rate	Thyroxine	Thyroid
Uterine contraction	Oxytocin	Hypothalamus
Heartbeat and blood pressure	Adrenaline, nor adrenaline	Adrenal medulla
Secretion of growth hormone	Somatotropin	Pituitary
Maturation of Graafian follicle	LH (Luteinizing hormone)	Pituitary

8.Explain the role of the hypothalamus and pituitary as a coordinated unit in maintaining homeostasis?

SOLUTION

i. The hypothalamus controls the secretory activity of the pituitary gland (anterior pituitary) by producing, releasing, and inhibiting hormones.

ii. Anterior pituitary and intermediate lobes are connected to the hypothalamus through the hypophyseal portal system. Various hormones secreted by the hypothalamus reach the pituitary gland through the hypophyseal portal system.

iii. The portal vein collects blood from various parts of the hypothalamus and opens into the anterior lobe of the pituitary. From the pituitary, the vein finally carries the blood into the superior vena cava. It helps in the feedback mechanism for hormonal control.

iv. Also, a negative feedback mechanism takes place in the form of hormones released by the target glands to decrease the secretion of the pituitary gland.

v. In such a negative feedback mechanism, the secretion of ACTH, TSH, and gonadotropins (FSH and LH) decreases when their target gland hormone levels rise.

e.g. Adrenocorticotropic hormone (ACTH) stimulates the cortex of the adrenal gland to secrete glucocorticoids, mainly cortisol. In turn, an elevated blood level of cortisol decreases secretion of both corticotropin and corticotropin-releasing hormone (CRH) by suppressing the activity of the anterior pituitary corticotrophs and neurosecretory cells.

9.What is adenohypophysis? Name the hormones secreted by it?

SOLUTION

Adenohypophysis: It is an outgrowth from the roof of the buccal cavity. This outgrowth is called Rathke’s pouch. It grows upwards towards the brain. It is the larger lobe of the pituitary gland. It is a highly cellular and vascular part of the pituitary gland. It contains various types of epitheloid secretory cells, acidophils, basophils, chromatophores. It is further divided into three parts – Pars distalis, pars tuberalis, and pars intermedia. Pars intermedia is poorly developed in human beings. It is a small reduced part lying in the cleft between the anterior and posterior lobe. It secretes Melanocyte Stimulating Hormone (MSH) in some lower vertebrates. MSH stimulates the dispersion of melanin granules in melanocytes and is responsible for skin pigmentation.

The pituitary gland located just below the hypothalamus.

Hormones secreted by anterior pituitary:

i. Somatotropic Hormone (STH) / Somatotropin / Growth Hormone (GH): The secretion of GH is high till puberty later its secretion becomes low. However, it is continuously secreted throughout life for repair and replacement of body tissue or cells.

Functions:

a. It stimulates the growth of the body and the development of all tissues.

b. It accelerates protein synthesis and cell division.

c. It stimulates the release of growth hormone.

ii. Thyroid Stimulating Hormone (TSH) / Thyrotropin:

Function:

It stimulates the thyroid gland to produce the hormone thyroxine.

iii. Adrenocorticotropic Hormone (ACTH) / Adrenocorticotropin:

Functions:

a. It stimulates the adrenal cortex to produce its hormones.

b. It maintains the functioning of the adrenal cortex.

iv. Prolactin / Luteotropin/ Mammotropin: The secretion of this hormone is regulated by PIF (Prolactin inhibiting factor) of the hypothalamus.

Functions:

a. Activates the growth of mammary glands during pregnancy (mammotropin).

b. Stimulates milk production and secretion of milk (lactogenic) by the mammary gland after childbirth.

v. Gonadotropins:

a. Follicle Stimulating Hormone (FSH): In males, it stimulates the development of seminiferous tubules. In females, it stimulates the growth of ovarian follicles.

b. Luteinizing hormone (LH): LH induces the ruptured follicles to develop into corpus luteum and for the production of progesterone FSH and LH are responsible for the stimulation of ovaries to produce estrogen.

c. ICSH: In males, it stimulates the testes to produce the androgen called testosterone. Testosterone is responsible for the development of secondary sexual characters.

10.Describe in brief, an account of disorders of the adrenal gland.

SOLUTION

Disorders of the adrenal gland:

i. Addison’s disease: It is caused due to hypersecretion of glucocorticoids (hormone secreted adrenal cortex). It is characterized by low blood sugar, low Na⁺, and high K⁺ concentration in plasma increased loss of Na⁺ and water in urine. It leads to weight loss, weakness, nausea, vomiting, and diarrhoea.

ii. Cushing’s disease: It is caused due to Hyposecretion of mineralocorticoids (a hormone secreted by the adrenal cortex) It leads to high blood sugar levels, excretion of glucose in the urine, rise in Na⁺ level in the blood, high blood pressure, obesity, and wasting of limb muscles.

11.Explain the action of steroid hormones and proteinous hormones.

SOLUTION

Hormones are released in very minute quantities. They produce their effect on the target cells by binding to hormone receptors. The hormone receptors are present on the cell membrane (i.e. membrane receptors) or maybe intracellular receptors.

i. Mode of hormone action through membrane receptors:

a. Hormones like catecholamines, peptide, and polypeptide hormones are not lipid-soluble and they cannot enter their target cells through the plasma membrane. These non-steroid water-soluble hormones interact with surface receptors and initiate metabolic activity.

b. Molecules of amino acid derivatives, peptide hormones bind to specific receptor molecules located on the plasma membrane.

c. The hormone-receptor complex causes the release of an enzyme adenylate cyclase from the receptor site. This enzyme forms cyclic AMP from ATP of the cell.

d. cAMP activates enzymatic actions. The hormone acts as the first messenger and the cAMP acts as the second messenger.

e. Some other secondary messengers are Ca++, cGMP, and IP3 (Inositol triphosphate), etc.

ii. Mode of action through intracellular receptors:

a. Steroid and thyroid hormones are lipid-soluble and can easily pass through the plasma membrane of the target cell into the cytoplasm.

b. In the cytoplasm, they bind to specific intracellular receptors proteins forming a hormone-receptor complex that enters the nucleus.

c. The hormone-receptor complex binds to a specific regulatory site of DNA, in the nucleus.

d. The activated genes transcribe mRNA which directs protein synthesis and enzymes in the cytoplasm.

e. The action of lipid-soluble hormones is slow but long-lasting,

Mechanism of hormone action through membrane receptor

Mechanism of hormone action through intracellular receptor

12.Describe in brief an account of disorders of the thyroid.

SOLUTION

Disorders of the thyroid gland are caused due to hypersecretion and hyposecretion of thyroid hormones.

i. Hypersecretion of thyroid hormones: It is caused by an increase in the levels of thyroid hormones. This increases metabolic rate, sensitivity, sweating, flushing, rapid respiration, bulging of eyeballs, and affects various physiological activities.

Graves’ disease: Hyperthyroidism in adults results in this disorder. It is characterised by protruding eyeballs, increased BMR, and weight loss. Increased BMR produces a range of effects like increased heartbeat, increased B.P., higher body temperature, nervousness, irritability, and tremor of fingers.

ii. Hyposecretion of thyroid hormone: It is caused by a deficiency of thyroid hormones or removal of the thyroid gland (Thyroidectomy).

a. Cretinism: It is caused due to deficiency of thyroid hormones in infants. A cretin (individual suffering from cretinism) has reduced BMR and low oxidation. They are short-statured because the skeleton fails to grow. They are mentally retarded, show stunted growth and delayed puberty. They show dry skin, thick tongue, prolonged neonatal jaundice, lethargy and constipation. This can be treated by early administration of thyroid hormones.

b. Myxoedema: It is caused due to the deficiency of thyroid hormones in adults. It is characterised by a peculiar thickening and puffiness of skin and subcutaneous tissue particularly of the face and extremities. The patient lacks alertness, intelligence. The patient suffers from slow heart rate, low B.P., low body temperature (feels cold) and stunted sexual development.

c. Simple goitre: It is iodine deficiency goitre. Iodine is required for the synthesis of thyroid hormone and if there is a deficiency of iodine in the diet, it causes enlargement of the thyroid gland leading to simple goitre. This disease is common in hilly areas. Addition of iodine to table salt prevents this disease. The size of the thyroid gland is increased but the total output of thyroxine is decreased.

iii. Hyposecretion of thyroid hormones in pregnant females causes defective development and maturation of growing baby.

COMPLETED

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Chapter 8: Respiration and Circulation, hsc, biology, maharashtra board, 12th std, balbharathi solution,

Chapter 8: Respiration and Circulation

Choose the correct alternative.

1.The muscular structure that separates the thoracic and abdominal cavity is _______.

OPTIONS

• pleura
• diaphragm
• trachea
• epithelium

2.What is the minimum number of the plasma membrane that oxygen has to diffuse across to pass from air in the alveolus to haemoglobin inside a RBC?

• Two
• Three
• Four
• Five

3.______ is a sound-producing organ.

OPTIONS

• Larynx
• Pharynx
• Tonsils
• Trachea

4.The maximum volume of gas that is inhaled during breathing in addition to T.V is _______.

OPTIONS

• residual volume
• I.R.V.
• G.R.V.
• Vital capacity

5._______ muscles contract when the external intercostal muscles contract.

OPTIONS

• Internal abdominal
• Jaw
• Muscles in bronchial walls
• Diaphragm

6.Movement of cytoplasm in unicellular organisms is called __________.

OPTIONS

• diffusion
• cyclosis
• circulation
• thrombosis

7.Which of the following animals do not have closed circulation?

OPTIONS

• Earthworm
• Rabbit
• Butterfly
• Shark

8.Diapedesis can be seen in _________ cell.

OPTIONS

• RBC
• WBC
• Platelet
• Neuron

9.Pacemaker of heart is _________.

OPTIONS

• SA node
• AV node
• His bundle
• Purkinje fibers

10.Which of the following is without a nucleus?

OPTIONS

• Red blood corpuscle
• Neutrophil
• Basophil
• Lymphocyte

11.Cockroach shows which kind of circulatory system?

OPTIONS

• Open
• Closed
• Lymphatic
• Double

12.Diapedesis can be seen in _________ cell.

OPTIONS

• RBC
• WBC
• Platelet
• Neuron

13.Opening of inferior vena cava is guarded by _______.

OPTIONS

• bicuspid valve
• tricuspid valve
• Eustachian valve
• Thebesian valve

14._______ wave in ECG represent atrial depolarization.

OPTIONS

• P
• QRS complex
• Q
• T

15.The fluid seen in the intercellular spaces in human is _________

OPTIONS

• blood
• lymph
• interstitial fluid
• water
• www.asterclasses.com

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1.Match the Respiratory surface to the organism in which it is found.

Respiratory surface	Organism
Plasma membrane	Insect
Lungs	Salamander
External gills	Bird
Internal gills	Amoeba
Trachea	Fish

SOLUTION

Respiratory surface	Organism
Plasma membrane	Amoeba
Lungs	Bird
External gills	Salamander
Internal gills	Fish
Trachea	Insect

Very short answer question.

1.Why does trachea have ‘C’ shaped rings of cartilage?

SOLUTION

The trachea has ‘C’ shaped rings of cartilage as they prevent the trachea from collapsing.

2.Why is respiration in insect called direct respiration?

SOLUTION

In insects, the respiratory system is independent of its circulatory system as blood does not play a direct role in oxygen transport but the tracheal tubes directly transport oxygen to the entire body. Therefore, respiration in the insect is called direct respiration.

3.Why is a gas exchange very rapid at the alveolar level?

SOLUTION

Alveoli are lined by layer of simple squamous epithelial epithelium. This thin, single layer of epithelium allows the rapid exchange of gases in alveolar region.

4.Name the organ which prevents the following the entry of food into the trachea while eating.

SOLUTION

Epiglottis

Short answer question.

1.Why is it advantageous to breathe through the nose than through the mouth?

SOLUTION

i. Nose filters and warms the inhaled air. Hair in the nose prevents the entry of microbes, dust, and other impurities which may harm the lungs.

ii. Mouth lacks any such structures for filtering and warming the air that is inhaled during inspiration. Hence, it is advantageous to breathe through the nose than through the mouth.

2.Identify the incorrect statement and correct it,

a. A respiratory surface area should have a large surface area.

b. A respiratory surface area should be kept dry.

c. A respiratory surface area should be thin, maybe 1mm or less.

SOLUTION

b. A respiratory surface should be kept dry.

Correct statement – A respiratory surface should be moist in order to facilitate the exchange of gases.

3.Given below is the characteristic of some modified respiratory movement. Identify them.

Spasmodic contraction of muscles of expiration and forceful expulsion of air through nose and mouth.

SOLUTION

Sneezing

4.Given below is the characteristic of some modified respiratory movement. Identify them.

An inspiration followed by many short convulsive expirations accompanied by facial expression.

SOLUTION

Crying

5.Write a note on blood plasma.

SOLUTION

Human blood consists of plasma and blood corpuscles or blood cells

Plasma: It is a straw-coloured, slightly alkaline, viscous fluid. It constitutes 55% of the blood.

Plasma consists of water, proteins (albumin, globulin, properdin, prothrombin, fibrinogen), inorganic salts (Na, K, Mg, Ca, Fe, Mn and Cl^–, HCO3- , and PO43-), food (glucose, amino acids, fatty acids, triglycerides), wastes (urea, uric acid and creatinine), regulators (hormones, enzymes, vitamins), anticoagulants (heparin), cholesterol and antibodies, dissolved gases (O₂, CO₂, N₂) Plasma contains 90% water, 7-8% proteins, inorganic salts – 1% and other substances 1-2%.

Blood Corpuscles: It constitutes 44% of the blood. Blood corpuscles are of three types as given below:

1. RBC (Red Blood Corpuscles) or Erythrocytes:

1. Erythrocytes are the most abundant cells in the human body.

2. They are circular, biconcave and enucleated (in camel and llama they are nucleated).

3. The red colour or RBCs is due to an oxygen-carrying pigment, the haemoglobin, in their cytoplasm.

4. In males, the RBC count is about 5.1–5.8 million/mm³ (per μL) and in females about 4.3–5.2 million/mm³.

5. The average life span of RBCs is 120 days.

6. The process of formation of RBCs is called erythropoiesis.

7. RBCs are produced from haemocytoblasts/reticulocytes.

8. The erythropoeitic organ of the foetus is the liver and spleen and in the adult, it is mainly the red bone marrow.

9. Vitamin B₁₂, folic acid and heme protein are required for the production of RBCs. The old and worn-out RBCs are destroyed in the liver and spleen (graveyard of RBCs).

10. Polycythemia is the condition in which the number of RBCs increase and erythrocytopenia is a decrease in number of RBCs.

11. The hormone erythropoietin produced by the kidney cells stimulates the bone marrow for the production of RBCs.

12. Mature erythrocyte is devoid of nucleus, mitochondria or other membrane bound cell organelles. Its cytoplasm (stroma) is rich in haemoglobin and O₂ carrying proteinaceous pigment that gives red colour to the RBCs and blood. It also contains an enzyme, carbonic anhydrase.

13. Erythrocytes are responsible for the transport of respiratory gases O₂ and CO₂, maintaining pH and viscosity of blood. They also contribute in the process of blood clotting.

14. The ratio of the volume of RBCs to the total blood volume of blood is hematocrit. It is different for men and women.

2. WBC (White Blood Corpuscles) or Leucocytes:

1a. Leucocytes are colorless, nucleated and amoeboid cells larger than RBCs.

2. These are colorless, irregular nucleated cells and show polymorphism (exist in variable forms)

3. Due to their amoeboid movement, they can move out of the capillary walls by a process called diapedesis.

4. A normal adult has on average, 5000-11000 WBCs per mm³ of blood.

5. Decrease in the number of WBCs (<4000) is called leucopenia (common in HIV, AIDS, and TB patients or those exposed to radiations, shock, etc.). Temporary increase in the number of WBCs is called as leucocytosis. It is due to infection. It also occurs during pregnancy and in newborn babies. An uncontrolled increase in the number of WBCs is a type of blood cancer called leukemia. WBCs are mainly concerned with defense mechanisms i.e. protection.

3. Blood Platelets or Thrombocytes:

1. Thrombocytes are cellular fragments formed from the large cells called megakaryocytes.

2. These are produced in the bone marrow. They are very small, oval shaped cell fragments without a nucleus.

3. Normal count of thrombocytes in human blood is about 2.5 – 4.5 lakh / mm³ of blood. If the number of thrombocytes decreases than normal, the condition is called thrombocytopenia. This condition causes internal bleeding (hemorrhage).

4. Platelets secrete platelet factors which are essential in blood clotting. They also seal the ruptured blood vessels by the formation of platelet plug/ thrombus. They secrete serotonin a local vasoconstrictor.

5. Functions of Blood: Blood perform various functions like transport, homeostasis and protection.

6.Explain blood clotting in short.

SOLUTION

1. Clotting or coagulation is the process of converting liquid blood into a solid form. This process may be initiated by contact of blood with any foreign surface (intrinsic process) or with damaged tissue (extrinsic process).
2. Intrinsic and extrinsic processes involve the interaction of various substances called clotting factors by a stepwise or cascade mechanism.
3. There are in all twelve clotting factors numbered as I to XIII (factor VI is not in active use). Interaction of these factors occurs in a cascade manner leading to the formation of the enzyme thrombin.
4. Thromboplastin helps in the formation of enzyme prothrombinase. This enzyme inactivates heparin and it also converts inactive prothrombin into its active thrombin.
5. Thrombin converts soluble blood protein fibrinogen into insoluble fibrin. Fibrin forms a mesh in which platelets and other blood cells are trapped to form the clot.

7.Describe pericardium.

SOLUTION

The heart is enclosed in a membranous sac called the pericardium. The pericardium is formed of two main layers – outer fibrous and inner serous pericardium. Serous pericardium is soft, moist, and elastic. It is formed of squamous epithelium and is further divisible into two layers as a parietal and visceral layer. Parietal and visceral layers of serous pericardium are separated by a pericardial space. This space is filled with pericardial fluid (about 50ml) which acts as a shock absorber and protects the heart from mechanical injuries. It also keeps the heart moist and acts as a lubricant.

8.Describe valves of human heart.

SOLUTION

Both the atria open into ventricles of their respective sides by atrioventricular apertures. The atrio-ventricular apertures are guarded by cuspid valves.

i. Cuspid valves: 

These are bicuspid and tricuspid valves. The bicuspid valve also known as the mitral valve is present in the left atrio-ventricular aperture. Tricuspid valve is present in the right AV aperture.

ii. Eustachian valve: 

It is present on the opening of the post-caval vein (inferior vena cava).

iii. Thebesian valve: 

It guards the opening of the coronary sinus into the right atrium.

iv. Semilunar valves: 

These three valves guard the opening between the right ventricle and pulmonary artery and left ventricle and aorta.

9.What is the role of papillary muscles and chordae tendinae in the human heart?

SOLUTION

The bicuspid and tricuspid valves are connected to chordae tendineae which in turn are connected to the papillary muscles present on the ventricular wall. Chordae tendineae and papillary muscles regulate the opening and closing of valves.

10.Explain in brief the factors affecting blood pressure..

SOLUTION

The factors affecting blood pressure are:

1. Cardiac output:

The normal cardiac output is 5 litres/min. An increase in cardiac output increases systolic pressure.

2. Peripheral resistance: 

It depends upon the diameter of blood vessels. A decrease in the diameter of arterioles and capillaries under the effect of vasoconstrictors like vasopressin or ADH cause increase in peripheral resistance and thereby increase in blood pressure.

3. Blood volume: 

Blood loss in accidents decreases blood volume, and thus the blood pressure.

4. Viscosity of blood: 

Blood pressure is directly proportional to the viscosity of blood.

5. Age: 

Blood pressure increases with age due to the increase in inelasticity of blood vessels.

6. Venous return: 

The amount of blood brought to the heart via the veins per unit time is called the venous return. It is directly proportional to blood pressure.

7. Length of blood vessel: 

Blood pressure is also directly proportional to the total length of the blood vessel. Blood pressure can also be affected by vasoconstriction or vasodilation.

8. Gender: 

Females have slightly lower BP than males before the age of menopause. However, the risk of high B. P. increases in the females after menopause sets in.

Give scientific reason.

1.Closed circulation is more efficient than open circulation.

SOLUTION

1. In open circulation, blood is not enclosed in blood vessels but pumped directly into the cavity called haemocoel whereas, in the closed type of circulation, blood flows within the blood vessels and does not come in direct contact with cells and body tissues.

2. Therefore, in closed blood circulation blood flows under high pressure and allows the blood to pass faster and achieve a high level of distribution within the body.

Thus, closed circulation is more efficient than open circulation.

2.Human heart is called as myogenic and autorhythmic.

SOLUTION

1. The human heart is capable of generating a cardiac contraction independent of the nervous system. It can generate its own rhythm due to the presence of nodal tissues.
2. The nodal tissue SA node (Sinoatrial node) is capable of generating the wave on contraction and making the pace of contraction.

Thus, human heart is myogenic and autorhythmic.

3.Person who has undergone a heart transplant needs lifetime supply of immunosuppressants.

SOLUTION

a. Immunosuppressants are the drugs that reduce the level of immune activity and the risk of rejection of foreign bodies such as transplant organs.

b. After transplantation, there is a risk of graft rejection as the body may recognize the transplanted organ/tissue as foreign and may trigger an immune response thereby damaging the transplanted organ. 

Therefore, the heart recipient has to rely upon lifetime supply of immunosuppressants.

4.Arteries are thicker than veins.

SOLUTION

i. Arteries carry oxygenated blood away from the heart to the body.

ii. The blood pumped out by the heart is under high pressure and to withstand this pressure arteries are thick-walled.

iii. Veins carry deoxygenated blood from the body back to the heart.

iv. They are thin-walled as the blood that flows through veins is under low pressure. Hence, arteries are thicker than veins.

5.Left ventricle is thick than all other chambers of heart.

SOLUTION

i. The thickness of the myocardium of the four chambers varies according to the functions of each chamber.

ii. The thin-walled atria deliver blood into adjacent respective ventricles.

iii. As compared to the right ventricle, the left ventricle pumps blood at great distances to all other parts of the body at higher pressure, and resistance to blood flow is larger. Therefore, the left ventricle is thick as it requires strength to withstand the high pressure.

Distinguish between

1. open and closed circulation.

SOLUTION

Open circulation	Closed circulation
1. In open circulation, blood is circulated through the body cavities (haemocoels).	1. In closed circulation, blood circulates the blood vessels and does not come in direct contact with cells and body tissues.
2. The blood flows with low pressure.	2. The blood flows with high pressure.
3. Exchange of material takes place directly between blood and cells or tissues of the body.	3. Exchange of material between blood and body tissues is through intermediate fluid called lymph.
4. It usually does not contain any respiratory pigment like haemoglobin so it does not transport respiratory gases	4. It contains respiratory pigments like haemoglobin for transportation of respiratory gases.
e.g. Arthropods and molluscs	e.g. All vertebrates, higher molluscs and annelids

2.Distinguish between Arteries and Veins

SOLUTION

Arteries	Veins
1. They carry blood away from the heart to various parts/organs of the body.	1. The carry blood towards the heart from various parts /organs of the body.
2. Blood flows under great pressure.	2. Blood flows under less pressure.
3. They are thick-walled.	3. They are thin-walled.
4. Arteries branch into arterioles and further into fine capillaries	4. Venules are small vessels that continue from capillaries and merge to form veins.
5. These are deeply situated except a few like the radial, brachial, femoral, etc. which are superficially located.	5. Mostly superficial in location.
6. They carry oxygenated blood, except pulmonary artery.	6. They carry deoxygenated blood, except pulmonary vein.
7. Tunica media is comparatively thicker.	7. Tunica media is comparatively thinner.
8. They do not have valves.	8. They have valves to prevent the backflow of the blood.

3.Distinguish between Blood and lymph.

SOLUTION

Blood	lymph
1. It is reddish in colour.	1. It is pale yellow in colour.
2. It has two main components – fluid plasma and formed elements (blood cells).	2. It has almost similar composition to the blood except for RBCs, platelets, and some proteins.
3. It flows through blood vessels.	3. It flows through lymph vessels.
4. It transports materials from one organ to another.	4. It transports material from tissues cells to blood and vice-versa.

4.Distinguish between Blood capillary and lymph capillary

SOLUTION

Blood capillary	lymph capillary
1. Its diameter is smaller than lymph capillary.	1. Its diameter is larger than blood capillary.
2. It contains blood.	2. It contains lymph.
3. It is less permeable than lymph.	3. It is more permeable than blood capillary.
4. Blood capillaries provide oxygen and other substances to the tissues.	4. Lymph capillaries absorb the excess of tissue fluid.

5.Distinguish between Intrinsic and extrinsic process of clotting

SOLUTION

Intrinsic pathway	Extrinsic pathway
1. It is stimulated by damage to blood vessels.	1. It is stimulated by damage to tissue outside the vessel.
2. It is more complex and takes more time than the extrinsic pathway.	2. It occurs rapidly as it has fewer steps as compared to the intrinsic pathway.
3. Tissue factor is not involved in the activation of the intrinsic pathway.	3. Tissue factor also known as thromboplastin activates extrinsic pathways.
4. It involves factor VIII, IX, XI, and XII.	4. It involves factors VII, X, and V.

Long answer question:

1.Smita was working in a garage with the doors closed and automobiles engine running. After some time she felt breathless and fainted. What would be the reason? How can she be treated?

SOLUTION

1. Smita felt breathless and fainted due to presence of an excess carbon monoxide released from automobile engines.

2. Carbon monoxide can be fatal if not treated. The affected person can be treated by administering pure oxygen. This will speed up the separation of carbon monoxide from hemoglobin.

2.Shreyas went to a garden on a wintry morning. When he came back, he found it difficult to breath and started wheezing. What could be the possible condition and how can he be treated?

SOLUTION

1. The symptoms – difficulty in breathing, wheezing indicate that Shreyas could be suffering from asthma.

2. Inhalers in which open-air passage ways are used to treat asthma.

3.Why can you feel a pulse when you keep a finger on the wrist or neck but not when you keep them on a vein?

SOLUTION

1. Pulse is the series of pressure waves that travel through arteries due to ventricular systole.
2. It is strongest in the arteries closer to the heart and gradually becomes weak in arteries away from the heart and will be the weakest till it reaches the vein.

4.A man’s pulse rate is 68 and cardiac output is 5500 cm3 . Find the stroke volume.

SOLUTION

Cardiac output = Stroke volume × Heart rate

∴ 5500 = Stroke volume × 68

Stroke volume = 550       

                   ————–

                        68

= 80.882 ≈ 80.88mL

5.Which blood vessel of the heart will have the maximum content of oxygen and why?

SOLUTION

1. Pulmonary vein carries the maximum content of oxygen.
2. Pulmonary circulation moves deoxygenated blood from the heart to the lungs for oxygenation and it returns to the heart as oxygenated blood. Systemic circulation pushes the oxygenated blood from the heart towards various body parts (except lungs) and returns back to the heart as deoxygenated blood.
3. Pulmonary vein is the only blood vessel that carries freshly oxygenated blood from the lungs to the heart for distribution to the body.

6.If the duration of the atrial systole is 0.1 sec and that of complete diastole is 0.4 sec, then how does one cardiac cycle complete in 0.8 sec?

SOLUTION

1. One cardiac cycle includes atrial systole, ventricular systole, and joint/complete diastole.

2. The duration for atrial systole is 0.1 sec, duration for complete diastole is 0.4 sec, which means if one cardiac cycle completes in 0.8 sec then the duration for ventricular systole is 0.3 sec.

3. Therefore, the duration of one cardiac cycle

= Atrial systole + Ventricular systole + Complete diastole

= 0.1 sec +0.3 sec + 0.4 sec = 0.8 sec

4. Also the relaxation period shortens as the heart beats faster whereas the durations of atrial systole and diastole shortens slightly. Hence, one cardiac cycle completes in 0.8 sec.

7.How blood is kept moving in the large veins of the legs?

SOLUTION

1. The blood in the large veins of legs is kept moving by the means of azygos system (located on either side of the vertebral column and drains the viscera within the mediastinum, as well as the back and thoracoabdominal walls).

2. It serves as a bypass for the inferior vena cava that drains blood from the lower body.

3. Several small veins link the azygos system directly with inferior vena cava. Large veins drain the lower limbs and abdomen, conducts blood into the azygos system.

4. If the inferior vena cava or hepatic portal vein becomes obstructed, the azygos system returns blood from the lower body to the superior vena cava.

8.Describe the histological structure of the artery, vein, and capillary.

SOLUTION

The three structural layers of a generalized blood vessel from innermost to outermost are the tunica interna (intima), tunica media, and tunica externa. Modifications in this basic design account for the five types of blood vessels and the structural and functional differences among the various vessel types. In the transverse section of an artery, three layers can be seen. They are:

T. S. of Artery, Vein and Capillary-

1. Tunica externa or tunica adventitia: 

It is a thick, tough layer of collagen fibers.

2. Tunica media: 

It is the middle layer made up of smooth muscle fibers and a network of elastic fibers. This thick muscular and elastic layer makes the arterial wall pulsatile.

3. Tunica interna or intima: 

The innermost tunica interna is a single layer of flat compact endothelial cells surrounding the lumen. The angular margin around the lumen shows tessellations. Arterial lumen is devoid of valves and blood flows through it rapidly and with high pressure.

9.What is blood pressure? How is it measured? Explain the factors affecting blood pressure.

SOLUTION

The pressure exerted by blood on the wall of the blood vessels is called blood pressure.

It is measured by the sphygmomanometer. It is usually measured from the arteries.

The factors affecting blood pressure are:

1. Cardiac output: 

The normal cardiac output is 5 litres/min. An increase in cardiac output increases systolic pressure.

2. Peripheral resistance: 

It depends upon the diameter of blood vessels. A decrease in the diameter of arterioles and capillaries under the effect of vasoconstrictors like vasopressin or ADH causes an increase in peripheral resistance and thereby increase in blood pressure.

3. Blood volume: 

Blood loss in accidents decreases blood volume, and thus the blood pressure.

4. Viscosity of blood: 

Blood pressure is directly proportional to the viscosity of blood.

5. Age: 

Blood pressure increases with age due to an increase in the inelasticity of blood vessels.

6. Venous return: 

The amount of blood brought to the heart via the veins per unit time is called the venous return. It is directly proportional to blood pressure.

7. Length of blood vessel: 

Blood pressure is also directly proportional to the total length of the blood vessel. Blood pressure can also be affected by vasoconstriction or vasodilation.

8. Gender: 

Females have slightly lower BP than males before the age of menopause. However, the risk of high B. P. increases in the females after menopause sets in.

10.Describe human blood and give its functions.

SOLUTION

Human blood consists of plasma and blood corpuscles or blood cells

Plasma: It is a straw-coloured, slightly alkaline, viscous fluid. It constitutes 55% of the blood. Plasma consists of water, proteins (albumin, globulin, properdin, prothrombin, fibrinogen), inorganic salts (Na, K, Mg, Ca, Fe, Mn and Cl^–, HCO3- and PO43-), food (glucose, amino acids, fatty acids, triglycerides), wastes (urea, uric acid and creatinine), regulators (hormones, enzymes, vitamins), anticoagulants (heparin), cholesterol and antibodies, dissolved gases (O₂, CO₂, N₂) Plasma contains 90% water, 7-8% proteins, inorganic salts – 1% and other substances 1-2%.

Blood Corpuscles: It constitutes 44% of the blood. Blood corpuscles are of three types as given below:

1. RBC (Red Blood Corpuscles) or Erythrocytes:

a. Erythrocytes are the most abundant cells in the human body.

b. They are circular, biconcave and enucleated (in camel and llama they are nucleated).

c. The red colour or RBCs is due to an oxygen-carrying pigment, the haemoglobin, in their cytoplasm.

d. In males, the RBC count is about 5.1–5.8 million/mm³ (per μL) and in females about 4.3–5.2 million/mm³.

e. The average life span of RBCs is 120 days.

f. The process of formation of RBCs is called erythropoiesis.

g. RBCs are produced from haemocytoblasts/reticulocytes.

h. The erythropoeitic organ of the foetus is the liver and spleen and in the adult, it is mainly the red bone marrow.

i. Vitamin B₁₂, folic acid and heme protein are required for the production of RBCs. The old and worn-out RBCs are destroyed in the liver and spleen (graveyard of RBCs).

j. Polycythemia is the condition in which the number of RBCs increase and erythrocytopenia is a decrease in the number of RBCs.

k. The hormone erythropoietin produced by the kidney cells stimulates the bone marrow for production of RBCs.

l. Mature erythrocyte is devoid of nucleus, mitochondria or other membrane-bound cell organelles. Its cytoplasm (stroma) is rich in haemoglobin and O₂ carrying proteinaceous pigment that gives the red colour to the RBCs and blood. It also contains an enzyme, carbonic anhydrase.

m. Erythrocytes are responsible for the transport of respiratory gases O₂ and CO₂, maintaining pH and viscosity of blood. They also contribute in the process of blood clotting.

n. The ratio of the volume of RBCs to the total blood volume of blood is hematocrit. It is different for men and women.

2. WBC (White Blood Corpuscles) or Leucocytes:

a. Leucocytes are colorless, nucleated, and amoeboid cells larger than RBCs.

b. These are colourless, irregular nucleated cells and show polymorphism (exist in variable forms)

c. Due to their amoeboid movement they can move out of the capillary walls by a process called diapedesis.

d. A normal adult has on average, 5000-11000 WBCs per mm3 of blood.

e. Decrease in the number of WBCs (<4000) is called leucopenia (common in HIV, AIDS, and TB patients or those exposed to radiations, shock, etc.). A temporary increase in the number of WBCs is called leucocytosis. It is due to infection. It also occurs during pregnancy and in newborn babies. An uncontrolled increase in the number of WBCs is a type of blood cancer called leukemia. WBCs are mainly concerned with defense mechanisms i.e. protection.

3. Blood Platelets or Thrombocytes:

a. Thrombocytes are cellular fragments formed from the large cells called megakaryocytes.

b. These are produced in the bone marrow. They are very small, oval-shaped cell fragments without a nucleus.

c. Normal count of thrombocytes in human blood is about 2.5 – 4.5 lakh / mm³ of blood. If the number of thrombocytes decreases than normal, the condition is called as thrombocytopenia. This condition causes internal bleeding (haemorrhage).

4. Platelets secrete platelet factors which are essential in blood clotting. They also seal the ruptured blood vessels by the formation of platelet plug/ thrombus. They secrete serotonin a local vasoconstrictor.

5. Functions of Blood: Blood perform various functions like transport, homeostasis, and protection.

COMPLETED

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Chapter 7, Plant Growth and Mineral Nutrition, hsc, biology, maharashtra board, 12th std, balbharathi solution,

Multiple choice question.

1.Which of the hormones can replace vernalization?

OPTIONS

• Auxin
• Cytokinin
• Gibberellins
• Ethylene

2.The principle pathway of water translocation in angiosperms is ______

OPTIONS

• Sieve cells
• Sieve tube elements
• Xylem
• Xylem and phloem

3.Abscisic acid controls ______.

OPTIONS

• cell division
• leaf fall and dormancy
• shoot elongation
• cell elongation and wall formation

4.Which is employed for the artificial ripening of banana fruits?

OPTIONS

• Auxin
• Ethylene
• Cytokinin
• Gibberellin

5.Which of the following is required for stimulation of flowering in the plants?

OPTIONS

• Adequate oxygen
• Definite photoperiod
• Adequate water
• Water and minerals

6.For short-day plants, the critical period is

OPTIONS

• light
• dark/ night
• UV rays
• both light and UV rays

7.Which of the following is day-neutral plant?

a. Tomato

b. Cotton

c. Sunflower

d. Soybean

SOLUTION

a. Tomato and c. Sunflower 

8.Essential macroelements are ____________.

OPTIONS

• manufactured during photosynthesis
• produced by enzymes
• absorbed from soil
• produced by growth hormones

9.Function of Zinc is ______.

OPTIONS

• closing of stomata
• biosynthesis of 3-IAA
• synthesis of chlorophyll
• oxidation of carbohydrates

10.Necrosis means ______.

OPTIONS

• yellow spots on the leaves
• death of tissue
• darkening of green colour in leaves
• wilting of leaves

11.Conversion of nitrates to nitrogen is called ______

OPTIONS

• ammonification
• nitrification
• nitrogen fixation
• denitrification

12.How many molecules of ATP are required to fix one molecule of nitrogen?

OPTIONS

• 12
• 20
• 6
• 16

Very Short Answer Question:

1.Enlist the phases of growth in plants?

SOLUTION

Three phases of growth in plants are:

i. Phase of cell division/ formation

ii. Phase of cell enlargement/ elongation

iii. Phase of cell maturation/ differentiation.

2.Give the full form of IAA?

SOLUTION

IAA: Indole-3-acetic acid

3.What does it mean by ‘open growth’?

SOLUTION

The form of growth where in new cells are being constantly added to the plant body by the activity of the meristem is called the open growth.

4.Which is the plant stress hormone?

SOLUTION

Abscisic acid (ABA) is the plant stress hormone

5.What is denitrification?

SOLUTION

i. Denitrification is the process in which anaerobic bacteria convert soil nitrates back into nitrogen gas.

ii. Denitrifying bacteria removes fixed nitrogen i.e. nitrates from the ecosystem and returns it to the atmosphere in an inert form.

iii. Denitrifying bacteria include Bacillus spp., Paracoccus spp. and Pseudomonas denitrificans. They transform nitrates to nitrous and nitric oxides and ultimately to gaseous nitrogen.

2NO₃ → 2NO₂ → 2NO → N₂

6.Name the bacteria responsible for conversion of nitrite to nitrate.

SOLUTION

Chemoautotrophs like Nitrobacter are responsible for conversion of nitrite to nitrate.

7.What is role of gibberellin in rosette plants?

SOLUTION

Gibberellin promotes bolting i.e. elongation of internodes just prior to flowering in plants with rosette habit e.g. beet, cabbage.

8.Define vernalization.

SOLUTION

The low-temperature treatment or chilling treatment of germinating seeds or seedlings to promote early flowering in plants is called vernalization. It was evidenced by Klippart (1918).

9.Define photoperiodism.

SOLUTION

The relative length of the day which is crucial in the growth and development of flowers is termed as photoperiodism.

OR

The response of plants to the relative length of light and dark periods with reference to the initiation of flowering is called photoperiodism.

10.What is a grand period of growth?

SOLUTION

The total time (period) required for all phases to occur, is called Grand Period of Growth.

Short Answer Question:

1.Write a short note on Differentiation.

SOLUTION

1. It is maturation of cells derived from the apical meristem of root and shoot.

2. Permanent change in structure and function of cells leading to maturation is called differentiation.

3. During cell differentiation, the cell undergoes few to major anatomical and physiological changes.

4. For e.g. Parenchyma in hydrophytes develops large schizogenous interspaces for mechanical support, buoyancy and aeration.

5. Cells lose the capacity to divide and redivide and mature.

2.Write a short note on Re-differentiation.

SOLUTION

1. The cells produced by dedifferentiation once again lose the capacity to divide and mature to perform a specific functions. This is called a re-differentiation.

2. For e.g. secondary xylem and secondary phloem are formed from dedifferentiated cambium present in the vascular bundle.

3.Differentiate between Arithmetic and Geometric growth.

SOLUTION

Arithmetic growth	Geometric growth
After mitosis one of the daughter cell continues to divide and the other cell takes part in the differentiation and maturation.	After mitosis both the daughter cells continue to divide and re-divide repeatedly.
On plotting the growth against time, a linear curve is obtained.	On plotting the growth against time, a sigmoid curve is obtained.

4.Enlist the role and deficiency symptoms of Nitrogen.

SOLUTION

Nitrogen:

a. Role: Constituent of proteins, nucleic acids, vitamins, hormones, coenzymes, ATP, chlorophyll.

b. Deficiency symptom: Stunted growth, chlorosis

5.Enlist the role and deficiency symptoms of Phosphorus.

SOLUTION

Phosphorus:

a. Role: Constituent of cell membrane, certain proteins, all nucleic acids, and nucleotides required for all phosphorylation reactions.

b. Deficiency symptom: Poor growth, leaves dull green.

6.Enlist the role and deficiency symptoms of Potassium.

SOLUTION

Potassium:

a. Role: Helps in determining anion- cation balance in cells involved in protein synthesis, involved in the formation of the cell membrane and in opening and closing of stomata; increases hardness; activates enzymes, and helps in the maintenance of turgidity of cells.

b. Deficiency symptom: Yellow edges to leaves, premature death.

7.What is short day plant? Give any two examples.

SOLUTION

1. Critical photoperiod is the length of photoperiod above or below which flowering occurs. Short Day Plants usually flower during winter and late summer when day length is shorter than the critical photoperiod.

2. These are called long night plants because they require long uninterrupted dark period/ night for flowering.

3. If the dark period is interrupted even by a flash of light, SDP will not flower.

4. Some of the short-day plants are Dahlia, Tobacco, Chrysanthemum, Soybean (Glycine max), Cocklebur (Xanthium), cotton, etc.

8.Define vernalization.

SOLUTION

The low-temperature treatment or chilling treatment of germinating seeds or seedlings to promote early flowering in plants is called vernalization. It was evidenced by Klippart (1918).

9.Give its significance of vernalization.

SOLUTION

1. Chouard (1960) defined vernalization as the acceleration of the ability to flower by chilling treatment.

2. The term vernalization was coined by T.D Lysenko (1928) for the effect of low temperature on flowering in plants.

3. It is an influence of temperature on development and flowering.

4. Many plants such as cereals, crucifers require a period of cold treatment for flowering.

5. It is the method of inducing early flowering in the plants by pretreatment to their seeds/ seedlings at low temperatures (1-6ºC for one to one and half months’ duration).

6. The site of vernalization is believed to shoot apical meristem.

7. Generally, vernalization is effective at the seed stage in annual plants.

8. It was suggested by Melchers (1939) that vernalization initiates a stimulus for the formation of a hormone called vernalin.

9. Significance of vernalization:

a. Crops can be produced earlier.

b. Crops can be cultivated in regions where they do not grow naturally.

Long Answer Question:

1.Explain sigmoid growth curve with the help of diagram.

SOLUTION

1. The curve obtained when a graph of growth rate against time is plotted for three phases of growth is called as a sigmoid curve.

2. Growth rate differs with three distinct phases of growth.

3. In the Lag phase, the growth rate is slow.

4. In Exponential (Log) phase, growth rate is faster and reaches its maximum.

5. In Stationary phase, growth rate gradually slows down.

2.Describe the types of plants on the basis of photoperiod required, with the help of suitable examples.

SOLUTION

Based on the photoperiodic response, plants were classified into three categories viz. Short Day Plants (SDP), Long Day Plants (LDP) and Day Neutral Plants (DNP).

1. Short Day Plants (SDP):

i. Critical photoperiod is the length of photoperiod above or below which flowering occurs. Short Day Plants usually flower during winter and late summer when day length is shorter than the critical photoperiod.

ii. These are called long night plants because they require long uninterrupted dark period/ night for flowering.

iii. If dark period is interrupted even by a flash of light, SDP will not flower.

iv. Some of the short-day plants are Dahlia, Tobacco, Chrysanthemum, Soybean (Glycine max), Cocklebur (Xanthium), cotton, etc.

2. Long Day Plants (LDP):

i. Plants that flower during summer are called long-day plants.

ii. They require a longer duration of light than the critical photoperiod, for flowering.

iii. They are called short night plants as they require a short dark period.

iv. When long dark period is interrupted by a brief flash of light, LD plants can flower e.g. pea, radish, sugar beet, cabbage, spinach, wheat, poppy, etc.

3. Day Neutral Plants (DNP):

a. These plants flower throughout the year-round, independent of the duration of light (photoperiod).

b. They do not require specific photoperiod to flower.

c. Therefore, they are called day-neutral plants e.g. Cucumber, tomato, sunflower, maize, balsam, etc.

Photoperiodism-

3.Explain biological nitrogen fixation with example.

SOLUTION

Biological nitrogen fixation:

i. It is carried out by prokaryotes called as ‘Nitrogen fixers’ or Diazotrophs’.

ii. It accounts for nearly 70% of natural nitrogen fixation.

iii. Nitrogen fixers are either symbiotic or free living.

iv. Symbiotic N₂ fixation: 

The best-known nitrogen-fixing symbiotic bacterium is Rhizobium. This soil living/ dwelling bacterium forms root nodules in plants belonging to the family Fabaceae e.g. beans, gram, groundnut etc.

v. Azotobacter, Azospirillum is free-living nitrogen-fixing bacteria.

vi. The cyanobacteria fix a significant amount of nitrogen in specialized cells called heterocysts.

vii. Nitrogen fixation is high energy-requiring process and nitrogen fixers use 16 molecules of ATP to fix each molecule of nitrogen to form ammonia.

N₂ + 8H⁺ + 8e^– + 16ATP → 2NH₃ + H₂ + 16ADP + 16Pi

               COMPLETED

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Balbharathi solution, chapter 6, plant water relation, science, biology, maharashtra board, hsc, full solution, latest edition,

Multiple Choice Question:

1.In soil, water available for absorption by root is ______.

OPTIONS

• gravitational water
• capillary water
• hygroscopic water
• combined water

2.The most widely accepted theory for ascent of sap is ______.

OPTIONS

• capillarity theory
• root pressure theory
• diffusion
• transpiration pull theory

3.Water movement between the cells is due to ______.

OPTIONS

• T.P.
• W.P.
• DPD
• incipient plasmolysiS

4.In guard cells, when sugar is converted into starch, the stomatal pore ______.

OPTIONS

• closes almost completely
• opens partially
• opens fully
• remains unchanged

5.Surface tension is due to ____________.

OPTIONS

• diffusion
• osmosis
• gravitational force
• cohesion

6.Which of the following type of solution has a lower level of solutes than the solution?

OPTIONS

• Isotonic
• Hypotonic
• Hypertonic
• Anisotonic

7.During rainy season wooden doors warp and become difficult to open or to close because of ______

OPTIONS

• plasmolysis
• imbibition
• osmosis
• diffusion

8.Water absorption takes place through ______.

OPTIONS

• lateral roots
• root cap
• root hair
• primary root

9.Due to low atmospheric pressure the rate of transpiration will ____________.

OPTIONS

• increase
• decrease rapidly
• decrease slowly
• remain unaffected

10.Osmosis is a property of ______.

OPTIONS

• solute
• solvent
• solution
• membrane

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Very short answer question.

1.What is osmotic pressure?

Explain the term osmosis.

SOLUTION

i. The pressure exerted due to osmosis is called osmotic pressure.

ii. Osmotic pressure is a pressure of the solution, which is required in opposite direction, so as to stop the entry of solvent molecules into the cell.

OR

Osmotic pressure of a solution is equivalent to the pressure which must be exerted upon it to prevent flow of solvent across a semipermeable membrane.

2.Name the condition in which protoplast of the plant cell shrinks.

SOLUTION

Plasmolysis

3.What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?

SOLUTION 1

The water potential of pure water or a solution increases on the application of pressure values more than atmospheric pressure. For example: when water diffuses into a plant cell, it causes pressure to build up against the cell wall. This makes the cell wall turgid. This pressure is termed as pressure potential and has a positive value.

SOLUTION 2

If a pressure greater than atmospheric pressure is applied to pure water or a solution, its water potential increases. It is equivalent to pumping water from one place to another. Pressure can build up in a plant system when water enters a plant cell due to diffusion causing a pressure built up against the cell wall, it makes the cell turgid.

4.Which type of solution will bring about deplasmolysis?

SOLUTION

Hypotonic solution can bring about deplasmolysis.

5.Which type of plants have negative root pressure?

SOLUTION

The plants in which transpiration occurs rapidly especially during midsummer shows negative root pressure.

6.In which conditions transpiration pull will be affected?

SOLUTION

For transpiration pull to operate, the water column should be unbroken and continuous. However, due to temperature fluctuations during day and night, gas bubbles may enter in water column breaking the continuity.

7.Mention the shape of guard cells in Cyperus.

SOLUTION

In Cyperus, both kidney-shaped and dumbbell-shaped guard cells are present.

8.Why do diurnal changes occur in osmotic potential of guard cells?

SOLUTION

1. According to Steward, diurnal changes occur in the osmotic potential of guard cells due to starch-sugar inter-conversion.

2. Whereas according to Levitt active transport of potassium ions into the guard cells and out of them causes diurnal changes in the osmotic potential of guard cells.

3. Endo-osmosis and exo-osmosis occur due to diurnal changes in osmotic potential of guard cells.

9.What is the symplast pathway?

SOLUTION

When water passes across from one living cell to another living cell through plasmodesmata, then it is called the symplast pathway. It is also called the trans-membrane pathway.

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Answer the following question.

1.Describe the mechanism for absorption of water.

SOLUTION

A mechanism for absorption of water:

1. In plants, water is absorbed mainly by two processes: Passive absorption and Active absorption

2. Passive absorption:

a. About 98% of the total water absorbed in plants occurs passively.

b. In passive absorption, living cells of the root do not play an important role in water absorption.

c. The driving force is transpiration pull and it thus proceeds through the DPD gradient.

d. There is no expenditure of energy (ATP) as water moves in accordance with the concentration gradient. Hence, it is passive absorption.

e. Passive absorption occurs during day time when transpiration is in progress. It stops at night when transpiration stops.

f. Rapid transpiration creates tension in the xylem vessel due to negative water potential. This tension is transmitted to xylem in the roots. Consequently, water is pulled upwards passively.

g. During passive absorption, no ATP is utilized. Thus, the rate of respiration is not affected.

3. Active absorption:

a. In this water is absorbed due to the activity of roots.

b. Root cells play an active role in the absorption of water.

c. The driving force is the root pressure developed, in the living cells of the root.

d. Active absorption occurs usually at night when transpiration stops due to closure of stomata.

e. As water absorption is against the DPD gradient, there is an expenditure of ATP (energy) generated through the respiratory activity of cells.

2.Discuss theories of water translocation.

SOLUTION

Theories of water translocation:

i. Various theories have been put forth to explain the mechanism of translocation of water. These theories include Vital force theory, Relay pump theory, Physical force theory, Root pressure theory, etc.

ii. Root Pressure Theory (Vital Theory): This theory was proposed by J. Priestley. According to this theory, the activity of living cells of the root is responsible for the translocation of water.

iii. Capillarity theory (physical force theory): This theory was put forth by Boehm in (1863). According to this theory, physical forces and dead cells are responsible for the ascent of sap.

iv. Cohesion- tension theory (Transpiration pull theory): This theory was put forth by Dixon and Jolly (1894). This is presently a widely accepted theory explaining the ascent of sap in plants. This theory is based on two principles i.e. Cohesion and adhesion, and transpiration pull.

3.What is transpiration?

SOLUTION

Transpiration:

The loss of water in the form of vapor is called transpiration that occurs through leaves, stem, flowers, and fruits. 

4.Describe the mechanism of opening and closing of stomata.

SOLUTION

Mechanism of opening and closing of stomata:

1. The opening and closing of stoma is controlled by turgor of guard cells.

2. During day time, guard cells become turgid due to endo-osmosis.

3. Thus turgor pressure is exerted on the thin walls of guard cells.

4. Being elastic and thin, lateral walls are stretched out.

5. Due to kidney or dumb-bell like shape, inner thick walls are pulled apart to open (widen) the stoma.

6. During night time, guard cells become flaccid due to exo-osmosis.

7. Flaccidity closes the stoma almost completely.

8. Endo-osmosis and exo-osmosis occur due to diurnal changes in the osmotic potential of guard cells.

9. According to starch-sugar inter-conversion theory (Steward 1964), during day time, enzyme phosphorylase converts starch to sugar, thus increasing the osmotic potential of guard cells causing entry of water, thereby guard cells are stretched and stoma widens. The reverse reaction occurs at night bringing about the closure of the stoma.

Starch(Stomata open)⇌(Night)Phosphorylase (Day)Sugar(Stomata close)

10. According to the theory of proton transport (Levitt-1974), stomatal movement occurs due to the transport of protons H⁺ and K⁺ ions. During the daytime, starch is converted into malic acid. Malic acid dissociates to form malate ions and protons. Protons are transported to subsidiary cells and K⁺ ions are imported from them.

Potassium Malate is formed that increases osmolarity and causes endoosmosis. Uptake of K⁺ ions is always accompanied by Cl^– ions. At night, uptake of K⁺ and Cl^– ions is prevented by abscisic acid, changing the permeability of guard cells. Due to this guard cells become hypotonic and thereby become flaccid.

5.What is transpiration?

SOLUTION

Transpiration:

The loss of water in the form of vapor is called transpiration that occurs through leaves, stem, flowers, and fruits. 

6.Explain the role of transpiration.

SOLUTION

Role of transpiration:

i. It removes excess of water.

ii. It helps in the passive absorption of water and minerals from the soil.

iii. It helps in the ascent of sap.

iv. As stomata are open, gaseous exchange required for photosynthesis and respiration is facilitated.

v. It maintains the turgor of the cells.

vi. Transpiration helps in reducing the temperature of leaf and in imparting a cooling effect.

7.What is the significance of transpiration?

SOLUTION

Significance of transpiration:

1. It removes excess of water.

2. It helps in the passive absorption of water and minerals from the soil.

3. It helps in the ascent of sap.

4. As stomata are open, gaseous exchange required for photosynthesis and respiration is facilitated.

5. It maintains the turgor of the cells.

6. Transpiration helps in reducing the temperature of leaf and in imparting a cooling effect.

8.Explain the root pressure theory and its limitations.

SOLUTION

Root pressure theory (Vital theory):

1. This theory was proposed by J. Priestley.

2. According to this theory, the activity of living cells of root is responsible for the translocation of water.

3. When a stem of a potted plant is cut few inches above the soil by a sharp knife, xylem sap is seen flowing out/ oozing out through the cut end.

4. This exudation at the cut end of the stem is a good proof for the existence of root pressure.

5. As water absorption by roots is a constant and continuous process, hydrostatic pressure is developed in the living cells of cortex of the root. This is termed as root pressure (coined by S. Hales).

6. It is due to root pressure water along with dissolved minerals is not only forced into xylem but it is also conducted upwards against the gravity.

7. Root pressure seems to be largely an osmotic phenomenon and its development is an active process.

8. The value of root pressure is +1 to +2 bars which is enough to pump water to a height of 10 to 20 meters.

9. The factors like oxygen, moisture, the temperature of the soil, salt contents, etc. influence the root pressure.

Limitations of root pressure theory:

Although ascent of sap takes place due to root pressure, there are certain objections raised, such as;

1. It is not applicable to plants taller than 20 meters.

2. Ascent of sap can also occur even in the absence of a root system.

3. Root pressure value is almost nearly zero in taller gymnosperm trees.

4. In actively transpiring plants, no root pressure is developed.

5. Xylem sap under normal condition is under tension i.e. it shows negative hydrostatic pressure or high osmotic pressure.

Thus, root pressure is not the sole mechanism explaining the ascent of sap in all plants of varying heights.

9.Explain capillarity theory of water translocation.

SOLUTION

Capillarity theory of water translocation:

1. This theory was put forth by Boehm in (1863).

2. According to this theory, physical forces and dead cells are responsible for the ascent of sap. For e.g. Wick dipped in an oil lamp, shows capillarity due to which oil is raised upwards. The conduction of water in a straw dipped in water is raised to a certain height because of capillarity. The height to which water is raised depends on the diameter of the straw.

3. Capillarity is because of surface tension, and forces of cohesion (attraction between like molecules) and adhesion (attraction between unlike molecules).

4. Xylem vessel/ tracheid with its lumen can be compared with straw.

5. Water column exists because of combined cohesive and adhesive forces of water and xylem wall, due to capillarity.

6. Due to capillarity, water is raised or conducted upwards against gravity, to few centimeters only.

10.Why is transpiration is called ‘a necessary evil’?

SOLUTION

Curtis (1926) regarded transpiration as ‘a necessary evil’, because;

1. For stomatal transpiration to occur, stoma must remain open, during day time.

2. When stomata are open then only the gaseous exchange needed for respiration and photosynthesis will take place.

3. If stomatal transpiration stops, it will directly affect the productivity of the plant through the loss of photosynthetic and respiratory activity.

4. Hence for productivity, stomata must remain open.

5. Consequently transpiration cannot be avoided.

11.Explain the movement of water in the root.

SOLUTION

Journey of water from soil to xylem in roots (from epiblema upto xylem in the stelar region):

1. Water is absorbed by root hair cells through processes like imbibition, diffusion, osmosis which occur sequentially.

2. Water passes through the epidermal cell (epiblema), cortex, endodermis, Casparian strip, pericycle, and then to protoxylem.

3. When the root hair cell absorbs water it becomes turgid. Its turgor pressure increases, but its DPD value decreases.

4. However, the immediately adjacent cortical cell inner to it, has more DPD value, because its O. P. is more.

5. Therefore, cortical cells will absorb water from the turgid root hair cell. It then becomes turgid.

6. The flaccid root hair cell now absorbs water from the soil.

7. Water from the turgid cortical cell is absorbed by the inner cortical cell and the process goes on.

8. Thus, a gradient of suction pressure (DPD) is developed from cells of epiblema to the cortex of the root.

9. Consequently water moves rapidly across the root through loosely arranged living cells of cortex, followed by passage cells of endodermis and finally into the cell of pericycle.

10. Protoxylem is in close proximity to the pericycle.

11. It is due to root pressure, water from pericycle is forced into the xylem.

12. Pathway of water across the root occurs in two types: Apoplast pathway and Symplast pathway

13. Apoplast pathway: When some amount of water passes across the root through the cell wall and the intercellular spaces of cortical cells of the root, it is then called the apoplast pathway. This pathway occurs up to endodermis.

14. Symplast pathway: When water passes across from one living cell to another living cell through plasmodesmata, then it is called the symplast pathway. It is also called the trans-membrane pathway.

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Very short answer question.

12.What is osmotic pressure?

Explain the term osmosis.

SOLUTION

i. The pressure exerted due to osmosis is called osmotic pressure.

ii. Osmotic pressure is a pressure of the solution, which is required in opposite direction, so as to stop the entry of solvent molecules into the cell.

OR

Osmotic pressure of a solution is equivalent to the pressure which must be exerted upon it to prevent flow of solvent across a semipermeable membrane.

13.Define and or explain the term:

Diffusion

SOLUTION

1. Diffusion means to disperse.
2. Diffusion can be defined as the movement of ions/ atoms/ molecules of a substance from the region of their higher concentration to the region of their lower concentration till equilibrium is reached.
3. The movement is due to the kinetic energy of the molecules.
4. Water passes into the cell by diffusion through a freely permeable cell wall.

Define and or explain the term:

14.Plasmolysis

SOLUTION 1

Plasmolysis – The shrinkage of cytoplasm of a living cell as a result of exosmosis is known as plasmolysis.

SOLUTION 2

1. When a living cell is placed in a hypertonic solution, exo-osmosis occurs. This is called plasmolysis.
2. During plasmolysis, the protoplast of the cell shrinks and recedes from the cell wall due to which cell becomes flaccid. Such a cell is called a plasmolysed cell.
3. In a plasmolyzed cell, a gap is developed between the cell wall and the protoplast. This gap is filled up by the outer solution.

15.imbibition

SOLUTION

1. Imbibition is swelling up of hydrophilic colloids due to the adsorption of water.

OR

The adsorption of water by hydrophilic compounds is called imbibition.

2. Substance that adsorbs water/liquid is called imbibant and water/ liquid that gets imbibed is called imbibate.

3. The root hair cell wall is made up of pectic compounds and cellulose which are hydrophilic colloids.

4. During imbibition, water molecules get tightly adsorbed without the formation of a solution.

5. Imbibition continues until the equilibrium is reached. In other words, water moves along the concentration gradient.

6. Imbibition is significant in soaking of seeds, swelling up of dried raisins, kneading of flour etc.

16.Guttation

SOLUTION

1. The loss of water in the form of liquid is called guttation.
2. It occurs through special structures called water stomata or hydathodes.

17.Transpiration

SOLUTION

i. The loss of water in the form of vapour is called transpiration that occurs through leaves, stem, flowers and fruits.

ii. Transpiration occurs through three main sites – cuticle, stomata, and lenticels.

Define and or explain the term:

18.Ascent of sap

SOLUTION

The transport of water with dissolved minerals from the root to other aerial parts like stem and leaves, against the gravity, is called translocation or ascent of sap.

19.Active absorption

SOLUTION

1. In this water is absorbed due to the activity of roots.
2. Root cells play an active role in the absorption of water.
3. The driving force is the root pressure developed, in the living cells of the root.
4. Active absorption occurs usually at night when transpiration stops due to the closure of stomata.
5. As water absorption is against the DPD gradient, there is an expenditure of ATP (energy) generated through the respiratory activity of cells.

Answer the following question.

1.Define and or explain the term:

DPD

SOLUTION

1. Diffusion pressure of pure solvent (pure water) is always more than the diffusion pressure of the solvent in a solution. The difference in the diffusion pressures of pure solvent and the solvent in a solution is called Diffusion Pressure Deficit (DPD) or Suction Pressure (SP).
2. The term DPD was coined by B.S. Meyer (1938). Nowadays, term water potential is used for DPD.
3. In colloquial language, the term DPD is actually the thirst of a cell with which it absorbs water from the surroundings.
4. The water around the cell wall has more diffusion pressure than cell sap. Due to this, water moves in the cell by diffusion. 

2.Define and or explain the term:

Turgor pressure

SOLUTION

Turgor pressure (T.P) is the pressure exerted by turgid cell sap on to the cell membrane and cell wall.

3.Define and or explain the term:

Water potential

SOLUTION

i. Chemical potential of water is called water potential.

ii. It is represented by Greek letter psi (ψ).

iii. The unit of measurement of water potential is bars/ pascals/ atmospheres.

iv. Water potential of protoplasm is equal but opposite in sign to DPD. It has a negative value.

v. Water potential of pure water is always zero. The addition of any solute in it decreases its psi (ψ) value. Therefore, it has a negative value.

vi. Difference between water potential of the adjacent cells decides the movement of water through plasmodesmata across the cells.

vii. Water always flows from less negative potential to more negative water potential (i.e. from high water potential area to low water potential area).

4.Define and or explain the term:

Wall pressure

SOLUTION

The cell wall is thick and rigid, exerts a counter pressure on the cell sap. This is called Wall pressure (W. P).

5.Define and or explain the term:

Root pressure

SOLUTION

1. During the absorption of water, the continuous flow of water develops hydrostatic pressure in living cells of the root. This is called root pressure.
2. Root pressure causes water to flow from pericycle into the xylem. It also causes upward conduction of water against gravity.
3. A manometer is used to measure the root pressure.

6.Distinguish between Osmotic pressure and Turgor pressure.

SOLUTION

Osmotic pressure	Turgor pressure
Osmotic pressure is a pressure of the solution, which is required in opposite direction, so as to stop the entry of solvent molecules into the cell.	Turgor pressure is the pressure exerted by turgid cell sap on to the cell membrane and cell wall.

7.Distinguish between Diffusion and Osmosis

SOLUTION

Diffusion	Osmosis
1. It takes place in solid, gas, or liquid medium.	1. It takes place only in a liquid medium.
2. It does not require the presence of a semi-permeable membrane.	2. It requires the presence of a semipermeable membrane.
3. In diffusion, the movement of ions/atoms/molecules from a region of higher concentration to the region of lower concentration takes place.	3. In osmosis, diffusion of the only solvent from a lower concentration of solution to a higher concentration of solution occurs
4. It is influenced by the diffusion pressure	4. It is only influenced by the turgor pressure.

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8.Enlist macronutrients and micronutrients required for plant growth.

Long answer question.

Write on macro- and micro nutrients required for plant growth.

SOLUTION

1. Macronutrients:
   Some minerals like C, H, O, P, N, S, Mg, K, Ca required in large quantity for normal growth of the plant, are called macro elements. Macronutrients are required in large quantities. They mainly play nutritive and structural roles.
2. Micronutrients:
   Some minerals like Cu, Mo, Mn, Cl, Bo, Zn required in small quantities for the growth of a plant, are called microelements.
   Micronutrients are required in traces because they function in the catalytic role as co-factors.

9.How are the minerals absorbed by the plants?

SOLUTION

i. The analysis of plant ash demonstrates that minerals are absorbed by plants from soil and surroundings.

ii. Minerals are absorbed by plants in the ionic (dissolved) form, mainly through roots and then transported.

iii. Mineral ion absorption is independent of water absorption.

iv. It can occur in two ways i.e. active and passive absorption.

v. In passive absorption, the movement of mineral ions into root cells occurs as a result of diffusion. Mineral ions diffuse from a region of their higher concentration to a region of their lower concentration without the expenditure of energy.

vi. Most minerals in the soil are charged particles hence, they cannot pass across the cell membranes. Hence most of the minerals are absorbed actively with the expenditure energy.

vii. Inactive absorption, minerals are absorbed against the concentration gradient with the expenditure energy.

viii. Absorbed mineral ions are pulled in an upward direction along with xylem sap because of transpiration pull.

ix. Hence, mineral ions are pulled from the source (root) and are transported ascendingly through the sap to the needed areas like apical, lateral, young leaves, developing flowers, fruits, seeds, and storage organs.

x. Mineral ions get unloaded by fine veins through the process of diffusion in the vicinity of cells. Cells uptake them actively.

Long answer question.

1.Describe structure of root hair.

SOLUTION

1. Root hair is a cytoplasmic extension (prolongation) of epiblema cell.

2. Each root hair may be approximately 1 to 10 mm long and tube-like structure.

3. It is colourless, unbranched, short-lived (ephemeral), and very delicate.

4. It has a large central vacuole surrounded by a thin film of cytoplasm, plasma membrane and thin cell wall, which is two-layered.

5. Outer layer is composed of pectin and the inner layer is made up of cellulose.

6. Cell wall of a root hair is freely permeable but the plasma membrane is selectively permeable.

2.Write on journey of water from soil to xylem in roots.

SOLUTION

1. Water is absorbed by root hair cells through processes like imbibition, diffusion, osmosis which occur sequentially.

2. Water passes through the epidermal cell (epiblema), cortex, endodermis, casparian strip, pericycle and then to protoxylem.

3. When root hair cell absorbs water it becomes turgid. Its turgor pressure increases, but its DPD value decreases.

4. However, the immediately adjacent cortical cell inner to it, has more DPD value because its O. P. is more.

5. Therefore, the cortical cells will absorb water from the turgid root hair cell. It then becomes turgid.

6. The flaccid root hair cell now absorbs water from the soil.

7. Water from the turgid cortical cell is absorbed by the inner cortical cell and the process goes on.

8. Thus, a gradient of suction pressure (DPD) is developed from cells of epiblema to the cortex of the root.

9. Consequently water moves rapidly across the root through loosely arranged living cells of cortex, followed by passage cells of endodermis and finally into the cell of pericycle.

10. Protoxylem is in close proximity with pericycle.

11. It is due to root pressure, water from pericycle is forced into the xylem.

12. Pathway of water across the root occurs in two types: Apoplast pathway and Symplast pathway

13. Apoplast pathway: When some amount of water passes across the root through the cell wall and the intercellular spaces of cortical cells of the root, it is then called the apoplast pathway. This pathway occurs up to endodermis.

14. Symplast pathway: When water passes across from one living cell to another living cell through plasmodesmata, then it is called the symplast pathway. It is also called the trans-membrane pathway.

3.Explain cohesion theory for translocation of water.

SOLUTION

1. This theory was put forth by Dixon and Jolly (1894).

2. This is presently a widely accepted theory explaining the ascent of sap in plants.

3. This theory is based on two principles i.e. Cohesion and adhesion, and transpiration pull.

4. Cohesion and adhesion:

a. A strong force of attraction between water molecules is called cohesive force.

b. While a strong force of attraction between water molecules and the lignified wall of the lumen of the xylem vessel, is called adhesive force.

c. Due to combined cohesive and adhesive forces a continuous water column is developed (formed) in the xylem right from root up to the tip of the topmost leaf in the plant.

5. Transpiration pull:

a. The transpiration pull developed in the leaf vessel is transmitted down to the root and thus accounts for the ascent of sap.

b. Excess water is lost in the form of vapour, mainly through the stomata found on a leaf.

c. This water loss increases the D.P.D. of mesophyll cells. These cells withdraw water ultimately from the xylem in the leaf.

d. In other words, due to continuous transpiration, a gradient of suction pressure (i.e. D.P.D.) is developed right from guard cells up to the xylem in the leaf. This will create a tension (called a negative pull or transpiration pull) in the xylem.

e. Consequently, the water column is pulled out of xylem. Thus, water is pulled upwards passively against the gravity leading to the ascent of sap.

4.Write mechanism of opening and closing of stoma.

SOLUTION

1. Opening and closing of stoma is controlled by the turgor of guard cells.

2. During day time, guard cells become turgid due to endo-osmosis.

3. Thus turgor pressure is exerted on the thin walls of guard cells.

4. Being elastic and thin, lateral walls are stretched out.

5. Due to kidney or dumb-bell like shape, inner thick walls are pulled apart to open (widen) the stoma.

6. During night time, guard cells become flaccid due to exo-osmosis.

7. Flaccidity closes the stoma almost completely.

8. Endo-osmosis and exo-osmosis occur due to diurnal changes in the osmotic potential of guard cells.

9. According to starch-sugar inter-conversion theory (Steward 1964), during day time, enzyme phosphorylase converts starch to sugar, thus increasing the osmotic potential of guard cells causing entry of water, thereby guard cells are stretched and stoma widens. The reverse reaction occurs at night bringing about the closure of the stoma.

Starch(Stomata open)⇌(Night)Phosphorylase (Day)Sugar(Stomata close)

10. According to the theory of proton transport (Levitt-1974), stomatal movement occurs due to the transport of protons H⁺ and K⁺ ions. During the daytime, starch is converted into malic acid. Malic acid dissociates to form malate ions and protons. Protons are transported to subsidiary cells and K⁺ ions are imported from them. Potassium Malate is formed that increases osmolarity and causes endosmosis. The uptake of K⁺ ions is always accompanied by Cl^– ions. At night, uptake of K⁺ and Cl– ions is prevented by abscisic acid, changing the permeability of guard cells. Due to this guard cells become hypotonic and thereby become flaccid.

5.What is hydroponics? How is it useful in identifying the role of nutrients?

SOLUTION

1. Hydroponics is a technique in which plants are grown in nutrient solutions in absence of soil. Roots are immersed in an adequately aerated, dilute, and defined solution of nutrients. Purified water and mineral salts are used in the nutrient medium.
2. In hydroponics, the concentration of a particular mineral in a solution of nutrients in which roots are immersed can be increased or decreased. By this method, essential elements can be identified and their deficiency symptoms can be discovered.
   Thus it helps to identify the role of nutrients in plant growth.

6.Explain the active absorption of minerals.

SOLUTION

1. Uptake of mineral ions against the concentration gradient is called active absorption.

2. Such movement requires an expenditure of energy by the absorbing cell. This energy is derived from respiration and is supplied through ATP.

3. The rate of active absorption of minerals depends upon respiration.

4. When the roots are deprived of oxygen, they show a sudden drop in the active absorption of minerals. The mineral ions accumulated in the root hair pass into the cortex and finally reach the xylem.

5. The minerals in the xylem are then carried along with water to other parts of the plant along the transpiration stream and are subsequently assimilated into organic molecules and then redistributed to other parts of the plant through the phloem.

7.Enlist macronutrients and micronutrients required for plant growth.

Long answer question.

Write on macro- and micro nutrients required for plant growth.

SOLUTION

1. Macronutrients:
   Some minerals like C, H, O, P, N, S, Mg, K, Ca required in large quantity for normal growth of the plant, are called macro elements. Macronutrients are required in large quantities. They mainly play nutritive and structural roles.
2. Micronutrients:
   Some minerals like Cu, Mo, Mn, Cl, Bo, Zn required in small quantities for the growth of a plant, are called microelements.
   Micronutrients are required in traces because they function in the catalytic role as co-factors.

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