Prasanna

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 16 Environmental Issues Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 16 Environmental Issues

PSEB 12th Class Biology Guide Environmental Issues Textbook Questions and Answers

Question 1.
What are the various constituents of domestic sewage? Discuss the effects of sewage discharge on a river.
Answer:
Domestic sewage is the waste originating from the kitchen, toilet, laundry, and other sources. It contains impurities such as suspended solid (sand, salt, clay), colloidal, material (faecal matter, bacteria, plastic and cloth fibre), dissolved materials (nitrate, phosphate, calcium, sodium, ammonia), and disease-causing microbes. When organic wastes from the sewage enter the water bodies, it serves as a food source for micro-organisms such as algae and bacteria. As a result, the population of these micro-organisms in the water body increases.

Here, they utilise most of the dissolved oxygen for their metabolism. This results in an increase in the levels of Biological oxygen demand (BOD) in river water and results in the death of aquatic organisms. Also, the nutrients in the water lead to the growth of planktonic algal, causing algal bloom. This causes deterioration of water quality and fish mortality.

Question 2.
List all the wastes that you generate, at home, school or during your trips to other places. Could you very easily reduce the generation of these wastes? Which would be difficult or rather impossible to reduce?
Answer:
Wastes generated at home include plastic bags, paper napkins, toiletries, kitchen wastes (such as peelings of vegetables and fruits, tea leaves), domestic sewage, glass, etc.

Wastes generated at schools include waste paper, plastics, vegetable and fruit peels, food wrappings, sewage etc.
Wastes generated at trips or picnics include plastic, paper, vegetable and fruit peels, disposable cups, plates, spoons etc.

Yes, wastes can be easily reduced by the judicious use of the above materials. Wastage of paper can be minimised by writing on both sides of the paper and by using recycled paper. Plastic and glass waste can also be reduced by recycling and re-using.

Also, substituting plastic bags with biodegradable jute bags can reduce wastes generated at home, school, or during trips. Domestic sewage can be reduced by optimising the use of water while bathing, cooking, and other household activities. Non-biodegradable wastes such as plastic, metal, broken glass, etc. are difficult to decompose because micro-organisms do riot have the ability to decompose them.

Question 3.
Discuss the causes and effects of global warming. What measures need to be taken to control global warming?
Answer:
Global warming is defined as an increase in the average temperature of the Earth’s surface. Causes of Global Warming: Global warming occurs as a result of the increased concentration of greenhouse gases in the atmosphere. Greenhouse gases include carbon dioxide, methane, and water vapour. These gases trap solar radiations released back by the Earth. This helps in keeping our planet warm and thus, helps in human survival. However, an increase in the amount of greenhouse gases can lead to an excessive increase in the Earth’s temperature, leading to global warming. Global warming is a result of industrialisation, burning of fossil fuels, and deforestation.

Effects of Global Warming: It has been observed that in the past three decades, the average temperature of the Earth has increased by 0.6°C. As a result, the natural water cycle has been disturbed resulting in changes in the pattern of rainfall. It also changes the amount of rainwater. Also, it results in the melting of Polar ice caps and mountain glaciers, which has caused a rise in the sea level, leading to the inundation of coastal regions.

Control Measures for Preventing Global Warming:

• Reducing the use of fossil fuels
• Use of bio-fuels
• Improving energy efficiency
• Use of renewable source of energy such as CNG etc.
• Reforestation.
• Recycling of materials

Question 4.
Match the items given in column A and B:

Answer:

Question 5.
Write critical notes on the following:
(a) Eutrophication
(b) Biological magnification
(c) Groundwater depletion and ways for its replenishment
Answer:
(a) Eutrophication:
It is the natural ageing process of a lake caused due to nutrient enrichment. It is brought down by the runoff of nutrients such as animal wastes, fertilisers, and sewage from land which leads to an increased fertility of the lake. As a result, it causes a tremendous increase in the primary productivity of the ecosystem. This leads to an increased growth of algae, resulting, into, algal blooms. Later, the decomposition of these algae depletes the supply of oxygen, leading to the death of other aquatic animal life.

(b) Biological Magnification: Unknowingly some harmful chemicals enter our bodies through the food chain. We use several pesticides and other chemicals to protect our crops from diseases and pests. These chemicals are either washed down into the soil or into the water bodies. From the soil, these are absorbed by the plants along with water and minerals, and from the water bodies, these are taken up by aquatic plants and animals. This is one of the ways in which they enter the food chain. As these chemicals are not degradable, these get accumulated progressively at each trophic level. As human beings occupy the topmost level in any food chain, the maximum concentration of these chemicals get accumulated in our bodies. This phenomenon is known as biological magnification.

(c) Ground Water Depletion and Ways for its Replenishment: The level of groundwater has decreased in the recent years. The source of water supply is rapidly diminishing each year because of an increase in the population and water pollution. To meet the demand of water, water is withdrawn from water bodies such as ponds, rivers etc. As a result, the source of groundwater is depleting.

This is because the amount of groundwater being drawn for human use is more than the amount replaced by rainfall. Lack of vegetation cover also results in very small amounts of water seeping through the ground. An increase in water pollution is another factor that has reduced the availability of groundwater.

Measures for Replenishing Ground Water:

• Preventing over-exploitation of groundwater
• Optimising water use and reducing water demand
• Rainwater harvesting
• Preventing deforestation and plantation of more trees.

Question 6.
Why does ozone hole form over Antarctica? How will enhanced ultraviolet radiation affect us?
Answer:
The ozone hole is more prominent over the region of Antarctica. It is formed due to an increased concentration of chlorine in the atmosphere. Chlorine is mainly released from chlorofluorocarbons (CFC’s) widely used as refrigerants. The CFC’s magnate from the troposphere to the stratosphere, where they release chlorine atoms by the action of UV rays on them.

The release of Chlorine atoms causes the conversion of ozone into molecular oxygen. One atom of chlorine can destroy around 10,000 molecules of ozone and causes ozone depletion. The formation of the ozone hole will result in an increased concentration of UV – B radiations on the Earth’s surface. UV -B damages DNA and activates the process of skin ageing. It also causes skin darkening and skin cancer. High levels of UV -B cause corneal cataracts in human beings.

Question 7.
Discuss the role of women and communities in protection and conservation of forests.
Answer:
Women and communities have played a major role in environmental conservation movements.
(i) Case Study of the Bishnoi Community: The Bishnoi community in Rajasthan strictly believes in the concept of living peacefully with nature. In 1731, the king of Jodhpur ordered his ministers to arrange wood for the construction of his new palace. For this purpose, the minister and 1116 workers went to Bishnoi village.

There, a Bishnoi woman called Amrita Devi along with her daughter and hundreds of other Bishnois showed the courage to step forward and stop them from cutting trees. They embraced the trees and lost their lives at the hands of soldiers of the king. This resistance by the people of the village forced the king to give up the idea of cutting trees.

(ii) Chipko Movement: The Chipko movement was started in 1974 in the Garhwal region of the Himalayas. In this movement, the women from the village stopped the contractors from cutting forest trees by embracing them.

Question 8.
What measures, as an individual, would you take to reduce environmental pollution?
Answer:
The following initiatives can be taken to prevent environmental pollution:
Measures for Preventing Air Pollution

• Planting more trees
• Use of clean and renewable energy sources such as CNG and bio-fuels
• Reducing the use of fossil fuels
• Use of catalytic converters in automobiles

Measures for Preventing Water Pollution:

• Optimising the use of water
• Using kitchen wastewater in gardening and other household purposes

Measures for Controlling Noise Pollution:

• Avoid burning crackers on Diwali
• Plantation of more trees

Measures for Decreasing Solid Waste Generation:

• Segregation of waste
• Recycling and reuse of plastic and paper
• Composting of biodegradable kitchen waste
• Reducing the use of plastics.

Question 9.
Discuss briefly the following:
(a) Radioactive wastes
(b) Defunct ships and e-wastes
(c) Municipal solid Wastes
Answer:
(a) Radioactive Wastes: Radioactive wastes are generated during the process of generating nuclear energy from radioactive materials. Nuclear waste is rich in radioactive materials that generate large quantities of ionising radiations such as gamma rays. These rays cause mutation in organisms, which often results in skin cancer. At high dosage, these rays can be lethal.

Safe disposal of radioactive wastes is a big challenge. It is recommended that nuclear wastes should be stored after pre-treatment in suitable shielded containers, which should then be buried in rocks.

(b) Defunct Ships and E-wastes: Defunct ships are dead ships that are no longer in use. Such ships are broken down for scrap metal in countries such as India and Pakistan. These ships are a source of various toxicants such as asbestos, lead, mercury etc. Thus, they contribute to solid* wastes that are hazardous to health.

E-waste or electronic wastes generally include electronic goods such as computers etc. Such wastes are rich in metals such as copper, iron, silicon, gold etc. These metals are highly toxic and pose serious health hazards. People of developing countries are involved in the recycling process of these metals and therefore, get exposed to toxic substances present in these wastes.

(c) Municipal Solid Wastes: Municipal solid wastes are generated from schools, offices, homes, and stores. It is generally rich in glass, metal, paper waste, food, rubber, leather, and textiles. The open dumps of municipal wastes serve as a breeding ground for flies, mosquitoes, and other disease-causing microbes. Hence, it is necessary to dispose of municipal solid waste properly to prevent the spreading of diseases. Sanitary landfills and incineration are the methods for the safe disposal of solid wastes.

Question 10.
What initiatives were taken for reducing vehicular air pollution in Delhi? Has air quality improved in Delhi?
Answer:
Delhi has been categorised as the fourth most polluted city of the world in a list of 41 cities. Burning of fossil fuels has added to the pollution of air in Delhi.
Various steps have been taken to improve the quality of air in Delhi.

(a) Introduction of CNG (Compressed Natural Gas): By the order of the supreme court of India, CNG-powered vehicles were introduced at the end of year 2006 to reduce the levels of pollution in Delhi. CNG is a clean fuel that produces very little unburnt particles.
(b) Phasing out of old vehicles
(c) Use of unleaded petrol
(d) Use of low-sulphur petrol and diesel
(e) Use of catalytic converters
(f) Application of stringent pollution-level norms for vehicles
(g) Implementation of Bharat stage I, which is equivalent to euro II norms in vehicles of major Indian cities.

The introduction of CNG-powered vehicles has improved Delhi’s air quality, which has lead to a substantial fall in the level of CO₂ and SO₂. However, the problem of suspended particulate matter (SPM) and respiratory suspended particulate matter,(RSPM) still persists.

Question 11.
Discuss briefly the following:
(a) Greenhouse gases
(b) Catalytic converter
(c) Ultraviolet B
Answer:
(a) Greenhouse Gases: The greenhouse effect refers to an overall increase in the average temperature of the Earth due to the presence of greenhouse gases. Greenhouse gases mainly consist of carbon dioxide, methane, and water vapour. When solar radiations reach the Earth, some of these radiations are absorbed.

These absorbed radiations are Released back into the atmosphere. These radiations are trapped by the greenhouse gases present in the atmosphere. This helps in keeping our planet warm and thus, helps in human survival.
However, an increase in the amount of greenhouse gases can lead to an excessive increase in the Earth’s temperature, thereby causing global warming.

(b) Catalytic Converter: Catalytic converters are devices fitted in automobiles to reduce automobile or vehicle pollution. These devices contain expensive metals such as platinum, palladium, and rhodium that act as catalysts.
As the vehicular discharge -passes through the catalytic converter, the unburnt hydrocarbons present in it get converted into carbon dioxide and water. Carbon monoxide and nitric oxide released by catalytic converters are converted into carbon dioxide and nitrogen gas respectively.

(c) Ultraviolet B: Ultraviolet-B is an electromagnetic radiation which has a shorter wavelength than visible light.
It is a harmful radiation that comes from sunlight and penetrates through the ozone hole onto the Earth’s surface.
It induces many health hazards in humans. UV -B damages DNA and activates the process of skin ageing. It also causes skin darkening and skin cancer. High levels of UV -B cause corneal cataracts in human beings.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 14 Respiration in Plants Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 14 Respiration in Plants

PSEB 11th Class Biology Guide Respiration in Plants Textbook Questions and Answers

Question 1.
Differentiate between:
(a) Respiration and Combustion
(b) Glycolysis and Krebs’ cycle
(c) Aerobic respiration and Fermentation
Answer:
(a) Differences between Respiration and Combustion

(b) Differences between Glycolysis and Krebs’ Cycle

(c) Differences between Aerobic Respiration and Fermentation

Question 2.
What are respiratory substrates? Name the most common respiratory substrate.
Answer:
The compounds oxidized during the process of respiration are called respiratory substrates. Carbohydrates, especially glucose, act as respiratory substrates. Fats, proteins, and organic acids also act as respiratory substrates.

Question 3.
Give the schematic representation of glycolysis.
Answer:
Glycolysis

• The scheme of reactions was given by Embden, Meycrhof and Pumas, hence it is also called EMP Pathway; it occurs in the cytoplasm of cells.
• In this process, glucose is partially oxidized/ converted into two molecules of pyruvic acid.
• Glucose is phosphorylated with the help of ATP into Glucose-6-Phosphate, catalyzed by the enzyme hexokinase.
• Glucose-6-Phosphate is converted into its isomer, Fructose-6-Phosphate, catalyzed by isomerase.
• Fructose-6-Phosphate is then phosphorylated with the use of ATP into Fructose 1,6-bisphosphate, catalyzed by
  phosphofructokinase.
• Fructose 1,6-bisphosphate is split into one molecule of 3-pho sphoglyceraldchyd and one molecule of
  dihydroxy-acetone phosphate; these two products are interconvertible.
• 3-phosphoglyceraldehyde is oxidized to 1, 3 bi phosphoglycerate, where NAD is reduced to NADH.
• 1,3-bisphosphoglycerate is split into 3-phosphoglycerate along with the formation of ATP, catalyzed by phosphoglycerate kinase.

• 3-Phosphoglycerate is subsequently converted into 2-phosphoglycerate and then into Phosphoenol Pyruvate (PEP).
• PEP is converted into pyruvate along with the formation of ATP, catalyzed by the enzyme pyruvate kinase.

In this pathway, ATP molecules are formed in two ways :

1. Direct/substrate phosphorylation of ADP to ATP.
2. Oxidation of NADH to NAD ’ and ATP formation through electron transport (oxidative phosphorylation).

Four molecules of ATP are formed directly in the following steps:

• When 1, 3 bi phosphoglyceric acids is converted into 3-phosphoglyceric acid and
• When phosphoenolpyruvate is converted into pyruvic acid.

Two molecules of ATP are consumed, one in each of the following steps:

1. Phosphorylation of glucose to glucose-6- phosphate and
2. Conversion of fructose-6-phosphate into fructose 1, 6 bisphosphates.

When 3-phosphoglyceraldehyde (PGAL) is converted into 1, 3 bi phosphoglycerates, two redox equivalents are removed in the form of two hydrogen atoms from PGAL and transferred to NAD⁺, which is reduced to NADH ⁺ H⁺.
Oxidation of two molecules of NADH to NAD yields six molecules of ATP during electron transport. The end products of glycolysis are two molecules each of pyruvic acid, NADH, and ATP.

Question 4.
What are the main steps in aerobic respiration? Where does it take place?
Answer:
The main steps in aerobic respiration are as follows :

• Glycolytic breakdown of glucose into pyruvic acid.
• Oxidative decarboxylation of pyruvic acid to acetyl Co-A (acetyl coenzyme-A).
• Krebs’ cycle.
• Terminal oxidation and phosphorylation in respiratory chain.
  It occurs inside the mitochondrial matrix.

Question 5.
Give the schematic representation of an overall view of Kreb’s cycle.
Answer:
Tricarboxylic Acid Cycle or Kreb’s Cycle:

• It is known as tricarboxylic acid cycle, since the first product, citric acid (hence citric acid cycle) is a tricarboxylic acid (has three COOH groups).
• The sequence of reactions in citric acid cycle was first followed by Sir Hans Kreb; hence it is also known as Kreb’s cycle.
• During this cycle of reactions, 3 molecules of NAD and one molecule of FAD are reduced to NADH and FADH respectively.
• Respiration in Plants 115:
• One ATP molecule is produced directly from GTP formed by substrate-level phosphorylation.

The reactions of aerobic oxidation of Pyruvic acid can be summed up as :
Pyruvic acid + 4NAD+ + FAD+ + 2H₂O + ADP + Pi → 3CO₂ + 4NADH + 4H⁺ + FADH₂ + ATP

Question 6.
Explain ETS.
Answer:
Electron Transport System (ETS):

• ETS occurs in the electron transport particles (ETP) on the inner surface of the inner membrane of mitochondria.
• NADH formed in glycolysis and citric acid cycle are oxidized by NADH-dehydrogenase (complex I) and the electrons are transferred to ubiquinone, via FMN.
• Ubiquinone also receives reducing equivalents via FADH generated during the oxidation of succinate by succinate dehydrogenase (complex II).
• The reduced ubiquinone called ubiquinol is then oxidized by transfer of electrons to cytochrome c, via cytochrome be bc₁ -complex (complex III).
• Cytochrome c acts as a mobile carrier between complex III and complex IV.
• Complex IV refers to cytochrome c oxidase complex containing cytochromes a and a3 and two copper centres.
• When the electrons are shunted over the carriers via complex I to IV in the electron transport chain, they are coupled to ATP synthetase (complex V) for the formation of ATP from ADP and IP.
• Oxygen functions as the terminal acceptor of electrons and is reduced to water along with the hydrogen atoms.

Question 7.
Distinguish between the following:
(a) Aerobic respiration and anaerobic respiration.
(b) Glycolysis and fermentation.
(c) Glycolysis and citric acid cycle.
Answer:
(a) Differences between Aerobic and Anaerobic Respiration

(b) Differences between Glycolysis and Fermentation:

(c) Differences between Glycolysis and Krebs’ Cycle

Question 8.
What are the assumptions made during the calculation of net gain of ATP?
Answer:
The calculations of net gain of ATP can be made only on certain assumptions :

• There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
• The NADH synthesized in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
• None of the intermediates in the pathway are utilised to synthesize any other compound.
• Only glucose is being respired-no other alternative substrates are entering in the pathway at any of the intermediary stages.

But this kind of assumptions are not really valid in a living system. All pathways work simultaneously and do not take place one after another. Substrates enter the pathways and are withdrawn from it as end when necessary; ATP is utilized as and when needed; enzymatic rates are controlled by multiple means. In overall steps, there is a net gain of 36 ATP molecules during aerobic respiration of one molecule of glucose.

Question 9.
Discuss ‘the respiratory pathway is an amphibolic pathway.’
Answer:
Aniphibolic Pathway: Glucose is the favored substrate for respiration. All carbohydrates are usually first converted into glucose before they are used for respiration. Other substrates can also be respired but then they do not enter the respiratory pathway at the first step.

Since respiration involves breakdown as well as synthesis of substrates, the respiratory process involves both catabolism and anabolism. That is why respiratory pathway is considered to be an amphibolic pathway rather than as a catabolic one.

Question 10.
Define RQ. What is its value for fats?
Answer:
Respiratory quotient (RQ) or respiratory ratio can be defined as the ratio of the volume of CO₂ evolved to the volume of O₂ consumed during respiration. The value of respiratory quotient depeñds on the type of respiratory substrate. Its value is one for carbohydrates. However, it is always less than one for fats as fats consume more oxygen for respiration than carbohydrates.

It can be illustrated through the example of tripalmitin fatty acid, which consumes 145 molecules of O₂ for respiration while 102 molecules of CO₂ are evolved.The RQ value for tripalmitin is 0.7.

Question 11.
What is oxidative phosphorylation?
Answer:
It refers to the formation of ATP in the mitochondria, utilizing the energy obtained by the oxidation of organic molecules.

Question 12.
What is the significance of step-wise release of energy in respiration?
Answer:
During oxidation within a cell, all the energy contained in respiratory substrates is not released free in a single step. It is released in a series of slow step-wise reactions controlled by enzymes and it is trapped as chemical energy in the form of ATP. The significance of this step-wise release of energy is that some energy is used to synthesize ATP. This ATP is stored for the later utilization wherever required. Hence, ATP acts as the energy currency of cell. The energy

stored in ATP can be utilized for following:

• In various energy-requiring processes of organisms.
• The carbon skeleton produced during respiration is used as precursors for the synthesis of other molecules.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 1 Reproduction in Organisms Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 1 Reproduction in Organisms

PSEB 12th Class Biology Guide Reproduction in Organisms Textbook Questions and Answers

Question 1.
Why is reproduction essential for organisms?
Answer:
Reproduction is a fundamental feature of all living organisms. It is a biological process through which living organisms produce offsprings (young ones) similar to them. Reproduction ensures the continuance of various species on the Earth. In the absence of reproduction, the species will not be able to exist for a long time and may soon get extinct.

Question 2.
Which is a better mode of reproduction sexual or asexual? Why?
Answer:
Sexual reproduction is a better mode of reproduction. It allows the formation of new variants by the combination of the DNA from two different individuals, typically one of each sex. It involves the fusion of the male and the female gamete to produce variants, which are not identical to their parents and to themselves. This variation allows the individual to adapt constantly changing and challenging environment. Also, it leads to the evolution of better suited organisms which ensures greater survival of a species. On the contrary, asexual reproduction allows very little or no variation at all. As a result, the individuals produced are exact copies of their parents and themselves.

Question 3.
Why is the offspring formed by asexual reproduction referred to as clone?
Answer:
A clone is a group of morphologically and genetically identical individuals. In the process of asexual reproduction, only one parent is involved and there is no fusion of the male and the female gamete. As a result, the offsprings so produced are morphologically and genetically similar to their parents and are thus, called clones.

Question 4.
Offspring formed due to sexual reproduction have better chances of survival. Why? Is this statement always true?
Answer:
Sexual reproduction involves the fusion of the male and the female gamete. This fusion allows the formation of new variants by the combination of the DNA from two (usually) different members of the species. The variations allow the individuals to adapt under varied environmental conditions for better chances of survival.

However, it is not always necessary that the offspring produced due to sexual reproduction has better chances of survival. Under some circumstances, asexual reproduction is more advantageous for certain organisms. For example, some individuals who do not move from one place to another and are well settled in their environment. Also, asexual reproduction is a fast and’ a quick mode of reproduction which does not consume much time and energy as compared to sexual reproduction.

Question 5.
How does the progeny formed from asexual reproduction differ from those formed by sexual reproduction?
Answer:

Question 6.
Distinguish between asexual and sexual reproduction. Why is
vegetative reproduction also considered as a type of asexual reproduction?
Answer:

Vegetative reproduction is a process in which new plants are obtained without the production of seeds or spores. It involves the propagation of plants through certain vegetative parts such as the rhizome, sucker, tuber, bulb, etc. It does not involve the fusion of the male and the female gamete and requires only one parent. Hence, vegetative reproduction is considered as a type of asexual reproduction.

Question 7.
What is vegetative propagation? Give two suitable examples.
Answer:
Vegetative propagation is a mode of asexual reproduction in which new plants are obtained from the vegetative parts of plants. It does not involve the production of seeds or spores for the propagation of new plants. Vegetative parts of plants such as runners, rhizomes, suckers, tubers, etc. can be used as propagules for raising new plants.

Examples of vegetative reproduction are given below:
1. Eyes of potato: The surface of a potato has several buds called eyes. Each of these buds when buried in soil develops into a new plant, which is identical to the parent plant.

2. Leaf buds of Bryophyllum: The leaves of Bryophyllum plants bear several adventitious buds on their margins. These leaf buds have the ability to grow and develop into tiny plants when the leaves get detached from the plant and come in contact with moist soil.

Question 8.
Define:
(a) Juvenile phase,
(b) Reproductive phase,
(c) Senescent phase.
Answer:
(a) Juvenile Phase: All organisms have to reach a certain stage of growth and maturity in their life, before they can be reproduce sexually. This phase of growth is called the juvenile phase or vegetative phase in plants.

(b) Reproductive Phase: When the juvenile phase is over the organisms enter the period of reproductive phase or sexual maturity. It is indicated by showing various morphological and physiological changes such as development of secondary sexual characters in animals and by flowering in plants. This is the actual period of the life span of any organism when it is capable of producing offsprings. This phase is of variable duration in different organisms.

(c) Senescent Phase : This is the final and third stage of growth cycle. It can be considered as the end of reproductive phase. It is accompanied by reduction in functional capacity and increase in cellular break down and metabolic failures. It ultimately leads to death.

Question 9.
Higher organisms have resorted to sexual reproduction in spite of its complexity. Why?
Answer:
Although sexual reproduction involves more time and energy, higher organisms have resorted to sexual reproduction in spite of its complexity. This is because this mode of reproduction helps introduce new variations in progenies through the combination of the DNA from two (usually) different individuals. These variations allow the individual to cope with various environmental conditions and thus, make the organism better suited for the environment. Variations also lead to the evolution of better organisms and therefore, provide better chances of survival. On the other hand, asexual reproduction does not provide genetic differences in the individuals produced.

Question 10.
Explain why meiosis and gametogenesis are always interlinked?
Answer:
Meiosis is a process of reductional division in which the amount of genetic material is reduced. Gametogenesis is the process of the formation of gametes. Gametes produced by organisms are haploids (containing only one set of chromosomes), while the body of an organism is diploid. Therefore, for producing haploid gametes (gametogenesis), the germ cells of an organism undergo meiosis. During the process, the meiocytes of an organism undergo two successive nuclear and cell divisions with a single cycle of DNA replication to form the haploid gametes.

Question 11.
Identify each part in a flowering plant and write whether it is haploid (n) or diploid (2n).
(a) Ovary …………………………
(b) Anther ……………………..
(c) Egg ………………………….
(d) Pollen ……………………..
(e) Male gamete ……………….
(f) Zygote ………………..
Answer:
(a) diploid (2n)
(b) diploid (2n)
(c) haploid (n)
(d) haploid (n)
(e) haploid (n)
(f) diploid (2n)

Question 12.
Define external fertilisation. Mention its disadvantages.
Answer:
External fertilisation is the process in which the fusion of the male and the female gamete takes place outside the female body in an external medium, generally water. Fish, frog, starfish are some organisms that exhibit external fertilisation.

Disadvantages of external fertilisation
In external fertilisation, eggs have less chances of fertilisation. This can lead to the wastage of a large number of eggs produced during the process.
Further, there is an absence of proper parental care to the offspring, which results in a low rate of survival in the progenies.

Question 13.
Differentiate between a zoospore and a zygote.
Answer:

Question 14.
Differentiate between gametogenesis from embryogenesis.
Answer:

Question 15.
Describe the post-fertilisation changes in a flower.
Answer:
As a result of double fertilisation in flowering plants, zygote (2n) and the primary endosperm nucleus (3n) is produced. The calyx, corolla, stamens and style wither away. The calyx may persist or even show growth in certain cases. The post fertilisation changes which take place are

• Endosperm formation
• Embryo formation
• Seed formation and
• Fruit formation.

The primary endosperm nucleus becomes active and forms a nutritive vegetative tissue. The endosperm at the expense of food present in the nucellus, Endosperm may be completely used up by the developing embryo (non-endospermic seeds e.g., pea) or may persist in the seed (endospermic seed e.g., castor). The zygote, waits for sometime till the formation of endosperm and then develops into embryo, by withdrawing nutrition from the endosperm. Ultimately the ovules are transformed into seeds and the ovary becomes a fruit. The formation of fruit helps in the nourishment and protection to the developing seeds and later helps in seed dispersal. Under favourable conditions the seeds germinate to form new plants.

Question 16.
What is a bisexual flower? Collect five bisexual flowers from your neighbourhood and with the help of your teacher find out their common and scientific names.
Answer:
A flower that contains both the male and female reproductive structure (stamen and pistil) is called a bisexual flower.
Examples of plants bearing bisexual flowers are:

1. Water lily {Nymphaea odorata)
2. Rose (Rosa multiflora)
3. Hibiscus (Hibiscus Rosa-sinensis)
4. Mustard (Brassica nigra)
5. Petunia (Petunia hybrida)

Question 17.
Examine a few flowers of any cucurbit plant and try to identify the staminate and pistillate flowers. Do you know any other plant that hears unisexual flowers?
Answer:
Cucurbit plant bears unisexual flowers as these flowers have either the stamen or the pistil. The staminate flowers bear bright, yellow coloured petals along with stamens that represent the male reproductive structure. On the other hand, the pistillate flowers bear only the pistil that represents the female reproductive structure.
Other examples of plants that bear unisexual flowers are corn, papaya, cucumber, etc.

Question 18.
Why are offspring of oviparous animals at a greater risk as compared to offspring of viviparous animals?
Answer:
Oviparous animals lay eggs outside their body. As a result, the eggs of these animals are under continuous threat from various environmental factors. On the other hand, in viviparous animals, the development of the egg takes place inside the body of the female. Hence, the offspring of an egg-laying or oviparous animal is at greater risk as compared to the offspring of a viviparous animal, which gives birth to its young ones.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 9 Biomolecules Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 9 Biomolecules

PSEB 11th Class Biology Guide Biomolecules Textbook Questions and Answers

Question 1.
What are macromolecules? Give examples.
Answer:
Chemical compounds, which are found in the acid insoluble fraction are called macromolecules or biomacromolecules. For example, proteins, lipids and carbohydrate, etc.

Question 2.
Illustrate a glycosidic, peptide and a phosphodiester bond.
Answer:
Glycosidic Bond: A glycosidic bond is a type of functional group that joins a carbohydrate (sugar) molecule to another group, which may or may not be another carbohydrate.

Peptide Bond: A peptide bond (amide bond) is a chemical bond formed between two molecules when the carboxyl group of one molecule reacts with the amine group of the other molecule, thereby releasing a molecule of water (H20).
H₂O.

Phosphodiester Bond: A phosphodiester bond is a group of strong covalent bonds between a phosphate group and two other molecules over two ester bonds. In DNA and RNA, the phosphodiester bond is the linkage between the 3′ carbon atom of one sugar molecule and the 5’carbon of another, deoxyribose in DNA and ribose in RNA.

Question 3.
What is meant by tertiary structure of proteins?
Answer:
Tertiary Structure of Protein; The overall shape of a single protein molecule; the spatial relationship of the secondary structures to one another. Tertiary structure is generally stabilized by non-local interactions, most commonly the formation of a hydrophobic core, but also through salt bridges, hydrogen bonds, disulfide bonds,1 and even post-translational modifications. The term “tertiary structure” is often used as synonymous with the term fold. The tertiary structure is what controls the basic function of the protein.

Question 4.
Find and write down structures of 10 interesting small molecular weight biomolecules. Find if there is any industry which manufactures the compounds by isolation. Find out who are the buyers?
Answer:

Question 5.
Proteins have primary structure. If you are given a method to know which amino acid is at either of the two termini (ends) of a protein, can you connect this information to purity or homogeneity of a protein?
Answer:
The sequence of amino acids, i.e., the positional information in a protein which is the first amino acid, which is second and so on is called the primary structure of a protein. The first amino acid is also called as N-terminal amino acid. The last amino acid is called the C-terminal amino acid. Yes, we can connect this information to purity or homogeneity of a protein. Based on number of amino and carboxyl groups, there are acidic (e.g., glutamic acid), basic (lysine) and neutral (valine) amino acids, proteins may be acidic, basic and neutral.

Question 6.
Find out and make a list of proteins used as therapeutic agents. Find other applications of proteins (e. g, cosmetics etc.)
Answer:
Some proteins and their functions are as follows:

Question 7.
Explain the composition of triglyceride.
Answer:
Triglycerides are composed of two types of molecules, i.e., glycerol i (3 carbon molecules) and fatty acids which attach to the glycerol at the alcohol unit. The following is a structural representation of a triglyceride at the molecular level.

Fatty acids are chains of hydrocarbons 4-22 (or more) carbons a long with a carboxyl group at one end. If each carbon has two hydrogen atoms, the fatty acid is saturated. If two carbon atoms are double-bonded, so that there is less hydrogen in the fatty acid, it is unsaturated (monounsaturated). If more than two carbon atoms are unsaturated, the fatty acid is polyunsaturated.

Question 8.
Can you describe what happens when milk is converted into curd or yoghurt, from your understanding of proteins.
Answer:
Milk contains a protein called casein. This protein gives milk its characteristic white colour. It is of high nutritional value because it contains all the essential amino acids required by man’s body. The curd forms because of the chemical reaction between lactic acid bacteria and casein. When curd is added to milk, the lactic acid bacteria present in it cause coagulation of casein and thus, convert it into curd.

Question 9.
Can you attempt building models of biomolecules using commercially available atomic models (Ball and Stick models).
Answer:
Yes, we can make models of biomolecules using commercially available atomic models.

Question 10.
Attempt titrating an amino acid against a weak base and discover the number of dissociating (ionisable) functional groups in the amino acid.
Answer:
When an amino acid is titrated against a weak base, it dissociates and gives two functional groups:
(i) -COOH group (carboxylic group)
(ii) Amino group (NH_2/sub>)

Question 11.
Draw the structure of the amino acid, alanine.
Answer:

Question 12.
What are gums made of? Is fevicol different?
Answer:
Gums are made of carbohydrates, i. e., L-rhamnose, D-galactose and D-galacturonic acid, etc. Fevicol is different from natural gums. It is a synthetic product.

Question 13.
Find out a qualitative test for proteins, fats and oils, amino acids and test any Fruit juice, saliva, sweat and urine for them.
Answer:

• Test for Proteins: Biuret test if Biuret’s reagent added to protein, then the colour of the reagent changes light blue to purple.
• Test for Fats and Oils: Grease or test.
• Test for Amino Acids: Ninhydrin test. If Ninhydrin reagent is added to the solution, then the colourless solution changes to pink, blue or purple depending upon the amino acid.

Question 14.
Find out how much cellulose is made by all the plants in the biosphere and compare it with how much of paper is manufactured by man and hence what is the consumption of plant material by man annually. What a loss of vegetation!
Answer:
Approximately, 100 billion tonnes of cellulose are made per year by all the plants in the biosphere and it takes 17 full grown trees to make one ton of paper. Trees are also used to fulfil the other requirements of man such as for timber, food, medicines, etc. Hence, it is difficult to calculate the annual consumption of plant material by man.

Question 15.
Describe the important properties of enzymes.
Answer:
Important properties of enzymes are given below :

• Enzymes are proteins which catalyse biochemical reactions in the cells.
• They are denatured at high temperatures.
• Enzymes generally function in a narrow range of temperature and pH. Each enzyme shows its highest activity at a particular temperature and pH called the optimum temperature and optimum pH.
• With the increase in substrate concentration, the velocity of the eyzymatic reaction rises at first. The reaction ultimately reaches a maximum velocity (V_max) which is not exceeded by any further rise in concentration of the
  substrate.
• The activity of an enzyme is also sensitive to the presence of specific chemicals that bind to the enzyme.
• Enzymes are substrate specific in their action.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 5 Principles of Inheritance and Variation Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation

PSEB 12th Class Biology Guide Principles of Inheritance and Variation Textbook Questions and Answers

Question 1.
Mention the advantages of selecting pea plant for experiment by Mendel.
Answer:
Mendel selected pea plants to carry out his study on the inheritance of characters from parents to offspring. He selected a pea plant because of the following features:

• Peas have many visible contrasting characters such as tall/dwarf plants, round/wrinkled seeds, green/yellow pod, purple/white flowers, etc.
• Peas have bisexual flowers and therefore undergo self pollination easily. Thus, pea plants produce offsprings with same traits generation after generation.
• In pea plants, cross pollination can be easily achieved by emasculation in which the stamen of the flower is removed without affecting the pistil.
• Pea plants have a short life span and produce many seeds in one generation.

Question 2.
Differentiate between the following:
Dominance and Recessive Homozygous and Heterozygous
(a) Dominance and Recessive
(b) Homozygous and Heterozygous
(c) Monohybrid and Dihybrid.
Answer:
(a) Dominance and Recessive

(b) Homozygous and Heterozygous

(c) Monohybrid and Dihybrid

Question 3.
A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?
Answer:
Locus is a fixed position on a chromosome, which is occupied by a single or more genes. Heterozygous organisms contain different alleles for an allelic pair. Hence, a diploid organism, which is heterozygous at four loci, will have four different contrasting characters at four different loci.

For example, if an organism is heterozygous at four loci with four characters, say Aa, Bb, Cc, Dd, then during meiosis, it will segregate to form 8 separate gametes.

If the genes are not linked, then the diploid organism will produce 16 different gametes. However, if the genes are linked, the gametes will reduce their number as the genes might be linked and the linked genes will be inherited together during the process of meiosis.

Question 4.
Explain the Law of Dominance using a monohybrid cross.
Answer:
Mendel’s law of dominance states that a dominant allele expresses itself in a monohybrid cross and suppresses the expression of recessive allele. However, this recessive allele for a character is not lost and remains hidden or masked in the progenies of F₁ generation and reappears in the next generation.

For example, when pea plants with round seeds (RR) are crossed with plants with wrinkled seeds (rr), all seeds in F₁ generation were found to be round (Rr). When these round seeds were self fertilised, both the round and wrinkled seeds appeared in F₂ generation in 3 : 1 ratio. Hence, in F₂ generation, the dominant character (round seeds) appeared and the recessive character (wrinkled seeds) got suppressed, which reappeared in F₂ generation.

Question 5.
Define and design a test-cross.
Answer:
Test cross is a cross between an organism with unknown genotype and a recessive parent. It is used to determine whether the individual is homozygous or heterozygous for a trait.
If the progenies produced by a test cross show 50% dominant trait and 50% recessive trait, then the unknown individual is heterozygous for a trait. On the other hand, if the progeny produced shows dominant trait, then the unknown individual is homozygous for a trait.

Question 6.
Using a Punnett Square, workout the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.
Answer:
In guinea pigs, heterozygous male with black coat colour (Bb) is crossed with the homozygous female having white coat colour (bb). The male will produce two types of gametes, B and b, while the female will produce only one kind of gamete, b. The genotypic and phenotypic ratio in the progenies of Fx generation will be same i.e., 1:1.

Question 7.
When a cross is made between tall plant with yellow seeds (TtYy) and tall plant with green seed (Ttyy), what proportions of phenotype in the offspring could be expected to he
(a) tall and green.
(b) dwarf and green.
Answer:
A cross between tall plant with yellow seeds and tall plant with green seeds will produce
(a) three tall and green plants
(b) one dwarf and green plant

Question 8.
Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F₁ generation for a dihybrid cross?
Answer:
When two individual heterozygous for two loci (Yy Rr) are crossed and the two loci are linked, the distribution of the phenotypic feature of F₁ generation will be in the ratio of 3:1 \(\frac{3}{4}\) of the individuals will show
both the dominant traits and \(\frac{1}{4}\) of the individuals will show both the
recessive traits. It is because the genes for both the traits are present on the same chro

Question 9.
Briefly mention the contribution of T.H. Morgan in genetics.
Answer:
Morgan’s work is based on fruit flies (Drosophila melanogaster). He formulated the chromosomal theory of linkage. He defined linkage as the co-existence of two or more genes in the same chromosome and performed dihybrid crosses in Drosophila to show that linked genes are inherited together and are located on X-chromosome. His experiments have also proved that tightly linked genes show very low recombination while loosely linked genes show higher recombination.

Question 10.
What is pedigree analysis? Suggest how such an analysis, can be useful.
Answer:
Pedigree analysis is a record of occurrence of a trait in several generations of a family. It is based on the fact that certain characteristic features are heritable in a family, for example, eye colour, skin colour, hair form and colour, and other facial characteristics. Along with these features, there are other genetic disorders such as Mendelian disorders that are inherited in a family, generation after generation. Hence, by using pedigree analysis for the study of specific traits or disorders, generation after generation, it is possible to trace the pattern of inheritance. In this analysis, the inheritance of a trait is represented as a tree, called family tree. Genetic counselors use pedigree chart for analysis of various traits and diseases in a family and predict their inheritance patterns. It is useful in preventing haemophilia, sickle cell anaemia, and other genetic disorders in the future generations.

Question 11.
How is sex determined in human beings?
Answer:
Human beings exhibit male heterogamy. In humans, males (XY) produce two different types of gametes, X and Y. The human female (XX) produces only one type of gametes containing X chromosomes. The sex of the baby is determined by the type of male gamete that fuses with the female gamete. If the fertilising sperm contains X chromosome, then the baby produced will be a girl and if the fertilising sperm contains Y chromosome, then the baby produced will be a boy. Hence, it is a matter of chance that determines the sex of a baby. There is an equal probability of the fertilising sperm being an X or Y chromosome. Thus, it is the genetic make up of the sperm that determines the sex of the baby.

Question 12.
A child has hlood group O. If the father has hlood group A and mother blood group B, work out the genotypes of the parents and the possible genotypes of the other offsprings.
Answer:
The blood group characteristic in humans is controlled by three set of alleles, namely, I^A,I^B and i. The alleles, I^A and I^B, are equally dominant whereas allele, i, is recessive to the other alleles. The individuals with genotype, I^A I^A and I^A i, have blood group A whereas the individuals with genotype, I^B I^B and I^B i, have blood group B. The persons with genotype I^A I^B have blood group AB while those with blood group O have genotype ii.
Hence, if the father has blood group A and mother has blood group B, then the possible genotype of the parents will be
Father
I^AI ^{or A}I^Ai

Mother
I^BI^B or I^Bi
A cross between homozygous parents will produce progeny with AB blood group.

A cross between heterozygous parents will produce progenies with AB blood group (I^A I^B) and O blood group (ii).

Question 13.
Explain the following terms with example
(a) Co-dominance
(b) Incomplete dominance
Answer:
(a) Co-dominance: Co-dominance is the phenomenon in which both the alleles of a contrasting character are expressed in heterozygous condition. Both the alleles of a gene are equally dominant. ABO blood group in human beings is an example of co-dominance. The blood group character is controlled by three sets of alleles, namely, I^A, I^B, and i. The alleles, I^A and I^B, are equally dominant and are said to be co-dominant as they are expressed in AB blood group. Both these alleles do not interfere with the expression of each other and produce their respective antigens. Hence, AB blood group is an example of co-dominance.

(b) Incomplete Dominance: Incomplete dominance is a phenomenon in which one allele shows incomplete dominance over the other member of the allelic pair for a character. For example, a monohybrid cross between the plants having red flowers and white flowers in Antirrhinum species will result in all pink flower plants in Fj generation. The progeny obtained in Fx generation does not resemble either of the parents and exhibits intermediate characteristics. This is because the dominant allele, R, is partially dominant over the other allele, r. Therefore, the recessive allele, r, also gets expressed in the generation resulting in the production of intermediate pink flowering progenies with Rr genotype.

What is point mutation? Give one example.
Answer:
Point mutation is a change in a single base pair of DNA by substitution, deletion, or insertion of a single nitrogenous base. An example of point mutation is sickle cell anaemia. It involves mutation in a single base pair in the beta-globin chain of haemoglobin pigment of the blood. Glutamic acid in short arm of chromosome II gets replaced with valine at the sixth position.

Question 15.
Who had proposed the chromosomal theory of the inheritance?
Answer:
Sutton and Boveri proposed the chromosomal theory of inheritance in 1902. They linked die inheritance of traits to the chromosomes.

Question 16.
Mention any two autosomal genetic disorders with their symptoms.
Answer:
Two autosomal genetic disorders are as follows:
1. Sickle Cell Anaemia: It is an autosomal linked recessive disorder, which is caused by point mutation in the beta-globin chain of haemoglobin pigment of the blood. The disease is characterised by sickle shaped red blood cells, which are formed due to the mutant haemoglobin molecule. The disease is controlled by Hb^A and Hb^S allele. The homozygous individuals with genotype, Hb^SHb^S, show the symptoms of this disease while the heterozygous individuals with genotype, Hb^A Hb^S, are not affected. However, they act as carriers of the disease.

Symptoms: Rapid heart rate, breathlessness, delayed growth and puberty, jaundice, weakness, fever, excessive thirst, chest pain, and decreased fertility are the major symptoms of sickle cell anaemia disease.

(b) Down’s Syndrome: It is an autosomal disorder that is caused by the trisomy of chromosome 21.
Symptoms : The individual is short statured with round head, open mouth, protruding tongue, short neck, slanting eyes, and broad short hands. The individual also shows retarded mental and physical growth.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 12 Mineral Nutrition Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 12 Mineral Nutrition

PSEB 11th Class Biology Guide Mineral Nutrition Textbook Questions and Answers

Question 1.
‘All elements that are present in a plant need not be essential to its survival’. Comment.
Answer:
All elements that are present in a plant need not be essential to its survival because they do not directly involved in the composition of their body. However, if the concentration of micronutrients such as Fe, Mn, Cu, Zn, Cl, etc., rise above their critical values, they appear to be toxic for the plant.

Question 2.
Why is purification of water and nutrient salts so important in studies involving mineral nutrition using hydroponics?
Answer:
It is to know the essentiality of a mineral element in the life cycle of a plant. Further, it helps in improving the deficiency symptoms of the plants. The nutrient solution must be adequetly aerated to obtain the optimal growth.

Question 3.
Explain with examples : macronutrients, micronutrients, beneficial nutrients, toxic elements and essential elements.
Answer:
(i) Macronutrients: These are generally present in plant tissues in large amount (in excess 10 m mole kg’1 of dry matter). The macronutrients include carbon, hydrogen, oxygen, nitrogen, phosphorus, sulphur, potassium, calcium and magnesium.

(ii) Micronutrients: Micronutrients or trace elements, are needed in very small amount (less than 10m mole kg~: of dry matter). These include iron, manganese, copper, molybdenum, zinc, boron, chlorine and nickel.

(iii) Beneficial Nutrients: The elements which are not essential for plants, but their presence are beneficial for the growth and development. Such, elements are called beneficial elements.

(iv) Toxic Elements: Any mineral ion concentration in tissues, that reduces the dry weight of tissues by about 10 % is considered toxic. For example, Mn inhibit the absorption of other elements.

(v) Essential Elements: The macronutrients including carbon, hydrogen, oxygen, nitrogen, phosphorus, sulphur, potassium, calcium and magnesium, which are require directly for the growth and metabolism of the plants and whose deficiency produces certain symptoms in the plants are known as essential elements.

Question 4.
Name at least five different deficiency symptoms in plants. Describe them and correlate them with the concerned mineral deficiency.
Answer:
The kind of deficiency symptoms shown in plants include chlorosis, necrosis, stunted plant growth, premature fall of leaves and buds, and inhibition of cell division.

• Chlorosis is the loss of chlorophyll leading to yellowing in leaves. This symptom is caused by the deficiency of elements N, K, Mg, S, Fe, Mn, Zn and Mo.
• Necrosis or death of tissue, particularly leaf tissue, is due to the deficiency of Ca, Mg, Cu, K.
• Lack or low level of N, K, S, Mo causes an inhibition of cell division.
• Some elements like N, S, Mo delay flowering if their concentration in plants is low.

Question 5.
If a plant shows a symptom which could develop due to deficiency of more than one nutrient, how would you find out experimentally, the real deficient mineral element?
Answer:
Every element shows certain characteristic deficiency symptoms in the plants. The deficiency of any one element cannot be met by supplying some other element. So, by absorbing the type of deficiency symptom, we can determine the real deficient mineral element.

Question 6.
Why is that in certain plants deficiency symptoms appear first in younger parts of the plant, while in others they do so in mature organs?
Answer:
For elements that are actively mobilised within the plants and exported to young developing tissues, the deficiency symptoms tend to appear first in the older tissues. For example, the deficiency symptoms of nitrogen, potassium and magnesium are visible first in the senescent leaves. In the older leaves, biomolecules containing these elements are broken down, making these elements available for mobilising to younger leaves.

The deficiency symptoms tend to appear first in the young tissues, whenever the elements are relatively immobile and are not transported out of the mature organs. For example, elements like sulphur and calcium are a part of the structural component of the cell and hence are not easily released.

Question 7.
How are the minerals absorbed by the plants?
Answer:
Mechanism of Absorption of Minerals: The process of absorption can occur into following two main phases :
(i) In the first phase, an initial rapid uptake of ions into the ‘free space’ or ‘outer space’ of cells the apoplast is passive.

(ii) In the second phase of uptake, the ions are taken in slowly into the ‘inner space’ the symplast of the cells. The passive movement of ions into the apoplast usually occurs through ion-channels, the trans-membrane proteins that function as selective pores. On the other hand, the entry or exit of ions to and from the symplast requires the expenditure of metabolic energies. The movement of ions is usually called the inward movement into the cells is influx and the outward movement, efflux.

Question 8.
What are the conditions necessary for fixation of atmospheric nitrogen by Rhizobium ? What is their role in Nitrogen-fixation?
Answer:
The first essential condition for nitrogen fixation is legume-bacteria relationship. Rhizobium bacteria cause nodule formation for this association. The enzyme nitrogenase is highly sensitive to the molecular oxygen. The nodules protect these enzymes by an oxygen scavenger called leghaerrloglobin.
Rhizobium bacteria are free living in soil. They are symbionts, which can fix atmospheric nitrogen for plants.

Question 9.
What are the steps involved in formation of a root nodule?
Answer:
Steps in Nodule Formation: Nodule formation involves a sequence of multiple interactions between Rhizobium and roots of the host plant. Principal stages in the nodule formation are given below:

• Rhizobia multiply and colonise the surroundings of roots and get attached to epidermal and root hair cells.
• The root-hairs curl and the bacteria invade the root-hair.
• An infection thread is produced carrying the bacteria into the cortex of the root, where they initiate the nodule formation in the cortex of the root. Then the bacteria are released from the thread into the cells which leads to the differentiation of specialised nitrogen fixing cells.
• The nodule thus formed, establishes a direct vascular connection with the host for exchange of nutrients.

Question 10.
Which of the following statements are true? If false, correct them:
(a) Boron deficiency leads to stout axis.
(b) Every mineral element that is present in a cell is needed by the cell.
(c) Nitrogen as a nutrient element, is highly immobile in the plants.
(d) It is very easy to establish the essentiality of micronutrients because they are required only in trace quantities.
Answer:
(a) True
(b) All the mineral elements present in a cell are not needed by the cell. For example, plants growing near radioactive mining sites tend to accumulate large amounts of radioactive compounds. These compounds are not essential for the plants.

(c) Nitrogen as a nutrient element is highly mobile in plants. It can be mobilised from the old and mature parts of a plant to its younger parts.
(d) True