Plus One Economics Notes Chapter 9 Environment Sustainable Development

Students can Download Chapter 9 Environment Sustainable Development Notes, Plus One Economics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Economics Notes Chapter 9 Environment Sustainable Development

Environment
The environment is defined as the total planetary inheritance and the totality of all resources. It includes all the biotic and abiotic factors that influence each other. While all living elements – the birds, animals and plants, forests, fisheries, etc. are biotic elements and abiotic elements include air, water, land, etc. Rocks and sunlight are all examples of abiotic elements of the environment. A study of the environment then calls for a study of the interrelationship between these biotic and abiotic components of the environment.

Functions of the Environment
The environment performs four vital functions

  • It supplies resources: resources here include both renewable and non-renewable resources.
    Renewable resources are those which can be used without the possibility of the resource becoming depleted or exhausted. That is, a continuous supply of the resource remains available.
  • It assimilates waste
  • It sustains life by providing genetic and bio-diversity
  • It also provides aesthetic services like scenery, etc.

Global Warming and Ozone Depletion
Two important issues faced by our environment is global warming and ozone depletion.

Global warming: Global warming is a gradual increase in the average temperature of the earth’s lower atmosphere as a result of the increase in greenhouse gases since the Industrial Revolution. Much of the recent observed and projected global warming is human-induced. It is caused by man-made increases in carbon dioxide and other greenhouse gases through the burning of fossil fuels and deforestation.

Ozone Depletion: Ozone depletion refers to the phenomenon of reductions in the amount of ozone in the stratosphere The problem of ozone depletion is caused by high levels of chlorine and bromine compounds in the stratosphere The origins of these compounds are chlorofluorocarbons (CFC), used as cooling substances in airconditioners and refrigerators, or as aerosol propellants, and bromo fluorocarbons (halons), used in fire extinguishers. As a result of depletion of the ozone layer, more ultraviolet (UV) radiation comes to Earth and causes damage to living organisms. UV radiation seems responsible for skin cancer in humans; it also lowers production of phytoplankton and thus affects other aquatic organisms. It can also influence the growth of terrestrial plants.

State of India’s Environment
India has abundant natural resources in terms of rich quality of soil, hundreds of rivers and tributaries, lush green forests, plenty of mineral deposits beneath the land surface, vast stretch of the Indian Ocean, ranges of mountains, etc. The black soil of the Deccan Plateau is particularly suitable for the cultivation of cotton, leading to a concentration of textile industries in this region. The Indo-Gangetic plains – spread from the Arabian Sea to the Bay of Bengal – are one of the most fertile, intensively cultivated and densely populated regions in the world.

India’s forests, though unevenly distributed, provide green cover for a majority of its population and natural cover for its wildlife. Large deposits of iron ore, coal and natural gas are found in the country. India alone accounts for nearly 20% of the world’s total iron-ore reserves. Bauxite, copper, chromate, diamonds, gold, lead, lignite, manganese, zinc, uranium, etc. are also available in different parts of the country. However, the developmental activities in India have resulted in pressure on its finite natural resources, besides creating impacts on human health and well-being.

The threat to India’s environment poses a dichotomy -threat of poverty-induced environmental degradation and, at the same time, threat of pollution from affluence and a rapidly growing industrial sector. Air pollution, water contamination, soil erosion, deforestation and wildlife extinction are some of the most pressing environmental concerns of India.
The priority issues identified are :

  • land degradation
  • biodiversity loss
  • air pollution with special reference to vehicular pollution in urban cities
  • management of fresh water and
  • Solid waste management. Land in India suffers from varying degrees and types of degradation stemming mainly from unstable use and inappropriate management practices.

Sustainable Development
The concept of sustainable development was emphasized by the United Nations Conference on Environment and Development (UNCED), which defined it as: ‘Development that meets the need of the present generation without compromising the ability of the future generation to meet their own needs’.

Strategies for sustainable development
Strategies for sustainable development include the following.

  • Use of Non-conventional Sources of Energy: LPG, Gobar Gas in Rural Areas:
  • CNG in Urban Areas
  • Wind Power
  • Solar Power through Photovoltaic Cells
  • Mini-hydel Plants
  • Traditional Knowledge and Practices
  • Bio-composting
  • Bio-pest Control

Plus One Economics Notes Chapter 8 Infrastructure

Students can Download Chapter 8 Infrastructure Notes, Plus One Economics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Economics Notes Chapter 8 Infrastructure

Infrastructure
Infrastructure means some kinds of permanent installation, which are used over a long period of time for the supply of basic inputs like railway lines, schools, colleges, universities, hospitals, etc. Infrastructural facilities are often referred to as economic and social overheads.

  1. The economic infrastructure consists of energy, transport, communication
  2. The social infrastructure consists of education, health, and housing.

Relevance of Infrastructure
Infrastructure plays important role in economic growth and development. Developed nations have good record of social and economic infrastructure. The contributions of infrastructure are:
• It invites investment which leads to growth.
• It enhances productivity.
• It improves the quality of life of people.

State of Infrastructure in India
Two important infrastructures in India are energy and health. We shall examine their details below:

Energy: Energy is very vital for rapid economic growth. There is a big gap between consumer demand and the supply of electricity in India. Energy is a critical aspect of the development process of a nation. It is, of course, essential for industries. Now it is used on a large scale in agriculture and related areas like the production and transportation of fertilizers, pesticides, and farm equipment. It is required in houses for cooking, household lighting, and heating.

Sources of Energy: There are commercial and non-commercial sources of energy. Commercial sources are coal, petroleum, and electricity as they are bought and sold. Non-commercial sources of energy are firewood, agricultural waste, and dried dung. These are non-commercial as they are found in nature/ forests. While commercial sources of energy are generally exhaustible, non-commercial sources are generally renewable.

Non-conventional Sources of Energy: Both commercial and non-commercial sources of energy are known as conventional sources of energy. There are three other sources of energy which are commonly termed as non-conventional sources – solar energy, wind energy and tidal power.

Power/Electricity: The most visible form of energy, which is often identified with progress in modern civilization, is power, commonly called electricity; it is one of the most critical components of infrastructure that determines the economic development of a country.

Health: Health is an essential element of human resource development. Health is a holistic process related to the overall growth and development of the nation. WHO defines health as Economists judge the health conditions of the people of a country by looking at the following indicators.

  • Infant mortality
  • Maternal mortality
  • Life expectancy
  • Nutritional levels
  • Incidence of communicable and non-communicable diseases
  • Health infrastructures

Health System in India: India’s health infrastructure consists of a three-tier system such as primary, secondary, and tertiary. Primary health care includes health education, health problems, prevention and control of diseases, promotion of nutrition, issues relating to potable water and sanitation, maternal and child health care, immunization against major infectious diseases like Polio, T.B, diphtheria, promotion of mental health and provision of essential drugs, etc.

Primary health care is provided through, sub-centers catering to a population of about 5000, Primary health care centres (PHCs) at block level and community health centres (CHCs) at the district level. The primary health care centres have only limited facilities. When the patient need advanced health care they are referred to secondary or tertiary hospitals.

Secondary care institutions are those which have facilities for clinical investigations like X-ray, clinical laboratory, scanning, etc., specialist doctors like a surgeon, gynecologists, pediatricians, etc. It is mostly available in district headquarters and big towns and cities.

Tertiary health care institutions these are health care institutions at the top of the three-tier system, devoted in health care, health education and research: Medical colleges, super-specialty hospitals and multi-specialty hospitals. All India Institute of Medical Sciences (AIIMS) Delhi, National Institute of Mental Health and Neuro Sciences (NIMHANS) Bangalore, Sree Chithra Institute of Medical Science (SCIM) Trivandrum, etc., are tertiary health care institutions.

Indian Systems of Medicine (ISM): Natural systems of medicine have to be explored and used to support public health. There is a great scope of advancement of medical tourism in India. It includes six systems: Ayurveda, Yoga, Unani, Siddha, Naturopathy and Homeopathy (AYUSH). At present, there are 3,004 ISM hospitals, 23,028 dispensaries and as many as 6,11,431 registered practitioners in India. But little has been done to set up a framework to standardize education orto promote research. ISM has huge potential and can solve a large part of our health care problems because they are effective, safe and inexpensive.

Plus One Economics Notes Chapter 7 Employment-Growth, Informalisation and Related Issues

Students can Download Chapter 7 Employment-Growth, Informalisation and Related Issues Notes, Plus One Economics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Economics Notes Chapter 7 Employment-Growth, Informalisation and Related Issues

Workers and Employment
Those activities that contribute to Gross National Product (GNP) or national income are known as economic activities. When farmers work in a field or a labour work in a factory, or doctor works in a hospital they produce goods or services. All those who are engaged in economic activities including self-employed are called workers. The employment situation is diverse and complex in India. It is due to the developing nature of the economy and the socio-economic and demographic factors that influence it.

Participation of People in Employment
The worker-population ratio is an important indicator of the employment situation in an economy.
Worker population-ratio refers to the ratio of workers to the population. It is computed by dividing the number of workers(W) by the total population (P) and express it in terms of percentage (W/P)*100. This ratio is useful in knowing the proportion of the population actively contributing to the production of goods and services in a country. The worker-population ratio in India for 2011 was 39.3%.

Employment in Firms, Factories, and Offices
In the course of the economic development of a country, labour flows from agriculture and other related activities to industry and services. Generally, we divide all economic activities into eight different industrial divisions. They are:

  1. Agriculture
  2. Mining and quarrying
  3. Manufacturing
  4. Electricity, gas and water supply
  5. Construction
  6. Trade
  7. Transport and storage
  8. Services.

For simplicity, all the working persons engaged in these divisions can be clubbed into three major sectors, viz.

  • Primary sector includes (1) and (2)
  • Secondary sector which includes (3), (4), and (5)
  • Service sector which includes divisions (6), (7), and (8).

Unemployment and Types of Unemployment
The unemployment situation in India is highly complex. There are different types of unemployment in our country like open unemployment, disguised unemployment, seasonal unemployment, etc. Unemployment is a situation in which people are willing to work at the prevailing wage rate, but do not get any work.

Open Unemployment: Open unemployment is the situation in which people above a certain age who are able to work and willing to work at the prevailing wage remain unemployed. Open unemployment is involuntary in nature. They are willing to work, but employment opportunities are not available to them. People standing in some selected areas waiting to be recruited as the hired worker is a case of open unemployment.

Disguised Unemployment: When more persons are working in a job than actually required, the situation is termed as disguised unemployment or hidden unemployment. If some workers are withdrawn from work, either total production or productivity falls. This type of unemployment is prominent in Indian agriculture.

Seasonal Unemployment: The type of unemployment caused by a change in seasons is termed as seasonal unemployment. This is normally found in the agricultural sector of India. Agriculture normally provides only seasonal employment and people are employed during the busy sowing and harvesting seasons. Seasonal unemployment could also be found in agro-based industries such as sugar mills, rice mills, cotton-spinning mills, etc.

Plus One Economics Notes Chapter 6 Rural Development

Students can Download Chapter 6 Rural Development Notes, Plus One Economics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Economics Notes Chapter 6 Rural Development

Rural Development
Rural development is quite a comprehensive term but it essentially means a plan of action for the development of areas which are lagging behind in socio-economic development. It essentially focuses on the action for the development of areas that are lagging behind in the overall development of the village economy. Some of the areas which are challenging and need fresh initiatives for development in India include:

  • Development of human resources including- literacy, more specifically, female literacy, education, and skill development-health, addressing both sanitation and public health
  • Land reforms
  • Development of the productive resources of each locality
  • Infrastructure development like electricity, irrigation, credit, marketing, transport facilities including construction of village roads and feeder roads to nearby highways, facilities for agriculture research and extension, and information dissemination
  • Special measures for alleviation of poverty and bringing about significant improvement in the living conditions of the weaker sections of the population emphasizing access to productive employment opportunities.

Credit and Marketing in Rural Areas
Credit: Growth of rural economy depends primarily on the infusion of capital, from time to time, to realize higher productivity in agriculture and non-agriculture sectors. As the time gestation between crop sowing and realization of income after production is quite long, farmers borrow from various sources to meet their initial investment on seeds, fertilizers, implements and other family expenses of marriage, death, religious ceremonies.

A major change occurred after 1969 when India adopted social banking and a multi-agency approach to adequately meet the needs of rural credit. Later, the National Bank for Agriculture and Rural Development (NABARD) was set up in1982 as an apex body to coordinate the activities of all institutions involved in the rural financing system.

The institutional structure of rural banking today consists of a set of multi-agency institutions, namely, commercial banks, regional rural banks (RRBs), co-operatives and land development banks. Recently, Self-Help Groups (henceforth SHGs) have emerged to fill the gap in the formal credit system because the formal credit delivery mechanism has not only proven inadequate but has also not been fully integrated into the overall rural social and community development. By March end 2003, more than seven lakh SHGs had reportedly been credit linked. Such credit provisions are generally referred to as micro-credit programmes.

Agricultural Market System
Agricultural marketing is a process that involves the assembling, storage, processing, transportation, packaging, grading and distribution of different agricultural commodities across the country. Let us discuss four such measures that were initiated to improve the marketing aspect.

1. The first step was regulation of markets to create orderly and transparent marketing conditions.

2. Second component is provision of physical infrastructure facilities like roads, railways, warehouses, godowns, cold storages and processing units.

3. Co-operative marketing, in realizing fair prices for farmers’ products, is the third aspect of a government initiative.

4. The fourth element is the policy instruments like

  • assurance of minimum support prices (MSP) for 24 agricultural products
  • maintenance of buffer stocks of wheat and rice by Food Corporation of India and
  • distribution of food grains and sugar through PDS.

These instruments are aimed at protecting the income of the farmers and providing food grains at subsidized rate to the poor.

Diversification into Productive Activities
Diversification of farm products has two aspects: The first one relates to the diversification of crop production. The second one relates to the shift of the workforce from agriculture to other allied activities such as livestock, poultry, fishers, etc., and to non-farm sectors like food processing. Diversification of agriculture helps to provide alternative employment opportunities in the non-farm sector and will minimize the risk of depending exclusively on agriculture. These activities related to diversification are given below:

  • Animal husbandry
  • Fisheries
  • Horticulture

Sustainable Development and Organic Farming
In recent years, awareness of the harmful effect of chemical-based fertilizers and pesticides on our health is on a rise. Conventional agriculture relies heavily on chemical fertilizers and toxic pesticides, etc., which enter the food supply, penetrate the water sources, harm the livestock, deplete the soil and devastate natural eco-systems. Efforts in evolving technologies which are eco-friendly are essential for sustainable development and one such technology which is eco-friendly is organic farming.

Benefits of Organic Farming:
1. Organic agriculture offers a means to substitute costlier agricultural inputs (such as HYV seeds, chemical fertilizers, pesticides, etc.) with locally produced organic inputs that are cheaper and thereby generate good returns on investment.

2. Organic agriculture also generates income through international exports as the demand for organically grown crops is on a rise.

3. Studies across countries have shown that organically grown food has more nutritional value than chemical farming thus providing us with healthy foods.

4. Since organic farming requires more labour input than conventional farming, India will find organic farming an attractive proposition.

5. Finally, the produce is pesticide-free and produced in an environmentally sustainable way.

Plus One Economics Notes Chapter 4 Poverty

Students can Download Chapter 4 Poverty Notes, Plus One Economics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Economics Notes Chapter 4 Poverty

Poverty and Poverty Line
Poverty is a multi-dimensional concept. It is defined as a situation in which a section of society is unable to fulfill even the basic necessities of life or deprived of basic necessities of life. Poverty is measured on the basis of the poverty line. One way is to determine it by the monetary value (per capita expenditure) of the minimum calorie intake that was estimated at 2,400 calories for a rural person and 2,100 for a person in the urban area.

Causes of Poverty
Important causes of poverty in India are:

  • Low income
  • Lack of assets
  • Unemployment
  • Inequality
  • Exploitation
  • Population explosion
  • Undesirable economic growth
  • Inflation
  • Absence of industrialisation.

Poverty Eradication Programmes
Poverty eradication programmes in India are classified as follows:

Self-employment Programmes
Integrated Rural Development Programme (IRDP): The IRDP was introduced in 2nd October 1980. This programme has been renamed as
Swarnajayanthi Grama Swarojgar Yojana (SGSY) from 1st April 1999. The SGSY is a very holistic programme compared to IRDP. SGSY is developed by merging various programmes such as TRYSEM (Training of Rural Youth for Self-employment), DWCRA (Development of Women and Children in Rural Areas), GKY (Ganga Kalyan Yojana), MSW (Million Well Scheme), and SITRA (Supply of Improved Tool Kits to Rural Artisans). It forms SHG’s (Self-Help Groups) of poor people and formulates self-employment programmes under their leadership. It includes development of infrastructure, technology, credit and marketing managements, etc. Unlike other programmes, the priority of the employment programme can be fixed by the beneficiaries.

Wage Employment Programmes
1. National Rural Emptoyment Programme (NREP): NREP was the new name given to Food for Work Programme. It was launched in 1980 as a centrally sponsored scheme with state and central governments sharing equal amounts. The aim of this programmer the development of community assets like drinking water wells, irrigation wells, rural roads, schools, etc.

2. The Rural Landless Employment Guarantee Programme (RLEGP): This programme was launched on 15th August 1983 to supplement NREP. This is a centrally sponsored scheme with 100 percent fund by the union government.

3. Jawahar Rozgar Yojana (JRY) and Nehru Rozgar Yojana (NRY): NREP and the RLEGP were merged and renamed into a single rural employment programme known as Jawahar Rozgar Yojana. JRY came into effect from 1st April 1999. The aim of the programme was to provide gainful employment for unemployed rural areas. The urban version of JRY is known as Nehru Rozgar Yojana (NRY).

4. The Million Well Scheme (MWS) was to provide open irrigation well, free of cost, to poor small and marginal farmers belonging to SC/ST category.

5. Indira Awas Yojana (IAY) was introduced for providing houses, free of cost to SC/ST families. Now this facility is extended to other poor families too.

6. Pradan Manthri Gramodaya Yojana: Gramin Awas (PMRY)
The PMRY: GAwas launched on 1st April, 2000. The programme aims at providing the housing needs of the rural people.

7. National Rural Employment Guarantee Programme (NREGP) 2005: In August 2005, the Parliament has passed a new Act known as National Rural Employment Guarantee Act 2005. The act provides guaranteed wage employment to every household whose adult volunteer is to do unskilled manual work for a minimum of 100 days in a year. Thj act came into force from 2nd February 2006 and implemented in India’s 200 most backward districts. Later on it was extended to all over the country in two phases. The programme was later on renamed as Mahatma Gandhi National Rural Employment Guarantee Programme (MGNREGP) or commonly called Employment Assurance Scheme.

Social Security Programmes: There are not many programmes for the social security of the poor. However, there are some centrally sponsored schemes. They are as follows:

  • Old-age pension for the elderly who are without support.
  • Financial support in the event of the death of the breadwinner.
  • Support for women of poor households on pregnancy.

Food Security Programmes: It is essential to ensure food security to the masses. As we know, though there js sufficient production of food grains, millions of people are starving in the country. The problem is mainly of distribution. To overcome these several measures are taken by the government. They are as follows:

1. Public distribution system (PDS): Foodgrains are made available at cheaper prices and distributed through Fair Price Shops, ration shops, Maveli Stores, Neethi Stores etc.

2. Targeted Public Distribution System (TPDS): The TPDS initiated in June 1997 aims at ensuring the availability of food grains to BPL families.

3. Integrated Child Development Schemes (ICDS): A nutrition programme meant for children below 6 years of age, pregnant and lactating women.

4. Mid-day Meal at School: Mid-day Meal at School is in operation in several states. The programme was launched in all India level on 15th August 1995.

5. Annapurna Scheme: This programme was commenced from 2000-01. Poor old people who are not getting old-age pensions are covered under this scheme.

6. Antyodaya Anna Yojana: This scheme is launched for the poorest of the poor. Under this scheme, food grains are made available to very poor families at a highly subsidised price.

Plus Two Physics Previous Year Question Paper Say 2018

Kerala State Board New Syllabus Plus Two Physics Previous Year Question Papers and Answers.

Kerala Plus Two Physics Previous Year Question Paper Say 2018 with Answers

BoardSCERT
ClassPlus Two
SubjectPhysics
CategoryPlus Two Previous Year Question Papers

Time: 2½ Hours
Cool off time : 15 Minutes

General Instructions to candidates

  • There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2½ hrs.
  • Your are not allowed to write your answers nor to discuss anything with others during the ‘cool off time’.
  • Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
  • Read questions carefully before you answering.
  • All questions are compulsory and only internal choice is allowed.
  • When you select a question, all the sub-questions must be answered from the same question itself.
  • Calculations, figures and graphs should be shown in the answer sheet itself.
  • Malayalam version of the questions is also provided.
  • Give equations wherever necessary.
  • Electronic devices except non programmable calculators are not allowed in the Examination Hall.

You may use following Physical constants wherever necessary.

Mass of proton 1.66 × 10-27 kg
Mass of Electron 9.11 × 10-31 kg
Elementary charge, e = 1.6 × 10-19 C
Velocity of light in vacuum c = 3 × 10m/s
Permittivity of free space ε0 = 8.85 × 10-12 F/m

Question Nos. 1 to 7 carry 1 score each. Answer 6 questions. (6 × 1 = 6)

Question 1.
Write the physical quantities having the following SI unit.
i) cm
ii) Ωm
Answer:
i) Electric dipole moment
ii) Electrical resistivity

Question 2.
A uniform wire of resistance 40Ω is cut into four equal parts and they are connected in parallel. The effective resistance of the combination is
i) 40Ω
ii) 10Ω
iii) 2.5Ω
iv) 4Ω
Answer:
Plus Two Physics Previous Year Question Paper Say 2018, 1

Question 3.
A particle of charge q is moving through a uniform magnetic field of intensity B with a velocity v. Write an expression for the force acting on the particle in vector form.
Answer:
\(\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})\)

Question 4.
The minimum distance between the object and its real image for a concave mirror of focal length f is
i) f
ii) 2f
iii) 4f
iv) zero
Answer:
iv) zero

Question 5.
Variation of photocurrent with collector plate potential for different intensities I1, I2, and I3 of incident radiation is shown below.
Plus Two Physics Previous Year Question Paper Say 2018, 2
Arrange these intensities in the accending order.
Answer:
I3 > I2 > I1

Question 6.
An electron is revolving around the nucleus of a hydrogen atom in an orbit of radius nine times the radius of the first orbit. Angular momentum of the electron in this orbit is.
i) \(\frac{h}{2 \pi}\)
ii) \(\frac{9 h}{2 \pi}\)
iii) \(\frac{3 h}{\pi}\)
iv) \(\frac{3 h}{2 \pi}\)
Answer:
radius of the nucleus r = n2r0
In this case
∴ n2 = 9, n= 3
Angular momentum L = \(\frac{\mathrm{nh}}{2 \pi}\)
for third orbit L = \(\frac{3 h}{2 \pi}\)

Question 7.
The voltage-current characteristics of an optoelectronic junction device is shown below.
Plus Two Physics Previous Year Question Paper Say 2018, 3
Identify the device.
Answer:
Solar cell

Question Nos. 8 – 15 carry 2 scores each. Answer any 7 questions. (7 × 2 = 14)

Question 8.
What is the radius of the circular path of an electron moving at a speed of 3 × 107 m/s in a magnetic field of 6 × 10-4 T perpendicular to it?
Answer:
radius r = \(\frac{m v}{q B}\)
v = 3 × 107 m/s, m = 9.1 × 10-31kg
B = 6 × 10-4, q = 1.6 × 10-19
r = \(\frac{9.1 \times 10^{-37} \times 3 \times 10^{7}}{1.6 \times 10^{-19} \times 6 \times 10^{-4}}\) = 0.285 m

Question 9.
a) “Parallel currents attract, and antiparallel currents repel”. State whether this statement is true or false.
b) Define the SI unit of current in terms of force between two current-carrying conductors.
Answer:
a) True
b) Ampere is defined as that current which is maintained in two straight parallel conductors of infinite length placed one meter apart in the vacuum will produce between a force of 2 × 10-7 Newton per meter length.

Question 10.
A circular metallic loop and a current-carrying conductor are placed as shown in the figure.
Plus Two Physics Previous Year Question Paper Say 2018, 4
If the current through the conductor is increasing from A to B, find the direction of the induced current the loop.
Answer:
Current flows in a clockwise direction

Question 11.
a) The electric field vector of an electromagnetic wave is represented as Ex = Em Sin(kz – ωt). Write the equation for the magnetic field vector.
b) The ratio of the intensity of the magnetic field to intensity of the electric field has the dimensions of
i) velocity
ii) acceleration
iii) reciprocal of velocity
iv) reciprocal of acceleration
Answer:
a) By = Bm Sin (Kz – ωt)
b) reciprocal of velocity

Question 12.
Write the equations for the following nuclear reactions:
a) β+decay of \({ }_{6}^{11} c\) to Boron (B)
b) β decay of \({ }_{15}^{32} c\) to Sulphur (S)
Answer:
Plus Two Physics Previous Year Question Paper Say 2018, 5

Question 13.
Write the truth table of the circuit shown below.
Plus Two Physics Previous Year Question Paper Say 2018, 6
Answer:

ABY
000
011
101
111

Question 14.
A TV transmitting antenna is 100 m tall. How much service area can it cover if the receiving antenna is at the ground level? Radius of earth is 6400 km.
Answer:
Hight of antenna, h = 100 m = 0.1 km
Radius of earth = 6400 km
Transmission range d = \(\sqrt{2 R h}\)
∴ Maximum area covered A = πd2
= π(\(\sqrt{2 R h}\))2
= 3.14 × 2 × 6400 × 0.1
= 4019 km2

Question 15.
Two magnetic dipoles P and Q are placed in a uniform magnetic field \(\vec{B}\) gas shown.
Plus Two Physics Previous Year Question Paper Say 2018, 7
a) Both the dipoles do not experience any torque. Why?
b) Identify the dipole which is in most stable equilibrium.
Answer:
a) Dipole moment is parallel to magnetic field.
b) Dipole Q

Question Nos. 16 to 22 carry 3 scores each. Answer any 6 questions. (6 x 3 = 18)

Question 16.
The experimental setup for the comparison of two resistances is shown below.
Plus Two Physics Previous Year Question Paper Say 2018, 8
a) The working principle of the above device is
i) Ohm’s law
ii) Kirchhoffs second law
iii) Wheatstone’s principle
iv) Point Rule
b) In figure, Let X is the effective resistance of series combination of two 3Ω resistors and R is the effective of a parallel combination of two 3Ω resistors. The balance point is obtained at C. If the length AB is 100 cm. find the length AC of the wire.
Answer:
a) Wheatston’s principle
b) x = 6 Ω
R = 1.5 Ω
Plus Two Physics Previous Year Question Paper Say 2018, 9

Question 17.
A magnetic needle has a magnetic moment 6.7 × 10-7 Am2 and moment of inertia 7.5 × 10-6 kgm2. In a uniform magnetic field, it performs 10 complete oscillations in 6.70s. What is the magnitude of the magnetic field?
Answer:
magnetic moment m = 6.7 × 10-2 Am2
moment of inertria, I = 7.5 × 10-6 kgm2
Plus Two Physics Previous Year Question Paper Say 2018, 10

Question 18.
With the help of a ray diagram, show the formation of image of a point object by refraction of light at a spherical surface separating two media of refractive indices n1 and n2. Using the diagram derive the relation
\(\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{n_{2}-n_{1}}{R}\)
Answer:
Plus Two Physics Previous Year Question Paper Say 2018, 11
Consider a convex surface XY, which separates two media having refractive indices n1 and n2. Let C be the centre of curvature and P be the pole. Let an object is placed at ‘O’, at a distance ‘u’ from the pole. I is the real image of the object at a distance ‘V’ from the surface. OA is the incident ray at angle ‘i’ and AI is the refracted ray at an angle ‘r’. OP is the ray incident normally. So it passes without any deviation. From snell’s law,
sin i/sin r = \(\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}\)
If ‘i’ and ‘r’ are small, then sin i » i and sin r » r.
Plus Two Physics Previous Year Question Paper Say 2018, 12
From the Δ OAC, exterior angle = sum of the interior opposite angles
i.e., i = α + θ ………(2)
Similarly, from Δ IAC,
α = r + β
r = α – β ………..(3)
Substituting the values of eq(2) and eq(3)in eqn.(1)
we get,
n1(α + θ) = n2(α – β)
n1α + n1θ = n2α – n2β
n1θ + n2β = n2α – n1α
n1θ + n2β = (n2 – n1)α ………..(4)
From OAP, we can write,
Plus Two Physics Previous Year Question Paper Say 2018, 13
According to New Cartesian sign convection, we can write,
OP = -u, PI = +v and PC = R
Substituting these values, we get
Plus Two Physics Previous Year Question Paper Say 2018, 14

Question 19.
a) Draw a graph showing the variation of stopping potential with the frequency of incident radiation. Mark the threshold frequency in the figure.
b) Using Einstein’s photoelectric equation, show that photoelectric emission is not possible if the frequency of incident radiation is less than the threshold frequency.
Answer:
a)
Plus Two Physics Previous Year Question Paper Say 2018, 15
b) According to Einsten’s Equation
hν = hν0 + KE
if = ν < ν0, kinetic energy of electron becomes negative.

Question 20.
a) The energy levels of an atom are as shown in the figure.
Plus Two Physics Previous Year Question Paper Say 2018, 16
Which transition corresponds to emission of radiation of maximum wavelength?
b) Sketch the energy level diagram for hydrogen atom and mark the transitions corresponding to Balmer series.
Answer:
a) In between 0 to -2ev
b)
Plus Two Physics Previous Year Question Paper Say 2018, 17

Question 21.
The decay rate of a radioactive ample is called its activity.
a) What is the SI unit of activity?
b) The half life of \({ }_{92}^{238} U\) against alpha decay is 1.5 × 10-17 s. Calculate the activity of a sample of \({ }_{92}^{238} U\) having 25 × 1020 atoms.
Answer:
a) becquerel
b) N = 25 × 1020, half life T1/2 = 1.5 × 1017s
activity, R = λN
R = \(\frac{0.693}{\mathrm{~T}_{1 / 2}} \times \mathrm{N}=\frac{0.693}{1.5 \times 10^{17}} \times 25 \times 10^{20}\)
R = 11550 Bq

Question 22.
a) Draw the circuit diagram where a zener diode is used as a direct voltage regulator.
b) In a zener regulated power supply a zener diode with zener voltage 4V is used for regulation. The load current is to be 4 m A and zener current is 20 mA. If the unregulated input is 10V, What should be the value of resistor that is to be connected in series with the diode?
Answer:
a)
Plus Two Physics Previous Year Question Paper Say 2018, 18
b) Zener voltage
V2 = 4V, IL = 4 × 10-3 A, I2= 20 × 10-3 A
Unregulated input = 10 v
Voltage across resister = 10 – 4 = 6v
Current through the resister = IL+ I2
= (4 + 20) mA
= 24 × 10-3 A
∴ resistance R = \(\frac{V}{I}\)
R = \(\frac{6}{24 \times 10^{-3}}\) = 250 Ω

Question Nos. 23 to 26 carry 4 scores each. Answer any 3 questions. (3 × 4 = 12)

Question 23.
Coulomb’s Law is a quantitative statement about the force between two point charges.
a) Write the mathematical expression of the above law.
b) Two ions carrying equal charges repel with a force of 1.48 × 10-8 N when they are separated by a distance of 5 × 10-10 m. How many electrons have been removed from each iron?
Answer:
Plus Two Physics Previous Year Question Paper Say 2018, 19

Question 24.
a) Potential energy of a system of charges is directly proportional to the product of charges and inversely to the distance between them.
a) Prove the above statement.
b) Two-point charges 3 × 10-8 C and -2 × 10-8 C are separated by a distance of 15cm. At what point on the line joining the charges the potential is zero?
Answer:
a) Potential of a system is the work done to assemble the system,
∴ w = qV
= \(\mathrm{q}_{1} \times \frac{1}{4 \pi q_{0}} \frac{\mathrm{q}_{2}}{\mathrm{r}}\)
In the above equation it is dear that energy is di-rectly proportial to charges and inversely proportional to the distance between them.
b) q1 = 3 × 10-8
Plus Two Physics Previous Year Question Paper Say 2018, 20

Question 25.
The circuit shown can be analysed using Kirchhoff’s rules.
Plus Two Physics Previous Year Question Paper Say 2018, 21
a) Apply Kirchhoffs first law to the point B.
b) State Kirchhoffs second law.
c) Apply Kirchhoffs second lawtothe mesh ABFGA.
Answer:
a) At the point B,
I1 = I2 = I3
b) In any closed circuit emf is eqaual to sum of potential drops.
c) -I2 R3 – I1 R2 – I1R1 + E1 = 0

Question 26.
Biot – Savart’s law relates current with the magnetic field produced by it.
a) Write the mathematical expression of the above law in vector form.
b) Using the law derive an expression for intensity of magnetic field at a point at distance x from the centre and on the axis of a circular current loop.
Answer:
Plus Two Physics Previous Year Question Paper Say 2018, 22
Consider a circular loop of radius ‘a’ and carrying current “I’. Let P be a point on the axis of the coil, at distance x from A and r from ‘O’. Consider a small length dl at A.
The magnetic field at ‘p’ due to this small element dI,
Plus Two Physics Previous Year Question Paper Say 2018, 23
The dB can be resolved into dB cos Φ (along with Py) and dBsinΦ (along with Px).
Similarly consider a small element at B, which produces a magnetic field ‘dB’ at P. If we resolve this magnetic field we get.
dB sinΦ (along px) and dB cosΦ (along py1)
dBcosΦ components cancel each other because they are in opposite direction. So only dB sinΦ components are found at P, so the total filed at P is
Plus Two Physics Previous Year Question Paper Say 2018, 24
Let there be N turns in the loop then,
B = \(\frac{\mu_{0} \text { NIa }^{2}}{2\left(\mathrm{r}^{2}+\mathrm{a}^{2}\right)^{3 / 2}}\)

Question Nos. 27 to 29 carry 5 scores each. Answer any 2 questions. (2 × 5 = 10)

Question 27.
A series LCR circuit shows the phenomenon called resonance.
a) Write the condition for resonance and obtain an equation for resonant frequency.
b) Obtain the Q value of a series LCR circuit with L = 2.0 H, c = 32µ F and R = 10 Ω.
c) Complete the following table using the suitable words from the bracket for two series LCR circuits.
(Current and applied voltage are in the same phase, currently leads the applied voltage, currently lags the applied voltage)
Plus Two Physics Previous Year Question Paper Say 2018, 25
Answer:
a) Condition for resonance XL = XC
Plus Two Physics Previous Year Question Paper Say 2018, 26
c) i) Cuurrent lags the applied voltage
ii) Current and applied voltage are in same phase.

Question 28.
Telescope is used to provide angular magnification of distant objects.
a) If f0 is the focal length of the objective and fe that of the eyepiece, the length of the telescope tube is ………..
b) Draw the ray diagram of a refracting type telescope when it is in normal adjustment.
c) Write any two advantages of reflecting type telescope over refracting type telescopes.
Answer:
a) f0 + fe
b)
Plus Two Physics Previous Year Question Paper Say 2018, 27
c) i) There is no chromatic aberration
ii) There is no spherical aberration

Question 29.
A wavefront is defined as a surface of constant phase.
a) The energy of the wave travels in a direction …….. to the wavefront.
b) Explain the Huygen’s principle.
c) Using Huygen’s Principle, prove that angle of incidence is equal to angle of reflection.
Answer:
a) Perpendicular

b) 1) Every point in a wavefront acts as a source of secondary wavelets.
2) The secondary wavelets travel with the same velocity as the original value.
3) The envelope of all these secondary wavelets gives a new wavefront.

c)
Plus Two Physics Previous Year Question Paper Say 2018, 28
AB is the incident wavefront and CD is the reflected wavefront. T is the angle of incidence and Y is the angle of reflection. Let c, be the velocity of light in the medium. Let PO be the incident ray and OQ be the reflected ray.

The time taken for the ray to travel from P to Q is
Plus Two Physics Previous Year Question Paper Say 2018, 29
Plus Two Physics Previous Year Question Paper Say 2018, 30
O is an arbitrary point. Hence AO is a variable. But the time to travel for a wavefront from AB to CD is a constant. So eq.(2) should be independent of AO. i.e., the term containing AO in eq.(2) should be zero.
∴ \(\frac{A O}{C_{1}}\) (sin i – sin r) = 0
sin i – sin r = 0
sin i = sin r
i = r
This is the law of reflection.

Plus Two Physics Previous Year Question Paper March 2019

Kerala State Board New Syllabus Plus Two Physics Previous Year Question Papers and Answers.

Kerala Plus Two Physics Previous Year Question Paper March 2019 with Answers

BoardSCERT
ClassPlus Two
SubjectPhysics
CategoryPlus Two Previous Year Question Papers

Time: 2½ Hours
Cool off time: 15 Minutes

General Instructions to candidates

  • There is a ‘cool off-time’ of 15 minutes in addition to the writing time of 2½ hrs.
  • You are not allowed to write your answers nor to discuss anything with others during the ‘cool off time’.
  • Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
  • Read questions carefully before you answering.
  • All questions are compulsory and the only internal choice is allowed.
  • When you select a question, all the sub-questions must be answered from the same question itself.
  • Calculations, figures, and graphs should be shown in the answer sheet itself.
  • Malayalam version of the questions is also provided.
  • Give equations wherever necessary.
  • Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

The given value of constants can be used wherever necessary.

Charge of proton = 1.6 x 10-19C
Mass of proton = 1.67 x 10-27kg

Answer any three questions from 1 to 4. Each carries 1 score. (3 × 1 = 3)

Question 1.
A charged particle enters a uniform magnetic field at an angle of 40°. It’s path becomes …………
Answer:
Helical

Question 2.
Figure shows the symbolic representation of ……….
Plus Two Physics Previous Year Question Paper March 2019, 1
i) OR gate
ii) NAND gate
iii) NOR gate
iv) NOT gate
Answer:
iii) NOR gate

Question 3.
Write the unit of mobility.
Answer:
Plus Two Physics Previous Year Question Paper March 2019, 2

Question 4.
If ‘h’ is Planck’s constant, the momentum of a photon of wavelength 1 A° is
i) h
ii) 10-10
iii) 1010 h
iv) 10 h
Answer:
Plus Two Physics Previous Year Question Paper March 2019, 3

Answer any six questions from 5 to 11. Each carries 2 scores. (6 × 2 = 12)

Question 5.
a) The ratio of electric field on the equatorial point and at the axial point at equal distances from the centre of a short electric dipole is ………..
b) A closed surface encloses an electric dipole. What is the electric flux through the surface?
Answer:
Plus Two Physics Previous Year Question Paper March 2019, 4

Question 6.
A series LCR circuit connected to an ac source is shown below:
Plus Two Physics Previous Year Question Paper March 2019, 5
a) Write an expression for impedance offered by its circuit.
b) Under what condition this circuit is used for tuning radio?
Answer:
a) Impedence Z = \(\sqrt{R^{2}+\left(x_{L}-x_{C}\right)^{2}}\)
b) at XL = XC this circuit is used as tuner circuit.

Question 7.
Which electromagnetic waves are used for the following purposes?
i) Diagnostic tool in medicine.
ii) Kill germs in water purifiers.
iii) Cellular phones.
iv) In remote switches of household electronic systems.
Answer:
i) X-rays
ii) Uv rays
iii) Radio Waves
iv) IR rays

Question 8.
Calculate the effective capacitance between ’a’ and ‘b’ from the figure given below:
C1 = C3 = 100 µF, C2 = C4 = 200 µF
Plus Two Physics Previous Year Question Paper March 2019, 6

Question 9.
Write any two uses of polaroids.
Answer:

  1. Polaroids are used in sunglasses
  2. Polaroids are used to produce 3D motion films

Question 10.
The temperature dependence of resistivity of a material is shown below:
Plus Two Physics Previous Year Question Paper March 2019, 7
a) Identify the type of material.
b) Write the relation between resistivity and average collision time for electron.
Answer:
a) Semi conductor
b) ρ = \(\frac{\mathrm{m}}{\mathrm{ne}^{2} \tau}\)

Question 11.
What is meant by the half-life of a radioactive substance? Write its relation with a decay constant.
Answer:
Half-life is the time taken for a radioactive substance to reduce half of its initial value.
\(T_{1 / 2}=\frac{0.693}{\lambda}\)

Answer any six questions from 12 to 18. Each carries 3 scores. (6 × 3 = 18)

Question 12.
A spherical shell of radius R’ is uniformly charged with charge +q. By Gauss’stheorem, find the electric field intensity at a point ‘p’.
a) Outside the spherical shell and
b) Inside the spherical shell.
Answer:
Field Due To A Uniformly Charged Thin Spherical Shell: Consider a uniformly charged hollow spherical conductor of radius R. Let ‘q’ be the total charge on the surface.
Plus Two Physics Previous Year Question Paper March 2019, 8
To find the electric field at P (at a distance r from the centre), we imagine a Gaussian spherical surface having radius ‘r’.

Then, according to Gauss’s theorem we can write,
\(\int \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{s}}=\frac{1}{\varepsilon_{0}} \mathrm{q}\)
The electric field is constant,at a distance T. So we can write,
Plus Two Physics Previous Year Question Paper March 2019, 9
b) E = 0.

Question 13.
The equipotential surface through a point is normal to the electric field at that point.
a) What is meant by equipotential surface?
b) What is the work done to move a charge on an equipotential surface?
c) Draw the equipotential surfaces for a uniform electric field.
Answer:
a) The surface over which potential is constant is equipotential surface.
b) Workdone = pd × charge
= 0 × q = 0
c)
Plus Two Physics Previous Year Question Paper March 2019, 10

Question 14.
The elements of earth’s magnetic field at a place are declination, dip and horizontal intensity.
a) A magnetic needle free to move in horizontal plane is shown below:
Plus Two Physics Previous Year Question Paper March 2019, 11
Which element of earth’s magnetic field is represented by θ in the figure?
b) The vertical component of earth’s magnetic field at a given place is \(\sqrt{3}\) times its horizontal component. If total intensity of earth’s magnetic field at the place is 0.4 G find the value of horizontal component of earth’s magnetic field.
Answer:
a) Declination
b) tan θ = \(\frac{B_{V}}{B_{H}}=\sqrt{3}\)
∴ θ = 60°
BH = B cos θ, B = 0.4 G
BH = 0.4 × cos 60 = 0.2 G
= 0.2 × 10-4 T

Question 15.
A transformer is used to change the alternating voltage to a high or low value.
a) What is the principle of a transformer?
b) A power transmission line feeds input power of 2300 V to a stepdown transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230 V?
Answer:
a) Mutual Induction
b) Vp = 2300 v, Np = 4000 turns
Vs = 230 v
Plus Two Physics Previous Year Question Paper March 2019, 12

Question 16.
Describe Young’s double-slit experiment and derive an expression for the bandwidth of the interference band.
Answer:
Expression for bandwidth
Plus Two Physics Previous Year Question Paper March 2019, 13
S1 and S2 are two coherent sources having wave length λ. Let ‘d’ be the distance between two coherent sources. A screen is placed at a distance D from sources. ‘O’ is a point on the screen equidistant from S1 and S2.

Hence the path difference, S1O – S2O = 0
So at ‘O’ maximum brightness is obtained.
Let ‘P’ be the position of nth bright band at a distance xn from O. Draw S1A and S2B as shown in figure.
From the right angle ΔS1AP
Plus Two Physics Previous Year Question Paper March 2019, 14
But we know constructive interference takes place at P, So we can take
(S2P – S1P) = nλ
Hence eq (1) can be written as
Plus Two Physics Previous Year Question Paper March 2019, 15
Let xn + 1 be the distance of (n + 1)th bright band from centre O, then we can write
Plus Two Physics Previous Year Question Paper March 2019, 16
This is the width of the bright band. It is the same for the dark band also.

Question 17.
The schematic diagram of an experimental setup to study the wave nature of electron is shown below:
Plus Two Physics Previous Year Question Paper March 2019, 17
a) Identify the experiment.
b) Explain how this experiment verified the wave nature of electrons.
Answer:
a) Davisson and Germer Experiment
b)
Plus Two Physics Previous Year Question Paper March 2019, 18
Experimental setup: The Davisson and Germer Experiment consists of filament ‘F’, which is connected to a low tension battery. The Anode Plate (A) is used to accelerate the beam of electrons. A high voltage is applied in between A and C. ’N’ is a nickel crystal. D is an electron detector. It can be rotated on a circular scale. Detector produces current according to the intensity of incident beam.

Working: The electron beam is produced by passing current through filament F. The electron beam is accelerated by applying a voltage in between A (anode) and C. The accelerated electron beam is made to fall on the nickel crystal. The nickel crystal scatters the electron beam to different angles. The crystal is fixed at an angle of Φ = 50° to the incident beam. The detector current for different values of the accelerating potential ‘V’ is measured. A graph between detector current and voltage (accelerating) is plotted. The shape of the graph is shown in figure.

Analysis of graph:
Plus Two Physics Previous Year Question Paper March 2019, 19
The graph shows that the detector current increases with accelerating voltage and attains maximum value at 54V and then decreases. The maximum value of current at 54 V is due to the constructive interference of scattered waves from nickel crystal (from different planes of the crystal). Thus wave nature of electrons is established.

The experimental wavelength of electron: The wavelength of the electron can be found from the formula
2d sinθ = nλ …….. (1)
From the figure, we get
θ + Φ + θ = 180°
2θ = 180 – Φ, 2θ = 180 – 50°
θ = 65°
for n = 1
equation (1) becomes
λ = 2d sinθ ……….. (2)
for Ni crystal, d = 0.91 A°
Substituting this in eq. (2), we get
wavelength λ = 1.65 A°

The theoretical wavelength of the electron:
The accelerating voltage is 54 V
Energy of electron E = 54 × 1.6 × 10-19 J
Plus Two Physics Previous Year Question Paper March 2019, 20

Discussion: The experimentally measured wave-length is found in agreement with de-Broglie wavelength. Thus wave nature of electron is confirmed.

Question 18.
The energy required to separate all the nucleons inside a nucleus is called binding energy of the nucleus.
a) Write an expression for binding energy in terms of mass defect.
b) Draw the graph showing the variation of binding energy per nucleon as a function of mass number.
c) Which nucleus possess maximum binding energy per nucleon?
Answer:
a) BE = Δ mc2 or
BE = (Zmp + (A – Z) mn – M)c2
b) Graph
Plus Two Physics Previous Year Question Paper March 2019, 21
c) Fe (nucleus of iron)

Answer any three questions from 19 to 22. Each carries 4 scores. (3 × 4 = 12)

Question 19.
Niels Bohr made certain modification in Rutherford’s model by adding the ideas of quantum hypothesis.
a) State Bohr’s second postulate of quantisation of angular momentum.
b) Derive an expression for the radius and energy of the electron in the nth orbit of hydrogen atom.
Answer:
a) The orbital angular momentum of electron is an integral multiple of \(\frac{\mathrm{h}}{2 \pi}\)

b) Radius of the hydrogen atom: Consider an electron of charge ‘e’ and mass m revolving round the positively charged nucleus in circular orbit of radius ‘r’.
The force of attraction between the nucleus and the electron is
Plus Two Physics Previous Year Question Paper March 2019, 22
This force provides the centripetal force for the orbiting electron
Plus Two Physics Previous Year Question Paper March 2019, 23
According to Bohr’s second postulate, we can write
Angular momentum, mvr = \(\frac{\mathrm{nh}}{2 \pi}\)
i.e., v = \(\frac{\mathrm{nh}}{2 \pi \mathrm{mr}}\)
Substituting this value of V in equation (2), we get
Plus Two Physics Previous Year Question Paper March 2019, 24
Energy of the hydrogen atom:
The K.E. of revolving electron is
Plus Two Physics Previous Year Question Paper March 2019, 25
Substituting the value of equation (5) in equation (9)
we get
Plus Two Physics Previous Year Question Paper March 2019, 26

Question 20.
Two long co-axial solenoids of same length are shown below:
Plus Two Physics Previous Year Question Paper March 2019, 27
a) Define mutual inductane of the pair of coils.
b) Derive an expression for mutual inductance of two co-axial solenoids.
c) Write the dimension of mutual inductance.
Answer:
a) Φ = MI, when I = 1 A, Φ = M
The mutual inductance of a pair of coils is numerically equal to the magnetic flux linked with one coil when unit current flows through the other.

b) Consider a solenoid (air core) of cross sectional area A and number of turns per unit length n. Another coil of total number of turns N is closely wound over the first coil. Let I be the current flow through the primary.
Flux density of the first coil B = μ0nI
Flux linked with second coil, Φ = BAN
Φ = μ0nIAN ………. (1)
But we know Φ = MI ………. (1)
From eq (1) and eq (2) weget
∴ MI = μ0nIAN
M = μ0nAN
If the solenoid is covered over core of relative per-meability μr
then M = μrμ0nAN
c) ML2T-2A-2

Question 21.
A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5 cm. What is the magnifying power of the telescope for viewing distant objects when
a) the telescope is in normal adjustment?
b) the final image is formed at the least distance of distinct vision.
Answer:
Plus Two Physics Previous Year Question Paper March 2019, 28

Question 22.
In Amplitude Modulation, the amplitude of the carrier wave is varied in accordance with the information signal.
a) What is meant by modulation index?
b) A message signal of frequency 10 kHz and peak value of 10 V used to modulate a carrier of frequency 1 MHz and peak voltage of 20 V. Determine the modulation index.
c) The block diagram of a transmitter is shown below. Identify the elements labelled X and Y.
Plus Two Physics Previous Year Question Paper March 2019, 29
Answer:
a) Modulation index, μ = \(\frac{A_{m}}{A_{c}}\)
b) Am = 10 V, Ac = 20 V
∴ μ = \(\frac{A_{m}}{A_{c}}=\frac{10}{20}\) = 0.5

Answer any three questions from 23 to 26. Each carries 5 scores. (3 × 5 = 15)

Question 23.
The cyclotron is a device used to accelerate charged particles.
a) With a suitable diagram briefly explain the working of a cyclotron and obtain an expression for cyclotron frequency.
b) Acyclotron oscillator frequency is 10 MHz. What should be the operating magnetic field for accelerating protons?
Answer:
a) Principles: Cyclotron is based on two facts

  1. An electric field can accelerate a charged particle.
  2. A perpendicular magnetic field gives the ion a circular path.

Constructional Details: Cyclotron consists of two semicircular dees D1 and D2, enclosed in a chamber C. This chamber is placed in between two magnets. An alternating voltage is applied in between D1 and D2. An ion is kept in a vacuum chamber.

Working: At certain instant, let D1 be positive and D2 be negative. Ion (+ve) will be accelerated towards D2 and describes a semicircular path (inside it). When the particle reaches the gap, D1 becomes negative and D2 becomes positive. So ion is accelerated towards D1 and undergoes a circular motion with larger radius. This process repeats again and again.

Thus ion comes near the edge of the dee with high K.E. This ion can be directed towards the target by a deflecting plate.
Plus Two Physics Previous Year Question Paper March 2019, 30

Mathematical expression: Let ‘v’ be the velocity of ion, q the charge of the ion and B the magnetic flux density.
If the ion moves along a semicircular path of radius Y, then we can write
Plus Two Physics Previous Year Question Paper March 2019, 31
Eq. (2) shows that time is independent of radius and velocity.

Resonance frequency (cyclotron frequency): The condition for resonance is half the period of the accelerating potential of the oscillator should be ‘t’. (i.e., T/2 = t or T = 2t). Hence period of AC
T = 2t
Plus Two Physics Previous Year Question Paper March 2019, 32
Plus Two Physics Previous Year Question Paper March 2019, 33

Question 24.
The experimental set up to find an unknown resistance using a metre bridge is shown below:
Plus Two Physics Previous Year Question Paper March 2019, 34
a) What is the principle of a metre bridge?
b) If the balance point is found to be at 39.5 cm from the end ‘A’, the resistor ‘S’ is of 12.5 Ω. Determine the resistance ‘R’. Why are the connections between resistors in a metre bridge made of thick copper strips?
c) If the galvanometer and cell are interchanged at the balance point of the bridge would the galvanometer show any current?
Answer:
Plus Two Physics Previous Year Question Paper March 2019, 35
The resistors in metre bridge are made of thick copper strips to minimise the resistance of connection.
c) No. The galvanometer will not show any current.

Question 25.
The circuit used to change alternating voltage to direct voltage is called rectifier.
a) With a neat diagram, explain the working of a full wave rectifier having two diodes.
b) What is the output frequency of a full wave rectifier if the input frequency is 50 Hz?
c) Draw the output wave form across the load resistance connected in the full wave rectifier circuit.
Answer:
a) Full wave rectifier:
Circuit details
Plus Two Physics Previous Year Question Paper March 2019, 36
Full wave rectifier consists of transformer, two diodes and a load resistance RL. Input a.c signal is applied across the primary of the transformer. Secondary of the transformer is connected to D1 and D2. The output is taken across RL.

Working: During the +ve half cycle of the a.c signal at secondary, the diode D1 is forward biased and D2 is reverse biased. So that current flows through D1 and RL.

During the negative half cycle of the a.c signal at secondary, the diode D1 is reverse biased and D2 is forward biased. So that current flows through D1 and RL. Thus during both the half cycles, the current flows through RL in the same direction. Thus we get a +ve voltage across RL for +ve and -ve input. This process is called full wave rectifcation.

b) 100 Hz

Plus Two Physics Previous Year Question Paper March 2019, 37

Question 26.
A ray of light parallel to the principal axis of a spherical mirror falls at a point M as shown in the figure below:
Plus Two Physics Previous Year Question Paper March 2019, 38
a) Identify the type of mirror used in the diagram.
b) By drawing a suitable ray diagram, obtain the mirror equation.
c) If the mirror is immersed in water, its focal length ………….
Answer:
a) Concave mirror

b)
Plus Two Physics Previous Year Question Paper March 2019, 39
Let points P, F, C be pole, focus and centre of curvature of a concave mirror. Object AB is placed on the principal axis. A ray from AB incident at E and then reflected through F. Another ray of light from B incident at pole P and then reflected. These two rays meet at M. The ray of light from point B is passed through C. Draw EN perpendicular to the principal axis.
Δ IMF and Δ ENF are similar.
Plus Two Physics Previous Year Question Paper March 2019, 40
but IF = PI – PF and NF = PF (since aperture is small)
hence eq. (1) can be written as
Plus Two Physics Previous Year Question Paper March 2019, 41
This is called mirror formula or mirror equation,

c) Remain the same

Plus One Botany Notes Chapter 11 Plant Growth and Development

Students can Download Chapter 11 Plant Growth and Development Notes, Plus One Botany Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Botany Notes Chapter 11 Plant Growth and Development

Growth
Growth is defined as an irreversible permanent increase in the size of an organ or its parts of an individual cell.
It is accompanied by metabolic processes (both anabolic and catabolic), that occur at the expense of energy.
Eg: expansion of a leaf.

Plant Growth Generally is Indeterminate
Plant growth is unlimited growth due to the presence of meristems.
Root apical meristem and the shoot apical meristem are responsible for the primary growth of the plants and contribute to the elongation of the plants along their axis.

Role of lateral meristem in plants
In dicotyledonous plants and gymnosperms, the lateral meristems, (vascular cambium and cork-cambium) cause an increase in the girth of the organs. This is known as secondary growth.
Plus One Botany Notes Chapter 11 Plant Growth and Development 1

Growth is Measurable

  • Growth is measured in terms of increase in fresh weight, dry weight, length, area, volume, and cell number.
  • One single maize root apical meristem can give rise to more than 17,500 new cells per hour, cells in a watermelon increase in size by up to 3,50,000 times.
  • In the former, growth is expressed as an increase in cell number.
    latter expresses growth as an increase in the size of the cell.
  • While the growth of a pollen tube is measured in terms of its length, an increase in surface area denotes the growth in a dorsiventral leaf.

Phases of Growth
The period of growth is generally divided into three phases.

  1. Meristematic: The constantly dividing cells, both at the root apex and the shoot apex, represent the meristematic phase of growth.
  2. Elongation: The cells proximal to the meristematic zone represent the phase of elongation.
  3. Maturation: Proximal to the phase of elongation represents the phase of maturation.

 

Plus One Botany Notes Chapter 11 Plant Growth and Development 2

Growth Rates
The increased growth per unit time is termed as growth rate. The growth rate may be

  1. Arithmetic
  2. Geometrical

 

Plus One Botany Notes Chapter 11 Plant Growth and Development 3

On plotting the length of the organ against time, a linear curve is obtained, it is expressed as
Plus One Botany Notes Chapter 11 Plant Growth and Development 4

Lt = L0 + rt
Lt = length at time ‘t’
L0 = length at time ‘zero’.
r = growth rate/elongation per unit time.

Different phases of the Sigmoid curve

  1. lag phase
  2. log or exponential phase
  3. stationary phase

In most systems, the initial growth is slow (lag phase), and it increases rapidly at an exponential rate (log or exponential phase)
In the end, due to the limited nutrient supply, the growth slows down leading to a stationary phase. It is the typical sigmoid or S-curve.
Plus One Botany Notes Chapter 11 Plant Growth and Development 5
The exponential growth can be expressed as
W1 = W0 e rt
W1 = final size (weight, height, number etc.)
W0 = initial size at the beginning of the period
r = growth rate
t = time of growth
e = base of natural logarithms

Quantitative comparisons between the growth of a living system can also be made in two ways:

  • Measurement and the comparison of total growth per unit time is called the absolute growth rate.
  • The growth of the given system per unit time expressed on a common basis.

 

Plus One Botany Notes Chapter 11 Plant Growth and Development 6
In Figure two leaves, A and B, are drawn that are of different sizes but show an absolute increase in area in the given time to give leaves, A1 and B1.

Conditions for Growth
Water, oxygen, and nutrients as very essential elements for growth.
The plant cells grow in size by cell enlargement it requires water.
Turgidity of cells helps in extension growth. Water also provides the medium for enzymatic activities
Oxygen helps in releasing metabolic energy essential for growth activities.
Nutrients (macro and micro essential elements) are required by plants for the synthesis of protoplasm and act as a source of energy.
An optimum temperature range is best suited for plant growth.
Environmental signals such as light and gravity also affect certain phases/stages of growth.

Differentiation, Dedifferentiation, and Redifferentiation
1. The cells derived from root apical and shoot-apical meristems and cambium differentiate and mature to perform specific functions. This is termed as differentiation.
For example, during differentiation, tracheary elements lose their protoplasm and develop a very strong, elastic, lignocellulosic secondary cell wall, to carry water too long distances.

2. The living differentiated cells, that have lost the capacity to divide can regain the capacity of division This phenomenon is termed dedifferentiation.
For example, interfascicular cambium and cork cambium is formed from fully differentiated parenchyma cells.

3. Meristems are able to divide and produce cells that once again lose the capacity to divide but mature to perform specific functions. This is called a redifferentiation.
For example, secondary tissues develop from vascular cambium and cork cambium

Development
It is the stage of the life cycle in which germination of the seed to senescence.
Plus One Botany Notes Chapter 11 Plant Growth and Development 7
The plant shows a response to the environment to form different kinds of structures. This ability is called plasticity
Heterophylly is an example of plasticity

Types of Heterophylly
1. The leaves of the juvenile plant are different in shape from those in mature plants.
e.g cotton, coriander, and larkspur.
2. Shapes of submerged leaves are different from those produced in the air.
Eg buttercup.
Plus One Botany Notes Chapter 11 Plant Growth and Development 8
Development is considered as the sum of growth and differentiation.
Development in plants is under the control of intrinsic and extrinsic factors.

A) Intrinsic factors

  1. Intracellular (genetic)
  2. Intercellular factors (chemicals such as plant growth regulators)

B) Extrinsic factors
Light, temperature, water, oxygen, nutrition, etc.

Plant Growth Regulators

Characteristics
The plant growth regulators (plant hormones or phytohormones) include

  1. Indole compounds (indole-3-acetic acid, IAA);
  2. Adenine derivatives (N6-furfurylamino purine, kinetin),
  3. Derivatives of carotenoids (abscisic acid, ABA);
  4. Terpenes (gibberellic acid, GA3) or
  5. Gases (ethylene, C2H4).

The PGRs are divided into two groups based on their functions in a living plant body.
One group of PGRs are involved in growth-promoting activities, e.g., auxins, gibberellins, and cytokinins.
The other group mainly involved in growth-inhibiting activities such as dormancy and abscission. Eg-abscisic acid and ethylene.
Plus One Botany Notes Chapter 11 Plant Growth and Development 9

The Discovery of Plant Growth Regulators
1. Auxin
At first, Charles Darwin and his son Francis Darwin observed that the coleoptiles of canary grass responded to unilateral illumination by growing towards the light source.
After a series of experiments, it was concluded that the tip of the coleoptile was the site of transmittable influence that caused the bending of the entire coleoptile.
Auxin was isolated by F. W. Went from tips of coleoptiles of oat seedlings.

2. Gibberellin
The ‘balance’ (foolish seedling) disease of rice seedlings, was caused by a fungal pathogen Gibberalla fujikuroi.
In this experiment, the uninfected rice seedlings were treated with sterile filtrates of the fungus. It led to the development of the disease. The active substance was gibberellic acid.
It was demonstrated by E. Kurosawa.

3. Cytokinin
The internodal segments of tobacco stems- the callus proliferated in the presence of auxins along with the extracts of vascular tissues, yeast extract, coconut milk or DNA.
Skoog and Miller later identified and crystallized the cytokinesis promoting active substances that they termed kinetin.

4. Abscisic acid(ABA)
During the mid-1960s inhibitory hormones were identified: inhibitor-B, abscission II and dormin.
Later all the three were named abscisic acid (ABA).

5. Ethylene
Ripened oranges that hastened the ripening of stored unripened bananas. Later this volatile substance was identified as ethylene, a gaseous PGR.

Physiological Effects of Plant Growth Regulators
Auxins
Auxins were first isolated from human urine.
They are generally produced by the growing apices of the stems and roots, from where they migrate to the regions of their action.

Two types of auxins

  1. Natural (IAA and indole butyric acid (IBA)
  2. Synthetic. NAA (naphthalene acetic acid) and 2, 4-D (2, 4-dichlorophenoxyacetic)

Plus One Botany Notes Chapter 11 Plant Growth and Development 10

  1. They help to initiate rooting in stem cuttings
  2. Auxins promote flowering e.g. in pineapples.
  3. They help to prevent fruit and leaf drop at early stages but promote the abscission of older mature leaves and fruits.
  4. In most higher plants, the growing apical bud inhibits the growth of the lateral (axillary) buds, a phenomenon called apical dominance.
  5. Removal of shoot tips (decapitation) usually results in the growth of lateral buds Hence it is widely applied in tea plantations, hedge-making, etc.
  6. Auxins also induce parthenocarpy.
  7. They are widely used as herbicides. 2, 4-D is used to kill dicotyledonous weeds, So it is used to prepare weed-free lawns by gardeners.
  8. Auxin controls xylem differentiation and helps in cell division.

Gibberellins
Gibberellic acid (GA3) was one of the first gibberellins to be discovered and remains the most intensively studied form.
GA3 is acidic.

  1. GA3 causes an increase in the length of grapes stalks. .
  2. Gibberellins, cause fruits like apple to elongate and improve its shape
  3. They delay senescence. Hence the fruits are keeping as fresh.
  4. GA3 is used to speed up the malting process in the brewing industry.
  5. Spraying sugarcane crop with gibberellins increases the length of the stem, thus increasing the yield by as much as 20 tonnes per acre.
  6. Spraying juvenile conifers with GAs hastens the maturity period, thus leading to early seed production.
  7. Gibberellins also promote bolting(internode elongation just prior to flowering) in beet, cabbages and many plants with rosette habit.

Cytokinins
Cytokinins were discovered as kinetin (a modified form of adenine, a purine) from the autoclaved herring sperm DNA.
Naturally occurring cytokinin-zeatin was isolated from corn-kernels and coconut milk.
Natural cytokinins are synthesised in regions where rapid cell division occurs, for example, root apices, developing shoot buds, young fruits, etc.

  1. It helps to produce new leaves, chloroplasts in leaves, lateral shoot growth and adventitious shoot formation.
  2. Cytokinins help to overcome apical dominance.
  3. They promote nutrient mo8/+9bilisation which helps in the delay of leaf senescence.

Ethylene
The most widely used compound as a source of ethylene is ethephon.
It is readily absorbed and transported within the plant and releases ethylene slowly.
Ethephon hastens fruit ripening in tomatoes and apples and accelerates abscission in flowers and fruits

  1. Ethylene is a gaseous hormone that promotes senescence and ripening fruits.
  2. It promotes horizontal growth of seedlings, swelling of the axis, and apical hook formation in dicot seedlings.
  3. Ethylene promotes senescence and abscission of plant organs, especially of leaves and flowers.
  4. Ethylene is highly effective in fruit ripening. It enhances the respiration rate during the ripening of the fruits (respiratory climactic).
  5. Ethylene breaks seed and bud dormancy and initiates germination in peanut seeds, sprouting of potato tubers.
  6. Ethylene promotes rapid internode/petiole elongation in deepwater rice plants.
  7. Ethylene also promotes root growth and root hair formation, thus helping the plants to increase their absorption surface.
  8. Ethylene is used to initiate flowering and fruit-set in pineapples.
  9. It also induces flowering in mango.
  10. It promotes female flowers in cucumbers

Abscisic acid

  1. It promotes abscission and dormancy.
  2. It acts as an inhibitor of plant metabolism.
  3. ABA inhibits seed germination.
  4. ABA stimulates the closure of stomata and increases the tolerance of plants to various kinds of stresses. Hence it is called the stress hormone.
  5. ABA plays an important role in seed development, maturation, and dormancy.
  6. ABA helps the seeds to withstand desiccation
  7. ABA acts as an antagonist to GA3.

Photoperiodism
It is the phenomenon of relative day and night length for the initiation of flowering.
Plus One Botany Notes Chapter 11 Plant Growth and Development 11

Based on the exposure to photoperiod there are three types of plants

  1. Long day plants: They require exposure to light for a period greater than critical duration (12 hr).
  2. Short-day plants: They require less than critical duration before flowering.
  3. Day-neutral plants: In this type, there is no such correlation between exposure to light duration and induction of flowering response.

Which is the organ of a plant perceives light for photoperiodism?
The site of perception of light/dark duration is the leaves. After receiving the required photoperiod, the hormonal substance migrates from leaves to shoot apices for inducing flowering. The shoot apices become changed into flowering apices prior to flowering.

Vernalisation
It is the phenomenon of exposure of low temperature for the initiation of flowering
Some important food plants, wheat, barley, rye have two kinds of varieties: winter and spring varieties.

Nature of spring and winter varieties
The ‘spring’variety are normally planted in the spring and come to flower and produce grain before the end of the growing season.
Winter varieties, planted in spring fail to flower or produce mature grain within a span of a flowering season.
If they are planted in autumn .they germinate and overwinter come out as small seedlings, resume growth in the spring, and are harvested usually around mid-summer.

Biennials and low-temperature treatment
Biennials are monocarpic plants that normally flower and die in the second season. Biennial plants are subjected to a cold treatment, it stimulates photoperiodic flowering response. Sugarbeet, cabbages, carrots are some of the common biennials.

NCERT Supplementary Syllabus

Seed Germination
The seeds germinate under favourable conditions after the period of dormancy.
After dormancy embryo becomes metabolically active and starts growing. This process is known as seed germination.
The conditions necessary for seed germination are the availability of water and oxygen.

A physiological phenomenon in seed germination
The physical phenomenon associated with seed germination is imbibition. It causes the swelling of seed then rupturing of the seed coat, through which radical emerges out.
It develops into a root system but the shoot system arises from the plumule of another end of the embryonal axis.
The metabolic activities require oxygen for breaking down the food reserves such as polysaccharides, proteins and lipid.
It is converted into soluble materials with the help of enzymes and mobilized to the embryonal axis.
The growth of radical and plumule is due to cell extension, cell division, and several biochemical processes.
The seed also needs a suitable temperature (optimum between 25 to35). The rate of respiration increases rapidly during seed germination.

What is the Viviparous type of germination?
Vivipary is the germination of a seed while it is still attached to the parent plant and is nourished by it. The plants grow in marshy land such as Rhizophora and Sonneratia (halophytes)show this type of germination.
During germination, radical elongates, and the weight of the germinating seed increases. As a result, the seedling separates and fail down vertically into the mud and grow into a new plant.

Seed Dormancy
It is the period of rest or a period of suspended growth due to this

  1. water content, the metabolic activities become extremely low.
  2. the seed coat becomes impermeable to oxygen and moisture and hardens.

The suspension of growth is due to exogenous (environmental conditions) or endogenous control during which metabolic activity of the seed is greatly reduced.

Causes of Dormancy

  1. Impermeable or mechanically resistant seed coats.
  2. Rudimentary or physiologically immature embryos or
  3. Due to the presence of germination inhibitors such as abscisic acid, phenolic acid, short-chain fatty acids, and coumarin.

How can overcome seed dormancy?

  1. Mechanical or chemical scarification of the seed coat (scratching of seed coat or seeds soaked in chemicals to break the dormancy)
  2. Stratification of seeds or changing environmental conditions such as temperature, light, and pressure. Stratification of seeds is subjecting the moist seeds to oxygen for variable periods of low or high temperatures.

Plus One Botany Notes Chapter 10 Respiration in Plants

Students can Download Chapter 10 Respiration in Plants Notes, Plus One Botany Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Botany Notes Chapter 10 Respiration in Plants

How are respiration and photosynthesis-related?
Green plants and cyanobacteria can prepare their own food by the process of photosynthesis, they trap light energy and convert it into chemical energy that is stored in the bonds of carbohydrates (Macromolecules) like glucose, sucrose, and starch.
By Cellular respiration, food materials undergo breakdown that release energy and the trapping of this energy for the synthesis of ATP.

ATP is called the energy currency of the cell why?
ATP is broken down whenever (and wherever) energy needs to be utilised. Hence, ATP acts as the energy currency of the cell.

Seat of photosynthesis and respiration
Photosynthesis takes place within the chloroplasts whereas the breakdown of complex molecules to yield energy takes place in the cytoplasm and in the mitochondria (also only in eukaryotes).
The compounds that are oxidized during respiration are known as respiratory substrates.
Eg. Proteins, fats, and even organic acids.

Do Plants Breathe?
Plants require O2 for respiration and give out CO2 and H2O as end products and release energy most of which is given out as heat.

Respiration is least important to plants than animals
Roots, stems, and leaves respire at rates far lower than animals When cells photosynthesize O2 is released within the cell.
But some cells live where oxygen may or may not be available.
All living organisms retain the enzymatic machinery to partially oxidise glucose without the help of oxygen. This breakdown of glucose to pyruvic acid is called glycolysis.

Glycolysis
The term glycolysis -(Greek words, glycols for sugar, and lysis for splitting).
The scheme of glycolysis was given by Gustav Embden, Otto Meyerhof, and J. Parnas, Hence glycolysis is called an EMP pathway.
In anaerobic organisms, it is the only process in respiration.
Glycolysis occurs in the cytoplasm of the cell and is present in all living organisms.
In this process, Glucose undergoes partial oxidation to form two molecules of pyruvic acid.

Steps lead to end products of glycolysis
Plus One Botany Notes Chapter 10 Respiration in Plants 1
1. Glucose and fructose are phosphorylated to give rise to glucose-6-phosphate by the activity of the enzyme hexokinase.

2. This phosphorylated form of glucose then isomerises to produce fructose-6-phosphate.

3. In this pathway, ATP is utilised at two steps: first in the conversion of glucose into glucose 6-phosphate and second in the conversion of fructose 6-phosphate to fructose 1,6 diphosphate).

4. The fructose 1,6-diphosphate is split into dihydroxyacetone phosphate and 3 phosphoglyceraldehydes (PGAL). In this step NADH +H+ is formed from NAD+.

5. 3-phosphoglyceraldehyde (PGAL) is converted to 1,3 bisphosphoglycerate (DPGA).

6. The conversion of DPGA to 3-phosphoglyceric acid (PGA), is also an energy-yielding process; this energy is trapped by the formation of ATP.

7. 3-phosphoglyceric acid (PGA) is converted into 2 phosphoglycerates.

8. 2 phosphoglycerates are converted into 2 phosphoenol pyruvic acid. ATP is synthesized during the conversion of PEP to pyruvic acid.

9. 2 phosphoenol pyruvic acid undergoes dephosphorylation to form 2 molecule of pyruvic acid

The fate of pyruvic acid
It involves

  1. Lactic acid fermentation
  2. Alcoholic fermentation
  3. Aaerobic respiration.

Fermentation takes place under anaerobic conditions in many prokaryotes and unicellular eukaryotes.

The complete oxidation of glucose to CO2 and H2O occurs in organisms that adopt Krebs’ cycle which is also called as aerobic respiration. This requires an O2 supply.

Fermentation
In fermentation, glucose undergoes incomplete oxidation and forms CO2 and ethanol
The enzymes, pyruvic acid decarboxylase, and alcohol dehydrogenase catalyze these reactions.
Fermentation occurs in the presence of yeast
Yeasts poison themselves to death when the concentration of alcohol reaches about 13 percent. Some organisms like bacteria produce lactic acid from pyruvic acid.

The lactic acid in eukaryotic cell
In animal muscle cells during exercise, when oxygen is inadequate for cellular respiration, pyruvic acid is reduced to lactic acid by lactate dehydrogenase.
The reducing agent is NADH+H* which is re oxidised to NAD+ in both the processes.
In both lactic acid and alcohol fermentation, less than seven percent of the energy in glucose is released.
In eukaryotes second step after glycolysis take place within the mitochondria and this requires O2.
It is aerobic respiration leads to complete oxidation of carbohydrate in the presence of oxygen and releases CO2, water and a large amount of energy.
This type of respiration is most common in higher organisms.

Aerobic Respiration
The second step of Aerobic respiration takes place within the mitochondria.
The product of glycolysis- pyruvate is transported from the cytoplasm into the mitochondria.
Plus One Botany Notes Chapter 10 Respiration in Plants 2

First step of oxidation of pyruvic acid
In the mitochondrial matrix, pyruvate undergoes oxidative decarboxylation by pyruvic dehydrogenase. The reactions require the participation of several coenzymes, including NAD+ and Coenzyme A.
Plus One Botany Notes Chapter 10 Respiration in Plants 3

During this process, two molecules of NADH are produced from the metabolism of two molecules of pyruvic acid.
The acetyl CoA then enters a cyclic pathway, tricarboxylic acid cycle(Krebs’ cycle) after the scientist Hans Krebs who first elucidated it.

Tricarboxylic Acid Cycle
The TCA cycle starts with the condensation of acetyl group with oxaloacetic acid (OAA) and water to yield citric acid.
The reaction is catalysed by the enzyme citrate synthase and a molecule of CoA is released.
It is followed by two successive steps of decarboxylation, leading to the formation of alpha-ketoglutaric acid and then succinyl-CoA. In the remaining steps, Succinic acid is oxidised to OAA.
Plus One Botany Notes Chapter 10 Respiration in Plants 4

Which step of the Krebs cycle substrate-level phosphorylation occurs?
During the conversion of succinyl-CoA to succinic acid, a molecule of GTP is synthesised. This is substrate-level phosphorylation.

At three sites in the cycle where NAD+ is reduced to NADH + H+ and one site where FAD+ is reduced to FADH2.
Plus One Botany Notes Chapter 10 Respiration in Plants 5
In the mitochondrial matrix, pyruvate is broken down to release.
8 molecules of NADH + H+
2 molecules of FADH2
2 molecules of GTP and
3 molecules of CO2

Electron Transport System (ETS) and Oxidative Phosphorylation
Plus One Botany Notes Chapter 10 Respiration in Plants 6
The metabolic pathway through which the electron passes from one carrier to another is called the electron transport system (ETS).
It is present in the inner mitochondrial membrane.
Reduced coenzyme like NADH(complex) in the mitochondrial matrix is oxidised and release 2 electrons and 2protons
Electrons and protons are transferred to FMN, it reduced to FMNH2
It breaks and releases protons and electrons .protons go to intermembrane space but electrons reach ubiquinone.
Ubiquinone also receives reducing equivalents via FADH2 (complex II).
The reduced ubiquinone is then oxidised with the transfer of electrons to cytochrome c via cytochrome bc1 complex (complex III).

Electron Transport System (ETS)
Cytochrome c acts as a mobile carrier for the transfer of electrons between complex III and IV.
Complex IV refers to cytochrome c oxidase complex containing cytochromes a and a3.

Oxidation of one molecule of NADH gives rise to 3 molecules of ATP, while that of one molecule of FADH2 produces 2 molecules of ATP.
Oxygen acts as the final hydrogen acceptor.

Oxidative phosphorylation in mitochondria
In ETS the energy of oxidation-reduction is utilised for the production of proton gradient required for phosphorylation. This process is called oxidative phosphorylation.

Chemiosmosis (proposed by peter Mitchel)
The energy released during the electron transport system is utilised in synthesizing ATP with the help of ATP synthase (complex V) called chemiosmosis.
Plus One Botany Notes Chapter 10 Respiration in Plants 7
F1 – F0/exosomes
This complex consists of two major components, F1 and Fo.
The F1 headpiece is a site for synthesis of ATP from ADP and inorganic phosphate.
F0 is an integral membrane protein complex act as a channel through which protons cross the inner membrane.
For each ATP produced, 2H+ passes through F0 from the intermembrane space to the matrix down the electrochemical proton gradient.

The Respiratory Balance Sheet
How many ATP molecules are produced in Aerobic respiration?
In aerobic respiration, the number of ATP molecules produced or utilized in glycolysis, TCA cycle and ETS gives the net gain of 36 ATP molecules
Fermentation accounts for only a partial breakdown of glucose whereas in aerobic respiration it is completely degraded to CO2 and H2O.

How many ATP molecules are produced in Fermentation?
In fermentation there is a net gain of only two molecules of ATP for each molecule of glucose degraded NADH is oxidized to NAD+ rather slowly in fermentation.

Amphibolic Pathway
It involves two processes anabolism and catabolism.
For example, fats is broken down into glycerol and fatty acids. Then fatty acids degraded to acetyl CoA and enter the pathway.
Glycerol enters the pathway after being converted to PGAL.
The proteins are degraded by proteases and the individual amino acids enter the pathway at some stage within the Krebs’cycle as pyruvate or acetyl CoA.

Is it true both catabolism and anabolism occur in fat metabolism?
Fatty acids( substrate) are broken down to acetyl CoA before entering the respiratory pathway. But when the organism needs to synthesize fatty acids, acetyl CoA withdrawn from the respiratory pathway for it.
Hence, the respiratory pathway involves the breakdown and synthesis of fatty acids, i.e catabolism, and anabolism respectively. Hence it is considered as an amphibolic pathway.
Plus One Botany Notes Chapter 10 Respiration in Plants 8

Respiratory Quotient
Definition: The ratio of the volume of CO2 evolved to the volume of O2 consumed in respiration is called the respiratory quotient (RQ) or respiratory ratio.
Plus One Botany Notes Chapter 10 Respiration in Plants 9

Respiratory quotient of some respiratory substrates
1. Carbohydrates: When carbohydrates are completely oxidised, the RQ is 1, because equal amounts of CO2 and O2 are evolved and consumed, respectively
Plus One Botany Notes Chapter 10 Respiration in Plants 10

2. Fats: If fats are used in respiration, the RQ is less than 1.
Plus One Botany Notes Chapter 10 Respiration in Plants 11

3. Proteins: When proteins are respiratory substrates the ratio is 0.9.

4. Organic acids: When organic acids are respiratory substrates, the ratio is more than one.

Plus One Botany Notes Chapter 9 Photosynthesis in Higher Plants

Students can Download Chapter 9 Photosynthesis in Higher Plants Notes, Plus One Botany Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Botany Notes Chapter 9 Photosynthesis in Higher Plants

What Do We Know?
Role of light, CO2, H2O, and Chlorophyll
Actually, chlorophyll (green pigment of the leaf), light, and CO2 are required for photosynthesis. A variegated leaf ora leaf that was partially covered with black paper, and one that was exposed to light. On testing these leaves for starch it was clear that photosynthesis occurred only in the green parts of the leaves in the presence of light.

Half leaf experiment and the importance of CO2 in photosynthesis
In this, a part of a leaf is enclosed in a test tube containing some KOH soaked cotton (which absorbs CO2), while the other half is exposed to air. The set up is then placed in light for some time. Then conducted the starch test, showed that the exposed part of the leaf tested positive for starch while the portion that was in the tube, tested negative. This showed that CO2 is required for photosynthesis.

Early Experiments

Historical aspects of photosynthesis
1. Priestley
He observed that a candle burning in a closed space – a bell jar, soon gets extinguished. Similarly, a mouse would soon suffocate in a closed space.
Plus One Botany Notes Chapter 9 Photosynthesis in Higher Plants 1
He concluded that a burning candle or an animal that breathes the air, both damage the air. But when he placed a mint plant in the same bell jar, he found that the mouse stayed alive and the candle continued to burn.
Plus One Botany Notes Chapter 9 Photosynthesis in Higher Plants 2

2. Jan Ingenhousz
He showed that sunlight is essential to the plant process that purifies the air fouled by burning candles or breathing animals. In aquatic habitat, during bright sunlight, small bubbles were formed around the green parts while in the dark they did not. Later he identified these bubbles are oxygen. So the green part of the plants could release oxygen.

3. Julius von Sachs
Glucose is usually stored as starch. He found that the green parts in plants where glucose is made.

4. T.W Engelmann
By using a prism he split light into its spectral components and then illuminated a green alga, Cladophora, placed in a suspension of aerobic bacteria. The bacteria were used to detect the sites of O2 evolution. He observed that the bacteria accumulated mainly in the region of blue and red light of the split spectrum.

An empirical equation for photosynthesis
Plus One Botany Notes Chapter 9 Photosynthesis in Higher Plants 3
[CH2O] represent a carbohydrate (e.g., glucose, a six-carbon sugar).

Hydrogen donor in bacteria and green plants
Some organisms do not release O2 during photosynthesis
When H2S, instead is the hydrogen donor for purple and green sulphur bacteria, the ‘oxidation’ product is sulphur or sulphate depending on the organism and not O2. In the green plants, the O2 evolved from H2O, not from carbon dioxide.

The modern equation for photosynthesis
Plus One Botany Notes Chapter 9 Photosynthesis in Higher Plants 4
C6H12O6 represents glucose. The O2 released is from water

Where Does Photosynthesis Take Place?
It takes place in the chloroplast of leaves that contain grana, the stroma lamellae, and the fluid stroma.

Where is the energy production site in chloroplast?
The energy-rich molecules like ATP and NADPH are synthesized in grana and stroma lamellae by light reactions.

Where is the Glucose production site in chloroplast?
In stroma by dark reactions, CO2 fixation leading to the synthesis of glucose, which in turn forms starch

Structure of chloroplast
Plus One Botany Notes Chapter 9 Photosynthesis in Higher Plants 5
Diagrammatic representation of an electron micrograph of a section of chloroplast

How Many Pigments are Involved in Photosynthesis?
Chromatographic separation of the leaf pigments shows that different types of pigments in leaves i.e
Chlorophyll a (bright or blue-green in the chromatogram)
chlorophyll b (yellow-green)
xanthophylls (yellow)
carotenoids (yellow to yellow-orange)

a) Graph showing the absorption spectrum of chlorophyll a, b, and the carotenoids.
Plus One Botany Notes Chapter 9 Photosynthesis in Higher Plants 6
b) Graph showing the action spectrum of photosynthesis.
Plus One Botany Notes Chapter 9 Photosynthesis in Higher Plants 7
c) Graph showing action spectrum of photosynthesis superimposed on the absorption spectrum of chlorophyll a.
Plus One Botany Notes Chapter 9 Photosynthesis in Higher Plants 8

Wavelengths of light absorbed by pigments
Chlorophyll pigments absorb light, at specific wavelengths of blue and the red regions while carotenoids absorb the blue and green wavelength
Chlorophyll is the major pigment responsible for trapping light, other thylakoid pigments like chlorophyll b, xanthophylls, and carotenoids, which are called accessory pigments, also absorb light and transfer the energy to chlorophyll a. but also protect chlorophyll a from photo-oxidation.

What is Light Reaction?
It is the photochemical phase include

  1. light absorption
  2. water splitting
  3. oxygen release
  4. Formation of high-energy rich molecules ATP and NADPH.

The pigments are organised into two light-harvesting complexes(LHC)

  1. Photosystem I (PS I)/P700
  2. Photosystem II (PS II)/P680

Each photosystem has single chlorophyll a molecule forms the reaction centre, all the pigments except chlorophyll-a forming a light-harvesting system also called antennae.

In PS I the reaction centre, chlorophyll a has an absorption peak at 700 nm while in PS II it has absorption maxima at 680 nm, and is called P680.
Plus One Botany Notes Chapter 9 Photosynthesis in Higher Plants 9

The Electron Transport
How does electron flow in electron carriers that connect two photosystems?
Initially, excitation of chlorophyll molecule occurs due to light, then electrons are emitted from Ps II (uphill) that are accepted by electron acceptor, electron flows through electron carriers cytochromes, (downhill) and (Loss of electrons of PSII is compensated by electrons coming from water and loss of electrons of PS I is compensated by electrons coming from PS II).
Plus One Botany Notes Chapter 9 Photosynthesis in Higher Plants 10
PS I is also excited due to light and electrons are emitted (uphill), it transfers an electron to another accepter, and finally down the hill to NADP+ causing it to be reduced to NADPH + H+ is called the Z scheme.

Result of Z-scheme

  1. production of ATP and NADPH
  2. O2 evolution

Splitting of Water
Photolysis
It is the splitting of water into H+, [O] and electrons in the presence of light and these electrons are available to PSII.
This process takes place on the inner side of the membrane of the thylakoid.
Oxygen released is one of the net products of photosynthesis.
2H2O → 4H+ + O2 + 4e

Cyclic and Non-cyclic Photo-phosphorylation
Phosphorylation
The process of which ATP from ADP and inorganic phosphate in the presence of light (in mitochondria and chloroplasts) is named phosphorylation.

Electron in a cyclic process
When only PS I is functional, the cyclic flow of electrons within the photosystem and the phosphorylation occurs in the stroma lamellae.
Cyclic photophosphorylation also occurs when only light of wavelengths beyond 680 nm are available for excitation i.e at 700 nm
Plus One Botany Notes Chapter 9 Photosynthesis in Higher Plants 11

Result of cyclic photophosphorylation
ATP is produced.

Where does the light-harvesting complex work for acyclic and noncyclic processes?
The membrane or lamellae of the grana have both PS I and PS II so a noncyclic process occurs,
The stroma lamellae membranes lack PS II as well as NADP reductase enzyme So a cyclic process occurs.

Chemiosmotic Hypothesis
It is the ATP synthesis linked to the development of a proton gradient across a membrane
In chloroplast, the proton accumulation is towards the inside of the membrane, i.e., in the lumen. In respiration, protons accumulate in the intermembrane space of the mitochondria when electrons move through the ETS.
The proton gradient develops due to,

a. Splitting of the water molecule takes place on the inner side of the membrane, the protons accumulate within the lumen of the thylakoids

b. As electrons move through the photosystems, protons are transported across the membrane moves into the lumen side of the membrane

c. The NADP reductase enzyme is located on the stromal side of the membrane. Along with electrons that come from the accepter of electrons of PS I, protons are necessary for the reduction of NADP+ to NADPH+ H+. These protons are also removed from the stroma.

In chloroplast, protons in the stroma decrease in number, while in the lumen there is an accumulation of protons. This creates a proton gradient across the thylakoid membrane The gradient is broken down due to the movement of protons across the membrane to the stroma through the transmembrane channel of the F0 of the ATP.
ATPase have a channel that allows diffusion of protons back across the membrane; this releases enough energy to activate the ATPase enzyme that catalyses the formation of ATP.
Plus One Botany Notes Chapter 9 Photosynthesis in Higher Plants 12
Where are the ATP and NADPH Used?
It is used in the biosynthetic phase of photosynthesis. This process does not directly depend on the presence of light but is dependent on the products of the light reaction, i.e., ATP and NADPH.
Melvin Calvin studied the algal photosynthesis by using radioactive 14C led to the discovery that the first CO2 fixation product was identified as 3-phosphoglyceric acid or PGA.

The Primary Acceptor of CO2
The studies showed that the accepter molecule was a 5-carbon sugar -ribulose bisphosphate (RuBP) in the Calvin cycle.

The Calvin Cycle
It involves three stages:

  1. carboxylation
  2. reduction
  3. regeneration

1. Carboxylation:
Carboxylation is the fixation of CO2 into a stable compound catalysed by the enzyme RuBisCO that results in the formation of two molecules of 3-PGA.

2. Reduction: These are a series of reactions that lead to the formation of glucose.
The steps involve the utilization of 3 molecules of ATP for phosphorylation and two NADPH for reduction per CO2 molecule fixed. For the fixation of six molecules of CO2, 6 turns of the cycle are required and one molecule of glucose is generated

3. Regeneration: Regeneration of the CO2 acceptor molecule require one ATP for phosphorylation to form RuBP.
Hence for every CO2 molecule entering the Calvin cycle, 3 molecules of ATP and molecules of NADPH are required
Plus One Botany Notes Chapter 9 Photosynthesis in Higher Plants 13
The Calvin cycle proceeds in three stages:
1. carboxylation, during which CO2 combines with ribulose- 1, 5- bisphosphate
2. reduction, during which carbohydrate is formed at the expenses of the photochemically made ATP and NADPH; and
3. regeneration during which the CO2 acceptor ribulose- 1, 5-bisphosphate has formed again so that the cycle continues
Plus One Botany Notes Chapter 9 Photosynthesis in Higher Plants 14

The C4 Pathway (Hatch and Slack Pathway)
Plants that are adapted to dry tropical regions have the C4 pathway, C4 plants have a special type of leaf anatomy. They tolerate higher temperatures. They lack a process called photorespiration and have greater productivity of biomass.

Special leaf anatomy-kranz anatomy
Large cells around the vascular bundles are centripetally arranged bundle sheath cells such anatomy is called ‘Kranz’ anatomy. Eg- maize or sorghum

Primary CO2, accepter, first stable product and Enzyme of C4 Pathway
The primary CO2 acceptor is a 3-carbon molecule- phosphoenolpyruvate (PEP) present in the mesophyll cells. The enzyme responsible for this fixation is PEP carboxylase or PEPcase.
The first stable product C4 acid OAA is formed in the mesophyll cells.
Plus One Botany Notes Chapter 9 Photosynthesis in Higher Plants 15
Diagrammatic representation of a Hatch arid Slack Pathway
OAA converted into 4-carbon compounds like malic acid or aspartic acid in the mesophyll cells .which are transported to the bundle sheath cells. In the bundle sheath cells, these C4 acids are broken down to release CO2 and a 3-carbon molecule. The 3-carbon molecule is transported back to the mesophyll where it is converted to PEP again, thus, completing the cycle.
Thus the basic pathway that results in the formation of the sugars, the Calvin pathway, is common to the C3 and C4 plants.

Photorespiration
In C3 plants, under high concentration of O2 and low CO2 concentration, RUBP binds with O2 to form one molecule of PGA and phosphoglycolate and a large quantity of CO2 is released.

Can you say photorespiration is a wasteful process?
This process utilise ATP but neither synthesis of sugars, nor of ATP. Hence photorespiration is a wasteful process.
Plus One Botany Notes Chapter 9 Photosynthesis in Higher Plants 16

The specialty of C4 plants to avoid Photorespiration
In C4 plants photorespiration does not occur because C4 acid from the mesophyll is broken down in the bundle cells to release CO2 – this results in increasing the intracellular concentration of CO2. Here RuBisCO functions as a carboxylase minimizing the oxygenase activity, productivity, and yields are better in these plants.

Factors Affecting Photosynthesis
Photosynthesis is influenced by several factors, both internal (plant) and external. The plant factors include the number, size, age, and orientation of leaves, mesophyll cells and chloroplasts, internal CO2 concentration, and the amount of chlorophyll. The external factors include the availability of sunlight, temperature, CO2 concentration, and water.

Blackman s Law of Limiting Factors
If a chemical process is affected by more than one factor, then its rate will be determined by the factor which is nearest to its minimal value: it is the factor that directly affects the process if its quantity is changed.
For example, In the green leaf, the light and CO2 conditions are optimum but the plant does not photosynthesize if the temperature is very low.

Light
The availability of light shows a direct relationship with CO2 fixation rates at low light intensities At higher light intensities the rate does not show further increase because other factors are not in optimal amount. The intensity of light beyond a point causes the breakdown of chlorophyll and a decrease in photosynthesis.
Plus One Botany Notes Chapter 9 Photosynthesis in Higher Plants 17

Carbon dioxide Concentration
The concentration of CO2 is very low in the atmosphere (between 0.03 and 0.04 percent). An increase in concentration up to 0.05 percent can cause an increase in CO2 fixation rates. The C3 and C4 plants respond differently to CO2 concentrations.

Graph of light Intensity on the rate of photosynthesis

C4 plants show saturation at about 360µL-1 while C3 responds to increased CO2 concentration and saturation is seen only beyond 450µL-1 Thus, the current availability of CO2 levels is limiting to the C3 plants.

C3 plants respond to higher CO2 concentration by showing increased rates of photosynthesis leading to higher productivity The above concept is used for some greenhouse crops such as tomatoes and bell pepper.

Temperature
The dark reactions that take place in stoma are enzymatic and temperature controlled. C4 plants respond to higher temperatures and show a higher rate of photosynthesis while C3 plants have a much lower temperature optimum.

Water
Water stress causes the closure of stomata and it is difficult to receive CO2 for photosynthesis. The stress condition also makes leaves wilt and reducing the surface area of the leaves and their metabolic activities.

Plus One Botany Notes Chapter 8 Mineral Nutrition

Students can Download Chapter 8 Mineral Nutrition Notes, Plus One Botany Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Botany Notes Chapter 8 Mineral Nutrition

Methods to Study the Mineral Requirements of Plants
In 1860, Julius von Sachs a German botanist demonstrated that plants could be grown in a nutrient solution in the complete absence of soil.

Hydroponics and its Importance

  • This technique of growing plants in a nutrient solution is known as hydroponics.
  • The nutrient solutions must be aerated to obtain optimum growth.
  • In this method, essential elements are used and their deficiency symptoms can be studied.
  • Hydroponics is used in the commercial production of vegetables such as tomato, seedless cucumber, and lettuce.

 

Plus One Botany Notes Chapter 8 Mineral Nutrition 1

Hydroponic plant production. Plants are grown in a tube or trough placed on a slight incline. A pump circulates a nutrient solution from a reservoir to the elevated end of the tube. The solution flows down the tube and returns to the reservoir due to gravity. Inset shows a plant whose roots are continuously bathed in the aerated nutrient solution. The arrows indicate the direction of the flow.

Essential Mineral Elements

Some minerals are not essential to plants

  • More than sixty elements are found in different plants.
  • Some plant species absorb selenium, some others gold, while some plants growing near nuclear test sites take up radioactive strontium.

Criteria for Essentiality
The criteria for the essentiality of an element are given below:

  • The element must be supporting normal growth and production.
    In the absence of the element, the plants do not complete their life cycle or set the seeds.
  • The requirement of the element must be specific and not replaceable by another element.
  • The element must be directly involved in the metabolism of the plant.

Based upon the above criteria 17 elements are essential for plant growth and metabolism. They are
i. Macronutrients:
Carbon, hydrogen, oxygen, nitrogen, phosphorous, sulphur, potassium, calcium, and magnesium
They are present in plant tissues in large amounts(in excess of 10 m mole/ Kg of dry matter).

ii. Micronutrients:
Iron, manganese, copper, molybdenum, zinc, boron, chlorine, and nickel
They are needed in very small amounts (less than 10 m mole /Kg of dry matter).

In addition to the essential elements, sodium, silicon, cobalt, and selenium are required by higher plants. Essential elements are grouped into four broad categories on the basis of their diverse functions.
i. Essential elements as components of biomolecules (e.g., carbon, hydrogen, oxygen, and nitrogen).

ii. Essential elements that are components of energy-related chemical compounds (e.g, magnesium in chlorophyll and phosphorous in ATP).

iii. Essential elements that activate or inhibit enzymes, (Mg2+ is an activator for both ribulose bisphosphate carboxylase oxygenase and phosphoenolpyruvate carboxylase, both of which are critical enzymes in photosynthetic carbon fixation
Zn2+ is an activator of alcohol dehydrogenase and Mo of nitrogenase during nitrogen metabolism.

iv. Essential elements alter the osmotic potential of a cell.
Potassium plays an important role in the opening and closing of stomata.

Role of Macro- and Micro-nutrients
Essential elements participate in various metabolic processes in the plant cells. The various forms and functions of mineral elements are given below.

Nitrogen

  • It is absorbed mainly as NO3 Some taken up as NO2 or NH4+
  • Nitrogen is required in meristematic tissues and the metabolically active cells.
  • It is one of the major constituents of proteins, nucleic acids, vitamins and hormones

Phosphorus

  • It is absorbed in the form of phosphate ions (either as HPO42- or H2PO4)
  • Phosphorus is a constituent of cell membranes, certain proteins, all nucleic acids, and nucleotides.
  • It is required for all phosphorylation reactions.

Potassium

  • It is absorbed as a potassium ion (K+).
  • It is required in abundant quantities for meristematic tissues, buds, leaves, and root tips.
  • Potassium helps to maintain an anion-cation balance in cells
  • It is involved in protein synthesis
  • It is involved in the opening and closing of stomata and activation of enzymes
  • It helps in the maintenance of the turgidity of cells.

Calcium

  • It is absorbed in the form of calcium ions (Ca2+).
  • Calcium is required by meristematic and differentiating tissues.
  • It is important in the formation of calcium pectate in the middle lamella.
  • It is also needed during the formation of the mitotic spindle.
  • It activates certain enzymes and plays an important role in regulating metabolic activities.

Magnesium

  • It is absorbed by plants in the form of divalent Mg2+
  • It activates the enzymes of respiration, photosynthesis, and are involved in the synthesis of DNA and RNA.
  • Magnesium is a constituent of the ring structure of chlorophyll
  • It helps to maintain the ribosome structure.

Sulphur

  • It is absorbed in the form of sulphate (SO42-)ion.
  • Sulphur is present in two amino acids – cysteine and methionine
  • It is the main constituent of several coenzymes, vitamins (thiamine, biotin, Coenzyme A), and ferredoxin.

Iron

  • It is absorbed in the form of ferric ions (Fe3+)
  • It is an important constituent of proteins involved in the transfer of electrons like ferredoxin and cytochromes.
  • It activates the catalase enzyme and is essential for the formation of chlorophyll.

Manganese

  • It is absorbed in the form of manganous ions (Mn2+).
  • It activates many enzymes involved in photosynthesis, respiration, and nitrogen metabolism.
  • It is also involved in the splitting of water to liberate oxygen during photosynthesis.

Zinc

  • Plants obtain zinc as Zn2+ ions.
  • It activates various enzymes, especially carboxylases.
  • It is also needed in the synthesis of auxin.

Copper

  • It is absorbed as cupric ions (Cu2+).
  • It is essential for the certain enzymes involved in redox reactions

Boron

  • It is absorbed as BO33- or B4O72-
  • It is required for uptake and utilisation of Ca2+
  • it helps in membrane functioning
  • it helps pollen germination
  • it helps cell elongation and cell differentiation
  • it is involved in carbohydrate translocation.

Molybdenum

  • Plants obtain it in the form of molybdate ions (MoO22+).
  • It is a component of nitrogenase and nitrate reductase both of which participate in nitrogen metabolism.

Chlorine

  • It is absorbed in the form of chloride anion (Cl).
  • Along with Na+ and K+, it helps in determining the solute concentration and the anion cation balance in cells.
    It is essential for the water-splitting reaction in photosynthesis, a reaction that leads to oxygen evolution.

Deficiency Symptoms of Essential Elements
If the concentration of the essential element below the critical concentration plants shows certain morphological changes. These are indications of deficiency symptoms.

Mobility of element determines deficiency symptoms
Deficiency symptoms in older tissues
Deficiency symptoms also depend on the mobility of the element in the plant. It first appears 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 and available for mobilising to younger leaves.

Deficiency symptoms in younger tissues
Sometimes the deficiency symptoms appear first in the young tissues. If the elements are immobile, they are not transported from mature organs to younger organs.
For example, Elements like sulphur and calcium are structural components of the cell and hence are not easily released.

The deficiency symptoms are

  1. Chlorosis
  2. Necrosis
  3. stunted plant growth
  4. premature fall of leaves and buds
  5. and inhibition of cell division.

Chlorosis is the loss of chlorophyll leading to yellowing in leaves. It is due to the deficiency of elements like N, K, Mg, S, Fe, Mn, Zn, and Mo.
Necrosis, or death of tissue, particularly leaf tissue. It is due to the deficiency of Ca, Mg, Cu, K.
Lack or low level of N, K, S, Mo causes inhibition of cell division.
Deficiency of elements like N, S, Mo delay flowering

Toxicity of Micronutrients
If the supply of micronutrients at a moderate decreased level shows deficiency symptoms but the moderate increase causes toxicity, i.e the excess of an element inhibits the uptake of another element.

Symptoms and other effects of Manganese toxicity

  • Symptom of manganese toxicity is the appearance of brown spots surrounded by chlorotic veins.
  • Manganese competes with iron and magnesium for uptake and for binding with enzymes.
  • Manganese also inhibits calcium translocation in the shoot apex.
  • Symptoms of manganese toxicity induce
  • Deficiency symptoms of iron, magnesium, and calcium.

Mechanism of Absorption of Elements
The process of absorption occurs in two main phases-

  1. Apoplast (passive). The passive movement of ions into the apoplast occurs through ion- channels and the transmembrane proteins.
  2. Symplast(active) The inward movement of ions into the cells is called influx and the outward movement efflux. This movement occurs by using metabolic energy.

Translocation of Solutes

  • Mineral salts are pulled up through the plant by the transpirational pull.
  • Analysis of xylem sap shows the presence of mineral salts in it.
  • Radioisotopic studies support the xylem transport of mineral elements.

Soil as a Reservoir of Essential Elements

  • Soil consist of a variety of minerals, nitrogen-fixing bacteria, and other microbes holds water and supplies air to the roots, and acts as a matrix that stabilises the plant.
  • If the amount of nutrients in the soil is decreased, it is supplied from outside as fertilizers in the form of macronutrients (N, P, K, S, etc.) and micronutrients (Cu, Zn, Fe, Mn, etc.)

Metabolism of Nitrogen
Nitrogen Cycle
Nitrogen is a constituent of amino acids, proteins, hormones, chlorophyll, and many vitamins.
Atmospheric nitrogen consists of two nitrogen atoms joined by a very strong triple covalent bond main nitrogen pools-atmospheric soil, and biomass.

Plus One Botany Notes Chapter 8 Mineral Nutrition 7

1. N2 Fixation: The process of conversion of atmospheric nitrogen (N2) to ammonia is termed nitrogen fixation.

2. Nitrification:

  • Ammonia is converted into nitrate.
  • Ammonia is first oxidized to nitrite by Nitrosomonas or Nitrococcus.
  • The nitrite is further oxidized to nitrate with the help of the bacterium Nitrobacter

3. Ammonification: Decomposition of organic nitrogen of dead plants and animals into ammonia is called ammonification

4. Denitrification: It is the conversion of soil nitrate into molecular N2 by Thiobacillus and pseudomonas

Formation of nitrogen oxides

  • In nature, lightning and UV provide energy to convert nitrogen to nitrogen oxides (NO, NO2, N2O).
  • Industrial combustions, forest fires, automobile exhausts, and power generating stations are also sources of atmospheric nitrogen oxides.

Biological Nitrogen Fixation
The nitrogen-fixing microbes are free-living or symbiotic. ‘Free-living nitrogen-fixing aerobic microbes are Azotobacter, Beijernickia Rhodospirillum Bacillus Anabaena Nostoc.

Plus One Botany Notes Chapter 8 Mineral Nutrition 3

Development of root nodules in soyabean

Plus One Botany Notes Chapter 8 Mineral Nutrition 4

Development of root nodule sin soyabean:

  • Rhizobium bacteria contact susceptible root hair, divide near it.
  • Upon successful infection of the root hair cause it to curl.
  • Infected thread carries the bacteria to the inner cortex. The bacteria get modified into rod-shaped bacteroids and cause inner cortical and pericycle cells to divide. Division and growth of cortical and pericycle cells lead to nodule formation.
  • A mature nodule is complete with vascular tissues continuous with those of the root.

Basic steps are given below

  • Rhizobium bacteria attach the root hair.
  • Root hair curls.
  • Infected thread carries the bacteria to the inner cortex.

The bacteria get modified into rod-shaped bacteroids and cause inner cortical and pericycle cells to divide. Division and growth of cortical and pericycle cells lead to nodule formation, d) A mature nodule is complete with vascular tissues continuous with those of the root.

Overall equation for N2 fixation
N2 + 8e + 8H+ +16ATP → 2NH3 + H2 + 16ADP + 16Pt

Fate of ammonia

  • At first, ammonia protonated to form NH4+.
  • This ammonium ion is used to synthesise amino acid in plants

There are two ways for the synthesis of amino acids in plants

1. Reductive animation
In this, ammonium ion reacts with alpha-ketoglutaric acid and forms glutamic acid.

Plus One Botany Notes Chapter 8 Mineral Nutrition 5

2. Transamination
It involves the transfer of an amino group from one amino acid to the keto group of a keto acid.
Glutamic acid is the main amino acid from which the transfer of amino groups takes place and other amino acids are formed in the presence of transaminase.

Plus One Botany Notes Chapter 8 Mineral Nutrition 6

Amides

  • The important amides are asparagine and aspartate.
  • Amide is formed when the hydroxyl group of one amino acid is replaced by an amino group.
  • Since amide contains more nitrogen than amino acids. They are transported through xylem vessels.

Plus Two Zoology Chapter Wise Previous Questions Chapter 8 Biodiversity and Conservation

Kerala State Board New Syllabus Plus Two Zoology Chapter Wise Previous Questions and Answers Chapter 8 Biodiversity and Conservation.

Kerala Plus Two Zoology Chapter Wise Previous Questions Chapter 8 Biodiversity and Conservation

Question 1.
Gopalan cultivated a variety of fruit crops and plants in this field and Raman destroyed the fruit crops and plants and cultivated rubber trees which are double in number. (MARCH-2010)
a) In your opinion which method is suitable for the ecosystem? Give reason.
b) Mention any three factors for the extinction of species.
Answer:
Method by Gopalan
Biodiversity depends on variety of species.  Extinction of species is due to

  1. Over exploitation
  2. Habitat loss and fragmentation
  3. Co-extinction

Question 2.
a) The “Evil Quartet” is the nickname used to describe the causes of biodiversity losses. Explain the reason leading to accelerated rates of extinction of flora and fauna.  (MAY-2010)
b) Philosophically or spiritually, we need to realize that every plant or animal species has an intrinsic value and we have a moral duty to protect them. Justify the statement and write down the protective measures.
Answer:
a) (1) Habitat loss and fragmentation
(2) Over-exploitation
(3) Alien species invasions
(4) Co-extinction
b) Agreed with the statement as we have a moral duty to care for their well-being and pass on our biological legacy in good order to future generations.
Protective measures-
Insitu conservation – Biosphere reserve, National park, Wild Life Sanctuary etc.
Exsitu conservation – Zoo, Botanical gardens, seed bank etc.

Question 3.
The year 2010 has been declared as the International Biodiversity Year by United Nations (UN)  (MARCH-2011)
a) Point out the levels of diversity in nature.(1 Score)
b) Give a brief description of The Evil Quartet.
Answer:
a) Genetic, Species, Ecosystem level diversity.
b) i) Habitat loss and fragmentation
ii) Overexploitation
iii) Alien species Invasion
iv) Co-extinction etc.

Question 4.
The given graph shows the distribution of insects in different latitudes of earth.  (MARCH-2012)
Plus Two Zoology Chapter Wise Previous Questions Chapter 8 Biodiversity and Conservation 1
a) What is your observation?
b) List the three reasons for greater biodiversity in tropical regions.
c) Write two causes of biodiversity lossess.
Answer:
a) Species richnes decreases from equator to poles.
b) Climate is constant and predictable Glaciations were absent. Tropical region get more solar energy.
c) Habital loss and fragmentation Invansion of alien species.

Question 5.
Last twenty years alone have witnessed the disappearance of 27 animal species from earth.  (MAY-2012)
a) Name an animal disappeared recently.
b) What may be the causes for this loss?
c) How can we conserve biodiversity?
Answer:
a) Out of Syllabus
b) i) Habitat loss and fragmentation
ii) Over-exploitation
iii) Alien species invasions
iv) Co-extinctions
c) It is through in situ (on site) conservation and ex situ (offsite) conservation

Question 6.
Gopalan cultivated a variety of fruit crops and plants . in this field and Raman destroyed the fruit crops and plants and cultivated rubbertrees which are double in number.  (MARCH-2013)
a) In your opinion which method is suitable for the ecosystem? Give reason.
b) Mention any three factors for the extinction of species.
Answer:
Method by Gopalan
Biodiversity depends on variety of species. Extinction of species is due to
1. Over exploitation
2. Habitat loss and fragmentation
3. Co-extinction

Question 7.
While preparing the species area relationship graph of 4 areas, the following z values are obtained.  (MAY-2013)
Area A = 0.1 Area B = 0.8 Area C = 1.2 Area D = 0.3
a) Which area shows maximum species richness?
b) What are the expected reasons for the loss of biodiversity in areas with low species richness?
Answer:
a) Area a = 0.1
b) (i) Habitat loss and fragmentation
(ii) Over-exploitation
(iii) Alien species invasions
(iv) Co-extinctions

Question 8.
“Nature provides all for the need of man but not for his greed.” (MARCH-2014)
a) Do you agree with this statement? Justify your answer.
b) Distinguish between two types of biodiversity conservations.
Answer:
a) yes, For example forest is used for some basic needs of a man but not for clearing of trees .
b) Exitu conservation – conservation of flora and fauna outside the natural habitat
Eg- Botanical garden
Insitu conservation – conservation of flora and fauna in the natural habitat
Eg-wild life sanctuaries and national parks

Question 9.
a) Variety of species are present around us, what they constitute and comment.  (MAY-2014)
b) Comment on in situ conservation and ex situ conservation.
c) In these aspects explain biodiversity hot spots with example – give importance to recent issues with regard to Western Ghats.
Answer:
a) Biodiversity
b) Exitu conservation – conservation of flora and fauna outside the natural habitat Insitu conservation- conservation.of flora and fauna in the natural habitat
c) Biodiversity hotspots are regions with very high levels of species richness and high degree of endemism Western ghats is the hotspots where accelerated habitat loss occurs.

Question 10.
We have a moral responsibility to take good care of earth’s biodiversity and pass it on in good order to next generation.  (MARCH-2015)
a) Define Biodiversity.
b) Write causes for biodiversity losses.
c) Name two types of biodiversity conservation.
Answer:
a) It is the variety within and between all species of plants, animals and micro-organisms and the ecosystems within which they live and interact.
b) i) Habitat loss and fragmentation
ii) Over-exploitation
iii) Alien species invasions
iv) Coextinctions
c) Insitu conservation, Exitu conservation

Question 11.
Two approaches for the conservation of biodiversity is shown as A and B.  (MAY-2015)
Plus Two Zoology Chapter Wise Previous Questions Chapter 8 Biodiversity and Conservation 2
a) Identify the type of biodiversity conservation shown in A and B.
b) Write the difference between the two, types of biodiversity conservation shown in A and B.
c) Which of the above approach is more desirable’ when there is an urgent need to save a species?
Answer:
a) A-Insitu conservation B-Exitu conservation
b) A- It is the conservation of animals in natural habitat
B- It is the conservation of plants outside the natural habitat
c) exitu conservation

Question 12.
Observe the concept diagram of the Evil Quartet of biodiversity loss. (MARCH-2016)
Plus Two Zoology Chapter Wise Previous Questions Chapter 8 Biodiversity and Conservation 3
a) Write A and B
b) What is Co-Extinction?
Answer:
a) A- Habitat loss and fragmentation
B -Alien species invasions
b) When a species becomes extinct, the plant and animal species associated with it also become extinct

Question 13.
Read the statement and choose the correct option:   (MARCH-2016)
A: Sacred grooves are examples of in situ conservation
B: Biodiversity hotspots have low degree of endemism.
C: Biodiversity increases when number of organisms in a particular species increases.
a) Statement ‘A’ alone is correct
b) Statement ‘A’ and ‘B’ are correct
c) Statement ‘A’ and ‘C’ are correct.
d) Statement ‘C’ alone is correct
Answer:
Statement ‘A’ alone is correct.

Question 14.
a) “When we conserve and protect the whole ecosystem, its biodiversity at all levels is protected.” Based on this statement explain the strategies of biodiversity conservation.  (MAY-2016)
OR
b) “When need turns to greed, it leads to biodiversity loss.” Substantiate this statement by explaining two causes of biodiversity loss.
Answer:
a) Insitu conservation- It is the conservation of plant and animal sp. in their natural habitat.
Eg-biosphere reserves, national parks and sanctuaries.
Exitu conservation- It is the conservation of plant and animal sp outside the natural habitat.
Eg- Zoological parks, botanical gardens and wildlife safari parks.
OR
b) i) Habitat loss and fragmentation: The degradation of many habitats by pollution affects the survival of many species. It results large habitats are broken up into small fragments.
Amazon rain forest is cleared for cultivating soya beans or for conversion to grasslands for raising beef cattle
ii) Over-exploitation : It leads to the over-exploitation of natural resources. For example the extinction of Steller’s sea cow, passenger pigeon was due to humans.
iii) Alien species invasions : The introduction of foreign species cause the reduction or extinction of indigenous species.
The Nile perch introduced into Lake Victoria in east Africa led to the extinction of more than 200 species of cichlid fish in the lake
iv) Co-extinctions : If two species are in obligatory relationship the extinction of one species affect the other.
Eg – coevolved plant-pollinator mutualism where extinction of one species leads to the extinction of the other.

Question 15.
Z – values of a frugivorous bat species are given below. Which value is not applicable to continents?  (MARCH-2017)
1) 0.6
2) 0.65
3) 0.20
4) 0.68
Answer:
Incorrect option

Question 16.
Distinguish in situ conservation from ex situ conservation with one example each.  (MARCH-2017)
Answer:
Insitu conservation- Ms the conservation of plant and animal sp. in natural habitat.
Eg- biosphere reserves, national parks and sanctuaries.
Exitu conservation- It is the conservation of plant and animal sp outside the natural habitat.
Eg- Zoological parks, botanical gardens and wildlife safari parks.

Question 17.
What are the advantages of biofertilizers over chemical fertilizers? Give an example for biofertilizer.  (MARCH-2017)
Answer:
a) 1) It prevents pollution
2) It improves soil structure and function
b) biofertilizer- Mycorrhiza

Question 18.
Explain the three levels of biodiversity. (MAY-2017)
OR
Explain different types of biodiversity conservation with example.
Answer:
1. Genetic diversity – A single species might show high diversity at the genetic level over its distributional range
2. Species diversity – Diversity at species level
3. Ecological diversity – Diversity at ecosystem level
OR
In situ conservation – the species are protected in their natural habitat.
example: National Park, Wildlife sanctuaries etc.
ex situ conservation – threatened animals and plants are protected outside their natural habitat.
example: Zoological Park, botanical gardens etc.

Plus Two Zoology Chapter Wise Previous Questions Chapter 7 Microbes in Human Welfare

Kerala State Board New Syllabus Plus Two Zoology Chapter Wise Previous Questions and Answers Chapter 7 Microbes in Human Welfare.

Kerala Plus Two Zoology Chapter Wise Previous Questions Chapter 7 Microbes in Human Welfare

Question 1.
Ramu cultivated pea plants as an intercrop in his paddy field. After harvesting, he allowed the roots of the pea plants remain in the soil for some period. (MARCH-2010)
a) Do you think the action of Ramu is reasonable?
b) Justify your answer.
Answer:
Yes.
Rhizohium found in the root nodules in the pea plant can fix atmospheric nitrogen into nitrates and in-creases fertility of soil.

Question 2.
Match column I with II. (MAY-2010)

Column IColumn II
1. Methano bacteria1. Plague
2. Bacillusthuringenesis2. Cyclosporin A
3. Azo Spirillum3. Gobar gas production
4. Trichoderma poly4. Bio Control sporum
5. Citric acid production
6. Bio fertilizer

Answer:

Column IColumn II
1. Methano bacteria1. Gobar gas production
2. Bacillus thuringenesis2. Bio Control
3. Azo Spirillum3. Bio fertilizer
4. Trichoderma4. Cyclosporin A poly sporum

Question 3.
A bacterial infection was effectively controlled by using a specific anitibiotic for a long time. But nowadays this antibiotic is not found to be so effective to control the said infection. (MARCH-2011)
Give a scientific explanation for this phenomenon based on evolution.
Answer:
Evidence for Natural selection and explanations like origin of antibiotic resistant varieties or elimination of sensitive varieties or Natural selection by Anthropo-genic action.

Question 4.
Rearrange the columns B & C with respect to A. (MARCH-2012)

ABC
Monascus

purpureus

StreptokinaseAntibiotic
StreptococcusStatinImmuno

suppressant

Penicillium

notatum

Cyclosporin-AClot buster
Trichoderma

polysporum

PenicillinCholesterol

lowering

agent

Answer:
Monascus – Statin – Cholesterol
Streptococcus – Streptokinase – Cholestrol lowering
Pencillium – Penicillin – Antibiotic
Trichoderma – Cyclosporin-A – Immuno suppressant

Question 5.
Match the following (MAY-2012)
Plus Two Zoology Chapter Wise Previous Questions Chapter 7 Microbes in Human Welfare 2
Answer:
(A) – (3)
(B) – (2)
(C) – (1)
(D) – (5)

Question 6.
Some bioactive molecules, their source and their medical impotance are given in the table below. Fill up the missing parts. (MARCH-2013)

Bioactive moleculeSourceMedical importance
astreptococcusRemoves clots from blood vessels
Cyclosporin Abc
dMonascus purpureusBlood cholesterol lowering agent

Answer:
a) streptokinase
b) Trichoderma polysporum
c) Immunosuppressant in organ transplant patients
d) Statins

Question 7.
Complete the illustration appropriately. (MAY-2013)
Plus Two Zoology Chapter Wise Previous Questions Chapter 7 Microbes in Human Welfare 1
Answer:
a) Biofertilisers
b) Yeast,ethanol production,
c) (1) Lady bird and dragon flies are useful in the elimination of aphids and mosquitoes.
(2) Bacillus thuringiencis
d) Dough for making dosa and idly is fermented by Bacteria.

Question 8.
The meaning of ‘antibiotics’is ‘against life’, whereas with reference to human beings they are ‘pro life’. Substantiate the statement with suitable example. (MARCH-2014)
Answer:
Penicillin was widely used to kill bacteria during infection.
During second world war,it was widely used to treat soldiers against bacterial infection

Question 9.
In our state waste managenfient is a problem. Government promotes and give subsidy to Biogas plants. Comment the functioning of biogas plants with the help of microbes. (MAY-2014)
Answer:
Methanobacterium is used to produce biogas and can be used as source of energy as it is inflammable. It is an anaerobic bacterium used in sludge digesters.

Question 10.
Microbes can also be used as a source of energy. (MARCH-2015)
Substantiate with suitable examples.
Answer:
Methanobacterium is used to produce biogas and can be used as source of energy as it is inflammable.

Question 11.
BOD of some water samples are given below: (MAY-2015)
A. Sample 1 – 200 mg/L
B. Sample 2 – 80 mg/L
C. Sample 3 – 300 mg/L
D. Sample 4 – 25 mgIL
a) Which of the above water sample is most polluted?
b) What is meant by floes ? What is its role in sewage treatment?
Answer:
a) Sample 4
b) Floes- masses of bacteria associated with fungal filaments to form mesh like structures.
They consume the major part of the organic matter in the effluent. This is significantly reduces the BOD of the effluent.

Question 12.
“BOD is commonly calculated as an index of water pollution”. (MARCH-2016)
a) Do you agree with this statement? Why?
b) Expand BOD.
Answer:
a) Yes, if pollution load increases ,BOD increases
b) biochemical oxygen demand

Question 13.
Choose the correct answer from the bracket. (MAY-2016)
Cyclosporin A is produced by ______
a) Aspergellus
b) Clostridium
c) Trichoderma
d) Acetobacter
Answer:
Trichoderma

Question 14.
Select a bio-control agent from the given microbes: (MAY-2016)
a) Baculo virus
b) Rhino virus
c) Picrona virus
d) Adenovirus
Answer:
a) Baculovirus

Question 15.
Complete the table by filling A, B, C and D using hints from the bracket:  (MAY-2017)
(Gobargas, Biological Control, Anabaena, Saccharomyces cerevisiae, Propionibacterium
sharmanii)
Plus Two Zoology Chapter Wise Previous Questions Chapter 7 Microbes in Human Welfare 3
Answer:
A -Gobargas
B – Saccharomyces cerevisiae
C – Anabaena
D-Biological control

Plus Two Zoology Chapter Wise Previous Questions Chapter 6 Human Health and Disease

Kerala State Board New Syllabus Plus Two Zoology Chapter Wise Previous Questions and Answers Chapter 6 Human Health and Disease.

Kerala Plus Two Zoology Chapter Wise Previous Questions Chapter 6 Human Health and Disease

Question 1.
“More and more children in Metro cities of India suffer from allergies and asthma.” (MARCH-2010)
a) Do you agree with this statement?
b) Justify your answer.
Answer:
a) Yes
b) Absence of previous encounter with the allergies in the over protected environment reduces immunity and children become more prone to allergy/ over exposure to allergens or pollutants in the metro cities can be considered.

Question 2.
Widespread incidence of diseases like H1H1, Chikungunya, dengue fever etc. are reported recently. As a science student, prepare an action plan in your school to control those diseases. (MARCH-2010)
Answer:
Measures to control mosquitoes, precaution to prevent the spread of disease. Awareness campaign for students etc.

Question 3.
The flow chart given below represents the life cycle of malarial parasite. Complete the flow chart and answer the following: (MAY-2010)
Plus Two Zoology Chapter Wise Previous Questions Chapter 6 Human Health and Disease 1
a) Which species of malarial parasite causes malignant malaria ?
b) Can you suggest two methods to control malaria?
Answer:
2 – Liver, 4 – RBC
a) Plasmodium falci parum
b) 1. Public hygiene and proper waste disposal
2. Periodic cleaning and disinfection of water reservoirs
3. Control of mosquitoes and their breeding grounds (any methods to control insect vectors can be added)

Question 4.
The list given below includes various stages of HIV infection. Arrange them according to correct sequence. (MAY-2010)
Answer:
Viral replication – Macrophage – Helper T cell – Formation of DNA – Progeny Virus – Viral particle – Release of progeny virus
Plus Two Zoology Chapter Wise Previous Questions Chapter 6 Human Health and Disease 2

Question 5.
Lakshmi and Sujitha are two Nursery students. Sujitha gets common cold very often. Lakshmi not (MARCH-2011)
a) How do you interpret this in an immunological aspects?
b) What are the common barriers protecting Lakshmi from cold?
Answer:
a) Low level of innate or weak or poor immunity to Sujitha / High immunity to Lakshmi.
b) Two correct points (Physical, Physiological, Cellular and Cytokine barriers) like mucous membrane / Nostril hairs / skin / Antibodies / Interferons.

Question 6.
Arrange the following diseases in the following columns in a meaningful order.  (MARCH-2012)
Typhoid, Ringworms, Amoebiasis, AIDS, Malaria, Pneumonia, Common cold.
Plus Two Zoology Chapter Wise Previous Questions Chapter 6 Human Health and Disease 3
Answer:
Plus Two Zoology Chapter Wise Previous Questions Chapter 6 Human Health and Disease 4

Question 7.
In a classroom discussion a student argues that allergic diseases are more common in children of metrocities than in villages. ,  (MARCH-2012)
a) Do you agree with this statement.
b) Which type of immunoglobulin is responsible for allergic reactions?
c) Suggest two drugs which reduce allergic symptoms.
Answer:
a) Yes
b) IgE
c) Antilistamine and Adrenalin drugs

Question 8.
Note the relation between first two terms and suggest a suitable term for the fourth place. (MAY-2012)
a) Erythroxylum coca : cocaine :
Papaversomniferum : _______
b) Salmonella typhi: typhoid :
Plasmodium falciparum: ______
Answer:
a) morphine
b) malaria

Question 9.
Nature has as many varieties of plants which give drugs for abuse, as there are medicinal plants which give medicines. Substantiate with two examples. (MARCH-2013)
Answer:
a) Cocaine is obtained from coca plant Erythroxylum coca. It affects the transport of the neuro-transmitter dopamine.
b) Opioids are the drugs, receptors present in our central nervous system and gastrointestinal tract Heroin (smack the depressant,it is prepared by the acetylation of morphine which is extracted from the latex of poppy plant Papaversomniferum.

Question 10.
“Prevention is better than cure”. This statement is true in the case of AIDS as well as immunisation. Substantiate. (MAY-2013)
Answer:
Diseases like AIDS are not curable even after the treatment for long period. This is the immuno deficiency disease affect the resistant power of the body. Medicine for these diseases are not discovered so far. Likewise immunization is done in earlier period of growth is good .Hence prevention is better than cure.

Question 11.
Classify the diseases given in the box as two groups based on their causative organisms. Specify the type of causative organism for each group. (MARCH-2014)
Plus Two Zoology Chapter Wise Previous Questions Chapter 6 Human Health and Disease 5
Answer:
protozoans – Malaria and amoebiasis Bacterial — diphtheria, typhoid and pneumonia

Question 12.
Prepare a pamphlet for an awareness programme in your school about the measures to prevent and control alcohol and drug abuse in adolescents. (MARCH-2014)
Answer:
Education and counseling, moral education, Avoid undue peer pressure, visual publicity through TV, seeking professional and medical help etc.

Question 13.
Briefly describe the characteristics of cancer cells. (MAY-2014)
Answer:
1. lose of contact inhibition- Normal cells are contact with other cells inhibits their uncontrolled growth. Cancer cells lost this property.
2. Metastasis- Cells sloughed from tumors reach distant sites through blood, and wherever they get lodged in the body, they start a new tumor there.

Question 14.
It is said that “Chikunguinea” once affected will not affect a person in the next half of his life. Justify this statement. (MAY-2014)
Answer:
This is due to acquired immunity. Due to infection body first time produces a response called primary response which is of low intensity. Subsequent encounter with the same pathogen a highly intensified secondary response. This is appears to have memory of the first encounter. The primary and secondary immune responses are carried out with the help of two special types of lymphocytes present in our blood, i.e. B-lymphocytes and T- lymphocytes.

Question 15.
Mother’s milk is considered essential for new born infants. (MARCH-2015)
a) Name the fluid secreted by mother from breast during the initial days of lactation.
b) What type of immunity it provides?
Answer:
a) Colostrum
b) IgA provides passive immunity

Question 16.
Cancer is one of the most dreaded diseases of human beings, and is a major cause of death all over the globe. (MARCH-2015)
a) What are the causes of cancer?
b) What are the methods of detection of cancer?
c) What are the types of treatment for cancer?
Answer:
a) Ionising radiations like X – rays and gamma rays and non-ionizing radiations like UV cause DNA damage leading to neoplastic transformation. The chemical carcinogens present in tobacco smoke have been identified as a major cause of lung cancer.

b) Biopsy and histopathological studies of the tissue and blood and bone marrow tests for increased cell counts in the case of leukemias Radiography (use of X-rays), CT (computed tomography) and MRI (magnetic resonance imaging) are very useful to detect cancers of the internal organs

c) 1. Surgery, radiation therapy and immuno therapy
2. Chemotherapeutic drugs are used to kill cancerous cells

Question 17.
Match the terms given in the three columns of the table correctly: (MAY-2015)
Plus Two Zoology Chapter Wise Previous Questions Chapter 6 Human Health and Disease 6
Answer:
Haemophilus Influenzae – Bacteria – pneumonia
Plasmodium Vivax – protozoa – malaria Wuchereria
Bancrofli – flat worm – filarisis
Trichophyton – fungus – ringworm

Question 18.
Identify the disease shown in the following figure and write the causative organism of the disease.(MARCH-2016)
Plus Two Zoology Chapter Wise Previous Questions Chapter 6 Human Health and Disease 7
Answer:
Elephantiasis orfilariasis

Question 19.
“Blood of a man is tested positive for cannabinoid.” (MARCH-2016)
a) What are these?
b) From where these are extracted naturally?
c) Which part of the body is affected by these?
Answer:
a) Drug obtained from plants
b) Natural cannabinoids are obtained from the inflorescences of the plant Cannabis sativa
c) Cardiovascular system of the body

Question 20.
Breast feeding during initial period of infant growth is necessary to develop immunity of new born babies. Why? (MARCH-2016)
Answer:
Initial days of lactation contains colostrum which is rich in antibodies and provide immunity to new borne babies.

Question 21.
Answer the questions about the given figure: (MAY-2016)
Plus Two Zoology Chapter Wise Previous Questions Chapter 6 Human Health and Disease 8
a) Identify the parts X and Y.
b) Name any two types of this molecule.
Answer:
a) X-antigen binding site
Y-Heavy chain
b) lgA, lgG, lg E

Question 22.
Select the odd one out and justify your selection. (MAY-2016)
Malaria, Gonorrhea, Amoebiasis, Filariasis
Answer:
Gonorrhoea – It is the only sexually transmitted disease

Question 23.
Complete the table by filling a, b, c and d. (MAY-2016)
Plus Two Zoology Chapter Wise Previous Questions Chapter 6 Human Health and Disease 9
Answer:
a) pneumonia
b) Rhino virus
c) malaria
d) Inflammation of lymphatic vessels of lower limb/ genital organs

Question 24.
Feeding _______ in the first few days is essential for preventing infections in a newly born baby. (MARCH-2017)
Answer:
colostrum

Question 25.
Morphine is said to be an abused drug. Discriminate the terms ‘use’ and ‘abuse’ of drugs based on this example. (MARCH-2017)
Answer:
For medical purpose it is used as painkiller or sedative but it is abused by some individuals as narcotic drug.

Question 26.
Differentiate Active immunity from Passive immunity. Give an example for Passive immunity. (MARCH-2017)
Answer:
When the antigens are coming in the form of living or dead microbes, the body of organism produce antibodies. This type of immunity is called active immunity.
When ready – made antibodies are directly injected to protect the body against foreign agents, it is called passive immunity.
Eg-Anti-venom.

Question 27.
Prepare a brief note to be presented in an awareness programme for adolescents about AIDS, their causes and preventive measures. (MAY-2017)
Answer:
AIDS-it is acquired immuno deficiency syndrome caused by virus called HIV It is mainly prevented by

Avoid sex with multiple partners
Use disposable needles and syringes
Ensure safe blood transfusion

Question 28.
Fill the boxes A, B, C and D. (MAY-2017)
Plus Two Zoology Chapter Wise Previous Questions Chapter 6 Human Health and Disease 10
Answer:
A – Acquired immunity
B – Physiological barrier
C -Cytokin barrier
D-passive immunity

Plus Two Zoology Chapter Wise Previous Questions Chapter 5 Evolution

Kerala State Board New Syllabus Plus Two Zoology Chapter Wise Previous Questions and Answers Chapter 5 Evolution.

Kerala Plus Two Zoology Chapter Wise Previous Questions Chapter 5 Evolution

Question 1.
Arrange the following in a hierarchical manner based on the period of their evolution. (MARCH-2010)
Homoerectus, Ramapithicus, Australopithicines, Homosapiens, Neanderthal man.
Answer:
Ramapithicus -> Australopithicines -> Homoerectus -> Neanderthalman -> Homosapiens

Question 2.
Fill the columns A and B using the items given below: (MARCH-2010)

Column AColumn B
Developmenf of DDT resistant ants
Adaptive radiation
Genetic drift
Inheritance of acquired characters

(Lamarkism, Evolution of anthropogenic action, Geneflow by chance, varieties of marsupials in Australia, De Vries)
Answer:

Column AColumn B
Development of DDT resistant antsEvolution of anthropogenic action
Varieties of Marsupials in AustraliaAdaptive radiation
Gene flow by chanceGenetic drift
Inheritance of acquired charactersLamarkism

Question 3.
The diagrams shown below represents the operation of natural selection on different traits. Observe the diagrams and answer the following : (MAY-2010)
Plus Two Zoology Chapter Wise Previous Questions Chapter 5 Evolution 1
a) Why graph C shows a marked difference from graph A.
b) What is the evolutionary significance of Directional Selection?
c) Mention the factors affecting gene frequency.
d) What is meant by founder effect?
Answer:
a) A is stabilizing selection and C is disruptive selection.
b) Peak shifted to one direction.
c) Gene migration
Genetic drift
Mutation
Genetic recombination
Natural selection
d) The effect where original drifted population becomes founders is called founder effect.

Question 4.
a) Arrange the given chemical compound in the sequential order as per the concept of origin of life. (Ammonia, Hydrogen, Protein, Nucleic acid, Amino acid) (MARCH-2011)
b) Correlate Miller’s experiment with this.
Answer:
a) Hydrogen -> Ammonia -> Amino acid -> Protein Nucleic acid
OR
b) Brief account of Miller’s Experiment / Labelled diagram of the experimental setup.

Question 5.
Note the relationship between the first pair and complete the second pair. (MARCH-2012)
a) Natural selection : Drawing ; Inheritance of acquired characters : ________.
b) Heart of vertebrates: homologous organs; Flippers of Penguin and Dolphin : _______.
Answer:
a) Lamarck
b) Analogous organs

Question 6.
An evolutionary process occurred in the evolution of marsupial mammals in Australia is given below. (MARCH-2012)
Plus Two Zoology Chapter Wise Previous Questions Chapter 5 Evolution 2
a) Name this evolutionary process.
b) Suggest another example for this phenomenon.
Answer:
a) Adaptive radiation
b) Darwins finches

Question 7.
Arrange the following examples under two heads viz- homologous organs and analogous organs. (MARCH-2013)
Forelimb of whale and bat,
Wings of butterfly and bat,
Heart of man and cheetah,
Eyes of Octopus and mammals
Answer:
Homologous organs
Forelimb of whale and bat Heart of man and cheetah
Analogous organs
Wings of butterfly and bat Eyes of octopus and mammals

Question 8.
Theory of chemical evolution is a version of theory of abiogenesis. Analyze the statement. (MARCH-2013)
Answer:
Oparin and Haldane proposed that the first form of life that arose from preexisting non-living organic molecules (e.g. RNA, protein, etc.) and it is followed by chemical evolution

Question 9.
A specific rat population was controlled for about a decade by a poison. After apopulation decline for about 10 years, the rat population was increased stabilized. (MAY-2013)
Plus Two Zoology Chapter Wise Previous Questions Chapter 5 Evolution 3
Resistance to poison is governed by a dominant autosomal gene ‘R’. In 1975 majority of the resistant animals are heterozygous at this locus (Rr).
a) What was the major genotype of the rat population before 1961 ?
(A) RR, (B) Rr, (C) rr, (D) R is absent as it is produced by a mutation.
b) What explanation you give for the development of resistance against poison in these rats?
c) This illustration can be utilized to explain a theory of Evolution” Substantiate.
Answer:
a) R is absent as it is produced by mutation
b) Mutation leads to genetic variation and which in turn lead to evolution
c) Evolution by anthropogenic action

Question 10.
Given below is the diagrammatic representation of the operation of Natural Selection on different traits. (MARCH-2014)
Plus Two Zoology Chapter Wise Previous Questions Chapter 5 Evolution 4
a) Identify the type of natural selection A, B and C with explanation of each.
b) Define Hardy – Weinberg Principle.
Answer:
a) A- Stabilising selection-More individuals acquire mean character value
B – Directional selection – More individuals ac-quire value other than mean character
C – Disruptive selection – More individuals acquire peripheral character value at both ends.
b) According to Hardy-Weinberg principle allele frequencies in a population are stable and is con-stant from generation to generation. This is called genetic equilibrium. Sum total of all the allelic frequencies is 1.
Hence,p2 + 2pq + q2 = 1 Disturbance in Hardy- Weinberg equilibrium, i.e., change of frequency of alleles in a population affected by five factors These are gene migration or gene flow, genetic drift, mutation, genetic recombination and natural selection.

Question 11.
Arrange the following in a hierarchical manner in ascending order based on the period of their evolution.  (MAY-2014)Homoerectus, Ramapithecus, Australopithecus, Homo sapiens.
Answer:
Ramapithecus —> Australopithecus—>Homoerectus —> Homo sapiens

Question 12.
a) The diagram given below shows a particular type of evolutionary process in Australian marsupials. Identify the’ evolutionary phenomenon and comment on. (MAY-2014)
Plus Two Zoology Chapter Wise Previous Questions Chapter 5 Evolution 5
b) Give another example for such a type of evolutionary process and explain.
Answer:
a) convergent evolution
Number of marsupials are different from the other evolved from an ancestral stock, but all within the Australian island continent. When more than one adaptive radiation appeared to have occurred in an isolated geographical area(representing different habitats), it is called as convergent evolution.
b) Placental mammals in Australia also exhibit adaptive radiation in evolving into varieties of such placental mammals each of which appears to be ‘similar’ to a corresponding marsupial (e.g., Placental wolf and Tasmanian wolf marsupial).

Question 13.
Match the following: (MARCH-2015)

a) Natural selection1) Convergent evolution
b) Inheritance of acquired2) Genetic drift characters
c) Analogous structures3) Charles Darwin
d) Gene flow by chance4) Lamarkism

Answer:

a) Natural selection3) Charles Darwin
b) Inheritance of acquired4) Lamarkism
c) Analogous structures1) Convergent evolution
d) Gene flow by chance2) Genetic drift characters

Question 14.
Plus Two Zoology Chapter Wise Previous Questions Chapter 5 Evolution 6
The above shown pictures are beaks of a particular type of bird seen in an island during Darwin’s journey.  (MARCH-2015)
a) Identify the bird and name the island.
b) Write the significance of this process in evolution.
Answer:
a) Darwin’s Finches, Galapagos Islands
b) From the original seed-eating features, many other forms with altered beaks arose, enabling them to become insectivorous and vegetarian finches.

Question 15.
Four groups of organs are given below:  (MAY-2015)
Read them carefully and answer the questions:
A. Thorns of bougainvilla and Tendrils of cucurbita
B. Eyes of octopus and mammals
C. Flippers of penguin and dolphin
D. Forelimbs of cheetah and man
(a) Categorise the four groups of organs as homologous organs and analogous organs.
(b) Based on each group of organs differentiate convergent evolution and divergent evolution.
(c) Illustrate homologous and analogous organs as evidences for evolution.
OR
Observe the diagrammatic representation and answer the questions:
Plus Two Zoology Chapter Wise Previous Questions Chapter 5 Evolution 7
a) Explain the phenomenon shown in the figure.
b) How can it be considered as an evidence of evolution?
c) Write any other example for this phenomenon. Explain.
Answer:
a) Homologous organs
Thorns of bougainvilla and tendrils of cucurbits , fore limbs of cheetah and man Analogous organs: Eye of octopus and mammals, flippers of penguin and dolphin
b) Convergent evolution – different structures evolving for the same function
Divergent evolution – same structure developed for different function
c) In convergent evolution, the similar habitat that has resulted in selection of similar adaptive fea-tures in different groups of organisms but toward the same function:
In divergent evolution ,same structure developed along different directions due to adaptations to different needs
OR
a) Adaptive radiation
b) A number of marsupials, each different from the other evolved from an ancestral stock, but all within the Australian island continent. When more than one adaptive radiation appeared to have occurred in an isolated geographical area (representing different habitats), it represents convergent evolution.
c) Darwin’s finches represent one of the best examples of this phenomenon. From the original seed-eating features, many other forms with altered beaks arose, it helps them to become insectivorous and vegetarian finches.

Question 16.
Which theory talks about the huge explosion that leads to origin of universe?  (MARCH-2016)
Answer:
Big Bang theory

Question 17.
Read the principle and answer the questions: “Allele frequencies in a population are stable and constant from generation to generation called genetic equilibrium.’ (MARCH-2016)
a) Name the principle mentioned here.
b) Mention any two factors affecting the equilibrium.
c) What is the significance of disturbances occur in the genetic equilibrium?
OR
’Natural selection can lead to stabilisation, directional change and disruptive changes.’
Explain the terms stabilization, directional change and disruptive change mentioned above.
Answer:
a) Hardy-Weinberg principle
b) gene flow and genetic drift
c) they become a different species
OR
Natural selection can lead to 1 stabilisation (in which more individuals acquire mean character value)
2. directional change (more individuals acquire value other than the mean character value)
3. disruption (more individuals acquire peripheral character value at both ends of the distribution curve)

Question 18.
Observe the diagram and answer the questions below: (MAY-2016)
Plus Two Zoology Chapter Wise Previous Questions Chapter 5 Evolution 8
a) Identify the types of evolution in the concept diagrams A and B.
b) Write example pair each for homologous and analogous organs.
Answer:
A – Divergent evolution
B – Convergent evolution
Homologous organs- Cheetah and human – bones of forelimbs, thorn of Bougainvillea and tendrils of cucurbits.
Analogous organs – eye of the octopus and of mammals and flippers of Penguins and Dolphins.

Question 19.
Statement below show the features of some human fossils. Read carefully and identify the fossil. (MAY-2016)
a) Human like beings with brains capacities between 650 – 800 cc
b) Lived in East and Central Asia with brain capacity of 1400 cc.
Answer:
a) Homohabilus
b) Neaderthal man

Question 20.
A population of 208 people of MN blood group was sampled and it was found that 119 were MM group, 76 MN group and 13 NN group. Answer the following questions: (MARCH-2017)
a) Determine the gene frequencies of M and N alleles in the population.
b) How does the above frequencies affect evolution?
OR
Examine the pictures of Darwin’s Finches given below and answer the following questions:
a) What phenomenon in evolution is represented in the picture?
b) Explain the phenomenon with the help of an additional example.
Plus Two Zoology Chapter Wise Previous Questions Chapter 5 Evolution 9
Answer:
a) Adaptive radiation –In this evolution starting from a point and radiating to other areas of geography
b) A number of different marsupials evolved from an ancestral stock within the Australian island.
Plus Two Zoology Chapter Wise Previous Questions Chapter 5 Evolution 10

Question 21.
Which of the following sets of gases were used in Mliller’s experiment? (MARCH-2017)
1) \(\mathrm{CH}_{4}, \mathrm{NO}_{2}, \mathrm{H}_{2} \mathrm{O}, \mathrm{CO}_{2}\)
2) \(\mathrm{NH}_{3}, \mathrm{CH}_{3}, \mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2}\)
3) \(\mathrm{NH}_{3}, \mathrm{CH}_{3}, \mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2}\)
4) \(\mathrm{H}_{2} \mathrm{O}, \mathrm{N}, \mathrm{CH}_{4}, \mathrm{H}_{2}\)
Answer:
2) \(\mathrm{NH}_{3}, \mathrm{CH}_{3}, \mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2}\)

Question 22.
Diagrammatic representation of the operation of Natural Selection on different traits is given. Observe it and answer the questions: (MAY-2017)
Plus Two Zoology Chapter Wise Previous Questions Chapter 5 Evolution 11
a) What do B and C represent?
b) Explain the process shown in B and C.
Answer:
a) B is directional selection C is disruptive selection
b) B natural selection leads to directional change in which more individuals acquire value other than the mean character value.
C natural selection leads to disruption in which more individuals acquire peripheral character value at both ends of the distribution curve.

Question 23.
Rearrange the following in the order of their evolution period (MAY-2017)
Australopithecines
Neanderthal man
Homo sapiens
Homo erectus
Dryopithicus
Answer:
Dryopithecus —> australopithecus —> Homo erectus —> neanderthal man —> Homo sapiens