Prasanna

Punjab State Board PSEB 12th Class Physics Book Solutions Chapter 15 Communication Systems Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Physics Chapter 15 Communication Systems

PSEB 12th Class Physics Guide Communication Textbook Questions and Answers

Question 1.
Which of the frequencies will be suitable for beyond-the-horizon communication using sky waves?
(a) 10 kHz
(b) 10 MHz
(C) 1 GHz
(d) 1000 GHz
Answer:
(b) 10 MHz.
For beyond-the-horizon communication, it is necessary for the signal waves to travel a large distance. 10 kHz signals cannot be radiated efficiently because of the antenna size. The high-energy signal waves (1 GHz – 1000 GHz) penetrate the ionosphere. 10 MHz frequencies get reflected easily from the ionosphere. Hence, signal waves of such frequencies are suitable for beyond-the-horizon communication.

Question 2.
Frequencies in the UHF range normally propagate by means of:
(a) Ground waves
(b) Sky waves
(c) Surface waves
(d) Space waves
Answer:
(d) Space waves
Owing to its high frequency, an ultra-high frequency (UHF) wave can neither travel along the trajectory of the ground nor can it get reflected by the ionosphere. The signals having UHF are propagated through line-of-sight communication, which is nothing but space wave propagation.

Question 3.
Digital signals
(i) do not provide a continuous set of values,
(ii) represent values as discrete steps,
(iii) can utilize binary system, and
(iv) can utilize decimal as well as binary systems.
Which of the above statements are true?
(a) (i) and (ii) only
(b) (ii) and (iii) only
(c) (i), (ii) and (iii) but not (iv)
(d) All of (i), (ii), (iii) and (iv)
Answer:
(c) A digital signal uses the binary (0 and 1) system for transferring message signals. Such a system cannot utilize the decimal system (which corresponds to analogue signals). Digital signals represent discontinuous values.

Question 4.
Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line-of-sight communication? A TV transmitting antenna is 81 m tall. How much service area can it cover if the receiving antenna is at the ground level?
Answer:
Line-of-sight communication means that there is no physical obstruction between the transmitter and the receiver.
In such communications, it is not necessary for the transmitting and receiving antennas to be at the same height.
Height of the given antenna, h = 81 m
Radius of earth, R = 6.4 x 10⁶ m

For range, d = 2Rh, the service area of the antenna is given by the relation
A = πd² = π(2Rh)
= 3.14 x 2 x 6.4 x 10⁶ x 81
= 3255.55 x 10⁶ m²
= 3255.55 ~ 3256 km².

Question 5.
A carrier wave of peak voltage 12 V is used to transmit a message signal.
What should be the peak voltage of the modulating signal in order to have a modulation index of 75%?
Answer:
Amplitude of the carrier wave, A_c =12,
Modulation index, m = 75% = 0.75
Amplitude of the modulating wave = A_m

Using the relation for modulation index,
m = \(\frac{A_{m}}{A_{c}}\)
∴ A_m =mA_c
= 0.75 x 12 = 9 V
Hence, amplitude of the modulating wave is 9 V.

Question 6.
A modulating signal is a square wave, as shown in Fig. 15.14.
The carrier wave is given by c(t) = 2 sin (8πt) volts.

(i) Sketch the amplitude modulated waveform.
(ii) What is the modulation index?
Answer:
Given, the equation of carrier wave
c(t) = 2sin(8πt) ……………………………… (i)
(i) According to the diagram.
Amplitude of modulating signal, A_m = 1V
Amplitude of carrier wave,
A_c = 2V [By eq.(1)]
T_m = 1s (From diagram)
From eq.(1) ω_m = \(\frac{2 \pi}{T_{m}}=\frac{2 \pi}{1}\) = 2π rad/s ………………. (2)
c(t) = 2sin8πt = Ac sinω_c t
ω_c =8π
So,
From Eq.(2)
So, ω_c = 4ω_m
Amplitude of modulated wave
A = A_m +A_c =2+1 =3V
The sketch of the amplitude modulated waveform is shown below :
For carrier signal, ω = 8π,T = \(\frac{2 \pi}{\omega}=\frac{2}{8}=\frac{1}{4}\) = 0.25s

(ii) Modulation index, µ = \(\frac{A_{m}}{A_{c}}=\frac{1}{2}\) = 0.5

Question 7.
For an amplitude modulated wave, the maximum amplitude is found to be 10 V while the minimum amplitude is found to be 2 V. Determine the modulation index, p.
What would be the value of p if the minimum amplitude is zero volts?
Answer:
Maximum amplitude, A_max = 10 V
Minimum amplitude, A_min = 2 V
Modulation index µ, is given by the relation
µ = \(\frac{A_{\max }-A_{\min }}{A_{\max }+A_{\min }}\)
= \(\frac{10-2}{10+2}=\frac{8}{12} \) = 0.67
If A_min=0,
Then µ = \(\frac{A_{\max }}{A_{\max }}=\frac{10}{10}\) = 1
Hence, µ = 1, if the minimum amplitude is zero volts.

Question 8.
Due to economic reasons, only the upper sideband of an AM wave is transmitted, but at the receiving station, there is a facility for generating the carrier. Show that if a device is available which can multiply two signals, then it is possible to recover the modulating signal at the receiver station.
Answer:
Let ω_c be the angular frequency of carrier waves and com by the angular frequency of signal waves.
Let the signal received at the receiving station be
e = E₁ cos(ω_c +ω_m )t
Let the instantaneous voltage of carrier wave
e_c = E_c cosω_c t
is available at receiving station.
Multiplying these two signals, we get
e x e_c = E₁E_c coscω_c t cos(ω_c+ω_m)t

Now, at the receiving end as the signal passes through filter, it will pass the high frequency (2ω_c + ω_m) but obstruct the frequency ω_m. So, we can \(\frac{E_{1} E_{c}}{2} \cos \omega_{m_{s}} t\) record the modulating signal frequency ω_m.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 13 Organisms and Populations Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 13 Organisms and Populations

PSEB 12th Class Biology Guide Organisms and Populations Textbook Questions and Answers

Question 1.
How is diapause different from hibernation?
Answer:
Diapause is a stage of suspended development to cope with unfavourable conditions. Many species of Zooplankton and insects exhibit diapause to tide over adverse climatic conditions during their development. Hibernation or winter sleep is a resting stage wherein animals escape winters of cold) by hiding themselves in their shelters. They escape the winter season by entering a state of inactivity by slowing their metabolism. The phenomenon of hibernation is exhibited by bats, squirrels, and other rodents.

Question 2.
If a marine fish is placed in a freshwater aquarium, will the fish be able to survive? Why or why not?
Answer:
If a marine fish is placed in a freshwater aquarium, then its chances of survival will diminish. This is because their bodies are adapted to high salt concentrations of the marine environment. In freshwater conditions, they are unable to regulate the water entering their body (through osmosis). Water enters their body due to the hypotonic environment outside. This results in the swelling up of the body, eventually leading to the death of the marine fish.

Question 3.
Define phenotypic adaptation. Give one example.
Answer:
Phenotypic adaptation involves changes in the body of an organism in response to genetic mutation or certain environmental changes. These responsive adjustments occur in an organism in order to cope with environmental conditions present in their natural habitats. For example, desert plants have thick cuticles and sunken stomata on the surface of their leaves to prevent transpiration. Similarly, elephants have long ears that act as thermoregulators.

Question 4.
Most living organisms cannot survive at temperatures above 45°C. How are some microbes able to live in habitats with temperatures exceeding 100°C?
Answer:
Archaebacteria (Thermophiles) are ancient forms of bacteria found in hot water springs and deep-sea hydrothermal vents. They are able to survive in high temperatures (which far exceed 100°C) because their bodies have adapted to such environmental conditions. These organisms contain specialised thermo-resistant enzymes, which carry out metabolic functions that do not get destroyed at such high temperatures.

Question 5.
List the attributes that populations but not individuals possess.
Answer:
A population can be defined as a group of individuals of the same species residing in a particular geographical area at a particular time and functioning as a unit. For example, all human beings living at a particular place at a particular time constitute the population of humans.

The main attributes or characteristics of a population residing in a given area are as follows :
(a) Birth Rate (Natality): It is the ratio of live births in an area to the population of an area. It is expressed as the number of individuals added to the population with respect to the members of the population.

(b) Death Rate (Mortality): It is the ratio of deaths in an area to the population of an area. It is expressed as the loss of individuals with respect to the members of the population.

(c) Sex Ratio: It is the number of males or females per thousand individuals.

(d) Age Distribution: It is the percentage of individuals of different ages in a given population. At any given time, the population is composed of individuals that are present in various age groups. The age distribution pattern is commonly represented through age pyramids.

(e) Population Density: It is defined as the number of individuals of a population present per unit area at a given time.

Question 6.
If a population growing exponentially double in size in 3 years, what is the intrinsic rate of increase (r) of the population?
Answer:
A population grows exponentially if sufficient amounts of food resources are available to the individual. Its exponential growth can be calculated by the following integral form of the exponential growth equation:
N_t = N₀e^rt
where,
N_t = Population density after time t
N₀= Population density at time zero
r = Intrinsic rate of natural increase
e = Base of natural logarithms (2.71828)

From the above equation, we can calculate the intrinsic rate of increase (r) of a population.
Now, as per the question,
Present population density = x
Then, population density after two years = 2x
t = 3 years
Substituting these values in the formula, we get
⇒ 2x = x e^3r
⇒ 2 = e^3r

Applying log on both sides,
⇒ log2 = 3r log e
⇒ \(\frac{\log 2}{3 \log e}\) = r

Hence, the intrinsic rate of increase for the above-illustrated population is 0.2311.

Question 7.
Name important defence mechanisms in plants against herbivory.
Answer:
Several plants have evolved various defence mechanisms both morphological and chemical to protect themselves against herbivory,
(1) Morphological Defence Mechanisms

• Cactus leaves (Opuntia) are modified into sharp spines (thorns) to deter herbivores from feeding on them.
• Sharp thorns along with leaves are present in Acacia to deter herbivores.
• In some plants, the margins of their leaves are spiny or have sharp edges that prevent herbivores from feeding on them.

(2) Chemical Defence Mechanisms

• All parts of Calotropis weeds contain toxic cardiac glycosides, which can prove to be fatal if ingested by herbivores.
• Chemical substances such as nicotine, caffeine, quinine, and opium are produced in plants as a part of self-defence.

Question 8.
An orchid plant is growing on the branch of mango tree. How do you describe this interaction between the orchid and the mango t tree?
Answer:
An orchid plant growing on the branch of a mango tree is an epiphyte. f Epiphytes are plants growing on other plants which however, do not derive nutrition from them. Therefore, the relationship between a mango tree and an orchid is an example of commensalisms, where one species gets benefited while the other remains unaffected. In the above interaction, the orchid is benefited as it gets support while the mango tree remains unaffected.

Question 9.
What is the ecological principle behind the biological control method of managing with pest insects?
Answer:
The basis of various biological control methods is on the concept of predation. Predation is a biological interaction between the predator and the prey, whereby the predator feeds on the prey. Hence, the predators regulate the population of preys in a habitat, thereby helping in the management of pest insects.

Question 10.
Distinguish between the following:
(a) Hibernation and Aestivation
(b) Ectotherms and Endotherms
Answer:
(a) Hibernation and Aestivation

Hibernation	Aestivation
1. Hibernation is a state of reduced activity in some organisms to escape cold winter conditions.	Aesrivarion is a state of reduced activity in some organisms to escape desiccation due to heat in summers.
2. Bears and squirrels inhabiting cold regions are examples of animals that hibernate during winters.	Fishes and snails are examples of organisms aestivating during summers.

(b) Ectotherms and Endotherms

Ectotherms	Endotherms
1. Ectotherms ate cold-blooded animals. Their temperature varies with their surroundings.	Endotherms are warm-blooded animals. They maintain a constant body temperature.
2. Fishes, amphibians, and reptiles are ectothermic animals.	birds and mammals are endothermal animals.

Question 11.
Write a short note on
(a) Adaptations of Desert Plants and Animals
(b) Adaptations of plants to water scarcity
(c) Behavioural adaptations in animals
(d) Importance of light to plants
(e) Effect of temperature or water scarcity and the adaptations of animals.
Answer:
(a) Adaptations of Desert Plants and Animals
(i) Adaptations of Desert Plants: Plants found in deserts are well adapted to cope with harsh desert conditions such as water scarcity and scorching heat. Plants have an extensive root system to tap underground water. They bear thick cuticles and sunken stomata o.i the surface of their leaves to reduce transpiration.

In Opuntia, the leaves are entirely modified into spines and photosynthesis is carried out by green stems. Desert plants have special pathways to synthesise food, called CAM (C₄ pathway). It enables the stomata to remain closed during the day to reduce the loss of water through transpiration.

(ii) Adaptations of Desert Animals: Animals found in deserts such as desert kangaroo rats, lizards, snakes, etc. are well adapted to their habitat. The kangaroo rat found in the deserts of Arizona never drinks water in its life. It has the ability to concentrate its urine to conserve water. Desert lizards and snakes bask in the sun during early morning and burrow themselves in the sand during afternoons to escape the heat of the day. These adaptations occur in desert animals to prevent the loss of water.

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

(c) Behavioural Adaptations in Animals: Certain organisms are affected by temperature variations. These organisms undergo adaptations such as hibernation, aestivation, migration, etc. to escape environmental stress to suit their natural habitat. These adaptations in the behaviour of an organism are called behavioural adaptations. For example, ectodermal animals and certain endotherms exhibit behavioural adaptations. Ectotherms are cold-blooded animals such as fish, amphibians, reptiles, etc.

Their temperature varies with their surroundings. For example, the desert lizard basks in the sun during early hours when the temperature is quite low. However, as the temperature begins to rise, the lizard burrows itself inside the sand to escape the scorching sun. Similar burrowing strategies are exhibited by other desert animals. Certain endotherms (warm-blooded animals) such as birds and mammals escape cold and hot weather conditions by hibernating during winters and aestivating during summers. They hide themselves in shelters such as caves, burrows, etc. to protect against temperature variations.

(d) Importance of Light to Plants: Sunlight acts as the ultimate source of energy for plants. Plants are autotrophic organisms, which need light for carrying out the process of photosynthesis. Light also plays an important role in generating photoperiodic responses! occurring in plants. Plants respond to changes in intensity of light
during various seasons to meet their photoperiodic requirements for flowering. Light also plays an important role in aquatic habitats for ‘ vertical distribution of plants in the sea.

(e) Effect of Temperature or Water Scarcity and the Adaptations of Animals: Temperature is the most important
ecological factor. Average temperature on the Earth varies from one place to another. These variations in temperature affect the distribution of animals on the Earth. Animals that can tolerate a wide range of temperature are called eurythermal. Those which can tolerate a narrow range of temperature are called stenothermal animals.

Animals also undergo adaptations to suit their natural habitats. For example, animals found in colder areas have shorter ears and limbs that prevent the loss of heat from their body. Also, animals found in Polar regions have thick layers of fat below their skin and thick coats of fur to prevent the loss of heat.

Some organisms exhibit various behavioural changes to suit their natural habitat. These adaptations present in the behaviour of an organism to escape environmental stresses are called behavioural adaptations. For example, desert lizards are ectotherms. This means that they do not have a temperature regulatory mechanism to escape temperature variations.

Question 12.
List the various abiotic environmental factors.
Answer:
All non-living components of an ecosystem form abiotic components. It includes factors such as temperature, water, light, and soil.

Question 13.
Give an example for:
(a) An endothermic animal
(b) An ectothermic animal
(c) An organism of benthic zone
Answer:
(a) Endothermic Animal: Birds such as crows, sparrows, pigeons, cranes, etc. and mammals such as bears, cows, rats, rabbits, etc. are endothermic animals.

(b) Ectothermic Animal: Fishes such as sharks, amphibians such as frogs, and reptiles such as tortoises, snakes, and lizards are ectothermic animals.

(c) Organism of Benthic Zone: Decomposing bacteria is an example of an organism found in the benthic zone of a water body.

Question 14.
Define population and community.
Answer:
Population: A population can be defined as a group of individuals of the same species residing in a particular geographical area at a particular time and functioning as a unit. For example, all human beings living at a particular place at a particular time constitute the population of humans.

Community: A community is defined as a group of individuals of different species, living within a certain geographical area. Such individuals can be similar or dissimilar, but cannot reproduce with the members of other species.

Question 15.
Define the following terms and give one example for each:
(a) Commensalism
(b) Parasitism
(c) Camouflage
(d) Mutualism
(e) Interspecific competition
Answer:
(a) Commensalism: Commensalism is an interaction between two species in which one species gets benefited while the other remains unaffected. An orchid growing on the branches of a mango tree and barnacles attached to the body of whales are examples of commensalisms.

(b) Parasitism: It is an interaction between two species in which one species (usually smaller) gets positively affected, while the other species (usually larger) is negatively affected. An example of this is liver fluke. Liver fluke is a parasite that lives inside the liver of the host body and derives nutrition from it. Hence, the parasite is benefited as it derives nutrition from the host, while the host is negatively affected as the parasite reduces the host fitness, making its body weak.

(c) Camouflage: It is a strategy adopted by prey species to escape their predators. Organisms are cryptically coloured so that they can easily mingle in their surroundings and escape their predators. Many species of frogs and insects camouflage in their surroundings and escape their predators.

(d) Mutualism: It is an interaction between two species in which both species involved are benefited. For example, lichens show a mutual symbiotic relationship between fungi and blue-green algae, where both are equally benefited from each other.

(e) Interspecific Competition: It is an interaction between individuals of different species where both species get negatively affected. For example, the competition between flamingoes and resident fishes in South American lakes for common food resources i.e., zooplankton.

Question 16.
With the help of suitable diagram describe the logistic population growth curve.
Answer:
The logistic population growth curve is commonly observed in yeast cells that are grown under laboratory conditions. It includes five phases: the lag phase, positive acceleration phase, exponential phase, negative acceleration phase, and stationary phase.
(a) Lag Phase: Initially, the population of the yeast cell is very small. This is because of the limited resource present in the habitat.

(b) Positive Acceleration Phase: During this phase, the yeast cell adapts to the new environment and starts increasing its population. However, at the beginning of this phase, the growth of the cell is very limited.

(c) Exponential Phase: During this phase, the population of the yeast cell increases suddenly due to rapid growth. The population grows exponentially due to the availability of sufficient food resources, constant environment, and the absence of any interspecific competition. As a result, the curve rises steeply upwards.

(d) Negative Acceleration Phase: During this phase, the
environmental resistance increases and the growth rate of the population decreases. This occurs due to an increased competition among the yeast cells for food and shelter.

(e) Stationary Phase: During this phase, the population becomes stable. The number of cells produced in a population equals the number of cells that die. Also, the population of the species is said to have reached nature’s carrying capacity in its habitat.

A Verhulst-pearl logistic curve is also known as an S-Shaped growth curve.

Question 17.
Select the statement which explains best parasitism.
(a) One organism is benefited.
(b) Both the organisms are benefited.
(c) One organism is benefited, other is not affected.
(d) One organism is benefited, other is affected.
Answer:
(d) One organism is benefited, other is affected.
Parasitism is an interaction between two species in which one species (parasite) derives benefit while the other species (host) is harmed. For example, ticks and lice (parasites) present on the human body represent this interaction wherein the parasites receive benefit (as they derive nourishment by feeding on the blood of humans). On the other hand, these parasites reduce host fitness and cause harm to the human body.

Question 18.
List any three important characteristics of a population and explain.
Answer:
A population can be defined as a group of individuals of the same species, residing in a particular geographical area at a particular time and functioning as a unit. For example, all human beings living at a particular place at a particular time constitute the population of humans.
Three important characteristics of a population are as follows :
(a) Birth Rate (Natality): It is the ratio of live births in an area to the population of an area. It is expressed as the number of individuals added to the population with respect to the members of the population.

(b) Death Rate (Mortality): It is the ratio of deaths in an area to the population of an area. It is expressed as the loss of individuals with respect to the members of the population.

(c) Age Distribution: It is the percentage of individuals of different ages in a given population. At any given time, a population is composed of individuals that are present in various age groups. The age distribution pattern is commonly represented through a^e pyramids.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 14 Ecosystem Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 14 Ecosystem

PSEB 12th Class Biology Guide Ecosystem Textbook Questions and Answers

Question 1.
Fill in the blanks.
(a) Plants are called as ……………………………. because they fix carbon dioxide.
(b) In an ecosystem dominated by trees, the pyramid (of numbers) is ………………….. type.
(c) In aquatic ecosystems, the limiting factor for the productivity is ………………………. .
(d) Common detritivores in our ecosystem are …………………………. .
(e) The major reservoir of carbon on earth is …………………………….. .
Answer:
(a) autotrophs
(b) inverted
(c) light
(d) earthworms
(e) oceans.

Question 2.
Which one of the following has the largest population in a food chain?
(a) Producers
(b) Primary consumers
(c) Secondary consumers
(d) Decomposers
Answer:
(a) Producers (decomposers can be maximum but they are excluded from the food chain ).

Question 3.
The second trophic level in a lake is
(a) Phytoplankton
(b) Zooplankton
(c) Benthos
(d) Fishes
Answer:
(b) Zooplankton
Zooplankton are primary consumers in aquatic food chains that feed upon phytoplankton. Therefore, they are present at the second trophic level in a lake.

Question 4.
Secondary producers are
(a) Herbivores
(b) Producers
(c) Carnivores
(d) None of the above
Answer:
(d) None of the above
Plants are the only producers. Thus, they are called primary producers. There are no other producers in a food chain.

Question 5.
What is the percentage of photosynthetically active radiation (PAR) in the incident solar radiation?
(a) 100%
(b) 50 %
(c) 1-5%
Answer:
(b) 50%
Out of total incident solar radiation, about fifty percent of it forms photosynthetically active radiation or PAR.

Question 6.
Distinguish between
(a) Grazing food chain and detritus food chain
(b) Production and decomposition
(c) Upright and inverted pyramid
(d) Food chain and Food web
(e) Litter and detritus
(f) Primary and secondary productivity
Answer:
(a) Grazing food chain and detritus food chain

Grazing food chain	Detritus food chain
1. In this food chain, energy is derived from the Sun.	In this food chain, energy comes from organic matter (or detritus) generated in trophic levels of the grazing food chain.
2. It begins with producers, present at the first trophic level. The plant biomass is then eaten by herbivores, which in turn are consumed by a variety of carnivores.	begins with detritus such as dead bodies of animals or fallen leaves, which are then eaten by decomposers or detrivores. These detritivores are in turn consumed by their predators.
3. This food chain is usually large.	It is usually smaller as compared to the grazing food chain.

(b) Production and decomposition

Production	Decomposition
1. It is the rare of producing organic matter (food) by producers.	It is the process of breaking down of complex organic matter or biomass from the body of dead plants and animals with the help of decomposers into organic raw material such as CO2, H2O and other nutrients.
2. It depends on the photosynthetic cápacity of the producers.	It occurs with the help of decomposers.
3. Sunlight is required by plants for primary production.	Sunlight is not required for decomposition by clecomposers.

(c) upright and inverted pyramid

Upright pyramid	Inverted pyramid
1. The pyramid of energy is upright.	always The pyramid of biomass and the pyramid of, numbers can be inverted.
2. In the upright pyramid, the number and biomass of organisms in the producer level of an ecosystem is the highest, which keeps on decreasing at each trophic level in a food chain.	In an inverted pyramid, the number and biomass of organisms in the producer level of an ecosystem is the lowest, which keeps on increasing at each tropic level.

(d) Food chain and food web

Food chain	Food web
1. The transfer of energy from producers to top consumers through a series of organisms is called food chain.	A number of food chain inter-connected with each other forming a web-like pattern is called food web.
2. One organism holds only one position.	One organism can hold more than one position.
3. The flow of energy can be easily calculated.	The flow of energy is very difficult to calculate.
4. It is always straight and proceed in a progressive straight line.	Instead of straight line, it is a series of branching lines.
5. Competition is limited to members of same trophic level.	Competition is amongst members of same and different trophic levels.

(e) Litter and detritus

Litter	Detritus
l. It is made of dried fallen plant matter.	It is freshly deposited organic matter, i. e. remains of plants and animals.
2. It is found above the ground.	It is found both above and below the ground.

(f) Primary and secondary productivity

Primary productivity	Secondary_productivity
1. Primary productivity is the amount of energy accumulation or amount of biomass produced per unit area over a time period.	Secondary productivity is the rate of formation of new organic matter by consumer.
2.  It is of two types, gross primary productivity (GPP) and net primary productivity (NPP). They are related as: GPP – R = NPP, where R is respiratory losses.	It is also of two types gross secondary productivity (GSP) and net secondary productivity (NSP). They are related as: NSP = GSP – R Where R is respiratory losses.

Question 7.
Describe the components of an ecosystem.
Answer:
An ecosystem is defined as an interacting unit that includes both the biological community as well as the non-living components of an area. The living and the non-living components of an ecosystem interact amongst themselves and function as a unit, which gets evident during the processes of nutrient cycling, energy flow, decomposition, and productivity. There are many ecosystems such as ponds, forests, grasslands, etc.

The two components of an ecosystem are as follows :
(a) Biotic Component: It is the living component of an ecosystem that includes biotic factors such as producers, consumers, decomposers, etc. Producers include plants and algae. They contain chlorophyll pigment, which helps them carry out the process of photosynthesis in the presence of light.

Thus, they are also called converters or transducers. Consumers or heterotrophs are organisms that are directly (primary consumers) or indirectly (secondary and tertiary consumers) dependent on producers for their food. Decomposers include micro-organisms such as bacteria and fungi. They obtain nutrients by breaking down the remains of dead plants and animals.

(b) Abiotic Component: They are the non-living component of an ecosystem such as light, temperature, water, soil, air, inorganic nutrients, etc.

Question 8.
Define ecological pyramids and describe with examples, pyramids of number and biomass.
Answer:
An ecological pyramid is a graphical representation of various ecological parameters such as the number of individuals present at each trophic level, the amount of energy, or the biomass present at each trophic level.
Ecological pyramids represent producers at the base, while the apex represents the top-level consumers present in the ecosystem.
There are three types of pyramids:

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

1. Pyramid of Numbers: It is a graphical representation of the number of individuals present at each trophic level in a food chain of an ecosystem. The pyramid of numbers can be upright or inverted depending on the number of producers. For example, in a grassland ecosystem, the pyramid of numbers is upright.

In this type of a food chain, the number of producers (plants) is followed by the number of herbivores (mice), which in turn is followed by the number of secondary consumers (snakes) and tertiary carnivores (eagles). Hence, the number of individuals at the producer level will be the maximum, while the number of individuals present at top carnivores will be least.

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

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

Question 9.
What is primary productivity? Give brief description of factors that affect primary productivity.
Answer:
It is defined as the amount of organic matter or biomass produced by producers per unit area over a period of time. Primary productivity of an ecosystem depends on the variety of environmental factors such as light, temperature, water, precipitation, etc. It also depends on the availability of nutrients and the availability of plants to carry out photosynthesis.

Question 10.
Define decomposition and describe the processes and products of decomposition.
Answer:
Decomposition is the process that involves the breakdown of complex organic matter or biomass from the body of dead plants and animals with the help of decomposers into inorganic raw materials such as carbon dioxide, water, and other nutrients. The various processes involved in decomposition are as follows :
1. Fragmentation: It is the first step in the process of decomposition. It involves the breakdown of detritus into smaller pieces by the action of detritivores such as earthworms.

2. Leaching: It is a process where the water-soluble nutrients go down into the soil layers and get locked as unavailable salts.

3. Catabolism: It is a process in which bacteria and fungi degrade detritus through various enzymes into smaller pieces.

4. Humification: The next step is humification which leads to the formation of a dark-colored colloidal substance called humus, which acts as reservoir of nutrients for plants.

5. Mineralisation: The humus is further degraded by the action of microbes, which finally leads to the release of inorganic nutrients into the soil. This process of releasing inorganic nutrients from the humus is known as mineralization.

Decomposition produces a dark-colored, nutrient-rich substance called humus. Humus finally degrades and releases inorganic raw materials such as C02, water, and other nutrients in the soil.

Question 11.
Give an account of energy flow in an ecosystem.
Answer:
Energy enters an ecosystem from the Sun. Solar radiations pass through the atmosphere and are absorbed by the Earth’s surface. These radiations help plants in carrying out the process of photosynthesis.
Also, they help maintain the Earth’s temperature for the survival of living organisms. Some solar radiations are reflected by the Earth’s surface.

Only 2-10% of solar energy is captured by green plants (producers) during photosynthesis to be converted into food.
The rate at which the biomass is produced by plants during photosynthesis is termed as ‘gross primary productivity.
When these green plants are consumed by herbivores, only 10% of the stored energy from producers is transferred to herbivores. The remaining 90 % of this energy is used by plants for various processes such as respiration, growth, and reproduction. Similarly, only 10% of the energy of herbivores is transferred to carnivores. This is known as ten percent law of energy flow.

Question 12.
Write important features of a sedimentary cycle in an ecosystem.
Answer:
Sedimentary cycles have their reservoirs in the Earth’s crust or rocks. Nutrient elements are found in the sediments of the Earth. Elements such as sulphur, phosphorus, potassium, and calcium have sedimentary cycles.
Sedimentary cycles are very slow. They take a long time to complete their circulation and are considered as less perfect cycles. This is because during recycling, nutrient elements may get locked in the reservoir pool, thereby taking a very long time to come out and continue circulation. Thus, it usually goes out of circulation for a long time.

Question 13.
Outline salient features of carbon cycling in an ecosystem.
Answer:
The carbon cycle is an important gaseous cycle which has its reservoir pool in the atmosphere.
All living organisms contain carbon as a major body constituent. Carbon is a fundamental element found in all living forms. All biomolecules such as carbohydrates, lipids, and proteins required for life processes are made of carbon.
Carbon is incorporated into living forms through a fundamental process called ‘photosynthesis’.

Photosynthesis uses sunlight and atmospheric carbon dioxide to produce a carbon compound called ‘glucose’.
This glucose molecule is utilised by other living organisms. Thus, atmospheric carbon is incorporated in living forms.
Now, it is necessary to recycle this absorbed carbon dioxide back into the atmosphere to complete the cycle.

There are various processes by which carbon is recycled back into the atmosphere in the form of carbon dioxide gas.
The process of respiration breaks down glucose molecules to produce carbon dioxide gas. The process of decomposition also releases carbon dioxide from dead bodies of plants and animals into the atmosphere. Combustion of fuels, industrialization, deforestation, volcanic eruptions and forest fires act as other major sources of carbon dioxide.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 12 Biotechnology and its Applications Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 12 Biotechnology and its Applications

PSEB 12th Class Biology Guide Biotechnology and its Applications Textbook Questions and Answers

Question 1.
Crystals of Bt toxin produced by some bacteria do not kill the bacteria themselves because
(a) bacteria are resistant to the toxin,
(b) toxin is immature,
(c) toxin is inactive,
(d) bacteria encloses toxin in a special sac.
Answer:
Toxin is inactive: In bacteria, the toxin is present in an inactive form, called prototoxin, which gets converted into active form when it enters the body of an insect.

Question 2.
What are transgenic bacteria? Illustrate using any one example.
Answer:
Transgenic bacteria contain foreign gene that is intentionally introduced into its genome. They are manipulated to express the desirable gene for the production of various commercially important products.

An example of transgenic bacteria is E.coli. In the plasmid of E.coli, the two DNA sequences corresponding to A and B chain of human insulin are inserted, so as to produce the respective human insulin chains. Hence, after the insertion of insulin gene into the bacterium, it becomes transgenic and starts producing chains of human insulin. Later on, these chains are extracted from E.coli and combined to form human insulin.

Question 3.
Compare and contrast the advantages and disadvantages of production of genetically modified crops.
Answer:
The production of genetically modified (GM) or transgenic crops has several advantages.

• Most of the GM crops have been developed for pest resistance, which increases the crop productivity and therefore, reduces the reliance on chemical pesticides.
• Many varieties of GM food crops have been developed, which have enhanced nutritional quality. For example, golden rice is a transgenic variety of rice, which is rich in vitamin A.
• These plants prevent the loss of fertility of soil by increasing the ‘
  efficiency of mineral usage.
• They are highly tolerant to unfavourable abiotic conditions.
• The use of GM crops decreases the post harvesting loss of crops.

However, there are certain controversies regarding the use of genetically modified crops around the world. The use of these crops can affect the native biodiversity in an area. For example, the use of Bt toxin to decrease the amount of pesticide is posing a threat for beneficial insect pollinators such as honey bee. If the gene expressed for Bt toxin gets expressed in the pollen, then the honey bee might be affected. As a result, the process of pollination by honey bees would be affected. Also, genetically modified crops are affecting human health. They supply allergens and certain antibiotic resistance markers in the body. Also, they can cause genetic pollution in the wild relatives of the crop plants. Hence, it is affecting our natural environment.

Question 4.
What are Cry proteins? Name an organism that produce it. How has man exploited this protein to his benefit?
Answer:
Cry proteins are encoded by cry genes. These proteins are toxins, which are produced by Bacillus thuringiensis bacteria. This bacterium contains these proteins in their inactive form. When the inactive toxin protein is ingested by the insect, it gets activated by the alkaline pH of the gut. This results in the lysis of epithelial cell and eventually the death of the insect. Therefore, man has exploited this protein to develop certain transgenic crops with insect resistance such as Bt cotton, Bt corn, etc.

Question 5.
What is gene therapy? Illustrate using the example of adenosine deaminase (ADA) deficiency.
Answer:
Gene therapy is a technique for correcting a defective gene through gene manipulation. It involves the delivery of a normal gene into the individual to replace the defective gene, for example, the introduction of gene for adenosine deaminase (ADA) in ADA deficient individual. The adenosine deaminase enzyme is important for the normal functioning of the immune system. The individual suffering from this disorder can be cured by transplantation of bone marrow cells. The first step involves the extraction of lymphocyte from the patient’s bone marrow. Then, a functional gene for ADA is introduced into lymphocytes with the help of retrovirus. These treated lymphocytes containing ADA gene are then introduced into the patient’s bone . marrow. Thus, the gene gets activated producing functional T-lymphocytes and activating the patient’s immune system.

Question 6.
Diagrammatically represent the experimental steps in cloning and expressing an human gene (say the gene for growth hormone) into a bacterium like E. coli ?
Answer:
DNA cloning is a method of producing multiple identical copies of specific template DNA. It involves the use of a vector to carry the specific foreign DNA fragment into the host cell. The mechanism of cloning and transfer of gene for growth hormone into E.coli is represented below:

Question 7.
Can you suggest a method to remove oil (hydrocarbon) from seeds based on your understanding of rDNA technology and chemistry of oil?
Answer:
Recombinant DNA technology (rDNA) is a technique used for manipulating the genetic material of an organism to obtain the desired result. For example, this technology is used for removing oil from seeds. The constituents of oil are glycerol and fatty acids. Using rDNA, one can obtain oilless seeds by preventing the synthesis of either glycerol or fatty acids. This( is done by removing the specific gene responsible for the synthesis.

Question 8.
Find out from internet what is golden rice.
Answer:
Golden rice is a genetically modified variety of rice, Oryza sativa, which has been developed as a fortified food for areas where there is a shortage of dietary vitamin A. It contains a precursor of pro-vitamin A, called beta-carotene, which has been introduced into the rice through genetic engineering. The rice plant naturally produces beta-carotene pigment in its leaves. However, it is absent in the endosperm of the seed. This is because beta-carotene pigment helps in the process of photosynthesis while photosynthesis does not occur in endosperm.

Since beta-carotene is a precursor of pro-vitamin A, it is introduced into the rice variety to fulfil the shortage of dietary vitamin A. It is simple and a less expensive alternative to vitamin supplements. However, this variety of rice has faced a significant opposition from environment activists. Therefore, they are still not available in market for human consumption.

Question 9.
Does our blood have proteases and nucleases?
Answer:
No, human blood does not include the enzymes, nucleases and proteases. In human beings, blood serum contains different types of protease inhibitors, which protect the blood proteins from being broken down by the action of proteases. The enzyme, nucleases, catalyses the hydrolysis of nucleic acids that is absent in blood.

Question 10.
Consult internet and find out how to make orally active protein pharmaceutical. What is the major problem to be encountered?
Answer:
Orally active protein pharmaceutical can be made by lining it with a substance that will dissolve after it has passed through the stomach.
The major problem encountered is that the stomach enzymes and acids may denature the therapeutic protein and render it ineffective.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 11 Biotechnology: Principles and Processes Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

PSEB 12th Class Biology Guide Biotechnology: Principles and Processes Textbook Questions and Answers
Question 1.
Can you list 10 recombinant proteins which are used in medical ; practice? Find out where they are used as therapeutics (use the internet).
Answer:

Recombinant proteins	Therapeutic uses
(a) Insulin	Used for diabetes mellitus
(b) OKT-3	Therapeutic antibody, used for reversal
(c) DNase	Treatment of cystic fibrosis
(d) Reo Pro	Prevention of blood clots
(e) Blood clotting factor VIII	Treatment of haemophilia A
(f) Blood clotting factor IX	Treatment of haemophilia B
(g) Tissue plasminogen activator	For acute myocardial infarction
(h) Interferon alpha (INF alpha)	Used for hepatitis C
(i) Interferon beta (INF beta)	Used for multiple sclerosis
(j) Interferon gamma (INF gamma)	Used for granulomatous disease

Question 2.
Make a chart (with diagrammatic representation) showing a restriction enzyme, the substrate DNA on which it acts, the site at which it cuts DNA and the product it produces.
Answer:

Question 3.
From what you have learnt, can you tell whether enzymes are bigger or DNA is bigger in molecular size? How did you know?
Answer:
The molecular size of DNA molecules is more than that of enzymes. It is because an enzyme (protein) is synthesised from a segment of DNA called the gene.

Question 4.
What would be the molar concentration of human DNA in a human cell? Consult your teacher.
Answer:
The molar concentration of human DNA in a human diploid cell is as follows:
Total number of chromosomes × 6.023 × 10²³
= 46 × 6.023 × 10²³
= 2.77 × 1018moles

Question 5.
Do eukaryotic cells have restriction endonucleases? Justify your answer.
Answer:
No. Eukaryotic cells do not have restriction endonucleases. All the restriction endonucleases have been isolated from the various strains of bacteria and they are also named according to the genus and species of prokaryotes. The first letter of the enzyme comes from the genus and the second two letters come from the species of the prokaryotic cell from which they have been isolated. For example, EcoRI comes from Escherichia coliRY 13. In EcoRI, the letter ‘R’ is derived from the name of the strain. Roman numbers following the names indicate the order in which the enzymes were isolated from that strain of bacteria.

Question 6.
Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors have over shake flasks?
Answer:
Shake flasks are used for growing microbes and mixing the desired materials on a small scale in the laboratory. However, the large-scale production of a desired biotechnological product requires large stirred tank bioreactors.

Besides better aeration and mixing properties, bioreactors have the following advantages:

• It has an oxygen delivery system.
• It has a foam control, temperature and pH control system.
• Small volumes of culture can be withdrawn periodically.

Question 7.
Collect 5 examples of palindromic DNA sequences by consulting your teacher. Better try to create a palindromic sequence by following base-pair rules.
Answer:
Some palindromic DNA sequences and the restriction enzymes which act on them are as follows :

• 5′-AGCT-3′ Alul (Arthrobacter luteus)
  3′-TCGA-5′
• 5′-GAATTC-3′ EcoRI (Escherichia coli)
  3′-CTTAAG-5′
• 5′-AAGCTT-3′ Hindlil (Haemophilus influenzae)
  3′-TTCGAA-5′
• 5′-GTCGAC-3′ Sail (Streptomyces albus)
  3′-CAGCTG-5′
• 5′-CTGCAG-3′ PstI (Providencia stuartii)
  3′-GACGTC-5′

Question 8.
Can you recall meiosis and indicate at what stage a recombinant DNA is made?
Answer:
A recombinant DNA is made in the pachytene stage of prophase I by f crossing over.

Question 9.
Can you think and answer how a reporter enzyme can be used to
monitor transformation of host cells by foreign DNA in addition to a selectable marker?
Answer:
Reporter enzyme can differentiate recombinants from non-recombinants on the basis of their ability to produce a specific colour in the presence of a chromogenic substrate. DNA is inserted within the coding sequence of the enzyme p-galactosidase. This results into inactivation of the enzyme which is referred to as insertional inactivation.

The presence of a chromogenic substrate gives blue-coloured colonies if the plasmid in the bacteria does not have an insert. The presence of the insert results in insertional inactivation of α-galactosidase and the colonies do not produce any colour. These are identified as recombinant colonies.

Question 10.
Describe briefly the following:
(a) Origin of replication
(b) Bioreactors
(c) Downstream processing
Answer:
(a) Origin of Replication: This is a sequence from where replication starts and any piece of DNA when linked to this sequence can be made to replicate within the host cells. This sequence is also responsible for controlling the copy number of the linked DNA. So, if one wants to recover many copies of the target DNA it should be cloned in a vector whose origin supports a high copy number.

(b) Bioreactors: Bioreactors are vessels in which raw materials are biologically converted into specific products, individual enzymes etc. using microbial plant, animal or human cells. A bioreactor provides the optimal conditions for achieving the desired product by providing optimum growth conditions (temperature, pH, substrate, salts, vitamins, oxygen). The most commonly used bioreactors are of stirring type. A biogas plant is a good example of a bioreactor.

(c) Downstream Processing: After completion of the biosynthetic stage, the product is subjected through a series of processes before it is ready for marketing as a finished product. The processes include separation and purification, which are collectively referred to as downstream processing. The product has to be formulated with suitable preservatives. Such formulation has to undergo thorough clinical trials as in the case of drugs. Strict quality control testing for each product is also required. Downstream processing and quality control testing vary from product to product.

Question 11.
Explain briefly
(a) PCR
(b) Restriction enzymes and DNA
(c) Chitinase
Answer:
(a) PCR: PCR stands for Polymerase Chain Reaction. In this reaction multiple copies of the gene (or DNA) of interest is synthesised in vitro using two sets of primers and the enzyme DNA polymerase. The enzyme extends the primers using the nucleotides provided in the reaction and the genomic DNA as template. The segment of DNA can be amplified to approximately billion times if the process of replication of DNA is repeated many times.

(b) Restriction Enzymes and DNA: Restriction enzymes are synthesised by microbes as a defence mechanism and are specifically endonucleases which cleave the double-stranded DNA with the desired genes. This activity occurs at a limited number of sites depending on the number of recognition sequences in DNA. Lysing enzymes, synthesising enzymes (DNA polymerase and reverse transcriptase) and ligases are also tools of genetic engineering.

(c) Chitinase: During the isolation of DNA in the processes of recombinant DNA technology, the fungal cell is heated with an enzyme called chitinase. The chitinase enzyme dissolves the chitin membrane to open the cell for release of DNA along with other macromolecules such as RNA, proteins, polysaccharides and lipids.

Question 12.
Discuss with your teacher and find out how to distinguish between
(a) Plasmid DNA and Chromosomal DNA
(b) RNA and DNA
(c) Exonuclease and Endonuclease
Answer:
(a)

Plasmid DNA	Chromosomal DNA
Plasmid DNA is the naked double-stranded DNA which forms a circle with no free ends. It is associated with few proteins but contains RNA polymerase enzyme. They are smaller than the host chromosomes and can be easily separated.	Chromosomal DNA is a double-stranded linear DNA molecule associated with large proteins. This DNA exists in relaxed and supercoiled forms and provides a template for replication and transcription. It has free ends represented as 3′-5′.

(b)

DNA	RNA
(i) It is mainly confined to the nucleus. A small quantity occurs in mitochondria and chloroplasts.	It mainly occurs in the cytoplasm. A small quantity is found in the nucleus.
(ii) Its quantity is constant in each cell of a species.	Its quantity varies in different cells.
(iii) It contains deoxyribose sugar.	It contains ribose sugar.
(iv) Its pyrimidines are adenine and thymine.	Its pyrimidines are adenine and uracil.
(v) The amount of adenine is equal to the amount of thymine. Also, the amount of cytosine is equal to the amount of guanine.	Adenine and uracil are not necessarily in equal amounts, nor are cytosine and guanine necessarily in equal amounts.
(vi) It can replicate itself.	It cannot replicate itself. It is formed by DNA. Some RNA viruses (paramyxo virus) can produce RNA from an RNA template.

(c) Exonucleases are nucleases which cut off the nucleotides from the 5′ or 3′ ends of a DNA molecule, whereas endonucleases are nucleases which cleave the DNA duplex at any point except at the ends.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 10 Microbes in Human Welfare Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 10 Microbes in Human Welfare

PSEB 12th Class Biology Guide Microbes in Human Welfare Textbook Questions and Answers

Question 1.
Bacteria cannot be seen with the naked eyes, but these can be seen with the help of a microscope. If you have to carry a sample from your home to your biology laboratory to demonstrate the presence of microbes under a microscope, which sample would you carry and why?
Answer:
Curd can be used as a sample for the study of microbes. Curd contains numerous lactic acid bacteria (LAB) or Lactobacillus. These bacteria produce acids that coagulate and digest milk proteins. A small drop of curd contains millions of bacteria, which can be easily observed under a microscope.

Question 2.
Give examples to prove that microbes release gases during metabolism.
Answer:
The examples of microbes that release gases during metabolism are as follows :
(a) Bacteria and fungi carry out the process of fermentation and during this process, they release carbon dioxide. Fermentation is the process of converting a complex organic substance into a simpler substance with the action of bacteria or yeast. Fermentation of sugar produces alcohol with the release of carbon dioxide and very little energy.

(b) The dough used for making idli and dosa gives a puffed appearance. This is because of the action of bacteria which releases carbon dioxide. This CO₂ released from the dough gets trapped in the dough, thereby giving it a puffed appearance.

Question 3.
In which food would you find lactic acid bacteria? Mention some of their useful applications.
Answer:
Lactic acid bacteria can be found in curd. It is this bacterium that promotes the formation of milk into curd. The bacterium multiplies and increases its number, which converts the milk into curd. They also increase the content of vitamin B₁₂ in curd.
Lactic acid bacteria are also found in our stomach where it keeps a check on the disease-causing micro-organisms.

Question 4.
Name some traditional Indian foods made of wheat, rice and Bengal gram (or their products) which involve use of microbes.
Answer:

1. Wheat Product: Bread, cake etc.
2. Rice Product: Idli, dosa etc.
3. Bengal Gram Product: Dhokla, khandvi etc.

Question 5.
In which way have microbes played a m^jor role in controlling diseases caused by harmful bacteria?
Answer:
Several micro-organisms are used for preparing medicines. Antibiotics are medicines produced by certain micro-organisms to kill other disease-causing micro-organisms. These medicines are commonly obtained from bacteria and fungi. They either kill or stop the growth of disease-causing micro-organisms. Streptomycin, tetracycline, and penicillin are common antibiotics. Penicillium notatum produces chemical penicillin, which checks the growth of Staphylococci bacteria in the body. Antibiotics are designed to destroy bacteria by weakening their cell walls. As a result of this weakening, certain immune cells such as the white blood cells enter the bacterial cell and cause cell lysis. Cell lysis is the process of destroying cells such as blood cells and bacteria.

Question 6.
Name any two species of fungus, which are used in the production of the antibiotics.
Answer:
Antibiotics are medicines that are produced by certain micro-organisms to kill other disease-causing micro-organisms. These medicines are commonly obtained from bacteria and fungi.
The species of fungus used in the production of antibiotics are as follows:

Antibiotic	Fungus source
1. Penicillin	Penicillium notatum
2. Cephalosporin	Cephalosporium acremonium

Question 7.
What is sewage? In which way can sewage be harmful to us?
Answer:
Sewage is the municipal waste matter that is carried away in sewers and drains. It includes both liquid and solid wastes, rich in organic matter and microbes. Many of these microbes are pathogenic and can cause several water borne diseases. Sewage water is a major cause of polluting drinking water. Hence, it is essential that sewage water is properly collected, treated, and disposed. If untreated sewage is disposed into rivers and streams, it will pollute the water bodies.

Question 8.
What is the key difference between primary and secondary sewage treatment?
Answer:

Primary sewage treatment	Secondary sewage treatment
1. It is a mechanical process involving the removal of coarse solid materials.	It is a biological process involving the action of microbes.
2. It is inexpensive and relatively less complicated.	It is a very expensive and complicated process.

Question 9.
Do you think microbes can also be used as source of energy? If yes, how?
Answer:
Yes, microbes can be used as a source of energy. Bacteria such as Methane bacterium is used for the generation of gobar gas or biogas.

The generation of biogas is an anaerobic process in a biogas plant, which consists of a concrete tank (10-15 feet deep) with sufficient outlets and inlets. The dung is mixed with water to form the slurry and thrown into the tank. The digester of the tank is filled with numerous anaerobic methane-producing bacteria, which produce biogas from the slurry. Biogas can be removed through the pipe which is then used as a source of energy, while the spent slurry is removed from the outlet and is used as a fertiliser.

Question 10.
Microbes can be used to decrease the use of chemical fertilisers and pesticides. Explain how this can be accomplished.
Answer:
Microbes play an important role in organic farming, which is done without the use of chemical fertilisers and pesticides. Bio-fertilisers are living organisms which help increase the fertility of soil. It involves the ’ selection of beneficial micro-organisms that help in improving plant growth through the supply of plant nutrients. Bio-fertilisers are introduced in seeds, roots, or soil to mobilise the availability of nutrients. Thus, they are extremely beneficial in enriching the soil with organic nutrients. Many species of bacteria and cyanobacteria have the ability to fix free atmospheric nitrogen. Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants. Azpirillum and Azotobacter are free living nitrogen-fixing bacteria whereas Anabaena, Nostoc and Oscillatoria are examples of nitrogen-fixing cyanobacteria. Bio-fertilisers are cost effective and eco-friendly.

Microbes can also act as bio-pesticides to control insect pests in plants.
An example of bio-pesticides is Bacillus thuringiensis, which produces a toxin that kills the insect pests. Dried bacterial spores are mixed in water and sprayed in agricultural fields. When larvae of insects feed on crops, these bacterial spores enter the gut of the larvae and release r toxins, thereby it. Similarly, Trichoderma are free living fungi. They live in the roots of higher plants and protect them from various pathogens.

Baculoviruses is another bio-pesticide that is used as a biological control I agent against insects and other arthropods.

Question 11.
Three water samples namely river water, untreated sewage water and secondary effluent discharged from a sewage treatment plant were subjected to BOD test. The samples were labelled A, B and C; but the laboratory attendant did not note which was which. The BOD values of the three samples A, B and C were recorded as 20mg/L, 8mg/L and 400mg/L, respectively. Which sample of the water is most polluted? Can you assign the correct label to each assuming the river water is relatively clean?
Answer:
Biological oxygen demand (BOD) is the method of determining the amount of oxygen required by micro-organisms to decompose the waste present in the water supply. If the quantity of organic wastes in the water supply is high, then the number of decomposing bacteria present in the water will also be high. As a result, the BOD value will increase. Therefore, it can be concluded that if the water supply is more polluted, then it will have a higher BOD value. Out of the above three samples, sample C is the most polluted since it has the maximum BOD value of 400 mg/L.

After untreated sewage water, secondary effluent discharge from a sewage treatment plant is most polluted. Thus, sample A is secondary effluent discharge from a sewage treatment plant and has the BOD value of 20 mg/L, while sample B is river water and has the BOD value of 8 mg/L.
Hence, the correct label for each sample is:

Label	BOD value	Sample
A.	20 mg/L	Secondary effluent discharge from a sewage treatment plant
B.	8 mg/L	River water
C.	400 mg/L	Untreated sewage water

Question 12.
Find out the name of the microbes from which Cyclosporin A (an immunosuppressive drug) and Statins (blood cholesterol lowering agents) are obtained.
Answer:

Drug	Function	Microbe
1. Cyclosporin-A	Immuno suppressive drug	Trichoderma polysporum
2. Statin	Blood cholesterol lowering agent	Monascus purpureus

Question 13.
Find out the role of microbes in the following and discuss it with your teacher.
(a) Single cell protein (SCP)
(b) Soil
Answer:
(a) Single Cell Protein (SCP): A single cell protein is a protein obtained from certain microbes, which forms an alternate source of proteins in animal feeds. The microbes involved in the preparation of single cell proteins are algae, yeast, or bacteria. These microbes are grown on an industrial scale to obtain the desired protein. For example, Spirulina can be grown on waste materials obtained from molasses, sewage, and animal manures. It serves as a rich supplement of dietary nutrients such as proteins, carbohydrate, fats, minerals, and vitamins. Similarly, micro-organisms such as Methylophilus and methylotrophus have a large rate of biomass production. Their growth can produce a large amount of proteins.

(b) Soil: Microbes play an important role in maintaining soil fertility. They help in the formation of nutrient-rich humus by the process of decomposition. Many species of bacteria and cyanobacteria have the ability to fix atmospheric nitrogen into usable form. Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants. Azospirillum and Azotobacter are free living nitrogen-fixing bacteria, whereas Anabaena, Nostoc, and Oscillatoria are examples of nitrogen-fixing cyanobacteria.

Question 14.
Arrange the following in the decreasing order (most important first) of their importance, for the welfare of human society. Give reasons for your answer.
Biogas, Citric acid, Penicillin and Curd
Answer:
The order of arrangement of products according to their decreasing importance is :
Penicillin > Biogas > Citric acid > Curd
Penicillin is the most important product for the welfare of human society. It is an antibiotic, which is used for controlling various bacterial diseases. The second most important product is biogas. It is aneco-friendly source of energy. The next important product is citric acid, which is used as a food preservative. The least important product is curd, a food item obtained by the action of Lactobacillus bacteria on milk.

Question 15.
How do biofertilisers enrich the fertility of the soil?
Answer:
Bio-fertilisers are living organisms which help in increasing the fertility of soil. It involves the selection of beneficial micro-organisms that help in improving plant growth through the supply of plant nutrients. These are introduced to seeds, roots, or soil to mobilise the availability of nutrients by their biological activity. Thus, they are extremely beneficial jf in enriching the soil with organic nutrients. Many species of bacteria and cyanobacteria have the ability to fix free atmospheric nitrogen. Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants. Azospirillum and Azotobacter are free living nitrogen-fixing bacteria, whereas Anabaena, Nostoc, and Oscillatoria are examples of nitrogen-fixing cyanobacteria. Bio-fertilisers are cost effective and eco-friendly.