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Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 6 Molecular Basis of Inheritance Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance

PSEB 12th Class Biology Guide Molecular Basis of Inheritance Textbook Questions and Answers

Question 1.
Group the following as nitrogenous bases and nucleosides:
Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Answer:
Nitrogenous Bases: Adenine, thymine, uracil, and cytosine.
Nucleosides: Cytidine and guanosine.

Question 2.
If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.
Answer:
According to Chargaffs rule, the DNA molecule should have an equal ratio of pyrimidine (cytosine and thymine) and purine (adenine and guanine). It means that the number of adenine molecules is equal to thymine molecules and the number of guanine molecules is equal to cytosine molecules.
% A = % T and % G = % C
If double stranded DNA has 20% of cytosine, then according to the law, it would have 20% of guanine.
Thus, percentage of G + C content = 40%
The remaining 60% represents both A + T molecule. Since adenine and guanine are always present in equal numbers, the percentage of adenine molecule is 30%.

Question 3.
If the sequence of one strand of DNA is written as follows: 5-ATGCATGCATGCATGCATGCATGCATGC-3′
Write down the sequence of complementary strand in 5′ → 3′ direction.
Answer:
The DNA strands are complementary to each other with respect to base sequence. Hence, if the sequence of one strand of DNA is
5′- ATGCATGCATGCATGCATGCATGCATGC – 3′
Then, the sequence of complementary strand in 5′-3′ direction will be
3′- TACGTACGTACGTACGTACGTACGTACG – 5′
Therefore, the sequence of nucleotides on DNA polypeptide in 5′-3′ direction is
5′- GCATGCATGCATGCATGCATGCATGCAT – 3′

Question 4.
If the sequence of the coding strand in a transcription unit is written as follows:
5′ – ATGCATGCATGCATGCATGCATGCATGC-3 Write down the sequence of mRNA.
Answer:
If the coding strand in a transcription unit is
5′ – ATGCATGCATGCATGCATGCATGCATGC-3′
Then, the template strand in 3′ to 5′ direction would be
3′ – TACGTACGTACGTACGTACGTACGTACG-5′
It is known that the sequence of mRNA is same as the coding strand of DNA.
However, in RNA, thymine is replaced by uracil.
Hence, the sequence of mRNA will be
5′ – AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Answer:
Watson and Crick observed that the two strands of DNA are f anti-parallel and complementary to each other with respect to their base sequences. This type of arrangement in DNA molecule led to the hypothesis that DNA replication is semi-conservative. It means that the double stranded DNA molecule separates and then, each of the separated strand acts as a template for the synthesis of a new complementary strand. As a result, each DNA molecule would have one parental strand and a newly synthesised daughter strand.

Since only one parental strand is conserved in each daughter molecule, it is known as semi-conservative mode of replication.

Quetion 6.
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.
Answer:
There are two different types of nucleic acid polymerases.

1. DNA-dependent DNA polymerases
2. DNA-dependent RNA polymerases

The DNA-dependent DNA polymerases use a DNA template strand for synthesising a new strand of DNA, whereas DNA-dependent RNA polymerases use a DNA template strand for synthesising a new strand of RNA.

Question 7.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer:
Hershey and Chase worked with bacteriophage and E.coli to prove that DNA is the genetic material. They used different radioactive isotopes to label DNA and protein coat of the bacteriophage.

They grew some bacteriophages on a medium containing radioactive phosphorus (³²) to identify DNA and some on a medium containing radioactive sulphur (³⁵S) to identify protein. Then, these radioactive labelled phages were allowed to infect E.coli bacteria. After infecting, the protein coat of the bacteriophage was separated’from the bacterial cell by blending and then subjected to the process of centrifugation. Since the protein coat was lighter, it was found in the supernatant while the infected bacteria got settled at the bottom of the centrifuge tube. Hence, it was proved that DNA is the genetic material as it was transferred from virus to bacteria.

Question 8.
Differentiate between the following :
(a) Repetitive DNA and Satellite DNA
(b) mRNA and tRNA
(c) Template strand and Coding strand Arts,
Answer:
(a) Repetitive DNA and Satellite DNA

Repetitive DNA	Satellite DNA
Repetitive DNA are DNA sequences that contain small segments, which are repeated many times.	Satellite DNA are DNA sequences that contain highly repetitive DNA.

(b) mRNA and tRNA

mRNA	tRNA
1. mRNA or messenger RNA acts as a template for the process of transcription.	tRNA or transfer RNA acts as an adaptor molecule that carries a specific amino acid to mRNA for the synthesis of polypeptide.
2. It is a linear molecule.	It has clover leaf shape.

(c) Template strand and Coding strand

Template strand	Coding strand
1. Template strand of DNA acts as a template for the synthesis of mRNA during transcription.	Coding strand is a sequence of DNA that has the same base sequence as that of mRNA (except thymine that is replaced by uracil in DNA).
2. It runs from 3′ to 5′.	It runs from 5′ to 3′.

Question 9.
List two essential roles of ribosome during translation.
Answer:
The important functions of ribosome during translation are as follows :
(a) Ribosome acts as the site where protein synthesis takes place from individual amino .acids. It is made up of two subunits.
The smaller subunit comes in contact with mRNA and forms a protein synthesising complex whereas the larger subunit acts as an amino acid binding site.

(b) Ribosome acts as a catalyst for forming peptide bond. For example, 23s r-RNA in bacteria acts as a ribozyme.

Question 10.
In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?
Answer:
Lac operon is a segment of DNA that is made up of three adjacent structural genes, namely, an operator gene, a promoter gene, and a regulator gene. It works in a coordinated manner to metabolise lactose into glucose and galactose.

In lac operon, lactose acts as an inducer. It binds to the repressor and inactivates it. Once the lactose binds to the repressor, RNA polymerase binds to the promoter region. Hence, three structural genes express their product and respective enzymes are produced. These enzymes act on lactose so that lactose is metabolised into glucose and galactose. After sometime, when the level of inducer decreases as it is completely metabolised by enzymes, it causes synthesis of the repressor from regulator gene. The repressor binds to the operator gene and prevents RNA polymerase from transcribing the operon. Hence, the transcription is stopped. This type of regulation is known as negative regulation.

Question 11.
Explain (in one or two lines) the function of the following:
(a) Promoter
(b) tRNA
(c) Exons
Answer:
(a) Promoter: Promoter is a region of DNA that helps in initiating the process of transcription. It serves as the binding site for RNA polymerase.

(b) tRNA: tRNA or transfer RNA is a small RNA that reads the genetic code present on mRNA. It carries specific amino acid to mRNA on ribosome during translation of proteins.

(c) Exons: Exons are coding sequences of DNA in eukaryotes that transcribe for proteins.

Question 12.
Why is the Human Genome project called a mega project?
Answer:
Human genome project was considered to be a mega project because it had a specific goal to sequence every base pair present in the human genome. It took around 13 years for its completion and got accomplished in year 2006. It was a large scale project, which aimed at developing new technology and generating new information in the field of genomic studies. As a result of it, several new areas and avenues have opened up in the field of genetics, biotechnology, and medical sciences. It provided clues regarding the understanding of human biology.

Question 13.
What is DNA fingerprinting? Mention its application.
Answer:
DNA fingerprinting is a technique used to identify and analyse the variations in various individuals at the level of DNA. It is based on variability and polymorphism in DNA sequences.
Applications

1. It is used in forensic science to identify potential crime suspects.
2. It is used to establish paternity and family relationships.
3. It is used to identify and protect the commercial varieties of crops and livestock.
4. It is used to find out the evolutionary history of an organism and trace out the linkages between groups of various organisms.

Question 14.
Briefly describe the following:
(a) Transcription
(b) Polymorphism
(c) Translation
(d) Bioinformatics
Answer:
(a) Transcription: It is the process of synthesis of RNA from DNA template. A segment of DNA gets copied into mRNA during the process. The process of transcription starts at the promoter region of the template DNA and terminates at the terminator region. The segment of DNA between these two regions is known as transcription unit. The transcription requires RNA polymerase enzyme, a DNA template, four types of ribonucleotides, and certain cofactors such as Mg²⁺.
The three important events that occur during the process of transcription are as follows:

1. Initiation
2. Elongation
3. Termination

The DNA-dependent RNA polymerase and certain initiation factors bind at the double stranded DNA at the promoter region of the template strand and initiate the process of transcription. RNA polymerase moves along the DNA and leads to the unwinding of DNA duplex into two separate strands. Then, one of the strands, called sense strand, acts as template for mRNA synthesis. The enzyme, RNA polymerase, utilises nucleoside triphosphates (dNTPs) as raw material and polymerises them to form mRNA according to the complementary bases present on the template DNA«. This process of opening of helix-and elongation of polynucleotide chain continues until the enzyme reaches the terminator region. As RNA polymerase reaches the terminator region, the newly synthesised mRNA transcripted along with enzyme is released. Another factor called terminator factor is required for the termination of the transcription.

(b) Polymorphism: It is a form of genetic variation in which distinct nucleotide sequence can exist at a particular site in a DNA molecule. This heritable mutation is observed at a high frequency in a population. It arises due to mutation either in somatic cell or in the germ cells. The germ cell mutation can be transmitted from parents to their offsprings. This results in accumulation of various mutations in a population, leading to variation and polymorphism in the population. This plays a very important role in the process of evolution and tracing human history.

(c) Translation: It is the process of polymerising amino acid to form a polypeptide chain. The triplet sequence of base pairs in mRNA defines the order and sequence of amino acids in a polypeptide chain.
The process of translation involves following three steps:

1. Initiation
2. Elongation
3. Termination

During the initiation of the translation, tRNA gets charged when the amino acid binds to it using ATP. The start (initiation) codon (AUG) present on mRNA is recognised only by the charged tRNA. The ribosome acts as an actual site for the process of translation and contains two separate sites in a large subunit for the attachment of subsequent amino acids. The small subunit of ribosome binds to mRNA at the initiation codon (AUG) followed by the large subunit. Then, it initiates the process of translation. During the elongation process, the ribosome moves one codon downstream along with mRNA so as to leave the space for binding of another charged tRNA. The amino acid brought by tRNA gets linked with the previous amino acid through a peptide bond and this process continues resulting in the formation of a polypeptide chain. When the ribosome reaches one or more STOP codon (VAA, UAG, and UGA), the process of translation gets terminated. The polypeptide chain is released and the ribosomes get detached from mRNA.

(d) Bioinformatics: It is the application of computational and statistical techniques to the field of molecular biology. It solves the practical problems arising from the management and analysis of biological data. The field of bioinformatics developed after the completion of human genome project (HGP). This is because enormous amount of data has been generated during the process of HGP that has to be managed and stored for easy access and interpretation for future use by various scientists. Hence, bioinformatics involves the creation of biological databases that store the vast information of biology.

It develops certain tools for easy and efficient access to the information and its utilisation. Bioinformatics has developed new algorithms and statistical methods to find out the relationship between the data, to predict protein structure and their functions, and to cluster the protein sequences into their related families.

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 11th Class Biology Book Solutions Chapter 11 Transport in Plants Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 11 Transport in Plants

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

Question 1.
What are the factors affecting the rate of diffusion?
Answer:
Factors affecting the rate of diffusion are as follows:

• Gradient of concentration
• Permeability of membrane
• Temperature
• Pressure

Question 2.
What are porins? What role do they play in diffusion?
Answer:
The porins are proteins that form huge pores in the outer membranes of the plastids, mitochondria and some bacteria allowing molecules up to the size of small proteins to pass through. Porins facilitate diffusion.

Question 3.
Describe the role played by protein pumps during active transport in plants.
Answer:
Protein pumps use energy to carry substances across the cell membrane. These pumps can transport substances from a low concentration to a high concentration (‘uphill’ transport). Transport rate reaches a maximum when all the protein transporters are being used or are saturated. Like enzymes, the carrier protein is very specific in what it carries across the membrane. These proteins are sensitive to inhibitors that react with protein side chains.

Question 4.
Explain why pure water has the maximum water potential.
Answer:
Water molecules possess kinetic energy. In liquid and gaseous form, they are in random motion that is both rapid and constant. The greater the concentration of water in a system, the greater is its kinetic energy or ‘water potential’. Hence, pure water will have the greatest water potential. Water potential is denoted by the Greek symbol psi or \p and is, expressed in pressure units such as pascals (Pa).

Question 5.
Differentiate between the following:
(a) Diffusion and Osmosis
(b) Transpiration and Evaporation
(c) Osmotic pressure and Osmotic Potential
(d) Imbibition and Diffusion
(e) Apoplast and Symplast Pathways of Movement of Water in Plants
(f) Guttation and Transpiration

(a) Differences between Diffusion and Osmosis

Diffusion	Osmosis
1. It is a movement of molecules from high concentration to low concentration.	It is a movement of molecules from high concentration to low concentration
2. It does not require any driving force.	It occurs in response to a driving force.

(b) Differences between Transpiration and Evaporation

Transpiration	Evaporation
1. It is the loss of water through the aerial parts of plants.	It is the loss of water from free surface of water.
2. It occurs in living tissues.	It occurs in non-living surfaces.
3. It is both physical and physiological process.	It is only a physical process, controlled by environmental factors.

(c) Differences between Osmotic Pressure and Osmotic Potential

Osmotic Pressure	Osmotic Potential
1. It is the pressure required to stop the movement of water molecules through a semipermeable membrane.	It is the amount by which water potential is reduced by the presence of solute.
2. Osmotic pressure is the positive pressure.	Osmotic potential is negative.

(d) Differences between Imbibition and Diffusion

Imbibition	Diffusion
It is a special type of diffusion, where water is absorbed by solids-colloids causing them to increase in volume. For example, absorption of water by dry seeds and dry wood.	In diffusion, molecules move in a random fashion. It is not dependent on a living system.

(e) Differences between Apoplast and Symplast Pathways of Movement of Water in Plants

Apoplast	Symplast
1. It is the system of adjacent cell walls that is continuous throughout the plant except casparian strips of the endodermis of the roots.	It is the system of interconnected protoplast.
2. Water moves through the intercellular spaces and the walls of cells.	Water travels through the cytoplasm
3. Movement does not involve crossing the cell membrane.	Water has to move in cells through the cell membrane.

(f) Differences between Guttation and Transpiration

Guttation	Transpiration
1. It occurs through hydathodes, present at the vein ends.	It occurs through general surface stomata and lenticles.
2. It occurs in leaves only.	It can occur through all aerial parts.
3. It does not occur in deficient water conditions and never leads to wilting.	It can occur in water deficient conditions leading to wilting.
4. It is regulated by humidity, temperature and presence of water in soil.	It is regulated by a number of external and internal factors such as relative humidity, temperature, opening and closing of stomata, etc.

Question 6.
Briefly describe water potential. What are the factors affecting it?
Answer:
Water potential is the potential energy of water relative to pure free water (e.g., deionised water). It quantifies the tendency of water to move from one area to another due to osmosis, gravity, mecanical pressure or matrix effects including surface tension. Water potential is measured in units of pressure and is commonly represented by the Greek letter (psi). This concept has proved especially useful in understanding water movement within plants, animals and soil.

Water potential of a cell is affected by both solute and pressure potential. The relationship between them is as follows:
Ψ_w = Ψ_s + Ψ_p

Question 7.
What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?
Answer:
If a pressure greater than atmospheric pressure is applied to pure water or a solution its water potential increases. It is equivalent to pumping water from one place to another. Pressure can be build up in a plant system when water enters a plant cell due to diffusion causing a pressure build up against the cell wall. It makes the cell turgid, this increases the pressure potential. Pressure potential is usually positive. It is denoted by Ψ_s.

Question 8.
(a) With the help of well-labelled diagrams, describe the process of plasmolysis in plants, giving appropriate examples.
(b) Explain what will happen to a plant cell if it is kept in a solution having higher water potential.
Answer:
(a) Plasmolysis occurs when water moves out of the cell and the cell membrane of a plant cell shrinks away from its cell wall. This occurs when the cell is kept in a solution that is hypertonic (has more solutes) to the protoplasm. Water moves out from the cell through diffusion and causes the protoplasm to shrink away from the walls. In such situation, cell becomes plasmolysed.

When the cell is placed in an isotonic solution. There is not flow of water towards the inside or outside. If the external solution balances the osmotic pressure of the cytoplasm, it is said to be isotonic. When the water flow into the cell and out of the cells are in equilibrium the cell is called flaccid.

(b) When the plant cell is kept in a solution having high water potential (hypotonic solution or dilute solution as compared to cytoplasm), water diffuses into the cell causing the cytoplasm to build up a pressure against the wall, called turgor pressure. The pressure exerted by the protoplasts due to entry of water against the rigid walls is called pressure potential (Ψ_p). Because of the rigidity of the cell wall, the cell does not rupture. This turgor pressure is ultimately responsible for enlargement of cells.

Question 9.
How is the mycorrhizal association helpful in absorption of water and minerals in plants?
Answer:
A mycorrhiza is a symbiotic association of a fungus with a root system. The fungal filaments form a network around the young root or they penetrate the root cells. The hyphae have a very large surface area that absorb mineral ions and water from the soil from a much larger volume of soil that perhaps a root cannot do. The fungus provides minerals and water to the roots, in turn the roots provide sugars and N-containing compounds to the mycorrhizae.

Question 10.
What role does root pressure play in water movement in plants?
Answer:
Root pressure can provide only modest push during water transport in plants. The main role of root pressure is in re-establishing the continuous chain of water molecules in the xylem. The continuous chain often breaks due to enormous tension created by transpiration pull.

Question 11.
Describe transpiration pull model of water transport in plants. What are the factors influencing transpiration? How is it useful to plants?
Answer:
Transpiration occurs mainly through the stomata in the leaves. As water evaporates through the stomata, since the thin film of water over the cells is continuous, it results in pulling of water, molecule by molecule, into the leaf from the xylem. Also, because of lower concentration of water vapour in the atmosphere as compared to the substomatal cavity and intercellular spaces, water diffuses into the surrounding air. This creates a transpiration pull.

Factors Affecting Transpiration: Temperature, light, humidity and wind speed.
Importance of Transpiration: Transport of liquids and minerals is facilitated because of transpiration.

Question 12.
Discuss the factors responsible for ascent of xylem sap in plants.
Answer:
The transpiration driven ascent of xylem sap depends mainly on the following physical properties of water:

• Cohesion: Mutual attraction between water molecules.
• Adhesion: Attraction of water molecules to polar surfaces (such as the surface of tracheary elements).
• Surface Tension: Water molecules are attracted to each other in the liquid phase more than to water in the gas phase.

These properties give water high tensile strength, i.e., an ability to resist a pulling force and high capillarity, i.e., the ability to rise in thin tubes. In plants, capillarity is aided by the small diameter of the tracheary elements, the tracheids and vessel elements.

Question 13.
What essential role does the root endodermis play during mineral absorption in plants?
Answer:
The endodermis of roots have many transport proteins embedded in their plasma membrane. They let some solutes cross the membrane but not all. Transport proteins in endodermis cells enable plant cells lo adjust the quantity and types of solutes to be absorbed from the soil. It regulates the quantity and type of minerals and ions, that reach the xylem tissue of plants.

Question 14.
Explain why xylem transport is unidirectional and phloem transport bi-directional?
Answer:
The source sink (food making tissue-tissue which stores food) relationship is variable in plants so, the direction of movement in the phloem can be upwards downwards, i.e., bi-directional. It is opposite to xylem, where the movement is always unidirectional. Hence, unlike one-way flow of water in transpiration, food in phloem sap can be transported in any required direction so long there is a source of sugar and a sink is able to use, store or remove the sugar. Here, in case of unidirectional flow in xylem tissue, it is important to note that root endodermis because of the layer of suberin has the ability to actively transport ions in one direction only.

Question 15.
Explain pressure flow hypothesis of translocation of sugars in plants.
Answer:
The Pressure Flow or Mass Flow Hypothesis: The accepted mechanism used for the translocation of sugars from source to sink is called the pressure flow hypothesis. As glucose is prepared at the source (by photosynthesis) it is converted to sucrose (a disaccharide). The sugar is then moved in the form of sucrose into the companion cells and then into the living phloem sieve tube cells by active transport.

As osmotic pressure builds up the phloem sap will move to areas of lower pressure. At the sink, osmotic pressure must be reduced. Again active transport is necessary to move the sucrose out of the phloem sap and into the cells which will use the sugar converting it into energy, starch or cellulose. As sugars are removed, the osmotic pressure decreases and water moves out of the phloem.

Hydrostatic pressure in the phloem sieve tube increases, pressure flow begins and the sap moves through the phloem. Meanwhile, at the sink, incoming sugars are actively transported out of the phloem and removed as complex carbohydrates. The loss of solute produces a high water potential in the phloem and water passes out, returning eventually to xylem.

Question 16.
What causes the opening and clog” T of guard cells of stomata during transpiration?
Answer:
The immediate cause of the opening or closing-of the stomata is a change in the turgidity of the guard cells. The inner wall of each guard cell, towards the pore or stomatal, aperture, is thick and elastic. When, turgidity increases within the two guard cells flanking each stomatal aperture or pore, the thin outer walls bulge out and force the inner walls into a crescent shape. The opening of the stoma is also aided due to the orientation of the microfibrils in the cell walls of the guard cells. Cellulose microfibrils are oriented radially rather than longitudinally making it easier for the stoma to open. When the guard cells lose turgor, due to water loss (or water stress) the elastic inner walls regain their original shape, the guard cells become flaccid and the stoma closes.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 4 Reproductive Health Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 4 Reproductive Healths

PSEB 12th Class Biology Guide Reproductive Health Textbook Questions and Answers

Question 1.
What do you think is the significance of reproductive health in a society?
Answer:
Reproductive health is the total well being in all aspects of reproduction. It includes physical, emotional, behavioural, and social well being. Sexually transmitted diseases such as AIDS, gonorrhoea, etc. are transferred from one individual to another through sexual contact. It can also lead to unwanted pregnancies.

Hence, it is necessary to create awareness among people, especially the youth, regarding various reproduction related aspects as the young individuals are the future of the country and they are most susceptible of acquiring sexually transmitted diseases. Creating awareness about the available birth control methods, sexually transmitted diseases and their preventive measures, and gender equality will help in bringing up a socially conscious healthy family. Spreading. awareness regarding uncontrolled population growth and social evils among young individuals will help in building up a reproductively healthy society.

Question 2.
Suggest the aspects of reproductive health which need to be given special attention in the present scenario.
Answer:
Reproductive health is the total well being in all aspects of reproduction. The aspects which have to be given special attention in the present scenario are as follows:
1. Counselling and creating awareness among people, especially the youth, about various aspects of reproductive health, such as sexually transmitted diseases, available contraceptive methods, case of pregnant mothers, adolescence, etc.

2. Providing support and facilities such as medical assistance to people during pregnancy, STDs, abortions, contraceptives, infertility, etc. for building a reproductively healthy society.

Question 3.
Is sex education necessary in schools? Why?
Answer:
Yes, introduction of sex education in schools is necessary. It would provide right information to young individuals at the right time about various aspects of reproductive health such as reproductive organs, puberty, and adolescence related changes, safe sexual practices, sexually transmitted diseases, etc.

The young individual or adolescents are more susceptible in acquiring various sexually transmitted diseases. Hence, providing information to them at the right time would help them to lead a reproductively healthy life and also protect them from the myths and misconceptions about various sex related issues.

Question 4.
Do you think that reproductive health in our country has improved in the past 50 years? If yes, mention some such areas of improvement.
Answer:
Yes, the reproductive health has tremendously improved in India in the last 50 years. The areas of improvement are as follows:

1. Massive child immunisation programme, which has lead to a decrease in the infant mortality rate.
2. Maternal and infant mortality rate, which has been decreased drastically due to better post natal care.
3. Family planning, which has motivated people to have smaller families.
4. Use of contraceptive, which has resulted in a decrease in the rate of sexually transmitted diseases and unwanted pregnancies.

Question 5.
What are the suggested reasons for population explosion?
Answer:
The human population is increasing day by day, leading to population explosion. It is because of the following two major reasons:
(a) Decreased death rate
(b) Increased birth rate and longevity
The death rate has decreased in the past 50 years. The factor leading to decreased death rate and increased birth rate are control of diseases, awareness and spread of education, improvement in medical facilities, ensured food supply in emergency situation, etc. All this has resulted in an increase in the longevity of an individual.

Question 6.
Is the use of contraceptives justified? Give reasons.
Answer:
Yes, the use of contraceptives is absolutely justified. The human population is increasing tremendously. Therefore, to regulate the population growth by regulating reproduction has become a necessary demand in the present times. Various contraceptive devices have been devised to reduce unwanted pregnancies, which help in bringing down the increased birth rate and hence, in checking population explosion.

Question 7.
Removal of gonads cannot be considered as a contraceptive option. Why?
Answer:
Contraceptive devices are used to prevent unwanted pregnancy and to prevent the spreading of STDs. There are many methods, such as natural, barrier, oral, and surgical methods, that prevent unwanted pregnancy. However, the complete removal of gonads cannot be a contraceptive option because it will lead to infertility and unavailability of certain hormones that are required for normal functioning of accessory reproductive parts. Therefore, only those contraceptive methods can be used that prevent the chances of fertilisation rather than making the person infertile forever.

Question 8.
Amniocentesis for sex determination is banned in our country. Is this ban necessary? Comment.
Answer:
Amniocentesis is a pre-natal diagnostic technique that is used to determine the sex and metabolic disorders of the developing foetus in the mother’s uterus through the observation of the chromosomal patterns. This method was developed so as to determine any kind of genetic disorder present in the foetus. However, unfortunately, this technique is being misused to detect the sex of the child before birth and the female foetus is then aborted. Thus, to prevent the increasing female foeticides, it is necessary to ban the usage of amniocentesis technique for determining the sex of a child.

Question 9.
Suggest some methods to assist infertile couples to have children.
Answer:
Infertility is the inability of a couple to produce a baby even after unprotected intercourse. It might be due to abnormalities present in either male or female, or might be even both the partners. The techniques used to assist infertile couples to have children are as follows:
(a) Test Tube Babies: This involves in-vitro fertilisation where the sperms meet the egg outside the body of a female. The zygote, hence produced, is then transferred in the uterus or fallopian tube of a normal female. The babies produced from this method are known as test tube babies.

(b) Gamete Intra Fallopian Transfer (GIFT): It is a technique that involves the transfer of gamete (ovum) from a donor into the fallopian tube of the recipient female who is unable to produce eggs, but has the ability to conceive and can provide right conditions for the development of an embryo.

(c) Intra Cytoplasmic Sperm Injection (ICSI): It is a method of injecting sperm directly into the ovum to form an embryo in laboratory.

(d) Artificial Insemination: Artificial insemination is a method of transferring semen (sperm) from a healthy male donor into the vagina or uterus of the recipient female. It is employed when the male partner is not able to inseminate the female or has low sperm counts.

Question 10.
What are the measures one has to take to prevent from contracting STDs?
Answer:
Sexually transmitted diseases (STDs) get transferred from one individual to the other through sexual contact. Adolescents and young adults are at the greatest risk of acquiring these sexually transmitted diseases. Hence, creating awareness among the adolescents regarding its after-effects can prevent them from contracting STDs. The use of contraceptives, such as condoms, etc. while intercourse, can prevent the transfer of these diseases. Also, sex with unknown partners or multiple partners should be avoided as they may have such diseases. Specialists should be consulted immediately in case of doubt so as to assure early detection and cure of the disease.

Question 11.
State True/False with explanation.
(a) Abortions could happen spontaneously too. (True/False)
(b) Infertility is defined as the inability to produce a viable offspring and is always due to abnormalities/defects in the female partner. (True/False)
(c) Complete lactation could help as a natural method of contraception. (True/False)
(d) Creating awareness about sex related aspects is an effective method to improve reproductive health of the people. (True/False)
Answer:
(a) False
Abortion is term given for medical termination of pregnancy.

(b) False
Infertility is defined as the inability of the couple to produce baby even after unprotected coitus. It might occur due to abnormalities/defects in either male or female or both.

(c) False
Complete lactation or lactational amenorrhea is a natural method of contraception. Flowever, it is limited till lactation period, which continues till six months after parturition.

(d) True.

Question 12.
Correct the following statements:
(a) Surgical methods of contraception prevent gamete formation.
(b) All sexually transmitted diseases are completely curable.
(c) Oral pills are very popular contraceptives among the rural women.
(d) In E. T. techniques, embryos are always transferred into the uterus.
Answer:
(a) Surgical methods of contraception prevent the flow of gamete during intercourse.
(b) Some of the sexually transmitted diseases are curable if they are detected early and treated properly. AIDS is still an incurable disease.
(c) Oral pills are very popular contraceptives among urban women.
(d) In embryo transfer technique, 8 celled embryos are transferred into the fallopian tube while more than 8 celled embryos are transferred into the uterus.