Author Archives: Rohit

Question 1.
How is diapause different from hibernation ?
Solution:
Diapause is different from hibernation. The table below shows the differences between them :

Question 2.
If a marine fish is placed in a freshwater aquarium/will the fish be able to survive? Why or why not?
Solution:
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 in the marine environment. In freshwater conditions, they are unable to regulate the water entering the 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.
Solution:
Phenotypic adaptation involves non-genetic changes in individuals such as physiological modifications like acclimatization or behavioural changes.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?
Solution:
organisms survive at a temperature range of 0° to 40°C or less. However, there are some notable exceptions. Certain microorganisms live in hot springs and deep-sea hydrothermal vents where temperature far exceeds 100°C. They survive at the high temperature due to the occurrence of branched-chain lipids in their cell membrane that reduces the fluidity of cell membranes and the occurrence of the minimum amount of free water in their cells that provides resistance to high temperature

Question 5.
List the attributes that populations but not individuals possess.
Solution:

1. Natality
2. Mortality
3. Growth forms
4. Population density
5. Population dispersion
6. Population age distribution

Question 6.
If a population growing exponentially double in size in 3 years, what is the intrinsic rate of increase (r) of the population?
Solution:
The intrinsic rate of increase(r), can be calculated by the following exponential growth equation:

Question 7.
Name important defence mechanisms in plants against herbivory.
Solution:

1. Modification of leaves into thorns.
2. Development of spiny margins on leaves.
3. Development of sharp silicated edges on leaves.

Question 8.
An orchid plant is growing on the branch of the mango tree. How do you describe this interaction between the orchid and the mango tree?
Solution:
An orchid growing as an epiphyte on a branch of mango tree is an example of commensalism. Commensalism is the relationship between individuals of two species of which one is benefited and the other is almost unaffected, i.e., neither benefited nor harmed. A commensal may get shelter (protection), or ride, or support instead of or in addition to food. Epiphytes are space parasites, they use trees only for attachment and manufacture their own food by photosynthesis. In Vanda, an epiphytic orchid, a special kind of aerial roots (hanging roots) hang freely in the air and absorb moisture with the help of their special absorptive tissue called velamen.

Question 9.
What is the ecological principle behind the biological control method of managing pest insects?
Solution:
Predation is the means of biological control to manage pest insects where predators prey upon pests and regulate their numbers in the habitat.

Question 10.
Distinguish between the following:

1. Hibernation and Aestivation
2. Ectotherms and Endotherms

Solution:

1. Differences between hibernation and aestivation are as follows :
   
2. Differences between ectotherms and endotherms are as follows:
   

Question 11.
Write a short note on :
(a) Adaptations of desert plants and animals
(b) Adaptations of plants to water scarcity
(c) Behavioral adaptations in animals
(d) Importance of light to plants
(e) Effect of temperature or water scarcity and the adaptations of animals.
Solution:
a. Desert plants are called xerophytes. They have adaptations for increased water absorption, reduction in transpiration and water storage. Many desert plants have a thick cuticle on their leaf surfaces and have their stomata arranged in deep pits to minimise water loss through transpiration. They also have a special photosynthetic pathway that enables their stomata to remain closed during day time. In desert plants like Opuntia, leaves are reduced to spines. Animals of dry areas may use metabolic water and reduce water loss bypassing nearly solid faeces and urine.

b. Xerophytes have special adaptations to withstand prolonged periods of drought. These are of four types – ephemerals, annuals, succulents and non-succulent perennials.

• Ephemerals (drought escapers): Plants which live for a brief period and complete their life cycle during the rains.
• Annuals (drought evaders): Plants which continue to live for a few
  months even after rains in hot dry conditions. They have modifications to reduce transpiration.
• Succulents (drought resistants): Plants have fleshy organs to store large amounts of water. They have a very thick cuticle, sunken stomata which open during night only.
• Non-succulent perennials: These are true xerophytes. They have an extensive root system to absorb the maximum amount of water. They possess waxy coatings on leaves, sunken stomata, reduced leaf blades etc. to reduce transpiration.

c. The animals with variable temperatures called poikilotherms are affected by temperature variations. They are also called ectotherms. They show different adaptations like hibernation, aestivation, periodic activity, winter eggs, and migration.

d. Sun is the ultimate source of energy for most of the organisms on this earth. Light is the visible range of the electromagnetic spectrum. Light (400 nm-700nm) is effective in photosynthesis and is called photo-synthetically active radiation or PAR. The intensity of light, duration of light, etc. are also influencing the growth of plants.

e. Animals live in arid regions show two kinds of adaptations

1. Reducing loss of water from their bodies.
2. Ability to tolerate arid conditions.

Question 12.
List the various abiotic environmental factors.
Solution:
Abiotic factors are non-living factors and conditions of the environment which influence the survival, function and behaviour of organisms. Various abiotic factors are :

(i) Temperature – Temperature is one of the most important environmental factors. The average temperature varies seasonally. It ranges from subzero level in polar areas and high altitudes to more than 50°C in tropical deserts in summer and exceeds 100°C in thermal springs and deep-sea hydrothermal vents.

(ii) Water – Next to temperature, water is the most important factor which influences the life of organisms. The productivity and distribution of land plants are dependent upon the availability of water. Animals are adapted according to water availability. E.g., aquatic animals are ammonotelic while xerophytic animals excrete dry feces and concentrated urine.

(iii) Light – Plants produce food through photosynthesis for which sunlight is essential to the source of energy. Light intensity, light duration and light quality influences the number of life processes in organisms, such as – photosynthesis, growth, transpiration, germination, pigmentation, movement and photoperiodism.

(iv) Humidity – Humidity refers to the moisture (water vapour) content of the air. It determines the formation of clouds, dew and fog. It affects the land organisms by regulating the loss of water as vapour from their bodies through evaporation, perspiration and transpiration.

(v) Precipitation – Precipitation means rainfall, snow, sleet or dew. Total annual rainfall, seasonal distribution humidity of the air and amount of water retained in the soil are the main criteria that limit the distribution of plants and animals on land.

(vi) Soil – The soil is one of the most important ecological factor called the edaphic factor. It comprises of different layers called horizons. The upper weathered humus containing part of soil sustains terrestrial plant life.

Question 13.
Give an example for:

1. An endothermic animal
2. An ectothermic animal
3. An organism of the benthic zone.

Solution:

1. Hedgehog
2. Frog
3. Sponges

Question 14.
Define population and community.
Solution:
Population: A population is a group of individuals of the same species, which can reproduce among themselves and occupy a particular area in a given time.

Community: It is an assemblage of several populations in a particular area and time and exhibits interaction and interdependence through trophic relationship.

Question 15.
Define the following terms and give one example for each.
(a) Commensalism
(b) Parasitism
(c) Camouflage
(d) Mutualism
(e) Interspecific competition
Solution:

a. Commensalism is an interspecific interaction between individuals of two species where one species is benefitted and the other is not affected.
e. g. Orchid and mango tree.

b. Parasitism is an interspecific interaction between individuals of two species where generally small species is benefitted and the large species are affected, e.g. Malarial parasite and human beings.

c. Camouflage: It is the ability of the animals to blend with the surroundings or background. In this way, animals remain unnoticed for protection or aggression. An example is a stick insect.

d. Mutualism is an interspecific interaction between individuals of two species where both the interacting species are benefitted in an obligatory way. e.g. Pollination in plants by animals.

e. Interspecific competition: It is an interaction between individuals of two species where both the interacting species are affected, e.g. Monarch butterfly and Queen monarch.

Question 16.
With the help of a suitable diagram describe the logistic population growth curve.
Solution:
Logistic population growth curve or S-shaped or sigmoid growth curve is shown by the populations of most organisms. It has the following phases: lag phase, log phase, exponential phase and stationary phase. In lag phase there is little or no increase in population. In log phase increase in population starts and occurs at a slow rate in the beginning. During exponential phase, increase in population becomes rapid and soon attains its full potential rate. This is due to the constant environment, availability of food and other requirements of life in plenty, absence of predation and interspecific competition and no serious intraspecific competition so that the curve rises steeply upward. The growth rate finally slows down as environmental resistance increases.

Finally, the population becomes stable during the stationary phase because now the number of new cells produced almost equals to the number of cells that die. Every population tends to reach a number at which it becomes stabilized with the resources of its environment. A stable population is said to be in equilibrium, or at saturation level. This limit in population is a constant K and is imposed by the carrying capacity of the environment. The sigmoid growth form is represented by the following equation :

r = intrinsic rate of natural increase
N = population density at time t; K = carrying capacity.

Question 17.
Select the statement which best explains 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.
Solution:
(d) One organism is benefited, other is affected,

Question 18.
List any three important characteristics of a population and explain.
Solution:
The three important characteristics of a population are:

1. Birth and death rate
2. Age structure
3. Sex ratio

(i) The birth rate (natality) of a population refers to the average number of young ones produced per unit time (usually per year). In the case of humans, it is commonly expressed as the number of births per 1,000 individuals in the population per year. The death rate (mortality) of a population is the average number of individuals that die per unit time (usually per year). In humans, it is commonly expressed as the number of deaths per 1,000 persons in a population per year.

(ii) The age structure of a population is the percentage of individuals of different ages such as young, adult and old. Age structure is shown bv organisms in which individuals of more than one generation coexist. The ratio of various age groups in a population determines the current reproductive status of the population. It also indicates what may be expected in the future. The population is divided into three age groups; pre-reproductive, reproductive and post-reproductive.

(iii) The sex ratio of a population refers to the number of females per thousand male individuals. There were 933 females per 1,000 males in our country in the 2001 census. The number of females in a population is very important as it is often directly related to the number of births. The number of males may be less significant because in many species a single male can mate with several females.

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.
Solution:
(c) Toxin is inactive: In bacteria, the toxin is present in an inactive form called prototoxin. This gets converted into the active form when it enters the salivary gland of insects having an alkaline medium.

Question 2.
What are transgenic bacteria? Illustrate using any one example.
Solution:
Transgenic bacteria are one that carries a transgene or a foreign gene of interest introduced using recombinant DNA technology. e. g., bacteria carrying the genes for human insulin.

In 1983, Eli Lilly an American company prepared two DNA sequences corresponding to A and B, chains of human insulin and introduced them in plasmids of E. coli to produce insulin chains. Chains A and B were produced separately, extracted and combined by creating disulfide bonds to form human insulin.

Question 3.
Compare and contrast the advantages and disadvantages of the production of genetically modified crops.
Solution:
Advantages of genetically modified crops or transgenic crops are as follows :

• They are resistant to pests, herbicides and diseases.
• They help to reduce post-harvest losses.
• They enhance the nutritional value of food, e.g., a transgenic variety of rice (golden rice) is rich in vitamin A content.
• Some transgenic plants, e.g., poplar trees are used to clean up heavy metal pollution from contaminated soil.
• They are efficient in mineral usage and thus prevent early exhaustion of fertility of the soil.

Transgenic crops have several disadvantages also which are mentioned below:

• Bt toxins expressed in pollen grains of transgenic crops are harmful for useful varieties of insects, e.g., honey bees and butterflies.
• The foods produced by transgenic crops might cause toxicity and might result in allergies.
• The bacteria present in human alimentary canal can become resistant to concerned antibiotic by taking up antibiotic resistance gene present in genetically modified food and become difficult to manage.

Question 4.
What are Cry proteins? Name an organism that produces it. How has man exploited this protein to his benefit?
Solution:
Cry proteins are a group of toxic protein which are highly poisonous to deficient types of insects. It is produced by a soil bacterium Bacillus thuringiensis. The genes controlling their formation are called cry genes eg:- Cry I Ab, Cry I Ac, Cry II Ab, The bacterium produces a protein in the crystal form of protoxin. Two cry genes have been incorporated in cotton (Bt cotton) while one has been introduced in corn (Bt corn) As a result Bt Cotton was disease resistant to bollworm and Bt corn was resistant to corn borer.

Question 5.
What is gene therapy? Illustrate using the example of adenosine deaminase (ADA) deficiency.
Solution:
Gene therapy is the technique of genetic engineering used to replace a faulty gene with a normal, healthy functional gene. The first clinical gene therapy was given in 1990 to a 4 years old girl with adenosine deaminase deficiency (ADA deficiency). This enzyme is very important for the immune system to function. Severe combined immunodeficiency (SCID) is caused due to a defect in the gene for the enzyme adenosine deaminase. SCID patient lacks functional T-lymphocytes and, therefore, fails to fight the infecting pathogens.

To perform gene therapy, lymphocytes are extracted from the patient’s bone marrow and a normal functional copy of human gene coding for ADA is introduced into these lymphocytes with the help of a retroviral vector. The cells so treated are reintroduced into the patient’s bone marrow. The lymphocytes produced by these cells contain functional ADA genes which reactivate the victim’s immune system. But, as these lymphocytes do not divide and are short-lived, so periodic infusion of genetically engineered lymphocytes is required. This problem can be overcome if stem cells are modified at an early embryonic stage.

Question 6.
Diagrammatically represent the experimental steps in cloning and expressing a human gene (say the gene for growth hormone) into a bacterium like E.coli?
Solution:
The given diagram represents the experimental steps in cloning and expressing a human gene for growth hormone into a bacterium E. coli.

Question 7.
Can you suggest a method to remove oil (hydrocarbon) from seeds based on your understanding of rDNA technology and the chemistry of oil?
Solution:
rDNA technology is a technique of genetic engineering that involves combining DNA from two different sources to produce recombined or recombinant DNA (rDNA). Oils are composed of glycerol and fatty acids. Thus, to produce oil-free seeds genes coding for glycerol or fatty acids should be identified and nucleotide sequences complementary to the sequence of these genes should be inserted adjacent to these genes in the early cells of the endosperm. During transcription, these complementary sequences will produce anti-sense RNAs to the RNAs produced by glycerol or fatty acids gene and will silence these genes. As a result, oil-free seeds will be produced.

Since glycerol is a common component of all the oils whereas various fatty acids combine with glycerol to form oils, thus it will be easier if we silence the gene for glycerol synthesis.

Question 8.
Find out from the internet what is golden rice.
Solution:
Golden rice is a GM rice with increased vitamin A content.

Question 9.
Does our blood have proteases and nucleases?
Solution:
Proteases occur naturally in all organisms. These enzymes are involved in a multitude of physiological reactions from simple digestion of food proteins to highly-regulated cascades (e.g., the blood-clotting cascade, the complement system, apoptotic pathways, and the invertebrate prophenoloxidase activating cascade). Proteases present in blood serum (thrombin, plasmin, Hageman factor, etc.) play important role in blood clotting, as well as in lysis of the clots, and the action of the immune system. Other proteases are present in leukocytes (elastase, cathepsin G) and play several different roles in metabolic control. Nucleases, such as deoxyribonucleases and ribonucleases are found in the blood which helps in the degradation of exogenous deoxyribonucleic acid and ribonucleic acid circulating in the blood.

Question 10.
Consult the internet and find out how to make orally active protein pharmaceuticals. What is the major problem to be encountered?
Solution:
The problem is stomach enzymes and acids. Once you orally ingest a protein, the proteases in your stomach juices (trypsin, chymotrypsin, pepsin) will cleave the holy-hell out of your therapeutic protein and the acids will denature whatever’s left beyond all recognition. This is why proteins like insulin have to be injected.

Page No: 205

Exercises

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

Page No: 189

Question 1.
What are the various public health measures, which you would suggest as a safeguard against infectious diseases?
Solution:
Prevention and control of infectious diseases

I. For water-borne diseases like typhoid, amoebiasis, etc.
Practice personal and public hygienic measures.

a. Personal hygienic measures

• Keeping the body clean
• Consumption of clean drinking water
• Eating fresh food

b. Public hygienic measures

• Proper disposal of waste and excreta
• Periodic cleaning and disinfection of water reservoirs, pool, tank etc.

II. For air-borne diseases like common cold, pneumonia

• Avoid close contact with infected persons.
• Avoid the use of belongings of the infected persons.

III. For vector-borne diseases like malaria

• Control and eliminate the vectors and their breeding places
• Introducing larvivorous fishes like Gambusia in ponds that feed on the larvae of the mosquito
• Avoid stagnation of water around the residential area.
• Spraying of insecticides in ditches, drainage areas, etc.
• Protection from a mosquito bite. Use mosquito nets in the doors and windows to prevent the entry of mosquitoes. It is very important in the light of recently widespread diseases like dengue fever, chikungunya etc.

The use of vaccines and immunization programmes has enabled us to eradicate smallpox. Diseases like polio, diphtheria, tetanus etc. have been controlled to an extent by the use of vaccines. Nowadays biotechnology is focussing on the preparation of newer and safer vaccines. A large number of antibiotics are available to treat many infectious diseases.

2. In which way has the study of biology helped us to control infectious diseases?

Answer

→ Various advancements that have occurred in the field of biology have helped us gain a better understanding to fight against various infectious diseases.
→ Biology has developed as we have come to know about the life cycle of various parasites, pathogens, and vectors along with the modes of transmission of various diseases and the measures for controlling them.

→ Vaccination programmes against several infectious diseases such as small pox, chicken pox, tuberculosis, etc. have helps us to eradicate these diseases.
→ Biotechnology has helped in the preparation of developed and safe drugs and vaccines.
→ Antibiotics have also played  a major role in  the treatment of various infectious diseases.

3. How does the transmission of each of the following diseases take place?
(a) Amoebiasis          (b) Malaria           (c) Ascariasis         (d) Pneumonia

Answer

(a) Amoebiasis: It is a vector transmitted disease that spreads by the means of contaminated food and water. The vector involved in the transmission of this disease is the housefly. Its mode of transmission is Entamoeba histolytica.

(b) Malaria: It is a vector transmitted disease that spreads by the biting of the female Anopheles mosquito. Its mode of transmission is Ascaris lumbricoides.

(c) Ascariasis: It spreads through contaminated food and water. Its mode of transmission is Ascaris lumbricoides

(d) Pneumonia: It spreads by the sputum of a diseased  person. Its mode of transmission is Streptococcus pneumoniae,

Question 4.
What measure would you take to prevent water-borne diseases?
Solution:
Water-borne diseases can be prevented by drinking clean water. Water should be free from contamination, suspended and dissolved substances. If water is contaminated it should be boiled and filtered before drinking. Periodic cleaning and disinfection of water reservoirs, pools, and tanks should be done.

Question 5.
Discuss with your teacher what does ‘a suitable gene’ means, in the context of DNA vaccines.
Solution:
‘A suitable gene’ means the gene which is able to produce antigenic polypeptides of the pathogen in bacteria and yeast. Using recombinant DNA technology, it is possible to produce vaccines in large scale for immunisation. Hepatitis B vaccine is produced using this technology.

Question 6.
Name the primary and secondary lymphoid organs.
Solution:
Primary lymphoid organs are bone marrow and thymus. Secondary lymphoid organs are the spleen, lymph nodes, tonsils, Peyer’s patches of the small intestine, and mucosa-associated lymphoid tissues (MALT).

Question 7.
The following are some well-known abbreviations, which have been used in this chapter. Expand each one to its full form.

1. MALT
2. CMI
3. AIDS
4. NACO
5. HIV

Solution:

1. MALT – Mucosal-associated lymphoid tissue.
2. CMI – Cell-Mediated Immunity
3. AIDS – Acquired Immuno Deficiency Syndrome
4. NACO – National AIDS Control Organisation
5. HIV – Human immunodeficiency virus.

8. Differentiate the following and give examples of each:
(a) Innate and acquired immunity    (b) Active and passive immunity

Answer

(a) Innate and acquired immunity

Innate immunity	Acquired immunity
(i) It is a non-pathogen specific type of defense mechanism.	(i) It is a pathogen specific type of defense mechanism.
(ii) It is inherited from parents and protects the individual since birth.	(ii) It is acquired after the birth of an individual.
(iii) It operates by providing barriers against the entry of  pathogenic  agents.	(iii) It produces primary and secondary responses, which are mediated by B-lymphocytes and T-lymphocytes.
(iv) It does not  specific memory.	(iv) It is observed by an immunological memory.

(b) Active and passive immunity

Active immunity	Passive immunity
(i) It is a type of acquired immunity in which the body produces its own antibodies against disease-causing antigens.	(i) It is a type of acquired immunity in which readymade antibodies are transferred from one person to another.
(ii) It shows long lasting effect.	(ii) It does not have long lasting effect.
(iii)It is slow. It takes time in producing antibodies and giving responses.	(iii) It is fast. It provides immediate respose.
(iv) Injecting microbes through vaccination inside the body is an example of active immunity.	(iv) Transfer of antibodies present in the mother’s milk to the infant is an example of passive immunity.

9. Draw a well-labelled diagram of an antibody molecule.

Answer

Question 1.
Group the following as nitrogenous bases and nucleosides:
Adenine, cytidine, thymine, guanosine, uracil, and cytosine.
Solution:
Adenine, Guanosine, Thymine, Uracil, and Cytosine are nitrogenous bases. (Adenine and Guanosine → Purine, Thymine, Uracil and Cytosine → Pyrimidine) Cytidine is a nucleoside.

Question 2.
If a double-stranded DNA has 20 percent of cytosine, calculate the percent of adenine in the DNA.
Solution:
According to Chargaff’s rule, in a double-stranded DNA, the total number of cytosine molecules will be equal to the number of guanine molecules and the number of adenine molecules will be equal to the number of thymine molecules. Therefore, if a double-stranded DNA has 20 percent of cytosine then the guanine will also be 20 per cent. The remaining 60% will consist of adenine and thymine in equal amount. Thus adenine will be 30%.

Question 3.
If the sequence of one strand of DNA is written as follows:
5′-ATGCATGCATGCATGCATGCA
TGCATGC-3′
Write down the sequence of complementary strand in 5′ -> 3′ direction.
Solution:
5′-GCATGCATGCATGCATGCAT G C ATG CAT-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.
Solution:
If the sequence of coding strand is :
5′ – ATGCATGCATGCATGCATGCATGCATGC – 3′
Then template strand is :
3′ – TACGTACGTACGTACGTACGTACGTACG – 5′
The mRNA will be formed on the template strand in 5′ —> 3’ direction. Thus mRNA sequence will be:
5′-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′
Thymine in DNA is substituted by uracil in RNA.

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesise a semi-conservative mode of DNA replication? Explain.
Solution:
The two strands of DNA show complementary base pairing. This property of DNA led Watson and Crick to suggest a semi-conservative mechanism of DNA replication in which one strand of a parent is conserved while the other complementary strand formed is new.

Question 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
Solution:
DNA dependent DNA polymerases and DNA dependent RNA polymerases.

Question 7.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic
material?
Solution:
They raised 2 types of bacteriophages

• On radioactive phosphorous (³²P)
• On radioactive sulphur (³⁵S).

³⁵S gets into protein and ³²P into DNA When both bacteriophages infected bacteria differently and by shaking them, the viral protein coat was separated

After raising these bacteria it was found that those infected with ³²P bacteriophage → radioactivity were found. But with ³⁵S → no radioactivity was found.

Question 8.
Differentiate between the following:

1. Repetitive DNA and Satellite DNA
2. Template strand and Coding strand
3. mRNA and tRNA

Solution:

1. Differences between repetitive DNA and satellite DNA are as follows:
   
   
2. Differences between template strand and coding strand are as follows:
   
   
3. Differences between mRNA and tRNA are as follows:
   
   
   
   

Question 9.
List two essential roles of ribosome during translation
Solution:
Two essential roles of the ribosome during translation are:

1. One of the RNA acts as a peptidyl transferase ribozyme for the formation of peptide bonds.
2. The ribosome provides sites for attachment of mRNA and charged tRNA for polypeptide synthesis.

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?
Solution:
The lac operon is regulated by the amount of lactose in the medium where the bacteria are grown. When the amount of lactose is exhausted in the medium, the lac operon shuts down.

Question 11.
Explain (in one or two lines) the function of the followings:

1. Promoter
2. tRNA
3. Exons

Solution:

1. Promoter: It is located at the 5′ end of the transcription unit and provides site for attachment of transcription factors (TATA Box) and RNA polymerase.
2. tRNA: It takes part in the transfer of activated amino acids from cellular pool to ribosome so that they can take part in protein formation.
3. Exons: In eukaryotes, DNA is mosaic of exons and introns. Exons are coding sequences of DNA which are both transcribed and translated.

Question 12.
Why is the Human Genome Project called a mega project?
Solution:
The human genome was a megaproject that aimed to sequence every base in the human genome. The estimated cost of the project would be a billion (1 billion = 100 crores) US dollars.

Question 13.
What is DNA fingerprinting? Mention its application.
Solution:
DNA fingerprinting is the identification of differences in specific regions of DNA sequences based on DNA polymorphism, repetitive DNA, and satellite DNA.
Application of DNA fingerprinting: Settling, paternity disputes and identity of criminal by different DNA profiles in forensic laboratories.

Question 14.
Briefly describe the following:

1. Transcription
2. Polymorphism
3. Translation
4. Bioinformatics

Solution:
1. Transcription – It is the process of copying genetic information from the anti-sense or template strand of the DNA into RNA. It is meant for taking the coded information from DNA in nucleus to the site where it is required for protein synthesis. Principle of complementarity is used even in transcription. The exception is that uracil is incorporated instead of thymine opposite adenine of template. The segment of DNA that takes part in transcription is called transcription unit. It has three components

• • a promoter,
  • the structural gene and
  • a terminator.

2. Polymorphism – It is the variation at genetic level, arisen due to mutations. Such variations are unique at particular site of
DNA. They occur approximately once in every 500 nucleotides or about 107 times per genome. These are due to deletions, insertions, and single-base substitutions. These alterations in healthy people, occur in non-coding regions of DNA and do not code for any protein but are heritable. The polymorphism in DNA sequences is the basis of genetic mapping of human genome as well as DNA fingerprinting.

3. Translation – It is the mechanism by which the triplet base sequence of mRNA guides the linking of a specific sequence of amino acids to form a polypeptide chain (protein) on ribosomes in the cell cytoplasm. All the protein that a cell needs are synthesised by the cell within itself.
The raw materials required in protein synthesis are ribosomes, amino acids, mRNA, tRNAs and amino acyl tRNA synthetase. Mechanism of protein synthesis involves following steps:

• • Activation of amino acids
  • Charging or aminoacylation of tRNA
  • Initiation
  • Elongation (Polypeptide chain formation)
  • Termination

The ribosomes move along the mRNA ‘reading’ each codon in turn. Molecules of transfer RNA (tRNA), each bearing a particular amino acid, are brought to their correct positions along the mRNA, molecule base pairing occurs between the bases of the codons and the complementary base triplets of tRNA. In this way, amino acids are assembled in the correct sequence to form the polypeptide chain.

4. Bioinformatics – Bioinformatics is the combination of biology, information technology and computer science. Basically, bioinformatics is a recently developed science which uses information technology to understand biological phenomenon. It broadly involves the computational tools and methods used to manage, analyse and manipulate volumes of biological data.

Question 1.
What do you think is the significance of reproductive health in a society?
Solution:
Significance of reproductive health in society are:

• Control over the transmission of STDs.
• Less death due to reproduction-related diseases like-AIDS, cancer of the reproductive tract.
• Control in a population explosion.
• Not only the reproductive health of men and women affects the health of the next generation.

Question 2.
Suggest the aspects of reproductive health which need to be given special attention in the present scenario.
Solution:
Providing medical facilities and care to the problems like menstrual irregularities, pregnancy related aspects, delivery, medical termination of pregnancy, STDs, birth control, infertility. Post-natal child maternal management is another important aspect of the reproductive and child health care programme.

Question 3.
Is sex education necessary in schools? Why?
Solution:
Yes, sex education is necessary for schools because:

• It will provide proper information about reproductive organs, adolescence, safe, hygienic sexual practices, and Sexually Transmitted Diseases (STDs).
• It will provide the right information to avoid myths and misconceptions about sex-related queries.

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.
Solution:
The reproductive health in our country has improved in the last 50 years. Some areas of improvement are :

• Massive child immunization.
• Increasing use of contraceptives.
• Better awareness about sex related matters.
• Increased number of medically assisted deliveries and better post-natal care leading to decreased maternal and infant mortality rates.
• Increased number of couples with small families.
• Better detection and cure of STDs and overall increased medical facilities for all sex related problems.

Question 5.
What are the suggested reasons for the population explosion?
Solution:

• Improved medical facilities
• Decline in death rate, IMR, MMR
• Slower decline in birth rate.
• Longer life span.
• Lack of 100% family planning and education among the village.

Question 6.
Is the use of contraceptives justified? Give reasons.
Solution:
Yes, the use of contraceptives is justified: To overcome the population growth rate, contraceptive methods are used. It will help in bringing birth rate down & subsequently curb population growth. With the rapid spread of HIV/ AIDS in the country, there is now a growing realization about the need to know about contraception & condoms.

Question 7.
Removal of gonads cannot be considered as a contraceptive option. Why?
Solution:
Removal of gonads not only stops the production of gametes but will also stop the secretions of various important hormones, which are important for bodily functions. This method is irreversible and thus, can not be considered as a contraceptive method.

Question 8.
Amniocentesis for sex determination is banned in our country. Is this ban necessary? Comment.
Solution:
Amniocentesis is a prenatal diagnostic technique to find out the genetic disorders and metabolic disorders of the foetus. Unfortunately, the useful technique of amnio-centesis had been misused to kill the normal female foetuses as it could help detect the sex of foetus also. Hence, this technique is now banned in our country. This ban is necessary as this technique was promoting female foeticide in our country.

Question 9.
Suggest some methods to assist infertile couples to have children.
Solution:
If the couples are enabled birth to the children and corrections are not possible, the couples could be assisted to have children through certain special techniques, commonly known as Assisted Reproductive Technologies (ART). Some methods are given as:

1. In Vitro Fertilization (IVF): In this method, ova from the female and the sperm from the male are collected and induced to form a zygote under simulated conditions in the laboratory. This process is called In Vitro Fertilization (IVF). Some method is given as follows:

• Zygote Intrafallopian Transfer (ZIFT): The zygote or early embryo with up to 8 blastomeres is transferred into the fallopian tube.
• Intra-Uterine Transfer (IUT): Embryo with more than 8 blastomeres is transferred into the uterus in females who cannot conceive embryos formed by the fusion of gametes in another female are transferred.
• Test tube baby: In this method, ova from the donor (female) and sperm from the donor (male) are collected and are induced to form a zygote under simulated conditions in the laboratory. The zygote could then be transferred into the fallopian tube and embryos transferred into the uterus, to complete its further development. The child born from this method is called a test-tube baby.

2. Gamete Intra Fallopian Transfer (GIFT): It is the transfer of an ovum collected from a donor into the fallopian tube 8 another female who cannot produce one, but can provide a suitable environment for fertilization and further development of the embryo.

3. Intra Cytoplasmic Sperm Injection (ICSI) : It is a procedure to form an embryo HI* the laboratory by directly injecting the sperm into an ovum.

4. Artificial Insemination (AI): In this method, the semen collected either from the husband or a healthy donor is artificially introduced into the vegina or into the uterus (Intra Uterine Insemination, IUI). This technique is used in cases where the male is unable to inseminate sperms in the female reproductive tract or due to very low sperm counts in the ejaculation.

5. Host Mothering: In this process, the embryo is transferred from the biological mother to a surrogate mother. The embryo then develops till it is fully developed or partially developed. It is then transferred to the biological mother or into any other. This technique is useful for females in which embryo forms but is not able to develop.

Question 10.
What are the measures one has to take to prevent contracting STDs?
Solution:
Diseases or infections which are transmitted through sexual intercourse are collectively called sexually transmitted diseases (STDs) or reproductive tract infections (RT), e.g., gonorrhea, syphilis, genital herpes, AIDS, etc. The measures that one has to take to prevent from contracting STDs are:

• Avoid sex with unknown partners/multiple partners.
• use condoms during coitus.
• In case of doubt, go to a qualified doctor for early detection and get complete treatment if diagnosed with the disease.

Question 11.
State True/False with an explanation.

1. Abortions could happen spontaneously too.
2. Infertility is defined as the inability to produce viable offspring and is always due to abnormalities/defects in the female partner.
3. Complete lactation could help as a natural method of contraception.
4. Creating awareness about sex related aspects is an effective method to improve the reproductive health of people.

Solution:

1. True: One-third of all pregnancies abort spontaneously (called miscarriage) within four weeks of conception and abortion passes unrecognized with menses.
2. False: Infertility is defined as the inability of the couple to produce viable offspring. It is due to abnormalities/defects in either male or female or both.
3. True: Complete lactation is a natural method of contraception as during this period ovulation does not occur, but this is limited to a period of 6 months after parturition.
4. True: Creating awareness in people about sex-related aspects like right information about reproductive organs, accessory organs of reproduction, safe and hygienic sexual practices, birth control methods, care of pregnant women, post-natal care of mother and child, etc., can help in improving the reproductive health of people.

Question 12.
State True/False with an explanation.
(a) Abortions could happen spontaneously too. (True/False)
Answer:
False, Abortion does not happen under normal conditions. It happens accidentally or under the will of Parents.

(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)
Answer:
False, Sterility always does not occur due to females sometimes. Males are also responsible for this.

(c) Complete lactation could help as a natural method of contraception. (True/False)
Answer:
True, the Menstrual cycle does not occur after parturition which can act as natural
contraception but this method is functional for a period of six months from parturition.

(d) Creating awareness about sex related aspects is an effective method to improve the reproductive health of dead people. (True/False)
Answer:
True, this creates better reproductive health among people.