Category: KSEEB Important Questions

Karnataka 2nd PUC Biology Important Questions Chapter 9 Strategies for Enhancement in Food Production

Question 1.
Define Animal husbandry?
Answer:
“It is the agricultural practice of breeding and raising live stock”. Includes Caring and breeding Buffaloes, cows, pigs etc.,
Poultry farming and fisheries Apiculture (bee keeping) for honey.
For food and eggs. Molluscs, prawns,crabs. India and china contributes about 70% of world live stock population.

Question 2.
Define dairying?
Answer:
Dairying is “the management of animals for milk and its products for human consumption.”

Question 3.
Explain the processes of dairy farm management?
Answer:
Processes of dairy farm management.

• Selection of good breeds having high yielding potential with disease-resistant characters.
• Cattles should be housed well, with adequate waters
• Providing correct quantity and quality of fodder.
• Maintaining hygienic conditions while milking and storage.
• Regular visits by a veterinary doctor.

Question 4.
Define Poultry?
Answer:
Poultry refers to “the class of domesticated fowl (birds) like chicken, ducks, turkeys and geese for their meat and eggs.

Question 5.
Write few sentences on successful poultry management?
Answer:
Management:

• Selection of disease-free and suitable breeds.
• Proper feed and water with hygienic conditions
• Healthcare.

Question 6.
Define the term Breed?
Answer:
Breed: A group of animals having the same ancestry characters appearance, size, configuration, etc.

Question 7.
Differentiate between Inbreeding and Outbreeding?
Answer:
Systems of breeding.

Inbreeding	Outbreeding
1. Mating of more closely related individuals. With in the same breed for 4-6 generations	1. Mating of unrelated animals within the same breed but having no common ancestors upto 4-6 generations.
2. Superior males and females of the same breed are identified and mated in pairs
3. Progeny obtained are evaluated selected used for further mating.

Question 8.
Significance of Inbreeding?
Answer:

1. Inbreeding is homozygosity
2. It exposes harmful recessive genes which can be eliminated by selection.
3. Inbreeding helps in the accumulation of superior genes.

Question 9.
Differentiate between outbreeding and outcrossing.
Answer:

• Outcrossing – Mating between unrelated members of the same breed.
• Outbreeding – Superior males of one breed in mated with superior females of another breed.

Question 10.
Define artificial insemination?
Answer:
Artificial insemination: The semen is collected from the male that is chosen as a parent and injected into the reproductive tract of the selected female by the breeder.

(Semen can be frozen and can be used at a later date, and also can be transported in a frozen form to the place where the male is housed). A.I. helps us overcome several problems of normal matings.

Question 11.
Explain MOET?
Answer:
MOET – Multiple Ovulation Embryo transfer technology for herd improvement (cows)
(a) Induction of superovulation in donor cow.
Answer:
A cow is administered with hormones with FSH-like activity, which results in production of 6-8 eggs instead of one egg, which they normally yield per cycle. “Production of more eggs is called superovulation”.

(b) Artificial insemination.
Answer:
The superovulated cow is artificially inseminated with semen collected from superior quality (Elite bull)

(c) Transfer of Embryos.
Answer:
The fertilized eggs at the 8-32 cells stage are transferred to surrogate mother cow which is in the same reproductive stage as the donor.

The genetic mother of is available for another round of super ovulation. This technology has been used for production high milk-yielding breeds, O⁺, high-quality meat-yielding bulls (o), and also in sheep, rabbits, buffaloes etc.

Question 13.
Define Inbreeding depression?
Answer:
Continuous inbreeding reduces fertility and even productivity. This is called inbreeding depression.

Question 14.
How can Be Inbreeding depression be overcome?
Answer:
To solve this problem, selected animals of the breeding population should be mated with unrelated superior
animals of the same breed or restore fertility and yield.

Question 15.
Define Out Crossing?
Answer:
Out Crossing: Mating between unrelated members of the same breed”.

Question 16.
Define Crossbreeding?
Answer:
Crossbreeding: “Superior males of one breed are mated with superior females of another breed”.

Question 17.
Give an example for cross breeded variety?
Answer:
Hisardale – new breed of sheep developed in Punjab by crossing Bikaneri eves and Marinorams.

Question 18.
Define Interspecific hybridisation with examples?
Answer:
“Male and female animals of two different related species are mated” Progeny may combine desirable features of both the parents.
Ex: MULE-cross between 2 different species, the female horse and the male donkey, which is stronger, disease, resistant, hard working.

Question 19.
Define apiculture?
Answer:
Apiculture – is maintenance of hives of honeybees for the production of honey.

Question 20.
Write Significance of apiculture?
Answer:
Significance:

1. Honey is a food of high nutritive value used in medicine .
2. Honey bees produces beeswax which is used in preparation of cosmetics and polishes of various kinds.

Question 21.
Write a note on management for successful Bee Keeping?
Answer:
Practice: Bee Keeping can be practiced in the area where sufficient bee pastures of shrubs, fruit orchards and cultivated crops,are present Many species of bees, the most commons are Apis indica.

The following points are important for successful bee-keeping.
(a) Knowledge of the nature and habits of bees.
(b) Suitable location, selection.
(c) Catching and hiving of swarms (group of bees).
(d) Management during different seasons.
(e) Handling and collection of honey and bees wax.

Question 22.
Define fishery? Give any 2 examples of fresh water and marine water fishes.
Answer:
Fishery is an industry devoted to the rearing, breeding, catching of fishes, shell fish, prawns, crabs, oysters, lobsters etc scientifically by man in ponds, tanks, etc, for Food.
Ex: Freshwater fishes: Catla, Rohu, Common Carp. Marine fishes: H-ilsa, Sardines, Maekeedl and Pomfrets.

Question 23.
Define Aquaculture?
Answer:
Aquaculture: Culturing economically important aquatic organisms. Harvesting, Processing, marketing in fresh, brackish and marine water.

Fisheries have an important place in the Indian Economy.

• Provides income and employment to millions of fishermen and farmers in coastal states.
• It is the only source of their livelihood.

Question 24.
Define Blue Revolution?
Answer:
Modem techniques in pisciculture and aquaculture have been able to increase production which is termed as ‘Blue Revolution’.

Question 25.
Define Green revolution?
Answer:
Green Revolution: Revolution in the breeding techniques in different crop plants for the development of high yielding and disease resistant, drought-resistant varieties.

Question 26.
Define Plant breeding?
Answer:
Plant breeding: It is the purposeful manipulation of plant species in order to create desired plant types that are better suited for cultivation.

Question 27.
Describe the steps involved in plant breeding?
Answer:
Following steps are involved in breeding a new genetic variety of a crop.
1. Collection of Variability: In many crops pre-existing genetic variability is available from wild varieties, where
these variations are inheritable. “The entire collection of plants and seeds having all the diverse alleles for all genes in a given crop is called germplasm collection.

2. Evaluation and selection of parents: The germplasm is evaluated to identify plants with a desirable combination of characters. Selected plants are multiplied, and pure lines are created where ever desirable and possible.

3. Cross hybridization among the selected parents: Desired characters are combined from two different plants
Eg: the high protein quality of one parent combined with disease-resistant variety from another parent. But it is time-consuming, and only one in 100/1000 shows the desirable combination.

4. Selection and testing of superior recombinants: Progeny among hybrids are selected which have the desired
character combination. These plants then self-pollinated of 6-8 generations, till they reach the state of homozygosity.

5. Testing, release and commercialization of new cultivars: The newly selected lines are tested for their yield and other characters, which is done in research fields by recording their performance under ideal crop management practices. The material is evaluated in comparison to the best available local crop cultivar which is a reference cultivar.

Note: Agriculture accounts for 33% of India’s GDP and employs nearly 62% of the population.
Examples of some Indian hybrids crops of high-yielding varieties.

Question 28.
Briefly explain on high yielding crops of India?
Answer:
Wheat and Rice: Wheat – wheat production increased from 11 to 75 million tonnes. Nobel Laureate Norman E. Borlang at “International centre for Wheat and Maize Improvement in Mexico” developed Semi-dwarf wheat.

Question 29.
Name two high-yielding wheat varieties?
Answer:
Sonka, Kalyan sona – high yielding and disease-resistant varieties.

Question 30.
Give 2 examples of Rice vairities?
Answer:
Rice: IR-8 (IRRI) Philippines. (International Rice Research Institute) Taichung Native -1 (Taiwan). These are semi-dwarf rice varieties. Later better yielding semi dwarf varieties Jaya and Ratna were developed in India.

Question 31.
Mention 2 sugarcane varieties used for breeding programme?
Answer:
Sugarcane:

1. Saccharum barberi – North India – Poor sugar content and yield.
2. Saccharum officinarum – South India – Thicker stems and higher sugar content.

(These 1 × 2 species ever successfully crossed, to get sugarcane varieties that have ability to grow in north India). Millets: Hybrid maize, jowar, and bajra have been successfully developed in India, high yielding, resistant to water stress.

Question 32.
Name fungal bacteria and viral diseases of plants?
Answer:

• Fungal diseases are Rusts → brown rust of wheat, red rot of sugarcane, late blight of potato.
• Bacterial diseases → black rot of crucifers, citrus canker.
• Viral disease → Tobacco mosaic and Turnip mosaic etc.

Question 33.
Explain the methods of plant breeding for disease resistance with examples.
Answer:
Methods of breeding for disease resistance:
The conventional method of breeding for disease resistance is – hybridization and selection

• Screening germplasm for resistance.
• Hybridization of selected parents.
• Selection and evaluation of the hybrids.
• Testing and release of new varieties.

Examples of disease-resistant varieties.

Question 34.
Give any two examples of disease-resistant variety?
Answer:

Question 35.
What is mutation breeding? Give example.
Answer:
Mutation breeding: “Improvement of crops by changing the genotype of plants through induced mutations “ (It is possible to induce mutations through use of chemicals or radiations (V radiations), the selection is made which plant have desirable Character, which sever as a source.)

Ex: Resistance to the yellow mosaic virus in bhendi (Abelmoschus esculentus) was transferred from a wild species, resulting in a new variety of bhendi known as “Parbhani Kranti.”

Note: Transfer of resistance gene is achieved by sexual hybridization between the target and the source plant followed by selection.

Question 36.
Explain how crop plants have evolved with natural resistant to insect pests along with examples?
Answer:

1. Cotton resistance to Jassids
2. Wheat resistance to cereal leaf beetle. In wheat solid stems, lead to non-preferences by stem sawfly.
3. Smooth leaved and nectar-less cotton varieties keep away Bollworm.
4. High aspartic acid, low N2 content, sugar content keep away maize stem borers.

Question 37.
Write a note on hidden hunger?
Answer:
Diet lacking essential micronutrients, Vitamins proteins leads to deficiency symptoms in man, i.e. ‘Hidden
hunger’, because of poor economic conditions, Plant breeding techniques have been used to improve nutrition quality of crops, vegetable, fruits etc.

Question 38.
Define Biofortification. Mention few examples of biofortified crops?
Answer:
Biofortification: “Breeding crops with higher levels of vitamins and minerals or higher protein and health; fats” The most practical means to improve public health.

Question 39.
Give few examples of vegetable crops released by IARI?
Answer:

1. Vitamin A enriched Carrots, Spinach, Pumpkin
2. Vitamin C enriched bitter gourd, mustard, tomato
3. Calcium enriched spinach
4. Protein-enriched -Beam, Peas etc.

Question 40.
Define SCP /Expand SCP?
Answer:
“Unicellular micro organisms, dried and uses as protein supplement in human food” They are alternate source of proteins for animal and plant nutrition.

Question 41.
Name few microbes usfed as SCP?
Answer:

• Bacteria- Methylophilus (250g) methylotrophs – its high rate of biomass production and growth, produces 25 tonnes of protein. (Where 250 kg cow gives 200g protein)
• Cyanobacteria – Spirulina
• Yeast – Saccharomyces cerevisiae, Mushrooms.
• Algae – Chlorella etc.

Question 42.
Why mircrobes are preferred for SCP culture?
Answer:
These microbes can be grown easily on waste water from potato processing plants, molasses, sewage, to produce large quantities such utilisation reduces environmental pollution.

Question 13.
Give a brief account of Tissue culture?
Answer:
“A branch of biotechnology where cells, tissues or organs of the plant body are isolated and grown on nutrient medium under aseptic conditions”.

Question 44.
Define Explant?
Answer:
Explant- The plant tissue or organ excised and used for invitro culture.

Question 45.
Define Totipotency?
Answer:
Totipotency – Capacity/ability of any cell to produce the entire plant of its own type in invitro culture.
(Nutrient) Medium – has sucrose as carbon source, vitamins amino acids and growth regulators like Auxin, Cytokinin etc.

Question 46.
What is Micropropagation?
Answer:
Micropropagation – Method of producing thousands of plants through tissue culture.

Question 47.
What are Somaclones?
Answer:
Somaclones Each of the plant resulted fram tissue culture is genetically identical to the original plant from which it is derived.

Question 48.
What is Meristem culture? Why they are prepared in tissue culture. .
Answer:
Meristem Culture – Culture of shoot apical meristems, produces large number of virus free microshoots.

Question 49.
What is Protoplast culture?
Answer:
Protoplast culture – The plants cells are digested with lytic enzymes like cellulase and pectinases and protoplast is obtained, which are cultured to get callus or microshoots. If protoplasts of same family are fused to hybrid protoplast which form new plant – Somatic hybrids, and process is known as Somatic hybridisation.

Question 50.
Write a note on applications of tissue culture?
Answer:
Applications of Tissue culture:

1. Micropropagation
2. Production of haploids (anther culture)
3. Virus-free plants
4. Production of secondary metabolites
   Ex: Rauwolfia, serpentine etc.

→ Strategies for Enhancement in Food Production:
Enhancement of food production is a major necessity to meet the demand for food of the ever-increasing population of the world.

Animal husbandry and plant breeding are the two important biological principles practiced to enhance food production.

→ Animal Breeding:
Animal breeding aims at increasing the yield of animals and increases desirable qualities of the breed.

→ Plant Breeding For Disease Resistance:
Cultivation of disease-resistant varieties enhances food production, which helps to reduce the dependence on use of fungicides and bacteriocides. Before breeding is undertaken, importance is given to know about the causative organisms and the mode of transmission.

→ Other Breeding Methods:

• Somaclonal variation: (Genetic variability which is regenerated during tissue culture)
• Genetic Engineering: Breeding organisms with manipulated genes.

→ Plant Breeding For Developing Resistance to Insect pests
Hairy leaves in several plants are associated with resistance to insect pests.

→ Objectives of Improving:

• Protein, Oil, Vitamin content, and quality.
• Micronutrient and mineral content.

Examples of biofortified crops.

1. Maize-amino acid, lysine, and tryptophan-rich.
2. Wheat-variety Atlas 66-high protein content, which is used as a donor of improving cultivated wheat.
3. Rice-with Iron-fortified (strengthen)
   Indian Agricultural Research Institute-New Delhi has released several vegetable crops i.e.

2nd PUC Biology Important Questions with Answers

Karnataka 2nd PUC Biology Important Questions Chapter 8 Human Health and Disease

Question 1.
What is disease? Mention two types with examples.
Answer:
Disease: Impaired physiological functioning of the body due to impairment, characterised by various signs and symptoms.

1. Infectious diseases: Transmitted easily from one person to another. Ex: viral, Bacterial, Fungal etc.
2. Non-infectious diseases: Diseases which do not transmit. Ex: Allergies, Genetic disorders, Cancer etc.

Question 2.
Define pathogen
Answer:
Pathogen → Disease causing organisms. They make entry into host, withstand hosts defence mechanism multiply and survive.

Question 3.
Name the causative orgapism of Typhoid?
Answer:
Pathogen – Salmonella typhii, Host – Children, Man’s intestine

Question 4.
Mention the symptoms of typhoid?
Answer:
Symptoms: High fever (39-40° C) head ache, weakness, liver and spleen enlargement. Intestinal perforation leads to death.

Question 5.
Name the confirmatory test for typhoid?
Answer:
Confirmed by test – Widal test.

Question 6.
Write the mode of transmission of typhoid.
Answer:

• Transmission: Through contaminated water and food, insects, house fly etc.,
• Classic Case: Mary Mallon – She was a typhoid carrier, who continued to spread typhoid through food she served.
• Prevention: By good sanitation, cleanliness.

Question 7.
Name the pathogen of pneumonia? Mode of infection.
Answer:
Pathogen: Streptococcus pneumoniae. Haemophilus influenza. Host: Human, (in alveoli of lungs)

Question 8.
Write the symptoms of pneumonia?
Answer:

• Symptoms: Persistent dry cough, high fever, chest pain, head ache, alveoli get filled with fluid. Breath
• Shortness. In severe cases, lips and finger nails turn grey to bluish in colour, which is known as Cyanosis.
• Transmission: Spreads by sputum of the patient, A healthy person acquires the infection by inhaling air born pathogens released by infected person.

Question 9.
How Pneumonia is prevented?
Answer:
Prevention: Vaccination (Pneumococcal), hygienic habits.

Question 10.
Name the disease caused by Rhinovirus and write its symptoms?
Answer:

• Disease: Common Cold.
• Host: Man (Nose and respiratory passage)
• Symptoms: Sore throat, hoarseness, cough nasal congestion,
• Transmission: Inhalation of droplets from cough and sneezes of an infected person, Contact with mucous contaminated objects like pen, book, keyboard, mouse etc,
• Prevention: Hygienic habits.

Question 11.
Name the causative organisms of Malaria?
Answer:

• Pathogen: Plasmodium vivax.
• P. falciparum causes malignant malaria which is fatal

Question 12.
Name the two hosts of Plasmodium?
Answer:
Host: Man and Mosquito

Question 13.
Write the symptoms of Malaria?
Answer:
Symptoms: Fever every 48/72 hrs, Cold, hot and sweating stages. On .chronic cases patients become anaemic and because of RBC destruction enlargement of liver and spleen may occur.

Question 14.
Name the insect vector which transmit malaria?
Answer:
Transmission: Female Anopheles mosquito acts as Vector which spreads disease from unhealthy to healthy person.

Question 15.
Give schematic representation of life cycle of Plasmodium?
Answer:
The life cycle of plasmodium (Malarial Parasite):

• Plasmodium enters the human body when mosquito (infected) of Anopheles) bites, as sporozoites (infection form).
• Parasites (Merozoites) multiply increase their no in liver cells and attack RBCs and results in rupturing of RBC, with the release of toxic substance Haemazoin which leads to the chills and fevers every 3 to 4 hours.
• These parasites enter the mosquito’s body when female Anopheles bites an infected person.
• Fertilisation and development of parasites takes place in mosquito gut and
• Matured sporozoites from gut migrate to salivary glands.
• When these mosquitoes bite human, sporozoites are injected to the body.

Question 16.
Name two hosts of plasmodium?
Answer:
Man Mosquitos.

Question 17.
Name the infectious stage of plasmodium?
Answer:
sporozoites.

Question 18.
Name the site of sexual reproduction of plasmodium?
Answer:
Female Anopheles mosquito.

Question 19.
Name the disease caused by Entamoeba histolytica?
Answer:
Disease: AMGEBIASIS (Amoebic dysentery).

Question 20.
Name the pathogen which causes Amoebic dysentry?
Answer:

• Pathogen: Entamoeba histolytica
• Host: Large intestine of man.

Question 21.
Write the symptoms of Amoebiasis?
Answer:
Symptoms: Constipation, abdominal pain, Cramps, stools with excess mueous and blood clots.

Question 22.
Write the mode of infection of Amoebiasis?
Answer:
Transmission: Through house flies, which carry parasite by faeces of infected person to food products and water.

Question 23.
How do you prevent Amoebiasis?
Answer:
Prevention: Hygienic habits.

Question 24.
Name the causative organism of Ascariasis?
Answer:
Ascaris lumbricoides.

Question 25.
Name the disease caused by Ascaris lumbricoides?
Answer:
Pathogen: Ascariasis Host: Man’s intestine.

Question 26.
Mention symptoms of Ascariasis?
Answer:

• Symptoms: Internal bleeding, muscular pain, fever, blockage of intestinal passage, anaemia.
• Transmission: Through faeces of infected persons which contaminate soil, water, plants etc. A healthy person acquires this infection through unwashed vegetables, fruits etc.

Question 27.
Name the pathogen, vector and a symptom of filariasis
Answer:

• Pathogen: Wuchereria bancrofti
• Host: Man.
• Disease: Elephantiasis/Filariasis
• Symptoms: Chronic inflammation of the organs in which they live. The lymphatic vessels of the lower limbs, genital organs are affected resulting in gross deformities.

Question 28.
Name the pathogen which causes elephantiasis?
Answer:

• Pathogen: Wuchereria bancrofti
• Host: Man.

Question 29.
Name the disease caused by Wuchereria Bancroft?
Answer:
Disease: Elephantiasis/Filariasis

Question 30.
Mention a few symptoms of Filariasis?
Answer:
Symptoms: Chronic inflammation of the organs in which they live. The lymphatic vessels of the lower limbs, genital organs are affected resulting in gross deformities.

Question 31.
Write the mode of infection of Filariasis?
Answer:
Transmission: Through female culex mosquito vectors.

Question 32.
Name the disease caused by Trichophyton?
Answer:
Disease: Ringworm (infectious disease).

Question 33.
Name the pathogen which causes Ring worm disease?
Answer:
Pathogen: Microsporum, Trichophyton, Epidermophyton are pathogens.
Host: Man

Question 34.
Mention symptoms of ring worm disease?
Answer:
Symptoms: Dry scaly lesions on skin, nails and scalp. Itchy reddish rashes, in shape of ring, Heat and moisture help these fungi to grow.

In Face and neck – Barbar’s itch. Foot-Athlet’s foot.

Transmission: Heat and moisture help these fungi to grow between toes, skin folds, groin. Ringworms are acquired by soil or using towels clothes, combs of infected individuals.

Question 35.
Write a note on public health measures?
Answer:
1. Maintenance of public hygiene and personal hygiene

Public Hygiene	Personal Hygiene
Proper disposal of waste and excreta. Periodic cleaning of water reservoirs, pools, tanks.	Keeping the body clean, consumption of clean drinking water, food, vegetables, fruits etc. Close contact with infected persons are avoided.

2. Awareness: Quare of infectious diseases – transmission, infection and preventive measures.

3. For air born diseases: like pneumonia and cold contact with infected persons should be avoided.

4. For vector-mediated diseases like Malaria and Filariasis

• Control and eliminate breeding places (stagnant water) in and around residential areas.
• Mosquito nets, spraying insecticides
• Introducing fishes like Gambusia which feeds on mosquito larvae.
• Doors and windows should be provided with mesh.

5. Vaccination: Vaccines and immunisation – to eradicate diseases complete eradication done – smallpox.
Polio, pneumonia, tetanus is controlled.
Biotechnology newer and safer vaccines.

Question 36.
Mention the preventive measures for vector-borne diseases?
Answer:
“The ability of the host to fight against the disease-causing organisms due to immune system in the body”,
i. e, resistance to disease.

Question 37.
Define Immunity and mention its types?
Answer:
Immunity- 2 types

1. Innate Immunity.
2. Acquired Immunity.

Question 38.
What is Innate immunity? Explain the 4 types?
Answer:
It is the non-specific type of defence, present by birth. It provides different types of barriers to the entry of foreign agents to our body. They are of 4 types of Barriers.

1. Physical Barriers
2. Physiological Barriers
3. Cellular Barriers
4. Cytakine Barriers

Question 39.
What is Acquired immunity?
Answer:
Acquired immunity: “It is pathogen-specific immunity developed by our body in response to diseases by microbes. It is characterised by memory”.

The body will be having a memory of the first encounter called primary response with low intensity, subsequent attacks by the same pathogen will encounter with the secondary response with high intensity, because of memory or first encounter.

Question 40.
Difference between innate immunity and acquired immunity?
Answer:

Innate Immunity	Acquired Immunity
1. It is called inborn immunity	1. It is called Adaptive immunity.
2. It is present since birth.	2. It is developed during life time.
3. Remains throughout life.	3. Short lived or life long

Question 41.
Differentiate between B lymphocytes and T lymphocytes?
Answer:

B-lymphocytes	T-lymphocytes
1. Originate in the bone marrow and differentiate there itself and gets matured into B-lymphocytes	1. Originate in the bone marrow and migrate to thymus gland and mature into T- lymphocytes.
2. Produce antibodies in the blood in response to pathogens. (Antibody-mediated immune system -AMIS)	2. do not produce antibodies but help B-lymphocytes to produce anti Bodies. (Cell-mediated immune system-CMlS).
3. Operate through humoral immune response (i.e., thro1 blood)	3. Operate through cells.

Question 42.
Which type of immune system is responsible for graft rejection?
Answer:
Cell-mediated immune system.

Question 43.
Draw the well-labelled diagram of an antibody molecule?
Answer:
Structure of Antibody Molecule:
Each antibody molecule has four peptide chins – 2 small Light Chains. -2 Longer heavy chains, so represented as H₂ L₂ these 4 chains are lead together by disulphide bonds. This antibody is a ‘ Y’ shaped protein molecule whose surface ‘shape’ reacts with antigen.

Question 44.
Differentiate between Active and Passive immunity?
Answer:

Active immunity	Passive immunity
1. Developed when host is exposed to antigens, and antibodies are produced naturally.	1. Developed when ready made antibodies are given to protect the body against pathogens.
2. Slow and takes time to give its full response and lasts longer.	2. Fast gives short lived immunity.
3. Antibody produced harmless	3. Harmful
4. Effective on large no of infections	4. Effective only on limited infections.

Question 45.
What is colostrum? Mention its importance.
Answer:
Ex: In Mother initial days of lactation, contain Colostrum, which has (IgA) antibodies to protect the infant.

Question 46.
Name the antibody passed from mother to foetus through placenta? (1M)
Answer:
Foetus receives antibodies (IgG) from mother through placenta.

Question 47.
Differentiate between Vaccination and Immunisation?
Answer:

Vaccination	Immunisation
1. The process of introduction of inactivated/ weakened pathogen into a healthy person to produce immunity.	1. The process by which body produces antibodies against the vaccine (Primary response) and develop the ability to neutralise pathogens during actual infection (Secondary response)
2. Vaccines generate memory B and T cells and produce quick immune response and increase antibody production

Question 48.
Passive immunization is required for snake venom. Why?
Answer:
If quick response is required like tetanus antibodies and antitoxins are directly injected. Snake bites-injection contain preformed antibodies against snake venom. —> passive immunisation.

Question 49.
Name vaccine produced in micobes thro Rec DNA technology.
Answer:
Recent Rec. DNA teck, allowed production of antigens of pathogen in bacteria or yeast.
Ex: hepatitis B vaccine produced from yeast.

Question 50.
Define Allergy?
Answer:
Allergy: The exaggerated response of the immune system to certain antigen (foreign substance) percent in the environment.

Question 51.
What are allergens?
Answer:
Allergens: The substances in response to immune system.

Question 52.
Name the antibody produced during Allergy?
Answer:

• Antibodies produce for allergy are of IgE type, for example dust pollens, animal dander etc. .
• Allergy is due to the production of chemicals like Histamine and serotonin from the mast calls.
• Determining cause of allergy, patient is exposed to allergens in tiny, increasing doses and reactions are studied.

Question 53.
Name the drugs for Allergy?
Answer:
Drugs: Antihistamine, Adrenalin and Steroids.

Question 54.
Why children in metro city are sensitive to allergies?
Answer:
More children in metro-cities sensitivity to allergens due to protected environment provided early in life.

Question 55.
Define Auto immunity. Give an example?
Answer:
Our own immune system attacks healthy cells in our body by mistake, (attack self cells), due to genetic or unknown reasons, which results in damage to the body.
Ex: Rheumatoid Arthritis.

Question 56.
What are lymphoid organs?
Answer:
Lymphoid organs: Organs where origin, maturation and proliferation of lymphocytes occur.
Primary lymphoid organs- Bone marrow and Thymus, these are the organs where immature lymphocytes undergo differentiation and maturation. After maturation they migrate to secondary lymphoid organs (spleen, lymphnodes, tonsils., Peyer’s patches of small intestine and appendix) and become effector cells. (Secondary’ . lymphoid organs).

Question 57.
Name the main lymphoid organ where lymphocytes are produced?
Answer:
Bone marrow → Main lymphoid organ – Lymphocytes and blood cells produced.

Question 58.
Define Thymus gland?
Answer:
Thymus → lobed organ located near the heart, beneath breast bone.
(Larger in size (birth time) grow (puberty) later degenerates.)

Question 59.
Name the graveyard of blood?
Answer:
Spleen: Graveyard of blood – filter blood from microbes. It acts by phagocytosis.

Question 60.
What are lymphnodes? Mention its function.
Answer:
Lymphnodes: Small nodes, solid structures located at different points along lymphatic system. Filters micro- organisms and antigens which activate lymphocytes.

Lymphoid tissue present in lining of major respiratory, digestive and urogenital tracts called as

Question 61.
Expand MALT and GALT?
Answer:
Mucus Associated LymphatimTissue (MALT); GALT → Gut Associated Lymphoid Tissue.

Question 62.
Expand AIDS?
Answer:
Acquired Immune Deficiency Syndrome. First reported in 1981.

Question 63.
Name the pathogen which causes AIDS?
Answer:
Pathogen → Human Immuno Deficiency Virus (HIV).

Question 64.
Name the virus group to which the AIDS belongs to?
Answer:
Belongs to Retroviruses (RNA is genetic material).

Question 65.
Write the mode of transmission of AIDS?
Answer:
Transmission: Sexual contact with infected person, multiple see partners Blood transfusion, injected needles (drug addicts) Injected mother to her child through placenta, spread through body fluids.

Question 66.
Mention the symptoms of AIDS?
Answer:
Symptoms: 5-10 years after infection. Fever, diarrhoea, weight loss couldn’t overcome infections like viruses ” fungi mycobacterium and parasite like Toxoplasma. Immuno defecient.

Question 67.
Draw a neat labelled diagram of RNA replication in humans?
Answer:

Question 68.
Explain mode of replication of HIV in humans?
Answer:

1. When virus enters the body it enters into macrophages, where viral RNA is converted into DNA by reverse transcriptase.
2. Viral DNA gets incorporated into host cell’s DNA and directs host DNA to produce viral RNA copies are released as viral particles.
3. The infected cells break open and released virus infect new cells (Tlym)
4. This process is repeated and leads to deerese in no of helper flymphin infected person.

Question 69.
Name the diagnostic test for AIDS?
Answer:
ELISA – (Enzyme Linked Immuno Sorbant Assay) Test Anti retro viral drugs ate partially effective, which can prolong life, but cannot prevent death.

Question 70.
Write the preventive measures of AIDS?
Answer:
No cure, prevention is the best option national AIDS control organisation (NGO) people world hea’ organisation (WHO) – programmes. Prevention by using disposable syringes, needles, free distribution e condoms, controlling drug abuse, regular check-ups etc.

Question 71.
Define Cancer?
Answer:
“Condition in which there is uncontrolled cell division resulting in the abnormal growth of excess tissue”

Question 72.
List any three types of cancer cells?
Answer:
Characters of Cancer cells:

1. Absence of contact inhibition.
2. Uncontrolled cell division.
3. Give rise to tumours.
4. It may show metastasis.

Question 73.
Differentiate between Benign and Malignant tumours?
Answer:

Benign	Malignant
1. Confined to original location do not spread to other parts	1. Proliferating neoplastic/tumour cells spread rapidly and damage surrounding tissues (normal)
2. Non-Fatal	2. Fatal

Question 74.
Define Metastasis? (1M) Why it is fatal.
Answer:
Metastasis- Character of malignant tumours. The process of carrying malignant tumour cells, times to other parts along with blood and where ever they are lodged they produce a new tumour.

Question 75.
What are Carcinogens?
Answer:
Cancer-causing agents – Carcinogens

Question 76.
Classify carcinogens for example?
Answer:

Oncogenic viruses have genes called viral oncogenes.

Question 77.
Describe the methods of detection of Cancer?
Answer:

1. Biopsy-where piece of suspected tissue cut into thin section, stained and examined.
2. Histopathological studies-tissues, blood, bone marrow
3. Techniques- Xrays, CT, MRT for detection in Internal organs. CT uses 3D image MRI uses strong
   magnetic fields and non-ionising radiations to accurately detection.
4. Antibodies against Cancer-specific antigens
5. Molecular biology-to detect genes.

Question 78.
Write a note on the treatment of Cancer?
Answer:
Treatment of Cancer

• Surgery (Benign tumours)
• Radiation therapy-tumour cells are irradiated lethally.

Chemotherapy – kill cancerous cells.
The majority of drugs have side effects – hair loss, anaemia, etc. Since tumour cells avoid the immune system. Helps in activating the immune system and destroy tumours

Question 79.
Name the biological response Immuno therapy modifier in treatment of Cancer?
Answer:
Patients are given biological response modifiers-a interferon.

Question 80.
Define drug and drug abuse?
Answer:
Drug: Any chemical compound that alters the biochemical or physiological process of tissue or organisms. When these drugs taken neither approved nor supervised by medical professionals – it is known as Drug abuse.
Commonly abused drugs.

Question 81.
Name the source of Opioids?
Answer:
Opioids: Extracted from the latex of poppy plant-Papaver Somniferum. These are Drugs that bind to receptors in our CNS and gastrointestinal tract.

Question 82.
What is Morphine?
Answer:
Morphine is a main alkaloid or Opium, by (Morphine is used as painkiller and sedative)acetylation of this.

Question 83.
What is SMACK? Give its ill-effects of drug abuse.
Answer:
Heroin (SMACK or brown sugar) diacetylmorphine, is produced which is a bitter, crystalline compound. It acts as a depressant and slows down body function. Taken by snorting and injection.

Question 84.
What are cannabinoids and name the products?
Answer:
2 Cannabinoids: → Group of chemicals, interact with cannabinoid receptors in the brain.
Extracted from the inflorescence of Cannabis sativa. Marijuana, Hashish, (Charas) and Ganja.
Effect on the cardiovascular system of the body.
Taken by inhalation and orally.

Question 85.
Name the source of cocaine?
Answer:
3 COCAINE → Extracted from Erythroxylum coca.

Question 86.
Write a note on action of cocaine on human body?
Answer:
Interferes with the transport of neurotransmitter Dopamine.
Stimulate CNS Known as Coke or Crack.
Taken by snorting Excess dosage-Hallucination

Question 87.
Name the plants which have hallucinogenic properties?
Answer:
Atropa belladonna and Datura – Hallucinogenic properties.

Drugs like Barbiturates amphetamines (LSD) Lysergicacid Diethylamides

Medicines are given for mental illnesses like depression and insomnia.

Question 88.
Name the source of Nicotine?
Answer:
NICOTINE → from Tobacco plant-Nicotiana tabaecum

Question 89.
Write the effect of nicotine on human body?
Answer:

• Stimulates adrenal gland to release adrenaline and noradrenaline into blood which increace heart rate, and BP.
• Smoking → Cancers of lung, throat, urinary bladder causing → oral cancer. Smoking increases carbon monoxide in blood and causes deficiency of O2 in the body.

Question 90.
Write the source and effect on the human body of the following drugs? Morphine, Cocaine, Marijuana?
Answer:
Adolescence and Drug/Alcohol Abuse
Child → Maturity process and period – leads to adolescence. Vulnerable phase of mental and psychological development, with curiosity, adventure, excitement and experimentation push towards abuses. Less supportive family → leads to abuse. If under pressure and stress in studies – it leads persuade to try alcohol and drugs.
Media promote this perception.
Addiction and Dependence
Because of known benefits – drugs are frequently used

Question 91.
Define addiction?
Answer:
Addiction: A sense of craving for anything which interferes with a person’s ability to function normally.
It is a psychological attachment to certain effects. When their effects (Temporary well being or euphoria) and their use become self-destructive.
Even once the use of these drugs leads to greater addiction, one becomes dependent on their use.

Question 92.
What is withdrawal syndrome?
Answer:
Dependence-Tendency of the body to experience unpleasant symptoms when dose of drugs/alcohol (regular) is abruptly discontinued is with drawl syndrome.

But relieved when use is resumed again. This withdrawal syndrome is characterised by-Anxiety shakiness, Nausea, sweating etc. and the person may need medical supervision.

Patient suffers from severe withdrawal syndrome, which makes him/her to ignore all social norms in order to get needs.

Question 93.
Mention the harmful effects of drugs/Alcohol?
Answer:

• Drug: Violence, Damage heart, liver, brain, Overdose leads to death. Intravenous injection-AIDS and Hep B.
• Alcohol: Liver’ disease-Cirrhoses (stores fat instead of glycogen) BP, Heart attack, Nervous system and circulatory system is affected.

Question 94.
Write a note of misuse of drugs?
Answer:
Misuse of Drugs
Sportspersons to enhance their performance.
Steroids in females-depression, menstrual cycle impaired, masculine characters.
Males: It makes acne, causes aggressiveness, decreased sperm production, Baldness and enlargement of prostate gland stunted growth in males and females.

Question 95.
Write a note on preventive and control measures of drug and alcohol abuse? (March 2014)
Answer:
Prevention and Control:

1. Avoid undue peer pressure: No child should be pressurised to perform beyond his/her limits in studies/sports/ activities
2. Education and Counselling!’ Channelise child’s energy into healthy pursuits like sports, reading, music and Yoga.
3. Seeking help from parents and peers: Guidance, advice, keeping away from friends who indulges them for abuse.
4. Looking for danger signs: Parents and teachers should watch, friends should bring notice to parents.
5. Seeking professional and medical help: help available from Qualified psychologists, psychiatrists and rehabilitation programmes. Detoxification programmes should be made available. With such help, the affected person get rid of problems completely to lead a normal and healthy life.

→ Health: State of complete physical, mental and social well being and not merely an absence of disease or infirmity

• People – Healthy – Efficient work → productivity increases → Economic prosperity.
• Health is affected by Impairment due to genetic disorders, infections lifestyles.

→ Immune System in The Body:
The immune system has the capacity to recognise the entry of antigens through all possible routes into the body and mobilise its cells to destroy them.

2nd PUC Biology Important Questions with Answers

Karnataka 2nd PUC Biology Important Questions Chapter 7 Evolution

Question 1.
Define evolution?
Answer:
It is descent with modification.

Question 2.
What does Big bang theory states?
Answer:
It states that single large explosion created the universe.

Question 3.
Who proposed the chemical evolution of life?
Answer:
Oparin of Russia and Haldane of England gave the chemical evolution of life.

Question 4.
Define the theory of A biogenesis?
Answer:
It states that life originated from non living material spontaneously and continuously.

Question 5.
Name the Biologist who demonstrated experiments and proved that life comes from pre existing life?
Answer:
Louis Pasteur.

Question 6.
Give the diagrammatic representation of Stanley Millers Experiment
Answer:
Stanley Miller and Urey created electric discharge in a closed flask containing CH₄, N₂, NH₃ and water vapour at 800C. He observed formation of amino acids.

This experiments justifies that organic molecules could be formed from the primitive gases on the pri mordial earth.

Question 7.
What are homologous organs? Give an example each from plants and animals?
Answer:
Homologous organs are organs whose origin or structure is same but they are functionally different. Ex. Forelimbs of whales, bats, cheetah share similarity in pattern of forelimbs bones but perfonn different functions. Ex. Thoms and tendrils of Bougainvillea cucurbita and cucurbit have same origin but different functions – Thoms are defensive in function where as tendrils help the plant in climbing.

Question 8.
What are analogous organs? Give an example from plants and animals?
Answer:
Analogous organs are organs whose origin is different but they perform same function.
Ex. Flippers of penguin and dolphins have different structure but perform the function of flight Ex. Wings of bird and butterfly are not anatomically similar but perform similar function.

Question 9.
What is Adaptive radiation?
Answer:
Process of evolution of different species in a given geographical area starting from a point and literally radiating to other areas of geography (habitats) is called adaptive radiation. Ex. Darwinian finches.

Question 10.
Industrial melanism in peppered moth is an excellent example of natural selection. Justify the statement.
Answer:
Evidence From Natural Selection:
A supporting evolution for natural selection comes from England. In a collection of moths made in 1850’s i.e., before industrialization sets in, it was observed that those were more white winged moths on trees than dark winged or raelanised moths.

But after industrialization it was found that there were more dark winged moths in the same area so the proportion was reversed.

Predators will spot a moth against a contrasting background. Before industrial revolution the grey colored or white moth were more because lichens are light in color and they got deposited on trees so white winged moth got camouflaged.

After Industrial revolution due to soot deposition population of dark winged or melanised moth increased as white winged moth were traced

Excess use of herbicides, pesticides only resulted in selection of resistant varieties. It is true for microbes against which we employ antibiotics or drugs against eukaryotic organisms/cells.

Question 11.
State Hardy Weinberg principle?
Answer:
It states that the allelic frequencies in a population are stable and constant from one generation to another generation.

Question 12.
Name the five factors which effect Hardy Weinberg equilibrium?
Answer:
The factors which affect the Hardy Weinberg equilibrium are gene flow, gene migration, genetic drift, mutation, genetic recombination, natural selection.

Question 13.
What is saltation
Answer:
Single large step in mutation is saltation.

Question 14.
Mention the time of origin, nature, food habits, habitats, cranial capacities of following hominids?
Answer:

1. Drvopithclus
2. RamapIhecus
3. Australopithecus
4. Homo hairs
5. Homo erectus
6. Neanderthal man
7. Homo sapiens.

→ Evolution:
Biological evolution is descent with modification.

→ Origin of Universe:

1. Universe is very old almost 20 Billion years.
2. Huge clusters of galaxies comprise the universe.
3. Galaxies contain stars and clouds of gas and dust.
4. According to Big bang theory single large explosion created the universe. .
5. As the universe expanded and cooled materials condensed under the influence of gravity to form stars and planets.
6. Earth is among the 8 planets revolving round the sun
7. Earth is formed 4.5 billion years back.

→ Theory of Panspermia:
Some scientists believed that life came from outside. Early Greek thinkers thought units of life called spores were transferred to different planets including Earth. ‘Panspermia’ is a favourite idea of astronomers.

→ Theory of a Biogenesis:

1. It states that life originated from non living material continuously and spontaneously.
2. This theory states that life came out of decaying and rotting matter like straw, mud etc.

→ Theory of Biogenesis:
It states that life comes from pre existing life. This theoiy was given by a scientist Louis Pasteur. He showed that in pre sterilized flasks, life did not come from killed yeast while in another flask open to air new organisms arose from killed yeast.

→ Theory of Chemical Evolution of Life:nd. This theory states that life was preceded by chemical evolution i.e., formation of divers organic molecules from inorganic constituents.

→ Evolution of Life Forms – A Theory:
Charles Darwin made a sea voyage in a ship called H.M.S Beagle. Any population has built in variation in characteristics. Those characteristics which enable some to survive better in natural conditions would out breed others that are less endowed to survive under natural conditions.

Fitness as per Darwin is reproductive fitness which means those who are better fit in an environment leave more progeny than others. They will survive more and are selected by nature. This is called as selection.

→ Evidence of Evolution:
Comparative anatomy and Morphology- It includes both homologous and Analogous organs.
Homologous organs have same origin but perform different functions. Whales, bats, cheetah and human (all mammals) share similarities in the pattern of bones of forelimbs but function is different. In whales forelimbs helps in swimming, in bats forelimbs helps in fight. In cheetah and human forelimbs help, in running. Thus homologous organs show divergent evolution.

Analogous organs have different origin but perform same function. Wings of butterfly and birds look alike perform same function but their structure is different. Analogous structures are a result of convergent evolution. Other examples of analogy are eye of octopus and of mammals or the flippers of penguins and dolphins.

→ Fossil Evidence / Evidence From Paleontology:
Fossils are remains of hard parts of life forms found in rocks. Different aged rock sediments contain fossils of different life forms that died during the formation of particular sediment. A study of fossils in different sedimentary layers indicates geological period in which they existed.

→ Evidence From Natural Selection:
A supporting evolution for natural selection comes from England. In a collection of moths made in 1850’s i.e., before industrialization sets in, it was observed that those were more white winged moths on trees than dark-winged or raelanised moths.

But after industrialization it was found that there were more dark winged moths in the same area so the proportion was reversed.

Predators will spot a moth against a contrasting background. Before industrial revolution the grey colored or white moth were more because lichens are light in color and they got deposited on trees so white winged moth got camouflaged.

After Industrial revolution due to soot deposition population of dark winged or melanised moth increased as white-winged moth were traced

Excess use of herbicides, pesticides only resulted in selection of resistant varieties. It is true for microbes against which we employ antibiotics or drugs against eukaryotic organisms/cells.

→ Biological Evolution:
Branching descent and natural selection are the two key concepts of Darwinian theory of evolution.
Lamarck gave the theory of use and disuse of orgHe gave the example of Giraffe. Giraffe used to raise its neck to reach the leaves of trees. Slowly giraffe had a long neck.

→ Mechanism Of Evolution:
Hugo Devries based on his work on evening primrose. Evolution of Darwin is gradual while Devries believed mutation caused speciation and hence it is saltation (single length step in mutation).

→ Hardy Weinberg Principle:
It states that allelic frequencies in a population are stable and constant from one generation to another. It is represented by binomial equation (p + q)² = p² + 2pq + q². Five factors are known to affect Hardy Weinberg equilibrium. These are gene migration or gene flow, gehetic drift mutation, genetic recombination, natural selection. When migration of a section of population to another place occurs gene frequencies change in the original as well as in the new population. New genes 7 alleles are added to the new population and lost from the old population. Sometimes the change in allelic frequency is so different in the new population that they become a different species. The original drifted population becomes founders and the effect is founder’s effect.

→ Origin And Evolution of Man:
About 15 millidn years ago primates called Dryopithecus and Ramapithecus were existing. They were hairy and walked like gorillas and chimpanzees.

Ramapithecus were more man like while Dryopithecus were ape like. Few fossils of man like bones have been discovered in Ethiopia and Tanzania. These revealed hominid features leading to the belief that about 3-4 mya ago, man like primates walked in Eastern Africa. They were not taller than 4 feet but walked upright.

Two million years ago Australopithecines lived in east African grasslands. Evidence showed they hunted with stone weapons but ate fruit. This creature was called hominid and was called Homo habilis. Brain capacity was between 650 – 800 CC. they did not eat meat.

Fossils discovered in Java in 1891 revealed the next stage i.e. Homo erectus about 1.5 million years ago. Homo erectus had a large brain around 900 CC. they ate meat.

Meanwhile Neanderthal man had a brain size of 1400CC and lived in east and central Asia between 1,00,000 – 40,000 years back. They used hides to protect their body and buried their dead.

Homo sapiens arose in Africa. During ice age between 75,000-10,000 years ago modem Homo sapiens arose. Pre historic cave art developed 15000 years ago. Agriculture and human settlement started.

2nd PUC Biology Important Questions with Answers

Karnataka 2nd PUC Biology Important Questions Chapter 6 Molecular Basis of Inheritance

Question 1.
Explain the structure of Nucleosome
Answer:
In prokaryotes, such as E coli through they do not have a defined nucleus the DNA is not scattered throughout the cell. DNA being negatively charged is held with some proteins that have positive charges in a region termed as nucleoid DNA in nucleoid is organized in large loops held by proteins.

In eukaryotes this organization is more complex. There is a set of positively changed basic proteins termed as histones. Histones are rich in amino acids arginine and lysine, which carry the charges in their side chains. There is an octamer of histone mojcehle and negatively charged DNA is wrapped and histone octamer is formed.

Nucleosome: A typical nucleosome contains 200bp of DNA helix. The nucleosomes in chromatin are seen as beads on string when seen in electron microscope. In a typical nucleus some regions of chromatin are loosely packed and light and are termed as euchromatin. Densely packed region of chromatin are heterochromatin which are dark in color. Euchromatin is active and heterochromatin is inactive, transcriptionally.

Question 2.
Name a nitrogenous base found only in DNA but absent in RNA?
Answer:
Thymine.

Question 3.
Name the nitrogenous base present in RNA but absent in DNA?
Answer:
Uracil.

Question 4.
Define nucleoside?
Answer:
A nucleoside is a combination of base and sugar.

Question 5.
Define nucleotide?
Answer:
It is a combination of Base, Sugar and phosphate.

Question 6.
What is Chargaff’s base equivalent rule?
Answer:
Total number of purines is equal to the total number of pyrimidine’s and their ratio is constant.

Question 7.
What do you mean by euchromatin?
Answer:
Lightly stained chromatin is euchromatin.

Question 8.
What do you mean by heterochromatin?
Answer:
Darkly stained chromatin is heterochromatin.

Question 9.
List the purines and pyrimidines in DNA?
Answer:
Purines Pyrimidines
Adenine and Guanine Cytosine Thymine.

Question 10.
List the nucleosides in DNA?
Answer:

1. Deoxyadenosine
2. Deoxyguanosine

Question 11.
List the nucleotides in DNA?
Answer:

1. Deoxyadenylic acid
2. Deoxyguanylic acid
3. Deoxythymidylic acid
4. Deoxycytidylic acid.

Question 12.
List the nucleosides of RIN A?
Answer:

1. Adenosine,
2. Guanosine
3. Uridine
4. Cytidine,

Question 13.
List the nucleotides of RNA?
Answer:

1. Adenylic acid
2. Guanylic acid.
3. Uridylic acid
4. Cytidylic acid.

Question 14.
Describe the double-helical structure of DNA with a labeled diagram?
Answer:

Question 15.
What is a nucleosome? Explain the structure of nucleosome with a neat labeled diagram?
Answer:
In prokaryotes, such as E coli through they do not have a defined nucleus the DNA is not scattered throughout the cell. DNA being negatively charged is held with some proteins that have positive charges in a region termed as nucleoid DNA in nucleoid is organized in large loops held by proteins.

In eukaryotes this organization is more complex. There is a set of positively changed basic proteins termed as histones. Histones are rich in amino acids arginine and lysine, which carry the charges in their side chains. There is an octamer of histone mojcehle and negatively charged DNA is wrapped and histone octamer is formed.

Nucleosome: A typical nucleosome contains 200bp of DNA helix. The nucleosomes in chromatin are seen as beads on string when seen in electron microscope. In a typical nucleus some regions of chromatin are loosely packed and light and are termed as euchromatin. Densely packed region of chromatin are heterochromatin which are dark in color. Euchromatin is active and heterochromatin is inactive, transcriptionally.

Question 16.
How did Avery, Macleod and Mccarty proved biochemical nature of transforming principle?
Answer:
Avery, Maccleod and Mccavty’s work stated the genetic material was thought to be a protein. They purified biochemical (Proteins, DNA, RNA etc) from the heat killed S cells to see which ones could transform R cells into S cells. They discovered that DNA from S bacteria caused R bacteria to become transformed.

They also discovered that protein digesting enzymes (proteases) and RNA digesting enzymes (RNases) did not affect transformation so the transforming substance was not protein or RNA. Digestion with DNase inhibited transformation suggesting that DNA caused the transformation.

Thus DNA is the hereditary material in all organisms.

Question 17.
Differentiate eukaryotic genes from prokaryotic genes?
Answer:

1. Eukaryotic gene are splits genes	1. Prokaryotic genes contain only exons contain both introns and exons
2. RNA Polymerase is of 3 types in Eukaryotic	2. It is of type only in prokaryotes Gene
3. It produces monocistronic m RNA	3. It produces polycistronic m RNA

Question 18.
Describe Griffith’s experiment on transformation to prone DNA is the genetic material?
Answer:

Question 19.
Describe Hershey and Chase’s experiment to prove DNA is the genetic material?
Answer:
Alfred Hershey and Martha Chase also gave experiments to prove DNA is the genetic material. They worked on viruses infecting bacteria called bacteriophages.

Bacteriophage attaches to bacteria and its genetic material enters bacterial cell. The bacterial cell treats the viral genetic material as if it was its own and manufactures more virus particles.

Hershey and Chase worked to discover whether it was protein or DNA from the viruses that entered the bacteria.
They grow some viruses on medium that contained radioactive phosphorus and some others on medium that contained radioactive sulfur.

Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not.

Similarly viruses growing on radioactive sulfur contained radioactive protein but not radioactive DNA because DNA does not contain sulfur.
Radioactive phage were allowed to attach to E. coli bacteria. As the infection proceeded viral coats were removed from the bacteria agitating them in ablender. Virus particles were separated from bacteria by spinning them in a centrifuge.

Bacteria infected with viruses that had radioactive DNA were radioactive indicating that DNA was the material that passed from virus to bacteria.

Bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicates that proteins did not enter the bacteria from the viruses DNA is the enetic material that is. passed from virus to bacteria.

Question 20.
What are split genes?
Answer:
Eukaryotic genes are termed as split genes as they contain both introns and exons.

Question 21.
Define transcription. Describe the process of transcription in prokaryotes?
Answer:
Transcription Unit: A transcription unit in DNA is defined primarily by three regions in DNA.

1. A promoter
2. Structural gene
3. A terminator

Since the two strands have opposite polarity and the DNA dependent RNA polymerase also called as polymerization in only one direction that is 5’ – 3’ the strand that has polarity 3’ – 5’ – acts a template and is referred to as template strand. The strand which does not code for anything is coding strand.

The promoter and terminator flank the structural gene in transcription unit. The promoter is said to be located towards 5’ end of the structural gene. It is a DNA sequence that provides binding site for RNA polymerase. Terminator is located towards 3 ’ end of the coating strand and it defines the end of the process of transcription.

Transcription Unit and Gene – Cistron is a segment of DNA Coding for a Polypeptide. Structural gene in a transcription unit could be monocistronic (mostly in eukaryotes) or polycistronic(bacteria or prokaryotes)

In Eukaryotes monocistronic’ structural genes are split. Coding sequences are defined as exons. Exons are interrupted by introns. Introns do not appear in mature or processed RNA.

Question 22.
Describe mechanism of semiconservative DNA replication?
Answer:
DNA replication: The parental DNA molecule produces two identical copies of daughter DNA molecules and it is called DNA replication.

1. The process of replication begins at a specific point on DNA strand called initiation site or site of origin (ORI).
2. The hydrogen bonds present between two strands are dissolved by DNA polymerase I.
3. The double stranded DNA is now getting separated to form single strands.
   This mechanism is called unzipping of DNA.
4. As the double helix is separated it gets into Y shaped appearance called replication fork. Consequently on one strand (the template with polarity (3’ – 5’) the replication is continuous while on the other (polarity 5’ – 3 ’) it is discontinuous. Discontinuously synthesized fragments are joined together by DNA ligase.

With reference to transcription define:
(a) Splicing – In enparyotes pritranscripts contain both exons and introns. Introns is non functional and it is subjected to splicing.
(b) Capping – Introns are removed and oxensone joined in a definite order in part.

Question 23.
Expand VNTR?
Answer:
Variable Number of Tandem Repeats.

Question 24.
Which is the largest human gene?
Answer:
Dystrophin.

Question 25.
Why t RNA is termed as adaptor molecule?
Answer:
It helps in transfer of amino acids to the site of protein synthesis hence it is termed as an adaptor molecule.

Question 26.
Name the initiator codon and terminator codons?
Answer:
AUG is termed as initiator codon and terminator codons are UAA, UAG, UGA.

Question 27.
Explain any two properties of genetic code
Answer:

1. Genetic code is universal: It is common and found in all living organisms.
2. Triplet codon: Genetic code is a triplet code and is represented by three-nucleotide sequences present on m RNA strand.
3. Initiator codon is AUG which codes for methionine. In case AUG is absent GUG acts as an initiator Codon which codes for amino acid valine.

Question 28.
Give any four important characteristic features of genetic code?
Answer:
Features:

1. Genetic code is universal: It is common and found in all living organisms.
2. Triplet codon: Genetic code is a triplet code and is represented by three nucleotide sequences present on m RNA strand.
3. Initiator codon is AUG which codes for methionine. In case AUG is absent GUG acts as an initiator Co don which codes for amino acid valine.
4. The process of translation is stopped by terminator Codons UAA, UAG, UGA which do not code for any amino acid.
5. Draw a neat labeled diagram of t RNA molecule.

Question 29.
Explain process of translation in detail.
Answer:
Translation: The process of polymerization of amino acids to form a polypeptide is termed as translation. The process of translation takes place in the cytoplasm of the cell where ribosome is located and it takes place in 4 levels
1. Activation or charging of amino acid: Amino acids before they attach to t RNA should be activated by using ATP molecules.
Amino acid + ATP- Activated amino acid.
Activated amino acid will be picked up by at RNA molecule and form amino a CY1 tRNA.
This process is called charging of amino acid.

2. Chain initiation:

• The formation of polypeptide chain is initiated by the initiator codon AUG.
• First mRNA binds to smaller subunit of ribosome ie (3os) and later on larger submit joins.
• Larger sub unit (5os) has a site (Aminoacyl site), P site peptidyl site (enzyme peptidyl transfer are and an E site or exit site.
  Image

3. Chain elongation: Ribosomal subunit shifts from one codon to another codon on the m RNA strand. T RNA carrying amino acid shifts from A site to P site. At P site of 5os subunit peptide bond formation takes place between 2 amino acids and it is catalyzed by peptidyl transfer enzyme and elongation factos EF , EF„ EF,. chain keeps elongating.

4. Chain termination: Codons present on m RNA strand are translated by ribosomes continuously to form a long polypeptide chain until the ribosomes reach the terminator codons (UAA, UAG, UGA).

Question 30.
With a neat labeled diagram explain Lac operon concept.
Answer:
Lac operon concept: The lac operon concept was given by Jacob and Monod. This concept was found in bacteria.

Lac operon consists of one regulatory’ gene (I gene) and three structural gene (Z, y; a) promoter gene gene Z codes for enzyme B galactosidase which is responsible for hydrolysis of disaccharide into its monomeric units galactose and glucose. Gene Y codes for the enzyme permease which increases permeability of all to lactose and gene a codes for the enzyme transacetylase.

In switch on mechanism the regulator gene synthesizes a repressor protein and forms inducer repressor complex to transcribe the 3 structural genes and by translation all 3 enzymes are synthesized which are needed for lactose metabolism.

In the switch off mechanism in the absence of inducer the repressor binds to operator gene and RNA polymerase is not allowed to transcribe the operon and thus no translation and no synthesis of enzymes takes place.

Question 31.
What is DNA finger printing? Mention steps involved in DNA finger printing technique.
Answer:
DNA finger printing:
It is a comparison of DNA from different sources to establish the identity.
This technique involves VNTR ie variable number of tandem repeats which are different between two individuals and are used in identifying different human beings.

Steps involved in DNA finger printing:
(a) Collection of samples: A sample of blood, semen (sperm) saliva, hair follicle, skin, bone or any other tissue even us dried condition is collected.

(b) Extraction of DNA: Sample is centrifuged at 200 P PM to spin off and separate the cells. Sample of DNA is treated with REN (restriction endonuclease enzyme) to obtain fragments of DNA of different length.

(c) Separation of DNA fragments: The DNA fragments are allowed to separate on agarose gel Electro phoresis. DNA fragments are separated on the gel plate according to their density gradient where heavier DNA stands are settling at the bottom of the gel plate and lighter fragments towards the top. These DNA fragments are treated with alkaline solution to dissolve H2 bonds to produce single stranded DNA.

(d) Extraction of DNA fragments: Single stranded DNA fragments on the gel plate are extracted on to the nylon mesh or nitro cellulose membrane by the process called southern blotting technique.

(e) Tagging Radio active probes: The nylon mesh containing single stranded DNA fragments are incubated with radio active probe. Probe is a radio active synthetic DNA of known nucleotide sequence.

(f) Auto radiography: An x-ray film is placed on the nylon mesh to obtain an autoradiography print. X-ray film is then processed to obtain visible patterns of bonds of DNA which is DNA fingerprint. A comparison of two different DNA finger prints can provide information about degree of similarities and differences.

Question 32.
Mention any five goals and salient features of HGP.
Answer:
Human Genome Project (HGP) is a mega project. It has some goals and salient feature.
Goals of HGP: Some of the important goals of HGP were as follows:

1. Identify all the approximately 20000 – 25000 genes in human.
2. Determine the sequences of the 3 million chemical base pains which make up in databases.
3. Store the information in databases.
4. Improve tools for data analysis
5. Transfer related technologies to other sectors such as industries.
6. Address the ethical, legal, social issues that may arise from the project.

Nucleic acids are macromolecules found in the nucleus as well as cytoplasm. Nucleic acids are broadly of two types Deoxyribonucleic acid (DNA) and Ribonucleic acid (RNA). DNA acts as the genetic material in most of the organisms. RNA acts as the genetic material in some viruses. RNA functions as adapter, structured and in some cases as a catalytic molecule.

DNA: It is a long polymer of deoxyribonucleotides. It is characteristic of an organism.

Structure of Polynucleotide chain.: A nucleotide has three components a nitrogenous base, pentose sugar (ribose and case of RNA And deoxyribose for DNA) and a phosphate group.

There are two types of nitrogenous bases – purine (Adenine and Guanine) and pyrimidine’s (Cytosine, Uracil and Thymine). Cytosine is common for both DNA and RNA and thymine is present in DNA. Uracil is present in RNA at the place of thymine.

Nucleoside consists of nitrogenous base and pentose sugar through glycosidic linkage. Nucleosides of DNA are deoxyadenosine, deoxyguanosine, deoxycytidine, deoxythymidine.

When a phosphate group is linked to a 5 OH group of nucleoside through phosphodester linkage a nucleotide is formed. More nucleotides form a polynucleotide chain.

In RNA pyrimidine base found is uracil instead of thymine.
Freidrich Meischer in 1969 noticed an acidic substance present in nucleus termed as nuclein. Watson and Crick in 1953 gave the double helical structure of DNA. . .

One of the hall marks of this proposition was base pairing between the two strands of polynucleotide chains.

Chargaff found out that for a double stranded DNA the ratios between adenine and Thymine and guanine and cytosine are constant and equals one.

The salient features of double helical structure of DNA are as follows.

1. It is made of two polynucleotide chains, where the backbone is constituted by sugar-phosphate.
2. The two chains have anti parallel polarity. It means if one chain has the polarity 5’-3’ other has 3’-5’.
3. The bases in two strands are paired through hydrogen bonds forming base pairs. Adenine forms two bonds with thymine and guanine forms three H bonds with cytosine. As a result always purine comes opposite to pyrimidine.
4. The two chains are coiled in a right handed fashion. Thtch of helix is 3.4mm and there are roughly 10BP in each turn. Diameter of the helix is 0.34nm

→ Search For Genetic Material:
Transforming principle: In 1928, Frederick Griffith in a series of experiments with Streptococcus pneumonia witnessed a transformation in bacteria.

When streptococcus pneumonia bacteria are grown on a culture plate, some produce smooth shiring colonies (S) while others produce rough colonies (R). This is because the S strain bacteria have a mucous polysaccharide coat, while (R) strain does not. Mice infected with S strain (Virulent) die from pneumonia infection but mice infected with R strain do not develop pneumonia.

• S Strain – Inject into Mice – Mice die
• R Strain – Inject into Mice – Mice lives
• S Strain (heat-killed) – Inject into mice – mice live.
• S Strain (heat-killed) + R strain (live) – Inject into Mice – Mice die.

He observed that when heat-killed S strain bacteria was injected with a mixture of heat killed S bacteria and live R bacteria the mice died.

He concluded that the R strain bacteria had some how been transformed by the heat killed S strain bacteria. This transforming principle is DNA.

→ Biochemical Characterization of Transforming Principle:
Avery, Maccleod and Mccavty’s work stated the genetic material was thought to be a protein. They purified biochemical (Proteins, DNA, RNA etc) from the heat killed S cells to see which ones could transform R cells into S cells. They discovered that DNA from S bacteria caused R bacteria to become transformed.

They also discovered that protein digesting enzymes (proteases) and RNA digesting enzymes (RNases) did not affect transformation so the transforming substance was not protein or RNA. Digestion with DNase inhibited transformation suggesting that DNA caused the transformation.

Thus DNA is the hereditary material in all organisms.

→ Genetic Material is DNA:
Alfred Hershey and Martha Chase also gave experiments to prove DNA is the genetic material. They worked on viruses infecting bacteria called bacteriophages.

Bacteriophage attaches to bacteria and its genetic material enters bacterial cell. The bacterial cell treats the viral genetic material as if it was its own and manufactures more virus particles.

Hershey and Chase worked to discover whether it was protein or DNA from the viruses that entered the bacteria.
They grow some viruses on medium that contained radioactive phosphorus and some others on medium that contained radioactive sulfur.

Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not.

Similarly viruses growing on radioactive sulfur contained radioactive protein but not radioactive DNA because DNA does not contain sulfur.
Radioactive phage were allowed to attach to E. coli bacteria. As the infection proceeded viral coats were removed from the bacteria agitating them in ablender. Virus particles were separated from bacteria by spinning them in a centrifuge.

Bacteria infected with viruses that had radioactive DNA were radioactive indicating that DNA was the material that passed from virus to bacteria.

Bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicates that proteins did not enter the bacteria from the viruses DNA is the enetic material that is. passed from virus to bacteria.

→ Properties of Genetic Material (DNA Vs RNA)

• It should be able to generate its replica.
• It should be chemically and structurally stable.
• It should provide for slow changes, (mutation) required for evolution.
• Protein does not fulfill the criteria hence it is not the genetic material.
• RNA and DNA fulfill the criteria.

→ RNA is Unstable:

• 2 – 0 H group present at every nucleotide (ribose sugar) in RNA is a reactive group and makes RNA liable and
  easily degradable.
• RNA is known as catalyst hence reactive RNA is Unstable and mutates faster.

ONA is More Stable:

• Stability is one of the properties of genetic material.
• DNA is chemically less reactive and structurally more stable when compared to RNA.
• Presence of thymine in place of uracil confers additional stability to DNA.
• Two strands being complimentary if separated by heating come together when appropriate conditions are provided,

→ Better Genetic Material (DNA OR RNA):

• Presence of thymine in place of uracil confers more stability to DNA.
• Both DNA, RNA are able to mutate.
• RNA can directly code for synthesis of proteins.
• For transmission of genetic information RNA is better.
  Hence DNA is better genetic material than RNA.

→ DNA replication: The parental DNA molecule produces two identical copies of daughter DNA molecules and it is called DNA replication.

1. The process of replication begins at a specific point on DNA strand called initiation site or site of origin (ORI).
2. The hydrogen bonds present between two strands are dissolved by DNA polymerase I.
3. The double stranded DNA is now getting separated to form single strands. This mechanism is called unzipping of DNA.
4. As the double helix is separated it gets into Y shaped appearance called replication fork. Consequently on one strand (the template with polarity (3’ – 5’) the replication is continuous while on the other (polarity 5’ – 3 ’) it is discontinuous. Discontinuously synthesized fragments are joined together by DNA ligase.

→ The Experimental Proof:
Matthew Messelson and Franklin Stahl performed the following experiment in 1958:
1. They grew E coli in a medium containing NH4 Cl is the heavy isotope of nitrogen. The result was that N was incorporated into newly synthesized DNA as well as other nitrogen containing compounds. This heavy DNA molecule could be distinguished from normal DNA by centrifugation in a calcium chloride density gradient (Cscl).

2. Then they transferred cells into a medium with normal NH4 Cl and took samples at various definite time intervals as cell multiplied.

3. Thus DNA that was extracted from culture one generation after the transfer from N to N medium had a hybrid or intermediate density. DNA extracted from the culture after another generation ie after 40 minutes was composed of equal amount of hybrid DNA and light DNA.

The experiments proved that DNA in chromosomes also replicate semi conservatively.

Transcription; Process of copying genetic information from one strand of DNA into RNA is termed as transcription. Here also the principle of complementarity governs the process of transcription except that adenosine forms base pair with uracil instead of thymine.

Transcription Unit: A transcription unit in DNA is defined primarily by three regions in DNA.

1. A promoter
2. Structural gene
3. A terminator

Since the two strands have opposite polarity and the DNA dependent RNA polymerase also called as polymerization in only one direction that is 5’ – 3’ the strand that has polarity 3’ – 5’ – acts a template and is referred to as template strand. The strand which does not code for any thing is coding strand.

The promoter and terminator flank the structural gene in the transcription unit. The promoter is said to be located towards 5’ end of the structural gene. It is a DNA sequence that provides binding site for RNA polymerase. Terminator is located towards 3 ’ end of the coating strand and it defines the end of the process of transcription.

Transcription Unit and Gene – Cistron is a segment of DNA Coding for a Polypeptide. Structural gene in a transcription unit could be monocistronic (mostly in euparyotes) or polycistronic(bacteria or prokaryotes)

In Eukaryotes monocistronic’ structural genes are split. Coding sequences are defined as exons. Exons are interrupted by introns. Introns do not appear in mature or processed RNA.

→ Types of RNA And The Process of Transcription:
In bacteria there arc major types of RNA’s mRNA tRNA and rRNA. All 3RNA’s are needed to synthesize a protein in the cell m RNA provides template, TRNA brings amino acids and reads the genetic code, or RNA’s play structural and catalytic role during translation.

There is a single DNA dependent RNA polymerase that catalyses transcription of all types of RNA in bacteria. RNA polymerase binds to promoter and initiates transcription. It facilitates opening of helix and continuous elongation. Once polymerases reaches the terminator region nascent RNA falls of and so also RNApolymerases. This results in termination of transcription.

In eukaryotes there are 3 types of RNApolymerases in the nucleus. RNA polymerase I transcribes or RNA where as RNA polymerase III is responsible for transcription of t RNA, 5 Sr RNA, SnRNA. The RNA polymerase II transcribes precursor of in RNA (the heterogeneous nuclear RNA (HnRNA).

In eukaryotes primary transcripts contains both exons and introns and is non functional. Hence it is subjected to a process called splicing where the introns are removed and exons are joined in a defined order hn RNA under¬goes additional processing called as capping and tailing. In capping an unusual nucleotide (methyl guanosine triphosphate) is added to 5’end of hn RNA.

In tailing adenylate residues (200 – 300) are added at 3’ in a template independent manner.

→ Genetic Code:
Features:

1. Genetic code is universal: It is common and found in all living organisms.
2. Triplet codon: Genetic code is a triplet code and is represented by three-nucleotide sequences present on m RNA strand.
3. Initiator codon is AUG which codes for methionine. In case AUG is absent GUG acts as an initiator Codon which codes for amino acid valine.
4. Process of translation is stopped by terminator codons UAA, UAG, UGA which do not code for any amino acid.
5. Genetic code is commaless
6. It is sensible
7. It is non-overlapping.

→ Mutations and Genetic Code:
Mutations can be sudden and heritable change in an offspring. Mutation can be of 2 types

1. Point Mutation.
2. Frameshift Mutation.

Point Mutation.	Frameshift Mutation.
1. Change of singe base pair in the gene	1. It occurs due to insertion or deletion of base
Ex: Sickle called anemia is due to change of pairs in the gene	Ex: Thalassemia amino acid residue glutamate to valine.

→ TRNA the adapter molecule:

T RNA is termed as adaptor molecule as it would read the code on one hand and on the other hand would bind to specific amino acids.

Transfer RNA or T RNA has an anticodon loop that has bases complimentary to the code. It has amino acid acceptor end to which it binds to amino acids. T RNA’s are specific to each amino acid. For initiation there is initiator t RNA. It has clover leaf structure.

Translation: The process of polymerisation of amino acids to form a polypeptide is termed as translation. The process of translation takes place in the cytoplasm of the cell where ribosome is located and it takes place in 4 levels

1. Activation or charging of amino acid: Amino acids before they attach to t RNA should be activated by using ATP molecules
Amino acid + ATP- Activated amino acid.
Activated amino acid will be picked up by at RNA molecule and form amino a CY1 tRNA.
This process is called charging of amino acid.

2. Chain initiation:

• The formation of polypeptide chain is initiated by the initiator codon AUG.
• First mRNA binds to smaller subunit of ribosome ie (3os) and later on larger submit joins.
• Larger sub unit (5os) has a site (Aminoacyl site), P site peptidyl site (enzyme peptidyl transfer are and an E site or exit site.

3. Chain elongation: Ribosomal subunit shifts from one codon to another codon on the m RNA strand. T RNA carrying an amino acid shifts from A site to P site. At P site of 5os subunit peptide bond formation takes place between 2 amino acids and it is catalyzed by peptidyl transfer enzyme and elongation factos EFp EF2, EF,. chain keeps elongating.

4. Chain termination: Codons present on m RNA strand are translated by ribosomes continuously to form a long polypeptide chain until the ribosomes reach the terminator codons (UAA, UAG, UGA).
The long polyptide chain so formed is dissociated from the ribosomeand released into the cytoplasm of a cell that becomes into a type of protein.

Chain termination is characterized by release factors RF₁, RF₂ RF₃.
Image
Lac operon concept: The lac operon concept was given by Jacob and Monod. This concept was found in bacteria.

Lac operon- consists of One regulatory gene (I gene) and three structural gene (Z, y, a) promoter gene gene Z codes for enzyme B galactosidase which is responsible for hydrolysis of disaccharide into its monomeric units galactose and glucose. Gene Y codes for the enzyme permease which increases permeability of cell to lactose and gene a codes for the enzyme transacetylasd.

In switch on mechanism the regulator gene synthesizes a repressor protein and forms inducer repressor com¬plex to transcribe the 3 structural genes and by translation all 3 enzymes are synthesized which are needed for lactose metabolism.

In the switch off mechanism in the absence of inducer the repressor binds to operator gene and RNA polymerase is not allowed to transcribe the operon and thus no translation and no synthesis of enzymes takes place.

Human Genome Project (HGP) is a mega project. It has some goals and salient feature.

→ Goals of HGP: Some of the important goals of HGP were as follows.

• Identify all the approximately 20000 – 25000 genes in humans.
• Determine the sequences of the 3 million chemical base pairs which make up in databases.
• Store the information in databases.
• Improve tools for data analyses
• Transfer related technologies to other sectors such as industries.
• Address the ethical, legal, social issues that may arise from the project.

→ Salient features of Human Genome:
Some of the salient features of HGP

• The human genome contains 3164.7 million nucleotides.
• The average gene consists of 3000 bases largest known human gene is dystrophin which has 2.4 million bases.
• Less than 2% of the genome codes for proteins.
• Repeated sequences make up a large portion of the human genome.
• Chromosome I has most genes (2968) and Y has fewest 231.
• Scientists have identified about 1.4 million locations were single base DNA differences occur in human.

→ Applications:

• Helps in diagnosing and treatment for genetic.
• DNA finger printing.
• It is a comparison of DNA from different sources to establish the identity.
• This technique involves VNTR ie variable number of tandem repeats which are different between two individuals and are used in identifying different human beings.

→ Steps Involved in DNA Finger Printing:
(a) Collection of samples: A sample of blood, semen (sperm) saliva, hair follicle, skin, bone or any other tissue even us dried condition is collected.

(b) Extraction of DNA: Sample is centrifuged at 200 or PM to spin off and separate the cells. Sample of DNA is treated with REN (restriction endonuclease enzyme) to obtain fragments of DNA of different length.

(c) Separation of DNA fragments: The DNA fragments are allowed to separate on agarose gel Electro phoresies. DNA fragments are separated on the gel plate according to their density gradient where heavier DNA stands are settling at the bottom of the gel plate and lighter fragments towards the top. These DNA fragments are treated with alkaline solution to dissolve H2 bonds to produce single-stranded DNA.

(d) Extraction of DNA fragments: Single stranded DNA fragments on the gel plate are extracted on to the nylon mesh or nitro cellulose membrane by the process called southern blotting technique.

(e) Tagging Radio active probes: The nylon mesh containing single stranded DNA fragments are incubated with radio active probe. Probe is a radio active synthetic DNA of known nucleotide sequence.

(f) Autoradiography: An X-ray film is placed on the nylon mesh to obtain an autoradiography print. X-ray film is then processed to obtain visible patterns of bonds of DNA which is DNA fingerprint. A comparison of two different DNA finger prints can provide information about degree of similarities and differences.

→ Applications of DNA Finger Printing:

1. This technique is used in identification of criminals like murderers, robber rapist etc.
2. It is employed in solving parental disputes.
3. It is used in identifying and reunion of lost children with their parents during natural disasters like earthquake, flood etc.
4. It is useful in identifying the victims of plane crash, building collapse, fire breakouts etc.
5. Useful in determination of genetic diversity and evolutionary biology.

→ Salient Features of Human Genome:
Some of the salient features of HGP

1. The human genome contains 3164.7 million nucleotides.
2. The average gene consists of 3000 bases largest known human gene is dystrophin which has 2.4 million bases.
3. Less than 2% of the genome codes for proteins.
4. Repeated sequences make up a large portion of the human genome.
5. Chromosome I has most genes (2968) and Y has fewest 231.
6. Scientists have identified about 1.4 million locations where single-base DNA differences occur in human.

2nd PUC Biology Important Questions with Answers

Karnataka 2nd PUC Biology Important Questions Chapter 5 Principles of Inheritance and Variation

Question 1.
Define test cross?
Answer:
Test cross is a cross between F; hybrid and its recessive parent.

Question 2.
What are alleles? .
Answer:
Genes which code for a pair of contrasting traits are termed as alleles.

Question 3.
Mention the phenotypic ratio of Monohybrid cross?
Answer:
3:1

Question 4.
Mention the phenotypic ratio of dihybrid cross?
Answer:
9:3:3:1

Question 5.
State the law of dominance?
Answer:
It states that in a dissimilar pair of factors one member of the pair dominates the other.

Question 6.
Write any three reasons for selection of pea plants by Mendel for his hybridization experiment?
Answer:

1. It is a true-breeding plant.
2. It is a self-pollinated crop plant.
3. It has a short life span.

Question 7.
Give the schematic representation of one gene inheritance with respect to law of dominance?
Answer:
It is a cross between two parents which differ from one another with respect to single pair of contrasting characters or single character. Mendel took height of pea plant as an example and made a cross between a pure or homozygous tall pea plant and a pure or homozygous dwarf pea plant. In the generation, all plants obtained were tall since T is dominant over t (dwarfness).

When F₁ plants were self-pollinated then in the F2 generation phenotypic ratio obtained was 3:1 and genotypic ratio 1:2:1.
Parents Pure Tall Pure Dwarf
TT tt -P₁

Gametes (T)(T) (t)(t)
Self-pollination Tt × Tt.

3:1 Phenotypic ratio
ITT Pure or homozygous tall.
2Tt Hybrid tall ltt Homozygous dwarf
Phenotypic ration = 3:1
Genotypic ratio = 1:2:1

Question 8.
Describe Monohybrid cross with an example?
Answer:
It is a cross between two parents which differ from one another with respect to single pair of contrasting characters or single character.

Mendel took height of pea plant as an example and made a cross between a pure or homozygous tall pea plant and a pure or homozygous dwarf pea plant.

In the F₁ generation all plants obtained were tall since T is dominant over t (dwarfness)

When F₁ Plants were self pollinated then in the F₂ generation phenotypic ratio obtained was 3:1 and genotypic ratio 1:2:1.
Parents Pure Tall Pure Dwarf
TT tt P₁
Gametes (T)(T) (t)(t)
Self pollination Tt X Tt.

Question 9.
State and explain the law of independent assortment with the help of inheritance of 2 genes?
Answer:
When two or more than two alleles are brought together in F₁ hybrids they assort independently of one another in F₁ generation during the formation of gametes.

When a plant producing round and yellow coloured seeds is crossed with a plant producing Wrinkled and green seeds all the F, hybrids produce round and yellow coloured seeds as Yellow (Y) is dominant over green (y) and allele for round (R) is dominant over wrinkled (r).

When the F, hybrids are self crossed to obtain F₂ generation all four types of plants are obtained such as round yellow (9), round and green (3), wrinkled and Yellow’ (3), Wrinkle and green (1) in the ratio of 9:3:3:1. Genotypic ratio is 1: 2: 2: 4: 1: 2: 1: 2: 1
(Phenotypic ratio) Parental at generation.

Phenotype Round and Yellow X Wrinkled and Green
Genotype RRYY rrYY
Gametes RY rY
RrYy F₁ Generation.
All offsprings are found and yellow
RrYy X RrYy

Round and yellow – 9, Round and Green – 3 Wrinkled and yellow – 3 Wrinkled and Green – 1

Test cross (Dihybrid):
When the F₁ hybrid obtained by crossing two parents differing in 2 characters are again crossed with double recessive parent it is test cross

Genotype RrYy X rryy
Gametes RYRy ry
rY ry

Round and yellow RrYy- 4 Round and Green Rryy – 4 Wrinkled and Yellow rryy – 4 Wrinkled and green rryy – 4. Test cross-ratio. Phenotypic and Genotypic ) = 4: 4: 4: 4 or 1: 1: 1: 1.

Question 10.
Explain the following terms with an example?
Answer:
(a) Codominance: When both the alleles are equally dominant it is termed as codominance.
AB blood group consists of two alleles IA and IB and they are both present together. Both express their own types of sugar because of codominance.

(b) Incomplete dominance: When the dominant gene is not able to mask the effect of recessive gene completely it results in blending of characters. This is termed as incomplete dominance.

When a homozygous red flowered plant is crossed with homozygous white flowered plant then in F, generation all flowers are pink.

When F, plants are self pollinated then in F2 generation both phenotypic and genotypic ratios obtained are 1:2:1.
RR homozygous red X rr – P,
Gametes (R) (P) (r) (r)
Rr = F₁
Pink.

Phonotypic and genotypic ration = 1:2:1
1- Pure Red, 2 – Pink, 1- Pure white.

Question 11.
Briefly mention the contribution of T.H Morgan in genetics?
Answer:
(a) They could be grown easily in the laboratory.
(b) They have a short life span
(c) A single mating can produce a large members of progenies.
(d) They exhibit sexual dimorphism i.e., clear differentiation between male and female sexes.

Question 12.
Who had proposed the chromosomal theory of inheritance?
Answer:
This theory was proposed by waiter Sutton and Theodore Bovery in 1902.
(a) A child has blood group O if the father has blood group A and mother blood group B, work out the genotypes of parents and genotypes of other offsprings.
Father × Mother
A B
Genotype of father I^A I^A I^B I^B
I^A I^B AB

Heterozygous state
Father I^Ai × I^Bi
Mother. I^AI^B, I^Ai, I^Bi, ii O
A B
Other offsprings can have genotype AB, A B, other than O.

Question 13.
How is sex determined in human beings?
Answer:
In humans, XX-Xy type of Sex determination is found. Females are homogametic with XX chromosome. Ma- leas are heterogametic as 50% of gametes carry the ‘Y’ chromosome and other 50% of gametes carry Y chromosome.

Parents Female Male
AA+XX AA+XY
Gametes A+X A+X A+Y
Offspring AA+XX AA+XY
Female Male.

Question 14.
What is male heterogamety? Give an example.
Answer:
In humans XX-XY type of sex determination is found. Females are homogametic with XX chromosome. Males are heterogametic as 50% of gametes carry ‘ Y’ chromosome and other 50% of gametes carry X chromosome.

Question 15.
What is pointmutation? Give on example?
Answer:
When mutation arises due to change in a single pair of DNA it is termed as point mutation Ex;- Sickle cell anaemia, Genetic disorders.

Pedigree analysis: Analysis of traits in several of generations of a family is called pedigree analysis. In human . genetics pedigree study provides a strong tool which is utilized to trace the inheritance of a specific trait, abnormality or disease. Some of the standard symbols used in pedigree analysis are as follows Image

Broadly genetic disorders are of 2 types.

1. Mendelian disorders.
2. Chromosomal disorders.

Mendelian disorders are mainly determined by alteration or mutation in a single gene. They are Haemophilia, cystic fibrosis, sickle celled anemia, color blindness, phenylketonuria. By pedigree analysis, one can understand whether the trait is dominant or recessive.

X-linked recessive gene shows transmission from carrier female to male progeny.

Haemophilia: It is also called bleeder’s disease. It is a sex linked recessive disease which shows transmission from female which is involved in clotting of blood.
In this particular disease a simple cut will result in continuous oozing of blood. The heterozygous female carrier will transmit thetdisease to her sons.
Female becoming haemophiliac is rare Queen victoria was carrier of the disease.

Sickle celled: This is a autosome-linked recessive trait transmitted from parents to the offspring when both partners are carrier for the gene or heterozygous. The disease is controlled by a single pair of alleles HbA and Hbs. Only homozygous individuals show the diseased phenotype. The defect is caused by the substitution of glutamic acid by valine at the sixth position of B globin chain of the haemoglobin molecule.
The mutant hemoglobin molecule undergoes polymerization under low O₂, tension causing change in the shape of RBC from biconcave disc to elongated sickle like structure.

Phenylketonuria: This error of metabolism is also due to autosomal recessive trait. The affected individual lacks an enzyme that convents amino acid phenylalanine to tyrosine. As a result phenyl alanine is accumulated and converted into phenyl pyruvic acid and other derivatives accumulation of these in brain results in mental retardation. They are excreted through urine because of poor absorption by kidney.

Chromosomal disorders: The chromosomal disorders are caused due to absence or excess or abnormal arrangement of one or more chromosomes.
Aneuploidy is the failure of segregation of chromatids during cell division cycle resulting in the gain or loss of a chromosome.

1. Down’s syndrome: This is caused due to the presence of an additional copy of chromosome no. 21 (Trisomy of 21). This disorder was described by Langdori down. The affected individual is short statured with small round head, furrowed tongue and partially open mouth, palm is broad with characteristic palm crease, physical, psychomotor, mental development is retarded.
2. Klinefelter’s syndrome: It is due to the presence of an additional copy of X chromosome resulting into karyotype of 47, XXY. Such an individual has overall masculine development (development of breast, i.e. Gynaecomastia is also expressed. Such individuals are sterile.
3. Turner’s Syndrome: This is caused due to absence of one of the X chromosome i.e 45 with XO, such females are sterile as ovaries are rudimentary besides other features including lack of secondary sexual charaters.

Question 16.
Which amino acid replaces glutamine at the 6th position of beta chain of haemoglobin?
Answer:
Valine.

Question 17.
Explain: (1) Down’s syndrome, (2) Klinefleter’s?
Answer:
1. Down’s syndrome: This is caused due to the presence of an additional copy of chromosome no. 21 (Trisomy
of 21). This disorder was described by Langdori down. The affected individual is short statured with small round head, furrowed tongue and partially open mouth, palm is broad with characteristic palm crease, physical, psychomotor, mental development is retarded.

2. Klinefelter’s syndrome: It is due to the presence of an additional copy of X chromosome resulting into karyotype of 47, XXY. Such an individual has overall masculine development (development of breast, i.e. Gynaecomastia is also expressed. Such individuals are sterile.

Question 18.
What is pedigree analysis?
Answer:

Question 19.
Write short note a linkage and recombination.
Answer:

• Physical association of genes on a chromosome is termed as linkage.
• The more closely the genes are found on chromosames greater is the linkage less closely the genes are found on chromosome lesser is the linkage.
• Original needed (O.S of Human Health and dieases strategies in improvement of crop).

Question 20.
Define the terms linkage and recombination.
Answer:
Physical association of genes on a chromosomes is termed as linkage.

The more closely the genes are found on chromosomes greater is the linkage less closely the genes are found on chromosome lesser is thf linkage.

Original needed (O.S of Human Health and dieases strategies in improvement of crop).
Recombination describes non parental gene combination.

→ Principles of Inheritance of Variation:

• Genetics is a branch of biology that deals with inheritance as well as variation of characters from parents to offspring.
• Gregor Mendel is also known as the father of genetics. He conducted hybridization experiments in garden pea plants (possum sativum).

He selected pea as his experimented plants because.

1. It is a true-breeding plant.
2. It is a self-pollinated crop plant.
3. It has several contrasting forms.
4. It has a short life span.

Pea plant exists in several contrasting states and they are as follows.

→ Some Important Terminologies in Genetics
(a) Gene – Gene is a basic unit of heredity.
(b) Allele – Genes that code for a pair of contrasting traits are known as alleles.
(c) Homozygous – When identical genes control a pair of characters it is termed as homozygous.
Ex: With respect to height of plant pure tall is represented as TT and pure dwarf TT.
(d) Heterozygous – When unidentical genes control a pair of contrasting characters it is heterozygous EX: Tt.
(e) Phenotype – The external morphological appearance of a character is termed as phenotype.
(f) Genotype – The genetic make up or genetic constitution of a particular individual is genotype.
(g) Test cross is a cross between Fj hybrid and its recessive parent.
(h) Pack cross is a cross between Fj hybrid with any one of its parents.
(i) Dominant – In a pair of dissimilar factor one dominates the other as in Fj it is called as dominant factor.
(j) Recessive – In a pair of dissimilar factor the factor which remains suppressed it termed as recessive.
(k) Co-dominant alleles,: When two different alleles are together both the alleles express their character side by side.

→ Monohybrid Cross:

• It is a cross made between two parents with respect to a single character.
• When a tall plant is crossed with a dwarf plant, all the F₁ hybrids are tall because tall is dominant.
• When the F₁ hybrids are self crossed in the F₂ generation there are 3/4th (75%) of tall plants and 1 /4th (25%) of the dwarf plants in the ration of 3:1 which is the phenotypic ratio.
• The genotypic ratio is 1:2:1 where 1 plant is pure or homozygous tall, 2 plants are heterozygous tall and 1 is pure dwarf. Parent Generation (P) Homozygous (Pure) × Homozygous (Pure)
• Tail Pea plant Dwarf pea plant

Image

→ Phenotypic Ratio:
3 Tall Plants: 1 Dwarf plant
Genotypic ration
1 Pure tall: 2 heterozygous tall: 1 pure dwarf.

Mendel Gave 2 laws.

1. Law of Dominance.
2. Law of Segregation.

→ Law of Dominance:

1. Characters are controlled by discrete units called factors.
2. Factors occur in pairs.
3. In a dissimilar pair of factors, one member of the pair dominates (dominant) the other (recessive).

Law of Segregation: This law is based on the fact that alleles do not show any blending both the characters are recovered as such in the F₂ generation.
The factors or alleles of a pair segregate from each other such that a gamete receives only one of the two factors.

Test Cross: It is the cross made between the F₁ hybrid, and its recessive parent to confirm whether the hybrid is homozygous or heterozygous.

When the test cross is conducted 50% of the plants are tall and other 50% of the plants are dwarf in the ration of 1:1 this confirms that F₁ hybrids are heterozygous.

Tall Dwarf
(Tt) 1:1 (tt)

→ Significance of Test Cross:
1. It helps to know whether the organism is homozygous or heterozygous for a trait.
Law of Independent Assortment: When two pairs of traits are combined in a hybrid segregation of one pair of characters is independent of other pair of characters.

→ Dihybrid Cross:
When a plant producing round and yellow colored seeds is crossed with a plant producing wrinkled and green seeds all the F, hybrids are round and yellow only. This indicates that alleles for yellow (Y) is dominant over green (y) and allele for round (R) is dominant over wrinkled.
When the hybrids are self ’crossed to obtain F₂ generation all four types of plants are obtained such as round yellow (9), and green (3), wrinkle and Yellow (3), Wrinkle and green (1) in the ratio of
9:3:3:1. Genotypic ratio is 1: 2: 2: 4: 1: 2: 1: 2: 1 (Phenotypic ratio) Parent at generation.

Phenotype Round and Yellow X Wrinkled and Green
Genotype RRYY rrYY
Gametes RY rY
RrYy F₁ Generation.
All off springs are found and yellow

RrYy X RrYy

Round and yellow – 9, Round and Green – 3 Wrinkled and yellow – 3 Wrinkled and Green – 1
Phenotypic ratio: 9:3:3:1
Genotypic ratio: – 1:2:2:4:1:2:1:2:1

→ Phenotypic Ratio:
ITT Pure or homozygous tall
2Tt Hybrid tall
1tt Homozygous dwarf
Phenotypic ration = 3:1
Genotypic ratio = 1:2:1

→ Deviations From Mendelian Principle Incomplete Dominance:
When the dominant gene is not completely dominant over the recessive gene it results in production of intermediate character or blended inheritance. Dog flower, or Antirrhinum is an example of incomplete dominance. When the red-colored flower were crossed with the white colored flower it resulted in production of pink colored flowers in the F₁ generation.

When F₁ plants were self-pollinated than in F₂ generation both phenotypic and genotypic ratios obtained are 1:2:1.
Parents RR homozygous red X rr – P,
Gametes (R) (P) (r) (r)
Rr = F1
Pink.

Phonotypic and genotypic ration =1:2:1
1- Pure Red, 2 – Pink, 1- Pure white.

→ Codominance:
When the alleles for 2 contrasting characters are brought together in heterozygous conditions both alleles express their character together and it is termed codominance.
Ex: ABO blood group in man

ABO blood group is controlled by gene 1. The plasma membrane of red blood cells has sugar polymers that protrude from its surface and the kind of sugar is controlled by the gene.

Gene I has 3 alleles I^A, I^B, and i. The alleles I^A and I^B produce a different type of sugar while it does not produce any sugar. When IA and IB are both present they express their own types of sugars because of codominance Hence red blood cells have both A and B types of sugars.

Multiple alleles – More than 2 i.e., the alleles govern the same character it is termed as Multiple alleles.

A single gene product may produce more than one effect. For example starch synthesized in pea seeds is controlled by one gene which has two alleles (B and b). Starch is synthesized effectively by homozygotes BB and therefore large starch grains are produced. On the other hand, bb homozygotes have lesser efficiency in starch synthesis and produce smaller starch grains. After maturation of the seeds BB seeds are round bb are wrinkled. Starch grains produced is of intermediate size in Bb seeds.

→ Chromosomal Theory of Inheritance:

1. This theory was proposed by Walter Sutton and Theodore Bovery in 1902.
2. They worked out the chromosome movement during meiosis.
3. Chromosome movement was parallel to the behavior of genes and he used chromosome movements to explain Mendel’s Law.
4. Sutton and Boveri argued that the pairing and separation of a pair of homologous chromosomes would lead to the segregation of a pair of factors’ they carried.

Thomas Hunt Morgan and his colleagues gave the experimental clarification of the chromosomal theory of inheritance. Morgan worked with tiny fruit flies drosophila melanogaster. He selected Drosophila for his experiments as:

1. They complete their life cycle in about two weeks.
2. A single mating could produce a large number of progeny flies.
3. Clear cut differentiation of sexes – male and female flies are easily distinguishable.
4. It shows lot of varations.

→ Linkage and Recombination:
Morgan carried out several dihybrid crosses in Drosophila to study sex linked genes. Morgan hybridized yellow bodied, white eyed females to brown bodied red eyed males and inter crossed their progeny he observed that the two genes did not segregate independently of each other and the F₂ ratio deviated significantly from the 9:3:3:1 ratio.

Morgan and his group knew that the genes were located on the X chromosome and they saw that when the two genes in the dihybrid cross are situated on the same chromosome the proportion of parental gene combinations were much higher than the non-parental type.

The term linkage was coined to describe the physical association of genes on a chromosome.

Recombination describes non-parental gene combinations. Morgan and his group found that when genes were grouped on the same chromosome some were tightly linked (i.e. they showed low recombination while others were loosely linked i.e. showed high recombination.

When genes of white and yellow were tightly linked they showed only 1.3% recombination while white and miniature showed 372% recombination.

→ Sex Determination:
Chromosomal mechanism of sex determination can be traced to some experiments carried out in insects. Henking (1891) found out a specific nuclear structure in spermato genesis in a fe insects.

50% of sperm received this structure after spermatogenesis where as the other 50% of sperm did not receive at. Henking gave the term ‘X’ body to it X body was later termed as X chromosome.

→ Grass Hopper XO Sex Determination:
Here males have only one X chromosome besides the auto somes where as females have a pair of X chromosomes. Xx – Xy type of sex determination is found in insects and man. Among males an X chromosome is but its counterpart is distinctly smaller and callec Y chromosome. Females have a pair of X chromosomes. B oth males and females bear same member of autosomes plusXX.

In human beings also XX – Xy type is found. Male heterogamety is observed where in some gametes have only X chromosome and some have Y chromosome.
XX XY Parents
Gametes (x) (x) (x) (y)
(xx) (xy)

→ Female off spring Male offspring:
In humans out of 23 pairs of chromosomes present 22 pairs are exactly same in both males and females and are autosomes.

A pair of X chromosmes are present in the female where as presence of X and Y chromosome are determinant of male characteristics.

During spermatogenesis among males, two types of gametes are produced 50% of the total sperm produced . carry X chromosome where as rest 50% has Y chromosome besides the autosomes.

Females however produce only one type of ovum with an X chromosome. In case the ovum fertilizes with a – sperm carrying X chromosome the zygote develops into a female (XX) and the fertrilization of ovum with Y chromosome carrying sperm results in male offspring.

Thus it is evident that it is the genetic make up of sperm that determines the sex of the child.

It is thus unfortunate in our society women are blamed for giving birth to female child.

Mutation: It is a phenomenon which results in alteration of DNA sequences and results in changes in the genotype and phenotype of the organism. Mutation also arises due to change in a single base pair of DNA. This is known as point mutation. A classic example of such mutation is sickle celled anaemia.

Deletions and insertion of base pairs of DNA causes frame shift mutation.
Sex determination in Birds.
Here it is ZZ-ZW type.
Here females have two sex chromosome one is of Z type and another is of W type males have a pair of Z chromosomes besides autosomes.

→ Sex Determination in Honeybee:

1. The female is diploid having 32 chromosomes.
2. Male is haploid having 16 chromosomes.
3. An offspring formed from fertilization of a sperm and egg develops into either queen or worker which are fe-male.
4. An infertilized egg develops into a male (drone) i.e. by means of parthenogenesis.
5. This method of sex determination is called haplodiploid sex determination.
6. Male produce sperms by mitosis and gemale produce egg by meioses.
7. Hence male honey bee donot have father and thus cannot have sons but have grandson.

Parents Female × Male
32 × 16
Meiosis Mitosis
16 16 16
16 32
Male Female.

2nd PUC Biology Important Questions with Answers

Karnataka 2nd PUC Biology Important Questions Chapter 4 Reproductive Health

Question 1.
Overpopulation India is the first country which initiates plans and programs at national level some of them are-
Answer:
(a) Family planning program – 1951.
(b) Reproduction and child health care program -1997(RCH).

Question 2.
Define Reproductive health according to WHO?
Answer:
According to W.H.O reproductive health means total well-being in all aspects of reproduction physical,
emotional, behavioural and social.

Question 3.
When was family planning programmes initiated at the national level in India?
Answer:
Family planning programmes were initiated in India in the year 1951.

Question 4.
Expand R.C.H?
Answer:
Reproductive and child health care programme.

Question 5.
What is R.C.H? Mention its goals?
Answer:
R.C.H is reproductive and child health care programme and its goals are-

1. 1 To create awareness among people about various reproductive related aspects.
2. To provides facilities and support for building a reproductively healthy society.

Question 6.
Name male and female sterilization techniques in humans?
Answer:
Male sterilization technique is vasectomy and female sterilization technique is tubectomy.

Question 7.
Name anyone intrauterine device?
Answer:
CuT (Copper T)

Question 8.
Expands M.M.R?
Answer:
Maternal mortality rate.

Question 9.
Give the full form of I.M.R?
Answer:
Infant Mortality Rate.

Question 10.
What is intrauterine devices?
Answer:
These devices are inserted by doctors or expert nurses in the uterus through vagina.

Question 11.
Name any three natural methods of contraception?
Answer:
The three natural methods of contraception are-

1. Periodic abstinence.
2. Coitus interruptus or withdrawal method.
3. Lactational amenorrhea

Question 12.
What is oral contraception? Name oral contraception.
Answer:
Oral administration of small doses of progestagens- estrogen combination is a contraceptive method used by females. They are used in the form of tablets called pills Saheli is an oral contraceptive.

Medical Termination of pregnancy MTP.

Intentional or voluntary termination of pregnancy before full term is called medical termination of pregnancy (MTP) or induced abortion.

1. Nearly 45-50 million MTPs are performed in a year all over world.
2. MTP can decrease the population to some extent.
3. In many countries, the debate was whether to accept/legalize MTP or not.

Question 13.
Why MTP?
Answer:

1. To get rid of unwanted pregnancies either due to casual unprotected intercourse or failure of the contraceptive used during coitus or rapes.
2. Useful in continued pregnancies which could be harmful or even fatal to mother or foetus or both.

Question 14.
When are MTPs safe?
Answer:
They are safe during the first trimester upto 12 weeks of pregnancy, second trimester abortions are much riskier.

Question 15.
Write any two preventive measures to control STDs.
Answer:

1. Avoid sex with unknown persons/multiple partners.
2. Always use condoms during intercourse.
3. Avoid sex with unknown persons/multiple partners.
4. Always use condoms during intercourse.
5. In case of doubt one must consult a qualified doctor without delay.

Question 16.
Define MTP?
Answer:
Intentional or voluntary termination of pregnancy before full term is called medical termination of pregnancy (MTP) or induced abortion.

Question 17.
What is the safe period of MTP?
Answer:
It is considered safe during the first trimester (Up to 12 weeks of pregnancy)

Question 18.
List any four symptoms of sexual transmitted diseases?
Answer:
Itching, fluid discharge, slight pain, swelling in the genital region.

Question 19.
What are sexually transmitted diseases?
Answer:
Disease or infections which are transmitted through sexual intercourse with an infected person are called STDs or venereal disease. Ex. Gonorrhoea, Syphilis, Genital herpes, AIDS.

Infertility – A large number of couples over the world including India are infertile i e they are unable to produce children in spite of unprotected sexual cohabitation.

The reasons could be physical, congenital diseases, drugs, immunological or psychological. Hence couple. , could be assisted to have their own child through special techniques termed as Assisted Reproductive Technologies (ART)
1. In – vitro Fertilization and Embryo Transfer (I.V.F-E.T) or test tube baby. In vitro fertilization takes place outside the body followed by embryo transfer. This method popularly termed as a Test tube baby programme involved a procedure in which ova from the wife and sperms from husband are collected and induced to form a zygote under simulated conditions in the laboratory.

2. ZIFT – Zygote intrafallopian transfer Zygote or early embryos (with upto 8 blastomeres) is then transferred into fallopian tube and embryos with more than 8 blastomeres into the uterus (Intrauterine transfer) or IUT.

3. (GIFT- Transfer of an ovum collected from a donor into the fallopian tube of another female who cannot produce one but can provide suitable environment for fertilisation is GIFT.

4. ICSI – Intracytoplasmic sperm injection is another procedure to form an embryo in the laboratory m which a sperm is directly injected into as ovum), (March 2016)

5. AI (Artificial insemination) is a technique in which the semen is collected either from husband Or a healthy donor and is artificially introduced either into the vagina or uterus of the female. All these techniques require extremely high precision handling by special professional

Question 20.
What is infertility?
Answer:
Inability to produce a child or inability to conceive (pregnancy) is called infertility.

Question 21.
Expand the following terms?
Answer:

• GIFT-Gamete Intrafallopian transfer
• ZIFT- Zygote Intrafallopian transfer
• IVF-ET-Invitro fertilization-embryo transfer technique
• ICSI- Intra Cytoplasmic sperm injection

Question 22.
What is population explosion? Mention the causes, effects, and control of overpopulation?
Answer:
Population explosion is the sudden increase in population.
Causes:

1. The decline in death rate.
2. The decline in Maternal Mortality Rate (MMR) and Infant Mortality Rate (IMR).
3. Increase in the number of people in reproducible age.

Effects:

1. Absolute scarcity can occur of the basic requirements i.e., food, shelter and clothing.
2. 2 It might also lead to poor conditions of living.

Control of Population:

1. The motivation of smaller families by using several contraceptive methods.
2. To have two children. Every couple must follow the slogan Hum Do Hamare Do.
3. The marriageable age should be raised and in female it should be 18years and in case of males it should be 21 years.
4. Incentives should be given to couple with small families.

→ Definition:
W.H.O (World Health Organisation) defines reproductive health as a means of total well-being in all aspects o Reproduction i.e. physical, emotional, social and behavioral
Reproductive Health – Problems and strategies various problem and strategies of reproductive health are

• Overpopulation
• Sex education
• Knowledge about birth control methods and care of mother and child
• Awareness about social evils.

→ Sex Education:

1. The introduction of sex education in schools and colleges is a step to disseminate right information to adolescents about reproductive organs, secondary sexual characters.
2. Adolescence and related changes safe and hygienic sexual practices, sexually transmitted diseases (STDs) etc.
3. This knowledge will save the young people from myths and misconceptions about sex related aspects and help them to lead a reproductively healthy life later.
4. Awareness About Social Evils: Creating awareness about consequences of uncontrolled population growth and social evils like sex abuse and sex related crimes is another important aspects of RCH programme.

→ Implementation of Reproductive Health:

1. Successful implementation of various action plans for maintaining reproductive health requires strong infrastructural facilities, professional expertise and material support.
2. Medical assistance and care should be provided to people in reproductive age.
3. Related problems, pregnancy, delivery, STDs, abortions, Contraception, menstrual problems, infertility etc.
4. Statutory ban should be there on amniocentesis for sex determination to legally check increasing female foeticide.
5. Research on various reproduction-related areas are encouraged and supported by Governmental and non- Governmental agencies to find new methods.
6. Better awareness about sex-related matters increased no. of medically assisted deliveries, better post natal care lead to decreased maternal and infant mortality rate. Increased no. of couples with small families’ better detection and care of STDs and overall increase medical facilities for all sex-related problems.

→ Population Explosion And Birth Control:

1. Population increased significantly due to increased health condition. World population which was around 2 billion in 1900 increased to 6 Billon by 2000.
2. A rapid decline in death rate maternal mortality rate (MMR) and infant mortality rate (IMR) and an increase in number of people in reproducible age are reasons for this.
3. Most important step to overcome this problem is to motivate smaller families by using contraceptive methods.
4. Advertisements in the media, as well as posters, show a happy couple with a slogan Hum do Hamaare do.
5. Rise of marriageable age of female to 18years and males to 21 years and incentives given to couples with small families are two of other measures taken to solve this problem.

→ Family Planning:
Planning for a small family by adopting scientific method of birth control is called family planning Contraceptive methods can be broadly divided into

1. Temporary method
2. Permanent method

Natural methods are of 3 types

1. Periodic abstinence.
2. Withdrawal or coitus interruptus.
3. Lactational amenorrhoea.

1. Periodic abstinence is a method in which couples avoid or abstain from coitus from 10-17* of the menstrual cycle when ovulation is expected.

2. Withdrawal or coitus interruptus is another method in which the partner withdraws his penis from vagina just before ejaculation to avoid insemination.

3. Lactational Amenorrhea :
1. It is an absence of menstruation and the fact that menstrual cycle does not occur during the period of intense lactation following parturition. As long as mother breastfeeds the child the chances of conception are almost nil.

2. In barrier methods ovum and sperms are prevented from physically meeting with the help of barrier methods available for both males and females. Condoms are the barriers made of thin latex sheath and used to cover the penis in male, vagina and cervix in females just before coitus so that the ejaculated semen would not enter into reproductive tract.

3. Condoms are also used to protect the user from contracting STD’s and AIDS. Diaphragms cervical caps and vaults are also barriers made of rubber capsand insertedinto female reproductive tract to cover the cervix during coitus. They block entry of sperms through cervix.

4. I.U.D are intra uterine devices inserted in the uterus through the vagina. I.U.D’s are available as non-medicated IUDs (Lippe’s loop) Cu releasing IUDs (Cu T, Cu 7, Multiload 375), hormone-releasing IUDs (Progestasert, LNG-20). IUDs increase phagocytosis of sperms within the uterus and Cu ions suppress sperm motility. Hormone releasing IUD’s make the uterus unsuitable for implantation and cervix hostile to the sperms.).

5. Oral Contraception – It includes administration of small doses of progestogens or progestogen/estrogen combinations and it is a contraceptive method used by females. They are used in the form of tablets called pills. Pills have to be taken for a period of 21 days starting from within first five days of menstrual cycle. After 7days during which menstruation occurs the pattern has to be repeated till female’s desires to prevent conception. Sahctius is a famous oral pill.

6. Progesterone alone or in combination with estrogen can be used by females as injection or implants under the skin.

→ Permanent Methods/Surgical Methods:
(Surgical methods also called sterilisation are terminal methods to prevent pregnancies. Sterilisation procedure in male is termed as vasectomy and in females it is tubectomy. In vasectomy a small part of the vas deferens is removed or tied up through a small incision on the scrotum. In tubectomy a small part of the fallopian tube is removed or tied up through a small incision in the abdomen or through the vagina.)

→ Sexually Transmitted Disease (Stds):
Diseases or infections which are transmitted through sexual intercourse with an infected person are collectively called STDs or venereal disease or reproductive tract Infections (RTI).

Some of the STDs are
(a) Gonorrhoea
(b) Syphilis
(c) Genital herpes
(d) Chlamydiosis
(e) Genital warts
(f) Trichomoniasis
(g) Hepatitis
(h) AIDS.

→ Mode Of Transmissions:

1. Sexual contact with infected persons.
2. Sharing of infected needles, surgical instruments with infected persons.
3. Transfusion of contaminated blood.
4. STDs can be transmitted from an infected mother to the foetus.

→ Common Symptoms of Stds:

1. Itching, fluid discharge, slight pain, swelling etc, in the genital region
2. Infected females may be asymptomatic for a long time.
3. Social stigma attached to STDs deters the infected persons from going for proper treatment.
4. Complications that can occur later includes pelvic inflammatory disease, abortion, ectopic pregnancy, infertility, cancer of reproductive tract.

2nd PUC Biology Important Questions with Answers

Karnataka 2nd PUC Biology Important Questions Chapter 3 Human Reproduction

Question 1.
What are viviparous organisms?
Answer:
Organisms in which the young one develops within the body of female are termed as viviparous organisms.

Question 2.
List out the reproductive events in humans. Any four?
Answer:
It includes the following.

1. Gametogenesis
2. Insemination
3. Fertilization
4. Implantation
5. Gestation or Pregnancy
6. Parturition.

Question 3.
Define gametogenesis.
Answer:
Formation of haploid male gametes called sperm and female gamete called ovum is termed as gametogenesis.

Question 4.
What is implantation?
Answer:
Attachment of embryo to the endometrial wall of uterus is termed as implantation

Question 5.
Define Parturition?
Answer:
Delivery of thr baby is termed as parturition.

Question 6.
Draw a neat labelled diagram male reproductive system in humans.
Answer:

Question 7.
Describe the male reproductive system?
Answer:
It is located in the pelvic region and consists of

1. A pair of testes
2. Accessory ducts
3. Glands
4. External genitalia.

1. The testes are situated outside the abdominal cavity within a pouch termed as scrotum. Scrotum helps in maintaining low temperature of the testes (2-2.5°c lower than normal internal body temperature. The testis is oval in shape, contains 250 compartments called testicular lobules. Each lobule contains 1-3 highly coiled seminiferous tubule in which sperms are produced. Each seminiferous tubule is lined on inside by 2 types of cells male germ cells, spermatogonia, sertoli cells.

Male germ cells undergo meiotic division to form sperms where as sertoli cells provide nourishment to germ cells.
Region outside the seminiferous tubule is termed as interstitial spaces which contains blood vessels, interstitial cells or leydig cells.

Leydig cells synthesise testicular hormones termed as androgens.

2. Male sex accessory ducts include rete testis, vasa efferentia, epididymis and vas deferens. Seminiferous tubules open into rete testis which leads to vasa efferentia. Vasa efferentia opens into epididymis which leads to was deferens which loops over urinary bladder. Vas deferens receives a duct from seminal vesicle and opens into urethra as ejaculatory duct.
The ducts store and transport sperms from testis to outside through urethra opens up into urethral meatus.

3. Male accessory glands include paired seminal vesicles, a prostrate gland and a paired bulbo urethral glands secretions of which comprise seminal plasma which is rich in fructose, calcium and certain enzymes.

4. External genitalia is penis which is made up special tissue which helps in erection of penis to facilitate insemination. Enlarged end of penis is glans penis opened by foreskin.

Question 8.
Which cells of testes secrete androgens?
Answer:
Leydig cells of testes secrete androgens.

Question 9.
What is the function of sertoli cell?
Answer:
The function of sertoli cells is to provide nourishment to germ cells during spermatogenesis.

Question 10.
Give the function of scrotum?
Answer:
It helps in maintaining low temperature of testes (2-2.5°c lower than normal internal body temperature).

Question 11.
What is the function of bulbourethral glands?
Answer:
The secretions of the bulbourethral glands help in the lubrication of the penis.

Question 12.
Draw a neat labeled diagram of sectional view of the female reproductive system?
Answer:

Question 13.
Describe female reproductive system in detail?
Answer:
Female reproductive system consists of a pair of ovaries, a pair of oviducts, uterus cervix, vagina, external genitalia in pelvic region.

Ovaries are the primary female sex organ which produces the female gamete or ovum and several steroid hormones. Each ovary is 2 – 4cm in length and connected to the pelvic wall and uterus by ligaments.

Oviducts (fallopian tubes), uterus, vagina constitute the female accessory ducts. Each fallopian tube is 10-12 cm long and it extends from periphery of each ovary to uterus.

The part closer to the ovary is infundibulum which contains finger like projections termed as fimbriae which helps in collection of ovum after ovulation. Infundibulum leads to a wider ampulla which leads to a narrow Isthmus which is the last part of oviducts and it joins the uterus. Uterus is single and also called womb shape is like inverted pear. Uterus opens into vagina through narrow cervix. Cavity of cervix along with vagina forms birth canal. Uterus has 3 layers outer thin perimetrium, middle muscular myometrium and inner glandular layer endometrium.

Female external genitalia include mons, pubis, labia majora, labia minora, hymen, clitoris.
Some important one mark questions from this topic:

Question 14.
Name the finger like projections present in the ovary at the end of infundibulum?
Answer:
Fimbriae.

Question 15.
What is the other name for oviduct?
Answer:
Fallopian tube.

Question 16.
What are the 3 parts of fallopian tube?
Answer:
Infundibulum, Ampulla, Isthmus.

Question 17.
Give the other name for uterus?
Answer:
It is termed as womb.

Question 18.
What forms birth canal?
Answer:
Cervical canal along with vagina forms birth canal.

Question 19.
Mention any two hormones secreted by ovary.
Answer:
Progesterone, Estrogen.

Question 20.
Draw a sectional view of the mammary gland?
Answer:

Mammary Gland — Diagrammatic Sectional View

Question 21.
Write a note on mammary gland?
Answer:

1. It is characteristic of all female mammals.
2. They are paired structures that contain glandular tissue and variable amount of fat.
3. Glandular tissue of each breast is divided into 10-20 mammary lobes which contain alveoli which secrete milk
4. Mammary lobe join to form mammary duct. Mammary duct is connected to laticiferous duct through mammary ampulla.
5. Through laticiferous duct milk is sucked out.

Question 22.
What are the 2 types of gametogenesis?
Answer:
They are:

1. Spermatogenesis
2. Oogenesis

Question 23.
What is Spermato genesis?
Answer:
The process of production of sperms in the male gonad or testes is spermatogenesis.

Question 24.
Define oogenesis?
Answer:
The process of formation of mature female gamete or ovum in the female gonad or ovary is oogenesis.

Question 25.
Define spermiogenesis?
Answer:
Spermatids are transformed into spermatozoa (sperms) by the process called spermiogenesis.

Question 26.
What is Spermiation?
Answer:
Sperm heads become embedded in sertoli cells and are released from seminiferous tubules by the process of Spermiation.

Question 27.
What is semen?
Answer:
Seminal plasma along with sperms constitute semen.

Question 28.
Give a detailed account of spermatogenesis?
Answer:
The spermatogonia which is found on inner wall of seminiferans tubule multiply by mitotic division and contain 46 chromosomes. It enlarges to form primary spermatocyte which undergoes Ist Meiosis to form 2 secondary spermatocytes which are equal in size and contains 23 chromosomes.

The secondary spermatocytes undergo the second meiotic division to form four equal haploid spermatids. Spermatids are transformed into sperms by spermiogenesis. Sperms are released from seminiferous tubules by the process called spermiation.

Question 29.
Give an account of hormonal control of spermatogenesis?
Answer:

1. Spermatogenesis starts at the age of puberty due to significant increase in the secretion of gonadotropin-releasing hormone.
2. The increased level of GnRH acts at the anterior pituitary gland and stimulates secretions of two gonadotrophins – Luteinising hormone and Follicle-stimulating hormone.
3. L-H acts at Leydig cells and stimulates synthesis and secretion of androgens. Androgens stimulate spermato genesis.
4. F.S.H acts on sertoli cells and stimulates the secretion of some factors which stimulate the process of spermiogenesis.

Question 30.
Draw a neat labeled diagram of human sperm.
Answer:
Any four labels:

Question 31.
Describe the structure of human sperm
Answer:
It is a microscopic structure and it is composed of a head, neck a middle piece and tail. Plasma membrane envelops whole body of sperm. Sperm head consists of a haploid nucleus and it is covered by cap like structure termed acrosome. Acrosome is filled with enzymes which help in the fertilisation of ovum. Middle piece possess mitochondria which produce energy for movement of tail. Human male ejaculates about 200 to 300 million sperms during coitus. Secretions of epididymis, vas deferens, seminal vesicle and prostrate are essential for maturation and motility of sperms.

Question 32.
Describe the process of Oogenesis with a schematic representation.
Answer:
The process of formation of mature female gamete is termed as oogenesis, Oogonia are formed within each foetal ovary no more oogonia are formed and added after birth. These cells enter prophase I of meiosis I and get arrested at primary oocyte stage, primary oocyte gets surrounded by layer of granulosa cells and is called primary follicle. The primary follicle gets surrounded by more layer of granulosa cells and forms secondary follicle. Secondary follicle gets converted to tertiary follicle which has a fluid filled cavity called antrum.

Theca layer is organised into outer theca externa, inner theca interna. Primary oocytae within the tertiary follicle grows in size and completes its first meiotic division to form a large haploid secondary oocyte and a tiny first polar body. Secondary oocyte retains bulk of nutrient-rich cytoplasm. Tertiary follicle forms a mature Grafian follicle. Secondary oocyte forms a new membrane zona pellucida Grafian follicle rypturcs to release secondary oocyte from ovary – process is ovulation.

The reproductive cycle in the female primates (Eg. monkeys, apes and human beings is termed as menstrual cycle.

Question 33.
Define menarche?
Answer:
First menstruation begins at puberty and is termed as menarche.

Question 34.
Define ovulation?
Answer:
Release of ovum/secondary oocyte from the Graafian follicle is termed as ovulation.

Question 35.
What is the function of corpus luteum?
Answer:
It secretes large amount of progesterone which is essential for maintenance of endometrium.

Question 36.
Ovulation takes place on the 14th day of menstrual cycle why?
Answer:
Graafian follicle matures and ruptures usually on the 14th day due to rapid secretion of L H or L.H surge.

Question 37.
Define menopause?
Answer:
In human beings menstrual cycle ceases around 50 years of age and it is termed as menopause.

Question 38.
Name the hormones secreted by Leydig cells and corpus luteum.
Answer:
Progesterone is the hormone secreted by corpus luteum. Androgen hormone is secreted by Leydig cells.

Question 39.
Name the four major events of menstrual cycle
Answer:
The four events of menstrual cycle are:

1. Menstrual phase
2. Follicular phase

Question 40.
What is menstrual cycle. Explain each events?
Answer:
The cycle of events starting from one menstruation till the next one is called as the menstrual cycle. Cycle starts with menstrual phase, where menstrual flow occurs and it lasts for 3-5 days. Menstrual flow is due to the breakdown of endometrial lining of uterus and its blood vessels which forms liquid that comes out through vagina. Menstruation occurs only if the released ovum is not fertilised. Lack of menstruation indicates pregnancy but stress, poor health may also lead to it.

The second phase is follicular phase during which the primary follicles in the ovary grow to become fully mature Grafian follicle and the endometrial lining of uterus regenerates. These changes in ovary, uterus are accompanied by changes in pituitary and ovarian hormones. L.H and F.S.H hormone increase in levels during the follicular phase and help in follicular development.

During ovulatory phase which occurs at the mid-cycle there is a surge in L.H. hormone secretion which induces rupture of Graafian follicle and the release of ovum (Ovulation).

After ovulatory phase is the luteal phase where the remaining parts of Graafian follicle transforms as corpus luteum. Corpus luteum secretes large amount of progesterone which is essential for maintenance of endometrium.

Endometrium is essential for implantation of fertilised ovum and other events of pregnancy.

During pregnancy menstrual cycle stops.

In the absence of fertilisation corpus luteum degenerates, endometrium disintegrates leading to menstruation. Thus there is marking of a new cycle.

During copulation semen is released by penis into vagina of the female. This process is termed as insemination. Motile sperms swim pass through cervix, then enter uterus and finally reach the ampullary isthmus junction of fallopian tube. Ovum released by ovary also reaches the ampullary isthmic junction and this is the place where fertilisation takes place. Process of fusion of sperm with an ovum is called fertilisation. During fertilisation, a sperm comes in contact with the zone Pellucida layer of the ovum. A change in membrane occurs and entry of more sperms is blocked. Thus it ensures only 1 sperm can fertilise an ovum. The acrosome of sperm secretes certain enzymes which helps the sperm to reach the cytoplasm of the ovum. The meiotic division of secondary oocyte is completed and there is formation of a haploid ovum and a second polar body.

Sperm and ovum fuse to form a diploid zygote.

Sex determination of a child:

1. The sex of the child is determined at the stage of fertilization.
2. Chromosome pattern in human female is XX and in the human male it is XY.
3. When the sperm carrying the X chromosome fuses with the egg carrying an X chromosome, the result is a female child with XX chromosome.
4. When sperm carrying the Y chromosome fuses with an ovum carrying the X chromosome the result is a male child with XX chromosome.
5. Thus human males are heterogametic with XY chromosome and human females are homogametic with XX chromosome.
6. The pattern of sex determination is XY-XY type.

Question 41.
What is the process of sex determination in humans termed as?
Answer:

1. It is termed as XX-XY type of sex determination.
2. When the sperm carrying the X chromosome fuses with an ovum carrying X chromosome the result is a female child with XX chromosome.
3. When the sperm carrying the Y chromosome fuses with an ovum carrying the X chromosome the result is a male child with an XY chromosome.

The mitotic division starts as the zygote moves through the isthmus of the oviduct called cleavage towards the uterus and forms 2, 4, 8, 16 daughter cells called blastomeres.

The chorionic villi and uterine tissue become interdigitated with each other and jointly form a structural and functional unit between the developing embryo (foetus) and maternal body called the placenta.

The average duration of human pregnancy is about 9 months which is called the gestation period. Vigorous contraction of the uterus at the end of pregnancy causes expulsion/delivery of the foetus. This process of delivery of the foetus (childbirth) is called parturition.

Question 42.
Define cleavage.
Answer:
The zygote undergoes repeated mitotic division at the ampullary-isthmus junction of the oviduct and it is termed as cleavage.

Question 43.
Name the two layers of a blastocyst?
Answer:
The two layers of the blastocyst are

1. outer trophoblast
2. Inner cell mass.

Question 44.
What is implantation?
Answer:
Blastocyst becomes embedded to the endometrium of uterus termed as implantation.

Question 45.
What is morula?
Answer:
Embryo with 8-16 blastomere is called morula.

Question 46.
Define placenta?
Answer:
The placenta is an organic connection between maternal tissue and foetus through which physilogical exchange takes place.

Question 47.
Write four hormones secreted by the placenta?
Answer:
The four hormones secreted by placenta are

1. Human chorionic Gonadotrophin (LCG).
2. Human placental lactogen (LPL).
3. Estrogen.
4. Progestogen.

Question 48.
What are the functions of placenta 2?
Answer:

1. It facilitates the supply of O₂ and nutrients to the embryo and helps in the removal of CO₂ and excretory material produced by the embryo.
2. The placenta is connected to the embryo through an umbilical cord which helps in the transport of substances to and from the embryo.
3. Placenta also acts as an endocrine tissue and produces several hormones like human chorionic gonadotropin (hCG), human placental lactogen(hPL), estrogen, progestogen.

Question 49.
Define parturition and Lactation?
Answer:
Vigorous contraction of uterus’ at the end of pregnancy causes expulsion/delivery of the foetus the process of delivery of the foetus is called childbirth.

Question 50.
Which hormone hastens childbirth 2?
Answer:
Oxytocin.

Question 51.
Define lactation?
Answer:
The mammary glands of females undergo differentiation during pregnancy and start producing milk towards the end of pregnancy termed lactation,

Question 52.
What is colostrum 2?
Answer:
Milk produced during the initial few days of lactation is termed as colostrum which contains several antibodies essential to develop resistance in newborns.

1. The zygote undergoes repeated mitotic division at the ampullary isthmic junction of oviduct and it is cleavage.
2. Zygote forms 2,4,8,16 daughter cells called blastomeres. Embryo with 8-16 blastomere is-called morula.
   Morula continues to divide and forms blastocyst.
3. The blastocyst has 2 layers outer layer called trophoblast and inner cell mass. Trophoblast gets attached to the endometrium of the uterus.
4. Inner cell mass forms embryo.
5. Blastocyst beocmes embedded to the endometrium of uterus termed as implantation.

Pregnancy:

1. After implantation finger-like projections appear on the trophoblast called chorionic Villi which is surrounded by uterine tissue and maternal blood.
2. Chorionic Villi and uterine tissue become linked with one another and jointly form a structural and functional unit between the developing embryo and maternal body called the placenta.
3. The placenta is thus an organic connection between maternal tissue and fetus through which physiological exchange takes place.

Question 53.
Name the hormones secreted by Leydig cells and corpus luteum.
Answer:
Progesterone is the hormone secreted by the corpus luteum. Androgen hormone is secreted by Leydig cells.

1. It facilitates the supply of oxygen and nutrients to the embryo and helps in the removal of CO₂ and excretory material produced by the embryo.
2. The placenta is connected to the embryo through an umbilical cord which helps in the transport of substances to and from the embryo.
3. Placenta also acts as an endocrine tissue and produces several hormones like human chorionic gonadotropin (hCG), human placental lactogen (hPL), estrogen, progestogen.
4. Relaxin is a hormone secreted in the later stage of pregnancy and supports fetal growth.

Question 54.
Define Cleavage?
Answer:
The zygote undergoes repeated mitotic division when it is at the isthmus of the oviduct and it is termed as cleavage.

Question 55.
Define Placenta?
Answer:
The placenta is an organic connection between maternal tissue and fetus through which physiological exchange takes place.

Question 56.
What is Colostrum?
Answer:
Milk produced during the initial few days of lactation is colostrum which contains several antibodies necessary to develop resistance against diseases.

Question 57.
Write four hormones secreted by the placenta?
Answer:
The four hormones secreted by the placenta are:

1. Human Chorionic Gonadotropin (hCG)
2. Human Placental lactogen (hPL)
3. Estrogen
4. Progestogen

Question 58.
Define fertilisation.
Answer:
The process of fusion of a sperm with an ovum is called fertilisation.

Question 59.
Name the hormone secreted by the Corpus luteum and Leydig cells.
Answer:
Progesterone is the hormone secreted by corpus lutcum and Androgen hormone is seended by and Leydigeells

Question 60.
Name the hormones secreted by Leydig cells and corpus luteum
Answer:
Progesterone is the hormone secreted by corpus luteum. Androgen hormone is secreted by Leydig cells.

Question 61.
Refine foetal ejection reflex.
Answer:
The signal of parturition originates from the fully developed fetus and placenta which induce mild uterine contractions called fetal ejection reflex.

Question 62.
What is parturition?
Answer:
Rigorous contraction of the uterus at the end of pregnancy leading to delivery of the fetus is called parturition.

Question 63.
What are stem cells.
Answer:
Inner cell mass contains cells called stein cells which have the potency to give rise to all tissues and organs.

→ Embryonic Development:
Immediately after implantation the inner cell mass differentiates into outer layer called ectoderm and inner layer called endoderm. Mesoderm appears in between outer ectodeim and inner endoderm and all these three layers gives rise to all tissues in adults.

1. Human pregnancy lasts 9 months. After the first month of pregnancy the embryo heart is formed.
2. By second month of pregnancy, foetus develops limbs and digits.
3. By 12 weeks (the first trimester) limbs and external genital organs are developed
4. Head develops during fifth month.
5. By second trimester body is covered with fine hair, eyelids separate, eyelashes are formed.
6. By end of nine months foetus is fully developed and is ready for delivery.

→ Parturition and Lactation:

1. The average duration of human pregnancy is 9 months and it is called gestation period.
2. Vigorous contraction of uterus at the end of pregnancy causes expulsion/delivery of foetus. The delivery of foetus is Parturition.

(March 2015) Signal of parturition originate from fully developed foetus and placenta induces mild uterine contractions termed as foetal ejection reflex and triggers the release of oxytocin from maternal pituitary oxytoin causes rapid uterine contractions and is preferred for parturtean.
The mammary glands of the female undergo differentiation during pregnancy and produce milk towards the end of pregnancy termed lactation. Milk produced during initial few days of lactation is termed as colostrum which contains several antibodies essential to develop resistance for new bom babies.

2nd PUC Biology Important Questions with Answers

Karnataka 2nd PUC Biology Important Questions Chapter 2 Sexual Reproduction in Flowering Plants

Question 1.
Draw a neat labelled diagram of longitudinal section of a typical flower?
Answer:

Question 2.
Explain the T. S. of mature anther with neat labelled diagram?
Mention the function of each? (a) Endothecium (b) Middle Layer (c)Tapetum.
Answer:
In transverse section a typical microsporangium is circular in outline, which is surrounded by four wall layers.
(a) Epidermis: Single layer of parenchyma, protective in function. It becomes stretched and shrivel off at
maturity.
(b) Endothecium: Single layered cells, radially elongated, with fibrous thickenings. They remain thin walled in the line of dehiscence (stomium) in between two microsporangia. They help in hygroscopic movement which help in dehiscence.
(c) Middle layers: Three layers of cells generally disintegrate in mature anthers and provide nourishment to sporogenous cells.
(d) Tapetum: It is a layer of nutritive tissue present around microspore mother cell. Each cell has dense cytoplasm, multinucleated due to endomitosis [because karyokinesis does not followed by cytokinesis]
(e) Sporogenous tissues ace a group of compactly arranged homogenous cells, present in the centre of each microsporangium,which diffrentiates into spores.

Question 3.
Define microsporogenesis?
Answer:
It is the process of “formation of microspores from microspore mother cell through meiosis”.

Question 4.
Explain the process of microspore formation?
Answer:

1. The cells of sporogenous tissue become round and loosely arranged cells known as pollen mother cell or microspore mother cell.
2. These cells undergo meiosis to form microspores, which are arranged in cluster of four cells, so known as microspore tetrads or pollen tetrads.
3. As the anthers mature and dehydrate, pollen tetrads, separate from one another and develop into “pollen grains”. They are released with dehiscence of anther.

Question 5.
Explain the structure of pollen grain? or Briefly explain the structure of pollen grain.
Answer:

Structure of Mature Pollen Grain

Pollen grain (Microspore): It is a haploid, uninucleate minute spores produced in large numbers as a result of meiosis in microspore mother cell inside the microsporangia.

Structrure: Pollen grains are generally spherical shaped measuring 25 to 50 pm in diameter. The outer layer is made up of two layered walls, outer exine and inner intine.

Exine: It is a hard protective outer layer made up of a highly resistant organic material sporopollenin, which preserve pollen grains as fossils.

It bears spines, ridges and other types of surface outgrowths which are of systemic value. Exine has prominent apertures called germ pores [where exine is thin, the sporopollenin is absent] through which pollen tube emerges out after pollination.

Intine: It is a inner layer made up of cellulose and pectin and has plates of enzymatic proteins in the vicinity of germpores.

Cytoplasm: It is surrounded by plasma membrane and uninucleate (nucleus) which divides mitotically. Matured pollen grain contains two cells, bigger vegetative cell and smaller generative cell.

Vegetative cell or Tube cell: It is larger, remains at the centre and have abundant reserve food and a large irregularly shaped nucleus.

Generative cell: Small and floats in the cytoplasm of vegetative cell. Spindle shape with dense cytoplasm and a. nucleus.

The pollen grains are shed at this two celled stage in 60% of angiosperms. In others 3-celled stage, where generative cell divide by mitosis and give rise to two male gametes.

Question 6.
Draw labelled diagram of pollen grain?
Answer:

Structure of Mature Pollen Grain

Question 7.
Write importance of pollen grains?
Answer:
Important in life cycle of angiosperms.

• Highly nutritious (certain plants) used as supplements (in the form of tablets and syrups)
• Increases the performance of athletes and racehorses.
• Gauses allergic diseases and respiratory disorders in some people. (Hay fever)

Question 8.
Write a note on viability of pollen grains?
Answer:
Pollen Viability is defined as the ability to live, germinate when conditions are favourable on the stigmatic surface. It differs from species to species, which depends on temperature and humidity. Ex: Rice and Wheat – viable only for 30 min.

Members of Rosaceae, Leguminosae, Solanaceae are viable for several months.

Question 9.
What is cryopreservation? or How pollen grains are preserved.
Answer:
Cryopreservation: Storing viable pollen grains in liquid Nitrogen (at – 196° C), which can be used as pollen banks in crop breeding programmes.

Question 10.
Explain the structure of Megasporangium or Ovule with labelled diagram? or Describe the .structure of Anatropus Ovule?
Answer:

• Funicle: Stalk of an ovule, which is attached to the placenta.
• Raphe: Long ridge formed by lengthwise fusion of funicle with the body of ovule.
• Hilum: It is a junction between ovule and funicle and even extends to form raphe.

Structure of an Ovule

• Integuments: Multicellular, one or two-layered which surrounds ovule and protective in function.
• Nucellus: These are the cells with abundant reserve food material enclosed with in integuments. It nourishes the developing Embryo cells.
• Micropyle: Ovule is surrounded by integuments except at the tip, resulting in a small opening known as micropyle. Through this pore pollen tube enters into the ovule. ,
• Chalaza: The basal part of an ovule opposite to micropyle, swollen part of the nucellus.
• Embryosac: It is the female gametophyte which contains Egg apparatus, Antipodal cells synergids and secondary nucleus.

Question 11.
Give any one function of the Nucellus in an Ovule? (1M)
Answer:
Nourishes developing embryo cells.

Question 12.
Explain megasporogenesis? or Write the events of development of female gametophyte incorrect order?
Answer:
The process of formation of megaspore from Megaspore Mother Cell, (MMC), through meiosis is known as megasporogenesis. MMC is large nucleated with dense cytoplasm. It undergoes meiosis, which results in. the production of 4 megaspores. Out of 4, 3 degenerates and one megaspore is functional which developes into female gametophyte. “Method of embryo sac formation from a single megaspore is known as Monosporic type of development”.

Developments of Female Gametophyte
Megaspore mother cell undergoes 3 successive mitotic divisions to form 8 nucleated embryo sac.

Question 13.
Name the components of egg apparatus in an Embryo sac?
Answer:
The egg cell, Synergids, Antipodals, Polar nuclei

Question 14.
What is monosporic development of female gametophyte?
Answer:
Method of embryo sac formation from a single megaspore.

Question 15.
Name the cells found at the chalazal end of embryo sac?
Answer:
Antipodal cells.

Question 16.
Mention the function of filiform apparatus?
Answer:
Filiform apparatus guides the entry of pollen tube towards Egg cell.

Question 17.
Name the female gametophyte of Angiosperms?
Answer:
Embryo sac.

Question 18.
How many nuclei and cells are present in female gametophytes?
Answer:
8 Nuclei and 7 cells.

Question 19.
Draw a neat labelled diagram of embryo sac?
Answer:

Question 20.
Define pollination?
Answer:
Pollination is the transfer of pollen grain from the anther to the stigma of a pistil.
Types – depending on the source of pollen 3 types

1. Autogamy
2. Geitonogamy
3. Xenogamy

Question 21.
Define Autogamy?
Answer:
Autogamy (Self-pollination): Transfer of pollen grain from anther to the stigma of the same flower.

Question 22.
What are Chasmogamous flowers? Give examples.
Answer:
Chasmogamous are normal flowers with exposed anthers and stigma which favours cross-pollination.
Ex: Oxalis, Catharanthus

Question 23.
What are Cleistogamous flowers?
Answer:
Cleistogamous are Flowers that do not open at all, which favours self-pollination in the absence of pollinators.
Ex: Oxalis, Commelina.

Question 24.
Explain the factors which favours self-pollination?
Answer:

1. Homogamy – Anthers and stigma of some bisexual flowers mature at same time. It requires synchrony in pollen release and receptivity of stigma and close nearer position of anther and stigma like bending and folding. Ex: Vinca, Potato, Sunflower etc.
2. Autogamy
3. Cleistogamous flower.

Question 25.
Mention the type of pollination which ensures genetic recombination?
Answer:
Cross-pollination.

Question 26.
What type of pollination is seen in Cleistogamous flowers?
Answer:
Self-pollination.

Question 27.
Explain types of cross-pollination? or What is Xenogamy/Geitonogamy? Give examples.
Answer:

• Geitonogamy: Transfer of pollen grain from tjie anther to the stigma of another flower of the same plant. (It has genetically similar to autogamy). It involves pollinators. Ex: Hibiscus.
• Xenogamy: Transfer of pollen grain from the anther to the stigma of a flower on different plant. (It has genetically variability). Ex: Calotropis, maize, cucurbita etc.
• Agents of Pollination: Cross-pollination involved plants have abiotic ,and biotic agents as pollinators.

Question 28.
What are the characters of wind-pollinated flowers? Give examples.
Answer:
(a) Anemophily: Pollination by wind, most. common amongst abiotic pollinators.

• Characteristics of anemophilous flowers.
• Flowers are small, colourless, odourless, nectarless.
• Flowers are light, non-sticky, winged.
• Weil exposed long stamens, long styles and feathery stigma.
  Ex; Tassels of com cob in which have clusters of style and stigma which wave in the wind to trap pollen grain. – Ex: Wheat, Paddy, Grasses.

(b) Hydrophily: Pollination by water: Ex: Monocots, Algae, Bryophytes, Pteridophytes etc.

Question 29.
How is pollination affected in Vallisneria and Zostera? or How does pollination is achieved in Vallisneria and Sea grass.
Answer:
Freshwater hydrophytes like Vallisneria is submerged hydrophyte, Dioecious. Female flowers reach the surface of the water with their long stalk. Pollen grain released on surface of water, which are carried passively by water current.

Marine water Hydrophytejike Zostera (sea grass) have female flower submerged in water and the pollen grains are released inside the water.

Question 30.
Write the characteristic features of water pollinated plant’s, pollen grains?
Answer:
Hydrophytes have pollen grains – needle-like without exine, long, ribbon, like, mucilage covering protects pollen grain.

Question 31.
Write the characters of entomophilous flowers or insect pollinated flowers?
Answer:
(a) Adaptations of entomophilous flowers pollinated by Bat:

• Large, brightly coloured bracts and with Sticky pollens, rich in nectar and scents.
• Rafflesia: The largest flower produces a foul smell to attract flies and beetles.
• Amorphophallus: Tallest flower, safe place to lay eggs for insects.
• Gestmm Noctorum: Nocturnal flower attracts nocturnal insects by its scent.

(b) Chiropterophily: Pollination by Bat.

Question 32.
List adaptation features of Bat pollinated flowers? Give examples.
Answer:
Large, freely exposed, tough, strong scent, open in the evening or night.
Ex: Kigelia pinnata, Anthocephalus, Adansonia etc.

Question 33.
Explain relationship between yucca and moth?
Answer:
Moth cannot complete its life cycle without Yucca and vice versa. Moth deposits eggs in the locule of the Ovary, larvae come out of eggs as the seeds start developing in turn Yucca flower gets pollinated.

Question 34.
What are pollen robbers?
Answer:
Many insects just consume nectar and pollen without bringing pollination – known as pollen or nectar robbers.
[Note; Omithophily: Pollination by birds

• Flowers attract birds by their colour and nectar.
  Ex: Bombax, Butea, Agave.
• Malacophily: Pollination by snails and slugs, small shrubs flowers are pollinated.
  Ex: Strawberry, Chrysanthemum etc.
• Myrmecophily: Pollination by Ants.
• Tree species flowers are pollinated by the activities of ants.
  Ex: Mango, Litchi, Acacia, Pongomia etc.]

Question 35.
Differences between self and cross-pollination?
Answer:

Self Pollination	Cross-Pollination
1. Maintains the purity of race from generation to generation like improved and high yielding character can be maintained.	1. Brings genetic recombination and produces variation Healthier and stronger offsprings
2. Disadvantage: Lack genetic diversity, lead to degeneration of race i.e. inbreeding depression.	2. Disadvantage: If unwanted disadvantage characters are introduced it leads to loss of species.

Question 36.
Mention two strategies evolved to prevent self pollination in flower? or What are outbreeding devices? Mention its significance.
Answer:
Since self pollination results in inbreeding depression, plants have many devices to discourage self pollination and encourage cross pollination.

• Pollen release and stigma receptivity are not synchronised.
  Ex: Stigma ripens earlier in wheat and paddy (protogyny). Anthers ripens before stigma (protoandry) in sunflower and .Oeimum.
• Production of unisexual flowers Ex: Castor, Maize.
• Anthers and stigma placed at different positions.
• Self-incompatibility and self sterility.

Question 37.
What is self-incompatibility?
Answer:
“The ability of a plant to reject its own pollen and pollens of closely related species” is known as seif incompatibility.

Question 38.
Write a note on pollen pistil interaction?
Answer:
The pistil (gynoecium ) has the ability to recognise the pollen whether it is of compatible (right type) or incompatible (wrong type) i.e. self-incompatibility.

This recognising capacity is mediated by chemical components of the pollen interacting with of pistil. If poll en grain has allele which matches an allele of stigma then pollen tube fails to grow.

Germination of pollen grain: After compatible pollination, pollen grain,

• Germinates on the stigma to produce pollen tube through one of the germ pores.
• The contents of pollen grain move into pollen tube, which grows through the tissues of stigma, style and reaches the ovary.
• Pollen tube enters the “-ovule through micropyle and enters one of the synergids through filiform apparatus. (which guides the pollen tube to entry towards egg).

All these events are referred to as pollen-pistil interaction.
[This knowledge gained in this area help the plant breeders in manipulating pollen pistil interaction].

Question 39.
Draw a neat labelled diagram to show pollen germination on stigma?
Answer:

Question 40.
Define Artificial Hybridisation, Emasculation, Bagging? or What is Artificial hybridisation? By which
technique it is achieved.
Answer:
[Hybridisation: Crossing two plants differing from each other genotypically in one or more characters].
Artificial Hybridisation is an approach for crop improvement programme – Crossing two plants, i.e. artificially pollinating female plant with desired pollen grain and female flowers stigma is protected from contamination. This is achieved by techniques – Emasculation and bagging.

Emasculation: Removal of stamens or anthers of a bisexual flower without affecting the female reproductive organs.

Bagging: Covering emasculated flower with a polythene bag immediately to prevent cross-pollination by unwanted pollen.

[During artificial hybridisation when the stigma of bagged flowers attain receptivity mature pollen grain from the anthers are dusted on the stigma, pollinated flowers are rebagged and fruits are allowed to develop.]

Question 41.
How is bagging useful in plant breeding programme?
Answer:
It prevents cross-pollination by unwanted pollen.

Question 42.
What is meant by Emasculation? When and why does a plant breeder employ this technique. (2M) Ans. Removal of anthers of a bisexual flower is known as emasculation. During performing artificial hybridisation to avoid self pollination .

Question 43.
Define Double fertilisation?
Answer:
Double fertilisation – “is a fertilisation process in flowering plants, in which one of the male gamete fertilises an egg cell to form zygote y/hile (second) the other fuses with two polar nuclei to form endopserm”. [Involves two types of fusions Syngamy + Triple fusion in an embryo sac.]

Question 44.
Explain Double fertilisation? or What is double fertilisation? Describe fertilised Embryo sac with a neat labelled diagram.
Answer:
Pollen tube with male gametes enter into embryo sac through micropyle.
After entering into one of the synergids, pollen tube release two male gametes into the cytoplasm of the synergid.

One of the male gamete moves towards the egg cell and fuses with its nucleus, completing syngamy, which results in formation of diploid zygote.

Other male gamete moves towards the two polar nuclei fuses.with them to produce triploid primary endosperm nucleus.

It involves fusion of three haploid nuclei – Triple fusion. (vegetative fertilisation).

Fertilized Embryo Sac

Question 45.
Give an account of significance of double fertilisation?
Answer:

• Stimulates embryo double fertilisation.
• Zygote maintain constant chromosome number in the life cycle of plant.
• Recombination of genes, leading to new characters in offsprings.

Question 46.
Difference between male gametophyte and female gametophyte?
Answer:

Male gametophyte	Female gametophyte
1. It is derived from microspores inside another.	1. It is derived from a megaspore formed inside the ovule.
2. Micropores shed at maturity.	2. Megaspores do not shed.
3. Male gametes 3 celled	3. Female gamete – 7 celled.

Question 47.
Differentiate between self and cross-pollination.
Answer:

Self Pollination	Cross-Pollination
1. Pollination in the same flower, or two flowers on the same plant.	1. Pollination between two flowers on different plants.
2. Monoecious condition.	2. Dioecious condition.
3. Pollinating agents are not required.	3. Require pollinating agents.
4. No variation. Ex: Wheat, Rice, Pea, Tomato etc.	4. Variation produced. Ex: Maize, Coconut, Papaya, Sugar cane etc.

Question 48.
Define Endosperm?
Answer:
“It is a nutritive tissue which provides essential nutrients to the developing embryo and to young seedlings during germination.”

Question 49.
Explain endosperm formation in Coconut?
Answer:
It develops from the primary endosperm nucleus (PEN) which has undergone vegetative fertilisation.
PEN undergoes first free nuclear division, later on cell wall formation and becomes cellular endosperm.
Ex: Coconut water-free nuclear division.
White kernel – Cellular.

Question 50.
Which cell of fertilised embryo sac forms the kernel of the coconut?
Answer:
An Endosperm cell forms kernel of coconut.
” [Based on the presence or absence (used up).
Endosperm seeds are classified as Endospermous / albuminous – Not completely Consumed endosperm → Monocots seeds. Nonendospermous / Exalbuminous – Completely consumed endosperm → Dicots seeds.]

Question 51.
Define Embryogenesis?
Answer:
Embryogenesis: “ The process of the development of embryo from the zygote”.

Question 52.
Explain Embryogenesis in Dicot along with a neat labelled diagram?
Answer:

The Embryogeny process are similar in both monocots, and dicots, but in dicots, after globular stage, it ends up with single terminal cotyledon whereas in dicots. 2 cotyledons preceded by heart-shaped embryo formation. Zygotes enlarge in size and divide transversely to form larger basal suspensor cell and smaller terminal embryo cells. The basal cell divides transversely and forms 6-8 suspensor cells, the last cell in hypophysis. The terminal cell divides longitudinally and transversely to form a globular, heart-shaped and mature embryo subsequently.

Question 53.
What is Epicotyl?
Answer:
The portion of the embryonal axis above the level of cotyledons is the Epicotyl, which terminates with plumule.

Question 54.
What is Hypocotyi?
Answer:
The cylindrical portion below the level of cotyledons is the Hypocotyi, terminates with radicle covered with a root cap.

Question 55.
Draw a neat labelled diagram of dicot embryo.
Answer:

Question 56.
ExplaIn the structure of Monocot Embryo?
Answer:
Monocot embryo has only one cotyledon known as Scutellum. Which is situated towards one side of embryonal axis. The embryonal axis above the level of attachment of scutellum is Epicotyl, which has shoot apex with leal pnrnordia enclosed with Coleoptile.

At its lower end, embryonal axis is Hypocotyl which has radicle and root cap enclosed with Coleorrhiza.

Question 57.
Define Scutellum?
Answer:
Single cotyledon in monocot seed.
Seed: After fertilisation the seed is the final product formed inside fruits, “it is a fertilised ovule”.
[Seeds typically consists of seed coats, cotyledons and an embryo axis].

Question 58.
Explain the structure of Dicot seed?
Answer:
The two integuments of the ovule becomes seed coat, outer leathery testa, and inner thin tegmen.
The micropyle remains as small pore, which helps in entry of O₂ and H₂O during germination.
Two Cotyledons are large and thick which store food.
One end of embryo axis in between cotyledons is plumule, other end posses Radicle.

Question 59.
Explain the structure of Monocot seed along with neat labelled diagram?
Answer:

1. Single seeded fruit – Caryopsis (Maize).
2. Endospermic, so endosperm has protein rich Aleurone layer.
3. Single cotyledon known as Scutellum.
4. Plumule covered with coleoptile and radicle covered with coleorrhiza.

L. S. of Maize grain (MONOCOT SEED)

Question 60.
Mention 2 types of Seeds with examples?
Answer:

1. Albuminous seeds: with Endosperm. Ex: Wheat, Com etc.
2. Exalbuminous seeds: without or (used up) endosperm. Ex: Bean, Pea etc.

Question 61.
Define perisperm give 2 examples.
Answer:
In some seeds nucellus remains even after seed formation, it is called as Perisperm Ex: Black pepper, Nymphea.

Question 62.
What is seed Dormancy?
Answer:
As the seed matures, water content is reduced, embryo slow down its activity, and enter a state of inactivity seed Dormancy, because seeds prefer to germinate under favourable conditions.

Question 63.
Differentiate between True and false fruits?
Answer:

1. True fruits: Fruits derived from ovary Ex: Tomato, Grapes, Coconut etc.
2. False fruits: Fruits derived from ovary along with accessory floral parts like thalamus. Ex: Apple, Strawberry, Cashew etc.
3. Parthenocarpic fruits (Seedless fruits)

Question 64.
What are Parthenocarpic fruits? Give examples.
Answer:
Fruits develop without fertilisation. Ex: Banana, Pineapple, Grapes etc.
[Artificially induced by spraying Auxin, Gibberllin etc.]

Question 65.
What are the advantages of seed?
Answer:
Advantage of seed production in Angiosperms:

• Seeds remain viable for several years.
• Seed coat protects embryo.
• Seedlings are nourished.
• Better adaptive strategies.
• Genetic recombinations leading to Variations.
• Basis of our agriculture – used as food, to raise crop in the next season.

Question 66.
Write a note on Seed Viability?
Answer:
Seed Viability is for several months to years. Oldest seed from Arctic Tundra in Lupinus arcticus plant, germinated after 10,000 years of dormancy.

Recently from Dead sea, in Phoenix dactylifera plant germinated after 2000 years of dormancy.

Question 67.
What is Apomixis?
Answer:
“Production of seed along with embryo without fertilisation”. (Agamospermy) it is a form of asexual reproduction Ex: Mango, Orange, Brinjal, Grasses etc.
[causes → Diploid egg develop into embryo without reduction division.
→ Embryo sac cells develop into embryo.]

Question 68.
Write a note on Apomixis significance?
Answer:

• Shortens the time required to produce new individuals and genetically controls.
• Assures total transfer of characters to the progeny and maintains hybrid vigour.
• Apomixis gained importance for problems of hybrids.

[Note: Hybrid seeds have to produced every year, if they are sown, genes segregate and do not maintain hybrid characters.

Hybrid seeds are expensive, so if apomictics is preferred there is no separation of characters, so need not to buy new hybrid seeds every years. Research studies are aiming towards transfer of apomictic genes into hybrid varieties.]

Question 69.
Define polyembryoy? Give any 2 examples.
Answer:
Phenomenon of development of more than one embryo in the seed. Ex: Citrus fruits, Mango, Onion etc.
[Causes: Cleavage of zygote into more than one unit, each develops into embryo.
Cells of embryo sac → haploid embryos.
Outside embryo sac cells like nucellus and integuments.]

Question 70.
Write significance of polyembryony?
Answer:
Plant breeding and horticulture.
Haploid embryos are used to raise homozygous diploids. Genetically uniform seedlings in fruit trees.

Question 71.
Mention any five differences between microsporogenesis and megasporogenesis?
Answer:

Microsporogenesis	Megasporogenesis
1. It is the formation of microspore from microspore mother cell through meosis.	1. It is the formation of megaspore from megaspore mother cell through meosis.
2. It results in the formation of four microspores all of which are functional.	2. It results in the formation of four megaspores out of which three degenerate and only one is functional.
3. The microspore develops wall layers and forms pollen grains.	3. The single functional megaspore forms embryosac.
4. This occurs in the sporogenous tissue with in the anther.	4. This occurs in the nucellus which is found with in the ovule.
5. The four microspores are arranged in the form of tetrad.	5. The four megaspores are arranged in linear form

→ Flower:
According to biologists, flowers are morphological embryological structures and sites of sexual reproduction. Diversity in their structure is due to range of adaptations ensures sexual reproduction and assures fruits and seeds set.

→ Pre Feitiusation Structures And Events:
Flowers – are modified shoots. Structural and hormonal changes act on vegetative buds, leads to differentiation and development of floral buds.
A flower morphologically ha&4 whorls.
Two accessory whorlš are Calyx and Corolla.
Two essential whorls are Androecium contain male reproductive structure namely Stamens and Gynocicurn containam female reproductive structure namely Carpets.

→ Stamen, Microsoprangium and Pollen Grain:

• Each stamen (microsporophyll) has terminal spore bearing organ the Anther, and the long slender stalk filament.
• Each anther consists of two anther lobes joined together by connective, each anther lobe consists of pollen sacs.
• Each pallen sac contains numerous pollen grains (microspores) and a 4 to 5 layered anther wall.

Stamens showing different parts

→ Structure of Microspohangium/Anther

T.S of young anther

→ The Pistil, Megasporangiúm And Embryo Sac:
The gynoecium or pistil is the female reproductive part of the flower consisting of carpels [Megasporophyll].

• A Pistil with one carpel – Monocarpellary
  Ex. Fabaceae
• rwt cirpeIs Bicarpellary
  Ex. Solanaceae
• Three earpcls – Tricarpellary
  Ex. Orchidaceae
• Five camels – Pentacarpellary
  Ex. Cucurbitaceae,
• If carpcs fused together – Syncarpous
  Ex. Hibiscus, papayer
• Carpels free – Apocarpous
  Ex. Vinca rosea, champak

→ Each gynoecium has 3 parts

1. Stigma,
2. Style,
3. Ovary.

Stigma: Serves as landing platform for pollen grains vary in shapes. It has sticky surface where pollen grains geminate after pollination.

Style: Elongated, slender part connects stigma with ovary.

Ovary: It is the basal swollen part of pistil, that contains ovules and eggs in its cavity. The placenta a ridgeftissu which is a place for attachment of ovules is located inside the locule.

→ Structure of Embryo Sac:
First the functional megaspore nucleus divides mitoticafly to form 2 nuclei which move towards opposite pole forming 2- Nucleate embryo sac. The developing embryo sac enlarges in size due to formation of vacuole in the centre.

Each 2 nulci, divides and redivides, forming 4 nculei at micropylar end and 4 nuclei at chalazal end, here they under go free nuclear division. At 8 celled stage cell wall formation.

Out of 4 nuclei at micropylar end, 3 differentiate to produce Egg apparatus = 1 Egg + 2 Synergid cells. 4 nuclei at chalazal, 3 remains at, polar region and foi-ni Antipodals.

One nucleus from micropylar end, one from chalazal end migrate towards the centre and fuse to form (2n) diploid secondary nucleus.

Thus a typical angiosperm embryosac at maturity though 8 nucleate, it is 7 celled.

→ Double Fertilisation:
Nawaschin discovered double fertilisation in Angiosperms. It is the process of forming Zygote and Endosperm.

→ Post Fertilisation Events:
Includes – development of Endosperm, Embryo mattifation of Ovary into fruit. Ovule into seed.

→ Embryo:
After fertilisation the zygote grows and give rise to the embryo. It is a post fertilisation product.

→ Development of Fruits:
The ovary develops into fruit.
The wall of the ovary is the fruit wall known as Pericarp.

Based on the type of pericarp fruits arc classified
(A) (1) Fleshy (Ex: Mango, Orange, Tomato etc.)
(2) Dry (Ex: Groundnut. Mustard etc.)

(B) Based on floral parts involved in fruit formation, classified as true and false fruits.

2nd PUC Biology Important Questions with Answers

Karnataka 2nd PUC Biology Important Questions Chapter 1 Reproduction in Organisms

Question 1.
Define Lifespan?
Answer:
The period from birth to the natural death of an organism.
Note: The life span of an organism varies greatly.
Ex: Crow – 15 Years, Tortoise – 100-150 Years.

Question 2.
Why unicellular organisms are considered as immortal?
Answer:
In single celled organisms, the parent continues to live as daughter cells.

Question 3.
Define Reproduction?
Answer:
It is a biological process in which organism gives rise to young ones similar to itself.

Question 4.
Write the Significance of reproduction or Why is reproduction essential for organisms? (2M)
Answer:

• The offspring grow mature and in turn produce new offspring.
• It is a process of species maintenance.
• If reproduction is absent, a species may soon get extinct.

Question 5.
Define asexual reproduction?
Answer:
It is a mode of reproduction in which a single parent produces young ones without the formation and fusion of gametes.

Question 6.
Mention the features of asexual reproduction?
Answer:

• It is uniparental.
• Occurs in unicellular organisms, plants and animals with simple organisation.
• It involves mitotic cell division.
• Offsprings are identical to one another and exact copies of their parents.

Question 7.
Why is the offspring formed by asexual reproduction referred to as clones? or Define clone.
Answer:
Morphologically and genetically similar off springs identical to their parents are known as clones.

Question 8.
Explain the different methods of asexual reproduction with examples?
Answer:
(a) Binary Fission: The body of a matured parent cell divides into two halves, each half grows into an adult.
Eg: Bacteria, Amoeba, Paramoecium.

(b) Cell division: The parent cell divides into two give rise to new individuals.
Eg: Protists, Monera.

(c) Budding: A new individual develops as a small outgrowth on the surface of the parent, later gets separated and matures into a new organism.
Ex: Yeast, Hydra. Sponges → Gemmules [internal buds or Internal asexual reproductive units]

(d) Spores: Specialised microscopic unicellular asexual reproductive structures to overcome unfavourable climatic conditions.
Ex: Zoospores (motile spores) – Chlamydomonas.Conidia – Penicillium.

Question 9.
Mention the asexual reproductive structures of Penicillium?
Answer:
Conidia.

Question 10.
Define vegetative reproduction?
Answer:
It is a process of formation of new plants from vegetative parts of the plant body.

Question 11.
What are vegetative propagules?
Answer:
Vegetative reproductive structures, having the capacity of giving rise to new offspring.

Question 12.
Name the vegetative propagules of the following? or Mention the units of vegetative propagation in plants, or Name the types of vegetation in propagules in the following plants (a) Potato (b) Ginger (c) Bryophyllum
Answer:

1. By Roots: Root tubers of sweet potato, Dahlia etc.
2. By Stem:
   (a) Tuber of Potato.
   (b) Rhizome of Ginger, Banana.
   (c) Sucker of Mint and Chrysanthemum.
   (d) Runner of Oxalis.
   (e) Offset of Eichhomia (Water Hyacinth).
   (f) Bulb of Onion.
3. By Leaves: Bryophyllum – leaf buds.
4. Inflorescence: Bulbils part of the inflorescence of plant Agave.

Question 13.
Name asexual reproductive structures in
(i) Hydra
Answer:
Budding

(ii) Chlamydomonas
Answer:
Zoospores

(iii) Pencillium.
Answer:
Conidia

Question 14.
Name the vegetative propagules of the following plant?
(a) Agave
Answer:
Bulbil

(b) Ginger
Answer:
Rhizome

(c) Bryophyllum.
Answer:
(c) Leaf buds

Question 15.
Why Eichhornia is known as Terror of Bengal?
Answer:
Eichhomia is an aquatic invasive weed,found growing in standing water. It drains oxygen from the water which leads to death of fishes and other aquatic organisms.

Question 16.
Write the importance of asexual reproduction in plant propagation?
Answer:
Asexual reproduction ability of plants is fully exploited for commercial propagation of economically important plants. Methods: Cutting, Grafting, Tissue Culture (Artificial Vegetative Reproduction).

Question 17.
Define sextjal reproduction?
Answer:
It is a mode of reproduction in which young ones are produced through formation and fusion of gametes by the same individual or by two different individuals of different sex.

Question 18.
Write any four differences between sexual and Asexual reproduction? (2M)
Answer:

Asexual Reproduction	Sexual Reproduction
1. Occurs in lower organisms.	1. Lower and Higher organisms.
2. Uniparental.	2. Generally biparental.
3. No formation of gametes.	3. Involves formation of gametes
4. Mitotic cell division.	4. Meiotic cell division.
5. Variations absent	5. Variations are produced.
6. Offspring genetically similar to parents	6. Offspring genetically dissimilar to parents.

Question 19.
Explain the different stages in the life cycle of living organisms?
(a) Define Juvenile Phase?
Answer:

1. Juvenile Phase: It is a period of growth of an organisms before reproductive maturity. It is called as a vegetative phase in plants.
2. Reproduction Phase: It is a period of growth of an organism where they start reproducing sexually.

(b) Define Senescent Phase?
Answer:
Senescent Phase (Ageing Process): It is the end of reproductive phase, which gradually results in slowing of physiological functions leading towards death of an organism.

Question 20.
What are Annuals/Biennials/Perennial plants give examples?
Answer:

Question 21.
Name the plant which flowers once in its lifetime?
Answer:
Bamboo [50-100 years].

Question 22.
Name the plant that flowers once in 12 years?
Answer:
Strobilanthus Kunthiana (Neelakuranji).
[Plants reproductive phase is marked by flowering, based on flowering capacity, they care classified into Mono and Polycarpic plants.]

Question 23.
Difference between Monocarpic plants and Polycarpic plants?
Answer:

Monocarpic Plants	Polycarpic Plants
1. These plants flower only once in their lifetime.	1. These plants flower many times a year throughout their life cycle.
2. After flowering and fruiting they die. Ex: Bamboo, Rice, Wheat, Carrot etc.	2. Flowering and Fruiting continues till the completion of their life span. Ex: Mango, Apple, Orange, Jackfruit etc.
3. Strobilanthes Kunthiana.

Question 24.
Differentiate between seasonal and continuous breeders?
Answer:

Seasonal Breeders	Continuous Breeders
These animals are reproductively active during favourable seasons. Ex: Frog, Lizards, Birds etc.	These animals are reproductively active throughout their reproductive phase. Ex: Rabbit, Primates etc.

Question 25.
What is reproduction cycle?
Answer:
Reproductive cycles are the cyclic changes that occur in the female reproductive systems of Mammals.

Question 26.
Differentiate between the Menstrual and Oestrus cycles.
Answer:

Menstrual Cycle	Oestrus Cycle
1. Females of placental mammals exhibit cyclical changes in the activities of ovaries during the reproductive phase of Primate mammals. Ex: Monkeys, Humans.	1. Females of placental mammals exhibit cyclical changes in the activities of ovaries during Apes and reproductive phase of Non-Primate mammals. Ex: Cow, Dog, Sheep.
2. Heat period is absent.	2. Heat period at the time of ovulation.
3. Breeding occurs during any season and any part of the reproductive cycle.	3. Non-breeding season, [anaestrous period]

Question 27.
Write 3 Phases of sexual reproduction?
Answer:

1. Pre fertilisation Events – (a) Gametogenesis, (b) Gamete transfer.
2. Fertilisation – Fusion of male and female gametes.
3. Post fertilisation Events – Zygote formation, Embryogenesis.

Question 28.
Define gametogenesis?
Answer:
The process of formation of gametes or sex cells within the sex organs or gonads.

Question 29.
Differentiate between isogamy and anisogamy? or Distinguish between homogametes and heterogametes?
Answer:

Homogametes (Isogametes)	Heterogametes (Anisogametes)
Male and female gametes which are morphologically similar Ex: Algae, Protozoa.	Male and Female gametes are morphologically dissimilar. Ex: Fucus, Human.

Question 30.
Defferentiate between Monoecious and Dioecious plants? Give examples.
Answer:
(a) Bisexual or Monoecious /Homothallic plants are the plants with both male and female reproductive structures on same plant. Ex: Cucurbit, coconut, Maize, Chara (Algae).
(b) Unisexual or Dioecious/Heterothallic plants are the plants with male and female reproductive structure on different plants. Ex: Papaya, Marchantia (Bryophyte), Date Palm, Cycas etc.

Question 31.
Why papaya is considered as dioecious plants?
Answer:
It bears male and female reproductive structures on different plants.

Question 32.
‘Papaya plants exhibit xenogamy only, why?
Answer:
Papaya plants are dioecious in nature, thus they exhibit xenogamy.

Question 33.
What is heterothallic condition? Differentiate between staminate and pistillate flowers.
Answer:

• If plant possess male and female reproductive structures on different plants is termed as heterothallic condition,
• If a flowering plant possess unisexual male flowers with only androecium it is called as staminate flower.
• If a flowering plant possess unisexual female flower with only gynoecium it is called as Pistillate flower. (Cell division during gamete formation
• All gametes are haploids, but parent body may be diploid or haploid. The type of cell division differs during gamete formation).

Question 34.
What are Diploids and Haploids? Give examples.
Answer:

1. Diploids: In these organisms, diploid parents produces haploid gametes by meiosis [gamete mother cells
   = Meiocytes],
   Eg: Pteridophytes, Gymnosperms, Angiosperms, Animals including human beings.
2. Haploids: In these organisms, Haploid parents produces Haploid gametes by mitosis (meiosis occurs during zygote formation). Eg: Algae, Fungi.

Question 35.
Define gamete transfer?
Answer:
It is the transfer of male gamete towards female gametes for the purpose of fertilisation.

Question 36.
Define fertilisation?
Answer:
“The process of fusion of haploid sperm with haploid egg to produce diploid Zygote” (syngamy).

Question 37.
Differentiate between external and internal fertilisation?
Answer:

External Fertilisation	Internal Fertilisation
1. Syngamy occurs outside the body of the parents in an external medium.	1. Syngamy occurs within the female body in the genital tract.
2. Copulatory organs are absent.	2. Males are provided with copulatory organs
3. Ex: Fishes, Amphibians, Algae.	3. Ex: Reptiles, Mammals, Gymnosperms, Angiosperms.

Question 38.
Define parthenogenesis with any 2 Examples? or Define parthenogenesis.
Answer:
Parthenogenesis: The female gamete develops into Embryo to form a new organism without fertilisation.
Ex: Rotifers, Insects, Drones, Lizards, Turkey etc.

Question 39.
Define Zygote?
Answer:
Fusion of male and female haploid gametes results in formation of zygote.
[Note: Development of zygote varies with organisms and environment.

• In External fertilisation it is formed in External medium (water) and formed inside the body of the organism in Internal fertilisation.
• In fungi and Algae Zygote develops thick wall and undergoes dormancy before germination to overcome unfavourable conditions.

Question 40.
What are haplontic and diplontic life cycles?
Answer:

• Haplontic life cycle – Zygote divides by meiosis to form haploid spores and form haploid individuals.
• Diplontic life cycle – Zygote divides mitotically to form diploid individuals.

Question 41.
Define Embryogenesis?
Answer:
“Development of Embryo from the zygote”.

Question 42.
Define organogenesis?
Answer:
The zygote grows by repeated mitotic division form cells and undergo cell differentiation to become tissues and organs to form an organism. This is known as organogenesis.
[Note: 1. In flowering plants, Zygote is formed inside the ovule and developes into Embryo. Ovary developes into fruit and Ovule developes into seed.]

Question 43.
Define Pericarp?
Answer:
The ovary wall or fruit wall develops into ‘Pericarp’.
[Note: 2. In Animals, based on the development of zygote, they are classified into Oviparous, Viviparous, animals.]

Question 44.
Define Ovipary? Give any two examples.
Answer:
Egg-laying animals. Eg. Reptiles, Birds.

Question 45.
Differentiate between Oviparous and Viviparous animals?
Answer:

Oviparous	Viviparous
1. These animals lay fertilised eggs, and the development of the embryo takes place outside the body of the female.	1. These animals give birth to young ones, and the development of the embryo takes place inside the body of the female.
2. After a period of incubation the young ones hatch out. Ex: Reptiles, birds.	2. After a period of growth young ones are delivered out. Ex: Mammals.

Question 46.
How Vivipary is advantageous to young ones.
Answer:
Because of proper embryonic care and protection,the chances of survival is greater in viviparous animals,than oviparous animals.

Question 47.
(a) What are hermaphrodites? Mention two example.
Answer:
Animals which possess both male and female reproductive organs are termed as hermaphrodites.
Ex: Leech, tapeworm, earthworm.

(b) Offsprings of asexual reproduction are called clones. Why?
Answer:
Morphologically and genetically similar offspring identical to their parents are known as clones.

Question 48.
Mention the asexual reproductive unit in sponges.
Answer:
Gemmules.

→ Introduction:
Reproduction is the production of new individuals, physically independent of their parents. It is one of the important characteristics of living organisms.

→ Types of Reproduction:

1. Asexual Reproduction.
2. Sexual Reproduction.

→ Phases in Sexually Reproducing Organisms:
All organisms have to reach a certain stage of growth and maturity in their life before they attain capable of sexual reproduction.

→ Gamete Transfer Types:

1. Male and female gametes are motile. Ex: lower Algae and fungi
2. Male Motile and female gamete stationary Ex: higher organisms.
3. In Angiospcms pollens are transferred to the stigmatic surface by pollination, which leads to fertilisation.
4. In invertebrates swimming movement of sperm enables it to reach egg.
5. In vertebrates sperms are transported from the region of deposition to the region of fertilisation.

2nd PUC Biology Important Questions with Answers

Karnataka 2nd PUC Biology Chapter Wise Important Questions and Answers

1. Reproduction in Organisms Important Questions
2. Sexual Reproduction in Flowering Plants Important Questions
3. Human Reproduction Important Questions
4. Reproductive Health Important Questions
5. Principles of Inheritance and Variation Important Questions
6. Molecular Basis of Inheritance Important Questions
7. Evolution Important Questions
8. Human Health and Disease Important Questions
9. Strategies for Enhancement in Food Production Important Questions
10. Microbes in Human Welfare Important Questions
11. Biotechnology: Principles and Processes Important Questions
12. Biotechnology and its Applications Important Questions
13. Organisms and Populations Important Questions
14. Ecosystem Important Questions
15. Biodiversity and Conservation Important Questions
16. Environmental Issues Important Questions