Important Questions for Class 12 Biology Chapter 8 – Microbes in Human Welfare

Microbes In Human Welfare Short Question And Answers

Question 1. What for are Cyclosporin A anil Streptokinase bioactive molecules prescribed by a doctor
Answer:

  • Cyclosporin A – It is used as an immunosuppressive agent.
  • Streptokinase – It is used as a clot bluster for removing clots from blood vessels.

Question 2. Assertion: Large holes in ‘Swiss cheese’ are due to the production of a large amount of carbon dioxide by a specific microbe

Reason: The specificity of the characteristic texture, flavour and taste of Swiss cheese is due to the use of the bacterium Propionibacterium Sherman.

  1. Both Assertion and Reason are true, and Reason is the correct explanation of the Assertion.
  2. Both Assertion and Reason are true, but Reason is not the correct explanation of the Assertion.
  3. The assertion is true, but Reason is false
  4. Both Assertion and Reason are false.

Answer: 1. Both Assertion and Reason are true, and Reason is the correct explanation of the Assertion

Question 3. Some cyanobacteria in aquatic and terrestrial environments that enrich the soil by fixing atmospheric nitrogen are

  1. Rhizobium and Azotobacter
  2. Azospirillum and Glomus
  3. Anabaena and Nostoc
  4. Azospirillumand Azotobacter

Answer: 3. Anabaena and Nostoc

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Question 4. The microbes commonly used in kitchens are

  1. Lactobacillus and Yeast
  2. PeniciiHum and Yeast
  3. Microspora and E.coli
  4. Rhizopus and

Answer: 1. Lactobacillus and

Microbes In Human Welfare Short Question And Answers:

Question 1. Fanners are often suggested to use the following organisms in their cropland to improve soil fertility. Explain.

  1. Rhizobium
  2. Anabaena

Answer:

  1. Rhizobium is a bacterium found in soil that helps in fixing nitrogen in leguminous plants
  2. It attaches to the roots of the leguminous plant and produces nodules. These nodules fix atmospheric nitrogen and convert it into ammonia that can be used by the plant for its growth and development.
  3. Anabaena plays a significant role in farming where it is used as a biofertilizer and soil stabilizer.

Question 2. Organic farmers use Trichoderma and liacuas as biological control agents. Explain.
Answer:

  1. Trichoderma – species are free-living fungi that are very common in the root ecosystems. They are effective biocontrol agents of several plant pathogens Baculoviruses are pathogens that attack Insects and other arthropods.
  2. Baculovirus – The majority of baculovirus ruses used as biological control agents are in the genus Nucleopolyhedrovtrus. These viruses are excellent candidates for species-specific, narrow-spectrum insecticidal applications. They have been shown to have no negative impacts on plants, mammals, birds, fish or even on non-target insects.

Question 3. A particular cyanobacterium is spread by (lie farmers in their fields while growing paddy). Name the cyanobacteria used and give two advantages of it.
Answer:

Cyanobacteria like Anaebena. Nosiocetc can fix atmospheric No, decompose organic wastes and residues, detoxify heavy metals, pesticides, and other xenobiotics, catalyze nutrient cycling, suppress the growth of pathogenic microorganisms in soil and water, and produce some bioactive compounds that contribute to plant growth.

Question 4. Study the given diagrams of Sewage Treatment Plants (STP) and answer the questions that follow:

Microbes In Human Welfare Sewage Treatment Plants

  1. Which of the two Sewage Treatment Plants? (A) or (B), will be more effective in treating human excreta in municipal waste?
  2. How is the primary effluent treated in the aeration tanks till there is a significant reduction in the BOD of (the effluent?

Answer:

  1. Sewage Treatment Plants (A) will be more effective in treating human excreta in municipal waste.
  2. The primary effluent is taken to aeration tanks, where it is constantly agitated mechanically. Air is pumped into it. periodically A large number of aerobic heterotrophic microbes grow in the aeration tank to form floes.
  3. Due to microbial activity, the organic matter gets digested The microbes convert it into microbial biomass and release the minerals Due to the breakdown of organic matter, the BOD (biochemical oxygen demand) of the wastewater reduces to about 10 to 15% of raw sewage.

Question 5. Write the different components of activated sludge. Explain the different ways it can be used further in the sewage treatment process.
Answer:

  • Dining secondary treatment BOD- of sewage or wastewater is reduced significantly the effluent is then passed into a settling tank where the bacterial “floes are allowed to sediment called activated sludge.
  • The floes are masses of bacteria associated with fungal filaments to form mesh-like structures.

Question 6.

1. Name the genus of baculovirus that acts as a biological control agent despite being a pathogen. Justify by giving three reasons that make it an excellent candidate for the job,
Answer:

Genus – Nucleopolyhedrovirus, species-specific, Narrow spectrum. No negative impact on (plants or mammals or birds or fish) non-target organisms.

2. ‘’Micro-organisms play an important role in the biological treatment of sewage.” Justify
Answer:

Primary effluent is passed into aeration tanks, constantly agitated and the air is pumped in. This allows the growth of useful aerobic microbes into flow or masses of bacteria and fungal filaments), these microbes consume organic matter and reduce the Biochemical oxygen demand (BOD) of effluent.

Question 7.

  1. Mention the difference in the level of BOD before and after the secondary treatment of sewage water.
  2. Write the importance of floes during the secondary treatment of sewage.

Answer:

1. Before – High. After- Low

2. Reduces BOP Biochemical Oxygen Demand

Question 8. Expand ‘LAB’. How are LABs beneficial to humans? (Write any two benefits).
Answer:

Lactic Acid Bacteria,

  1. They produce acid which partially digests the milk protein or sets milk into curd.
  2. They improve nutrition and quality by producing Vitamin B.
  3. Check disease-causing microbes in our stomach

Question 9. Your advice is sought to improve the nitrogen content of the soil to be used for the cultivation of a noil-leguminous terrestrial crop.

  1. Recommend two microbes that can enrich the soil with nitrogen.
  2. Why do leguminous crops not require such enrichment of the soil?

Answer:

  1. Azospiri/fum Azotobacter Anabaena Nosioc Oscillaloriu rankia
  2. They can fix atmospheric nitrogen, due to the presence of Rhizobium which is a V-fixing bacteria in their root nodules.

Question 10. Why are microbes like Spirulina being produced on a commercial scale? Mention its two advantages.
Answer:

  • As a source of food protein.
  • Reduces environmental pollution or solves the problem of hunger and malnutrition or a rich source of protein or low-cost production.

Question 11.Name (the microbes that help the production of the following products commercially:

  1. Statin
  2. Citric acid
  3. Penicillin
  4. Butyric acid

Answer:

  1. Monascus purpureas
  2. Aspergillus
  3. PenicilHum notatum
  4. Clostridium butylicum

Question 12. Name the first antibiotic discovered and by whom.
Answer: Penicillin by A. Fleming.

Question 13. What is the pathogenic property of baculovirus, used as a biological agent? Name the genus of these organisms.
Answer:

  • Baculoviruses are pathogens that attack insects and other arthropods.
  • The majority of baculoviruses used as biological control agents are in the genus

Question 14. Explain the changes that mills undergo when a suitable starter or inoculum is added to it. flow does the end product formed prove to be beneficial for human health?
Answer:

Lactobacillus lactic acid bacteria (LAB) present in inoculum, grows in milk at a suitable optimum temperature, multiplies converting milk to curd,-and produces acids that coagulate and partially digest the milk proteins. This improves its nutritional quality by increasing Vitamin B12, and LAB checks diseases causing.

Question 15. Why does an organic farmer intentionally not use toxic chemicals to kill the pests which damage the last crops? Explain giving three reasons. 
Answer:

  • Toxic chemicals can have adverse side effects cause biomagnification,
  • it kills both useful and harmful life forms indiscriminately,
  • It eradicates pests not control pests,
  • Beneficiary predatory and parasitic insects which depend upon them as food or hosts would not be able to survive,
  • It disturbs the food chain food webs or vibrant ecosystems.

Question 16.

1. Organic farmers prefer biological control of diseases and pests to the use of chemicals for the same purpose. Justify.
Answer:

  • Reduces dependence on toxic chemicals.
  • Protects our ecosystem or environment.
  • Protects and conserves non-target organisms they are species-specific.
  • These chemicals being non-biodegradable may pollute the Permanently.
  • These chemicals being non-biodegradable may cause biomagnification.

2. Give an example of a bacterium, a fungus and an insect that are used as biocontrol agents.
Answer:

  • Bacteria Bacillus thuringiensis Fungus Trichodcnva.
  • Insect Lady bird or Dragonfly or Moth.

Question 17. The three microbes are listed below. Name the product produced by each one of them and mention their use.

  1. spergillusniger
  2. Trichodermtt polysporum
  3. Monascus purpure

Answer:

  1. Aspergillus niger – Citric Acid, natural preservative or flavouring agent.
  2. Trichodennapolysporum – Cyclosporin A, immunosuppressive agent.
  3. Monascus purpureas – Statin, blood cholesterol-lowering agent.

Question 18. Baculoviruses are good examples of biocontrol agents. Justify by giving three reasons.
Answer:

  • Species-specific narrow-spectrum insecticidal application
  • They have no negative impact on plants mammals or birds or fish or non-target insects
  • They are beneficial for 1 PM (Integrated Pest Management) or the Pest Management Programme.

Question 19. Secondary treatment of the sewage is also called Biological treatment. Justify this statement and explain the process.
Answer:

Involves biological organisms such as aerobic and anaerobic microbes or bacteria and fungi to digest or Consume organic waste Primary effluent is passed into an aeration tank where vigorous growth of aerobic microbes (floes) takes place. BOD is reduced (microbes consume a major part of organic matter), effluent is passed to a settling tank where floe sediment produces activated sludge, and sludge is pumped to an anaerobic sludge digester to digest bacteria and fungi.

Question 20. Identify A, B, C, D, and E and find the following table :
Answer:

Microbes In Human Welfare Cyclosporin

  1. Streptokinase
  2. ‘clot buster’
  3. cyclosporin A
  4. Trichoderma  polysporrum.
  5. Monascus  purpureus
  6. Blood-cholesterol-lowering agents.

Microbes In Human Welfare Long Question And Answers

Question 1. Describe the process of secondary treatment given to municipal wastewater (sewage) before it can be released into fresh waterbodies. Mention another benefit provided by this process.
Answer:

Process of secondary treatment Passing of primary effluent into a large aeration tank which is constantly agitated mechanically and air is pumped into it allowing vigorous growth of useful aerobic microbes into floes

Microbes consume a major part of organic matter in effluent which significantly reduces BOD

Now effluent is passed into a settling tank where floes are allowed to settle or sediment called activated sludge

Digestion of activated sludge by anaerobic microbes and effluents from secondary treatment can be released into river or stream

This resulted in the production of Biogas (CH4, IBS and CO2) which can be used as a source of energy.

Class 12 Biology Chapter 7 Human Health and Disease

Human Health And Disease Very Short Question and Answers

Question 1. write the scientific name of the causative Lent of pneumonia in humans and mention one specific symptom of the disease.
Answer:

Bacteria like Streptococcus pneumoniae and Haemophilus influenzae are responsible for die disease pneumonia. The symptoms of pneumonia include fever, chills, cough and headache.

Question 2. Expand MALT and mention any one location of it in the human body.
Answer:

The mucosa-associated lymphoid tissue (MALT), also called mucosa-associated lymphatic tissue, is a diffuse system of small concentrations of lymphoid tissue found in various submucosal membrane sites of the body, such as the gastrointestinal tract, nasopharynx, thyroid, breast, lung, salivary glands, eye, and skin.

Question 3. For early detection of cancer, 3-D images of tissues are essential. Name the technique and the basis on which it can generate a three-dimensional image of changes in the IMtig tissue.
Answer:

Medical imaging is the technique and process of imaging the interior of a body for clinical analysis and medical intervention, as well as visual repair of the juice the organs
Of tissues (physiology). Medical imaging seeks to reveal internal! structures are hidden by the skin and bones, as well as to diagnose and treat disease.

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Question 4. The main barrier that prevents the entry of micro-organisms into our body is

  1. Antibodies
  2. Macrophages
  3. Monocytes
  4. skin

Answer: 4. Skin

Question 5. Colostrum provides passive immunity to human infants as it contains antibody

  1. IgA
  2. IgM
  3. tgE
  4. igG

Answer: 1. IgA

Question 6. Name any two physiological barriers that provide innate immunity. 
Answer:

Acid in Stomach, Saliva in mouth

Question 7. Name the pathogen which causes Typhoid. Name the test that confirms the disease.
Answer:

Salmonella typhl,w idal test

Question 8. How does the body respond when harmony in produced hy is released in its blood?
OR
Write the role of interferons.
Answer:

Chill and high fever occur, in regular intervals every 3 to 4 days.
OR
Virus-infected cells secrete proteins called interferons, which protect non-infected cells (from further viral infection )

Question 9. Name two diseases whose spread can be controlled by the eradication of Aedes mosquitoes.
Answer:

Dengue and Chikunguniya.

Question 10. How do cytokine barriers promote innate immunity in humans?
Answer:

Interferon (proteins) which is secreted by virus-infected cells protect non-infected cells from further viral infection

Question 11. Flow do monocytes act as a cellular barrier in humans to provide innate immunity?
Answer:

Phagocytosis of microbes or destroy microbes.

Question 12. Name the condition in vertebrates where the body attacks self-cells.
Answer:

Autoimmunity.

Human Health And Disease Short Questions and Answers

1. A body developed some allergic reactions when he straight entered his air-conditioned room after a game of football outside his house. Write any two symptoms that could be noticed in such a condition. How does our body combat such conditions?
Answer:

Symptoms of allergic reactions include sneezing, watery eyes, runny nose and difficulty in breathing. The immune system overreacts by producing antibodies called Immunoglobulin E (IgE). These antibodies travel to cells that release chemicals, causing an allergic reaction

Question 2.

1. Write the Scientific name of the plant from where natural cannabinoids are obtained.

2. Mention the parts of the plant that are used for extracting the drug.

3. How does the drug affect the human body?
OR
The epithelial lining of our intestine is considered a secondary lymphoid organ. Justify the statement.
Answer:

1. Condition saliva

2. Cannabinoids are obtained from the inflorescence of the plant.

3. These drugs affect the cardiovascular system of the body. They affect brain areas that influence pleasure, memory, thinking, concentration, movement and coordination.
OR
There is lymphoid tissue also located within the lining of the major tracts (respiratory, digestive and urogenital tracts) called mucosal-associated lymphoid tissue (MAtT) It constitutes about 50 per cent of the lymphoid tissue in the human body

Question 3. The major cause of death among infants is because of Vcute Respirators Infection ( XRI) of the lower respiratory tract, which further affects the alveoli. Name the associated disease, and a causative bacteria and give any two symptoms of the disease.
Answer:

The associated disease is pneumonia and the most common bacteria causing pneumonia is Streptococcus pneumoniae.

Symptoms:-

Fast or Laboured Breathing – Breathing patterns would be rapid but shallow, directed from the stomach instead of the chest, accompanied by wheezing.

Pale Skin – The skin around the lips and face starts turning blue (a sign of decreased oxygen in the bloodstream).

Pain – Depending on the infected part, he will experience pain in the lungs or the abdomen. Especially when coughing or breathing deeply

Question 4.

1. In our country YVC’O and other Non-Government Organisations (NGOs) are doing a lot to educate people about AIDS. Expand ’NACO’. Enlist any three ways by which transmission of HIV- AIDS infection occurs.
Answer:

National Aids Control Organization

Ways of Transmission

  • Mother-to-child transmission is the most common way that children get HIV & This is called perinatal transmission It is less common because of advances in HIV prevention and treatment,
  • Used needles, syringes, and other injection equipment may have someone raise blood on them, and blood can carry HIV  with someone who has HIV
  • The primary way in which HIV’ is transmitted from person to person is when a partner has unprotected vaginal or anal sex with an HIV-positive partner.

2. write the scientific name of the plant and the part from which opioids are extracted. How does it affect our body functions?
Answer:

  • Opioids are extracted from the late plant called tie opium poppy, whose scientific name is Papaver souvenir.
  • Receptors are present m the central nervous system, cardiomiscible system and gastrointestinal tract.
  • Opioids act as a depressant and slow down body functions.

Question 5. It is often observed that the chances of a person suffering from measles in his or her lifetime are low if he or she has suffered from the disease in their early childhood. Justify the statement.
Answer:

Memory cells develop during measles in early childhood, subsequent encounters with the same pathogen elicit a highly intensified secondary anamnestic response.

Question 6. A student on a field trip suddenly felt breathlessness and started to sneeze very badly. Name this response and explain what it is due to.
Answer:

Allergy Allergens (dust or pollen mites etc.) are due to the release of chemicals like histamine serotonin (from mast cells).

Question 7. Causative organisms of some diseases gain entry into the human body through mosquito bites and make humans suffer from the disease. Name one such: 

  1. protozoan disease along with the scientific name of the causative organism.
  2. Helminths disease along with the scientific name of the causative organism.

Answer:

  1. Malaria → Plasmodium vivax Plasmodium malaria Plasmodium falciparum
  2. Elephantiasis/Filariasis → Wuchereria bancrofti Wuchcicria malayi

Question 8. Write the functions of bone marrow as the primary lymphoid organ and lymph nodes as the secondary lymphoid organs.
Answer:

Bone Marrow-lymphocytes are produced here, develop and mature into antigen-sensitive lymphocytes

Lymph nodes – trap the microorganism or antigens from the tissue fluid, the trapped antigens activate the lymphocytes (present in lymph nodes) to cause an immune response.
OR
What is a vaccine? State the type of immunity that it induces.
Answer:

The vaccine is a weakened/inactivated pathogen, or its antigenic protein, Active immunity

Question 9.

1. Name the source plant of heroin drug. Mow is it obtained from the plant?
Answer:

Papaver somtiifermnlPoppy plant.

Extracted from the latex of the plant or acetylation of morphine (obtained from the latex of the plant)

2. Write the effects of heroin on the human body.
Answer:

Depressant slows down body function

Question 10. Why is the structure of an antibody molecule represented as ILL2? Name any two types of antibodies produced in the human body.
Answer:

L 2 – Two light or small polypeptide chains,

Ha,” two heavy or longer polypeptide chains.

I gA / I gM / I gE / I gG

Question 11. Mention one application for each of the following :

  1. Passive immunization
  2. Antihistamine
  3. Colostrum
  4. Cytokinin-barrier

Answer:

  1. Provide preformed antibodies anti-toxins for quick response in ca microbes! tetanus) or snake bite.
  2. Reduces symptoms of allergy
  3. Provides passive immunity, antibodies or IgV to newborns.
  4. Protection of non-infected cells from further viral infection

Question 12.

1. Write the complete name of the diagnostic test for AIDS. Explain the principle it works on.
Answer:

ELISA Enzyme Linked Immune Sorbent Assay. It is based on the principle of antigen-antibody interaction where a pathogen can be detected by the presence of antibodies (proteins, glycoproteins, etc.) on it

2. Name the type of genetic material present in the AIDS-causing pathogen.
Answer:

MIX is a retrovirus, which means it carries single-stranded RNA as its genetic material rather than the double-stranded DNA human cells carry.

Question 13. A patient complains of suffering from constipation, stomach ache, stool with blood clots and excess mucous. The physician diagnosed it as amoebiasis, after a stool test.

  1. Write the scientific name of the microbe identified in the stool sample.
  2. How do you think, the patient must have contracted it?
  3. Write your suggestions to the patient to avoid infection in future.

Answer:

  1. Entamoeba histolytica
  2. I houseflies act as mechanical carriers and serve to transmit the parasite from the faeces of infected persons to food and food products, thereby contaminating them. Drinking water and food contaminated by faecal matter are the main sources of infection.
  3. Perform hand hygiene frequently, especially before handling food or eating, and after using the toilet or handling faecal matter Wash hands with liquid soap and water, and rub for at least 20 seconds.

Question 14. Name any two techniques which can be used to generate a three-dimensional image to detect cancers of the internal organs and explain any one of the techniques.
Answer:

Techniques like radiography (use of X-rays), CT (computed tomography) and MRI (magnetic resonance imaging) are very useful in detecting cancers of the internal organs.

Magnetic resonance imaging (MRI): Magnetic resonance imaging (MRI)  is a spectroscopic imaging technique used in medical settings to produce images of the inside of the human body MRI is based on the principles of nuclear magnetic resonance (NMR). which is a spectroscopic technique used to obtain microscopic chemical and physical data about molecules.

Magnetic resonance imaging is accomplished through the absorption and emission of energy of the radio frequency (RF) range of the electromagnetic spectrum

MRI Uses

  1. Utilizes non-ionizing radiation (unlike X-rays)
  2. Ability to image in any plane, (unlike CT scans).
  3. Very low incidents of side effects.
  4. Ability to diagnose, visualize, and evaluate various illnesses.

Question 15. Given Below Is A Diagrammatic Representation Of the Immune System Of The Human Body:

Hunan Health And Disease Diagrammatic Representation Of Immune System

  1. Identify ‘ X’and ’Y’ in the given diagram.
  2. Explain two major functions of the organs that you have identified,

Answer:

1. X – Lymph Nodes

Y – Thymus

2. Lymph Nodes –

  1. The removal of excess fluids from body tissues.
  2. Absorption of fatty acids and subsequent transport of fat, and chyle, to the circulatory system
  3. Production of immune cells (such as lymphocytes, monocytes, and antibody-producing cells called plasma cells).
  4. There is lymphoid tissue also located within the lining of the major tracts (respiratory, digestive and urogenital tracts) called mucosa-associated lymphoid tissue (MALT) It constitutes about 50 per cent of the lymphoid tissue in the human body.

Thymus – The thymus gland is a lobular structure located between the lungs behind the sternum on the ventral side of the aorta.

  • The thymus plays a major role in the development of the immune system.
  • This gland secretes the peptide hormones called thymosins.
  • Thymosins play a major role in the differentiation of T – T-lymphocytes, which provide cell-mediated immunity.
  • Thymosins also promote the production of antibodies to provide humoral immunity.
  • Thymus is degenerated in old individuals resulting in a decreased production of thymosins.
  • As a result, the immune responses of old persons become weak.

Question 16. Answer the following questions concerning “opioids”, the commonly abused drug :

  1. Where in our body are the specific opioid receptors present?
  2. What is heroin chemically known as?
  3. Write the scientific name of the plant from which opioids are extracted.

Answer:

  1. Opioids are drugs, which bind to specific opioid receptors present in our central nervous system and gastrointestinal tract.
  2. Heroin commonly called smack is chemically known as diacetylmorphine.
  3. Papaver sonmifentni.

Question 17. Name the two primary lymphoid organs in humans. Explain their functions in providing immunity.
Answer:

Bone Marrow: The main lymphoid organ where all blood cells including lymphocytes are produced

Thymus:

  1. A lobed organ located near the heart arid beneath the breastbone
  2. Quite large at the time of birth but keeps reducing in size with age and by the time puberty is attained it reduces to a very small size

Question 18.

  1. Name the causative agents of pneumonia and the common cold.
  2. How do these differ in their symptoms?
  3. Mention two symptoms common to both.

Answer:

1. Streptococcus pneumoniae Haemophilus influenza, rhinoviruses

2. Different symptoms

Human Health And Disease Different Symptoms

3. Common symptoms

Hunan Health And Disease Common Symptoms

Or

  1. Write the scientific names of the causative agent and vector of malaria, and write its symptoms.
  2. Name any two diseases spread by Aedes sp.

Answer:

  1. Plasmodium vivax P. falciparum P.malariae,vector-female Anopheles mosquito Symptoms chills, high fever
  2. Dengue, Chikungunya

Question 19. Name is a human disease, its causal organism, symptoms (any three) and sector, spread by intake of water and food contaminated by human faecal- matter. 
Answer:

Amoebiasis (Amoebic dysentery). Entamoeba histolytica, constipation abdominal pain cramps or stools with excess mucus or blood clots & causing agent is Housefly.
OR
Ascariasis. A scar is, internal bleeding muscular pain fever anaemia or blockage of intestinal passage and The causing agent is Housefly.
OR
Typhoid, Salmonella n phi, high fever weakness stomach pain or constipation headache or loss of appetite and the causing agent is Housefly.
OR

  1. Why is there a fear among the guardians that their adolescent wards may get trapped in drug or alcohol abuse?
  2. Explain ‘addiction” and ‘dependence’ in respect of drug/alcohol abuse in youth.

Answer:

  1. Adolescents are easily affected by this. (vulnerable to) peer pressure adventure or curiosity excitement or experimentation or media
  2. Addiction- Psychological attachment to certain effects such as Euphoria or temporary feeling of well-being Dependence- Tendency of the body to show withdrawal syndrome or symptoms if regular doses of drug/alcohol are abruptly discontinued.

Question 20.

1. What is an “allergic reaction”?
Answer:

The exaggerated response of the immune system to certain antigens present in the environment is called allergy The substances to which such an immune response is produced are called allergens.

2. Name any two drugs used to quickly reduce the symptoms of allergy.
Answer:

Anti-histamine, adrenalin and steroids

3. Why do more and more children in metro cities of India suffer from allergies and asthma?
Answer:

More and more children in metro cities of India suffer from allergies and asthma due to sensitivity to the environment. This could be because of the protected environment provided early in life.

Human Health And Disease Long Question And Answer

Question 1. It is commonly observed that parents feel embarrassed to discuss freely with their adolescent children about sexuality and reproduction. The result of Ibis parental inhibition is that the children go astray sometimes. 

  1. Explain the reasons that you feel are behind such embarrassment amongst some parents to freely discuss such issues with their growing children.
  2. By taking one example of a local plant and animal, how would you help these parents overcome such inhibitions about reproduction and sexuality?

Answer:

  1. Illiteracy or conservative attitudes or misconceptions social myths etc. are the main reasons.
  2. If a student gives the clarity of the concept of reproduction and sexuality by taking an example of a plant and an animal to reproductive organs, gamete formation fertilization, sexual behaviour etc.

Question 2.

1. Differentiate between active and passive immunity.
Answer:

Human Health And Disease Difference between Active Immunity And Passive Immunity

2. Comment on the role of vaccination and immunization in keeping the human population healthy.
Answer:

Role of vaccination or immunization:

  • Antibodies produced in the body against antigens neutralize pathogenic agents.
  • Vaccines also generate memory cells (B and T cells) that recognize quickly subsequent exposure and control the growth of pathogens with massive production of antibodies.
  • Preformed antibodies or antitoxin protect our body from deadly microbes like tetanus and against snake venom.

Class 12th Biology Important Questions Chapter 6 Evolution

Evolution Very Short  Question And Answers

Question 1. Mention one example each from plants and animals exhibiting divergent evolution.
Answer:

Thorn of Bougaimilira and tendrils of Cucurbita, forelimbs of whales bats, cheetahs,s and humans (all mammals or vertebrate hearts or vertebrate drains

Question 2. Write the names of the following:

  1. A 15 ape-like mya primate
  2. A 2 mya primate that In ed in East African grasslands

Answer:

  1. Dryopithecus
  2. Australopithecines or Australopithecus or Homo habilis

Question 3. State two postulates of Oparin and Haldane concerning the origin of life.
Answer:

  1. The first form of life could have come from existing non-living organic molecules RNA And Protein
  2. Formation of life was preceded by chemical evolution or formation of diverse organic, molecules from inorganic coast entrants

Evolution Short Question And Answers

Question 1. Wings of birds and wings of butterflies contribute to locomotion, and the evolution of such organs is a result.
Answer:

They are not anatomically similar in structure but perform the same function, hence these are analogous structures type of evolution is convergent evolution similar habitats of birds and butterflies have resulted in. the selection of similar adaptive features (wings.) in different groups of organisms, but towards the same function convergent evolution.

Read and Learn More Important Questions for Class 12 Biology Chapter Wise

Question 2. According to the Darwinian theory of natural selection, the rate of appearance of new forms is linked to the life cycle or the life span of an organism. Explain with the help of an example.
Answer:

A colony of bacteria (say A) growing in a given medium has built-in variation in terms of the ability to utilize a feed component, a change in the medium composition would bring out only that part of the population B) that can survive under the new conditions In due course of time this variant population outgrows the others and appears as new species thus organisms with shorter life-cycle or life-span will undergo evolution faster or for the same thing to happen in fish or fowl would take millions of years as life spans of these animals are in years.

Question 3. Explain the Ilardy-Weinberg principle with the help of an algebraic equation.
Or
With the help of an algebraic equation, how did Ilardy-Weinberg explain that in a given population the frequency of occurrence of alleles of a gene is supposed to remain the same through generations?
Answer:

The Principle says that allele frequency in a population is stable and is constant from generation to generation, the gene pool remains constant and expressed as p2 +2pq+ q2/ ( p+q)2

  • Where p2 = frequency of individuals with A A genotype
  • Where q2= frequency of individuals with a genotype
  • Where 2pq = frequency of individuals with Aa genotype

Question 4. What is adaptive radiation? How did Darwin explain it?
Answer:

The process of the evolution of different species in a given geographical area starting from a point and radiating to other areas of geography (habitats) is called adaptive radiation He observed that there were many different varieties of finches (Darwin’s finches) in the same island, and all those varieties evolved ou the island itself, from the original seed-eating features many other forms with altered beaks arose to become insectivorous or vegetarian finches.

Question 5.

1. Differentiate between analogous and homologous structures.
Answer:

  1. Aitalogous-Anatomically embryonic development, origin, and structure are not similar though perform similar functions and are a result of convergent evolution.
  2. Honiologus Anatomically embryonic development, origin, and structure are similar (but perform different functions) and are a result of divergent evolution.

2. Select and write analogous structures from the list given below :

  1. Wings of butterflies and birds
  2. Vertebrate hearts
  3. Tendrils of bougainvillea and cucurbita
  4. Tubers of sweet potato and potato

Answer: 1 Or 4

Question 6. How can the Hardy- Weinberg Equilibrium be affected? Explain giving three reasons.
Answer:

  1. Gene Migration Or Gene Flow: When the migration of a section of the population occurs to another place and gene frequencies change in the original as well as in the new population.
  2. Genetic drift: If the same change occurs by chance or new genes or alleles are added to the new population and these are lost from the old population.
  3. Mutation: Pro existing advantageous mutations when selected will result in new phenotypes.
  4. Genetic recombination: Variation in characteristics will be there because of genetic recombination, during meiosis and also due to random fusion of gametes.
  5. Natural selection: Heritable variations enabling better survival enabled organisms to reproduce and leave a greater number of progeny.

Question 7. Write the characteristics of Raniapithecus, Drvopitheeiis, and Neanderthal man.
Answer:

  • Raniapithecus: I fairy Or walked like gorillas and chimpanzees, more man-like.
  • Drvopitheeiis: Hairy or walked like gorillas and chimpanzees, more aped ike.
  • Neanderthal man: brain size is 1400cc, used hides to protect their body or buried their dead,

Question 8. Excessive and continuous use of pesticides has resulted in the evolution of some new species of pests. Explain what must have led to this. What is this type of evolution called?
Answer:

  • Excess use of herbicides, pesticides, etc. only resulted in the selection of resistant varieties in a much lesser time scale.
  • Hence, resistant organisms seel Is appear in a time scale of months or years and not centuries.
  • These are examples of evolution by anthropogenic action.
  • This also tells us that evolution is not a directed process in die sense of determinism.
  • It is a stochastic process based on chance events in nature and chance mutation in the organisms.

Question 9. What type of organs eye of an Octopus and that of a human called? Give another example from the animal group and one from the plants of such organs. Name and explain the evolutionary process they exhibit.
Answer:

  • The eyes of the octopus and mammals are an example of analogous organs. Sweet potato (root modification) and potato.
  • Wings of butterflies and birds.
  • The similar habitat has resulted in the selection of similar adaptive features in different groups of organisms but toward the same function, analogous structures are a result of convergent evolution – different structures evolving for the same function and hence having similarity.

Evolution Long Question And Answers

Question 1.

1. Describe the observations made on the collection of white-winged moths and dark-winged moths in England between the years 1850 and 1920. What did these observations lead to?
Answer:

Before industrialization tree bark was covered with light-colored lichens, In this background white-winged moth survived but dark colored moth was picked out. by predators, post-industrialization tree trunks became- dark due to industrial smoke and soot, under this condition, and the white-winged moth did not survive due to predators, while the dark-winged/melanized moth survived, this showed that organism that is better adapted to survive are selected by Nature/Natural selection.

2. How is the use of herbicides, pesticides, and antibiotics by humans for various purposes, comparable with the observations made on moths in the above question? What is this type of phenomenon called?
Answer:

Excessive use of these chemicals has resulted in the selection of resistant varieties, in u much lesser time (scale).

These are examples Of evolution by anthropogenic action.

Question 2. Describe S.E. Miller’s experiment. Comment on the observations he made and his contribution towards the origin of life on Earth.
Answer:

High temperature (800″C), high energy radiation, reducing atmosphere created, by electric discharge in a closed flask, containing CH4, IE. NIC. and water vapors in the experimental setup.

Evolution Diagrammatic Representation Of Miller's Experiment

Observation and Contribution

  • Formation of amino acids
  • The first form of life arose slowly through evolutionary forces in front of living molecules or abiogenesis.

Question 3. Taking an example of white-winged moths and dark-winged moths of England in pre and post-industrialised eras, explain evolution by natural selection. 
Answer:

An interesting observation supporting evolution by natural selection comes from England In a collection of moths made in the 1850s, therefore before industrialization set in, it was observed that there were more white-winged moths on trees than dark-winged or melanic moths.

  • However, in the collection carried out from the same area, but after industrialization, in 1920, there were more dark-winged moths in the same area, therefore the proportion was reversed.
  • The explanation put forth for this observation was that ‘predators will spot a moth against a contrasting background’.
  • During the post-industrialization period, the tree trunks became dark due to industrial smoke and soot. Under this condition, the white-winged moth did not survive due to predators, dark winged or melanised moth survived.
  • Before industrialization set in, thick growth of almost white-colored lichen covered the trees – in that background, the white-winged moth survived but the Clark color. red moth was picked out by predators.
  • Lichen will not grow in areas that are polluted. Hence, moths that were able to camouflage themselves, therefore, hide in the background, survived.
  • This understanding is supported by the fact that in areas where industrialization did not occur example in rural areas, the count of matric moths was low.
  • This showed that in a mixed population, those that can better adapt, survive and increase in population size. Remember that no variant is completely wiped out.

Important Questions for Class 12 Biology Chapter 5 – Molecular Basis of Inheritance

Molecular Basis Of Inheritance Very Short Question And Answers

Question 1. Given the low are the observations drawn in HGP. Select the options that show the correct observations.

  1. The human genome contains 3164.7 billion base pairs.
  2. The average gene consists of 3000 bases.
  3. Less than 2% of the genome codes for proteins.
  4. Chromosome one has the most genes ( 2698 )
    1. 1 and 2
    2. 2 and 3
    3. 3 and 4
    4. 1 and 3

Answer: 2. 2 and 3

Question 2. The phosphodiester linkage in the formation of a nucleotide involves the bonding between

  1. Phosphate group and OH of 3’C of a nucleoside
  2. Phosphate group and OH of 5’C of a nucleoside
  3. Phosphate group and H of 3’C of a nucleoside
  4. Phosphate group and H of 5’C of a nucleoside

Answer: 2. Phosphate group and OH of 5’C of a nucleoside.

Question 3. The switching ‘on’ and ‘off of the lac operon in prokaryotes is regulated by

  1. Glucose
  2. Galactose
  3. Lactose
  4. Fructose

Answer: 3. Lactose.

Question 4.For ‘in-vitro’ DNA replication, which one of the following substrates needs to be added along with the necessary enzymes the DNA template, and specific conditions?

  1. Ribonucleotide triphosphate
  2. Deoxyribonucleoside triphosphate
  3. Deoxyribonucleotide triphosphate
  4. Ribonucleoside triphosphate

Answer: 2. Deoxy r i bo nu Pelosi triphosphate

Read and Learn More Important Questions for Class 12 Biology Chapter Wise

Question 5. Which one of the following factors will associate transiently with RNA polymerase to terminate transcription in prokaryotes?

  1. Sigma factor
  2. Rho factor
  3. Delta factor
  4. Theta factor

Answer: 2. Rho factor

Question 6.Choose the correct pair of codons with their corresponding amino acids from the following list:

  1. UAG: Glycine
  2. AUG: Arginine
  3. UUU: Phenylalanine
  4. UGA: Methionine

Answer: 3. UUU: Phenylalanine

Question 7. During the elongation process of translation, the peptide bond formation between amino acids is catalyzed

  1. Ribosomal RNA
  2. Protein in the small subunit of the ribosome
  3. Protein to the large subunit of the ribosome
  4. Transfer RNA

Answer: 1 . Ribosomal RNA

Question 8. A region of the coding strand of DNA has the following nucleotide sequence: 5-TGCGCCA – 3’ The sequence of bases on niRNA transcribed by this DNA strand would be:

  1. 3′ – ACGCGGT – 5′
  2. 5′ – ACGCGGT – 3′
  3. 5′ – UGCGCCA – 3’
  4. 3′ – UGCGCCA – 5′

Answer: 3.  5′ – UGCGCCA – 3′

Question 9. A DNA molecule is 160 base pairs long. It has 20# adenine. How many cytosine bases are present in this DNA molecule?

  1. 192
  2. 96
  3. 64
  4. 42

Answer: 2. 96

Question 10. A template strand in a bacterial DNA has the following base sequence:
5′ – TTTAACGAGG – 3′

  1. 5′ – AAATTGCTCC- 3′
  2. 3′ -AATTGCTCC-5′
  3. 3′ – AAAUUGCUCC – 3’
  4. 5’ – CCUCGUUAAA – 3′

Answer: 3. 5′ – CCUCGUUAAA – 3′

Question 11. tRNA has an that has bases complementary to the codon. Its actual structure is a compact molecule which looks like. Select the option that has the correct choices for the two ‘blanks’

  1. Amino acid acceptor end, clover-leaf
  2. Anticodon loop, clover-leaf
  3. Amino acid acceptor end, inverted L
  4. Anticodon loop, inverted L

Answer: 3. Anticodon loop, inverted L

Question 12. Which type of RNA is correctly paired with its function? 

  1. Small nuclear RNA – Processes rRNA
  2. Transfer RNA: attaches to amino acid
  3. Ribosomal RNA: involved in transcription
  4. Micro RNA: involved in translation

Answer: 2. Transfer RNA: attaches to an amino acid.

Question 13. The figure given below has labeling (1), (2), and (3), which two labeling in the given figure are components of a nucleosome? Select the correct option.

Molecular Basis Of Inheritance Nucleotide Histone

  1. 1 – H1 histone, 2 – DNA
  2. 1 – DNA, 2 – Histone Detainer
  3. 2 – DNA, 3 – HI Histone
  4. 2 – Histone octamer, 3 – DNA

Answer: 4. 2 – Histone octamer, 3 – DNA

Question 14. Which one of the following diagrams is a correct depiction of a Polynucleotide chain To DNA?

Molecular Basis Of Inheritance Polynucleotide Chain To DNA

Answer:  2

Question 15. In molecular biology who proposed that genetic information flows in one direction?

  1. Hargobind Khoraim
  2. Francis Crick
  3. Watson mid-Crick
  4. Marshall Nirenhcrg

Answer: 2. Francis Crick

Watson and Crick proposed that genetic information flows in one direction.

Question 16. Meseison and Stahl carried out centrifugation in CsC12 density gradient to separate :

  1. DNA from RN A
  2. DNA from protein
  3. The normal DNA from rN-DNA
  4. DNA from tRNA

Answer: 3. The normal DNA from rN-DNA

Question 17. Write the dual purpose served by Deoxyribonudeoside triphosphates in polymerization.
Answer:

Acts as a substrate towards the reaction and provides energy to the poh memorization ie energy source (from the initial two phosphates’)

Question 18. Name one amino add, which is coded by only one codon.
Answer:

Methionine Tryptophan

Question 19. Write the conclusion Griffith arrived at the end of his experiment with Streptococcus pneumoniae.
Answer:

He concluded that 11 ml of the R .sbaiu bacteria had somehow been transformed h\ the local ‘killed S strain bacteria.

Molecular Basis Of Inheritance Short Question And Answers

Question 1. Draw a labeled schematic representation of the Central Dogma of Molecular Biology as proposed by Francis Crick.
Answer:

Francis Crick proposed the Central dogma in molecular biology, which states that the genetic information flows from DNA → RNA → Protein

Molecular Basis Of Inheritance Central Dogma

Question 2. State four salient observations drawn from the Human Genome Project.
Answer:

  1. Some of the salient observations drawn from the human genome project are as follows :
  2. The human genome contains 3164.7 million bp.
  3. The average gene consists of 3000 bases, but sizes vary greatly, with the largest known human gene being dystrophin at 2.4 million bases.
  4. The total number of genes is estimated at 30,000-much lower than previous estimates of 80.000 to 1.40.000 genes. Almost: all (99.9 percent) nucleotide bases are the same in all people.

The functions are unknown for over 50 percent of the discovered genes.

Question 3. Given below is one of the strands of a DNA segment :

Molecular Basis Of Inheritance DNA Segment

1. Write its complementary strand.
Answer:

Molecular Basis Of Inheritance Polarity Nucleotide Sequance(Polarity, Nucleotide sequence)

2. Write a possible RNA strand that can be transcribed from the above DNA molecule formed.
Answer:

Molecular Basis Of Inheritance Polarity Nucleotide Sequance DNA(Polarity, Nucleotide sequence)

Question 4. Draw a schematic diagram of a transcription unit with the polarity of the DNA strands and label the coding strand, template strand, and terminator.
Answer:

Molecular Basis Of Inheritance Polarity Of The DNA Strands

Question 5. A segment of a DNA molecule comprises 546 nucleotides. How many cytosine nucleotides would be present in it if the number of adenine nucleotides is 96?
Answer:

\(\mathrm{A}+\mathrm{T}=\mathrm{C}+\mathrm{G} \text {, Given } \mathrm{A}=96 \text { so } \mathrm{T}=96 \text {, and } \mathrm{A}+\mathrm{T}=192\)

⇒\(\begin{aligned}
& \text { Given total Nucleotides }=546 \\
& \mathrm{G}+\mathrm{C}=546-192=354 \text { because } \mathrm{G}=\mathrm{C} \text { so } \mathrm{C}=354 / 2, \\
& \text { Cytosine }=177
\end{aligned}\)

6. Although a prokaryotic cell has no defined nucleus, DNA is not scattered throughout the cell ‘Explain.
Answer:

DNA is negatively charged & positively charged proteins and can hold it in places and large loops (in a region termed as nucleoid)

Question 7. Differentiate between the genetic codes given below :

  1. Unambiguous and Universal
  2. Degenerate and Initiator

Answer:

Molecular Basis Of Inheritance Universal And Unambiguous

Question 8.

1. Draw a polynucleotide chain (four nucleotides long) of DNA having four variable
nitrogenous bases.
Answer:

Molecular Basis Of Inheritance Nitrogenous Bases

2. Draw a neat labeled diagram of the nucleosome. Name the basic amino acid residues present mainly in the nucleosome.
Answer:

Molecular Basis Of Inheritance Nucleotide

Amino Acid: Lysine and Arginine

Question 9.

1. Explain the events occurring in a ‘Replicating Fork” during the replication of DNA.
Answer:

DNA-dependent DNA polymerase catalyzes the polymerization, of deoxynucleotides Or deoxyribonucleotide triphosphates.

  • The polymerization takes only in one direction 5’→3‘, on one strand (the template with polarity 3′ .5) the replication is continuous,  while on the other (the template with polarity 5″ →3′) it is discontinuous, – the discontinuous)’ synthesized fragments are later joined by the enzyme DNA Haase.

2. Name the different types of RNA polymerases in a eukaryotic cell. Write their roles in transcription.
Or
3. Name the three RNA polymerases found in eukaryotic cells and mention their functions.
Answer:

RNA polymerase 1 →transcribes rRNAs(28S or 18S or 5.8S).

RNA polymerase 2→ transcribes precursor of mRNA or hnRNA or heterogenous nuclear RNA

RNA polymerase 3 → transcribes tRNA or 5srRNA or snRNAs or small nuclear RNAs

Question 10. Explain the mechanism of translation that occurs in the ribosomes in a prokaryote.
Answer:

  1. Charging of tRNA Or aminoacylation of tRNA. the small subunit of ribosome binds to RNA
    (5′ end), for initiation, the ribosome binds to the mRNA at the start codon (AUG) that is recognized only by initiator tRNA
  2. In the elongation phase, an amino acid with tRNA sequentially binds to the appropriate codon on mRNA(forming complimentary base pairs with tRNA anticodon),
  3. Ribosome moves from codoil to codon along the mRNA and amino acids are added one by one in the two sites of the large subunit joined by a peptide bond.
  4. Termination occurs when a release factor binds to the stop codon and releases the complete po 1 ypepti de.

Question 11. Explain the role of regulatory genes in a lac operon. Why is the regulation of lac operon called negative regulation?
Answer:

  • Regulatory gene-gene codes for the repressor of the lac operon, the repressor protein (synthesized by the I gene, binds to the operator site of the operon, and prevents the RNA polymerase from transcribing the operon
  • The repressor of the lac operon is synthesized constitutively all 1 the time, and thus the operon is in a switched-off position generally, it is switched on only when lactose is present in the culture medium of the E.coli when the operon gets ‘switched on’

Question 12.

1. Expand VNTR and describe its role in DNA fingerprinting.
Answer:

VNTR -Variable Number of Tandem Repeat(s) -used as a probe (because of its high degree of polymorphism)

2. List any two applications of DNA fingerprinting techniques.
Answer:

Forensic science or criminal investigation (any point related to forensic science (determining population and genetic diversity paternity testing or maternity testing or study of evolutionary biology.

13. Why is the DNA molecule a better hereditary material than the RNA molecule?
Answer:

The DNA molecule is a better hereditary material.

  1. It is more stable (due to the presence of thymine and not uracil as in RNA).
  2. Less reactive than RNA (as RNA has 2″ – Oil making it more reactive).
  3. Being less reactive, DNA is not easily degradable (RNA being more reactive is easily degradable).
  4. The rate of mutation is slow (The rate of mutation in RNA is faster)

Question 14. Explain the post-transcriptional modifications the hn-RNA undergoes in eukaryotic cells.
Answer:

  1. Splicing → lustrous are removed and exons are joined
  2. Capping → Methyl guanosine triphosphate Or map is added to the 5’ end of hnRNA
  3. Tailing → Polyadenylate residues are added to 3’etui in a template-independent manner

Question 15.

1. List the two methodologies involved in the human genome project. Mention how they were used.
Answer:

Expressed Sequence Tags, Identifying all the genes that are expressed as RNA Sequence Annotation, sequencing the whole set of genome coding or noncoding sequences, and later assigning different regions with functions.

2. Expand ’YAC and mention what was it used for.
Answer:

Yeast Artificial Chromosomes used as cloning vectors (cloning/amplification)

Question 16. Identify, A, B, C, D, E, and F In the table given below :
Answer:

Molecular Basis Of Inheritance Nitrogenous Base

  1. Nitrogenous base
  2. OH of PC pentose sugar
  3. N-glycosidic linkage
  4. phosphate group
  5. phosphodiester linkage
  6. 3′-5’ phosphodiester linkage

Molecular Basis Of Inheritance Long Question and Answers

Question 1.

1. How did Matthew Meselson and Franklin Stall? experimentally prove that DNA replication
Answer:

Matthew Meselson and Franklin Stahl performed the following experiment in 1958 : (They grew E.coli in a medium containing NiliCl ( ‘.NHs the heavy isotope of nitrogen) as the only nitrogen source for many generations. The result was that is N was incorporated into newly synthesized DNA (as well as other nitrogen-containing compounds). This heavy DNA molecule could be distinguished from the normal DNA by centrifugation in a cesium chloride (CsCl) density gradient (Please note that 15N is not a radioactive isotope, and it can be separated from 14N only based on densities).

2. The use of heavy isotope of nitrogen by Meselson and Stahl demonstrated a semi-conservative mode of replication of a DNA molecule.” Explain how they arrived at this conclusion.
Answer:

Then they transferred the cells into a medium with normal “14NH4Cl and took samples at various definite time intervals as the cells multiplied, and extracted the DNA that remained as double-stranded helices. The various samples were separated independently by oil CsCl gradients to measure the densities of DNA.

Molecular Basis Of Inheritance Meselson And Stahl Experiment

Question 2.

1. Name and describe the technique which is an important tool of forensic science.
Answer:

The technique is DMA fingerprinting DNA profiling DNA typing Or DNA test.

  1. It is a technique to identify a person based on his/her DNA specificity.
  2. This technique was invented by Sir Alec Jeffery (1984).
  3. In India, DNA Fingerprinting has been started by Dr. V.K. Kashyap & Dr. Lai Ji Singh.
  4. DNA of humans is almost (99.9%) the same but a very small amount (0.1 that differs from person to person,
  5. The human genome has 3.3 x 10 bp
  6. 0.1 % different = 3.3 x 1 bp
  7. These differences are mainly due to – Repetitive DNA sequences.
  8. These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation.
  9. Differences in DNA sequences.
  10. Arises due to mutations.
  11. Allelic sequence variation has traditionally been described as a DNA polymorphism if more than one variant (allele) at a locus occurs in a human population with a frequency greater than 0.01.
  12. In simple terms, if an inheritable mutation is observed in a population at high frequency, it is referred to as DNA polymorphism.

Molecular Basis Of Inheritance Steps Of DNA Finger Printing

2. Mention any two applications of this technique other than its use in forensic studies.
Answer:

To determine the population and genetic diversities of the population.

Question 3. Describe the experiment carried out by Mershey and Chase. Write the conclusion they arrived at.
Or
1. Hershey and Chase carried out their experiment in three steps: infection, blending, and centrifugation. Explain each step.
2. Write the conclusion and interpretation of the result they obtained.

Answer:

They grew viruses on a medium containing radioactive phosphorus r P) and some on radioactive sulfur ( “Si radioactive D.NA contained in viruses grown on radioactive phosphorus, the radioactive protein contained in virus grown in radioactive sulfur. Radioactive phages were allowed to attach to E.rolibacteria, the virus coats were removed from bacteria by agitating in a blender, and the virus particles were separated from the bacteria on centrifugation. Bacteria infected with viruses containing radioactive DNA were radioactive, whereas bacterial cells infected with viruses containing radioactive protein did not show radioactivity.

Molecular Basis Of Inheritance The Hershey Chase Experiment

Conclusion – DNA is therefore the genetic material that is passed from virus to bacteria.

Question 4. List the different components of a Lac Operon. Explain the role of these components, when the operon is in an ‘open state’.
Answer:

Components: Regulatory gene/the T gene (inhibitor), and three structural genes i.e. z, y, and a Repressor, which is synthesized (all-the-time -constitutively from the I gene, is inactivated by interaction with the inducer (lactose or also lactose), this allows RNA polymerase access to die promoter, turning on the transcription of these three genes in the lac operon, which in turn produces enzymes responsible for digestion of lactose (ß – galactosidase break lactose to glucose and galactose)

Molecular Basis Of Inheritance The Lac Operon

Question 5. Write the five important goals of the Human Genome Project (HGP).
Answer:

Goals of HGP

  1. Identify all the approximately 20.000-25.000 genes in human DNA.
  2. Determine the sequences of the 3 billion chemical base pairs that make up human DNA:
  3. Store this information in databases
  4. Improve tools for data analysis
  5. Transfer related technologies to other sectors, such as industries:
  6. Address the ethical, legal, and social issues (El. SI) that may arise front the project.

Question 6. Compare the processes of DN A replication and transcription in prokaryotes.
Answer:

Similarities:

Both the processes involve –

  1. Unwinding of the helix and separating the two DNA strands
  2. Breaking the hydrogen bonds between the bases/pairs
  3. Follow the complementary base pair rule
  4. Polymerization occurs in a 5′ → 3′ direction
  5. linking Polvmeriaction of nucleotides

Dissimilarities:

  1. DNA replication Transcription
  2. DNA nucleotides added are RNA nucleotides added are ATP, GTP, CTP, TTP ATP, GTP, CTP. UTP
  3. Deoxyribose sugar is the part of Ribose sugar is the part of nucleotide Adenine pairs with Thymine Adenine with Uracil
  4. Both strands copied Only one strand copied.
  5. Resulting in two DNA molecules resulting in the formation of an RNA molecule (Any other correct dissimilarity)

Question 7.

1. Explain Griffith’s ‘transforming principle’ experiment.
Answer:

When Streptococcus pneumoniae (pneumococcus) bacteria are grown on a culture plate produce smooth shiny colonies (S) because the S strain bacteria have a mucus (polysaccharide) coat, Mice infected with the S strain(x indent) die from pneumonia infection while others produce rough colonies (R), but mice infected with the R strain do not develop pneumonia. Griffith observed that heat-killed S strain bacteria when injected into mice did not kill them, When he injected a mixture of heat-killed S and live R bacteria, the mice died. Moreover, he recovered living S bacteria from the dead mice.

2. In the above experiment, ‘’heat which killed one type of bacteria, did not destroy the properties of genetic material.’” Justify
Answer:

The two DN A strands get separated by heating and come together, when appropriate conditions are provided heat does not destroy the genetic properties.

Question 8.

1. State the ‘Central dogma’ as proposed by Francis (“rick. Are there any exceptions to it? Support your answer with a reason and an example.
Answer:

Molecular Basis Of Inheritance Central Dogma

Yes, in some viruses flow of information is in reverse direction/reverse transcription e.g. Retrovirus/HIV.

2. Explain how the biochemical characterization (nature) of the ‘Transforming Principle’ was determined, which was not defined in Griffith’s experiments.
Answer:

Protein DNA and RNA were purified from heat-killed S strain/smooth Streptococci! s/Diploco( cits pneumoniae

  1. Protein + Protease → transformation secured (R cell to S type)
  2. RNA + RNA use → transformation occurred (R cell to S type)
  3. DNA + DNA use → transformation inhibited

Hence, DNA alone is the transforming material

Question 9.

1. Why does DNA replication occur in small replication forks and not in its entire
length?
Answer:

DNA being very long, requires high energy for opening along its entire length

2. Why is DNA replication continuous and discontinuous in a replication fork?
Answer:

DNA-dependent DNA polymerase catalyzes polymerization only in one direction, i.e.Two strands of DNA are anti-parallel and have opposite polarity

3. State the importance of the origin of replication in a replication fork.
Answer:

The site where replication originates

Molecular Basis Of Inheritance Replication Originates

4. What is an operon? Explain the functioning of the lac option when in an open state.
Answer:

The arrangement where a (Polycistronic) structural gene is regulated by a common promoter and regulatory genes.

Lactose acts as an inducer and binds with repressor protein, RNA polymerase freely moves over the structural genes, and transcribes lac miRNA, which in turn produces enzymes-transacetylase. permease, ß-galactosidase (by lac z), is responsible for the digestion of lactose.

Molecular Basis Of Inheritance Digestion Of Lactose

Question 10.

1. Describe the structure and function of a t-RNA molecule. Why is it referred to as an adapter molecule?
Answer:

Clover-leaf shaped/inverted L-shaped molecules have an anti-codon loop with bases complementary to specific codons. has an amino acid acceptor end. It reads the code on one hand and binds with the specific amino acid on the other hand.

Molecular Basis Of Inheritance The Adapter Molecule

2. Explain the process of splicing of hn-RNA in a eukaryotic cell.
Answer:

Introns are removed, and exons are joined in a definite order called splicing.

Molecular Basis Of Inheritance Process Of Trascription In Eukaryotes

Class 12 Biology Chapter 4 Principles of Inheritance and Variation Important Questions

Principles Of Inheritance And Variation Very Short Questions And Answers

Question 1. The case of Down’s syndrome in humans is;

  1. Extra copy of an autosome
  2. Extra copy of a sex chromosome
  3. Absence of an autosome
  4. Absence of a sex chromosome

Answer: 1. Extra copy of an autosome

Question 2. Which of the following features shows the mechanism of sex determination in honeybees’

  1. An offspring formed from the union of a sperm and egg develops as a male
  2. Males have half the number of chromosomes than that of female
  3. The females are diploid having 32 chromosomes ”
  4. Males have fathers and can produce sons-.

Answer: 2. Males have half the number of chromosomes than females.

Question 3. Select the incorrect pair:

  1. Sickle-cell anaemia: Autosomes like recessive
  2. Haemophilia Autosome linked recessive trait,
  3. Colourblindness Sex sex-linked recessive trait
  4. Thalassemia: Autosome-linked recessive trait

Answer: 2. Haemophilia: Autosome-linked recessive trait

Read and Learn More Important Questions for Class 12 Biology Chapter Wise

Question 4. An example of a human trait where a single gene can exhibit multiple phenotypic expression is

  1. Phenylketonuria
  2. Cystic fibrosis
  3. Thalassemia
  4. Haemophilia

Answer: 1. Phenylketonuria

Question 5. The life cycle of Drosophila melanogaster is completed in

  1. 7 days
  2. 14 days
  3. 21 days
  4. 28 days

Answer: 2. 14 days

Question 6. How many types of gametes would developed by an organism with genotype AaBBCcDD?

  1. 1
  2. 2
  3. 3
  4. 4

Answer:  4. 4

Question 7.Assertion (A): There is an expression of only one gene of the parental character in a Mendelian Monohybrid cross in Ft generation

Reason (R): In a dissimilar pair of factors one member of the pair dominates the other

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.

Answer: 1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Question 8. In Pisum sativum, the flower colour may be Violet (V) or White (v). What proportion of the offspring in a cross of VV vv would be expected to be violet 

  1. 25%
  2. 50%
  3. 75%
  4. 100%

Answer: 4. 100%

Question 9. Which one of the gene pairs is expected to give a ratio of 1 1: I I in the progeny of a Mendelian Dihybrid cross

  1. AaBb×AbBb
  2. A ABB × AaBb
  3. AaBb × aabb
  4. A ABB × aabb

Answer: 3. AaBh x aabb

Question 10. The progeny of a cross between two snap-dragon plants heterozygous for flower colour, bearing different coloured flowers would be:

  1. 25% red, 50%
  2. pink, 25% white
  3. 50% red, 50% white
  4. 75% red, 25% white

Answer: 2. 25% red. 50% pink, 25% white.

Question 11. Study the given pedigree of a family and select the trait that shows this pattern of inheritance

Principles Of Inheritance And Variation Autosomal Recessive

  1. Autosomal recessive, Phenylketonuria
  2. Sex-linked recessive. Colour-blindness
  3. Autosomal dominant. Myotonic dystrophy
  4. Sex-linked dominant. Vitamin I) resistant rickets

Answer: 1. Autosomal recessive. Phenylketonuria

Question 12. A child with blood group A has a father with blood group B and a mother with blood group AB What would be the possible genotypes of the parents and the child Choose the correct option:

Principles Of Inheritance And Variation Phenylketonuria

Answer: 3. Father – IBi, Mother – IAiB, Child – I A i

Question 13. In a hybrid Mendelian cross, garden pea plants heterozygous for violet flowers and round seeds are crossed with homozygous white flowers and wrinkled seeds. The genotypic and phenotypic ratio of F, progeny would be:

  1. 9:3:3: 1
  2. 1:2:2: 1
  3. 1:1.1.1
  4. 3: 1

Answer: 3. 1: 1: 1: 1

Question 14. Colour blindness is a sex-linked recessive trait in humans. A man with normal colour vision marries a woman who is colourblind. What would be the possible genotypes of the parents, the son and the daughter of this couple?

Principles Of Inheritance And Variation Genotype

Answer: 1. Mother- XX, Father – XCY, Daughter- XCX, Son – XY

Question 15. Given below are the pairs of contrasting traits in Pisum sativum as studied by Mendel Select the incorrectly mentioned option from the table given below:

Principles Of Inheritance And Variation Karyotype Analysis

Answer: 1 Or 4

Question 16. How many types of gametes can be produced in a diploid organism which is heterozygous for 4 loci?

  1. 4
  2. 8
  3. 16
  4. 32

Answer: 3.  16

Question 17. Given below is a Karyotype obtained after analysis of foetal cells for a probable genetic disorder.

Principles Of Inheritance And Variation Human Karyotype Obtained Analysis Of foetal Cells For probable Genetic disorder

Based on the above Karyotype, the chromosomal disorder detected in the unborn foetus and the consequent symptoms the child may suffer from are

  1. Down’s syndrome: Gynecomastia, overall masculine development
  2. Down’s syndrome: Furrowed tongue, short stature
  3. Klinefelter’s syndrome Gynecomastia, Masculine development.
  4. Klinefelter’s syndrome: Rudimentary ovaries, short stature

Answer: 3. Klinefelter’s syndrome: Gynecomastia, Masculine development.

Question 18 The recombinant Frequency between the four linked genes is as follows:

  1. Between X and Y is 40%.
  2. Between Y and Z is 30%.
  3. Between Z and W is 10%.
  4. Between W and X. is 20%.

Select the option that shows the correct order of the position of W, X. Y and Z genes on the chromosome:

  1. Y – X -Z -W
  2. Y – W – Z – X
  3. X- Y -Z -W
  4. Z – X – Y – W

Answer: 2. Y – W – Z – X

Question 19. Write the possible genotypes of a person with blood group ‘B
Answer:

The genotype for blood group B is H1 in homozygous condition and H is in heterozygous condition

Question 20. Write the dominant traits in pea plants observed by Mendel concerning:

  1. colour of a pea pod.
  2. flower position.

Answer:

  1. Green pod colour was the dominant
  2. The terminal flower position was recessive to the axial position

Question 21. Write the symbolic representation used in a pedigree chart showing

  1. a carrier mother and
  2. a sufferer son, concerning haemophilia.

Answer:

A carrier mother→Principles Of Inheritance And Variation Carrier Mother

A sufferer’s son → Principles Of Inheritance And Variation Sufferer Son

Question 22. Assert ion: The progenies of a test cross can be easily analysed to predict the genotype of the test organism.

Reason: In a typical test cross, an organism showing a recessive phenotype is crossed with a recessive parent instead of self-crossing 

  1. Both Assertion and Reason are true, and Reason is the correct explanation of the Assertion
  2. Both Assertion and Reason are true, but Reason is not the correct explanation of the Assertion
  3. The assertion is true, but Reason is false
  4. Both Assertion and Reason are false.

Answer: 3. Assertion is true, but Reason is false.

Question 23.

1. Name a human genetic disorder due to the following :

  1. An additional X-chromosome in a male
  2. Deletion of one X-chromosome in a female

Answer:

  1. Klinefelter’s Syndrome
  2. Turner’s Syndrome

2. State what does aneuploidy lead to.
Answer:

Abnormal number of chromosomes in a cell. Down’s Syndrome or Turner’s Syndrome or Klinefelter’s Syndrome

Question 24. State Mendel’s Law of Independent Assortment.
Answer:

When two pairs of traits (characters) are combined in a hybrid segregation of one pair of characters is independent of the other pair of characters.

Question 25. Write one example of each of the organisms exhibiting

  1. Male heterogamety, and
  2. Female heterogamety.

Answer:

  1. Human or Drosophila or Grasshopper
  2. Birds Chicken

Question 26. Write the sex of a human having XXY chromosomes with 22 pairs of autosomes. Name the disorder this human suffers from.
Answer:

Mate, Klinefelter’s syndrome

Question 27. Name the type of cross that would help to find the genotype of a pea plant bearing violet flowers.
Answer:

Test cross

Question 28. A colour-blind bos is born to a couple with normal colour vision. Write the genotype of the parents.
Answer:

Mother-Xc
X Father-XY

Principles Of Inheritance And Variation Short Question And Answers

Question 1.

1. Write two closely linked genes that control a-Thalassemia.
Answer:

Thalassemia is controlled by two closely linked genes HBA1 and HBA2 on chromosome 16 of each parent and it is observed due to mutation or deletion of one or more of the four genes

2. Differentiate between Thalassemia and Sickle cell anaemia based on their effect on the globin molecule of haemoglobin.
Answer:

Thalassemia differs from sickle-cell anaemia in. that the former is a quantitative problem of synthesising too few globin molecules while the latter is a qualitative problem of synthesising an incorrectly functioning globin

Question 2.

1. Mendel did not explain the expression of incomplete dominance in plants, (give an example of a flower exhibiting incomplete dominance. Name and state the Law of Mendel that the genes which exhibit incomplete dominance follow.
Answer:

  • Antirrhinum Snapdragon or Dog flower or Four o’clock plant or A Law of segregation.
  • Allele or factors of a pair segregate from each other such that a gamete receives only one of the two factors.

2. Your teacher gave you a tall pea plant and asked you to find out whether the plant is homozygous or heterozygous. Mow will you proceed to find the genotype of the given plant?
Answer:

The Genotype was found by test cross. The Crossing of Unknown plant with recessive parent.

Principles Of Inheritance And Variation Genotype Plant With Recessive Parent

Question 3. Why is the frequency of red-green colour blindness more in human males than in females? Explain.
Answer:

The gene for colour blindness is located on the X chromosome in humans, it is a recessive gene, since human males have a single X chromosome the recessive gene always expresses when present, whereas in human females as they have two X chromosomes (the trait is expressed only if both the sex chromosomes have this reason e gene

Question 4. A lipophilic father can never pass the gene for haemophilia to his son. Explain.
Answer:

It is a sex-linked recessive disorder in which the X chromosome has the haemophilic gene, the Son inherits an ‘S’ chromosome from the father and the gene for haemophilia is not present on the S chromosome

Question 5.

1. What happens when chromatids fail to segregate during cell division cycle? Explain your answer with an example.
Answer:

Failure of segregation of chromatids during the cell division cycle results in the gain or loss of a chromosome (s), called aneuploidy. For example, Down’s syndrome results in the gain of an extra copy of chromosome 21

2. ABO blood groups are a good example of co-dominance. Justify.
Answer:

In blood groups I ‘ and I1’ are present together they both express their types of sugars this is because of co-dominance, lienee red blood cells have both A and B types of sugar polymers

Question 6.

1. Generally it is observed that human males suffer more than human females, who rarely suffer from it. Explain giving reasons.
Answer:

This is a sex-linked X chromosomes recessive disease, the heterozygous female carrier for haemophilia may transmit the disease to sons (mate progeny), and the possibility of a female becoming a haemophimophilia is extremely rare haemophilia because the mother of such a female has to be at least carrier and the father should be haemophilic.

2. F1 progeny of pea plant bearing violet flowers and snapdragon plant bearing red flowers were soiled to produce their respective F2 progeny. Compare the phenotypes, the genotypes and the pattern of inheritance of their respective F1 progeny.
Answer:

Principles Of Inheritance And Variation Progeny Of Pea Plant Bearing

Question 7.

1. Differentiate between pleiotropy and polygenic inheritance by taking one example of each.

2. How is polygenic inheritance different from pleiotropic? Give one example of each.

Answer:

Principles Of Inheritance And Variation difference Between Pleiotropy And Polygenic Inheritance

Question 8.

1. How does mutation occur?
Answer:

Eosstdeletion or gain (insertion duplication or addition) or change in position of DNA segments/chromosome

2. Differentiate between point mutation and frameshift mutation.
Answer:

Mutation due to a change in a single base pair of DNA is point mutation. Insertion or deletion of one or two bases changes the reading frame from the point, of insertion or deletion.

Question 9. Explain the mechanism of ‘sex determination’ in birds. How does it differ from that of human beings?
Answer:

In birds, sex determination takes place by the ZW -method;

Principles Of Inheritance And Variation Sex Determination

  • In Birds, female heterogamety female produces (Z) type and (W) type of gametes in the case of bird males in hornogamety (ZZ) and female is heterogamety.
  • In 11 humans being male heterogamety male produces (X) type and (Y) type of gametes.

Question 10. Both Haemophilia and Thalassemia are blood-related disorders in humans. Write their causes and the difference between the two. Name the category of genetic disorder they both come under.
Answer:

Principles Of Inheritance And Variation difference Between Haemophilia And Thalassemia

Mendelian disorder

Principles Of Inheritance And Variation Long Question And Answer:

Question 1. The cytological observations made in several insects led to the development of the concept of the genetic or chromosomal basis of the sex-determination mechanism. The honey bee is an interesting example to study the mechanism of sex determination. Study the schematic cross between the male and the female honey bees given below and answer the questions that follow :

Principles Of Inheritance And Variation Development Of The Concept Of Genetic And Chromosomal

1. Identify the cell divisions V and ‘IT’ that lead to gamete formation in female and male honey bees respectively.
Answer:

  • ‘A’-Meiosis
  • ‘B’-Mitosis

2. Name the process ( that leads to the development of the male honey bee (drone).
Answer:

‘C’-Parthenogenesis

Question 2. T.H. Morgan carried out a cross on Drosophila involving genes for body colour (y+/Y) and genes for eye colour (w +/w). Study the schematic representation of the cross-opto I I generation and answer the questions that follow :

Principles Of Inheritance And Variation T.H Morgan

  1. Name the kind of cross it represents.
  2. Identify and write the dominant phenotype concerning eye colour.
  3. What are these genes located on the chromosome shown referred to as?

Answer:

  1. Dihybrid cross
  2. The red-eye phenotype of w
  3. Linked genes or sex-linked genes

Question 3. Mendel crossed a homozygous pea plant having yellow and round seeds with another pea plant hearing green and wrinkled seeds, lie found that in some of the IS populations new combination of parental characters was observed. How will you explain the appearance of a new combination of parental characters in IS offspring? Support your answer with the help of Punned Square.

Answer: When two pairs of traits are combined in a hybrid segregation of one pair of characters is independent of the other pair of characters.

Principles of Inheritance And Variation Dihybrid

 Phenotypic ratio

Principles Of Inheritance And Variation Phenotypic Ratio

Genotypic ratio

Principles Of Inheritance And Variation Genotype Ratio

Question 4.

1. Write the scientific name of the organism Thomas Hunt Morgan and his colleagues worked with for their experiments. Explain the correlation between linkage and recombination concerning genes as studied by them.
Answer:

Trosophikttnehmogasier
They observed that two genes (located closely on a chromosome) did not segregate independently of each other (Fe ratio deviated very significantly from 9:3:3: 1) lightly linked genes tend to show very fewer (lesser) recombinant frequency of parental traits/show higher (more) frequency of parental type loosely linked genes show higher percentage (more) of recombinant frequency of parental traits Or lower frequency percentage of parental type genes present on the same chromosome are said to be linked and the recombinant frequency depends on their relative distance on the chromosome.

2. How did Sturtevant explain gene mapping while working with Morgan?
Answer:

He used the frequency of recombination between gene pairs on the same chromosome, as a measure of the distance between genes and ‘mapped’ their position on the chromosome.

Important Questions for Class 12 Biology Chapter 3 – Reproductive Health 

Reproductive Health Very Short Questions and Answers

Question 1. Assertion (A): Through Reproductive and Child Health (RCI) programs in India; we could bring down the population growth rate.

Reason (R): A rapid increase in MMR and IMR were the reason, along with other reasons for this.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion t A) is false, but Reason (R) is true.

Answer: Assertion (A) is true, but Reason (R) is false.

Question 2. Assertion (A): Sterilisation methods are generally advised for male or female partners as a terminal method to prevent any more pregnancies.

Reason (R): These techniques are less effective and have high reversibility.

  1. Both Assertion (A) and Reason (R) are true and Reason (R)is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R ) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.

Answer: 3. Assertion (A) is true, but Reason (R) is false.

Read and Learn More Important Questions for Class 12 Biology Chapter Wise

Question 3. A specialized procedure to form an embryo in the laboratory in which sperm is directly, injected into the ovum is:

  1. HIT
  2. IUI
  3. ICSI
  4. ZIFT

Answer: 3. ICSI

Question 4. Listed below are all reproductive tract infections except

  1. Genital herpes
  2.  Filariasis
  3. Trichomoniasis
  4. Syphilis

Answer: 2. Filariasis

Question 5. Write the full name of the technique used for the transfer of early embryos in the uterus of the mother for further development. Write the minimum number of blastomeres the embryo must have before being transferred.
Answer:

The zygote or early embryo with up to 8 blastomeres is transferred into the Fallopian tube and the process is called Zygote Intra Fallopian Transfer or ZIFT.

Question 6.

1. Assertion: A statutory ban on amniocentesis for sex determination is to legally check increasing female foeticide.

Reason: In amniocentesis, some of the amniotic fluid that has the developing fetus is taken to analyze the chromosomes in the fetal cells.

  1. Both Assertion and Reason are true, and Reason is the caned explanation of the Assertion.
  2. Both Assertion and Reason are true, but Reason is not the correct explanation of the Assertion.
  3. The assertion is true, but Reason is false.
  4. Both Assertion and Reason are false.

Answer: 3. Assertion is true, but Reason is false.

 2. Assertion: Our laws permit legal adoption and it is as yet, one of the test methods for couples looking for parenthood.

Reason: Emotional, religious, and social factors are also no deterrents to the legal adoption of orphaned and destitute children in India.

  1. Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
  2. Both Assertion and Reason are true, but Reason is not the correct explanation of the Assertion.
  3. The assertion is true, but Reason is false.
  4. Both Assertion and Reason are false.

Answer: 1. Both Assertion and Reason are true, and Reason is the correct explanation of the Assertion.

Question 7. Give one reason to justify a statutory ban on amniocentesis.
Answer:

Prevent female foeticide.

Reproductive Health Short Question And Answers

Question 1. Why do doctors suggest some monied couples go for ‘IVF’? Explain the steps carried out in the process of ‘IVF’.
Answer:

Childless couples could be assisted in having children through IVF

Ova from the wife/donor (female) and sperm from the husband or donor(male) are collected and are induced to form a zygote under simulated conditions in the laboratory, the zygote or early embryos (with up to 8 blastomeres) could then be transferred into the fallopian tube(ZIFT), and embryos with more than 8 blastomeres transferred into the uterus(lUT)

Question 2. A doctor after conducting certain tests on a pregnant woman advised her to undergo M.T.P., as the fetus she was earning showed trisomy of the 21st chromosome.

1. State the cause of trisomy of the 21st chromosome.
Answer:

Cause non-disjunction or failure of segregation of chromatids of the 21st chromosome during gamete formation, leading to a gain of a chromosome.

2. Why was the pregnant woman advised to undergo M.T.P. and not to complete the full term of her pregnancy?
Answer:

Mother was advised to undergo MTP because Trisomy of the 21st Chromosome would lead to s, Down’s syndrome or an individual is short-statured with furrowed tongue or broad palm with characteristic palm crease or retarded physical or mental and psychomotor development.

Question 10. List any two types of IEDs that are available for human females and state their mode of action.
Answer:

  1. Copper releasing fUDs (CuT, Cu7, Multiload 375)
  2. Cu ions released suppress sperm motility and. the fertilising capacity of sperms,
  3. The hormone-releasing TUDs (Progcstascrt, LNG-20)
  4. The hormone-releasing JUDs. in addition, make the uterus unsuitable for implantation and the cervix hostile to the sperm.

Reproductive Health Long Question And Answer

Question 1. IVF is a very popular method this day that is helping childless couples to bear a child. Describe The different steps that are carried out in this technique. Would you consider Gamete Intrafallopian Transfer (GIFT) as an IN F1 Give a reason in support of your answer.
Answer:

  1. Ova from the wife or donor (female) and sperms from the husband or donor (male) are collected and induced to form a zygote, in simulated conditions in the laboratory, the zygote or early embryos up to 8 blastomeres) then transferred into the fallopian tube(ZIFT), and embryos with more than 8 blastomeres. into the uterus (JUT) to complete its further development.
  2. No, GIFT cannot be considered an IVF technique because fertilization takes place in the female body,

Question 2.

1. Explain the steps involved in in vitro fertilization popularly known as the test tube baby program.
Answer:

In vitro fertilization followed by embryo transfer (ET) is one such method. In this method, popularly known as the test tube baby program, ova from the wife or donor (female) and sperms from the husband or donor (male) are collected and induced to form a zygote under simulated conditions in the laboratory. The zygote or early embryos (with up to 8 blastomeres) could then be transferred into the fallopian tube (ZIFF-zygote intrafallopian transfer) and embryos with more than 8 blastomeres, into the uterus (IUT intra uterine transfer), to complete its further development.

2. State the importance of this program.
Answer:

  • A large number of couples all over the world including India are infertile, i.e., they are unable to produce children despite unprotected sexual cohabitation.
  • The reasons for this could be many physical, congenital, diseases, drugs, immunological or even psychological,
  • In India, often the female is blamed for the couple being childless, but more often than not, the problem lies in the male partner.
  • Specialized health care units (infertility clinics, etc.) could help in the diagnosis and corrective treatment of some of these disorders and enable these couples to have children.

Important Questions of Biology Class 12 Chapter 1 Sexual Reproduction in Flowering Plants

Sexual Reproduction In Flowering Plants Very Short Question And Answers

Question 1. The hilum in a typical angiosperm ovule represents the junction between –

  1. Integuments and the embryo sac.
  2. Embryo sac and its ocellus
  3. The body of the ovule and the funicle
  4. Nocellus and the funicle

Answer: 3. Body of the ovule and the funicle.

Question 2. In the given diagram of a transverse section of a young amber Choose the labelling showing the correct placement of the wall layers from the table given below.
Answer:

Sexual Reproduction In Flowering Plants

Sexual Reproduction In Flowering Plants Ovule And The Funicle

  1. Tapetum
  2. Endotliecium
  3. Epidermis
  4. Middle layers

Read and Learn More Important Questions for Class 12 Biology Chapter Wise

Question 3. The term used for the embryo entering into the state of inactivity as the seed matures is

  1. Quiescent
  2. Parthenogenesis
  3. Parthenocarpy
  4. Donnancy

Answer: 4. Dormancy

Question 4. The ploidy of the apomictic embryo developed from the integument cells and megaspore mother a cell without reduction division respectively will be

  1. 2n and 2n
  2. n and n
  3. 2n and n
  4. 3n and 2n

Answer: 1. 2n and 2n

Question 5. Below is a diagrammatic representation of a mature embryo sac of a typical angiosperm plant. Choose the option showing the correct labelling for the parts W, X, Y and Z from the table below.
Answer:

Sexual Reproduction In Flowering Plants Mature Of Embryo Sac

Sexual Reproduction In Flowering Plants Mature Of Embryo

Answer: 3. Y – Micropylar end. X – Antipodals, Y-Synergids, Z – Central cell

Question 6. Given below is a figure of an angiosperm plant showing two different types of flowers ‘X’ and the possible type of pollination in them :

Sexual Reproduction In Flowering Plants Angiosperm diagram Two Different Types of Flowers

Select the correct option for the flower (X) and flower (Y) and the possible type of pollination from the given table:

Sexual Reproduction In Flowering Plants Pollination

Answer: 3. Flower X Chasmogamous self Or cross-pollination Flower Or Cleistogamous, self-pollination

Question 7. An undifferentiated sheath covering the root cup of a monocotyledonous embryo is:

  1. Scutellum
  2. Coleorhiza
  3. Coleoptile
  4. Epi blast

Answer: 2. Coleorhiza

Question 8. Residual persistent nucellus in black pepper is known as :

  1. Perisperm
  2. Pericarp
  3. Putvinus
  4. Perianth

Answer: 1. Perisperm

Question 9. Amongst tile insects, the dominant biotic pollinating agents are –

  1. Ants
  2. Wasps
  3. Beetles
  4. Bees

Answer: 4. Bees

Question 10. A genetic mechanism which prevents inbreeding depression in the majority of atigiospenu plants is

  1. Parthenogenesis
  2. Parthenoeurpy
  3. Mutation
  4. Self-incompatibility

Answer: 3. Self-incompatibility

Question 11. Self-pollination is fully ensured if

  1. The flower is bisexual.
  2. The style is longer than the filament.
  3. The flower is cleistogamous.
  4. The time of pistil and another maturity is different.

Answer: 3. The flower is cleistogamous.

Question 12. Some flowers, selected for artificial hybridization, do not require emasculation but bagging is essential for them. Give a reason.
Answer:

As some Bowers are unisexual, to prevent contamination of their stigma with unwanted pollen grains.

Question 13. Write one advantage and one disadvantage of cleistogamy to flowering plants.
Answer:

Cleistogamy Advantage:

Assured seed set maintains pure lines.

Cleistogamy Disadvantage:

No variation or only parental characters are preserved or it can lead to inbreeding depression.

Sexual Reproduction In Flowering Plants Short Question And Answers

Question 1:

1. Write the two crucial changes the seeds undergo while reaching maturity that enable them to be in a viable state until the onset of favourable conditions.
Answer:

Seeds remain in a state of dormancy-induced desiccation and the hormone abscisic acid until conditions for growth become favourable.

2. Name the oldest viable seed excavated from the Arctic Tundra as per the records.
Answer:

The oldest is that of a lupine, Lupinus an excavated front Arctic Tundra.

Question 2. Explain the mechanism of pollination in marine seagrasses like Zostcra.
Answer:

In marine Seagrasses, female flowers remain submerged in water and the pollen grains are released inside the water. Pollen grains in many such species are long, ribbon and they are carried passively inside the water; some of them reach the stigma and achieve pollination.

Question 3. What is ‘bagging’? State its importance in the artificial hybridization of flowering plants.
Answer:

Bagging – Process of covering emasculated or female or artificially pollinated flowers with a bag of suitable size generally made up of butter paper (to prevent contamination of stigma with unwanted pollen).

Importance – The desired pollen grains are used for pollination or the stigma is protected from contamination from unwanted pollen. It combines desirable characteristics to produce commercially superior varieties.

Question 4:

1. You are given castor and bean seeds. Which one of the two would you select to observe the endosperm?
Answer:

Castor

2. The development of endosperm precedes that of embryos in plants. Justify.
Answer:

Endosperm stores reserve food materials or provide nutrition to the developing embryo.

Question 5. Write any two ways by which apomictic seeds may be developed in angiosperms.
Answer:

  • Develops from a diploid egg cell (formed without reduction division) which grows into an embryo without fertilization.
  • Develops from the nuclear cell which divides and protrudes into the embryo sac and develops into an embryo

Question 6. A pollen grain in angiosperm at the time of dehiscence from an anther could be 2-celled or 3-celled. Explain. Mow arc the cells placed within the pollen grain when shed at a 2-celled stage?
Answer:

  • In the 2-celled stage, the mature pollen grain contains a generative and vegetative cell, whereas in the 3- 3-celled stage one vegetative cell and two male gametes are present.
  • The generative cell floats in the cytoplasm of the vegetative cell.

Question 7:

1. When a seed of orange is squeezed, many embryos, instead of one are observed. Explain how it is possible.
Answer:

Polyemhryony and nucellar cells surrounding the embryo sac start dividing, protrude into the embryo sac and develop into many embryos.

2. Are these embryos genetically similar or different? Comment. 
Answer:

These embryos are genetically similar, as produced from nucellar cells by mitotic division/formed without fertilisation (but different from the embryo formed by fertilization)

Question 8. Draw a schematic transverse section of a mature anther of an angiosperm. Label its epidermis, middle layers, tapetum, eudothecium, sporogenous tissue and the connective.
Answer:

Sexual Reproduction In Flowering Plants Transverse Section of Mature Anther Of An Angiosperm

Question 9. Differentiate between wind-pollinated and insect-pollinated flowers.
Answer:

Wind pollinated: Pollen grains are light sticky or well-exposed stamens or Large feathery stigmas or single ovule in each flower or Small flowers packed in inflorescence.

Insect pollinated: Pollen grains sticky or Moral rewards or Fragraut or Neetar rich or Large Flower.

Question 10:

1. When and where do tapetum and synergids develop in (lowering plants? Mention their functions.
Answer:

  1. Tapetum – Mierosporogenesis. Microsporang (Anther), nourishes the developing pollen grains.
  2. Synergids – Megusporogenesis. Megasporangiumtovule), synergids have filiform apparatus to guide the pollen tube into it.

2. Where are the following structures present in a male gametophyte of an angiosperm? Mention the function of each one of them.

  1. Germ pore
  2. Sporopollenin
  3. Generative cell

Answer:

  1. Germ pore- Pollen grain exiting, site from where pollen tube emerges.
  2. Sporopollenin – Exine of the pollen grain, protects the pollen grains from high temperature and strong acids And alkalis or enzymes or adverse conditions.
  3. Generative cells – Pollen grains give rise to two male gametes.

Question 11. Draw L.S. of an embryo of grass and label its parts.
Answer:

Sexual Reproduction In Flowering Plants L S Of An Embryo Of Grass

Question 12. Draw a labelled diagram of a mature male gametophyte of an angiosperm.
Answer:

Sexual Reproduction In Flowering Plants Structure Of Mature Gametophyte

Question 13. Do you think apomixis can be compared with asexual reproduction? Support your answer, by giving one reason. Flow is apomixis beneficial to farmers? Explain.

Answer:

Yes, seeds are produced without fertilisation.

Production of hybrid seeds is costly, if hybrids with desirable characteristics can be made into apomicts, there is no segregation of characters in the hybrid progeny, and farmers can continue using hybrid seeds year after year and do not have to buy hybrid seeds.

Question 14:

1. Can a plant flowering in Mumbai be pollinated In pollen grains of the same species growing in New Delhi? Provide explanations for your answer.
Answer:

Yes. Artificial hybridisation in pollen grains of the flower is introduced artificially on the stigma of another flower.

2. Draw the diagram of a pistil where pollination has successfully occurred. Label the parts involved in reaching the male gametes to their desired destination.
Answer:

Diagram with following labelling Stigma, Pollen tube, Synergid or Filiform Apparatus. Micropyle

Sexual Reproduction In Flowering Plants Longitidinal Section Of A flower Showing Growth Of Pollen Tube

Question 15:

1. Draw a diagram of Pistil showing pollen tube growth in angiosperm and label

  1. Stigma
  2. male gametes
  3. micropyle end
  4. Ovule.

Answer: 1.

Sexual Reproduction In Flowering Plants Longitidinal Section Of A flower Showing Growth Of Pollen Tube.

2. Write the function of micropyle.
Answer:

The pollen tube, after reaching the ovary, enters the ovule through the micropyle and then enters one of the synergids through the filiform apparatus.

Sexual Reproduction In Flowering Plants Long Question And Answers

Question 1. Study the figures given below of the development of megaspores in an angiosperm and answer the questions that billow :

Sexual Reproduction In Flowering Plants Megaspore

1. Describe the developmental events in the nucellus of the ovule. What is this type of development of megaspore referred to as?
Answer:

  • Enclosed within the integuments is a mass of cells called the nucellus.
  • Cells of the nucellus have abundant reserve food materials.
  • Located in the nucleus is the embryo sac or female gametophyte.
  • This method of embryo sac formation from a single megaspore is termed monosporic development.

 2. How many free nuclear mitotic divisions will the functional megaspore undergo to form a mature embryo sac?
Answer:

3 mitotic division

3. Describe the structure of a typical female gametophyte of a flowering plant.
Answer:

  • A characteristic distribution of the cells within the embryo sac.
  • Three cells are grouped at the micropylar end and constitute the egg apparatus.
  • The egg apparatus, in turn, consists of two synergids and one egg cell.
  • The synergids have special cellular thickenings at the micropylar tip called filiform, apparatus, which plays an important role in guiding the pollen tubes into the syttergid.
  • Three cells are at the chalazal end and are called the antipodal.
  • The large central cell, as mentioned earlier, has two polar nuclei.
  • Thus, a typical angiosperm embryo sac. at maturity, though 8-nucleate is 7-celled.

Question 2:

1. Name the specific part of the anther and the process responsible for the development of a male gametophyte in an angiosperm.
Answer:

porogenous tissue or Microsporungium Microspore mother cell Pollen mother cell or PMC Microsporogenesis.

2. Draw a labelled diagram of a mature male gametophyte (3-celled) of an angiosperm. Write the functions of each labelled part
Answer:

Sexual Reproduction In Flowering Plants Structure Of Mature Gametophyte.

  • Vegetative cell – It has an abundant food reserve
  • Male gametes – Participate in double fertilisation or one male gamete fuses with the egg and the other fuses with two polar nuclei or secondary nucleus (any two)
  • Exilic – Made up of the most resistant organic material sporopollenin or can withstand high temperatures or strong acids or alkalis or no enzyme can degrade it.
  • In tine – Contributes pollen tube formation.
  • Germ pore – Region from where pollen tubes arise.

Question 3. Explain the post-pollination events up to double fertilisation, that occur in an angiosperm.
Answer:

The Pollen grain germinates on the stigma to produce a pollen tube through one of the germ pores.

  • The contents of the pollen grain or vegetative cell generative cell or two male gametes move into the pollen tube.
  • The pollen tube grows through the tissues of the stigma and pollen tube to reach the ovary.
  • The pollen tube enters (through micropyle) the synergids through the filiform apparatus, pollen tube releases two male gametes in the cytoplasm of the synergids.
  • One of the male gametes fuses with the egg cell or female gamete completing syngamy. to form (diploid) zygote.
  • The other male gamete fuses with two polar nuclei in the (central cell) to produce (a triploid) primary endospermic cell, three haploid cells fuse called triple fusion, and two types of fusion syngamy and triple fusion are called double fertilisation.

Question 4:

1. Describe any two devices in a flowering plant which pretend both autogamy and geitonogamy.
Answer:

Dioecy/production of unisexual flowers (in different plants) Self sterility or self-incompatibility or intraspecific incompatibility lists is a genetic mechanism and prevents self pollen (front the same flower or another flower of the same plant) from fertilising the ovules by inhibiting pollen germination or pollen tube growth in the pistil.

2. Explain the events up to double fertilisation after the pollen tube enters one of the synergids in an ovule of an angiosperm.
Answer:

Pollen tube releases 2 male gametes in the cytoplasm of synergid

  • One male gamete fuses with egg cells or syngamy. resulting in a diploid zygote
  • Other male gamete fuses with polar nuclei or triple fusion, to form diploid PEN (Primary Endosperm Nucleus) or PEC (Primary Endosperm Cell)

Question 5:

1. Draw a diagrammatic sketch of a transverse section of an anther of an angiosperm. Label its different walls and the tissue forming microspore mother cells.
Answer:

Sexual Reproduction In Flowering Plants Structure Of Mature Gametophyte.

2. Describe the process of microsporogenesis up to the formation of a microspore.
Answer:

Sporogenous tissue → Microspore mother cell → Microspore tetrad

3. Write the function of germ pore1 in a pollen grain of an angiosperm.
Answer:

Germ pores allow the germinating or growing pollen tube with contents of the pollen grain or male gametes + vegetative cell to come out of the pollen grains

Question: 6

1. State one difference and one similarity between geitonogamy and xenogamy.
Answer:

GeitOnogamy – Transfer of pollen grains from the anther to the stigma of another flower of the same plant. Although is functionally cross-pollination involves 1 a pollination. agent, genet it ally it m similar to autogamy since the pollen grains come Item the sunn plant

Xenogamy – transfer of pollen grains from the anther to the stigma of a different plant This is the only tv[x of pollination which during pollination brings genetically different types of pollen grains to the stigma.

2. Explain any three devices developed in flowering plants to discourage self-pollination and encourage cross-pollination.
Answer:

Flowing plants have Haw developed many devices to discourage sell-pollination and to encourage cross-pollination.

  • At some speed. pollen and base and stigma negativity are not synchronised.
  • Lit her the pollen is released before the stima becomes receptive or stigma becomes receptive much before the release of pollen.
  • In some other species, the anther and stigma are placed at different positions so that the pollen cannot come in contact with the stigma of the same flower.

NEET Class 12 Biology MCQ – Environmental Issues

Question 1. Carbon monoxide is a major pollutant of 

  1. Water
  2. Air
  3. Noise
  4. Soil

Answer: 2. Air

Carbon monoxide (CO) is major pollutant in air, exhausted by various automobiles.

Question 2. Air pollution is caused by excess of 

  1. Dinitrogen
  2. Hydrogen
  3. Water vapour
  4. None of these

Answer: 1. Dinitrogen

Nitrogen form various oxides such as N2O, NO, NO N2O5, and N2O5 play an important role in the formation of photochemical smog and acid rains. Thus, excess of dinitrogen cause air pollution.

Question 3. Statements given below are pertaining to air pollutants.

  1. They cause injury to all living organisms.
  2. They reduce the growth and yield of crops and cause premature death of plants.
  3. They affect the respiratory system of humans and animals.

Choose the option containing the correct statements.

  1. 1 and 2
  2. 1 and 3
  3. 2 and 3
  4. 1, 2 and 3

Answer: 4. 1,2 and 3

  • All the given statements are correct as Air pollution has several effects on all living organisms and on climate. Diseases like bronchitis, lung cancer and emphysema are caused by air pollution.
  • NO2 causes bronchitis and lowers the resistance to influenza.
  • SO2 obstructs breathing and irritates the eyes.
  • Nitric acid, nitrous acid and sulphuric acid cause respiratory diseases. Air pollutants reduce the growth and yield of crops and cause premature death of plants.

Biology MCQs with answers for NEET

Question 4. The Air Prevention and Control of Pollution Act came into force in the year …A… It was amended in …B… to incorporate …C… as an air pollutant.

  1. A–1980, B–1986, C–water
  2. A–1981, B–1987, C–noise
  3. A–1982, B–1988, C–radioactive
  4. A–1983, B–1989, C–soil

Answer: 2. A–1981, B–1987, C–noise

Question 5. Which of the following is a primary pollutant?

  1. CO
  2. HNO3
  3. H2SO4
  4. O3

Answer: 1. CO

A primary pollutant is that which persists in the form in which it is released in the environment, e.g. CO.

Question 6. Assertion Secondary air pollutants are formed by interaction among the primary pollutants and are more toxic. Reason (R) DDT is a secondary air pollutant.

  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true, but R is not the correct explanation of A
  3. A is true, but R is false
  4. Both A and R are false

Answer: 3. A is true, but R is false

A is true, but R is false because Secondary air pollutants are produced photochemically from primary pollutants. Smog, photochemical smog, ozone and PAN all are secondary air pollutants. DDT is a primary soil pollutant.

Question 7. Which of the following is a secondary pollutant?

  1. CO
  2. O3
  3. SO2
  4. CO2

Answer: 2. O3

O3 (ozone) is a secondary pollutant. These are formed by the reaction of primary pollutants. CO is a quantitative pollutant. CO2 and SO2 are primary pollutants.

Biology MCQs with answers for NEET

Question 8. One of the following acts as a secondary pollutant.

  1. Br2
  2. CI2
  3. NO2
  4. HNO2

Answer: 3. NO2

Secondary pollutants are not emitted directly in the air, water or soil, but are synthesised by chemical reactions and the best examples are oxides of nitrogen. They further form ozone, PAN and aldehydes.

Question 9. The secondary pollutant which stops Hill reaction is 

  1. Sulphuric Acid
  2. Nitric Acid
  3. Peroxyacetyl Nitrate (Pan)
  4. Aldehydes
  5. None of the above

Answer: 3. Peroxyacetyl Nitrate (Pan)

PAN prevents the photolysis of water in photosynthesis or Hill reaction.

Question 10. Match the following columns.

NEET Biology Air Pollution and Noise Pollution Match the columns Q 40

Answer:

1–1, 2, 4

2–3, 5

Question 11. Match the following columns.

NEET Biology Air Pollution and Noise Pollution Match the columns Q 41

Answer:

1–3, 4, 5, 6, 7

2–1, 2, 8, 9

Question 12. Consider the following statements about pollution.

  1. Pollution is defined as an undesirable change in physical, chemical or biological characteristics of air, land, water or soil.
  2. The Air (Prevention and Control of Pollution) Act was amended in 1987.
  3. The Environment Protection Act, of 1976 was formed to bring into effect the parameters of the United Nations Conference on the Human Environment.

Biology MCQs with answers for NEET

Choose the option containing the correct statements.

  1. 1 and 2
  2. 1 and 3
  3. 2 and 3
  4. 1, 2 and 3

Answer: 1. 1 and 2

Statements 1 and 2 are correct, but 3 is incorrect because In order to control environmental pollution, the government of India has passed the Environment Protection Act, 1986 to protect and improve the quality of our environment (air, water and soil). The air act was amended in 1987 to include noise as air pollution.

Question 13. Which of the following is a weightless pollutant?

  1. Sewage and exhaust emissions
  2. SO2 and NO2
  3. Photochemical smog
  4. Heat, sound and radioactive wastes

Answer: 4. Heat, sound and radioactive wastes

Heat, sound and radioactive wastes are weightless pollutants.

Question 14. A scrubber in the exhaust of a chemical industrial plant removes CBSE AIPMT, Karnataka CET

  1. Gases Like Sulphur Dioxide
  2. Particulate Matter Of The Size 5 Micrometer Or Above
  3. Gases Like Ozone And Methane
  4. Particular matter of the size 2.5 micrometre or less

Answer: 1. Gases Like Sulphur Dioxide

A scrubber can remove gases like sulphur dioxide. In a scrubber, the exhaust is passed through a spray of water or lime

Question 15. In the scrubber, the exhaust is passed through a

  1. Spray Of Water
  2. Spray Of Lime
  3. Both 1 And 2
  4. Spray Of Electrons

Answer: 3. Both 1 And 2

A scrubber can remove gases like sulphur dioxide. In a scrubber, the exhaust is passed through a spray of water or lime

Biology MCQs with answers for NEET

Question 16. Consider the following statements about scrubbers.

  1. It removes gases like sulphur dioxide from industrial exhaust.
  2. The exhaust passes through a spray of water or lime.
  3. Sulphur dioxide reacts with lime to form a precipitate of calcium sulphate and water dissolves gases.

Choose the option containing the correct statements.

  1. 1 and 2
  2. 1 and 3
  3. 2 and 3
  4. 1, 2 and 3

Answer: 4. 1,2 and 3

All the given statements are correct for the scrubber. A scrubber can remove gases like sulphur dioxide. In a wet scrubber, a fine spray of water or alkaline fluid like lime is allowed to fall over exhaust emissions. The particles also become heavy and fall down. The lime reacts with sulphur dioxide to produce a precipitate of calcium sulphate or calcium sulphide. Water is used to remove soluble gases and particles.

Question 17. The below diagram shows a scrubber. Identify A, B, C and D.

NEET Biology Air Pollution and Noise Pollution scrubber

  1. A–Particulate matter, B–Clean air, C–Dirty air, D–A dust particle
  2. A–Dirty air, B–Clean air, C–Water lime spray, D–Slurry outlet
  3. A–Slurry outlet, B–Dirty air, C–Particulate matter, D–Clean air
  4. A–Dust particle, B–Clean air, C–Particulate matter, D–Water spray

Answer: 2. A– Dirty air, B–Clean air, C–Water spray, D–Slurry outlet.

Question 18. Cyclone collectors are nowadays commonly used to control 

  1. Air Pollution With Special Reference To Dust Particles
  2. Radioactive Pollution
  3. Water Pollution In General
  4. Water pollution with special reference to Ganga Action Plan

Answer: 1. Air Pollution With Special Reference To Dust Particles

The cyclone collector is the device that is used for minimising air pollution by trapping the suspended particles like coarse dust generated due to activities like mining, cement and wood factories. These are dry scrubbers that work on the principle of inertia. Cyclone collector work similar to a centrifuge, but with a continuous feed of dirty air.

Thus, option 1 is correct.

Question 19. Gaseous pollutants can be controlled by

  1. Arrestors
  2. Electrostatic Precipitators
  3. Pyrolysis
  4. Incineration

Answer: 2. Electrostatic Precipitators

A gaseous pollutant can be controlled by Electrostatic Precipitator (ESP). It is an electrical device that removes particulate matter present in the exhaust of a thermal power plant. More than 99% of particulate matter can be removed by this method.

Question 20. In a coal-fired power plant, electrostatic precipitators are installed to control the emission of 

  1. NOx
  2. SPM
  3. CO
  4. SO2

Answer: 2. SPM

The electrostatic precipitators are installed to control the emission of Suspended Particulate Matter (SPM) as these can cause various respiratory disorders in humans if released into the air.

Question 21. Which one of the following is the most efficient device to eliminate particulate matter from industrial emission?

  1. Cyclonic separators
  2. Trajectory separators
  3. Pyrolysis
  4. Incineration
  5. Electrostatic precipitator

Answer: 5. Electrostatic precipitator

An electrostatic Precipitator (ESP) is the most efficient device to eliminate submicron particulates from industrial emissions. It removes the impurities (dust, fibres) by applying a high-voltage electrostatic charge which precipitates the impurities on the charged plates.

Question 22. Which of the following is made use in an electrostatic precipitator?

  1. Catalysts
  2. Absorbers
  3. Electrodes
  4. Chemicals

Answer: 3. Electrodes

An electrostatic precipitator is used to remove particulate matter present in the exhaust of a thermal power plant. It has electrode wires that are maintained at several thousand volts which produce a corona that releases electrons.

NEET Biology Mcq Chapter Wise

Question 23. In an electrostatic precipitator, very high voltage of electricity is passed through electrodes that produce a corona, which emits …A… . The dust particles present inside the precipitator gains a …B… charge instantly.

  1. A–electron, B–positive
  2. A–neutron, B–negative
  3. A–electron, B–negative
  4. A–proton, B–positive

Answer: 3. A–electron, B–negative

In an electrostatic precipitator, electrode wires are provided with an electric current of several thousand volts, which produces a corona that releases electrons (A). These electrons attach to dust particles and gives them a negative charge within a very small fraction of a second.

Question 24. Electrostatic Precipitator (ESP)

  1. Removes particulate matter from the exhausts of thermal power
    plants.
  2. It has a removal efficiency of 99%.
  3. ESP has thin wires and stack of thick metal plates.

Choose the option containing correct statements.

  1. 1 and 2
  2. 1 and 3
  3. 2 and 3
  4. 1, 2 and 3

Answer: 4. 1, 2 and 3

All the given statements are correct.

Question 25. The below diagram shows an electrostatic precipitator. Identify A, B, C and D and select the correct option.

NEET Biology Air Pollution and Noise Pollution electrostatic precipitator

  1. A–Dust particles, B–Negatively charged wire, C–Discharge corona, D–Collection plate grounded
  2. A–Discharge corona, B–Collection plate grounded, C–Dust particles, D–Negatively charged wire
  3. A–Discharge corona, B–Negatively charged wire, C–Dust particles, D–Collection plate grounded
  4. A–Discharge corona, B–Dust particles, C–Negatively charged wire, D–Collection plate grounded

Answer: 3. A–Discharge corona, B–Negatively charged wire, C–Dust particles, D–Collection plate grounded

NEET Biology Air Pollution and Noise Pollution Electrostatic precipitator.

Question 26. Choose the correct statement regarding the catalytic converters.

  1. Vehicles fitted with catalytic converters must use leaded petrol
  2. Platinum, palladium and rhodium are used as catalysts in catalytic converters
  3. Catalytic converters reduce the particulate matter
  4. Nitrogen gas in the exhaust is converted to nitric oxide by the catalytic converter

Answer: 2. Platinum, palladium and rhodium are used as catalysts in catalytic converters

The statement in option 2 is correct. Other options are incorrect and can be corrected as Catalytic converters are fitted in vehicles for reducing the emission of poisonous gases. As the exhaust passes through, the catalytic converter, unburnt hydrocarbons are converted into carbon dioxide and water. Carbon monoxide and nitric oxide are changed to carbon dioxide and nitrogen gas, respectively.

Question 27. Catalytic converters are fitted into automobiles to reduce the emission of harmful gases. Catalytic converters convert leftover hydrocarbons into

  1. Carbon Dioxide And Water
  2. Carbon Monoxide
  3. Methane
  4. Carbon dioxide and methane

Answer: 1. Carbon Dioxide And Water

The statement in option 1 is correct. Other options are incorrect and can be corrected as Catalytic converters are fitted in vehicles for reducing the emission of poisonous gases. As the exhaust passes through, the catalytic converter, unburnt hydrocarbons are converted into carbon dioxide and water. Carbon monoxide and nitric oxide are changed to carbon dioxide and nitrogen gas, respectively.

NEET Biology Mcq Chapter Wise

Question 28. Catalytic converters

  1. Fitted into automobiles for reducing emission of poisonous gases.
  2. Metals like platinum, palladium and rhodium are used as catalysts.
  3. On passing through a converter, the nitric oxide in the exhaust splits into nitrogen and oxygen. Carbon monoxide is oxidised to carbon dioxide.
  4. Leaded petrol is harmful to catalytic converters as they can inactivate the catalyst.

Choose the option containing correct statements about converters.

  1. 1, 2 and 3
  2. 2, 3 and 4
  3. 1, 3 and 4
  4. 1,2,3, and 4

Answer: 4. 1,2,3, and 4

All given statements are correct. Catalytic converters are fitted into automobiles for reducing the emission of poisonous gases. Rhodium and platinum, palladium are examples of catalysts used in catalytic converters. They convert unburnt hydrocarbons into CO2 and HO2 and CO and nitric oxide to CO2 and N2 gas, respectively. Use of unleaded petrol is however recommended as the lead petrol causes inactivation of the catalyst.

Question 29. Match the following columns.

NEET Biology Air Pollution and Noise Pollution Match the columns Q 59

Answer: 2. A–2, B–1, C–4, D–3

Question 30. Which of the following is not an air pollutant?

  1. NO2
  2. SO2
  3. Hydrocarbons
  4. CO2

Answer: 4. CO2

Carbon dioxide occurs naturally in the atmosphere. It is an essential ingredient in photosynthesis.

Thus, it is not an air pollutant.

NEET Biology Mcq Chapter Wise

Question 31. Air pollution causing photochemical oxidant production includes

  1. Carbon Monoxide, Sulphur Dioxide
  2. Nitrous Oxide, Nitric Acid Fumes, Nitric Oxide
  3. Ozone, Peroxyacetyl Nitrate, Aldehyde
  4. Oxygen, chlorine, fuming nitric acid

Answer: 3. Ozone, Peroxyacetyl Nitrate, Aldehyde

Photochemical smog is formed at high temperatures over cities and towns due to still air, emission of oxides of nitrogen and sulphur and hydrocarbon from automobile exhausts and solar energy. Nitrogen dioxide splits into nitric oxide and nascent oxygen. Nascent oxygen combines with molecular oxygen to form ozone. Ozone reacts with hydrocarbon to form aldehyde and ketone. Nitrogen oxides, oxygen and ketones combine to form Peroxyacyl Nitrates (PAN). In areas with intense solar radiation, photoelectrical smog forms brown air.

Question 32. Smog is a combination of

  1. Fire And Water
  2. Smoke And Fog
  3. Smoke And Water
  4. Air and water

Answer: 2. Smoke And Fog

Smog is basically derived from the merging of two words; smoke and fog. Smog occurs mainly because of air pollution and can also be defined as a mixture of various gases with dust and water vapour. Smog also refers to hazy air that makes breathing difficult.

Question 33. The basic component of smog may be

  1. PAN
  2. ozone
  3. PBN
  4. All of the above

Answer: 4. All of the above

Smog is a highly oxidising polluted atmosphere comprising O3, NO2, PAN (Peroxyacetyl Nitrate) and PBN (Peroxybenzoyl Nitrate).

Question 34. Smog is commonly formed due to JIPMER

  1. Mixing Of Ozone With NO2
  2. Deposition Of CO2 On Condensed Vapour Particles
  3. Greenhouse Effect And Global Warming
  4. The reaction of oxides of N and S, etc., with sunlight and volatile organic compounds

Answer: 4. Reaction of oxides of N and S, etc., with sunlight and volatile organic compounds

Photochemical smog is formed at high temperatures over cities and towns due to still air, emission of oxides of nitrogen and sulphur and hydrocarbon from automobile exhausts and solar energy. Nitrogen dioxide splits into nitric oxide and nascent oxygen. Nascent oxygen combines with molecular oxygen to form ozone. Ozone reacts with hydrocarbon to form aldehyde and ketone. Nitrogen oxides, oxygen and ketones combine to form Peroxyacyl Nitrates (PAN). In areas with intense solar radiation, photoelectrical smog forms brown air.

Question 35. Assertion Smog is commonly formed at places having low temperatures and high pollution of aerosol. Reason (R) It is very common in metropolis cities of India.

  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true, but R is not the correct explanation of A
  3. A is true, but R is false
  4. Both A and R are false

Answer: 1. Both A and R are true and R is the correct explanation of A

Both A and R are true and R is the correct explanation of A. Smog is a common problem in many large urban areas, where human activity gives rise to particulates in the air. It can be exacerbated significantly by weather conditions and local geography, which can prevent particulate from being dispersed naturally by wind. It is more closely associated with automobile and industrial emissions, which interact frequently to create the particles that cause smog. Smog is the interaction of light with fine solid or liquid particles in the air, known as aerosols. The most common cause of photochemical smog is the interaction of nitrogen oxides and volatile compounds.

Question 36. Which are the primary constituents of photochemical smog?

  1. Carbon dioxide and NO2
  2. Hydrocarbons and CFCs
  3. SO2 and CO
  4. NO2 and hydrocarbons

Answer: 4. NO2 and hydrocarbons

Photochemical smog is formed at high temperatures over cities and towns due to still air, emission of oxides of nitrogen and sulphur and hydrocarbon from automobile exhausts and solar energy. Nitrogen dioxide splits into nitric oxide and nascent oxygen. Nascent oxygen combines with molecular oxygen to form ozone. Ozone reacts with hydrocarbon to form aldehyde and ketone. Nitrogen oxides, oxygen and ketones combine to form Peroxyacyl Nitrates (PAN). In areas with intense solar radiation, photoelectrical smog forms brown air.

NEET Biology Mcq

Question 37. Which is always present in photochemical smog?

  1. Ozone
  2. CO2
  3. SO2
  4. CH4

Answer: 1. Ozone

Ozone is generated in the lower atmosphere during the formation of photochemical smog when nitrogen dioxide splits to produce reactive oxygen atoms, which combine with molecular oxygen. Oxygen molecules split under ultraviolet radiations to produce oxygen atoms which combine with molecular oxygen to form ozone. It is always present in photochemical smog.

Question 38. Photochemical smog affects plants by

  1. Bleaching Of Leaves And Excess Production Of Pigments
  2. Shedding Of Foliage And Increase Of Transpiration
  3. Decolourising And Curling Of Leaves
  4. Degradation of chlorophyll by ozone

Answer: 4. Degradation of chlorophyll by ozone

Ozone is formed by resultant secondary pollutants in smog and degrades the photosynthetic pigments, i.e. chlorophyll of leaves.

Question 39. Choose the correct sequence of air pollution and its components with the effect it produces.

  1. Chemical factory → NO2 → ozone hole
  2. Automobile exhausts → NO2 → greenhouse effect
  3. Heavy industry → CO2 → acid rain
  4. Incinerators → NOx gases → photochemical smog

Answer: 4. Incinerators → NOx gases → photochemical smog

Nitrogen oxides are produced naturally through biological and non-biological activities. Human activity forms nitrogen oxides in the combustion process of industries, automobiles, incinerators and nitrogen fertilisers. Nitrogen oxides give photochemical smog. Thus, the correct sequence of air pollution and its component with the effect it produces is shown in option 4

Question 40. Match the following columns.

NEET Biology Air Pollution and Noise Pollution Match the columns Q 70

Answer: 4. A–3, B–2, C–1

Question 41. Burning of plastics releases

  1. CFC
  2. SO2
  3. Benzopyrene
  4. PCB

Answer: 4. PCB

Polychlorinated Biphenyls (PCB) are released from the burning of plastics.

NEET Biology Mcq

Question 42. The pollutants emitted by jet aeroplanes in the outer atmosphere are known as

  1. Smog
  2. Photochemical Oxidants
  3. Aerosols
  4. Loess

Answer: 4. Loess

Chemicals released in the atmosphere with force in the form of mist or vapours are called aerosols. Jet aeroplanes release aerosols which contain CFC.

Question 43. In the ‘wet system’ of removing gaseous pollutants

  1. Alkali Fluid Is Used
  2. Acidic Fluid Is Used
  3. Neutral fluid is used
  4. None of the above

Answer: 1. Alkali Fluid Is Used

In wet scrubbing processes for gaseous control, a liquid is used to remove pollutants from an exhaust stream. The removal of pollutants in the gaseous stream is done by absorption by the use of scrubbing reagents such as lime or sodium hydroxide (alkaline nature). Thus, the correct answer is 1.

Question 44. Which of the following statements is true?

  1. Benzene hexachloride is a non-biodegradable pollutant.
  2. Anthropogenic air pollutants are natural in origin.
  3. Carbon monoxide is a primary air pollutant.
  4. Sulphur dioxide causes brown air effects during traffic congestion in cities.

Choose the correct option.

  1. 1 and 3
  2. 1 and 2
  3. 2 and 3
  4. 2 and 4
  5. 1 and 4

Answer: 1. 1 and 3

Statements I and III are true, but II and IV are false because Non-biodegradable pollutants either do not degrade or degrade extremely slowly in the natural environment, e.g. DDT, BHC (Benzene Hexachloride). Anthropogenic pollutants are man made pollutants, e.g. sewage, pesticides, fertilisers, etc. Primary air pollutants are those which enter the air directly from the source, e.g. carbon monoxide. In traffic congested cities, the brown air effect is caused due to oxides of nitrogen.

Question 45. Which of the following is most poisonous?

  1. CO
  2. CO2
  3. C
  4. SO2

Answer: 1. CO

Carbon monoxide is produced due to incomplete combustion, metallurgical operations and naturally by plants as well as animals. CO combine with haemoglobin and produce carboxy haemoglobin. At 50 ppm carbon monoxide convert 7.5% of haemoglobin into carboxy haemoglobin. It impair oxygen transport resulting in giddiness, headache, decreased vision, cardiovascular malfunctioning and asphyxia. Thus, CO is most poisonous.

NEET Biology Mcq

Question 46. CO is more toxic than CO2 because 

  1. It Affects The Nervous System
  2. It Damages Lungs
  3. It Reduces The Oxygen Carrying Capacity Of Haemoglobin
  4. It forms acid with water

Answer: 3. It Reduces The Oxygen Carrying Capacity Of Haemoglobin

CO is more toxic because it reduces the oxygen carrying capacity of haemoglobin.

Question 47. According to the Central Pollution Control Board (CPCB), particles that are responsible for causing great harm to human health are of diameter 

  1. 2.5 micrometres
  2. 5.00 micrometres
  3. 10.00 micrometres
  4. 7.5 micrometres

Answer: 1. 2.5 micrometres

According to Central Pollution Control Board (CPCB), particulate size 2.5 micrometres or less in diameter (PM 2.5) are responsible for causing the greatest harm to human health. These fine particulates can be inhaled deep into the lungs and can cause breathing and respiratory symptoms, irritation inflammations and damage to the lungs and premature deaths.

Question 48. Examples of regional pollution are 

  1. Acid Rain
  2. Smog
  3. Both 1 and 2
  4. None of these

Answer: 3. Both 1 and 2

Regional pollutants are those which have been emitted from all sources in a region and have had time to mix, diffuse from their peak concentration and undergo physical, chemical and photochemical reactions. The size of a region is indeterminate, but usually incorporates one or more cities and is on the order of 100 to 10, 000 km2. Examples of regional pollution are acid rain, smog, etc.

Question 49. Rain is called acid rain when its pH is below

  1. 7
  2. 6.5
  3. 6
  4. 5.6

Answer: 4. 5.6

Acid rain refers to precipitation with a pH of less than 5.

Question 50. Acid rain is caused by increase in the atmospheric concentration of 

  1. CO2 and CO
  2. O2 and dust
  3. SO2 and NOv
  4. SO3 and CO

Answer: 3. SO2 and NO2

Acid rain is rainfall and other forms of precipitation with a pH of less than 5. Acid rain is caused by large scale emission of acidic gases into the atmosphere from thermal power plants, industries and automobiles. The common ones are sulphur dioxide, nitrogen oxides (NO2), Volatile Organic carbon (VOCs) and hydrogen chloride.

Question 51. The acid rain destroys vegetation because it contains

  1. Nitrates
  2. H2SO4
  3. O2
  4. CO

Answer: 2. H2SO4

Acid rain contains acidic components, such as sulphuric acid or nitric acid. The type of acid rain that contains water is called wet deposition. Acid rain formed with dust or gasses is called dry deposition. When these rain falls on the soil the soil become acidic. Also the leaves

Question 52. Acid rain occurs above the areas where 

  1. There Is Excess Production Of Nh3 And Coal Gas
  2. Factories Are Expelling SO2 In Air
  3. There Is an Excess Release Of CO2 Due To Increase In Combustion And Respiration
  4. Excess production of gaseous hydrocarbons

Answer: 2. Factories Are Expelling SO2 In Air

Acid rain occurs above the area where factories are expelling of sulphur dioxide into the air. These oxides react with water and precipitate in the form of the acid rain. For example, around the Taj Mahal the presence of the Mathura refinery has caused multiple events of acid rain.

NEET Biology MCQ Chapter Wise

Question 53. In acid rain, sulphuric acid accounts for 

  1. 100% of acid
  2. 70% of acid
  3. 50% of acid
  4. 30% of acid

Answer: 2. 70% of acid

Acid rain is caused by a chemical reaction. Sulphur dioxide on reaction with water and oxygen forms sulphur trioxide which is rapidly converted into sulphuric acid in the presence of water. It accounts for 70% volume in acid rain. Hence, the correct answer is option 2.

Question 54. It is said that Taj Mahal is getting destroyed due to 

  1. Flood In Yamuna River
  2. Decomposition Of Marble As A Result Of High Temperature
  3. Air pollutant released from oil refinery of Mathura
  4. All of the above

Answer: 3. Air pollutant released from oil refinery of Mathura

Threat to Taj Mahal from the Mathura refinery is due to pollutant gases comprising SO2, H2S and nitrogen oxides. They would convert CaCO3 (marble) into calcium sulphate and calcium nitrate.

NEET Biology Mcq

Question 55. Which of the following is the most serious pollutant among sulphur compounds?

  1. Oxides of sulphur
  2. Carbonyl sulphide
  3. Carbon disulphide
  4. Sulphates

Answer: 1. Oxides of sulphur

Oxides of sulphur react with water in the atmosphere and form sulphuri cacid which cause rainwater to get acidic and thus oxides of sulphur cause acid rain. Carbonyl sulphide is a colourless gas which decomposes into carbon dioxide and carbon disulphide. Carbon disulphide is a colourless volatile liquid in its pure form. It is not considered to cause any type of pollution. Sulphates are polyatomic ions derived from sulphur and oxygen. Sulphates generally cause water pollution. So, the correct answer is oxides of sulphur.

Question 56. SO2 pollution affects 

  1. Nucleus
  2. Mitochondria
  3. Ribosome
  4. Vacuole

Answer: 2. Mitochondria

SO2 pollution affects the respiratory system mostly, thus it affects the mitochondria. Higher concentrations of SO2 can result in temporary breathing impairment for children and adults who are active outdoors. Short-term exposures of individuals to elevated SO2 levels while at moderate exertion may result in reduced lung function that may be accompanied by such symptoms as wheezing, chest tightness or shortness of breath.

Question 57. A component of living cell affected by pollutant SO2 is

  1. Nucleus
  2. All Cell Membrane Systems
  3. Cell wall
  4. None of the above

Answer: 2. All Cell Membrane Systems

The component of living cell affected by pollutant SO2 is cell membrane. SO2 replaces oxygen in cellular materials. It affects structural proteins in the membrane and these changes the membrane permeability.

Question 58. The ciliary passage of the respiratory tract gets damaged by

  1. Oxides Of Sulphur
  2. Oxides Of Nitrogen
  3. Oxides of carbon
  4. None of the above

Answer: 1. Oxides Of Sulphur

Oxides of sulphur irritate the respiratory epithelium and impair normal breathing or cause asthma. Thus, oxides of sulphur can damage ciliary passage of respiratory tract.

Question 59. Which of the following are the most suitable indicators of SO2 pollution in the environment?

  1. Algae
  2. Fungi
  3. Lichens
  4. Conifers

Answer: 3. Lichens

Lichens are very sensitive to SO2 pollution. They are completely destroyed at places where there is SO2 pollution in the atmosphere. Therefore, they act as very good indicators of SO2 pollution.

Question 60. SO2 of the atmosphere combines with water to form H SO2 4. When this reaches the soil

  1. It Increases Of Ph Of The Soil
  2. It decreases the pH of the soil
  3. pH remains the same
  4. Soil texture changes

Answer: 2. It decreases the pH of the soil

When sulphur dioxide combines with water and air, it forms sulphuric acid, which is the main component of acid rain. When the rain falls on the soil, the soil become acidic resulting in the lower pH of the soil.

Question 61. Which one of the following statements is incorrect in case of the Bhopal tragedy?

  1. Methyl isocyanate gas leakage took place
  2. Thousands of human beings died
  3. Radioactive fallout engulfed Bhopal
  4. It took place in the night of December 2/3, 1984

Answer: 3. Radioactive fallout engulfed Bhopal

All options are correct except option 3 because the Bhopal gas tragedy took place in the night of December 2/3, 1984, which was worst chemical disaster in history. People started dying within hours and more than 2000 lives were lost in first few days. Methyl isocyanate gas was the main chemical released that engulfed Bhopal that day.

Question 62. The effect of hydrogen sulphide on man leads to which of the following disease?

  1. Paralysis
  2. Sterility
  3. Blindness
  4. None of these

Answer: 4. None of these

HS2 does not accumulate in the body, but repeated/prolonged exposure to moderate levels can cause low blood pressure, headache, loss of appetite and weight. Repeated exposure over time to high levels of HS2 may cause convulsions, coma, brain and heart damage or even death. So, option 4 is correct.

Question 63. In 1984, Bhopal gas tragedy was caused due to leakage of 

  1. Potassium Isocyanate
  2. Methyl Isocyanate
  3. Sodium monoxide
  4. None of the above

Answer: 2. Methyl Isocyanate

Bhopal gas tragedy occurred on 2/3 December, 1984 in which a storage tank containing 36 tonnes of Methyl Isocyanate (MIC) burst in the pesticide manufacturing plant of Union Carbide in Bhopal. MIC is one of the deadliest toxins which when inhaled even in very small doses can kill the animal. It is highly irritating to skin, eyes or mucous membranes and causes death by lung oedema. It is also a carcinogenic agent.

Question 64. What were the effects of the Bhopal gas tragedy?

  1. Death of newborn babies
  2. Newborn babies were born with deformities
  3. Vegetation in the radius of 3.5 sq km around the Union Carbide factory was affected
  4. All of the above

Answer: 4. All of the above

Approximately 3598 deaths have resulted from the Bhopal Gas tragedy. Chronic inflammatory damage to the eye and lungs is the main cause. Other effects of this tragedy are the death of newborns and babies were born with deformities and the vegetation in the radius of 3.5 sq. km around the Union Carbide factory was affected.

Thus, option 4 is correct.

Question 65. The TLY (Threshold Limit Value) of methyl isocyanate responsible for the Bhopal gas tragedy is 

  1. 0.002 ppm
  2. 0.02 ppm
  3. 0.2 ppm
  4. 0.005 ppm

Answer: 2. 0.02 ppm

Methyl isocyanate is a volatile, toxic chemical used to manufacture carbamate pesticides. Bhopal gas tragedy (2/3 December 1984) is a case of air pollution in which MIC (Methyl Isocyanate) gas released from a fertiliser manufacturing plant of Union Carbide caused the death of approximately 3598 persons. MIC causes irritation in the eyes which may result in blindness, various lung diseases may result in death. Workers exposed to the MIC 8-hours threshold limit value of 0.02 ppm (46) are exposed to approximately 460 mg MIC in a workday.

Question 66. Various atmospheric air borne pollutants may cause

  1. Pneumonia
  2. Diarrhoea
  3. Respiratory Diseases And Allergies
  4. Cholera

Answer: 3. Respiratory Diseases And Allergies

Airborne pollutant enters via the respiratory tract in humans and cause respiratory diseases and allergies on the skin.

Question 67. Sulphur dioxide causes

  1. Emphysema
  2. Bronchitis
  3. Asthma
  4. All of the above

Answer: 4. All of the above

Sulphur dioxide causes respiratory tract diseases like asthma, bronchitis, cancer, emphysema, etc.

Question 68. Accumulation of fluoride in plant leaves causes

  1. Chlorosis
  2. Stem Rot
  3. Tip burn
  4. None of the above

Answer: 3. Stem Rot

Fluoride is an accumulative poison in plant foliage. Accumulation may be gradual over time and causes tip burn in leaves.

Question 69. Which of the following is pollution-related disorder?

  1. Silicosis
  2. Pneumoconiosis
  3. Fluorosis
  4. Leprosies

Answer: 1. Silicosis and 3. Fluorosis

(1 & 3) Fluorides are given out during the refining of metals. Fluorides cause fluorosis. This is a pollution-related disorder. When the level of fluoride content in drinking water becomes as high as 3-12 mg/L, the water becomes polluted. It affects teeth as well as bones. Silicosis is caused by inhalation of dust containing free silica or silicon dioxide, especially by workers engaged in mining, pottery, ceramic industry, sandblasting, and building and construction industries. So, both fluorosis and silicosis are pollution-related disorders.

Question 70. Bagassosis is

  1. Hypersensitivity Pneumonitis Due To Exposure To Fibrous Cellulose Part Of Crushed Sugarcane
  2. Is An Asthma Like Condition Produced Due To Exposure To Flax Or Cotton Fibres
  3. Hypersensitivity Pneumonitis Due To Exposure To Damp Hay Or Grain
  4. Hypersensitivity pneumonitis due to exposure to paprika

Answer: 1. Hypersensitivity Pneumonitis Due To Exposure To Fibrous Cellulose Part Of Crushed Sugarcane

Bagassosis is an interstitial lung disease, is a type of hypersensitivity pneumonitis due to exposure to the fibrous cellulose part of crushed sugarcane.

Question 71. Which of the following diseases is caused or aggravated by pollution?

  1. Bronchitis
  2. Rheumatism
  3. Scurvy
  4. Haemophilia

Answer: 1. Bronchitis

Inflammation of the lining of bronchial tubes causes bronchitis. Particulate matter, a prominent source of air pollution causes pneumoconiosis, byssinosis, emphysema, bronchitis and severe coughing. Rheumatism is a disease of joints. Scurvy is a deficiency disease caused due to the deficiency of vitamin C. Haemophilia is a genetic sex-linked disorder in which blood fails to clot normally.

Question 72. Black foot disease occurs due to 

  1. Cadmium
  2. Mercury
  3. Arsenic
  4. Copper

Answer: 3. Arsenic

Black foot disease is caused by excessive consumption of arsenic in water. Black foot disease, with gangrene-like symptoms, affects the feet and sometimes the fingers. It is a rare peripheral vascular disease found mainly in the Province of Taiwan (China). The best prevention for this disease is avoidance of exposure to arsenic by ingesting clean drinking water.

Question 73. ‘Knock knee’ syndrome occurs due to the pollution of 

  1. Nitrates
  2. Phosphates
  3. Fluorides
  4. Heavy metals

Answer: 3. Fluorides

Fluorides are given out during the refining of minerals (e.g. aluminium) and can contaminate groundwater. Excess fluorides in drinking water causes teeth deformity, hardened bones and stiff and painful joints (skeletal fluorosis) or knock knee disease.

NEET Biology MCQ Chapter Wise

Question 74. Black lung disease is common in

  1. Coal Miners Odisha Jee 2005
  2. Refinery Workers
  3. Petrochemical Industry
  4. Farmers

Answer: 1. Coal Miners Odisha Jee 2005

Black lung is a man-made, occupational lung disease that is contracted by prolonged breathing of coal mine dust. The disease is called ‘black lung’ because it causes the lungs to look black instead of pink. Medically, it is a type of pneumoconiosis called coal worker’s pneumoconiosis.

Question 75. Assertion Fluorides destroy vegetation. Reason (R) Fluorosis hinders respiration.

  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true, but R is not the correct explanation of A
  3. A is true, but R is false
  4. Both A and R are false

Answer: 3. A is true, but R is false

A is true, but R is false because, In plants, fluorides combine chemically with Mg 2+ of chlorophyll and hence inhibit photosynthesis. It causes the abscission of leaf and fruit and hence destroy vegetation. In human beings, the typical symptom of excess fluorine (fluorosis) is the mottling of teeth.

Question 76. In a small town, pasture is contaminated with airborne fluorides. Grazing animals in that area will suffer from

  1. Dental And Bone Diseases
  2. Goitre Disease
  3. Thyroid Disease
  4. Beriberi

Answer: 1. Dental And Bone Diseases

Fluorides cause fluorosis. It is a disease which is defined by mottling of teeth, and abnormal bones that are liable to fracture because fluorine replaces Ca²+ and makes the bones brittle. Fluoride is light airborne particulate matter.

Question 77. Fly-ash is a/an 

  1. Insectivorous Plant
  2. Light Airborne Particulate Matter
  3. New Name Of Orchid Plant
  4. Causal organisms of various diseases

Answer: 2. Light Airborne Particulate Matter

Fly ashes are finely divided residues resulting from the combustion of ground or powdered coal. It is discharged as an airborne emission. It is a light airborne particulate matter.

Question 78. Assertion COPD disease is very common in Kanpur. Reason (R) It is disease of respiratory tract occurring due to pollution.

  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true, but R is not the correct explanation of A
  3. A is true, but R is false
  4. Both A and R are false

Answer: 1. Both A and R are true and R is the correct explanation of A

Both A and R are true and R is the correct explanation of A. Kanpur is the most polluted city in the country, with high quantities of Respirable Suspended Particulate Matter (RSPM) in the air according to the results of a survey carried out by the Central Pollution Control Board (CPCB) in 2002. Chronic Obstructive Pulmonary Disease (COPD) is a condition characterised by narrowing of the air ways, but these changes are permanent rather than reversible. COPD is caused by exposure to pollutant that produce inflammation.

Question 79. Fibrosis of lung is due to

  1. Asbestos And Cement Particles
  2. Mercury
  3. China Clay
  4. Rubber Particles

Answer: 1. Asbestos And Cement Particles

The asbestos and cement particles from industries are added to the air and these are inhaled by workers and people living nearby. These particles form a coating on the internal lining of lungs and bronchioles causing very serious fibrosis which may later lead to cancer.

Question 80. In silicosis, the silica particles can be seen

  1. As Needle Shaped Crystal By Light Microscopy
  2. As Needle Shaped Crystals Under Polarised Light
  3. As Maltese Cross Shaped Crystal By Light Microscopy
  4. As Maltese Cross Shaped Crystals Under Polarised Light

Answer: 2. As Needle Shaped Crystals Under Polarised Light

Silicosis is a lung disease caused by breathing-in tiny bits of silica. It is characterised by the fibrotic nodules with the concentric onion skinned arrangement of collagen fibres, central hyalinisation and a cellular peripheral zone with light birefringent particle seen under polarised light.

Question 81. Which of the following ailments is a result of inhalation of fine particulate matter?

  1. Irritation
  2. Inflammation
  3. Damage of lungs and premature deaths
  4. All of the above

Answer: 4. Damage of lungs and premature deaths

Fine particulates can be inhaled deep into the lungs and can cause breathing and respiratory symptoms, irritation, inflammations and damage of the lungs and premature deaths.

Question 82. 3, 4-benzopyrene causes

  1. Leukaemia
  2. Cytosilicosis
  3. Lung Cancer
  4. Tuberculosis

Answer: 3. Lung Cancer

There are a number of chemicals, which causes lung cancer and one of them is 3,4-benzopyrene. Benzopyrene mutates three positions is the p53 gene, which leads to lung cancer. Hence, the correct option is 3.

NEET Biology MCQ Chapter Wise

Question 83. What is true about the Euro II norms? It is laid down to

  1. Control Sulphur At 350 Ppm In Diesel And 150 Ppm In Petrol
  2. Reduce Sulphur Level To 50 Ppm In Gasoline
  3. Reduce Sulphur Level To 200 Ppm In Diesel And Petrol
  4. Reduce sulphur level to 200 ppm in diesel and 100 ppm LPG

Answer: 1. Control Sulphur At 350 Ppm In Diesel And 150 Ppm In Petrol

Euro 2 norms were stipulated to control sulphur content at 350 ppm in diesel and 150 ppm in petrol and aromatic hydrocarbons are to be contained at 42%.

Question 84. All automobiles and fuel (petrol and diesel) were to have met the Euro III emission specification in eleven Indian cities from 1 April 2005 and have to meet the Euro IV norms by

  1. 1 April 2007
  2. 1 April 2008
  3. 1 April 2009
  4. 1 April 2010

Answer: 4. 1 April 2010

All automobiles and fuel were to have met the Euro III emission specification in eleven Indian cities from 1 April, 2005 and have to meet the Euro IV norms by 1 April 2010.

Question 85. Air (Prevention and Control of Pollution) Act was ammended in 1987 to include among pollutants

  1. Vehicular Exhaust Neet 2020
  2. Allergy Causing Pollen
  3. Noise
  4. Particulates of size 2.5 micrometers or below

Answer: 3. Noise

Air (Prvention and Control of Pollution) Act was ammended in 1987 to include noise as pollutant.

Question 86. Noise is

  1. Loud Sound
  2. Sound Of High Frequency
  3. Unwanted Sound
  4. Constant sound

Answer: 3. Unwanted Sound

Noise is defined as an undesired high level of sound. It is a physical form of pollution that affects the receiver directly. Noise or unwanted sound has a value of 80 dB and above.

Question 87. What is the intensity of sound in normal conversation?

  1. 10-20 dB
  2. 30-60 dB
  3. 70-90 dB
  4. 120-150 dB

Answer: 3. 70-90 dB

The unit of sound level is decibel. Moderate conversation produces 30–60 dB sound.

NEET Biology MCQ Chapter Wise

Question 88. During day time sound level in the silent zone is about 

  1. 20 dB
  2. 30 dB
  3. 40 dB
  4. 50 dB

Answer: 4. 50 dB

Sound is expressed in decibels (dB). A decibel is one-tenth of bel, a unit of measuring electric or acoustic power. The World Health Organisation (WHO) has recommended a noise of less than 75 decibels for industries. The National Pollution Control Board has recommended the following maximum permissible noise level in Indian cities

NEET Biology Air Pollution and Noise Pollution National Pollution Control Board

Question 89. Noise pollution is measured in

  1. Decibels
  2. Amperes
  3. Fathoms
  4. Ohm

Answer: 1. Decibels

Noise pollution is measured in decibels (dB).

Question 90. Sound above what level is considered hazardous noise pollution?

  1. Above 80 dB
  2. Above 120 dB
  3. Above 150 dB
  4. Below 30 dB

Answer: 1. Above 80 dB

Sound above 80 dB becomes hazardous to humans.

Question 91. Consider the following statements regarding noise.

  1. Causes psychological disorders in humans.
  2. Causes physiological disorders in humans.
  3. Unit of measurement of noise dB.
  4. 150 dB is tolerable for humans.

Choose an option containing correct statements.

  1. 1 and 4
  2. 1 and 2
  3. 1, 2 and 3
  4. 1 and 3

Answer: 3. 1,2, and 3

All statements are correct except IV because Noise pollution causes psychological and physiological disorders in humans. Noise is measured in dB unit. Above 80 dB noise became hazardous for human.

Question 92. The green muffler is used against which type of pollution?

  1. Air
  2. Soil
  3. Water
  4. Noise

Answer: 4. Noise

Green plants are being planted along road side to check noise pollution. This is known as green muffler.

Question 93. The anxiety and stress reactions are caused by 

  1. Air Pollution
  2. Noise Pollution
  3. Water Pollution
  4. Nuclear pollution

Answer: 2. Noise Pollution

NEET Biology MCQ Chapter Wise

Noise pollution has several adverse effects on human health as The first effects of noise are anxiety and stress however, in extreme cases it may lead to fright. Noise causes headache by dilating blood vessels of the brain, eye strain by dilating the pupil, digestive spasms through anxiety and high blood pressure by increasing cholesterol level in the blood. Noise can impair the development of nervous system of unborn babies which leads to abnormal behaviour in later life.

Question 94. Assertion Inhabitants close to very busy airports are likely to experience health hazards. Reason (R) Sound level of jet aeroplanes usually exceeds 160 dB.

  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true, but R is not the correct explanation of A
  3. A is true, but R is false
  4. Both A and R are false

Answer: 1. Both A and R are true and R is the correct explanation of A

Both A and R are true and R is the correct explanation of A. Noise level up to 65 dB (decibel) in a commercial area and up to 75 dB in industrial areas is permitted as per law. Prolonged exposure to noise level to 80 dB or more leads to loss of hearing ability, fatigue, nervousness, fever, hypertension, gastric disorder, increase in cholesterol level and dilation of pupil of the eye. As the jet aeroplanes have the noise up to 150-160 dB or more, the inhabitants in the vicinity of busy airports are likely to experience above health hazards. Maximum noise level is recorded in rockets, i.e. 180 dB.

Question 95. Which of the following physiological manifestations is the result of noise pollution?

  1. Dilation of pupil
  2. Increase in the rate of heartbeat
  3. Constriction of blood vessels
  4. All of the above

Answer: 4. All of the above

Because of noise pollution nervous tension, blood pressure and heart trouble increases. It causes constriction of blood vessels and increases the rate of heartbeat. Noise pollution also increases digestive spasms and causes the dilation of the pupils of the eyes. It causes impairment of night vision and decreases the rate of colour perception.

Class 10 Biology Nutrition Question and Answers

Class 10 Biology  Life Processes Nutrition Short Answer Type Questions

Question 1. Where do plants get each of the raw materials required for photosynthesis?
Answer:

There are three raw materials besides sunlight. They are carbon dioxide, water and mineral salts.

  1. Carbon Dioxide. Atmosphere through stomata.
  2. Water. Soil through roots.
  3. Minerals. Soil with the help of roots.

Question 2. What is the role of acid in our stomach?
Answer:

Acid (HCl) is present in our gastric juice. It has several functions:

  1. Killing of germs entering along with food,
  2. Softening of food.
  3. Stoppage of action of salivary amylase.
  4. Conversion of proenzyme pepsinogen into enzyme pepsin,
  5. Acidify food for the action of the enzyme pepsin.

Question 3. What is the function of digestive enzymes?
Answer:

Digestive enzymes are hydrolytic biocatalysts (hydrolases) which bring about the break up of complex substances of food into simpler, soluble and absorbable components.

Question 4. How is the small Intestine designed to absorb digested food?
Answer:

  1. The small intestine is the longest part of the alimentary canal.
  2. It is the area where all components of food are completely digested.
  3. The small intestine has a number of transverse folds or villi or increasing internal surface area.
  4. Villi are richly supplied with blood capillaries and a lymph vessel lacteal for quick removal of absorbed food.
  5. Enterocytes or surface cells of the small intestine have a brush border or microvilli. Microvilli increase the surface area of the cells several times for the absorption of food.

Class 10 Biology  Life Processes Nutrition Multiple Choice Questions And Answers

Question 1. Most of the digestion and absorption of the food takes place in the

  1. Small intestine
  2. Liver
  3. Stomach
  4. Large intestine

Answer: 1. Small intestine

Question 2. Pseudopodia are

  1. Small hair-like structures present in unicellular organisms
  2. False feet developed in some unicellular organisms
  3. Long tube-like structures coming out of the mouth
  4. Suckers which are attached to the walls of the intestine.

Answer: 2. False feet developed in some unicellular organisms

Class 10 Biology  Life Processes Nutrition Short Answer Type Questions

Question 1. Explain the significance of peristaltic movements that occur all along the gut during digestion.
Answer:

They are alternate contraction and expansion movements from the oesophagus to the large intestine. It moves the food forward automatically in a regulated manner depending upon the time required for the action of digestive enzymes and absorption of the digested materials.

Question 2. Leaves of a healthy potted plant were coated with Vaseline. Will the plant remain healthy for long? Give a reason for your answer.
Answer:

No. Due to the covering of the leaf with vaseline, the stomata become blocked. The plant can neither receive carbon dioxide for photosynthesis nor oxygen for respiration. There is no cooling by transpiration. The effect of sunburn can also occur. Later the plant with wither and die.

Question 3. Tooth enamel is the hardest substance in our body. Name the compound of which it is made up of At which pH of the mouth it gets eroded. State the role of bacteria present in the mouth. Suggest a method to prevent tooth decay.
Answer:

Enamel consists of calcium phosphate (95%) and proteins amelogenin and enamelin (5%). Enamel gets corroded in acidic conditions as created by plaque-forming bacteria Streptococcus mutans.

Brushing the teeth after each meal or at least twice a day will prevent plaque formation and tooth decay.

Question 4. Name the cells that control the opening and closing of the stomatal pore. How do they perform their function?
Answer:

Guard cells. Guard cells bring about the opening and closing of stomata by gain or loss of turgidity. When the stomata are open, guard cells withdraw K” ions from the surrounding epidermal cells. They also obtain water from the surrounding cells due to endosmosis.

Their elastic outer wall bends outwardly pulling the thickener wall also. This creates a pore in between the two guard cells. During closure movement, guard cells lose k ions and water to the surrounding epidermal cells. They become flaccid. The inner walls come closer and closer to the stomatal pore.

Question 5.

  1. Name two organisms that obtain food through parasitic mode.
  2. How do fungi obtain their food?

Answer:

  1. Parasitic Mode of Nutrition. Cuscuta (plant), Ascaris (animal).
  2. Fungi are generally saprophytes. They obtain nourishment from organic remains by first pouring digestive enzymes over them and then absorbing the dissolved materials.

Question 6. List two different functions performed by the pancreas in our body.
Answer:

The pancreas is a heterochrony gland

  1. Its exocrine part secretes digestive juice called pancreatic juice.
  2. Its endocrine region secretes hormones, insulin and glucagon.

Question 7. List in tabular form two differences between pepsin and trypsin.
Answer:

Nutrition Differences Between Pesin And Trypsin

Question 8.

  1. Name the process by which autotrophs prepare their own food.
  2. List three events which occur during this process.
  3. State two sources from which plants obtain nitrogen for the synthesis of proteins and other compounds.

Answer:

  1. Photosynthesis.
  2. Events In Photosynthesis:
    1. Photolysis of water.
    2. Absorption of solar energy and production of assimilatory power (ATP + NADPH).
    3. Reduction of carbon dioxide.
  3. Nitrate and Ammonium from soil.

Question 9. What are the Final products formed after the digestion of carbohydrates, proteins and fats?
Answer:

  1. Carbohydrates – Glucose.
  2. Proteins – Amino acids.
  3. Fats – Fatty acids and Glycerol.

Question 10. Mention the role of the following in digestion

  1. Pepsin
  2. Saliva
  3. Villi
  4. Bile juice.

Answer:

  1. Pepsin. Breaks down protein into peptones and proteoses in an acidic medium.
  2. Saliva.
    1. Moistening the food
    2. Sterilising the food by lysozyme
    3. Partial digestion of starch into maltose.
  3. Villi. Absorption of digested food by increasing surface area and having microvilli over their cells.
  4. Bile Juice.
    1. Neutralizing the acidity and making the chyme coming from the stomach alkaline.
    2. Emulsification of fat.

Question 11. List the role of each of the following in the digestive system:

  1. Muscles of stomach
  2. Hydrochloric acid
  3. Mucus.

Answer:

  1. Muscles of Stomach.
    • Mixing of gastric juice with food,
    • Breaking food into smaller pieces by churning movement.
  2. Hydrochloric Acid. It sterilizes, softens and acidifies the food in the stomach for the action of enzyme peps
  3. Mucus.
    • It lubricates the food for easy passage of food in the alimentary canal.
    • The mucus protects the wall of the alimentary canal from its own digestive enzymes.

Question 12.

  1. Why does a piece of bread taste sweet when chewed for some time?
  2. “Cellulose acts as a roughage in man but serves as a source of nutrients in cow”. Justify the statement.

Answer:

  1. In the mouth, food mixes with saliva. Saliva contains the enzyme ptyalin or salivary amylase that hydrolyses star to sweet sugar maltose.
  2. In human beings, cellulose cannot be digested as they do not have any enzyme or bacteria to metabolize Cellulose functions as roughage in the human digestive system. In cows, the stomach contains cellulose ferment bacteria that convert cellulose to glucose state.

Question 13. What is photosynthesis?
Answer:

Photosynthesis is the process of the formation of organic food from inorganic raw materials (CO2, H2O) with the he! of sunlight inside chlorophyll-containing cells.

Question 14. Complete the following flow chart as given instructions
Answer:

Nutrition Flowchart For Instructions

  1. Mucus— (d) Moistening of food, protection of the inner lining of the stomach from HCl.
  2. HCI—(e) Killing of germs, softening of food and providing acidic medium.
  3. Pepsin—(f) Digests proteins in an acidic medium.

Question 15. State the role played by the following in the process of digestion

  1. Explain
    1. Enzyme trypsin
    2. Enzyme lipase
  2. List two functions of finger-like projections present in the small intestine.

Answer:

  1. That is
    1. Trypsin. Digestion of proteins.
    2. Lipase. Digestion of emulsified fat.
  2. Villi.
    1. Brings intestinal wall in contact with food
    2. Increasing surface area for digestion and absorption of food.

Class 10 Biology  Life Processes Nutrition Long Answer Type Questions And Answers

Question 1. List two differences between holozoic nutrition and saprophytic nutrition. Give two examples of each of them.
Answer:

Nutrition Differences Between Holozoic Nutrition And Saprophytic Nutrition

State the roles of the liver and pancreas.
Answer:

  1. Role Of Liver. Storage of glycogen, formation of urea, formation of prothrombin, fibrinogen and heparin, secretion of bile, detoxification, and elimination of pathogens.
  2. Role Of Pancreas. Secretion of hormones insulin and glucagon, secretion of pancreatic juice for digestion-pancreatic amylase, pancreatic lipase and trypsin.

Name the organ which performs the following functions in humans:

  1. Absorption of digested food
  2. Absorption of water.

Answer:

  1. Absorption Of Digested Food Mostly in the ileum part of the small intestine.
  2. Absorption Of Water. Partly in the ileum and partly in the large intestine.

Explain the statement, “Bile does not contain any enzyme but it is essential for digestion”.
Answer:

Role Of Bile In Digestion.

  1. Neutralization and alkalisation of chyme coming from the stomach.
  2. Emulsification of fat.

Question 2. Name the glands, associated with the digestion of starch in the human digestive tract and mention their role.
Answer:

  1. Salivary Glands. Their secretion called saliva contains the enzyme ptyalin or salivary amylase which converts part of starch into maltose and dextrins.
  2. Pancreas. Pancreatic amylase changes starch and dextrins into maltose.
  3. Intestinal Glands. They secrete disaccharidases which hydrolyse disaccharides into monosaccharides:
    • Sucrose-Glucose + Fructose.
    • Lactose-Glucose + Galactose.
    • Maltose-Glucose + Glucose.

How is the required pH maintained in the stomach and small intestine?
Answer:

  1. In the stomach, acidic pH is maintained by HCl.
  2. In the intestine, alkaline pH is maintained by bile and pancreatic juice.

Question 3. Mention the location of four major glands associated with the digestive system of humans and explain the functions of each.
Answer:

Four Major Glands. Salivary glands, liver, pancreas, gastric glands.

  1. Salivary Glands. Three pairs— parotid below the ear, submaxillary at an angle of the lower jaw and sublingual below the tongue.
    • Salivary glands secrete saliva. Saliva moistens the food, disinfects it with lysozyme and partly digests starch/glycogen into maltose.
  2. Liver It occurs on the right upper side of the abdomen below the diaphragm. The liver secretes bile for neutralization an alkalisation of chyme, emulsification of fat, storage of glycogen and many nondigestive functions.
  3. Pancreas. It lies behind the stomach with its head connected to the duodenum by the pancreatic duct. Digestive secretion of the pancreas, called pancreatic juice contains lipolytic, amylolytic and proteolytic enzymes for the digestion of all major components of food.
  4. Gastric Glands. They occur inside the stomach. The secretion called gastric juice contains HCl and the proteolytic enzyme pepsin which breaks down protein in an acidic medium to form peptones and proteoses.

Question 4. What function is served by:

  1. Pyloric sphincter
  2. Anal sphincter.

Answer:

Pyloric Sphincter — Release of chyme into duodenum.

Anal Sphincter — Release of faeces through the anus.

Question 5. What is the nutrition in

  1. Fungi
  2. Amarbel?

Answer:

  1. Nutrition In Fungi. It is commonly saprotrophic, for example, Rhizopus and mushroom. A few fungi are parasitic.
  2. Amarbel (Cuscuta). Heterotrophic, holoparasite.

Name the part of the alimentary canal where:

  1. Food is completely digested
  2. Secrete juice that has trypsin
  3. Secrete bile
  4. Absorb water from unabsorbed food.

Answer:

Mention the names of any two secretions by the gastric glands and state one role played by each in our body.
Answer:

  1. Gastric Glands. HCl and Pepsin.
  2. Hydrochloric acid (HCl).
  3. Pepsin.

Question 6. Write the function of the following in the human alimentary canal:

  1. Saliva
  2. HCl in stomach
  3. Bile juice
  4. Villi.

Answer:

  1. Function of Saliva,
    • Moistening the food
    • Conversion of starch into maltose by salivary amylase.
  2. Function of HCl.
    • Disinfection of food
    • Acidification of food for the action of pepsin.
  3. The function of Bile Juice:
    • Deacidification of food and making it alkaline
    • Emulsification of fat.
  4. The function of Villi. Increase the surface area of the small intestine for absorption.

Write one function of the following enzymes:

  1. Pepsin
  2. Lipase.

Answer:

  1. Function of Pepsin. Digestion of proteins to form peptones and proteoses.
  2. Function of Lipase. Digestion of fats into fatty acids and glycerol.

Question 7. A gas is released during photosynthesis. Name the gas and also state the way by which the gas is evolved.
Answer:

Oxygen. It is formed during photosynthetic splitting or photolysis of water.

Question 8. Why is nutrition necessary for the human body?
Answer:

Nutrition Is Required For

  1. Obtaining energy
  2. Materials for the formation of new cells
  3. Repair of damaged cells and tissues
  4. Development of immunity.

Class 11 Chemistry Classification of Elements and Periodicity in Properties Question and Answers

Chemistry Classification of Elements and Periodicity in Properties Question and Answers

Question 1. Write the name of the element which is diagonally related to the elementberyllium. Three elements A, B, and C have atomic numbers 11, 14, and 17 respectively. State the blocks in the periodic table to which elements A and C belong. Write the formulas ofthe compounds formed between B and C and A and C. State the nature ofthe bonds.
Answer: Aluminium (Al)

11A: 1, 14B: ls22s22p63s23p2,

17C: ls22s22p63s23p5

From the electronic configuration is evident that A is an s -a block element whereas C is a p -a block element.

The compound formed between B and C has the formula BC4 and the nature of the bond is covalent. A compound formed by a combination of A and C is AC (electrovalent).

Question 2. Classify as basic, amphoteric or acidic: BeO, A1203, CaO, Si02 State the modem periodic law.
Answer: BeO, A1203: amphoteric; CaO: basic; Si02: acidic

Question 3. Write the name of the element which is diagonally related to the elementberyllium.

Three elements A, B, and C have atomic numbers 11, 14, and 17 respectively.

State the blocks in the periodic table to which elements A and C belong. Write the formulas ofthe compounds formed between B and C and A and C. State the nature ofthe bonds.

Symbol of one transition element — Fe.

Symbol of one inner-transition element — Ce

Ionic radius is highest for O2- for the given case.

Electronegativity is lowest for Na for the given case.

Ionization energy is highest for Ar for the given case.

Question 4. State the group number in the modem periodic table where solid, liquid, and gaseous elements are present at room temperature. Identify solid, and liquid elements. Indicate the given elements as alkali metal, alkaline earth metal, coinage metal, or chalcogen: Li, Ca, S, Cu.
Answer: Group-17 of modem periodic table contains solid (iodine, I2), liquid (bromine, Br2), and gaseous (chlorine, Cl2) elements at the same time

Group 17 of modern periodic table contains solid

(iodine, I2), liquid (bromine, Br2) and gaseous

(chlorine, Cl2) elements at the same time.

Question 5. Among which of the following pairs of elements, the first has lesser ionization energy than the second—

  1. Na, K
  2. N, O
  3. B, Be
  4. Br, I

Answer: 3. B, Be

Question 6. Name a pair of elements that exhibit a diagonal relationship.
Answer: Li and Mg.

Question 7. Name a chalcogen and an alkaline earth metal.
Answer: Oxygen (O) is an example of chalcogen and magnesium (Mg) is an alkaline earth metal

Question 10. Mention the names of the most electropositive and most
electronegative stable elements in the periodic table. 2
Answer:  

Question 11. What is a metalloid? Give one example. 
Answer: The most electropositive stable element of the periodic table is cesium (Cs) and the most electronegative stable element of the periodic table is fluorine (F).

Question 11. Which has greater ionization energy and why—S or P?
Answer: Phas greater ionization energy than S. According to Hund’s rule the half-filled electronic configuration is highly stable and the outermost 3p -orbital of the P atom is half-filled.

To produce P+ ions by removing an electron from the 3p orbital ofPrequireshuge energyresulting a very high value of ionization energy.

On the other hand, the S atom has a partially filled 3p -orbital which attains a stable half-filled electronic configuration in the S+ ion.

Therefore, to produce S to S+ comparatively lower energy is required resulting in a lower value of ionization energy.

Question 13. Arrange as directed: 1. Si02, N02, A1203, C102 (Increasing acidic property) 2. Na+, Al3+, F-, Cl- (Increasing ionic radius) Arrange the following ions in the increasing order of their ionic radius. F-, Mg2+, Al3+, O2-
Answer: A1203 < Si02 < NOz < C102

Al3+ < Na+ < F- < Cl-

Question 14. Arrange the following ions in the increasing order of their ionic radius. F-, Mg2+, Al3+, O2-
Answer: Al3+ < Mg2+ < F- < O2-

Question 15. Which ofthe given elements has the lowest electron affinity

  1. C
  2. P
  3. O
  4. S

Answer: 2. P

Question 16. Arrange the following elements according to their decreasing oxidising ability: I, Br, F, Cl.
Answer: F > Cl > Br >I

Question 17. Arrange as per instruction: Na20, B203, A1203, MgO (increasing order of basicity) Be, 0, Cl, I (increasing order of electro-negativity)
Answer: B203 < A1203 < MgO < Na20

Be <I < Cl < O

Question 19. Two atoms X and Y have electronic configurations [He] 2sz2p3 and [Ne] 3s23p2 respectively. Which period and group of the periodic table do they belong to? The second electron affinity of an element is always endothermic. Explain.
Answer: X —1 Period 2, Group-15. Y —1 Period 3, Group-14.

Question 20. The Electron affinity of Cl is greater than that of. Explain. Indicate the basic/acidic behavior of the hydrides of the following elements: F, C, O, and N Metallic properties of elements in the periodic table.
Answer: HF: acidic, CH4: neutral, H20: neutral, NH3: basic (From left to right along a period the relative acidity of the hydrides ofthe elements increase).

Question 21. Metallic property of elements down the group in the periodic table
Answer: HF: acidic, CH4: neutral, H20: neutral, NH3: basic (From left to right along a period the relative acidity of the hydrides ofthe elements increases).

Question 22. Arrange the following in increasing order of ionic radius: Na+, F-, O2-, Al3+, N3-.
Answer: Al3+ < Na+ < F- < O2- < N3

Question 23. Why does nitrogen have a higher ionization enthalpy than that of Oxygen? Arrange the following in increasing order of acidity: N02, A1203, Si02, C102
Answer: Na < A1 < Mg < Si

Question What will be the order of, Mg, A1, and Sn terms of the first ionisation enthalpy?
Answer: Mg2+ < Na+ < F- < O2-

Question 25. Arrange the following ions in order of increasing ionic radii: Na+, F-, O2-, Mg2+.
Answer: Mg2+ < Na+ < F- < O2-

Question 26. Why is the electron-gain enthalpy of oxygen less than
that of sulfur? Arrange the following metal oxides in terms of ascending order of basicity: ZnO, MgO, CaO, CuO.
Answer: Electron-gain enthalpy of O is less than that of S as the 2p -orbital of the outermost shell of O-atom is much smaller in size than the 3p -orbital of the outermost shell of the S-atom.

So, the additional electron-electron repulsive force produced due to the addition of one electron to the 2p -orbital of the oxygen atom from outside is more than the additional electron-electron repulsive force developed by the addition of one electron to the 3p -orbital of the S-atom.

So the electron-gain enthalpy of the O-atom becomes less than that S-atom.

\(\mathrm{CuO}<\mathrm{ZnO}<\mathrm{MgO}<\mathrm{CaO}\)

Question 27. Why is the first ionization enthalpy of helium maximum among all the elements? Arrange the given compounds in terms of ascending order of oxidizing property: HC1, HBr, HI, HF
Answer: Configuration of He is Is2, i.e., all electrons of He are presenting Is -orbital. These electrons are attracted very strongly by the nucleus.

Further, there are no inner electronic orbitals to shield these electrons from the pull ofthe nucleus.

So removal of an electron from the Is -orbital requires a large amount of energy. Therefore, the first ionization enthalpy ofHe is maximum.

Hydrogen halides do not show oxidizing properties. However, the reducing power of the hydrogen halides follows the sequence: HF <HCl < HBr < H

Question 28. Determine the position of an element in the long form of the periodic table if its electronic configuration is [18Ar]3d104s2.
Answer: It is a d -block element. Hence period no. = principle quantum no. ofthe outermost shell = 4 and its group no. = total electronin 3d and 4s -orbital = 10 + 2 = 12.

Question 29. Mention the name and the position of two elements, one of which is the most electronegative and the most electropositive periodic table.
Answer: Most electronegative element =F (group 17, period-2)

Most electropositive element = Cs (group 1, period-6)

Question 30. Which of the following two elements has a diagonal relationship? Li, Be, A1, and Si Between 2gCu and igK which one has higher ionization enthalpy and why do Be and Al
Answer:

  1. Electronic configuration of 19K: ls22s22p63s23p64s1
  2. Electronic configuration of 2gCu: ls22s22p63s23p63d104s1

The nucleus of the Cu-atom contains 10 more protons than that of the K-atom.

Additional nuclear pull on the outermost electron (4s1) in Cu-atom is not counter-balanced by the shielding effect of ten 3d-electrons because orbitals have poor screening effects.

Thus the effective nuclear charge acting on the 4s-electron of copper is greater than that acting on the 4s-electron of

Question 31. Arrange the following ions in ascending order of radius: Na+, F-, O2-, Mg2+
Answer: Mg2+ < Na+ <F- < 02-

Question 32. Is the electronegativity of Sn2+ and Sn4+ equal or different?
Answer: Elements in a higher oxidation state have higher electronegativity compared to the elements in a lower oxidation state. Thus Sn4+ is more electronegative than Sn2

Question 33. What is the oxidation state of TI in the compound t1i3? Which is the stronger oxidizing agent between CO2 and PbO2 and why?
Answer: +1

PbO2. Due to the inert pair effect, Pb is very unstable in the +4 oxidation state. Thus Pb4+ can be easily reduced to Pb2.

Question 34. Which iw is more stable between BC13 and T1C13 and why? What is the oxidation state of Zn in Zn-Hg Arrange the following elements in the increasing order of their first ionisation enthalpy.IX Be. Na, Mg 56 Arrange the following elements in the decreasing order of their electro-negativity. Si N. F. Cl
Answer: Because of the poor shielding effect by the inner d – and/- electrons, the inert pair effect is maximum for Tl.

Thus the most stable oxidation state of TL is +1 and not +3. Therefore TIClg is unstable. B does not exhibit an inert pair effect and thus BC13 is stable.

Question 37. The outermost electronic configuration of the atom of an element is 3s-3p3. Mention the position of the element in the long periodic table. Why is electron gain enthalpy of oxygen less than that of sulfur?
Answer: Because of the poor shielding effect by the inner d – and f- electrons, the inert pair effect is maximum for Tl. Thus the most stable oxidation state of TL is +1 and not +3. Therefore TIClg is unstable. B does not exhibit an inert pair effect and thus BC13 is stable.