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Selina Concise Biology Class 10 ICSE Solutions Pollution- A Rising Environmental Problems

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Selina Concise Biology Class 10 ICSE Solutions Pollution- A Rising Environmental Problems

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 15 Pollution A Rising Environmental Problems. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 15 Pollution – A Rising Environmental Problems

Exercise 1

Solution A.1.
(d) The dust raised during road-cleaning

Solution A.2.
(b) Disposing of corpses in rivers

Solution A.3.
(b) Ozone

Solution B.1.
(i) SO2
(ii) Bromochlorodifluoromethane and chlorofluoromethane
(iii) Mercury

Solution B.2.

Column I

 Column II (Answers)
(i) Chlorofluocarbons (CFCs)

(ii) Flyash

(iii) Cow dung

(iv) CO2 and methane

(v) Sulphur dioxide

(vi) Iodine – 131

(f) Ozone depletion

(e) Industrial Waste

(b) Biodegradable

(a) Global Warming

(d) Acid Rain

(c) Nuclear Radiation Pollutant

Solution B.3.
(i) vehicular air
(ii) X-ray
(iii) hot
(iv) domestic activities

Solution C.1.

(i) Rivers contaminated with sewage:

  • A number of waterborne diseases are produced by the pathogens present in polluted water, affecting humans as well as animals.
  • The flora and fauna of rivers, sea and oceans is adversely affected.

(ii) Too much gaseous exhausts containing CO2 and SO2:

  • The high concentration of CO2 in atmosphere has been the main component of the green house effect that has caused global warming i.e. the rise of atmospheric temperature in recent years. Global warming causes melting of snow caps rise in sea levels.
  • SO2 is poisonous and irritates the respiratory system of animals and humans. A continuous exposure to SO2 has been reported to damage the lungs and increase the rate of mortality.
  • SO2 is also responsible for acid rain

(iii) Pesticides such as DDT used in agriculture:

  • Pesticides kill soil microbes which are responsible to recycle the nutrients in the soil.
  • Pesticides can enter the food chain and affect the health of humans as well as animals. It can cause damage to the lungs and central nervous system, failures of reproductive organs and dysfunctions of the immune system, endocrine system, and exocrine system, as well as potential cancer risks and birth defects.

(iv) Prolonged noise such as the one produced by crackers throughout night:

  • Prolonged exposure to the high decibel noise damages ear drums and can bring permanent hearing impairment.
  • Noise pollution can lead to high blood pressure (hypertension), constant headache, lack of concentration.

Solution C.2.
Three major constituents of sewage:

  1. Kitchen wastes
  2. Sanitary waste
  3. Waste from agricultural lands

Solution C.3.

  • The common sources of oil spills are: The overturned oil tankers, offshore oil mining and Oil Refineries.
  • The sea birds and sea animals sometimes get thick, greasy coating on their bodies due to oil spills.
  • Sea birds may ingest their oil coated. This may irritate their digestive system, may damage liver and kidney.
  • Oil spills lead to the death of sea birds as well as sea animals.

Solution C.4.
Measures to minimise noise pollution:

  1. Use of loud speakers should be banned.
  2. Airports should be located away from the residential area.

Solution D.1.

  1. Industrial Waste:
    Large number of industries produces waste water which contains various types of chemical pollutants. Such wastes are commonly discharged into the rivers. These chemicals cause irritation to the body systems of fish.
  2. Thermal Pollution:
    Many industries such as thermal power plants, oil refineries, nuclear plants use water for cooling their machinery. This hot waste water may be 8-10oC warmer than the intake water. This hot water is released into the nearby streams, rivers or the sea and causes warming. The sudden fluctuation in the temperature of water kills the fishes and harms the plant life growing in it.

Solution D.2.
(i) Noise Pollution
(ii) Industrial machines, workshops, trains, loud conversation, loudspeakers, etc.
(iii) Effects of noise pollution:

  1. It lowers efficiency of work.
  2. It disturbs sleep and leads to nervous irritability.

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Selina Concise Biology Class 10 ICSE Solutions Health Organisations

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Selina Concise Biology Class 10 ICSE Solutions Health Organisations

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 14 Health Organisations. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 14 Health Organisations

Exercise 1

Solution A.1.
(a) May 8

Solution A.2.
(c) Geneva

Solution B.1.

  1. A
  2. B
  3. A
  4. B
  5. A
  6. A
Column I
(Activity)		 Column II
(Organisation)
(i)To extend relief to victims of earthquake. Red Cross
(ii)To lay pharmaceutical standards for Important drugs. WHO
(iii)Arranging ambulance in emergencies. Red Cross
(iv)To suggest quarantine measures. WHO
(v)Training of midwives. Red Cross
(vi)Procuring and supplying blood for transfusion. Red Cross

Solution B.2.
Geneva

Solution B.3.
(a) WHO: World Health Organisation
(b) UNO: United Nations Organisation

Solution C.1.
(i) Sanitation – Removal and proper disposal of garbage, sewage and other wastes, elimination of breeding places of flies, mosquitoes, etc.
(ii) Supply of safe drinking water.
(iii) Keeping statistical records – Apart from the registration of births and deaths, to maintain the information about the health and diseases of the people in their area need regularly.

Solution C.2.

  1. Food and water borne diseases:
    The contaminated food and water cause several diseases. Water borne diseases occur due to contaminated water from hand pumps or mixing of untreated sewage with river water.
  2. Insect and air-borne diseases:
    Lack of cleanliness leads to breeding of houseflies, mosquitoes which are the carries of certain diseases.
  3. Lack of medical facilities:
    Lack of medical facilities especially in rural areas, leads to unavoidable deaths and damage to health. Lack of knowledge and superstitions beliefs also delay timely treatment which may result in serious consequences.

Solution C.3.
Functions of WHO:

  • To promote and support projects for research on diseases.
  • To collect and supply information about the occurrence of diseases of epidemic nature such as cholera, plague, yellow fever, etc.

Solution C.4.
Functions of Red Cross:

  • To extend relief and help to the victims of any calamity – flood, fire, famine, earthquakes, etc.
  • To procure and supply blood for the needy victims of war and other calamities.
  • To extend all possible first-aid in any accident.
  • To arrange for ambulance services in all emergencies.

Solution C.5.
Functions of World Health Organisation (WHO):

  • To promote and support projects for research on diseases.
  • To collect and supply information about the occurrence of diseases of epidemic nature such as cholera, plague, yellow fever, etc.
  • To lay pharmaceuticals standards for important drugs, to ensure purity and size of the dose.
  • To organize campaigns for the control of epidemic (widespread) and endemic (local) diseases.

Solution C.6.

  1. To extend relief and help to victims of any calamity
  2. To procure and supply blood for the needy victims of war or calamity
  3. To extend all possible first aid in any accident
  4. To educate people in accident prevention
  5. To arrange for ambulance services in emergencies
  6. To look after maternal and child welfare centres
  7. To train midwives
    (any four)

Solution D.1.

  • The World Health Organization (WHO) is a specialized agency of the United Nations (UN) that is concerned with international public health.
  • It was established on 7 April 1948, with headquarters in Geneva, Switzerland and is a member of the United Nations Development Group.
  • There were several reasons for the formation of WHO:
  • Member countries of the UNO focused on the need for creating an international body to look after the health problem of people of the world.
  • This was particularly felt in the field of research on the causes and cures of the diseases.
  • The combined efforts in this direction were to give better and faster results.
  • The poor and developing countries were to benefit quickly.

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Selina Concise Biology Class 10 ICSE Solutions Aids To Health

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Selina Concise Biology Class 10 ICSE Solutions Aids To Health

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 13 Aids To Health. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 13 Aids To Health

Exercise 1

Solution A.1.
(c) An antibiotic

Solution A.2.
(c) Tetanus

Solution A.3.
(c) April 7

Solution B.1.
(a) Salvarson
(b) Penicillin
(c) Passive Acquired Immunity
(d) Antiseptics – Lysol, Iodine, Boric acid and Carbolic acid;
Disinfectants – Cresol and Phenol;
Antibiotics – Ampicillin and Penicillin
(e) Oral polio vaccine (OPV)

Solution B.2.
(i) Acquired Immuno Deficiency Syndrome
(ii) Bacillus Calmette Guerin
(iii) Diphtheria, Pertussis and Tetanus
(iv) World Health Organisation

Solution B.3.
Antibodies which are immunoglobulins are produced in the blood to fight and destroy harmful microbes.

Solution C.1.
(a) False
(b) True
(c) False
(d) False
(e) False
(f) False

Solution C.2.
(a) Antiseptic is a mild chemical substance which, when applied on the body, kills germs whereas an antibiotic is a chemical substance produced by a micro-organism, which can kill or inhibit the growth of some other disease producing microorganisms.

(b) Antiseptic is a mild chemical substance which, when applied on the body, kills germs whereas a disinfectant is a strong chemical, which is applied on spots and places on the body where germs thrive and multiply.

(c) Disinfectant is a strong chemical, which is applied on spots and places on the body where germs thrive and multiply whereas deodorants are neither antiseptics nor disinfectants; they aerosols used to mask a bad smell.

(d) Vaccination is the introduction of any kind of dead or weakened germs into the body of a living being to develop immunity (resistance) against the respective disease or diseases whereas sterilization is a process of eliminating or killing all the microbes present on a surface, contained in a fluid, in medication, or in a compound such as biological culture media.

(e) Active immunity is the immunity developed by an individual due to a previous infection or antigen which enters his body naturally whereas passive immunity is the immunity provided to an individual from an outside source in the form of ”readymade” antibodies.

(f) Innate immunity is the immunity by the virtue of genetic constitutional makeup i.e. it is inherited from parents. It is present in the body without any external stimulation or a previous infection whereas acquired immunity is the resistance to a disease which an individual acquires during his lifetime. It may be the result of either a previous infection or readymade antibodies supplied from outside.

Solution C.3.

  1. TAB vaccine for typhoid
  2. BCG vaccine for measles
  3. DTP vaccine for diphtheria, tetanus and whooping cough

Solution C.4.
(a) Lysol, Benzoic acid, DDT, mercurochrome
Antiseptics. DDT is a wrong example for this category as it belongs to disinfectant which is not good for human skin.

(b) Formalin, iodine, Lysol, phenol.
Disinfectants. Iodine is a wrong example as it is an antiseptic.

(c) BCG, DTP, ATP.
Vaccines. ATP is a wrong example as it is used as an energy carrier in the cells of all known organisms.

(d) Tears, skin, nasal secretion, HCl (in stomach).
Germ Killing Secretions. Skin is a wrong example as it is a protective mechanical barrier. It prevents the entry of microorganisms at first place.

Solution C.5.

Vaccine

 Disease (s) The Nature of Vaccine
TAB Typhoid

Killed germs

Salk’s Vaccine

 poliomyelitis Killed germs
BCG tuberculosis

Living weakened germs

Vaccines for Measles

 Measles Living weakened germs
Cowpox Virus small pox

Living fully poisonous germs

Toxoids

 Diphtheria Extracts of toxins
Tetanus

Secreted by bacteria

Solution C.6.

  1. Innate Immunity
  2. Acquired Immunity
  3. Specific Immunity
  4. Active Acquired Immunity
  5. Passive Acquired Immunity
  6. Natural Acquired Active Immunity
  7. Artificial Acquired Active Immunity
  8. Natural Acquired Passive Immunity
  9. Artificial Acquired Passive Immunity

Solution C.7.
(a) antibiotics have a wide use in medicine to fight infections.
(b) Certain antibiotics are used as food preservatives, especially for fresh meat and fish.
(c) Some antibiotics are used in treating animal feed to prevent internal infections.
(d) Some antibiotics are used for controlling plant pathogens.

Solution C.8.

Merits of the Local Defence Systems:

  1. Local defence systems start working instantaneously.
  2. These systems are not dependent on previous exposure to infections.
  3. They are effective against a wide range of potentially infectious agents.

Solution C.9.

  • Diphtheria is a serious bacterial infectious disease. It leads to cold, coughing, sneezing and in severe cases if undiagnosed it might result in heart failure or paralysis.
  • Treatment includes a combination of medications and supportive care. The most important step is prompt administration of diphtheria toxoid which is made harmless is given intravenously. The harmless toxoid once administered in patient’s body triggers the production of antibodies against the pathogens causing diphtheria.

Solution C.10.

(a) Bleeding from a cut in the skin:

  • In case of bleeding, raise the affected part to minimize the blood flow.
  • Wash the cut surface with clean water.
  • Press the area with a piece of clean cotton and apply some antiseptic.

(b) A fractured Arm:

  • Lay the victim comfortably, loosen or remove the clothes from the affected part.
  • Do not move the part fractured.
  • If the affected limb is an arm, then tie a sling around the neck to rest the arm in it.

(c) Stoppage of breathing due to electrical shock:

  • Lay the victim flat on his back and put a pillow or folded towel under his shoulders in a way that his chest is raised and the head thrown back.
  • Hold and draw his arms upwards and backwards. This will cause his chest to expand and draw the air.
  • Next, fold the victim’s arms and press them against the ribs. The air will now be expelled.
  • Repeat the two steps at the rate of about 15 times per minute. Continue till the victim starts breathing without any extra help or till the doctor arrives.

Solution D.1.
Vaccination is the practice of artificially introducing the germs or the germ substance into the body for developing resistance to particular diseases. Scientifically, this practice is called prophylaxis and the material introduced into the body is called the vaccine. The vaccine or germ substance is introduced into the body usually by injection and sometimes orally (e.g. polio drops). Inside the body, the vaccine stimulates lymphocytes to produce antibodies against the germs for that particular disease. Antibodies are the integral part of our immunity. Their function is to destroy unwanted particles entered in the body. Vaccines give our immunity a signal to produce specific antibodies. Hence, the principle of vaccination is to produce immunity against a disease.

Solution D.2.
Whenever a germ or infection invades the body. A signal is sent to the immune system to produce specific antibodies. In order to cope up with the number of germs being multiplied inside the body, white blood cells start multiplying rapidly. This enables them to produce more number of antibodies and invade infection in time. Therefore, ”Abnormally, large numbers of WBCs in the blood are usually an indication of some infection in our body”.

Solution D.3.
(a) Antiseptics:  Antiseptics are mild chemical substances applied to the body, which prevent the growth of some bacteria and destroy others.
Example: Lysol and Iodine

(b) Disinfectants:  Disinfectants are chemicals which will kill all micro-organisms they come in contact with. Disinfectants are usually too strong to be used on body.
Example: Cresol and Phenol

(c) Vaccines: Vaccines are the materials used to administer in the body to provide passive immunity. The materials are generally germs or the substances secreted by the germs.
Example: OPV (Oral polio vaccine) and DTP (Diphtheria, Tetanus and Pertussis)

Solution D.4.
First aid is the immediate care given to a victim of an accident, sudden illness or other medical emergency before the arrival of an ambulance, doctor or other qualified help.

(a) Little toe in the foot is pierced by a thorn and is bleeding:

  • In the case of bleeding, raise the affected part to minimise gravitational outflow of blood.
  • Wash the cut surface with clean water, press the area with a piece of clean cotton wool, and if possible, apply some mild antiseptic.

(b) An elderly woman walking on the footpath during a hot mid-day has fallen unconscious:

  • Immediately lay the woman comfortably on a side of the road.
  • Loosen the clothes.
  • Let fresh air be around the woman.
  • Give some fluids to drink to the woman.

(c) A young boy has burnt his finger tip while firing crackers:

  • Immediately wash his burnt finger with sufficiently cold water for a few minutes.
  • Do not rub the burnt region.
  • Apply creams/ointments specially recommended, in case they are readily available.

(d) Your gardener has been bitten by a snake while digging soil in the flower bed:

  • Immediately squeeze out some blood from the wound.
  • Tie a tourniquet above the site to prevent spreading of venom into the body.

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Selina Concise Biology Class 10 ICSE Solutions Population- The Increasing Numbers and Rising Problems

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Selina Concise Biology Class 10 ICSE Solutions Population- The Increasing Numbers and Rising Problems

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 12 Population The Increasing Numbers and Rising Problems. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 12 Population – The Increasing Numbers and Rising Problems

Exercise 1

Solution A.1.
(d) Use of antibiotics and prophylactic vaccinations

Solution A.2.
(a) Per 1000 people per year

Solution B.1.
Demography

Solution B.2.
(i) Tubectomy (For Female)
(ii) Vasectomy (For Male)

Solution C.1.
(a) True
(b) False

Solution C.2.
It is approximately 7 billion.

Solution C.3.

(a) Birth Rate: It is the number of live births per 1000 people of population per year.

(b) Death Rate: It is the number of deaths per 1000 of population per year.

(c) Rate of growth of population: It is the difference between the birth rate and the death rate. As long as the birth rate exceeds the death rate, the population grows. If the birth rate is lower than the death rate, the population declines.

(d) Population density: It is the number of individuals per square kilometre (Km2) at any given time.

(e) Exhaustible resource: Exhaustible resources are the once which are perished by the prolonged us and they can never be renewed or replaced.

Solution C.4.
(a) False.
Dog was the first domesticated animal.

(b) True.
Rapidly growing industries made human life more and more comfortable, with greater opportunities of jobs and with more production of food. All this favoured population rise.

(c) False.
Present human population growth is following geometrical progression.

(d) False.
Birth Rate (natality) is the number of live births per 1000 people of population per year.

(e) False.
Vasectomy is the surgical method of contraception in human males while tubectomy is the surgical method used in females.

Solution C.5.
The rate of growth of population of the world is 1.092% (this rate results in about 145 net additions to the worldwide population every minute or 2.4 every second {2011 estimates}) and for India is 1.344%.

Solution C.6.
By law the minimum age is 21 years for boys and 18 years for girls.

Solution C.7.
Two advantages of small family are:

  1. Parents can give more attention to their children.
  2. Small family helps every country in controlling the growth of population.

Solution C.8.

  1. Illiteracy:
    Most of the rural population which forms the bulk of our society is still illiterate, ignorant and superstitious.
    They also do not know the functioning of the human reproductive system.
  2. Traditional Beliefs:
    Among the people from lower strata of the society, children are regarded as a gift of God and a sign of prosperity.
    Therefore, they make no effort to avoid pregnancy.

Solution C.9.
Population growth is not the only threat humanity is facing, but it will be a major contributor to the crises that await us and the planet in the coming century. Overpopulating the planet puts us all at risk of extreme environmental and social consequences that we are beginning to witness today. The extreme growth in human population is mortally taxing the Earth and its resources. Each individual person has a unique impact on the planet’s environment. Some people may be relatively less damaging than others, but no living individual is without an ecological footprint. In other words, each person needs basic resources and almost all people aspire to utilize significantly more resources than are required by their basic needs. As a result, the Earth is attempting to impose its own checks on human population. We can witness these “checks” in the form of widespread disease and the emergence of new disease strains, food and water shortages, poor harvests and violent and destructive weather caused by climate change. While it should be obvious that the Earth is a finite sphere and cannot endure infinite growth by any single species, we should also remember that Earth’s current web-of-life is the result of billions of years of complex evolution. It is irreplaceable. When we look forward to the next 40 years, the most significant population increases will take place in the areas of our world where natural resources and the infrastructure of modernity are already the scarcest. 95% of the human population growth is occurring in countries already struggling with poverty, illiteracy and civil unrest. It will further stress, the already strained ecological systems and worsen poverty in much of the developing world, thus aggravating threats to international security.

Thus, the statement made by an author ‘some great author has said that a population explosion is far more dangerous than an atomic explosion’ is true.

Solution C.10.
Poverty and population have been closely linked ever since the world faced changes due to the major revolutions. Poverty has its own effects on the population and vice versa. Poverty prevails because of illiteracy and traditional beliefs in the economically weaker strata. Since illiteracy and traditional beliefs prevail the people from this stratum, they regard children as gift of God and a sign of prosperity. They consider children to be helping hands in increasing the family income, hence they keep producing more children forgetting that their current situation would do no good for the children and they would add more to this already overburdened poverty strata. Hence, the population keeps on rising and so does poverty. As the population increases the quality of life goes down.

Solution C.11.

  1. Tool making revolution.
  2. Agricultural revolution.
  3. Scientific industrial revolution.

Solution C.12.
According to census, the Indian population in 1981 was 685 million and it was 846 million in 1991.

Solution C.13.
Yes, there could be a corresponding operation made in women. The name of the surgical procedure in females is ‘tubectomy’. In tubectomy, the abdomen is opened and the fallopian tubes (oviducts) are cut or ligated i.e. tied with nylon thread to close the passage of the egg.
Selina Concise Biology Class 10 ICSE Solutions Population- The Increasing Numbers and Rising Problems image - 1

Solution C.14.
Family welfare centres are those places where any help or advice about family planning is available free of cost. These places could be any hospitals, dispensaries, etc. The inverted red triangle is the symbol of family welfare in India.
Selina Concise Biology Class 10 ICSE Solutions Population- The Increasing Numbers and Rising Problems image - 2

Solution C.15.
Below are some of the advantages of having a small family:

  1. Financial condition of family is deeply related to the size of the family. A living cost of a large family is surely much higher than a small family. A large family has more expenses on cloth, toys, education and food whereas expenses in small family are very low.
  2. Parents can easily fulfill the needs of one or two children. They can provide them best education and look after them very well whereas when there are many children to look after parents just cannot fulfill even the basic needs of the children properly. Therefore, as a result, children suffer, the parents suffer and the nation suffers.
  3. A child in a small family receives more support from their parents than in a large family. In large family, parents have many children to look after, so they cannot give their best support to everyone whereas in small family parents can give more support to children as they have only one or two children to look after.
  4. Family size also affects the health, especially that of the mother and the child. Frequent pregnancies can cause illness to both the mother and the children. It can disrupt the health of the women. It puts mother and baby’s health at risk. So having a small family definitely leads to healthy and happy family.

Solution D.1.

  1. Food: The first and most important need of the humans (or any living organism) is food. But with the production of food rising by arithmetic progression and population growing by geometric progression i.e. the number at each step is being multiplied. At the same time growing population is increasing the use of more and more agricultural land to build houses. Thus it is evident that food would be running short for the unchecked rising population.
  2. Water: Availability of clean and germ-free water for drinking purposes would be more and more scarce with increase in population; the reason would be mainly, the pollution of rivers, ponds, lakes etc.
  3. Land: Man is bringing more and more land under cultivation and also using up land for building more residential colonies, factories and industries. Usable land would thus become less and less available.

Solution D.2.

  1. The orthodox view, to have at least one son especially in Indian society, should be modified with education. People should be educated that their greed for a son can lead to numerous children in the household which would worsen both their family’s health and wealth. They should focus on proper upbringing of the child, be it a son or a daughter.
  2. Married couples should be educated to delay the birth of their first child, to space the second with a sufficient interval for proper upbringing and to stop the third. They should also be educated to adopt family planning methods by which they can prevent pregnancy after two children. These include devices for both men and women, for example: Condoms, intrauterine devices (IUD) and oral pills.

Solution D.3.
For developing countries like India, population explosion is a curse and is damaging the development of the country and its society. The developing countries are already facing a lack in their resources, and with the rapidly increasing population, the resources available per person are reduced further, leading to increased poverty, malnutrition, and other large population-related problems. The literal meaning of population is “the whole number of people or inhabitants in a country or region”, and the literal meaning of population explosion is “a pyramiding of numbers of a biological population”. As the number of people in a pyramid increases, so do the problems related to the increased population. Some of the reasons for this population explosion are poverty, better medical facilities, and immigration from the neighboring countries. The population in India continues to increase at an alarming rate. The effects of this population increase are evident in the increasing poverty, unemployment, air and water pollution, and shortage of food, health resources and educational resources.

Solution E.1.
(a) B; 1981; 1991.
(b) B; 1981; 1991.
(c) B; 1971; 1981.

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Selina Concise Biology Class 10 ICSE Solutions The Reproductive System

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Selina Concise Biology Class 10 ICSE Solutions The Reproductive System

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 11 The Reproductive System. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 11 The Reproductive System

Exercise 1

Solution A.1.
(d) Epididymis → vas deferens → urethra

Solution A.2.
(d) 28 days

Solution A.3.
(d) About seven days

Solution B.1.
(a) Scrotum
(b) Seminiferous Tubules
(c) GraafianFollicle
(d) Seminal vesicle
(e) Epididymis

Solution B.2.
(a) Testosterone
(b) Ureter
(c) Ovum
(d) After birth

Solution B.3.
(a) Testes → Sperms → Sperm duct → Semen → Penis
(b) Menarche Puberty → Reproductive age → Menstruals → Menopause
(c) Graafian follicle → Ostium → Fallopian tube → Uterus

Solution B.4.
Seminiferous tubule → Epididymis → Vas deferens → Penis

Solution C.1.
Semen is the mixture of sperms and secretions from seminal vesicles, prostate gland and Cowper’s (bulbo-urethral gland).

Solution C.2.
(a) Inguinal canal: It is the canal which allows the descent of testes along with their ducts, blood vessels and nerves into the abdomen.
(b) Prostate gland: It is a bilobed structure which surrounds the urethra and pours an alkaline secretion into the semen.
(c) Testis: Testis is a male reproductive organ. There a pair of testes present in a scrotal sac descended outside the body cavity. Testes produce sperms which are the male gametes.
(d) Ovary: Ovary is a female reproductive organ.It produces ova i.e. female gametes.
(e) Oviduct: A pair of oviduct is present on either side of the uterus. Oviduct carries the released ovum from the ovary to the uterus.

Solution C.3.
Secondary sexual characters in males:

  1. Beard and moustache
  2. Stronger muscular built
  3. Deeper voice

Secondary sexual characters in females:

  1. Breasts in females
  2. Large hips
  3. High pitched voice

Solution C.4.
The accessory reproductive organs include all those structures which help in the transfer and meeting of two kinds of sex cells leading to fertilization and growth and development of egg up to the birth of the baby.
For example: uterus in females, penis in males.

Solution C.5.

Primary Reproductive Organs

 Accessory Reproductive Organs
The primary reproductive organs produce sex cells.

The accessory reproductive organs help in the transfer and meeting of two kinds of sex cells leading to fertilization.

The primary reproductive organs do not help in the development of baby.

 The accessory organs help in the growth and development of egg up to the birth of baby.
Example: Testes in males and ovaries in females.

Example: penis in males, Uterus, vagina in female.

Solution C.6.
Hymen is a thin membrane which partially covers the opening of the vagina in young females.

Solution C.7.
(a) Hernia: It is an abnormal condition which is caused when the intestine due to the pressure in abdomen bulges into the scrotum through the inguinal canal.
(b) Ovulation: It is the release of the mature ovum by the rupture of the Graafian follicle.
(c) Puberty: It is the period during which immature reproductive system in boys and girls matures and becomes capable of reproduction.

Solution C.8.
Changes in human male:

  1. Development of Beard and moustache
  2. Voice becomes deeper

Changes in human female:

  1. Development of Breasts in females
  2. Development of high pitched voice

Solution C.9.
(a) Menarche is the onset of menstruation in young females at about 13 years of age whereas menopause is the permanent stoppage of menstruation at about 45 years of age.

(b) Cowper’s gland opens into urethra in human males and its secretion serves as a lubricant whereas the prostate gland surrounds the urethra in males and its alkaline secretion neutralizes acid in female’s vagina.

(c) Hymen is a thin membrane that partially covers the opening of vagina in young females whereas clitoris is a small erectile structure located in the uppermost angle of vulva in front of the urethral opening.

(d) Uterus is a hollow, pear shaped muscular organ located in the pelvic cavity. It is the site of implantation for the embryo after fertilisation whereas the vagina is the muscular tube extending from the cervix to the outside. At the time of sexual intercourse, the vagina receives the male penis and provides entry for the sperms.

(e) Efferent ducts join to form the epididymis whereas the epididymis is continued by the side of the testes to give rise to the sperm duct or vas deferens.

Solution D.1.

  • Testes are responsible for the production of male gametes i.e. sperms. The normal body temperature does not allow the maturation of the sperms. Being suspended outside the body cavity, the temperature in the scrotal sac is 2 to 3oC which is the suitable temperature for the maturation of the sperms.
  • When it is too hot, the skin of the scrotum loosens so that the testes hang down away from the body. When it is too cold, the skin contracts in a folded manner and draws the testes closer to the body for warmth.
  • In an abnormal condition, in the embryonic stage, the testes do not descend into the scrotum. It can lead to sterility or incapability to produce sperms.

Solution D.2.
Testosterone is the male reproductive hormone produced by the interstitial cells or the Leydig cells. These cells are located in the testes. They serve as a packing tissue between the coils of the seminiferous tubules. Therefore, it can be said that the testes produce the male hormone testosterone.

Solution D.3.
otal reproductive period = 45 – 13 = 32 years
Total eggs produced = 32 x 12 = 384 eggs approximately

Solution E.1.
(a) Excretory system and Female Reproductive system

(b)

  1. Kidney
  2. Ureter
  3. Fallopian Tube
  4. Infundibulum
  5. Ovary
  6. Uterus
  7. Urinary Bladder
  8. Cervix
  9. Vagina
  10. Vulva

(c)

  • Function of Fallopian Tube (part 3): The fallopian tubes carry the ovum released from the ovary to the uterus.
  • Function of Infundibulum (part 4): Infundibulum is the funnel shaped distal end of the ovary which picks up the released ovum and pushes it further on its passage into the fallopian tube.
  • Function of Ovary (part 5): Ovary produces female gametes i.e. ova.
  • Function of Uterus (part 6): Uterus allows the growth and development of the embryo.

Solution E.2.
(a)

  1. Fallopian Tube
  2. Infundibulum
  3. Ureter
  4. Vagina
  5. Ovary
  6. Uterus
  7. Urinary Bladder
  8. Urethra

(b) Oestrogen secreted by the corpus luteum secrets oestrogen. Oestrogen stimulates the thickening of the endometrial wall of the uterus. The uterine wall becomes thickened and is supplied with a lot of blood to receive the fertilized egg.
(c) If fertilization fails to take place, the endometrial lining of the uterus starts shedding on the 28th day of the menstrual cycle. Finally it is discharged out along with the unfertilised ovum as the menstrual flow.

Solution E.3.
a.

  1. Seminal vesicles
  2. Prostate gland
  3. Bulbo-urethral gland
  4. Epididymis
  5. Testis
  6. Scrotum
  7. Urinary bladder
  8. Vas deferens
  9. Erectile tissue
  10. Penis
  11. Urethra

b. Functions of

  1. Seminal vesicles
    They produce the fluid which serves as the transporting medium for sperms.
  2. Prostate gland
    It produces an alkaline secretion which mixes with the semen and helps neutralise the vaginal acids.
  3. Bulbo-urethral gland
    It produces a secretion which serves as a lubricant for the semen to pass through the urethra.
  4. Testis
    It produces the male gamete sperm and the male sex hormone testosterone.
  5. Vas deferens
    They carry the sperms from the epididymis to the urethra.
  6. Urethra
    It serves as an outlet for delivering the sperms into the vagina.

Exercise 2

Solution A.1.
(c) fallopian tube

Solution A.2.
(a) energy

Solution A.3.
(c) 280 days

Solution B.1.
(a) Amniotic fluid
(b) Uterus
(c) Amniotic membrane
(d) Inguinal canal

Solution B.2.
(a) Sperm
(b) Follicle

Solution B.3.
(a) Ovulation → fertilization → implantation → gestation → child birth
(b) Sperm → sperm duct → urethra → coitus → vagina → ovum

Solution B.4.
(a) Menarche
(b) Ovulation
(c) Menstruation
(d) Fertilization
(e) Implantation

Solution B.5.

Column I

 Column II
(a) Acrosome

(v) spermatozoa

(b) Gestation

 (vii) Time taken by a fertilized egg till the delivery of baby
(c) Menopause

(vi) complete stoppage of menstrual cycle

(d) Foetus

 (i) An embryo which looks like human baby
(e) Oogenesis

(iii) ovum producing cells

(f) Ovulation

(ii) Luteinizing hormone

 

Solution C.1.
(a)

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

(b)

  1. Fertilization occurs in the fallopian tube.
  2. Vagina is also known as the birth canal.
  3. Nutrition and oxygen diffuse from the mother’s blood into the foetus’s blood through placenta.
  4. Gestation period in humans is about 280 days.

Solution C.2.

Structure

 Function
1. Corpus luteum

1. secretes progesterone & other hormones to prepare the uterine wall for the receival of the embryo.

2. Testes

 2. produces male gametes in mass
3. Placental disc

3. supplies oxygen and nutrients to embryo

4. Oxytocin

 4. increases the force in uterine contractions during child birth
5. Umbilical cord

5. connects placenta with foetus

6. Fallopian tube

6. The site of fertilization for the sperm and ovum

Solution C.3.

(a) Foetus:

  1. It is contained in the uterus.
  2. In foetus, limbs have appeared and resembles the humans unlike the embryo which is a growing or dividing zygote.

(b) Hyaluronidase:

  1. Enzyme
  2. It is an enzyme secreted by the sperm that allows the sperm to penetrate the egg.

(c) Morula:
It is the stage in the development of human embryo which consists of a spherical mass of cells. Blastocyst

(d) Amniotic fluid:

  1. Between amnion and embryo
  2. It protects the embryo from physical damage, keeps the pressure all around embryo and prevents sticking of foetus to amnion.

(e) Gestation:
Gestation is the full term of the development of an embryo in the uterus. 280 days in humans.

(f) Placenta:

  1. Placenta is formed by two sets of minute finger like processes called the villi. One set of villi is from the uterine wall and the other set is from the allantois.
  2. Oxygen and amino acids.
  3. Progesterone and oestrogen.

(g) Implantation:

  1. Blastocyst
  2. It occurs in about 5-7 days after ovulation.

Solution D.1.

(a) Sperm is the male gamete produced by the testes. Semen on the other hand is the mixture of sperms and alkaline secretions from the seminal vesicle, prostate gland and Cowper’s gland.

(b) Implantation is the fixing of embryo in the wall of uterus. The state that implantation produces is known as pregnancy.

(c) Follicle is the cellular sac containing a maturing egg. Corpus luteum on the other hand is the remnant of the follicle the release of ovum during ovulation.

(d) Amnion is a sac which develops around the embryo whereas allantois is an extension from the embryo which forms villi of placenta.

(e) Sterility is the incapability to produce sperms whereas impotency is the inability to copulate.

(f) Prostate gland pours alkaline secretions into the semen to neutralize the acid in female’s vagina whereas the secretion of Cowper’s gland serves as a lubricant.

(g) Identical twins are produced from one ovum i.e. one developing zygote splits and grows into two foetuses whereas fraternal twins are produced when two ova get fertilized at a time.

Solution D.2.

  1. After fertilization zygote is formed inside the fallopian tube.
  2. The zygote then divides repeatedly to form a spherical mass of cells known as ‘Morula’.
  3. The morula then develops into a hollow sphere of cells with a surrounding cellular layer and an inner cell mass projecting from it centrally. This stage is known as the ‘blastocyst’. It implants itself into the uterine wall.
    Selina Concise Biology Class 10 ICSE Solutions The Reproductive System image -1
  4. From the blastocyst arises an embryo which is around 3 weeks old. It is a tiny organism that hardly resembles human being.
  5. By the end of 5 weeks, the embryo is with a develoed heart and blood vessels.
  6. By the end of 8 weeks, limbs are developed. This stage is known as ‘foetus’.
  7. At the end of nearly 40 weeks i.e. end of gestation period, the infant is born.

Solution D.3.
(a) Amnion:

  1. Amnion contains the amniotic fluid which surrounds the embryo.
  2. This fluid protects the embryo from physical damage.
  3. It maintains even pressure all around the embryo.
  4. It also prevents sticking of foetus to amnion.

(b) Placenta:

  1. The placenta allows the diffusion of oxygen and nutrients such as glucose, vitamins and amino acids from mother to foetus.
  2. Similarly, it also allows the diffusion of carbon dioxide, urea and waste products from foetus to mother.
  3. Placenta also acts as an endocrine tissue. It secretes oestrogen and progesterone.

Solution E.1.
a. A – ovum
B – sperm

b. Sperms are produced in the testis.
The ovum is produced in the ovary.

c. The reproductive cells unite in the fallopian tubes of the female reproductive system.

d. Ovary – Oestrogen and progesterone
Testis – Testosterone

e. Accessory glands:

  • Seminal vesicle – Seminal fluid
  • Prostate gland – Alkaline secretion
  • Bulbo-urethral gland – Lubricant

Solution E.2.
(a)

  1. umbilical cord,
  2. placenta,
  3. amnion,
  4. mouth of uterus,
  5. muscular wall of uterus

(b) Gestation
(c) 280 days
(d) Placenta provides the foetus with oxygen and nutrients. In addition, the placenta also removes carbon dioxide and waste products of the foetus.
(e) Progesterone

Solution E.3.

Selina Concise Biology Class 10 ICSE Solutions The Reproductive System image -2

Solution E.4.
(a) A – Muscular wall of uterus,
B – Oviduct,
C – Ovary,
D – Cervix
(b) If part B will get blocked, ovum released from the ovary will not get fertilized by the sperm and hence pregnancy will be prevented.

Solution E.5.

  1. Prostate gland
  2. Bulbo-urethral gland
  3. Urethra
  4. Vas deferens
  5. Testis

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Selina Concise Biology Class 10 ICSE Solutions Endocrine Glands

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Selina Concise Biology Class 10 ICSE Solutions Endocrine Glands

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 10 Endocrine Glands. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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ICSE SolutionsSelina ICSE Solutions

Selina ICSE Solutions for Class 10 Biology Chapter 10 Endocrine Glands

Exercise 1

Solution A.1.
(c) pancreas

Solution A.2.
(b) thyroid

Solution A.3.
(c) alcohol

Solution B.1.
(a) Insulin, glucagon, somatostatin
(b) Adrenaline
(c) Hypoglycemia
(d) Insulin
(e) Adrenaline
(f) Anti-diuretic hormone (Vasopressin)
(g) Adrenaline

Solution B.2.
If there was hyposecretion of the thyroid gland in a child; the child will suffer from cretinism. The symptoms of cretinism are dwarfism, mental retardation, etc.

Solution B.3.
(a) Prostate
(b) Scurvy
(c) Cretinism
(d) Cortisone

Solution B.4.
(a) Larynx
Reason- Larynx is the sound box while the rest three i.e. glucagon; testosterone and prolactin are hormones.

(b) Penicillin
Reason – Penicillin is an antibiotic while adrenaline; insulin; thyroxine are hormones.

(c) Adrenaline
Reason – Adrenaline is a hormone while the stomach, ileum and liver are the organs of the digestive system.

(d) Insulin
Reason- Insulin is secreted by the pancreas while TSH, GH, ADH are the hormones secreted by the pituitary gland.

(e) Iodine
Reason- Iodine is required for the synthesis of thyroxine hormone. While cretinism, goitre, myxoedema are the deficiencies occur due to the deficiency of thyroxine.

Solution B.5.

Column I

 Column II
1. Beta cells of islets of Langerhans

(g) Insulin

2. Thyroid

 (c) Exophthalmic goitre
3. Cretinism

(h) Under secretion of thyroxine in a child

4. Addison’s disease

 (b) Glucocorticoids
5. Hypothyroidism

(e) Thyroxine

6. Myxoedema

 (a) condition due to under     secretion of thyroxine in adults
7. Adrenaline

(d) Increases heart beat

8. Cortisone

(f) Adrenal cortex

Solution B.6.

A (Condition)

 B (Cause)
(a) Dwarfism and mental retardation

v. Hypothyroidism

(b) Diabetes mellitus

 i. Excess of glucose in blood
(c) Shortage of glucose in blood

iii. Insulin shock

(d) Gigantism

 ii. Over secretion of growth hormone
(e) Enlargement of breasts in adult males

vi. Over secretion of cortical hormones

(f) Exophthalmic goitre

iv. Over secretion of thyroxine

Solution C.1.
(a) True
Reason- Adrenaline is described as emergency hormone because during any emergency situation more adrenaline is secreted which makes the heart beat faster, increases the breathing, releases more glucose into the blood stream to fulfill the energy requirement.

(b) False
Reason- The two different kinds of diabetes are diabetes insipidus caused due to insufficient secretion of vasopressin and the other is ‘diabetes mellitus’ caused due to hyposecretion of insulin but they cannot be described as mild and severe.

(c) True
Reason-Iodine is an active ingredient in the production of the thyroxine hormone.

(d) True
Reason- Pituitary gland controls the functioning of all the other endocrine glands.

(e) True
Reason- Hormones are poured directly into blood the blood stream and control physiological processes by chemical means. Their action depends on the feedback mechanism.

(f) True
Reason- Gigantism and dwarfism are controlled by the growth hormone from the pituitary gland. Growth hormone is much more active in children for their normal body growth along with which necessary substance required for the synthesis of growth hormone need to be consumed.

Solution C.2.
Endocrine glands are ductless glands, means they pour their secretion i.e. hormones directly into the blood stream while the other glands are exocrine glands which have ducts. Through ducts they pour their secretions (not hormones) into the blood stream.

Solution C.3.
Hormones unlike enzymes are secreted by the endocrine glands only. Also the hormones unlike the enzymes are poured directly into the blood. Hormones can be peptides, steroids, amine but all enzymes are proteins.

Solution C.4.
Chemically hormones are peptides, amines or steroids. They are involved in regulating the metabolism of the body. They can bring about specific chemical changes during metabolic process. Therefore hormones can be termed as ‘chemical messengers’.

Solution C.5.
Iodine is an active ingredient in the production of the thyroxine hormone secreted by the thyroid gland. Thyroxine hormone is a very essential hormone for our body. In case of its abnormal secretions a person may suffer certain sever disorders. Therefore, it is an important nutrient for our body.

Solution C.6.
Adrenaline is the hormone which prepares the body to meet any emergency situation. Adrenaline makes the heart beat faster. At the same time, it stimulates the constriction of the arterioles of the digestive system reducing the blood supply of the digestive system which makes the mouth dry.

Solution C.7.
If one adrenal gland is removed, the other one gets enlarged. This is to meet the requirement of hormones produced by the body.

Solution C.8.

  1. Diabetes mellitus:
    Cause – under secretion of Insulin hormone
    Symptoms – excretion of great deal of urine with sugar, Person feels thirsty and loss of weight. In severe cases, the person may lose the eye sight.
  2. Diabetes insipidus:
    Cause – Under secretion of Anti-diuretic hormone
    Symptoms – frequent urination resulting in loss of water from body and the person feels thirsty.

Solution C.9.
The Himalayan soil is deficient in iodine. Thus, the food grown in such soil also becomes iodine deficient. Due to this reason, when Himalayan people consume iodine deficient food, they do not get the proper intake of iodine. Therefore, people living in the low Himalayan hilly regions often suffer from goitre.

Solution C.10.

S.No.

 Source

Gland cells

 Hormone produced Chief function Effect of over secretion

Effect of under secretion

1.

 Thyroid thyroxine Regulates basal metabolism Exophthalmic goiter Simple goiter, cretinism in children and myxoedema in adults
2. Beta cells of Islets of Langerhans Insulin Promotes glucose utilization by the body cells Hypoglycemia

Diabetes mellitus

3.

 Anterior pituitary Growth hormone Promotes growth of the whole body Gigantism Dwarfism
4. Posterior pituitary Vasopressin Increases reabsorption of water from kidney tubule More concentrated and less amount of urine

Diabetes insipidus

Solution C.11.

Gland

 Hormone secreted

Effect on body

Thyroid

 Thyroxine Regulates basal metabolism
Pancreas (“beta” cells) Insulin

Controls blood sugar level

Adrenal gland

 Adrenaline Increases heart beat
Anterior pituitary Thyroid stimulating hormone

Stimulates thyroxine secretion

Solution C.12.

Gland

 Secretions

Effect on body

Ovary

 oestrogen development of secondary sexual characteristics
Alpha cells of islets of Langerhans Glucagon

Raises blood sugar level

Thyroid

 Hypersecretion of thyroxine Protruding eyes
Anterior pituitary Hypersecretion of Growth hormone

Gigantism

Solution D.1.

Hormonal Response

 Nervous Response
Hormonal response is slow.

Nervous response is immediate.

Hormones are chemical messengers transmitted through blood stream.

 Nerve impulses are transmitted in the form of electro-chemical responses through nerve fibres.
This response brings about a specific chemical changes. Therefore it regulates the metabolism.

This response does not bring any chemical change during metabolism.

Solution D.2.

Action of Hormones

 Action of Nerves
The effect of hormones is wide spread in the body. They can show their effect on more than one target site at a time.

The nerve response affects only particular glands.

The effect of hormones can be short-lived or long lasting.

 The effect of nervous response is always short-lived.
Cannot be modified by the previous learning experiences.

Can be modified by the previous learning experiences.

Solution E.1.
a. Glucagon: Alpha cells of the islets of Langerhans
Insulin: Beta cells of the islets of Langerhans
b. Insulin: It maintains the levels of glucose (sugar) in the blood.
Glucagon: It raises the blood glucose levels by stimulating the breakdown of glycogen to glucose in the liver.
c. An endocrine gland is one which does not pour its secretions into a duct, while an exocrine gland is a gland which pours its secretions into a duct. Because the pancreas produces hormones such as insulin, glucagon and somatostatin directly into the blood and not into a duct, it functions as an endocrine gland. Because it secretes the pancreatic juices for digestion via a duct, it functions as an exocrine gland. Hence, the pancreas is an exo-endocrine gland.
d. Insulin is not administered orally because the digestive juices degrade insulin, and thus the insulin is ineffective in the body.
e. Islets of Langerhans
f. The pancreas is located in the abdomen behind the stomach.

Solution E.2.
(a) This portion is located in the neck region above the sternum.
(b) 1- Larynx, 2 – Thyroid gland, 3 – Trachea
(c) Larynx is the voice box containing vocal cords. It helps in producing sound.
Thyroid gland produces thyroxine and calcitonin which are essential hormones.
Trachea is the wind pipe that helps in passing air to and from the respiratory system while breathing.
(d) Structure 2 is the thyroid gland. It is an endocrine gland, so it is ductless and pours its secretions directly into the blood. Hence, there is no duct.

Solution E.3.
(a) 1- Pituitary gland, 2 – thyroid gland, 3 – pancreas, 4 – adrenal glands
(b) All the glands shown in the above diagram are endocrine glands. They secrete essential hormones and pour their secretions directly into the blood.
(c) Iodine is essential for the normal working of thyroxine.

Solution E.4.
(a) A
(b) Hormone secreted by the endocrine gland is shown in the image A to be moving only in one direction i.e. towards the target organ. But actually the hormones poured into the blood stream may have one or more target sites at a time. The arrows shown are carried to all parts by the blood and their effect is produced only in one or more specific parts.

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Selina Concise Biology Class 10 ICSE Solutions The Nervous System

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Selina Concise Biology Class 10 ICSE Solutions The Nervous System

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 9 The Nervous System. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 9 The Nervous System

Exercise 1

Solution A.1.
(b) neurolemma

Solution A.2.
(d) Pons – consciousness

Solution A.3.
(b) Contains both sensory and motor fibres

Solution B.1.
(a) Cerebrospinal fluid
(b) Synapse
(c) Cerebrum
(d) Hypothalamus

Solution B.2.
(a) Stimulus: Receptor:: Impulse: Effectors
(b) Cerebrum: Diencephalon:: Cerebellum: Medulla oblongata
(c) Receptor: Sensory nerve:: Motor nerve: Effector

Solution B.3.
(a) Sensory
(b) Maintaining posture and equilibrium
(c) Spinal cord

Solution C.1.
(a) Corpus Callosum – It is located located in the forebrain. It connects two cerebral hemispheres and transfers information from one hemisphere to other.
(b) Central canal – It is located in centre of the spinal cord. It is in continuation with the cavities of the brain. It is filled with cerebrospinal fluid and acts as shock proof cushion. In addition, it also helps in exchange of materials with neurons.

Solution C.2.
(a) False
(b) False
(c) True
(d) True

Solution C.3.
(a)

Cerebrum Cerebellum
The cerebrum controls all voluntary actions. It enables us to think, reason, plan and memorize. The cerebellum on the other hand maintains balance of the body and coordinates muscular activity.

(b)

Sympathetic Nervous System Parasympathetic Nervous System
Sympathetic nervous system prepares the body for violent action against the abnormal condition. Parasympathetic nervous system is concerned with re-establishing normal conditions after the violent act is over.

(c)

Sensory Nerve Motor Nerve
Sensory nerve brings impulses from the receptors i.e. sense organs to the brain or spinal cord.

Motor nerve carries impulse from the brain or spinal cord to effector organs such as muscles or glands.

(d)

Medulla Oblongata Cerebellum
Medulla oblongata controls the  activities of internal organs and many other involuntary actions The cerebellum on the other hand maintains balance of the body and coordinates muscular activity.

(e)

Cerebrum Spinal Cord
The grey matter containing cytons lies in the cortex (outer region) while the white matter containing axons lies in the medullary region (inner region). The grey matter containing cytons lies in the medullary region i.e. inner side while the white matter containing axons lies in the cortex i.e. the outer region.

Solution C.4.
(a) Cerebellum maintains balance of the body and coordinates muscular activity.
(b) Myelin sheath acts like an insulation and prevents mixing of impulses in the adjacent axons.

Solution C.5.
(a) Synapse: It is a gap between the axon terminal of one neuron and the dendrites of the adjacent neuron. It transmits nerve impulse from one neuron to another neuron.
(b) Association Neuron: It interconnects sensory and motor neurons.
(c) Medullary sheath: It provides insulation and prevents mixing of impulses in the adjacent axons.
(d) Medulla Oblongata: It controls activities of internal organs such as peristalsis, breathing and many other involuntary actions.
(e) Cerebellum: It maintains balance of the body and coordinates muscular activity.
(f) Cerebrospinal Fluid: It acts like a cushion and protects the brain from shocks.

Solution C.6.
(a) Sensory, motor and mixed nerves
(b) Somatic and autonomic nervous system
(c) Natural and conditioned reflexes
(d) Sensory, motor and association neurons
(e) Gray and white matter

Solution C.7.
(a) Stimulus — receptor — sensory neuron — central nervous system — motor neuron — effector — response
(b) Resting — depolarization — repolarization
(c) Dendrites — Dendron — perikaryon — nucleus — axon — axon endings
(d) Cerebrum — diencephalon — mid-brain — cerebellum — pons — medulla oblongata

Solution D.1.
(a) Reflex action  is an autonomic, quick and involuntary action in the body brought about by a stimulus.
(b)

Example Type of Reflex
(i) Sneezing Simple
(ii) Blushing Simple
(iii) Contraction of eye pupil Simple
(iv) Lifting up a book Conditioned
(v) Knitting without looking Conditioned
(vi)  Sudden application of brakes of the cycle on sighting an obstacle in front Conditioned

Solution D.2.
The advantages of having a nervous system are as follows:

  1. Keeps us informed about the outside world through sense organs.
  2. Enables us to remember, think and reason out.
  3. Controls and harmonizes all voluntary muscular activities such as running, holding, writing
  4. Regulates involuntary activities such as breathing, beating of the heart without our thinking about them.

Solution D.3.
The brain and the spinal cord lie in the skull and the vertebral column respectively. They have an important role to play because all bodily activities are controlled by them. A stimulus from any part of the body is always carried to the brain or spinal cord for the correct response. A response to a stimulus is also generated in the central nervous system. Therefore, the brain and the spinal cord are called the central nervous system.

Solution D.4.
Reflex actions are involuntary actions which occur unknowingly. Voluntary actions on the other hand are performed consciously.
Picking up an apple and eating it is an example of voluntary action whereas withdrawal of hand on touching a hot object is an example of reflex action.

Reflex Action Voluntary Action
Reflex actions are involuntary actions which occur unknowingly. Voluntary actions on the other hand are performed consciously.
Commands originate in the spinal cord, autonomic nervous system and a few in the brain as well. Commands originate in the brain.

Solution D.5.
Selina Concise Biology Class 10 ICSE Solutions The Excretory System (Elimination of Body Wastes) image - 1

Solution D.6.

Organ

 Sympathetic System Parasympathetic System
e.g. Lungs Dilates bronchi and bronchioles

Constricts bronchi and bronchioles

1. Heart

 Accelerates heartbeat Retards heartbeat
2. Pupil Dilates

Constricts

3. Salivary gland

 Inhibits the secretion of saliva causing the drying of the mouth

Stimulates the release of saliva

Solution E.1.
Salivation is an example of conditioned reflex that develops due to experience or learning. Saliva starts pouring when you chew or eat food. Therefore, this reflex will occur not just on the sight or smell of food. The brain actually needs to remember the taste of food. Boy B started salivating because he must have tasted that food prior unlike boy A.

Solution E.2.

Situation

 Organ/body part Change/action Part of autonomic nervous system involved
1. You have entered a dark room Eye Pupil dilates

Sympathetic
2. Your body is consuming lot of glucose while running a race Liver Glycogen is converted into glucose in liver Sympathetic
3. You are chewing a tasty food Salivary gland Salivation increases

Parasympathetic
4. You are running a race Adrenal gland Release of adrenaline and noradrenaline increases Sympathetic
5. You are retiring to bed for sleep Heart Heart rate slows down

Parasympathetic
6. You are shivering in intense cold Body hairs Hair raised

Sympathetic

Solution E.3.
Fill in the following information in the diagram.

  1. Central Nervous System
  2. Autonomic
  3. 12
  4. spinal
  5. 31
  6. dilates
  7. constricts
  8. liver

Exercise 2

Solution A.1.
(b) Cornea

Solution A.2.
(b) Cochlea

Solution A.3.
(c) Eustachian tube, tympanum and utriculus

Solution A.4.
(a) Retina

Solution B.1.
(a) Rhodopsin
(b) Eustachian tube
(c) Hammer
(d) Dura mater
(e) Eustachian tube
(f) Cornea
(g) Auditory nerves
(h) Rods and cones
(i) Hypermetropia

Solution B.2.
(a) Cones: Iodopsin:: rods: rhodopsin
(b) Sound: ear drum:: dynamic balance: semi-circular canals

Solution B.3.

Column I Column II
i. The blind spot (h) no sensory cells
ii. The yellow spot (g) centre of the retina
iii. Ciliary muscle (b) Shape of the lens
iv. Iris (e) free of rod cells, (a) colour of the eye
v. Meninges (c) Protective covering of the brain

Solution C.1.
(a) Myopia results when the eye ball is lengthened from front to back or the lens is too curved.
Hyperopia results from either too shortening of the eyeball from front to back or when the lens is too flat.
(b) Rods are sensitive to dim light but do not respond to colour.
cones are sensitive to bright light and are responsible for colour vision.
(c) cochlea is responsible for hearing; it can perceive the senses of hearing.
Semicircular canals are responsible for perceiving the senses to maintain the body balance.
(d) Rod cells contain rhodopsin whereas the cone cells contain iodopsin.
(e) Dynamic balance is when the body is in motion whereas static balance is positional balance with respect to gravity.

Solution C.2.
(a) False
Correct statement: Deafness is caused due to rupturing of the eardrum.
(b) False
Correct statement: Semicircular canals are concerned with dynamic balance.

Solution C.3.
(a) Fovea centralis is located at the back of the eye almost at the centre of the eyeball. It is the region of the brightest vision and also of the colour vision.
(b) Organ of corti is located in the inner ear. It contains sensory cells which process hearing.

Solution C.4.
(a) True
(b) False/ Ciliary muscles regulate the size of the lens.
(c) True
(d) False/The auditory nerve responsible for sound as well as for the body balance.
(e) True
(f) False/ flavour is a combination of taste and smell.
(g) False/ short-sightedness is myopia and hyperopia is long-sightedness.
(h) True

Solution C.5.
(a) Auditory canal, tympanum, ear ossicles, oval window, cochlea
(b) Conjunctiva, cornea, lens, retina, optic nerve

Solution C.6.
(a) Organ of Corti and hearing
(b) Olfactory nerve and smell
(c) Retina and vision
(d) Taste bud and taste

Solution C.7.
(a) Lacrimal gland is a tear gland located at the upper sideward portion of the eye orbit. Its secreation lubricates the surface of the eye, washes aways the dust particles and kills germs
(b) Yellow spot is the region of brightest vision and contains maximum sensory cells whereas a blind spot contains no sensory cells and this is the point of no vision.
(c) Presbyopia is an age-old eye defect. In this condition, the lens loses flexibility resulting in far-sightedness.
Cataract is also very common in old people, the cornea becomes opaque and the vision is cut down even to blindness.
(d) The process of focusing the eye at different distances is called the power of accommodation.
(e) The image formed on the retina is inverted and real.

Solution C.8.
An optical illusion is the life-like continuous movement on the screen. Television is an example of optical illusion, where the scanning beam of a picture frame of the TV camera moves so rapidly on the viewing screen of the TV set that our eyes cannot keep pace with it.

Solution C.9.
(a) Oval window is located in the middle ear. It helps in setting the fluid in the cochlear canals into vibration.
(b) Cochlea is located in the inner ear. It helps in transmitting impulses to the brain via the auditory nerve.
(c) Semicircular canals are located in the inner ear. These help in maintaining the dynamic equilibrium of the body.
(d) Utriculus is located in the inner ear. It joins the semi-circular canals to cochlea. It also helps in maintaining static balance of the body.

Solution C.10.
Selina Concise Biology Class 10 ICSE Solutions The Excretory System (Elimination of Body Wastes) image - 2

Solution C.11.

Structure

 Function
Yellow Spot

Region of the brightest vision

Auditory nerve

 Transfers impulse from inner ear to brain
Ciliary muscle

Helps to change the focal length of the eye lens

Spinal cord

 Conducts impulses
Oval window

Sets fluid in cochlear canal into vibration

Semicircular canals

Dynamic equilibrium

Solution D.1.
While reading a book, the lens is more convex or rounded due to contraction of ciliary muscles because the book is usually read from a short distance. When we raise our head and look at a distant object, the ciliary muscles relax to build the tension on the suspensory ligament so that they can stretch the lens. This change in the curvature of the lens makes us focus on distant object.
Selina Concise Biology Class 10 ICSE Solutions The Excretory System (Elimination of Body Wastes) image - 3

Solution D.2.
The brain sees the vivid picture of the dream through the eyes. Our eyes have actually never seen the vivid picture. This is an example of optical illusion. The area of dream is controlled by the cerebrum of the central nervous system. So sometime we can remember the vivid picture seen in the dream.

Solution D.3.
If we look at a bright object and then close our eyes, the sensation of light persists for a short period. This is known as persistence image or the after image. It lasts for one-tenth of a second. Therefore by closing the eyes and gently pressing them with your palms, you see some specs of brilliant light.

Solution D.4.
Adaptation is the ability to adjust vision in bright and dark areas. When we enter a dark room from bright light, the rhodopsin pigment broken down in bright light is regenerated. It dilates the pupil and allows more light to enter the eyes. This is called dark adaptation. On the other hand, if we enter bright area from a dark room, the rhodopsin pigment is bleached. This constricts the pupil and reduces the light entering the eyes. This is called ‘light adaptation‘

Accommodation is the process of focusing the eye at different distances. This is mainly brought about by a change in the curvature of the lens. When the ciliary muscles contract, the lens becomes thicker and we are able to focus a nearby object. On the other hand when the ciliary muscles relax, the lens remains stretched i.e. the normal condition and we are able to focus on distant object.

Solution D.5.
Our eyes are designed to focus at a great variety of distances. To focus constantly at a short distance can make the lens focusing muscles fatigued. Therefore, we do not enjoy watching a movie from a very short distance from the screen in cinema hall.

Solution D.6.

Defect of vision

 Cause Corrective measure
Myopia Lengthening of eye ball from front to back or the lens is too curved.

Using suitable concave lens

Hyperopia

 Shortening of eye ball from front to back or the lens is too flat. Using suitable convex lens
Astigmatism Uneven curvature of the cornea

Using suitable cylindrical lenses

Presbyopia

 Loss of flexibility of lens Using suitable convex lens
Cataract Lens turning opaque

Surgery or use of convex lens or implantation of plastic lens.

Colour blindness

 Genetic defect No control measure
Squint Formation of cross-eye

Surgery and suitable exercise

Solution D.7.
The three ear ossicles are: Malleus (hammer), Incus (anvil) and Stapes (stirr up).
The last ear ossicle, stapes, vibrates and transmits the vibration to the oval window.
The role of other two ear ossicles is to magnify the vibration of stapes as a result of their lever like action.

Solution D.8.
The process of focusing the eye at different distances is called the power of accommodation. The ciliary muscles are responsible for the power of accommodation.

Solution E.1.
a. The ability of the eye to focus sharply on things which are near to the eye as well as far off is known as the power of accommodation.
b. Shape of the eye:
Near vision – flattened
Distant – rounded or more convex
c. Ciliary muscles and suspensory ligament
d. In the dark: Cells – rod cells, Pigment – rhodopsin
In the light: Cells – cone cells, Pigment – iodopsin

Solution E.2.
a. The middle ear or membranous labyrinth has two structures inside it, the cochlea and the semi-circular canals.
b. Malleus, incus and stapes
c. Static balance – Utriculus and sacculus (inner ear)
Hearing – Internal ear
Dynamic balance – Semi-circular canals (inner ear)
d. Collectively they are termed as ossicles.

Solution E.3.
(a) Cornea is comparable to the lens cover of the camera.
The iris and pupil act like the aperture of a camera.
(b) The cornea is the eye’s main focusing element. It takes widely diverging rays of light and bends them through the pupil; the rays are further converged by the lens.

Solution E.4.
(a) Myopia
(b) The two possible reasons for myopia are either the eye ball is lengthened from front to back or the lens is too curved.
(c) 1 – vitreous humour, 2 – blind spot, 3-lens, 4-pupil
(d) Concave lens
Selina Concise Biology Class 10 ICSE Solutions The Excretory System (Elimination of Body Wastes) image - 4

Solution E.5.
(i) Ear
(ii) m – malleus, i – incus and s – stapes respectively. These are collectively called as ear ossicles.
(iii) Cochlea. The vibrating movements in the hair of the sense cells of cochlea transmit the impulse for hearing to the brain via auditory nerve.
(iv) Tympanic membrane. It vibrates and then sets the ear ossicles into vibration in the process of hearing.

Solution E.6.
(i) Ear ossicles
(ii) A – Cochlea, B – Semicircular canals, C – Ear ossicles
(iii) Cochlea helps in transmitting impulses to the brain via the auditory nerve. Semicircular canals help in maintaining dynamic equilibrium of the body.
(iv) Organ of Corti

Solution E.7.
Selina Concise Biology Class 10 ICSE Solutions The Excretory System (Elimination of Body Wastes) image - 5

Solution E.8.
(a) Myopia
(b) A-Normal eye, B-Myopia
(c) Looking glasses with the concave lens are required here.

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 8 The Excretory System (Elimination of Body Wastes). You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 8 The Excretory System (Elimination of Body Wastes)

Exercise 1

Solution A.1.
(c) Removal of nitrogenous wastes

Solution A.2.
(a) Proximal convoluted tubule

Solution A.3.
(c) Sweating

Solution A.4.
Liver

Solution B.1.
(a) Liver
(b) Cortex
(c) Glomerulus
(d) Collecting duct
(e) Renal artery (Renal vein has urea but renal artery has higher concentration of urea as compared to renal vein).

Solution B.2.
(a) Afferent arteriole, glomerulus, efferent arteriole, capillary network, renal vein
(b) Renal artery, kidney, ureter, urinary bladder, urethra

Solution B.3.
(a) Ultrafiltration
(b) Excretion
(c) Osmoregulation
(d) Excretion

Solution C.1.
(a) Glomerulus is involved in the process of ultrafiltration.The liquid part of the blood which is plasma including urea, salts, glucose filters out from the glomerulus into the renal tubule.
(b) Henle’s loop is involved in reabsorption of water and sodium ions.
(c) Ureter carries urine to the urinary bladder by ureteral peristalsis.
(d) Renal artery supplied blood to the kidney.
(e) Urethra is involved in the process of micturition i.e. expelling urine out of the body.

Solution C.2.
Excretion helps in removing toxic wastes from our body and it also plays an important role osmoregulation i.e. the maintenance of the homeostasis of the body.
Carbon dioxide, water, nitrogenous compounds such as urea, uric acid and excess salts are some common excretory products.

Solution C.3.
A uriniferous tubule also known as the kidney tubule is the structural and functional unit of the kidney.
It takes in impure blood from the renal artery and removes wastes in the form of urine. It also provides a larger surface area for reabsorption of salts and water.

Solution C.4.
Maintaining a normal osmotic concentration in the body means regulating the percentage of water and salts. If this regulation mechanism fails we either end up losing vital salts and water or may accumulate unwanted salts and excess water in our body.

Solution C.5.
If one kidney is donated to a needy patient, the other kidney alone is sufficient for removing wastes or excretion. Thus, the donor can live a normal life.

Solution C.6.
During summer, a considerable part of water is lost through perspiration so the kidneys have to reabsorb more water from the urine. This makes the urine thicker in summer than in winters.

Solution C.7.

(a) Bowman’s capsule is a thin walled cup containing the glomerulus. This Bowman’s capsule along with the glomerulus is known as malpighian capsule.

(b) The renal cortex is the outer darker region of the kidney whereas the renal medulla is the inner lighter region of the kidney.

(c) Renal pelvis is the expanded front end of the ureter in the kidney whereas the renal papilla is the apex of the renal pyramid which projects into the pelvis.

(d) Urea is the chief excretory product which is excreted in the form of urine whereas urine is the filtrate left after reabsorption and tubular secretion which contains 95% water and 5% solid wastes.

(e) Excretion is the process of removal of chemical wastes especially nitrogenous wastes from the body.
Catabolism on the other hand is the set of metabolic pathways which break down molecules into smaller units and release energy.

Solution C.8.
Urea, creatinine, uric acid

Solution C.9.

Column I Column II 
(a)   Bowman’s Capsule Glomerulus
(b)   Contains more CO2 and less urea Renal Vein
(c)    Anti-diuretic hormone Regulates amount of water excreted
(d)   Contains more urea Renal artery

Solution C.10.
In a nephron, the blood flows through the glomerulus under great pressure. The reason for this great pressure is that the efferent (outgoing) arteriole is narrower than the afferent arteriole (incoming). This high pressure causes the liquid part of the blood to filter out from the glomerulusinto the renal capsule.

Solution D.1.

(a) Ultrafiltration – The process of the filtration of blood in the glomerulus under great pressure during which the liquid part of the blood i.e. plasma along with urea, glucose, amino acids and other substances enter the renal tubule.

(b) Micturition – The process of expelling urine out of the body through urethra by opening the sphincter muscles passing of urine involving relaxation of sphincter muscles between the urinary bladder and urethra.

(c) Renal pelvis – Renal pelvis is the expanded front end of the ureters into the kidney.

(d) Urea – A nitrogenous waste produced primarily in the liver due to the break down dead protein remains and extra amino acids.

(e) Osmoregulation – It is a process of maintaining the blood composition of the body i.e. the normal osmotic concentration of water and salts in the body.

Solution D.2.

Ultrafiltration – Ultrafiltration involves filtration of the blood which takes place in the glomerulus. The blood containing urea from the afferent arteriole enters the glomerulus under high pressure. The high pressure is created because the efferent arteriole is narrower than the afferent arteriole. The high pressure causes the liquid part of the blood to filter out from the glomerulus into the renal tubule. This filtrate is known as ‘glomerular filtrate’.
Glomerular filtrate consists of water, urea, salts, glucose and other plasma solutes. Blood corpuscles, proteins and other large molecules remain behind in the glomerulus. Therefore the blood which is carried away by the efferent arteriole is relatively thick.

Selective absorption – The Glomerular filtrate entering the renal tubule contains a lot of usable materials such as glucose and sodium. As this filtrate passes down the renal tubule, a lot of water along with these usable materials is reabsorbed. Such reabsorption is called ‘selective absorption’. The reabsorption occurs only to the extent that the normal concentration of the blood is undisturbed.

Solution D.3.
Dialysis involves the use of artificial kidney or a dialysis machine. The patient’s blood is from the radial artery is led through the machine where excess salts and urea is removed. The purified blood is then returned to a vein in the same arm.
Dialysis is carried out in case of failure of both the kidneys. In case there is a permanent damage, then the dialysis is to be repeated for about 12 hours twice a week.

Solution E.1.
(a) The image shown can be left or right kidney. As the right kidney is slightly lower than the left one, so we need to have the images of both the kidneys for comparison.
(b) It is a longitudinal section of the kidney.
(c) 1-renal artery, 2-renal vein, 3-ureter, 4-cortex, 5-pelvis
(d) (i) 4/cortex
(ii) medulla
(iii) 5/pelvis

Solution E.2.
(a) Excretory system and Circulatory system.
(b) 1-kidney, 2-renal artery, 3-ureter, 4-urinary bladder, 5-urethera
(c) Nephron
(d) Urea and ammonia
(e) Ultrafiltration and selective reabsorption

Solution E.3.
(a) 4/Glomerulus
(b) 2/Efferent arteriole
(c) 1/ Afferent arteriole from renal artery
(d) 7/Collecting tubule
(e) 5/ Proximal convoluted tubule with blood capillaries

Solution E.4.
(a) The process of removal of chemical wastes especially nitrogenous waste from the body is known as excretion.
(b) Nephrons
(c) As the cortex region contains numerous nephrons or kidney tubules, therefore, it shows a dotted appearance.
(d) Kidneys help in removing wastes or excretion and osmoregulation.
(e) The blood vessel ‘B’ is renal artery and the blood vessel ‘A’ is renal vein.

So the blood vessel ‘B’ contains oxygenated blood with high concentration of urea and glucose whereas the blood vessel ‘A’ contains deoxygenated blood with low concentration of urea and glucose as compared to renal artery.

Solution E.5.
a. The structure is a Bowman’s capsule, which is part of the nephron. The Bowman’s capsule is found in the cortex of the kidney.

b.

  1. Afferent arteriole
  2. Glomerulus
  3. Bowman’s capsule
  4. Efferent arteriole

c. Urine formation occurs in two steps – ultrafiltration and reabsorption.

d. The process occurring in 2 and 3 is known as ultrafiltration.
In the glomerulus, the blood flows under high pressure because of the narrow lumen of the capillary network of the glomerulus. This forces most of the components (both waste and useable materials) of the blood out of the capillaries. This process of the filtration of blood under high pressure in the Bowman’s capsule is known as ultrafiltration.

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 7 The Circulatory System. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 7 The Circulatory System

Exercise 1

Solution A.1.
(a) lymphocytes and monocytes

Solution A.2.
(b) phagocytosis

Solution B.1.
(a) Blood platelets and blood coagulation
(b) Neutrophils and phagocytosis
(c) Erythrocytes and transportation of gases
(d) Lymphocytes and Produce antibodies
(e) Bone marrow and destruction of old and weak RBC’s/production of RBCs and WBCs.

Solution B.2.
(a) Red Blood Cells
(b) Blood Platelets

Solution C.1.
Structural Differences between White Blood Cells and Red Blood Cells:

White Blood Cells Red Blood Cells
1. White blood cells are amoeboid. Red blood cells are minute biconcave disc-like structures.
2. They are nucleated cells. They anucleated cells.
3. Haemoglobin is absent in red blood cells. Haemoglobin is present in red blood cells.

Solution C.2.
During blood transfusion it is necessary to know the blood groups before transfusion because it is important that the blood groups of the donor and the recipient are compatible. In case of an incompatible blood transfusion, the recipient develops antibodies that attack the antigens present on the RBCs of the donor causing the blood cells to clump together which may result in death.

Solution C.3.
(a) Antibodies are produced by lymphocytes in response to the entry of pathogens in the blood stream.
Antibiotics are the medicines extracted from some bacteria and fungi. Antibiotics destroy or inhibit the growth of pathogens.

(b) RBC: RBC is enucleated, biconcave, disc-like structure, flat in the centre while thick and rounded at the periphery.
WBC: WBC is nucleated and amoeboid in shape.

(c) Serum: The plasma from which the protein fibrinogen has been removed is called serum.
Vaccine: Vaccine is killed or living weakened germs which are introduced in the body to stimulate the production of antibodies against pathogens for a particular disease.

Solution C.4.
Heparin

Solution D.1.
The functions of blood plasma are:

  • Transports of digested food from the alimentary canal to tissues.
  • Transports excretory materials from tissues to excretory organs.
  • Distributes hormones from the glands to their target site.
  • Distributes heat in the body to maintain the body temperature.

Solution D.2.
Blood clotting or coagulation occurs in a series of the following steps:

  • The injured tissue cells and the platelets disintegrate at the site of wound to release thromboplastin.
  • The thromboplastin with the help of calcium ions converts inactive prothrombin into active thrombin.
  • Thrombin in the presence of calcium ions converts soluble fibrinogen into insoluble fibrin which forms a mesh or network at the site of wound.
  • The blood cells trapped in this network shrink and squeeze out the plasma to leave behind a solid mass known as the clot.
    Selina Concise Biology Class 10 ICSE Solutions The Circulatory System image -1

Solution D.3.

(a) Rh factor – It is an inherited antigen often found on the blood cells. Some individuals have these antigens and are thus Rh positive (Rh+) while others who do not have this antigen are Rh negative (Rh-).

(b) Universal donor – The person with blood group O is a universal donor as this type of blood can be given to persons with any blood group i.e. O, A, B, AB.

(c) Diapedesis – It is the squeezing of leucocytes through the wall of capillaries into the tissues.

Solution D.4.
Blood clotting is not dependent on the exposure of blood to air. In fact, clotting can be caused by the movement of blood over a rough surface such as on cholesterol deposit inside of a blood vessel of the skin.

Solution D.5.
The functions of the blood are:

  1. Transport of digested food from the alimentary canal to tissues. These substances are simple sugars like glucose, amino acids, vitamins, mineral salts, etc.
  2. Transport of oxygen in the form of an unstable compound ‘oxyhaemoglobin’ from the lungs to the tissues.
  3. Transport of carbon dioxide from the tissues to the lungs.
  4. Transport of excretory materials from the tissues to the liver, kidney or the skin for elimination.
  5. Distribution of hormones from glands to the target sites.
  6. Distribution of heat to keep the body temperature uniform.
    (Any five)

Solution E.1.
(a)

  1. Red Blood Cell (RBC),
  2. White Blood Cell (WBC),
  3. Blood Platelet
  4. Blood Plasma.

(b) The red blood cells are minute biconcave disc-like structures whereas the white blood cells are amoeboid.

(c) Function of part 1 (RBC): Transport of respiratory gases to the tissues and from the tissues, transport of nutrients from the alimentary canal to the tissues.
Function of part 2 (WBC): WBCs play major role in defense mechanism and immunity of the body.
Function of part 3 (Blood Platelet): Blood platelets are the initiator of blood clotting.

(d) The average life span of a red blood cell (RBC) is about 120 days.

(e) Thromboplastin

Exercise 2

Solution A.1.
(d) Heart itself

Solution A.2.
(c) artery

Solution A.3.
(a) tricuspid valve

Solution A.4.
(b) renal artery

Solution A.5.
(b) inadequate supply of oxygen to the heart muscle

Solution A.6.
(d) destroy pathogens

Solution A.7.
(a) closure of tricuspid and bicuspid valves
(b) closure of aortic and pulmonary valves
(c) rushing of blood through valves producing turbulence

Solution B.1.
The average values of blood pressure in a normal adult human are 100-140 mm for systolic pressure and 60-80 mm for diastolic pressure.

Solution B.2.
Yes, the heart beats approximately 1, 03,680 times in a day.

Solution B.3.
(a) Hepatic portal vein
(b) Blood Capillaries
(c) Pulmonary artery
(d) White blood cells
(e) Venules
(f) Portal vein
(g) Atrial systole
(h) Tricuspid valve
(i) Atrial systole
(j) Pericardial fluid

Solution B.4.

(a) Blood platelets are involved in blood clotting or coagulation. Blood platelets integrate at the site of injury to release thromboplastin which initiates the process of blood clotting.

(b) Neutrophils perform phagocytosis i.e. they engulf pathogens that enter the blood stream and destroy them.

(c) Erythrocytes transport oxygen from the lungs to the tissues in the form of an unstable compound oxyhaemoglobin and transport carbon dioxide from the tissues to the lungs.

(d) Lymphocytes produce antibodies against pathogens which enter the blood stream. In some cases they also perform phagocytosis.

(e) Bone marrow is involved in formation of RBCS and WBCs. It is also involved in the destruction of old and weak RBCs.

Solution B.5.
(a) The blood vessel that begins and ends in capillaries is the hepatic portal vein.
(b) A blood vessel which has small lumen and thick wall is artery.
(c) The valve which prevents the back flow of blood in the veins and lymph vessels is semilunar valve.

Solution B.6.
(a) Lubb: Atrio-ventricular valve:: Dup: Semilunar valves
(b) Coronary artery: Heart::Hepatic artery: Liver

Solution B.7.
A matured mammalian erythrocyte lacks a nucleus and mitochondria. The lack of a nucleus increases the surface area-volume ratio of RBCs, thus increasing the area for oxygen absorption. Also, the lack of a nucleus reduces the size of the cell, making it easy to flow through the blood vessels and more cells can be accommodated in a small area.
The lack of mitochondria implies that the cell does not use any oxygen absorbed for respiration, thus increasing the efficiency of the cell to transport oxygen as all the oxygen absorbed is transported without any loss.

Solution C.1.
Blood flows twice in the heart before it completes one full round. The full round thus includes pulmonary and systemic circulation. In pulmonary circulation, blood enters the lungs through pulmonary arteries. Pulmonary veins collect the blood from the lungs and carry it back to the left atrium.
In systemic circulation, blood from the left ventricle enters the aorta through which the blood is sent to the body parts. From the body parts blood is collected by veins and sent back to the heart. Therefore, the blood circulation in the human body is called double circulation.

Solution C.2.
The first sound LUBB is produced when the atrio-ventricular valves i.e. tricuspid and bicuspid valves close at the start of ventricular systole.
The second sound DUP is produced at the beginning of ventricular diastole, when the pulmonary and aortic semilunar valves close.

Solution C.3.
People have a common belief that the heart is located on the left side of the chest because the narrow end of the roughly triangular heart is pointed to the left side and during its working the contraction of the heart is more powerful on the left side which can be felt.

Solution C.4.
(a)

Erythrocytes Leucocytes
They function in the transport of oxygen throughout the body and in the removal of carbon dioxide from the body. They help in the defense of the body against disease-causing pathogens.

(b) An artery carries blood away from the heart whereas a vein brings blood towards the heart.
(c) An artery generally contains oxygenated blood whereas a vein generally carries deoxygenated blood.
(d) Tricuspid valve is located between the right atrium and right ventricle of the heart whereas a bicuspid valve is located between the left atrium and left ventricle of the heart.

Solution C.5.

Column A Column B
SA node Pacemaker
Defective hemoglobin in RBC Sickle cell anemia
Muscle fibres located in the heart Purkinje fibres
The liquid squeezed out of blood during clotting Serum
Never tires, keep on contracting and relaxing Cardiac muscles
Cardiac cycle 0.85 sec
Liquid part of the blood without corpuscles Plasma

Solution C.6.

Substance From To
Oxygen Lungs Whole body
Carbon dioxide Whole body Lungs
Urea Whole body Kidneys
Digested carbohydrates Intestines Whole body
Hormones Endocrine glands Target organs

Solution D.1.
(a) Endothelium– It is the innermost layer of the muscular wall of an artery or a vein which faces the lumen.
(b) Lymph nodes– The structures from which fresh lymph channels arise which pour the lymph into major anterior veins.
(c) Venule– The smallest common blood vessel formed by the union of capillaries.
(d) Diastole– The relaxation of muscles of ventricles or atria.

Solution D.2.

Artery Vein
An artery is a vessel which carries blood away from the heart towards any organ. A vein is a vessel which conveys the blood away from an organ towards the heart.
Artery has thick muscular walls. Vein has thin muscular walls.
It has a narrow lumen. It has a broad lumen.
There are no valves. Thin pocket-shaped valves are present in the veins.
Arteries progressively decrease in size and branch to form arterioles. Arterioles further breaks up to form capillaries. Capillaries unite to form branches called Venules. Venules further unite to form veins.

Solution D.3.
Tonsils: Tonsils are lymph glands located on the sides of the neck. They tend to localize the infection and prevent it from spreading it in the body as a whole.
Spleen: The spleen is a large lymphatic organ. The spleen acts as a blood reservoir in case of emergency such as haemorrhage, stress or poisoning. It produces lymphocytes and destroys worn out RBCs.

Solution D.4.
(a) The left ventricle pumps blood to the farthest points in the body such as the feet, the toes and the brain against the gravity while the right ventricle pumps the blood only up to the lungs. Therefore, the left ventricle has thicker walls than the right ventricle.
(b) The right ventricle pumps blood to the lungs for oxygenation whereas the right auricle receives the blood from vena cavae and passes it to the right ventricle. Therefore, walls of the right ventricle are thicker than those of the right auricle.

Solution D.5.
(a) The left ventricle pumps blood to the farthest points in the body such as the feet, the toes and the brain against the gravity. Thus, it requires greater force to push the blood. In order to with stand with the force applied the walls of the left ventricle are thicker than the walls of all the chambers.

(b) The blood from stomach and intestines enters the liver via hepatic portal vein because the liver monitors all the substances that have to be circulated in body. The excess nutrients such as glucose, fats are stores in the liver. Excess amino acids are broken down by the process deamination. Toxic substances are detoxified.

(c) During blood transfusion it is important that the blood groups of the donor and the recipient are compatible. In case of an incompatible blood transfusion, the recipient develops antibodies that attack the antigens present on the RBCs of the donor causing the blood cells to clump together which may result in death. The examination of Rh factor is also necessary for the blood transfusion. Therefore, the blood groups of both the donor and recipient must be known before transfusing blood.

(d) Veins carry the blood from the body part towards the heart while the arteries carry the blood from the heart. Veins carry the blood against the force of gravity. Therefore, only the veins and not the arteries are provided with valves.

(e) Atrial wall is less muscular than the ventricular wall because the major function of atria is to receive blood from the body and pump in into very next ventricles. While the ventricles pump the blood out of the heart. Right ventricle to the lungs and the left ventricle to all the body parts.

(f) Arteries are responsible to carry oxygenated blood from the heart to the tissues. The blood flows in the artery under high pressure and in spurts. If arteries are located superficially then there is a high possibility of their damage which could lead to a lot of blood loss. To prevent this damage and blood loss, the arteries are deep seated in the body.

Solution E.1.
(a) A is artery, B is vein.
(b)

  1. Endothelium of the artery,
  2. Middle layer of smooth muscles and elastic fibres of the artery,
  3. External layer of connective tissue of the artery,
  4. Endothelium of the vein,
  5. Middle layer of smooth muscles and elastic fibres of the vein,
  6. External layer of connective tissue of the vein.

(c) An artery has thick muscular walls and a narrow lumen. It carries blood away from the heart towards any organ.
A vein on the other hand has thin muscular walls and a wider lumen. It carries blood away from an organ towards the heart.

Solution E.2.
(a) The structure 3 represents the heart. It forms the centre of double circulation and is located between the liver and the head (as per the diagram). Also the blood circulation (indicated by 1) begins from heart to lungs.
(b)

Aorta 5
Hepatic portal vein 7
Pulmonary artery 1
Superior vena cava 9
Renal vein 8
Stomach 10
Dorsal aorta 11

Solution E.3.
(a)

1 Systemic Circulation
2 Vena Cava
3 Aorta
4 Right Ventricle
5 Left Atrium
6 Pulmonary Artery
7 Pulmonary Vein
8 Pulmonary Circulation

(b) Blood flows twice in the heart before it completes one full round. The full round thus includes pulmonary and systemic circulation. For this reason, the blood circulation in the human body is called double circulation.
(c) The relaxation of muscles of ventricles or auricles is known as diastole.

Solution E.4.
(a) Tissue Fluid
(b) Red blood cells
(c) Lymph
(d) The lymph supplies nutrition and oxygen to those parts where blood cannot reach. The lymph drains away excess tissue fluids and metabolites and returns proteins to the blood from tissue spaces.

Solution E.5.
(a) Hepatic portal vein (4)
(b) Hepatic portal vein (4)

Solution E.6.
(a) A- Vein, B-Artery, C-Capillary
(b)

  1. External layer made of connective tissue
  2. Lumen
  3. Middle layer of smooth muscles and elastic fibres
  4. Endothelium

(c) An artery has thick muscular walls and a narrow lumen. It does not have any valve.
A vein on the other hand has thin muscular walls and a wider lumen. It has valves to prevent back flow of blood.

(d) A (Vein)- deoxygenated blood, B (Artery)- oxygenated blood

(e) At the capillary level the actual exchange of gases takes place.

Solution E.7.
(a) Atrial Diastole and Ventricular Systole

(b) Ventricular muscles are contracting during this phase because the valves between the two ventricles and pulmonary artery and aorta are open while the atrio-ventricular valves are closed.

(c)

1 Pulmonary Artery
2 Aorta
3 Pulmonary Vein
4 Left Atrium
5 Bicuspid Valve
6 Right Ventricle

(d) Part 1 (Pulmonary artery) – Deoxygenated blood
Part 2 (Aorta) – Oxygenated Blood

(e) Two i.e., bicuspid and tricuspid valves are closed in this phase.

Solution E.8.
a.

  1. Arteriole
  2. Artery
  3. Venule
  4. Capillaries
  5. Vein

b. Such an arrangement can be observed in the lungs.

Solution E.9.
a. 1 – Red blood cell
b. Diapedesis
c.

RBC WBC
They lack a nucleus. They have a nucleus.
They are biconcave and disc-shaped. They are spherical and have different sizes.

d. The process which occurs in B and C is phagocytosis. In this process, the WBCs engulf the foreign particles and destroy them, thus preventing the occurrence of disease.

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Selina Concise Biology Class 10 ICSE Solutions Photosynthesis: Provider of Food for All

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Selina Concise Biology Class 10 ICSE Solutions Photosynthesis: Provider of Food for All

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 6 Photosynthesis: Provider of Food for All. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 6 Photosynthesis: Provider of Food for All

Exercise 1

Solution A.1.
b) glucose formed in photosynthesis soon gets converted into starch

Solution A.2.
b) twelve

Solution A.3.
b) humidity

Solution A.4.
c) trapping light energy

Solution A.5.
a) continue to live, but will not be able to store food

Solution A.6.
a) Carbon dioxide is reduced and water is oxidised

Solution A.7.
c) activate chlorophyll

Solution A.8.
d) ensure that the leaves are free from starch

Solution A.9.
a) CO2

Solution B.1.
(a) Producers / Autotrophs
(b) Chloroplasts
(c) ATP (Adenosine triphosphate)
(d) Glucose
(e) Green plants
(f) Carbon dioxide dissolved in water
(g) Stroma
(h) Phloem

Solution C.1.

(a)

Respiration Photosynthesis
The gas released during respiration is carbon dioxide. The gas released during photosynthesis is oxygen.

(b)

Light Reaction Dark Reaction
Hydrogen and oxygen are produced here, along with release of electrons, which converts ADP into ATP. Glucose is the main product formed during dark reaction.

(c)

Producers Consumers
Producers show autotrophic mode of nutrition i.e. they are able to produce their own food from basic raw materials.
For example: green plants

Consumers show heterotrophic mode of nutrition i.e. they depend directly or indirectly on the producers for their food.
For example: Animals

(d)

Grass Grasshopper
Green grass being a producer is capable of producing its own food by photosynthesis. Grasshopper is a primary consumer (herbivore) and directly feeds on producers like grass.

(e)

Chlorophyll Chloroplast
Chlorophyll is the green pigment present in cell organelles called chloroplasts. Chloroplasts are cell organelles, situated in the cytoplasm of plant cells. They are present mainly in the mesophyll cells and in the guard cells of stomata.

Solution C.2.
(a) False
Correct Statement: Dark reaction of photosynthesis is independent of light and occurs simultaneously with light reaction.

(b) True

(c) False
Correct Statement: Starch produced in a leaf is stored temporarily in the leaf until the process of photosynthesis. At night it is converted back into soluble sugar and translocated to different part of the body either for the utilization or for the storage.

(d) True

(e) False
Correct Statement: Green plants are producers.

(f) False
Correct Statement: Respiration results in loss of dry weight of the plants.

(g) False
Correct Statement: Photosynthesis stops at a temperature of above 40oC.

(h) True
(i) True
(j) True

Solution C.3.
(a) grana
(b) iodine solution
(c) chloroplast
(d) Calvin cycle
(e) Sucrose

Solution C.4.

(a) False
Photosynthesis increases with the light intensity up to a certain limit only and then it gets stabilized.

(b) False
The atmospheric temperature is an important external factor affecting photosynthesis. The rate of photosynthesis increases up to the temperature 35oC after which the rate falls and the photosynthesis stops after 40oC.

(c) False
Ice cold water will hamper the process of photosynthesis in the immersed leaf, even if there is sufficient sunshine because the temperature is an important factor for the rate of photosynthesis.

(d) False
For destarching, the potted plant can kept in a dark room for 24-48 hours.

(e) False
There is no start point or end point in the carbon cycle, the carbon is constantly circulated between the atmosphere and the living organisms.

(f) False
If a plant is kept in bright light all the 24 hours for a few days, the dark reaction (biosynthetic phase) will continue to occur because the dark reaction is independent of light and it occurs simultaneously with the light dependent reaction.

(g) True

Solution C.5.
Photons, grana, water molecules, hydrogen and hydroxyl ions, oxygen

Solution C.6.

Photosynthesis Respiration
Carbon dioxide is used up and oxygen is released. Oxygen is used up and carbon dioxide is released.
Photosynthesis occurs in plants and some bacteria. Respiration occurs in all living organisms.
Photosynthesis results in gain of dry weight of the plants. Respiration results in loss of dry weight of the plants.
Glucose is produced which is utilized by the plants. Glucose is broken down to obtain energy.
The raw materials for the photosynthesis are water, carbon dioxide and sunlight. The raw material for respiration is glucose.

(Any 4)

Solution C.7.
Oxygen is released during photosynthesis. Some of this oxygen may be used in respiration in the leaf cells, but the major portion of it is not required and it diffuses out into the atmosphere through the stomata. However, in a sense, even this oxygen is not a waste because all organisms require it for their existence including the plants.

Solution C.8.
The presence of starch is regarded as evidence of photosynthesis. Hence before starting an experiment on photosynthesis, the plant should be placed in the dark for 24-48 hours to destarch the leaves. During this period, all the starch from the leaves will be sent to the storage organs and the leaves will not show the presence of starch. So the various experiments on photosynthesis can be carried out effectively.

Solution C.9.
Destarching means removal of starch. Destarching is carried out so that all the starch from the leaves will be sent to the storage organs. Hence all the leaves will not show the presence of starch and photosynthesis can be studied. Destarching ensures that any starch present after the experiment has been formed under experimental conditions.

Solution C.10.
If a green plant is kept in bright light, it tends to use up all the CO2 produced during respiration, for photosynthesis. Thus, the release of CO2cannot be demonstrated. Hence, it is difficult to demonstrate respiration as these two processes occur simultaneously.

Solution C.11.
The chloroplasts are concentrated in the upper layers of the leaf which helps cells to trap the sunlight quickly. Also the epidermis is covered by a waxy, waterproof layer of cuticle. This layer is thicker on the upper surface than the lower one. Hence most leaves have the upper surface more green and shiny than the lower one.

Solution C.12.

  • Place hydrilla plant (a water plant) in a beaker containing pond water and cover it by a short-stemmed funnel. (Make sure the level of water in the beaker is above the level of the stem of the funnel)
  • Invert a test tube full of water over the stem of the funnel.
  • Place the set up in the sun light for a few hours.

Observation:
Bubbles appear in the stem which rise and are collected in the test tube. When sufficient gas gets collected, a glowing splinter will be introduced in the test tube, which will burst into flames.

Inference:
The splinter glows due the presence of oxygen in the test tube which proves that the gas collected in the test is released by hydrilla during photosynthesis.
Selina Concise Biology Class 10 ICSE Solutions Photosynthesis Provider of Food for All image - 1

Solution C.13.

(i) Light Reaction: 
The light reaction occurs in two main steps:

  1. Activation of chlorophyll – On exposure to light energy, chlorophyll becomes activated by absorbing photons.
  2. Splitting of water – The absorbed energy is used in splitting the water molecule into hydrogen and oxygen, releasing energy. This reaction is known as photolysis of water.

Selina Concise Biology Class 10 ICSE Solutions Photosynthesis Provider of Food for All image - 2
The fate of H+, e– and (O) component are as follows:
The hydrogen ions (H+) obtained from above are picked up by a compound NADP (Nicotinamide adenine dinucleotide phosphate) to form NADPH.
Selina Concise Biology Class 10 ICSE Solutions Photosynthesis Provider of Food for All image - 3
The oxygen (O) component is given out as molecular oxygen (O2).
2O → O2
The electrons (e–) are used in converting ADP into energy rich ATP by adding one inorganic phosphate group Pi.
ADP + Pi → ATP
This process is called photophosphorylation.

(ii) Dark reaction: The reactions in this phase does not require light energy and occur simultaneously with the light reaction. The time gap between the light and dark reaction is less than one thousandth of a second. In the dark reaction, ATP and NADPH molecules (produced during light reaction) are used to produce glucose (C6H12O6) from carbon dioxide. Fixation and reduction of carbon dioxide occurs in the stroma of the chloroplast through a series of reactions. The glucose produced is either immediately used up by the cells or stored in the form of starch.
Selina Concise Biology Class 10 ICSE Solutions Photosynthesis Provider of Food for All image - 4

Solution C.14.
Complete the following food chains by writing the names of appropriate organisms in the blanks:
(i) Grass → Rabbit. → Snake → Hawk
(ii) Grass/Corn → Mouse → Snake → Peacock

Solution C.15.
Non-green plants such as fungi and bacteria obtain their nourishment from decaying organic matter in their environment. This matter comes from dead animals and plants. Fungi and bacteria break down the organic matter to obtain the nourishment and they release carbon dioxide back in the atmosphere.

Solution C.16.
Chlorophyll is the foundation site for the photosynthesis in green plants. The initiation of photosynthesis takes place when the chlorophyll molecule traps the light energy. The light energy is then converted into chemical energy in the form of glucose using carbon dioxide (CO2) from the atmosphere, and water (H2O) from the soil. All other organisms, directly or indirectly depend on this food for their survival. The starting point of any food chain is always a plant. If green plants were to suddenly disappear, then so would virtually all life on Earth. Thus, we can say that all life owes its existence to chlorophyll.

Solution C.17.
To test the leaf for starch, the leaf is boiled in water to kill the cells. It is next boiled in methylated spirit to remove chlorophyll. The leaf is placed in warm water to soften it. It is then placed in a dish and iodine solution in added. The region, which contains starch, turns blue-black and the region, which does not contain starch, turns brown.

Solution D.1.

a. The student wanted to show that sunlight is necessary for photosynthesis. / The role of sunlight in photosynthesis is being investigated.

b. Yes. The other uncovered leave of the potted plant act as a control.

c. Destarching ensures that any starch present after the experiment has been formed under experimental conditions. Therefore, the plant was kept in the dark before the experiment.

d.

  • The student dipped the leaf in boiling water for a minute to kill the cells.
  • Then he boiled the leaf in alcohol/methylated spirit over a water bath to remove chlorophyll. The leaf becomes hard and brittle.
  • He then places the leaf in hot water to soften it.
  • Next the student spreads the leaf in a dish and pours iodine solution on it. The presence of starch is indicated by a blue-black colour.
  • The uncovered portion (exposed to sunlight) turned blue-black colour and the covered portion showed brown colour. The difference in the colours of covered and uncovered part of leaves indicates the importance of sunlight in photosynthesis.

Solution D.2.
(a) Guard cells: They regulate the opening and closing of stomata and thus regulate the entry of carbon dioxide through the stomata.

(b) Cuticle: Cuticle is transparent and water proof due to which light can penetrate this later easily.

(c) Mesophyll cells: Mesophyll cells are the main sites for photosynthesis. Chloroplasts are mainly contained in the mesophyll cells. When sunlight falls on the leaf, the light energy is trapped by the chlorophyll of the upper layers of mesophyll, especially the palisade cells.

(d) Xylem Tissue in the Leaf Veins: Water is essential for photosynthesis to occur. Water is taken up by the roots from the soil, sent up through the stem and finally brought to the leaves (site of photosynthesis) through the xylem tissue. The water is then distributed in the mesophyll tissue.

(e) Phloem Tissue in the Leaf Veins: The prepared food is transported from leaves to all parts of the plant by the phloem tissue. The glucose is converted into insoluble starch and later into soluble sugar i.e. sucrose, which is transported in solution through the phloem in the veins of the leaf and down through the phloem of the stem.

(f) Stoma: The main function of stoma is to let in carbon dioxide from the atmosphere for photosynthesis. Also most of the oxygen produced during photosynthesis diffuses out into the atmosphere through the stomata.

Solution D.3.
a.

  1. Sunlight
  2. Oxygen
  3. Glucose
  4. Xylem

b. A – Transpiration
B – Translocation

Solution D.4.
a. Food chain
b. Hawk, eagle
c. Photosynthesis
d. Carbon

Solution D.5.
Test to determine the presence of starch in a leaf:

  • Dip a leaf in boiling water for a minute to kill the cells.
  • Boil the leaf in methylated spirit in a water bath to remove the chlorophyll, till the leaf turns pale blue and becomes hard and brittle.
  • Now place the leaf in hot water to soften it.
  • Place the leaf in a Petri dish and pour iodine solution over it.
  • The appearance of a blue-black colour on the leaf is indicative of the presence of starch.
  • The absence of starch is indicated by a brown colouration.

Solution D.6.
a. To demonstrate the importance of carbon dioxide in photosynthesis
b. No, the experiment will not work satisfactorily, as the beaker contains lime water and not potassium hydroxide to absorb CO­2.
c. Place potassium hydroxide in the beaker instead of lime water
d. Before starting the experiment, it is necessary to destarch the leaves of the plant by keeping the plant in complete darkness for 48 hours. This is because if the plant is not destarched, then the experiment will give false results because starch stored previously may be detected in the leaf placed in the beaker even if no starch is produced during the experiment.

Solution D.7.
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Selina Concise Biology Class 10 ICSE Solutions Transpiration

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Selina Concise Biology Class 10 ICSE Solutions Transpiration

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 5 Transpiration. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 5 Transpiration

Exercise 2

Solution A.1.
(a) Open stomata, dry atmosphere and moist soil

Solution A.2.
(a) increase

Solution A.3.
(b) temperature is high

Solution A.4.
(c) sunken stomata

Solution A.5.
(d) hydathodes

Solution A.6.
(d) transpiration

Solution A.7.
(d) hot, dry and windy

Solution A.8.
(b) Lenticels

Solution A.9.
(b) evaporation of water from the aerial surfaces of a plant

Solution B.1.
(a) Lenticels
(b) Guttation
(c) Potometer
(d) Nerium
(e) Ganong’s photometer
(f) Stomata and cuticle
(g) Hydathodes
(h) Guttation

Solution B.2.
(a) vapour, aerial
(b) stomata, transpiration
(c) suction, water (heat)

Solution C.1.
(a) guttation
(b) protection and reduced transpiration
(c) transpiration
(d) reduced transpiration

Solution C.2.
(i) False
(ii) True
(iii) True
(iv) False
(v) Most transpiration occurs at mid-day.
(vi) Potometer is an instrument used for measuring the rate of transpiration in green plants.

Solution C.3.
(a) Transpiration increases with the velocity of wind. If the wind blows faster, the water vapours released during transpiration are removed faster and the area surrounding the transpiring leaf does not get saturated with water vapour.

(b) When the rate of transpiration far exceeds the rate of absorption of water by roots, the cells lose their turgidity. Hence, excessive transpiration results in wilting of the leaves.

(c) Plants absorb water continuously through their roots, which is then conducted upwards to all the aerial parts of the plant, including the leaves. Only a small quantity of this water i.e. about 0.02% is used for the photosynthesis and other activities. The rest of the water is transpired as water vapour. Hence water transpired is the water absorbed.

(d) There are more stomatal openings on the lower surface of a dorsiventral leaf. More the number of stomata, higher is the rate of transpiration. Hence more transpiration occurs from the lower surface.

(e) Cork and Bark of trees are tissues of old woody stems. Bark is thick with outermost layer made of dead cells and the cork is hydrophobic in nature. These properties make them water-proof and hence they prevent transpiration.

(f) In both perspiration and transpiration, water is lost by evapouration from the body of the organism as water vapour. This evaporation reduces the temperature of the body surface and brings about cooling in the body of the organism.

(g) On a bright sunny day, the rate of transpiration is much higher than any other days. The leaves of certain plants roll up on a bright sunny day to reduce the exposed surface and thus reduce the rate of transpiration.

Solution C.4.
(a) False
Reason: Potometer is used to measure the rate of transpiration in a plant. Demonstration of transpiration occurring from the lower surface of a leaf is done by analyzing the changes in colour of pieces of dry cobalt chloride paper attached (and held in place) to the two surfaces of a leaf.

(b) True
Reason: Transpiration carried out by the large number of trees in a forest. This increases the moisture in the atmosphere and brings rain.

(c) False
Reason: Hydathodes are special pores present on the ends of leaf veins through which guttation occurs and water droplets are given out. Their openings cannot be regulated. Stomata on the other hand are minute openings in the epidermal layer of leaves through which exchange of gases as well as transpiration occurs. Water is given out as water vapour. Stomatal opening is regulated by guard cells.

(d) False
Reason: Transpiration is reduced during high atmospheric humidity. High humidity in the air reduces the rate of outward diffusion of the internal water vapour across stomata, thereby reducing the rate of transpiration.

(e) True
Reason: Desert plants need to reduce transpiration as much as possible so as to survive in the hot and dry environment. Hence some of them have sunken stomata as an adaptation to curtail transpiration.

(f) True
Reason: During the day, the stomata are open to facilitate the inward diffusion of carbon dioxide for photosynthesis. During mid-day, the outside temperature is higher, due to which there is more evaporation of water from the leaves. Therefore more transpiration occurs during mid-day.

Solution C.5.

Guttation Bleeding
It is the removal of excess of water from the plants because of excess water buildup in the plant. It is the removal of water from the plant because of injury.
Water escapes from specialisedstructures called hydathodes. Water escapes in the form of sap from the injured part of the plant.

Solution D.1.
Wilting refers to the loss of cellular turgidity in plants which results in the drooping of leaves or plant as a whole because of lack of water.
During noon the rate of transpiration exceeds the rate of absorption of water by roots. Due to the excessive transpiration, the cells of leaves lose their turgidity and wilt.

Solution D.2.
The lower surface of leaf is sheltered from direct sunlight. If more stomata are on the upper surface of a leaf, then excessive transpiration would occur, resulting in quick wilting of the plant. Hence most plants have more numerous stomata on the lower surface of a leaf to control the rate of transpiration.

Solution D.3.
Take the small potted rose plant and cover it with a transparent polythene bag. Tie its mouth around the base of the stem. Leave the plant in sunlight for an hour or two.
Selina Concise Biology Class 10 ICSE Solutions Transpiration image - 1
Drops of water will soon appear on the inner side of the bag due to the saturation of water vapour given out by the leaves. A similar empty polythene bag with its mouth tied and kept in sunlight will show no drops of water. This is the control to show that plants transpire water in the form of water. If tested with dry cobalt chloride paper, the drops will be confirmed as water only.

Solution D.4.
Selina Concise Biology Class 10 ICSE Solutions Transpiration image - 2
Potometer is a device that measures the rate of water intake by a plant. This water intake is almost equal to the water lost through transpiration. Potometers do not measure the water lost due to transpiration but measure the water uptake by the shoot.

Solution D.5.

  • Transpiration occurring through lenticels i.e. minute openings on the surface of old stems is called lenticular transpiration.
  • Stomatal transpiration is controlled by the plant by altering the size of the stoma, where as this does not happen in case of lenticular transpiration. This is because the lenticels never close, but remain open all the time.
  • The amount of stomatal transpiration is much more than the amount of lenticular transpiration.

Solution D.6.
The factors that accelerate the rate of transpiration are:

  • High intensity of sunlight
  • High temperature
  • Higher wind velocity
  • Decrease in atmospheric pressure
    (Any three)

Solution D.7.
Forests have large number of plants especially trees. Each plant loses water in the form of water vapour everyday into the atmosphere through transpiration. A large apple tree loses as much as 30 litres of water per day. So huge amount of water is escaped into the atmosphere by forests. This increases the moisture in the atmosphere and brings more frequent rains.

Solution D.8.
The advantages of transpiration to the plants are:

  • Transpiration brings about a cooling effect to the plant body since evaporation of water reduces the temperature of leaf surface.
  • Transpiration helps in the ascent of sap by producing a suction force acting from the top of the plant.
  • Transpiration helps in distributing water and mineral salts throughout the plant body.
  • Transpiration helps in eliminating excess water.

Solution D.9.

  1. If the water content of the leaves decreases due any reason, the guard cells turn flaccid, thereby closing the stomatal opening and transpiration stops.
  2. Some plants have sunken stomata whereas others have reduced number of stomata to reduce transpiration.
  3. In some plants, leaves may be dropped or may be absent or changed into spines as an adaptation to reduce transpiration.
  4. The leaves may be covered by thick cuticle such as in Banyan tree, so as to reduce transpiration.

Solution D.10.
No, they are not dew drops.

This is water given out by the plant body through guttation. Since the banana plant is growing in humid environment, transpiration is hampered. But the roots continue to absorb water from the soil. This builds up a huge hydrostatic pressure within the plant and forces out the excess water from the hydathodes, which are pores present at the tips of veins in the leaf. This is observed especially during the mornings.

Solution D.11.
(a) Intensity of light – During the day, the stomata are open to facilitate the inward diffusion of carbon dioxide for photosynthesis. At night they are closed. Hence more transpiration occurs during the day. During cloudy days, the stomata are partially closed and the transpiration is reduced.
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(b) Humidity of the atmosphere – When the air is humid; it can receive very less water vapour. Thus, high humidity in the air reduces the rate of outward diffusion of the internal water vapour across stomata, thereby reducing the rate of transpiration.

Solution E.1.

(i) The leaf D would become most limp. This is because water would be lost through transpiration from upper as well as the lower surface of leaf D since it is uncoated.

(ii) The least limping would be shown by leaf C since its upper and lower surfaces have been coated with vaseline. So no water is lost from the leaf through transpiration since the stomatal openings get blocked by vaseline.

Solution E.2.
(a)
Guard Cell
Inner wall of the Guard Cell
Stoma/Stomatal Aperture
(b) Open state
(c) The structure of stoma remains same in monocots as well as in dicots. Hence, the stoma from the diagram can be of a monocot leaf or of a dicot leaf.
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Solution E.3.
(a) Transpiration
(b) Oil is put on the surface of water to prevent loss of water by evaporation.
(c) Yes, the transpiration rate will increase. Transpiration would occur faster. The observable changes will occur in less time.
(d) The spring balance progressively measures the change in weight of the set-up. This because as the plant transpires, it creates the suction force in plant which allows roots to absorb more water from the test tube. Hence, the water in the test will get reduced. Thus, the weight of the entire set will decrease.

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Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots : The Processes Involved

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Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots : The Processes Involved

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 4 Absorption by Roots : The Processes Involved. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 4 Absorption by Roots : The Processes Involved

Exercise 1

Solution A.1.
c) Imbibition

Solution A.2.
c) Hypertonic salt solution

Solution A.3.
b) Turgidity

Solution A.4.
d) Grow downward into the soil

Solution A.5.
b) Diffusion

Solution A.6.
c) a selectively permeable membrane in between

Solution A.7.
a) Pure water

Solution A.8.
d) water

Solution A.9.
b) Root pressure

Solution A.10.
d) it allows a solvent to pass through freely but prevents the passage of the solute

Solution B.1.
(a) Turgidity
(b) Guttation
(c) Osmosis
(d) Xylem
(e) Endosmosis
(f) Diffusion
(g) Root pressure

Solution B.2.
(a) Turgor pressure
(b) Flaccidity
(c) Bleeding

Solution B.3.
(a) the fluids inside
(b) transported inside against their concentration gradient
(c) turgor movements

Solution B.4.
(a) shrink
(b) water
(c) opposite

Solution B.5.

Column I Column II
a Xylem (iv) upward flow of water
b Phloem (iii) downward flow of sap
c Cell membrane (i) semi-permeable
d Root pressure (v) guttation
e Cell wall (ii) permeable

Solution C.1.
(a)

Plasmolysis Deplasmolysis
1. It refers to the shrinkage of the cytoplasm and withdrawal of the plasma membrane from the cell wall caused due to the withdrawal of water when placed in a hypertonic solution.

2. In Plasmolysis, the cell becomes flaccid.

1.Deplasmolysis is the recovery of a plasmolysed cell when it is placed in water, wherein the cell’s protoplasm again swells up due to the re-entry of water.

2. In deplasmolysis, the cell becomes turgid.

(b)

Turgor pressure Wall pressure
Turgor pressure is the pressure of the cell contents on the cell wall. Wall pressure is the pressure exerted by the cell wall on the cell content.

(c)

Guttation Bleeding
Guttation is the process by which drops of water appear along leaf margins due to excessive root pressure. Bleeding is the loss of cell sap through a cut stem.

(d)

Turgidity Flaccidity
1. It is the state of a cell in which the cell cannot accommodate any more water and it is fully distended. 1. It is the condition in which the cell content is shrunken and the cell is not tight.

Solution C.2.
(a)

  1. False
  2. False
  3. False
  4. True
  5. False
  6. False

(b)

  1. A plant cell placed in hypotonic solution gets turgid.
  2. Addition of salt to pickles prevents growth of bacteria because they turn flaccid.
  3. Cells that have lost their water content are said to be plasmolysed.
  4. The shrinkage of protoplasm, when a cell is kept in hypertonic solution.

Solution C.3.
The cell is said to be turgid when the plant cell wall becomes rigid and stretched by an increase in the volume of vacuoles due to the absorption of water when placed in hypotonic solution. On the other hand, the cell is said to be flaccid when the cell contents get shrunken when the cell is placed in hypertonic solution and the cell is no more tight. Flaccidity is the reverse of turgidity.

Example: Weeds can be killed in a playground by sprinkling excessive salts around their base.
Or
A plant cell when immersed in hypertonic solution like salt solution for about 30 minutes will become flaccid or limp.

Solution C.4.

(a) Common salt when sprinkled on the grass causes the Plasmolysis of grass cell ultimately leading them to death. Hence, if we sprinkle some common salt on grass growing on a lawn, it is killed at the spot.

(b) If a plant is uprooted, the leaves continue losing water by transpiration, but there is no more water absorbed the roots. This does not allow the compensation for the loss of water by transpiration; hence the leaves of the uprooted plant wilt soon.

(c) Transplantation causes stress to the seedlings. If the seedlings are transplanted in the morning, they would have to immediately bear the additional stress of excessive transpiration occurring during the hot afternoon. Transplantation in the evening helps the seedlings to adjust for a longer time during the night (cooler temperatures) because the quantity of water absorbed exceeds the loss of water through transpiration. Therefore, it is better to transplant seedling in a flower bed in the evening and not in the morning.

(d) In a hypertonic solution, the solution outside the cell has higher solute concentration than the fluids inside the cell. Therefore, water flows out from the plant cell due to exosmosis. The cytoplasm shrinks and the plasma membrane withdraws away from the cell wall and this the cell becomes flaccid. Hence a plant cell when kept in a hypertonic salt solution for about 30 minutes turns flaccid.

(e) Potato cubes contain excess of salts and sugars as compared to the water in which the cubes are placed. Hence, due to endosmosis, water from the surrounding enters the potato cubes making them firm and increasing their size.

Solution C.5.

(a) True.
Plasmolysis occurs due to outflow of water from the cell when placed in hypertonic solution due to which the cytoplasm shrinks away from the cell wall. On the other hand, deplasmolysis is the result of the re-entry of water into the plasmolysed cell when placed in hypotonic solution due to which the protoplasm again swells up pressing tight against the cell wall.

(b) False.
Guttation is the process by which drops of water appear along leaf margins due to excessive root pressure whereas bleeding is the loss of cell sap through a cut stem.

(c) False.
There is only one seed coat in a seed.

(d) False.
The leaves of the twig remain turgid since its xylem is intact and xylem is responsible for water conduction in plants.

(e) False.
Guttation occurs due to excessive root pressure. It is maximum when root pressure is maximum which occurs in the early mornings or at night. This is because during these times, transpiration is very low and water absorption is very high.

(f) False.
Dry seeds when submerged in water swell up due to imbibitions. On contact with water dry seeds imbibe water and swell up.

Solution D.1.
Examples of turgor movements in plants:

  1. In Mimosa pudica, a sensitive plant, the stimulus of touch leads to loss of turgor at the base of the leaflets and at the base of the petioles called pulvinus. This causes the folding and drooping of leaves of the plant.
  2. The leaves of insectivorous plants close up to entrap a living prey. When the insect come in contact with the leaf, it loses it turgor hence closing the leaves of the plant.
  3. The bending movements of certain flowers towards the sun.
    (Any two)

Solution D.2.
The closing and opening of the stomata depends on the turgidity of the guard cells. Each guard cell has a thicker wall on the side facing the stoma and a thin wall on the opposite side. Guard cells contain chloroplasts. As a result of the synthesis of glucose during photosynthesis and some other chemical changes, the osmotic pressure of the contents of the guard cells increases and they absorb more water from the neighbouring cells, thus becoming turgid. Due to turgor, the guard cells become more arched outwards and the aperture between them widens, thereby opening the stoma.
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At night or when there is shortage of water in the leaf, the guard cells turn flaccid and their inner rigid walls become straight, thus closing the stomatal aperture.

Solution D.3.
If the concentration of mineral nutrient elements is higher inside the root-hairs than in the surrounding soil, then roots take them in from the soil by ‘active transport’. In active transport, the mineral ions are forcibly carried from the surrounding soil i.e. the region of their lower concentration into the roots i.e. the region of their higher concentration through the cell membrane by expenditure of energy. This energy is supplied by the cell in the form of ATP.

Solution D.4.
When soaked in water, the seeds swell up due to imbibition and endosmosis. During these two processes water enters the cell. Due to endosmosis, at some point, the seed coat is unable to bear the turgor pressure and hence, the seed coat bursts.

Solution D.5.
Leaves of the sensitive plant wilt and droop down on a slight touch due to turgor movement. Petiole of sensitive plant is held up by turgid pulvinus tissue. The stimulus of touch leads to loss of turgor at the base of the leaflets and at the base of the petioles i.e. pulvinus. The cells of the lower side of pulvinus lose water and the petiole collapses. This causes the wilting and drooping of the leaves.
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Solution D.6.
As water is lost from the leaf surface by transpiration, more water molecules are pulled up due to the tendency of water molecules to remain joined i.e. cohesion. This produces a continuous column of water throughout the stem which is known as ‘transpiration pull’. A negative pressure or tension is produced in the xylem that pulls the water from the roots and soil. Transpirational pull is an important force which causes the ascent of sap.
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Solution E.1.
(a) The cell is flaccid i.e. it is plasmolysed.
(b) Plasma Membrane
(c) Plasmolysis would not occur and flaccidity would not be seen i.e. the protoplasm would not have shrunken away from the cell wall.
(d) Cell Wall is absent in animal cell.

Solution E.2.
(a) Flaccid Cell
(b) The liquid is hypertonic solution. It has higher solute concentration outside the cell than the fluids inside the cell.
(c)
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Solution E.3.

(a) Osmosis

(b) Osmosis is the diffusion of water molecules across a semi-permeable membrane from a more dilute solution (with a lower solute concentration) to a less dilute solution (with a higher solute concentration).

(c) After an hour or so, the level of sugar solution in the thistle funnel will rise and the level of water in the beaker will drop slightly.

(d) For control experiment, the beaker will contain the water. At the same time, instead of the sugar solution; the thistle funnel with the cellophane paper tied on its mouth and inverted in the beaker will also contain water.

(e)

  1. concentrated sugar solution – Cell sap (of higher concentration than that of the surrounding water) within the root hair.
  2. parchment paper – cell membrane of root hair.
  3. water in the beaker – water in soil.

(f) cellophane paper, egg membrane, animal bladder (any one)

(g)

  1. The roots of plants absorb water and minerals from surrounding soil due to osmosis.
  2. Osmosis allows plants to absorb water from the soil which helps plants to keep cells alive in roots, stems and leaves.
  3. Osmosis is also important in the opening and closing of stomata which is an important feature for the processes like transpiration and photosynthesis. (Any two)

Solution E.4.

a.
A – Cell wall
B – Cell membrane
C – Cytoplasm
D – Nucleus

b. A root hair gets turgid because of the absorption of water from the surrounding. Absorption of water by root hair is achieved by the process of osmosis. The concentration of water in the surrounding is more than that of the interior of the cell; this causes the water from the surrounding to move in because of endosmosis.

c.

Cell wall Cell membrane
The cell wall of a root hair is freely permeable and allows both salt and water to pass through. The cell membrane of a root hair is semi-permeable and does not allow large dissolved salt molecules to pass through.

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Solution E.5.

(a) Water is hypotonic to the potato cells, due to which endosmosis occurs and water enters the potato cells. The protoplasm swells up pressing tight against the cell wall. The cells are fully distended i.e. turgid. This causes the firmness and increase in the size of the potato cubes when placed in water.

(b) Sugar solution is hypertonic to the potato cells, due to which exosmosis occurs and water flows out of the potato cells. The potato cell loses its distended appearance, the cytoplasm shrinks and the plasma membrane withdraws from the cell wall. The cells become limp or flaccid. This causes the softness and decrease in size of the potato cubes when placed in sugar solution.

(c) The process being investigated is osmosis. Osmosis is the diffusion of water molecules across a semi-permeable membrane from a more dilute solution (with a lower solute concentration) to a less dilute solution (with a higher solute concentration).

Solution E.6.
(a) It is the diagrammatic cross-section of a part of a root.

(b)

  1. Root hair
  2. Epidermis
  3. Cortex
  4. Endodermis
  5. Phloem
  6. Xylem

(c) Cortex (label 3) is the ground tissue and is active in the uptake of water and minerals. It also helps in storage of photosynthetic products.
Phloem (label 5) helps in transporting the prepared food from leaves to different parts of the plant.

Solution E.7.
(a) The process of water absorption by plant roots through osmosis is being studied here.

(b) A root-hair contains cell sap which contains higher concentration of salts as compared to outside soil water. This difference sets off osmosis and outside water diffuses into the root-hair. From the cell bearing root-hair, water passes into adjoining cells one after another to finally the xylem vessels.

(c) The surface of water was covered with oil to prevent any loss of water by evaporation.

Solution E.8.
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