Bhagya

MP Board Class 11th Maths Solutions Chapter 11 शंकु परिच्छेद Ex 11.3

निम्नलिखित प्रश्नों 1 से 9 तक प्रत्येक दीर्घवृत्त में नाभियों और शीर्षों के निर्देशांक, दीर्घ और लघु अक्ष की लंबाइयाँ, उत्केंद्रता तथा नाभिलंब जीवा की लम्बाई ज्ञात कीजिए।
प्रश्न 1.
\(\frac{x^{2}}{36}+\frac{y^{2}}{16}\) = 1
हल:

प्रश्न 2.
\(\frac{x^{2}}{4}+\frac{y^{2}}{25}\) = 1
हल:

दीर्घ अक्ष की लंबाई = 2a = 2 × 5 = 10
लघु अक्ष की लंबाई = 2b = 2 × 2 = 4
उत्केंद्रता = e = \(\frac{c}{a}=\frac{\sqrt{21}}{5}\)
नाभिलंब जीवा की लंबाई = \(\frac{2 b^{2}}{a}=\frac{2 \times 4}{5}=\frac{8}{5}\).

प्रश्न 3.
\(\frac{x^{2}}{16}+\frac{y^{2}}{9}\) = 1
हल:

प्रश्न 4.
\(\frac{x^{2}}{25}+\frac{y^{2}}{100}\) = 1
हल:
दीर्घवृत्त का समीकरण \(\frac{x^{2}}{25}+\frac{y^{2}}{100}\) = 1
∴ a² = 100, b² = 25
∴ a = 10, b = 5
∴ c² = a² – b² = 100 – 25 = 75
∴ c = 5\(\sqrt{3}\)
नाभि के निर्देशांक (0, ± c) या (0, ± 5\(\sqrt{3}\))
शीर्षों के निर्देशांक (0, ± a) या (0, ± 10)
दीर्घ अक्ष की लंबाई = 2a = 2 × 10 = 20
लघु अक्ष की लंबाई = 2b = 2 × 5 = 10

प्रश्न 5.
\(\frac{x^{2}}{49}+\frac{y^{2}}{36}\) = 1
हल:
दीर्घवृत्त का समीकरण \(\frac{x^{2}}{49}+\frac{y^{2}}{36}\) = 1
∴ a² = 49, b² = 36
∴ a = 7, b = 6
दीर्घ अक्ष, x-अक्ष के अनुदिश है
c² = a² – b² = 49 – 36 = 13
c = \(\sqrt{13}\)
नाभियों के निर्देशांक (± c, 0) या (± \(\sqrt{13}\), 0)
शीर्षों के निर्देशांक (± a, 0) या (± 7, 0)
दीर्घ अक्ष की लंबाई = 2a = 2 × 7 = 14
लघु अक्ष की लंबाई = 2b = 2 × 6 = 12

प्रश्न 6.
\(\frac{x^{2}}{100}+\frac{y^{2}}{400}\) = 1
हल:
दीर्घवृत्त का समीकरण \(\frac{x^{2}}{100}+\frac{y^{2}}{400}\) = 1
∴ a² = 400, b² = 100
∴ a = 20, b = 10
c² = a² – b² = 400 – 100 = 300
∴ c = 10\(\sqrt{3}\)
दीर्घ अक्ष, y- अक्ष के अनुदिश है
नाभियों के निर्देशांक (0, ± c) या (0, ± 10\(\sqrt{3}\))
शीर्षों के निर्देशांक (0, ± a) या (0, ± 20)
दीर्घ अक्ष की लंबाई = 2a = 2 × 20 = 40
लघु अक्ष की लंबाई = 2b = 2 × 10 = 20

प्रश्न 7.
36x² + 4y² = 144.
हल:
दीर्घवृत्त का समीकरण 36x² + 4y² = 144
या \(\frac{x^{2}}{4}+\frac{y^{2}}{36}\) = 1
∴ a² = 36, b² = 4
∴ a = 6, b = 2
∴ c² = a² – b² = 36 – 4 = 32
∴ c = 4\(\sqrt{2}\)
दीर्घवृत्त का अक्ष, y-अक्ष के अनुदिश है
नाभियों के निर्देशांक (0, ± c) या (0, ± 4\(\sqrt{2}\))
शीर्षों के निर्देशांक (0, ± a) या (0, ± 6)
दीर्घ अक्ष की लंबाई = 2a = 2 × 6 = 12
लघु अक्ष की लंबाई = 2b = 2 × 2 = 4

प्रश्न 8.
16x² + y² = 16.
हल:
दीर्घवृत्त का समीकरण 16x² + y² = 16
या \(\frac{x^{2}}{1}+\frac{y^{2}}{16}\) = 1
∴ दीर्घवृत्त का अक्ष, y-अक्ष के अनुदिश है।
a² = 16, b² = 1
∴ a = 4, b = 1
c² = a – b² = 16 – 1 = 15
∴ c = \(\sqrt{15}\)
नाभियों के निर्देशांक (0, ± c) या (0, ± \(\sqrt{15}\))
शीर्षों के निर्देशांक (0, ± a) या (0, ± 4)
दीर्घ अक्ष की लंबाई = 2a = 2 × 4 = 8
लघु अक्ष की लंबाई = 2b = 2 × 1 = 2

प्रश्न 9.
4x² + 9y² = 36.
हल:
दीर्घवृत्त का समीकरण 4x² + 9y² = 36
या \(\frac{x^{2}}{9}+\frac{y^{2}}{4}\) = 1
दीर्घ अक्ष, x-अक्ष के अनुदिश है।
∴ a² = 9, b² = 4
∴ a= 3, b = 2
c² = a² – b² = 9 – 4 = 5
∴ c = \(\sqrt{5}\)
नाभियों के निर्देशांक (± c, 0) या (± \(\sqrt{5}\), 0)
शीर्षों के निर्देशांक (± a, 0) या (± 3,0)
दीर्घ अक्ष की लंबाई = 2a = 2 × 3 = 6
लघु अक्ष की लंबाई = 2b = 2 × 2 = 4

निम्नलिखित प्रश्नों 10 से 20 तक प्रत्येक में, दिए प्रतिबंधों को संतुष्ट करते हुए दीर्घवृत्त का समीकरण ज्ञात कीजिए।
प्रश्न 10.
शीर्षों (± 5, 0), नाभियाँ (± 4, 0).
हल:
a = 5, c = 4, c² = a² – b²
या 16 = 25 – b²
∴ b² = 25 – 16 = 9
और a² = 25
दीर्घवृत्त का समीकरण,
\(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1

प्रश्न 11.
शीर्षों (0, ± 13), नाभियाँ (0, ± 5).
हल:
दीर्घ अक्ष, y-अक्ष के अनुदिश है।
∴ c = 5, a = 13, c² = a² – b²
∴ 25 = 169 – b²
∴ b² = 169 – 25 = 144,
और a² = 13² = 169
∴ दीर्घवृत्त का समीकरण
\(\frac{x^{2}}{144}+\frac{y^{2}}{169}\) = 1

प्रश्न 12.
शीर्ष (± 6, 0), नाभियाँ (± 4, 0).
हल:
दीर्घ अक्ष x-अक्ष के अनुदिश है।
a= 6, ∴ a² = 36, c = 4
c² = a² – b² या 16 = 36 – b²
∴ b² = 36 – 16 = 20
∴ दीर्घवृत्त का समीकरण
\(\frac{x^{2}}{36}+\frac{y^{2}}{20}\) = 1.

प्रश्न 13.
दीर्घ अक्ष के अंत्य बिन्दु (± 3, 0), लघु अक्ष के अंत्य बिन्दु (0, ± 2).
हल:
दीर्घ अक्ष x-अक्ष के अनुदिश है।
a = 3, b = 2, ∴ a² = 9, b² = 4
दीर्घवृत्त का समीकरण,
\(\frac{x^{2}}{9}+\frac{y^{2}}{4}\) = 1

प्रश्न 14.
दीर्घ अक्ष के अंत्य बिन्दु (0, ± \(\sqrt{5}\)), लघु अक्ष के अंत्य बिन्दु (± 1, 0).
हल:
दीर्घ अक्ष, y-अक्ष के अनुदिश है।
a = \(\sqrt{5}\), b = 1, ∴ a² = 5, b² = 1
दीर्घवृत्त का समीकरण,
\(\frac{x^{2}}{1}+\frac{y^{2}}{5}\) = 1

प्रश्न 15.
दीर्घ अक्ष की लंबाई = 26, नाभियाँ (45, 0).
हल:
दीर्घ अक्ष, x-अक्ष के अनुदिश है।
और 2b = 26, ∴ b = 13 या a² = 169,
c = 5, c² = 25 = a² – b² = 169 – b²
∴ b² = 169 – 25 = 144
अतः a² = 169, b² = 144
∴ दीर्घवृत्त का समीकरण,
\(\frac{x^{2}}{169}+\frac{y^{2}}{144}\) = 1.

प्रश्न 16.
दीर्घ अक्ष की लंबाई = 16, नाभियाँ (0, ± 6).
हल:
दीर्घ अक्ष, y-अक्ष के अनुदिश है।
2b = 16, ∴ b = 8 या b² = 64,
c = 6, c² = 36,
c² = a² – b²
या 36 = a² – 64
∴ a² = 64 + 36 = 100
∴ दीर्घवृत्त का समीकरण
\(\frac{x^{2}}{64}+\frac{y^{2}}{100}\) = 1

प्रश्न 17.
नाभियाँ (± 3, 0), a = 4.
हल:
दीर्घ अक्ष, x-अक्ष के अनुदिश है।
∴ c = 3, a = 4
अब c² = a² – b²
या 9 = 16 – b²
∴ b² = 16 – 9 = 7
∴ दीर्घवृत्त का समीकरण,
\(\frac{x^{2}}{16}+\frac{y^{2}}{7}\) = 1

प्रश्न 18.
b = 3, c = 4, केन्द्र मूल बिन्दु पर, नाभियाँ x-अक्ष पर है।
हल:
दीर्घ अक्ष, x-अक्ष के अनुदिश है
c² = a² – b²
16 = a² – 9
a² = 16 +9 = 25
∴ दीर्घवृत्त का समीकरण,
\(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1

प्रश्न 19.
केंद्र (0, 0) पर, दीर्घ अक्ष y-अक्ष पर और बिन्दुओं (3, 2) और (1, 6) से जाता है।
हल:

प्रश्न 20.
दीर्घ अक्ष,x-अक्ष पर और बिन्दुओं (4, 3), (6, 2) से जाता है।
हल:
मान लीजिए दीर्घवृत्त का समीकरण \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1

∴ दीर्घवृत्त का समीकरण,
\(\frac{x^{2}}{52}+\frac{y^{2}}{13}\) = 1

MP Board Class 11th Maths Solutions

MP Board Class 12th Biology Solutions Chapter 12 Biotechnology And its Application

Biotechnology And its Application NCERT Text Book Questions and Answers

Question 1.
Crystals of Bt toxin produced by some bacteria do not kill the bacteria themselves because:
(a) Bacteria are resistant to the toxin,
(b) Toxin is immature
(c) Toxin is inactive
(d) Bacteria enclose toxin in a special sac
Answer:
(c) Toxin is inactive: In bacteria, the toxin is present in an inactive form called prototoxin. This gets converted into the active form when it enters the salivary gland of insects having alkaline medium.

Question 2.
What are transgenic bacteria? Illustrate using any one example.
Answer:
The becteria whose DNA is manipulated to carry and express a foreign DNA is called transgenic bacteria. These microbes are used for producing important bio-chemicals. They have been synthesizing alcohol, enzymes, steroids and antibiotics. Example: Bacillus thuringiensis for Bt cotton, hirudin from transgenic Brassica napus seed. Hirudin is a protein which prevents blood clotting. Its gene was chemically synthesized and introduced in Brassica napus, in which hirudin accumulates in the seed from where it is extracted, purified and used as a medicine.

Question 3.
Compare and contrast the advantages and disadvantages of production of genetically modified crops.
Answer:
Advantages of GM crops :

• Genetic modification has made crops more tolerant to abiotic stresses (cold, drought, heat, salt.)
• Viral resistance can be introduced.
• Over ripening losses can be reduced. Example: Flavr Savr Tomato.
• Enhanced nutritional value of food. Example: Golden Rice.
• Reduced reliance on chemical pesticides.

Disadvantages of GM crops :

• Transgenes in crop plants can endanger native species. Example: The gene for Bt toxin expressed in pollen may end natural pollinators such as honeybees.
• Weeds also become resistant.
• Products of transgenes may be allergic or toxic.
• They cause damage to the natural environment.

Question 4.
What are Cry proteins? Name an organism which produces it How has man exploited this protein to his benefit?
Answer:
Cry proteins are toxic proteins (insecticidal proteins) secreted by Bacillus thuringiensis in crystal form during a particular phase of their growth. The toxin is coded by a gene called cry.

The genes encoding cry proteins called Bt toxin genes were isolated from B. thuringiensis and incorporated into several crop plants such as Bt cotton, Bt com etc. to provide resistance against insect pests.

Question 5.
What is gene therapy? Illustrate using the example of adenosine deaminase (ADA) deficiency.
Answer:
It is a collection of methods which allows correction of a gene defect that has been diagnosed in a child or embryo. In gene therapy, normal genes are inserted into a person’s cells or tissues to treat a hereditary defect. Gene therapy is being tried for sickle cell anaemia and Severe Combined Immuno Deficiency (SCID).

In some children, ADA deficiency can be cured by bone marrow transplantation. In others, it can be treated by enzyme replacement therapy, in which functional ADA is given to the patient by injection. However, both of these approaches are not completely curative.

In gene therapy, lymphocytes from the blood of the patient are grown in culture outside the body. A functional ADA cDNA (using a retroviral vector) is then introduced into these lymphocytes, which are subsequently returned to the patient. Because these cells are not immortal, the patient requires periodic infusion of such genetically engineered lymphocytes. However, if the gene isolated from marrow cells producing ADA is introduced into cells at early embryonic stages, the disease could be cured permanently.

Question 6.
Diagrammatically represent the experimental steps in cloning and expressing a human gene (say the gene for growth hormone) into a bacterium like E. coli.
Answer:

Question 7.
Can you suggest a method to remove oil (hydrocarbon) from seeds based on your understanding of rDNA technology and chemistry of oil?
Answer:
To remove oil from seeds using recombinat DNA technology would involve:

• Identifying the genes that code for oil production.
• Deleting these genes from the seed genome.
• Splicing back together the remaining DNA.
• Putting it back into the cell.

It will be not be very easy because the oils are made up of fatty acids and glycerol. Since, fatty acids are important components of cell membrane system, deleting or switching off of its genes might affect the cell structure itself.

Question 8.
Find out from internet what is golden rice.
Answer:
Golden rice is a variety of rice produced through genetic engineering to biosynthesize β -carotene, a precursor of vitamin ‘A’ in the edible parts of rice. It is intended to produce a fortified food to be grown and consumed in areas with a shortage of dietary vitamin ‘A’.

Question 9.
Does our blood have proteases and nucleases?
Answer:
No.

Question 10.
Consult internet and find out how to make orally active protein pharmaceutical. What is the major problem to be encountered?
Answer:
For making any oral drug or nutritional supplement, the action of digestive enzymes has to be taken into account. Most of the antibiotics and vitamin supplements are made in capsule form to prevent the action of HC1 in the stomach. For protein preparation, the major source is groundnut shells. The protein extracted from the source is predigested so, as to make it absorbable by the digestive system.

Biotechnology And its Application Other Important Questions and Answers

Biotechnology And its Application Objective Type Questions

1. Choose the Correct Answer:

Question 1.
Triticum aestivum wheat is:
(a) Haploid
(b) Diploid
(c) Tetraploid
(d) Hexaploid.
Answer:
(c) Tetraploid

Question 2.
Man-made cereal is:
(a) Potato
(b) Triticale
(c) Triticum
(d) Sugarcane.
Answer:
(b) Triticale

Question 3.
Wheat grain is a:
(a) Fruit
(b) Seed
(c) Embryo
(d) Glume.
Answer:
(b) Seed

Question 4.
Removal of stamens from the flower during hybridization is called:
(a) Cutting
(b) Self-fertilization
(c) Emusculation
(d) Topnin.
Answer:
(c) Emusculation

Question 5.
New crop is:
(a) Triticale
(b) Rye
(c) Winged bean
(d) Wheat.
Answer:
(a) Triticale

Question 6.
Wheat used in bread is:
(a) Triticum aestivum
(b) Triticale
(c) All species of triticum
(d) Secale.
Answer:
(a) Triticum aestivum

Question 7.
Sonera-64 and Lerma roja-64 A are the varieties of:
(a) Wheat
(b) Rice
(c) Pea
(d) Maize.
Answer:
(a) Wheat

Question 8.
Haploid male plants can be produced by the culturing of:
(a) Filament
(b) Pollen grains
(c) Stamens
(d) Androecium.
Answer:
(b) Pollen grains

Question 9.
Ti plasmid which is used in genetic engineering, is found in:
(a) Escherichia coli
(b) Bacillus thuringiensis
(c) Agrobacterium rhizogenes
(d) Agrobacterium tumefaciens.
Answer:
(d) Agrobacterium tumefaciens.

Question10.
Bt-Toxin is:
(a) Intercellutar lipid
(b) Intercellular crystal protein
(c) Extracellular crystal protein
(d) Lipid.
Answer:
(c) Extracellular crystal protein

Question 11.
The function of Bacillus thuringiensis is:
(a) Bio-metallurgy technique
(b) Bio-insecticides plant
(c) Bio-fertilizers
(d) Bio-mineralization process.
Answer:
(b) Bio-insecticides plant

Question 12.
Non-toxic crystal of Bt are made by bacteria but bacterias are not killed by own because:
(a) Non-toxic is immature
(b) Bacteria is immune resistant
(c) Non-toxic is inactive
(d) Bacteria has non-toxic sac.
Answer:
(c) Non-toxic is inactive

Question 13.
Genetic transfer through viruses is called:
(a) Sexduction
(b) Transduction
(c) Conjugation
(d) Transformation.
Answer:
(b) Transduction

Question 14.
Biopiracy is realated with:
(a) Discovery of biomolecules and genes
(b) Cultural knowledge
(c) Bio-research
(d) All of these.
Answer:
(d) All of these.

Question 15.
Golden rice is rich in which vitamin:
(a) Vitamin A
(b) Vitamin B
(c) Vitamin C
(d) Vitamin D.
Answer:
(a) Vitamin A

2. Fill in the Blanks:

1. …………………………. is the appropriation of another’s knowledge of use of biological resources.
2. A ……………………………. is a patent granted by the government to the inventor for biological entities.
3. ………………………………….. is a loosely used term for molecules that are present in organisms.
4. …………………………………… is a group of standards which is used in control of relations between our work and bio-diversity.
5. The production of product, its extraction and process is called ……………………………….

Answer:

1. Biopiracy
2. Biopatent
3. Bio-molecules
4. Bio-code of conduct
5. Downstream.

3. Match the Following:

Answer:

1. (b)
2. (d)
3. (a)
4. (c).

4. Answer in One Word/Sentence:

1. Give the name of first transgenic crop.
2. What is the name of insect resistant protein which is transferred in Bt Cotton?
3. What is the name of first man-made insulin?
4. In which organism nif-genes are found?
5. Give the name of an antiviral protein.

Answer:

1. Tobacco
2. Cry protein
3. Humulin
4. Rhizobium
5. Interferon.

Biotechnology And its Application Very Short Answer Type Questions

Question 1.
Name the organism whose genetic material has been altered using genetic engineering techniques.
Answer:
Genetically Modified Organisms (GMO).

Question 2.
Name the crops which are prepaired with the help of biotechnology.
Answer:
Bt cotton, Bt maize, paddy, tomato, potato and soya been.

Question 3.
Through whom Bt toxin protein originates?
Answer:
By Bacillus thuringiensis.

Question 4.
By which Bt toxin is coded?
Answer:
By cry genes Bt toxin is coded.

Question 5.
Full form of RNAi.
Answer:
RNA interference (RNAi).

Question 6.
Name the therapy which is help of missing of defective ones in order to correct genetic disorders.
Answer:
Gene therapy.

Question 7.
Name the scientific name of bacteria in which be form organism toxin.
Answer:
Bacillus thuringiensis.

Biotechnology And its Application Short Answer Type Questions

Question 1.
What is genetically modified (GM) food ? Give two examples.
Answer:
Genetically modified food (GM food): The food substances produced from ge-netically modified crops or transgenic crops is called GM food. This food differ from conventionally developed varieties in the following aspects :

• GM food contains antibiotic resistance gene itself.
• It contains protein produced by transgene, e.g. Cry protein in insect resistance varieties.
• These GM foods contain enzyme produced by the antibiotic resistance gene that was used during gene transfer by recombinant DNA technology.

Examples of GM Crops, Food and Fruits:

1. Flavr Savr Tomato : It is the first food containing genetically engineered DNA. . These tomatoes contain genes for antibiotic resistance for kanamycin.

2. Maize : GM maize has a bacterial gene which increases its resistance to pests and
diseases. It also has a gene for ampicillin resistance which is harmful for us, therefore introduction of GM maize is opposed by many European countries. ,

3. Rape oil seed : It is a new type of plant that contain genes for resistance to the herbicide Basta. It has for more potential, dangers and can become a weed and would be impossible to control with Basta. It could cross fertilize with relatives such as wild mustard, thus, spreading the resistance to wild plants. Such type of environmental risks could occur with genetically modified rapeseed crop. They might also effect food chains in unpredictable ways.

Question 2.
Write down the advantage of GM Crops.
Answer:

Advantages of GM crops :

• Genetic modification has made crops more tolerant to abiotic stresses (cold, drought, heat, salt.)
• Viral resistance can be introduced.
• Over ripening losses can be reduced. Example: Flavr Savr Tomato.
• Enhanced nutritional value of food. Example: Golden Rice.
• Reduced reliance on chemical pesticides.

Disadvantages of GM crops :

• Transgenes in crop plants can endanger native species. Example: The gene for Bt toxin expressed in pollen may end natural pollinators such as honeybees.
• Weeds also become resistant.
• Products of transgenes may be allergic or toxic.
• They cause damage to the natural environment.

Question 3.
What is perfect agriculture? How is this method better than traditional method? Explain.
Answer:
Perfect agriculture is a method of agriculture which is sustainable, perfect and harmless. Green revolution and there after the production of agricultural crops has definitely increased due to use of new and high yielding varieties, development of irrigation facilities, increased irrigated area, use of fertilizers etc. but it results many problems such as loss of soil fertility, pollution of food and water and diseases. The resistance power of plants and human heings falls slowly. Food and water borne diseases affecting the health of human beings and animals.

All of these conditions and events taking place due to the modem commercial agriculture. Therefore, it would become necessary to develop a method of agriculture which would be free from above mentioned demerits. This kind of agriculture is called to be as perfect or sustainable agriculture. Organic agriculture is the best example of perfect agriculture.

Question 4.
What is organic cropping? What are its basis?
Answer:
Organic agriculture: Organic agriculture is a method of agriculture which does not allow the use of synthetic fertilizers, pesticides, insecticides, weedicides, plant growth regulators, substances of animal origin and genetically modified bacteria. In this method biofertilizers, crop rotation methods are used to increase crop production and biopesticides are used to control insects and weeds.

Thus, organic farming is a holistic way of agriculture which tries to bridge the widening gap between man and nature. If has the commitment of meeting production needs on one hand and sustaining resources tand ecosystem function on the other hand. Thus, organic farming is an alternative agriculture production system which avoid or largely excludes the use of synthetic chemicals, fertilizers, pesticides and growth regulating hormones and live stock additives.

Basics of Organic agriculture :

• Organic agriculture is based on improvement of soil, plants, animals, man and global scinery and make it sustainable.
• Organic agriculture is based on those ecosystems and biocycles which utilize that organisms which would be promoted.
• It is based on the principle which are related with making pollution free environment and possibilities of life.
• It is also based on saving environment and health of present and future generations.

Question 5.
What is gene library?
Answer:
Gene library: Several clones of cells, each clone containing one or a few foreign genes representing almost all the genes of an organism is referred to as genes library. From this gene library it is possible to identify a clone containing gene of interest. In order to obtain gene library of an organism, its genome is first cut into smaller DNA fragments containing one or a few genes such as fragments can be cloned into a cell which may brfthat of bacteria, yeast, insects, plant or animal cell.

When such a cell multiplies to form a group of cells, all cells will contain the same foreign DNA fragment which was introduced initially. These cells which have similar foreign DNA fragment are referred to as a clone of cells. Several clones of cells each clone containing one or a few foreign genes are finally obtained and is called gene library.

Biotechnology And its Application Long Answer Type Questions

Question 1.
What is eugenics? Write importance of eugenics.
Answer:
Eugenics: The branch of biology which deals with the study of improvements of human race is called eugenics.
Importance:

• Development of selective reproduction in similar species.
• Transfer of genetic materials in various organisms.
• Development of GM food and GM crops.
• Gene cloning
• Gene therapy, etc.

Question 2.
Explain the following in brief:

1. Biopiracy,
2. Biopatent.

Answer:
1. Biopiracy : Some organisations and multinational companies exploid biological resources and genetical resources indegenous to a country without proper authorisation. This is called biopiracy. In fact it is illegal removal of biological material. The process of biopyracy involves collection of samples of biological sources, which can be done unnoticed. This biological material is then subjected to product development for use on a commercial scale.

Today a range of biological resources are facing biopiracy. It includes plants and animals, micro-organisms genetic materials etc. Western companies are getting great benefits from using the knowledge and biological resources of the third world communities.While the companies stand to make huge revenue from this process, the local communities are unrewarded and infact, may have to buy the products of these companies at high prices.
To check illegal exploitation of biological resources Government of India has signed the General Agreement on Tariffs and Trade (GATT), which opens country natural resources for foreign exploitation.

2. Biopatent : The protection given by government to an inventor of biological material to secure him for a specific time the exclusive right of manufacturing, exploiting, using and selling of an invention is called biopotent.

Today manufacturing companies are being granted patents for products and technologies that make of biological resources, such as plants and animals, genetic materials which was identified developed and used by farmers and indegenous peoples.

There is growing worldwide opposition to the granting of patents on biological materials such as genes, plants, animals and human. Farmers and indegenous peoples are outaged that’ plants that they developed are being ‘hijacked’ by companies. Groups are diverse as religious leaders, parliamentarians and environment NGOs are intensifying campaign against corporate patenting of living things.

Question 3.
Describe the application of genetic engineering in the field of Agriculture and Medicine.
Answer:
(A) Application of Genetic engineering or Biotechnology in Agriculture : Genetic engineering is found to be very beneficial in agriculture. Its important use in agriculture are:
1. Increase in photosynthetic efficiency: An increase in photosynthetic efficiency of crop plants can be achieved by introducing suitable Carbon dioxide Fixation Gene (cfx) from any plant into the crop plants.

2. Transfer of nitrogen fixing ability: Number of symbiotic and non-symbiotic micro-organism have capacity of fixing atmospheric nitrogen. Nitrogen fixers are found to possess nitrogen fixing gene (nif genes) which are located on chromosomes or plasmids. Introduction of nif gene in crop plants results in ability in crop plants to fix atmospheric nitrogen and reduction in the use of chemical nitrogen fertilizers.

3. Disease resistance in crop plants : Plant breeders at present are developing high yield varieties by transferring gene for disease resistance through conventional breeding.

4. Plant tissue in crop improvement: Some of the areas of plant improvement where tissue culture has been applied with success are as follows :

• Rescuing hybrids through embryo culture.
• Multiplication of germplasm.
• Production of disease free plants.
• Production of haploid through another culture.
• Somaclonal variation.
• Somatic hybridization.
• Cryopreservation of germplasm.

5. VAM (Vesicular-Arbuscular Mycorrhiza) fungi with Rhizobium can boost the yields: Recently there has been a new dimension to this farm practice by the way of increasing Rhizobium inoculation effect by simultaneous inoculating seeds with VAM as well as Rhizobium culture.VAM are structural modification of hyphae helping in absorption and storage of phosphorus.

(B) Application of Genetic engineering in Medical field :
1. The hereditary diseases like colour-blindness, haemophilia which are caused by recessive genes and also many inborn metabolic disorders due to defective genes as alkaptonuria, phenylketonuria can be cured with the gene therapy.

2. Substances like vitamins, hormones, amino acids and antibodies can be synthesized in bacteria by introducing the genes which code these substances. In this way bacteria can be used as biofactories for the synthesis of these substances.

3. Production of insulin: Insulin is medicine used for the treatment of diabetes. Initially it is derived from animals (pig and cows) but today it is produced by gene splicing.

4. Hepatitis-B vaccine: Hepatitis-B is a viral disease of liver. Today this vaccine is prepared with the help of genetic engineering.

MP Board Class 12th Biology Solutions

MP Board Class 12th Biology Solutions Chapter 4 जनन स्वास्थ्य

जनन स्वास्थ्य NCERT प्रश्नोत्तर

प्रश्न 1.
समाज में जनन स्वास्थ्य के महत्व के बारे में अपने विचार प्रकट कीजिए।
उत्तर
जनन स्वास्थ्य का तात्पर्य जनन के सभी पहलुओं जैसे शारीरिक, भावनात्मक, व्यावहारिक तथा सामाजिक स्वास्थ्य से है। दुनिया में भारत पहला ऐसा देश है जिसने राष्ट्रीय स्तर पर जननात्मक स्वस्थ समाज को प्राप्त करने की कार्य योजनाएँ बनाई हैं। इन कार्यक्रमों को परिवार कल्याण के नाम से जाना जाता है। इनकी शुरूआत 1951 में (परिवार नियोजन के नाम से) हुई थी। जनन संबंधित और आवधिक क्षेत्रों को इसमें सम्मिलित करते हुए बहुत उन्नत व व्यापक कार्यक्रम फिलहाल ‘जनन एवं बाल स्वास्थ्य सेवा कार्यक्रम (आ.सी.एच.)’ के नाम से प्रसिद्ध है। इन कार्यक्रमों के अंतर्गत् जनन संबंधी विभिन्न पहलुओं के बारे में लोगों में जागरूकता पैदा करते हुए और जननात्मक रूप से संबंध समाज तैयार करने के लिए अनेक सुविधाएँ एवं प्रोत्साहन दिए जा रहे हैं।

प्रश्न 2.
जनन स्वास्थ्य के उन पहलुओं को समझाइए, जिस पर आज के परिदृश्य में विशेष ध्यान देने की जरूरत है
उत्तर
जनन स्वास्थ्य प्राप्ति के लिए विभिन्न कार्य-योजनाओं के सफलतापूर्वक क्रियान्वयन के लिए मजबूत संरचनात्मक सुविधाओं, व्यावसायिक विशेषज्ञता तथा भरपूर भौतिक सहारों की आवश्यकता होती है। लोगों को जनन संबंधी समस्याओं जैसे-सगर्भता (Pregnancy), प्रसव (Parturition), यौन संचारित रोगों, गर्भपात (Abortion), गर्भ निरोधकों, आर्तव चक्र संबंधी समस्याओं, बांझपन (Infertility) आदि के बारे में चिकित्सकीय सहायता देने के लिए बेहतर तकनीकों एवं नई कार्य योजनाओं को क्रियान्वित करने की भी आवश्यकता है ताकि लोगों की अधिक सुचारु रूप से देखभाल और सहायता की जा सके।

बढ़ती “मादा भ्रूण हत्या” की कानूनी रोक के लिए ऐम्नियोसेंटेसिस (Amniocentesis) जैसे लिंग परीक्षण पर वैधानिक प्रतिबंध तथा व्यापक बाल प्रतिरक्षीकरण (टीकाकरण) आदि कुछ महत्वपूर्ण कार्यक्रमों को शामिल किया गया है। जनन संबंधी विभिन्न अनुसंधानों को बढ़ावा देने के लिए हमारे देश की सरकारी एजेंसियाँ सतत् क्रियाशील हैं। लखनऊ स्थित केन्द्रीय औषध अनुसंधान संस्थान (Central Drug Research Institue CDRI) ने “सहेली” नामक गर्भनिरोधक गोली का निर्माण किया है। यौन संचारित रोगों की सही जाँच तथा देखभाल और लगभग सभी जनन स्वास्थ्य समस्याओं हेतु विकसित चिकित्सा सुविधाओं के होने से बेहतर समाज एवं जनन स्वास्थ्य के संकेत प्राप्त हो रहे हैं।

प्रश्न 3.
क्या विद्यालयों में यौन शिक्षा आवश्यक है ? यदि हाँ तो क्यों ?
उत्तर
हाँ, विद्यालयों में यौन शिक्षा आवश्यक है, ताकि छात्र/छात्राओं को यौन संबंधी विभिन्न पहलुओं के बारे में फैली हुई भ्रान्तियों एवं यौन संबंधी गलत धारणाओं से छुटकारा मिल सके। बच्चों को जनन अंगों, किशोरावस्था एवं उससे संबंधित परिवर्तनों, सुरक्षित और स्वच्छ यौन क्रियाओं, यौन संचारित रोगों एवं एड्स की जानकारी देना विशेष रूप से किशोर आयु वर्ग में जनन संबंधी स्वस्थ जीवन बिताने में सहायक होती है।

प्रश्न 4.
क्या आप मानते हैं कि पिछले 50 वर्षों के दौरान हमारे देश के जनन स्वास्थ्य में सुधार हुआ है? यदि हाँ तो इस प्रकार के सुधार वाले कुछ क्षेत्रों का वर्णन कीजिए।
उत्तर
जी हाँ, पिछले 50 वर्षों के दौरान हमारे देश के जनन स्वास्थ्य क्षेत्र में सुधार हुआ है। निम्न क्षेत्रों में हुए सुधार को आसानी से देखा जा सकता है

1. यौन संबंधी मामलों के बारे में बेहतर जागरुकता आयी है।
2. चिकित्सकीय देख-रेख में होने वाले प्रसवों की संख्या में वृद्धि हुई है।
3. प्रसवोत्तर देखभाल बेहतर हुई है।
4. मातृ मृत्यु-दर (MMR) में गिरावट आई है।
5. शिशु मृत्यु-दर (IMR) में गिरावट आई है।
6. यौन संचारित रोगों की सही जाँच-पड़ताल तथा देखभाल व उपचार बेहतर हुआ है।
7. जनसंख्या वृद्धि दर पर रोक लगी है।
   कुल मिलाकर सभी यौन समस्याओं हेतु बढ़ी हुई चिकित्सा सुविधाओं का होना समाज के बेहतर जनन स्वास्थ्य की ओर संकेत देता है।

प्रश्न 5.
जनसंख्या विस्फोट के कौन-से कारण हैं ?
उत्तर
शीघ्र एवं अनियमित जनसंख्या वृद्धि को जनसंख्या विस्फोट कहा जाता है।इसके प्रमुख कारण निम्नलिखित हैं

1. रोगों व महामारियों पर नियंत्रण
2. कृषि का विकास
3. संचार व आवागमन के साधन
4. उच्च जन्मदर
5. निम्न मृत्युदर
6. मनुष्य का वर्ष भर मैथुनकाल होना।

प्रश्न 6.
क्या गर्भ-निरोधकों का उपयोग न्यायोचित है ? कारण बताइए।
उत्तर
भारत में जनसंख्या वृद्धि-दर अत्यधिक होने के कारण राष्ट्रीय संकट उत्पन्न हो गया है। अतः गर्भ निरोधकों का उपयोग न्यायोचित है। इसके उपयोग से परिवार को सीमित किया जा सकता है एवं उनकी सुविधाओं में वृद्धि की जा सकती है। इन गर्भ निरोधकों के उपयोग से यौन संचारित रोगों (STDs) से बचा जा सकता है। इसके साथ ही दो संतानों के बीच अन्तराल भी रखा जा सकता है।जनसंख्या वृद्धि को कम करके परिवार, समाज व देश की समृद्धि में सहयोग कर सकते हैं।

प्रश्न 7.
जनन ग्रंथि को हटाना, गर्भ निरोधकों का विकल्प नहीं माना जा सकता है, क्यों?
उत्तर
जनन ग्रंथियाँ सन्तान उत्पन्न करने वाले अंग हैं । वृषण में शुक्राणुओं तथा अण्डाशय में अण्डों का निर्माण होता है। गर्भ निरोधक के लिए स्वस्थ अंगों को शरीर से हटाना उचित नहीं है। इससे मानसिक, शारीरिक स्वास्थ्य संबंधी समस्याएँ उत्पन्न हो सकती हैं। जबकि गर्भ निरोधक विधियाँ सुलभ होती हैं एवं उनका कोई बुरा प्रभाव भी शरीर पर नहीं पड़ता है। इनका आसानी से प्रयोग भी किया जा सकता है। जरूरत पड़ने पर इसे हटाया भी जा सकता है। जनन ग्रंथि को हटाना एक जटिल प्रक्रिया है, कानून हमें इसकी आज्ञा भी नहीं देता है।

प्रश्न 8.
उल्बवेधन एक घातक लिंग निर्धारण (जाँच) प्रक्रिया है, जो हमारे देश में निषेधित है। क्या यह आवश्यक होना चाहिए ? टिप्पणी कीजिए।
उत्तर
बढ़ती मादा भ्रूण हत्या की कानूनी रोक के लिए उल्बवेधन (Amniocentesis) जाँच (भ्रूणीय लिंग निर्धारण), लिंग परीक्षण पर वैधानिक प्रतिबंध उचित है क्योंकि यह एक खतरनाक प्रवृत्ति है। कई बार ऐसा देखा जाता है कि यह पता चलने पर कि भ्रूण मादा (लड़की) है, चिकित्सीय सगर्भता समापन (Medical termination of pregnancy) कराया जाता है, सामान्य भाषा में इसे गर्भपात (Abortion) कहा जाता है। यह पूरी तरह गैर-कानूनी है। इस प्रकार की प्रवृत्ति से बचना चाहिए क्योंकि यह माता एवं बच्चा (भ्रूण) दोनों के लिए खतरनाक है । इससे समाज में पुरुष एवं महिलाओं की संख्या का अनुपात भी बिगड़ सकता है। जिससे वैवाहिक तथा स्वास्थ्य संबंधी समस्याएँ भी उत्पन्न हो सकती हैं।

प्रश्न 9.
बन्ध्य दम्पत्तियों को संतान पाने हेतु सहायता देने वाली कुछ विधियाँ बताइए।
उत्तर
यदि दंपत्ति बच्चा पैदा करने में अक्षम है तथा ऐसे दोष को ठीक करने का इलाज संभव न हो तो कुछ विशेष तकनीकों के द्वारा उन्हें बच्चा पैदा करने में सहायता की जा सकती है, ये तकनीकें सहायक जनन प्रौद्योगिकी (ART) कहलाती हैं।
1. पात्रे निषेचन (Invitro fertilization IVF)-
इस विधि में अण्डाणु का निषेचन शरीर से बाहर लगभग शरीर के भीतर जैसी स्थिति में कराया जाता है। इसके पश्चात् स्त्री में भ्रूण स्थानान्तरण (Embryo transfer, ET) कराया जाता है।

• युग्मनज डिंबवाहिनी स्थानान्तरण (Zygote intrafallopian transfer ZIFT)-इस प्रक्रिया में प्रारंभिक भ्रूण को 8 ब्लास्टोमियर तक की अवस्था में स्त्री की डिंबवाहिनी या फैलोपियन नलिका में अग्रिम परिवर्धन के लिए स्थानांतरित किया जाता है।
• आन्तर गर्भाशयी स्थानान्तरण (Intrauterine transfer IUT)-जब भ्रूण को 8 ब्लास्टोमियर से अधिक अवस्था पर परिवर्तन हेतु स्त्री के गर्भाशय में स्थानान्तरित किया जाता है तो इसे IUT कहते हैं । यदि किसी स्त्री में गर्भधारण नहीं हो पाता है तो उसकी सहायता के लिए पात्रे निषेचन अर्थात् अन्य स्त्री के भीतर ही निषेचन कराने के बाद भ्रूण को उस स्त्री में स्थानान्तरित किया जा सकता है।
• टेस्ट ट्यूब बेबी (Test tube baby)-इस विधि में दाता स्त्री के अण्डे को दाता पुरुष शुक्राणु से प्रयोगशाला में परखनली के भीतर (स्त्री शरीर के बाहर) निषेचन कराया जाता है तथा निश्चित समय तक अनुरूपी परिस्थितियों में परिवर्धन के बाद अग्रिम परिवर्धन हेतु स्त्री के गर्भाशय में स्थानान्तरित कर दिया जाता है। इस विधि द्वारा जन्मे बच्चे को “टेस्ट ट्यूब बेबी” कहते हैं।

2. युग्मक आन्तर फैलोपीयन स्थानान्तरण (Gamete intrafallopian transfer GIFT)-
जिन स्त्रियों में अण्डाणु उत्पन्न नहीं होता है किन्तु इनमें निषेचन और भ्रूण परिवर्धन के लिए उचित वातावरण होता है उनमें यह विधि अपनायी जाती है। इसमें एक दाता स्त्री से अण्डाणु लिया जाता है तथा उस स्त्री को फैलोपीयन नलिका में स्थानांतरित कर दिया जाता है।

3. आन्तर कोशिकीय शुक्राणु निक्षेपण (Intra-cytoplasmic sperm injection)-
प्रयोगशाला में भ्रूण बनाने के लिए इस प्रक्रिया में शुक्राणु को सीधे ही अण्डाणु में अंत:क्षेपित कर दिया जाता है।

4. कृत्रिम वीर्य सेचन (Artificial insemination AI)-
यदि पुरुष-स्त्री को वीर्य सेचित करने में अक्षम हो अथवा उसके वीर्य में शुक्राणुओं की संख्या कम हो तो ऐसी अवस्था में यह तकनीक अपनायी जाती है। इस विधि में पति अथवा एक स्वस्थ दाता का वीर्य कृत्रिम रूप से स्त्री की योनि में प्रविष्ट कर दिया जाता है। जब कृत्रिम विधि से वीर्य को गर्भाशय में प्रविष्ट किया जाता है तो इसे गर्भाशय वीर्य सेचन (Intra-uterine insemination IUI) कहते हैं।

5. परपोषी मातृत्व (Host mothering)-
इस विधि में भ्रूण को प्राकृतिक माता से निकालकर एक अन्य स्त्री धात्रेय माता (Foster mother) में रोपित कर दिया जाता है। धात्रेय माता में यह भ्रूण जन्म तक अथवा अस्थायी रूप से निश्चित समय तक वर्धित होता है। जिसके बाद इन्हें पुनः मूल माता में या अन्य किसी स्त्री में रोपित कर दिया जाता है। यह तकनीक ऐसी स्त्रियों के लिए लाभदायक है जिनमें भ्रूण बन तो जाता है किन्तु पूर्ण परिवर्धन के समय तक ये इसे रख नहीं पाती हैं।

प्रश्न 10.
किसी व्यक्ति को यौन संचारित रोगों की चपेट में आने से बचने के लिए कौन-से उपाय अपनाने चाहिए?
उत्तर
यौन संचारित रोगों से बचाव (Protection from Sexually Transmitted Diseases)यौन संचारित रोगों से बचाव हेतु निम्न बातों का ध्यान रखना चाहिए

1. किसी अनजान व्यक्ति या बहुत से व्यक्तियों के साथ यौन संबंध न रखें।
2. मैथुन के समय हमेशा कंडोम का इस्तेमाल करें।
3. संक्रमित व्यक्ति का रुधिर किसी अन्य व्यक्ति को न चढ़ाया जावे।
4. पहले से उपयोग किये गये इंजेक्शन व सुईयों का प्रयोग न करें।
5. समलैंगिकता से बचें।
6. यदि कोई आशंका है तो तुंरत ही प्रारंभिक जाँच के लिए किसी योग्य चिकित्सक से मिलें और रोग की पहचान होने पर पूरा इलाज करावें।

प्रश्न 11.
निम्न वाक्य सही है या गलत, व्याख्या सहित बताइए

(क) गर्भपात स्वतः भी हो सकता है। (सही/ गलत)
उत्तर
गलत, सामान्य परिस्थितियों में गर्भपात नहीं होता। किसी दुर्घटनावश या स्वैच्छिक रूप से गर्भसमापन (गर्भपात) होता है।

(ख) बंध्यता को जीवनक्षम संतति न पैदा कर पाने की अयोग्यता के रूप में परिभाषित किया गया है और यह सदैव स्त्री की असामान्यताओं/दोषों के कारण होती है।(सही/गलत)
उत्तर
गलत, बंध्यता हमेशा स्त्री की असामान्यताओं/ दोषों के कारण नहीं होती, कभी-कभी पुरुष भी बंध्यता के लिए दोषी होता है।

(ग) एक प्राकृतिक गर्भ निरोधक उपाय के रूप में शिशु को पूर्णरूप से स्तनपान कराना सहायक होता है। (सही/गलत)
उत्तर
सही, शिशु को पूर्णरूप से स्तनपान कराने से अण्डोत्सर्ग नहीं होता है। अतः आर्तव चक्र (Menstrual cycle) भी नहीं होता है, जिसके कारण गर्भ की संभावनाएँ समाप्त हो जाती हैं। किन्तु यह विधि शिशु के जन्म के अधिकतम 6 माह तक कारगर है।

(घ) लोगों के जनन स्वास्थ्य के सुधार हेतु यौन संबंधित पहलुओं के बारे में जागरुकता पैदा करना एक प्रभावी उपाय है।(सही/गलत)
उत्तर
सही, क्योंकि ऐसा करने से लोगों की जनन स्वास्थ्य की समस्याएँ समाप्त अथवा कमतर हो जाती हैं।

प्रश्न 12.
निम्नलिखित प्रश्नों को सही कीजिए

(क) गर्भ निरोधक के शल्य क्रियात्मक उपाय युग्मक बनने को रोकते हैं।
उत्तर
युग्मक बनने से नहीं, बल्कि युग्मकों के परिवहन को रोकते हैं।

(ख) सभी प्रकार के यौन संचारित रोग पूरी तरह उपचार योग्य हैं।
उत्तर
हिपेटाइटिस-B, एड्स एवं जननिक हर्पिस का उपचार नहीं होता।

(ग) ग्रामीण महिलाओं के बीच गर्भ निरोधक के रूप में गोलियाँ (पिल्स) बहुत अधिक लोकप्रिय
उत्तर
ग्रामीण महिलाओं में गर्भ निरोधक के रूप में गोलियाँ (पिल्स)लोकप्रिय नहीं हैं । ग्रामीण महिलाओं को यौन शिक्षा की आवश्यकता है।

(घ) ई. टी. तकनीकों में भ्रूण को सदैव गर्भाशय में स्थानान्तरित किया जाता है।
उत्तर
ई. टी. तकनीकों में 8 ब्लास्टोमियर से ज्यादा अवस्था वाले भ्रूण को गर्भाशय में स्थानान्तरित किया जाता है। जबकि 8 ब्लास्टोमियर से कम अवस्था वाले भ्रूण को अण्डवाहिनी में स्थानान्तरित किया जाता है।

जनन स्वास्थ्य अन्य महत्वपूर्ण प्रश्नोत्तर

जनन स्वास्थ्य वस्तुनिष्ठ प्रश्न

1. सही विकल्प चुनकर लिखिए-

प्रश्न 1.
स्त्रियों के शरीर से फैलोपियन ट्यूब को अलग करना कहलाता है
(a) वैसेक्टोमी
(b) ट्यूबेक्टोमी
(c) ओवरीक्टोमी
(d) कैस्ट्रेशन।
उत्तर
(b) ट्यूबेक्टोमी

प्रश्न 2.
आबादी की सर्वाधिक वृद्धि का प्रमुख कारण है
(a) कम मृत्यु दर
(b) जन्मदर में वृद्धि
(c) अकाल न पड़ना
(d) युद्ध कम होना।
उत्तर
(a) कम मृत्यु दर

प्रश्न 3.
बड़े शहरों में अधिक जनसंख्या का कारण है
(a) शिक्षा के अवसर
(b) उपलब्ध भौतिक सुविधाएँ
(c) अधिक आय के स्रोत
(d) उपर्युक्त सभी।
उत्तर
(d) उपर्युक्त सभी।

प्रश्न 4.
जनसंख्या घनत्व अधिक है____
(a) यू.एस.ए. में
(b) भारत में
(c) चीन में
(d) जापान में।
उत्तर
(c) चीन में

प्रश्न 5.
AIDS रोग फैलता है
(a) बैक्टीरिया से
(b) प्रोटोजोआ से
(c) वाइरस से
(d) फंगस से।
उत्तर
(c) वाइरस से

प्रश्न 6.
यौन संक्रमित रोगों के कारक होते हैं
(a) विषाणु
(b) जीवाणु
(c) प्रोटोजोआ
(d) उपर्युक्त सभी।
उत्तर
(a) विषाणु

प्रश्न 7.
भारत में जनसंख्या की गणना की गई थी
(a) सन् 1891
(b) सन् 1947
(c) सन् 1950
(d) सन् 1961.
उत्तर
(a) सन् 1891

प्रश्न 8.
निम्न में से जन्मदर को नियंत्रित करने की विधि
(a) TUD
(b) GIFT
(c) MIF
(d) IVEET
उत्तर
(a) TUD

प्रश्न 9.
एल्कोहॉल से सर्वाधिक प्रभावित अंग है
(a) यकृत
(b) सेरीब्रम
(c) सेरीबेलम
(d) हृदय।
उत्तर
(a) यकृत

प्रश्न 10.
विश्व की जनसंख्या वृद्धि दर है
(a) 2.4%
(b) 2%
(c) 3%
(d) 4%.
उत्तर
(b) 2%

2. रिक्त स्थानों की पूर्ति कीजिए

1. मानव जनसंख्या का सांख्यिकीय अध्ययन ………………. कहलाता है।
2. माल्थस के अनुसार जनसंख्या में ………………. वृद्धि होती है जबकि खाद्य उत्पादन में ……………….. वृद्धि होती है।
3. किसी समष्टि में एक निश्चित अवधि में जीवों का आना और बाहर जाना ………………. कहलाता
4. शुक्राणु नली को काटकर बन्द करना ………………. कहलाता है।
5. स्थलीय जीवों के जनसंख्या घनत्व के लिये स्थान के क्षेत्रफल को ……………… से प्रदर्शित करते हैं।
6. जन्मदर एवं अप्रवासन जनसंख्या घनत्व में ………………. करते हैं।
7. संशोधित जन्मदर ……………… जन्मदर से अधिक होती है।
8. जनसंख्या आकार को नियन्त्रित करने वाले कारक को …………….. कहते हैं।
9. जैव सूचकांक = ( ………………. / मृत्युदर) x 1000.
10. बालिकाओं में बालकों की अपेक्षा ……………. परिपक्वता आती है।
11. भारत पहला देश है जिसने …………………… स्वास्थ्य देखभाल को ………………….में सामाजिक लक्ष्य के रूप में लाया।
12. RCH भारत में .. ……………. में लाया गया।
13. एम्नियोसेन्टेसिस में लिंग निर्धारण …………. गुणसूत्र के निरीक्षण द्वारा लाया गया।
14. प्रसव के ठीक बाद महिलाएँ …… अनार्तव (मासिक धर्म amoeriorrhoea) का अनुभव करती हैं।
15. LNG-20 ……………….. स्रावी IUD है।
उत्तर

1. डेमोग्राफी
2. बीजगणितीय, अंकगणितीय
3. प्रवासन
4. वैसेक्टोमी
5. द्विविमा (m²)
6. वृद्धि,
7. अशोधित
8. धनात्मक अवरोधक
9. जन्मदर
10. जल्दी
11. प्रजनन, सन् 1951
12. सन् 1997,
13. लिंग
14. स्तनपान संबंधी
15. हॉर्मोन।

3. सही जोड़ी बनाइए

‘A’ – ‘B’

1. S-आकृति वक्र – (a) जन्म नियंत्रण विधि
2. kसीमा स्थिरांक – (b) जनसंख्या
3. शुक्राणु नाशक रसायन – (c) लॉजिस्टिक वक्र
4. माल्थस 2009 – (d) D/Px1000
5. वास्तविक मृत्युदर – (e) संतृप्ति अवस्था।
उत्तर
1.(c), 2.(e), 3.(a), 4. (b), 5.(d).

‘A’ – ‘B’

1. माल्थस – (a) कॉपर-टी
2 वैसेक्टोमी – (b) लिंग परीक्षण
3. ट्यूबेक्टोमी – (c) सेन्सस
4. जनगणना – (d) जनसंख्या पर निबंध
5. एम्निओसेण्टेसिस – (e) पुरुष
6. आई.ग.सी.डी. – (f) स्त्री या जनसंख्या नियंत्रण।
उत्तर
1.(d),2.(e),3. (1), 4.(c), 5.(b), 6.(a).

4. एक शब्द में उत्तर दीजिए

1. दो ऐसे कारण दीजिए जिनसे जनसंख्या अनियन्त्रित रूप से बढ़ जाती है।
2. IUCD का पूरा नाम लिखिए।
3. मानव जनसंख्या का अध्ययन क्या कहलाता है?
4. माला D और N में कौन-सा रासायनिक गर्भ-निरोधक पाया जाता है ?
उत्तर

1. (i) कम उम्र में विवाह, (ii) शिक्षा
2. इन्ट्रायूरिक कन्ट्रासेप्टिव डिवाइसेस
3. डेमोग्राफी (Demography)
4. एस्ट्रोजन एवं प्रोजेस्टीरॉन हॉमोन।

जनन स्वास्थ्य अति लघु उत्तरीय प्रश्न

प्रश्न 1.
परिवार नियोजन को अब किस नाम से जाना जाता है ?
उत्तर
परिवार नियोजन को अब “परिवार कल्याण” के नाम से जाना जाता है।

प्रश्न 2.
सेन्ट्रल ड्रग रिसर्च इन्स्टीट्यूट (CDRI) कहाँ अवस्थित है।
उत्तर
लखनऊ (उत्तरप्रदेश) में स्थित है।

प्रश्न 3.
कंडोम के उपयोग के कोई एक लाभ लिखिए।
उत्तर
कंडोम यौन संचरित रोगों से बचाव करता है।

प्रश्न 4.
हफ्ते में एक बार लेने वाली गर्भनिरोधक गोली का नाम लिखिए।
उत्तर
“सहेली”

प्रश्न 5.
IUCD काशब्द विस्तार बताइए।
उत्तर
इन्ट्रा यूटेराइन कॉन्ट्रासेप्टिव डिवाइस।

प्रश्न 6.
STD का सम्पूर्ण रूप लिखिये।
उत्तर
यौन संचारित रोग (Sexually Transmitted Diseases)।

प्रश्न 7.
विवाह की वैधानिक आयु स्त्री और पुरुष के लिए क्या सुनिश्चित है ?
उत्तर
स्त्री की आयु 18 वर्ष तथा पुरुष के लिए 21 वर्ष सुनिश्चित है।

प्रश्न 8.
दो STD के नाम लिखिए जो संदूषित रक्त से संचारित होते हैं।
उत्तर
(a) एड्स
(b) हिपेटाइटिस-BI

प्रश्न 9.
लैंगिक संपर्क से होने वाले कोई दो रोग बताइए।
उत्तर

• सूजाक (Gonorrhoea)
• सिफलिस (Syphilis)

प्रश्न 10.
HIV एवं AIDS का सम्पूर्ण रूप लिखिए।
उत्तर

• HIV—ह्यूमन इम्यूनोडेफीसिएन्सी वाइरस।
• AIDS- एक्वायर्ड इम्यूनोडेफीसिएन्सी सिन्ड्रोम।

प्रश्न 11.
ZIFT का पूरा नाम लिखिए।
उत्तर
युग्मनज अन्तः डिम्बवाहिनी स्थानान्तरण (Zygote Intra-fallopian Transfer)।

प्रश्न 12
पुरुष में नसबंदी की शल्य क्रिया का क्या नाम है ?
उत्तर
वैसेक्टोमी (Vasectomy)।

प्रश्न 13.
दो ऐसे कारण दीजिये जिनसे जनसंख्या अनियन्त्रित रूप से बढ़ जाती है।
उत्तर
रोगों का नियंत्रण, कृषि का विकास।

प्रश्न 14.
मानव जनसंख्या का अध्ययन क्या कहलाता है ?
उत्तर
डेमोग्राफी।

प्रश्न 15.
माला D और N में कौन-सा रासायनिक गर्भ निरोधक पाया जाता है ?
उत्तर
एस्ट्रोजन एवं प्रोजेस्ट्रॉन।

प्रश्न 16.
एम्नियोसेण्टेसिस विधि द्वारा किसका परीक्षण किया जाता है ?
उत्तर
भ्रूण का लिंग परीक्षण।

प्रश्न 17.
किसी इकाई क्षेत्रफल में पाई जाने वाली मानव संख्या कहलाती है ?
उत्तर
जनसंख्या घनत्व।

जनन स्वास्थ्य लघु उत्तरीय प्रश्न

प्रश्न 1.
परखनली शिशु कैसे प्राप्त किये जा सकते हैं ?
उत्तर
स्त्रियाँ जब गर्भधारण करने योग्य नहीं होती हैं, उस स्थिति में परखनली शिशु तकनीक का उपयोग किया जाता है। स्त्रियों के अनिषेचित अण्डाणु को ऐण्टिसेप्टिक स्थिति में निकाला जाता है। इस अण्डाणु को परखनली में लेकर शुक्राणु द्वारा निषेचन (Fertilization) की क्रिया कराते हैं । निषेचित अण्डाणु से विदलन क्रिया द्वारा 32 कोशिकीय अवस्था वाला भ्रूण बनता है, जिसे ब्लास्टोसिस्ट (Blastocyst) कहते हैं। इस 32 कोशिकीय भ्रूण का रोपण स्त्रियों के गर्भाशय में कर दिया जाता है जहाँ पर इस भ्रूण का अगला विकास होता है और शिशु बनता है जिसे ‘टेस्ट ट्यूब बेबी’ कहते हैं।

प्रश्न 2.
गिफ्ट (GIFT) क्या है ?
उत्तर
GIFT-इसका पूरा नाम गैमीट इन्ट्रा फैलोपियन ट्रान्सफर (Gamete Intra Fallopian Transfer) है । यह गर्भ निरोधन की आधुनिक तकनीक है। इस तकनीक में शुक्राणु कैथेटर (Catheter) में संगृहीत कर लिया जाता है। वाश स्विम तकनीक (Wash swim up technique) द्वारा शुक्राणुओं को अण्डाणु के पीछे कैथेटर में संगृहीत किया जाता है। यह सम्पूर्ण क्रिया स्त्रियों की फैलोपियन नलिका में होती है। निषेचन क्रिया शुक्राणुओं की संख्या कम होने से पूर्ण नहीं हो पाती है।

प्रश्न 3.
ट्यूबेक्टोमी और वैसेक्टोमी को समझाइये।
उत्तर
ट्यूबेक्टोमी-इसमें स्त्री की अण्डवाहिनी को काटकर बाँध दिया जाता है। जिससे अण्डाणु गर्भाशय में नहीं आ पाते और निषेचन नहीं हो पाता। इसे ट्यूबेक्टोमी कहते हैं। यह परिवार नियोजन की एक विधि है। वैसेक्टोमी-यह भी परिवार नियोजन की एक विधि है जो पुरुषों द्वारा अपनाई जाती है। इसमें पुरुष की शुक्राणु नलिका को काटकर बाँध दिया जाता है जिससे शुक्राणु स्खलन के समय नहीं निकलते और निषेचन नहीं हो पाता। यह क्रिया वैसेक्टोमी कहलाती है।

प्रश्न 4.
एम्नियोसेण्टेसिस क्या है ?
उत्तर
एम्नियोसेण्टेसिस भ्रूण परीक्षण की एक तकनीक है जिसमें सर्जिकल सुई द्वारा मादा के गर्भाशय से एम्नियोटिक द्रव को शरीर से बाहर निकाला जाता है और एम्नियोटिक द्रव में उपस्थित फोयटस कोशा का संवर्धन किया जाता है और इसका गुणसूत्रीय परीक्षण करके निम्न बातों का पता लगाया जाता है

• गुणसूत्रीय असामान्यता, जैसे-डाउन सिण्ड्रोम, फिलोडेल्फिया सिण्ड्रोम एवं एडवर्ड सिण्ड्राम।
• उपापचयी अनियमितताएँ, जैसे- PKU, क्रिटेनिज्म, एल्केप्टोन्यूरिया।
• लिंग भ्रूण के परीक्षण में इसका उपयोग किया जाता है।

प्रश्न 5.
मानव जनसंख्या वृद्धि के सामाजिक कारण समझाइये।
उत्तर
मानव जनसंख्या वृद्धि के प्रमुख सामाजिक कारण निम्नलिखित हैं

• समाज का अशिक्षित होना ।
• निम्न सामाजिक स्तर का होना।
• समाज में विभिन्न प्रकार की कुरीतियाँ होना।
• विभिन्न प्रकार की सामाजिक मान्यताएँ ।
• कम उम्र में विवाह करना, क्योंकि उम्र के पूर्वार्द्ध में प्रजनन क्षमता अधिक होती है।
• निम्न सामाजिक स्तर के कारण मनोरंजन का अभाव होना।
• सामाजिक मान्यताएँ जैसे-पुत्र रत्न की प्राप्ति से मोक्ष मिलता है
• सामाजिक पिछड़ापन।

MP Board Class 12th Biology Solutions

MP Board Class 12th Biology Solutions Chapter 7 Evolution

Evolution NCERT Textbook Questions and Answers

Question 1.
Explain antibiotic resistance observed in bacteria in light of Darwinism selection theory.
Answer:
Darwinism theory of natural selection states that environment selects organisms with favourable variations and these organisms thus, survive and reproduce. It is observed when bacterial populations are exposed to certain antibiotic, the sensitive bacteria could not tolerate and hence, died due to the adverse environment. Whereas some bacteria that developed mutation became resistant to the particular antibiotic and survived. As a result such resistant bacteria survive and multiply quickly as compared to other sensitive bacteria. So, the whole population is regained by multiplication of resistant variety and antibiotic resis¬tant gene becomes widespread in the bacterial population.

Question 2.
Find out from newspapers and popular science articles any new fossil dis-coveries or controversies about evolution.
Answer:
Fossils of dinosaurs have revealed the evolution of reptiles in Jurassic period. As a result of this evolution of other animals such as birds and mammals has also been discovered. However, two unusual fossils recently unearthed in China have ignited a controversy over the evolution of birds confiiciusomis is one such genus of primitive birds that were crow sized and lived during the Cretaceous period in China.

Question 3.
Attempt giving a clear definition of the term species.
Answer:
A species generally includes similar organism. Members of this group can show interbreeding. Similar group of genes are found in the members of same species and this group has capacity to produce new species. Every species has some cause of isolation which intruped the interbreeding with nearest reactional species which is refer as reproductively isolated.

Question 4.
Try to trace the various components of human evolution (Hint: Brain size and function, skeletal structure, dietary preference, etc.)
Answer:
The various components of human evolution are as follows:

1. Brain capacity.
2. Posture
3. Food/Dietary preference and their important features.

Name brain capacity, posture and food features :

• Dryopithecus Africans : Knuckle walker, walked similar to Gorillas and Chim-panzees (was more apelike) soft fruit and leaves; canines large, arm and legs are of equal size.
• Ramapithecus: Semi-erect (more manlike) seeds, nuts canines were small while molars were large.
• Australopithecus africanus: Australopithecus africanus 450 cm3 full erect prosture, height (1 -05m) Herbivorous (ats fruits). Hunted with stone weapons, lived as trees, canines and incisors were small.
• Homo habilus: Homo habilus 735 cm3 fully erect posture, height (1 -5m) carnivo¬rous canines were small. They were first tool makers.
• Homo erectus: Homo erectus 800-1100 cm3 fully erect posture, height (1-5- l-8m) omnivorous. They used stone and bone tools for hunting games.
• Homo neanderihalensis : Homo neanderthalensis 1300 – 1600 cm3 fully erect posture, height (1-5 – l-6m) omnivorous cave dwellers, used hiles to protect their bodies and buried their dead.
• Homo sapiens fossils : Homo sapiens fossils 1650 cm3 fully erect posture with height (l -8m) omnivorous. They had strong jaw with teeth close together. They were cave dwellers, made painting and carvings in the caves. They developed a culture and were called first modem man.
• Homo sapiens sapiens : Homo sapiens sapiens 1200 – 1600 cm3 fully erect pos¬ture, height (1-5 – l-8m) omnivorous. They are the living modem men with high intelligence. They developed art, culture, language, speech, etc. They cultivated crops and domes¬ticated animals.

Question 5.
Find out through internet and popular science articles whether animals other than man has self-consciousness.
Answer:
There are many animals other than humans, which have self-consciousness. An example of an animal being self-conscious is dolphins. They are highly intelligent. They have a sense of self and, they also recognize others among themselves and others. They communicate with each other by whistles, tail-slapping, and other body movements, not dolphins, there are certain other animals such as Crow, Parrot, Chimpanzee, Gorilla, Orangutan, etc., which exhibit self-consciousness.

Question 6.
List 10 modern day animals and using the internet resources link it to a corresponding ancient fossil. Name both.
Answer:
Modern and Ancient corresponding animals :

Question 7.
Practise drawing various animals and plants.
Answer:

Question 8.
Describe one example of adaptive radiation.
Answer:
Darwin’s finches in the Galapagos Island once had a common ancestor but with evolution, they modified into different types according to their food habitat.

Question 9.
Can we call human evolution as adaptive evolution ?
Answer:
No, we can not be called human evolution as adaptive evolution.

Question 10.
Using various resources such as your school Library or the internet and discussions with your teacher, trace the evolutionary stages of any one animal, say horse.
Answer:
The evolution of horse started with Eohippus during Eocene period. It involved the following evolutionary stages :

• Gradual increase in body size
• Elongation of head and neck region
• Increase in the length of limbs and feet
• Gradual reduction of lateral digits
• Enlargement of third functional toe
• Strengthening of the back
• Development of brain and sensory organs
• Increase in the complexity of teeth for feeding on gras

• Eohippus : It had a short head and neck. It had four functional toes and a splint of 1 and 5 on each hind limb and a splint of 1 and 3 in each forelimb. The molars were short crowned that were adapted for grinding the plant diet.
• Mesohippus : It was slightly taller than Eohippus. It had three toes in each foot.
• Merychippus : It had the size of approximately 100 cm. Although it still had three toes in each foot, but it could run on one toe. The side toe did not touch the ground. The molars were adapted for chewing the grass.
• Pliohippus : It resembled the modern horse and was around 108 cm tall. It had a single functional toe with splint of 2nd and 4th in each limb.

Equus : Pliohippus gave rise to Equus or the modern horse with one toe in each foot. They have incisors for cutting grass and molars for grinding food.

Evolution Other Important Questions and Answers

Evolution Objective Type Questions

1. Choose the Correct Answer.

Question 1.
These are found at the atmosphere of earth before evolution:
(a) Water vapour, CH₄, NH₃ and Oxygen
(b) CO₂, NH₃, H₂ and Water vapour
(c) CH₄, NH₃, H₂ and Water vapour
(d) CH₄, O₃, O₂ and Water vapour.
Answer:
(c) CH₄, NH₃, H₂ and Water vapour

Question 2.
Which gas is absent at the atmosphere of earth at the time of origin of earth:
(a) NH₃
(b) H₂
(c) O₃
(d) CH₄
Answer:
(c) O₃

Question 3.
Which gaseous mixture are used and found the amino acid by Miller:
(a) Methane, Ammonia, Hydrogen and Water vapour
(b) Methane, Ammonia, Nitrogen and Water vapour
(c) Methane, Nitrogen, Hydrogen and Water vapour
(d) Ammonia, Carbon dioxide, Nitrogen and Water vapour.
Answer:
(a) Methane, Ammonia, Hydrogen and Water vapour

Question 4.
Who gave the theory “Survival of the fittest” :
(a) Charles Darwin
(b) Herbert Spencer
(c) Lamarck
(d) Hugo de Vries.
Answer:
(b) Herbert Spencer

Question 5.
Who gave the theory of inheritance of acquired, characters :
(a) Charles Darwin
(b) Lamarck
(c) Valles
(d) de Vries.
Answer:
(b) Lamarck

Question 6.
Who wrote origin of species :
(a) Oparin
(b) Beajman
(c) Lamarck
(d) Darwin.
Answer:
(d) Darwin.

Question 7.
What is the name of Darwin’s ship:
(a) Gangotri
(b) Beagle
(c) Atlantic
(d) Seagull.
Answer:
(b) Beagle

Question 8.
‘Atmosphere is a factor of variation’ who gave this theory :
(a) Mendel
(b) Darwin
(c) Lamarck
(d) Laplace.
Answer:
(c) Lamarck

Question 9.
Darwinism explain it:
(a) Characters are grown by heredity
(b) Species are changed into structural form with time .
(c) Nature selected the animals which are adapted
(d) Origin of evolution due to effect of environment.
Answer:
(c) Nature selected the animals which are adapted

Question 10.
Which antibiotic is used to Replica planting experiment of Ladderberg:
(a) Penicillin
(b) Streptomycin
(c) Erythromycin
(d) Neomycin.
Answer:
(a) Penicillin

Question 11.
Which is the basic unit of natural selection:
(a) Species
(b) Community
(c) Genous
(d) Solitary organism.
Answer:
(c) Genous

Question 12.
Who gave the theory of natural selection :
(a) Lamarck
(b) de Vries
(c) Darwin
(d) Mendel.
Answer:
(d) Mendel.

Question 13.
Darwinism is based on:
(a) Segregation
(b) Independent assortment
(c) Quantitative heredity
(d) Natural selection.
Answer:
(b) Independent assortment

Question 14.
Who discover the uses and experiment of organs:
(a) Valles
(b) Lamarck
(c) Darwin
(d) de Vries
Answer:
(a) Valles

Question 15.
The unit of natural selection :
(a) Solitary animal
(b) Family
(c) Community
(d) Species.
Answer:
(d) Species.

Question 16.
Mule is a product of:
(a) Mutation
(b) Reproduction
(c) Inter-species hybridization
(d) Intra-species hybridization
Answer:
(b) Reproduction

Question 17.
Who use first the word ‘species’:
(a) Linnaeus
(b) John
(c) Aristotle
(d) Darwin
Answer:
(b) John

Question 18.
Homologus organs are :
(a) Similar in origin
(b) Similar in function
(c) Similar in evolution
(d) Similar in behaviour.
Answer:
(a) Similar in origin

Question 19.
Which era are called as golden period of Reptilia or Dinosaurs :
(a) Mesozoic
(b) Cenozoic
(c) Palaeozoic
(d) Cenozoic.
Answer:
(a) Mesozoic

Question 20.
Dinosaurs are distinct in this era:
(a) Jurassic
(b) Triassic
(c) Cretaceous
(d) Permian.
Answer:
(c) Cretaceous

Question 21.
Vestigial organs of human’ are:
(a) Wisdom teeth, Coccyx, Nail, Eyelids, Vermiform appendix
(b) Wisdom teeth, Coccyx, Vermiform appendix, Pancreas
(c) Wisdom teeth, Coccyx, Vermiform appendix, Nictating membrane
(d) Coccyx, Wisdom teeth, Nail, Auricular muscle.
Answer:
(c) Wisdom teeth, Coccyx, Vermiform appendix, Nictating membrane

Question 22.
How made fossils :
(a) Animals are naturally buried in land after death
(b) Animals are decomposed by decomposers
(c) Animals are eaten by their hunter species
(d) Animals are destroyed due to environmental conditions.
Answer:
(d) Animals are destroyed due to environmental conditions.

Question 23.
Species which are similar in shape and different in reproduction are called :
(a) Sub-species
(b) Sibling
(c) Isotopic
(d) Allopetric species.
Answer:
(c) Isotopic

Question 24.
Name the ship which had been used by Darwin for journey :
(a) Ciloge
(b) Beagle
(c) Seagull
(d) Atlantic.
Answer:
(b) Beagle

Question 25.
Life is not in this era :
(a) Mesozoic
(b) Palaeozoic
(c) Cenozoic
(d) Azoic.
Answer:
(d) Azoic.

Question 26.
Which factor is responsible for evolution due to New-darwinism theory :
(a) Mutation
(b) Usefull variation
(c) Hybridization
(d) Mutation and Natural selection.
Answer:
(d) Mutation and Natural selection.

Question 27.
Found the age of fossils by which :
(a) By quantity of calcium ions
(b) By quantity of organic radioactive components
(c) By struggle from other mammals
(d) By structure of bones.
Answer:
(d) By structure of bones.

Question 28.
Darwin finches are related to which of the following evidences :
(a) Fossils
(b) Embryology
(c) Anatomy
(d) Geographical distribution.
Answer:
(d) Geographical distribution.

Question 29.
Who gave Recapitulation theory :
(a)Weizmann
(b) Haeckel
(c) Darwin
(d) Malthus.
Answer:
(b) Haeckel

Question 30.
Which definition of organic evolution is correct:
(a) Evolutionary history of a species
(b) History of species with the variations is the species
(c) Embryonal history of species
(d) Development of species.
Answer:
(a) Evolutionary history of a species

Question 31.
What is the perfect sequence of development of human:
(a) Peking man, Heidelberg man, Neanderthal Cro-Magnon
(b) Peking man, Homo sapiens, Cro-Magnon, Neanderthal
(c) Peking man, Neanderthal,.Homo sapiens, Heidelberg
(d) Peking man, Cro-Magnon, Homo sapiens, Neanderthal.
Answer:
(a) Peking man, Heidelberg man, Neanderthal Cro-Magnon

Question 32.
Which of the following is the most primitive ancestor of man :
(a.) Australopithecus
(b) Ramapithecus
(c) Homo habilis
(d) Homo neanderthalensis.
Answer:
(b) Ramapithecus

Question 33.
Ancestral amphibians were tetrapods that evolved during :
(a) Jurassic period
(b) Cretaceous period
(c) Devonian period
(d) Carboniferous period.
Answer:
(c) Devonian period

2. Fill in the Blanks :

1. Earth is a member of …………….
2. ……………. wrote ‘Origin of life’.
3. Oxygen is find the earth because photo-synthetic organisms are present on the earth, this phenomenon is called …………….
4. ………….. is the connecting link between Reptiles and Aves.
5. …………………. is the ancesters of horse.
6. The evolution of ……………. molecule set the stage for evolution of autotrophs.
7. Evolution of birds and mammals were ……………. era.
8. Golden era of Dinosaurs is ……………. period.
9. ……………. fossil man has been known from Shivalik hills in India.
10. Book of Charles Darwin “Origin of life” has been explained of …………….
11. ……………. be change of hareditary characters.

Answer:

1. Solar system
2. Oparin
3. Oxygen revolution
4. Archaeopteryx
5. Eohippus
6. Chlorophyll
7. Jurassic
8. Mesozoic
9. Ramapithecus
10. Natural selection
11. Mutation.

Question 3.
Match the Following :

Answer:

1. (d)
2. (a)
3. (e)
4. (c)
5. (b)

Answer:

1. (d)
2. (c)
3. (e)
4. (a)
5. (b)
6. (f).

4. Answer in One Word / Sentence:

1. The matter which gave rise to the universe 15 billion years ago.
2. The matter which is a link between acellular and cellular system.
3. The island at which Darwjn studies the organisms for his theory of evolution.
4. Origin of two or more unlike species from a common ancestor.
5. The whole process of development at changes from embryo to adult organism.
6. The process of changes in the course of origin of a new species.
7. The theory of organic evolution proposed by Darwin.
8. The permanent, heritable and spontaneous changes in organisms.
9. The remains of ancient organisms.
10. The organs which are different in their origin but have similar function.

Answer:

1. Ylem
2. Coacervates
3. Galopegos
4. Adaptive radiation
5. Ontogeny
6. Phytogeny
7. Natural selection
8. Mutation
9. Fossils
10. Homologous organs.

Evolution Very Short Answer Type Questions

Question 1.
Name the non-cellular organism.
Answer:
Virus.

Question 2.
Name the theory which tell about the milky way and formation of star.
Answer:
Big Bang.

Question 3.
Releasing of 02 by the evolution of photosynthesis an ancient time is called?
Answer:
Oxygen revolution.

Question 4.
Which planet has the possibility of origin of life other than earth?
Answer:
ars.

Question 5.
Where did life originate ?
Answer:
In water.

Question 6.
What is the art of limitation, which is practiced the world over from annoying siblings or plant resembles another creature to gain other advantages?
Answer:
Mimicry.

Question 7.
Name the method of the changing of the structure of a gene resulting in a variant form that may be transmitted to.
Answer:
Mutation.

Question 8.
What is the collection of different genes within an inter breading population?
Answer:
Gene pool.

Question 9.
Who wrote origin of species?
Answer:
Charles Darwin.

Question 10.
Who proposed the theory of natural selection?
Answer:
Darwin.

Question 11.
Who proposed the Recapitulation theory or Biogenetic law?
Answer:
Haeckel.

Question 12.
Astronomical distance measured in?
Answer:
Astronomical distance measured in Light years.

Question 13.
Name the theory by which earth is said to originate.
Answer:
The big bang theory.

Question 14.
What is fossil ?
Answer:
Fossils are the remains or impressions of ancient organisms preserved in sedimentary rocks or other media.

Question 15.
What is mutation ?
Answer:
New species originate due to changes of hereditary characters are called mutation.

Question 16.
Name the scientist who tell the spontaneous theory is wrong.
Answer:
Louis Pasteur’s.

Question 17.
Which era is called golden period of Dinosaurs?
Answer:
Mesozoic period is called golden period of Dinosaurs.

Question 18.
In which ship Darwin studied the nature ?
Answer:
Beagle.

Question 19.
Name any two vertibrates body organ which are homologus organs of human forelimb.
Answer:
(1) Flipper of Whale,
(2) Wing of birds.

Question 20.
What is the scientific name of modern man ?
Answer:
Homo sapiens.

Question 21.
Who is the early man of the modern human?
Answer:
Cro-Magnon peoples are early human of modern human.

Question 22.
Who are the early human and sub-human.
Answer:
Ramapithecus is early human and Australopithecus is early human.

Question 23.
Which human form first started to walk on two legs?
Answer:
Australopithecus form first started to walk on two legs.

Question 24.
Which type of human was ‘Cro-Magnon’ on the basis of food in take?
Answer:
Cro-Magnon was carnivorous.

Question 25.
On the basis of evolution which human had brain size of 1400cc?
Answer:
Neanderthal.

Question 26.
Give the name of apelike ancestors of humans
Answer:
Apelike ancestors of human’s are Dryopithecus.

Question 27.
Differentiate between Dryopithecus and Ramapithecus.
Answer:
Dryopithecus were apelike but Ramapithecus were mostly human like.

Evolution Short Answer Type Questions

Question 1.
What is virus ? Why is it treated as a link between living and non-living ?
Answer:
Viruses are simplest organism of the earth, which consists of nucleic acid (DNA or RNA) surrounded by protein cover. It shows characteristics of living as well as non-living organisms.
(A) Living characters of virus:

• Virus shows structural differentiations.
• They contain hereditary material.
• They exhibit mutation.
• They spread plant and animal diseases.
• Growth and development present.
• They exhibit adaptaion.
• They possess sensitivity.

(B) Non-living characters of virus:

• Lack protoplasm and cell organelles.
• Can be crystallized.
• No metabolic activities seen.
• Cannot reproduce outside living cells.
• They lack enzymes.

Due to above reason, viruses are considered as link between living and non-living organisms, thus, it is the first life originated in the earth.

Question 2.
What is oxygen revolution? Explain.
Answer:
Oxygen revolution : Evolution of 02 in photosynthesis during primitive environmental conditions is very important because, it is required in the evolution of organism and conversion of reducing environment into oxidizing environment hence, it is called oxygen revolutioa Oxygen evolution should cause the following changes in the environment:

• Oxygen evolution should cause the conversion of reducing environment into oxidizing environment
• Ozone layer is formed 15 miles above from the earth which absorbs the ultraviolet light of the sunlight and thus, prevents the entry of uv light in the atmosphere.
• O₂ present in the environment dissociates methane (CH₄) into CO₂ and O₂. This CO₂ is used in photosynthesis.
• NH₃ of the primitive environment is dissociated into H₂O.and nitrogen.
  CH₄ + 2O₂2 → CO₂ + 2H₂O
  4NH₃ + 3O₂ → 2N₂ + 6H₂O.

Question 3.
Explain the origin of the earth.
Answer:
Origin of the earth:

• Earth was formed 4-5 billion years back.
• Initially, die surface was covered with water vapOur, methane, C02 and NH3.
• The UV rays of the sun broke water into hydrogen and oxygen.
• Hydrogen escaped and oxygen combined with NH3 and CH4 to form water, C02 and other gases, also forming die ozone layer.
• Cooling of water vapour led to rain which filled the depressions on earth’s surface, forming water bodies.

Question 4.
Mention the names of discoveries and principles given by the following scientists:

1. Louis Pasteur
2. A. I. Oparin
3. Urey and Miller
4. Francesco Redi
5. Faux.

Answer:

1. Louis Pasteur : He proved that air contains spores of microorganism and Biogenesis theory was supported by him.
2. A. I. Oparin: He presented the biochemical explanation of origin of life in his book “The Origin of Life on Earth”.
3. Urey and Miller : They supported the evidence of Oparin-Haldane theory of Origin of life.
4. Francesco Redi: He by conducting experiments proved that abiogenesis cannot exist but bioginesis theory can exist i.e., Life arises from pre-existing life.
5. Faux: He has been experimentally supporting the organic substances as described by Oparin.

Question 5.
Write down the difference between :

1. Ozone and Oxygen
2. Micro-molecules and Macro-molecules.

Answer:
1. Differences between Ozone and Oxygen :

2. Differences between Micro-molecules and Macro-molecules :

Question 6.
What are homologous organs ?
Or
What is homology ?
Answer:
Organs which are similar in structure and origin but different in appearance and functions are called homologous organs and the phenomenon is called homology.
Examples : Forelimbs of bat, wings of bat, hands of man, forelimbs of horse. These are the examples of homologous organs because, they are made up of similar bones, hu¬merus, radius-ulna, carpals, metacarpals and fingers.

Question 7.
What do you mean by analogous ?
Answer:
Analogous organs: Organs which are different in origin and structure but per¬forming similar functions are known as analogous organs and the phenomenon is called as analogy. Analogous organs do not indicate phylogeny.
Examples : Wings of butterflies are made up of chitin, wings of birds made by production of feathers on forelimbs and skin present between the fingers of bat are the examples of analogous organs.

Question 8.
What is the difference between homologous and analogous organs ? Give two examples of each of them.
Answer:
Differences between Homologous and Analogous organs :

Question 9.
What do you mean by vestigial organs ?
Or
Write the two names of vestigial organs of man.
Or
What are vestigial organs ? Explain. Write four vestigial organs of the human body.
Answer:
Vestigial organs : Organs that are reduced and have become functionless in an organism but were functional in their ancestors are called as vestigial organs :
Examples :

• Vermiform appendix
• Coccygial vertebrae
• Nictitating membrane in the eyes of human
• Muscles of external ear (Pinna).

Question 10.
What is connecting link ? Also explain the significance of connecting links.
Answer:
Connecting link : Certain organisms which share characters with two different groups. Such organisms are called connecting links.
Examples : Some fossils present in nature like Archaeopteryx is known as fossils connecting link and it is the connecting link of birds and reptiles. Neopilina, Platyptus, Protopterus, etc. Neopilina is a mollusc and is the connecting link of mollusca and annelida.
Significance of Connecting links : It proves the process of organic evolution and helps in the identification of organism closely related.

Question 11.
Write down the demerits of Darwinism.
Answer:
Some of the demerits of Darwinism are :

• Darwinism stresses upon small fluctuating variations which has no role in evolution.
• Does not satisfactorily explain effect of use and disused and presence of vestigial organs:
• It did not differentiate somatic and germinal variations.
• It explains survival of the fittest but not arrival of the fittest. .

Question 12.
Differentiate between Lamarckism and Darwinism.
Answer:
Differences between Lamarckism and Darwinism:

Question 13.
Explain Lamarckism in short
Or
Explain Lamarckism of organic evolution in brief.
A ns. Lamarckism : In 1809, Lamarck has proposed a theory to explain organic evolution, which is known, as the “Theory of inheritance of acquired characters.” According to Lamarck, organisms acquire certain characters during their lifetime due to changes in environment and these acquired characters are heritable. According to this theory, new species are originated as follows :

• New requirements and wills are produced in the organisms due to the effect of changed environment
• New requirements and wills of organisms resulting in the production of new habits.
• Changes in the habit bringing about modifications of the organ.
• New habits resulting in the use or disuse of the organs:
• Use of organ resulting in the development of acquired characters.
• These acquired characters are heritable.
• Inheritance of acquired characters resulting in the development of new species.

Evolution Long Answer Type Questions

Question 1.
Draw a well-labelled diagram of Miller and Urey’s experiment
Answer:

Experimental evidence of Chemical evolution or Miller’s experiment:

• Experiment was performed by S.L. Miller and H.C. Urey in 1953.
• Experimental set-up: In a closed flask containing CH4, H2, NH3 and water vapoiy at 800°C, electric discharge was create. The conditions were similar to those in primitfve atmosphere.
• Observations: After a week, they observed presence of amino acids and complex molecules like sugars, nitrogen bases, pigments and fats in the flask.
• Conclusions : (i) It provides experimental evidence for the theory of chemical origin.
  (ii) It showed that the first non-cellular form of life was created about 3 billion years ago.
  (iii) It showed that non-cellular biomolecules exist in the form of DNA, RNA, polysaccharides and protein.

Question 2.
Write an essay on modern concepts of origin of life.
Or
Explain the role of non-living in origin of life.
Answer:
Modern concept of origin of life : The modem concept of origin of life was postulated by a Russian biochemist A.I. Oparin in 1936. According to this theory, after the formation of earth various chemicals played important role in the formation of atmosphere. Life originated and first organism came into existence from certain molecules when atmospheric conditions became suitable. According to Oparin, life originated in the following steps:

1. Formation of earth and its atmosphere: Earth is believed to be originated some 4,500 million years ago by the condensation and cooling of the clouds of cosmic dust and gases called ylem. The heavier elements collected at the core and lighter elements around the core. Outermost layer contains H, C, O and N. Oxygen was found only in combination of other elements. These four elements reacted with each other forming H₂, H₂O, CH₄, NH₃, CO₂ and HCN.

2. Formation of small organic molecules : The mixture of methane, ammonia, water and hydrogen comes in contact of solar energy. Cosmic rays and electric discharge could produce some simple organic compounds. These simple organic compounds formed in such a way and accumulated in primitive atmosphere and oceans were responsible for synthesis of complex micro molecules as follows :

3. Formation of polymers : It is clearly understood from the above description that a large number of micro molecules such as hydrocarbons, amino acids, fatty acids, purine and pyrimidines and simple sugars accumulated in the oceans. When atmospheric water condensed on further cooling, the inorganic precursors collided, reacted and aggregated to form new molecules of increasing size and complexity. Thus, by polymerization macromolecules were formed. The chemical reactions for the formation of macro-molecules can be summarized as follows:

(a) Sugar + Sugar → Polysaccharides
(b) Fatty acid + Glycerine → Lipids
(c) Amino acid + Amino acid → Protein
(d) Nitrogenous base (Adenine) + Sugar + Phosphate → Adenosine phosphate
(e) Nitrogenous base + Sugar → Nucleoside
(f) Nucleoside + Phosphate → Nucleotide
(g) Nucleotide + Nucleotide → Nucleic acid.

4. Formation of molecular aggregates and primitive cells : Over a vast of time, these molecules became associated with one temporary complex. Ultimately, it leads to the formation of a coacervate. A coacervate is a solution of high molecular weight of chemicals, i.e., proteins and carbohydrates, which become bounded by lipid membrane, which is selectively permeable. The coacervate grows by absorbing molecules from their environment. The substances which got accumulated in the coacervates underwent reactions and resulted in the molecular reorganization of some proteins into enzymes. A coacervate having nucleoprotein surrounded by various nutritive organic substances and covered by surface membrane is considered to be the precell, which got later transformed into first living cell. The coacervate can reproduce by budding.

5. Evolution of complex biochemical reactions : Primitive organism utilize chemi¬cal substances present in the environment as food hence, they are:

(a) Heterotrophic, chemosynthetic organisms appeared due to mutation and natural selection in heterotrophs.
(b) Blue-green algae evolved from chemosynthetic organisms by mutation and natural selection.
(c) The liberation of free oxygen into the atmosphere produced by the blue-green algae due to the process of photosynthesis. It finally changed the reducing atmosphere into an oxidizing one and therefore, all possibilities of further chemical evolution were finished.
Free living eukaryotes originated in the ocean from blue-green algae.

6. The origin of well-developed organisms : From the simple eukaryotes which were like unicellular organisms of today various forms of life evolved during passage of time.

Question 3.
Write the process of formation of organic molecules in sea water on earth with the help of Miller and Urey’s experiment
Answer:
The work of A.I. Oparin (1938-1965), HLUrey and Stanley Miller (1959) pro¬vided evidences in the favour of biochemical origin of life. They had prepared the atmos¬phere like that of primitive earth and as described by Oparin, they made the synthesis of organic compounds by the following methods:

(i) Four elements H, C, O (not free O₂) and N react .with each other to form H₂O, CH₄, NH₃, CO₂ and HCN on primitive earth.

(ii) From these four elements following organic molecules were formed in sea water of earth:
(a)

(b) CH₄, H₂O, NH₃ → Amino acids.
(c) CH₄, HCN, H₂O, NH₃ → Nitrogenous bases.

(iii) Macro-molecules of organic compounds were synthesized by these above pre¬pared organic compounds.

(a) Sugar + Sugar → Polysaccharides (Carbohydrate).
(b) Fatty acids + Glycerol → Fats.
(c) Amino acid + Amino acid → Protein.
(d) Nitrogenous base + Sugar → Nucleoside.
(e) Nucleoside + Phosphoric acid → Nucleotide.
(f) Nucleotide + nucleotide → Nucleic acid.
(g) Nitrogenous base (Adenine) + Sugar + Phosphate →Adenosine phosphate.
(iv) The above organic compounds and salts together constituted the first living being.

Question 4.
Name connecting link of reptiles and birds. Also write their characters.
Answer:
Archaeopteryx is the connecting link of birds and reptiles. Archaeopteryx was a bird. It is regarded as the connecting link between reptiles and birds, which suggests the path of evolution of the latter from the former. It is found as fossils. They are found during Jurassic period 140 million years. Archaeopteryx exhibits both reptiles and birds like characters.

1. Reptiles like characters:

• Bones were similar to that of reptiles in which air sacs were absent.
• Tail bearing vertebra.
• It had teeth in jaws, scales were present on the body.
• Metacarpals were free.
• Pelvic girdle recombines with the pelvic girdle of reptiles.

2. Birds like characters :

• Presence of feathers on body. .
• Forelimbs were modified in wings.
• Skull large and monocondylar.
• Jaws were modified into beak.
• Hallux was backwarded and pointed.

Question 5.
Give a detailed account of theory of natural selection.
Or
Describe the Darwin’s opinion about the origin of new species of organisms.
Answer:
Charles Darwin (1809-1882) explained the theory of evolution in his book “Origin of species by natural selection.” Darwin undertook a long voyage for five years in the capacity of a naturalist on a British warship ‘Beagle’. He travelled to islands of Galapagos and collected evidences to explain evolution.
To explain origin of species he gave theory of natural selection as a mechanism for evolution.

The main points of Darwinism are given below:

1. Over production of offsprings : Every living being has an inherent tendency to produce more offspring than that can survive.

2. Struggle for existence : Though the offsprings are produced in large number yet their production remain almost constant. This is because of struggle for existence. There is struggle for food, space, breeding, etc. Moreover death of individuals due to diseases an

3. Survival of the fittest: Darwin believed that any individual is successful in struggle for existence if it survives long enough to produces offsprings. Individuals who are fit in a particular environment can only survive.

4. Variation: Due to constant struggle, the organisms change themselves in accordance with the new needs.

5. Natural selection : In the struggle, for existence organisms having variations favourable to the environment would have more chance to survive and reproduce their own kind. Those which do not possess favourable variations would die or fail to reproduce.

6. Origin of species : Any changes in environmental conditions cause natural selection to act upon the population and select the well adapted individuals. It results in changes of characters of the populations. By the inheritance of these changes in successive generations, new species are formed.

Question 6.
Organic Evolution is a continued process, explain it in favour of it giving any three evidences.
Answer:

Evolution is a complex phenomenon accounting for the present day diversity among organisms. But it has clearly maintained the basic unity among them since it occurred over a period of millions of years, no one would have seen/recorded evolution and hence scientists have provided various evidences to prove evolution.

Some of the evidences of organic evolution are described below :

I. Evidences from Embryology :

• Important activities that occur various animals are:
• For the survival, all animals get energy and various substances from environment.
• In all organisms energy is produced from ATP.
• In all organisms, the duplication of DNA is similar.
• In all organisms, protein synthesis is same and it is produced from ribosomes.
• In all organisms respiration and steps of respiration is same.
• All organisms multiplicate and reproduce, due to which they have basic similarities.
• All organisms conduct hereditary characters on similar principles.

II. Evidence from Anatomy: Anatomy of living organisms will be explained with different examples:

• Homologous organs: Organs which are similar in structure and origin but diffeifent in function and appearance is known as homologous organs.
• Analogous organs : Organs which are different in origin and structure but performing similar functions are known as analogous organs.
• Vestigial organs : Organs that are reduced and have become functionless in an organism, but were functional in their ancestors are called vestigial organ.

III. Evidences from vestigial organs: Organs of the body which are non-functional but they are functional in some other organisms are called vestigial organs:

Morphological evidence of evolution is provided by the presence of vestigial organs of body which are often undesired, degenerated and non-functional. These might have been large and functional in some other animals or in ancestors of those which now possess it in rudimentary forms. e.g„ vermiform appendix in man, muscles of external ear (pinna) in man, nictitating membrane or plica, semilunaris in human eye, wisdom teeth (third pair of molars) tail bone (coccyx) in man, wings of ostrich, hindlimbs in snakes, etc.

Question 7.
What do you mean by organic evolution? How do fossils exhibit evidence to prove organic evolution?
Answer:
Descent with modification in organism is known as organic evolution.

Evidences of organic evolution from fossils record : Fossils are treated as significant evidence of organic evolution. Fossils are the remains or impressions of ancient organisms preserved in the layers of rock and soil. Fossils only do not prove the theory of organic evolution, yet it evidently prove that gradually complexity increased in body organization. The complexity in the body of organization can be noticed as we study the upper layers. Thus, it can be concluded from above observations :

• The crust of the earth and the organisms living on it underwent change in the course of time.
• The organisms with simple structural organization originated earlier than the complex ones.
• Some of the organisms lived on the earth for short time and became extinct. This was a result of drastic changes in the climate on the earth.

Hence, forth fossils produce bonafide record of such plants and animals which had shown their existence once upon a time and now are extinct or not present exactly in the same form, thus, producing strong evidence in favour of organic evolution.

MP Board Class 12th Biology Solutions

MP Board Class 11th Maths Solutions Chapter 11 शंकु परिच्छेद Ex 11.2

निम्नलिखित प्रश्न 1 से 6 तक प्रत्येक में नाभि के निर्देशांक, परवलय का अक्ष, नियता का समीकरण और नाभिलंब जीवा की लंबाई ज्ञात कीजिए।
प्रश्न 1.
y² = 12x.
हल:
परवलय का समीकरण, y² = 12x
∴ y² = 4ax से तुलना करने पर
4a= 12 या a= 3
(i) नाभि के निर्देशांक (a, 0) या (3, 0)

(ii) परवलय का अक्ष OX
इसका समीकरण y = 0
(iii) नियता का समीकरण : x = – a अर्थात् x = – 3
(iv) नाभिलंब जीवा की लंबाई = 4a = 12.

प्रश्न 2.
x² = 6y.
हल:
परवलय का समीकरण x² = 6y
∴ 4a = 6 या a = \(\frac{3}{2}\)

इसका अक्ष y-अक्ष है जिसका
(i) समीकरण x = 0 है।
(ii) नाभि F (0, a) के निर्देशांक \(\left(0, \frac{3}{2}\right)\) है।
(iii) नियता y = – a का समीकरण y = – \(\frac{3}{2}\)
(iv) नाभिलंब जीवा की लम्बाई 4a = 6.

प्रश्न 3.
y² = – 8x.
हल:
परवलय का समीकरण y² = – 8x
∴ 4a = 8, a = 2
(i) नाभि F(- a, 0) के निर्देशांक (- 2, 0)

(ii) परवलय का अक्ष x-अक्ष
इसका समीकरण y = 0
(iii) नियता x = a का समीकरण x = 2.
(iv) नाभिलंब जीवा की लंबाई = 4a = 8.

प्रश्न 4.
x² = – 16y.
हल:
परवलय का समीकरण x² = – 16y
∴ 4a = 16 या a = 4
नियता
(i) नाभि F (0, – a) के निर्देशांक (0, – 4)
(ii) परवलय अक्ष का समीकरण x = 0.
(iii) नियता y = 0 का समीकरण y = 4.
(iv) नाभिलंब जीवा की लंबाई 4a = 16.

प्रश्न 5.
y² = 10x.
हल:
परवलय का समीकरण y² = 10x (आकृति प्रश्न 1 में देखें)
4a = 10 या a = \(\frac{5}{2}\)
(i) नाभि F (a, 0) के निर्देशांक \(\left(\frac{5}{2}, 0\right)\)
(ii) परवलय का अक्ष : x-अक्ष, समीकरण y = 0
(iii) नियता x = – a का समीकरण x = – \(\left(\frac{5}{2}, 0\right)\)
(iv) नाभिलंब जीवा की लंबाई 4a = 10.

प्रश्न 6.
x² = – 9y.
हल:
परवलय का समीकरण x² = – 9y (आकृति प्रश्न 4 में देखें)
4a = 9 या a = \(\frac{9}{4}\)
(i) नाभि (0, – a) के निर्देशांक \(\left(0,-\frac{9}{4}\right)\)
(ii) परवलय का अक्ष : y-अक्ष, समीकरण x = 0
(iii) नियता y = a का समीकरण y = \(\frac{9}{4}\)
(iv) नाभिलंब जीवा की लंबाई 4a = 9.

निम्नलिखित प्रश्न 7 से 12 तक प्रत्येक में परवलय का समीकरण ज्ञात कीजिए जो दिए प्रतिबंध को संतुष्ट करता है।
प्रश्न 7.
नाभि (6, 0), नियता x = – 6.
हल:
परवलय का अक्ष : x-अक्ष, y = 0

शीर्ष (0, 0) है, नाभि के निर्देशांक (6, 0)
परवलय का अक्ष, धन x-अक्ष के अनुदिश है।
परवलय का समीकरण y² = 24x.

प्रश्न 8.
नाभि (0, – 3), नियता y = 3.
हल:
परवलय का अक्ष y-अक्ष है।
शीर्ष (0, – 3), (0, 3) का मध्य बिन्दु (0, 0) है। नाभि (0, – 3) से स्पष्ट होता है कि परवलय की अक्ष OY के अनुदिश है।

∴ परवलय के समीकरण का रूप x² = – 4ay
यहाँ पर a = 3, ∴ 4a = 12
∴ परवलय का समीकरण x² = – 12y.

प्रश्न 9.
शीर्ष (0, 0), नाभि (3, 0) (आकृति प्रश्न 7 की देखिए)
हल:
परवलय का अक्ष OX के अनुदिश हैं।
∴ परवलय के समीकरण का रूप y² = 4ax
नाभि (3, 0) है। ∴ a = 3
4a = 4 x 3 = 12
∴ परवलय का समीकरण y² = 12x.

प्रश्न 10.
शीर्ष (0, 0), नाभि (-2, 0).
हल:
परवलय का अक्ष OX’ के अनुदिश
नाभि (- 2, 0) है तो a= 2

∴ 4a = 8
परवलय का रूप y² = – 4ax
परवलय का समीकरण y² = – 8x.

प्रश्न 11.
शीर्ष (0, 0), (2, 3) से जाता है और अक्ष, x-अक्ष के अनुदिश है।
हल:
परवलय का शीर्ष (0, 0) है और अक्ष : x-अक्ष है।
∴ परवलय के समीकरण का रूप y² = 4ax
यह बिन्दु (2, 3) से होकर जाता है
∴ 9 = 4a.2 या 4a = \(\frac{9}{2}\)
अतः परवलय का समीकरण y² = \(\frac{9}{2}\)x या 2y² = 9x.

प्रश्न 12.
शीर्ष (0, 0), (5, 2) से जाता है और y-अक्ष के सापेक्ष सममित है।
हल:
शीर्ष (0, 0), परवलय y-अक्ष के सापेक्ष सममित है।
समीकरण का रूप x² = 4ay है। यह बिन्दु (5, 2) से गुजरता है।
∴ 25 = 4a × 2
∴ 4a = \(\frac{25}{2}\)
∴ परवलय का समीकरण, x² = \(\frac{25}{2}\)y या 2x² = 25y

MP Board Class 11th Maths Solutions

MP Board Class 12th Biology Solutions Chapter 14 Ecosystem

Ecosystem NCERT Text Book Questions and Answers

Question 1.
Fill in the Blanks:

1. Plants are called as …………….. because they fix carbon dioxide.
2. In an ecosystem dominated by trees, the pyramid (of numbers) is ………………… type.
3. In aquatic ecosystem, the limiting factor for the productivity is …………………….
4. Common detritivores in our ecosystem are ………………………..
5. The major reservoir of carbon on earth is ………………………
Answer:

1. Producer
2. Inverted
3. Light
4. Earthworm
5. Ocean and Biosphere.

Question 2.
Which one of the following has the largest population in the food chain:
(a) Producers
(b) Primary consumers
(c) Secondary consumers
(d) Decomposers.
Answer:
(d) Decomposers.

Question 3.
The second trophic level in a lake is:
(a) Phytoplankton
(b) Zooplankton
(c) Benthos
(d) Fishes.
Answer:
(b) Zooplankton

Question 4.
Secondary producers are:
(a) Herbivors
(b) Producers
(c) Carnivors
(d) None of the above.
Answer:
(a) Herbivors

Question 5.
What is the percentage of photosynthetically active radiation (PAR) in the incident solar radiation:
(a) 100%
(b) 50%
(c) 1-5%
(d)2 -10%.
Answer:
(b) 50%

Question 6.
Distinguish between:
(a) Grazing food chain and detritus food chain
(b) Production and decomposition
(c) Upright and inverted pyramid
(d) Food chain and Food web
(e) Litter and Detritus
(f) Primary and Secondary productivity.
Answer:
(a) Grazing food chain and detritus food chain

(b) Production and decomposition

(c) Upright and inverted pyramid

(d) Food chain and Food web

(e) Litter and Detritus

(f) Primary and Secondary productivity.

Question 7.
Describe the components of an ecocystem.
Answer:
Ecosystem: The system resulting from the interaction between organisms and their environment is called as ecosystem.

(a) Producers: Organism, which can synthesize their own food are included under producers, e.g., Volvox, Pandorina, Oedogonium, Saggitaria, Utricularia, Azolla, Trapa, Lemna, Typha, Nymphaea etc. form the producer class of the pond ecosystem.

(b) Consumers:

• Primary consumer: Animals, which feed on producers are included into this category e.g., Daphnia, Cyclops, Paramoecium, Amoeba and small fishes.
• Secondary consumers: Primary consumers also serve as food for water sn’akdls, a few tortoise, few types of fish etc. hence, these are carnivores.
• Tertiary consumers: Secondary consumers also serve as food for aquatic birds like kingfisher, cranes, big fish and these together form a top class carnivorous group and called as tertiary consumers.

(c) Decomposers: All producers and consumers die and accumulate on the floor of the pond. Even the waste material and faeces of these animals get accumulated on the floor of the pond. Similarly, the floor of pond is also occupied by decomposers, which include bacteria and fungi. These decomposers decompose complex organic compounds of then- bodies into simpler forms which are finally mixed with soil of floor of ponds. These are again absorbed by the roots of producer plants and thus matter is recycled.

Question 8.
Define ecological pyramids and descirbe with examples, pyramids of number and biomass.
Answer:
Ecological Pyramid : The relation between producers and consumers in a ecosystem can be graphically represented in the form of a pyramid called ecological pyramid. The base always represent the producers or the first trophic level and the apex represents top level consumer or the lost trophic level ecological pyramids are of three types :

1. Pyramid of number
2. Pyramid of biomass
3. Pyramid of energy.

The biomass, i.e., the living weight of the organisms in the food-chain present at different trophic levels in an ecosystem forms the pyramid of biomass.

When biomass of consumers is greater than biomass of producer then pyramid is called as inverted pyramid of biomass.eg., pyramid of biomass of pond ecosysyem is always inverted.

Ecosystem: The system resulting from the interaction between organisms and the environment is called as ecosystem.

Question 9.
What is primary productivity ? Give brief description of factors that affect primary productivity.
Answer:
The rate of biomass production is called productivity.
It is expressed in terms of g^-2yr^-1  or(Kcal-m^-2) yr^-1 to compare the productivity of ecosystems.
It can be divided into Gross Primary Productivity (GPP) and Net Primary Productivity (NFP).

Gross Primary Productivity of an ecosystem is the rate of production of organic matter during photosynthesis. A considerable amount of GPP is utilized by plants in respiration.

Gross primary productivity minus respiration losses (R), is the Net Primary Productivity (NPP). GPP – R=NPP.
Primary productivity depends on :

• The plant species inhabiting a particular area.
• The environmental factors.
• Availability of nutrients.
• Photosynthetic capacity of plants.

Question 10.
Define decomposition and describe the processes and products of decomposition.
Answer:
Decomposition is the process that involves the breakdown of complex organic matter or biomass from the body of dead plants and animals with the help of decomposers into inorganic raw materials such as carbon dioxide, water, and other nutrients.
The various processes involved in decomposition are as follows:
1. Fragmentation: It is the first step in the process of decomposition. It involves the breakdown of detritus into smaller pieces by the action of detritivores such as earthworms.

2. Leaching: It is a process where the water soluble nutrients go down into the soil layers and get locked as unavailable salts.

3. Catabolism: It is a process in which bacteria and fungi degrade detritus through various enzymes into smaller pieces.

4. Humification: The next step is humification which leads to the formation of a dark coloured colloidal substance called humus, which acts as reservoir of nutrients for plants.

5. Mineralization: The humus is further degraded by the action of microbes, which finally leads to the release of inorganic nutrients into the soil. This process of releasing inorganic nutrients from the humus is known as mineralization. Decomposition produces a dark coloured, nutrient-rich substance called humus. Humus finally degrades and releases inorganic raw materials such as CO₂, water, and other nutrient in the soil.

Question 11.
Given account of Explain energy flow in ecosystem.
Answer:
Energy flow: In die ecosystem, energy is transferred in an orderly sequence. The flow of solar energy from producers to consumers and to decomposers subsequently in an ecosystem is known as energy flow. Energy flow is an ecosystem is always unidirectional. Sun is the sole source of solar energy in an ecosystem. Green plants utilize this energy in photosynthesis and convert it in the form of chemical energy and store it. Plants utilize maximum part of this energy to do its biological functions. Some of it is converted into heat and released in the environment. Remaining part of the energy is stored in various components of the body.

When a consumer eats these producer plants, the energy is then transferred into its body. In any food-chain energy flows from primary producers to primary consumers, from primary consumers to secondary consumers and secondary consumers to tertiary consumers and so on. Because every organism of a trophic level continuously converts chemical energy into heat, there is always a loss of energy with each step in a food-chain. According to an estimate only 10% of the total energy obtained is transferred from one trophic level to another.

Question 12.
Write important features of a sedimentary cycle in an ecosystem.
Answer:
Sedimentary cycles have their reservoirs in the earth’s crust or rocks. Nutrient elements are found in the sediments of the earth. Elements such as sulphur, phosphorus, potassium, and calcium have sedimentary cycles.

Sedimentary cycles are very slow. They take a long time to complete their circulation and are considered as less perfect cycles. This is because during recycling, nutrient elements may get locked in the reservoir pool, thereby taking a very long time to come out and continue circulation. Thus, it usually goes out of circulation for a long time.

Question 13.
Outline salient features of carbon cycling in an eosystem.
Answer:
Carbon constitutes 49 percent of dry weight Of an organisms.

• 71 percent of the carbon is found dissolved in oceans which is responsible for its regulations in atmosphere.
• The carbon cycle occurs through atmosphere, oceans and through living and dead organisms.
• It is estimated that 4xl013 kg of Carbon is fixed in the biosphere through photosynthesis annually.
• Carbon is returned to atmosphere as CO₂ by animals and plants through respiration and the activities of decomposers.
• Some amount of fixed carbon is lost as sediments and removed from circulation.
• Burning of wood, forest fire, volcanic activity and combustion of organicjnatter and fossil fuels are some essential sources for releasing CO₂ in the atmosphere.
• Human activities like deforestation and vehicular burning or fossil fuels has caused in increase in the amount of CO₂ in atmosphere.

Ecosystem Other Important Questions and Answers

Ecosystem Objective Type Questions

Choose the Correct Answer

Question 1.
Position of man in food chain of all ecosystem is as :
(a) Consumer
(b) Producer
(c) Producer and consumer
(d) Decomposer.
Answer:
(a) Consumer

Question 2.
Flow of energy in an ecosystem is :
(a) Unidirectional
(b) Bi-directional
(c) Three directional
(d) Quadri-directional.
Answer:
(a) Unidirectional

Question 3.
The word “ Ecosystem” was first used by :
(a) Tansley
(b) Odum
(c)Writier
(d) Mishra and Puri.
Answer:
(a) Tansley

Question 4.
Source of energy in an ecosystem is:
(a) Solar energy
(b) Green plants
(c) Food substances
(d) All of the above.
Answer:
(a) Solar energy

Question 5.
Pyramid of number of trees in an ecosystem is always:
(a) Inverted
(b) Upward
(c) Both (a) and (b)
(d) None of these.
Answer:
(a) Inverted

Question 6.
Correct food chain is:
(a) Grass → Grasshopper → Frog Snake → Hawk
(b) Grass → Frog → Snake → Peacock
(c) Grass → Peacock → Grasshopper → Hawk
(d) Grass → Snake → Rabbit.
Answer:
(a) Grass → Grasshopper → Frog Snake → Hawk

Question 7.
Pyramid of biomass of an ecosystem of lake is:
(a) Upward
(b) Inverted
(c) Inverted and Upward both
(d) None of these.
Answer:
(b) Inverted

Question 8.
Man is:
(a) Producer
(b) Carnivore
(c) Herbivore
(d) Omnivorus.
Answer:
(d) Omnivorus.

Question 9.
Decomposer are:
(a) Rats
(b) Algae
(c) Bacteria and Fungi
(d) Goats.
Answer:
(c) Bacteria and Fungi

Question 10.
Man made ecosystem is:
(a) Pond
(b) Aquarium
(c) Forest
(d) Lake.
Answer:
(b) Aquarium

Question 11.
Pyramid of energy is a forest ecosystem is:
(a) Always Inverted
(b) Always upward
(c) First upward than Inverted
(d) None of the above.
Answer:
(b) Always upward

Question 12.
Study of ecosystem of species is called:
(a) Ecology
(b) Autecology
(c) Synecology
(d) None of these.
Answer:
(b) Autecology

Question 13.
Food chain starts from:
(a) Respiration
(b) Photosynthesis
(c) Decomposition
(d) Nitrogen fixation.
Answer:
(b) Photosynthesis

Question 14.
Flow of energy and food in a food web is:
(a) Unidirectional
(b) B i-directional
(c) Multidirectional
(d) None of these.
Answer:
(c) Multidirectional

Fill in the Blanks:

1. The transitional zone between two adjacent communities is called ………………………
2 The term ecosystem was proposed for the first time by ………………………..
3. The brief source of energy in ecosystem is …………………….
4. Bacteria which fix the nitrogen is called ……………………
5. All ecosystem are dependent of energy on …………………….
6. Balance of ecosystem is called ………………………
7. Pyramid of energy is always …………………………
8. All the plant of a particular geographic area constitute ………………………. of the place.
Answer:

1. Ecotone,
2. Tansley,
3. Sunlight,
4. Rhizobuim,
5. Solar energy,
6. Ecological balance,
7. Upright,
8. Flora.

3. Match the Following:


Answer:

1. (b)
2. (e)
3. (d)
4. (a)
5. (c).

Answer:

1. (b)
2. (e)
3. (d)
4. (a)
5. (c)

Question 4.
Answer in One Word/Sentence:

1. Mention the two name of biotic components of ecosystem.
2. Who proposed the term ecosystem?
3. Which ecosystem is more stable?
4. Name the relatively loss stable ecosystem.
5. Name the ecosystem showing maximum stratification.
6. Give two examples of artificial ecosystem.
7. Give the name of any two food chains.
8. Write the name of basic unit of ecology.
9. Name the organisms found on the surface of water bodies.
10. Name the transitional zone of two adjacent communities.
11. Name the process of establishment of organisms is any new area.
12. From where xerosere started?
13. Who has proposed the term succession?
14. Name the succession started from naked rock. ‘

Answer:

1. Biotic, Aboitic
2. Tansley
3. Ocean
4. Desert
5. Aquatic
6. Aquarium, crop¬land
7. Grazing, Detritus
8. Producer
9. Benthos
10. Ecotone
11. Succession
12. Rocks
13. Hult(1885)
14. Lithosere.

Ecosystem Very Short Answer Type Questions

Question 1.
Name the process of soil formation.
Answer:
Pedogenesis.

Question 2.
What is the orientation of pyramid of energy?
Answer:
Upright.

Question 3.
Which ecosystem is most established?
Answer:
Complex ecosystem.

Question 4.
When many food chain operate simultaneously and interlock such pattern is formed.
Answer:
Food web.

Question 5.
Name the ecosystem which shows most productivity.
Answer:
Tropical ecology.

Question 6.
What are fungi and bacteria called in an ecosystem?
Answer:
Micro consumer or Decomposer.

Question 7.
Which part of the energy is transferred from one trophic level to other in ecosystem?
Answer:
10 %.

Question 8.
Name the type of chemosynthetic bacteria.
Answer:
Autotrophic.

Question 9.
Name the word which is similar to ecosystem given by Pro. R. Mishra.
Answer:
Ecocosm.

Question 10.
Name the trophic level in which green plants are found.
Answer:
Primary trophic level.

Question 11.
Who gave the word transformer for producer?
Answer:
By E. J. kormondy.

Question 12.
Name any two sedimentary cycle.
Answer:

1. Phosphorus cycle
2. Sulpher cycle.

Question 13.
The energy pyramids are always.
Answer:
Upright.

Question 14.
Give examples of decomposers.
Answer:
Bacteria and Fungi.

Question 15.
Who gave 10% rule of energy?
Answer:
Lindeman.

Question 16.
Which form of nitrogen is absorbed by plants?
Answer:
In the form of nitrate ion (No₃^–).

Question 17.
Name two types of food chain.
Answer:

1. Grazing food chain,
2. Detritus food chain.

Ecosystem Short Answer Type Questions

Question 1.
Differentiate between food chain and food web.
Answer:

Question 2.
Write the names of important components of atmosphere.
Answer:
(A) Abiotic components:

• Energy: Light, temperature, energy of chemical substances.
• Materials: Water, soil, salts etc.

(B) Biotic components:

• Producers: e.g., green plants.
• Consumers: Primary, secondary and tertiary.
• Decomposers: Bacteria, fungi.

Question 3.
Write the name and ratio of different components of biosphere.
Answer:
Name and ratio of different components of biosphere is :

Some gases are also found in biosphere e.g., Helium, Neyon and Crypton are found in less quantity.

Question 4.
Differentiate between detritivore and decomposer.
Answer:
Detritivore are organisms which feed on detritus and break them into smaller particles, e.g., earth worm. And decomposer are organisms which by secreting enzyme’s break down complex organic matter into in organic substance e.g. some bacteria and fungi.

Question 5.
Explain consumers of ecosystems.
Answer:

1. Producers: All the green plants.
2. Consumers: Depends on others for food.
   (i) Primary consumer: Depends on plants called herbivores.
   (ii) Secondary consumers: Depends on herbivores for food.
   (iii) Tertiary consumers: Depends on secondary consumers.
3. Decomposers: They decomposed dead organic matter.

Question 6.
Differentiate between biome and ecosystem.
Answer:
An ecosystem is the interaction of living and non-living things in an environment. A biome is a specific geographic area notable for the species living there.

Question 7.
Draw a pyramid of energy of grass land in ecosystem.
Answer:

Question 8.
Explain nitrogen cycle in nature.
Answer:

Question 9.
Explain sulphur cycle by diagram.
Answer:

Question 10.
Explain the effect of light on plants.
Answer:
Effect of light on plants: Light is the source of energy. It is essential for life. It is an important factor of an ecosystem. The existence of life on earth is because of light obtained from sun. Sunlight is essential for photosynthesis, help in preparation of food for the whole living world. Light effects biological activities of plants by its intensity, period and duration. Plants are classified into following two categories on the basis of requirement of light intensity:

• Heliophytes: Plants, which can grow better in bright light are called heliophytes.
• Sciophytes: Plants, which require relatively less of light and they can grow better in shades are called sciophytes.

Question 11.
Explain the meaning of food web and draw its diagram.
Answer:
Food Web: In nature, food-chains are not isolated sequences, but are interrelated and interconnected with one another. When many food-chains operate simultaneously and interlock such pattern is termed as food web. Thus, the food web is a description of feeding connections between the organisms which make up a community. Energy passes through one trophic level to next via these food web links, e.g., a rat feeds on various kinds of grains, fruits, stems, roots, etc.

A rat in its turn is consumed by a snake which is eaten by a falcon. The snakes feed on both frogs and rats. Thus, a network of food-chains exists and this is called food web. The food web gets more complicated because of variability in taste and preference availability and compulsion and several other factors at each level. For example, tigers normally do not feed on fishes or crabs but in Sunderbans they are forced to eat them.

Question 12.
Explain Calcium cycle with well labelled diagram.
Answer:

Question 13.
Draw ecological pyramid of number of a tree ecosystem and grassland ecosystem.
Answer:

Question 14.
What do you mean by ecosystem? Describe the important components of a pond ecosystem.
Or
Write about role of decomposers in an ecosystem with example.
Answer:
Ecosystem: The system resulting from the interaction between organisms and their environment is called as ecosystem.

Components of a pond ecosystem:
A pond ecosystem should contain all components of ecosystems like:

(a) Producers: Organism, which can synthesize their own food are included under producers, e.g., Volvox, Pandorina, Oedogonium, Saggitaria, Utricularia, Azolla, Trapa, Lemna, Typha, Nymphaea etc. form the producer class of the pond ecosystem.

(b) Consumers:

• Primary consumer: Animals, which feed on producers are included into this category e.g., Daphnia, Cyclops, Paramoecium, Amoeba and small fishes.
• Secondary consumers: Primary consumers also serve as food for water sn’akdls, a few tortoise, few types of fish etc. hence, these are carnivores.
• Tertiary consumers : Secondary consumers also serve as food for aquatic birds like kingfisher, cranes, big fish and these together form a top class carnivorous group and called as tertiary consumers.

(c) Decomposers: All producers and consumers die and accumulate on the floor of the pond. Even the waste material and faeces of these animals get accumulated on the floor of the pond. Similarly, the floor of pond is also occupied by decomposers, which include bacteria and fungi. These decomposers decompose complex organic compounds of then- bodies into simpler forms which are finally mixed with soil of floor of ponds. These are again absorbed by the roots of producer plants and thus matter is recycled.

Question 15.
Explain pyramid of biomass of pond ecosystem.
Answer:
The biomass, i.e., the living weight of the organisms in the food-chain present at different trophic levels in an ecosystem forms the pyramid of biomass.

When biomass of consumers is greater than biomass of producer then pyramid is called as inverted pyramid of biomass. eg. pyramid of biomass of pond ecosysyem is always inverted.

Ecosystem: The system resulting from the interaction between organisms and the environment is called as ecosystem.

Ecosystem Long Answer Type Questions

Question 1.
What are ecological pyramids? Explain various types of ecological pyramids.
Answer:
Trophic level: In an ecosystem, the producer consumer arrangement is a kind of structure known as trophic structure and each food level in the food-chain is called as trophic level or energy level. In other, words each level of food in food-chain is called its trophic level. The first trophic level (T,) in an ecosystem is occupied by producers. Herbivores (primary consumers) form second trophic level (T2), secondary consumers form third trophic level (T3), tertiary consumers form fourth trophic level (T4) and decomposers form fifth trophic level (T5) in an ecosystem. .

Food or Ecological Pyramids: If we express the organisms of various trophic levels according to their number, biomass and ratio of energy stores within it, then we obtain ac6ne or pyramid like structure which is known as food or ecological pyramid. Ecological pyramids represent the trophic structure and function of an ecosystem. In base and successive trophic levels the tiers which make up the apex. Ecological pyramids are of the following three types :

1. Pyramid of Biomass
2. Pyramid of number
3. Pyramid of energy.

1. Pyramid of Biomass : Biomass is the dry weight of living organisms per unit of space. The ecological pyramid, which shows the quantitative relationship of the standing crop at each trophic level. The pyramid of biomass shows gradual reduction in biomass at each trophic level from base to apex. The pyramid of biomass may be:

• Upright: e.g., all terrestrial ecosystems.
• Inverted: e.g., all aquatic ecosystems.

2. Pyramid of Number : The ecological pyramid which shows the number of individual organisms at each trophic level. It represents numerical relationship between different trophic level of a food-chain. In this pyramid more abundant species from the first trophic level and from the base of pyramid and the less abundant species remain near the top. The pyramid of number may be:

• Upright: e.g. grassland, pond, forest ecosystem.
• Inverted :e.g. ecosystem of tree.

3. Pyramid of Energy : It indicates the total amount of energy at each trophic level of the food-chain. At each producer level, the total energy available is relatively more than at the higher trophic levels because of the loss of the energy at each trophic level. Thus, there is a gradual loss of energy at each trophic level. The pyramid of energy of each types of ecosystem is always upright.

Question 2.
What is meant by terrestrial biomes? What are its types? Explain any one biomes in detail.
Answer:
Terrestrial Biome : Large area occupying ecosystems in nature are called biomes. If biomes are on land than they are called Terrestrial biomes.
Terrestrial biomes may be:
(a) Forest Biomes : They may be as below;

• Topical rain forest
• Cold tropical forest
• Taiga forest.

(b) Grassland biomes : They may be as below:

• Tropical rain forest
• Cold tropical forest.

(c) Desert Biomes
(d) Tundra Biomes.
Grassland Biome or Ecosystem has long grasses, Its, land is fertile. It receives approximately 25 to 75 cm average rainfall. Its component are:

(a) Abiotic component: All organic, inorganic substances and climatic factors together form abiotic component.

(b) Biotic component :

• Producers: Grasses, herb, shrubs.
• Primary consumers: Herbivore like cow, buffalo, goats, sheep, deer, rabbit, rat insect.
• Secondary consumers: Carnivore animals which eat primary consumers-like snake, birds, foxes, jackal etc.
• Tertiary consumers: These organism which eat secondary consumers because no other one eats them like Hawk, Peacock, etc.
• Decomposers: Micro-Fungus, Bacteria, actinomycetes are decomposer of grassland biomes and recycle the material back to soil and used by producers.

Question 3.
What are Biogeochemical cycles? Write in short sulphur and calcium cycle.
Answer:
All living organisms get matter from the biosphere components i.e„ lithosphere, hydrosphere and atmosphere. Essential elements or inorganic substances are provided by earth and are required by organisms for their body building and metabolism, they are known as Biogeochemicals or biogenetic nutrients.

Sulphur Cycle : Producers (green plants) need sulphur in the form of sulphates from soil or from water (aquatic plants). The animals get sulphur through food. Some animals get sulphur from water also. Sulphur is found in three amino acids hence, sulphur is component of most proteins, some vitamins and enzymes. Plants pick up sulphur in the form of sulphates. They are converted to organic form mostly as component of some amino acids. It is found in nature as element and also as sulphates in soil, water and rocks. After the death of plants and animals, they are decomposed by microbes like Asperigillus, Neurospora and Escherichia releasing hydrogen sulphide.

Calcium cycle : Calcium is slowly released from the rocks by water and wind action. These are either blown into the air or absorbed by plants through their roots. Animals obtain it directly as compounds and also from plants. Calcium is released from plant and animal bodies by decomposition after death. Molluscs and corals deposit a large quantity of calcium in their shells and skeletons making it unavailable for quick cycling.

MP Board Class 12th Biology Solutions

MP Board Class 11th Maths Solutions Chapter 11 शंकु परिच्छेद Ex 11.1

निम्नलिखित प्रश्न 1 से 5 तक प्रत्येक में वृत्त का समीकरण ज्ञात कीजिए :
प्रश्न 1.
केंद्र (0, 2) और त्रिज्या 2 इकाई।
हल:
यहाँ h = 0, k = 2 तथा r = 2 रखने पर,
वृत्त का समीकरण, (x – 0)² + (y – 2)² = 2²
या x² + y² – 4y + 4 =4
अतः वृत्त का अभीष्ट समीकरण, x² + y² – 4y = 0.

प्रश्न 2.
केंद्र (- 2, 3) और त्रिज्या 4 इकाई।
हल:
∴ वृत्त का समीकरण (x + 2)² + (y – 3)² = 4²
या (x² + 4x + 4) + (y² – 6y + 9) = 16
या x² + y² + 4x – 6y – 3 = 0.

प्रश्न 3.
केंद्र \(\left(\frac{1}{2}, \frac{1}{4}\right)\) और त्रिज्या \(\frac{1}{12}\) इकाई।
हल:
यहाँ h = \(\frac{1}{2}\), k = \(\frac{1}{4}\), तथा r = \(\frac{1}{12}\) हो, तब
वृत्त का समीकरण,

प्रश्न 4.
केंद्र (1, 1) और त्रिज्या -2 इकाई।
हल:
यहाँ h = 1, k = 1 तथा r = \(\sqrt{2}\) हों, तब
वृत्त का समीकरण,
(x – 1)² + (y – 1)² = (\(\sqrt{2}\))²
या (x² – 2x + 1) + (y² – 2y + 1) = 2
या x² + y² – 2x – 2y= 0.

प्रश्न 5.
केंद्र (- a, – b) और त्रिज्या \(\sqrt{a^{2}-b^{2}}\) इकाई।
हल:
वृत्त का समीकरण,
(x + a)² + (y + b)² = \((\sqrt{a^{2}-b^{2}})\)²
या x² + 2ax + a² + y² + 2by + b² = a² – b².
या x² +y² + 2ax + 2by + 2b² = 0.

निम्नलिखित प्रश्न 6 से 9 तक में प्रत्येक वृत्त का केन्द्र और त्रिज्या ज्ञात कीजिए :
प्रश्न 6.
(x + 5)² + (y – 3)² = 36.
हल:
वृत्त (x + 5)² + (y – 3)² = 36 की (x – h)² + (y – k)² = r² से तुलना करने पर,
– h = 5, – k = – 3, r² = 36
∴ h = – 5, k = 3, r= 6
∴ केन्द्र (- 5, 3), त्रिज्या = 6.

प्रश्न 7.
x² +y² – 4x – 8y – 45 = 0.
हल:
(x² – 4x) + (y² – 8y) = 45
या (x² – 4x + 4) + (y² – 8y + 16) = 45 + 4 + 16 = 65
(x – 2)² + (y – 4)² = 65
∴ h = 2, k = 4, r = \(\sqrt{65}\)
⇒ केन्द्र (2, 4) और त्रिज्या = \(\sqrt{65}\)
दूसरी विधि : 2g = – 4, 2f= – 8, c = – 45
g = – 2, f = – 4, r = \(\sqrt{g^{2}+f^{2}-c}\)3
r = \(\sqrt{4+16+45}\)
= \(\sqrt{65}\)
केन्द्र (- g, – f) अर्थात् (2, 4)
त्रिज्या = r = \(\sqrt{65}\).

प्रश्न 8.
x² + y² – 8x + 10y – 12 = 0.
हल:
(x² – 8x) + (y² + 10y) = 12
या (x² – 8x + 16) + (y² + 10y + 25) = 12 + 16 + 25
(x – 4)² + (y + 5)² = 53.
∴ केन्द्र (4, – 5), त्रिज्या= \(\sqrt{53}\).

प्रश्न 9. 2x² + 2y² – x= 0.
हल:

प्रश्न 10.
बिन्दुओं (4, 1) और (6, 5) से जाने वाले वृत्त का समीकरण ज्ञात कीजिए जिसका केन्द्र रेखा 4x + y = 16 पर स्थित है।
हल:
वृत्त का व्यापक समीकरण
x² + y² + 2gx + 2fy + c = 0
बिन्दु (4, 1) इस पर स्थित है
∴ 16 + 1 + 8g + 2f + c = 0
∴ 8g + 2f + c = – 17 …..(1)
बिन्दु (6, 5) वृत्त पर स्थित है
∴ 36 + 25 + 12g + 10f + c = 0
∴ 12g + 10f + c = – 61 …..(2)
केंद्र (- g, – 1) रेखा 4x + y = 16 पर स्थित है
∴ – 4g – f = 16
या 4g + f = – 16 …..(3)
समीकरण (1) को (2) में से घटाने पर
4g + 8f = – 44
समीकरण (3) को (4) में से घटाने पर
7f = – 44 + 16 = – 28
f = – 4
समीकरण (3) में f का मान रखने पर
4g – 4 = – 16 या 4g = – 12
∴ g = – 3
f और g का मान समी (1) में रखने पर
– 24 – 8 + c = – 17
c = 32 – 17 = 15
अतः वृत्त का समीकरण
x² + y² – 6x – 8y + 15 = 0.

प्रश्न 11.
बिन्दुओं (2, 3) और (- 1, 1) से जाने वाले वृत्त का समीकरण ज्ञात कीजिए जिसका केंद्र रेखा x – 3y – 11 = 0 पर स्थित है।
हल:
मान लीजिए वृत्त का समीकरण
x² + y² + 2gx + 2fy + c = 0 ….(1)
इस पर बिन्दु (2, 3) स्थित है।
∴ 4 + 9 + 4g + 6f + c = 0
या 4g + 6f + c = – 13 ……(2)
इसी प्रकार (- 1, 1) भी वृत्त (1) पर स्थित है।
जब p = – 2, वृत्त का समीकरण
(x + 2)² + y² = 25
या x² + y² + 4x – 21 = 0
जब p = 6, वृत्त का समीकरण
(x – 6)² + y² = 25
x² + y² – 12x + 36 – 25 = 0
या x² + y² – 12x + 11 = 0
∴ वृत्त के अभीष्ट समीकरण
x² + y² + 4x – 21 = 0 और x² + p² – 12x + 11 = 0^π

प्रश्न 13.
(0, 0) से होकर जाने वाले वृत्त का समीकरण ज्ञात कीजिए जो निर्देशांक्षों पर a और b अंत: खण्ड बनाता है।
हल:
वृत्त मूल बिन्दु से होकर जाता है और अक्षों पर अंत:खण्ड a, b बनाता है।
OA = a, ∴ A के निर्देशांक (a, 0)
OB = b, ∴ B के निर्देशांक (0, b)

या x² + y² – ax – by = \(\frac{a^{2}+b^{2}}{4}-\frac{a^{2}+b^{2}}{4}\)
∴ वृत्त का अभीष्ट समीकरण
x² + y² – ax – by = 0.

प्रश्न 14.
उस वृत्त का समीकरण ज्ञात कीजिए जिसका केंद्र (2, 2) हो तथा (4, 5) से जाता है।
हल:
वृत्त की त्रिज्या = केंद्र (2, 2) और बिन्दु (4, 5) के बीच की दूरी

प्रश्न 15.
क्या बिन्दु (- 2.5, 3.5) वृत्त x² + y² = 25 के अंदर, बाहर या वृत्त पर स्थित है।
हल:
वृत्त का केंद्र O(0, 0) है।
दिया हुआ बिन्दु P(- 2.5, 3.5) है।
OP= \(\sqrt{(-2.5)^{2}+(3.5)^{2}}\)
= \(\sqrt{6.25+12.25}\)
= \(\sqrt{18.50}\)
= 4.25 (लगभग)
यह त्रिज्या जो 5 इकाई से कम है
अतः बिन्दु (- 2.5, 3.5) वृत्त के अंदर स्थित होगा।

MP Board Class 11th Maths Solutions

MP Board Class 12th Biology Solutions Chapter 4 Reproductive Health

Reproductive Health NCERT Textbook Questions and Answers

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

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

Question 2.
Suggest the aspects of reproductive health which need to be given special attention in the present scenario.
Answer:
Special attention need to be given to the following aspect :

• Introduction of sex education in school that to helps in eradicating myths and mis-conceptions regarding sex-related aspects.
• Proper information about reproductive organs, safe and hygienic sexual practices and sexually transmitted diseases.
• Awareness of problems due to uncontrolled population growth, social evils like sex abuse and sex-related crimes etc.
• Strong infrastructural facilities, professional expertise and material support to provide medical assistance and care to people in reproduction related problems.
• Educating people about available birth control options, care of pregnant mothers, post-natal care of mother and child, importance of breast feeding, equal opportunities for the male and female child.

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

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

Question 4.
Do you think that reproductive health in our country has improved in the past 50 year’s ? If yes, mention some such areas of improvement
Answer:
Yes, Reproductive health in our country has improved in the last 50 year’s.

Some areas of improvement are :

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

Question 5.
What are the suggested reasons of population explosion?
Answer:
The situation where population exceeds productive capacity is known as population explosion. In nature, the amount of resources are limited hence, if the population increases in the present rate and increasing beyond the limit, they (Resources) would get exhausted.
Reasons of population explosion : Following are the reasons of population explosion:

• Increasing birth rate.
• Decreasing death rate.
• Higher rate of reproduction.
• Medical services have brought down mortality due to fatal diseases and epidemics.
• Lack of predator in the civilized world today, the only predator of man is man himself.

Question 6.
Is the use of contraceptives justified ? Give reasons.
Answer:
Yes, use of contraceptive is justified because it helps to control the rapid growth of human population. It will also help in preventing unwanted pregnancies and STDs. Contraceptive also help in controlling the population growth rate.

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

Question 8.
Amniocentesis for sex determination is banned in our country. Is this ban necessary? Comment
Answer:
Yes, the ban is necessary because amniocentesis is misused for determining the sex of the fetus and then aborting the child of it is a female. This process is illegal it causes harm to the fetus as well as mother it can also disturb the sex ratio.

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

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

(i) Zygote Intrafallopian Transfer (ZlFT) : The zygote or early embryo with up to 8 blastomeres is transferred into the fallopian tube.

(ii) Intra Uterine Transfer (IUT) : Embryo with more than 8 blastomeres is transferred into the uterus in females who cannot conceive embryos formed by fusion of gametes in another female are transfered.

(iii) Test tube baby : In this method, ova from the donor (female) and sperm from the donor (male) are collected and are induced to form zygote under stimulated conditions in the laboratory. The zygote could then be transferred into the fallopian tube and embryos transferred into the uterus, to complete its further development. The child born from this method is called test tube baby.

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

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

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

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

Question 10.
What are the measures one has to take to prevent from contracting STDs ?
Answer:
STDs can be prevented by the following methods :

• Avoid sex with unknown partners/multiple partners.
• Always use condoms during coitus.
• Always contact a qualified doctor for any doubt in early stage of infection and get complete treatment if diagnosed with disease.

Question 11.
State True/False with explanation.
(a) Abortions could happen spontaneously too. (True/False)
Answer:
False, Abortion does not happen under normal conditions. It happens accidently or under the will of Parents.

(b) Infertility is defined as the inability to produce a viable offspring and is always due to abnormalities/defects in the female partner. (True/False)
Answer:
False, Sterility always does not occur due to female sometimes. Males are also responsible for this.

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

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

Question 12.
Correct the following statements :
(a) Surgical methods of contraception prevent gamete formation.
Answer:
It prevents the transportation of gametes not their formation.

(b) All sexually transmitted diseases are completely curable.
Answer:
Hepatitis-Band AIDS are not curable.

(c) Oral pills are very popular contraceptives among the rural women.
Answer:
Oral pills are not popular among rural women. They require sex education.

(d) In E.T. techniques, embryos are always transfered into the uterus.
Answer:
In E.T. techniques 8-called blastomere is transfered into the fallopian tube. While more than 8-celled blastomere is transfered into the uterus.

Reproductive Health Other Important Questions and Answers

Reproductive Health Objective Type Questions

1. Choose the Correct Answer
Question 1.
Removal of Fallopian tube in females are called :
(a) Vasectomy
(b) Tubectomy
(c) Ovariectomy
(d) Castration (Gonadectomy).
Answer:
(b) Tubectomy

Question 2.
Main region ofthe population growth are :
(a) Less death rate
(b) Growth of birth rate
(c) Drought
(d) Less war.
Answer:
(a) Less death rate

Question 3.
The causes of popülatîon explosion in large cities is :
(a) Chances of education
(b) Available facilities
(c) Sources of income
(d) All of these.
Answer:
(d) All of these.

Question 4.
Population density is more in :
(a) USA
(b) India
(c) China
(d) Japan.
Answer:
(c) China

Question 5.
AIDS disease caused by :
(a) From bacteria
(b) From protozoa
(c) From virus
(d) From fungus.
Answer:
(c) From virus

Question 6.
Sex transmitted diseases are caused by :
(a) Virus
(b) Bacteria
(c) Protozoa
(d) All of these.
Answer:
(a) Virus

Question 7.
Population calculated in India :
(a) In 1891
(b) In 1947
(c) In 1950
(d) In 1961.
Answer:
(a) In 1891

Question 8.
The technique of the control of birth rate is called :
(a) TUD
(b) GIFT
(c) MIF
(d) IVEET.
Answer:
(a) TUD

Question 9.
Alcohol affected organ is :
(a) Liver
(b) Cerebrum
(c) Cerebellum
(d) Heart
Answer:
(a) Liver

Question 10.
Population growth rate of world is :
(a) 2-4%
(b) 2%
(c) 3%
(d) 4%.
Answer:
(b) 2%

2. Fill in the Blanks :

1. Statistical study of population is called ……………….
2. According to MALTHUS, Population grows ……………….whereas the means of its subsistences grows ……………….
3. The entry and exit of member in any population is called ……………….
4. Cutting and ligating ends of segments of vas deferens is called ……………….
5. The population density of terrestrial animals, the area is represented by ……………….
6. Birth rate and migration ………………. the population density.
7. Revised birth fate is always greater than ………………. birthrate.
8. Vital index = ………………. /Mortality x 100.
9. Population ………………. with time and place.
10. From the population point of view, every ………………. person in the world, is an Indian.

Answer:

1. 1. Demography
   2. Geometrically
   3. Migration
   4. Vasectomy
   5. Meter2 (m2)
   6. Growth
   7. Unrevised
   8. Natality
   9. Changes
   10. Sixth.

3. Match the Following :

Answer:

1. (c)
2. (e)
3. (a)
4. (b)
5. (d).

Answer:

1. (c)
2. (d)
3. (e)
4. (b)
5. (a).

III. Answer in One Word/Sentence:

1. Write two such causes by which unchecked growth takes place. (SSCE 1992)
2. Write the full name of IUCD. (SSCE 1992!)
3. What we called the study of human population?
4. What type of chemical contraceptive is found in Mala-D and N?

Answer:

1. Early marriage, illiteracy
2. Intra-Uterine Contraceptive Devices
3. Demography
4. Oestrogen and Progesteron hormone.

Reproductive Health Very Short Answer Type Questions

Question 1.
What is the other name of family planning?
Answer:
The other name of family planning is “Parivar Kalyan”.

Question 2.
Where is Central Drug Research Institute located ?
Answer:
Central Drug Research Institute (CDRI) is located in Lucknow (Uttar Pradesh).

Question 3.
Write one benefit of condom.
Answer:
Condom provided protection from Sexually Transmitted Diseases (STDs).

Question 4.
What is the name of contraceptive pills which is taken only in once in a week ?
Answer:
“Saheli.”

Question 5.
Write the full form of IUCD.
Answer:
Intra Uterine Contraceptive Device.

Question 6.
Write the full form of STDs.
Answer:
Sexually Transmitted Disease.

Question 7.
What is the legal age for marriage of male and female in India ?
Answer:
For male 21 Year and for female 18 years.

Question 8.
Write the name of two STDs which are spread through infected blood.
Answer:
AIDS, Hepatitis-B.

Question 9.
Write the name of two disease which are caused by sexual contact
Answer:
Gonorrhoea, Syphilis.

Question 10.
Write the full form of HIV and AIDS.
Answer:
HIV : Human Immunodeficiency Virus.
AIDS : Acquired Immuno Deficiency Syndrome.

Question 11.
Write the full form of ZIFT.
Answer:
Zygote Intra Fallopian Transfer.

Question 12.
What is the process for surgical sterilization in males called ?
Answer:
Vasectomy.

Reproductive Health Short Answer Type Questions

Question 1.
What is test-tube baby ?
Answer:
In some cases, a woman is unable to have a normal fertilization to bear the child. In such cases, test-tube technique may be successful. In this technique, the unfertilized eggs of such woman is isolated in aseptic condition and fertilized it in test-tube by the sperms of her husband. The fertilized egg or blastocyst can be maintained in vitro till it gets 32 celled stage. It can be implanted in the uterus of the female. The female remains under the supervision of doctor till completion of gestation period of 280 days. The baby produced in such a way is called a test-tube baby.

Question 2.
What is GIFT ? Explain in brief.
Answer:
GIFT (Gametes Intra Fallopian Transfer): It is a latest technique to produce a child. In this technique, the gametes are kept separately in a catheter and injected directly into the fallopian tube of the woman using laproscopy. Thus, in this case, fertilization occurs inside the body of woman. Prior to injection of gametes, the mother would be given hormones for about a week to stimulate follicle formation. This causes development of several eggs i.e,, super ovulation. The first GIFT in India was done in Mumbai and twins were born in August, 1990.

Question 3.
Explain the following:

1. Tubectomy
2. Vasectomy.

Answer:

1. Tubectomy : The surgical removal of a segment of oviduct and then ligating the cut end is called tubectomy. It is applied in females to check the pregnancy.
2. Vasectomy : Surgical cutting and ligating ends of segments of vas deferens is called vasectomy.

Question 4.
What is amniocentesis ? Write its ill effects and two advantages.
Answer:
Amniocentesis : It is prenatal diagnostic technique in which amniotic fluid of uterus is isolated by a surgical needle and foetus cells are cultured on a culture medium and chromosomes are examined. This technique is used to understand the following things or Advantages :

• Chromosomal abnormalities like that of Down’s syndrome, Philadelphia syndrome and Edward’s syndrome.
• Metabolic disorders like that of PKV, Cretinism, Alkaptonuria, etc.
• Sex of the embryo can be examined by this technique.
• Effects of amniocentesis: Due to this type of test, the female embryos are being eradicated. This leads into the
• decrease in number of females which may cause a serious problem.

Question 5.
Explain the social causes of human population growth.
Answer:
Social causes of human population growth rate are:

• Illiteracy in society.
• Low level of society.
• Various orthodox tradition like son leads to progeny.
• Social barriers.
• Early child marriage presuming that after mid-age marriage reproductive capacity degenerates.
• Social backwardness.
• Social set-up that “Putra Ratan se Moksh milta hai”.

MP Board Class 12th Biology Solutions

MP Board Class 11th Maths Solutions Chapter 10 सरल रेखाएँ विविध प्रश्नावली

प्रश्न 1.
k के मान । ज्ञात कीजिए जब कि रेखा (k – 3)x – (4 – k²)y + K² – 7k + 6 = 0
(a) x-अक्ष के समान्तर है।
(b) y-अक्ष के समान्तर है।
(c) मूल बिन्दु से जाती है।
हल:
(i) x-अक्ष के समान्तर y = a
∴ प्रश्न में दिए गए समीकरण में x का गुणांक = 0 या k – 3 = 0 अर्थात् k = 3.
(ii) xy-अक्ष के समान्तर रेखा x =q
दिए गए समीकरण में y का गुणांक = 0 या 4 – k² = 0 अर्थात् k = ± 2.
(iii) यदि रेखा मूल बिन्दु से जाती है तो (0, 0) इसके समीकरण को संतुष्ट करेगा।
0 – 0 + K² – 7k + 6 = 0 या (k – 6) (k – 1) = 0 या k = 1, 6.

प्रश्न 2.
θ और p के मान ज्ञात कीजिए यदि समीकरण x cos θ + y sin θ = p रेखा \(\sqrt{3}\)x + y + 2 = 0 का लम्ब रूप है।
हल:

प्रश्न 3.
उन रेखाओं के समीकरण ज्ञात कीजिए जिनके अक्षों से कटे अंतःखण्डों का योग और गुणनफल क्रमशः 1 और – 6 हैं।
हल:
मान लीजिए अक्षों पर कटे अतः खण्ड a और b हैं।
दिया है : a + b = 1, ab = – 6
b = 1 – a
∴ a(1 – a) = – 6
या a – a₂ = – 6
a₂ – a – 6 = 0
या (a – 3) (a + 2) = 0
∴ a = 3, – 2
∴ b = – 2, 3
3, – 2 अंत:खण्ड वाली रेखा का समीकरण,
\(\frac{x}{3}+\frac{y}{-2}\) = 1
या 2x – 3y = 6.
और – 2, 3 अंत:खण्ड वाली रेखा का समीकरण,
\(\frac{x}{-2}+\frac{y}{3}\) = 1
या – 3x + 2y = 6.

प्रश्न 4.
y-अक्ष पर कौन से बिन्दु ऐसे हैं, जिनकी रेखा \(\frac{x}{3}+\frac{y}{4}\) = 1 से दूरी 4 इकाई है।
हल:
मान लीजिए y-अक्ष पर बिन्दु (0, y₁) है।
(0, y₁) की रेखा 4x + 3y = 12 से दूरी

प्रश्न 5.
मूल बिन्दु से बिन्दुओं (cos θ, sin θ) और (cos ϕ, sin ϕ) को मिलाने वाली रेखा की लांबिक दूरी ज्ञात कीजिए।
हल:
(cos θ, sin θ), (cos ϕ, sin ϕ) को मिलाने वाली रेखा का समीकरण,

प्रश्न 6.
रेखाओं x – 7y + 5 = 0 और 3x + y = 0 के प्रतिच्छेद बिन्दु से खीची गई और y-अक्ष के समान्तर रेखा का समीकरण ज्ञात कीजिए।
हल:
दी गयी रेखाएँ
x – 7y + 5 = 0 …(1)
3x + y = 0 …(2)
समी (2) से,
y = – 3x
y का मान समी (1) में रखने पर,
x – 7(- 3x) + 5 = 0
या 22x + 5 = 0
x = \(\frac{-5}{22}\)
अब y = – 3x = – \(3\left(\frac{-5}{22}\right)\) = \(\frac{15}{22}\)
वह रेखा का जो \(\left(-\frac{5}{22}, \frac{15}{22}\right)\) से जाती है और y-अक्ष के समान्तर है, समीकरण x = – \(\frac{5}{22}\) या 22x + 5 = 0.

प्रश्न 7.
रेखा \(\frac{x}{4}+\frac{y}{6}\) = 1 पर लंब उस बिन्दु से खींची गई रेखा का समीकरण ज्ञात कीजिए जहाँ यह रेखा y-अक्ष से मिलती है।
हल:
रेखा AB का समीकरण,
\(\frac{x}{4}+\frac{y}{6}\) = 1 या 3x + 2y = 12,
2y = – 3x + 12
y = \(\frac{-3}{2} x+\frac{12}{2}\)

∵ रेखा y-अक्ष पर प्रतिच्छेद करती है, इसलिए बिन्दु B(0, 6) है।
∴ BC रेखा का समीकरण
y – 6 = \(\frac{2}{3}\)(x – 0)
या 3y – 18 = 2x
या 2x – 3y + 18 = 0.

प्रश्न 8.
रेखाओं y – x = 0, x + y = 0, और x – k = 0 से बने त्रिभुज का क्षेत्रफल ज्ञात कीजिए।
हल:
y – x = 0 और y + x = 0 बिन्दु (0, 0) पर मिलते हैं।
x = k को y – x = 0 में रखने से, y – k = 0 या y = k
x – k = 0 और y – x = 0 बिन्दु (k, k) पर मिलते हैं।
x = k को y + x = 0 में रखने से,
y + k = 0 या y = – k
x = k और y + x = 0 बिन्दु (k, – k) पर मिलते हैं।
अब बिन्दु (0, 0), (k, k) और (k, – k) से बने त्रिभुज का क्षेत्रफल
= \(\left|\frac{1}{2}[0 .(-2 k)+k(-k)+k(-k)]\right|\)
= \(\left|\frac{1}{2}\left(-k^{2}-k^{2}\right)\right|\)
= k₂ वर्ग इकाई।

दूसरी विधि : त्रिभुज OPQ का क्षेत्रफल
= 2 × क्षेत्रफल ∆OAP
= 2 × [\(\frac{1}{2}\) × k × k] = k₂ वर्ग इकाई।

प्रश्न 9.
p का मान ज्ञात कीजिए जिससे तीन रेखाएँ 3x + y – 2 = 0, px + 2y – 3 = 0 और 2x – y – 3 = 0 एक बिन्दु पर प्रतिच्छेद करें।
हल:
दी गयी रेखाएँ
3x + y = 2 …(1)
2x – y = 3 …(2)
समी (1) और (2) को जोड़ने पर,
5x = 5 या x = 1
∴ y = 2 – 3x = 2 – 3 = – 1
∴ समी (1) और (2) वाली रेखाएँ बिन्दु (1, – 1) पर प्रतिच्छेद करती हैं।
तीसरी रेखा px + 2y – 3 = 0 भी इसी बिन्दु से जाती है इसलिए (1, – 1) इस समीकरण को संतुष्ट करेगा।
p × 1 + 2 × (- 1) – 3 = 0
p – 2 – 3 = 0
∴ p = 5
अतः दी गयी रेखाएँ संगामी हैं यदि p = 5.

प्रश्न 10.
यदि तीन रेखाएँ जिनके समीकरण y = m₁x + c₁, y = m₂x + c₂ और y =m₃x + c₃ हैं, संगामी हैं, तो दिखाइए कि m₁(c₂ – c₃) + m₂(c₃ – c₁) + m₃(c₁ – c₂) = 0.
हल:

प्रश्न 11.
बिन्दु (3, 2) से जाने वाली उस रेखा का समीकरण ज्ञात कीजिए जो रेखा x – 2y = 3 से 45° का कोण बनाती है।
हल:
माना रेखा AB का समीकरण : x – 2y = 3
या y = \(\frac{1}{2}\)x – 3
तब रेखा AB की ढाल = \(\frac{1}{2}\)

+ve चिन्ह लेने पर, 1 = \(\frac{2 m-1}{m+2}\)
या 2m – 1 = m + 2
∴ m = 3
2m -1
– ve चिन्ह लेने पर, – 1 = \(\frac{2 m-1}{m+2}\)
या 2m – 1 = – m – 2
∴ 3m = – 1
या m = \(\frac{-1}{3}\)
अतः रेखा PA का समीकरण जहाँ बिन्दु P = (3, 2) हो और m = \(\frac{-1}{3}\) हो।
y – 2 = – \(\frac{-1}{3}\)(x – 3)
3y – 6 = – x + 3
या x + 3y – 9 = 0
अब जब ढाल m = 3 हो, तब बिन्दु P(3, 2) से रेखा का समीकरण,
y – 2 = 3(x – 3)
y – 2 = 3x – 9
या 3x – y – 7 = 0.

प्रश्न 12.
रेखाओं 4x + 7y – 3 = 0 और 2x – 3y + 1 = 0 के प्रतिच्छेद बिन्दु से जाने वाली रेखा का समीकरण ज्ञात कीजिए जो अक्षों से समान अंतः खण्ड बनाती हैं।
हल:
4x + 7y = 3 …(1)
2x – 3y = – 1 …(2)
समी (2) को 2 से गुणा करने पर,
4x – 6y = – 2 …(3)
समी (3) को (1) में से घटाने पर
13y = 5
∴ y = \(\frac{5}{13}\)

जो रेखा अक्षों पर बने अंतः खण्ड समान हैं तो वह धन x-अक्ष के साथ 45° या 135° का कोण बनाती हैं। इसलिए उसकी ढाल ±1 होगी।
∴ PA और PB रेखाओं के समीकरण
y – y₁ = m(x – x₁)
(i) जब m = 1 हो, तब y – \(\frac{5}{13}\) = 1 \(\left(x-\frac{1}{13}\right)\)
या 13y – 5 = 13 x – 1 या 13x – 13y + 4 = 0.
(i) जब m = – 1 हो, तब y – \(\frac{5}{13}\) = 1 \(\left(x-\frac{1}{13}\right)\)
13y – 5 = – 13x + 1
∴ 13x + 13y – 6 = 0.

प्रश्न 13.
दर्शाइए कि मूल बिन्दु से जाने वाली और रेखा y = mx + c से θ कोण बनाने वाली उस रेखा का समीकरण \(\frac{y}{x}=\pm \frac{m+\tan \theta}{1-m \tan \theta}\) हैं।
हल:
रेखा PA का समीकरण y = mx + c है
यह रेखा OP के साथ कोण θ बनाती है।
रेखा PA की ढाल = m
मान लीजिए OP की ढाल = m₁ है।


या \(\frac{y}{x}\) = m₁
∴ अभीष्ट रेखाओं का समीकरण
\(\frac{y}{x}\) = \(\frac{m \pm \tan \theta}{1 \mp m \tan \theta}\)

प्रश्न 14.
(- 1, 1) और (5, 7) को मिलाने वाली रेखाखण्ड को रेखा x + y = 4 किस अनुपात में विभाजित करती है ?
हल:
मान लीजिए बिन्दु P रेखाखण्ड AB को k : 1 के अनुपात में विभाजित करता है। जबकि A और B के क्रमशः (- 1, 1) और (5, 7) निर्देशांक हैं।

अतः बिन्दु P रेखाखण्ड AB को 1 : 2 के अनुपात में विभाजित करता है। .

प्रश्न 15.
बिन्दु (1, 2) से रेखा 4x + 7y + 5 = 0 की 2x – y = 0 के अनुदिश दूरी ज्ञात करो।
हल:
माना रेखा PC का समीकरण, 2x – y = 0 जिस पर बिन्दु P(1, 2) स्थित है।

रेखा AB का समीकरण 4x + 7y + 5 = 0 और 2x – y = 0 को हल करने पर,
∴ x = \(\frac{-5}{18}\)
और y = \(-\frac{5}{9}\)

प्रश्न 16.
बिन्दु (- 1, 2) से खींची जा सकने वाली उस रेखा की दिशा ज्ञात कीजिए जिसका रेखा x + y = 4 से प्रतिच्छेदन बिन्दु दिए बिन्दु से 3 इकाई की दूरी पर है।
हल:
मान लीजिए अभीष्ट रेखा PQ की ढाल m है
रेखा PQ जो बिन्दु P(- 1, 2) से होकर जाती है और ढाल m है, का समीकरण
y – y₁ = m(x – x₁)
y – 2 = m(x + 1)
या mx – y + m + 2 = 0 …(1)

रेखा AB का समीकरण x+ y = 4
∴ y = 4 – x
y का मान समी (1) में रखने पर,
mx – (4 – x) + m + 2 = 0
या (m + 1) x + m – 2 = 0
∴ x = – \(\frac{m-2}{m+1}\)
अब y = 4 – x = 4 + \(\frac{m-2}{m+1}\)

अतः रेखा PQ की ढाल 0 है अर्थात् रेखा x-अक्ष के समांतर है।

प्रश्न 17.
समकोण त्रिभुज के कर्ण के अंत्य बिन्दु (1, 3) और (- 4, 1) हैं। त्रिभुज के पाद (leg) (समकोणीय भुजाओ) का एक समीकरण ज्ञात कीजिए।
हल:
माना त्रिभुज ABC एक समकोणीय त्रिभुज है जिसका कर्ण AB है। A और B के निर्देशांक क्रमशः (1, 3) और (- 4, 1) हैं।
मान लीजिए BC की ढाल m है।

AC ⊥ BC
∴ AC की ढाल = – \(\frac{1}{m}\)
रेखा BC का समीकरण,
y – y₁ = m(x – x₁)
y – 1 = m(x + 4)
या mx – y + 4m + 1 = 0 …(1)
रेखा AC का समीकरण
y – 3 = – \(\frac{1}{m}\)(x – 1)
या my – 3m = – x + 1
या x + my – 3m – 1 = 0 …(2)
यह दोनों रेखाएँ m के दिए मान से इन का समीकरण ज्ञात कर सकते हैं। यदि BC भुजा x-अक्ष के समांतर हो तो m = 0.
BC का समीकरण, y – 1 = 0
या y = 1
∴ AC, y-अक्ष के समांतर हो और यह A(1, 3) से जाती है। अतः AC का समीकरण x = 1
अत: BC और AC के समीकरण y = 1 और x = 1 हैं।

प्रश्न 18.
किसी बिन्दु के लिए रेखा को दर्पण मानते हुए बिन्दु (3, 8) का रेखा x + 3y = 7 में प्रतिबिंब ज्ञात कीजिए।
हल:
माना रेखा AB का समीकरण x + 3y = 7 है और बिन्दु P के निर्देशांक (3, 8) हैं।
y = – \(\frac{1}{3} x +\frac{7}{3}\)
बिन्दु P का प्रतिबिंब Q होगा यदि PQ ⊥ AB, PQ और AB बिन्दु M पर इस प्रकार प्रतिच्छेद करते हैं कि
PM = QM

रेखा AB की ढाल = – \(\frac{1}{3}\)
और PQ की ढाल = 3
∴ PQ रेखा का समीकरण,
y – 8 = 3(x – 3)
= 3x – 9
या 3x – y = 1 …(1)
AB का समीकरण x + 3y = 7 …(2)
समी (1) को 3 से गुणा करके समी (2) में जोड़ने पर,
10x = 10 या x = 1
समी (1) से y = 3x – 1 = 3 – 1 = 2
∴ बिन्दु M के निर्देशांक (1, 2) हैं।
मान लीजिए Q के निर्देशांक (x₁, y₁) हैं
बिन्दु M रेखाखण्ड PQ का मध्य बिन्दु है
∴ जबकि P(3, 8) है।
∴ \(\frac{x_{1}+3}{2}\) = 1 या x₁ = – 1
\(\frac{y_{1}+8}{2}\) = 2 या y₁ = – 4
∴ P का प्रतिबिंब (- 1, – 4) हैं।

प्रश्न 19.
यदि रेखाएँ y = 3x + 1 और 2y = x + 3, रेखा y = mx + 4 पर समान रूप से आनत हो तो m का मान ज्ञात कीजिए।
हल:
रेखा AB का समीकरण, y = 3x + 1 की ढाल = 3
रेखा BC का समीकरण, y = mx + 4 को ढाल = m


अतः m का अभीष्ट मान = \(\frac{1 \pm 5 \sqrt{2}}{7}\).

प्रश्न 20.
यदि एक वर बिन्दु P(x, y) की रेखाओं x + y – 5 = 0 और 3x – 2y + 7 = 0 से लांबिक दूरियों का योग सदैव 10 रहे तो दर्शाइए कि P अनिवार्य रूप से एक रेखा पर गमन करता है।
हल:
P(x, y) से रेखा x + y – 5 = 0 की दूरी
= \(\frac{x+y-5}{\sqrt{2}}\)
P(x, y) से रेखा 3x – 2y + 7 = 0 की दूरी
= \(\frac{3 x-2 y+7}{\sqrt{9+4}}\)
दोनों दूरियों का योग = 10 (दिया है)
∴ \(\sqrt{13}\)(x + y – 5) + \(\sqrt{2}\)(3x – 2y + 7) = 10\(\sqrt{26}\)
या \((\sqrt{13}+3 \sqrt{2}) x+(\sqrt{13}-2 \sqrt{2}) y-5 \sqrt{13}+7 \sqrt{2}-10 \sqrt{26}\) = 0
जो कि एक सरल रेखा का समीकरण है।
अत: P एक अनिवार्य रूप से एक रेखा पर गमन करता है।

प्रश्न 21.
समांतर रखाओं 9x + 6y – 7 = 0 और 3x + 2y + 6 = 0 से समदूरस्थ रेखा का समीकरण ज्ञात कीजिए।
हल:
दी गयी समांतर रेखाएँ
9x + 6y – 7 = 0 …(1)
और 3x + 2y + 6 = 0
या 9x + 6y + 18 = 0 …(2)
एक रेखा जो इसके समांतर है, उसका समीकरण
9x + 6y + c = 0 …(3)
रेखा (1) और (3) के बीच दूरी
\(\frac{(c+7)}{\sqrt{81+36}}=\frac{(c+7)}{\sqrt{117}}\) …(4)
रेखा (2) और (3) के बीच दूरी
\(\frac{(c-18)}{\sqrt{81+36}}=\frac{c-18}{\sqrt{117}}\) ….(5)
दूरियाँ (4) और (5) आपस में समान हैं।
\(\frac{(c+7)}{\sqrt{117}}=\pm \frac{c-18}{\sqrt{117}}\) [+ve चिन्ह मान्य नहीं है।]
c + 7 = – (c – 18)
= – c + 18
2c = 18 – 7 = 11
या c = \(\frac{11}{2}\)
c का मान समी (3) में रखने पर,
9x + 6y + \(\frac{11}{2}\) = 0
या 18x + 12y + 11 = 0.

प्रश्न 22.
बिन्दु (1, 2) से होकर जाने वाली एक प्रकाश किरण x-अक्ष के बिन्दु A से परावर्तित होती है और परावर्तित किरण बिन्दु (5, 3) से होकर जाती है। A के निर्देशांक ज्ञात कीजिए।
हल:
मान लीजिए BC, x-अक्ष के अनुदिश उस बिन्दु के निर्देशांक A (a, 0) है। AN इस पर लंब है। PA एक आपतित किरण है और AQ परावर्तित किरण है।

⇒ आपतित कोण PAN = परावर्तित कोण NAQ
⇒ ∠PAB = ∠QAC
⇒ यदि QA का झुकाव 0 हो तो PA का झुकाव 180 – θ होगा।
QA की ढाल जबकि Q(5, 3) और A(a, 0) हो, तो

प्रश्न 23.
दिखाइए कि (\(\sqrt{a^{2}-b^{2}}\), 0) और (-\(\sqrt{a^{2}-b^{2}}\),0) बिन्दुओं से रेखा \(\frac{x}{a}\)cosθ + \(\frac{y}{b}\)sinθ = 1 पर खींचे गए लम्बों की लंबाइयों का गुणनफल b₂ है।
हल:

प्रश्न 24.
एक व्यक्ति समीकरणों 2x – 3y + 4= 0 और 3x + 4y – 5 = 0 से निरूपित सरल रेखीय पथों के संधि बिन्दुओं (junction/crosing) पर खड़ा है और समीकरण 6x – 7y + 8 = 0 से निरूपित पथ पर न्यूनतम समय में पहुँचना चाहता है। उसके द्वारा अनुसरित पथ का समीकरण ज्ञात कीजिए।
हल:
AB और BC दो रेखीय पथ हैं। AB व BC रेखाओं के समीकरण
2x – 3y + 4 = 0 …(1)
और 3x + 4y – 5 = 0 …(2)

AB और BC बिन्दु B पर मिलते हैं।
समी (1) को 3 से तथा समी (2) को 2 से गुणा करने पर
6x – 9y = – 12 ….(3)
6x + 8y = 10 ….(4)
समी (3) को समी (4) में से घटाने पर,
17y = 10 + 12 = 22
∴ y = \(\frac{22}{17}\)
y का मान समी (1) में रखने पर,

B से AC तक न्यूनतम समय में पहुंचने के लिए कम से कम दूरी BD ( BD ⊥ AC) तय करनी है।
रेखा AC का समीकरण, 6x – 7y + 8= 0 की ढाल = \(\frac{6}{7}\)
BD की ढाल = – \(\frac{7}{6}\)
BD बिन्दु B \(\left(\frac{-1}{17}, \frac{22}{17}\right)\) से होकर जाती है।
∴ रेखा BD का समीकरण
y – y₁ = m(x – x₁)
\(y-\frac{22}{17}=-\frac{7}{6}\left(x+\frac{1}{17}\right)\)
102 से गुणा करने पर,
102y – 132 = – 119x – 7
119x + 102y – 125 = 0 .
अतः B से AC तक पहुँचने के लिए BD पथ अपनाना है जिसका समीकरण 119x + 102y – 125 = 0 है।

MP Board Class 11th Maths Solutions

MP Board Class 12th Biology Solutions Chapter 15 Biodiversity And Conversation

Biodiversity and Conversation NCERT Text Book Question and Answers

Question 1.
Name the three important components of biodiversity.
Answer:
Three important components of biodiversity are :

1. Genetic diversity.
2. Species diversity.
3. Ecosystem diversity.

Question 2.
How do ecologists estimate the total number of species present in the world?
Answer:
The diversity of living organisms present on the earth is very vast. According to an estimate by researchers, it is about seven millions. The total number of species present in the world is calculated by ecologists by statistical comparison between a species richness of a well studied group of insects of temperate and tropical regions. Then, these ratios are ‘ extrapolated with other groups of plants and animals to calculate the total species richness present on the earth.

Question 3.
Give three hypotheses for explaining why tropics show greatest levels of species richness.
Answer:
There are three different hypotheses proposed by scientists for explaining species richness in the tropics :

1. Tropical latitudes receive more solar energy than temperate regions, which leads to high productivity and high species diversity.
2. Tropical regions have less seasonal variations and have a more or less constant environment. This promotes the niche specialization and thus, high species richness.
3. Temperate regions were subjected to glaciations during the ice age, while tropical regions remained undisturbed which led to an increase in the species diversity in this region.

Question 4.
What is the significance of the slope of regression in a species-area relationship?
Answer:
The slope regression (z) has a great significance in order to find a species-area relationship, it has been found that in smallar areas (where the species-area relationship is analyzed), the value of slopes of regression is similar regardless of the taxonomic group or the region. However, when a similar analysis done in larger areas, then the slope of regression is much steeper.

Question 5.
What are the major causes of species losses in geographical regions?
Answer:
The following are the major causes for the loss of biodiversity around the world:
1. Habitat loss and fragmentation: Habitats of various organisms are altered or destroyed by uncontrolled and unsustainable human activities such as deforestation, slash and burn agriculture, mining, and urbanization. This results in the breaking up of the habitat into small pieces, which effects the movement of migratory animals and also, decreases the genetic exchange between populations leading to a declination of species.

2. Over-exploitation: Due to over-hunting and over-exploitation of various plants and animals by humans, many species have become endangered of extinct (such as the tiger and the passenger pigeon).

3. Alien species invasions: Accidental or intentional introduction of non-native species into a habitat has also led to the declination or extinction of indigenous species. For example, the Nile perch introduced in lake Victoria in Kenya led to the extinction of more than two hundred species of native fish in the lake.

4. Co-extinction: In a native habitat, one species is connected to the other in an intricate network. The extinction of one species causes the extinction of other species, which is associated with it in an obligatory way. For example, the extinction of the host will cause the extinction of its parasites.

Question 6.
How is biodiversity important for ecosystem functioning?
Answer:
An ecosystem with high species diversity is much more stable than an ecosystem with low species diversity. Also, high biodiversity makes the ecosystem more stable in productivity and more resistant towards disturbances such as alien species invasions and floods.

If an ecosystem is rich in biodiversity, then the ecological balance would not get affected. Various tropic levels are connected through food chains. If any one organism or all organisms of any one trophic level is illed, then it will disrupt the entire food chain. For example, in a food chain, if all plants are killed, then all deer’s will die due to the lack of food.

If all deer’s are dead, soon the tigers will also die. Therefore, it can be concluded that if an ecosystem is rich in species, then there will be other food alternatives at each trophic level which would not allow any organism to die due to the absence of their food resource. Hence, biodiversity plays an important role in maintaining the health and ecological balance of an ecosystem.

Question 7.
What are sacred groves? What is their role in conservation?
Answer:
Sacred groves are forest patches for worship in several parts of India. All the trees and wildlife in them are again treated and given total protection. They are found in khasi and jointia hills in Meghalaya. Westen Ghat regions of Karnataka and Maharashtra, etc. Tribals do not allow anyone to cut even a single branch of tree in these sacred groves thus sacred groves have been free form all types of exploitations.

Question 8.
Among the ecosystem services are control of floods and soil erosion, how is this achieved by the biotic components of the ecosystem?
Answer:

• Control of soil erosion: Plant roots hold the soil particles tightly and do not allow the top soil to be drifted away be winds or moving water. Plants increase the porosity and fertility of the soil.
• Control of floods: It is carried out by retaining water and preventing run off rain water. Litter and humus of plants function as sponges thus, retaining the water which percolates down and get stored as underground water. Hence, the flood is controlled.

Question 9.
The species diversity of plants (22 %) is much less than that of animals (72 %). What could be the explanations to how animals achieved greater diversifications?
Answer:
Animals have achieved greater diversification than plants due to following reasons:
1. They are mobile and thus can move away from their predators or unfavourable environments. On the other hand plants are fixed and have fewer adaptations to obtain optimum amount of raw materials and sunlight therefore, they show lesser diversity.

2. Animals have well developed nervous system to receive stimuli against external factors and thus can respond to them. On the other hand, plants do not exhibit any such mechanism, thus they show lesser diversity than animals.

Question 10.
Can you think of a situation where we deliberately want to make a species extinct? How would you justify it?
Answer:
Yes, there are various kinds of parasites and disease causing microbes that we deliberately want to eradicate from the earth. Since these micro-organisms are harmful to human beings, scientists are working hard to fight against them. Scientists have been able to eliminate smallpox virus from the world through the use of vaccinations. This shows that humans deliberately want to make these species extinct. Several other eradication programmes such as polio and hepatitis B vaccinations are aimed to eliminate these disease causing microbes.

Biodiversity and Conversation Other Important Questions and Answers

Biodiversity And Conversation Objective Type Questions

Choose the Correct Answer:

Question 1.
Wildlife conservation act was inforced in India in:
(a) 1883
(b) 1972
(c) 1973
(d) 1982.
Answer:
(b) 1972

Question 2.
Indian board of wildlife was established in:
(a) 1952
(b) 1981
(c) 1971
(d) 1972.
Answer:
(a) 1952

Question 3.
NBPGR is situated at:
(a) Delhi
(b) Kolkata
(c) Lucknow
(d) Mumbai.
Answer:
(a) Delhi

Question 4.
Number of biosphere in our country is :
(a) 73
(b) 7
(c) 416
(d) 23
Answer:
(b) 7

Question 5.
National park related with white tiger is:
(a) Kanker
(b) Satipada
(c) Bandhavgarh
(d) Kanha.
Answer:
(c) Bandhavgarh

Question 6.
The first national park of Madhya Pradesh is :
(a) Shivpuri
(b) Bandhavgarh
(c) Kanha
(d) Kanker.
Answer:
(c) Bandhavgarh

Question 7.
Plant fossils national park is situated in :
(a) Shivpuri
(b) Mandala
(c) Kanha
(d) Kanker.
Answer:
(b) Mandala

Question 8.
Kanha national park is famous for:
(a) Rhinoceros
(b) Tiger
(c) Birds
(d) Aligator.
Answer:
(b) Tiger

Question 9.
Bandipur national park is the habitat of:
(a) Deer Project
(b) Peacock Project
(c) Elephant Project
(d) Tiger Project
Answer:
(d) Tiger Project

Question 10.
Animal which is extinct in india:
(a) Hippopotamus
(b) National Park
(c) Cheetah
(d) Wolf
Answer:
(c) Cheetah

Question 11.
Which is in-situ conservation method:
(a) Tissue culture’
(b) National park
(c) Botanical garden .
(d)Cryotest.
Answer:
(b) National park

Question 12.
Which of the following are not done in a wild sanctuary:
(a) Conservation of flora
(b) Conservation of fauna
(c) Use of soil and flora
(d) Prohibition of hunting.
Answer:
(c) Use of soil and flora

Question 13.
In India maximum area covered by forest
(a) Odisha
(b) Arunachal Pradesh
(c) Madhya Pradesh
(d) All of the above
Answer:
(b) Arunachal Pradesh

Question 14.
Forest destruction:
(a) Destruction of natural sources
(b) Environmental Pollution
(c) Genetic disorder
(d) All of the above
Answer:
(b) Environmental Pollution

Question 15.
Forest Research Institute (FRI) is situated in:
(a) Shimla
(b) Chennai
(c) Dehradun
(d) Kolkata.
Answer:
(c) Dehradun

Question 16.
World environmental day is celebrated on :
(a) 15th March
(b) 15th April
(c) 4th May
(d) 5th June.
Answer:
(d) 5th June.

Question 17.
Species which are continuously distinct:
(a) Long lasting animal
(b) Dangerous animal
(c) Simple animal
(d) None of these.
Answer:
(c) Simple animal

Question 18.
Kaziranga national park is situated in:
(a) West Bengal
(b) Kerala
(c) Assam
(d) Umariya.
Answer:
(c) Assam

Question 19.
Bandhavgarh national park is situated in:
(a) Satana
(b) Shivpuri
(c) Mandla
(d) Umariya.
Answer:
(d) Umariya.

Question 20.
Kanha national park is located in :
(a) Uttar Pradesh
(b) Rajasthan
(c) Gujrat
(d) Madhya Pradesh.
Answer:
(a) Uttar Pradesh

Question 21.
National wildlife (Protection) Act was formulated during:
(a) 1972
(b) 1974
(c) 1976
(d) 1978.
Answer:
(a) 1972

Question 2.
Fill in the Blanks:

1. Planting of trees and forests on waste land is called ……………………..
2. Wildlife protection act is come in force in …………………….
3. UNESCO has started Man and Biosphere programme in ……………………..
4. ………………….. is the first Biosphere Reserve of India.
5. Kajiranga wildlife sanctuary is famous for ………………………
6. Red Data book is related with ……………………..
7. World Environment Day is celebrated on ………………………… of every year.
8. FRI is situated at ……………………..
9. …………………….. is the main source of energy.
10. National parks are made for the conservation of ……………………….
11. ………………………….. is the main region of forest depletion.
12. Indian Lion and Cheetah is ……………………….. animal.
13. ………………………….. is called the development of forest on barren land.
14. World environment day celebrated on ………………………

Answer:

1. 1. Afforestation
   2. 1972
   3. 1971
   4. Nilgiri
   5. Rhinoceros
   6. Endangered species
   7. 5th June
   8. Dehradun
   9. Sun
   10. Threatened species

1. Growing population
2. Rare
3. Forestration
4. 5th June.

Question 3.
Match the Following:

Answer:
1. (c)
2. (d)
3. (e)
4. (b)
5. (a).

Answer:
l.(e)
2. (d)
3. (a)
4. (b)
5. (c).

Question 4.
Answer in One Word/Sentence:

1. In which place of India, lion is found?
2. When does World Environment Day is celebrated?
3. When does Wildlife Protection Act was formed in India?
4. In which state Ghana bird Sanctuary is situated?
5. Where is Tiger Project situated?
6. Name any two National Parks of India.
7. Where does Kanha National Park situated?
8. Name any two threatened species.
9. Name the state having maximum forest area.
10. Name the National Animal and National Bird of India.
11. Name any one book associated with the conservation of wildlife.

Answer:

1. Gir forest, Gujarat
2. 5th June
3. 1972
4. Bharatpur, Rajasthan,
5. Kanha, M.P.
6. Rajaji Park and Gir National Park
7. M.P.
8. Lion, Wild buffalo
9. Chattisgarh (C.G.)
10. Tiger and Peacock
11. Red Data Book.

Biodiversity And Conversation Very Short Answer Type Questions

Question 1.
When does WLF-India was established?
Answer:
1962.

Question 2.
Where does Tiger project is established in Madhya Pradesh?
Answer:
Kanha national park.

Question 3.
Where does wild Ass is found?
Answer:
Runn of Kutch.

Question 4.
Where does great Indian bustards are found in India?
Answer:
Rajasthan, Punjab, Gujrat.

Question 5.
Write other name of biodiversity.
Answer:
Biological diversity.

Question 6.
Which part of the world are known as lesser biodiversity?
Answer:
Poles has lesser biodiversity.

Question 7.
What is the basic concept of biodiversity of biosphere?
Answer:
Gene.

Question 8.
Name the two plants which are worshiped any two states.
Answer:

1. Kadamba in Rajasthan.
2. Mango in Odisha.

Question 9.
Which part of the world which has more biodiversity?
Answer:
In Brazil.

Question 10.
Write the full form of IUCN.
Answer:
International Union for Conservation of Nature and Natural Resources.

Question 11.
Who perform important role in nature balancing?
Answer:
Forest animals.

Question 12.
Name the branch of science in which done the study of forest animal conservation.
Answer:
Forestry.

Question 13.
When, started the programme of social forestry?
Answer:
1976

Question 14.
Write the modern name of sanctuary.
Answer:
Shelter.

Biodiversity And Conversation Short Answer Type Questions

Question 1.
Explain conservation of biodiversity.
Answer:
Conservation of biodiversity: India is one of the richest (among the 12 mega centres of the world) countries in biological diversity. This rich biodiversity is due to a variety of climatic conditions prevailing on different ecological habits ranging from tropical, subtropical, temperate, alpine to desert. These varied conditions harbour a plethora of organisms, which forms an important natures wealth, responsible for socio-economic development of life in our country. But the biodiversity of organisms are under serious threat and there is an urgent need for biodiversity conservation on war foot level.

Question 2.
What is social forestry?
Answer:
The planning of social forestry started in India since 1976, which is related with the conservation of forests. This project is useful for local people in various ways, such as it fulfil their requirements, provides work to unemployed, use of wasteland and help to maintain O₂ and CO₂ balance in the atmosphere, etc. Thus, project is started by Indian government, the chief objectives of this project are as follows :

• Plantation of useful plants in the forest.
• Development of forests on personal lands by the cooperation of government.
• To prevent the harmful effects of pollution by development of artificial forest.
• Preservation of endangered wild animals.

Question 3.
Write importance of forests.
Answer:
Importance of forests :
1. Forests play a vital role in the life and economy of all tribes living in forests, by providing food, medicines and other products of commercial value.

2. Forests are large biotic communities. It provides shelter and sustenance to a larger number of diverse species of plants, animals and microorganisms.

3. Forests prevent soil erosion by wind and water. The trees provide shade which prevents the soil from drying during summer. Trees reduce the velocity of raindrops or wind striking the ground so that dislodging of the slil partiles is reduces. The root system of plants firmly binds the soil.

Question 4.
Write any five features of Indian forests.
Answer:
Indian forests are characterized by :

1. Indian forests are mainly tropical forests.
2. Himalayan forests are characterized by the presence of coniferous trees.
3. In few parts of our country having temperate forests.
4. Our forests contain a large number of useful varieties of plants and animals.
5. A great variations are present in indian forests.

Question 5.
Enumerate any five reasons for the destruction of wildlife (animals).
Answer:
The main reason of destruction of wildlife (animals) are:

• Entertainment, personal profit, earning money by wrong methods are the human illattitudes, unkindness toward wild animals has brought the animals at endanger level.
• Huge reduction in the natural habitat of wild animal so.it has reduced their living area due to urbanization, industrialization and deforestation.
• Exhorbitant extraction and consumption is harmful for wild animals like skin of animals, Teeth of elephant etc.
• Various types of pollution has force the reduction of wild animals. .
• Very loose and unpunishable, wildlife act which has increase poaching, huntidg of wild animals.

Biodiversity And Conversation Long Answer Type Questions

Question 1.
What do you understand by Threatened species? Explain its types.
Answer:
Species which have been greatly reduced in their number or whose natural habitats have been disturbed due to which these are near extinction and may become extinct if the causative factors continue are called threatened species. It is estimated that about 25,000 plant species and 1,000 vertebrate species and subspecies and many invertebrate species are threatened with extinction. It is believed that at least 10% of the living species are in danger.
The organisms which are near extinction are of following types :

1. Endangered (E) species: The species which are facing danger of extinction and whose survival is unlikely if the causal factors continue to operate. These are the species whose number have been reduced to a critical level or whose habitats have been so drastically reduced that they are deemed to be in immediate danger of extinction. For example, Indian rhinoceros, Asiatic lion and the great Indian bustard, snow leopard etc.

2. Vulnerable (V) species: These are the species having sufficient number of individuals in their natural habitats. However, in the near future, they might represent the category of endangered species if unfavourable factors in the environment continue to operate. e.g„ Musk deer, black buck, golden langur, etc.

3. Rare (R) species: These are species with small population in the world. At present these are not endangered and vulnerable but are at risk. These species are usually localize within geographical areas or habitats or are thinly scattered over a more extensive range. e.g., Indian elephant, Asiatic wildass, gharial, wild yak etc.

4. Threatened (T) species: The species which do not fall under the endangered or vulnerable categories but indications are available that such species may come under any of these two categories if appropriate measures are not taken to protect them.

Question 2.
What are National Parks? Explain any five National Parks found in India.
Answer:
Natinal park is an area which is strictly reserved for the betterment of the wildlife and where activities like forestry, grazing or cultivation are not permitted.
Five national parks of India are :

1. Shivpuri park: It is located at Shivpuri near Gwalior (Madhya Pradesh). Wildlife present in this park tiger, cheetal, sambhar.
2. Gundi deer park: It is locssssated near Chennai (Madras) in Tamil Nadu. Wildlife found here are Alvino deer, black buck, cheetal and famous snake park.
3. Betia national park: It is located in Palamu at Bihar. Wildlife found in this national park are elephant, tiger.
4. Dachigham national park: It is located at Srinagar in Jammu and Kashmir. Wildlife
   leopard, black beer, brown bear, musk deer, hangul, scrow, etc.
5. Bandhavgarh national park: It is located at Shahdol in Madhya Pradesh. Wildlife such as white tiger, panther, cheetal, bison, nilgai, barking deer, wild boar, etc. are found here.

Question 3.
Write in brief, the reasons necessary for conservation of wild species.
Answer:
Necessity for wildlife conservation : The conservation of wildlife is required . for the following reasons:

1. To maintain balance in nature: The wildlife helps us in maintaining the balance of nature. Once this equilibrium is disturbed it leads to many problems. The destruction of carnivores or insectivores often leads to an increase in the herbivores which in turn affects the forest vegetation or crop.

2. Economic value: The wildlife can be used commercially to earn money e.g., Animal products like hides, ivory, fur etc. are of tremendous economic value. The collection and supply of dead or living specimens of wildlife for museums and zoos fetches good amount of money. Wildlife can increase our earning of foreign exchange if tourism is promoted properly.

3. Scientific value: The preservation of wildlife helps many naturalists and behaviour biologists to study morphology, anatomy, physiology, ecology and behaviour biology of the wild animals under their natural surroundings.

4. Recreational value: The wildlife of any country provides best means of sports and recreation. Bird-watching is a hobby of many people all over the world. A visit to the parks and sanctuaries is an enjoyable proposition for children as well as for adults.

5. Cultural value: The wildlife of India is our cultural asset and has deep rooted impact on Indian art, sculpture, literature andreligion. Indus valley civilization shows the use of animals symbols in their seals.

6. Preservation of human race: The destruction of wildlife in an area may eventually lead to the end of human civilization.

Question 4.
Describe the National and International efforts prescribed for the conservation of forests.
Answer:
The forest conservation is started in India on national level since British government. In 1856, Lord Dalhousi had formulated a policy for the conservation of forest in Burma. In 1894, Indian government also prepared a forest policy on national level. The main points of this policy are:

• Forest management,
• Proper use of forest land,
• Policy for protected forests,
• Improved forest production.

Indian government established national parks, sanctuaries and zoological parks. The F.A.O. of United Nations is also functioning on forest conservation on international level. This organization also provides financial help for this purpose. In 1952, Indian government also prepared India’s New National Forest Policy under the direction of F.A.O. Forest policy has been planned for

• Prevention of deforestation of hill plants.
• Reforestation of grazing land.
• Development of grazing land,
• Plantation of economically useful forest trees.
• Increase in the profit of government from forests.

Question 5.
Write a short note on wild animals in India.
Answer:
India as a country has a diverse range of wildlife. India is home for many species of wild animals. More than 25% land are dense forest in India and around 400 national parks. Some of the most important and popular wild animals in India are as follows :
(A) Animals: In Indian forest, below mentioned animals are found :

1. Deer: Its many species are found in India-e.g. Musdeer, Sambhar deer, chital, etc.
2. Antelope: These are same as deer e.g.-Nilgai, Barasingha (Swamp deer), four- homed antelope (chousingha) etc.
3. Elephant: Elephants are large mammals of the family elephantidae. It is found in Kerela and North India.
4. Rhinoceros: It is found in Himalaya region and in the forest of Bengal and Assam. Humans are the biggest threat to
5. the Indian rhinoceros as the have been hunted to the brink of extinction for their horns.
6. Wild Ass: Wild asses are not found in any part of the world. Now in India, it is found in die little Rann of Kutch in the Gujrat state of India.

Carnivorous animals: Some India wild carnivores are :

(a) Indian lion (Asiatic lion): Now it is confined to forests in the fall.
(b) Cheetah: It is on the verge of extinction.
(c) Lion: Lion is a national animal, at present its population in our country are more than 3,000.
(d) Leopard: It is similar in cheetah but smaller than cheetah.

(B) Birds: Peacock, wild fowl, many types of duck, stork, pigeons, partridge, quail, vulture, kite, piquant, owl, indian paradise flycatcher (dudhraj) are found in forests of our country.

(C) Reptiles: Crocodiles, alligators, tortoise, lizards, snakes and other reptilians are found in Indian forest.
Many vertebrates and invertebrates are also found in Indian forest.

Question 6.
What is the main rules of Indian forest act?
Answer:
The Indian forest act, 1927 was largely based on previous Indian forest acts implemented under the British. Things which are included in this act are as follows :

• Forest arrangement: Due to this act give the protection and arrangement to forest and wildlife.
• Appropriate use of forest land: Uses of extra land for forest animals and grow some plant these are useful in wild animals.
• Act for protection of forest: This act, stop passion of deforestation and must be conservation of forest and wild animals.
• Increasing of forest product: Due to this act try to increasing of forest product and discovered new information which are better for wildlife.

Question 7.
Write an essay on measures of forest conservation.
Answer:
Forest conservation: Forest conservation and management are essential to maintain the forests in their natural state and also to prevent the depletion of wildlife and forest wealth. For the success of conservation it is necessary to know the cause of depletion and destruction of forests. Forests are generally destroyed by fire, improper cutting of trees and by animals.

Essentiality of forest conservation: Forest is a complex system which is responsible for the ecological balance in nature. Deforestation causing natural imbalance and affects the biotic components of the environment resulting floods, drought, epidemics, environmental pollution. Many ecologically important species of plants and animals are lost due to which economically important substance like wood, medicines, resin, lac and various food materials will not be available for us.

Measures of forest conservation: The following measures or efforts are prescribed for reforestation:

• Establishment of conserved forests and their conservation in proper way.
• Reforestation on deforested land. Old and damaged plants would be replaced by new plants.
• Proper management of forests.
• By promoting public awareness about forests.
• Replacement of burnt off areas of the forests.
• Plantation of trees that increase forest productivity.
• Forestation of plants on hills and wastelands and prevention of grazing by cattle.
• Prevent forests from fire, diseases and insects.
• Providing basic protection for all forests by law.
• Regulating human activity in the forest such as grazing by cattle and collection of firewood and fodder etc.
• Provide special attention for the conservation of endangered plant and animal species under the inspection of specialists.
• Removal of undesirable trees and vegetation for the better growth of desirable species.
• Forestation of industrially useful plants.
• The government will arrange the management of useful forests.

MP Board Class 12th Biology Solutions