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NCERT MCQ CLASS-11 CHAPTER-10 | MATH NCERT MCQ | STRAIGHT LINES | EDUGROWN

In This Post we are providing Chapter-10 Straight Lines NCERT MCQ for Class 11 Math which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MCQ ON STRAIGHT LINES

Question 1.
The locus of a point, whose abscissa and ordinate are always equal is
(a) x + y + 1 = 0
(b) x – y = 0
(c) x + y = 1
(d) none of these.

Answer: (b) x – y = 0

Question 2.
The equation of straight line passing through the point (1, 2) and parallel to the line y = 3x + 1 is
(a) y + 2 = x + 1
(b) y + 2 = 3 × (x + 1)
(c) y – 2 = 3 × (x – 1)
(d) y – 2 = x – 1

Answer: (c) y – 2 = 3 × (x – 1)

Question 3.
What can be said regarding if a line if its slope is negative
(a) θ is an acute angle
(b) θ is an obtuse angle
(c) Either the line is x-axis or it is parallel to the x-axis.
(d) None of these

Answer: (b) θ is an obtuse angle

Question 4:
The equation of the line which cuts off equal and positive intercepts from the axes and passes through the point (α, β) is
(a) x + y = α + β
(b) x + y = α
(c) x + y = β
(d) None of these

Answer: (a) x + y = α + β

Question 5.
Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are coincident if
(a) a₁/a₂ = b₁/b₂ ≠ c₁/c₂
(b) a₁/a₂ ≠ b₁/b₂ = c₁/c₂
(c) a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
(d) a₁/a₂ = b₁/b₂ = c₁/c₂

Answer: (d) a₁/a₂ = b₁/b₂ = c₁/c₂

Question 6:
The equation of the line passing through the point (2, 3) with slope 2 is
(a) 2x + y – 1 = 0
(b) 2x – y + 1 = 0
(c) 2x – y – 1 = 0
(d) 2x + y + 1 = 0

Answer: (c) 2x – y – 1 = 0

Question 7.
The slope of the line ax + by + c = 0 is
(a) a/b
(b) -a/b
(c) -c/b
(d) c/b

Answer: (b) -a/b

Question 8.
Equation of the line passing through (0, 0) and slope m is
(a) y = mx + c
(b) x = my + c
(c) y = mx
(d) x = my

Answer: (c) y = mx

Question 9.
The angle between the lines x – 2y = y and y – 2x = 5 is
(a) tan^-1 (1/4)
(b) tan^-1 (3/5)
(c) tan^-1 (5/4)
(d) tan^-1 (2/3)

Answer: (c) tan^-1 (5/4)

Question 10.
Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel if
(a) a₁/a₂ = b₁/b₂ ≠ c₁/c₂
(b) a₁/a₂ ≠ b₁/b₂ = c₁/c₂
(c) a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
(d) a₁/a₂ = b₁/b₂ = c₁/c₂

Answer: (a) a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Question 11.
The locus of a point, whose abscissa and ordinate are always equal is
(a) x + y + 1 = 0
(b) x – y = 0
(c) x + y = 1
(d) none of these.

Answer: (b) x – y = 0

Question 12.
In a ΔABC, if A is the point (1, 2) and equations of the median through B and C are respectively x + y = 5 and x = 4, then B is
(a) (1, 4)
(b) (7, – 2)
(c) none of these
(d) (4, 1)

Answer: (b) (7, – 2)

Question 13.
The length of the perpendicular from the origin to a line is 7 and the line makes an angle of 150 degrees with the positive direction of the y-axis. Then the equation of line is
(a) x + y = 14
(b) √3y + x = 14
(c) √3x + y = 14
(d) None of these

Answer: (c) √3x + y = 14

Question 14.
If two vertices of a triangle are (3, -2) and (-2, 3) and its orthocenter is (-6, 1) then its third vertex is
(a) (5, 3)
(b) (-5, 3)
(c) (5, -3)
(d) (-5, -3)

Answer: (d) (-5, -3)

Question 15.
The sum of squares of the distances of a moving point from two fixed points (a, 0) and (-a, 0) is equal to 2c² then the equation of its locus is
(a) x² – y² = c² – a²
(b) x² – y² = c² + a²
(c) x² + y² = c² – a²
(d) x² + y² = c² + a²Answer

Answer: (c) x² + y² = c² – a²

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CLASS 12TH CHAPTER -11 Biotechnology : Principles and Processes |NCERT BIOLOGY SOLUTIONS | EDUGROWN

Class 12 biology sets a strong basis for advanced studies. Subject experts clear all your queries during exam preparation by the explanation of the solutions in a conceptual way.

All chapter class 12 Biology NCERT Solutions are prescribed by the subject teachers based on the NCERT textbook questions and explained all solutions in a better way for easy understanding. In-depth knowledge of the biology subject can be attained by the students with the help of detailed NCERT Solutions of Class 12 Biology

NCERT Solutions for Class 12 Biology Chapter :10 Microbes in Human Welfare

Page No: 205

Exercises

1. Can you list 10 recombinant proteins which are used in medical practice? Find out where they are used as therapeutics (use the internet).

Answer

Recombinant proteins are obtained from the recombinant DNA technology. This technology involves the transfer of specific genes from an organism into another organism using vectors and restriction enzymes as molecular tools.

Ten recombinant proteins used in medical practice are:
(i) Insulin: used for the treatment of diabetes mellites
(ii) Interferon-α: Used for chronic hepatitis C
(iii) Interferon: Used for herpes and viral enteritis
(iv) Coagulation factor VII: Treatment of haemophilia A

(v) Coagulation factor IX: Treatment of haemophilia B

(vi) DNAase I: Treatment of cystic fibrosis
(vii) Anti-thrombin III: Prevention of blood clot
(viii) Interferon B: For treatment of multiple sclerosis
(ix) Human recombinant growth hormone: For promoting growth in an individual
(x) Tissue plasminogen activator: Treatment of acute myocardial infection

2. Make a chart (with diagrammatic representation) showing a restriction enzyme, the substrate DNA on which it acts, the site at which it cuts DNA and the product it produces.

Answer

3. From what you have learnt, can you tell whether enzymes are bigger or DNA is bigger in molecular size? How did you know?

Answer

Enzymes are smaller in size than DNA molecules. This is because DNA contains genetic information for the development and functioning of all living organisms. It contains instructions for the synthesis of proteins and DNA molecules.while, enzymes are proteins which are synthesized from a small strend of DNA known as ‘genes’, which are involved in the formatoin of the polypeptide chain.

4. What would be the molar concentration of human DNA in a human cell? Consult your teacher.

Answer

The molar concentration of human DNA in a human diploid cell is as follows:

⇒ Total number of chromosomes × 6.023 × 10²³

⇒ 46 × 6.023 × 10²³

⇒ 2.77 × 10²³ moles

Hence, the molar concentration of DNA in each diploid cell in humans is 2.77 × 10²³ moles.

5. Do eukaryotic cells have restriction endonucleases? Justify your answer.

Answer

No, eukaryotic cells do not have restriction endonucleases. This is because the DNA of eukaryotes is highly methylated by a modified enzyme, called methylase. Methylation protects the DNA from the activity of restriction enzymes .These enzymes are present in prokaryotic cells where they help prevent the invasion of DNA by virus.

6. Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors have over shake flasks?

Answer

The shake flask method is used for a small-scale production of biotechnological products in a laboratory. whereas stirred tank bioreactors are used for a large-scale production of biotechnology products.
Stirred tank bioreactors have several advantages over shake flasks:
(i) Small volumes of culture can be taken out from the reactor for testing.
(ii) It has a foam breaker for regulating the foam.
(iii) It has a control system that regulates the temperature and pH.

7. Collect 5 examples of palindromic DNA sequences by consulting your teacher. Better try to create a palindromic sequence by following base-pair rules.

Answer

The palindromic sequence is a certain sequence of the DNA that reads the same whether read from 5′ → 3′ direction or from 3′ → 5′ direction. They are the site for the action of restriction enzymes. Most restriction enzymes are palindromic sequences.
Five examples of palindromic sequences are-
(i) 5′-AGCT-3′
3′-TCGA-5′

(ii) 5′-GAATTC-3′
3′-CTTAAG-5′

(iii) 5′-AAGCTT-3′
3′-TTCGAA-5′

(iv) 5′-GTCGAC-3′
3′-CAGCTG-5′

(v) 5′-CTGCAG-3′
3′-GACGTC-5′

8. Can you recall meiosis and indicate at what stage a recombinant DNA is made?

Answer

Meiosis is a process that includes the reduction in the amount of genetic material. It is of two types, namely meiosis I and meiosis II. During the pachytene stage of prophase I, crossing over of chromosomes takes place where the exchange of segments between non-sister chromatids of homlogous chromosomes takes place. This results in the formation of recombinant DNA.

9. Can you think and answer how a reporter enzyme can be used to monitor transformation of host cells by foreign DNA in addition to a selectable marker?

Answer

A reporter gene can be used to monitor the transformation of host cells by foreign DNA.They act as a selectable marker to determine whether the host cell has taken up the foreign DNA or the foreign gene gets expressed in the cell. The researchers place the reporter gene and the foreign gene in the same DNA construct. Then, this combined DNA construct is inserted in the cell. then, the reporter gene is used as a selectable marker to find out the successful uptake of genes of interest .
Example of reporter genes – lac Z gene, which encodes a green fluorescent protein in a jelly fish.

Page No: 206

10. Describe briefly the following:
(a) Origin of replication
(b) Bioreactors
(c) Downstream processing

Answer

(a) Origin of replication -Origin of replication is defined as the DNA sequence in a genome from where replication initiates. The initiation of replication can be either uni-directional or bi-directional. A protein complex recognizes the ‘on’ site, unwinds the two strands, and initiates the copying of the DNA.
(b) Bioreactors – Bioreactors are large vessels used for the large-scale production of biotechnology products from raw materials. They provide optimal conditions to obtain the desired product by providing the optimum temperature, pH, vitamin, oxygen, etc. Bioreactors have an oxygen delivery system, a foam control system, a PH, a temperature control system, and a sampling port to obtain a small volume of culture for sampling.
(c) Downstream processing – Downstream processing is a method of separation and purification of foreign gene products after the completion of the biosynthetic stage. The product is subjected to various processes in order to separate and purify the product. After downstream processing, the product is formulated and is passed through various clinical trials for quality control and other test

11. Explain briefly
(a) PCR
(b) Restriction enzymes and DNA
(c) Chitinase

Answer

(a) PCR: – Polymerase chain reaction (PCR) is a technique in molecular biology to amplify a gene or a piece of DNA to obtain its several copies. It is extensively used in the process of gene manipulation. The process involves in-vitro synthesis of sequences using a primer, a template strand, and a thermostable DNA polymerase enzyme obtained from a bacterium, called Thermus aquaticus. The enzyme utilizes building blocks dNTPs (deoxynucleotides) to extend the primer. In the first step, the double stranded DNA molecules are heated to a high temperature so that the two strands separate into a single stranded DNA molecule. This process is called denaturation. Then, this ssDNA molecule is used as a template strand for the synthesis of a new strand by the DNA polymerase enzyme and this process is called annealing, which results in the duplication of the original DNA molecule. This process is repeated over several cycles to obtain multiple copies of the rDNA fragment.

(b) Restriction enzymes are molecular scissors used in molecular biology for cutting DNA sequences from a specific site. It plays an important role in gene manipulation. The enzymes recognize a specific six-box pair sequence known as the recognition sequence and cut the sequence at a specific site. For example, the recognition site for enzyme ECORI is as follows
Restriction enzyme are categorized into two types:
(i) Exonuclease: It is a type of restriction enzyme that removes the nucleotide from either 5′ or 3′ ends of the DNA molecule.
(ii) Endonuclease: It is a type of restriction enzyme that makes a cut within the DNA at a specific site. This enzyme acts as an important tool in genetic engineering. It is commonly used to make a cut in the sequence to obtain DNA fragments with sticky ends, which are later joined by enzyme DNA ligase.

(c) Chitinase – Chitinase is a class of enzymes used for the degradation of chitin, which forms a major component of the fungal cell wall. Therefore, to isolate the DNA enclosed within the cell membrane of the fungus, enzyme chitinase is used to break the cell for releasing its genetic material.

12. Discuss with your teacher and find out how to distinguish between
(a) Plasmid DNA and Chromosomal DNA
(b) RNA and DNA
(c) Exonuclease and Endonuclease

Answer

(a) Plasmid DNA and Chromosomal DNA

Plasmid DNA is an extra-chromosomal DNA molecule in bacteria that is capable of replicating, independent of chromosomal DNA.

Chromosomal DNA is the entire DNA of an organism present inside chromosomes.

(b) RNA and DNA

RNA is a single stranded molecule.

DNA is a double stranded molecule. It contains ribose sugar. It contains deoxyribose sugar. The pyrimidines in RNA are adenine and uracil. The pyrimidines in DNA are adenine and thymine. RNA cannot replicate itself. DNA molecules have the ability to replicate. It is a component of the ribosomes. It is a component of the chromosomes.
(c) Exonuclease and Endonuclease

It is a type of restriction enzyme that removes the nucleotide from 5′ or 3′ ends of the DNA molecule.

It is a type of restriction enzyme that makes a cut within the DNA at a specific site to generate sticky ends.

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CLASS 12TH CHAPTER -10 Microbes in Human Welfare |NCERT BIOLOGY SOLUTIONS | EDUGROWN

Class 12 biology sets a strong basis for advanced studies. Subject experts clear all your queries during exam preparation by the explanation of the solutions in a conceptual way.

NCERT Solutions for Class 12 Biology Chapter :10 Microbes in Human Welfare

Page No: 189

Exercises

1. Bacteria cannot be seen with the naked eyes, but these can be seen with the help of a microscope. If you have to carry a sample from your home to your biology laboratory to demonstrate the presence of microbes under a microscope, which sample would you carry and why?

Answer

Curd can be used as a sample for the study of microbes. Curd contains numerous lactic acid bacteria (LAB) or Lactobacillus. These bacteria produce acids that coagulate and digest milk proteins.
A small drop of curd is carried to the biology laboratory because it contains contains multitude of bacteria, which can be easily observed under a microscope which are of various shapes and sizes.

2. Give examples to prove that microbes release gases during metabolism.

Answer

The examples of bacteria that release gases during metabolism are:
(a) The dough used for making idli and dosa gives a puffed appearance. This is because of the action of bacteria which releases carbon dioxide. This CO₂ released from the dough gets trapped in the dough, thereby giving it a puffed appearance.

(b) During the digestion of sludge during waste water treatment, bacteria produce a mixture of gases such as methane, hydrogen sulphide and carbon dioxide.

3. In which food would you find lactic acid bacteria? Mention some of their useful applications.

Answer

Lactic acid bacteria can be found in curd.

Some of their useful applications are as follows:
(i) It is this bacterium that promotes the formation of milk into curd.
(ii) The bacterium multiplies and increases its number, which converts the milk into curd.

(iii) They also increase the content of vitamin B₁₂ in curd.
(iv) Lactic acid bacteria are also found in our stomach where it keeps a check on the disease-causing micro-organisms.

4. Name some traditional Indian foods made of wheat, rice and Bengal gram (or their products) which involve use of microbes.

Answer

(i) In the making dosa and idli, rice powder is fermented by bacteria and for making bread (from wheat), yeast (Saccharomyces cerevisiae) is used.
(ii) Gutta (made from black gram) also uses bacteria.

(ii) Microbes are also used to ferment fish, soyabean and bamboo shoots to make foods.

5. In which way have microbes played a major role in controlling diseases caused by harmful bacteria?

Answer

(i) Several micro-organisms are used for preparing medicines. Antibiotics are medicines produced by certain micro-organisms to kill other disease-causing micro-organisms.
(ii) These medicines are commonly obtained from bacteria and fungi. They either kill or stop the growth of disease-causing micro-organisms.
(iii) Antibiotics have been used to treat deadly diseases such as plague, whooping cough (kali khansi), diphtheria (galghotu) and leprosy (kusht rog) and many other common infections Streptomycin, tetracycline, and penicillin are common antibiotics.

(iv) Penicillium notatum produces chemical penicillin, which checks the growth of staphylococci bacteria in the body.
(v) Antibiotics are designed to destroy bacteria by weakening their cell walls. As a result of this weakening, certain immune cells such as the white blood cells enter the bacterial cell and cause cell lysis. Cell lysis is the process of destroying cells such as blood cells and bacteria.

6. Name any two species of fungus, which are used in the production of the antibiotics.

Answer

Penicillium notatum and Streptomyces are two species offungus, which are used in the production of the antibiotics.

7. What is sewage? In which way can sewage be harmful to us?

Answer

Sewage is the municipal waste matter that is carried away in sewers and drains.
It includes both liquid and solid wastes, rich in organic matter and microbes. Many of these microbes are pathogenic and can cause several water- borne diseases. Sewage water is a major cause of polluting drinking water. Hence, it is essential that sewage water is properly collected, treated, and disposed

8. What is the key difference between primary and secondary sewage treatment?

Answer

Primary treatment involves physical removal of large and small particles from the sewage through filtration and sedimentation.
Whereas, secondary sewage treatment involves biological digestion of organic matter by microbes.
Primary treatment is inexpensive and relatively less complicated where as secondary sewage is a very expensive and complicated process.

9. Do you think microbes can also be used as source of energy? If yes, how?

Answer

Yes, microbes can be used as a source of energy. Bacteria such as Methane bacterium is used for the generation of gobar gas or biogas.
(i) The generation of biogas is an anaerobic process in a biogas plant, which consists of a concrete tank (10-15 feet deep) with sufficient outlets and inlets.
(ii) The dung is mixed with water to form the slurry and thrown into the tank. The digester of the tank is filled with numerous anaerobic methane-producing bacteria, which produce biogas from the slurry.

(iii) Biogas can be removed through the pipe which is then used as a source of energy, while the spent slurry is removed from the outlet and is used as a fertilizer.

10. Microbes can be used to decrease the use of chemical fertilisers and pesticides. Explain how this can be accomplished.

Answer

Microbes can be used as biofertilisers, organisms that enrich the nutrient quality of the soil.
The main sources of bio-fertilisers are bacteria, fungi and cyanobacteria.They help in increasing the fertility of the soil in many ways
(i) Rhizobium that forms nodules on the roots of leguminous plants(a symbiotic association) fixes atmospheric nitrogen into organic
forms, which is used by the plant as nutrient.
(ii) Azospirillum and Azotobacter fix atmospheric nitrogen, while living freely, and enriching the nitrogen content of the soil.
(iii) Many members of the genus Glomus (fungi) form symbiotic associations with plant known as mycorrhiza that
(a) Absorption of phosphorus from soil and pass it to the plant.
(b) Help the plants to develop resistance to root-borne pathogens.
(c) increase their tolerance to salinity and drought and thus, help inoverall increase in plant growth and development.
(iv) Cyanobacteria autotrophic microbes, e.g., Anabaena, Nostoc,Oscillatoria can fix atmospheric nitrogen, in aquatic and terrestrial environment and also add organic matter to the soil and increase its fertility.

11. Three water samples namely river water, untreated sewage water and secondary effluent discharged from a sewage treatment plant were subjected to BOD test. The samples were labelled A, B and C; but the laboratory attendant did not note which was which. The BOD values of the three samples A, B and C were recorded as 20mg/L, 8mg/L and 400mg/L, respectively. Which sample of the water is most polluted? Can you assign the correct label to each assuming the river water is relatively clean?

Answer

Sample A (BOD 20mg/L) is secondary effluent discharged from a sewage treatment plant.
Sample B (BOD 8mg/L) is river water.
Sample C (BOD 400mg/L) is the untreated sewage water.
As BOD is the direct measure of the organic matter present in water, higher the BOD, more polluted the water.

Page No: 190

12. Find out the name of the microbes from which Cyclosporin A (an immunosuppressive drug) and Statins (blood cholesterol lowering agents) are obtained

Answer

(i) Cyclosporin-A is obtained from the fungus Trichoderma polysporum.
(ii) Statins is obtained from Monascus purpureus.

13. Find out the role of microbes in the following and discuss it with your teacher.
(a) Single cell protein (SCP)
(b) Soil

Answer

(a) Single Cell Protein (SCP) refers to harmless microbial cells that can be used as an alternate source of good protein.Just like mushrooms (a fungus) is eaten by many people and yeast isused by athletes as a protein source;similarly, other forms of microbial cells can also be used as food rich in protein, minerals, fats, carbohydrate and vitamins.Microbes like Spirulina and Methylophilus methylotrophus are being grown on an industrial scale on materials containing starch like wastewater from potato processing plants, straw,molasses, animal manure and even sewage. These single cell microbes can be used as source.

(b) Soil: Microbes play an important role in maintaining soil fertility. They help in the formation of nutrient-rich humus by the process of decomposition. Many species of bacteria and cyanobacteria have the ability to fix atmospheric nitrogen into usable form. Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants. Azospirillium and Azotobocter are free living nitrogen-fixing bacteria, whereas Anabena, Nostoc, and Oscillitoria are examples of nitrogen-fixing cyanobacteria.

14. Arrange the following in the decreasing order (most important first) of their importance, for the welfare of human society. Give reasons for your answer.
Biogas, Citric acid, Penicillin and Curd

Answer

Penicillin > Biogas > Curd > Citric acid
• Penicillin is an antibiotic that helps kill pathogens that cause infections and diseases and thus, saves lives.
• Biogas is a non-polluting clean fuel that is produced as a byproduct of sewage treatment. It is used for cooking and lighting up the homes in rural areas.
• Curd has good nutrient value, provides vitamin-B₁₂ and replaces harmful bacteria of the stomach with helpful ones.
• Citric acid it is used as preservative of food.

15. How do biofertilisers enrich the fertility of the soil?

Answer

Bio-fertilizers are living organisms which help in increasing the fertility of soil. It involves the selection of beneficial micro-organisms that help in improving plant growth through the supply of plant nutrients. These are introduced to seeds, roots, or soil to mobilize the availability of nutrients by their biological activity. Thus, they are extremely beneficial in enriching the soil with organic nutrients. Many species of bacteria and cyanobacteria have the ability to fix free atmospheric nitrogen. Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants. Azospirillium and Azotobocter are free living nitrogen-fixing bacteria, whereas Anabena, Nostoc, and Oscillitoria are examples of nitrogen-fixing cyanobacteria. Bio-fertilizers are cost effective and eco-friendly.

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NCERT MCQ CLASS-11 CHAPTER-9 | MATH NCERT MCQ | SEQUENCES AND SERIES | EDUGROWN

In This Post we are providing Chapter-9 Sequences and Series NCERT MCQ for Class 11 Math which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MCQ ON SEQUENCES AND SERIES

Question 1. If in A.P., S_n = qn2 and Sm = qm², where S_r denotes the sum of r terms of the A.P., then Sq equals
(a) q³/2
(b) mnq
(c) q³
(d) (m + n)q²

Answer : C

Question 2. If tn denotes the n th term of the series 2 + 3 + 6 + 11 + 18 + … then t50 is
(a) 492 – 1
(b) 492
(c) 502 + 1
(d) 492 + 2

Answer : D

Question 3. The sum to 200 terms of the series 1 + 4 + 6 + 5 +11 + 6 + ………. is
(a) 31,200
(b) 29,800
(c) 30,200
(d) None of these

Answer : C

Question 4. If the sum of the series 54 + 51 + 48 + ……….. is 513, then the number of terms are
(a) 18
(b) 20
(c) 17
(d) None of these

Answer : A

Question 5. The sum of 11 terms of an A.P. whose middle term is 30,
(a) 320
(b) 330
(c) 340
(d) 350

Answer : B

Question 6. If the sum of the first 2n terms of 2, 5, 8, ……. is equal to the sum of the first n terms of 57, 59,61……., then n is equal to
(a) 10
(b) 12
(c) 11
(d) 13

Answer : C

Question 7. There are four arithmetic means between 2 and –18. The means are
(a) –4, –7, –10, –13
(b) 1, –4, –7, –10
(c) –2, –5, –9, –13
(d) –2, –6, –10, –14

Answer : D

Question 8. 2, 3, 5 are the following terms of an A.P.:
(a) 2nd, 3rd, and 5th
(b) 4th, 9th and 25th
(c) 4th, 6th, and 10th
(d) none of the above

Answer : D

Question 9. The first term of an infinite G.P. is 1 and each term is twice the sum of the succeeding terms. then the sum of the series is
(a) 2
(b) 3
(c) 3/2
(d) 5/2

Answer : C

Question 10. The sum of the first nine terms of an arithmetic progression is 171. Which one of the following statements is not correct about this A.P.?
(a) The sum of the first and the ninth terms cannot be determined
(b) No term of the A.P. can be determined
(c) The first term of the A.P. cannot be determined
(d) The common difference cannot be determined

Answer : A

Question 11. If the nth term of an arithmetic progression is 3n + 7, then what is the sum of its first 50 terms?
(a) 3925
(b) 4100
(c) 4175
(d) 8200

Answer : C

Question 12. What is the value of 91/3. 91/19. 91/27…… ∞ ?
(a) 9
(b) 3
(c) 91/3
(d) 1

Answer : B

Question 13. The minimum value of the expression
3^x + 3^{1 – x}, x ∈ R, is
(a) 0
(b) 1/3
(c) 3
(d) 2√3

Answer : D

Question 14. At the end of each year the value of a certain machine has depreciated by 20% of its value at the beginning of that year. If its initial value was `₹1250, the value at the end of 5 years is
(a) 409.6
(b) 409
(c) 408
(d) 409.5

Answer : A

Question 15. An even number of AM are inserted between two numbers whose sum is 13/6. If the sum of means exceeds their number by 1, what is the number of means?
(a) 8
(b) 18
(c) 12
(d) 6

Answer : C

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CLASS 12TH CHAPTER -9 Strategies for Enhancement in Food Production |NCERT BIOLOGY SOLUTIONS | EDUGROWN

Class 12 biology sets a strong basis for advanced studies. Subject experts clear all your queries during exam preparation by the explanation of the solutions in a conceptual way.

NCERT Solutions for Class 12 Biology Chapter : 9 Strategies for Enhancement in Food Production Biology

Page No: 178

Exercises

Question 1.
Explain in brief the role of animal husbandry in human welfare.
Solution:

Animal husbandry evolves new techniques and technologies for the management of livestock like buffaloes, cows, pigs, horses, cattle, sheep, camels, goats, etc., that are useful to humans.
These methods can also be applied to rearing animals like bees, silkworms, prawns, crabs, fishes birds, pigs, cattle, sheep, and camels for their products like milk, eggs, meat, wool, silk, honey, etc.

Role of animal husbandry in human welfare is discussed as follows:

Milk is an important product of farm animals that are consumed as such, in the form of curd, cheese, butter, ice cream, etc. Milk is the only source of animal protein for vegetarians and is a complete food. Most of the milk is obtained from cows and buffalo. Other milk-yielding animals are goat, sheep, camel, and yak.
Egg, like milk, is also a complete food. Chicken and duck are the two major sources of the egg.
Meat is a protein-rich diet that is obtained from all types of livestock, e.g., goat, sheep, pig, cattle, chicken, fish, etc.
Honey is a sweet syrup obtained from the hives of the honey bee. Honey is used in sweetening various preparations.
Fibers like wool and silk are two high-quality fibres which we get from animals. Wool is the hair of sheep, some goats, and rabbits. Silk is a product of silkworms.
The skins of many animals are converted into hides and leather.
Drought animals are trained to carry men and materials besides other functions, e.g., buffalo, bullock, horse, camel, ass, elephant, reindeer, yak.
The rearing of animals provides employment to many persons.
Animal byproducts like horns, feathers, bone, dung, and droppings are all used in developing useful products.

Question 2.
If your family owned a dairy farm, what measures would you undertake to improve the quality and quantity of milk production?
Solution:
Some of the measures to be followed for proper management of dairy farm are :

Selection of good breeds having high milk yielding potential according to the climatic conditions of the area.
The shed under which the cattle are kept should be well ventilated with an adequate water supply for drinking as well as for washing. Shed should have pucca floor and proper drainage channel.
The feed of the animals should be a balanced diet with right proportions of carbohydrates, fats, proteins, and roughage and it should be given timely in good quantity.
Cleanliness and hygiene comes first for maintaining the livestock’s health and productivity. So, washing cattle and taking precautionary measures while milking are a must.
Inspection, keeping records of the activities and consulting a veterinary doctor for regular checkups of the livestock should be undertaken.

Question 3. What is meant by the term ‘breed’? What are the objectives of animal breeding?

Answer

A breed is a improved variety of animals within a species. It is similar in most characters such as general appearance, size, configuration, and features with other members of the same species.
For example- Jersey and Brown Swiss are foreign breeds of cattle. These two varieties of cattle have the ability to produce abundant quantities of milk which is nutritious with high protein content.
Objectives of animal breeding:
(i) To improve the desirable qualities of the animal produce
(i) To increase the yield of animals
(iii) To produce disease-resistant varieties of animals.

Question 4. Name the methods employed in animal breeding. According to you which of the methods is best? Why?

Answer

Animal breeding is the method of mating interrelated individuals. There are several methods involded in animals breeding, which can be classified into the following categories:

(i) Natural methods of breeding include inbreeding and out-breeding. Breeding between animals of the same breed is known as inbreeding, while breeding between animals of different breeds is known as out-breeding. Out-breeding of animals is of three types:

→ Out-crossing: In this type of out-breeding, the mating of animals occurs within the same breed.as they have no common ancestors up to the last 4-5 generations.
→ Cross-breeding: In this type of out-breeding, the mating occurs between different breeds of the same species, thereby producing a hybrid.
→ Interspecific hybridization: In this type of out-breeding, the mating occurs between different species.

(ii) Artificial methods of breeding include modern techniques of breeding. It involves controlled breeding experiments, which are of two types:-
→ Artificial insemination: It is a process of introducing the semen (collected from the male) into the oviduct or the uterus of the female body by the breeder. This method of breeding helps the breeder overcome certain problems faced in abnormal mating.
→ Multiple ovulation embryo technology (MOET): It is a technique for cattle improvement in which super-ovulation is induced by a hormone injection. Then, fertilization is achieved by artificial insemination and early embryos are collected. Each of these embryos are then transplanted into the surrogate mother for further development of the embryo.

The best method to carry out animal breeding is the artificial method of breeding, which includes artificial insemination and MOET technology. These technologies are scientific in nature. They help to minimize problems of normal mating and have a high success rate of crossing between mature males and females. Also, it ensures the production of hybrids with the desired qualities. This method is highly economical as a small amount of semen from the male can be used to inseminate several cattle as semen is not destroyed.

5. What is apiculture? How is it important in our lives?

Answer

(i) Apiculture is the practice of bee-keeping for the production of various products such as honey, bee’s wax, etc. Honey is a highly nutritious food source and is used as an indigenous system of medicines.
(ii) It is useful in the treatment of many diseases such as cold, flu, and dysentery.

(iii) Other commercial products obtained from honey bees include bee’s wax and bee pollen. Bee’s wax is used for making cosmetics, polishes, and is even used in several medicinal preparations.

(iv) As demand of honey is increasing , people have started practicing bee-keeping on a large scale. It has become an income generating activity for farmers since it requires a low investment and is labour intensive.

6. Discuss the role of fishery in enhancement of food production.

Answer

(i) Fishery is an industry devoted with catching, processing, and marketing of fishes and other aquatic animals that have a high economic value.
(ii) Some commercially important aquatic animals are prawns crabs, oysters, lobsters, and octopus.
(iii) Fisheries play an important role in the Indian economy. This is because a large part of the Indian population is dependent on fishes as a source of food, which is both cheap and high in animal protein.
(iv) A Fishery is an employment generating industry especially for people staying in the coastal areas. Both fresh water fishes (such as Catla, Rohu, etc) and marine fishes (such as tuna, mackerel pomfret, etc.) are of high economic value.

7. Briefly describe various steps involved in plant breeding.

Answer

Plant breeding is purposeful manipulation of plants species in order to create desired plants that are better suited for cultivation, give better yield and are disease restistant. various steps involved in plant breeding are as follows:
(i) Collection of genetic variability: Genetic variability from various wild relatives of the cultivated species are collected to maintain the genetic diversity of a species. The entire collection of the diverse alleles of a gene in a crop is called the germplasm collection.
(ii) Evaluation of germplasm and selection of parents: The germplasm collected is then evaluated for the desirable genes. The selected plants with the desired genes are then used as parents in plant breeding experiments and are multiplied by the process of hybridization.
(iii) Cross-hybridization between selected parents: The next step in plant breeding is to combine the desirable characters present in two different parents to produce hybrids. It is a tedious job as one has to ensure that the pollen grains collected from the male parent reach the stigma of the female parent.
(iv) Selection of superior hybrids:the selection process is crucial to the success of breeding objective and requires careful scientific evaluation of the progeny. The progenies of the hybrids having the desired characteristics are selected through scientific evaluation. The selected progenies are then self-pollinated for several generations to ensure homozygosity.
(v) Testing, release, and commercialization of new cultivars: the newly selected lines are evaluated for theire yield and other agronomic traits of quality, disease resistance, by growing them in research fields for at least three growing seasons in different parts of the country. After thorough testing and evaluation, the selected varieties are given to the farmers for growing in fields for a large-scale production.

8. Explain what is meant by biofortification.

Answer

(i) Biofortification is a process of breeding crops with higher levels of vitamins, minerals, proteins, and fat content.
(ii) This method is employed to improve public health. Breeding of crops with improved nutritional quality is undertaken to improve the content of proteins, oil, vitamins, minerals, and micro-nutrients in crops.
(iii) It is also undertaken to upgrade the quality of oil and proteins. An example of this is a wheat variety known as Atlas 66, which has high protein content in comparison to the existing wheat. In addition, there are several other improved varieties of crop plants such as rice, carrots, spinach etc. which have more nutritious value and more nutrients than the existing varieties.

9. Which part of the plant is best suited for making virus-free plants and why?

Answer

Apical and axillary meristems of plants is used for making virus-free plants. In a diseased plant, only this region is not infected by the virus as compared to the rest of the plasnt region. Hence, the scientists remove axillary and apical meristems of the diseased plant and grow it in vitro to obtain a disease-free and healthy plant. Banana, sugarcane, and potato have been obtained using this method by scientist are virus free plants.

Question 10.
What is the major advantage of producing plants by micropropagation?
Solution:
Micropropagation is the tissue culture technique used for rapid vegetative multiplication of ornamental plants and fruit trees by using small-sized explants. Because of the minute size of the propagules in the culture, the propagation technique is named micropropagation. This method of tissue culture produces several plants. Each of these plants will be genetically identical to the original plant from which explants were taken. Plants obtained by vegetative propagation of a single plant constitute a somaclonal. The members of a single somaclonal have the same genotype. It is the only process adopted by Indian plant biotechnologists in different industries mainly for the commercial production of ornamental plants like lily, orchids, Eucalyptus, Cinchona, blueberry, etc., and fruit trees like tomato, apple, banana, grapes, potato, Citrus, palm, etc.
Question 11. Find out what the various components of the medium used for propagation of an explant in vitro are?

Answer

The major components of medium used for propagation of explants in vitro are carbon sources such as sucrose, inorganic salts, vitamins, amino acids, water, agar-agar, and certain growth hormones such as auxins and gibberellins.

Question 12. Name any five hybrid varieties of crop plants which have been developed in India.

Answer

The five hybrid varieties of crop plants which have been developed in India are:

Crop Plant	Hybrid Variety
Wheat	Sonalika and kalian sona
Rice	Jaya and Ratna
Cauliflower	Pusa shubra and Pusa snowball K-1
Cowpea	Pusa komal
Mustard	Pusa swarnim

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CLASS 12TH CHAPTER -8 Human Health and Disease |NCERT BIOLOGY SOLUTIONS | EDUGROWN

Class 12 biology sets a strong basis for advanced studies. Subject experts clear all your queries during exam preparation by the explanation of the solutions in a conceptual way.

NCERT Solutions for Class 12 Biology Chapter : 8 Human Health and Disease

Page No: 164

Exercises

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

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

a. Personal hygienic measures

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

b. Public hygienic measures

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

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

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

III. For vector-borne diseases like malaria

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

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

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

Answer

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

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

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

Answer

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

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

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

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

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

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

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

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

MALT
CMI
AIDS
NACO
HIV

Solution:

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

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

Answer

(a) Innate and acquired immunity

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

(b) Active and passive immunity

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

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

Answer

10. What are the various routes by which transmission of human immuno-deficiency virus takes place?

Answer

AIDS (Acquired Immuno Deficiency Syndrome) is caused by the Human immunodeficiency virus (HIV).
This is transmitted by following modes –
(a) Unprotected sexual contact with a diseased person.
(b) Transfusion of blood from a healthy to a diseased person.
(c) Sharing infected needles or syringes.
(d) infected mother to a child through the placental connection.

11. What is the mechanism by which the AIDS virus causes deficiency of immune system of the infected person?

Answer

→ AIDS (Acquired Immuno Deficiency Syndrome) is caused by the Human immunodeficiency virus (HIV) via sexual or blood to blood contact.
→ After entering the human body, the HIV virus attacks and enters into the macrophages. Inside the macrophages, the RNA of the virus replicates with the help of enzyme reverse transcriptase and gives rise to viral DNA copy.
→ Then, this viral DNA incorporates into the host DNA and directs the synthesis of virus particles. → At the same time, HIV enters helper T- lymphocytes. It replicates and produces viral progeny.
→ These newly formed progeny viruses get released into the blood, attacking other healthy helper T-lymphocytes in the body.
→ As a result, the number of T-lymphocytes in the body of an infected person decreases in number, which causes decrease in immunity of person.

12. How is a cancerous cell different from a normal cell?

Answer

Normal cell	Cancerous cell
(i) Normal cells show the property of contact inhibition. Therefore, when these cells come into contact with other cells, they stop dividing.	(i) Cancerous cells lack the property of contact inhibition. Therefore, they continue to divide, thereby forming a mass of cells called tumor.
(ii) They undergo differentiation after attaining a specific growth.	(ii) They do not undergo differentiation.
(iii) These cells remain confined at a particular location.	(iii) These cells do not remain confined at a particular location. They move into neighboring tissues and disturbs the functioning.

13. Explain what is meant by metastasis.

Answer

The property of metastasis is perfomed by malignant tumors. These melingnent cells moves through different part of body by a pathological process. These cells divide uncontrollably, forming a mass of cells called tumor. From the tumor, some cells get shed off and enter into the blood stream. From the blood stream, these cells reach distant parts of the body and therefore, start the formation of new tumors by dividing actively.

14. List the harmful effects caused by alcohol/drug abuse.

Answer

Alcohol and drugs have several adverse effects on the individual, his family, and the society.
(i) Effects of alcohol:
→ Effects on the individual: Alcohol is injurious to the health of the individual. When an individual consumes excess alcohol, it causes damage to the liver and the central nervous system. As a result, other symptoms such as depression, fatigue, aggression, loss of weight and appetite may also be observed in the individual. Sometimes, extreme levels of alcohol consumption may also lead to heart failure, resulting coma and death. The immediate adverse effects of alcohol abuse are manifested in form of reckless bhehaviour, vandalism and voilence,Also, it is advisable for pregnant women to avoid alcohol as it may inhibit normal growth of the baby.
→ Effects on the family: Consumption of excess alcohol by any family member can have devastating effects on the family. It leads to several domestic problems such as quarrels, frustrations, insecurity, etc.
Effects of alcohol on society-
(a) Rash behaviour
(b) Malicious mischief and violence
(c) Disturbing social network
(d) Loss of interest in social activities, loss of interests in hobbies,change in eating and sleeping habites etc.

(ii) Effects of drugs: An individual who is addicted to drugs creates problems not only for himself but also for his family.
→ Effects on the addited individual: Drugs have an adverse effect on the central nervous system of an individual. This leads to the malfunctioning of several other organs of the body such as the kidney, liver, etc. The spread of HIV is most common in these individuals as they share common needles while injecting drugs in their body. Drugs have long-term side effects on both males and females. These side effects include increased aggressiveness, mood swings, and depression
→ Effects on the family and society: A person addicted to drugs creates problems for his family and society. A person dependant on drugs becomes frustrated, irritated, and anti-social.At the time of drug addicted becomes mental and financial distress.

15. Do you think that friends can influence one to take alcohol/drugs? If yes, how may one protect himself/herself from such an influence?

Answer

Yes, friends can influence one to take drugs and alcohol. A person can take the following steps for the prevention of themself against drug abuse:
→ by avoiding undue peer pressur as everyone has theire own field of interest which shouid be respected by theire teachers and family. One should not experiment with alcohol for curiosity and fun.
→ Avoid the company of friends who take drugs.
→ Seek help from parents and peers.a child should not pushed beyond his/her threshold limits.
→ Take proper knowledge and counseling about drug abuse. Devote your energy in other extra-curricular activities.
→ Seek immediate professional and medical help from psychologists and psychiatrists if symptoms of depression and frustration become apparent.
→ Get rid of the problems completely and lead a perfectly normal life by increasing their will power.

16. Why is that once a person starts taking alcohol or drugs, it is difficult to get rid of this habit? Discuss it with your teacher.

Answer

This question should be discussed with your subject teacher.

17. In your view what motivates youngsters to take to alcohol or drugs and how can this be avoided?

Answer

→ Various factors are responsible for motivating youngsters towards alcohol or drugs. Curiosity, need for adventure and excitement, experimentation for fun are the initial causes for motivating youngsters.
→ Some youngsters start consuming drugs and alcohol in order to overcome negative emotions (such as stress, pressure, depression, frustration) and to excel in various fields..such youngsters use to abuse alcohol due to their family pressure for academics .
→ Several mediums like television, internet, newspaper, movies promote various brend by the brand ambassadors like celebrities . Celebrities are also ideal of youngsters so the get much infiuenced by them.
→ Amongst these factors, reasons such as unstable and unsupportive family structures and peer pressure can also lead an individual to be dependant on drugs and alcohol.
Preventive measures against addiction of alcohol and drugs:
(i) Parents should motivate and try to increase the will power of their child.
(ii) Parents should educate their children about the ill-effects of alcohol. They should provide them with proper knowledge and counselling regarding the consequences of addiction to alcohol.
(iii) It is the responsibility of the parent to discourage a child from experimenting with alcohol. Youngsters should be kept away from the company of friends who consume drugs.
(iv) Children should be encouraged to devote their energy in other extra- curricular and recreational activities.
(v) Proper professional and medical help should be provided to a child if sudden symptoms of depression and frustration are observed.

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CLASS 12TH CHAPTER – 7 Evolution |NCERT BIOLOGY SOLUTIONS | EDUGROWN

Class 12 biology sets a strong basis for advanced studies. Subject experts clear all your queries during exam preparation by the explanation of the solutions in a conceptual way.

NCERT Solutions for Class 12 Biology Chapter : 7 Evolution

Page No: 142

Exercises

1. Explain antibiotic resistance observed in bacteria in light of Darwinian selection theory.

Answer

In the presence of antibiotic, the bacteria that are sensitive to it will die. However, if there are any mutants in the population, that can somehow survive its effect, they will multiply and increase in numbers. After that, they will live as antibiotic resistant bacteria.

Question 2.
Find out from newspapers and popular science articles any new fossil discoveries or controversies about evolution.
Solution:
Chimps are more evolved than humans (The Times of India):
Chimpanzees are more evolved than humans, a study suggests. There is no doubt that humans are the more advanced species. But a comparison of 14,000 human and chimpanzee genes shows the forces of natural selection have and the greatest impact on our ape cousins.

The researchers’ discovery challenges the common assumption that our large brains and high intelligence were the gifts of natural selection. Humans and chimps followed different evolutionary paths from a common ape ancestor about 5 million years ago. Both underwent changes as the fittest survived to pass their genes on to future generations. But the US study shows that humans possess a ‘substantially smaller’ number of positively-selected genes than chimps.

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

Question 4.
Try to trace the various components of human evolution (hint: brain size and function, skeletal structure, dietary preference, etc.)
Solution:Answer

Human evolution shows the following trends:
A. Brain size: It increased gradually along with evolution. The brain capacity of Australopithecus africanus – 500 cc, Homo habilis – 700 cc, Homo eredus – 800 – 1300 cc, Homo sapiens sapiens – 1450 cc.

B. Skeletal structure:

Name	Features
Dryopithecus	Ape like, canines large, arms and legs are of equal size, ate soft fruits and leaves
Ramapithecus	More man-like, canines were small while molars were large, walked more erect, ate seeds and nuts
Australopithecus	Man-like, canines and incisors were small, walked upright, hunted with stone weapons, ate fruits, brain capacities were between 400-600cc.
Homo habilis	First human like being, canines were small, first tool makers, did not ate meats, brain capacities were between 650-800cc.
Homo erectus	Used stone and bone tools for hunting games, ate meat, brain capacity 900cc.
Homo neanderthalnsis	Cave dwellers, used hides to protect their bodies, and buried their dead, brain capacity 1400cc.
Homo sapiens (Modern human)	Modern man with high intelligence, developed art, culture, language etc., cultivated crops and domesticated animals.

Question 5.

Find out through the internet and popular science articles whether animals other than man have self-consciousness.
Solution:
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, chimpanzees, Gorilla, Orangutan, etc., which exhibit self-consciousness.

Question 6.
List 5-6 modern-day animals and using the internet resources link it to a corresponding ancient fossil. Name both.
Solution:
The list of few modern-day animals and their corresponding ancient fossils is as follows:

Question 7.
Describe one example of adaptive radiation.
Solution:
Adaptive radiation – Formation of different species from a common ancestor with new species adapting to different geological niches.
Example: Darwin’s finches are Galapagos island have wolves from mainland finches. They underwent changes in the shape, size of beaks, food habits, feathers.

Question 8.
Can we call human evolution adaptive radiation?
Solution:
No, we can not be called human evolution as adaptive evolution.

Question 9.
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.
Solution:
The evolutionary stages of the modern horse are listed in the table given below:

• Eohippus: It appeared in the Eocene period about 52 million years ago. It was approximately the size of a fox (0.4 m), with a relatively short head and neck and a springy, arched back. 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.

• Mesohippus: Approx, 40 million years ago in Oligocene period, Mesohippus which was slightly larger than Eohippus about 0.6 metre. It had three toes in each foot.

• Merychippus: In Miocene period the grazer Merychippus flourished. It had the size of approx 1m. 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: Around 12 million years in Pilocene period, modern horse Pilohippus emerged. It had a single functional toe with splint of 2nd and 4th in each limb.

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

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NCERT MCQ CLASS-11 CHAPTER-8 | MATH NCERT MCQ | BINOMIAL THEOREM| EDUGROWN

In This Post we are providing Chapter-8 Binomial Theorem NCERT MCQ for Class 11 Math which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MCQ ON BINOMIAL THEOREM

Question 1.
The coefficient of y in the expansion of (y² + c/y)⁵ is
(a) 10c
(b) 10c²
(c) 10c³
(d) None of these

Answer: (c) 10c³

Question 2:
(1.1)¹⁰⁰⁰⁰ is _____ 1000
(a) greater than
(b) less than
(c) equal to
(d) None of these

Answer: (a) greater than

Question 3.
The fourth term in the expansion (x – 2y)¹² is
(a) -1670 x⁹ × y³
(b) -7160 x⁹ × y³
(c) -1760 x⁹ × y³
(d) -1607 x⁹ × y³

Answer: (c) -1760 x⁹ × y³

Question 4.
If n is a positive integer, then (√3+1)²ⁿ⁺¹ + (√3−1)²ⁿ⁺¹ is
(a) an even positive integer
(b) a rational number
(c) an odd positive integer
(d) an irrational number

Answer: (d) an irrational number

Question 5.
If the third term in the binomial expansion of (1 + x)^m is (-1/8)x² then the rational value of m is
(a) 2
(b) 1/2
(c) 3
(d) 4

Answer: (b) 1/2

Question 6.
The greatest coefficient in the expansion of (1 + x)¹⁰ is
(a) 10!/(5!)
(b) 10!/(5!)²
(c) 10!/(5! × 4!)²
(d) 10!/(5! × 4!)

Answer: (b) 10!/(5!)²

Question 7.
The coefficient of xⁿ in the expansion of (1 – 2x + 3x² – 4x³ + ……..)^-n is
(a) (2n)!/n!
(b) (2n)!/(n!)²
(c) (2n)!/{2×(n!)²}
(d) None of these

Answer: (b) (2n)!/(n!)²

Question 8.
The value of n in the expansion of (a + b)ⁿ if the first three terms of the expansion are 729, 7290 and 30375, respectively is
(a) 2
(b) 4
(c) 6
(d) 8

Answer: (c) 6

Question 9.
If α and β are the roots of the equation x² – x + 1 = 0 then the value of α²⁰⁰⁹ + β²⁰⁰⁹ is
(a) 0
(b) 1
(c) -1
(d) 10

Answer: (b) 1

Question 10.
The general term of the expansion (a + b)ⁿ is
(a) T_r+1 = ⁿC_r × a^r × b^r
(b) T_r+1 = ⁿC_r × a^r × b^n-r
(c) T_r+1 = ⁿC_r × a^n-r × b^n-r
(d) T_r+1 = ⁿC_r × a^n-r × b^r

Answer: (d) T_r+1 = ⁿC_r × a^n-r × b^r

Question 11.
The coefficient of xⁿ in the expansion (1 + x + x² + …..)^-n is
(a) 1
(b) (-1)ⁿ
(c) n
(d) n+1

Answer: (b) (-1)ⁿ

Question 12.
If n is a positive integer, then (√5+1)²ⁿ + 1 − (√5−1)²ⁿ + 1 is
(a) an odd positive integer
(b) not an integer
(c) none of these
(d) an even positive integer

Answer: (b) not an integer

Question 13.
In the expansion of (a + b)ⁿ, if n is even then the middle term is
(a) (n/2 + 1)^th term
(b) (n/2)^th term
(c) n^th term
(d) (n/2 – 1)^th term

Answer: (a) (n/2 + 1)^th term

Question 14.
In the expansion of (a + b)ⁿ, if n is odd then the number of middle term is/are
(a) 0
(b) 1
(c) 2
(d) More than 2

Answer: (c) 2

Question 15.
if n is a positive integer then 2³ⁿn – 7n – 1 is divisible by
(a) 7
(b) 9
(c) 49
(d) 81

Answer: (c) 49

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NCERT MCQ CLASS-11 CHAPTER-7 | MATH NCERT MCQ | PERMUTATIONS AND COMBINATIONS | EDUGROWN

In This Post we are providing Chapter-7 Permutations and Combinations NCERT MCQ for Class 11 Math which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MCQ ON PERMUTATIONS AND COMBINATIONS

Question 1. In how many ways can a consonants and a vowel be chosen out of the word COURAGE?
(a) ⁷C₂
(b) ⁷P₂
(c) ⁴P₁× ³P₁
(d)⁴P₁+ ³P1

Answer : C

Question 2. How many words can be formed from the letters of the word DOGMATIC, if all the vowels remain together :
(a) 4140
(b) 4320
(c) 432
(d) 43

Answer : B

Question 3. Numbers lying between 999 and 10000 than can be formed from the digits 0, 2, 3, 6, 7, 8 (repetition of digits not allowed) are :
(a) 100
(b) 200
(c) 300
(d) 400

Answer : C

Question 4. How many numbers of 6 digits can be formed from the digits of the number 112233?
(a) 30
(b) 60
(c) 90
(d) 120

Answer : C

Question 5. The sum of all five digit numbers that can be formed using the digits 1, 2, 3, 4, 5 when repetition of digits is not allowed, is
(a) 366000
(b) 660000
(c) 360000
(d) 3999960

Answer : D

Question 6. How many 10 digit numbers can be written by using the digits 1 and 2 :
(a) 10C1 + 9C2
(b) 210
(c) 10C2
(d) 10!

Answer : B

Question 7. In a test there were n questions. In the test 2n – i students gave wrong answers to i questions i = 1, 2, 3, ……..n. If the total number of wrong answers given is 2047 then n is
(a) 12
(b) 11
(c) 10
(d) None of these

Answer : C

Question 8. A set contains (2n + 1) elements. If the number of subsets of this set which contain at most n elements is 4096, then the value of n is
(a) 6
(b) 15
(c) 21
(d) None of these

Answer : D

Question 9. The number of numbers of 9 different non-zero digits such that all the digits in the first four places are less than the digit in the middle and all the digits in the last four places are greater than the digit in the middle is
(a) 2 (4 !)
(b) (4 !) 2
(c) 8 !
(d) None of these

Answer : B

Question 10. The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines:
(a) 6
(b) 9
(c) 18
(d) 12

Answer : C

Question 11. Ten different letters of an alphabet are given, words with five letters are formed. The number of words which at least one letter repeated is :
(a) 69760
(b) 30240
(c) 99748
(d) 37120

Answer : A

Question 12. There are 10 points in a plane, out of which 4 points are collinear. The number of triangles formed with vertices as there points is:
(a) 20
(b) 120
(c) 40
(d) 116

Answer : D

Question 13. In a college of 300 students, every student read 5 newspapers and every newspaper is read by 60 students. The number of newspaper is:
(a) at least 30
(b) at most 20
(c) exactly 25
(d) none of these

Answer : C

Question 14. The number of ways in which 6 rings can be worn on four fingers of one hand is :
(a) 46
(b) 6C4
(c) 64
(d) 24

Answer : C

Question 15. In a chess tournament where the participants were to play one game with one another, two players fell ill having played 6 games each, without playing among themselves. If the total number of games is 117, then the number of participants at the beginning was :
(a) 15
(b) 16
(c) 17
(d) 18

Answer : A

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NCERT MCQ CLASS-11 CHAPTER-6 | MATH NCERT MCQ | LINEAR INEQUALITIES| EDUGROWN

In This Post we are providing Chapter-6 Linear Inequalities NCERT MCQ for Class 11 Math which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MCQ ON LINEAR INEQUALITIES

Question 1.
Sum of two rational numbers is ______ number.
(a) rational
(b) irrational
(c) Integer
(d) Both 1, 2 and 3

Answer: (a) rational

Question 2.
If x² = -4 then the value of x is
(a) (-2, 2)
(b) (-2, ∞)
(c) (2, ∞)
(d) No solution

Answer: (d) No solution

Question 3.
Solve: (x + 1)² + (x² + 3x + 2)² = 0
(a) x = -1, -2
(b) x = -1
(c) x = -2
(d) None of these

Answer: (b) x = -1

Question 4.
If (x + 3)/(x – 2) > 1/2 then x lies in the interval
(a) (-8, ∞)
(b) (8, ∞)
(c) (∞, -8)
(d) (∞, 8)

Answer: (a) (-8, ∞)

Question 5.
The region of the XOY-plane represented by the inequalities x ≥ 6, y ≥ 2, 2x + y ≤ 10 is
(a) unbounded
(b) a polygon
(c) none of these
(d) exterior of a triangle

Answer: (c) none of these

Question 6.
The interval in which f(x) = (x – 1) × (x – 2) × (x – 3) is negative is
(a) x > 2
(b) 2 < x and x < 1
(c) 2 < x < 1 and x < 3
(d) 2 < x < 3 and x < 1

Answer: (d) 2 < x < 3 and x < 1

Question 7.
If -2 < 2x – 1 < 2 then the value of x lies in the interval
(a) (1/2, 3/2)
(b) (-1/2, 3/2)
(c) (3/2, 1/2)
(d) (3/2, -1/2)

Answer: (b) (-1/2, 3/2)

Question 8.
The solution of the inequality |x – 1| < 2 is
(a) (1, ∞)
(b) (-1, 3)
(c) (1, -3)
(d) (∞, 1)

Answer: (b) (-1, 3)

Question 9.
If | x − 1| > 5, then
(a) x∈(−∞, −4)∪(6, ∞]
(b) x∈[6, ∞)
(c) x∈(6, ∞)
(d) x∈(−∞, −4)∪(6, ∞)Answer

Answer: (d) x∈(−∞, −4)∪(6, ∞)

Question 10.
The solution of |2/(x – 4)| > 1 where x ≠ 4 is
(a) (2, 6)
(b) (2, 4) ∪ (4, 6)
(c) (2, 4) ∪ (4, ∞)
(d) (-∞, 4) ∪ (4, 6)

Answer: (b) (2, 4) ∪ (4, 6)

Question 11.
If (|x| – 1)/(|x| – 2) ‎≥ 0, x ∈ R, x ‎± 2 then the interval of x is
(a) (-∞, -2) ∪ [-1, 1]
(b) [-1, 1] ∪ (2, ∞)
(c) (-∞, -2) ∪ (2, ∞)
(d) (-∞, -2) ∪ [-1, 1] ∪ (2, ∞)

Answer: (d) (-∞, -2) ∪ [-1, 1] ∪ (2, ∞)

Question 12.
The solution of the -12 < (4 -3x)/(-5) < 2 is
(a) 56/3 < x < 14/3
(b) -56/3 < x < -14/3
(c) 56/3 < x < -14/3
(d) -56/3 < x < 14/3

Answer: (d) -56/3 < x < 14

Question 13.
If x² = -4 then the value of x is
(a) (-2, 2)
(b) (-2, ∞)
(c) (2, ∞)
(d) No solution

Answer: (d) No solution

Question 14.
Solve: |x – 3| < 5
(a) (2, 8)
(b) (-2, 8)
(c) (8, 2)
(d) (8, -2)

Answer: (b) (-2, 8)

Question 15.
The graph of the inequations x ≥ 0, y ≥ 0, 3x + 4y ≤ 12 is
(a) none of these
(b) interior of a triangle including the points on the sides
(c) in the 2nd quadrant
(d) exterior of a triangle

Answer: (b) interior of a triangle including the points on the sides

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NCERT Solutions for Class 12 Biology Chapter :10 Microbes in Human Welfare

NCERT Solutions for Class 12 Biology Chapter :10 Microbes in Human Welfare

Page No: 189

NCERT Solutions for Class 12 Biology Chapter : 9 Strategies for Enhancement in Food Production Biology

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