CLASS 12TH CHAPTER -13 Organisms and Populations |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 :13 Organisms and Populations

Question 1.
How is diapause different from hibernation ?
Solution:
Diapause is different from hibernation. The table below shows the differences between them :
NCERT Solutions for Class 12 Biology Chapter 13 Organisms and Populations Q1.1

Question 2.
If a marine fish is placed in a freshwater aquarium/will the fish be able to survive? Why or why not?
Solution:
If a marine fish is placed in a freshwater aquarium, then its chances of survival will diminish. This is because their bodies are adapted to high salt concentrations in the marine environment. In freshwater conditions, they are unable to regulate the water entering the body (through osmosis). Water enters their body due to the hypotonic environment outside. This results in the swelling up of the body, eventually leading to the death of the marine fish.

Question 3.
Define phenotypic adaptation. Give one example.
Solution:
Phenotypic adaptation involves non-genetic changes in individuals such as physiological modifications like acclimatization or behavioural changes.Question 4.
Most living organisms cannot survive at temperatures above 45°C. How are some microbes able to live in habitats with temperatures exceeding 100°C?
Solution:
organisms survive at a temperature range of 0° to 40°C or less. However, there are some notable exceptions. Certain microorganisms live in hot springs and deep-sea hydrothermal vents where temperature far exceeds 100°C. They survive at the high temperature due to the occurrence of branched-chain lipids in their cell membrane that reduces the fluidity of cell membranes and the occurrence of the minimum amount of free water in their cells that provides resistance to high temperature

Question 5.
List the attributes that populations but not individuals possess.
Solution:

  1. Natality
  2. Mortality
  3. Growth forms
  4. Population density
  5. Population dispersion
  6. Population age distribution

Question 6.
If a population growing exponentially double in size in 3 years, what is the intrinsic rate of increase (r) of the population?
Solution:
The intrinsic rate of increase(r), can be calculated by the following exponential growth equation:
NCERT Solutions for Class 12 Biology Chapter 13 Organisms and Populations Q6.1

Question 7.
Name important defence mechanisms in plants against herbivory.
Solution:

  1. Modification of leaves into thorns.
  2. Development of spiny margins on leaves.
  3. Development of sharp silicated edges on leaves.

Question 8.
An orchid plant is growing on the branch of the mango tree. How do you describe this interaction between the orchid and the mango tree?
Solution:
An orchid growing as an epiphyte on a branch of mango tree is an example of commensalism. Commensalism is the relationship between individuals of two species of which one is benefited and the other is almost unaffected, i.e., neither benefited nor harmed. A commensal may get shelter (protection), or ride, or support instead of or in addition to food. Epiphytes are space parasites, they use trees only for attachment and manufacture their own food by photosynthesis. In Vanda, an epiphytic orchid, a special kind of aerial roots (hanging roots) hang freely in the air and absorb moisture with the help of their special absorptive tissue called velamen.

Question 9.
What is the ecological principle behind the biological control method of managing pest insects?
Solution:
Predation is the means of biological control to manage pest insects where predators prey upon pests and regulate their numbers in the habitat.

Question 10.
Distinguish between the following:

  1. Hibernation and Aestivation
  2. Ectotherms and Endotherms

Solution:

  1. Differences between hibernation and aestivation are as follows :
    NCERT Solutions for Class 12 Biology Chapter 13 Organisms and Populations Q10.1
  2. Differences between ectotherms and endotherms are as follows:
    NCERT Solutions for Class 12 Biology Chapter 13 Organisms and Populations Q10.2

Question 11.
Write a short note on :
(a) Adaptations of desert plants and animals
(b) Adaptations of plants to water scarcity
(c) Behavioral adaptations in animals
(d) Importance of light to plants
(e) Effect of temperature or water scarcity and the adaptations of animals.
Solution:
a. Desert plants are called xerophytes. They have adaptations for increased water absorption, reduction in transpiration and water storage. Many desert plants have a thick cuticle on their leaf surfaces and have their stomata arranged in deep pits to minimise water loss through transpiration. They also have a special photosynthetic pathway that enables their stomata to remain closed during day time. In desert plants like Opuntia, leaves are reduced to spines. Animals of dry areas may use metabolic water and reduce water loss bypassing nearly solid faeces and urine.

b. Xerophytes have special adaptations to withstand prolonged periods of drought. These are of four types – ephemerals, annuals, succulents and non-succulent perennials.

  • Ephemerals (drought escapers): Plants which live for a brief period and complete their life cycle during the rains.
  • Annuals (drought evaders): Plants which continue to live for a few
    months even after rains in hot dry conditions. They have modifications to reduce transpiration.
  • Succulents (drought resistants): Plants have fleshy organs to store large amounts of water. They have a very thick cuticle, sunken stomata which open during night only.
  • Non-succulent perennials: These are true xerophytes. They have an extensive root system to absorb the maximum amount of water. They possess waxy coatings on leaves, sunken stomata, reduced leaf blades etc. to reduce transpiration.

c. The animals with variable temperatures called poikilotherms are affected by temperature variations. They are also called ectotherms. They show different adaptations like hibernation, aestivation, periodic activity, winter eggs, and migration.

d. Sun is the ultimate source of energy for most of the organisms on this earth. Light is the visible range of the electromagnetic spectrum. Light (400 nm-700nm) is effective in photosynthesis and is called photo-synthetically active radiation or PAR. The intensity of light, duration of light, etc. are also influencing the growth of plants.

e. Animals live in arid regions show two kinds of adaptations

  1. Reducing loss of water from their bodies.
  2. Ability to tolerate arid conditions.

Question 12.
List the various abiotic environmental factors.
Solution:
Abiotic factors are non-living factors and conditions of the environment which influence the survival, function and behaviour of organisms. Various abiotic factors are :

(i) Temperature – Temperature is one of the most important environmental factors. The average temperature varies seasonally. It ranges from subzero level in polar areas and high altitudes to more than 50°C in tropical deserts in summer and exceeds 100°C in thermal springs and deep-sea hydrothermal vents.

(ii) Water – Next to temperature, water is the most important factor which influences the life of organisms. The productivity and distribution of land plants are dependent upon the availability of water. Animals are adapted according to water availability. E.g., aquatic animals are ammonotelic while xerophytic animals excrete dry feces and concentrated urine.

(iii) Light – Plants produce food through photosynthesis for which sunlight is essential to the source of energy. Light intensity, light duration and light quality influences the number of life processes in organisms, such as – photosynthesis, growth, transpiration, germination, pigmentation, movement and photoperiodism.

(iv) Humidity – Humidity refers to the moisture (water vapour) content of the air. It determines the formation of clouds, dew and fog. It affects the land organisms by regulating the loss of water as vapour from their bodies through evaporation, perspiration and transpiration.

(v) Precipitation – Precipitation means rainfall, snow, sleet or dew. Total annual rainfall, seasonal distribution humidity of the air and amount of water retained in the soil are the main criteria that limit the distribution of plants and animals on land.

(vi) Soil – The soil is one of the most important ecological factor called the edaphic factor. It comprises of different layers called horizons. The upper weathered humus containing part of soil sustains terrestrial plant life.

Question 13.
Give an example for:

  1. An endothermic animal
  2. An ectothermic animal
  3. An organism of the benthic zone.

Solution:

  1. Hedgehog
  2. Frog
  3. Sponges

Question 14.
Define population and community.
Solution:
Population: A population is a group of individuals of the same species, which can reproduce among themselves and occupy a particular area in a given time.

Community: It is an assemblage of several populations in a particular area and time and exhibits interaction and interdependence through trophic relationship.

Question 15.
Define the following terms and give one example for each.
(a) Commensalism
(b) Parasitism
(c) Camouflage
(d) Mutualism
(e) Interspecific competition
Solution:

a. Commensalism is an interspecific interaction between individuals of two species where one species is benefitted and the other is not affected.
e. g. Orchid and mango tree.

b. Parasitism is an interspecific interaction between individuals of two species where generally small species is benefitted and the large species are affected, e.g. Malarial parasite and human beings.

c. Camouflage: It is the ability of the animals to blend with the surroundings or background. In this way, animals remain unnoticed for protection or aggression. An example is a stick insect.

d. Mutualism is an interspecific interaction between individuals of two species where both the interacting species are benefitted in an obligatory way. e.g. Pollination in plants by animals.

e. Interspecific competition: It is an interaction between individuals of two species where both the interacting species are affected, e.g. Monarch butterfly and Queen monarch.

Question 16.
With the help of a suitable diagram describe the logistic population growth curve.
Solution:
Logistic population growth curve or S-shaped or sigmoid growth curve is shown by the populations of most organisms. It has the following phases: lag phase, log phase, exponential phase and stationary phase. In lag phase there is little or no increase in population. In log phase increase in population starts and occurs at a slow rate in the beginning. During exponential phase, increase in population becomes rapid and soon attains its full potential rate. This is due to the constant environment, availability of food and other requirements of life in plenty, absence of predation and interspecific competition and no serious intraspecific competition so that the curve rises steeply upward. The growth rate finally slows down as environmental resistance increases.

Finally, the population becomes stable during the stationary phase because now the number of new cells produced almost equals to the number of cells that die. Every population tends to reach a number at which it becomes stabilized with the resources of its environment. A stable population is said to be in equilibrium, or at saturation level. This limit in population is a constant K and is imposed by the carrying capacity of the environment. The sigmoid growth form is represented by the following equation :
NCERT Solutions for Class 12 Biology Chapter 13 Organisms and Populations Q16.1
r = intrinsic rate of natural increase
N = population density at time t; K = carrying capacity.

Question 17.
Select the statement which best explains parasitism.
(a) One organism is benefited.
(b) Both the organisms are benefited
(c) One organism is benefited, other is not affected
(d) One organism is benefited, other is affected.
Solution:
(d) One organism is benefited, other is affected,

Question 18.
List any three important characteristics of a population and explain.
Solution:
The three important characteristics of a population are:

  1. Birth and death rate
  2. Age structure
  3. Sex ratio

(i) The birth rate (natality) of a population refers to the average number of young ones produced per unit time (usually per year). In the case of humans, it is commonly expressed as the number of births per 1,000 individuals in the population per year. The death rate (mortality) of a population is the average number of individuals that die per unit time (usually per year). In humans, it is commonly expressed as the number of deaths per 1,000 persons in a population per year.

(ii) The age structure of a population is the percentage of individuals of different ages such as young, adult and old. Age structure is shown bv organisms in which individuals of more than one generation coexist. The ratio of various age groups in a population determines the current reproductive status of the population. It also indicates what may be expected in the future. The population is divided into three age groups; pre-reproductive, reproductive and post-reproductive.

(iii) The sex ratio of a population refers to the number of females per thousand male individuals. There were 933 females per 1,000 males in our country in the 2001 census. The number of females in a population is very important as it is often directly related to the number of births. The number of males may be less significant because in many species a single male can mate with several females.

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CLASS 12TH CHAPTER -12 Biotechnology and Its Applications |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 :12 Biotechnology and Its Applications

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 encloses toxin in a special sac.
Solution:
(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 an alkaline medium.

Question 2.
What are transgenic bacteria? Illustrate using any one example.
Solution:
Transgenic bacteria are one that carries a transgene or a foreign gene of interest introduced using recombinant DNA technology. e. g., bacteria carrying the genes for human insulin.

In 1983, Eli Lilly an American company prepared two DNA sequences corresponding to A and B, chains of human insulin and introduced them in plasmids of E. coli to produce insulin chains. Chains A and B were produced separately, extracted and combined by creating disulfide bonds to form human insulin.

Question 3.
Compare and contrast the advantages and disadvantages of the production of genetically modified crops.
Solution:
Advantages of genetically modified crops or transgenic crops are as follows :

  • They are resistant to pests, herbicides and diseases.
  • They help to reduce post-harvest losses.
  • They enhance the nutritional value of food, e.g., a transgenic variety of rice (golden rice) is rich in vitamin A content.
  • Some transgenic plants, e.g., poplar trees are used to clean up heavy metal pollution from contaminated soil.
  • They are efficient in mineral usage and thus prevent early exhaustion of fertility of the soil.

Transgenic crops have several disadvantages also which are mentioned below:

  • Bt toxins expressed in pollen grains of transgenic crops are harmful for useful varieties of insects, e.g., honey bees and butterflies.
  • The foods produced by transgenic crops might cause toxicity and might result in allergies.
  • The bacteria present in human alimentary canal can become resistant to concerned antibiotic by taking up antibiotic resistance gene present in genetically modified food and become difficult to manage.

Question 4.
What are Cry proteins? Name an organism that produces it. How has man exploited this protein to his benefit?
Solution:
Cry proteins are a group of toxic protein which are highly poisonous to deficient types of insects. It is produced by a soil bacterium Bacillus thuringiensis. The genes controlling their formation are called cry genes eg:- Cry I Ab, Cry I Ac, Cry II Ab, The bacterium produces a protein in the crystal form of protoxin. Two cry genes have been incorporated in cotton (Bt cotton) while one has been introduced in corn (Bt corn) As a result Bt Cotton was disease resistant to bollworm and Bt corn was resistant to corn borer.

Question 5.
What is gene therapy? Illustrate using the example of adenosine deaminase (ADA) deficiency.
Solution:
Gene therapy is the technique of genetic engineering used to replace a faulty gene with a normal, healthy functional gene. The first clinical gene therapy was given in 1990 to a 4 years old girl with adenosine deaminase deficiency (ADA deficiency). This enzyme is very important for the immune system to function. Severe combined immunodeficiency (SCID) is caused due to a defect in the gene for the enzyme adenosine deaminase. SCID patient lacks functional T-lymphocytes and, therefore, fails to fight the infecting pathogens.
NCERT Solutions for Class 12 Biology Chapter 12 Biotechnology and Its Applications Q5.1
To perform gene therapy, lymphocytes are extracted from the patient’s bone marrow and a normal functional copy of human gene coding for ADA is introduced into these lymphocytes with the help of a retroviral vector. The cells so treated are reintroduced into the patient’s bone marrow. The lymphocytes produced by these cells contain functional ADA genes which reactivate the victim’s immune system. But, as these lymphocytes do not divide and are short-lived, so periodic infusion of genetically engineered lymphocytes is required. This problem can be overcome if stem cells are modified at an early embryonic stage.

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?
Solution:
The given diagram represents the experimental steps in cloning and expressing a human gene for growth hormone into a bacterium E. coli.
NCERT Solutions for Class 12 Biology Chapter 12 Biotechnology and Its Applications Q6.1

Question 7.
Can you suggest a method to remove oil (hydrocarbon) from seeds based on your understanding of rDNA technology and the chemistry of oil?
Solution:
rDNA technology is a technique of genetic engineering that involves combining DNA from two different sources to produce recombined or recombinant DNA (rDNA). Oils are composed of glycerol and fatty acids. Thus, to produce oil-free seeds genes coding for glycerol or fatty acids should be identified and nucleotide sequences complementary to the sequence of these genes should be inserted adjacent to these genes in the early cells of the endosperm. During transcription, these complementary sequences will produce anti-sense RNAs to the RNAs produced by glycerol or fatty acids gene and will silence these genes. As a result, oil-free seeds will be produced.

Since glycerol is a common component of all the oils whereas various fatty acids combine with glycerol to form oils, thus it will be easier if we silence the gene for glycerol synthesis.

Question 8.
Find out from the internet what is golden rice.
Solution:
Golden rice is a GM rice with increased vitamin A content.

Question 9.
Does our blood have proteases and nucleases?
Solution:
Proteases occur naturally in all organisms. These enzymes are involved in a multitude of physiological reactions from simple digestion of food proteins to highly-regulated cascades (e.g., the blood-clotting cascade, the complement system, apoptotic pathways, and the invertebrate prophenoloxidase activating cascade). Proteases present in blood serum (thrombin, plasmin, Hageman factor, etc.) play important role in blood clotting, as well as in lysis of the clots, and the action of the immune system. Other proteases are present in leukocytes (elastase, cathepsin G) and play several different roles in metabolic control. Nucleases, such as deoxyribonucleases and ribonucleases are found in the blood which helps in the degradation of exogenous deoxyribonucleic acid and ribonucleic acid circulating in the blood.

Question 10.
Consult the internet and find out how to make orally active protein pharmaceuticals. What is the major problem to be encountered?
Solution:
The problem is stomach enzymes and acids. Once you orally ingest a protein, the proteases in your stomach juices (trypsin, chymotrypsin, pepsin) will cleave the holy-hell out of your therapeutic protein and the acids will denature whatever’s left beyond all recognition. This is why proteins like insulin have to be injected.

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NCERT MCQ CLASS-11 CHAPTER-13 | MATH NCERT MCQ | LIMITS AND DERIVATIVES| EDUGROWN

In This Post we are  providing Chapter-13 Limits and Derivatives 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 LIMITS AND DERIVATIVES

Question 1. limx→4 |x-4 | /x – 4 is equal to     
(a) 1
(b) –1
(c) does not exist
(d) None of these

Answer :  C

Question 2. If y = √x + 1/√x , then dy/dx at x = 1 is

(a) 1
(b) 1/2
(c) 1/√2
(d) 0

 Answer :  D

Question 3. limx→0 |Sinx | /x  is      
(a) 1
(b) –1
(c) does not exist
(d) None of these

Answer :  C

Question 4. If y =

Class 11 Mathematics Limits And Derivatives MCQs

(a) -4x / (x2 – 1)2

(b) -4x / (x2 – 1)
(c)  1 – x2 / 4x
(d) 4x / x2 – 1
 

Answer :  A

Question 5. limx→0 tan2x – x / 3x – Sinx      
(a) 2
(b) 1/2
(c) -1/2
(d) 1/4

Answer :  B

Question 6. The derivative of (3x + 5)(1 + tan x) is     
(a) 3x sec2 x 6 sec2 x 3 + 3 tan x
(b) 3x sec2 x  5 sec2 x 3 3 tan x
(c) 3x sec2 x + 5 sec2 x + 3 + 3 tan x
(d) None of the above

Answer :  C

Question 7. The derivative of (ax2 + cot x)( p + q cos x) is     
(a) ax2 cot (q sin ) + ( p+ q cosx )(2ax cosec2x )
(b) ax2 cot (q sin ) + ( p+ q cosx )(2ax cosec2x )
(c) ax2 cot (q sin ) + ( p q cosx )(2ax cosec2x )
(d) None of the above

Answer :  A

Question 8. The derivative of (x sinx cosx) 2 + 2 is      
(a) x2 cos x + 2x sin x sin 2x
(b) x2 cos x + 2x sin x sin 2x
(c) x2 cos x + x sin x sin 2x
(d) x2 cos x + 2x sin x + sin 2x

Answer :  B

Question 9. The derivative of sin3x cos3x xis       
(a) 3 /5 sin2 2x cos 2x
(b) 3 /4 sin2 2x cos 2x
(c) 3 /4 sin 2x cos2 2x 
(d) 3 /2 sin2 2x cos 2x

Answer :  B

Question 10. The derivative of x5 cos x /sinx is

Class 11 Mathematics Limits And Derivatives MCQs

Answer :  B

Question 11. Find the differentiation of x2/3 by using first principle.     
(a) 2/3 x-1/3
(b) -2/3 x-1/3
(c) 2/3 x 1/3
(d) None of these

Answer :  A

Question 12. Find the differentiation of cos (x2 + 1 ) by using first principle.      
(a) 2x sin (x2 + 1 )
(b) 2x sin (x2 + 1 )
(c) -2x sin (x2 – 1 )
(d) None of these

Answer :  B

Question 13. If  f (x )= x-4 /2√x then f ‘(1) is

(a) 5/4
(b) 4/5
(c) 1
(d) 0

Answer :  A

Question 14. Find the differentiation of x cos x by using first principle.   
(a) x sin x + cos x
(b) x sin x + cos x
(c) x cos x cos x
(d) None of these

Answer :  A

Question 15. If y = Sin (x+9) / Cos x , then dy/dx at x = 0 is    
(a) cos 9
(b) sin 9
(c) 0
(d) 1

Answer :  A

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NCERT MCQ CLASS-11 CHAPTER-12 | MATH NCERT MCQ | INTRODUCTION TO THREE DIMENSIONAL GEOMETRY| EDUGROWN

In This Post we are  providing Chapter-12 Introduction to Three Dimensional Geometry 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 INTRODUCTION TO THREE DIMENSIONAL GEOMETRY

Question 1. The coordinates of a point which is equidistant from the points (0, 0, 0), (a, 0, 0), (0, b, 0), (0, 0, c) are given by
(a) (a/2 , b/2 , c/2,)
(b)  ( -a/2 , -b/2 , c/2,)
(c)  (a/2 , -b/2 , c/2,)
(d) (-a/2 , b/2 , c/2,)

Answer :   A

Question 2. If x2+y2 = 1, then the distance from the point (x, y, 1-x2-y2 )  to the origin is 
(a) 1
(b) – 1
(c) 0
(d) 2

Answer :   A

Question 3. Three vertices of a parallelogram ABCD are A(1, 2, 3), B(-1, -2, -1) and C(2, 3, 2). Find the fourth vertex D. 
(a) (– 4, – 7, – 6)
(b) (4, 7, 6)
(c) (4, 7, – 6)
(d) None of these

Answer :   B

Question 4. If a parallelopiped is formed by planes drawn through the points (5, 8, 10) and (3, 6, 8) parallel to the coordinate planes, then the length of diagonal of the parallelopiped is 
(a)2 3
(b)32
(c) 2
(d) 3

Answer :   A

Question 5. The points (5, – 4, 2),(4,– 3, 1), (7, -6, 4) and (8, – 7, 5) are the vertices of 
(a) a rectangle
(b) a square
(c) a parallelogram
(d) None of these

Answer :   C

Question 6. If the coordinates of the vertices of a ΔABC are A(-1, 3, 2), B(2, 3, 5) and C(3, 5, – 2), then ? ∠A is equal to
(a) 45°
(b) 60°
(c) 90°
(d) 30°

Answer :   A

Question 7.  Find the ratio in which the YZ-plane divides the line segment formed by joining the points (– 2, 4, 7) and (3, – 5, 8).
(a) externally 2 : 3
(b) internally 2 : 3
(c) internally 3 : 2
(d) externally 3 : 2

Answer :   B

Question 8. Find the coordinates of the point which divides the line segment joining the points (– 2, 3, 5) and (1, -4, 6) in the ratio 2 : 3 externally.
(a) (– 8, – 17, 3)
(b) (– 8, 17, 3)
(c) (8, – 17, 3)
(d) None of these

Answer :   B

Question 9. Find the length of the medians of the triangle with vertices A(0, 0, 6), B(0, 4, 0) and C(6, 0, 0).
(a) 7, 7, 34
(b) 7, 8, 34
(c) 7, 9, 34
(d) None of these

Answer :   A

Question 10. Find the coordinates of the points which trisect the line segment joining the points P(4 , 2, -6) and Q(10, -16, 6).
(a) (6, – 4, – 2), (8, – 10, 2)
(b) (6, 4, – 2), (8, – 10, 2)
(c) (6, – 4, – 2), (8, 10, 2)
(d) None of these

Answer :   B

Question 11. Find the centroid of a triangle, the mid-point of whose sides are D(1, 2, -3), E(3, 0, 1) and F (-1, 1, -4).
(a) (1, 1, 2)
(b) (1, 1, – 2)
(c) (– 1, –1, –2)
(d) (1, –1, –2)

Answer :   B

Question 12. The points A(5, -1, 1), B(7, -4, 7), C(1, -6, 10) and D(-1, -3, 4) are vertices of a 
(a) square
(b) rhombus
(c) rectangle
(d) None of these

Answer :   B

Question 13. If vertices of a triangle are A(1, -1, 2), B(2, 0, -1) and C(0, 2, 1), then the area of a triangle is (a) √6
(b) 26
(c) 36
(d) 46

Answer :   B

Question 14. The point ( -2, -3, -4) lies in the  
(a) first octant
(b) seventh octant
(c) second octant
(d) eight octant

Answer :   B

Question 15. The mid-points of the sides of a triangle are (5, 7, 11), (0, 8, 5) and (2, 3, – 1). Then, the vertices are
(a) (7, 2, 5), (3, 12, 17), (– 3, 4, – 7)
(b) (7, 2, 5), (3, 12, 17), ( 3, 4, 7)
(c) (7, 2, 5), (– 3, 12, 17), (– 3, – 4, – 7)
(d) None of the above

Answer :   A

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NCERT MCQ CLASS-11 CHAPTER-11 | MATH NCERT MCQ | CONIC SECTIONS | EDUGROWN

In This Post we are  providing Chapter-11 Conic Sections 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 CONIC SECTIONS

Question 1.
The straight line y = mx + c cuts the circle x² + y² = a² in real points if

(a) √{a² × (1 + m²)} < c
(b) √{a² × (1 – m²)} < c
(c) √{a² × (1 + m²)} > c
(d) √{a² × (1 – m²)} > c

Answer: (c) √{a² × (1 + m²)} > c
The straight line y = mx + c cuts the circle x² + y² = a² in real points if
√{a² × (1 + m²)} > c

Question 2.
Equation of the directrix of the parabola x² = 4ay is

(a) x = -a
(b) x = a
(c) y = -a
(d) y = a

Answer: (c) y = -a
Given, parabola x² = 4ay
Now, its equation of directrix = y = -a

Question 3.
The equation of parabola with vertex at origin and directrix x – 2 = 0 is
(a) y² = -4x
(b) y² = 4x
(c) y² = -8x
(d) y² = 8xAnswer

Answer: (c) y² = -8x
Since the line passing through the focus and perpendicular to the directrix is x-axis,
therefore axis of the required parabola is x-axis.
Let the coordinate of the focus is S(a, 0).
Since the vertex is the mid point of the line joining the focus and the point (2, 0) where the directix is x – 2 = 0 meets the axis.
So, 0 = {a – (-2)}/2
⇒ 0 = (a + 2)/2
⇒ a + 2 = 0
⇒ a = -2
Thus the coordinate of focus is (-2, 0)
Let P(x, y) be a point on the parabola.
Then by definition of parabola
(x + 2)² + (y – 0)² = (x – 2)²
⇒ x² + 4 + 4x + y² = x² + 4 – 4x
⇒ 4x + y² = – 4x
⇒ y² = -4x – 4x
⇒ y² = -8x
This is the required equation of the parabola.

Question 4.
The perpendicular distance from the point (3, -4) to the line 3x – 4y + 10 = 0
(a) 7

(b) 8
(c) 9
(d) 10Answer

Answer: (a) 7
The perpendicular distance = {3 × 3 – 4 × (-4) + 10}/√(3² + 4²)
= {9 + 16 + 10}/√(9 + 16)
= 35/√25
= 35/5
= 7

Question 5.
The equation of a hyperbola with foci on the x-axis is

(a) x²/a² + y²/b² = 1
(b) x²/a² – y²/b² = 1
(c) x² + y² = (a² + b²)
(d) x² – y² = (a² + b²)

Answer: (b) x²/a² – y²/b² = 1
The equation of a hyperbola with foci on the x-axis is defined as
x²/a² – y²/b² = 1

Question 6.
If the line 2x – y + λ = 0 is a diameter of the circle x² + y² + 6x − 6y + 5 = 0 then λ =
(a) 5

(b) 7
(c) 9
(d) 11

Answer: (c) 9
Given equation of the circle is
x² + y² + 6x − 6y + 5 = 0
Center O = (-3, 3)
radius r = √{(-3)² + (3)² – 5} = √{9 + 9 – 5} = √13
Since diameter of the circle passes through the center of the circle.
So (-3, 3) satisfies the equation 2x – y + λ = 0
⇒ -3 × 2 – 3 + λ = 0
⇒ -6 – 3 + λ = 0
⇒ -9 + λ = 0
⇒ λ = 9

Question 7.
The number of tangents that can be drawn from (1, 2) to x² + y² = 5 is
(a) 0

(b) 1
(c) 2
(d) More than 2Answer

Answer: (b) 1
Given point (1, 2) and equation of circle is x² + y² = 5
Now, x² + y² – 5 = 0
Put (1, 2) in this equation, we get
1² + 2² – 5 = 1 + 4 – 5 = 5 – 5 = 0
So, the point (1, 2) lies on the circle.
Hence, only one tangent can be drawn.

Question 8.
The equation of the circle x² + y² + 2gx + 2fy + c = 0 will represent a real circle if

(a) g² + f² – c < 0
(b) g² + f² – c ≥ 0
(c) always
(d) None of these

Answer: (b) g² + f² – c ≥ 0
Given, equation of the circle is: x² + y² + 2gx + 2fy + c = 0
This equation can be written as
{x – (-g)}² + {y – (-f)}² + = √{g² + f² – c}²
So, the circle is real is g² + f² – c ≥ 0

Question 9.
The equation of parabola whose focus is (3, 0) and directrix is 3x + 4y = 1 is

(a) 16x² – 9y² – 24xy – 144x + 8y + 224 = 0
(b) 16x² + 9y² – 24xy – 144x + 8y – 224 = 0
(c) 16x² + 9y² – 24xy – 144x – 8y + 224 = 0
(d) 16x² + 9y² – 24xy – 144x + 8y + 224 = 0

Answer: (d) 16x² + 9y² – 24xy – 144x + 8y + 224 = 0
Given focus S(3, 0)
and equation of directrix is: 3x + 4y = 1
⇒ 3x + 4y – 1 = 0
Let P (x, y) be any point on the required parabola and let PM be the length of the perpendicular from P on the directrix
Then, SP = PM
⇒ SP² = PM²
⇒ (x – 3)² + (y – 0)² = {(3x + 4y – 1) /{√(3² + 4²)}²
⇒ x² + 9 – 6x + y² = (9x² + 16y² + 1 + 24xy – 8y – 6x)/25
⇒ 25(x² + 9 – 6x + y²) = 9x² + 16y² + 1 + 24xy – 8y – 6x
⇒ 25x² + 225 – 150x + 25y² = 9x² + 16y² + 1 + 24xy – 8y – 6x
⇒ 25x² + 225 – 150x + 25y² – 9x² – 16y² – 1 – 24xy + 8y + 6x = 0
⇒ 16x² + 9y² – 24xy – 144x + 8y + 224 = 0
This is the required equation of parabola.

Question 10.
If the parabola y² = 4ax passes through the point (3, 2), then the length of its latusrectum is
(a) 2/3
(b) 4/3
(c) 1/3
(d) 4

Answer: (b) 4/3
Since, the parabola y² = 4ax passes through the point (3, 2)
⇒ 2² = 4a × 3
⇒ 4 = 12a
⇒ a = 4/12
⇒ a = 1/3
So, the length of latusrectum = 4a = 4 × (1/3) = 4/3

Question 11.
The eccentricity of an ellipse is?

(a) e = 1
(b) e < 1
(c) e > 1
(d) 0 < e < 1

Answer: (d) 0 < e < 1
The eccentricity of an ellipse e = √(1 – a²/b²) and 0 < e < 1

Question 12.
If the length of the tangent from the origin to the circle centered at (2, 3) is 2 then the equation of the circle is
(
a) (x + 2)² + (y – 3)² = 3²
(b) (x – 2)² + (y + 3)² = 3²
(c) (x – 2)² + (y – 3)² = 3²
(d) (x + 2)² + (y + 3)² = 3²

Answer: (c) (x – 2)² + (y – 3)² = 3²
Radius of the circle = √{(2 – 0)² + (3 – 0)² – 2²}
= √(4 + 9 – 4)
= √9
= 3
So, the equation of the circle = (x – 2)² + (y – 3)² = 3²

Question 13.
If the length of the major axis of an ellipse is three times the length of the minor axis then its eccentricity is
(a) 1/3
(b) 1/√3
(c) 1/√2
(d) 2√2/√3

Answer: (d) 2√2/√3
Given, the length of the major axis of an ellipse is three times the length of the minor axis
⇒ 2a = 3(2b)
⇒ 2a = 6b
⇒ a = 3b
⇒ a² = 9b²
⇒ a² = 9a² (1 – e²) {since b² = a²(1 – e²)}
⇒ 1 = 9(1 – e²)
⇒ 1/9 = 1 – e²
⇒ e² = 1 – 1/9
⇒ e² = 8/9
⇒ e = √(8/9)
⇒ e = 2√2/√3
So, the eccentricity of the ellipse is 2√2/√3

Question 14.
The equation of parabola with vertex at origin and directrix x – 2 = 0 is

(a) y² = -4x
(b) y² = 4x
(c) y² = -8x
(d) y² = 8x

Answer: (c) y² = -8x
Since the line passing through the focus and perpendicular to the directrix is x-axis,
therefore axis of the required parabola is x-axis.
Let the coordinate of the focus is S(a, 0).
Since the vertex is the mid point of the line joining the focus and the point (2, 0) where the directix is x – 2 = 0 meets the axis.
So, 0 = {a – (-2)}/2
⇒ 0 = (a + 2)/2
⇒ a + 2 = 0
⇒ a = -2
Thus the coordinate of focus is (-2, 0)
Let P(x, y) be a point on the parabola.
Then by definition of parabola
(x + 2)² + (y – 0)² = (x – 2)²
⇒ x² + 4 + 4x + y² = x² + 4 – 4x
⇒ 4x + y² = – 4x
⇒ y² = -4x – 4x
⇒ y² = -8x
This is the required equation of the parabola.

Question 15.
In an ellipse, the distance between its foci is 6 and its minor axis is 8 then its eccentricity is
(a
) 4/5
(b) 1/√52
(c) 3/5
(d) 1/2

Answer: (c) 3/5
Given, distance between foci = 6
⇒ 2ae = 6
⇒ ae = 3
Again minor axis = 8
⇒ 2b = 8
⇒ b = 4
⇒ b² = 16
⇒ a² (1 – e²) = 16
⇒ a² – a² e² = 16
⇒ a² – (ae)² = 16
⇒ a² – 3² = 16
⇒ a² – 9 = 16
⇒ a² = 9 + 16
⇒ a² = 25
⇒ a = 5
Now, ae = 3
⇒ 5e = 3
⇒ e = 3/5
So, the eccentricity is 3/5



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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.

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: 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., Sn = qn2 and Sm = qm2, where Sr 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    
3x + 31 – 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.

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 : 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:

  1. 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.
  2. Egg, like milk, is also a complete food. Chicken and duck are the two major sources of the egg.
  3. Meat is a protein-rich diet that is obtained from all types of livestock, e.g., goat, sheep, pig, cattle, chicken, fish, etc.
  4. Honey is a sweet syrup obtained from the hives of the honey bee. Honey is used in sweetening various preparations.
  5. 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.
  6. The skins of many animals are converted into hides and leather.
  7. Drought animals are trained to carry men and materials besides other functions, e.g., buffalo, bullock, horse, camel, ass, elephant, reindeer, yak.
  8. The rearing of animals provides employment to many persons.
  9. 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 :

  1. Selection of good breeds having high milk yielding potential according to the climatic conditions of the area.
  2. 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.
  3. 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.
  4. 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.
  5. 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 PlantHybrid Variety
WheatSonalika and kalian sona
RiceJaya and Ratna
Cauliflower        Pusa shubra and Pusa snowball K-1
CowpeaPusa komal
MustardPusa swarnim


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