NCERT Class 11 Biology Solutions

NCERT Solutions for Class 11 Biology Chapter 15 Plant Growth and Development

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NCERT Solutions for Class 11 Biology Chapter 15 Plant Growth and Development

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 15 Plant Growth and Development.

Question 1.
Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem, and growth rate.
Solution:
(1) Growth: Growth is accompanied by metabolic processes (both anabolic and catabolic), that occur at the expense of energy. For example, the expansion of a leaf is growth.

(2) Differentiation: The cells derived from root apical and shoot apical meristems and cambium differentiate and mature to perform specific functions. This act leading to maturation is termed differentiation. They undergo a few or major structural changes both in their cell walls and protoplasm. For example, to form a tracheary element, the cells would lose their protoplasm.

(3) Development: Development is a term that includes all changes that an organism goes through during its life cycle from germination of the seed to senescence. It is also applicable to tissues/organs.

(4) Dedifferentiation: Plants show another interesting phenomenon. The living differentiated cells, that by now have lost the capacity to divide can regain the capacity of the division under certain conditions. This phenomenon is termed dedifferentiation. For example, interfascicular cambium and cork cambium.

(5) Redifferentiation: While doing so, such meristems/ tissues are able to produce cells that once again lose the capacity to divide but mature to perform specific functions, i.e., get redifferentiated. List some of the tissues in a woody dicotyledonous plant that are the products of redifferentiation.

(6) Determinate growth: The growth in plants is open i.e. it can be indeterminate or determinate. Even differentiation in plants is open because cells tissues arising out of the same meristem have different structures at maturity. The final structure at maturity of a cell/’ tissue is also determined by the location of the cell within. For example, cells positioned away from root apical meristems differentiate as rootcap cells, while those pushed to the periphery mature as the epidermis.

(7) Meristem growth: The constantly dividing cells, both at the root apex and shoot apex, represent the meristematic phase of growth.

(8) Growth rate: The increased growth per unit time is termed as growth rate. Thus, the rate of growth can be expressed mathematically. An organism or a part of the organism can produce more cells in a variety of ways. The growth rate shows an increase that may be arithmetic or geometrical.

Question 2.
Why is not any one parameter good enough to demonstrate growth throughout the life of a flowering plant?
Solution:
In plants, growth is said to have taken place when the number of protoplasm increases. Measuring the growth of protoplasm involves many parameters such as weight of fresh tissue sample, the weight of dry tissue sample, differences in length, area, volume and cell number measured during the growth period. Measuring the growth of plants using only one parameter does not provide enough information and hence, is insufficient for demonstrating growth.

Question 3.
Describe briefly:
(a) Arithmetic growth
(b) Geometric growth
(c) Sigmoid growth curve
(d) Absolute and relative growth rates
Solution:
(i) Arithmetic growth: In this type of growth after mitosis, only one daughter cell continues to divide while the others take part in differentiation and maturation e.g., root elongating at constant rate. Here a ‘linear curve is obtained.
(ii) Geometric growth : In most systems, the initial growth is slow (lag phase), and it increases rapidly thereafter – at an exponential rate (log or exponential phase), Here both the progeny cells following mitotic cell division divide continuously.
(iii) Sigmoid growth curve : Sigmoid or S-shaped growth curve consists of three phases i.e., lag phase, log phase and stationary phase. During lag phase plant i growth is slow (in phase of cell division), but increases at log or exponential phase , (due to cell enlargement). During stationary phase the growth again slows down due f to the limitation of nutrients.
(iv) Absolute and relative growth rates : Measurement and comparison of total growth per unit time is called the absolute growth rate. The growth of the given system per unit time expressed on a common basis e.g., per unit initial parameter is called the relative growth rate,

Question 4.
List flve main groups of natural plant growth regulators. Write a note on discovery, physiological functions and agricultural/ horticultural applications of any one of them.
Solution:
The five main groups of natural growth regulators are
(a) auxins
(b) gibberellins
(c) cytokinins
(d) ethylene
(e) abscisic acid
Gibberellins
Discovery: – They are another kind of promotory PGR. There are more than 100 gibberellins reported from different organisms such as fungi and higher plants. They are denoted as GA1, GA2, GA3. E. Kurosawa reported the symptoms of the disease r in infected rice seedings when they were treated with filtrates of the fungus. Gibberalla fujikuroi caused, ‘bakane’ (foolish seedling) a disease of rice seedlings. The active substances were later identified as gibberellic acid.
Physiological functions
(i) They cause an increase in length of axis is used to increase the length of grapes stalk.
(ii) Gibberellins cause fruit like apple to elongate and improve its shape.
(iii) They also delay senescence. Thus the fruits can be left on the tree longer so as to extend the market period.
Agricultural Applications
(i) Spraying sugarcane crop with gibberellins increases the length of stem. Thus increasing the yield as much as 20 tonnes per acre.
(ii) Spraying juvenile conifers with GAs hastens the maturity period, thus leading to early seed production.
(iii) Gibberellins also promotes bolting (internode elongation just prior to flowering) in beet, cabbages and many plants with rosette habit.

Question 5.
What do you understand by photoperiodism and vernalisation? Describe their significance.
Solution:
Photoperiodism refers to the response of plants with respect to the duration of light. On the basis of its response to the duration of light, a plant is classified as a short-day plant, a long-day plant or a day-neutral plant. Photoperiodism helps in studying the response of flowering in various crop plants with respect to the duration of exposure to light.

Vernalisation is the cold-induced flowering in plants. In some plants, exposure to low temperatures is necessary for flowering to be induced. The winter varieties of rye and wheat are planted in autumn. They remain in the seeding stage during winters and flower during summers. However, when these varieties are sown in spring, they fail to flower.

Question 6.
Why is abscisic acid also known as stress hormone?
Solution:
Abscisic acid is known as the stress hormone because it stimulates the closure of stomata in the epidermis and increases the tolerance of plants to various kinds of stresses.

Question 7.
‘Both growth and differentiation in higher plants are open’ Comment.
Solution:
Growth and development in higher plants are referred to as being open because various meristems, having the capacity for continuously dividing and producing new cells, are present at different locations in these plant bodies.

Question 8.
‘Both a short day plant and a long day plant can flower simultaneously in a given place’. Explain.
Solution:
There are two different plants one is Oat which is a long-day plant and the other one is Xanthium which is a short-day plant. Both have different photoperiods i.e. 9 hrs in Oat and 15.6 hrs in Xanthium. At 9.5 hrs both Oat and Xanthium will be flowering simultaneously.

Question 9.
Which one of the plant growth regulators would you use if you are asked to:
(a) induce rooting in a twig
(b) quickly ripen a fruit
(c) delay leaf senescence
(d) induce growth in axillary buds
(e) ‘bolt a rosette plant’
(f) induce immediate stomatal closure in leaves.
Solution:
(a) Auxin
(b) Ethylene
(c) Cytokinin
(d) Cytokinin
(e) Gibberellin
(f) Abscisic acid

Question 10.
Would a defoliated plant respond to the photoperiodic cycle? Why?
Solution:
The flowering in certain plants depends not only on a combination of light and dark exposures but also on their relative durations. This is a photoperiodic cycle. Because, while shoot apices modify themselves into flowering apices prior to flowering, they (i.e., shoot apices of plants) by themselves cannot perceive photoperiods. The site of perception of light/ dark duration is the leaves. It has been hypothesized that there is a hormonal substance(s) called florigen that is responsible for flowering. Florigen migrates from leaves to shoot apices for inducing flowering only when the plants are exposed to the necessary inductive photoperiod.

Question 11.
What would be expected to happen if:
(a) GA3 is applied to rice seedlings
(b) dividing cells stop differentiating
(c) a rotten fruit gets mixed with unripe fruits
(d) you forget to add cytokinin to the culture medium.
Solution:
(a) It causes elongation of stems and leaf sheaths.
(b) A callus of the undifferentiated cells will be produced.
(c) It stimulates the ripening of unripe fruits.
(d) It inhibits the growth of callus.

VERY SHORT ANSWER QUESTIONS

Question 1.
What is the full form of IAA?
Solution:
Indole acetic acid

Question 2.
Name the apparatus used In determining growth in the plant? (Oct. 85)
Solution:
Auxanometer

Question 3.
Name stress hormone in plants that functions during drought.
Solution:
Abscisic acid

Question 4.
Name the hormone that makes the plant more tolerant to various stresses.
Solution:
Abscisic acid

Question 5.
In a wheat field, some broad-leaved weeds were seen by a farmer. Which plant hormone would you suggest to get rid of them?
Solution:
2,4-dichloro phenoxy acetic acid (2,4-D)

Question 6.
A farmer grows cucumber plants in his field. He wants to increase the number of female flowers in them. Which plant growth regulator can be applied to achieve this?
Solution:
Ethylene

Question 7.
What is the result of the addition of gibberellins to plants? (Oct. 91)
Solution:
Bolting

Question 8.
Define growth rate.
Solution:
The increased growth per unit time is called a growth rate.

Question 9.
Who isolated auxin? Name the plant source.
Solution:
F. W. Went isolated auxin. He isolated it from tips of coleoptiles of oat seedlings.

Question 10.
Define photoperiodism. (Apr. 97)
Solution:
The response of a plant to varying photoperiods of light is called photoperiodism.

Question 11.
Name the causative agent of ‘bakane’ disease in rice seedlings.
Solution:
Gibberella fujikuroi

Question 12.
Define climacteric.
Solution:
Climacteric refers to the increased rate of respiration during the ripening of fruits.

Question 13.
What would happen when a branch from a short day plant after floral induction is grafted on a non-induced long day plant?
Solution:
The long day plant would start flowering because a short day plant is capable to induce flowering in the long-day plant.

Question 14.
What is the most abundant natural cytokinin that was isolated from com kernels and coconut milk?
Solution:
Zeatin is the most abundant natural Cytokinin that was isolated from com Kernels and Coconut milk

Question 15.
Which, plant hormone was first isolated from human urine?
Solution:
Auxin was first isolated from human urine.

SHORT ANSWER QUESTIONS

Question 1.
List some structural modifications which occur during cell differentiation.
Solution:
During differentiation, cells undergo few to major structural changes both in their cell walls and
protoplasm. For example, to form a tracheary element, the cells would lose their protoplasm. They also develop very strong, elastic, lignocellulosic secondary cell walls, to carry water to long distances even under extreme conditions.

Question 2.
How do you induce lateral branching in a plant which normally does not produce them? Give reasons in support of your answer.
Solution:
Apical bud checks the sprouting of lateral buds due to the presence of auxins. When the apical bud is removed, lateral branches are produced. Due to the removal of apical bud effect of auxins is destroyed inducing the lateral buds to grow rapidly.

Question 3.
Define growth regulators.
Solution:
The plant growth regulators (PGRs) are small, simple molecules of diverse chemical
composition. They could be indole compounds (indole-3-acetic acid, IAA); adenine derivatives (kinetin), derivatives of carotenoids (abscisic acid, ABA); terpenes (gibberellic acid, GA3) or gases (ethylene, C2H4). Plant growth regulators are variously described as plant growth substances, plant hormones or phytohormones.

Question 4.
Define the term Growth. Mention the phases of growth. (Oct. 1988, 2000, 2003, July 2006, March 2011)
Solution:
Growth is a permanent irreversible change brought about by an increase in size, weight or volume. The phases of growth are

  • Phase of cell division or formation
  • Phase of cell elongation or enlargement
  • Phase of cell maturation or differentiation.

Question 5.
Define plasticity.
Solution:
Plants follow different pathways in response to the environment or phases of life. It leads to formation of different structures. This ability is called plasticity.

Question 6.
What is growth? How will you measure the rate of growth?
Solution:

  1. Growth is defined as a permanent or irreversible increase in dry weight, size, mass or volume of a cell, organ or organism.
  2. Generally growth is accompanied by metabolic processes (both anabolic and catabolic). At the cellular level, growth is due to increase in amount of protoplasm.
  3. However, it is difficult to measure increase in protoplasm.
  4. Increase in protoplasm leads to increase in cell, cell number and cell size. This fact is used in calculating growth which, therefore, is a quantitive or measurable phenomenon.
  5. The parameters used for measuring growth increase in fresh weight, dry weight, length, area, volume and cell number.

Question 7.
Explain the different phases of growth with the help of a diagram.
Solution:
The period of growth is generally divided into three phases – meristematic, elongation, and maturation.
Meristematic phase: The constantly dividing cells, both at the root apex and the shoot apex, represent the meristematic phase of growth. The cells in this region are rich in protoplasm possess large conspicuous nuclei.

Their cell walls are primary in nature, thin, and cellulosic with abundant plasmodesmatal connections.

Elongation phase: The cells proximal to the meristematic zone represent the phase of elongation.

Increased vacuolation, cell enlargement, and new cell wall deposition are the characteristics of the cells in this phase.

Maturation phase: The cells of this zone, attain their maximal size in terms of wall thickening and protoplasmic modifications.
NCERT Solutions for Class 11 Biology Chapter 15 Plant Growth and Development 1

Question 8.
Define growth and describe the three phases of growth. (Oct. 83, 85)
Solution:
Growth is a permanent irreversible change brought about by an increase in height, weight, or volume. The three phases of growth are;

(a) Phase of cell division or cell formation:
This region is located at the tip of shoot and root. It is represented by the apical meristem capable of rapid cell division. The cells are undifferentiated, with a thin cell wall made of cellulose, with an active protoplasm and prominent nucleus. This region is mainly concerned with cell division.

(b) Phase of cell elongation or cell enlargement: This region lies next to the cell formation zone. The cells enlarge because of their elastic cell walls. Growth takes place during this stage either by apposition or intussusception. Cells are turgid.

(c) Phase of cell differentiation or cell maturation: This represents the last region and differentiation based on functions is seen here. Secondary walls are laid down where some have additional deposits of lignin, Suberin, and others lose their protoplast and become dead.

Question 9.
Where are auxins synthesized in plants? Mention any two of their functions.
Solution:
Auxins are produced in the growing shoot apices and root apices.
Functions of auxins are as follows :
(i) Auxins control apical dominance, i.e., they suppress the growth of lateral buds into branches.
(ii) They help to prevent fruit and leaf drop at early stages but promote the abscission of older mature leaves and fruits.

Question 10.
Where are cytokinins synthesised in plants? Mention any two of their functions.
Solution:
Cytokinins are synthesised in plant parts where rapid cell division occurs, like root apices, shoot buds, young fruits, etc. The functions of cytokines are as follows:
(i) Cytokinins influence cell division (cytokinesis), cell enlargement and differentiation.

Question 11.
Explain apical dominance. Name the hormone that controls it.
Solution:
Apical dominance is the phenomenon in which the apical bud suppresses the growth of lateral buds into branches. Auxin is the hormone that controls it.

Question 12.
How does abscisic acid act antagonistically to auxins and gibberellins?
Solution:
ABA induces the formation of the abscission layer, while auxins prevent the formation of the abscission layer.
ABA induces seed dormancy and bud dormancy, while gibberellins break seed dormancy and bud dormancy.

Question 13.
What is ethephon? How does it function in plants? Give any two of its functions.
Solution:
Ethephon:

  •  It is a compound used as a source of ethylene for plant growth.
  •  It is an aqueous solution that is easily absorbed by the plants and transported within the plant.
  •  It releases ethylene slowly.

Functions of ethephon are as follows:

  • It accelerates abscission in flowers and thinning in cotton, walnut, and cherry, etc.
  • It promotes the development of female flowers in cucumbers thereby increasing the yield.

Question 14.
Discuss the practical applications of auxins in Agriculture and Horticulture. (Oct. 96) OR What are auxins? Explain briefly uses of Auxins. (Oct. 99)
Solution:
Auxins are a group of plant growth substances, acidic in nature and bring about over-all growth.

  • Apical dominance: As long as the apical bud is present growth of lateral buds is pre-vented which is used in the long term storage of potato tubers.
  • Rooting: In low concentrations, auxins stimulate root formation which is used to propagate cuttings. When dipped in a dilute solution of auxins the root formation is initiated.
  • Flower initiation: Low Concentration of 2, 4-D and NAA are used to initiate flowering in a pineapple so that harvesting becomes easy.
  • Abscission: Application of auxins increase the concentration and thereby delays the development of abscission which prevents premature leaf and fruitful. This is used in orchards and prevention of defoliation in cabbages and cauliflower.
  • Parthenocarpy: This is the process of obtaining fruits without fertilization and gives rise to seedless varieties which is successfully used in citrus, dates.
  • Sex expression: Use of auxins on cucurbit plants increases the production of female flower.
  • Weedicide / Herbicide: 2-4-D is widely used as a herbicide on broad-leaved forms because of its non-toxic nature. Widely used in crop plant cultivation or lawns.
  • Tissue culture (organogenesis): In tissue culture where micropropagation is carried out the auxins are used to bring about organogenesis.

Question 15.
Differentiate between phototropism and Geotropism.
Solution:
NCERT Solutions for Class 11 Biology Chapter 15 Plant Growth and Development 2

Question 16.
What are Terpenoids?
Solution:
Terpenoids are derivatives of terpenes, includes abscisic acid and gibberellin and the carotenoid and chlorophyll pigments.

LONG ANSWER QUESTIONS

Question 1.
What is meant by vernalization? Explain the significance of vernalization.
Solution:

  • Vernalization may be defined as the method of inducing early flowering in plants by pretreatment of their seeds at low temperatures.
  • It is the acquisition or acceleration of the ability to flower by chilling treatment.
  • Some cereals such as wheat, barley, oat and rye have two kinds of varieties: winter and spring varieties.
  • The ‘spring’ variety are normally planted in the spring and come to flower and produce grain before the end of the growing season.
  • Winter varieties, however, if planted in spring would normally fail to flower or produce mature grain within a span of a flowering season.
  • Hence, they are planted in autumn. They germinate, and during winter come out as small seedling, resume growth in the spring, and are harvested usually around mid-summer.
  • Another example of vernalization is seen in biennial plants. Biennials are monocarpic plants that normally flower and die in the second season. Sugarbeet, cabbages, carrots are some of the common biennials.
  • Subj ecting the growing of a biennial plant to a cold treatment stimulates a subsequent photoperiodic flowering response.

The significance of vernalization is as follows:

  • It reduces the vegetative period of the plant.
  • It prepares the plants for flowering.
  • It increases yield, resistance to cold and diseases.
  • Vernalization is beneficial in reducing the period between germination and flowering.

Thus, more than one crop can be obtained during a year.

Question 2.
Discuss the role of auxins in plant growth.
OR
Describe any four physiological effects of auxins. (Oct. 89, 2001)
Solution:

  1. Cell division and Differentiation: Auxins promote cell division and their subsequent differentiation into tissues. They are used in cultures to bring about organogenesis.
  2. Apical dominance: Auxins are more concentrated in the terminal buds rather than lateral buds. Therefore the presence of the terminal buds inhibits the growth of lateral buds which is also true when auxins are applied to the cut surface of the stem. This is used in preventing the sprouting of potato buds (axillary buds).
  3. Root Initiation: Low concentrations of auxins promote rooting which can propagate more plants vegetatively.
  4. Abscission formation: The development of abscission is due to a decrease in auxin concentration resulting in fruit fall and defoliation. In young leaves and fruits, the concentration is high. Hence the external application of auxins helps to prevent premature fruit drop of apple, pear, and defoliation of cabbage.
  5. Parthenocarpy: A normal fruit develops after fertilization, during which the auxin concentration increases. The application of auxins stimulates fruit formation without fertilization and is called parthenocarpy.
  6. Herbicide: Synthetic auxins like 2, 4 – D and 2, 4, 5-T are toxic to broad-leaved plants and because of this used as selective herbicides in crop plants, lawn grass, etc.

Question 3.
What is meant by seed dormancy? Describe the methods to overcome seed dormancy.
Solution:
Seed dormancy

  • There are certain seeds which fail to germinate even when external conditions are favourable. Such seeds are undergoing a period of dormancy which is controlled not by the external environment but are under endogenous control or conditions within the seed itself.
  • Impermeable and hard seed coat; the presence of chemical inhibitors such as abscisic acids, phenolic acids, para-ascorbic acid; and immature embryos are some of the reasons which cause seed dormancy.
  • Seed dormancy, however, can be overcome through natural means and various other means e.g. the seed coat barrier in some seeds can be broken by mechanical abrasions using knives, sandpaper, etc. or vigorous shaking. In nature, these abrasions are caused by microbial action, and passage through the digestive tract of animals.
  • The effect of inhibitory substances can be removed by subjecting the seeds to chilling conditions or by application of certain chemicals like gibberellic acid and nitrates.
  • Changing the environmental conditions, such as light and temperature are other methods to overcome seed dormancy.

Question 4.
Describe the phenomenon of photoperiodism.
Solution:
The effect of photoperiods or day duration of light hours (and dark periods) on the growth and development of plants, especially flowering, is called photoperiodism. On the basis of photoperiodic response to flowering, plants have been divided into the following categories:

  • Short-day plants: They flower when the photoperiod or day length is below a critical period. Most winter flowering plants belong to this category, e.g., Xanthium, Chrysanthemum, rice, sugarcane, etc.
  • Long-day plants: These plants flower when they receive long photoperiods or light hours which are above a critical length, e.g., wheat, oat, sugar beet, spinach, radish, barley, etc.
  • Day-neutral plants: There are many plants, however, where there is no such correlation between exposure to the light duration and induction of flowering response; such plants are called day-neutral plants e.g. tomato, cucumber, etc.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 15 Plant Growth And Development, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 15 Plant Growth And Development, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules

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NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules.

Question 1.
What are macromolecules? Give examples.
Solution:
Biomolecules i.e. chemical compounds found in living organisms are of two types. One, those which have molecular weights less than one thousand and are usually referred to as macromolecules or simply as biomolecules while those which are found in the acid-insoluble fraction are called macromolecules or as biomacromolecules.

The molecules in the insoluble fraction with the exception of lipids are polymeric substances. Then why do lipids, whose molecular weights do not exceed 800, come under acid-insoluble fractions i.e., macromolecular fractions?

Question 2.
Illustrate a glycosidic, peptide and a phospho-diester bond.
Solution:
(a) Glycosidic bond: It is a bond formed between two monosaccharide molecules in a polysaccharide. This bond is formed between two carbon atoms of two adjacent monosaccharides.

(b) Peptide bond: Amino acids are linked by a peptide bond which is between the carboxyl (- COOH) group of one amino acid and the amino (- NH2) group of the next amino acid which is formed by the dehydration process.

(c) Phosphodiester bond: This is the bond present between the phosphate and hydroxyl group of sugar which is called an ester bond. As this ester bond is present on either side, it is called a phosphodiester bond.

Question 3.
What is meant by the tertiary structure of proteins?
Solution:
Tertiary structure of protein : When the individual peptide chains of secondary structure of protein are further extensively coiled and folded into sphere-like shapes with the hydrogen bonds between the amino and carboxyl group and various other kinds of bonds cross-linking on-chain to another they form tertiary structure. The ability of proteins to carry out specific reactions is the result of their primary, secondary and tertiary structure.
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 1

Question 4.
Find and write down structures of 10 interesting small molecular weight biomolecules. Find if there is any industry which manufactures the compounds by isolation. Find out who are the buyers?
Solution:
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 2
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 3
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 4
Fat is being manufactured by many companies in pharmaceuticals business as well as in food business. Vitamins come in many combination and are being used as supplementary medicines. Lactose is made by companies in manufacturing baby food. All of us are buyers of fat, protein and lactose.

Question 5.
Proteins have primary structures. If you are given a method to know which amino acid is at either of two termini (ends) of a protein, can you connect this information to purity or homogeneity of a protein?
Solution:
The primary structure of proteins is described as the type, number, and order of amino acids in the chain. A protein is imagined as a line whose left end represents the first and right end represents the last amino acid. But in fact, this is not so simple. Actually, the number of amino acids in between the two termini determines the purity or homogeneity of a protein.

Question 6.
Find out and make a list of proteins used as therapeutic agents. Find other applications of proteins (e.g., Cosmetics, etc.)
Solution:
Haemoglobin, Insulin, thyroxine, growth hormone, other hormones of the adenohypophysis, serum albumen, serum globulin, fibrinogen, etc. are used as the therapeutic agents. Proteins are also used for the synthesis of food supplements, film, paint, plastic, etc.

Question 7.
Explain the composition of triglyceride.
Solution:
Triglycerides are esters of three molecules of fatty acids and one molecule of glycerol.
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 5

Question 8.
Can you describe what happens when milk is converted into curd or yoghurt, from your understanding of proteins.
Solution:
Conversion of milk into curd is the digestion of milk protein casein. Semi digested milk is the curd. In the stomach, renin converts milk protein into paracasein which then reacts with Ca++ ion to form calcium paracaseinate which is called the curd or yoghurt.

Question 9.
Can you attempt building models of biomolecules using commercially available atomic models (Ball and Stick model)?
Solution:
Yes, the Three-dimensional structure of cellulose can be made using balls and sticks. Similarly, models of other bimolecular can be made
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 6

Question 10.
Attempt titrating an amino acid against a weak base and discover the number of dissociating (ionizable) functional groups in the amino acid.
Solution:
When an amino acid is titrated with weak base then its-COOH group also acts as weak acid. So it forms a salt with weak base then the pH of the resulting solution is near 7, so there is no sudden change. Number of dissociating functional groups are two, one is amino group (NH2) and another is carboxylic group ( – COOH). In the titration, amino acid acts as an indicator. Amino acids in solution acts as basic or acidic as situation demands. So these are also called amphipathic molecules.

Question 11.
Draw the structure of the amino acid, alanine.
Solution:
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 7

Question 12.
What are gums made of? Is fevicol different?
Solution:
Gums are categorized into secondary metabolites or biomolecules. Thousands of compounds one present in plant-fungal and microbial cells. They are derived from these things. But is different. Fevicol has not derived from paper written cells.

Question 13.
Find out a qualitative test for proteins, fats and oils, amino acid and test any fruit juice, saliva, sweat and urine for them.
Solution:
Qualitative Tests for proteins, amino acids, and fats:
Biuret Test: Biuret test for protein identifies the presence of protein by producing violet colour of solution. Biuret H2NCONHCONH2 reacts with copper ion in a basic solution and gives violet colour.
Liebermann-Burchard Test for cholesterol:
This is a mixture of acidic anhydride and sulphuric acid. This gives a green colour when mixed with cholesterol.
Grease Test for oil: Certain oils give a translucent stain on clothes. This tesi can be used to show presence of fat in vegetable oils. These tests can be performed to check presence of proteins and amino acids and fats in any of the fluid mentioned in the question.

Question 14.
Find out how much cellulose is made by all the plants in the biosphere and compare it with how much of paper is manufactured by man and hence what is the consumption of plant material by man annually. What a loss of vegetation?
Solution:
According to a 2006 report from the UN, forests store about 312 billion tons of carbon in their biomass alone. If you add to that the carbon in deadwood, litter, and forest soil, the figure increases to about 1.1 trillion tons! The UN assessment also shows that the destruction of forests adds almost 2.2 billion tons of carbon to the atmosphere each year, the equivalent of what the U.S. emits annually. Many climate experts believe that the preservation and restoration of forests offers one of the least expensive and best ways to fight against climate change.
Although it is difficult to get exact data about the quantum of cellulose produced by plants, but above information can give some idea. About 10% of cellulose is used in paper making. The percentage is less but wrong practice of cutting wood and re-plantation makes the problem complicated. Usually older trees are cut for large quantity of cellulose and re-plantation is limited to selected species of plants. Selected species disturb the biodiversity as it leads to monoculture.
Add to this the problem of effluents coming out of a paper factory and the problem further aggravates.

Question 15.
Describe the important properties of enzymes.
Solution:
Properties of enzymes

  • Enzyme catalysis hydrolysis of ester, ether, peptide, c-c, c-halids, or P-N bonds.
  • Enzymes catalysis removal of the group from the substrate by mechanisms other than hydrolysis of leaving double bonds.
  • Enzymes generally function in a narrow range of temperature and pH.
  • Activity declines both below and above optimum temperature and pH.
  • The higher the affinity of the enzyme for its substrate the greater is its catalytic activity.
  • The activity of an enzyme is also sensitive to the presence of specific chemicals that bind to the enzyme.
  • For eg: Inhibitors that shuts off enzyme activity and Co-factors that facilitate catalytic activity.
  • Enzymes retain their identity at the end of the reaction.

VERY SHORT ANSWER QUESTIONS

Question 1.
Which organic compound is commonly called animal starch?
Solution:
Glycogen

Question 2.
Name the biomolecules of life.
Solution:
Carbohydrates, Lipids, Proteins, Enzymes, and nucleic acids.

Question 3.
Name one basic amino acid.
Solution:
Lysine.

Question 4.
Name one heteropolysaccharide.
Solution:
Chitin

Question 5.
Name the biomolecules present in the acid-insoluble fraction.
Solution:
Protein, polysaccharide, nucleic acid, and lipids.

Question 6.
Name the bond formed between sugar molecules.
Solution:
Glycosidic bond.

Question 7.
Name three pyrimidines.
Solution:
Thymine, cytosine, and uracil

Question 8.
Which enzyme does catalyse covalent bonding between two molecules to form a large molecule?
Solution:
Ligases.

Question 9.
On reaction with iodine, starch turns blue-black, why?
Solution:
The appearance of blue colour with the addition of iodine is due to its reaction with amylose fraction of starch.

Question 10.
Which type of bonds are found in proteins and polysaccharides?
Solution:
Peptides bond in protein and glycosidic bonds in polysaccharides.

Question 11.
Name one neutral amino acid.
Solution:
Valine.

Question 12.
Where does histone occur?
Solution:
Chromosomes.

Question 13.
Name two different kinds of metabolism.
Solution:
Anabolism and catabolism.

SHORT ANSWER QUESTIONS

Question 1.
Which type of bonds are found in nucleic acids?
Solution:
Phosphodiester bond.

Question 2.
What are the monosaccharides present in DNA and RNA? (Chikmagalur 2004)
Solution:
Deoxyribose in DNA and Ribose in RNA.

Question 3.
What are fatty acids? Give two examples.
Solution:
Fatty acids are compounds which have a carboxyl group attached to an R-group, which could be a methyl (CH3), or ethyl (C2H5) group or a higher number of CH2 groups e.g., Linoleic acid, Palmitic acid.

Question 4.
What are co-enzymes? Give two examples.
Solution:
Coenzymes are the non-protein organic ^compounds bound to the apoenzyme in a conjugate enzyme, their association with the apoenzyme is only transient, e.g., Nicotinamide adenine dinucleotide (NAD). Flavin adenine dinucleotide (FAD), Nicotinamide adenine dinucleotide phosphate (NADP).

Question 5.
(i) What is meant by complementary base pairing?
(ii) What is the distance between two successive bases in a strand of DNA?
(iii) How many base pairs are present in one turn of the helix of a DNA strand?
Solution:
(i) Complementary base pairing is the type of
pairing in DNA, where a purine always pairs with a pyrimidine, i.e., adenine pairs with thymine (A=T) and guanine pairs with cytosine (G=C).
(ii) 0.34 nm or 34 A is the distance between two successive bases in the strand of DNA
(iii) 10 base pairs

Question 6.
Differentiate between DNA and RNA.
Solution:
The main differences between DNA add RNA are as following
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 8
Question 7.
What la a prosthetic group? Give an example.
Solution:
The non-protein part of a conjugated protein is called a prosthetic group. For example in a nucleoprotein (nucleic acid is the prosthetic group).

Question 8.
Differentiate between essential amino acids and non-essential amino acids.
Solution:
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 9

Question 9.
Differentiate between Structural Proteins and Functional Proteins.
Solution:
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 11

Question 10.
What is activation energy?
Solution:
Activation Energy: An energy barrier is required for the reactant molecules for their activation. So this energy with enzyme-substrate reaction is called Activation energy.
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 12

The activation energy is low for reactions with catalysts [enzymes] than those with Non enzymatic reactions.

Question 11.
What are the components of enzymes?
Solution:
Enzymes are made up of protein as well as non – protein parts. The protein part is called an apoenzyme and the non-protein part is a coenzyme. These two together are called a holoenzyme.

LONG ANSWER QUESTIONS

Question 1.
How many classes are enzymes divided into? Name all the classes.
Solution:
Enzymes are divided into 6 classes. Namely

  1. Oxidoreductases/dehydrogenases: Enzymes which catalyze oxidoreduction between two substrates
  2. Transferases: Enzymes catalyzing a transfer of group between a pair of substrates.
  3. Hydrolases: Enzymes catalyzing the hydrolysis of ester, ether, peptide, glycosidic, C-C-C-halide or P.N bonds.
  4. Lyases: Enzymes catalyze the removal of groups from – substrates by mechanisms other than hydrolysis leaving double bonds.
  5. Lyases: Enzymes catalyzing the interconversion of optical geometric or positional isomers.
  6. Ligases: Enzymes catalyzing the linking together of 2 compounds.

Question 2.
Distinguish between the primary, secondary, and tertiary structures of proteins.
Solution:
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 13

Question 3.
Explain the effect of the following factors on enzyme activity:
(i) Temperature
(ii) pH.
Solution:
Temperature: An enzyme is active within a narrow range of temperature. The temperature at which an enzyme shows its highest activity is called optimum temperature.

It generally corresponds to the body temperature of warm blood animals e.g., 37°C in human beings. Enzyme activity decreases above and below this temperature. Enzyme becomes inactive below minimum temperature and beyond maximum temperature.

Low temperature present inside cold storage prevents spoilage of food. High temperature destroys enzymes by causing their denaturation.
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 15

The relation between temperature and enzyme controlled reaction velocity

pH – Every enzyme has an optimum pH when it is most effective.

A rise or fall in pH reduces enzyme activity by changing the degree of ionisation of its side chains. A change in pH may also reverse the reaction.

Most of the intracellular enzymes function near-neutral pH with the exception of several digestive enzymes which work either in acidic range of pH or alkaline range of pH. pH for trypsin is 8.5.

Question 4.
Discuss the B-DNA helical structure with the help of a diagram.
Solution:

  • Watson & Crick suggested the double-helical structure of DNA in 1953.
  • The backbone of the DNA molecule is made up of deoxyribonucleotide units joined by a phosphodiester bond.
  • The DNA molecule consists of two chains wrapped around each other.
  • The two helical strands are bound to each other by Hydrogen Bonds.
  • Purines bind with pyrimidines A = T, C = G
  • The pairing is specific and the two chains are complementary.
  • One strand has the orientation 5’ → 3’ and other has 3’ → 5’.
  • Both polynucleotides strands remain separated with a 20A° distance.
  • The coiling is right-handed.

NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 16

Question 5.
What are different kinds of enzymes? Mention with enzyme examples.
Solution:
Enzymes with substrate bonds are broken and changed to different kinds as

  1. Oxidoreductases: eg Alcohol dehydrogenase, oxidation, Reduction occurs
  2. Transferases: transfer a particular group to another substrate, eg. transavninase
  3. Hydrolases: cleave their substrates by hydrolysis of a covalent bond e.g. Urease, amylase.
  4. Lyases: break the covalent bond eg. Deaminase
  5. Isomerase: by changing the bonds they make isomers. eg: Aldolase.
  6. Ligase: These bind two substrate molecules eg: DNA ligase, RNA ligase

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 9 Biomolecules, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 9 Biomolecules, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement

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NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement.

Question 1.
Draw the diagram of a sarcomere of skeletal.
Solution:
NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 1

Question 2.
Define sliding filament theory of muscle contraction.
Solution:
The mechanism of muscle contraction is best explained by the sliding filament theory which states that contraction of a muscle fiber takes place by sliding of the thin filaments over the thick filaments.

Question 3.
Describe the important steps in muscle contraction.
Solution:
Muscle contraction is initiated by a signal sent by the central nervous system (CNS) via a motor neuron. A motor neuron along with the muscle fibers connected to it constitutes a motor unit. The junction between a motor neuron and the sarcolemma of the muscle fiber is called the neuromuscular junction or motor-end plate. A neural signal reaching this junction releases a neurotransmitter (Acetylcholine) which generates an action potential in the sarcolemma. This spreads through the muscle fiber and causes the release of calcium ions into the sarcoplasm. An increase in Ca++ level leads to the binding of calcium with a subunit of troponin on actin filaments and thereby removes the masking of active sites for myosin.

Utilizing the energy from ATP hydrolysis, the myosin head now binds to the exposed active sites on actin to form a cross bridge. This pulls the attached actin filaments towards the center of the A-bonds. The Z-line attached to these actions is also pulled inwards thereby causing a shortening of the sarcomere i.e., contraction. It is clear from the above steps, that during shortening of the muscle i.e., contraction, the ‘I’ bonds are getting reduce whereas the A-bonds are retaining the length. The myosin, releasing the ADP and p1 goes back to its relaxed state.

A new ATP binds and the cross-bridge is broken. This causes the return of Z-lines back to their original position i.e., relaxation. The reaction time of the fibers can vary in different muscles. Repeated activation of the muscles can lead to the accumulation of lactic acid due to the anaerobic breakdown of glycogen in them, causing fatigue. Muscle contains a red-colored oxygen storing pigment called myoglobin. Myoglobin content is high in some of the muscles which gives a reddish appearance. Such muscles are called the Red muscles. These muscles also contain plenty of mitochondria which can utilize a large amount of oxygen stored in them for ATP production.

These muscles, therefore, can also be called aerobic muscles. On the other hand, some of the muscles possess a very little quantity of myoglobin and therefore, appear pale or whitish. These are the white fibers. The number of mitochondria is also few in them, but the amount of sarcoplasmic reticulum is high. They depend on the anaerobic process for energy.

Question 4.
Write true or false. If false, change the statement so that it is true.

  1. Actin is present in the thin filament.
  2. H-zone of striated muscle fibre represents both thick and thin filaments.
  3. The human skeleton has 206 bones.
  4. There are 11 pairs of ribs in man.
  5. The sternum is present on the ventral side of the body.

Solution:

  1. True
  2. False: The H-zone of striated muscle fibre represents only thick filaments.
  3. True
  4. False: There are 12 pairs of ribs in man.
  5. True

Question 5.
Write the differences between.
(a) Actin and Myosin
(b) Red and White muscles
(c) Pectoral and Pelvic girdle
Solution:
(a) Differences between actin and myosin are as following;
NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 2
(b) The main difference between red muscles and white muscles are as following :
NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 3
(c) The main difference between the pectoral girdle and pelvic girdle are as following:
NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 4

Question 6.
Match Column I with Column II
Column I                                            Column II
(a) Smooth muscle                         (i) Myoglobin
(b) Tropomyosin                            (ii) Thin filament
(c) Red muscle                               (iii) Sutures
(d) Skull                                         (iv) Involuntary
Solution:
(a)– (iv)
(b)-(ii)
(c) -(i)
(d) – (iii)

Question 7.
What are the different types of movements exhibited by the cells of the human body?
Solution:
Cells of the human body exhibit three main types of movements, namely, amoeboid, ciliary and muscular. Some specialized cells in our body like macrophages and leucocytes in blood exhibit amoeboid movement. It is affected by pseudopodia formed by the streaming of protoplasm (as in Amoeba). Cytoskeletal elements like microfilaments are also involved in the amoeboid movement.

Ciliary movement occurs in most of our internal tubular organs which are lined by ciliated epithelium. The coordinated movements of cilia in the trachea help us in removing dust particles and some of the foreign substances inhaled along with the atmospheric air. Passage of ova through the female reproductive tract is also facilitated by the ciliary movement. Movement of our limbs, jaws, tongue, etc, requires muscular movement. The contractile property of muscles is effectively used for locomotion and other movements by human beings and the majority of multicellular organisms. Locomotion requires a perfect coordinated activity of muscular, skeletal, and neural systems.

Question 8.
How do you distinguish between a skeletal muscle and a cardiac muscle?
Solution:
The main difference between skeletal muscle and cardiac muscle are as following:
NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 5

Question 9.
Name the type of joint between the following:
(a) Atlas/Axis
(b) Carpal/metacarpal of the thumb
(c) Between phalanges
(d) Femur/acetabulum
(e) Between cranial bones
(f) Between pubic bones in the pelvic girdle
Solution:
(a) Pivot joint
(b) Saddlejoint
(c) Gliding joint
(d) Ball and socket joint
(e) Fibrous joint
(f) Cartilagenousjoint

Question 10.
Fill in the blank spaces:
(a) All mammals (except a few) have ………….. cervical vertebra.
(b) The number of phalanges in each limb of a human is …………
(c) Thin filament of myofibril contains 2 ‘F’ actins and two other proteins namely ………. and ……………….
(d) In a muscle fibre Ca++ is stored in ……………
(e) …….. and ……….. pairs of ribs are called floating ribs.
(f) The human cranium is made of ……………. bones.
Solution:
(a) Seven
(b) Fourteen
(c) Troponin, tropomyosin
(d) Sarcoplasmic reticulum
(e) 11th, 12th
(f) Eight

VERY SHORT ANSWER QUESTIONS

Question 1.
What causes gouty arthritis in humans?
Solution:
Gouty arthritis (= Gout) is caused either due to excessive formation of uric acid or inability to excrete it.

Question 2.
How many tarsals are there in the ankle?
Solution:
Seven.

Question 3.
What are the bones of the heel called?
Solution:
Metatarsals.

Question 4.
How many types of movement shows by the human body?
Solution:
Three types of movements: amoeboid, ciliary, and muscular movement.

Question 5.
Name the lubricant which is responsible for the movable joint at the shoulder.
Solution:
Synovial fluid.

Question 6.
Give two disorders of skeleton and joints.
Solution:
Arthritis and Osteoporosis.

Question 7.
Mention two sites on all bodies where striated muscles are present.
Solution:
Limbs and tongue.

Question 8.
Name the two filaments which form the cross-bridges during muscle contraction?
Solution:
Actin and myosin.

Question 9.
Name the monomers of myosin.
Solution:
Meromyosins.

Question 10.
How many ribs are present in adult men?
Solution:
Twelve pairs.

Question 11.
Name the single U-shaped bone present at the base of the buccal cavity.
Solution:
Hyoid.

Question 12.
Name the location where Z-line is present in the sarcomere.
Solution:
Centre of I band.

Question 13.
What is the total number of bones present in the left pectoral girdle and the left arm respectively in a normal human?
Solution:
Left pectoral girdle – 2
Left-arm – 30

Question 14.
Name the kind of joint which permits movements in a single plane only.
Solution:
Hinge joint.

Question 15.
What are neuromuscular junctions?
Solution:
The junction between a motor neuron and the sarcolemma of a muscle fiber is known as the neuromuscular junction.

Question 16.
Why are the ribs described as bicephalic?
Solution:
Since each rib has two articulation surfaces on its dorsal end, it is described as bicephalic.

Question 17.
What is acromion?
Solution:
It is a flat expanded process projecting from the spine of the scapula; the clavicle articulates with it.

Question 18.
What is arthritis?
Solution:
Arthritis is painful stiffness and inflammation of joints.

Question 19.
What is sarcomere?
Solution:
A sarcomere is a structural unit within a microfibril bounded by Z lines that contain actin and myosin.

Question 20.
Which muscle protein acts as ATPase?
Solution:
Myosin.

SHORT ANSWER QUESTIONS

Question 1.
What causes osteoporosis?
Solution:
Osteoporosis is a disease in which bone loses minerals and fibres from its matrix. There are more chances of fractures. Decreased level of estrogen is a common cause.

Question 2.
Why a red muscle fibre can work for a prolonged period, while a white muscle fiber suffers from fatigue soon?
Solution:
Red muscle fibres contain myoglobin that stores oxygen in the form of oxymyoglobin.
Since, there is a continuous supply of oxygen; for the oxidation of food materials to release energy, the red muscle fibers retain energy and do not become fatigued and work for long periods whereas white muscle fibres lack myoglobin. At times they carry out anaerobic respiration and become fatigued.

Question 3.
Name the major components of the appendicular skeleton.
Solution:
It is situated at the lateral sides which actually extend outwards from the principal axis. It consists of pectoral and pelvic girdles and bones of arms and legs.

Question 4.
What is the sarcoplasmic reticulum? What is its function?
Solution:
The endoplasmic reticulum of muscle fiber is called the sarcoplasmic reticulum which acts as a storehouse of calcium ions.

Question 5.
Differentiate between A and I bands.
Solution:
The main differences between A-band and I-band are as following :
NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 6

Question 6.
Draw the labeled diagram of the pectoral girdle and upper arm.
Solution:
NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 7

Question 7.
Differentiate between bone and cartilage.
Solution:
The main differences between bone and cartilage are as following:
NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 8

Question 8.
Describe the vertebro-chondral ribs.
Solution:
Vertejno-chondral ribs

  • 8th, 9th and 10th pairs of ribs are called vertebro-chondral (false) ribs.
  • They remain attached dorsally to the respective thoracic vertebrae and vertrally to the sternum through the seventh rib by hyaline cartilage.

Question 9.
How muscular contraction is triggered?
Solution:
It is triggered by nerve releasing a neurotransmitter, which in turn triggeres the sarcoplasmic reticulum to release calcium ions into muscle interior. Where they bind to troponin, thus causing tropomyosin to shift from the face of the actin filament to which myosin heads need to produce a contraction.

Long ANSWER QUESTIONS

Question 1.
Draw a well diagram of human skull.
Solution:

NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 9

Question 2.
Write short notes on:
(a) Muscular dystrophy
(b) Tetany
(c) Myasthenia gravis
Solution:
Muscular dystrophy
The abnormality of muscles associated with dysfunction and ultimately deterioration is called muscular dystrophy. It is a genetic disorder caused by lack of dystrophin.
Myasthenia gravis: It is an auto-immune disorder that affecting neuromuscular junction and leads to fatigue, weakening, and paralysis of skeletal muscles. ‘
Tetany: The rapid spasm and (wild contractions) is called tetany. In this case, the muscles do not get a chance to relax at all. It is caused due to deficiency of parathyroid hormone and thus lowering Ca++ in blood fluid.

Question 3.
Give differences between movable and immovable joints?
Solution:
Differences between movable and immovable joint are tabulated below:
NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 10

Question 4.
Describe the structure of the rib cage of a human.
Solution:
Rib Cage: There are 12 pairs of ribs. Each rib is a thin flat bone connected dorsally to the vertebral column and ventrally to the sternum. It has two articulation surfaces on its dorsal end and is hence called bicephalic. First, seven pairs of ribs are called true ribs.

Dorsally, they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilage. The 8th, 9th, and 10th pairs of ribs do not articulate directly with the sternum but join the seventh rib with the help of hyaline cartilage. These are called vertebrochondral (false) ribs. The last 2 pairs (11th and 12th) of ribs are not connected ventrally and are, therefore, called floating ribs. Thoracic vertebrae, ribs, and sternum together form the rib cage.

NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 11

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 20 Locomotion and Movement, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 20 Locomotion and Movement, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption

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NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption.

Question 1.
Choose the correct answer among the following:
(a) Gastric juice contains
(i) pepsin, lipase, and rennin
(ii) trypsin, lipase, and rennin
(iii) trypsin, pepsin, and lipase
(iv) trypsin, pepsin, and rennin
(b) Succuss enterics is the name given to:
(i) a junction between the ileum and large intestine
(ii) intestinal juice
(iii) swelling in the gut
Solution:
(a) (i) Pepsin, lipase, and rennin
(b) (ii) Intestinal juice

Question 2.
Match column I with column II
Column I                                         Column II
(a) Bilirubin and biliverdin           (i) Parotid
(b) Hydrolysis of starch                (ii) Bile
(c) Digestion of fat                        (iii) Lipases
(d) Salivary gland                            (iv) Amylases
Solution:
Column I                                 Column II
(a) Bilirubin and biliverdin     (ii) Bile
(b) Hydrolysis of starch          (iv) Amylases
(c) Digestion of fat                 (iii) Lipases
(d) Salivary gland                   (i) Parotid

Question 3.
Answer briefly:
(a) Why are villi present in the intestine and not in the stomach?
(b) How does pepsinogen change into its active form?
(c) What are the basic layers of the wall of the alimentary canal?
(d) How does bile help in the digestion of fats
Solution:
(a) Villi increases surface area for absorption and maximum absorption takes place in the intestine.
(b) Coming in contact with hydrochloric acid in stomach proenzyme pepsinogen convert to its active form pepsin, the proteolytic enzyme of the stomach.
(c) There are four basic layers in the wall of alimentary canal i.e. serosa, muscularis, submucosa and mucosa.
(d) Bile helps in the emulsification of fats i.e. breakdown the fats into very small micelles. Bile also activates lipases.

Question 4.
State the role of pancreatic juice in the digestion of proteins.
Solution:
Pancreatic juice contains inactive enzymes like trypsinogen, chymotrypsinogen, procarboxypeptidases. Trypsinogen is activated by an enzyme enterokinase secreted by the intestinal mucosa into active trypsin which in turn activates other enzymes in the pancreatic juice. Proteins, proteases, and peptones in the chyme are digested by the proteolytic enzymes of pancreatic juice.

Question 5.
Describe the process of digestion of protein in the stomach.
Solution:
The mucosa of the stomach has gastric glands that secrete mucus, proenzyme pepsinogen, HCl, and intrinsic factor. Intrinsic factor is essential for the absorption of vitamin B12. The food mixes thoroughly with acidic gastric juice of the stomach and called the chyme. The proenzyme pepsinogen on exposure to HC1 gets converted to active enzyme pepsin. Pepsin converts proteins into proteases and peptones. Renin found in the gastric juice of infants also helps in the digestion of milk protein.

Question 6.
Give the dental formula of human beings.
Solution:
Dental formula of human beings is 2123/2123.

Question 7.
Bile juice contains no digestive enzymes, yet it is important for digestion. Why?
Solution:
The bile juice released into the duodenum contains bile pigments (bilirubin and biliverdin), bile salts, cholesterol, and phospholipids but no enzymes. Bile helps in the emulsification of fats i.e. break down the fats into very small micelles. Bile also activates lipases. Bile is alkaline (pH about 8). Bile pigments are excreted in faeces. In absence of HCl, the above-mentioned functions will not occur and digestion of food will be affected.

HCl of the gastric juice may cause gastric or duodenal ulcers and damage, the underlying blood vessels. These cause hemorrhage inside. Strong HCl is produced in stress conditions also. Parental or oxyntic cells which secrete HCl and intrinsic factor (Intrinsic factor is essential for the absorption of vitamin B2)

Question 8.
Describe the digestive role of chymotrypsin. Which two other digestive enzymes of the same category are secreted by their source gland?
Solution:
Chymotrypsin changes proteins into peptides and also milk protein, changes into paracasein (curd). The pancreatic juice contains inactive enzymes, trypsinogen, chymotrypsinogen, procarboxypeptidases, amylases, lipases and nucleases. Trypsinogen is activated by an enzyme, enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice.

Trypsin changes proteases and peptones into peptides and amino acids. Procarboxypetidases change peptides into small peptides and amino acids. Amylases change starch into maltose. Lipases change emulsified fat into fatty acids and glycerol. Nucleases changes nucleotides into phosphate, sugar and nitrogen bases.

Question 9.
How are polysaccharides and disaccharides digested?
Solution:
(a) Digestion of carbohydrates starts in the mouth cavity with the help of enzymes salivary amylase.
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 1

Question 10.
What would happen if HCl were not secreted in the stomach?
Solution:
HCl is secreted by oxyntic cells in the stomach wall and performs five functions:

  1. Kill bacteria and germs
  2. Loosens fibrous material of food.
  3. Activates proenzyme pepsinogen to its active form pepsin.
  4. Gives an acidic medium for action by pepsin.
  5. Curdles milk.  Pepsin changes proteins into proteases and peptones.

Question 11.
How does butter in your food get digested and absorbed in the body?
Solution:
Butter is a kind of fat. Fats are broken down by lipases with the help of bile into di- and monoglycerides:
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 2

Question 12.
Discuss the main steps in the digestion of proteins as the food passes through different parts of the alimentary canal.
Solution:
(a) In the stomach the proenzyme pepsinogen, on exposure to HC1 converted into active pepsin that converts proteins into proteases and peptones.
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 3
(b) Proteins, proteases and peptides in the chyme reaching the intestine are acted upon by proteolytic enzymes of pancreatic juice and converted to dipeptides.
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 4
(c) The enzymes in succuss entericus act on the end product to form amino acids.
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 5

Question 13.
Explain the terms thecodont and diphyodont.
Solution:
Thecodont: In human beings, teeth are
embedded in pits, the sockets of the jawbones. Such teeth are called the thecodont.
Diphyodont: The teeth that appear in two sets, i. e., milk-teeth which are later replaced by permanent teeth. This condition is called diphyodont.

Question 14.
Name different types of teeth and their number in an adult human.
Solution:
Incisors – 8
Canines – 4
Premolars – 8
Molars – 12

Question 15.
What are the functions of the liver?
Solution:

  • Secretion of bile
  • Synthesis of blood clotting factors
  • Regulating carbohydrate, protein, and lipid metabolism.
  • Synthesis of amino acids, plasma proteins, cholesterol, etc.
  • Stores glycogen, fats, fat-soluble vitamins, etc.
  • Detoxifies toxic substances
  • Elimination of foreign bodies
  • Produces heat through metabolism
  • Produces anticoagulant heparin
  • Synthesis of urea from ammonia.

VERY SHORT ANSWER QUESTIONS

Question 1.
What is digestion?
Solution:
Digestion is the mechanical, enzymatic, and biochemical transformation of complex (polymers) food molecules into simple molecules (monomers) which are suitable for absorption.

Question 2.
Which is the food constituent that bile helps to digest and absorb?
Solution:
Fats.
Question 3.
What is the function of enterokinase?
Solution:
Enterokinase of intestinal juice activates the inactive trypsinogen into trypsin which digests protein in the duodenum.

Question 4.
Which is a nondigestive activating enzyme?
Solution:
Enterokinase.

Question 5.
Mention the role of bile salt in the digestion of fats.
Solution:
Bile salts emulsify fat particles and reduce the surface tension of fat droplets to increase the action of enzyme lipase.5. Bile salts emulsify fat particles and reduce the surface tension of fat droplets to increase the action of the enzyme lipase.

Question 6.
What is the role of HCl in protein digestion?
Solution:
Role of HCl:

  • It activates pepsinogen into active pepsin.
  • It provides a suitable acidic medium for the action of proteases in the stomach.

Question 7.
Name the hardest substance in the body.
Solution:
Enamel.

Question 8.
What is a cystic duct?
Solution:
Duct of gall bladder.

Question 9.
Mention two functions of mucus.
Solution:
Role of mucus:
(a) Acts as a lubricant.
(b) Protects the epithelial surface of the stomach from the corrosive effect of hydrochloric acid and digestion by pepsin.

Question 10.
Name the secretion of goblet cells in the human stomach.
Solution:
Goblet cells secrete mucus.

Question 11.
Where the taste buds located?
Solution:
Taste buds are located in the papillae on the upper surface of the tongue.

Question 12.
What is the ‘Pancreatic Enzyme’ which acts on starch? (Oct. 83, 01)
Solution:
The pancreatic amylase(Amylopsin).

Question 13.
What is the function of epiglottis?
Solution:
Epiglottis prevents the entry of food into the trachea, by closing its opening called the glottis.

Question 14.
Which part of the stomach continues into the duodenum?
Solution:
Pyloric region

Question 15.
What name is given to the major lymph vessel present in the intestinal villi?
Solution:
Lacteal

Question 16.
Where are the crypts of Leiberkuhn located?
Solution:
Crypts of Lieberkuhn are located in between the bases of the villi in the intestine.

Question 17.
Name the structural and functional unit of the liver.
Solution:
Hepatic lobules

Question 18.
Mention the secretion of Goblet cells.
Solution:
Mucin (Mucus) (April 93)

Question 19.
What is a bolus?
Solution:
When the thoroughly masticated food mixes with the saliva, the food particles become adhered together by the mucus known as bolus.

Question 20.
What is chyme?
Solution:
After partial digestion in the stomach, the form of food is called chyme.

Question 21.
What is the meaning of deglutition?
Solution:
The act of swallowing is called deglutition.

Question 22.
Name the enzyme involved in the breakdown of nucleotides into sugars and bases.
Solution:
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 6

SHORT ANSWER QUESTIONS

Question 1.
Mention the juice secreted by the liver. State its function indigestion. (April 1985)
Solution:
The liver cells secrete a juice called the ‘Bile juice’. ‘Bile salts’ are one of the components of Bile juice. These are very important in digesting lipids or fats. Specifically ‘Bile salts’ aid in digestion than bringing about digestion of fat by emulsifying fats and making the fat molecules accessible to the action of the pancreatic Lipase (or Lipid digesting) enzyme. Apart from this the bile juice also creates an alkaline medium in the intestine so that food particles can be actively acted upon by the pancreatic enzymes.

Question 2.
Name the organs which secret carboxypeptidases and aminopeptidases respectively. Give the function performed by these enzymes
Solution:
Carboxypeptidases are secreted by the pancreas. Aminopeptidases are secreted by the intestine. Both the enzymes act on the terminal peptide bonds and release the terminal/last amino acids of the peptide chain.

Question 3.
State the Role of HCI indigestion. (Oct. 88, April 93, 98)
Solution:
The main role of HCI is to convert the inactive enzyme pepsinogen to pepsin which in turn helps in the digestion of proteins. It also creates an acidic medium in the stomach for the activity of pepsin.

Question 4.
What is succus entericus? Mention any two carbohydrate digesting enzymes present in it? (April 2006)
Solution:
Succus entericus intestinal juice is the secretion of intestinal glands in the ileum of the small intestine. The carbohydrate digesting enzymes are maltase, lactase, and sucrase.

Question 5.
Differentiate between micelles and chylomicrons.
Solution:
The differences between micelles and chylomicrons are
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 7

Question 6.
How is our gut lining protected from its own secretion of proteases?
Solution:
(i) Protease is secreted in an inactive form and poses no threat to the gut lining.
(ii) The mucus provides protection to the epithelial lining.

Question 7.
Name the organs which secrete carboxypeptidases and aminopeptidases respectively. Give the functions performed by these enzymes.
Solution:
Carboxypeptidases are secreted by the exocrine part of the pancreas. Aminopeptidases are secreted by the intestinal mucosa. These enzymes act on the terminal peptide bonds and release the last amino acid from the polypeptide chain, thereby progressively shortening the peptide chain.

Question 8.
Where is the ileocaecal valve present? What is its function?
Solution:
The ileo-caecal valve is present at the junction of the ileum of the small intestine and the caecum of the large intestine.
It prevents the backflow of the matter from the caecum into the ileum.

Question 9.
How does the nervous system control the activities of the gastro-intestinal tract?
Solution:
The sight, smell and presence of food in the oral cavity can stimulate the secretion of saliva. Gastric and intestinal secretions are also stimulated by similar neural signals. Muscular activities of the alimentary canal are coordinated by both local and CNS neural mechanisms. Hormonal control of secretion of digestive enzymes is carried out by local hormones.

Question 10.
Name the final products of the digestion of proteins. How and where are they absorbed from the alimentary canal?
Solution:
The final products of digestion of proteins are amino acids They are absorbed by an active process utilising energy against the concentration gradient in the ileum.

Question 11.
What is the pancreas? Mention the major secretions of the pancreas that are helpful in digestion.
Solution:
The pancreas is a carrot-shaped soft greyish pink gland that lies transversely below the stomach between the duodenum and spleen, which secretes digestive enzymes from its exocrine parts and hormones from its endocrine parts.
The pancreas secretes three enzymes in inactive proenzyme or zymogen state and three in active enzyme state.
Proenzymes. Trypsinogen, chymotrypsinogen, and procarboxypeptidases.
Active Enzymes. Amylase, lipase, and nucleases.

LONG ANSWER QUESTIONS

Question 1.
What is meant by vernalization? Explain the significance of vernalization.
Solution:
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 8

Question 2.
Write short notes on
(a) Liver
(b) Layers of the alimentary canal
Solution:
(a) Liver: The liver is the largest gland of the body. It is situated in the abdominal cavity, just below the diaphragm and has two lobes. The hepatic lobules are the structural and functional units of the liver containing hepatic cells arranged in the form of cords. Each lobule is covered by a thin connective tissue sheath called the Glisson’s capsule. The bile secreted by the hepatic cells passes through the hepatic ducts and is stored and concentrated in a thin muscular sac called the gall bladder. The duct of gall bladder (Cystic duct) along with the peptic duct from the liver forms the common bile duct. The bile contains pigments like bilirubin and biliverdin, bile salts, cholesterol and phospholipids but no enzymes. Bile helps in the emulsification of fats and activates lipases.

(b) Layers of the alimentary canal:
The wall of the alimentary canal possesses four layers namely serosa, muscularis, sub-mucosa, and mucosa. The serosa is the outermost layer and is made up of a thin mesothelium with some connective tissues. Muscularis is formed by smooth muscles usually arranged into an inner circular and an outer longitudinal layer. An oblique muscle layer may be present in some regions.

The sub-mucosal layer is formed of loose connective tissues containing nerves, blood, and lymph vessels. In the duodenum, glands are also present in the sub-mucosa. The innermost layer lining the lumen of the alimentary canal is the mucosa. This layer forms irregular folds in the stomach and small finger-like foldings called villi in the small intestine. The cells lining the villi produce numerous microscopic projections called microvilli giving a brush border appearance.

Villi are supplied with a network of capillaries and a large lymph vessel called the lacteal. The mucosal epithelium has goblet cells which secrete mucus that help in lubrication. Mucosa also forms glands in the stomach and crypts in between the bases of villi in the intestine (crypts of lieberkuhn).

Question 3.
Describe the major disorders of the human digestive system.
Solution:
Disorders of the digestive system:
(i) Indigestion:

  • It is the condition in which the food is not properly digested leading to a feeling of fullness.
  • It is caused by inadequate secretion of digestive enzymes, food poisoning, overeating or spicy food.

(ii) Constipation

  • It refers to the condition where the faeces are retained in the rectum for longer periods as the bowel movements occur irregularly.

(iii) Diarrhoea:

  • It refers to the abnormal frequency of bowel movement and increased liquidity of the faecal discharge; absorption of food is impaired.

(iv) Vomiting:

  •  It is the ejection of stomach contents through the mouth; this reflex action is controlled by the vomit centre in the medulla.
  • It is due to viral infection, where liver is affected and digestion of fats is impaired.
  • The eyes and skin turn yellow due to the deposit of bile pigments.

Question 4.
How is the DNA content in our diet digested in the body?
Solution:
DNA content is digested in the intestinal part of our alimentary canal by the enzymes present in pancreatic juice and succus entericus.
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 9
DNAase is found in pancreatic juice while nucleotidase and nucleosidase occur in succus entericus and hydrolyse the DNA content in our diet.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 16 Digestion and Absorption, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 16 Digestion and Absorption, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Biology Chapter 8 Cell: The Unit of Life

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NCERT Solutions for Class 11 Biology Chapter 8 Cell: The Unit of Life

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 8 Cell: The Unit of Life.

Question 1.
Which of the following is not correct?
(a) Robert brown discovered the cell.
(b) Schleiden and Schwann formulated the cell theory.
(c) Virchow explained that cells are from pre¬existing cells.
(d) A unicellular organism carries out its life activities within single cell.
Solution:
(a) Robert brown discovered the cell.

Question 2.
New cells generate from
(a) bacterial fermentation
(b) regeneration of old cells
(c) pre-existing cells
(d) abiotic material
Solution:
(c) pre-existing cells

Question 3.
Match the following
Column A Column B
(a) Cristae (i) Flat membranous sac in stroma
(b) Cisternae (ii) Infoldings in mitochondria
(c) Thylakoids (iii) Disc-shaped sacs in Golgi apparatus
Solution:
(a) Cristae (ii) Infoldings in mitochondria
(b) Cisternae (iii) Disc-shaped sacs in Golgi apparatus
(c) Thylakoids (i) Flat membranous sac in stroma

Question 4.
Which of the following is correct
(a) Cells of living organisms have a nucleus.
(b) Both animals and plant cells have a well defined cell wall.
(c) In prokaryotes, there are no membrane bound organelles.
(d) Cells are formed de novo from abiotic materials
Solution:
(c) In prokaryotes, there are no membrane-bound organelles.

Question 5.
What are mesosomes in a prokaryotic cells? Mention the function that it performs.
Solution:
A special membranous structure is a mesosome that is formed by the extensions of the plasma membrane into the cell. These extensions are in the form of vesicles, tubules, and lamellae. They help in cell wall formation, DNA replication, and distribution to daughter cells. They also help in respiration, in the secretion process increase plasma membrane surface and enzymatic content.

Question 6.
How do neutral solutes move across the plasma membrane? Can the polar molecules also move across it in the same way? If not, then how are these transported across the membrane?
Solution:
Neutral solutes may move across the membrane by the process of simple diffusion along the concentration gradient i.e. from higher concentration to the lower. Water may also move across this membrane from higher to lower concentrations. The movement of water by diffusion is called osmosis. As the polar molecules cannot pass through the non-polar lipid bilayer, they require a carrier protein of the membrane to facilitate their transport across the membrane.

A few ions or molecules are transported across the membrane against their concentration gradient i.e. from lower to the higher concentration. Such transport is an energy-dependent process, in which ATP is utilized and is called active transport. e,g., Na’/K* Pump.

Question 7.
Name two cell-organelles that are double membrane-bound. What are the characteristics of these two organelles? State their functions and draw labelled diagrams of both.
Solution:
Two double membrane-bound cell organelles:
(a) Mitochondria: It has finger-like folds in the inner membrane called cristae. Mitochondria is the place for aerobic respiration.
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1
(b) Chloroplast: Chloroplast is responsible for converting light energy into chemical energy. Chloroplast contains stacked thylakoid in its matrix.
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 2
Functions of mitochondria: The double membrane mitochondria are actively associated with aerobic respiration and the release of energy for cellular activity. The biological oxidation of the fats and carbohydrate release much amount of energy which is utilised by mitochondria for ATP synthesis. The required energy is released from ATP molecules for various cell processes in cells so they are termed as “the powerhouse of the cell”
Functions of chloroplast :
(i) Their main function of the chloroplast is to trap the sun’s energy and to convert it into chemical energy of food by photosynthesis.
(ii) Storage of starch.
(iii) Chloroplast in fruits and flowers changes into chromoplasts.

Question 8.
What are the characteristics of prokaryotic cells?
Solution:
The prokaryotic cells are represented by bacteria, blue-green algae, mycoplasma, and PPLO (Pleuro Pneumonia-like organisms). They are generally smaller and multiply more rapidly than the eukaryotic cells. They may vary greatly in shape and size. The four basic shapes of bacteria are bacillus (rod-like), coccus (spherical), vibrio (comma-shaped), and spirillum (spiral).

The organization of the prokaryotic cell is fundamentally similar even though prokaryotes exhibit a wide variety of shapes and functions. All prokaryotes have a cell wall surrounding the cell membrane. The fluid matrix filling the cell is the cytoplasm. There is no well-defined nucleus. The genetic material is basically naked being not enveloped by a nuclear membrane.

In addition to the genomic DNA (the single chromosome/ DNA circle), many bacteria have small DNA circles outside the genomic DNA. These smaller DNA circles are called plasmids, The plasmid DNA confers certain unique phenotypic to antibiotics. Prokaryotes have something unique in the form of inclusions.

A specialised differentiated form of the cell membrane called mesosome is the characteristic of prokaryotes. They are essentially infoldings of the cell membrane.

Question 9.
Multicellular organisms have a division of labour. Explain.
Solution:
In unicellular organisms, there is no division of labour.

  1. The single cell of the organisms is capable of performing all the vital activities of life i.e., respiration, movement, digestion and reproduction, etc. Respiration, nutrition, and excretion in most of these unicellular organisms takes place through the general body surface.
  2. No special organs for these are present in them because they are too small to need them. Most of these unicellular organisms reproduce by simple binary division, to maintain their continuity.
  3. However, in some, sexual reproduction has also been observed.

Question 10.
Cell is the basic unit of life. Discuss in brief.
Solution:
Cell: The Basic Unit of life: All living organisms are composed of small, tiny structures or compartments called cells. These cells are called the ‘building blocks’ of life.

  1. The cells in true sense are considered as the basic unit of life because all the life processes i.e., metabolism, responsiveness, reproduction are carried out by the cells.
  2. Respiration, nutrition, release of energy for the body are carried out within the cells only.
  3. Even the animals and plants reproduce because the cells reproduce individually.
  4. Growth occurs because cell grow and multiply.
    In Amoeba all the life processes are performed within the boundaries of the single cell.
  5. This is true of all other multicellular organisms. The only difference in the multicellular organisms is that the body of these organisms is made up of many cells.
  6. In these organisms, the cell do not behave independently but get organized into tissues. Each tissue is specialized to perform specific functions. Different tissues then get organised into tissues.
  7. Each tissue is specialized to perform specific functions. Different tissues then get organised into organs which perform certain specific functions.
  8. Different organs are finally organised to form organ systems. Now it must be very clear that the basic structure to tissues, organs and organ system are the cells only.
  9. These tissues, organs and organ system of the organisms work because the cells work.
  10. Thus “the cells are structural and functional unit of the living beings” hence it is the basic unit of life.

Question 11.
What are nuclear pores? State their function.
Solution:
At a number of places, the nuclear envelope is intercepted by minute pores which are called nuclear pores. These are formed by the fusion of two nuclear membranes. These nuclear pores are the passages through which the movement of RNA and protein molecules takes place in both directions between the nucleus and the cytoplasm.

Question 12.
Both lysosomes and vacuoles are * endomembrane structures, yet they differ in
terms of their functions. Comment.
Solution:
Lysosomes are filled with hydrolytic enzymes that are capable of digesting carbohydrates, proteins, lipids and nucleic acids whereas vacuoles contain water, sap, excretory product and other materials not useful for cell.

Question 13.
Describe the structure of the following with the help of labelled diagrams.
(i) Nucleus
(ii) Centrosome
Solution:
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 3
(i) Nucleus:

    1. The nucleus is a large organelle controlling all the activities of the eukaryotic cells. Some cells have more than one nucleus.
    2. Binucleate cells have 2 nuclei per cell eg. Paramoecium. Multinucleate cells have many nuclei e.g. Ascaris.
    3. Some cells lack nucleus (anucleate) at maturity. Examples: mammalian RBCs and sieve tube cells in vascular plants.
    4. The nucleus is bounded by two membranes, which make the nuclear envelope.
    5. The outer and inner membranes are separated by a narrow space, perinuclear space.
    6. The outer membrane remains in continuation with endoplasmic reticulum (ER) and the inner one surrounds the nuclear contents.
    7. At some points, the nuclear evelope is interrupted by the presence of small structures called nuclear pores.
    8. These pores help in exchange of materials between nucleoplasm and cytoplasm. Nuclear membrane dissappears during cell division. It reappears during nuclear reorganization in stage.
    9. The nucleoplasm contains chromatin and nucleolus. The nucleolus is a rounded structure. It is not separated from the rest of the nucleoplasm by membrane.
    10. It is associated with a specific nucleolar organizing region (NOR) of some chromosomes. Nucleolus is the “site for ribosomal RNA synthesis”.
    11. The cells which remain engaged in protein synthesis have larger and more numerous nuclei in their nucleoplasm.

NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 4

(ii) Centrosome:

  1. Under the electron microscope, each centriole is seen to be formed of nine sets of tubular structures arranged in a circular fashion.
  2. Each of these sets is a triplet composed of three microtubules. Each microtubule has a diameter of about 250A. The triplets are found in the matrix.
  3. Sometimes delicate strands appear to connect sets of the triplet to each other.
  4. Also can be seen radiating from the central core of the cylinder, delicate strands which connect sets of the triplets to each other giving a cartwheel appearance.
  5. Basal bodies are structures similar to the centrioles. They produce cilia and flagella.

Question 14.
What is a centromere? How does the position of the centromere form the basis of the classification of chromosomes? Support your answer with a diagram showing the position of the centromere on different types of chromosomes.
Solution:
Eukaryotic chromosomes: The chromosomes are uncoiled in a loose, indistinct network called the chromatin that contains DNA, RNA and protein in interphase.

The types of proteins present and associated with DNA are histone and non-histone proteins.

Chromosomes are thread-like structures. They become visible (under light microscope) during cell division.

In higher organisms, the well-developed nucleus contains a definite number of chromosomes of definite size and shape.
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 5
The shape of a chromosome is usually observable at metaphase and anaphase when the position of primary constriction {centromere) is clearly seen. Based on the position of the centromere, chromosomes are of 3 types:

  • telocentric – with terminal centromere,
  • the acrocentric – terminal centromere is capped by a telomere
  • submetacentric – the centromere is subterminal in position
  • metacentric – these have median centromere.

VERY SHORT ANSWER QUESTIONS

Question 1.
Who discovered the Golgi body?
Solution:
Camillo Golgi (1898).

Question 2.
Give the location of 70S ribosomes.
Solution:
Prokaryotic cells, plastids, and mitochondria

Question 3.
Name the cell organelle rich in acid hydrolases.
Solution:
Lysosomes

Question 4.
Who proposed the cell theory?
Solution:
Schleiden and Schwann.

Question 5.
Expand PPLO.
Solution:
PPLO (Pleuro Pneumonia Like Organisms)

Question 6.
Name the organelle responsible for protein synthesis in a cell.
Solution:
Ribosome

Question 7.
Give the full form of SER and RER.
Solution:

  • SER – Smooth endoplasmic reticulum.
  • RER – Rough endoplasmic reticulum.

Question 8.
Name the membrane which surrounds the vacuole in the cell.
Solution:
Tonoplast

Question 9.
Name two types of constituents of the plasma membrane.
Solution:
Proteins and lipids.

Question 10.
Name two processes of passive transport.
Solution:

  • Osmosis
  • Diffusion

Question 11.
What is plasmodesmata? What is its function?
Solution:
Plasmodesmata: Adjoining the cells and the linking gap of cytoplasmic protoplasmic presence is called Plasmodesmata. It links the neighbouring cells together.

SHORT ANSWER QUESTIONS

Question 1.
Why fluid-mosaic model is more accepted than other models of the plasma membrane?
Solution:
The fluid mosaic model explains
(i) quasifluid state of the plasma membrane,
(ii) It differentiates two 6. types of proteins
(iii) It explains functional specificity and variability in two surfaces of PM.

Question 2.
Name three types of elements in the Golgi body. List two major functions of the Golgi body.
Solution:
Three types of elements in golgi body are cistemae, vesicles and vacuoles. The main function of golgi bodies are cellular secretion and acrosome formation.

Question 3.
What is peculiar about mitochondrial DNA?
Solution:
Mitochondrial DNA is circular double-stranded and not associated with histone proteins.

Question 4.
How does cytokinesis take place in plant and animal cells?
Solution:
In plant cell cytokinesis take place by cell plate formation and in animal cells it occurs by constriction

Question 5.
Differentiate between prokaryotic and eukaryotic cells.
Solution:
The main differences between prokaryotic cell and eukaryotic cell are
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 6

Question 6.
Differentiate between active and passive transport across the membrane.
Solution:
The main differences between active transport and passive transport are
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 7

Question 7.
Describe the different methods of transport of nutrients in cell.
Solution:
Different methods of transport of nutrients in the cell are:
(i) Simple diffusion : It is the movement of ions/ molecules of any substance from a region of higher concentration to ‘the region of lower concentration, until equilibrium is reached. Many neutral solutes move by diffusion.
(ii) Osmosis : It is the movement of solvent molecules across a semipermeable membrane, from the region of higher concentration to the region of lower concentration, until equilibrium is reached. Water moves by osmosis from one cell to the other.
(iii) Facilitated diffusion : It refers to the movement of ions/molecules across the membrane with the help of transmembrane proteins.

Question 8.
Why are ribosomes of prokaryotes different from eukaryotes?
Solution:
The type of ribosomes of prokaryotes is different from eukaryotes because the prokaryotes were primitive, simpler and have remained intact during evolution while at the base level eukaryotes have adapted with the environment and are retaining their kind of entities for the complex structure. Prokaryotes have 70S ribosomes with 30S and 50S subunit and eukaryotes have 80S ribosome with 40S subunit and 60S sub unit.

Question 9.
What is the function of
(1) Nuclear Pores
(2) Slimy Capsule in Bacteria
(3) Golgi Bodies:
(4) Centrosome with Centrioles
Solution:
(1) Nuclear Pores: There is exchange of RNA and proteins through the nuclear pores
(2) Slimy Capsule in Bacteria : A slimy capsule is the outer covering of cell wall of bacteria and is an additional protection for the bacteria.
(3) Golgi Bodies :
(i) It takes part in packaging materials delivered either to the intra-cellular targets or secteted outside the cell.
(ii) It is also a important site of formation of glycoproteins and glycolipids.
(4) Centrosome with Centrioles
(i) Centrioles help in organising the spindle fibres and astral rays during cell division.
(ii) It also provides basal bodies which give rise to cilia and flagella.

LONG ANSWER QUESTIONS

Question 1.
Describe the structure of cell wall.
Solution:

  • A non-living rigid structure called the cell wall forms an outer covering for the plasma membrane of fungi and plants.
  • Cell wall does not only give shape to the cell and but protect the cell from mechanical damage and infection.
  • It also helps in cell-to-cell interaction and provides barrier to undesirable macromolecules.
  • Algae have cell wall, made of cellulose, galactans, mannans and minerals like calcium carbonate, while in other plants it consists of cehulose, hemicellulose, pectins and proteins.
  • The cell wall of a young plant cell, the primary wall is capable of growth, which gradually diminishes as the cell matures and the secondary wall is formed on the inner side of the cell.
  • The middle lamella is a layer mainly of calcium pectate which holds or glues the different neighbouring cells together.
  • The cell wall and middle lamellae may be transferred by plasmodesmata which connect the cytoplasm of neighbouring cells.

Question 2.
Give an account of prokaryotic cells.
Solution:

  • The prokaryotic cells are represented by bacteria, blue green algae, mycoplasma and PPLO (Pleuro Pneumonia Like Organisms). They are generally smaller and multiply more rapidly than the eukaryotic cells.
  • They may vary greatly in shape and size. The four basic shapes of bacteria are bacillus (rod like), coccus (spherical), vibrio (comma shaped) and spirillum (Spiral).
    The organisation of the prokaryotic cell is fundamentally similar even though prokaryotes exhibit a wide variety of shapes and functions. All prokaryotes have a cell wall surrounding the cell membrane.
  • The fluid matrix filling the cell is the cytoplasm. There is no well defined nucleus.
  • The genetic material is basically naked, not enveloped by a nuclear membrane.
  • In addition to the genomic DNA (the single chromosome/circular DNA), many bacteria have small circular DNA outside the genomic DNA. These smaller DNA is called plasmids.
  • The plasmid DNA confers certain unique phenotypic characters to such bacteria. One such character is resistance to antibiotics. Nuclear membrane is found in eukaryotes.
  • No organelles, like the ones in eukaryotes, are found in prokaryotic cells except ribosomes.
  • Prokaryotes have something unique in the form of inclusions. A specialised differentiated form of cell membrane called mesosome is the characteristic of prokaryotes which helps in respiration process.
  • They are essentially infoldings of cell membrane.

Question 3.
Give an ultrastructure of mitochondria.
Solution:

  • Mitochondria, unless specifically stained, are not easily visible under the microscope.
  • The number of mitochondria per cell is variable depending on the physiological activity of the cells.
  • In terms of shape and size also, considerable degree of variability is observed.
  • Typically it is sausage shaped or cylindrical having a diameter of 0.2-1.0 ft m (average 0.5 film) and length (1.0 -4.1 ft).
  • Each mitochondrion is a double membrane-bound structure with the outer membrane and the inner membrane dividing its lumen distinctly into two aqueous compartments, i.e. the outer compartment and the inner compartment.
  • The inner compartment is called the matrix. The outer membrane forms the continuous limiting boundary of the organehe.
  • The inner membrane forms a number of infoldings called the cristae. The cristae increase the surface area.
  • The two membranes have their own specific enzymes associated with the mitochondrial function. Mitochondria are the sites of aerobic respiration.
  • They produce cellular energy in the form of ATP, hence they are called. “Power houses” of the cell.
  • The matrix also possesses single circular DNA molecule, a few RNA molecules, ribosomes (70s) and the components required for the synthesis of proteins. The Mitochondria divide by fission.
    NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 8

Question 4.
Describe the fluid mosaic model of membrane.
Solution:
The characteristic features of fluid mosaic model
• This model was proposed by Singer and Nicholson.
• According to this model, there is a central bilipid layer (of phospholipids) with their polar head group toward the outside and the non-polar tails pointing inwards.
• Some proteins which are embedded in the lipid layer are called integral proteins and they cannot be separated from the membrane easily.
• Some large globular integral proteins which project beyond the lipid layer on both the sides are believed to have channels through which water soluble materials can pass across.
• Those proteins which are superficially attached are called peripheral (extrinsic) proteins and they can be easily removed.
• Some membrane lipids and integral proteins remain bound to oligosaccharides; such oligosaccharides project into the extracellular fluid and they influence the manner in which cells interact with the other cell.
• There are also certain specific proteins called membrane receptors, which mediate the flow of materials and information into the cell.

Question 5.
What are plastids? How are they classified on the basis of the type of pigments? Name them and their pigments and mention their functions.
Solution:
Plastids are double-membrane bound organelles
of different shapes, that are found only in plant
cells and contain pigments and storage products.
They are of three types :
(i) Leucoplasts
These are the oval, spherical, rod-like or filamentous colourless plastids which are found in storage organs. Their main function is to store reserve materials like starch (amyloplasts), proteins (aleuroplasts) and fats (elaioplasts).
(ii) Chromoplasts
• These are coloured plastids containing mainly the yellow, red and orange pigments (carotene and xanthophyll).
• These are found in petals of flowers and skin of fruits.
• They attracts agents for pollination and dispersal of fruits/seeds.
(iii) Chloroplasts
• These are the green plastids containing mainly chlorophylls and very little carotene and xanthophyll.
• Their main function is photosynthesis and the formation of starch.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 8 Cell: The Unit of Life, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 8 Cell: The Unit of Life, drop a comment below and we will get back to you at the earliest.