CBSE Class 11 – Page 24

NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants

January 5, 2022June 18, 2021 by Raju

NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants.

Question 1.
What are the factors affecting the rate of diffusion?
Solution:
Diffusion rates are affected by the gradient of concentration, the permeability of the membrane separating them, temperature, and pressure.

Question 2.
What are porins? What role do they play in diffusion?
Solution:
Porins are proteins that form huge pores in outer membranes of plastids, mitochondria and some bacteria, allowing molecules upto the size of small proteins to pass through. Thus porins facilitate diffusion.

Question 3.
Describe the role played by protein pumps during active transport in plants.
Solution:
Active transport uses energy to pump molecules against a concentration gradient. Active transport is carried out by membrane-proteins. Hence different proteins in the membrane play a major role in both active as well as passive transport. Pumps are proteins that use energy to carry substances across the cell membrane.

This pump can transport substances from a low concentration to a high concentration. The transport rate reaches a maximum when all the protein transporters are being used or are saturated. Like enzymes, the carrier protein is very specific in what it carries across the membrane. These proteins are sensitive to inhibitors that react with protein side chains.

Question 4.
Explain why pure water has maximum water potential?
Solution:
Water molecules possess kinetic energy. In liquid and gaseous form they are in random motion that is both rapid and constant. The greater the concentration of water in a system, the greater is its kinetic energy or water potential. Hence, it is obvious that pure water will have the greatest water potential.

Question 5.
Differentiate between the following:
(a) Diffusion and Osmosis
(b) Transpiration and Evaporation
(c) Osmotic Pressure and Osmotic Potential
(d) Imbibition and Diffusion
(e) Apoplast and Symplast pathways of movement of water in plants
(f) Guttation and Transpiration
Solution:
(a) The differences between diffusion and osmosis are as following:
NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 1
(b) The differences between transpiration and evaporation are as following:

(c) The main differences between osmotic pressure and osmotic potential are as following:

(d) The main differences between imbibition and diffusion are as following:
NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 4
(e) The main differences between apoplast and symplast pathways of movement of water in plants are as following :

(f) The main differences between guttation and transpiration are as following :

Question 6.
Briefly describe water potential. What are the factors affecting it?
Solution:
Water potential: Free energy per mole of water molecule is called water potential. It is represented by Greek letter ‘φ’ (Psi). Water potential of pure water is zero and addition of solute, decreases its free energy or water “‘potential.
Factors affecting water potential are as following:
(i) Matric potential (φ_m) causes by adsorbent or colloidal particles (always negative and almost negligible).
(ii) Solute potential (φ_s) caused by presence of solute particles (decreases the water potential).
(iii) Pressure potential (φ_p) caused by entry or exit of water (increases the water potential).
φ_w= φ_p + φ_s + φ_m

Question 7.
What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?
Solution:
If a pressure greater than atmospheric pressure is applied to pure water or a solution, its water potential increases. It is equivalent to pumping water from one place to another.

Question 8.
(a) With the help of well-labelled diagrams, describe the process of plasmolysis in plants, giving appropriate examples.
(b) Explain what will happen to a plant cell if it is kept in a solution having higher water potential.
Solution:
(a) Plasmolysis is the shrinkage of the protoplast of a cell from its cell wall under the influence of a hypertonic solution (solution of higher concentration).
NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 7
(b) The process of plasmolysis is usually reversible. When the cells are placed in hypotonic solution (solution with low concentration), water diffuses into the cell causing the cytoplasm to swell up. The swelling of shrunken protoplast is called as deplasmolysis.

Question 9.
How is the mycorrhizal association helpful in the absorption of water and mineral in plants?
Solution:
Some plants have additional structures associated with them that help in water (and mineral) absorption. A mycorrhiza is a symbiotic association of a fungus with a root system. The fungal filaments form a network around the young root or they penetrate the root cells. The hyphae have a very large surface area that absorbs mineral ions and water from the soil from a much larger volume of soil that perhaps a root cannot do.

The fungus provides minerals and water to the roots, in turn, the roots provide sugars and N-containing compounds to the mycorrhizae. Some plants have an obligate association with the mycorrhizae. For example, Pinus seeds cannot germinate and establish without the presence of mycorrhizae.

Question 10.
What role does root pressure play in water movement in plants?
Solution:
As various ions from the soil are actively transported into the vascular tissues of the roots, water flows and increases the pressure inside the xylem. This positive pressure is called root pressure and can be responsible for pushing up water to small heights in the stem.

Root pressure can only provide a modest push in the overall process of water transport. The greatest contribution is to re-establish the continuous chains of water molecules in the xylem which often break under the enormous tensions created by transpiration.

Question 11.
Describe the transpiration pull model of water transport in plants. What are the factors influencing transpiration? How is it useful to plants?
Solution:
Transpiration pull theory: The theory was put forward by Dixon and Jolly.
The main features of the theory are :
(i) There is a continuous column of water (present in the tracheary element) from the root through the stem and into the leaves.
(ii) Water molecules attached to one another, by cohesion force, adhesion force (between their walls and water molecules), and surface tension.
(iii) The water in the tracheary element would come under tension, due to transpiration also called transpiration pull.

Factor affecting transpiration: External factors e.g. atmospheric pressure, temperature, humidity, CO2, sunlight, wind velocity, etc. The thickness of cuticle, number, position and closing and opening of stomata, pH, and hormones are internal factors which affect transpiration. Significance of transpiration:

Ascent of sap
Removal of the excess water
Cooling effect
Distribution of minerals.
Transpiration supplies water for photosynthesis.

Question 12.
Discuss the factors responsible for the ascent of xylem sap in plants.
Solution:
The ascent of xylem sap is supported by the following factors.
(i) Root pressure (positive pressure that develops in the xylem sap of root).
(ii) Cohesion (adhesion of water molecules due to hydrogen bonding).
(iii) Adhesion (force between tracheary wall and water molecule).
(iv) Transpiration pull (tension develops due to transpiration) is the main cause of ascent of xylem sap.

Question 13.
What essential role does the root endodermis play during mineral absorption in plants?
Solution:
Water flowing through apoplast contains minerals useful to plants and also toxins. Endodermal cells have many transport proteins embedded in their plasma membrane; they let some solutes cross the membrane, but not others. Transport proteins of endodermal cells are control points, where a plant adjusts the quantity and types of solutes that reach the xylem.

Question 14.
Explain why xylem transport is unidirectional and phloem transport bidirectional.
Solution:

The plant part which synthesise the food, i.e. leaf known as source and part that needs or stores the food called as sink.
Food, primarily sucrose, is transported by vasculai tissue phloem, from a source to a sink. But depending on season or plants need, the source sink may be reversed.
Sugar stored in roots may be mobilised to become a source of food source in early spring when buds of trees require energy for their growth and development and act as sink.
Since the source – sink relationship is bidirection in phloem while in xylem it is unidirection.

Question 15.
Explain pressure flow hypothesis of translocation of sugars in plants.
Solution:
Mass flow or Pressure Flow Flypothesis : It was put forward by Munch (1927,1930).

According to this hypothesis organic substance move from region of high osmotic pressure to low osmotic pressure due to turgor pressure.
A high osmotic concentration is present in source e.g. mesophyll cells (due to photosynthesis).
Sugar present in them passed into sieve tube therefore high osmotic concentration develops and it absorbs water from xylem and develop a high turgor pressure. It causes flow of sugar towards area of low turgor pressure or sink.

Question 16.
What causes the opening and closing of guard cells of stomata during transpiration?
Solution:

The opening and closing of stomata depend upon the turgid or flaccid state of the guard cells.
The inner wall of guard cells (towards pore) is thick and outer wall (towards other epidermal cells) is thin.
When the turgor pressure of the guard cells in increased the outer thinner wall of the guard cell is pushed out (towards the periphery) thus pulling the inner thicker wall leading to the opening of stomatal pore.
When the guard cells are in a flaccid state the outer thinner wall of guard cells returns to original position (moves towards pore) due to which tension on the inner wall is released and stomatal aperture gets closed.

VERY SHORT ANSWER QUESTIONS

Question 1.
What fraction of soil water is readily available to plants?
Solution:
Capillary water.

Question 2.
What is a hypertonic solution? (Oct. 87, M.Q.P., Apr. 2000)
Solution:
The external solution which has a concentration higher than that of the cell sap is called a Hypertonic solution.

Question 3.
What is the chemical potential of pure water at normal temperature and pressure?
Solution:
Zero.

Question 4.
What is transmembrane pathway?
Solution:
The movement of water through the cell membrane is called transmembrane pathway or symplast pathway.

Question 5.
What happens to the plant cell when it is plasmolyzed? (Apr. 95)
Solution:
becomes flaccid

Question 6.
Identify a type of molecular movement which is highly selective and requires special membrane proteins, but does not require energy.
Solution:
Facilitated diffusion.

Question 7.
How can you revert a freshly plasmolysed plant cell to its normal state?
Solution:
By placing it in aqueous solution.

Question 8.
What is meant by uniport?
Solution:
A molecule moves across the membrane independent of other molecule called as uniport.

Question 9.
What is the use of imbibition pressure to plants?
Solution:
Responsible for the seedling to emerge out of the soil.

Question 10.
What is a casparian strip?
Solution:
Casparian strip is band of suberin on the tangential and radial walls of endodermal cells.

Question 11.
Name the form in which cabohydrates are transported in plants and the tissue through which it is transported.
Solution:
Sucrose, phloem.

Question 12.
What is the pressure potential of a plasmolysed cell?
Solution:
Pressure potential of a plasmolysed cell is negative.

Question 13.
Give reason:
Excessive use of chemical fertilizers results in wilting of plants. (June 2009)
Solution:
Excessive use of fertilizers increases the solute concentration of the soil solution causing exosmosis resulting in the wilting of plants.

Question 14.
If water enters a cell what is the pressure exerted by its swollen protoplasts?
Solution:
The pressure exerted will be turgor pressure.

SHORT ANSWER QUESTIONS

Question 1.
List any four mechanisms that contribute to the ascent of sap in tall trees.
Solution:
Mechanisms that contribute to the ascent of sap in tall trees are :

Adhesion
Cohesion
Transpirational pull
Root pressure.

Question 2.
Mention two advantage of transpiration.
Solution:
Advantages of transpiration:
(1) It helps in mineral absorption and ascent of sap.
(2) Transpiration, by evaporating of water causes a cooling effect of plant body.

Question 3.
How do rise in temperature and wind velocity affect transpiration? Write any three adaptations shown by plants to reduce transpiration.
Solution:
Rise in temperature, wind velocity increases the rate of transpiration.
Three adaptations shown by plants to reduce transpiration are:
(a) leaves with thick cuticle
(b) sunken stomata
(c) modification of leaves into scales or spines and stem into phylloclades.

Question 4.
What is meant by apoplast pathway? Why does it occur in cortex and not in endodermis?
Solution:
The movement of water exclusively through cell wall without crossing cell membrane, is called apoplast pathway. It occurs in cortex due to the absence of casparian strip and presence of loosely packed cells. It does not occur in endodermis due to the presence of suberin in casparian strip.

Question 5.
What is the significance of osmosis?
Solution:
Significance of osmosis:
(1) Osmosis is important in the absorption of water by plants.
(2) Helps in cell to cell movement of water.
(3) Provides rigidity to the plant organs.
(4) Helps in opening and closing of stomata.

Question 6.
Describe the apoplastic and symplastic movements of water in plants.
Solution:
Apoplastic movement of water occurs exclusively through the non-living cell wall without crossing any membrane. Symplastic movement of water occurs from one cell to the other through plasmodesmata it passes across membrane and i.e. living components.

Question 7.
Sugar crystals do not dissolve easily in ice cold water. Explain.
Solution:
Rate of diffusion is increased when temperature increase. So, sugar crystal do not dissolve easily in ice cold water.etrads of homo logous chromosomes and crossing as the genetic material exchange takes place. Thus this stage is marked with genetic recombination.

Question 8.
What is imbibition pressure? What is the usefulness of imbibition pressure to seed germination?
Solution:
Imbibition pressure is the pressure developed in an adsorbent due to diffusion of water into it.
This pressure makes the seedlings to emerge above the ground during seed germination.

Question 9.
Differentiate between active and passive transport.
Solution:
The main difference between active transport and passive transport are :
NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 10

Question 10.
Differentiate between diffusion and facilitated diffusion.
Solution:
The main difference between diffusion and facilitated diffusion are :
NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 11

Question 11.
What is meant by source and sink in plants, with regard to translocation?
Solution:
Source : Source is the part of the plant which synthesises the food, they are the leaves.
Sink: Sink is the part of the plant that needs or stores the food like roots, tubers, any storage organ etc.

Question 12.
Differentiate between transportation of water and translocation of Food.
Solution:

LONG ANSWER QUESTIONS

Question 1.
Describe with the help of labelled diagrams, the closing and opening of stomata in plants.
Solution:

Stomata are minute openings formed in the epidermis of leaves, stem and in (some cases) even flowers.
Each stomata contains pore surrounded by two guard cells. The guard cells are joined at both ends but separate in the mid-region of their length.
Stomata are mostly present on the lower epidermis of the leaf. The open stomata account for diffusion of water vapour through them.
During the day, the cell- sap concentration becomes high due to accumulation of sugar as a result of photosynthesis.
This results into endosmosis and tl\e water is withdrawn into guard cells from the neighboring cells.
This makes the guard cells turgid and stomata open. If the availability of water is reduced, the guard cells lose their turgidity and they become flaccid by exosmosis of water from guard cells.
This leads to the closing of stomata and transpiration stops.

Question 2.
Describe the root pressure theory and its demerits.
Solution:
Root pressure theory : The theory was put forward by Priestley (1916). Root pressure refers to positive hydrostatic pressure which is develops in xylem sap, when rate of water absorption is more than the rate of transpiration and as a result of which water is pushed up in tracheary element of root.
Demerits of root pressure concept :
(1) Magnitude of root pressure is maximum upto 2 atmosphere which can raise water upto 64ft. only. It can’t drive water in tall trees (300-400 feet) there is need of 20 atmosphere root pressure.
(2) No root pressure in conifers.
(3) Root pressure is low in summer when transpiration is high and high in spring when transpiration is low
(4) Root pressure is not seen in rapidly transpiring plants.

Question 3.
Discuss the factor affecting the water absorption.
Solution:
The factor affecting the water absorption are :
(1) Temperature : Optimum temperature for
absorption is 20-30°C. Absorption decrease at low temperature by decreasing permeability, retarding in growth of root and availability of water and by increasing viscosity of water.
(2) Concentration of soil solution: Absorption decreases with increase in concentration of soil solution, e.g. Halophytes (Rhizophora) are able to grow in swamps with high salt cont-entration because their cell sap concentration is also high.
(3) Soil aeration : Flooding of fields reduces soil air (0₂). Hence metabolic processes and absorption are also reduced.
(4) Increase in C0₂ : High C0₂ concentration lowers the rate of respiration and thus influence metabolic processes. These result low water absorption.
(5) Available soil water : Maximum absorption occurs when the available soil water is in the range of field capacity and permanent wilting coefficient.
(6) Root growth : Deep rooted plant absorb more water. Older roots are suberized and absorb slowly.

Question 4.
Differentiate the following
(a) Simple diffusion and active transport
(b) Turgid cell and flaccid cell
(c) Isotonic solution and hypotonic solution
Solution:
(a) The main difference between simple diffusion and active transport are :
NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 14
(b) The main difference between the turgid cell and flaccid cell are :

(c) The main difference between isotonic solution and hypotonic solution are :

Question 5.
What are the properties that favours the upward movement water through xylem?
Solution:
The properties are:
(1) Cohesion : mutual attraction between water molecules
(2) Adhesion : attraction of water molecules to polar surfaces
(3)Surface Tension : water molecules are attracted to each other and give water high tensile strength i.e., an ability to resist a pulling force
(4) Capillarity : ability to rise in their tubes. These are also related to the transpiration pull. With this pull water comes to leaves for photosynthesis as well.

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

NCERT Solutions for Class 11 English Hornbill Poem 2 The Laburnum Top

January 5, 2022June 18, 2021 by Prasanna

Here we are providing NCERT Solutions for Class 11 English Hornbill Poem 2 The Laburnum Top. Students can get Class 11 English The Laburnum Top NCERT Solutions, Questions and Answers designed by subject expert teachers.

The Laburnum Top NCERT Solutions for Class 11 English Hornbill Poem 2

The Laburnum Top NCERT Text Book Questions and Answers

Find out

Question 1.
What laburnum is called in your language.
(Answers will vary)

Question 2.
Which local bird is like the goldfinch.
Answer:
Birds which sing melodiously like the goldfinch in India include the Asian Koel, the Bulbul, the House Sparrow, the Himalayan Cuckoo, the Swallow and so on.

The Laburnum Top Think it out

Question 1.
What do you notice about the beginning and the ending of the poem?
Answer:
The beginning of the poem is about the silence and stillness of the laburnum. The end of the poem again drifts into silence and emptiness of the laburnum.

Question 2.
To what is the bird’s movement compared? What is the basis for the comparison?
Answer:
The bird’s movement is compared to that of a lizard. The basis for the comparison is the sleek, alert and abrupt movement with which the bird enters its nest. This movement of the bird is comparable to that of a lizard.

Question 3.
Why is the image of the engine evoked by the poet?
Answer:
The poet evokes the image of an engine because as soon as the goldfinch enters its nest on the laburnum top, the whole tree comes alive with the chirruping of the young ones of the bird, their twittering and the tremor of their wings. As an engine brings a seemingly dead machine to life, so the chittering family of the goldfinch brings the whole tree into life.

Question 4.
What do you like most about the poem?
Answer:
Open-ended question. Here is a suggested answer: I like the depiction and the imagery in the poem. The poet creates a mundane yet beautiful picture of a tree top that goes through stages from being silent to alive and drifts back into silence again. The imagery of an engine bringing alive a machine is also something that catches my fancy.

Question 5.
What does the phrase “her barred face identity mask” mean?
Answer:
The phrase refers to the face of the bird that is covered with fur of different colours and which looks like a mask making it the identity of the bird.

Note down

1. the sound words
Chirrup, machine starts up, chitterings, whistle-chirrup whisperings, trillings, engine, stokes it full, subsides to empty.

2. the movement words
Seeds fallen, sleek as a lizard, enters the thickness, tremor of wings, tree trembles and thrills, flirts out to a branch-end, she launches away.

3. the dominant colour in the poem.
The Laburnum tree has yellowing leaves, sitting still in the afternoon yellow sunlight. The goldfinch, a bird full of yellow feathers flies on to its branches to sit and sing. The dominant colour in all the descriptions in the poem is yellow.

The Laburnum Top List the following

Question 1.
Words which describe ‘sleek’, ‘alert’ and ‘abrupt’.
Answer:
Sleek – Smooth, Polished, Effortless
Alert – Watchful, Careful, Brisk, Active
Abrupt – Hasty, Quick, Sudden, Rushed

Question 2.
Words with the sound ‘ch’ as in ‘chart’ and ‘tr’ as in ‘trembles’ in the poem.
Answer:
Words with ‘Ch’ – Chirrup, Chitterings, Whistle-Chirrup
Words with ‘Tr’ – Tremor, Trillings, Tree, Trembles, Thrills

Question 3.
Other sounds that occur frequently in the poem.
Answer:
‘St’ – Still, Startlement, Starts up
‘-ings’ – Chitterings, Wings, Trillings, Whisperings

The Laburnum Top Thinking about language

Question 1.
Look for some other poem on a bird or a tree in English or any other language.
Answer:
Other poems on birds and trees include “Ode to a Nightingale” by John Keats which describes the beauty in the bird’s song, and “Hawk Roosting” by Ted Hughes which looks at a Hawk overlooking its domain to prey upon.
(Answers will vary)

Try this out
Write four lines in verse form on any tree that you see around you.
(Answers will vary)

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

January 5, 2022June 18, 2021 by Raju

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:

Growth is defined as a permanent or irreversible increase in dry weight, size, mass or volume of a cell, organ or organism.
Generally growth is accompanied by metabolic processes (both anabolic and catabolic). At the cellular level, growth is due to increase in amount of protoplasm.
However, it is difficult to measure increase in protoplasm.
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.
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:

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:

Cell division and Differentiation: Auxins promote cell division and their subsequent differentiation into tissues. They are used in cultures to bring about organogenesis.
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).
Root Initiation: Low concentrations of auxins promote rooting which can propagate more plants vegetatively.
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.
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.
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

January 5, 2022June 18, 2021 by Raju

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 3

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:

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 p^H.
Activity declines both below and above optimum temperature and p^H.
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:

Question 9.
Differentiate between Structural Proteins and Functional Proteins.
Solution:

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

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

Question 2.
Distinguish between the primary, secondary, and tertiary structures of proteins.
Solution:

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

Oxidoreductases: eg Alcohol dehydrogenase, oxidation, Reduction occurs
Transferases: transfer a particular group to another substrate, eg. transavninase
Hydrolases: cleave their substrates by hydrolysis of a covalent bond e.g. Urease, amylase.
Lyases: break the covalent bond eg. Deaminase
Isomerase: by changing the bonds they make isomers. eg: Aldolase.
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

January 5, 2022June 18, 2021 by Raju

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:

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.

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

Solution:

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

(c) The main difference between the pectoral girdle and pelvic girdle are as following:

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) 11^th, 12^th
(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:

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 8^th, 9^th, and 10^th 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 (11^th and 12^th) 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

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