Level - I
SECTION - A
School/Board Exam. Type Questions Very Short Answer Type Questions :
1. What is mycorrhiza?
Sol. A mycorrhiza is a symbiotic association of a fungus and the root system of a higher plant. 2. Explain plasmodesmata.
Sol. Plasmodesmata are the protoplasmic connections present between adjacent cells. They are large openings in the cell wall specifically of plants.
3. What is the membrane of vacuole called? Sol. Tonoplast
4. Define antiport.
Sol. In facilitated diffusion, when molecules move in opposite directions by transport proteins, it is called antiport. 5. Give an example of region in plants, where negative pressure potential is found.
Sol. Xylem
6. Give net direction of water movement in a plant cell in flaccid state.
Sol. There is no net movement of water occurs in flaccid cell. (i.e., water flowing into the cell and out of cell is equal).
7. What happens to the water potential of a solution when more solute is added to it. Sol. Its water potential decreases.
8. State the direction of movement of food in the phloem.
Sol. The direction of movement of food in the phloem can be upwards or downwards i.e., bidirectional? 9. Which means of transport occurs when sugar is loaded in phloem at the source?
Sol. Active transport
10. Give shape of guard cells in dicotyledonous plants. Sol. They are kidney-shaped in dicotyledonous plants. Short Answer Type Questions :
11. What is cytoplasmic streaming? Give an example of a plant where cytoplasmic streaming can be observed? Sol. It is an automatically vital movement that occurs continuously in the cytoplasmic matrix of eukaryotic cells
including plant cells.
For example, in cells of Hydrilla leaf, the movement of chloroplast due to streaming is easily visible.
Chapter
8
Transport in Plants
12. What are the two conditions for imbibition to occur?
Sol. Two conditions which are necessary for imbibition to occur are:
(a) Presence of water potential gradient between the surface of adsorbent and the liquid to be imbibed into it.
(b) Affinity between the adsorbent i.e., solid substances and imbibed liquid.
13. What is turgor pressure? Give its one significance to plants.
Sol. The hydrostatic pressure that develops in the protoplasm of a cell, due to entry of water into it, is called turgor pressure. This turgor pressure helps plants to transport water and solutes over long distances through xylem.
14. Name the two factors on which water potential depends? Also give its relationship with them.
Sol. The two factors are :
(a) Solute potential (b) Pressure potential
The equation which summarises this relationship between these terms is : Water potential (w) = Solute potential (s) + Pressure potential (p)
15. Give any two characteristic features of transport through special proteins in facilitated diffusion.
Sol. The two characteristic features of carrier proteins are: (a) Specific in nature
(b) Sensitive to inhibitors
16. What will be the water potential of solution B, if its solute potential S at atmospheric pressure is – 200 Pa?
Sol. Water potential equals to solute potential, at atmospheric pressure. So,
Water potential of solution B = Solute potential of B W = S (at atmospheric pressure) Therefore, W = –200 Pa
17. How solute potential, water potential and pressure potential of a cell (plant) vary when kept in hypotonic solution? Sol. When a plant cell is placed in hypotonic solution, it will gain water from outside as water moves from its higher concentration to lower concentration region. This causes central vacuole to expand and thus causing protoplast to swell. This increases turgor pressure or pressure potential. This pressure potential rises to become equal to solute potential S or osmotic potential. At this time, water potential becomes zero so that further entry of water into cell stops automatically.
18. How imbibition pressure helps in germination of seeds?
Sol. Dry seeds have a highly negative water potential and also have strong affinity for water. So when these seeds are placed in water (w = 0) they rapidly absorb water. The imbibition of water into seeds continues until an equilibrium between the water outside and inside the seeds is reached. This results in the breakage of seed coat in the process of seed germination.
19. Differentiate between active water absorption and passive water absorption.
Sol. Active water absorption Passive water absorption
1. It requires energy for water absorption. It does not require energy for water absorption.
2. It is a mode of water absorption in which force develops in the root itself.
It is a mode of water absorption in which the force develops in the shoot system where transpiration is going on.
3. In such absorption, positive pressure is developed in xylem due to continuous absorption of water in xylem which has high osmotic concentration (i.e., concentration of sugars and salts).
In this kind of absorption, a negative pressure or tension is developed in the xylem due to the loss of water from the aerial parts.
4. It is common in some plants like herbaceous plants or grasses.
It operates in all plants. 5. Only a fraction of water absorption by a plant
is through active method.
Most of the water absorption by a plant is through passive method.
6. Root pressure is essential in it. Transpiration is essential in it. 20. How can you demonstrate by an experiment that transpiration takes place in plants?
Sol. Transpiration can be easily demonstrated by placing a bell jar over a potted plant with the pot enclosed in a plastic bag to prevent water loss from the soil. As transpiration occurs, a fluid get collected on the inside walls of the bell jar which shown to contain water, when tested with cobalt (II) chloride paper. It turns blue to pink on absorption of this water.
Water Drops Bell Jar Plant Pot covered by oil cloth Glass Plate
Fig. : Demonstration of transpiration
21. Give difference between turgid and flaccid cells.
Sol. Turgid cells Flaccid cells
1. In such cells, there is entry of water i.e., endosmosis takes place.
No endosmosis or exosmosis takes place, in such cells.
2. These cells develop turgor pressure which rises and then becomes equal to wall pressure, in case of plant cells.
There is no such pressure inside flaccid cells.
3. The cell swells up. There is no swelling as equal amounts of water pass in and out of the cell.
22. Differentiate between hypertonic and hypotonic solution.
Sol. Hypertonic solution Hypotonic solution
1. It is a solution which is more concentrated, as compared to another solution.
It is a solution which is more dilute as compared to another solution.
2. The osmotic potential of such solution is lower than hypotonic solution i.e., more negative.
The osmotic potential is higher i.e., less negative than hypertonic solution.
3. Cell if placed in hypertonic solution will undergo plasmolysis.
23. Discuss four plant factors that influence the rate of transpiration.
Sol. The rate of transpiration is affected by a number of internal (plant) and external (environmental) factors. Some internal (plant) factors that affect transpiration include:
(i) Frequency of occurrence of stomata i.e., number and distribution of stomata: Usually, the greater is the number of stomata per unit area, the greater is the transpiration. However, much depends on the nature of stomata also. In general, xerophytes possess large number of stomata than mesophytes. (ii) Number of open stomata: Rate of transpiration depends on the number of stomata open which is in
turn dependent on factors like light duration, CO2 requirement etc.
(iii) Availability of soil water i.e., water status of the plant: A decrease in the amount of available soil water reduces absorption causing a reduction in the rate of transpiration also.
(iv) Canopy structure: Canopy refers to the shape of crown i.e., arrangement of branches and leaves on the top of main shoot, above ground. Transpiration also depends on arrangement and structure of leaves. 24. How can you say that transpiration is a compromise for plants in a way, to carry out photosynthesis? Sol. An actively photosynthesizing plant requires good quantity of water, minerals and cooler temperature. All these
requirements are meant by transpiration. As it creates a pull for absorption and translocation of water as well as minerals.
Photosynthesis is limited by available amount of water which is swiftly depleted by transpiration. Environment that reduces transpiration, keeps the stomata open lowers surface temperature and makes water available from soil in good quantity, ideal for photosynthesis. It is found in rain forests where humidity in air is kept high due to cycling of water from root to leaf and then to atmosphere, which is again put back to soil. Transpiration and photosynthesis occur simultaneously due to opening of stomata which leads to gaseous exchange but water vapour also escapes from it side by side.
25. Why are minerals remobilised in plants? Mention any two minerals which are remobilised.
Sol. Mineral ions are frequently remobilised in plants from senescent parts to younger growing parts because due to ageing older parts are removed from plants such as dry, yellow leaves fall, dried flowers also fall. As these parts have to be removed or fall from plants, so before falling these parts pass out their minerals to those parts which need these minerals in order to grow. Some minerals like nitrogen, sulphur, phosphorous and potassium are re-exported from older parts to younger parts like growing buds or leaves etc.
26. What is the principle of mass flow or pressure flow system of food translocation, in plants?
Sol. Pressure flow system of food translocation in plants depends on the principle that organic solutes tend to move from the region of high osmotic pressure to the region of low osmotic pressure. This is due to the development of a gradient of turgor pressure in different parts of plant. For example green leaves synthesize food so sieve tubes of this region are loaded with sugar and develop high osmotic concentration. Due to this water enters into it, consequently develop high turgor pressure. It causes flow of sugar and other solutes along with water towards the area of low turgor pressure which can be any other part of plant like branch, stem, root etc. 27. How are C4 plants more efficient than C3 plants?
Sol. C4 plants are twice as efficient as C3 plants in terms of fixing carbon (making sugar). As these C4 plants lose only half as much of water as C3 plants for the same amount of CO2 fixed. It is due to the different pathway followed by them in synthesis of sugar in leaves because they have peculiar leaf anatomy i.e., kranz anatomy. 28. What role does active transport play in transport of substances in living cell?
Sol. Active transport performs following major functions in cells and organelles :
It makes possible the uptake of essential nutrients from the environment or surrounding fluid even when their concentrations in the environment are much lower than inside the cell.
It allows various substances, such as secretory products and waste materials to be removed from the cell or organelle, even when the concentration outside is greater than the inside.
29. What will happen if we remove a ring of bark leaving a narrow strip in a plant?
Sol. When bark along with phloem is removed, nutrients get collected above the ring, so that the bark above the ring swells up and tissues present below the ring stop growing. Even roots of such plants can be damaged if ring is not heated after some time. This, in turn can kill the whole plant.
Long Answer Type Questions :
30. Why do well-watered plants transpire more during sunny and windy days than in cool and calm morning? Why it is important to plants?
Sol. Transpiration rate is indirectly affected by light because it regulates opening of stomata. In most plants, stomata open in presence of light and close in darkness. Thus the rate of transpiration increases in light and decreases in dark. So, in day time due to sunlight stomata remain open and transpiration continues to occur. Also due to increase in temperature, stomata tend to open and close with a decrease in temperature. Therefore, during sunny days, plants get both sunlight and high temperature, favourable conditions for transpiration.
Windy days also favour transpiration rate by removing the saturated air around the leaves and bringing in dry air. However, high-speed wind sometimes reduces transpiration due to closure of stomata and cooling of leaf surface. Advantages of transpiration are:
(a) Transpiration is important for plants because it directly influences the absorption of water from the soil. (b) Transpiration exerts a tension of pull on water column in xylem which is responsible for the ascent of sap. (c) Transpiration helps in the movement of water and minerals absorbed by the roots to the other parts of
the plant.
(d) The evaporation of water during transpiration contributes to the cooling of leaves (and also the surrounding air) and protects leaves from heat injury particularly under conditions of high temperature and intense sunlight. 31. (i) What is the name of phenomenon which occurs when any liquid particles diffuse in solid particles and
increase their volume? (ii) How such movement occurs?
(iii) Can you stop this movement from occurring?
(iv) What are the conditions necessary for such movement to occur? (v) Give its one application to man to illustrate your answer.
Sol. (i) Imbibition is the phenomenon which causes water (i.e., liquid) to diffuse, in solid particles of a substance. This increases the volume of the solid substance on which liquid is imbibed.
(ii) It occurs due to strong affinity of water molecules to hydrophilic colloids present in a dry substance like wood etc. Also water tends to move from its higher concentration to lower concentration region as water potential gradient is established between solid and liquid to be imbibed.
(iii) Naturally, we cannot stop such imbibition to occur. (iv) Two conditions are required which are:
(a) Presence of water potential gradient between surface of adsorbant and the liquid imbibed. (b) Affinity between the adsorbant and the imbibed liquid.
(v) Jamming of wooden doors and drawers of desks are the common examples of imbibition. A classical example of imbibition is the swelling of wood which had been used by prehistoric man to split rocks and boulders. As when imbibition occurs, it increases the volume but this increase is less than the water absorbed, so the absorbed water in the imbibant is held under pressure, which also releases energy in the form of heat. This pressure helps in breaking rocks. This imbibition pressure not only helps rocks to split but also causes seedlings to emerge out of the soil into the air.
32. Describe girdling experiment with diagram.
Sol. This experiment was conducted by Malpighi, to demonstrate the pathway of translocation of organic nutrient and also to identify tissue involved in it. In this experiment, following steps are involved:
On the trunk of a tree, a ring of bark upto a depth of the phloem layer is carefully removed. This narrow ring of bark consists of phloem, cortex and cork.
After a few weeks, the bark above the ring on the stem swells. The leaves do not wilt but growth below the ring was reduced.
This swelling of bark above the ring was due to the accumulation of nutrients in that region. As the translocation of sugars and other materials down the plant was stopped whereas the upward movement of water was not affected.
This experiment proves that phloem is the responsible tissue for translocation of food in one direction, i.e., towards the roots.
Bulge Ring of bark removed Adventitious root A B
Fig. : Girdling of tree trunk to show that organic nutrients accumulate in the bank above the girdle where a bulge is also produced
33. Compare different transport systems present in living cell. i.e., simple diffusion, facilitated diffusion and active transport on basis of their properties.
Sol. Diffusion and facilitated diffusion occurs as molecules move from an area of higher concentration to an area of lower concentration (i.e., downhill transport) while in active transport movement of molecules or ions occur against their concentration gradient (i.e., uphill transport). On the other hand, facilitated diffusion and active transport of molecules requires special membrane proteins and pumps respectively, while diffusion does not require any such transport proteins. An important distinction between active transport and simple or facilitated diffusion concerns the direction of transport. The essential characteristics of facilitated diffusion are specificity, passivity and saturation while that of active transport is specificity and saturation only.
34. What is osmotic pressure? How is it developed? How is it related to solute potential? How osmotic pressure of a solution is measured?
Sol. The osmotic pressure of a solution is equivalent to the pressure required to prevent the passage of pure water molecules (any solvent) into solution through a semipermeable membrane when both, the solution and solvent are separated from each other by means of that membrane. In other words, the pressure required to stop diffusion of water molecules from their higher concentration to lower concentration through semipermeable membrane is known as osmotic pressure.
In the thistle funnel experiment if external pressure is applied from the upper part of the funnel such that it prevents water diffusion into the funnel through that semipermeable membrane. Then this pressure will be eventually equal to the osmotic pressure.
Because, higher number of solute molecules in a solution lowers its free energy and this results in its increased osmotic pressure. In other words, the osmotic pressure is directly proportional to the concentration of the solutions, i.e., solute concentration.
Osmotic pressure (O.P.) Solute concentration
Numerically, osmotic pressure is equivalent to the osmotic potential, but the sign is opposite. Osmotic pressure is the positive pressure applied while osmotic potential (or solute potential) is negative.
Solute potential (s) or Osmotic potential = – OP (osmotic pressure)
35. What do you mean by active transport? Explain it with any example.
Sol. Cells can also move substances across a cell membrane up their concentration gradients (i.e., uphill transport). This process requires the expenditure of energy, typically from ATP, and is therefore called active transport. Like facilitated diffusion, active transport involves highly selective carrier proteins situated in the cell membrane. These membrane proteins bind to the specific transport substance, which could be an ion or a simple molecule, such as a sugar or amino acid, etc. and actively transport it across the membrane against their concentration gradient.
Active transport enables a cell to take up additional molecules of a substance that is already present in its cytoplasm in higher concentrations than in extracellular fluid. For example, without active transport, liver cells would be unable to accumulate glucose molecules from the blood plasma inside them because the glucose concentration is often higher inside the liver cells than it is in the plasma.
36. Explain the significance of : (i) Active transport (ii) Osmosis (iii) Turgor pressure
Sol. (i) Active transport performs following major functions in cells and organelles :
It makes possible the uptake of essential nutrients from the environment or surrounding fluid even when their concentrations in the environment are much lower than inside the cell.
It allows various substances, such as secretory products and waste materials to be removed from the cell or organelle, even when the concentration outside is greater than the inside.
(ii) Osmosis :
(1) Osmosis helps in absorption of water by plants.
(2) Movement of water from one cell to another is due to osmosis. (3) Opening and closing of stomata is brought about by osmosis. (4) Osmosis plays a key role in the growth or cell enlargement.
(iii) Turgor pressure: The process of plasmolysis is usually reversible i.e., if the plasmolysed cell is immediately placed in a hypotonic solution, water will enter the cell and plasmolysed cytoplasm will swell up and will also come in contact with the cell wall. In such condition, when water diffuses into the cell pressure is exerted by till contents i.e., cytoplasm against the cell wall and this pressure is known as Turgor pressure. This pressure which is exerted by protoplasts of a plant cell against its wall due to the entry of water into it, (called as pressure potential Yp). Hence, when a cell is placed in hypotonic solution, it absorbs water by osmosis and increases its volume, such swollen condition of the cell is known as turgid condition and cell is called turgid. Due to turgor pressure, the protoplast of a plant cell will press the cell wall to the outside. The cell wall being rigid presses back the protoplast with a pressure equal in magnitude but opposite in direction. This pressure (i.e., wall pressure) exerted by cell wall over the protoplast counter balances the turgor pressure. And prevents the plant cell from bursting. Thus, plant cell can maintain its normal size.
37. Discuss the forces which help in mass flow of water in apoplast pathway. Sol. The movement of water through this pathway is due to:
(i) Transpiration pull: When transpiration occurs i.e., water evaporates from intercellular spaces into atmosphere, a tension or negative pressure is generated in the xylem, which is transmitted down to the roots. As a result, whole the water column is lifted like a rope or cord, in apoplast pathway. The mass flow of water occurs, largely due to the transpiration pull.
(ii) Adhesion-cohesion forces: The forces of cohesion (i.e., attraction among water molecules and adhesion (i.e., attraction between water molecules and the walls of the root cells and xylem vessels), together help to form thin, unbroken column of water and leads to mass flow of water in such pathways. These forces act due to the cohesive and adhesive properties of water.
38. What is the significance of channel proteins in transport of substances across the membrane? Which kind of transport is this?
Sol. Significance of channel proteins in transport: In facilitated diffusion, movement across the membranes is achieved by special proteins i.e., channel proteins. The channel proteins from hydrophilic channels through the membrane so that it allows the passage of solutes without undergoing any change in their structure. Some channel proteins always remain open while some are controlled by stimulus so they are “gated”.
Some of these channels are relatively large and non-specific as pores found in the outer membrane of bacteria, mitochondria and chloroplasts. Such pores are formed by transmembrane proteins called “porins”. Such proteins allow passage of various hydrophilic solutes with the size limit for the solute molecules determined by the pore size of that particular porins. Such molecules generally allow small-sized proteins to pass through them.
Some channel proteins concerned specifically with inorganic ions transport are known as Ion channels; and some of which are specific for water transport as they form water channels. These channel proteins mediate water movement and are known as Aquaporins (AQPs). They facilitate the rapid movement of water molecules into or out of cells in specific tissues that require this capability. A water channel is made up of eight different types of aquaporins.
39. How is pressure potential related to water potential?
Sol. When a pressure greater than the atmospheric pressure is applied to pure water or a solution its water potential increases. This is because the pressure develops when the water molecules move from one place to another. Such a situation occurs in living cells. For example, when water enters plant cells due to diffusion, causing a pressure to build up inside that cell against the cell wall by protoplast. This makes the cell turgid and increases its pressure potential. The pressure developed in the cytoplasm and exerted on the cell wall is called pressure potential (p). Hence, the pressure potential or hydrostatic pressure develops in an osmotic system due to the osmotic entry or exit of water from it.
A positive pressure develops in a plant cell due to entry of water into it. Such positive hydrostatic pressure is also called turgor pressure. Loss of water from the cell also leads to the production of a negative hydrostatic pressure or tension. It develops in xylem due to loss of water from aerial part while transpiration. This is an important aspect in transport of water and other solutes over long distances in plants.
40. How is tension developed in xylem vessels for conduction of water? Explain its mechanism in water conduction upto shoot tips.
Sol. A tension or stress is developed in xylem vessels by transpiration which in turn helps in conducting water upto shoot tips. This can be explained as :
(i) Transpiration pull exerted on the water column.
(ii) Cohesive and adhesive properties of water molecules to form continuous water column in the xylem. The moist walls of mesophyll cells of leaves lose water vapour into the intercellular spaces which are inturn connected to the atmosphere through stomatal pores. This makes them highly saturated. The air in outside environment is comparatively dry (i.e., water concentration outside is low) and hence, a potential gradient is established between intercellular spaces of plant and air. As water moves from its higher concentration to its lower concentration, so water vapour moves out from intercellular spaces through stomata into the air. This in turn lowers the water potential of the surrounding mesophyll cells and they continue to lose water to intercellular spaces. Thus transpiration sets up a pressure gradient inside the various tissues in the plants. And this establishes pressure gradient in intercellular spaces and mesophyll cells which is transmitted into the photosynthetic cells (i.e., chloroplast) upto the water-filled xylem elements in the leaf vein. This causes movement of water, molecules by molecule from deeper layers of the leaf towards the outermost layer. Thus, during transpiration, a stress or pressure gradient is developed which causes movement of water from root tips to xylem in the stem and ultimately to the leaves, at the top. This pull or tension is known as transpiration pull. This tension or pull results in upward movement of water.
41. Represent diagrammatically:
(i) Apoplastic and symplastic pathway (ii) Water movement in a leaf.
Sol. (i)
Xylem Endodermis Pericycle
Cortex Plasmodesmata Plasma membrane
Casparian strip = apoplast
= symplast Epidermis
Fig. : Pathway of water movement in the root OR Symplastic path Apoplastic path Cortex Casparian strip Pericycle Phloem Xylem Endodermis
Fig. : Symplastic and apoplastic pathways of water and ion absorption and movement in roots (ii)
Xylem Phloem
Diffusion into surrounding air
Stomatal Guard Cell Palisade
pore
Fig. : Water movement in the leaf. Evaporation from the leaf sets up a pressure gradient between the outside air and the air spaces of the leaf. The gradient is transmitted into the photo synthetic cells and on the water-filled xylem in the leaf vein.
42. Explain diagrammatically the mechanism of translocation of food in plants.
Sol.
Sugars leave sieve tube for metabolism and storage; water follows by osmosis = High turgor pressure Phloem Root
Sugars enter sieve tubes; water follows by osmosis Sugar solution flows
to regions of low turgor pressure Tip of stem
Sugars leave sieve tubes; water follows by osmosis
Fig. : Diagrammatic representation of mechanism of translocation
43. Study the given figure, in which the two chambers A and B containing solutions are separated by a semipermeable membrane.
A B
low % of solutes
high % of solutes
(i) Which chamber has hypertonic solution?
(ii) Which chamber has higher solute potential?
(iii) In which direction would water movement occur?
(iv) Give the name of phenomenon occurring in this chamber.
(v) If external pressure greater than atmospheric pressure is applied on chamber B, what will happen?
Sol. (i) B chamber has hypertonic solution.
(ii) B has higher solute potential than A.
(iii) Water will move from A to B chamber across semipermeable membrane.
(iv) Osmosis.
(v) The water movement will stop hence, osmosis will stop occurring; if external pressure is applied greater than atmospheric pressure.
44. Study the following figure of adjacent cells carefully and answer the following: P = 10 atm = –15 atm S P S = 6 atm = –8 atm B A
(i) Calculate the water potential of both A and B cells in given figure. (ii) What is the value of osmotic pressure in both cells?
(iii) Give direction of net flow of water in these cells. Sol. (i) Water potential of A, w =s + p
w = –15 + 10 – 5 atm
Water potential of B, w =s + p w = –8 + 6
= –2 atm
(ii) As osmotic pressure is equal to solute potential but opposite sign In A : Osmotic pressure = 15 atm
In B : Osmotic pressure = 8 atm
(iii) Net movement occurs from chamber B to A.
45. (i) With the help of a well-labelled diagrams, describe the process of plasmolysis in plants. (ii) Is plasmolysis a reversible process? If yes, then how can it be reversed?
Sol. (i) Cell sap Plasmalemma Protoplast (central vacuole) Tonoplast Cytoplasm Cell wall Central vacuole Shrunken protoplast Central vacuole External solution
Turgid Cell Plasmolysed Cell
Fig. : Plasmolysis in a cell.
(ii) Yes, plasmolysis a reversible process, it can be deplasmolysed if immediately placed in a hypotonic solution. So that, water enters the cell and plasmolsed cytoplasm gains water and swells up.
SECTION - B
Model Test Paper Very Short Answer Type Questions :1. Give one example of symport.
Sol. Glucose and Na+ ions are transported by symport across membrane by symporters (i.e., carrier proteins). 2. Name the process which involves diffusion of water vapour from plants.
3. Name the process through which plants maintain their body temperature? Sol. Transpiration
4. Where are casparian strips located in plants?
Sol. They are found in the endodermis region of roots in plants.
5. Give example of a plant whose seeds cannot germinate without mycorrhizae. Sol. Pinus.
6. Name the phenomenon involved in water absorption by root hairs. Sol. Diffusion.
7. In which part or tissue of plants, quantity and types of solutes are selected for transport to all its parts. Sol. In endodermal cells of root hairs.
8. How are sugars removed at the sink region in plants? Sol. By active transport, sugars are removed at the sink.
Short Answer Type Questions :
9. When does pressure flow of phloem sap begins in plants?
Sol. When hydrostatic pressure in the phloem sieve tube increases, pressure flow begins and the sap moves through the phloem.
10. What happens in sieve tubes of phloem when sugars are removed, at the sink?
Sol. At the sink, when sugars are actively transported out of the phloem sap, osmotic pressure decreases and water moves out of the phloem.
11. What is sink? Give two factors on which source and sink part of the plant is adjusted.
Sol. Sink is the part of plant which needs or stores the food. This source and sink adjustment depends on-(a) Season
(b) Plant’s requirement
12. (i) Name any two chief sink regions for the mineral elements in plants.
(ii) Name two elements which are transported in small amounts as organic compounds. Sol. (i) (a) Apical and lateral meristems
(b) Fruits and seeds (ii) Phosphorous and sulphur
13. From where are mineral nutrients obtained by plants?
Sol. Plants obtain carbon and oxygen from CO2 in the atmosphere. Remaining mineral nutrients are absorbed by roots from soil.
14. What changes have to occur inside guard cells to facilitate pore opening?
Sol. Opening and closing of the stomata is brought about by the change in the turgidity of the guard cells. This happens, when guard cells, gain water, they become turgid. Their outer thinner walls are pushed out and this creates tension in inner thicker wall, which moves these wall towards periphery. This leads to the opening of pore i.e., stomatal aperture.
15. What happens when :
(i) A cell is kept in water and then transferred to salt solution.
(ii) External pressure is applied on thistle funnel filled with sucrose solution and is separated by a semipermeable membrane from water kept in a beaker.
Sol. (i) The cell will first swell up when kept in water and then shrinks when transferred to salt solution.
(ii) When external pressure is applied on a osmotic system comprising thistle funnel and water, the diffusion of water into the funnel stops.
Short Answer Type Questions :
16. Give one example to support the fact that the source and sink may be reversed depending on the season. How this relationship helps in food translocation?
Sol. Source is the region which has high concentration of organic substances while sink region has its low concentration. For example, roots behave as sink in summers and winters while they act as source in early springs because the food stored in them is frequently mobilised to growing buds. Storage organs function both as source and sink.
This source and sink relationship helps in translocation of food through vascular tissue, phloem. As source has higher concentration of organic substances, they tend to flow towards lower concentration areas i.e., sink. 17. Explain with figure, how molecules can be transported passively across the plasma membrane by help of
transporters?
Sol. The molecules present in extracellular matrix bind to the transport proteins i.e., porins which rotates and releases it inside the cell. This happens passively as the molecules are transported from their high concentration to their low concentration i.e., along the concentration gradient.
Outer side
of cell Outer side of cell
Membrane Membrane Transported molecule Transport protein Inner side of cell
Fig. : Facilitated Diffusion 18. How osmotic pressure is related to solute potential?
Sol. Osmotic pressure is the maximum pressure which can develop in an osmotically active solution when it is separated from its pure solvent by a semipermeable membrane under ideal conditions of osmosis. It is dependent upon the concentration of solutes.
More is the solute concentration, greater will be the pressure required to prevent water from diffusing in. Numerically, osmotic pressure is equivalent to the osmotic potential, but its sign is opposite. Osmotic pressure is the positive pressure applied while osmotic potential (or solute potential) is negative.
Osmotic pressure (O.P.) = – Solute potential (s) or Osmotic potential
The osmotic pressure of a solution can be measured with the help of apparatus called osmometer. 19. (i) When does plasmolysis occur?
(ii) How does a cell become turgid?
(iii) Which phenomenon will occur when grapes are added in 10% salt solution into a petridish?
Sol. (i) Plasmolysis occurs when cell is placed in hypertonic solution (i.e., in a solution of higher concentration as compared to the cytoplasm)
(ii) When a cell is placed in hypotonic solution, it absorbs water by osmosis and increases its volume to become turgid.
(iii) When grapes or swollen raisins are placed in 10% salt solution, they shrink. Because, they lose water to the salt solution as salt solution is more concentrated than the sap present in grapes. The phenomenon in which water is drawn out of the cell through diffusion into the extracellular fluid, is known as Exosmosis.
20. Why ultimately water moves through the root layers by symplast pathway?
Sol. In the root layers endodermis layers is present in the inner boundary of cortex which is impervious to water. This layer consists of a band of suberised matrix called the casparian strip which is impervious to water. Therefore, water molecules are unable to penetrate this layer, so they get directed to wall regions which are not suberised, into the cells proper through the membranes. Therefore, now water have to enter symplast pathway in the endodermis i.e., water enters into the cells through the cell membrane. In this pathway, water travels through the cell i.e., their cytoplasm and pass from one cell to other through the plasmodesmata. It is aided by cytoplasmic streaming.
21. (i) Do plants utilise the whole amount of water absorbed by them? (ii) If not then, why?
(iii) Give one example in support of your answer.
Sol. (i) No, plants cannot utilise the whole amount of water absorbed by them.
(ii) Because it evaporates from stem and leaves i.e., shoot system into the surroundings i.e., air. This loss of water in vapour form, from the aerial parts of plant is called transpiration.
(iii) For example, a plant of maize (corn) absorbs about 3 litres of water per day. A plant of mustard absorbs water equal to its own weight in about 5 hours. But the actual amount of water needed by plant is hardly 1-2% of what is absorbed by them.
Long Answer Type Questions :
22. Describe how minerals are absorbed and transported in plants.
Sol. Most minerals enter the root by active absorption into the cytoplasm of epidermal cells. This movement of ions from soil to interior of root is against concentration gradient and require energy. This energy is made available by plants in the form of ATP. The active uptake of ions is partly responsible for water potential gradient in roots and therefore helps in uptake of water by osmosis. Some ions can also be passively absorbed into the epidermal cells of roots by mass flow and diffusion.
After the ions, have reached xylem vessels through active or passive absorption or combination of the two, they are further transported upwards, to the stem and all other parts of plant through the transpiration stream. This allows movement of minerals from their conducting tissue towards the area of their sink. The area of sink for the mineral elements are the growing regions of the plants such as the apical and lateral meristems, young leaves, developing flowers, fruits and seeds and the storage organs.
Minerals flow towards these areas and are unloaded at fine vein endings through diffusion. They are picked up by cells through active uptake.
23. How is food transported in plants?
Sol. In plants, the most accepted mechanism used for the translocation of sugars from source to sink; is called the pressure flow hypothesis. According to mass flow or pressure flow hypothesis, organic substances (i.e., sucrose) flow in solution form in sieve elements due to development of an osmotically generated pressure gradient between source and sink ends.
In plants, the mesophyll cells of leaves synthesize carbohydrates (i.e., glucose) by process of photosynthesis. This glucose is readily converted into sucrose (a disaccharide sugar) Mesophyll cells are only few cells away from sieve tube elements. Therefore, this sucrose moves into the companion cells and then into the living sieve tube elements of phloem, by active transport. This transport requires energy. Now, sieve tubes become loaded with sucrose. This creates hypertonic condition in the phloem. Now these sieve tubes absorb water from the nearby xylem elements by osmosis. This in tern increases osmotic pressure inside sieve tubes and causes flow of phloem sap towards the region where osmotic pressure is low. Usually, a low osmotic pressure is required to be maintained at the sink.
Finally, at sink sucrose is moved out of the phloem sap by active transport and these cells use this sugar for obtaining energy or convert in into starch or cellulose. As sugars are removed the osmotic pressure decreases in sieve tube elements and water moves out of the phloem.
Sugars leave sieve tube for metabolism and storage; water follows by osmosis = High turgor pressure Phloem Root
Sugars enter sieve tubes; water follows by osmosis Sugar solution flows
to regions of low turgor pressure Tip of stem
Sugars leave sieve tubes; water follows by osmosis
Fig. : Diagrammatic representation of mechanism of translocation
Objective Type Questions
(Means of transport)
1. Diffusion is a ______process and is not dependent on _______.
(1) Slow, gradient of concentration (2) Slow, living system
(3) Rapid, temperature (4) Rapid, pressure
Sol. Answer (2)
As it does not require any carrier protein
2. Diffusion rates are affected by
(1) Pressure (2) Temperature
(3) Concentration gradient (4) All of these
Sol. Answer (4)
Factors affecting diffusion Pressure, temperature, concentration gradient, density etc. 3. Gaseous movement into and out of the plants occurs indirectly through
(1) Osmosis (2) Diffusion (3) Transpiration (4) Imbibition
Sol. Answer (2)
Diffusion is the only means of gaseous exchange.
4. Which of the following process requires membrane proteins?
(1) Simple diffusion (2) Imbibition
(3) Facilitated diffusion (4) More than one option is correct
Sol. Answer (3)
Facilitated diffusion involves protein channels.
5. Select the incorrect statement w.r.t. facilitated diffusion
(1) Highly selective (2) Uphill transport
(3) Requires special membrane proteins (4) Transport saturates Sol. Answer (2)
Facilitated diffusion occurs along with concentration gradient i.e., downhill.
6. Select the odd one out w.r.t. porins
(1) Not associated with the inner membrane of plastids (2) Associated with the outer membrane of mitochondria (3) Found in outer membrane of gram positive bacteria
(4) Allow movement of low molecular weight hydrophilic substances. Sol. Answer (3)
Found in outer membrane of gram negative bacteria. 7. In antiport,
(1) Two molecules are transported in same direction across the membrane (2) Only one molecule is transported across the membrane
(3) Two molecules are moved in opposite directions across the membrane (4) No transport occurs
Sol. Answer (3)
If same direction Symport. If occurs only in one direction Uniport 8. In active transport
(1) Energy is required (2) Membrane proteins are involved
(3) Pumps are present (4) All of these
Sol. Answer (4)
i.e., Energy is required in the form of ATP Membrane proteins are involved Pumps are present
(Plant water relations)
9. A cell when immersed in a solution, increases in volume, so the external solution is
(1) Hypertonic (2) Isotonic
(3) Hypotonic (4) Either hypertonic or hypotonic
Sol. Answer (3)
High concentration of water as a result endo-osmosis occurs, water moves from external medium into the cell as a result volume of cell increases.
10. A cell is placed in 0.4 M solution of sugar and no change in volume of cell is found. What is the concentration of the cell sap?
(1) 40 M (2) 4 M (3) 0.4 M (4) 0.20 M
Sol. Answer (3)
If there is no change in volume of cell then it means solution must be isotonic i.e., of same concentration to cell sap.
11. A plant cell if placed in distilled water will
(1) Shrink (2) Swell up
(3) Not change its shape or size (4) Burst immediately Sol. Answer (2)
Because concentration of water molecules will be high in distilled water as compared to cell as a result endo osmosis will occur – leading to increase in volume and swelling of cell.
12. Doors made up of wood, swell up in rainy season due to
(1) Transpiration (2) Imbibition (3) Exosmosis (4) Guttation
Sol. Answer (2)
Phycocolloids present in wood will absorb the moisture and increase the volume. Transpiration : Loss of water in the form of water vapour.
Exosmosis : Movement of water molecules out of the cell. Guttation : Loss of water in the form of droplets.
13. Select the incorrect statement w.r.t. imbibition (1) It is diffusion process
(2) Affinity between the adsorbent and the liquid is not a pre-requisite (3) It involves both capillary action and adsorption
(4) Phycocolloids are best imbibants Sol. Answer (2)
14. Concentration of water molecules in a system determines
(1) Uphill transport rate (2) Number of carrier proteins
(3) Water potential of the system (4) Membrane permeability Sol. Answer (3)
Water potential is kinetic energy of the water molecules. 15. Osmotic potential is
(1) Positive (2) Negative (3) Always zero (4) Greater than one
Sol. Answer (2)
s (osmotic potential) is always negative for a solution. 16. When solute potential increases then water potential would(1) Increase (2) Decrease
(3) Remain same (4) First increase then decrease
Sol. Answer (1) At atmosphere pressure p = 0 w = s +p w = s + 0 w = s It is directly proportional.
17. The hydrostatic pressure which develops due to entry of water into a plant cell is
(1) Positive (2) Negative (3) Zero (4) Undetermined
Sol. Answer (1)
Turgor pressure is positive for a system.
18. For a solution at atmospheric pressure, w is equivalent to
(1) p (2) s (3) Zero (4) TP Sol. Answer (2) At atmospheric pressure ( p= 0) s = s + p w = s + 0 w = s
19. When a cell is placed in a solution whose osmotic concentration is equal to cell sap then (1) Water moves inside the cell (2) Water moves outside the cell (3) No net movement of water occurs (4) Cell will be plasmolysed Sol. Answer (3)
Concentration inside cell and outside cell is same.
20. What would be p of cell sap in cell-A and cell-B respectively?
w = –900 kPa
s = –1200 kPa
w = –600 kPa
s = –1000 kPa
Cell-A Cell-B
Sol. Answer (1) Cell-A Cell-B w = –900 kPa w = –600 kPa s = –1200 kPa s = –1000 kPa w = s + p w = s + p p = w – s p = w – s = –900 – (–1200) = –600 – (–1000) = –900 + 1200 = –600 + 1000) p = 300 kPa p = 400 kPa
21. Turgor pressure is the
(1) Positive pressure (2) Negative pressure
(3) Atmospheric pressure (4) Imbibition pressure
Sol. Answer (1)
TP is exerted by the cell on to the external environment.
22. Plant seeds when sown in soil, germinate and come out of it; due to
(1) Turgor pressure (2) Imbibition pressure (3) Osmotic pressure (4) Atmospheric pressure Sol. Answer (2)
As seed absorbs moisture it exerts pressure and seed coat breaks.
23. In mass flow, various substances move independently according to their
(1) Size (2) Concentration gradient (3) Root pressure (4) Carrier proteins Sol. Answer (2)
In mass flow various substances move from high concentration to low concentration according to the concentration gradient and pressure gradient.
(Long distance transport of water) 24. In mycorrhiza, fungal filaments help in
(1) Water absorption (2) Food translocation
(3) Developing tension in xylem (4) Development of root pressure Sol. Answer (1)
In mycorrhiza fungal filaments increase the surface area for water absorption.
25. Non-living components of xylem tissues are involved in
(1) Symplast pathway (2) Apoplast pathway (3) Osmosis (4) Active transport Sol. Answer (2)
Apoplastic pathway is also known as dead channel as it involves all the dead components.
26. Water reaches xylem from root hairs by
(1) Apoplast pathway (2) Symplast pathway (3) Both (1) & (2) (4) Imbibition Sol. Answer (3)
Both Apoplast and symplast pathway. Apoplast occurs via cell wall and intercellular spaces and symplast via plasmodesmata and cell membrane.
27. Symplast pathway of water in plants is not related to
Sol. Answer (4)
Because cell wall is dead component while symplastic pathway is living channel i.e., it comprises living components only.
28. Guttation happens due to the development of
(1) Negative hydrostatic pressure in xylem (2) Positive hydrostatic pressure in xylem
(3) Intense transpiration pull (4) Low root pressure
Sol. Answer (2)
Due to excessive absorption of solute and water in xylem via root there is push force developing in xylem which results in guttation.
29. Hydathodes help in
(1) Bleeding (2) Guttation
(3) Protection against grazing (4) More than one option is correct Sol. Answer (2)
They are broad opening of vein endings which help in excudation of solutes along with water which is known as Guttation.
30. Select an incorrect statement
(1) Guard cells are dumb-bell shaped in monocots (2) Inner wall of the guard cell is thin and elastic in dicots (3) Cellulosic microfibrils are arranged radially in guard cells (4) Guard cells are surrounded by subsidiary cells
Sol. Answer (2)
Inner wall of guard cells is thick, outer wall is thin and elastic.
31. Water is mainly transported to shoot tips by the help of
(1) Capillarity (2) Root pressure (3) Transpiration pull (4) Canopy structure Sol. Answer (3)
Transpiration pull is mainly responsible for ascent of sap.
32. Xylem helps in translocation of
(1) Some hormones (2) Water and mineral salts
(3) Amides (4) More than one option is correct
Sol. Answer (4)
i.e. Water and minerals Some hormones Amides
33. The water rises in straw due to suction, it is due to
(1) Positive hydrostatic pressure (2) Negative hydrostatic pressure
(3) Zero hydrostatic pressure (4) Diffusion pressure
Sol. Answer (2)
Suction through straw creates a pull which is equivalent to transpiration pull.
34. Transport proteins of _________are control points, where a plant adjusts the quantity and types of solutes that reach the xylem
Sol. Answer (2)
Endodermis of root has a band of casparian strip which is of suberin, thus blocks the apoplastic pathway by making the space impermeable. lt acts as check point or barrier, So that, every molecule must pass through living membrane which is selective before reaching to xylem.
35. The casparian strips of root endodermis is made up of
(1) Suberin (2) Cellulose (3) Chitin (4) Keratin
Sol. Answer (1)
Xylem
Endodermis Pericycle
Cortex
Plasmodesmata Plasma membrane Casparian strip (Suberin)
= apoplast = symplast Epidermis
(Transpiration)
36. The outer wall of guard cell in sunflower is
(1) Thin and elastic (2) Thick and elastic (3) Thin and inelastic (4) Thick and inelastic Sol. Answer (1)
Sunflower is a dicot plant.
37. Stomata when open lead to
(1) Exchange of gases (2) Evaporation of water
(3) Uptake of carbon dioxide (4) All of these
Sol. Answer (4)
i.e., – Exchange of gases – Evaporation of water – Uptake of CO2
38. Rate of transpiration is increased by
(1) Sunlight (2) Darkness (3) High humidity (4) High speed winds
Sol. Answer (1)
Rest all cause the closing of stomata thus, transpiration rate decreases.
(Uptake and transport of mineral nutrients) 39. Ions are absorbed from the soil by
(1) Passive transport (2) Active transport (3) Both (1) & (2) (4) Imbibition Sol. Answer (3)
i.e., Passive transport Active transport
Along concentration gradient and no expense of energy
Uphill or against concentration gradient with expense of energy
40. Remobilised minerals become available to
(1) Chlorotic leaves (2) Dried leaves (3) Young leaves (4) Older parts of plant Sol. Answer (3)
Major sink of mineral elements is young and growing organs.
41. An analysis of the xylem exudates shows that much of the nitrogen travels as
(1) Inorganic ions (2) Nitrate and nitrites
(3) Organic form i.e., amino acids and amides (4) Molecular nitrogen Sol. Answer (3)
Much of nitrogen travels through xylem as organic compounds.
42. Find odd one w.r.t. chief sink for mineral elements.
(1) Apical and lateral meristem (2) Young leaves
(3) Fruits and seeds (4) Mature or older leaves
Sol. Answer (4)
Sink are the plant parts where concentration of food is low and requirement is very high.
Since, Apical and lateral meristem, young leaves and fruits and seeds are growing parts of plants.
Thus these regions require food, therefore they serve as sink. Mature and old leaves are about to fall they don’t have high requirement of food as their metabolic rate decreases.
(Phloem transport : Flow from source to sink)
43. A plant organ having high concentration of food, will serve as a
(1) Source (2) Sink (3) Conducting tissue (4) Plasmodesmata
Sol. Answer (1)
Source During translocation of food in phloem. 44. Sucrose moves into sieve tube elements by
(1) Diffusion (2) Endosmosis (3) Active transport (4) Exosmosis
Sol. Answer (3)
Phloem loading as well as unloading both require energy therefore it is an active transport.
45. Translocation of photosynthates occur in the form of
(1) Sucrose (2) Starch (3) Glucose (4) 3 PGA
Sol. Answer (1)
Sucrose is non-reducing sugar i.e., less reactive and more stable therefore this form of disaccharides sugar is used to be transported.
