NEET Biology Transport in Plants Important Questions Set 01

These Class 11 NEET Biology important questions on Transport in Plants give 45 questions with step-by-step answers, arranged as very short, short and long answer types, as asked in school and board examinations. They help NEET aspirants revise the definitions, structures, differences and key terms of the chapter.

Very Short Answer Type Questions

Question 1. What is mycorrhiza?
Answer: A mycorrhiza is a symbiotic association of a fungus and the root system of a higher plant.

Question 2. Explain plasmodesmata.
Answer: Plasmodesmata are the protoplasmic connections present between adjacent cells. They are large openings in the cell wall specifically of plants.

Question 3. What is the membrane of vacuole called?
Answer: Tonoplast

Question 4. Define antiport.
Answer: In facilitated diffusion, when molecules move in opposite directions by transport proteins, it is called antiport.

Question 5. Give an example of region in plants, where negative pressure potential is found.
Answer: Xylem

Question 6. Give net direction of water movement in a plant cell in flaccid state.
Answer: There is no net movement of water occurs in flaccid cell. (i.e., water flowing into the cell and out of cell is equal).

Question 7. What happens to the water potential of a solution when more solute is added to it.
Answer: Its water potential decreases.

Question 8. State the direction of movement of food in the phloem.
Answer: The direction of movement of food in the phloem can be upwards or downwards i.e., bidirectional?

Question 9. Which means of transport occurs when sugar is loaded in phloem at the source?
Answer: Active transport

Question 10. Give shape of guard cells in dicotyledonous plants.
Answer: They are kidney-shaped in dicotyledonous plants.

Short Answer Type Questions

Question 11. What is cytoplasmic streaming? Give an example of a plant where cytoplasmic streaming can be observed?
Answer: It is an automatically vital movement that occurs continually 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.

Question 12. What are the two conditions for imbibition to occur?
Answer: 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.

Question 13. What is turgor pressure? Give its one significance to plants.
Answer: 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.

Question 14. Name the two factors on which water potential depends? Also give its relationship with them.
Answer: The two factors are :
(a) Solute potential
(b) Pressure potential The equation which summarises this relationship between these terms is : Water potential (Ψw) = Solute potential (Ψs) + Pressure potential (Ψp)

Question 15. Give any two characteristic features of transport through special proteins in facilitated diffusion.
Answer: The two characteristic features of carrier proteins are:
(a) Specific in nature
(b) Sensitive to inhibitors

Question 16. What will be the water potential of solution B, if its solute potential ΨS at atmospheric pressure is –200 Pa?
Answer: Water potential equals to solute potential, at atmospheric pressure. So, Water potential of solution B = Solute potential of B ΨW = ΨS (at atmospheric pressure) Therefore, ΨW = –200

Question 17. How solute potential, water potential and pressure potential of a cell (plant) vary when kept in hypotonic solution?
Answer: 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 ΨS or osmotic potential. At this time, water potential becomes zero so that further entry of water into cell stops automatically.

Question 18. How imbibition pressure helps in germination of seeds?
Answer: Dry seeds have a highly negative water potential and also have strong affinity for water. So when these seeds are placed in water (Ψw = 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.

Question 19. Differentiate between active water absorption and passive water absorption.
Answer: 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 It is a mode of water absorption in which the develops in the root itself. force develops in the shoot system where transpiration is going on.
3. In such absorption, positive pressure is In this kind of absorption, a negative pressure developed in xylem due to continuous or tension is developed in the xylem due to absorption of water in xylem which has high the loss of water from the aerial parts. osmotic concentration (i.e., concentration of sugars and salts).
4. It is common in some plants like herbaceous It operates in all plants. plants or grasses.
5. Only a fraction of water absorption by a plant Most of the water absorption by a plant is is through active method. through passive method.
6. Root pressure is essential in it. Transpiration is essential in it.

Question 20. How can you demonstrate by an experiment that transpiration takes place in plants?

Answer:

A well-labelled diagram should show these parts: Water Drops; Bell Jar; Plant; Pot covered by oil cloth; Glass Plate.

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.

Question 21. Give difference between turgid and flaccid cells.
Answer: Turgid cells Flaccid cells
1. In such cells, there is entry of water i.e., No endosmosis or exosmosis takes place, in endosmosis takes place. such cells.
2. These cells develop turgor pressure which There is no such pressure inside flaccid cells. rises and then becomes equal to wall pressure, in case of plant cells.
3. The cell swells up. There is no swelling as equal amounts of water pass in and out of the cell.

Question 22. Differentiate between hypertonic and hypotonic solution.
Answer: Hypertonic solution Hypotonic solution
1. It is a solution which is more concentrated, as It is a solution which is more dilute as compared to another solution. compared to another solution.
2. The osmotic potential of such solution is lower The osmotic potential is higher i.e., less than hypotonic solution i.e., more negative. negative than hypertonic solution.
3. Cell if placed in hypertonic solution will Cell placed in hypotonic solution swells up. undergo plasmolysis.

Question 23. Discuss four plant factors that influence the rate of transpiration.
Answer: 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.

Question 24. How can you say that transpiration is a compromise for plants in a way, to carry out photosynthesis?
Answer: 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.

Question 25. Why are minerals remobilised in plants? Mention any two minerals which are remobilised.
Answer: Mineral ions are frequently remobilised in plants from sensing 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.

Question 26. What is the principle of mass flow or pressure flow system of food translocation, in plants?
Answer: 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.

Question 27. How are C₄ plants more efficient than C₃ plants?
Answer: 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.

Question 28. What role does active transport play in transport of substances in living cell?

Answer:

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.

Question 29. What will happen if we remove a ring of bark leaving a narrow strip in a plant?
Answer: 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

Question 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?

Answer:

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.

Question 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.
Answer: (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 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.

Question 32. Describe girdling experiment with diagram.

Answer:

A well-labelled diagram should show these parts: Bulge; Ring of bark removed; A; Adventitious; B root.

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:

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.

in the bank above the girdle where a bulge is also produced

Question 33. Compare different transport systems present in living cell. i.e., simple diffusion, facilitated diffusion and active transport on basis of their properties.
Answer: 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 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.

Question 34. What is osmotic pressure? How is it developed? How is it related to solute potential? How osmotic pressure of a solution is measured?

Answer:

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 (Ψs) or Osmotic potential = – OP (osmotic pressure)

The osmotic pressure of a solution can be measured with the help of an apparatus called osmometer.

Question 35. What do you mean by active transport? Explain it with any example.
Answer: 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.

Question 36. Explain the significance of :
(i) Active transport
(ii) Osmosis
(iii) Turgor pressure

Answer:

(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.

Question 37. Discuss the forces which help in mass flow of water in apoplast pathway.

Answer:

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.

Question 38. What is the significance of channel proteins in transport of substances across the membrane? Which kind of transport is this?

Answer:

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.

Question 39. How is pressure potential related to water potential?
Answer: 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 (Ψp). 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.

Question 40. How is tension developed in xylem vessels for conduction of water? Explain its mechanism in water conduction upto shoot tips.

Answer:

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.

Question 41. Represent diagrammatically:
(i) Apoplastic and symplastic pathway
(ii) Water movement in a leaf.

Answer:

A well-labelled diagram should show these parts: (i); Plasmodesmata; Plasma membrane; Xylem; = apoplast; Casparian strip; = symplast; OR; Endodermis; Symplastic path; Cortex; Casparian; Phloem; Apoplastic strip Pericycle path; (ii); Palisade; Guard Cell; Stomatal; Diffusion into pore surrounding air.

Epidermis Cortex Endodermis Pericycle 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.

Question 42. Explain diagrammatically the mechanism of translocation of food in plants.
Answer: A well-labelled diagram should show these parts: Sugars leave bysieve tubes; water follows osmosis; Tip of stem; Sugars enter sieve tubes;; High; Sugarto solutionof lowflows water follows by osmosis turgor= regions pressure turgor pressure; Phloem; Sugarsfor leave sieveand tube metabolism storage;by water follows osmosis; Root.

Question 43. Study the given figure, in which the two chambers A and B containing solutions are separated by a semipermeable membrane.
high % of solutes
low % of solutes A B
(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?
Answer: (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.

Question 44.

Study the following figure of adjacent cells carefully and answer the following:

(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.

Answer:

(i) Water potential of A, Ψw = Ψs + Ψp Ψw = –15 + 10 = – 5 atm Water potential of B, Ψw = Ψs + Ψp Ψw = –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.

Question 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?

Answer:

A well-labelled diagram should show these parts: Cell sap Plasmalemma; (i); Protoplast; (central vacuole); Shrunken protoplast; Central vacuole; Cell; External wall solution; Cytoplasm; Tonoplast; Turgid Cell; Plasmolysed 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.