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Transport in Plants Test - 4

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Transport in Plants Test - 4
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  • Question 1
    4 / -1

    A flowering plant is planted in an earthen pot and irrigated. Urea is added in high amounts to make the plant grow faster, but after some time the plant died. This may be due to

    Solution

    Urea solution is more concentrated than the sap present in plant cells, so plant cells lose water to the soil through the process of exosmosis. Loss of water caused plant to die.

  • Question 2
    4 / -1

    Absorption of water from soil by seeds increases the______ thus helping seedlings to come out of soil.

    Solution

    Imbibition is the initial step in the germination of the seeds. When mature seeds are sown in soil, they imbibe water and swell. The water is imbibed first by seed coat and then by other tissues of embryo and endosperm. The process of imbibition is important to initiate the seed germination. The pressure that an imbibant develops when submerged in pure imbibing liquid is called imbibition pressure.

  • Question 3
    4 / -1

    Cell wall of plant cells is

    Solution

    A membrane is said to be permeable to a substance, if it allows the passage of the substance through it.This property is termed as permeability. The plant cell wall is permeable because it allows both solvent and solute molecules to pass through it. On the other hand, cuticle layer is impermeable because it does not allow both the solvent and the solute particles to pass through it.

  • Question 4
    4 / -1

    Read the given statements and select the correct ones.
    (i) A membrane which permits the passage of pure solvent molecules to pass through it and not the solute particles is called semi-permeable.
    (ii) A membrane which allows some substances to pass through it more readily than others is known as selectively/ differentially permeable.
    (iii) All living biological membranes are perfectly semipermeable.

    Solution

    There are certain membranes, which allow penetration of only solvent molecules but not the solute particles. Such membranes are known as semi-permeable. Examples of such membranes are cellophane, fish and animal bladders, egg membrane, plasma membrane, etc. The biological membranes are not perfectly semi-permeable. They allow to pass solute molecules only up to a certain extent and are, therefore, selective in nature. They are known as selectively or differentially permeable membranes. All living biological membranes are differentially permeable membranes, which allow some substances to pass through it more readily than others.

  • Question 5
    4 / -1

    Movement of solvent molecule from a region of its higher concentration to a region of its lower concentration through a semipermeable membrane, is referred to as

    Solution

    When the two solutions having different osmotic concentrations are separated by means of a semipermeable membrane, the molecules of solvent or water move from the region of their higher diffusion pressure (or free energy) to the region of their lower diffusion pressure (or free energy). This movement of water or solvent is called osmosis.

  • Question 6
    4 / -1

    Read the following statements and select the correct option.
    (i) Pure water has the highest water potential i.e., zero.
    (ii) Process of diffusion does not require any input of energy.
    (iii) Water moves from the system containing water at higher water potential to the one having lower water potential.

    Solution

    • Pure water is with no dissolved solutes hence its potential is highest that is zero, as solute is added to water potential or energy of water becomes negative.
    • Diffusion is movement of substances from their higher potential/energy/concentration to places of lower potential /energy/concentration hence does not require input of energy.
    • Water always moves from its higher potential i.e with lower solute potential to its lower potential or higher solute potential by a process called osmosis.

    So, the correct answer is 'Statements (i), (ii) and (iii) are correct'

  • Question 7
    4 / -1

    If ψw = water potential; ψs = solute potential; ψρ = pressure potential, then select the correct equation showing their inter- relation

    Solution

    Ψw = Ψs + Ψ
    where Ψ= wster potential,
    Ψs = solute potential
    and Ψp = pressure potential

  • Question 8
    4 / -1

    If  DPD represents diffusion pressure deficit OP is the osmotic pressure and TP is the turgor pressure, then which of the following equation is correct?

    Solution

    The reduction in the diffusion pressure of water in a solution over its pure state is called diffusion pressure deficit or DPD. Diffusion pressure deficit is also called suction pressure. Its value is equal to the osmotic pressure or potential (OP) (positive value taken in bars or atm) of the solution in a cell or system minus the wall pressure (WP) (= turgor pressure, TP) which opposes the entry of water into it provided the external water is pure. DPD = OP − WP(=TP)
    Water is absorbed by a system having higher DPD from another system with lower DPD.

  • Question 9
    4 / -1

    Which of the following occupies the space between the cell wall and the shrunken protoplast in a plasmo lysed cell?

    Solution

    Hypertonic solution is more concentrated as compared to standard solution. If a plant cell is placed in highly concentrated salt or sugar solution (hypertonic solution), water from the cell sap flows out due to exosmosis through the plasma membrane outside the cell. Loss of water from the cell sap causes contraction or shrinkage of proptoplast. This is referred to as plasmolysis. Hypertonic solution occupies the space between the cell wall and shrunken protoplast in a plasmolysed cell.

  • Question 10
    4 / -1

    Two adjacent cells A and B are being studied. Cell A has OP of 10 atm and TP of 6 atm. Cell B has OP of 10 atm and TP of 4 atm.
    Movement of water will occur from

    Solution

    For Cell A : DPD = OP − TP = 10 − 6 = 4 atm
    For Cell B : DPD = OP − TP = 10 − 4 = 6 atm
    As water moves from a cell with lower DPD to a cell with higher DPD, therefore water will move from cell A to cell B.

  • Question 11
    4 / -1

    If a cell A with DPD = 5 bars is connected to cells B, C and D, whose OP and TP are respectively 5 and 5, 10 and 4, and 8 and 3, the flow of water will be

    Solution

    DPD of cell A = 5 bars
    DPD of cell B = 5 − 5 = 0 bars
    DPD of cell C = 10 − 4 = 6 bars
    DPD of cell D = 8 − 3 = 5 bars
    As cell B has the lowest value of DPD, therefore water will move from cell B to cells A, C and D.

  • Question 12
    4 / -1

    The plant cell cytoplasm is surrounded by both cell wall and cell membrane. The specificity of transport of substances is mostly across the cell membrane, because

    Solution

    The plant cell cytoplasm is surrounded by both cell wall and cell membrane. The specificity of transport of substances are mostly across the cell membrane, because cell membrane is selectively permeable.

  • Question 13
    4 / -1

    If some solute is dissolved in pure water, its water potential

    Solution

    The water potential of a solution can be determined using pure water as the standard of reference. The pure water, at atmospheric pressure, has a water potential of zero (0). The presence of solute particles reduces the free energy of water and thus decreases the water potential (negative value). Because of this, the water potential of a solution is always less than zero.

  • Question 14
    4 / -1

    A plasmolysed cell can be deplasmolysed by placing it in

    Solution

    The swelling up of a plasmolysed protoplast under the influence of hypotonic solution or water is called deplasmolysis. It is due to endosmosis. Deplasmolysis is possible only immediately after plasmolysis otherwise the cell protoplast becomes permanently damaged. During deplasmolysis water diffuses into protoplast.

  • Question 15
    4 / -1

    Water potential of a flaccid cell will be:

    Solution

    If plant cell happens to be bathed in hypertonic solution, it loses water through the process of exosmosis. The loss of water is first from cytoplasm and then central vacuole. As a result, the protoplast is reduced in size. This decreases turgor pressure or pressure potential (Ψp) and corresponding wall pressure. Solute potential becomes slightly more negative due to loss of water. The cell attains a minimum size when turgor pressure is zero/pressure potential (Ψp) is zero.
    Therefore, Ψw = Ψs + Ψp
    or Ψw = Ψs (as Ψp = 0)
    A cell which is deficient in turgor is called flaccid.

  • Question 16
    4 / -1

    Salt is added to preserve meat, pickles, etc. because salting kills bacteria by the process of

    Solution

    Plasmolysis is the characteristic feature of living walled cells. All living walled cells plasmolyse when kept in a hypertonic solution. Pickets, meat and fish are preserved by salting. Similarly, jams and jellies are preserved by sweetening with sugars. Salting and sweetening create hypertonic condition in which the fungi and bacteria get killed by plasmolysis.

  • Question 17
    4 / -1

    Select the incorrect statement regarding imbibition.

    Solution

    The adsorption of water by the solid particles or an adsorbent causing it to enormously increase in volume without forming a solution is called imbibition. Solid substances or adsorbent which take part in imbibition are called imbibants e . g , e.g, seeds dry wood the liquid (usually water) which is imbibed is known as imbibate. the imbibate is held in between and over the surface of particles of the solid substance through the processes of capillarity and adsorption . the swelling imbibant also develops a pressure called imbibition pressure (matric potential).

  • Question 18
    4 / -1

    Read the given statements and select the correct option.
    Statement 1 : Plant cells do not rupture when placed in distilled water.
    Statement 2 : Animal cells rupture when placed in distilled water.

    Solution

    When a plant cell is placed in pure distilled water, the water enters into the cell. Cell wall is outer rigid, protective and supportive covering of a plant cell, which prevents the bursting of plant cell by inhibiting excessive endosmosis. On the other hand animal cells rupture if they are placed in distilled water, due to the absence of cell wall.

  • Question 19
    4 / -1

    Which of the following statments is/are correct?

    Solution

    Following are the correct statements:
    (1) Bacteria do not survive in salted pickles because they get plasmolysed in hypertonic medium.
    (2) Osmotic potential is numerically equal to the osmotic pressure but is negative in sign.
    (3) Kneading of wheat flour is accompanied by the release of heat which is due to imbibition.
    So, the correct answer is (d).

  • Question 20
    4 / -1

    Water passes into a cell due to

    Solution

    Cells gain or lose water among themselves on the basis of their water potential or DPD and not their solute or osmotic potentials only. Movement of water (solvent) takes place from lower DPD to high DPD or from higher water potential to lower water potential.

  • Question 21
    4 / -1

    Which option is true for a fully turgid cell?

    Solution

    As the cell becomes fully turgid, the value of turgor pressure becomes equal to that of solute potential (Ψs) so that water potential (Ψw) or DPD becomes either zero or equal to that of external hypotonic solution.
    Ψw = Ψs + Ψp = 0
    Though there is no net movement of water between the cell and its environment, equilibrium is dynamic and not static. Equal exchange of water molecules continues between the cell and its environment.

  • Question 22
    4 / -1

    The cell A has an osmotic potential of −20 bars and a pressure potential of +6 bars. What will be its water potential?

    Solution

    Ψw = Ψs + Ψp
    where Ψw = water potential,
    Ψs = osmotic potential
    and Ψp = pressure potential
    Ψw = (−20)+(+6) = −14 bars

  • Question 23
    4 / -1

    Use of excessive fertilisers causes wilting due to

    Solution

    Excessive use of fertilisers causes exosmosis. In such condition, osmotic withdrawal of water from all parts of plant takes place. This kind of loss of water causes wilting in plant.

  • Question 24
    4 / -1

    The practice of breaking of rocks during rainy season by inserting wooden pegs in them is based on the phenomenon of

    Solution

    Imbibition pressure generated during germination of seeds and spores is so enormous that it can break asphalt roads and concrete pavements. In older times, the imbibition pressure was used in breaking the rocks and stones. Dry wooden stakes, if driven into a small crack in a rock and then soaked, can develop enough pressure to split the rock.

  • Question 25
    4 / -1

    Which out of the four plant cells (P, Q, R and S) would not exhibit any wall pressure?

    Solution

    Wall pressure is equal and opposite pressure of the cell wall against the cell content to the turgor pressure. In the given figure, cell walls of cells Q and S, generate resistance against the expanding protoplasm as they are shown to be fully turgid cells. But cells P and R are shown to be in the stage of plasmolysis, hence these cells would not exhibit any wall pressure.

  • Question 26
    4 / -1

    Given figure represents demonstration of osmosis by egg membrane osmoscope.

    After few days, which of the following would have occurred ?

    Solution

    Water enters from the beaker to the thistle funnel through the egg membrane (semi-permeable membrane) demonstrating the process of osmosis.

  • Question 27
    4 / -1

    The given figure shows set up of potato osmoscope experiment. Select the option that correctly identifies the labels A, B and C.

    Solution

    In the given figure we can see that endosmosis has taken place because the water level rises from the one which was marked initially with the pin.
    A represents a field potato where the potato acts as an osmoscope through which osmosis occurs. the potato SPL because the potato peel does not act as a semipermeable membrane and hence would prevent osmosis from occurring.
    For endosmosis to occur the solution kept inside the potato should be of higher concentration hence it is a sugar solution.
    Similarly for and osmosis to occur potato should be kept in a solution of low concentration or plain water. Water hence moves inside the potato as it contains a solution of higher concentration.

  • Question 28
    4 / -1

    Read the given statments and select the correct option.
    Statement 1: It becomes difficult to open and shut the wooden doors and windows during rainy season.
    Statement 2: Wooden doors and windows imbibe water in rainy season and thus their volume is increased.

    Solution

    Wooden doors and windows adsorb water in humid rainy season and increase in their volume so that they are hard to open or close. If a dry piece of wood is placed in water it swells and increases its volume During imbibition the water molecules get tightly adsorbed and become immobilised.

  • Question 29
    4 / -1

    The given table shows properties of four cells systems A, B, C and D. The maximum rate of inward diffusion of water will be observed in which of these systems?

    Solution

    Diffusion of water will occur from higher concentration to lower concentration. Maximum rate of inward diffusion of water will occur where concentration difference is maximum. System C has maximum concentration difference of 0.65M. Hence, maximum diffusion will occur in system C.

  • Question 30
    4 / -1

    The concentration of solute in four cells is 0.4M. They are placed in four separate containers I, II, III and IV, filled with saline water of concentrations 0.1M, 0.4M, 2M and 3M respectively. In which container will the cell swell?

    Solution

    Concentration of container-I with respect to the cell is hypotonic, therefore, endosmosis occurs and water will move from its higher concentration towards its lower Concentration (i.e., from container to cell), as a result of which cell in container-I will swell.

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