MASS TRANSPORT PLANTS Flashcards

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1
Q

Q1.(a) (i) Give two ways in which the structure of starch is similar to cellulose.

A
  1. Are polymers / polysaccharides / are made of monomers / of
    monosaccharides;
  2. Contain glucose / carbon, hydrogen and oxygen;
  3. Contain glycosidic bonds;
  4. Have 1−4 links;
    Neutral: references to ‘unbranched’, insoluble, formed by
    condensation, flexible and rigid
    Are made of the monomer glucose = MP 1 and 2 = 2 marks
  5. Hydrogen bonding (within structure).
    Ignore reference to H bonds between cellulose molecules
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2
Q

(ii) Give two ways in which the structure of starch is different from cellulose. (2)

A
(Starch)
1. Contains α / alpha glucose;
Assume ‘it’ refers to starch
Accept: converse arguments only if linked directly to
cellulose
Accept: forms α glycosidic bonds
2. Helical / coiled / compact / branched / not straight;
3. 1,6 bonds / 1,6 branching;
4. Glucoses / monomers same way up;
5. No H-bonds between molecules;
6. No (micro / macro) fibres / fibrils.
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3
Q

(b) In plants, mass transport of sugars takes place through columns of sieve cells in the
phloem. Other cells, called companion cells, transport sugars into, and out of, the
sieve cells.
The diagram shows the structure of phloem.

Structures I and J allow the transport of sugars between cells.

(i) Using the diagram, suggest and explain one other way in which sieve cells are
adapted for mass transport (2)

A
  1. No / few organelles / very little cytoplasm / cytoplasm at
    edge / more room / hollow / large vacuole / large space /
    thick walls;
    Accept strong walls for thick walls
  2. (So) easier / more flow / (thick / strong walls) resist pressure.
    Easier flow may be expressed in other ways e.g. lower
    resistance to flow
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4
Q

(ii) Using the diagram, suggest and explain one other way in which companion
cells are adapted for the transport of sugars between cells. (2)

A

) 1. Mitochondria release energy / ATP / site of respiration;
Q Reject: ‘produce energy’
but accept produce energy in form of ATP
2. For active transport / uptake against concentration gradient.

  1. Ribosomes / rough endoplasmic reticulum produce(s)
    proteins;
    Concept of making proteins needed
  2. (Proteins) linked to transport e.g. carrier proteins / enzymes.
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5
Q

PCMBS is a substance that inhibits the uptake of sucrose by plant cells.
Scientists investigated the effect of PCMBS on the rate of translocation in sugar
beet.
The figure below shows their results

(b) During their experiment, the scientists ensured that the rate of photosynthesis of
their plants remained constant.
Explain why this was important. (2)

A
  1. Rate of photosynthesis related to rate of sucrose production;
  2. Rate of translocation higher when sucrose concentration is higher.
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6
Q

(c) The scientists concluded that some translocation must occur in the spaces in the
cell walls.
Explain how the information in the figure above supports this conclusion. (2)

A
  1. Rate of translocation does not fall to zero / translocation still occurs after
    120 minutes;
  2. But sucrose no longer able to enter cytoplasm of phloem cells.
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7
Q

Q3.(a) Contrast the processes of facilitated diffusion and active transport.

A
  1. Facilitated diffusion involves channel or carrier proteins whereas active transport
    only involves carrier proteins;
  2. Facilitated diffusion does not use ATP / is passive whereas active
    transport uses ATP;
  3. Facilitated diffusion takes place down a concentration gradient whereas
    active transport can occur against a concentration gradient.
    Since ‘contrast’, both sides of the differences needed
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8
Q

Students investigated the uptake of chloride ions in barley plants. They divided the
plants into two groups and placed their roots in solutions containing radioactive
chloride ions.
• Group A plants had a substance that inhibited respiration added to the
solution.
• Group B plants did not have the substance added to the solution.
The students calculated the total amount of chloride ions absorbed by the plants
every 15 minutes. Their results are shown in the figure below.

(c) Explain the results shown in the figure above. (4)

A
  1. Group A – initial uptake slower because by diffusion (only);
  2. Group A – levels off because same concentrations inside cells and
    outside cells / reached equilibrium;
  3. Group B – uptake faster because by diffusion plus active transport;
  4. Group B fails to level off because uptake against gradient / no
    equilibrium to be reached;
  5. Group B – rate slows because few / fewer chloride ions in external
    solution / respiratory substrate used up.
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9
Q

Q4.(a) Describe the mass flow hypothesis for the mechanism of translocation in plants.
(4)

A
  1. In source / leaf sugars actively transported into phloem;
  2. By companion cells;
  3. Lowers water potential of sieve cell / tube and water enters by osmosis;
  4. Increase in pressure causes mass movement (towards sink / root);
  5. Sugars used / converted in root for respiration for storage.
    Accept starch
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10
Q

Scientists measured translocation in the phloem of trees. They used carbon dioxide
labelled with radioactive 14C.
They put a large, clear plastic bag over the leaves and branches of each tree and
added 14CO2. The main trunk of the tree was not in the plastic bag.
At regular intervals after adding the 14CO2 to the bag, the scientists measured the
amount of 14CO2 released from the top and bottom of the main trunk of the tree. On
the surface of the trunk of these trees, there are pores for gas exchange.
The following figure shows the scientists’ results.

(b) Name the process that produced the 14CO2 released from the trunk.

A

b) Respiration.
1
(c) 1. (About) 30 hours;
2. Time between peak 14C at top of trunk and bottom.
2
(d) Length of trunk (between top and bottom).

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11
Q

0 7 Ulva lactuca is an alga that lives on rocks on the seashore. It is regularly covered by
seawater.
Figure 6 shows a diagram of one Ulva lactuca alga.
Figure 6
0 7 . 1 Unlike plants, Ulva lactuca does not have xylem tissue.
Suggest how Ulva lactuca is able to survive without xylem tissue.
[1 mark]

A

Short diffusion pathway (to cells)
OR
It has a surface permeable (to water/ions into cells);

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12
Q

0 8 Scientists investigated the effect of a heat treatment on mass transport in barley box
plants.
• They applied steam to one short section of a leaf of the heat-treated plants. This
area is shown by the arrows in Figure 9.
• They did not apply steam to the leaves of control plants.
• They then supplied carbon dioxide containing radioactively-labelled carbon to each
plant in the area shown by the rectangular boxes in Figure 9.
• After 4 hours, they:
• found the position of the radioactively-labelled carbon in each plant. These
results are shown in Figure 9.
• recorded the water content of the parts of the leaf that were supplied with
radioactively-labelled carbon dioxide. These results are shown in Table 4.
Figure 9

8 box . 1 The scientists concluded that this heat treatment damaged the phloem.
Explain how the results in Figure 9 support this conclusion.
[2 marks]

A
  1. The radioactively labelled carbon is converted
    into sugar/organic substances during
    photosynthesis;
  2. Mass flow/translocation in the phloem
    throughout the plant only in plants that were
    untreated/B/control

OR

  1. Movement in phloem requires living
    cells/respiration/active transport/ATP

4.Heat treatment stops respiration/active
transport/ATP production so transport in the
phloem throughout the plant only in plants that
were untreated/B/control;

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13
Q

0 8 . 2 The scientists also concluded that this heat treatment did not affect the xylem.
Explain how the results in Table 4 support this conclusion.
[2 marks]

A
  1. (The water content of the leaves was) not
    different because (means ± 2) standard
    deviations overlap;
  2. Water is (therefore) still being transported in the
    xylem (to the leaf)
    OR
    Movement in xylem is passive so unaffected by
    heat treatment;
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14
Q

0 8 box . 3 The scientists then investigated the movement of iron ions (Fe3+) from the soil to old
and young leaves of heat-treated barley plants and to leaves of plants that were not
heat treated. Heat treatment was applied half way up the leaves. The scientists
determined the concentration of Fe3+ in the top and lower halves of the leaves of each
plant.
Their results are shown in Figure 10.

What can you conclude about the movement of Fe3+ box in barley plants?
Use all the information provided.
[4 marks]

A
  1. Heat treatment has a greater effect on young
    leaves than old;
  2. Heat treatment damages the phloem;
  3. Fe3+ moves up the leaf/plant;
  4. (Suggests) Fe3+ is transported in the xylem in
    older leaf;
  5. In young leaf, some in xylem, as some still
    reaches top part of leaf;
  6. (Suggests) Fe3+ is (mostly) transported in phloem
    in young leaf
    OR
    Xylem is damaged in young leaf
    OR
    Xylem is alive in young leaf;
  7. Higher ratio of Fe3+ in (all/untreated) old leaves
    than (all/untreated) young;
  8. All ratios show there is less Fe3+ in the top than
    the lower part of leaves;
  9. (But) no statistical test to show if the difference(s)
    is significant
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