XYLEM 2

Question 1

Q1- Using your knowledge of cohesion-tension theory
(i) describe and explain the changes in rate of flow of water in the branch over the
24 hour period;

Answer 1

rate of flow increases to max at 1200 and then decreases;
increasing transpiration / evaporation from leaves;
transpiration creates tension / increases transpirational pull;
water molecules are cohesive / stick together;
produces a water column;

Question 2

1.(ii) explain why the diameter of the branch decreased during the first 12 hours.

Answer 2

(ii) (increase transpiration) produce a higher tension / reduces the
pressure in the xylem reducing the diameter / adhesive forces
between xylem and water;

Question 3

1(b) A stem was cut from a transpiring plant. The cut end of the stem was put into a solution
of picric acid, which kills plant cells. The transpiration stream continued. Suggest an
explanation for this observation.

Answer 3

(b) water moves in dead cells / xylem is non-living tissue;

the process is passive / no energy is needed;

Question 4

Q2.​ (a) Describe how water is moved through a plant according to the cohesion-tension hypothesis. (4)

Answer 4

1. water evaporates / transpires from leaves;
2. reduces water potential in cell / water potential / osmotic gradient across
   cells (ignore reference to air space);
3. water is drawn out of xylem;
4. creates tension (accept negative pressure, not reduced pressure);
5. cohesive forces between water molecules;
6. water pulled up as a column;

Question 5

(b) The mass of water lost from a plant was investigated. The same plant was used in every treatment and the plant was subjected to identical environmental conditions. In some treatments, the leaves were coated with a type of grease. This grease provides a
waterproof barrier. The results of the investigation are given in the table.
Treatment
Mass​ ​lost​ ​in​ ​5​ ​days​ ​/​ ​g
No grease applied
10.0
Grease applied only to the upper surface of every leaf
8.7
Grease applied to both surfaces of every leaf
0.1

(i) What is the advantage of using the same plant in every treatment? (1)

Answer 5

(b) (i) same surface area of leaf / number of leaves / age / thickness of

cuticle;

Question 6

(b) The mass of water lost from a plant was investigated. The same plant was used in every treatment and the plant was subjected to identical environmental conditions. In some treatments, the leaves were coated with a type of grease. This grease provides a
waterproof barrier. The results of the investigation are given in the table.
Treatment
Mass​ ​lost​ ​in​ ​5​ ​days​ ​/​ ​g
No grease applied
10.0
Grease applied only to the upper surface of every leaf
8.7
Grease applied to both surfaces of every leaf
0.1
(b) The mass of water lost from a plant was investigated. The same plant was used in every treatment and the plant was subjected to identical environmental conditions. In some treatments, the leaves were coated with a type of grease. This grease provides a
waterproof barrier. The results of the investigation are given in the table.
Treatment
Mass​ ​lost​ ​in​ ​5​ ​days​ ​/​ ​g
No grease applied
10.0
Grease applied only to the upper surface of every leaf
8.7
Grease applied to both surfaces of every leaf
0.1
(ii) Why was it important to keep the environmental conditions constant? (1)

Answer 6

(ii) (environmental conditions) affect rate of transpiration / evaporation;

Question 7

(iii) What is the evidence that the grease provides a waterproof barrier? (1)

Answer 7

(iii) presence of grease reduces water loss;

1

Question 8

(ii) Use your knowledge of leaf structure to explain why less water is lost through the
upper surface of leaves than is lost through the lower surface. (2)

Answer 8

(ii) more stomata on the lower surface;

(thicker) waxy cuticle on the upper surface;

Question 9

3. ​ ​ ​ ​ ​ ​A student investigated the rate of transpiration from a leafy shoot. She used a potometer to measure the rate of water uptake by the shoot. The diagram shows the potometer used by the student.
   (a) Give one​ ​environmental factor that the student should have kept constant during this investigation.

Answer 9

M3.​(a) Light (intensity) / temperature / air movement / humidity;

Question 10

(b) The student cut the shoot and put it into the potometer under water. Explain why. (1)

Answer 10

(b) Prevent air entering / continuous water column;

Allow answer in context of shoot, xylem or potometer.

Question 11

(c) The student wanted to calculate the rate of water uptake by the shoot in cm

3 per minute.

What measurements did she need to make? (2)

Answer 11

(c) Distance and time;

Reject ‘amount bubble moves’

1

Radius / diameter / area (of capillary tube);

Question 12

(d) The student assumed that water uptake was equivalent to the rate of transpiration. Give two​ ​reasons why this might not​ ​be a valid assumption.
(2)

Answer 12

(used to provide) turgidity / support / description of;
(used in) photosynthesis / (produced in) respiration;
Apparatus not sealed / ’leaks’;

Question 13

(e) The student measured the rate of water uptake three times.
(i) Suggest how the reservoir allows repeat measurements to be made. (1)

Answer 13

Returns bubble (to start);

Question 14

(ii) Suggest why she made repeat measurements. (1)

Answer 14

(ii) Increases reliability (of results) / anomalous result can be identified;
Q​ Ignore references to validity / precision / accuracy etc.

Question 15

Q4.​ ​ ​ ​ ​ ​Organic compounds synthesised in the leaves of a plant can be transported to the plant’s roots.
This transport is called translocation and occurs in the phloem tissue of the plant.
(a) One theory of translocation states that organic substances are pushed from a high pressure in the leaves to a lower pressure in the roots.
Describe how a high pressure is produced in the leaves. (3)

Answer 15

1. Water potential becomes lower / becomes more negative (as sugar enters phloem);
2. Water enters phloem by osmosis;
3. Increased volume (of water) causes increased pressure.

Question 16

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.

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

Answer 16

(b) 1. Rate of photosynthesis related to rate of sucrose production;
2. Rate of translocation higher when sucrose concentration is higher.

Question 17

(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)

Answer 17

(c) 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.

Question 18

Q5.​ ​ ​ ​ ​ ​(a) (i) Give two​ ways in which the structure of starch is similar​ to cellulose. (2)

Answer 18

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

Question 19

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

Answer 19

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.

Question 20

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)

Answer 20

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

Question 21

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

Answer 21

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.
Note: no mark is awarded for simply naming an organelle