Exam-style Questions: Transport

1. The roots of two groups of pea plants were placed in solutions containing radioactive potassium ions. For the experimental plants a respiratory inhibitor was added to the solution. At regular intervals the solutions surrounding the roots were tested for radioactive potassium ions. The table shows the results of this investigation.

Time from placing roots in solution/minutes Concentration of radioactive potassium ions in the solutions surrounding the roots/arbitrary units
Experimental plants Control plants
0 7.5 7.5
15 6.6 3.3
30 6.4 2.9
60 6.3 2.4
120 6.3 1.2
240 6.3 0.6

a)

(i) The rate of uptake of potassium by the experimental plants in the first 15 minutes was 0.06 units per minute.

Calculate the rate of uptake of potassium by the control plants over the same time period.

(ii) Suggest an explanation for the difference between the rates of uptake of the experimental and control plants in the first 15 minutes.

(iii) The rate of potassium ion uptake in the control plants in the first hour was faster than in the second hour. Suggest why.

b)

At the end of the investigation sections were cut across the stems of the pea plants and the amount of radioactivity measured. The diagram shows a section across the stem of a pea plant.

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(i) Give one feature by which this section can be recognised as a stem.

(ii) Using a guideline, label and name the tissue in which you would expect to find the greatest amount of radioactivity.

(Marks available: 7)

Answer

Answer outline and marking scheme for question: 1

Give yourself marks for mentioning any of the points below:

a) (i) 0.28 (units per minute)

(ii) uptake in (control plants) by active transport;

Use energy/ATP from respiration;

Amount absorbed by experimental plants is due to diffusion.

(iii) Concentration falls therefore rate of diffusion falls;

Active transport involves carrier/membrane proteins;

More potassium ions so more chance of collision with carriers.

(5 marks)

b) (i) Cylindrical arrangement of vascular bundles/vascular tissues in bundles;

(ii) Correct label to Xylem.

(2 marks)

(Marks available: 7)

2. A biologist named Stephen Hales described how he carried out an investigation in 1727.

  • I cut a branch (b) off an apple tree about 1 metre long, then sealed the cut end (p) and tied a piece of wet bladder over it.
  • Then I cut off the other end of the branch at (i) and attached a glass tube (z) to it.
  • After filling the glass tube with water, I placed the lower end in a bowl of mercury (x).

Copyright S-cool

(Reproduced from Vegetable staticks, Bales, S., by permission of the Royal Botanic Gardens, Kew)

  • I left the apparatus outside on a warm afternoon.
  • By 3.00 p.m. the mercury had risen over 30 cm.
  • When the mercury reached the cut end of the stem (i) air bubbles appeared and the mercury slowly ran back into the bowl (x).

a) Suggest a hypothesis that was being tested in this investigation.

b) (i) Through which tissue in the stem is most water transported?

(ii) Give one structural feature of this tissue which enables it to transport water rapidly (as shown in Hales's demonstration).

c) Explain, in terms of the cohesion-tension theory,

(i) why the level of mercury rose during the investigation;

(ii) why the level of mercury fell towards the end of the investigation.

(Marks available: 7)

Answer

Answer outline and marking scheme for question: 2

Give yourself marks for mentioning any of the points below:

a) e.g. water can travel downwards/both ways in a stem/water movement through plant is passive;

(1 mark)

b)

(i) xylem

(ii) e.g. no cross walls/continuous tubes/hollow/lignified/perforated end walls.

(2 marks)

c)

(i)

  • water evaporates/leaves transpire;
  • lowers water potential in leaf cells;
  • reducing pressure in xylem/leaf cells 'pull' water out of xylem;

(ii)

EITHER:

air entered vessels;

no longer a continuous column held by cohesive forces/H bonds broken;

OR

mercury is heavy/dense/viscous;

cohesive forces insufficient to hold mercury up/mercury not cohesive to walls of xylem.

(4 marks)

(Marks available: 7)

3. Read the extract and then answer the questions which follow.

Fish breathe easy by putting the squeeze on blood cells
In the gills of a fish blood passes through structures called lamellae, two thin membranes held apart by cells which look like the pillars supporting a roof. Video recordings of this flow show that red blood cells become deformed as they pass between the pillar cells.

Normally the red blood cells of a trout are oval and measure 13.5 by 8.4 micrometres. As the cells flow through the lamellae, however, they stretch to more than 18 micrometres in length, and take the shape of a letter C or S. Some of the red blood cells get jammed between the pillar cells, blocking the progress of other blood cells.

This means that redblood cells passing through the gill lamellae travel about 50 per cent further than the shortest path. This helps to explain why fish gills are so good at picking up oxygen from water.

(Reproduced by permission of New Scientist)

a) The function of red blood cells is to transport oxygen. How is oxygen transported in red blood cells?

b) (i) Explain the advantage to the trout of the change in shape of its red blood cells as they pass through the gills.

(ii) Explain the advantage to the trout of some red cells getting jammed between the pillar cells.

(Marks available: 5)

Answer

Answer outline and marking scheme for question: 3

Give yourself marks for mentioning any of the points below:

a)

As oxyhaemoglobin/by haemoglobin/combined with haemoglobin.

(1 mark)

b)

(i) Greater surface area;

for oxygen to diffuse through.

(ii) Red cell in gill capillaries for longer period; haemoglobin more likely to be saturated with oxygen/more oxygen absorbed.

(4 marks)

(Marks available: 5)