TRANSPORT IN LIVING ORGANISMS

In living organisms; Transport refers to the movement of materials within

the body of organisms. Transport is necessary for the movement of

substances within, into and out of cells so as to enable vital life process to

occur

Importance of transport in organisms

❖ It enables distribution of food materials in the body of an

organism.

❖ It enables distribution of respiratory gases (oxygen and carbon

dioxide) to the body.

❖ It enables removal of metabolic waste products from the body

❖ It enables absorption of food nutrients by organisms

❖ It enables cooling of the body of organisms.

Movement of materials occurs through: -

❖ Diffusion

❖ Osmosis

❖ Active transport

❖ Mass flow

Diffusion

Diffusion This is the passive movement of particles from an area of high

concentration to an area of low concentration. The difference in the

concentration of a substance between two areas is known as a

concentration gradient. When the difference in concentration between the

two areas is great, the concentration gradient becomes Steep and the rate

of diffusion increases.

As the difference in concentration between the two areas decreases the

rate of diffusion slows down. This process continues until the substances

are distributed evenly throughout the two areas. When the particles are

evenly distributed. They move at the same rate in either direction.

Concentration gradient is the difference in concentration of a substance

between two regions.

Example:

1. when tea bag is dissolved in hot water and cold water

2. when potassium permanganate crystals are dissolved in water

Note; for both particles the aim of experiment will be Demonstration of

diffusion in liquids

Roles of diffusion in living organisms.

❖ Movement of oxygen gas from the alveoli to the blood capillaries;

❖ Movement of oxygen gas from the blood capillaries to the tissue

fluid;

❖ Movement of oxygen gas from the tissue fluid to the cell;

movement of carbon dioxide gas from the cell to the tissue fluid;

❖ Movement of carbon dioxide gas from the tissue fluid to the blood

capillaries; movement of carbon dioxide gas from the blood

capillaries to the alveoli

❖ Movement of air from the atmosphere to the leaves;

❖ Movement of carbon dioxide gas from the intercellular space to the

palisade cells;

❖ Movement

oxygen

gas

from

the

palisade

cells

the

intercellular spaces of spongy mesophyll;

❖ Movement of oxygen gas from the intercellular spaces to the

atmosphere;

❖ Movement of water vapour from the leaves to the atmosphere; and

❖ Movement of digested food substances from the ileum to the

circulatory system

Factors affecting the rate of diffusion

❖ Concentration gradient. When there is a big difference in

concentration of molecules between the two areas, high rate of

diffusion occurs.

❖ Surface area to volume ratio: The larger the surface area to

volume ratio, the larger the number of particles that will be able to

move in a given time hence, the higher the diffusion rate,

❖ Distance over which diffusion takes place: When the distance

over which the material is transported is long, the rate of diffusion

decreases. If the distance is short, diffusion occurs faster because

the materials do not have to travel far. For example, in a thin layer

of cells the rate of diffusion increases.

Osmosis

Osmosis is the passive diffusion of water through a semi-permeable

(partially permeable) membrane. It is regarded as a special form of

diffusion because it involves movement of water molecules through a

semi-permeable membrane. Osmosis is defined as the process by which

example

1. You are provided with two Irish potatoes, water troughs, watch glasses with

sample A (table salt) and boiling water. carry out experiments as directed by

procedures (i) – (xi), and then answer the questions that follow:

(i) Peel off the two Irish potatoes provided to remove the outer cover.

(ii) Label one of the Irish potatoes as specimen U and the other as

specimen V.

(iii) Put specimen V into boiling water for 2 minutes, then take it out and

cool.

(iv) Using a knife/scalpel, cut the cross section of the specimen U to

obtain two halves.

(v) Scoop out the central portion of one half of the specimen U to make a

hole of about 2.5 cm deep from the cut surface. The walls of the hole

must be thin (5-8 mm) thick but take care not to damage it.

(vi) Place a scooped specimen U in the trough. (vii) Put 3 g of sample A in

the hole of the specimen U.

(vii) Using a pipette or dropper, add 1 drop of water to dissolve

sample A in a hole of specimen U.

(viii) Put water in the trough until specimen U is half immersed.

Carefully observe the experiment and note the set up and the level of

water in the beginning.

(ix) Repeat step (iv) and (ix) for specimen V that has been boiled and

cooled.

(x) Leave the experiment for 40 minutes, there after observe the

experiment again and note the changes.

Questions

(a) What is the aim of the experiment?

(b) Draw

well

labeled

diagrams

indicate

the

setup

the

experiment;

(i) at the beginning

(ii) (ii) after 40 minutes

(d) Give a reason for the observed changes in the holes and the troughs

after 40 minutes of the experiment.

(e) Identify the specimen which acts as a control experiment.

(f) Give the biological terminologies used to identify the

concentration of the solution in each of the following:

(i) Hole of the specimens

(iii)Water troughs

(g) Based on the observation made from the experiment, why is it not

advised to urinate frequently nearby the plants in the dry season?

(h) What are the two benefits the plant gets by undergoing the process

you investigated in the experiment?

Answers

(a) The aim of the experiment was to investigate osmosis in living

organism using Irish potato

(b)

(ii) the volume of water in the trough of specimen V increased

(d) (i)The hole had high concentration of solute than water in the trough

therefore water moved from the trough to the potato hole through the

semipermiable membrane of raw potato.

(ii) The volume of water in the trough of specimen v remained the same

(e) The specimen wchich acted as the control experiment is specimen V

(f) (i) hypertonic solution in the potatoes

(ii) hypotonic solution in water trough

labeled by letters N

(mixture of distilled water and orange food

colour), N

(mixture of distilled water and orange food colour), N

(distilled/rain/tap water), and N

(mixture of distilled water and table

salt). They were required to follow the procedures as directed in (i)-

(vii) to demonstrate the process of capillarity and osmosis

experiments, and then answer the questions that follow.

(i)

Dip a delivery tube in the beaker containing solution N

and a

stirring rod in the beaker containing solution N

until it reaches 5

cm deep while holding them careful to ensure that they do not

touch the bottom of the beakers for 5 minutes.

(ii)

Remove the delivery tube and stirring rod from the beakers

containing solution N

and N

after 5 minutes and observe what is

happening.

(iii)

Peel the Irish potato to remove the outer cover and chip it to make

four (4) small bars of about 4 cm long and 3 mm thick.

(iv)

Place two bars in a test tube which is half filled with water. Boil

for 2 minutes and allow it to cool.

(v)

Place one boiled bar and one unboiled bar into the beaker containing

solution N

(vi)

Place one boiled bar and one unboiled bar into the beaker containing

solution N

(vii)

After 30 minutes remove the bars from solutions N

and

. Try to bend each bar and touch to feel its texture.

The questions asked were:

a) What was your observation after 5 minutes when a

delivering tube and stirring rod were deepened into solutions

and N

in procedure

b) (i)?

c) Account for observation made in procedure (i).

d) Which of the beakers N

and N

acted as a control of the

capillarity experiment? Give reason to support your answer.

e) Which tissue found in plants can perform the same function

as represented by delivery tube in this experiment?

f) How does the function of capillarity investigated in this

experiment important for the survival of the plants?

g) (i) What was the texture of each bar after removing them

from the solution N

(ii) Briefly explain the cause of each texture observation in

(f) (i)

h) (i) What was the texture of each bar after removing them

from the solution N

? Briefly explain the cause of each

texture observed in

i) (i). which of the bars in solutions N

and N4 acted as a

control of osmosis experiment? Give reason to support your

answer.

j) Which of the solutions N

and N

4 had

the following:

(i) Higher concentration of water molecules?

(ii) Higher concentration of solute molecules?

2) You have been provided with specimen X and two solutions S

and

• Using a razor blade or scalpel, cut a 6 cm long petiole from

specimen X. Use this same piece of petiole in all the 3 stages of the

experiment described below.

Stage I:

Using a razor blade or scalpel split the piece of petiole from specimen

X longitudinally, up to of its length, so as to produce 4 strips on one

end of the specimen, while the other end 2 remains intact as shown in

the diagram below.

Make a sketch drawing of the pet1ole and label it as sketch No. I.

Stage II:

Dip the piece of petiole m solution S1 for about I 0 minutes.

Remove it from the solution, observe and touch it gently to feel its

hardness or softness.

Make a sketch drawing of the petiole and label it as sketch No.2.

Stage Ill:

Dip the petiole in solution S3 for about 10 minutes.

Remove it from the solution, observe and touch it gently to feel its

hardness or softness.

Make a sketch drawing and label it sketches No. 3.

a) Record your observations and explanations for stages II and III of

the experiment as shown m table 2 below.

b) What was the aim of the experiment?

c) Give brief comments on the concentrations of solutions s

ands