REPUBLIC OF NAMIBIA
KAVANGO WEST REGIONAL COUNCIL
DIRECTORATE OF EDUCATION, ARTS AND CULTURE
PERFORMANCE
ENHARNCING NOTES
NSSCO BIOLOGY GRADE 10-11
PREPARED NOTES FOR GRADE 10-11 LEARNERS, BASED ON THE
SPECIFIC OBJECTIVES OF THE REVISED SYLLABUS. 2019
Compiled By:
1. Amutenya Bilha
2. Hausiku Andreas
3. Katega Maria
4. Nauyoma Elizabeth
5. Sifwaku Lydia
6. Simbombo Julius
7. Sindano Ambrosius
When Excellence Becomes A Tradition, Greatness Will Sky-Rocket.
Table of contents
Topic 1 Scientific processes
1.1. Mathematical requirements …………………………………………………………………............2-6
1.2. Scientific skills …………………………………………………………………………………………7-30
Topic 2 Classification and diversity of living organisms
2.1. Classification of living organisms………………………………………………………………......31-36
2.2. Diversity of living organisms ………………………………………………………………………..37-51
Topic 3 Organisation and maintenance of the organism
3.1. Microscope ……………………………………………………………………………………………52-55
3.2. Cell structure, organisation and levels of organisation ……………………...............................55-67
3.3. Movement of substances in and out of cells ………………………………………………………67-77
3.4. Biological molecules …………………………………………………………………………………77-86
3.5. Enzymes ………………………………………………………………………………………………86-93
3.6. Nutrition ……………………………………………………………………………………………...93-142
3.7. Transport in plants …………………………………………………………………………… ……143-162
3.8. Transport in animals ……………………………………………………………………………….162-191
3.9. Gas exchange in humans …………………………………………………………………………192-197
3.10. Respiration ……………………………………………………………………............................198-203
3.11. Excretion in humans ……………………………………………………………………………...203-208
3.12. Co-ordination in plants ……………………………………………………………………….......209-219
3.13. Co-ordination in humans …………………………………………………………………………219-238
3.14. Homeostasis ……………………………………………………………………………………….239-249
Topic 4 Development of the organism and the continuity of life
4.1. Cell division ………………………………………………………………………………………….249-254
4.2. Reproduction ……………………………………………………………………………………......254-261
4.3. Human reproductive system ………………………………………………………………………268-296
4.4. Inheritance ……………………………………………………………………………………….....296-316
Topic 5 Relationship of organisms with one another and with their environment
5.1. Energy flow, food chains and food webs ………………………………………………………..316-324
5.2. Biochemical cycling ………………………………………………………………………………..325-330
5.3. Population …………………………………………………………………………………………..331-337
5.4. Human influences on the ecosystem …………………………………………………………….337-356
SAMPLES OF EXAM BASED QUESTIONS ………………………………………………………. 357-372
1
Topic 1: Scientific processes
1.1 Mathematical requirements
Specific objective: perform simple arithmetical calculations (add, subtract, multiply
and divide)
(a) The plant lost 8.2 g of mass on day 6, 6.4 g on day 7 and on day 4.3 g on day 8.
What was the total mass lost from day 6 to 8?
Answer: 8.2 g + 6.4 g + 4.3 g = 18.9g
(b) A plant’s mass was 675 g on the first day of an experiment, and 649 g on the fifth
day.
How much mass did the plant lose?
Answer: 675 g – 649 g = 26 g
(c) Approximately 2000 dm3 of blood is pumped through your heart every day.
How much blood does your heart pump in a week?
Answer: 200 dm3 x 7 days = 14 000 dm3 per week
(d) Your heart beats about 4200 times an hour. What is the heart rate per minute?
Answer: 4200 beats /hour ÷ 60 minutes = 70 beats / min
Specific objective: use averages, decimals, fractions, percentages, ratios and
reciprocals
Averages
 An average is a measure of central tendency in a set of data.
 It is a representative value for the whole set of data.
 The median, mean and mode are three measures of central tendency.
Finding the median
 Organise (arrange) the data in order, from the smallest to largest.
 The value in the middle is the median. In cases of bimodal (two
values) in the middle, add the two values and divide by 2.
Finding the mean (average)

Add up all the values involved and then divide by the total number of
values in the set.
Finding the mode

Mode is the common value (the one appearing the most) or the value
with the highest frequency.
2
Decimals


A decimal refers to the notation of a number in the decimal system,
which contains a decimal marker (separator), for example 11. 00 or 4.
13249.
Decimal markers in examination papers will be a single dot on the line
(in the value).
Fractions

A fraction represents a part of a whole umber.
Improper fraction – when the top (numerator) is larger than the bottom
(denominator), e.g. 7
3
Proper fraction – the top is less than the bottom, e.g. 5  1 = 0.5
10
2
Mixed number – a whole number and a fraction, e.g. 4 2
3
Percentages



A percentage is a number or ratio expressed as a fraction of 100.
It is often denoted using the percent sign (%).
E.g. if a leopard kills 60 prey out of 80 attempts, his success
percentage is 75%
 60 x 100 = 75%
80
Ratios



A ratio is the relationship between two quantities, normally expressed as
the quotient of one divided by the other.
For an example if a dog has three white puppies and two brown puppies,
the ratio of white puppies to brown puppies is 3 to 2, also written as 3:2.
In monohybrid inheritance if two heterozygous tall plants are crossed, the
phenotypic ratio is 3 tall to 1 short (3 tall:1 short)
3
Reciprocals

A reciprocal of a number is 1 divided by the number.

e.g. the reciprocal of 4 is 1 and the reciprocal of 10 is 1
10
4
Specific objective: use usual mathematical instruments (ruler, compasses,
protractor, Set Square)
Ruler – is used to measure length
E.g. measure the length of an organism to calculate magnification
Compasses - are used to draw circles
- e.g. when drawing pie charts
Protractor - is used to measure angles (in degrees)
- e.g. measuring angles when drawing a pie chart
4
Set square – is used to draw a perpendicular line
Specific objective: recognise and use direct and inverse proportion

Proportion refers to the relationship between variables
Direct proportion
- Means that two variables increase at a constant rate
- E.g. temperature of a reptile such as a snake, increases as external
(environmental) temperature increases
Inverse (indirect) proportion
- Means that as one variable increases, a related variable decrease at a
constant rate
- E.g. rate of photosynthesis (number of bubbles produced per minute)
decreases as the distance of the lamp away from the plant increases
Specific objective: use positive, whole number indices in algebraic expressions
Using indices
- The index (plural indices) of a number (base) says how many times to use the
number in a multiplication.
- Indices are used to write large numbers in a simple manner.
- E.g. 26 = 2 x 2 x 2 x 2 x 2 x 2 = 64 ( 2 is the base and 6 is the index/exponent)
The exponent of 2 is 6 in this example and is pronounced as two is raised to
the power of 6.
5
Specific objective: solve equations of the form x = y + z and x = yz for any one term
when the other two are known
Examples:
(a) The form x = y + z
The total number of learners (x) in grade 10 at a particular school is 135. If there
are 65 girls(y), how many boys (z) are there?
x=y+z
135 = 65 + z
135 – 65 = z
70 = z
Therefore, number of boys is 70
(b) In the form x = yz
2480 stomata (x) were found in 80 leaves (y), with each leaf containing an equal
amount. How many stomata (z) were found in each leaf?
x = yz
2480 = 80z
2 480 = 80z
80
80
31 = z
Number of stomata in each leaf is 31
6
1.2 . Scientific skills
1.2.1 Planning and conducting investigations
Specific objective: make observations accurately; use appropriate techniques;
handle apparatus / material competently and have due regard to safety


Make observations accurately
Make precise observations by recording everything which happens in great
detail, including the timing of these events using instruments which enhance the
human senses.
Scientists must make accurate observations, because usually these results are
communicated to others that might want to do the same experiment.
Use appropriate techniques




Techniques are ways of carrying out a scientific procedure.
Scientific investigations can involve observation and experiment.
Techniques vary with the type of investigation that is being carried out.
Scientists record observations accurately in the form of drawings, photographs,
measurements or descriptions.
Handle apparatus/material competently and have due regard to safety









Biologists use many different types of apparatus for their observations and
experiments.
Many materials in the laboratory can cause injury if not handled carefully.
To protect yourself from possible injury, wear safety googles whenever working
with chemicals, burners or any substance that might get into eyes.
Wear a laboratory coat and tie back long hair whenever working with chemicals
or heated substances.
Follow instructions correctly.
Do not eat or drink in the laboratory or from laboratory glassware.
When you light a Bunsen burner, strike the match first, and then turn on the gas.
Be careful when handling sharp instruments such as scalpels and scissors.
Report any breakage or spills to your teacher immediately and ask about the
proper clean-up procedure.
7
Specific objective: distinguish between independent, dependent and constant
variables
Variables refers to all the conditions that might affect the outcome of an
experiment.

Independent variable
Is the variable that can be changed on purpose (manipulated) in an experiment.
Dependent variable


Is the variable that is being measured in an experiment
They are the changes observed due to changes in the independent variable.
Constant variable

Is the variable that remains unchanged (constant/same) throughout an
experiment, though other variables may change.
Specific objective: state the hypothesis or the aim of the investigation in relation to
dependent and independent variables
Hypothesis

Is an educated guess concerning the outcome of an investigation by stating the
relationship between the dependent and the independent variable.
-
To test a hypothesis, you need to design a fair test where you isolate the
factor influencing the hypothesis while keeping other variables the same
(constant)
8
1.2.2 Recording data
Specific objective: locate, select and organize information from a variety of sources




Data is any information gathered from experiments, observations, or from
reading about a topic.
It can be measurements, descriptions, drawings or information.
One can use a variety of sources such as the internet, books, scientific journals
and magazines to locate and select data.
The information can then be organised in a table or chart.
Specific objective: record results of experimental investigations in a logical manner
(tables or graphs) and explain the importance of units and recording results of
experimental investigations



After collecting information from your experiments or investigations, it needs to
be organised scientifically for you to analyse the results.
Recording the results of an experimental investigation may be done in a table or
graphs with specified units.
It is important to record results of experimental investigations as it helps to:
identify errors which may have occurred during the experiment, analyse and
draw conclusions from the results obtained or compare with previous findings
should the investigation be repeated.
9
1.2.3 Drawing graphs and tables
-
Data can be presented in the form of tables and graphs where:
Tables summarises the main findings and Graphs show results visually,
often making the trends easier to see
Specific objective: complete tables of data, and process data, using a calculator where
necessary





Constructing a table
A table must have an appropriate heading/title
A table consists of a number of different columns and rows
The independent variable usually appears in the left-hand column
The dependent variable usually appears in the right-hand column
The name of each variable must appear at the top of the column with the unit
written next to or below the heading e.g. time/s (units should be in the heading
column and not with measurements column).
Example:
The table shows heart rate during running and cycling
Heart rate/min
Time/min
Run
cycle
0
70
70
10
72
71
20
74
96
30
120
125
40
140
127
50
143
103
60
122
89
-
Has a heading/title
Time is the independent variable (appears in the left-hand column)
because you can change it on purpose
Heart rate is the dependent variable (appears in the right-hand
column) the readings depend on time
Variables have units next to them
10
Drawing graphs (bar graph, histogram and line graph)
The following should be considered when drawing any graph:
 select suitable scale and axes for graphs
 The scale is the number of units per major division on your graph.
 Space the units equally on both x-axis and the y-axis
 A suitable scale should cover 50% or more of the space provided on
both axes.
 label each axes with the physical quantity and the appropriate unit, e.g.
time/s
 plot the independent variable on the x-axis (horizontal axis) and plot
the dependent variable on the y-axis (vertical axis)
 label each graph with the appropriate heading (by convention always
the dependent versus independent variable)
Drawing bar graphs / charts
 The independent variable is usually non-numerical and the dependent
variable is a numerical (number) value
 Bars are drawn upward and must all have the same width
 Spaces in between bars must also be the same
 A bar graph is usually used to present discontinuous data e.g. blood
groups
 Only a sharp pencil is allowed when drawing graphs
Example of a bar graph/chart
11
Drawing histogram



Similar to bar graph but with no spaces in between bars / bars should
touch
Bars should be drawn in order of increasing or decreasing magnitude
A histogram is used to show the distribution of continuous data e.g. height
Example of a histogram
Drawing line graphs




Present points (plots) on the curve clearly marked as crosses (x) or
encircled dots (⊙).
If a further curve is included, vertical crosses (+) may be used to mark the
points
Join the plots using a ruler for precise reading. Always start from the first
plotted coordinate up to the last plot (no extrapolation).
When two or more line graphs are drawn on the same grid, use a key to
distinguish them clearly.
Drawing pie charts


Should be drawn with the sectors in rank order, largest first, beginning at
noon and proceeding clockwise
Pie charts should preferably contain no more than six sectors
12
Specific objective:
- draw charts and graphs from given data to include plotting of one or
several curves
- label each axis with the physical quantity and the appropriate unit
- plot the independent variable on the x-axis and depended variable
on the y-axis
- present points on the curve clearly marked as crosses or encircled
dots
- label each graph with the appropriate heading
Example: use the data in the table to draw a line graph on the
grid provided
pH
3
4
5
6
7
8
Time taken to break
down starch/minutes
20
13
3
7
12
22
13
Specific objective: interpret charts and graphs


Interpreting a chart or a graph means pointing out the trends that
the graph or chart shows.
The interpretation must point out the main trends and any
unexpected features of the graph or chart.
Activity:
From the graph drawn above, interpret the relationship between pH and
time taken to break down starch in minutes.
--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
14
1.2.4 Basic units and derived units
(Note: The solidus (/) will be used for a quotient and to indicate units in labels of
tables and graphs)
Specific objective: explain and use the relationship between length, surface
area and volume and their units on metric scales



-
Length
Is a measure of the distance between two points.
Metre (m) is the basic unit of length.
Metric unit of length symbols:
1 micrometre (µm) = 0.001 or 1/1000 millimetre
1 millimetre (mm) = 0.001 or 1/1000 metre
1 centimetre (cm) = 0.01 or 1/100 metre
1 metre = 1m
1 decimetre (dm) = 0.1 or 1/10 metre
1 kilometre (km) = 1000 metres
Surface area
 Is the size of the surface of an object.
 The units for measuring surface area are mm2, cm2 and m2.
Volume



Is a measure of the amount of space an object occupies.
The units for measuring volume are mm3, cm3 and m3.
Surface area : volume ratio is inversely proportional to the volume of
an object.
15
Class Activity
Complete the table that shows surface area, volume and surface area : volume
ratios of different sized cubes. One column has been completed for you. Use the
following to help you:
-
The formula for calculating the surface area of a cube is 6s2 (6 x s x s), where s
is the length of one side.
The formula for calculating the volume of a cube is s3 (s x s x s), where s is the
length of one side.
Calculation of surface area : volume ratios of cubes
Length of one side
of cube/ cm
Surface area/
cm2
Volume/cm3
Surface area : volume ratio
96
64
1.5:1
2
4
6
8
10
Specific objective: identify the correct SI unit and derived units
Quantity being measured
length
mass
Name of unit
Symbol of unit
kilometre
km
metre
m
centimetre
cm
millimetre
mm
micrometre
µm
tone (1000 kg)
(no symbol)
kilogram
kg
gram
g
milligram
mg
microgram
µg
16
time
temperature
amount of substance
year
y
day
d
hour
h
minute
min
second
s (not sec)
degree Celsius
oC
mole
mol
Derived SI units are listed below
energy
kilojoule
kJ
joule
J
Recommended units for area, volume and density area listed below
area
hectare = 104 m2
ha
volume
square metre
m2
square decimetre
dm2
square centimetre
cm2
square millimetre
mm2
cubic kilometre
km3
cubic metre
m3
cubic decimetre
dm3 (not l)
litre
density
(preferred to dm3 litre)
cubic centimetre
cm3
cubic millimetre
mm3
kilogram per cubic
metre
gram per cubic
centimetre
kg m-3
17
g cm-3
Specific objective: explain and use sub-multiple prefixes for units (kilo, deci,
centi, milli, micro)
-
The value of a unit measurement can be changed by a prefix to the unit
name.
The prefix shows the multiple of 10 of the unit or the sub-multiple of 10 of
a unit.
The table summarises the use of units and prefixes
Prefix
Meaning
Length
Megamillions of a unit
Kilothousands of a unit
kilometre (km)
Decione-tenth of a unit
decimetre (dm)
Centone-hundredth of a
centimetre (cm)
unit
Millione-thousandth of a
millimetre (mm)
unit
Microone-millionth of a unit
micrometer (µm)
Mass
tone (Mg)
kilogram (kg)
milligram (mg)
microgram
(µg)
Specific objective: use standard notation

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

Standard notation is the normal way of writing large and small numbers.
Scientific notation is a special way of writing numbers.
Scientific notation is sometimes called standard notation.
Scientific notation is represented in the form:
A x 10n, where A is a number between 1 and 10 inclusively and n is an
integer (positive or negative whole number).
e.g.(a) 900 000 = 9 x 105
(b) 35 000 = 3.5 x 104
(c) 0.00000054 = 5.4 x 10-7
18
Specific objective: use acceptable methods of stating units, e.g. metres per
second or m per s, or m/s or ms-1
Example:
- A young bull at full speed can run 6.7 metres per second, while a gazelle
can run 20 metres per second.
- Metres per second or m per second can be written as m/s or m/s -1.
1.2.5 Error, accuracy and uncertainty
Specific objective: identify sources of error and suggest possible improvements
in procedures
-
A source of error is any factor that may affect the outcomes of an
experiment.
Human error



-
Human error can occur when tools or instruments are used or read
incorrectly.
For example, a temperature reading from a thermometer in a liquid
should be taken after stirring the liquid and whilst the bulb of
thermometer is still in the liquid.
Never report the following as human error. They are mistakes that
should not have happened.
Spilling or sloppiness, dropping the equipment, etc.
Bad calculations, doing math incorrectly or using the wrong formula
Reading a measuring device incorrectly (thermometer, balance, etc.)
Not cleaning the equipment
Using the wrong chemical
Not following the planned procedure
Instrumental errors


Can occur when the tools are not functioning exactly as they should be.
An example is when a thermometer is not calibrated correctly, can cause
an error in the reading shown.
19
Observational error

Can occur if the experimenter did not read the instrument e.g.
thermometer or measuring cylinder correctly when recording results.
Systematic errors



These are errors caused by the way in which the experiment was
conducted.
They are caused by an inaccuracy in the design of the system or faulty
equipment.
Systematic errors cannot be eliminated by taking more measurements and
using averages.
Random errors



These are unknown and unpredictable errors.
They are chance variations in the measurements over which you as the
experimenter have little or no control.
Random errors can be reduced by averaging a sufficiently large number of
measurement results.
Specific objective: handle and process experimental observations and data,
including dealing with anomalous or inconsistent results
-
Anomalous results are those which deviate from what is usual, normal or
expected.
These are results which are not in line with the rest of the results or which
do not follow any of the connections between results.
If you repeat the experiment several times, you will usually find fewer
anomalous results than if you carry it out only once.
Anomalous results can be excluded when drawing up a conclusion, as
they may make your conclusion unreliable, especially when you are using
numbers.
20
Specific objective: evaluate presented results or experimental data by applying
scientific knowledge and interpret and draw appropriate conclusions from
practical observations and data in relation to hypotheses
-
-
When you interpret results, you look for trends and patterns in the data,
drawing comparisons between two or more sets of data, and writing about
these trends.
Look for relationship between the independent and the dependent
variables.
A conclusion is a statement, supported by data and based on sound
reasoning about the results of an appropriate investigation.
The conclusion must refer to the hypothesis you set out to test.
Specific objective: suggest possible measures to prevent errors
Following these guidelines may prevent errors:
-
Make sure you know how to operate the measuring instruments correctly
before you begin taking measurements.
Repeat the data collection or observations many times and always state
the number of observations or repeats of your experiment.
Make sure that the measuring equipment are working correctly before
using them in an experiment.
Use instruments that allow you to measure accuracy rather than
estimating.
21
1.2.6 Experimental techniques
Specific objective: plan an experiment or investigation, including making
reasoned predictions of expected results ad suggesting suitable apparatus and
techniques
Plan an experiment or investigation
Always keep the following in mind when planning any investigation:
- Aim (to determine)
- Hypothesis (what you think the outcome of your experiment will be)
- Apparatus (list of apparatus used)
- Method (the different types)
- Observation (if possible - what you see, feel, smell, hear and taste)
- Recording of results (table of your findings)
- Presentation of results (e.g. graphs)
- Conclusion (answer to aim in full sentence or paragraphs)
Steps used in a scientific investigation
1. Ask a question (formulate a testable question)
2. Make a hypothesis (make a prediction)
3. Plan the investigation (creating a plan)
4. Test the hypothesis (conducting a fair test)
5. Conclusion or explanation of result (analyzing and interpret results)
Specific objective: name appropriate apparatus for the measurement of time,
temperature, mass and volume, including burettes, pipettes and measuring
cylinders
Apparatus for measuring time
 Stop watches measure intervals of time in units called seconds and
minutes.
 The two main stop watches are analogue and digital.
Apparatus for measuring temperature


We use thermometers to measure how hot or cold things are in units
called degrees Celsius.
The symbol used is oC.
22
Apparatus for measuring mass



Balances are used to measure mass in units called grams.
The main two types are digital balances and triple beam balances.
All balances work best on firm, level surfaces.
Apparatus used for measuring volume



A measuring cylinder is used to measure the volume of liquids.
A pipette, sometimes called a dropper, is used to measure the volume of a
liquid, or transfer a particular quantity of liquid from one container to another.
A burette is a graduated glass tube, used for accurately measuring or measuring
out small quantities of liquid.
23
Laboratory apparatus
24
Specific objective: recall familiar techniques to record observations and make
deductions from them
In biology, we use various techniques to record observations. Here are some
examples.





Counting to find out how many times a particular event occurs.
Measuring is very important when you observe a specimen and when you
record dependent or independent variables in an experiment.
You can measure length, time, temperature, pH, mass, height, breadth, volume
and area.
Drawing or photographing an object records data about that object.
Sampling is a method of collecting data from a small part intended to show what
the whole area under investigation is like.
Specific objective: recall of simple chemical tests, e.g. for food substances and the
use of hydrogen carbonate indicator, litmus and universal indicator paper
Food tests
Test and reagent
used
Benedict’s test
(uses Benedict’s
solution)
What it tests
for
Reducing
sugars, e.g.
glucose,
maltose
Biuret test
Protein
(uses biuret solutions:
copper sulfate
(biuret A –blue),
sodium /potassium
hydroxide
(Biuret B - clear)
Initial
colour of
reagent
blue
What happens to show
positive result
blue
- Solution changes from blue
to purple
- Stays blue when protein are
absent
25
- Solution changes from blue
to green, yellow and
orange-red (brick –red)
- Remains blue when
reducing sugars are absent
Iodine test
(uses iodine solution)
Starch
Brown /
yellowbrown
- Solution changes from
brown to blue-black
- Stays brown when starch is
absent
Ethanol test (uses
ethanol and water)
Lipids (fats
and oils)
Clear
Solution becomes cloudy
DCPIP test (uses
DCPIP)
Vitamin C
(ascorbic
acid)
Blue
- Solution changes from blue
to clear
- Stays blue when vitamin C
is absent
Test for carbon dioxide




The most effective way to test for carbon dioxide (CO2) is to bubble the gas
through clear limewater.
The clear limewater turns milky or cloudy white in presence of CO2.
Hydrogen carbonate indicator is a pH indicator solution used to show carbon
dioxide concentration in a solution.
It becomes orange / yellow with increased carbon dioxide levels (acidic
conditions), it changes from red to purple if carbon dioxide is removed
(alkaline conditions) and stays red at atmospheric carbon dioxide (neutral
conditions).
Test for acids and bases
The main use of litmus paper is to test whether a solution is acidic or alkaline (basic).



Blue litmus paper is used to test an acid, the litmus paper turns from blue to red
in acids.
Red litmus paper is used to test an alkali (base), red litmus paper turns from red
to blue in alkaline.
Universal indicator paper or solution is used to ascertain the pH of a solution
by comparing the colour of the solution to it.
Test for oxygen

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A glowing splint is used to test for oxygen.
The splint will glow brighter or relights (reignites) if the gas is oxygen.
26
Test for water

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Cobalt chloride paper (blue), tests for the presence of water.
If water is present, cobalt chloride will change colour from blue to pink.
Specific objective: draw an appropriate conclusion, and justify it in line with the data
using an appropriate explanation
During practical and investigations, observations are recorded as data.
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Data can be qualitative (descriptive) observations or quantitative (numerical)
observations.
Qualitative observations, e.g. the feeding mechanism of bees when visiting
flowers, may be recorded in written observation notes, photographs or drawings.
Quantitative observations, e.g. the size (length or height of organism such as
leaves of a plant in sunny or shaded positions, must be recorded in specific units
like cm2 (surface area of leaf), and mm (length of leaves)
Specific objective: recognise, observe, record and measure images of familiar, and
unfamiliar, biological specimens
Record temperature from a thermometer

To increase the accuracy of a measurement of temperature, you may need to
take several measurements and calculate the mean (average).
27
Recording the volume of a measuring cylinder

The volume of liquid in the measuring cylinder above is 67 cm3.
Measuring the length of a biological object

If the grasshopper is drawn to scale of x1, the real measuremet would be 5 cm.
28
Specific objective: make a clear line drawing from an image of a specimen,
calculating the magnification and adding labels as required
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Rules for making scientific drawings
Draw exactly what you see.
Use a sharp pencil.
Make your drawing as large as possible.
The outline of the object must be clear and sharp, no gaps or fluffy lines.
Keep the proportions correct. Each part of the drawing must be the correct size
in relation to the other parts.
Do not shade your drawing or try to make it look artistic.
Rules for labeling drawings
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Write a title below the drawing.
Only label the parts that you have been asked to label.
Draw the label lines with a ruler, using a pencil.
Label lines should never cross over each other.
Label line should touch the structure you are labelling.
Print the label in pen at the end of the label line.
Write the labels horizontally underneath each other.
Calculating magnification
Magnification is how much larger or smaller a drawing is than the real thing.
To calculate magnification:


Measure and record the size of the specimen and that of your drawing
including units of measurement.
Use the formula:
Magnification = size (length) of drawing (image) ÷ actual size of object
Always put X before you write down your answer or after your answer, without
any unit.
29
Specific objective: record readings from diagrams of apparatus, including reading a
scale with accuracy and precision and taking repeated measurements, where
appropriate, to obtain an average value
Accuracy is how close measurements are to the true value.
 Accuracy depends on the quality of the measuring instrument and
the skills of the individual taking the readings.
Precision is the degree to which repeated measurements, under the same
experimental conditions, give the same result.
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Increase precision by repeating observations or measurements
several times and obtain average value.
Increase precision by making sure that the measuring equipment is
suitable for the task, working properly and that you can use it
correctly.
30
Topic 2: Classification and diversity of organisms
2.1 Classification of living organisms
Specific objective: state that organisms can be classified into groups by features that
they share
Organisms can be classified into groups by features that they share for an example:
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Features that are similar in fundamental structure
Features that are similar in position and development
Features that share a common ancestor
The importance of classification of living organisms
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To help when trying to identify unknown organisms
To make it easier to study organisms
Classification enables us to keep track of all organisms
To sort organisms in order
It enables global / international communication about organisms
Classification shows evolutionary relationship of organisms
Specific objective: describe the binomial system of naming species as a system in which
the scientific name of an organism is made up of two parts showing the genus and species
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Binomial nomenclature is a system of naming species of living things by giving
each a name in Latin known as the scientific name.
The name consist of two parts / two word name.
The first part of the name is the genus (generic) name.
The second part of the name is the species (specific) name.
The genus comes first and always begins with a capital letter.
Followed by the species name which is written with lowercase letters.
When printed, the name should be typed in italics.
When handwritten, the two words of the name should be underlined separately.
An example is Homo sapiens, the binomial name of humans.
Homo is the genus name and sapiens is the species name of the human.
31
Specific objective: outline the use of hierarchical classification system for living
organisms
Hierarchical system of classification
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Is when larger groups are progressively divided into smaller groups starting from
the most inclusive group to the smallest least inclusive / more specific group and
each group has a set of distinguished features.
The groups in classification are known as taxa (singular taxon).
The science of classification is called taxonomy. It is a branch of biology that deals
with identification and classification of living organisms.
The taxon at the very top is the kingdom which is the broadest category, followed by
phylum, class, order, family, genus and species.
Way to remember: King Peter Came Over For Ginger Snacks
Classification of the human
Taxa of human Explanation of taxon
classification
Kingdom
Animalia – includes all animals which are multicellular, eukaryotic
organisms, which are heterotrophic, meaning they obtain nutrition
from organic sources.
Phylum
Chordata – includes all the vertebrata, which are all animals
having a back bone or spinal column and internal skeleton
Class
Mammalia – includes all mammals, which are animals
characterised by a hairy body covering, external earflaps (pinnae)
and mammary glands
Order
Primates – animals having flexible hands with grasping fingers and
feet, good eye sight and in the higher apes, a highly developed
brain: includes monkeys, apes and man
Family
Hominidae –modern great apes and humans and a number of their
extinct ancestors and relatives with flat faces and three dimensional vision
Genus
Homo – includes modern humans and species closely related to
them, with upright posture
Species
sapiens – having higher and vertical forehead, large brain volume,
thin skull bones, smaller teeth and jaw and prominent chin
Scientific name Homo sapiens
(binomial name)
Common name human
32
Specific objective:
identifiable features
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
construct and use simple dichotomous keys based on easily
A dichotomous key is a tool that allows the user to determine the identity of items
in the natural world.
It consists of a series of two choices that leads the user to the correct name of a
given item.
The identification of organisms based on a series of choices between alternative
characters is described as dichotomous keys.
How to use dichotomous keys
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Dichotomous keys are based on pairs of descriptions.
It is a series of questions asking about features you can see on an organism.
In each case two opposite descriptions are given and you have to choose the one
that applies.
The key will tell you which pair to go to next.
The process repeats itself until eventually you arrive at the identity of the unknown
organism.
Start any key by finding the organism you want to identify, and read the first pair
of descriptions.
33
Construction of dichotomous keys
What to consider when constructing dichotomous keys:
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Start the key by writing 1(a) and 1(b).
Make sure you use features that can clearly divide your specimens into two groups.
Use visible and opposing differences.
Avoid colour when constructing keys.
Characteristics/features must be suitable and unambiguous.
Try not to use words indicating size (small, big, short, etc.)
- Size can only be used when it is qualified (e.g. legs longer than abdomen).
- Size can only be used when the correct scale is given for each organism.
A key should always have one set of questions less than it has organisms e.g. if
there are six organisms to classify, there will be five sets of questions or
statements.
Be precise and avoid statements such as ‘many’ or ‘few,’ but rather use ‘more than
ten’ or ‘less than five’.
Use distinct, clearly visible and recognizable features, e.g. number of legs rather
than shape of legs.
Avoid grouping more than one feature at a time.
Avoid repeating the same feature.
At the end of your key, there must be one organism in each group.
34
Activity 2.1
Fig. 2.1 shows different variations of the termites, found in the Kalahari Desert.
Fig. 2.1
Use Fig. 2.1 to construct a dichotomous key to identify the different variations.
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Activity2.2
Use the dichotomous key which would distinguish between the seven organisms in Fig.
2.2. Only use features that are visible in the drawings
Fig. 2.2
36
2.2 Diversity of living organisms
Specific objective: outline the structure of a virus, limited to protein coat and genetic
material, and consider the arguments for and against the classification of viruses as living
organisms.
Structure of a virus

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Viruses are non-cellular and have no cell contents (no cytoplasm and no
organelles like nucleus, mitochondria, cell membrane and ribosomes)
The core of a virus is a strand of genetic material which can either be DNA or RNA.
The core of genetic material is enclosed in a protein coat or shell known as the
capsid.
HIV has RNA as genetic material.
37
Arguments For the classification of viruses as living organisms

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Viruses are capable of reproduction (though only within the cells of host
organisms).
Viruses also have genetic material (DNA or RNA) which means that they can
evolve.
Arguments Against the classification of viruses as living organisms

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Viruses do not have cell structure – no cytoplasm and no organelles.
There are seven characteristics of living organisms. Of these, six do not apply to
viruses – respiration, excretion, movement, nutrition, growth and sensitivity.
Although they can reproduce, it happens only within the cells of host organisms
(they remain dormant outside a host).
Conclusion: Most Biologists therefore consider viruses to be non-living. Thus, viruses
cannot be classified into any of the five kingdoms of living things, so they are classified
separately.
Specific objective: list the main features used in the classification of the following
groups, using visible, external characteristics only: flowering plants (monocotyledons
and dicotyledons using seeds, flowers, leaves and roots).
Flowering plants
Flowering plants produce flowers and fruits that contain and protect the seeds.
The phylum of flowering plants is divided into two classes: Monocotyledons and
Dicotyledons
feature
Monocotyledons
Dicotyledons
Seeds
one cotyledon
two cotyledons
Leaves
long and narrow
short and broad
Veins
parallel
branching / network of veins
Roots
fibrous / adventitious
Taproot
Floral parts
multiples of 3
multiples of 2 or 5
38
Specific objective: list the main features used in the classification of the following
groups, using visible, external characteristics only and their adaptation to the
environment: Molluscs; Annelids; Arthropods (insects, arachnids, crustacean and
myriapods).
External diagnostic features of Annelids
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Have long segmented bodies
Have a fixed number of similar segments known as metameric segmentation
Move with bristles called chaetae
Their bodies are covered by flexible non-chitinous cuticle
39
Adaptations of annelids to the environment
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Fluid-like skeleton (hydrostatic skeleton) to keep their shape.
Mucus on their skin protects them from damage and helps them to move through
the soil.
Muscular structure of annelids gives them the needed strength to push through the
soil.
Chaetae which provide grip and traction allowing them to move through the soil
more easily. examples include, rag worms, earthworms, leeches, etc.
40
External diagnostic features of Molluscs
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Have soft, unsegmented bodies
Have a hump which is covered by a thin mantle which secretes a calcareous shell.
Some have one coiled shell (snail), some have two uncoiled shells (mussel), some
have no shell (slugs) and some have an internal shell (octopuses).
Molluscs have a large muscular foot that contains muscles used for locomotion,
suction, clinging to surfaces, burrowing and grasping.
Have a mouth with a rasping tongue-like radula used for feeding.
Adaptations of Molluscs to the environment
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A long, rough tongue (radula) helps with the scraping off pieces of food.
The head has two pairs of tentacles which are used for smell, taste and sight.
The slimy muscular foot that projects from the shell enables them to move easily.
Glands in the foot secrete sticky mucus for easy movement.
The skin is permanently moist and retraction into the shell protects it from drying
out in dry weather.
Arthropods
Classes of phylum Arthropoda
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Insecta
Crustacea
Arachnida
Myriapoda
41
Visible external diagnostic features of all Arthropods
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Have jointed limbs (appendages)
Have segmented bodies
Are covered with an exoskeleton (made up chitin)
Class Insecta (the insects)
External Diagnostic features of Insecta
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Body divided into three sections namely the head, thorax and abdomen.
Three pairs of jointed legs
Most have one or two pairs of wings and others have no wings.
One pair of compound eyes.
One pair of antennae on the head.
Adaptations of Insecta to their environment
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Compound eyes that can see nearly in every direction when objects are close
Wings to allow flight to avoid predators and to find food and detection of vibrations
Antennae to detect odours, tastes, wind speed, wind direction, heat, moisture and
touch
Hairs on feet or abdomen or antennae for taste
A hard, rigid exoskeleton protects insects against harsh environmental conditions
Exoskeleton covered with waterproof cuticle that prevents water loss from insect’s
body
Different colours camouflage them against their predators
42
Class Crustacea (the Crustaceans)
External diagnostic features of Crustacea
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Body divided into two sections – the cephalothorax (head and thorax fused) and
abdomen
Five pairs of legs (or more)
Two pairs of antennae
A hard exoskeleton hardened by calcium salts
Compound eyes on stalks
Cray fish
Adaptations of Crustacea to their environment
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Thick hard exoskeleton protects their bodies
Some, like lobsters have strong pincers or claws to fight off enemies and capture
prey.
Antennae are sense receptors that respond to touch and chemical stimuli. They
are used as sense organs to find food.
Some have modified gills, which allows them to live on land, e.g. the woodlouse,
43
Class Arachnida (the arachnids)
External diagnostic features of Arachnida
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Body is divided into two sections, namely the cephalothorax (head and thorax
fused) and abdomen.
Have four pairs of legs attached to cephalothorax
No antennae are present
Some have chelicerae to seize and poison their prey
Arachnids have eight simple eyes
They have a leathery exoskeleton
A spider
Adaptations of Arachnida to the environment
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Chelicerae carry poison fangs to seize and inject venom into prey which paralyses
it, especially in spiders.
Some (like scorpions) have strong pincers to catch and hold prey firmly
The eight legs are made up of seven segments each, therefore spiders can move
quickly.
Most spiders spin sticky webs which they use to trap and capture insects.
44
Class Myriapoda (the myriapods)
This is a class of arthropods containing the Millipedes and Centipedes
External diagnostic features of myriapoda
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Have bodies divided into two parts – the head and trunk
Have many jointed legs
Have one pair of antennae
Have a long segmented trunk which can be flattened (in centipedes) or cylindrical
(in millipedes)
Their body is divided into segments, most of which are completely identical
External diagnostic features of centipedes
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Have flattened bodies made up of about 18 or more segments
Each segment bears one pair of legs (two legs per segment)
Have one pair of long antennae
External diagnostic features of millipedes
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Have cylindrical rounded bodies divided into many segments
Each segment has two pairs of legs (four legs per segment)
Have one pair of short antennae
45
Adaptations of myriapods to their environment
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Centipedes are carnivores with poison claws to paralyse their prey
Centipedes have strong mouthparts to eat their prey
Millipedes are herbivores with strong mouthparts to feed on decaying plant matter
Strong mouthparts help to grind plant matter into fine pieces before swallowing
A hard exoskeleton reduces water loss from the bodies of myriapods
Specific objective: list the main features used in the classification of the following
groups, using visible, external characteristics only and their adaptation to the
environment: vertebrates (fish, amphibians, reptiles, birds and mammals).
Vertebrates
Vertebrates represent the overwhelming majority of phylum chordata.
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All vertebrates have a backbone (vertebral column)
Have an internal skeleton made up of bones
Have post-anal tail
The vertebrates includes the following classes:
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Fish / Osteichthyes
Amphibia
Reptilia
Birds /Aves
Mammalia
46
Class osteichthyes (the bony fish)
External diagnostic features of fish (the bony fish) / osteichthyes
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Have gills for gaseous exchange in water
Gills are covered on each side by the gill cover called the operculum
Have fins for swimming and balancing
Have a lateral line for detecting vibrations
Body covered by overlapping scales which protect their smooth skin
Adaptations of Fish /osteichthyes to their environment
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Have streamlined bodies which makes it easier to move and cut through water
Scales overlap and face backwards to reduce friction in water
Have fins for moving forward and backward and to help them turn in water
Have a swim bladder which helps keep fish buoyant and to be weightless in water
Have a flexible vertebral column which allows fish to manoeuvre from side to side
The lateral line helps fish to sense vibrations and water pressure
The operculum covers and protects the gills
47
Class Amphibia (the amphibians)
External diagnostic features of amphibians
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Have smooth, moist skin
Adults breathe with lungs and use their skin for gaseous exchange
Tadpoles have a tail, but adults have no tail
Tadpoles have gills only for gaseous exchange (do not have lungs)
The hind limbs have webbed feet
A frog
Adaptations of amphibians to their environment
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Smooth bodies with mucus to prevent drying out in hot conditions
Nostrils mounted high on top of head to breathe while the rest of the body is under
water
Amphibians use their skin for gaseous exchange to get dissolved oxygen when
under water
Tadpoles breathe with gills and undergo metamorphosis to develop lungs in order
to breathe on land
Eyes have transparent membranes to cover them and enable them to see when in
water
The hind limbs have webbed skin to help them when swimming
48
Class Reptilia (the reptiles)
The class Reptilia comprises of turtles, crocodiles, snakes, lizards, tortoises, etc.
External diagnostic features of reptiles
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Have tough, dry, scaly skins
Lay eggs with tough leathery shell on land
Have four limbs, but snakes have no limbs
Adaptations of reptiles to their environment
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Dry, scaly skin minimizes water loss from their bodies
Dry scaly skin also protects them from their predators
Eggs are tough and leathery and resistant to water loss
Leathery shells prevent damage, because reptiles do not look after their eggs after
laying them
Class Aves (the birds)
External diagnostic features of Aves /birds
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Skin is covered by feathers
Have scales covering the legs
Have toothless beaks
Lay hard-shelled eggs with food store (yolk) and calcareous shell
The fore limbs are modified as wings
49
A bird
Adaptations of Aves / birds to their environment
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Forelimbs modified into wings which enables flight (for most species)
Toothless beak is used to peck and tear food
Feathers trap air for insulation to keep them warm during winter
Have light weight, hollow bones that make them lighter for flying
Streamlined bodies to cut through air with minimum resistance when flying
Class Mammalia (the mammals)
External diagnostic features of mammals
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Have skin (body) covered with hair or fur
Have mammary glands
Have external earflaps called pinnae (absent in whales, dolphins and seals)
Have whiskers
50
Adaptations of mammals to their environment
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Bodies covered with hair or fur which traps air for insulation
Mammary gland with milk to feed young before they are able to eat an adult diet
Sweat glands in skin produce sweat to lower body temperature
Some have big ears for losing heat from body, e.g. elephants
Specific objective: observe and draw organisms found locally, concentrating on
diagnostic features and/ or features that adapt them to their environment.
Activity
-
Look for any organism of any class covered in this unit. It can be a plant or an
animal.
You do not need to catch the organism, it is better to observe it in its natural
environment.
Give your diagram a title.
Draw the organism, following instructions for drawing.
Label the diagnostic features of the organism.
Decide which phylum and/ or class your organism belongs to
51
Topic 3: Organisation and maintenance of the organism
3.1. The microscope
Specific objective: use and manipulate a simple light microscope
A microscope is
An apparatus used to enlarge very small objects that cannot be seen with a naked eye.
There are two types of microscopes:
-
A light microscope
An electron microscope
Light microscope
52
Functions of the parts of a light microscope
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Body tube/body
It connects the ocular (eyepiece) to the objective lenses.
The part through which an image is viewed with the aid of an eyepiece
Eyepiece (ocular lens)
The image projected by the objective lenses is further magnified when viewed
through the eyepiece lens. Allows us to observe the specimen in a microscope
more clearly
Arm
The central part of the microscope that supports the body tube and it is the part
to hold when carrying the microscope
Base / foot/ stand
Act as a stand for a microscope and gives it a steady support
Objective lens
Forms a magnified image of the object in the intermediate image plane.
There are three objective lenses to choose from (low power objective, medium
power and high power objective)
Revolving nose piece
It contains the objective lenses. It rotates to allow the user to examine the
specimen on the slide at different magnifications
Stage /stage large
Supports the slide. Stage has a hole that allows light to pass through
Stage clips
Holds the slide in position
Coarse adjustment knob
Moves the tube and lenses up and down faster so that the specimen is in focus.
This knob is used only with the low power objective lens
Fine adjustment knob
Moves the tube and lenses up and down slower to put the specimen at the right
position until image is perfectly focused. It gives fine focus with the high power
and medium power objectives
Light source
Bulb supplies light on to the stage
Iris diaphragm (condenser)
A hole under the stage that regulates the amount of light that goes through the
specimen on the stage
Mirror
It is flat on one side and concave on the other side in order to reflect light up
through the specimen on the stage
53
Specific objective: Calculating the magnification and size of biological
specimens (in millimetres and micrometres)
What is a biological specimen?
A small sample or part taken to show the nature of the whole for microscopic
studies
Magnification
The number of times an object or image is increased in size by a lens system
Calculating the magnification of the microscope
Formula used = Magnification of the objective lens X the magnification of the
eyepiece lens
(The magnifications are indicated on the lenses)
Calculating magnification and size of biological specimens
Magnification of the drawing formula = size of the drawing ÷ actual size
NB: answer shows how many times the drawing is magnified hence, an X is put
in front of the answer.
Example:
actual leaf size: 5mm
Length of drawing: 20mm
Magnification of drawing = length of drawing ÷ actual length
=
20mm ÷ 5mm
=
X4
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Calculating the actual size of a specimen
-
When a drawing or specimen has already been magnified, we can work out the
actual size of the specimen.
Formula used:
Actual size = size of image (drawing or photograph) ÷ magnification
Biological drawings
When making biological drawings:
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Drawings should be made using a sharp pencil
Make your drawing as large as possible to fill half the available space given
Make accurate observations, correct proportions, clear and sharp outlines
Lines should be clear, lines joined up smoothly
Do not shade or colour the drawing
Label features on the drawing and give the drawing a title
Label lines:
Must be in pencil
Must touch the part you want to identify
May not cross
Should be on one side of the drawing only, if possible
Must be straight lines with no arrows
Labels must be printed in ink
3.2. Cell structure, organisation and levels of organisation
Specific objective: describe and compare the structure of a plant cell (palisade cell)
and animal cell (liver cell) as seen under a light microscope, limited to the location of
the cell membrane, cell wall, cytoplasm, nucleus, vacuoles and chloroplasts
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A cell is the smallest biological unit that possess all the characteristics of a living
organism
The size and shape of a cell ranges from millimetres to microns depending on its
function
An individual cell have one or more cells performing several functions
55
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Cells can be prokaryotic or eukaryotic
Prokaryotic cell is a cell that lack a nucleus
Eukaryotic cell is a cell that possess a clearly defined nucleus
Living organisms can be made up of a single cell (unicellular) or many cells
(multicellular)
Cells contains structures called organelles
An organelle is a tiny cellular structure that performs specific functions within a
cell
Structure of a plant cell (e.g. palisade cells in the leaf)
As seen under a light microscope
Drawing of a plant palisade cell
Structure of an animal cell (e.g. liver cell)
As seen under a light microscope
Drawing of a liver cell
56
Functions of the parts of a plant and animal cell
Plant / animal
part
Cell wall
Description
Function (s)
- Structures located around
the cell surface membrane of
plant cells
- Cell walls are fully
permeable
Cell surface
membrane
- Structure that surrounds the
cytoplasm of all cells
- It is selectively permeable,
with receptors for cell
recognition
- Jelly-like fluid within a cell’s
membrane but not inside a
cell’s nucleus
- It contains all the contents
of the cell
Its location in a cell varies
(usually at the centre in
animal cells but can be at the
edge of plant cells)
Animal cells have small
temporary vacuoles while
plant cells have one large
vacuole
- Provides mechanical support
around the plant cell
- It gives plant cells a specific
shape
- It prevents bursting of plant
cells during endosmosis
- It controls the entry and exit of
substances into the cell
- It maintains a constant internal
environment in cells
Cytoplasm
Nucleus
Vacuole
Chloroplasts
Organelles found in the
cytoplasm of green plants
57
- The site for chemical reactions
in the cell
- Provides a physical structure
for the cell
- It controls the functions and
activities of the cell
- It carries genetic information of
a cell
- It stores salts, ions, sugars,
wastes and pigments in cell sap
- It supports the plant cells and
keeps the cell firm
- It controls the water content of
the plant cell
- The site of photosynthesis
- Contains chlorophyll the green
pigment that traps light needed
during photosynthesis
Comparing the structures of a plant (palisade) cell and animal (liver) cell
Plant cell
has cell wall of cellulose
Has chloroplasts with chlorophyll
Has large permanent vacuole
Mostly larger in size
Has a regular shape
Has a cytoplasm, cell surface
membrane and mitochondria present
Animal cell
no cell wall present
Has no chloroplasts with chlorophyll
Small temporary vacuoles present
Smaller in size
Irregular with many shapes
Has a cytoplasm, cell surface
membrane and mitochondria present
Specific objective: Making temporary slides of plant cells (e.g. epidermal cells from a
leaf or onion), make observations and draw cells as seen under a light microscope
Apparatus set up:
 A thin layer taken from the inside of an onion bulb
 A glass microscope slide and cover slip
 Tissue or blotting paper
 Methylene blue solution (1 part added to 4 parts of water) or Iodine solution
 Plastic dropper or pipette, scalpel and forceps
 A light microscope to view the prepared slide
Steps to follow:
 With clean hands and a clean working surface, cut a small piece from the onion
bulb with a scalpel
 Use forceps to peel off a small piece of thin skin from the inside (do not let it get
dry)
 Put a drop of distilled water on the centre of a slide and place the peeled piece of
epidermis onto it and spread it flat
 Gently lower a cover slip on top of the piece using a sharp pencil. Clean the slide
using the tissue or blotting paper
 Use a pipette to take up a small amount of the Methylene blue or Iodine solution
and drop it carefully next to the edge of the cover slip
 Soak up excess solution using a filter paper
 Place the prepared slide on the stage of the microscope and secure the slide
using the stage clips
 Start viewing the specimen starting with the low objective lens before moving
onto a higher magnification
58
Image of the onion cells seen under the microscope
59
Specific objective: draw prepared slides of animal tissues (for example: epithelium of
mammalian trachea, human cheek cells and muscle tissue)
Epithelium of mammalian trachea
As seen under the microscope:
Drawing the epithelium:
Human cheek cells
As seen under the microscope:
Drawing the human cheek cells:
60
Muscle tissue
As seen under the microscope:
Drawing the muscle tissue
Specific objective: state the functions and structures in the cytoplasm of a eukaryotic
cell limited to rough endoplasmic reticulum, ribosomes, vesicles and mitochondria (from
diagrams and images)
Eukaryotic cell:
-
-
A cell with a clearly defined nucleus.
It has a nuclear membrane that surrounds the nucleus containing chromosomes.
These are large cells containing numerous specialised organelles enclosed
within membranes e.g. endoplasmic reticulum, ribosomes, vesicles and
mitochondria
Examples of eukaryotic cells include plant, animal cells and fungi
61
Animal eukaryotic cell
Plant eukaryotic cell
62
Cell organelles found in eukaryotic cells
Rough endoplasmic reticulum (rough ER)
-
Has ribosomes attached to its surface, giving it a rough appearance
Functions:
It transports mRNA from the nucleus to the ribosomes
It synthesise ribosomes and proteins and transport the proteins to the rest of the
cell
Ribosomes
-
the site of protein synthesis
Vesicles
-
are secretory vesicles produced by the Golgi bodies
Functions:
stores, transport or digest cellular products and waste
contains digestive enzymes to break down and destroy bacteria
Mitochondria
-
the site of cellular respiration to release energy from food
ATP is produced during cell respiration
Specific objective: relate the number of mitochondria to the release of sufficient
energy
The human body has many cellular types performing a variety of functions and thus
require an abundant supply of energy




some cells have many mitochondria compared to other cells, depending on the
energy required
skeletal and heart muscle cells, liver cells and brain cells are examples of cells
containing more mitochondria
cells requiring less energy like nerve cells or skin cells have less mitochondria
more mitochondria is produced if a cell is not getting enough energy
63
Specific objective: identify different levels of organisation in drawings and diagrams
of familiar and unfamiliar material








Atoms and molecules form the most basic level of organisation
This includes the chemicals essential for manufacturing life e.g. glucose, amino
acids etc.
Most cells have variation of a basic structure to help it carry out their functions
Structure of different cells relate to its functions
A group of cells can group together to form another level of organisation
This structures can be grouped to form larger systems
Different systems performing different functions combine together to form an
organism
These levels of increasing complexity are called levels of organisations
64
A cell
The smallest basic structural and functional unit of a living organism. Each cell
type has different structure and function
Tissues
A group of similar cells working together to perform a particular function.
Plant tissue examples: Phloem tubes, Xylem vessels, Epidermis etc.
Root hair cells
Xylem vessels
Animal tissue examples: Liver tissue, Bone tissue, Nerve tissue, Red blood cells
Nerve tissue
Muscle tissue
65
Red blood cells
Ciliated cells
Organs
Are groups of different tissues working together to perform a particular function.
-
Organs are comprised of two or more tissues
Examples of animal organs
Examples of plant organs: Leaf, Flower, Root, Ovary etc.
Organ systems
A group or organs performing several closely related functions
Animal organ systems examples:
Support/muscle system, excretory system, reproductive systems, lymphatic
system etc.
Plant organ systems examples:
Reproductive systems, transport system, root system etc.
66
The organism
A complex, functioning whole that is the sum of all its component parts.
-
Some organisms have only one cell (unicellular) e.g. a Bacterium cell, Amoeba
species
Other organisms consist only of tissues e.g. Jelly fish
Other organisms have many cells (multi cellular)
3.3. Movement of substances in and out of cells
3.3.1 Diffusion
Specific objective: describe Process of diffusion
Definition: the movement of molecules, atoms or ions from a region of high
concentration to a region of low concentration, down a concentration gradient.
-
Example of diffusion:
Adding potassium permanganate crystals to pure water
The crystals start to dissolve in water and start to spread out until evenly
distributed and water becomes blue
Specific objective: state that the energy for diffusion comes from the kinetic energy
of random movements of molecules and ions
-
Diffusion does not require energy therefore there is a small amount of kinetic
energy involved
Particles always move randomly because they have kinetic energy
Particles in the air collide (bump) against each other and are pushed around by
other particles
67
Structure of a cell surface membrane
Specific objective: describe the factors that influence diffusion, limited to surface
area, temperature, concentration gradient and distance

-
Surface area
Refers to the total area of the surface of an object
The greater the surface area of the cell, the faster the rate of diffusion. As the
surface area increases, more particles can spread out faster on the large surface
area

-
Temperature
High temperature means molecules have more kinetic energy, and thus
increases rate of diffusion.
Temperature is directly proportional to rate of diffusion

-
Concentration gradient
The greater the difference in the concentration of molecules, the faster the rate of
diffusion.
Concentration gradient is directly proportional to rate of diffusion.

-
Distance
The shorter the travel distance for molecules, the faster the rate of diffusion.
Gas diffuses faster through a thin wall than a thick wall.
This is an inversely proportional relationship
68
Specific objective: describe the importance of diffusion of gases and solutes

-

-

-
Allows gaseous exchange in plants:
Carbon dioxide needed for photosynthesis inside the leaf is in higher
concentration outside the leaf (air in the atmosphere)
Hence, carbon dioxide diffuses from the air into the leaf through the stomatal
openings on leaves where there is a lower carbon dioxide concentration
Oxygen, a by-product of photosynthesis becomes highly concentrated in the leaf
then the surrounding air
Hence, oxygen diffuses from where it is in higher concentration in the leaf, to the
outside of the leaf where the concentration is low through the stomata
Importance in gaseous exchange in animals:
Oxygen is needed for respiration inside all living cells
The air inhaled into the lungs (alveoli) contains a higher concentration of oxygen
than the blood capillaries surrounding the alveoli
Hence, oxygen diffuses across the alveoli walls into the blood capillaries down
the concentration gradient
Blood capillaries transport the oxygen with other nutrients to cells and tissues
where it diffuses across to cells and tissues
Cells and tissues releases waste products like carbon dioxide and it diffuses into
the capillaries down the concentration gradient
Carbon dioxide highly concentrated in the blood capillaries surrounding the
alveoli diffuses across into the alveoli down the concentration gradient
Importance of diffusion of solutes
Some products of digestion are absorbed from the ileum into the blood capillaries
by diffusion
Such solutes include glucose and amino acids that are highly concentrated in the
ileum and diffuse into the capillaries where there is a lower concentration
69
Specific objective: investigating diffusion, for example, the rate at which
ammonia diffuses along a glass tube containing pieces of red litmus paper
Rate at which ammonia diffuses along a glass tube containing pieces of red
litmus paper
-
Apparatus set up:
Use forceps to place red litmus paper strips at equal intervals in the tube
Place a piece of cotton wool soaked in ammonia solution at one end of a tube
Seal the tube at both ends
Observe the changes on the red litmus papers, as it starts turning blue as the
ammonia gas reaches it
Observe that the litmus paper nearest to the cotton turns blue first
Ammonia gas makes red litmus paper turn blue
Record the time taken for each piece of red litmus to turn blue and plot the
results on a graph
70
Specific objective: investigate the factors that influence diffusion, limited to
surface area, temperature, concentration gradients and distance
Diffusion and temperature
Apparatus and methodology:
-
Using two large beakers of equal size, fill one beaker with the same amount of
cold tap water while pouring hot water in the second beaker
Use a dropper to add one drop of food colouring into each beaker at the same
time
Do not stir the mixtures to allow the food colourant to spread on its own
throughout the water
Observe the time taken for the food colourant to be distributed in the water
Notice that the food colouring diffuses faster in hot water as the molecules have
more kinetic energy at high temperature
3.3.2. Osmosis
Specific objective: describe the effects of osmosis on plant and animal tissues
(include reference to hypotonic, isotonic and hypertonic solutions)
Definition of osmosis
The movement of water molecules from a high water potential to a low water
potential through a partially permeable membrane down a water potential
gradient
-
Partially permeable membrane refers to a membrane that allows small molecules
to pass through but not larger molecules
Water potential is the force responsible for movement of water from one area to
another and is symbolised as ᴪ
A solution is made up of a solute (e.g. sugar or salt) mixed in a solvent (like
water)
A dilute solution has more water molecules and few solutes, therefore a high
water potential
A concentrated solution has few water molecules and more solutes, therefore a
low water potential
A visking tubing is a manufactured membrane that act like a cell membrane
Small atoms and molecules like water and gases can pass through the
membrane but large molecules like proteins and sugar cannot pass through
71
High water potential
low water potential
Effects of osmosis on plant and animal tissues
Types of environments (solutions) where osmosis occurs:
-
Hypotonic solution / dilute solution (high water potential)
A solution that has more free water molecules and less solutes than the cell, so
water molecules moves into the cell. E.g. distilled water
-
Isotonic solution
A solution where the water potential is the same inside and outside the cell,
water will than move across the membrane in both directions maintaining cell
size.
-
Hypertonic solution / concentrated solution (lower water potential)
A solution that has a higher solute concentration and less free water molecules
than the cell, hence water molecules will move out of the cell into the solution
Placing a plant and animal cells in a hypotonic solution
-
the solution has a high water pontential than the plant and animal cell, water
molecules move from the solution into the cells
the cells swells and become turgid
the animal cell will become turgid and eventually burst because it does not have
a protective cell wall like the plant cell
72
a plant cell:
an animal cell:
Placing plant and animal cells in an isotonic solution
-
the water potential is the same outside the cells as well as inside the cells, hence
water molecules moves in both directions and both cells maintain their shapes.
plant cell:
animal cell:
Placing plant and animal cells in a hypertonic solution
-
the solution have a higher water potential then the cells, hence water leaves the
plant and animal cells
this causes the cell membrane to shrink and the cell become flaccid and
plasmoysed
73
plant cell:
animal cell:
Specific objective: investigate and explain the effects of immersing plant tissues
in solutions of diferent concentrations by using the terms turgor pressure, turgid,
flaccid and plasmolysis

Practical experiments that can be carried out in class to demonstrate effects of
osmosis on plant tissues including using raw potato strips or dried raisins in
various salt solutions (see activity at back of booklet)
Placing plant and animal cells in hypotonic solutions (e.g. pure water/dilute solution)





the water will move into the cells by endosmosis
the pure water has a high water potential while the cell sap (cytoplasm) of the cell
has a low water potential
the plant cell expands and vacuole swell and cell becomes turgid
the cell wall of plant cell expands slightly but prevents the cell from bursting
the animal cell expands and bursts as there is no protective cell wall
Placing cells in hypertonic solutions




the water leaves the cell by exosmosis
the solution outside the cell has a low water potential while the cytoplasm has a
higher water potential
the cells cytoplasm and vacuole shrinks and becomes flaccid
a flaccid cell has its plasma membrane tear away from the cell wall and cell
becomes plasmolyzed
74
Specific objective: explain the importance of water potential and osmosis in the
uptake of water by plants





osmosis facilitates the movement of water molecules into and out of the cells
living cells need water all the time for many biological processes and osmosis
controls how much water to be kept in a cell
root hair cells on plant roots absorb water from the soil through osmosis
water potential between the soil particles and root hair cells makes plants to
absorb water
absorbing water to replace water used and lost by plants prevents plants from
wilting
Specific objective: outline how plants are supported by turgor pressure in cells, in
terms of water pressure acting against a cell wall
-
as water enters a cell, the vacuole swells and pushes the cytoplasm and cell
membrane up against the cell wall, the cell becomes turgid
turgor in plants cells is maintained because of a protective cell wall, that prevents
plant cells from bursting
turgor helps to support the stem (of unligified), leaves and flowers and keep them
firm
the leaves are well exposed to trap maximum sunlight for photosynthesis
75
3.3.3 Active transport
Specific objective: define active transport as:
The movement of particles (ions/ molecules) through a cell membrane from a
region of lower concentration to a region of higher concentration against a
concentration gradient, using energy from respiration
Specific objective: discuss importance of active transport as a process of
movement across membranes with reference to the uptake of ions by root hairs
and uptake of glucose by epithelial cells of villi and kidney tubules
In the uptake of mineral ions by root hairs:




mineral ions like phosphates and nitrates are absorbed by root hair cell from the
soil through active transport
these ions are often in higher concentration in the root hair cell than in the soil,
hence cannot easily diffuse across
the nutrients are absorbed from a low concentration in the soil against a
concentration gradient
active transport requires energy in the form of ATP which is used by carrier
proteins in the cell membrane to take the ions from a low concentration
In the uptake of glucose by epithelial cells of villi and kidney tubules:




glucose is absorbed from the epithelial cells of the villi in the small intestines and
transported to the body cells and tissues through the blood stream
cells may need glucose which is already in higher concentration in the cell
compared to the outside surrounding
hence, glucose moves against the concentration gradient by active transport
energy released during respiration is used by carrier proteins in the membrane to
absorb the glucose against the concentration gradient
Uptake of glucose by epithelial cells of kidney tubules


kidneys reabsorb useful substances such as glucose from the blood during
filtration
glucose is transported from the kidney tubules into the epithelial walls of
capillaries against a concentration gradient
76
Specific objective: how protein molecules move particles across a membrane
during active transport


the carrier proteins bind to the solute molecule, change shape and carry the
molecule across the membrane
it then reverts back to its original position
3.4. Biological molecules
Specific objective: describe the synthesis of large molecules from smaller basic
units, i.e. simple sugars to starch and glycogen; amino acids to proteins; fatty
acids and glycerol to fats and oils
77


-
a large molecule called a polymer is formed by bonding together many smaller
basic units called monomers
there are four major types of polymers:
polysaccharides that are made from many glucose molecules bonded together
by glycosidic bonds
lipids (fats and oil) made from many three fatty acids and glycerol bonded
together by ester bonds
polypeptide (protein) made from many amino acids bonded together by peptide
bonds
deoxyribose nucleic acid (DNA) made from nucleotides bonded together by
hydrogen bonds
Synthesis of large carbohydrates

larger molecules called polymers are made from smaller basic units called
monomers
Carbohydrates




are a large group of organic molecules containing the elements of Carbon
(C), Hydrogen (H) and Oxygen (O)
this chemical elements bind together to form different types of
carbohydrates
glucose, sucrose, lactose, starch, glycogen, cellulose are all examples of
types of carbohydrates
all carbohydrates are made up of units of a sugar molecule called
saccharides
Types of carbohydrates:



monosaccharides
disaccharides
polysaccharides
Monosaccharides
-
simple sugars made up of one glucose molecule
the general formula of a monosaccharide molecule is C6H12 O6
examples of monosaccharides include glucose, fructose, galactose
this are simple sugars that form the basic structure of complex carbohydrates
monosaccharides are small, simple, soluble and sweet tasting molecules
78
Disaccharides
-
are carbohydrates composed of two monosaccharides
when disaccharides are formed a glycosidic bond is formed between the two
monosaccharides
examples of disaccharides include sucrose, maltose, lactose
Polysaccharides
-
these are large complex insoluble and non-sweet tasting carbohydrates made up
of many monosaccharides bonded together
polysaccharides are a result of polymerisation
examples of polysaccharides are starch, glycogen, cellulose
79
Synthesis of proteins
-
protein molecules contain the chemical elements Carbon, Hydrogen, Oxygen and
Nitrogen (CHON)
sometimes Sulphur can be added to the CHO to make proteins
amino acids are the building blocks of proteins
proteins are assembled from only about 20 different naturally occurring amino
acids
one protein molecule can consist of many amino acids bonded together by
peptide bonds
Synthesis of lipids (fats and oils)
-
fats or oils are composed of Carbon, Hydrogen and Oxygen elements
a lipid molecule is made up of a glycerol and three fatty acids
fats do not dissolve in water
animal fats are made up of saturated fatty acids and are usually solids at room
temperature
plant fats are composed of unsaturated fatty acids and are usually liquids at
room temperature
the fatty acids and glycerol are bonded together by Ester bonds
Fatty acid and glycerol structure
80
Specific objective: outline role of carbohydrates, fats/oils and proteins in living
organisms
Role of carbohydrates





carbohydrates like glucose are broken down during respiration to release energy
sucrose is a transport form of carbohydrates from the leaves to other plant parts
in the phloem tubes
glycogen is a storage form of carbohydrates in animals
starch is storage form in plants
cellulose is a carbohydrate mainly used in the formation of cell walls
Role of proteins







proteins are mainly responsible for growth
proteins help to repair of damaged tissues
proteins also provide energy in the absence of carbohydrates and fats
enzymes are proteins that act as biological catalysts in chemical reactions
collagen is another type of protein needed in cartilage, tendons and ligaments
some hormones like insulin are proteins
proteins are part of the immune system as they are part of antibodies to help fight
against infections
Role of lipids (fats and oils)




fats are a major source of energy
fats acts as insulating layer against heat loss
fats help to form the water proof cuticle in leaves to reduce water loss
lipids are part of the structure of a cell membrane
Specific objective: describe the role of water as a solvent, in living organisms with
respect to digestion, excretion and transport






water is a solvent that dissolves solutes to make solutions
water dissolves more substances than any other liquid
water is a good solvent making it a transport medium for many blood molecules
water is part of blood plasma, in which soluble nutrients dissolve after digestion
to be transported to other body parts
water aids in removal of metabolic wastes, as the wastes dissolve in blood
plasma to be transported to excretory organs
water dissolves plant nutrients in the soil, so that it is absorbed by root hair cells
81
Specific objective: explain that different sequences of amino acids give different
shapes to protein molecules





there are 20 different types of amino acids in protein structures
long chains of amino acids linked together through peptide bonds make up
protein molecules
each protein chain has its own particular amino acids sequence
these sequences cause a folding of the chain to give proteins different shapes
and functions
the sequence of amino acids in a protein is determined by the sequence of
nucleotide bases in the DNA
Specific objective: describe the structure of DNA as: two strands coiled together to
form a double helix; each strand contains chemicals called bases; bases always pair up
in the same way: A with T, and C with G (no references to full names is required)



Deoxyribonucleic acid (DNA) is the heredity material in organisms as it carries
genetic instructions
DNA is located in the chromosomes in the nucleus of cells
DNA is a long molecule in the form of a double helix (two long thin strands
twisted around each other)
82








The DNA molecule is made up of subunits called nucleotides
Each nucleotide is made up of three chemical groups
o Group one: is a sugar molecule called deoxyribose sugar
o Group two: is a nitrogenous base and
o Group three: is a phosphate part
There are four nitrogenous bases: adenine (A), thymine (T), guanine (G) and
cytosine (C)
The nitrogenous bases have complementary pairing (adenine combines with
thymine while guanine combines with cytosine)
Between the complimentary base pairs are weak Hydrogen bonds
There are two Hydrogen bonds between A and T while there are three Hydrogen
bonds between G and C
Adenine and guanine are called purines (large bases) while cytosine and thymine
are called pyrimidines (small bases)
DNA is made up of many nucleotides joined together, hence it is a polynucleotide
Specific objective: describe the use of:
-
Benedict’s solution to test for reducing sugars
Iodine solution to test for starch
Biuret test for proteins
Ethanol test for fats and oils
DCPIP test for vitamin C
83
Benedict’s test




Used to test for reducing sugars
Simple sugars like glucose and maltose are reducing sugars
Benedict’s solution is a clear blue solution
In the presence of simple sugars, the blue solution changes colour to green,
yellow or red depending on the amount of sugar
Procedure followed:
-
-
Place 5 cm³ Benedict’s solution in a test tube
Shake the tube gently to obtain a clear blue colour
Crush the food sample to be tested and add it to the test solution
Shake the mixture thoroughly and heat the tube in a water bath over a Bunsen
burner
Result interpretation:
An orange-red precipitate indicates a large amount of sugar present
A greenish-yellow colour indicates a small amount of glucose present
It the solution remains blue, there is no reducing sugar present
Iodine test



A solution used to test for the presence of starch in food samples
Iodine solution is a brown solution that turns blue-black in the presence of starch
Iodine penetrates the starch more easily when the cell walls have been
destroyed
Procedure followed:
-
Crush the food to be tested into small pieces
Spread the crushed food on a white tile
Use a dropper or syringe to add a few drops of iodine solution to the food
Result interpretation:
-
A change from brown to blue/black shows there is starch present
If the solution remains brown, there is no starch present
84
Biuret test



Test that is used to detect protein in food
The biuret reagent is blue in colour and it’s a solution that consists of potassium
hydroxide or sodium hydroxide and copper sulfate
It detect peptide bonds in proteins
Procedure followed:
-
Crush the food sample if necessary and place it in a test tube
Add equal amounts (5cm³) of biuret reagent A (copper sulfate) and biuret B
potassium sulfate/ sodium hydroxide in the test tube containing the food sample
Add a little distilled water to the mixture and shake gently
Allow the mixture to settle for a few minutes and observe colour changes
Result interpretation:
-
If the blue reagent colour remain blue, there is no protein present
If it changes from blue to purple/violet protein is present
Ethanol test

A test used to find out if the food sample contains fat
Procedure followed:
-
Crush the food to be tested into small pieces and place it in a test tube
Dissolve the crushed food sample in 10cm³ of ethanol (alcohol)
Shake the mixture thoroughly and allow it to settle for 10 minutes for the food to
dissolve in the ethanol
Pour the clear liquid into another test tube and add 2cm³ of distilled water to it
Result interpretation:
-
If the sample contains fat, tiny globules will float in the water
A milky/white suspension indicates the presence of a lipid
If it stays clear, there is no fat present in the sample
DCPIP (dichlorophenolindolphenol)


A test used to detect the presence of vitamin C (ascorbic acid) in food
DCPIP solution is blue in colour
85
-
Procedure followed:
Place 5cm³ of blue DCPIP solution in a test tube
Use a dropper or graduated pipette to add the sample drop by drop to the DCPIP
test tube
Shake the tube gently after adding each drop
Record the number of food sample drops that are added to the DCPIP
Result interpretation:
-
The lesser the drops taken to change the blue DCPIP to clear means the sample
contains a lot of vitamin C
If the DCPIP remains blue, there is no vitamin C present
Specific objective: investigate the distribution of carbohydrates, fats and proteins in
different parts of a seed or fruit
Materials needed:
-
Fruits e.g. camel thorn pods, bananas, apples
Seeds e.g. sunflower seeds, water melon seeds
Benedicts solution, biuret solution, iodine solution, ethanol test and distilled water
Mortar and pestle
Test tubes
Procedure:
-
Follow the procedure for each test as described in the notes and test all various
food sources for starch, reducing sugar, fats and protein content separately
Draw a well labeled table and record your observations for each test
Make conclusion by comparing the various results for each test
3.5 Enzymes
Specific objective: Define the term catalyst as:
A substance that increase (speeds up) the rate of a chemical reaction and is not
changed by the reaction.
Metabolic reactions
Metabolic reactions are all chemical reactions that occur in cells. They include
reactions that are catalysed by enzymes.
86
Types of chemical reactions:
Anabolic reaction-is a reaction that build up complex compounds from simple
ones for, example photosynthesis
Catabolic reaction- is a reaction that break down complex compounds into
simple ones, example digestion of food.
Specific objective: Define enzymes as:
Proteins that function as biological catalysts
Specific objective: explain enzyme action with reference to the active site,
enzyme-substrate complex, substrates and product.
Properties of enzyme.
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Enzymes are proteins
Enzymes are biological catalyst
Enzymes can be used over and over again
Enzyme activity can be affected by temperature and pH.
Enzymes are specific in the reaction they catalyse.
Enzyme action
Only molecules with the correct shape can fit into the enzyme. Just like one key
can open a lock. Only one type of enzyme that can speed up a specific reaction.
This is called a lock and key model.
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Substrate: is the molecule or substance the enzyme acts on
Active site: the ‘dent’ on the enzyme which is exactly the correct shape for the
substrate to fit into
Enzyme-substrate complex: the substrates bonds with the enzyme’s active site
and an enzyme-substrate complex is formed
Products: ‘new’ molecules that leave the active site after the enzyme acts on the
substrate
Some enzymes, their substrates and products
enzyme
Amylase
Maltase
Lipase
Sucrase
Pepsin
Trypsin
substrate
starch
maltose
lipids (fats & oils)
sucrose
proteins
polypeptides
product
maltose
glucose
fatty acids & glycerol
fructose & glucose
polypeptides
amino acids
Specific objective: explain the effect of changes in temperature and pH on
enzyme activity in terms of shape, fit, and denaturalization.
The effect of temperature on enzymes activities
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Each enzyme has a temperature at which the rate of reaction is greatest this is
called an optimum (best) temperature.
The optimum temperature of an enzyme is the level where enzyme works most
effectively by forming
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The effect of low temperature on enzyme activity.
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Rate of enzyme activity is very slow as enzymes become inactive
Low temperature causes less kinetic energy, which causes the molecules to
move slowly and cause few collisions between enzymes and substrates.
Few ESC’s are likely to be formed
The effect of high temperature on enzyme activity
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High temperature gives more kinetic energy, which cause the enzyme to vibrate
too much.
This breaks the bonds of the enzyme and its structure changes
The active site changes shape
The substrate will not fit into the active site and no enzyme-substrate complex
(ESC) will form.
The enzyme will be denatured and the reaction rate will decrease or stop.
When enzyme denature it is a permanent change and the enzyme will not work
again.
The graph shows the effect of temperature on enzyme activity
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The effect of change in pH on Enzyme activity
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Different enzymes work best at different pH levels.
The pH where the rate of enzyme activity is the fastest is called the optimum pH.
If the pH is either too high or too low, the enzyme can denature irreversibly.
Some enzymes work best in neutral (pH 7-7.5) conditions like Amylase in saliva.
Some enzymes work best in acidic conditions like Pepsin in the stomach.
Some enzymes work best in alkaline conditions like Lipase in the duodenum.
Changes in pH also alter an enzyme’s shape and slow down its activity, but this
can usually be reversed if the optimum pH is restored.
An extreme pH level can denature enzymes – the active site gets deformed
permanently.
Specific objective: investigate the effect of changes in temperature and pH on
the rate of amylase and lipase activity
Investigation on how enzyme activity is affected by changes in temperature
Procedure:
1. Put 5cm3 of starch solution in three test tubes labelled 15°C, 35°C and 65°C
To make the starch solution:
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Dissolve 10g of soluble starch in 8cm3 of cold water
Pour the solution into 1litre (1000cm3) of boiling water and stir well
Boil the solution for 15 minutes, keep it covered to reduce evaporation
Allow to cool
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2. Add amylase to the starch solution
To make an amylase solution:
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Weigh out 0.5g of the enzyme and add to 80ml of distilled water at room
temperature in a beaker
Stir gently to dissolve and add water to a final volume of 100ml
Store at 4°C (fridge) for a short period of time or on ice during use
3. Put the three test tubes with the mixture into a water bath for 30 minutes at
temperatures 15°C (use ice cubes), 35°C and 65°C
4. Use a dropper pipette to put a small sample of the mixture (of what is in each
test tube after 30 minutes) onto a white tile and add iodine solution to it
5. Observe which solution turns blue-black and which remains yellow-brown.
Explain your results
Specific objective: describe the role of enzymes in the germination of seeds
and their uses in biological washing powders and in the food industry
In seed germination
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Seeds contain inactive enzymes, an embryo plant and insoluble nutrients
Maize seeds contain starch, bean seeds contain protein and sunflower seeds
contain lipids
This starch, proteins and lipids are insoluble, and the embryo plant cannot use it
unless it is digested into soluble molecules
Enzymes become activated when the seed absorb water during germination
A plant growth substance is formed in the embryo and it activates the enzymes to
digest insoluble nutrients into soluble forms
Enzyme amylase change the starch in the seed into maltose while protease
digests proteins into polypeptides and amino acids
Enzyme lipase digests lipids into fatty acids and glycerol
The soluble nutrients are used by the embryo plant during germination as a
source of energy for growth
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In biological washing powders
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Biological washing powder contain enzymes which breakdown insoluble
proteins, fats and starch stains that cannot be removed by detergents alone.
Once broken down, the stains disappears because the smaller molecules that
result from the enzyme actions are soluble and can be washed away easier
Protein stains like blood haemoglobin are digested by protease enzymes into
soluble amino acids
Fat stains are digested by lipase enzymes into soluble fatty acids and glycerol
Amylase digest starch stains into soluble sugars
In food industry
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The enzyme rennin or chymosin are added to milk to make cheese.
o The enzymes clot the milk to form milk curds, which are used to make
cheese
The enzyme pectinase is used to make fruit juice.
o Pectinase soften the fruit cell walls. This makes it easier to extract more
juice from fruits, such as apples and oranges .
o Pectinase also helps to make cloudy juice become clear
The enzyme amylase is added to bread dough in bread baking
o Amylase breaks down the starch in the dough into sugar
o When yeast is added to the dough, it can breakdown the sugar more
easily, releasing carbon dioxide that makes the bread dough to rise
Enzyme trypsin is used to predigest proteins in the manufacturing of baby food.
o This makes it easier for babies to digest their food
The enzyme papain is used to produce tenderised meat.
o Papain breaks down protein in the meat, making the meat easier to chew.
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Specific objective: Investigate the uses of biological powders that contain
enzymes.
Requirements
1. Two boiled eggs
2. Two teaspoons
3. Two glass beakers with water
4. Regular and biological detergent
Procedure
1. Boil two eggs for about five minutes
2. Put marks on the two spoons for identification.
3. Push the teaspoons into the eggs to collect some yolk on the spoons.
4. Dissolve equal amounts of ‘regular’ and 'biological' washing powders in two
separate beakers of water and leave a yolk stained spoon in each glass.
5. Observe what happens to the yolk on both spoons put in the two washing
powders. The spoon put in the ‘regular’ washing powder still has yolk on but the
yolk on the other spoon has been digested by the ‘biological’ washing powder
3.6 Nutrition
Specific objective: distinguish between autotrophic nutrition and heterotrophic
nutrition
Definition of nutrition:
Nutrition (feeding) is the intake of inorganic and organic substances from which
organisms obtain energy and raw materials for growth and development.
Autotrophic nutrition
Is a process where organisms use inorganic materials (carbon dioxide and water),
an external source of energy (sunlight) and chlorophyll to build organic molecules
during photosynthesis
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Examples of organic molecules are: glucose, starch, fats and proteins
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Heterotrophic nutrition
Is a process where organisms obtain and digest organic molecules during
nutrition and use it as a source of energy and for growth
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3.6.1 Plant nutrition
3.6.1. 1 Leaf Structure
Specific objective: Identify the cellular and tissue structure of a dicotyledonous
leaf, as seen in cross section.
External structure of a dicotyledonous leaf
Internal structure of a dicotyledonous leaf, as seen in cross section
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Specific objective: state the significance of these structures in terms of function,
i.e. distribution of chloroplasts for photosynthesis; stomata, (opening and closure)
and mesophyll cells for gaseous exchange and vascular bundles for transport.
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A leaf is made of different layers of cells (different tissues).
The upper epidermis is a layer of transparent cells (because they do not have
chloroplasts) which allows light to pass through to the mesophyll cells.
The upper epidermis covers the upper surface of the leaf to protect the inner
cells of the leaf and it produces the cuticle.
The cuticle is a waxy, waterproof layer covering the upper epidermis, to reduce
water loss from the leaf by evaporation.
The lower epidermis is a layer of transparent cells (because they do not have
chloroplasts) which covers the lower surface of the leaf to protect the inner cells
of the leaf and sometimes produces the cuticle.
The lower epidermis a layer of transparent cells that allows light to pass through
to mesophyll cells. The mesophyll layer is the middle layer of the leaf, situated
between the two epidermises.
The cells in the mesophyll layer just below the upper epidermis are called the
palisade mesophyll cells (layer).
The palisade mesophyll cells is where most of the photosynthesis takes place as
it have larger number of chloroplasts.
The cells in the mesophyll layer in the lower part of the leaf, just above the lower
epidermis, are called the spongy mesophyll cells (layer).
Cells of the spongy mesophyll are round, loosely arranged cells with large air
space between them (like a sponge), which allows for gaseous exchange by
diffusion between mesophyll cells and the air.
Photosynthesis also takes place in this layer.
The leaf contains vascular bundles (veins), which are made up of xylem and
phloem.
Xylem is made up of tiny tubes called xylem vessels (tubes) which transport
inorganic substances (water and mineral ions).
The phloem transports organic substances (sucrose and amino acids) by
translocation.
Stomata are found in lower epidermis. These are small pores (stomata
openings) with two guard cells, one on either side of the stomata openings. The
guard cells control the opening and closing of a stoma.
The stomata allow gaseous exchange, where carbon dioxide and oxygen can
diffuse into or out of the leaf, and allow transpiration to take place.
The guard cells contain few chloroplasts so some photosynthesis can take place
here.
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Specific objective: explain how the internal structure is adapted for photosynthesis
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A leaf has a very large surface area that can be exposed to absorb the
greatest (maximum) amount of light.
It is thin to allow light to penetrate to all cells.
It is thin to reduce the distance for diffusion.
A leaf is supported by a stem and petiole to expose as much of it as possible
to light and air.
The upper and lower epidermis are transparent to allow light through to the
mesophyll layer.
The palisade mesophyll cells are closely packed to absorb more incidental
light.
The palisade mesophyll cells are near the leaf surface to maximise light
interception.
Palisade mesophyll cells are arranged at right angles to the leaf surface to
reduce the number of cell walls for light to pass through.
Palisade mesophyll cells have large numbers of chloroplasts to maximise
light absorption.
Spongy mesophyll cells are round and loosely arranged to accommodate air
spaces, which act as reservoirs for gases and to assist in gaseous exchange.
Mesophyll cells have large vacuoles to push chloroplasts to the edge of the
cells.
Mesophyll cells have thin cell walls so there is a short diffusion pathway.
Chloroplasts can move within mesophyll cells towards light.
Chloroplasts can move away from high light intensity to avoid damage.
There are stomata in the lower epidermis for gases to enter and leave during
gaseous exchange.
Xylem vessels is present for the transport of water to the chloroplasts in the
mesophyll cells.
Phloem tube is there to translocate the products of photosynthesis away
from the leaf.
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Specific objective: make temporary mounts of the upper and lower epidermis
(using nail varnish) with emphasis on the distribution of stomata.
Making temporary slides of the epidermis
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Stomatal density varies between monocots and dicots, between plants species
and between the underside and top side of the leaves on a plant.
use clear nail varnish to make an impression of the epidermis
Materials needed:
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Different plant leaves.
Clear fingernail polish.
Clear cellophane tape (clear package sealing
tape).
Microscope and microscope slides.
Procedure:
1. Obtain a leaf from a plant, generally any plant will work for this procedure.
2. Paint a chick patch of clear nail polish on the leaf surface being studied. Make
the patch at least one square centimetre (cm2). Try not to paint on the large
veins, as this makes it difficult to remove the dried nail polish.
3. Allow the nail polish to dry completely.
4. Tape a piece of clear cellophane tape to the dried nail polish patch.
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5. Gently peel the nail polish patch from the leaf by pulling on a corner of the tape
and “peeling” the fingernail polish off the leaf. This is the leaf impression you will
examine.
6. Tape your peeled impression to a very clean microscope slide. Use scissors to
trim away any excess tape.
7. Scan the slide until you find a good area where you can see the stomata.
Examine the leaf impression under a light microscope at X 400. Search for areas
where there are numerous stomata and where there are no dirt, thumbprints,
damaged areas or large leaf veins.
8. Draw the leaf surface with stomata. Each stoma is bordered by two sausageshaped cells that are usually smaller than chloroplasts. Sketch this. Label the
stoma, guard cells, epidermal cells and chloroplasts.
9. Count all the stomata in one microscopic field. Record the number in your data
table.
10. Repeat counts for at least three other distinct microscopic fields. Record all the
counts. Determine an average number per microscopic field.
11. From the average number / X 400 microscopic field, calculate the stomata per
mm2 by multiplying by 8.
View of the stomata as seen under a microscope
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Specific objective: draw and interpret prepared slides of transverse sections
through a leaf.
3.6.1.2 Mineral requirements
Specific objective: state the importance and explain the effects of iron,
magnesium, phosphate and nitrate ions on plant growth.
Mineral requirements
The metabolites that plants require for growth, development and repair come
from photosynthesis and respiration. However, there are other ingredients
needed for the production of these essential metabolites. They are called
mineral ions
Importance of iron and effects on plant
growth
a micro nutrient required in small amounts
 Iron is needed for chlorophyll
synthesis.
 Iron is also a constituent of electron
carriers.
 It is needed for formation of some
enzymes.
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Deficiency effects on plant growth
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Iron chlorosis (yellowing of the
new upper leaves).
Leave become whitish and start
to die.
Stunted growth.
Importance of magnesium and effects on
plant growth
a macro nutrient needed in relatively large
amounts
 Magnesium ions are required for the
synthesis of chlorophyll in plants.
 Magnesium also acts as an enzyme
activator.
 It forms part of the middle lamellae
in plant cells.
Deficiency effects on plant growth
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Yellowing of the older (lower)
leaves will occur first as they
become yellow between the veins
and around the edges.
Eventually the leaf and the plant will
die.
Importance of phosphorus and effects on Deficiency effects on plant growth
plant growth
a micro nutrient absorbed in the form of
phosphate ions
 Used to synthesise nucleotides which
 It leads to stunted growth (plants
form nucleic acids (DNA and RNA).
are small), especially the roots.
 Used to form high energy compounds
 Brown areas occur on leaves and
like ATP,
the petioles.
 A component of cell membranes in
 they produce little or no flowers,
the form of phospholipids.
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Phosphate ions affects plant growth
by increasing early growth and root
formation.
Phosphorus also improves the plant’s
ability to absorb water and other
nutrients from the soil.
Phosphorus encourages flower or
fruit production.
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Have weak root systems or a bright
green or purplish
Importance of nitrates and effects on
Deficiency effects on plant growth
plant growth
 Nitrates are required for amino acid
 yellowing of all leaves will occur
synthesis (amino acids combine to
 Symptoms include stunted plant
form proteins).
growth
 They are used to synthesise
 Thin weak stems and leaves that
chlorophyll.
are pale green or yellow
 Nitrates are used to synthesise
nucleotides which form nucleic acids.
 Some plant hormones, like auxin,
contain nitrogen.
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Specific objective: Investigate the effect of mineral deficiency on the plant growth
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Here is a method to investigate mineral deficiency in a plant, using nitrate ions as
an example.
You could do the same experiment with any of the following minerals:
magnesium ions, phosphate ions and iron ions.
Materials needed:
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Nitrogen fertiliser (ammonium or nitrate fertiliser)
30 maize / mahangu (millet) seedlings
Soil for each pot
3 bottles with water to make the three liquid nutrient mediums
Procedure
1. Take 30 seedlings of the same plant (they should be the same age and height)
and plant them in separate pots.
2. Make up three nutrient mixtures (liquid nutrient medium) with varying
concentrations of nitrate ions. Make up one mixture with a high concentration,
one with a medium concentration and one with a low concentration of nitrate.
3. Split the plants into three groups. Each group should be given only one of the
three mixtures. Add about a cup of the nitrate liquid medium to each pot.
4. Record the height of each plant after seven weeks. Calculate the average height
of each group of plants.
5. During the experiment, it is important to keep all other variables the same, e.g.
the amount of sunlight and water the plant receive.
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Results
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The greater the concentration of nitrate, the more the plants will grow
(average heights of 12, 18 and 23 cm have been measured for plants given
low, medium and high concentrations respectively)
3.6.1.3 Photosynthesis
Specific objective: state the word and balanced chemical equation for
photosynthesis.
Photosynthesis
A method of nutrition in green plants, where light energy is trapped in the
chlorophyll and organic substances (carbohydrates / glucose) are produced by
using water and carbon dioxide, with the release of oxygen as a by-product.
Equations for photosynthesis
Word equation
Carbon dioxide + water + (chlorophyll & light energy) → glucose + oxygen
Balanced chemical equation
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Specific objective: Investigate the need for chlorophyll, light and carbon dioxide
for photosynthesis, using appropriate controls
Carry out starch tests on leaves
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Carbohydrates are the products of photosynthesis.
Simple sugars (glucose) is formed first but are immediately converted into
starch, the first visible product of photosynthesis.
It is therefore possible to find out if the plant has carried out photosynthesis by
testing for starch.
Presence of starch implies that photosynthesis has occurred.
Procedures (steps) when carrying out a starch test on leaves
Use a green leaf which was exposed to sunlight for a few hours (glucose is
produced which is converted into starch for storage).
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Step one: Boil the leaf in water for about a minute till it is flabby (cell walls
and membranes are broken down and enzymes and cytoplasm destroyed for
iodine solution to penetrate)
Step two: Place the leaf in a test tube with alcohol (ethanol) and place the
test tube in a beaker with boiling water (water bath) till the leaf is decolourised
(for chlorophyll to be dissolved and extracted).
Step three: Rinse the yellow – white coloured leaf in hot water (to soften it
after it was hardened by the alcohol).
Step four: spread the leaf in a shallow petri dish or on a white tile and cover it
with yellow – brown Iodine solution for a few minutes.
If the leaf turns blue- black, starch is present and photosynthesis took place.
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Removal of starch from a plant
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Place a potted plant in a dark cupboard for 48 hours.
This will ensure that no starch will be formed in its leaves during that period.
Any starch already present is converted into sugar (glucose) to be
transported out of the leaves to the rest of the plant.
When a leaf is now tested for starch, it will be noticed that no starch is
present.
Such a plant is said to be destarched.
To show that light is essential for photosynthesis:
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Destarch a potted geranium plant (as outlined above)
Do not remove any leaves, but select one at the top of a leafy shoot.
Wrap a strip of aluminium foil around a part of this leaf and press it as close as
possible to the surface of the leaf so that this part cannot received any light.
Expose the plant to sunlight for 4 to 6 hours.
Remove the leaf with the foil from the plant. Test the leaf for starch.
The exposed parts will turn blue- black.
The covered part of the leaf will stain brown.
Since starch has not formed in the part which receive no light, this shows that
light is necessary for starch formation and thus for photosynthesis.
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Investigation set up to show if light is essential for photosynthesis
To show that carbon dioxide is essential for photosynthesis
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Destarch a potted geranium plant.
Put a small container of soda lime (or potassium hydroxide / sodium hydroxide)
close to the plant.
Cover it with a transparent plastic bag and tightly seal with an elastic band so
that no carbon dioxide (air) can enter.
The soda lime absorbs all the carbon dioxide.
Expose the plant to sunlight for 4 to 6 hours.
Remove the leaves and test them for starch.
The leaves will stain brown.
Since starch has not formed in the leaves which receive no carbon dioxide, it
seems that carbon dioxide is necessary for starch formation and thus for
photosynthesis.
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To investigate if carbon dioxide is essential for photosynthesis
To show that chlorophyll is essential for photosynthesis
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Destarch a potted geranium plant which has variegated leaves (leaves with
parts that have chlorophyll and non- green parts without chlorophyll).
Expose the plant to sunlight for 4 to 6 hours.
Remove a leaf and make a careful drawing of its green and non- green parts.
Then test the leaf for starch.
The green parts (containing chlorophyll) turn blue- black to show that (starch is
present).
The non- green parts (lacking chlorophyll) stains brown (starch is absent).
Since starch is present only in the parts which originally contained chlorophyll, it
seems reasonable to assume that photosynthesis has taken place.
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To show that oxygen is released during photosynthesis
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Set up the experiment as shown below
Elodea can be used as a submerged plant.
The gas collected in the test tube can be tested with a glowing splint.
If it is oxygen, the glowing splint will glow brighter or even burst into a flame.
This indicates that oxygen has been given off by the splint.
To investigate if oxygen is released during photosynthesis
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Specific objective: describe the effects of varying light intensity, carbon dioxide
concentration and temperature on the rate of photosynthesis
Effects of varying light intensity, carbon dioxide concentration and temperature
on the rate photosynthesis
Light intensity
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At low light intensity, the rate of photosynthesis is lowered, because the energy
that the light provides is less and the reaction rate is slowed down.
A higher light intensity will enable photosynthesis to happen faster.
Carbon dioxide concentration
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When there is insufficient carbon dioxide concentration, a plant will not be able to
photosynthesise to its full potential. Because there is less carbon dioxide
concentration – less reactant – fewer products are made.
Temperature
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In cooler temperatures (low), the rate of photosynthesis will decrease.
At optimum temperature, the rate of photosynthesis become faster.
If temperature is too high, photosynthesis stop.
Specific objective: Describe the synthesis of carbohydrates.
The synthesis of carbohydrates
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Green plants synthesise carbohydrates during the process of photosynthesis.
Photosynthesis requires light energy, water and carbon dioxide.
The end product of photosynthesis is glucose (carbohydrate) and oxygen.
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Requirements for the synthesis of carbohydrates
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Light intensity is absorbed by the chloroplasts and trapped by the chlorophyll in
the chloroplasts of the mesophyll cells.
This light energy is converted into chemical energy.
This chemical energy is used to split water molecules into hydrogen ions and
oxygen.
The oxygen is released by diffusion into the air spaces of the leaf and from there
through the stomata into the air outside the leaf.
Carbon dioxide diffuses into leaf through the stomata and then into the air
spaces.
From the air spaces, it diffuses into the chloroplasts of the mesophyll cells.
Water is absorbed from the soil by the root hair cells by osmosis.
Water moves from the root hair cells through the cortex and endodermis into the
xylem and up to the leaf.
From the leaf xylem, water moves by osmosis into the mesophyll cells and into
the chloroplasts.
The carbon dioxide and hydrogen atoms (from the water) combine. Carbon
dioxide is converted into carbohydrates, using the energy from sunlight.
This forms an energy- rich carbohydrate called glucose.
Specific objective: describe the use and storage of carbohydrates made in
photosynthesis.
Use of carbohydrates (glucose)
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Glucose is used in the leaves as an energy source for respiration.
It is changed into sucrose for translocation in the phloem sieve tubes.
It is changed into cellulose to build and strengthen the cell walls.
It is converted into amino acids / proteins (by reaction with nitrates) for growth
and repair.
It is converted into chlorophyll (by reaction with magnesium and iron) to trap
light energy.
It is converted into different vitamins.
It is used to make nectar and fruits.
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Storage of carbohydrates (glucose)
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It is changed into insoluble starch to be stored in seeds, leaves and roots.
It is changed into fats and oils to be stored in seeds.
It is converted into amino acids / proteins to be stored.
Specific objective: define the term limiting factor as:
Something present in the environment in short supply that restricts life processes.
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Examples of limiting factors: temperature, availability of water, light intensity,
carbon dioxide concentration and amount of chloroplasts / chlorophyll.
Specific objective: identify and explain the limiting factors of photosynthesis, in
different environmental conditions.
Limiting factors of photosynthesis
Light as a limiting factor
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Light is a requirement for photosynthesis.
As light intensity increases, so does the rate of photosynthesis (e.g. at 50 au
(arbitrary units) of light the rate of photosynthesis is 70, and at 150 au it is 170).
It increases until the optimum light intensity is reached (at 230 au).
A further increase in light intensity does not bring about any further increase in
the rate of photosynthesis (after 230 au it remains constant).
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Graphical presentation of light intensity is a limiting factor
Carbon dioxide as a limiting factor
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Photosynthesis requires carbon dioxide.
At low carbon dioxide levels, the rate of photosynthesis will be slow.
The more carbon dioxide a plant is given, the faster it can carry out
photosynthesis.
It increases until the optimum is reached.
Graphical presentation of carbon dioxide concentration is a limiting factor
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Carbon dioxide is a limiting factor even though there is sufficient light
Temperature as a limiting factor
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Most plants function best at optimum temperatures between 20°C and 35°C
As the temperature increases, the rate of photosynthesis increases.
Any further increase in the temperature above the optimum temperature can
cause a drop in the rate of photosynthesis, because the enzymes become
denatured.
If the temperature is low, the enzymes become less active and even inactive and
the rate of photosynthesis is slowed down
Temperature as a limiting factor
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Specific objective: describe the use of CO2 enrichment, optimum light and optimum
temperatures in greenhouse systems and their importance to increase plant
productivity.
Greenhouse systems and plant productivity
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A greenhouse, also called a “glasshouse”, is a structure with walls and roof made
chiefly of transparent materials, such as glass, where the environment of plants
can be controlled and optimum conditions can be maintained to maximise growth
and yields.
Control of carbon dioxide (CO2) concentration
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Carbon dioxide can be increased (carbon dioxide enrichment) by using burning
of fuels, e.g. gas or paraffin.
Bottles of carbon dioxide can also be used to increase its concentration.
Ventilators can allow fresh air with carbon dioxide to enter the greenhouse.
As the CO2 concentration becomes higher, CO2 is no longer a limiting factor and
the rate of photosynthesis increases.
Control light intensity
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Position the greenhouse to receive maximum light.
Additional light can be supplied by artificial lights (bulbs).
Light which is too bright can be reduced by blinds which can be closed.
Reflectors inside the greenhouse can reflect light directly onto plants.
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Control of temperature
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Temperature can be increased by using heaters or by burning fuels.
Temperature can be reduced by using of fans and ventilators.
The temperature must be maintained because photosynthesis involves enzymes.
Control of humidity (water)
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Sprinkler irrigation systems can be used to control water and give continuous
supply of water to plants.
Hydroponics is a method of growing plants using mineral nutrient solutions, in
water, without soil. Terrestrial plants are grown with their roots in the solution
only.
Humidifiers can control the humidity to keep the air around plants humid, to
reduce transpiration and evaporation of water.
Greenhouse (glasshouse)
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Specific objective: investigate the effect of gas exchange on an aquatic plant
kept in the light and in the dark (use hydrogen carbonate indicators solution).
Procedure
1. Take two test- tubes label them P and Q
2. Fill up each test tube with a solution of red hydrogen carbonate indicator.
Hydrogen carbonate indicator turns yellow when the carbon dioxide
concentration increases and turns purple when the carbon dioxide concentration
decreases.
3. Place similar pieces of the same aquatic plant into tubes P and Q
4. Leave tube uncovered, but cover tube with tube Q with a black light – proof
cover.
5. Leave both tubes in a warm room in sunlight for four hours.
What would be the colour of the hydrogen carbonate indicator in the two tubes after four
hours?
Tube P: When the carbon dioxide level falls, the hydrogen carbonate indicator changes
from red to purple (the rate of photosynthesis exceeds the rate of respiration).
Tube Q: It changes from red to yellow as the carbon dioxide concentration increases
(the rate of respiration exceeds the rate of photosynthesis).
Conclusion: Tube P absorb light and carbon dioxide that will enable photosynthesis to
take place while Tube Q did not received light thus photosynthesis did not take place
and carbon dioxide accumulate.
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3.6.2 Human nutrition
3.6.2.1 Diet and nutrients
Specific objective: describe the causes and effects of vitamins A, C, D and
mineral salts (iodine and iron only) deficiencies.
Vitamins
Vitamins are a group of organic compounds that are required in small quantities
in the diet.
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Vitamin A (also known as retinol)
Vitamin A cannot be made by the human body and so it is an essential part of the
diet.
It is a fat- soluble Vitamin that is stored in the liver.
Examples of food sources rich in vitamin A are liver, milk, butter, cheese, yogurt,
egg yolk, oily fish and fish liver oils and green leafy vegetables
beta- carotene (which is found in green leafy vegetable and red, orange and
yellow fruits) can also be converted by the liver into vitamin A
Causes of vitamin A deficiency
Consuming a diet with very low fat.
Consuming a diet which lack food such as liver, dairy foods and dark green
vegetables
When rice is the main food in your diet, vitamin A deficiency usually occurs as
rice doesn’t contain any carotene
Excess alcohol consumption (alcoholism)
Effects of vitamin A deficiency
Night blindness (does not produce a substance called rhodopsin that aids in night
vision)
Poor complexion and dry skin
Weaken immune system
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Vitamin C (also known to as ascorbic acid)
Vitamin C is water soluble (your body doesn’t make or store it and you flush out
the excess daily via urine)
Thus, we need to eat vitamin C daily.
Food sources rich in vitamin C include mainly fruits and vegetables like
tomatoes
Causes of vitamin C deficiency
 Consuming diet that lack vitamin C like citrus fruits and vegetables
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Effects of vitamin C deficiency
Slow in wound healing due to iron deficiency
Weakened immune system unable to fight colds and flu
Bruising as small vessels break
Bone fractures
Scurvy (bleeding of the gums)
Vitamin D
Sometimes called the “sunshine vitamin” because it is produced in response to
sunlight.
It is fat- soluble vitamin
Our bodies can make most of what we need when we are directly exposed to
sun.
Vitamin D is stored in the liver and fatty tissue.
Sources of food rich in vitamin D include fortified foods, salmon, sardines, fish
liver oil, liver, raw milk, butter, cheese, margarine and eggs and few minutes of
exposure to the sun daily
Causes of vitamin D deficiency
 A strict vegetarian diet (not eating any animal products that are good sources of
vitamin D)
 Your exposure to sunlight is limited (you are homebound, wear long robes or
head coverings for religious reasons, or have an occupation that prevents
exposure to the sun)
 You have dark skin as the pigment melanin reduces the skin’s ability to make
vitamin D in response to sunlight exposure.
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Effects of vitamin D deficiency
Deficiency of vitamin D cause Rickets in children
Deficiency of vitamin D in adults cause Osteomalacia and Osteoporosis (weak
bones that can break easily)
Mineral salts
Mineral salts are naturally occurring inorganic substances
-
Iodine
Iodine is an essential mineral that is crucial for the thyroid of function properly.
Sources include some sea food like fish, lobster, shell fish, table salt and
vegetables grown on iodine rich soils
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Causes of iodine deficiency
Eating a diet that lack iodine rich food like seafood
It is also common in mountainous regions where food is grown in iodine poor soil.
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Effects of iodine deficiency
Leading to goitre (thyroid gland enlarges)
Lead to lethargy and fatigue
Slow metabolism and weight gain.
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Iron
Iron is a mineral that is naturally present in many foods or added to some food
products and available as a dietary supplement.
Iron is a mineral found in every cell of the body.
Foods (dietary sources) that contain iron include red meat, pork, liver, kidneys,
eggs, dark green leafy vegetables, such as spinach, cabbage, broccoli and
lettuce.
Dried fruits such as raisins and apricots, as well as iron fortified cereals and
bread are also good sources of iron
Causes of iron deficiency
Women, who lose a lot of blood during their monthly period, called menstruation,
are at risk of iron deficiency
The loss of blood associated with trauma or surgery can also lead to iron
deficiency called anaemia.
Effects of iron deficiency
anaemia, this reduces the number of red blood cells due to lack of iron
General objective
 Know the importance of a balanced diet for every person and understand their
different needs
 Know the effects of malnutrition and understand the problems related to famine
.
Definition of a balanced diet
A diet that contains all seven food nutrients in the correct quantity for growth,
repair of body cells and to stay healthy
Specific objective: describe the role of a balanced diet with reference to:
children, adult/workers, nursing mothers, athletes and people with HIV/ AIDS
 In children
o Breast milk provide antibodies to strengthen their immune system
o Vitamin D promote calcium absorption for the development of strong
bones
o More protein promotes rapid growth
o Children need more carbohydrates and fats for more energy.
121
 Adults/workers
o Balanced diet will help them maintain healthy weight and a strong immune
system
o Men need more energy than women because they have faster metabolism
o Minerals such as calcium will maintain their bone strength
o High energy intake and less activity will lead to weight gain and problems
associated with being overweight
 Pregnant and nursing mothers
o Pregnant women need more protein for foetal growth.
o Pregnant women need more iron for the formation of haemoglobin (red
blood cell) for the foetus.
o Fibre is important for pregnant women to prevent constipation
o Breastfeeding women need additional calcium for production of breast
milk.
 Athletes
o They need more energy-rich food for their physical activities
o More protein is required to build the muscle mass
 People with HIV/AIDS
o A healthy and balanced diet boosts their immune system
o For example, they should eat plenty of fresh fruits and vegetable to ensure
that they take in sufficient vitamins, especially vitamin C and D.
o They need more protein to gain weight or build muscle mass
Specific objective: explain how age, gender and activity affect the dietary needs of
humans, including during pregnancy and whilst breast-feeding
Babies, toddlers and young children (0- 10 years)
 Breast milk is the most natural food for new born babies as it provides all
essential nutrients for a baby up to the first six months.
 Carbohydrates provide more energy
 Proteins to make new cells to grow and repair tissues.
 Children need extra calcium and vitamin D for growth and development of their
bones and teeth
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Puberty and adolescence (11- 19 years).
Energy and protein requirements may be higher than in adults.
Females have an increased requirement for iron because of menstruation.
Males have an increased requirement for calcium for thickening bones.
Elderly persons
An older person’s energy needs decline due to decreased physical activity and
basic metabolic rate thus needs less energy giving food.
The ageing body replaces muscle with fat, which burns fewer kilojoules than
muscle does. Loss of muscle mass can also be due to inactivity therefore, they
need less fat-rich food intake.
Elderly persons may also need less protein, or even more if they have very
inactive lifestyles.
Elderly people, especially women, may be at risk of osteoporosis, so
supplementation of calcium and vitamin D is recommended.
A reduction in foods like fruits and vegetables can lead to an insufficient intake of
Vitamin C, so supplements can benefit them.
Gender (sex) affects dietary needs
Males need more energy giving food than females of the same age and even
when doing the same job, because males have more muscle mass and are
generally larger (broader muscles).
Men have more muscle mass and higher metabolic rates than women therefore
have higher requirements for protein.
Activities affect dietary needs
 Active persons require more energy than less active persons do.
 When you are active, you burn more kilojoules and if you burn kilojoules than you
eat, you lose more weight.
 If a child is extremely active, he/she will need more nutrients (i.e. carbohydrates,
fat, and protein).
 When athletes are doing sport, they require more proteins for building muscles.
 When doing sport, athletes require more carbohydrates for energy.
 A worker doing heavy work requires more energy than a worker doing light work
does.
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Pregnancy
The extra protein is needed for growth of the foetus.
Calcium is used to build a baby’s bones and teeth
Calcium is needed for the production of breast- milk.
Extra iron can prevent anaemia in the mother and essential for formation of foetal
haemoglobin in the blood.
Vitamin C assists in the absorption of iron and is necessary to make collagen, a
structural protein that is a component of cartilage, tendons and skin of the foetus.
Vitamin A is important for the baby’s embryonic growth, including the
development of the heart, lungs, kidneys and formation of the retina and visual
pigments of the eyes.
Vitamin D assists in the absorption of calcium for the formation of bones and
teeth.
Fibre is important of prevent constipation in the pregnant mother.
Reduced sugar intake is advised, because the mother is less active so less
energy is used. Eating sugar during pregnancy causes weight gain.
Folic acid (a B vitamin) supplements help to prevent birth defects in this foetal
brain and spinal cord, known as neural tube defects.
Specific objective: describe the causes and effects of malnutrition in relation to
starvation, constipation, coronary heart disease, obesity and scurvy

Starvation
Is when there is a severe lack of food or under nutrition e.g. Kwashiorkor and
marasmus
Cause
 Poverty
 Famine
Effects
 During the first stage of starvation, the body uses up all
available carbohydrates for energy followed by all fat
reserves
 After all the fats are exhausted, the body start using
protein, including the muscles
 Once all the muscle have been used up, the organs start
breaking down, which lead to death
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Constipation
 A condition where faeces or stool is hard and egestion becomes difficult and
infrequent
Cause
 Lack of fibre (roughage)
Effects
 Difficult and infrequent egestion
 Colon cancer
Coronary heart disease
 Refers to the condition which develop when the coronary artery, which supply the
heart muscles with nutrient (glucose) and oxygen, become narrowed
Cause
 Eating too much animal fats increase the level of cholesterol, which is deposited
in the coronary arteries and this lead to narrowing/blockage of the coronary
arteries.
Effects
 If the heart muscles do not receive adequate supply of glucose and oxygen this
can cause a heart attack, which can lead to death.
Obesity
Cause
 Eating too much sugary food, refined and fatty food
Effects
 Can lead to diabetes
 Can lead to coronary heart disease
Scurvy
Cause
 Lack of vitamin C (diet lacking fresh fruits and vegetable)
Effects
 Wall of blood vessels become weak and bruising on the skin can occur easily.
 Bleeding gums
 Poor healing of wounds
 Anaemia
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Specific objective: investigate health problems particular to Namibia associated
with the over-consumption of sugar, fat, meat and salt
Dietary health problems in Namibia
Health problems associated with overconsumption of sugar
 The extra sugar will be stored as fats and this can contribute to weight gain that
can lead to being overweight or obesity.
 Being overweight or obese can lead to coronary heart disease (CHD), high blood
pressure and diabetes Type 2.
 Tooth decay results from eating large amounts of sugar, especially in
concentrated forms found in cold drinks and sweets.
Health problems associated with the overconsumption of fat
 Extra fats are stored in the body and can contribute to weight gain and obesity
 Animal fats (saturated fats) contained cholesterol, which can cause hypertension
(high blood pressure) and coronary heart disease.
 Cholesterol leads to blockage of coronary arteries, causing heart attacks.
Health problems associated with the overconsumption of meat
 Excessive use of meat can lead to constipation.
 Overconsumption of meat can lead to weight gain. People who eat large
quantities of meat tend to be overweight or obese.
 Red meat contains saturated fats (cholesterol) which can lead to hypertension
and coronary heart disease.
Health problems associated with the overconsumption of salt
 Excessive use of salts can cause hypertension (high blood pressure)
 People who consume large amounts of salt run the risk of suffering from strokes
and coronary heart disease.
 Excessive use of salts can cause osteoporosis, because it can lead to the loss of
calcium.
 Excessive use of salts can also increase the risk of stomach cancer and kidney
disease.
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Specific objective: discuss the problems that contribute to famine (unequal
distribution of food, drought and flooding, increasing population).
Increasing population
 When the population increase, the availability of food resources fall
 This leads to an inability to grow sufficient food
Drought and flooding
 Lack of rain or unpredictable rainfall makes it difficult to grow crops and
rear animals
 For example:
 In Africa, droughts are responsible for many food shortages
 In Asia, floods are the main cause of shortage of food
Poverty
 Unequal distribution of wealth between the rich and poor also
contribute to famine
 The country may have enough food to feed its population, but poor
people do not have enough money to buy food
Unequal distribution of food
 Some countries faced with drought or wars do not receive sufficient
amounts of food relief
 This increases starvation and possible death due to starvation
3.6.2.2 Alimentary canal (gut):
Specific objective: define the following terms as:
Ingestion- the taking in of substances into the body through the mouth
Digestion- the breakdown of food into smaller pieces without chemical change to
the food molecules (mechanical digestion) and the breakdown of large, insoluble
molecules into smaller, soluble molecules (chemical digestion)
Absorption- the movement of small food molecules and ions through the wall of
the intestine into the blood
Egestion- the passing out of food that has not been digested or absorbed
through the anus
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Specific objective: describe the functions of the various parts of the alimentary
canal in relation to ingestion, digestion, absorption and egestion of food.
Mouth and mouth cavity (buccal cavity)
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The mouth, bounded by two lips, is the anterior (front) opening of the alimentary
canal.
The mouth cavity contains the tongue, teeth, and duct openings of the salivary
glands.
Chewing of food (mastication) occurs here, this is to make swallowing easier, to
increase the surface area for enzyme activity and to mix the food with saliva.
Chewing is mechanical digestion.
Six salivary glands produce saliva with water and mucus to soften, moisten and
lubricate dry food and the oesophagus.
Saliva, which is secreted by the salivary glands into the mouth, contains amylase
which digests (hydrolyses) insoluble starch into soluble maltose.
Masticated food is rolled into food bolus, by the tongue and pushed into the
pharynx.
128
Pharynx
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The region at the back of the mouth cavity where the windpipe, crosses the
oesophagus.

The tongue pushes masticated food into it and the pharynx then contracts by
muscular action to push the bolus into the oesophagus.
Oesophagus
-
The oesophagus is the part that starts in the pharynx and ends with an opening
in the stomach.

It allows the food bolus to move from the mouth to the stomach by peristalsis
(rhythmical contractions).
A sphincter muscle (cardiac sphincter) between the stomach and the
oesophagus stops food from passing back into the oesophagus.

Epiglottis
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It closes the trachea (windpipe) during swallowing to ensure that the food
passes into the oesophagus.
Stomach
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It is the part between the oesophagus and the duodenum. At the entry point of
the oesophagus into the stomach is the cardiac sphincter muscle and at the exit
point of the stomach, where the duodenum is connected, is the pyloric sphincter
muscles
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It holds and stores the food for a while.
The circular muscles causes wavelike contractions to churn up the food and mix
it with gastric juice in the stomach lining.
Gastric juice contains hydrochloric acid (HCI) that kills pathogens such as
bacteria in the food and provide a suitable pH for pepsin.
Pepsin breaks down proteins into polypeptides
HCI can convert sucrose into glucose and fructose.
Some absorption of soluble molecules like glucose, water, alcohol and minerals
occur in the stomach.
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129
The small intestine
The first part is the duodenum
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Chyme (mixture of food) from the stomach passes into it.
Most digestion is completed in the duodenum.
The intestinal glands in the wall of the duodenum release intestinal juice, which
contains the enzymes maltase, lipase and sucrase.
Digestion of fats by lipase starts in the duodenum
Pancreas release the pancreatic juice that contain hydrolytic (enzymes) to break
down polypeptides, starch and fats.
Bile secreted from the liver into the duodenum neutralises the stomach acid.
The second part is the ileum
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The end products of chemical digestion by enzymes end up here and are called
chyle.
Amino acids, glucose, fructose, water- soluble vitamins and minerals are
absorbed into the bloodstream by the villi of the small intestine.
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The large intestine
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The last part of the alimentary canal consisting of the caecum, appendix, colon,
rectum and anus.
It receives indigestible materials from the ileum
Minerals and some vitamins (like vitamin K and B) are absorbed from the large
intestine.
The colon is responsible for absorption of water from the undigested food.
The colon compacts the indigestible residue into semi- solid faeces, which moves
on to the rectum.
To be as released faeces during defecation through the anus, which is controlled
by the anal sphincter muscle.
The caecum is a pouch-like structure that receives the chyle from the ileum.
The appendix is a finger-like tube attached to the caecum that has no proven
function in humans
131
Specific objective: describe cholera as a disease caused by a bacterium.
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Cholera is an infectious disease, which is caused by a bacterium called Vibrio
cholera.
Cholera is contracted by eating food or drinking water that has been
contaminated with faeces from someone infected with this bacterium.
The symptoms of cholera include:
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vomiting
severe watery diarrhea, which can lead to dehydration and loss of salts
Death, if untreated.
Specific objective: explain that cholera bacterium produces a toxin that causes
secretion of chloride ions into the small intestine, causing osmotic movement of
water into the gut, causing diarrhea, dehydration and loss of salts from blood.
Process of how cholera bacteria cause diarrhea
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The cholera bacteria are ingested, live, and breed in the small intestine.
Cholera bacteria produce and release a toxic (poison) into the small intestine.
The toxin causes the epithelial cells of the small intestine to secrete chloride
(salt) ions, which diffuse into the lumen of the small intestine.
The presence of these chlorine ions results in an increased amount of water to
be released from epithelial cells of the small intestines.
This is because water moves by osmosis into the lumen of the intestine (as the
salt level is now higher and the water potential is lower in the lumen).
This increase in water causes watery faeces. It is discharged from the bowels
frequently and in a liquid form, resulting in diarrhea and dehydration.
However, as long as enough fluids are given to replace these water losses, a
person will eventually recover from cholera.
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3.6.2.3 Digestion
Specific objective: state the role of enzymes in digestion with reference to
dependence of enzyme activity on the conditions of pH and temperature.
Role of enzymes in digestion
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Enzymes are protein that function as biological catalysts to speed the rate of
reactions such as digestion.
Enzymes are necessary for digesting food nutrients
Enzymes works best at optimum temperature, which is around 37°C in human.
Each enzyme work best at its optimum pH.
Deviations in pH outside their range can cause the enzymes to denature
The pH differs in different parts of the alimentary canal
Salivary amylase best digests starch into maltose in the mouth at pH 6.5 to 7.2
Pepsin best digests proteins into polypeptides in the stomach at pH 2.5
Pancreatic amylase best digests starch in the small intestine at pH 8
Lipase best digests lipids (fats and oils) into fatty acids and glycerol in the small
intestine at pH 8
Trypsin best digests polypeptides into amino acids in the small intestine at pH 8
Sucrase best digests sucrose into glucose in the small intestine at pH 8
Maltase best digests maltose into glucose in the small intestine at pH 8
133
The diagram and graph showing pH requirements of different enzymes
Specific objective: state the importance of chemical digestion in the alimentary canal
Types of digestion
Mechanical digestion
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A physical break down of food into smaller pieces by chewing, churning,
chopping and smashing in order to provide large surface area for enzymes to act
on.
Chemical digestion
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Chemical digestion begins in the mouth where enzyme amylase breaks down
starch into maltose.
Food containing proteins are digested chemically by enzymes
Chemical digestion breaks large, complex, insoluble nutrients in food into small,
simple, soluble nutrient molecules that can be taken in by cells.
Acids, made by the stomach, and enzymes, secreted by the stomach, chemically
digest nutrients.
Enzymes from the pancreas further digest the carbohydrates, lipids and proteins
found in food in the duodenum into soluble form.
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Structures in the digestive system and types of digestion
Structure
mouth
stomach
small intestine
Function
Teeth grind food
amylase changes starch to
maltose
acids allow pepsin to break down
proteins
contractions smash and squeeze
food
enzymes break down food
nutrients
food mixed, rolled, chopped and
squeeze
Type of digestion
mechanical digestion
chemical digestion
chemical digestion
mechanical digestion
chemical digestion
mechanical digestion
Specific objective: describe the digestion of starch in the alimentary canal
Digestion of starch in the alimentary canal
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Saliva, which is secreted by the salivary glands into the mouth, contains salivary
amylase.
Amylase digests (hydrolyses) insoluble starch into soluble maltose. The digestion
of starch to maltose is the first stage of starch digestion.
Salivary amylase in the mouth starts the breakdown process and further
breakdown occurs in the small intestine, where the remaining starch is digested
by pancreatic amylase.
Once starch has been broken down to maltose, the maltase enzyme (which is
produced in the wall of the small intestine) continues the chemical digestion and
breaks the maltose down to glucose, i.e. a soluble, smaller molecule that can
easily be absorbed.
Maltase is responsible for the final stage of starch digestion.
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Specific objective: describe the functions of a typical amylase, protease and lipase,
listing the substrate and end- products.
Different enzymes and their functions
enzyme
function
substrate
salivary amylase
digests starch into
starch
maltose
Pancreatic
digests starch into
starch
amylase
maltose
maltase
digests maltose
maltose
into glucose
Pepsin (protease) digests proteins
protein
into polypeptides
trypsin (protease) digests
polypeptides
polypeptides into
amino acids
lipase
digests lipids into
lipids
fatty acids and
glycerol
product
maltose
maltose
glucose
polypeptides
amino acids
fatty acids and
glycerol
Specific objective: state the functions of hydrochloric acid in gastric juice, limited to
killing bacteria in food and giving an acid pH environment for enzymes.
Functions of hydrochloric acid in gastric juice (in the stomach)
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The hydrochloric acid in the gastric juice kills bacteria that enter the stomach,
protecting us from harm.
The hydrochloric acid also creates acidic conditions below (pH 2-3) in the
stomach.
These acidic conditions activate inactive pepsinogen into the digestive enzyme
pepsin, which begins protein digestion into polypeptides.
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Specific objective: outline the role of bile in neutralising acidic mixtures and in
emulsifying fats in food
Bile
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Is a dark green to yellowish-brown, alkaline solution produced by the liver and
stored in the gall bladder.
It enters the duodenum through the bile duct. Even in cases where the gall
bladder has been removed by surgery, the liver continues to manufacture bile.
Role of bile in digestion: Neutralising acidic mixtures
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Bile is alkaline, it helps to neutralize stomach acids (gastric acids).
Bile juice reduces the acidity of chyme (the pulpy acidic fluid that passes from the
stomach to the small intestine).
The food coming from the stomach has to be made alkaline to create optimum
conditions for the pancreatic enzymes, e.g. lipase, trypsin and amylase to act on
it.
Role of bile in digestion: Emulsification of fats
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Bile causes emulsification of fats. (Emulsification is the breakdown of large fat
globules into smaller fat droplets to increase its surface area for lipase enzyme to
act on it).
Fats is broken down by lipase into fatty acids and glycerol in order to pass
through the walls of the intestine to be absorbed.
Emulsification of fats
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Specific objective: investigate the effects of bile on fats
Effects of bile salts on fats
Bile is first used to emulsify fats entering the small intestine and lipase is used
afterwards to break down fats into fatty acids and glycerol.
Materials needed
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Four test tubes.
Bile solution.
Lipase solution.
Fat solution (oil).
Procedure
1. Label 4 test tubes A, B, C and D. Divide the fat solution equally into the 4 tubes.
2. In tube B, add about 2 cm3 of bile solution.
3. In tube C, add about 2 cm3 of lipase solution.
4. In tube D, add the 2 cm3 of bile and 2 cm3 of lipase solution.
5. Mix well until everything looks settled.
6. Tube A is your control
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Compare the 3 other tubes to the control and write down your observations.
You should be able to make conclusions about the role of bile in digesting a fatty
solution and the extent of digestion with and without the addition of lipase.
3.6.2.4 Absorption
Specific objective: identify the small intestine as the region for the absorption of
digested food.
The small intestine
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The small intestine is the main region for the absorption of nutrients. It is
responsible for the absorption of over 90% of nutrients from the food.
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Specific objective: describe the significance of villi in increasing the internal surface
area of the small intestine.
Significance of villi
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The primary function of the villi in the small intestine is to increase the
absorption of nutrients from food passing through the small intestine.
Intestinal villi (singular villus) are small, finger-like projections that extend into the
lumen of the small intestine.
Millions of villi increase the surface area.
Micro-villi further increase the surface area.
The ileum contains about 5 million villi
Specific objective: describe the structure of a villus, including the role of capillaries
and lacteals in it.
Structure of a villus
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Villi are finger-like projections that increases surface area for absorption.
Each villus is surrounded by a single layer of epithelium cells. One layer of
epithelial cells minimizes the distance for nutrients to travel. The epithelial cells
contain many mitochondria for active transport.
Each epithelial cells has micro-villi (a brush border) to further increase the
surface area and absorption of nutrients.
Between epithelial cells are goblets, which secrete mucus to keep the gut moist
and dissolve nutrients.
Inside each villus is a single central lacteal (lymph vessel) for absorption of fats,
fatty acid, glycerol and fat-soluble and vitamins.
139
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Inside each villus is a dense network of blood capillaries for the absorption of
digested nutrients, e.g. glucose, amino acids, water-soluble vitamins and
minerals.
Structure of a villus
Specific objective: state the role of the hepatic portal vein in the transport of
absorbed food to the liver.
Role of the hepatic portal vein
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The hepatic portal vein is a major blood vessel in the abdomen, transport
blood from the spleen, stomach, pancreas and intestines to the liver.
All capillaries and venules from these organs join together to form the hepatic
portal vein and transport nutrient-rich blood to the liver.
140
The position of hepatic portal vein in the digestive system
Specific objective: describe the role of the liver in the metabolism of excess
glucose and in the breakdown of excess amino acids.
The role of the liver in the metabolism of excess glucose
High glucose level
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It lowers the level of blood glucose by converting excess glucose into glycogen.
Low glucose levels
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It increases the glucose level by converting glycogen into glucose.
Role of the liver in the breakdown of excess amino acids
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Excess amino acids cannot be stored as amino acids.
In the liver, the nitrogen part from amino acids is removed. This is known as
deamination.
Ammonia is then formed which is poisonous.
The liver immediately converts ammonia into a harmless substance called urea
by combining ammonia with carbon dioxide (CO2)
The rest of the amino acids is converted into fats or carbohydrates for energy
(glucose) and storage (glycogen).
141
Specific objective: describe the role of fats as a storage substance
Fats (a lipid) has many uses in the body.
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Fat serves as the storage substance for the body’s extra kilojoules.
Fat act as a storage substance that fills the fat cells in adipose tissue
(subcutaneous fat) underneath the skin to help insulate the body.
Fats are important to store certain fat-soluble vitamins like vitamins A, D, E and
K, which are insoluble in water.
Fats are stored around some organs (visceral fat) in the abdominal cavity. It
helps protect vital organs such as the pancreas, kidney and liver against physical
trauma to the body.
Fats act as a storage substance
Specific objective: state that water is absorbed in the small intestine and colon.

Water is absorbed in the small intestine and colon into the blood stream
142
3.7 Transport in plants
Specific objective: Identify and describe the structure of a dicotyledonous root
and stem limited to epidermis with root hairs, cortex, phloem and xylem from
photomicrographs and as seen under the light microscope.
Dicotyledonous root
Main functions of roots
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Roots are responsible for the absorption of water and mineral salts/ions (inorganic
substances).
Roots firmly anchor the plant in the soil.
External structure of a dicotyledonous root
The internal structure of a dicotyledonous root
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The epidermis is a single-celled layer which surrounds the outer part of the root.
The epidermis protects the underlying tissues.
The root hairs are finger-like outgrowths of some epidermal cells. They increase
the absorption surface area for water and mineral salts.
An epidermal cell plus its root hair outgrowth is called a root hair cell.
The first layer of cells directly inside the epidermis is the broad cortex. The cortex
allows water and mineral ions to pass through to the xylem and it stores starch.
The innermost layer of the cortex is a single-celled layer called the endodermis.
The endodermis regulates the passage of water from the cortex to the xylem.
143
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The vascular bundle is a central column in the very centre of the root.
The endodermis is found around the central vascular bundle.
Inside the vascular bundle are the xylem vessels and the phloem sieve tubes.
The xylem is located in the centre of the vascular bundle and is arranged in the
form of a cross. The xylem transports water and mineral ions from the roots to the
rest of the plant.
The phloem is found closer to the endodermis between the “arms’ of the xylem
cross. The phloem transports manufactured organic substances from the
photosynthesizing part to the rest to the plant (during translocation).
Between the xylem and phloem is the cambium tissue.
Specific objective: draw and interpret prepared slides of roots and stems, including
the structure of xylem and phloem tissue
Cross and longitudinal section through the structure of a dicotyledonous root
144
Outline (drawing) of a cross section through the structure of a dicotyledonous root
Photomicrograph of a cross section of a root
145
Structure of a dicotyledonous stem
Main functions of stems
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Stems expose the leaves in a favourable position for the absorption of light for
photosynthesis.
They bear the flowers in a favourable position for pollination.
They bear fruits in a favourable position for the dispersal and distribution of seeds.
They transport water and mineral salts from roots to the rest of the plant.
They transport organic substances from the leaves to the rest of the plant.
They store reserve nutrients and water.
External structure of a dicotyledonous stem
The internal structure of a dicotyledonous stem
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The epidermis is a single-celled layer which surrounds the outer part of the stem.
Some epidermal cells are modified to form lenticels (stomata).
The epidermis protects the underlying tissues. The lenticels allow gaseous
exchange.
The region directly under the epidermis is the broad cortex. Intracellular spaces in
the cortex allow gaseous exchange.
The vascular bundles are mostly arranged towards the outside of the stem and
organised into a circle in the cortex,
The reason for this is so that the xylem can give support to the herbaceous stems.
The xylem is always located more on the inside of the vascular bundle, while the
phloem tissue is located on the outside of the vascular bundle.
The xylem and phloem are separated by the cambium.
The region internal to the cortex is called the pith.
146
Outline drawing of a cross section through the structure of a dicotyledonous stem
Photomicrograph of a cross section of a dicotyledonous stem
147
Specific objective: draw and interpret prepared slides of roots and stems, including
the structure of xylem and phloem tissue.
Diagrams of prepared slides of roots showing the xylem and phloem tissue
When making an outline drawing of the above slides, it must look like this:
148
Diagrams of prepared slides of stem showing the xylem and phloem tissue
When making an outline drawing of the above slides, it must look like this:
Specific objective: explain the mechanisms by which water enters a plant,
crosses the root, moves up through xylem vessels, enters leaf cells, and leaves
the plants through stomata
The pathway of water through a plant
Uptake of water by osmosis

Water is absorbed from the soil into the root hairs by osmosis.
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Water moves from a high water potential (Ψ) in the soil to a low water potential (Ψ)
in the root hair cells.
149
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Root hair cells are adapted for this by having a large surface area to speed up
osmosis.

Root hair cells have a partially permeable membrane that permits water molecules
(and minerals) to enter but not to leave.

A root hair cell has a large vacuole to increase the concentration of salts and
sugars, thus increasing the rate of absorption.
Water movement through a plant
Water movement through the root
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Water moves by osmosis from the soil into the root hairs, from where it moves
through the epidermis of the root to the cortex cells.
From the cortex cells, water moves through the endodermis into the xylem vessels.
Flow diagram of water movement through the root
150
Water moves through the root by following three pathways.

The apoplast pathway is where most water travels from cell to cell via the cell walls.
Most water moves this way because of little obstruction and it is the fastest way.

The symplast pathway is where water travels from cell to cell via the cytoplasm
and through the plasmodesmata. This is a slower way of movement.

The vacuolar pathway is where little water passes via vacuoles located in the cells.
This is a very slow movement of water and very little water flow occurs this way
Water movement up the xylem in the root and the stem

Water can move up the stem xylem vessel because of different forces.

Water which is absorbed from the soil into the roots creates a force known as root
pressure, which forces water up the xylem. (If a stem is cut near the roots, water
will seep from the cut surface).

Cohesion is the tendency of water molecules to attract one another. Water
molecules pull up other water molecules and thus moves up the stem xylem.

This pulling effect causes transpiration pull because of evaporation of water
during transpiration, which causes a sucking force for water up the xylem.

Adhesion is the tendency of water molecules to be attracted to molecules of a
different type. Adhesion forces between the water molecules and the walls of
xylem vessels help water to move upwards in the xylem.

This phenomenon is known as capillarity. Water automatically moves up a tube
and the smaller the diameter of the tube, the higher it rises. Xylem vessels have a
diameter of 2 mm, therefore has considerable capillarity force to move water up
the stem.
151
Demonstration of capillarity
Demonstration of root pressure and transpiration pull
Water movement through the leaf

Water moves by osmosis from the stem xylem into the leaf xylem.

From the leaf xylem, water moves to the mesophyll (palisade and spongy) cells by
osmosis.

Water evaporates from the mesophyll cell walls into the air spaces as water
vapour.
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Water vapour diffuses out of the stomata by transpiration which causes
transpiration pull
Movement of water through a leaf
152
Xylem
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Xylem transports water and mineral ions upwards in plants.
It provides support to plants
Xylem consists of long cylindrical hollow tubes connected end to end to form a
long continuous column.
It has no end walls, enabling it to form a continuous tube.
It contains no cytoplasm and organelles (cells are dead) so there is no
obstruction to the flow of water and mineral ions.
Its walls are thickened with lignin (lignified walls) to strengthen it and preventing
the cell from collapsing (it supports the plant)
Xylem walls have unlignified areas called pits for lateral movement of water.
Phloem sieve tubes
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Phloem transports sucrose and amino acid upwards and downwards in plants.
Phloem consists of phloem sieve tubes and companion cells.
Have living cells and contain cytoplasm and organelles.
The tubes are connected end to end for easy flow of sucrose and amino acids.
They contain very little cytoplasm and few organelles to transport sucrose and
amino acids with little resistance.
Organelles are at the edges to give more space for movement of sucrose and
amino acids.
Have end walls with sieve plates (plates with pores) to permit continuous
movement form cell to cell.
Specific objective: investigate, using a suitable stain, the pathway of water through
the above-ground parts of a plant.
For this experiment, you will need:
Three stalks of celery, water, food coloring, three clear plastic or glass cups/beakers
and a knife or scalpel.
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Cut the base (non-leafy end) of each of the fresh celery stalks under water.
Place the cut ends into the cups/beakers of water containing red, green and blue
food dye.
Leave the celery stalks in bright light at room temperature in a breeze (wind).
After a few hours, cut across the stalk with a sharp knife or scalpel and observe
the cut end.
153
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If you leave the celery stalks in the coloured water for a day, the leaves will turn
green, red and blue.
As leaves transpire and lose water to the air, the coloured water is sucked in at the
cut end of celery into the xylem vessel and appears as coloured dots.
The coloured water, which moves from the stem xylem into the xylem of the leaves
makes the leaves to appear green, red and blue.
A cross and longitudinal cut section of celery stalks showing the position of the xylem
(celery left standing in food colouring, causing leaves to turn green, red and blue as seen
below)
154
Specific objective: define transpiration as:
Loss of water vapour from plant leaves by evaporation of water at the surfaces of
mesophyll cells followed by diffusion of water vapour through the stomata.
A closed and open stomata
Specific objective: describe transpiration, and explain how environmental
conditions (temperature, wind speed, humidity and light intensity) affect the rate at
which water vapour diffuses out of stomata, and hence affects rate of water uptake.
Describe transpiration:
-
Transpiration occurs through the stomata of leaves and through lenticels of green
stems
Advantages of transpiration:
- It causes the transpiration pull
- It has a cooling effect on plants
- Transpiration pull brings mineral ions
with the water to the plant
Disadvantages of transpiration:
- The rate of water loss can exceed
the uptake of water and this can lead
to wilting and death of the plant
Environmental conditions which affect the rate of transpiration
Temperature
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An increase in temperature causes an increase in transpiration rate.
Higher temperature causes an increase in kinetic energy of water molecules and
thus increases the rate of transpiration.
Water molecules move faster when heated.
Air with a high temperature can hold more water molecules.
High temperature lowers the relative humidity of air.
A decrease in temperature decrease the rate of transpiration.
155
Wind speed
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Wind (air movement) will increase the rate of transpiration and water loss through
the leaves.
In the absence of any wind, the water vapour which diffuses out of the stomata
accumulates near the leaf surface.
The faster the wind speed, the more rapidly the moist air is removed and the
greater the rate of transpiration. This creates a water potential gradient between
the moist air on the leaf and outside the leaf.
Humidity
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Low humidity increases the rate of transpiration because there is less water vapour
outside the leaf and more water vapour inside the leaf.
This creates a water potential gradient between the moist air in the leaf and dry air
outside the leaf.
When the external air has a high humidity, the gradient is reduced and less water
is transpired.
Light intensity
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Stomata open in light and close in the dark.
Light stimulates the opening of the stomata.
An increase in light intensity increases the transpiration rate.
A decrease in light intensity decrease the transpiration rate.
Specific objective: investigate and compare the relative number of stomata on the
upper and lower epidermis of a leaf using clear nail varnish or water-based varnish
(Refer to previous chapter on procedure)
Another method is to use cobalt chloride paper to show that transpiration occurs
through the stomata
Materials needed:
-
Cobalt chloride paper (special paper that is blue when dry but turns pink when wet)
A healthy tree around the school yard
Tape/rubber band to hold the cobalt paper against the leaf
Stop watch
156
Procedure:
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Take the blue cobalt chloride paper and attach it on the lower surface of three
leaves
Take another cobalt chloride paper and attach it on the upper surface of three
identical leaves on the same tree
Use the rubber band or tape to keep the paper attached onto the leaf while
observing changes
Observation:
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The paper attached on the lower leaf surface will turn pink faster than the one
attached to the upper leaf surface
Suggest a reason for the results observed in terms of stomata distribution
Specific objective: investigate and describe the effects of variation of temperature
and humidity on transpiration rate (use a simple photometer).
How to use a simple potometer to investigate and describe the effects of
variation of temperature and humidity on transpiration rate
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The more water is lost during transpiration, the faster the uptake of water occurs.
Transpiration speed in a cut leafy shoot can be measured by means of a
potometer.
To be strictly accurate, the potometer measures the rate of water uptake of a shoot,
but in practice, this is almost the same as the rate of transpiration.
A little of the water taken up can be used by the plant for photosynthesis and other
metabolic processes, but most water is transpired.
What is a potometer?
This is an instrument used to determine the rate of transpiration in a plant by
measuring water uptake.
157
Precautions when setting up a potometer
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Use a healthy, undamaged non-wilted shoot.
Cut the shoot under water to make sure no air goes into the xylem.
Cut the shoot at a slant to increase the surface area.
Insert the shoot into the potometer under water.
Make sure the entire apparatus is full of water – no airlocks or bubbles.
Dry off the leaves.
Ensure air and watertight joints
Allow time for the shoot to acclimatize and become adjusted to the factor being
investigated.
Keep the environmental conditions constant.
Measure the distance travelled by the bubble in the calibrated tube in a given time
Move the air bubble back to the beginning by opening the tap.
A bubble potometer
Variation of temperature affects transpiration rate
-
Increasing the temperature is predicted to cause an increase in the rate of
transpiration.
The effect of temperature variation can be tested experimentally by using heaters.
158
Using a heater to show the effect of temperature on transpiration
Variation of humidity affects transpiration rate
-
Increasing humidity is predicted to cause a decrease in the rate of transpiration.
The effect of humidity can be tested experimentally by enclosing the plant in a
transparent plastic bag.
Using a transparent plastic bag to show the effect of humidity on transpiration
Specific objective: describe how and why wilting occurs.

Wilting occurs when the rate at which plant leaves lose moisture into air, in a process
called transpiration, exceeds the ability of plant roots to supply enough water.

Wilting refers to the loss of rigidity of non- woody parts of plants.

This occurs when the turgor pressure in non- lignified plant cells falls towards zero.

The cells lose their turgidity and become flaccid.

Wilting is one of the first signs of plant stress and most often indicates a lack of soil
moisture.

Mild wilting typically does not permanently harm a plant if it is corrected immediately.
159
Why wilting occurs in plants
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When the rate of water loss from the plant is greater than absorption of water in
the plant, wilting occurs.
If the soil around the roots of a plant lacks water, the rate of water loss is faster
than the rate of absorption of water.
A plant wilts because of shortage of water, or because there is no water in the
central vacuole over a period of time when the plant really needs water.
Specific objective: discuss ways in which xerophytes can reduce water loss, with
reference to two locally occurring examples (e.g. Aloe; Euphorbia; Quiver tree).
Xerophytes
Xerophytes are plants that need very little water and are adapted for growing in very
dry (arid) places, where water is very scarce.
Xerophyte plants
160
Adaptations of xerophytes to reduce water loss (xeromorphic features)
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Some xerophytes have thick succulent stems to store water, e.g. Euphorbia and
quiver tree.
Some xerophytes have thick succulent leaves to store water, e.g. Aloe.
Xerophytes have a reduced number of stomata, so there are fewer openings to
cause loss of water.
The stomata are situated in pits (sunken)
The surface (epidermis) of the leaf has a layer of hairs that can trap water vapour
to reduce water loss.
Leaves may be rolled up (curled) to expose the outer waterproof cuticle while
stomata are hidden inside.
Leaves may be reduced in size (smaller leaves) to have less surface area for loss
of water. Smaller leaves have fewer stomata.
Leaves may be reduced to spines (thorns), so there is less surface area for loss
of water
Some plants lose their leaves during the winter (deciduous)
Specific objective: define translocation in terms of:
The movement of sucrose and amino acids from the region of production or of storage
to the region of utilisation or demand in respiration or growth.
-
Translocation is the transport of organic molecules (sucrose and amino acids) in
the phloem from a region of production e.g. leaves to regions of utilization, like
areas that demand it for growth or respiration (e.g. flowers) or for storage (e.g.
roots.).
Specific objective: describe the translocation of applied systemic pesticides in
phloem throughout the plant.
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Systemic pesticides are sprayed onto plants.
The pesticides are absorbed into the leaf of the plant through the cuticle and
stomata.
The pesticides will be translocated in the phloem.
The pesticides move into the phloem by active transport.
Aphids and other sucking plant pests use their proboscis to suck plant sap (juices)
with amino acids and sucrose from the phloem.
The pests also remove poisonous sap from the phloem.
Once an insect have ingested enough pesticide it will die.
161
Figure above shows an aphid feeding on plant stem by inserting its tube-like mouthpart
(proboscis) into phloem sieve tubes and sucking out plant sap with sucrose and amino
acids:
3.8 Transport in animals
3.8.1. Heart and lymphatic vessels
Specific objective: describe the heart in terms of its gross structure and the
functions of the parts
-
The circulatory system, also called the cardiovascular system, is an organ
system that permits blood to circulate around the body.
The circulatory system is composed of the heart, arteries, capillaries, veins and
blood.
The circulatory system also includes the lymphatic system, which circulates
lymph. The lymph, lymph nodes and lymph vessels form the lymphatic system.
162
Gross structure of the heart
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The major part of the heart is made up of cardiac muscles called the
myocardium.
The heart has four chambers –two atria and two ventricles.
The atria are the receiving chambers
The ventricles are the discharging chambers.
The upper, thin-walled chambers are the left and right atria (singular-atrium).
The two atria are separated by the septum.
Each of the atria opens into a thick-walled chamber, the ventricle, below.
The left and right ventricles are separated by the interventricular septum
The blood is prevented from flowing backwards by four sets of valves.
Between the right atrium and left ventricle is the bicuspid or mitral (two-flap)
valve.
The bicuspid (mitral) valve prevents backflow of oxygenated blood from the left
ventricle to the left atrium.
The bicuspid and tricuspid valves are known as atrioventricular valves.
The flaps of these valves have strings called tendons or cords.
These tendons link the valves to little bumps of muscle protruding from the inner
surface of the ventricles, called papillae. The tendons prevent valves from turning
inside out.
At the start of the aorta is the aortic semi-lunar valve.
At the start of the pulmonary artery is the pulmonic semi-lunar valve.
The heart muscle is supplied with blood from coronary blood vessels.
The coronary arteries provide the heart muscle with glucose and oxygen from the
aorta.
163
Internal structure of the heart and direction of blood flow
Functions of the different parts of the heart.
The heart pumps blood through the circulatory system all around the body. Blood
enters the atria from large veins and leaves via the ventricles into arteries.
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The pulmonary veins carry oxygenated blood from the lungs to the left atrium.
The left atrium receives oxygenated blood from the pulmonary veins and pumps
it through the bicuspid (mitral) valve into the left ventricle.
The left ventricle receives oxygenated blood from the left atrium and pumps it
through the semi-lunar valve (aortic valve) into the aorta.
The aortic valve (semi-lunar valve) prevents backflow of blood from the aorta to
the left ventricle.
The aorta carries oxygenated blood from the left ventricle to the body.
The vena cava inferior carries deoxygenated blood from the lower body to the
right atrium.
The vena cava superior carries deoxygenated blood from the upper body to the
right atrium.
164
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The right atrium receives deoxygenated blood from the vena cava and pumps it
through the tricuspid valve into the pulmonary artery.
The pulmonary artery carries deoxygenated blood from the right ventricle to the
lungs.
The pulmonic valve (semi-lunar valve) prevents backflow of blood from the
pulmonary artery to the right ventricle.
Specific objective: draw the external structure of the mammalian heart
A photograph of the external structure of the mammalian heart
Class activity
Observe the heart on the photograph above and label the parts A to G below
165
Drawing the external structure of the heart should look like this:
Specific objective: outline how the structure of arteries, veins and capillaries are
adapted for their functions.
The structure of blood vessels and their functions
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Blood vessels are tubular structures that transport blood throughout the human
body.
There are three major types of blood vessels: arteries, capillaries and veins.
Arteries
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Arteries have a diameter of up to 3 cm (aorta) and are not permeable.
Blood flows through arteries under high pressure (10 – 16 kPa).
Blood moves in pulses and flows rapidly.
Arteries are capable of constriction.
Small arteries are known as arterioles
Arteries are the blood vessels that carry blood away from the heart to the body.
All arteries transport oxygenated blood, except the pulmonary artery and the
umbilical artery (in the umbilical cord) which transport deoxygenated blood.
166
An artery
Veins
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Veins have a diameter up to 2.5 cm (vena cava inferior) and are not permeable.
Small veins are known as venules.
Blood flows through veins under low pressure (1 kPa).
There are no pulses in veins and blood flows slowly.
Veins are the blood vessels that carry blood to the heart from the body.
All veins transport deoxygenated blood, except the pulmonary vein and the
umbilical vein (in the umbilical cord) which transport oxygenated blood.
A vein
167
Valves in veins
Capillaries
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Capillaries have an average diameter of 8 micrometer and are permeable.
Blood flows through capillaries under reduced pressure (4-1 kPa).
There is no pulse and blood flows slowly.
They are the smallest blood vessels and are considered to be the endings of
arteries and beginnings of veins
They link arteries and veins.
A capillary
168
The table Shows adaptations of blood vessels.
Structure of arteries
How structure adapts for function
Thick muscular walls
Single layer of epithelial cells inside lumen
Thick layer of smooth muscle and elastic
tissue
Small lumen
To withstand high blood pressure and
prevent bursting
Reduces friction inside
To stretch outwards and recoil with the
pressure of the heartbeat
To maintain high blood pressure
Structure of veins
How structure adapts for function
Strong walls with collagen
Large lumen, relative to thickness of wall
One-way valves
Thinner walls
Single layer of epithelial cells inside lumen
Prevent bursting
Blood flows easily and freely
Prevent backflow of blood
Allow skeletal muscles to squeeze vein
causing blood to move forward
Reduces friction inside
Structure of capillaries
How structure adapts for function
169
Numerous (many-100 000 km of
capillaries)
Very, very thin walls – one cell layer thick
Provide large surface area and are close
to all body cells
To provide short pathway for easy and
fast diffusion of different materials to and
from the blood
For exchange of materials to and from
blood, for white blood cells to squeeze
out and tissue fluid to leak out
Reduces friction inside
Small endothelial gaps between cells of
wall
Smooth inner endothelial surface
Changes in pressure of blood as it flows through blood vessels
Blood pressure is the pressure, which the blood exerts against the wall of the blood
vessel through which it flows. Arteries have the highest blood pressure and veins have
the lowest.
170
Specific objective: describe the double circulatory system of a mammal.

A double circulatory system refers to a system in which blood is pumped through
the heart twice during one complete journey around the body.
-
The deoxygenated blood arrives at the heart from the body, is then first pumped
into the lungs where it becomes oxygenated and then goes back into the heart,
before it is finally pumped into the rest of the body.
The advantage of a double circulation is that oxygenated and deoxygenated
blood is separated.
-
Follow the blood flow from the kidney back to the kidney, showing how blood travels
through the heart twice:
Kidney→ renal veins → vena cava inferior → heart→ pulmonary arteries →
lungs → Pulmonary veins→ heart →Aorta artery → renal arteries → kidney.
Remember: In blood flow around the body there is no short-cut, blood must flow
through the heart twice!
171
Double circulatory system
Specific objective: Investigate and state the effect of physical activity on the
pulse rate
Taking pulse rate (heart rate)
Your pulse is the rate at which your heart beats. As your heart pumps blood
through your body, you can feel a pulsing in some of the arteries close to the
skin’s surface.
The usual resting pulse for an adult is 60 to 100 beats per minute.
172
Checking your pulse in the radial artery (radial pulse).







Find a watch that shows seconds or a digital watch and put it on the table next to
you.
Turn your left hand so that your palm is face-up.
Find your pulse by placing the index finger, middle finger and sometimes even
the ring finger of your right hand in the dent at the wrist on the radial artery. Do
not press too hard, that will make the pulse go away. Use pressure. It can take
several seconds and several micro-adjustments in the placement of your fingers
on wrist to find your pulse.
Keep moving your fingers down or up your wrist in small increments (and
pausing a few seconds) until you find it.
After finding your pulse, count the number of beats for 20 seconds.
Multiply by 3 to get your pulse rate in beats per minute, e.g. 26 x 3 = 78
beats/min (or just count for 60 seconds for a math-free option).
Do not use your thumb if you take someone else’s pulse rate, because the thumb
has its own pulse that you may feel.
Taking the pulse rate at the radical and carotid arteries
Checking your pulse in the carotid artery in the neck




You can also check your pulse in the carotid artery.
This is located in your neck, on either side of your windpipe.
To check your pulse in your carotid artery, place your index and middle fingers
on your neck to the side of your windpipe. When you feel your pulse, follow the
same procedures as above.
Be careful when checking your pulse in this location. If you press to hard, you
may become lightheaded and fall down.
173
Investigate the effect of exercise on pulse rate of a human
TAKE NOTE: Do not attempt this activity if you have a health problem or if you are
recovering from an illness.
Materials
A stopwatch
Procedure







Read through and familiarize yourself with all procedures before starting.
Sit down comfortably on a chair. Take 5 minutes to settle. Locate your pulse.
Count the number of pulses per minute and record. Count the number of pulses
in 20 seconds and multiply this value by 3 to convert it to beats/ min (bpm).
Repeat twice and calculate the average number of pulses per minute and record.
This is called the resting pulse rate.
Stand up. Immediately measure your pulse rate and record. Do not count for
longer than 20 seconds, because your heart rate begins to slow down as soon as
you stop exercising.
Walk briskly for 5 minutes. Immediately measure the pulse rate and record. Do
not count your pulse rate for longer than 20 seconds.
Walk briskly for 5 minutes. Immediately measure the pulse rate and record.
Again, count for only 20 seconds.

Run for 5 minutes. Immediately measure the pulse rate and record. Remember,
count only for 20 seconds.




Compare the pulse rates after the different levels of exercise.
Replicate the investigation or cross-reference your results with other groups.
Record your results in a table and draw conclusions from the results
A bar chart of the results should be drawn.
Activity
Pulse in beats per minute
(bpm)
Sitting down
Stand up
Gentle walking
Brisk walking
Running
174
Specific objective: describe the likely causes of a heart attack limited to diet,
smoking, stress, age and gender.
Heart attack (cardiac arrest)
The terms “cardiac arrest” and “heart attack” are often used interchangeable, but
they are actually two distinct conditions.
-
-
A heart attack is when a coronary artery suddenly becomes blocked, stopping
the flow of blood to the heart muscle and damaging it. All or part of the heart
muscle becomes cut off from its oxygen supply. Left without oxygen, the heart
muscle is injured.
A cardiac arrest is when the heart malfunctions and suddenly stops beating. The
treatment for a cardiac arrest is to, as soon as possible, deliver a large electrical
shock to the heart.
Likely causes of heart attacks
Poor diet





Eating a diet with a high level of saturated animal fat results in high cholesterol in
the blood and body.
It can cause damage of the endothelium layer lining the inside of the artery walls.
Cholesterol is also deposited inside the walls in coronary arteries, which narrow
and eventually become blocked, leading to heart attacks.
Too much salt can raise blood pressure, which can increase a person’s risk of a
heart attack.
Obesity (being overweight) is also linked to heart attacks. The increased flow
often leads to high blood pressure (hypertension).
Stress (tension, anxiety, fear)





High levels of stress make other risk factors (such as high cholesterol or high
blood pressure) worse.
Intense anger and grief can cause a heart attack. It probably occurs from a
sudden increase in heart rate and blood pressure triggered by the emotion
Chronic stress that causes an increase in heart rate and blood pressure may
damage the artery walls.
During moments of high stress, your body releases stress hormones and if your
body is constantly exposed to stress hormones, it can contribute to heart attacks.
People may drink too much alcohol or smoke cigarettes to “manage” their chronic
stress however, these habits can increase blood pressure and may damage
artery walls.
175
Smoking (cigarettes, other tobacco products and passive smokers





Smoking decreases oxygen to the heart muscle.
It also increases blood pressure and heart rate.
Smoking damages endothelial cells that line the inside of the coronary arteries.
Smoking cause platelets to clump together, increasing the risk of a blood clot.
It can cause the coronary arteries to spasm, cutting off blood supply to the heart
muscle.
Age and gender



As you get older, your risk of a heart attack increases.
As you age, so do your blood vessels. They become less flexible, making it
harder for blood to move through them easily. Fatty deposits called plaques also
collect along your artery walls and slow the blood flow to the heart.
Men have a greater risk of heart attacks than women do and they have attacks
earlier in life.
Preventative measures for heart disease: (additional information)





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

Follow a low cholesterol diet (less food rich in saturated animal fats)
Avoid being obese (overweight).
Reduce stress levels.
Do not smoke.
Exercise regularly (walking cycling, swimming – active sports).
Reduce salt intake.
Avoid excessive sugar intake.
Reduce alcohol consumption.
Specific objective: state ways of treating coronary heart disease limited to drug
treatment with aspirin and surgery (stents, angioplasty and bypass).
Treating coronary heart disease
Your heart’s coronary arteries can become narrowed or blocked by the build-up of
the cholesterol-laden plaques that form due to atherosclerosis.
176
Treating coronary heart disease: Drug treatment with aspirin



People at high risk of a heart attack should take a daily low dose of Aspirin (if told
to by their doctor). Heart attack survivors also regularly take a low dose of
Aspirin.
Aspirin also helps prevent blood clots. Aspirin works by “thinning” the blood and
preventing the formation of clots. Some chemicals in the blood trigger events that
cause blood clots.
When Aspirin stops those chemicals, it helps slow the formation of the clots.
Treating coronary heart disease: Angioplasty


Angioplasty is a procedure in which a narrowed section of the coronary artery is
widened or a blocked coronary artery opened, by putting in a balloon stent to
restore normal blood flow to the heart muscle.
It is not a major surgery. Angioplasty can also be used during a heart attack to
quickly open a blocked artery and reduce the amount of damage to your heart.
Treating coronary heart disease: Heart bypass surgery



In heart bypass surgery, a surgeon takes blood vessels from another part of the
patient’s body to go around, or bypass, a blocked artery.
The result is that more blood and oxygen can flow to the heart muscle again,
lowering the risk of a heart attack.
The type of bypass surgery depends on how many of the arteries are blocked. A
single bypass (only one artery is blocked), a double bypass (two arteries are
blocked), a triple bypass (three arteries are blocked)
177
Specific objective: state that the lymphatic system consists of lymphatic vessels
and lymphatic nodes
The lymphatic system
-
-
The lymphatic system is part of the circulatory system and is a vital part of the
immune system, comprising a network of lymphatic nodes and lymphatic
vessels throughout the body.
The tonsils, spleen and thymus are all part of the lymphatic system.
Diagrams showing some lymph nodes
Specific objective: describe the functions of the lymphatic system limited to the
protection of the body from infection and circulation of body fluids.
Functions of the lymphatic system



The lymphatic system collects and returns tissue fluid to the blood circulation.
This prevents oedema (accumulation of too much tissue fluid).
Lymph nodes are the major site of lymphocyte production, which produces
antibodies to defend the body against infectious diseases.
The lymphatic system also absorbs and transports fats, fatty acids and
glycerol from the ileum (lacteals in ileum) to the blood circulation.
178
3.8.2 Blood
Specific objective: identify red and white blood cells, as seen under the light
microscope, on prepared slides, in diagrams and photomicrographs.
Bloods cells as seen under a light microscope
Blood cells as seen from prepared slides and drawings
179
Specific objective: state the functions of the components of blood limited to red
blood cells, white blood cells, platelets and plasma.


Blood is made up of different types of cells: red blood cells, white blood cells and
platelets.
All of these cells are floating in a liquid called plasma.
Functions of red blood cells


-
To transport oxygen to all the tissues from the lungs. Red blood cells contain
haemoglobin, which binds with the oxygen to form oxyheamoglobin, enabling the
transport of oxygen.
Red blood cells also transport carbon dioxide back to the lungs.
Some carbon dioxide produced in the tissue cells diffuses into the blood plasma.
The largest fraction of carbon dioxide diffuses into the red blood cells and
combines with haemoglobin.
Functions of white blood cells



White blood cells protect the body from infection.
Neutrophils are white blood cell forming an essential part of the body’s defense
system by carrying out phagocytosis. Phagocytosis is the ingestion and digestion
of pathogens
Lymphocytes are white blood cells that function as part of the immune system
which produces antibodies, which are proteins that specifically target bacteria,
viruses and other foreign materials.
Functions of platelets

The primary function of platelets is to aid in the blood clotting process (or
coagulation), by gathering at the site of an injury and forming a scab, which
covers the wound and prevents blood from leaking out
Plasma
-
Blood plasma is a yellowish, liquid component of blood and a mixture of many
different substances. Plasma the main transporter of different substances to and
from the cells.
180
Functions of plasma

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The plasma transport blood cells (white blood cells, platelets and red blood cells
suspended within it) throughout the body.
Plasma is the main transporter of nutrients to the cells. The soluble products of
digestion, e.g. glucose, amino acids, fatty acids and glycerol, ions (salts) and
vitamins are transported in the blood plasma.
Plasma also transports waste products like urea from the liver to the kidneys,
where they are excreted from the body in urine.
Plasma also transports waste products like carbon dioxide from the cells to the
lungs to be excreted during exhalation.
Plasma transports chemical messengers such as hormones from endocrine
glands to target organs.
Plasma also transports antibodies and blood clotting proteins (fibrinogen) around
the body.
Specific objective: State the function of lymphocytes and phagocytes as seen
under light microscope, and in diagram and images


Phagocytes (Neutrophils) are white blood cells forming an essential part of the
body’s defense system by carrying out phagocytosis. Phagocytosis is ingestion
and digestion of pathogens
Lymphocytes are white blood cells that functions as part of the immune system
by producing antibodies, which are proteins that specifically target bacteria
viruses and other foreign materials.
181
Specific objective: describe the process of clotting limited to the conversion of
fibrinogen to fibrin only.
Process of blood clotting


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


A blood vessel is cut or damaged. blood plasma, blood cells and platelets flow
from a cut vessel.
Platelets in the blood are activated by vitamin K and calcium ions.
Platelets release an enzyme which acts on soluble fibrinogen in the blood
plasma.
Soluble fibrinogen becomes insoluble threads of fibrin.
Red blood cells become trapped in the mesh of fibrin and dry out.
This causes a scab to form and blood is stopped from flowing out
How different parts of the blood help to form a clot when a tissue is damaged
Specific objective: state the role of blood clotting.
Role of blood clotting



If the blood vessels in the skin are damaged (cuts or extensive burns), the way is
open for bacteria, other pathogens and outside particles to enter the blood
stream, which could cause infectious diseases and infection.
Damaged blood vessels in the skin could endanger life due to extensive blood
loss, so clotting prevents blood loss after an injury to keep a person from
bleeding to death.
Blood clotting helps to seal wounds rapidly until a more permanent repair is
produced.
182
Specific objective: describe the transfer of nutrients between capillaries and tissue
fluid (details of roles of water potential and hydrostatic pressure are not required).
The transfer of nutrients between capillaries and tissue fluid
-
-




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


Tissue fluid is formed when blood plasma, the liquid part of blood inside blood
vessels, leaks out and escapes from capillaries to surround and bathe the cells
by filling the spaces between the cells.
Some of the tissue fluid returns directly back to the capillaries. The rest of the
tissue fluid returns to the circulation as lymphoma in the lymphatic vessels, to
become tissue fluid again.
The tissue fluid which leaks out of the capillaries contains amino acids, glucose,
minerals, vitamins and oxygen. It is the medium through which substances are
exchanged between the blood and the cells.
The fluid leaks from capillaries through tiny gaps between the cells of the wall of
the capillaries.
The gaps allow small molecules to pass through, but keep back red blood cells
and large plasma proteins.
Tissue fluid contains carbon dioxide and other waste products from cells which
are carried away in the bloodstream.
Tissue fluid maintains a relatively constant environment around body cells,
helping them to function efficiently.
Not all tissue fluid returns to the blood in the capillaries. Excess fluid is returned
to the blood via the lymphatic system as lymph.
The lymphatic system is a network of vessels called lymphatic or lymph vessels
which are spread all over the body.
It does to form a continuous circulatory system and lymph is carried in one
direction only, namely away from the tissues.
Lymph vessels contain one-way valves to prevent backflow of lymph.
Lymph moves because of muscle contractions which compress lymph vessels,
exerting pressure to move lymph forward.
183
Lymphatic vessels
184
3.8.3 Defence against Diseases
Specific objective: define pathogen as:
A disease- causing organism. For example bacterium, virus. Fungus or
protozoan
Specific objective: define active immunity as:
The defence against a pathogen by antibody production in the body
Immunity
Immunity is the state in which the human body is protected from the invasion of
pathogens and therefore infectious diseases.
Specific objective: state the body’s defence mechanisms limited to mechanical
barriers, chemical barriers, antibody production and phagocytosis (white blood
cells).
The body’s defence mechanisms
Mechanical barriers
Mechanical barriers represent the first line of defence for the body.

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The skin, when intact, is a very effective physical or mechanical barrier to block
the entry of pathogens into the body.
The mucous membrane of various body systems such as the respiratory, urinary,
digestive tracts and eyes prevent invasion by microorganisms with the help of
their epithelium and mucous secretions, which form a protective covering.
The conjunctiva of the eyes lines the interior surface of each eyelid and the
exposed surface of the eyeball.
The respiratory system also forms part of the mechanical barriers against
pathogens. The respiratory epithelium traps the microorganism less than 10 µm
in diameter and transports them by ciliary action away from the lungs.
Coughing and sneezing also help with the removal of microorganisms from the
respiratory tract.
185
Chemical barriers
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
Pathogens may manage to reach the stomach but many of them are
destroyed by the gastric juice of the stomach.
Kidneys, ureters and urinary bladder are sterile under normal conditions. The
lower urinary tract is flushed with urine which eliminates potential pathogens.
The acidic environment (pH3 to 5) of the vagina also provides defence, as it is
unfavourable and inhospitable for many pathogens.
Lysozyme, an enzyme produced in tears, perspiration (sweat) and saliva can
break down peptidoglycan cell walls of bacteria and thus acts as an antibiotic
(kills bacteria).
White blood cells
-
-
Cells of the immune system that originate from bone marrow, where stem cells
divide by mitosis to produce cells that differentiate into different white blood cells
called lymphocytes and phagocytes (neutrophils).
Some lymphocytes mature in the bone marrow (B lymphocytes) and some
mature in the thymus gland (T lymphocytes)
Antibody production

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
When lymphocytes come in contact with pathogens, e.g. bacteria and viruses,
some B lymphocytes differentiate into plasma cells, which produce antibodies.
Pathogens have antigens which cause an immune response in the host
organism.
Antigens are substances on pathogens that cause an immune system to produce
antibodies against it.
Antibodies combine with the antigen to form an antigen-antibody complex which
makes the pathogen harmless or inactivates it.
Antibodies have a Y–shaped molecule with a receptor site to bind a specific
antigen (like a lock and key).
Antibodies are specific to each antigen.
186
Lymphocytes produce antibodies

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
Some lymphocytes differentiate into memory cells to provide long-term immunity.
These memory cells remain in the lymph nodes after an infection, to produce
plasma cells very fast when re-infected
The plasma cells can then quickly produce antibodies before symptoms of the
disease are shown.
Phagocytosis



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


Is the process of engulfing an object, e.g. a pathogen, and digesting it.
Neutrophils (phagocytes) which are produced and matured in the bone marrow
carry out phagocytosis.
There are receptor proteins on the cell membrane of a neutrophil.
The neutrophil moves towards the pathogen, e.g. bacterium or virus, and when
the pathogen comes into contact with its cell membrane, it sticks to the receptor
protein.
The neutrophil membrane infolds, then flows around and engulfs the pathogen.
The cell membrane then pinches off and forms a vacuole (a vesicle called a
phagosome) inside the neutrophil.
Lysosomes then fuse with the vacuole and enzymes are secreted into the
vacuole by these.
The enzymes digest (dissolve) the pathogen into amino acids, glucose, fatty
acids and glycerol, which are absorbed into cytoplasm of the cell and are used.
187
Phagocytosis
Specific objective: explain the role of vaccination in controlling the spread of
disease.
Vaccination
-
-
Vaccination (inoculation) is the treatment with a vaccine to stimulate an
individual’s immune system to develop acquired active immunity against a
pathogen and therefore a disease.
A vaccine can be administered as an injection or given orally as drops.


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
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
A vaccine contains weakened or dead pathogens, e.g. bacteria or viruses,
or poisonous toxins (made harmless) produced by pathogens.
These pathogens and toxins do not cause infectious diseases
(occasionally they cause minor symptoms of diseases).
The surface antigens are still present on the weakened pathogens, or on
the harmless toxins.
This causes differentiation of lymphocytes into plasma cells.
This process causes antibody production by plasma cells.
Some plasma cells differentiate into memory cells.
Memory cells ‘remember’ the antigen and rapidly release high levels of the
correct antibodies when the real pathogen enters the body (secondary
response).
188

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
The antibodies are specific to the pathogen (antigen-antibody specificity).
The antibodies attach to the surface antigen of the pathogen, preventing it
from causing the disease.
The person is said to be immune to the disease.
Specific objective: define passive immunity as:
A short-term defence against a pathogen by antibodies acquired from another
individual.
Specific objective: describe the events leading to passive and active immunity.
Passive immunity:
-
Passive immunity gives a person short-term protection against some pathogens
and infectious diseases because lymphocytes cannot produce antibodies.
The antibodies must thus be acquired from another individual.
Events leading to passive immunity
Event one

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
During pregnancy, antibodies cross the placenta from the mother to the foetus.
These antibodies are present in the mother’s blood from previous infections and
immunisations she had.
After birth, the baby also gets antibodies when feeding on breast milk.
Babies acquire immunity from their mothers by getting ready-made antibodies.
The borrowed antibodies eventually disappear, so it gives a person only
temporary protection against diseases.
Event two




Horses are injected with small doses of snake venom, as they can withstand high
doses of these venoms.
The horses then produce antibodies against the snake venom. These antibodies
are then extracted (it is now known as antiserum).
Snakebite victims can be injected with these antibodies to give them temporary
immunity and time for their own immune system to respond to the venom.
The borrowed antibodies eventually disappear, so it gives a person only
temporary protection against venom.
189
Active immunity
-
Active immunity gives a person long-term protection against some infectious
diseases.
Some lymphocytes differentiate into memory cells so you can produce your own
antibodies before symptoms of a disease are shown.
Events leading to active immunity
Event one




A person is exposed to a live pathogen (e.g. bacterium or virus) as a child and
develops an infectious disease such as chicken pox, mumps or measles.
During the recovering period from a disease such measles, chicken pox or
mumps, the lymphocytes produce the correct antibodies.
Some lymphocytes become memory cells so that if the same pathogen enters
again, the lymphocytes will release the correct antibodies very quickly to prevent
the symptoms of the disease.
This immunity causes the body to make its own antibodies, is long lasting
Event two





A person is vaccinated (immunised or inoculated) against infectious diseases
such as polio, meningitis or tuberculosis with an oral vaccine or an injection.
A vaccine can be in a harmless, weakened form of a disease-causing pathogen,
or a killed pathogen, or the inactivated toxin from the pathogen.
Lymphocytes produce the correct antibodies in response to the vaccine the
person received.
Some lymphocytes become memory cells so that if the real pathogen enters, it
will release the correct antibodies very quickly to prevent the symptoms of the
disease.
This immunity causes the body to make its own antibodies, is long lasting
190
Specific objective: State that memory cells are not produced in passive immunity.




In passive immunity a person is given ready-made antibodies. The borrowed
antibodies eventually disappear. It only gives you temporary protection against
diseases.
Only antibodies can be passed across the placenta to the foetus during
pregnancy, and to the baby through breast milk while feeding.
Memory cells do not go to the baby, so immunity is temporary.
The body does not have memory cells to produce more antibodies.
Specific objective: explain the importance of passive immunity for breast-fed infants.
Importance of passive immunity for breast- fed infants



Passive immunity is useful, because the baby’s immune system is not yet well
developed for the first few months of its life, the mother’s antibodies can protect
the baby against any disease she is immune to.
The bulk of antibodies in breast milk the infant consumes adhere to the gut wall,
where they form a strong line of defence against a range of pathogens.
Antibodies can move throughout the body, searching for any pathogens that may
harm the body.
Specific objective: explain how passive immunity is a short-term defence against a
pathogen.



Passive immunity is a short-term defence against a pathogen through antibodies
acquired from another individual, such as from mother to infant.
However, the effects are immediate, but only last for a short time (a few weeks to
3 months)
The antibodies will eventually break down when the body does not produce its
own memory cells or lymphocytes to clone them.
191
3.9. Gas exchange in humans
Specific objective: distinguish between breathing and respiration


Breathing is a physical process of inhaling oxygen-rich air from the atmosphere
into the lungs and exhaling carbon dioxide-rich air from the lungs into the
atmosphere
Respiration is the release of energy from breaking down of food substances
(glucose) in all living cells.
Specific objective: list the features of gas exchange surfaces in humans limited to
large surface area, thin surface, good blood supply and good ventilation with air
Alveoli (alveolus) - Is where gas exchange takes places.
 Alveoli are the gas exchange surface in humans.
Its features:





Very large surface area
Thin walls (one cell thick)
Moist surface
Good blood supply
Good ventilation with air
Alveoli structure
192
Respiratory system in humans
Specific objective: state the functions of internal and external intercostal
muscles and cartilage in the trachea
External intercostal muscles


Contract to move the rib cage upwards and outwards in order to allow oxygenrich air to flow into the lungs.
It relax to move the rib cage inwards and downwards in order to allow carbon
dioxide-rich air to flow out of the lungs.
Internal intercostal muscles


Relax to move the rib cage upwards and outwards in order to allow oxygen richair to flow into the lungs.
It contracts to move the rib cage inwards and downwards in order to allow carbon
dioxide rich-air to flow out of the lungs.
Diaphragm


Contract and flattens to increase the volume of the chest cavity in order to allow
oxygen-rich air to flow into the lungs.
It relaxes and forms a dome-shape to decrease the volume of the chest cavity in
order to allow carbon dioxide-rich air to flow out of the lungs.
193
Cartilage in the trachea (cartilage ring in the trachea)

Keep the trachea (windpipe) open to prevent it from collapsing during breathing.
Breathing in
Breathing out
Specific objective: explain the differences in composition between inspired and
expired air.
Composition of inspired and expired air
Gas
nitrogen
% in inspired air
78 %
% in expired air
78 %
oxygen
21 %
16 %
Carbon dioxide 0.04 %
4%
Temperature
warmer
Variable
194
Reason for difference
It is insoluble, not
absorbed into the blood
and not needed for
respiration
It is soluble, absorbed
into the blood and used
in respiration
It is a waste product of
respiration which goes
from the alveoli out of the
lungs
The body heats up the
air to body temperature
Water vapour
variable
saturated
It is a product of
respiration which
evaporates from the
surface of the alveoli
Specific objective: investigate the differences in carbon dioxide concentration in
inspired and expired air, using limewater and or hydrogen carbonate indicator
solution


Lime water is used to test for the presence of carbon dioxide
Clear lime water turns milky/cloudy in the presence of carbon dioxide
Practical procedure
1. Put 50ml of clear lime water into two 100ml beakers
2. Bubble room air through one beaker using a pipette and pipette pump
3. Observe and record the results
4. With the other beaker, bubble exhaled air through the solution for one minute
5. After one minute, record your results
Findings:
Bubbling carbon dioxide into lime water calcium carbonate is produced which makes the
clear lime water appear cloudy or milky
195
Specific objective: explain the effects of physical activity on the rate and depth of
breathing.
Breathing rate is the number of times a person inhale or exhale per minute.



During vigorous exercise, muscles need more energy which increases the
breathing rate and depth.
The breathing rate and depth increases to allow more oxygen to be supplied to
the active muscle cells
More carbon dioxide to be breathed out of the body through the breathing
system.
Practical Activity (Investigate the effects of exercise on the rate and depth of
breathing)
You will need:

A stopwatch
Procedure
1. Work in pairs and record each other’s breathing rate.
2. Let your partner jump up and down twenty times and record the breathing rate.
3. Jump up and down twenty times and let your partner take your breathing rate.
4. After a rest, repeat step 2 and 3 in order to make sure your results are accurate.
Results
Draw up a table with the measurements you have taken.
Questions
1. What is the difference in breathing rate at rest and after exercise?
2. Explain your answer.
196
Specific objective: describe the effects of tobacco smoke on the gas exchange system
with reference to carbon monoxide, nicotine and tar








Tobacco smoke contains tar, nicotine and carbon monoxide, which are harmful to
gas exchange system.
Tar is a sticky substance that stimulates the production of extra mucus, which
narrows the air passages and restricts the flow of air.
Tar contains chemicals that are carcinogenic, which causes lung cancer in
smokers.
Tar damages the alveoli which results into emphysema (people who suffer from
emphysema are usually short of breathe and they lack oxygen in the blood since
the alveoli are broken down)
Nicotine is an addictive substance in tobacco smoke.
Nicotine can cause narrowing of arteries, causing high blood pressure
Nicotine increases chances of stickiness of platelets which may cause blood
clots
Carbon monoxide is a poisonous gas since it combines with haemoglobin in the
red blood cells to prevent oxygen from being transported in the body, this makes
the smoker to sometimes be short of breath.
Specific objective: state that tobacco smoking can cause chronic obstructive
pulmonary disease (COPD)



COPD is a lung disease which makes it difficult to breathe because airflow is
obstructed
Smoking increases the risk of getting COPD
Tobacco smoking cause COPD diseases such as lung cancer and emphysema,
which can cause death.
197
3.10. Respiration
There are two types of respiration
i.
ii.
Aerobic respiration
Anaerobic respiration
Specific objective: Define aerobic respiration as:
A chemical reaction in cells that use oxygen to break down nutrient molecules to
release energy.
Specific objective: state the equation of respiration.
In words:
Glucose + Oxygen  carbon dioxide + water
As a balanced molecular equation:
C6H12O6 + 6O2  6CO2 + 6H2O
Specific objective: state the uses of energy in the human body

Energy is used for muscle contraction, protein synthesis, cell division, active
transport, transmission (passage) of nerve impulses and homeostasis.
Specific objective: investigate the uptake of oxygen by germinating seeds and
the effect of temperature on the rate of respiration of germinating seeds
Practical activity (Investigate the uptake of oxygen by germinating seeds)
You will need:






20 germinating (live) seeds
20 dead seeds
Two gas jars with lids
A marker pen
Cotton wool
Deflagration spoon with tea candle , matches and a stop watch
198
Methods
1. Label the two jars A and B. line the bottom of the two jars with some cotton wool,
and moisten it with a little water.
2. Place the live seeds in jar A and dead seeds in jar B. seal each jar tightly with its
lid.
3. Leave the seeds in a sunny spot to germinate for 24 hours to a week.
4. After the seeds have been germinating for a day, loosen the lid of jar A (but do
not remove it). Light the tea candle with a match.
5. Remove the lid of jar A and, working as quickly as possible, lower the lit candle
into the jar, making sure that the deflagrating spoon cover seals the jar opening.
6. Use the stopwatch to time how many seconds pass before the flame goes out.
Record your observation.
7. Repeat steps 4-6 with jar B.
Questions
1. Identify the following:
a) The independent variable
b) The dependent variable
c) Two constant variables
2. Describe what happens when the lit candle is lowed into:
a) Jar A
b) Jar B
3. Formulate a conclusion based on your observations in questions 2.
4. Why is it necessary to have a jar with dead seeds when doing the experiment?
Explain your answer.
Specific objective: investigate the effect of temperature on the rate of respiration
of germinating seeds
Practical activity (Investigate the effect of temperature on the rate of
respiration of germinating seeds)
You will need:





Three petri dishes
Three pieces of cotton wool
15 bean seeds
A fridge
A heater or lamp
199
Methods
1. Moisten the cotton wool and wrap five beans in each piece of cotton wool. Place
the wrapped bean ‘packages’ in each of the petri dishes.
2. Place one petri dish in the fridge, leave another in the classroom (at room
temperature) and place the third petri dish next to a heater or under a lamp.
3. Ensure that you keep the cotton wool in each petri dish equally moist throughout
the investigation.
4. Check on the petri dishes every day at the same time until germination starts.
Note the day that germination started at each of the three temperatures.
Questions
1. Complete the following table:
Temperature/ °C
Germination time/ days
2. Write a conclusion based on your results in the table.
3. Suggest a reason why the rate of respiration is the lowest at the lowest
temperature.
3.10.2 Anaerobic respiration
Specific objective: define anaerobic respiration as:
The chemical reactions in cells that break down nutrient molecules to release a
relatively small amount of energy, in the absence of oxygen.
Specific objective: State the word equation for anaerobic respiration in muscles
during strenuous physical activity and a balanced chemical equation for yeast
Word equation in muscles:
Glucose  lactic acid
Balanced chemical equation for yeast.
C6H12O6  2C2H5OH + 2CO2
200
Specific objective: describe the production of lactic muscles during exercise and
outline how lactic acid is removed during recovery
Production of lactic acid





Lactic acid is produced during strenuous exercise when muscle cells use energy
fast that the circulatory system cannot supply enough oxygen to meet the
demand
Muscle cells start to break down glucose in the absence of oxygen by lactic acid
fermentation to form lactic acid.
This results in a built up of an oxygen debt as the body takes up insufficient
oxygen
Lactic acid causes stiffness in the muscles during and after exercise, therefore it
needs to be removed.
Blood becomes more acidic contributing to muscle fatigue and cramps
How lactic acid is removed during recovery



After exercise a person continues to breathe quickly and deeply to supply the
body with extra oxygen to repay the oxygen debt
The extra oxygen is transported to the liver with lactic acid
In the liver, lactic acid is oxidized (broken down) into glucose, carbon dioxide and
water
Specific objective: describe the role of anaerobic respiration in brewing and breadmaking
The role of anaerobic respiration in brewing



Yeast breaks down the sugars firstly aerobically until oxygen runs out
Then sugars are broken down anaerobically to release alcohol (ethanol) and
carbon dioxide
This is called anaerobic fermentation
The role of anaerobic respiration in bread-making




Bread is made from dough (mixture of flour, salt, water, sugar and yeast)
The yeast ferments the sugar to produce alcohol and bubbles of carbon dioxide.
Baking the bread causes the bubbles of carbon dioxide to expand, making the
bread to rise and give it a light porous texture
The alcohol evaporates during baking
201
Specific objective: compare aerobic respiration and anaerobic respiration in terms of
relative amounts of energy released
Feature
Substrates needed
Products produced
Anaerobic respiration
Glucose
Lactic acid
Number of molecules of ATP/
molecule of substrate
2 ATP
Aerobic respiration
Glucose
Carbon dioxide and
water
38 ATP
Specific objective: investigate the production of carbon dioxide by yeast in anaerobic
conditions
Practical activity (Investigate the production of carbon dioxide by yeast in
anaerobic condition)
Procedure:
1. Add glucose solution to a test tube
2. Add dried (killed) yeast to the test tube and stir
3. Add two drops of oil to ensure that oxygen cannot interfere with the reaction. The oil
prevent oxygen entering the solution, making the yeast respire anaerobically
4. Put the test tube in a water bath and heat the water to vary the temperature
5. Use a rubber bung and delivery tube to connect the test tube, with glucose and yeast,
to a test tube with clear lime water
Conclusions:
-
What changes happen to the lime water after carbon dioxide is passed through it
202
Apparatus set to test for anaerobic respiration in yeast
3.11. Excretion in humans
Excretion is the removal of toxic materials, the waste products of metabolism and
substances in excess of requirement from organisms.
Specific objective: define deamination as:
The removal of the nitrogen-containing part of amino acids to form urea.
Specific objective: describe the formation of urea and the breakdown of alcohol, drugs
and hormones in the liver
Formation of urea


The amino group is removed from excess amino acids (deamination) and
ammonia is formed.
Then, ammonia combines (react) with carbon dioxide to form urea.
Breaking down of alcohol


Alcohol is broken down by the liver into water and carbon dioxide.
The breaking down of alcohol by the liver is known as detoxification.
Breaking down of drugs

Drugs are broken down by the liver to less harmful substances and the end
products pass out of the body in urine.
203
Breaking down of hormones


Excess hormones are broken down by the liver and the end products are
expelled from the body in urine.
Hormones are broken-down (metabolised) into inactive molecules to prevent
accumulation in the tissues.
Specific objective: identify on drawings, diagrams and images the relative positions of
the ureter, bladder and urethra
Diagram showing the positions of the ureter, bladder and urethra
Specific objective: describe the functions of the kidney simply in terms of the removal
of urea and excess water and the re-absorption of glucose and some salts (details of
kidney structure and nephron are not required)
Removal of urea and excess water



The renal artery carries blood with urea, excess water and excess salts to the
kidney.
Blood is filtered as it passes through the kidneys to remove urea, excess water
and excess salts to form a solution called urine.
Water is passively re-absorbed by osmosis
204
Re-absorption of glucose and some salts.




During the filtration of urea and excess water, some useful substances like
glucose, amino acids and vitamins filtrate out of the blood.
As the filtrates moves through the kidneys, any glucose in it must be re-absorbed
back into the blood.
Most of the water, along with some of the salts is also re-absorbed and need to
be kept in the blood.
Glucose is re-absorbed by diffusion and active transport.
Specific objective: outline dialysis and discuss its application in kidney machines
Kidney failure can be caused by:



Infections of the kidneys.
Damage during accidents.
High blood pressure or diabetes mellitus.
There are two major treatments for patients with kidney failure:
a) Dialysis
b) Kidney transplant
Dialysis

It is the process of removing waste products and excess fluid from the body,
when the kidneys are not able to adequately filter the blood.
205
The diagram shows a kidney machine
Dialysis application in kidney machines (dialysis machine)










Blood enters machine from a patient’s artery in the arm
A pump push blood into the dialyser.
Blood passes along the dialyser tubing (visking tubing or cellophane tubing)
which is partially (selectively) permeable and acts as a filter.
The tubing is surrounded by a liquid called dialysis fluid.
The fluid always contains some salts, glucose and amino acids, but no urea or
uric acid.
The dialysis fluid has the same osmotic potential as blood.
Waste substances (urea and uric acid) and excess substances (water and salts)
pass from the blood to the dialysis fluid by diffusion.
A bubble trap or counter flow removes air bubbles and blood clots from the
blood, and re-warms the blood so that the body is not ‘shocked’ when the blood
returns.
‘Cleaned blood’ returns to the patient’s circulation via a vein.
The dialysis fluid must be replaced continuously to maintain the correct
concentration.
206
Specific objective: discuss the advantage and disadvantage of kidney transplants
compared with dialysis
Advantage
Disadvantage
Dialysis machine
- No donor is needed.
- The recipient cannot reject the
dialysis machine.
- It is a life saver if a transplant is
not possible.
Kidney transplants
- One can live a normal life after a
transplant if the kidney is not rejected.
- One is not dependent on machines
after a successful transplant.
- Continuous filtration of blood occurs
after a transplant.
- A kidney transplant is cheaper than
dialysis, which has to be done over a
long period of time.
- A patient has a better quality of life
and a longer life expectancy.
- time-consuming process, 3 to 4 - Rejection of the transplanted kidney
times a week for 6 to 10 hours at by the immune response of the
a time.
recipient can occur.
- One cannot live a normal life,
- The recipient must receive
because dialysis is a continuous
immunosuppressant drugs which can
process.
lead to loss of resistance to infections.
- One needs to be close to a
- In such cases, the recipient is
hospital at all time.
vulnerable to other diseases and
- There can be a shortage of
therefore must be kept in isolation.
machines.
- In some cases immunosuppressant
- Dialysis is very expensive.
drugs have to be taken for the rest of
- There is no continuous filtration the patient’s life.
of blood, therefore dangerous
- There are always risks in any
waste substances can
operation- bleeding, infection or
accumulate in the body before
breathing problems.
the next dialysis.
- The tissue type and blood group of
- The process can cause the loss donor and recipient must be closely
of some minerals like calcium and matched.
iron.
- Diet and fluid intake must be
controlled and maintained,
especially the intake of salts and
protein.
- A person cannot do sports.
207
Specific objective: use dialysis or Visking tubing to separate and identify
mixtures of glucose and starch or sodium chloride and starch.
An activity on dialysis tubing experiment.
-
After 20 minutes have passed, the solution inside the dialysis tubing and the
solution in the beaker will be tested for glucose and starch.
The presence of glucose will be tested with Benedict’s solution, the presence of
starch will be tested using the iodine solution
Sodium chloride will be tested using silver nitrate.
Which molecules or ions do you think were able to diffuse through the bag and
into the distilled water?
As the purpose of dialysis is to mimic the function of the kidney, it is designed to
allow small molecules such as salt, glucose and water through, while stopping big
molecules like starch.
-
Therefore, the starch will not pass through the distilled water, whereas the
glucose and salt will.
This is why a patient is given additional glucose while on a dialysis machine, as
the machine is unable to reabsorb the glucose in a way that the kidney can.
208
3.12 Co-ordination in plants
Co-ordination
Is the ability of an organism to detect and respond to internal and external stimuli.
Stimuli
Any changes that occurs in an environment that causes an organism to react or
respond.
For example:
Plants respond to light, gravity, water, and chemicals sound, touch, temperature
etc.
Importance of responding to stimuli
The environment of plants and animals is continuously changing so,
-
In order to survive, organisms have to detect and respond to stimuli.
This allows their bodies to adapt to the changing environmental conditions.
Coordination in plants is control by:
-
plant growth substances (plant hormones)
Specific objective: Define Plant growth substances/hormones as:
Chemical substances that affect the activities of a particular cell and organs in
plants.
Examples of plant growth substance
Auxins, Ethene, Abscisic acid, Gibberellin
209
Tropism is a growth response of a part of a plant to stimuli (light, water, gravity,
chemical and touch)
Positive tropism

a response of plant growing toward a stimuli
Negative tropism

A response of a part of a plant growing away from stimuli
Two types of growth /tropic response in plants

Phototropism

Gravitropism (geotropism)
Specific objective: define Gravitropism as:
A response in which parts of a plant grow towards or away from gravity.
Positive gravitropism (geotropism)
Radicles and the roots they develop into, always grow down towards gravity
Negative gravitropism (geotropism)
Shoots (coleoptiles) and the stem and leaves they develop into, always grow
away from gravity
Specific objective: define Phototropism
As a response in which parts of a plant grow towards or away from the direction
from which light is coming.
Positive phototropism
Shoots (coleoptiles), the stem and leaves that they develop into, always grow
upwards towards light
210
Negative phototropism
Radicles and the roots they develop into, always grow away from light
Specific objective: describe the chemical control of plant growth by Auxins in
gravitropism and phototropism
Auxins (plant growth substance)

Are plant growth substances, which control response in plants or that cause
differential growth
Where Auxin produced or made

By cells in tips of the shoots (coleoptiles) and roots.
Auxin diffuse to the nearby cells where they cause differential growth.
Differential growth

One side behaves differently from the other side.
Role of Auxin as plant growth substances
In shoots:


Promote cell elongation when in high concentration
Slow down the rate of growth when in low concentration.
In roots:


Inhibit or slow down the rate of growth when in high concentration.
Increase the rate of growth when in low concentration.
Chemical control of plant growth by Auxins in gravitropism (geotropism)





Auxins is produced near the tip of the root
Auxin diffuse away from the upper part of the root to the lower side.
More auxin accumulate on the lower side of the root (due to effect of gravity).
Inhibit elongation of the root
The root grows downward due to gravity resulting in differential growth.
211
The chemical control of plant growth by Auxins in Phototropism





Auxin is produced in the tip of the coleoptiles (shoot)
Auxin are light sensitive and thus diffuse to the shaded side away from light
More auxin collects and accumulate on shaded side
This stimulate cells on the shaded side to grow faster causing differential
growth.
Shoot grows toward light.
NB: when shoot receive light from both side, auxin diffuse and is distributed equally
thus the shoot grows upwards
NB: shoots are also referred to as coleoptiles
212
Specific objective: describe the effects of synthetic plant growth substances used as
weed killer
Synthetic plant growth substances /regulator
These are synthetic hormones sprayed onto weeds to encourage rapid growth that
cause the weed to die. This hormone is selective because it affect only some plants.




Weed killers (herbicides) contains synthetic plant hormones such as auxin.
Spray the selective weed-killer to dicot plants (broad leaved) so that it is
absorbed by the weed
The weed grow rapidly die while
Monocots grow at normal rate
213
Specific objective: distinguish between gravitropism and phototropism
Gravitropism
Growth of a part of a plant to the
stimulus of gravity
Auxin inhibits cell growth in roots if in high
concentrations
Roots always grow towards gravity
(positive gravitropism)
Stems always grow away from gravity
(negative gravitropism)
Differential growth can occur during
gravitropism
Phototropism
Growth of a part of a plant to the stimulus
of light
Auxin stimulates cell growth in stems if in
high concentrations
Roots always grow away from light
(negative phototropism)
Stems always grow towards light (positive
phototropism)
Differential growth can occur during
phototropism
Specific objective: explain gravitropism and phototropism in terms of auxins regulating
differential growth
-
Differential growth is when one part of a plant grow at a different rate (faster)
than the other side.
Phototropism





Auxins regulating differential growth during gravitropism and phototropism
At the top of the cell, there is a section where cells are being produced by
dividing therefore the shoot grows.
The growth of the cells is controlled by auxin, if you cut off the tip containing
auxin the plant will not grow.
If plant grows facing where light is coming in, auxin accumulates on the dark side
where there is no light causing cells to grow at a faster rate on that side.
The side where light is coming in grows normal. This happens because of
differential growth.
Gravitropism




Auxin likes to settle at the bottom of the root due to the pulling force of gravity
High concentration of auxin at the tip of the root slows down the growth
This causes the root to bend and grow down towards gravity into the soil causing
top side to grow different from the lower side.
This happens because of differential growth.
214
Specific objective: observe and interpret the appearance of seedlings grown in
uniform light, one-sided light and no light
Practical procedure:
 Take three pots filled with the same amount of soil.

Plant about 20 seeds in each pot and leave to germinate and grow.
 Then place one pot in normal growth conditions, leave another pot in a dark
cupboard for several days and put the third pot in a place with light only entering
from one side (a box opened on one side)
 Supply equal amounts of water to all.
A
B
C
 Plants under normal conditions will grow slower, stronger and will be green (A)
 Plants in the dark will have under developed smaller leaves which will lack
chlorophyll and will appear more yellow, pale and white (B)
 Stems in the dark will be longer, thinner and weaker, have longer internodes and
grow faster.
 Plants in light will bend and grow towards the light because of positive phototropism
(C)
215
Specific objective: investigate gravitropism and phototropism in shoots and roots
Gravitropism in shoots and roots
Use a Clinostat

An apparatus for studying tropism in plants,

A rotating disc to which the plant/seedling is attached so that it receives an equal
stimulus on all sides.
At rotating/when the clinostat rotates

gravity will act equally on all sides of the shoot/roots

Gravity have no influence on distribution of the auxins

Auxin concentration spread/distributed equally or evenly

Auxin cannot cause differential growth in the shoot/root

there is no response to gravity/light as radicles and shoots grow straight
At stationary /when clinostat is switched off

Gravity will act on one side of the root and shoot only

Gravity has an effect on the distribution of the auxins

More auxin will concentrate on lower side of the shoot and root tip

Stimulating cells to elongate faster in shoots and inhibit cell growth in roots

Shoots respond away from gravity due to negative gravitropism and roots
respond to gravity due to positive gravitropism
216
Rotating clinostat
Rotating clinostat
stationary clinostat
Stationary clinostat
Investigating phototropism in shoots

The tip of the shoot is the sensitive region of the shoot where the receptors that
detect light are found.

The parts of the shoot that respond to the stimulus is the part below the tip called
the effectors.

When auxins is produced at tip of the shoots, it diffuse downward causing the
cells behind the tip to get longer.

If there is no auxins present, the cells will not elongate and grow.
217
Tips of coleoptiles is cut off, and separated from the rest of the coleoptiles by a
piece of gelatin/agar jelly:

Growth continues toward the light

Auxin diffuses through the jelly from the tip

Cause more cell division or growth on the shaded side
If the gelatin block/ jelly jar with auxin is placed on one side of the tip

Shoot grow bend to the side where there is no jelly

Auxin diffuse from the jelly block and cell growth occurs on that side only.
Tip removed and mica substance inserted and tip replaced

Growth stops

Auxin cannot diffuse through the mica from the tip
Tip covered

The shoot still grows because auxin is still produced at the tip
218
Class activity
The figure above shows the coleoptiles A, B and C after a certain experiment.
a) Suggest the aim of the experiment?
b) Explain the appearance of each A, B and C after a week.
3.13 Coordination in humans
The coordination in humans is divided into two parts:
1. The Nervous control system

Is the main coordinating system of the body.

It is responsible for detecting and responding to internal and external changes of
the environment (stimuli).

Coordinate the body using electrical impulses through neurons (nerves),

Made up of receptors, neurons (nerves) effectors (peripheral system), the spinal
cord, brain (central nervous system)
2. The endocrine (hormonal) system:

Made of receptors, endocrine glands and hormones and effectors.
219
The human brain
3.13.1 Nervous control in humans
Specific objective: Identify motor, relay and sensory neurones from diagrams
Neurones/nerve cells
Three types of neurones (nerve cells):

Sensory

Inter (relay/connector)

Motor
Sensory neurone
220
Motor neurone
Function of Neurones/nerve cells
Sensory neurons
Conduct electrical impulses from the receptors (sense organs) to the CNS (brain
& spinal cord).
Inter (relay/connector) neurones)
Conduct electrical impulses inside the CNS from sensory to motor neurons
Motor neurons
Conduct electrical impulses from CNS to the effector (muscles and glands)
Specific objective: Describe effectors in terms of muscles and glands





Effectors are muscles and glands that respond to stimuli.
They receive impulse via motor neurons from the CNS.
Impulses are sent along motor neurons to effectors to ensure the correct
response will take place
Muscles contract and relax to cause the correct movement
Glands secrete correct hormones
221
Specific objective: define a synapse as:
A junction between two neurones or between a neurone and an effector.
It consist of the

presynaptic neuron,

synaptic gap

and post synaptic neuron
Specific objective: describe how nerve impulses are transmitted across the
synapse








A nerve impulse arrives at the end of the presynaptic neurone at the presynaptic
knob.
The calcium channels open and calcium ions enter the presynaptic neuron
knob
This causes the vesicles which contain neurotransmitter to move towards and
fuse with presynaptic membrane.
The vesicles release the neurotransmitter into synaptic cleft
Neurotransmitter diffuse across the synaptic cleft to the post synaptic membrane
The neurotransmitter attach itself to the receptor site on the postsynaptic
membrane
The sodium channel open
Then sodium ions enter the postsynaptic neurone
Diagram showing a synapse
222
Specific objective: describe the effects of alcohol on the release of the
neurotransmitter
Alcohol influences and affects the release of neurotransmitters by:



Slowing down the release of neurotransmitters
Less impulses cross the synapse and slowed from being passed to the next
neuron
Resulting in slower reaction or response of a person
Specific objective: Distinguish between voluntary and involuntary actions
Voluntary actions




A voluntary action can be controlled for example kicking the ball, bending the
arm
It involves a decision or thought-process
It is not automatic
Actions are made deliberately and consciously
Involuntary actions



Involuntary actions cannot be controlled for example heartbeat, blinking,
breathing and walking
It is automatic
No thought required as it is an unconscious action
Specific objective: describe a simple reflex arc in terms of sensory, relay and motor
neurons
Reflex Arc
Is the pathway a nerve impulse travels along from the receptor to the effector to
bring about a correct reflex action.

Reflex arc involve the sensory, relay and motor neuron.
223
Reflex arc
Reflex arc sequence

A stimuli is received by a receptor.

The stimuli is converted into a nerve impulse.

An impulse is conducted along a sensory neurone to the coordinator (CNS).

The impulse is conducted from the sensory neuron along the inter neurone to
the motor neurone.

The motor neurone conducts the impulse from the spinal cord to the effector.

The effector makes the correct response (reaction) to the original stimuli.
224
Specific objective: draw the structure of the transverse section of the spinal
cord, from photo micrographs and diagrams
Transverse section diagram of the spinal cord
Transverse section of the spinal cord as seen from photomicrographs
225
Specific objective: define a reflex action as:
A means of integrating and coordinating stimuli with responses of effectors

Reflex action is a rapid (very fast) response to a stimulus

An automatic response to a stimulus

An inborn response
Examples:

when you accidentally step on a pin/thorn and automatically jerk your foot

Swallowing, sneezing, constriction and dilation of pupil are all reflex actions
Specific objective: Define Sense organs as:
Groups of receptor cells responding to specific stimuli: light, sound, touch,
temperature and chemicals.
Sense
organ
sense
stimulus
SKIN
feeling of touch
Touch, heat, cold,
pain, pressure
NOSE
smell
Chemical in air
TONGUE
taste
Chemical in food
EYE
sight
Light, Colours
EAR
hearing and balance
Sound vibrations,
motion and gravity
226
Specific objective: identify the structure of the eye, limited to the cornea, iris,
pupil, lens, retina, optic nerve and blind spot and state the function of each part
Functions of the parts of the eye
Cornea


Allow light to enter into the interior of the eyeball
Helps to refract (bend) light rays and causes them to converge
Lens


Focuses light onto the retina at the back of eye.
Helps the eye to focus on objects at various distances by changing shape
227
Iris

Control the amount of light entering the eye, by changing the size of the pupil
through contraction and relaxation of radial and circular muscles
Pupil

It controls the amount of light entering the eye
Retina

The light-sensitive cells that releases impulses when stimulated by light. The
rods are concerned with black and white vision and cones with colour vision
Optic nerve

Transmit impulses from the retina to the visual cortex where they are interpreted
Blind spot

Blood vessels enter and leave at this point and no image formed
Specific objective: explain the pupil reflex in terms of light intensity and
antagonistic action of circular and radial muscles in the iris

The pupil reflex is the changing of the size of the pupil of the eye in dim or bright
light.
Pupil reflex in bright light



Circular muscles of the iris contract.
Radial muscles of the iris relax
Pupil constricts (smaller), less light enter the eye, preventing the retina from
damage.
228
Pupil reflex in dim light




Circular muscles of the iris relax
Radial muscles of the iris contract
Pupil dilate (bigger)
More light enter the eye, allowing better vision in dim light intensity
Antagonistic muscle action






Antagonistic muscles are those that work in pairs but in opposite direction.
The iris has radial and circular muscles which act as involuntary antagonistic
muscles.
When the radial muscles of the iris contract, the circular muscles of the iris relax.
This causes the pupil to dilate.
When the radial muscles of iris relax, the circular muscles of the iris contract.
This causes the pupil constrict.
229
Specific objective: explain accommodation to view near and distant objects in
terms of the contraction and relaxation of the ciliary muscles, tension in the
suspensory ligament, shape of the lens and refraction of light.
Accommodation (focusing) of an eye
Definition
Is the change in the shape of the lens to focus light coming from different distances
object (far or near).
Parts of the eye involved
in focusing of light rays
Ciliary muscle
Suspensory ligaments
Lens
Light rays from a near-by
object
Contracts
Becomes slackened/loose
Becomes more rounder of
fatter/more convex to
refract the light rays more
230
Light rays from a distant
object
Relaxes
Becomes taut/tight
Becomes less
convex/flatter/less round
as light rays do not need
more bending
Specific objective: distinguish between rods and cones, in terms of function and
distribution in the retina of a human eye



The retina contains two types of photoreceptors, rods and cones.
The cones are more sensitive than the rods.
The rods and cones are not evenly distributed across the retina.
Function of cones




Cones detect colour (red, green, and blue)
Each cone is linked to one neurone and therefore it gives a very sharp image.
Cones work in bright light
Cones are more sensitive to light than rods
Distribution of cones


Cones are found in the centre of the retina, mainly in the fovea centralis (yellow
spot)
There are no cones in the blind spot
Function of rods



Rods do not detect colour
Groups of rods are linked to one neurone, therefore they do not give as sharp an
image as cones
They work in dim light
Distribution of rods


Rods are found in the peripheral areas (edges) of retina
There are no rods in the blind spot
231
Specific objective: compare nervous and hormonal control systems in terms of
speed and longevity of action
Hormonal control
Message transmitted by hormones
Hormones are transmitted through blood
Hormones travel slowly and
reaction/response is slow.
Response/effect usually lasts longer and
may be permanent
Transported to the target organ
Response/effect is widespread
Originate in endocrine glands
Response may be voluntary or
involuntary.
Nervous control
Message transmitted by electrical impulses
Impulses transmitted by neurones/nerves.
Impulse travels faster and
reaction/response is very fast
Response/effect is short lived and is
temporary
Transmitted to the effector or CNS
Response/effect is localized
Originated in the sense organs/receptors
Response always involves involuntary.
3.13.2 Drugs
Specific objective: define a drug as:
A substance taken into the body that modifies or affects chemical reactions in the
body

Drugs include legal and illegal drugs.

Legal drugs are prescribed to a patient by their doctor or bought over the counter

Illegal drugs include prescription drugs that have been dangerously modified and
substances that are banned by law.
232
Specific objective: describe the use of antibiotics for the treatment of bacterial
infection



Antibiotics are drugs that are used to kill or stop the growth of disease-causing
bacteria.
Bacteria cause bacterial diseases like tuberculosis, syphilis, gonorrhea and
pneumonia.
Many bacterial diseases of humans and other animals can be successfully
treated with antibiotics
Use of antibiotics




Antibiotics weaken the cell walls of bacteria so that when substances enter the
bacterium, it cannot resist internal pressure that will lead to bursting of the
bacterium cell.
Antibiotics are used to damage the cell surface membrane of the bacterium
cell, which causes the cell contents to leak out, thus killing the cell
Antibiotics can also inhibit bacterial growth by binding to the ribosome of some
bacteria.
This stops protein synthesis therefore no proteins and enzymes can be
produced for the bacteria.
Specific objective: explain why antibiotics kill bacteria but do not affect viruses




Antibiotics cannot kill viruses because viruses have different structures and
replicate in a different way from bacteria.
Antibiotics weaken or prevent cell wall production and damage the cell surface
membranes.
Viruses do not have cell walls or cell surface membranes, they are surrounded
by a capsid (protein coat) which is unaffected by antibiotics.
Viruses also do not have cytoplasm therefore there is no internal pressure
which can cause the virus to burst
233
Specific objective: describe the effects of excessive alcohol consumption and
heroin, limited to:
- Powerful depressant drugs
- Effect on reaction time and self-control
- Liver damage
- Negative social implications, e.g. crime, rape etc.
- Cause infections such as HIV
Effects of excessive alcohol consumption
Alcohol is the intoxicating agent found in beer, wine and liquor made through a
process called fermentation.
Powerful depressant drugs





Alcohol is a depressant which in low doses can cause euphoria, sociability and
reduce anxiety.
in higher doses, alcohol causes intoxication and unconsciousness
Alcohol slows down the functions of the central nervous system.
Consuming alcohol prior to driving increases the risk of vehicle accidents and
deaths.
Alcohol affects safe driving skills such as judgements, concentration,
comprehension, coordination, visual acuity, and reaction time become impaired.
Reaction time and self-control




Reaction time is the amount of time taken to respond to a stimulus.
Alcohol slows down reaction time to external stimuli.
Alcohol dulls the brain signal that warns people they are making a mistake, while
drinkers are aware of their error, the alcohol causes them to care less about it.
Alcohol damages the part of the brain that governs self-control.
Consumption and liver damage


Alcohol related liver disease is caused by damage to the liver from excessive
drinking for many years.
Years of alcohol abuse can cause the liver to become inflamed and swollen that
can also cause cirrhosis.
234


Cirrhosis is a disease in which normal liver cells are replaced by scar tissue
which interferes with all of the liver’s important functions
Cirrhosis is the final stage of liver disease.
Addictions and withdrawal symptoms




Occurs when heavy drinkers suddenly stop or reduce their alcohol intake.
They may experience a combination of physical and emotional symptoms.
Moderate withdrawal symptoms include physical reactions like headache,
excessive sweating and increased heart rate.
Some symptoms are as severe as hallucinations.
Negative social implications



Alcohol plays a strong role in relation to crime.
40% of all violent crimes and 37% of convicted offenders in jail report that they
had been drinking at the time of their arrest.
Alcohol has been found to be closely associated with violent crimes, including
murder, rape, assault and child and spousal abuse.
Infections such as HIV




People who abuse alcohol are more likely to engage in behaviour that puts them
at risk of contracting HIV.
Alcoholics have a high rate of injection drug use.
Alcohol abusers have high tendency toward high-risk sexual behaviour,
including unprotected sexual intercourse.
Too much alcohol can weaken the immune system.
Effects of heroin
Powerful depressant drug



Is the most dangerous drug that reduces nervous activity and slows down brain
activity
Heroin also lowers neurotransmission levels
Abusers typically report feeling a surge of pleasurable sensation
Reaction time

A depressant drug, that slows down the transmission of impulses, thus
increasing the reaction time and slows down a person’s reactions.
235
Liver damage

Heroin has little effect on the liver. The major problems that result from its use
are not from the heroin, but from contaminants in the drug which can severely
damage the liver.
Addiction and withdrawal



When someone has a heroin dependency, they will feel uncomfortable and sick
without the substance in their system.
This occurs because the body has become so accustomed to the heroin that its
absence is unusual and may results in withdrawal symptoms.
Withdrawal symptoms of heroin include: restlessness and discomfort racing
heartbeat, anxiety, sweating, shivering, pain in the muscles and bones,
diarrhoea, vomiting and inability to sleep.
Negative social implications

Criminal behaviour and substance abuse manifests in the form of prostitution,
drug trafficking, theft and burglary, rape, domestic violence poor judgement of
behaviour, poor ability to work and emotional problems.
Infections such as HIV


Using heroin increases the risk of a person being exposed to HIV infection,
through contacting with infected blood or body fluids (e.g. semen, saliva)
resulting from the sharing of syringes
or through unprotected sexual intercourse with an infected person
3.13.3 Hormones in humans (endocrine system)
Specific objective: define a hormone as:
A chemical substance, produced by a gland and carried by the blood, which alters
the activity of one or more specific target organs
Endocrine glands: Are ductless glands that secrete hormones directly into the
bloodstream and not into a duct
Target organs: are organs or cells in the body which are affected by hormones
236
Specific objective: Identify specific endocrine glands and their secretions, limited to
adrenal glands and adrenaline, pancreas and insulin, testes and testosterone,
ovaries and oestrogen
Diagram showing some major endocrine glands in humans
Endocrine glands
Location
Hormone produced
2. Pancreas
Pancreas
Insulin and glucagon
3. Adrenal gland
Top of kidney
Adrenaline
4. Ovary
Ovary
Oestrogen and
progesterone
5. Testes
Testes
Testosterone
237
Specific objective: discuss the role of the hormone adrenaline in the chemical
control of metabolic activity, including increasing the blood glucose concentration
and pulse rate



Adrenaline is produced by the adrenal glands, which are located on top of each
kidney in the abdominal area.
Adrenaline prepares the body to cope with stress and to meet danger and deal
with emergencies
Fear or anger cause adrenaline to be released into the bloodstream.
Adrenaline causes an increase in the blood glucose concentration




Promote the release of glucose from the locations in the body where it is stored.
Adrenaline stimulates the breakdown and conversion of glycogen into glucose
in the liver, which sends it via the bloodstream to the muscles.
Stimulates the breakdown and conversion of glycogen into glucose in the
muscles, instead of releasing that glucose, the muscles hold on to it to use for
immediate release of energy.
This results in an increased glucose concentration in the blood and muscles and
can be used for energy to either combat or flee dangerous situation
Adrenaline causes an increase in the pulse rate






Pulse (heart) rate is a measurement of the number of times a person’s heart
beats in one minute.
Adrenaline increases the pulse (heart) rate , because it increases pressure in the
coronary arteries
It also increases the breathing rate and depth of breathing
This results in more oxygen being taken in faster
It also results in more glucose and oxygen being delivered to the muscles
To increase the rate of aerobic respiration
More energy is therefore released to combat or flee a potentially dangerous
situation
238
3.14 Homeostasis
-
Our body is equipped with a self-regulatory system that monitor and regulate
changes of factor in our body.
Specific objective: define Homeostasis as:
The maintenance of constant internal environment
Living organisms have two environments:
1. External environment: the surrounding in which the organism lives.
2. Internal environment: the space inside your body. I.e. the space between the
cells filled with tissue fluid.
Homeostasis helps to maintain:
-
A constant internal body temperature of around 37˚C, which helps enzymes to
work.
A constant amount of water, to prevent damage of body cells or losing too much
water by osmosis.
A constant blood glucose level, which means that there is always enough fuel for
respiration.
A constant blood pH at about 7.4 as variation in pH can lead to denaturation of
enzymes.
Oxygen and carbon dioxide concentrations in the body
Specific objective: explain the concept of control by Negative Feedback
Negative feedback is a response which automatically brings about a correction
through a certain condition that brings a situation back to normal.



Receptor cells detect changes and send a message (chemical or hormonal) to
the control centre (brain).
The control centre (the brain) Issues new instructions that are received by an
effector.
The effector responds by carrying out some corrective action to bring the
changed factor back to its normal state.
239





This is followed by feedback to the receptor cells about the effector’s corrective
action
If the factor is within its normal range, nothing more happens.
If the factor is still out of limits, the receptor cells send another message to the
brain and the effector responds again.
Another message, (a negative one) is now sent from the receptor to the effector
to stop its corrective action from going too far (to prevent overcorrection).
This is known as negative feedback, getting the body environment back to an
ideal balance
Principle of negative feedback
Specific objective: Describe temperature regulation in ectothermic and
endothermic animals
Exothermic Animals (cold-blooded animals)
-
Animals that cannot generate sufficient heat inside the body
Rely on external sources such as the sun for body heat
Ectothermic animals have low metabolic rates and their body temperature
changes as external temperature changes
Examples include: reptiles, insects
240
Endothermic Animals (warm-blooded animals)
-
Rely on heat produced within bodies by metabolic reactions (i.e. respiration) to
control their body temperature.
Endothermic animals include the mammals and the birds
Ways of regulating temperature in Ectothermic animals
In cold weather:




Position themselves to absorb solar radiation by climbing up vegetation and
stones (bask in the sun).
Use radiant energy from the sun to raise their body temperature.
Skin become darker in colour and absorbs more radiant energy
Goes into hibernation
In warm weather:



Escape to places of low heat (shade, underground burrow, under stones or
water).
Opening of mouth (panting) to lose more water by evaporation (leading to
cooling).
Aestivation a state of dormancy due to dry condition in the summer, also occurs
in some reptiles and toads
Ways of regulating Temperature in Endothermic animals


Endothermic have higher metabolic rates and effective physiological
mechanisms for controlling body temperature to keep it constant.
Endothermic animals maintain a constant body temperature of 37°C
When temperature is low



Cellular respiration which takes place in each living cell, is the main heat
generator. An increase in adrenaline causes an increase in metabolic rate,
which results in faster and more heat released
By consuming hot drinks and food or
By exposing the body to heat from the sun.
241
When temperature is high




Endotherms lose heat by radiation (when heat is lost from a warm body to a cold
environment)
Through Evaporation (when fluids absorb energy and change into gas therefore,
sweat evaporates, heat is lost)
Through Convection (when heat is lost in fluids)
Through Conduction (when heat is lost from touching warm objects to cold
objects)
Graphical presentation of temperature regulation by ectotherm and endotherms
Specific objective: describe the maintenance of a constant internal body temperature
in humans in terms of insulation, the effects of sweating, shivering related to
vasodilation and vasoconstriction only
Insulation
-
An insulator is something that does not easily allow heat to escape.
242




Fats underneath the skin provide humans with insulation by creating a fatty layer
in their bodies which stops internal body heat from escaping.
Wearing of more clothes when it is cold slow down the heat loss from your body
to the air around you.
When the external temperature is low (cold) hair erector muscle contracts, hairs
stand upright and trap a layer of air that acts as insulation against the cold
When the external temperature is high (hot) hair erector muscle relaxes, hairs
lies flat on skin surface and does not trap a layer of air and thus prevent
insulation
Sweating










When temperatures are high, the arterioles, which supply the sweat glands with
blood dilate
This causes more blood to flow to the sweat glands
More sweat is produced
More evaporation of sweat from the skin surface occurs.
More heat is lost through evaporation
When temperatures are low, the arterioles which supply the sweat glands with
blood constricts
Less blood flows to the sweat glands
Less sweat is therefore produced and secreted
Less evaporation of sweat from the skin surface occurs.
Less heat is lost by evaporation
243
A section through the skin
Shivering



When the external temperatures are low, the muscles in the skin start to
contract and relax uncontrollably.
This twitching exercise produces heat, which helps to raise body temperature
More heat is generated through respiration in the muscle cells during shivering
Vasoconstriction
Vasoconstriction refers to the narrowing of the blood vessels, in particular
arterioles.



When the external temperature is low and the body temperature falls below 37°C
the muscles around the arterioles contract.
The arterioles in the skin constrict and reduce in diameter resulting in less blood
flowing in the skin capillaries.
This makes the skin turn pale because less blood reaches the surface, so less
heat is lost through radiation from the skin.
244
Vasodilation
Vasodilation refers to the widening of the blood vessels, in particular arterioles.



When the external temperature is very high and your body temperature rises too
much above 37°C, the muscles around the arterioles in the skin relax.
The arterioles in the skin dilate and increase in diameter resulting in more blood
flowing in the skin capillaries.
This makes the skin turn darker because more blood reaches the surface, so
more heat is lost through radiation and convection from the skin.
The diagram shows vasoconstriction (cold environment) and vasodilation (warm
environment)
Specific objective: describe the control of the glucose content of the blood by the
liver, and the roles of insulin and glucagon from pancreas
245
Control of blood glucose level in the body
-
The normal concentration of glucose in the blood is 90-100mg glucose/ 100cm3
of blood (0.1%), but this can vary.
After a meal rich in carbohydrates, the blood glucose level may rise to 150mg.
After a prolonged and strenuous period of physical exercise, the blood glucose
levels may drop to as low as 70mg.
After an extended period without eating food, the blood glucose levels may also
drop to as low as 70mg.
Blood glucose levels need to be kept within narrow limits, because respiration is
a continuous process which needs a regular supply of glucose.
If the blood glucose falls below normal, a person may faint and go into a coma
which can be fatal.
If the blood glucose level is too high, water may be drawn out of cells and tissue
fluid by osmosis back into the blood.
This will cause kidney to excrete this water which can lead to dehydration.
The roles of insulin in the pancreas (high glucose levels)






The beta cells of the islets of Langerhans in the pancreas secrete the hormone
insulin into the bloodstream.
This stimulates cells all over the body to uptake glucose especially the skeletal
muscles
The bloodstream transports the insulin to the liver
The insulin instructs the liver cells to convert excess glucose into glycogen
Insulin can also increase the rate of respiration
Glycogen can be stored in the liver and the muscles
The roles of glucagon in the pancreas (Low glucose levels)






The alpha cells of the islets of Langerhans in the pancreas secrete the hormone
glucagon into the bloodstream.
The bloodstream transports the glucagon to the liver
The glucagon instructs the liver cells to convert glycogen into glucose
Glucagon can also decrease the rate of respiration
Adrenaline can also be secreted from the adrenal glands to stimulate the
conversion of glycogen to glucose
The entire activity is controlled by homeostasis.
246
Diagrammatic presentation of blood glucose control
Specific objective: outline the symptoms of and treatment of Type 1 diabetes
(details of β (beta) cells are not required)


Type 1 diabetes is a chronic auto-immune disease in which the immune system
attacks the insulin-producing beta cells in the pancreas.
The beta cells of the pancreas become incapable of secreting insulin and the
lack of insulin leads to increased glucose in the blood and urine.
Symptoms of Type 1 diabetes








Extreme constant thirst
Frequent urination
Unintended weight loss
Extreme hunger
Irritability and other mood changes
Fatigue and weakness
Blurry vision
In females, a vaginal yeast infection
247
Treatment of Type 1 diabetes




There is no cure for diabetes. Frequent blood sugar monitoring involves regular
measurement of blood glucose levels.
Too high blood glucose can be controlled by daily administrations of insulin
injection
Sufferers must change their diet by avoiding eating too much sugary food
(carbohydrates)
Exercise regularly and maintain a healthy weight
Specific objective: Investigate the rate of cooling of a warm body (represented by
tubes containing hot water) under different conditions (e.g. large or small surface
area, with wet or dry covering).
Materials
-
Hot water
4 glass beakers
4 rubber bands
4 thermometers
Stop watch
Piece of newspaper
Woolen sock (to represent feathers or fur)
Piece of cotton cloth
Piece of tinfoil
Procedure
1. Align the four glass beakers in a row on an even surface.
2. Pour the same amount of very hot water into each glass beaker.
3. Cover each glass beaker with one of the four materials from above (foil, cotton
cloth, woolen sock and newspaper)
4. Secure the materials over the top of each glass beaker with a rubber band.
5. Leave the glass beaker for half an hour.
6. Remove all the materials from the glass beakers.
7. Place a thermometer into each glass beaker. Leave it for two minutes and read the
temperature in °C. The one which has the warmest water had the best insulator as
a cover.
248
For our experiment a good insulator is a kind of material which keeps things warm for
longer, it help to stop heat from escaping.
Question
Which material do you think would make the best insulator? Foil, cotton cloth,
newspaper or a woolen sock? Can you explain why?
Topic 4 Development of the organism and the continuity of life
4.1 Cell division
- Growth and reproduction in animals is usually a result of cell divission.
- Cell division is the process by which cells produce new cells.
- All cells are derived from pre-existing cells.
Specific objective: define mitosis as:
Nuclear division, giving rise to genetically identical cells
Specific objective: describe mitosis simply, in term of the exact duplication of
chromosomes resulting in identical daughter nuclei (details of stages are not required




During mitosis, the cell nucleus divides into two daughter nuclei.
Each contain exactly the same number of chromosomes.
Before division, the nucleus makes copy of the chromosomes.
Mitosis form two daughter nuclei of identical cells with the full set of
chromosomes (diploid).
249
Simple presentation of the process of mitosis
Specific objective: State the role of mitosis in growth, repair of damage tissues,
replacement of cell and sexual reproduction.
Role of mitosis




-
Growth
Mitosis occur where new cells are needed for growth
Mitosis is the basic component of growth
A zygote will develop into a functioning multicellular organism because of mitosis
Repair of tissues
Mitosis multiply cells for the repair of damaged cells of multicellular organism
Skin cells go through many mitotic division during the repair of the epidermis
after an injury
Repair can also occur during regeneration of missing parts in some multicellular
organisms
Replacement of cells
Skin cells are shed daily and they are replaced by mitosis.
Red blood cells that are lost are replaced by mitosis
Cells lost in donated blood gets replaced by mitosis
Asexual reproduction
Mitosis is a form of asexual reproduction
Simple organisms make copies of themselves and split by binary fission during
mitosis
A wide range of organisms reproduce by asexual method involving Mitosis.
250
Specific objective: observe, and interpret from diagrams and photomicrographs,
mitosis in root tips.
Diagram of mitosis in root tips
Photomicrograph showing stages of mitosis
Specific objective: prepare and stain temporary slides of root tip squashes to show
mitosis
(Refer to page 337 NAMCOL Biology learner’s book for procedure)
Specific objective: define meiosis as:
The reduction division in which the chromosome number is halved from diploid to
haploid resulting in genetically different cells (details of stages are not required)
251
Specific objective: draw and interpret prepared slides showing meiosis in animal and
plant tissues.
Slides showing meiosis
Specific objective: state that meiosis is involved in production of gametes.
Meiosis and the production of gametes
-
Not any diploid cell produces gametes through meiosis.
Diploid cells destined to become haploids are cells in the ovaries in females and
testes in males or in ovule and anthers of a plant flower.

Meiosis result in the formation of four daughter cells, each with the half number
of chromosomes of the parents. This form the basis of sexual reproduction.
Reduction division reduces the diploid chromosome to a haploid number.
Gametes are formed during meiosis.
At fertilization, haploid male gametes fuse with female haploid gametes to form a
diploid zygote.
Genetic variation occurs because meiosis provide opportunity for new
combination of genes to occur in gametes.




252
The diagram shows the process of meiosis
Specific objective: explain how meiosis produces variation by forming new
combinations of maternal and paternal chromosomes (details are not required)
Genetic Variation.
-
Maternal chromosomes are the ones received from the mother and paternal ones
are received from the father.
Chromatids exchange genetic materials and different sections of chromosomes
are exchanged.
This major genes reshuffling take place during cross over.
Meiosis causes genetic variation (due to crossing over and reshuffling of genes)
253
Specific objective: distinguish between mitosis and meiosis
The table below shows the distinction between mitosis and meiosis
Features
mitosis
meiosis
4.2 Reproduction
4.2.1 Asexual and Sexual Reproduction
- Reproduction is the biological process by which new individual organisms are
produced from their parents.
- There are two forms of reproduction: Sexual and Asexual
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Specific objective: define asexual reproduction as:
A process resulting in the production of genetically identical offspring from one parent.
Specific objective: describe asexual reproduction in bacteria, spore production in fungi
and tuber formation in potatoes
Asexual reproduction in bacteria
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Bacteria are unicellular organisms that reproduce by a form of a cell division
called binary fission.
This occurs when duplicates of DNA divides into daughter cells which then also
divides and the life cycle continue
Binary fission in bacteria
Asexual reproduction in spore production in fungi
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Reproduction takes place by means of spores
Fungi produce genetically identical spores by mitosis.
Some of the hyphae become aerial
At the tip of these aerial hyphae, sporangia form where spores are produced and
stored.
When matured, sporangia burst open, spores are released into the air and
dispersed by wind.
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Once these spores land on a favourable environment (moist and warm) they
geminate, grow and develop again
All these offspring are genetically identical.
Spore production in fungi
Asexual reproduction in tuber formation in potatoes.
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Potato tubers have auxiliary buds on their surface.
In favourable conditions, new shoots and roots grow from these buds using
stored nutrients in the tuber.
The new roots and leaves grow by mitosis. The new plants are identical to the
parent plant.
Roots grow downwards and the shoots and leaves grow upwards.
The tips of the ground stem swell as they store nutrients.
These nutrients are translocated down the stem to be stored in tubers.
As more nutrients are stored, new tubers are formed
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Asexual reproduction in Irish potatoes
Specific objective: discuss the advantages and disadvantages of asexual reproduction
Advantages of asexual reproduction
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It’s a fast and simple process
Only one parent is needed or does not involve gametes
The offspring are identical to their parents and retain advantageous
characteristics of the parent
Rapid increase in number and spreading can out-compete competitors by
forming large colonies
No need for agents for pollination to carry pollen and seeds during seed
dispersal.
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Disadvantages of asexual reproduction
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All offspring are genetically identical
Has lesser chance of adopting to environmental changes
All offspring maybe affected or killed by the same diseases because they are
genetically identical.
Overcrowding may occur and resources can be used up.
Specific objective: define sexual reproduction as:
A process involving the fusion of the nuclei of two gametes (sex cells) to form a zygote
and the production of offspring that are genetically different from each other
Specific objective: define fertilization as the fusion of gamete nuclei
Fertilization in animals
Specific objective: discuss the advantages and disadvantages of sexual reproduction
Advantages of sexual reproduction
Results in genetic variation
Gives new offspring better chance of
survival in changing conditions
Seeds can be dispersed to other areas
Evolution and formation of a new species
can occur
Disadvantages of sexual reproduction
Two parents required
The new young ones are vulnerable
Produces fewer offspring than asexual
reproduction
It takes time and energy
Outside agents are needed to carry
pollen in flowering plants
Pollen and seeds are wasted
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4.2.2 Sexual reproduction in plants
Specific objective: define self-pollination as:
The transfer of pollen grains from the anther of a flower to the stigma of the same flower
or different flower on the same plant
Specific objective: define cross-pollination as:
The transfer of pollen grains from the anther of a flower to the stigma of a flower on a
different plant of the same species
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Specific objective: describe the different structural adaptations and functions of the
parts of insect-pollinated and wind-pollinated flowers
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Sexual reproduction in plants takes place in the flowers
Flowers carry the reproductive organs of a flower
Unisexual flowers are those that have the male reproduction organs (stamen)
and female reproduction organs (carpel) on separate flowers
Bisexual flowers have both stamen and carpels in the same flower
Structure of an insect pollinated flower
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Structure of a wind-pollinated flower
Functions of the parts of a flower
Flower part
Functions
Flower stalk (stem) Connects the flower to the stem; carries inorganic and organic
nutrients to the flower
Receptacle
Top thickened part of the flower stalk which bears the flower;
where the parts of the flower are attached to
Sepals (calyx)
Green outer part of the flower that enclose and protect a
developing flower bud
Carpel (pistil)
The female reproductive organs of the flower (the stigma, style
and ovary containing ovules)
Stigma
Receptive tip of the style where pollen is deposited during
pollination; it produces the nutritive juices for pollen to
germinate
Style
Part of the carpel that joins the ovary to the stigma; supports
the stigma to be exposed to agents of pollination; pollen tube
grows through the style
Ovary
Enlarged basal part of the carpel where ovules are produced; it
contains the ovules; it becomes the fruit after fertilization
Ovule
Structure inside the ovary which contains the female gametes;
it becomes the seed after fertilization
Female gamete
The haploid sex cell which fuses with the male gamete during
fertilization to form the diploid zygote
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Stamen
Anther
Pollen sacs
Pollen grains
Male gamete
Filament
Petals (corolla)
Nectary gland and
nectar
The male reproductive organs, each stamen consists of an
anther and a filament
The tip of the stamen which contains pollen sacs to produce
pollen grains
The chambers in the anther where the pollen grains are
produced
Pollen is a fine to coarse powdery substance comprising pollen
grains in which the male gametes are produced
The haploid sex cell produced in the pollen grains which fuses
with the female gamete during fertilisation to form a haploid
zygote
The stalk of the stamen which carries the anther and hold it to
be exposed to agents of pollination
The often conspicuous coloured and scented part of a flower
which attracts pollinators for pollination
A gland-like swelling found on the receptacle or other parts of
some flowers which produces nectar a sweet sugary fluid
collected by bees to make honey
Structural adaptations of an insect and wind pollinated flower
Insect pollinated flower
Large, brightly coloured and conspicuous
petals to attract insects for pollination
Flowers have nectaries to produce nectar
to attract pollinators
Scented flowers with attractive odours to
attract pollinaters
Firm anthers inside the flower to allow
visiting insects to brush against them
Wind pollinated flower
Small, dull and inconspicuous flowers
Nectaries absent, no need to attract
pollinators
flowers not scented (have no attractive
smell)
Anthers dangle outside the flower on
long, slender filaments for pollen to be
easily released into the air
Stigmas are enclosed within the flower for long filaments with anthers that hang
visiting insects to brush against them
outside the flower
when collecting nectar
Small anthers that produce less pollen
Large anthers to produce lots of pollen
as most pollen are wasted
Small stigmas to deposit pollen
Large, branched and feathery stigmas
accurately
with a large surface area to filter and
catch pollen grains from the air
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Pollen grains are sticky and spiky to help
it adhere to the body of the insects
Pollen grains are larger in size compared
to wind-pollinated pollen grains
Pollen grains are smooth, light and dry
to aid wind transport
Pollen grains are small in size
Specific objective: investigate and draw the anthers and stigmas of an insectpollinated flower using a hand lens
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A hand lens is a magnifying glass that is used to enlarge an object
Using a hand lens:
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First hold the instrument out at arm’s length
to get the best view of the object, hold the lens as close to your eye as possible
bring the object up in front of the lens until it is in focus
obtain your focal distance, which is the distance at which you must hold the
object in front of the lens for it to be in focus
the higher the magnification of the lens, the shorter the focal distance needs to
be and the smaller the field of view
e.g. a 10X hand lens has a focal distance of 25mm while a 20X hand lens has a
focal view of 12mm
Specific objective: observe, draw, interpret and compare the structure of windpollinated and insect-pollinated flowers
Activity
Collect a flower from a wind pollinated plant like grass and from an insect pollinated
plant and make an enlarged labelled drawing of the flowers.
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Specific objective: Name the agents of pollinated
Pollination:
Is the transfer of pollen grains from the anther of a flower to the stigma of a flower
Agents of pollination
Agent
Wind
method
Pollen is carried from one flower to the next by blowing wind
Insects
Pollen is carried from one flower to the next by insects like bees,
flies, ants
Water
Animals
Humans
Pollen is carried from one flower to the next by flowing water
Pollen is carried from one flower to the next by animals
Pollen is carried artificially from one flower to the next by humans
using a small brush
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Specific objective: discuss the implications to a species of self-pollination and crosspollination in terms of variation and reliance on pollinators
Implications to a species of self-pollination
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there is less genetic variation in the offspring than if there was cross-pollination
variation allows species to have a better chance of survival, with self-pollination,
species become less adaptive to changing conditions
if plants become susceptible to a disease, their resistance can be reduced
Self-pollination creates less variation, but can be advantageous if there are no
other plants nearby. Pollination can still occur in areas with scarce pollinators
less pollen is wasted as pollen transfer is more precise
Implications to a species of cross-pollination
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causes more genetic variation due to mixing of gametes
plants are more able to adapt to changing conditions creating better chances of
survival for a plant species
plants can develop more resistance to diseases
more pollen is wasted as pollen transfer is less precise
plants like maize avoid self-pollination because anthers mature before the
stigma
Specific objective: describe the growth of the pollen tube and its entry into the ovule
followed by fertilisation (details of production of endosperm and development are not
required)
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Pollen grains land on the stigma during pollination
The stigma produces a sticky nutritive fluid
Pollen grain absorbs this fluid and burst open
A pollen tube emerges from the pollen grain and grows through the stigma and
style to the ovary
Male nucleus moves down the pollen tube
The tip of the pollen tube ruptures and the male gamete enters the ovule
The male gamete fuses with the female gamete (egg cell) to form a diploid
zygote during fertilization
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The zygote then forms the embryo plant by mitosis
After fertilization, the petals and style dry and fall off
After fertilisation, the ovule stores nutrients and becomes the seed
After fertilisation the ovary becomes the fruit
Specific objective: investigate the formation of pollen tubes in sugar solutions
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Growth of a pollen tube can be observed under medium power of a microscope
Materials needed:
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Sucrose (a sugar) solution
Boric acid solution
Microscope slide
Light microscope
Fresh pollen grains
Procedure:
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Prepare a nutrient solution by dissolving 10g sucrose and 10mg boric acid in
100ml distilled water
Take a few drops of this solution on a clear slide and dust a few pollen grains
from the stamen of a flower in it
Observe the slide under a microscope
Observations:
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After every half hour, observe the slide under a simple microscope
Pollen grains with developing pollen tubes will be seen
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Most pollen grains germinate in a 10-20% sugar solution (you can use sugar
solution alone if there is no boric acid
Germinating pollen grains as seen through a light microscope
Specific objective: explain the importance of seed dispersal
Seed dispersal:
The spreading of seeds and fruits as far away as possible from the mother plant
Methods of seed dispersal:
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Wind
Water
Self-explosion
Animals
Importance of seed dispersal
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The offspring (seedlings) do not compete with parents for resources such as
water, sunlight, space and minerals
Avoid overcrowding of plants in a small space
Risk of soil depletion of its resources is reduced
Risk of spreading of diseases in a plant species is reduced
A natural disaster cannot destroy a whole community
Dispersal reduces inbreeding and thus promote outbreeding
Increases chances of plants becoming established in new habitats and to
colonise new areas
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4.3. Human reproductive system
4.3.1 Sex hormones in humans
Specific objective: describe the sites of oestrogen and progesterone in the
menstrual cycle and in pregnancy
Menstrual cycle:
Regular natural changes that occur in the female reproductive system and it makes
pregnancy possible
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the cycle is required for the production of the female gametes
it prepares the uterus for pregnancy
the cycle occurs over 28 days and includes ovulation (the release of an egg from
the ovary)
the cycle includes thickening of the uterus wall and menstruation if fertilization did
not occur
Sites of oestrogen and progesterone production in the menstrual cycle
Hormone
Oestrogen
Site of production:
- produced and secreted by the
Graafian follicle in the ovary
Progesterone
Site of production:
- produced and secreted by the
corpus luteum (yellow body) in
the ovary
Site of oestrogen and progesterone production during pregnancy
Hormone
Oestrogen
Site of production:
- produced by the placenta
Progesterone
Site of production:
- produced by the placenta
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Specific objective: explain the role of hormones in controlling the menstrual cycle
and pregnancy limited to FSH, LH, progesterone and oestrogen
Hormone
FSH (follicle stimulating
hormone) produced and
secreted by the anterior
lobe of the pituitary gland
LH (luteinizing
hormone) produced and
secreted by the anterior
lobe of the pituitary gland
Oestrogen
Progesterone
Roles in the menstrual cycle
- in the first 5 days of the cycle, it stimulates
the development of the Graafian follicle in
the ovary
- it stimulates the Graafian follicle in the
ovary to produce oestrogen
- its concentration rises after menstruation
and peaks on day 14
- a low FSH level after menstruation prevents
further follicle stimulation
- it peaks on day 14
- it controls and stimulates ovulation to occur
on day 14
- stimulates the corpus luteum In the ovary to
produce and secrete progesterone
- oestrogen rises from day 4/5 and peaks on
day 12/13
- high levels of oestrogen causes the cells in
the uterus lining after menstruation to
divide, repair, thicken and build up a new
uterus lining
- causes blood vessels and glands to
develop in the uterus lining
- stimulates the pituitary gland to produce
and secrete LH
- its production peaks a week after ovulation
to help maintain the thickness of the uterus
lining
- it prevents the uterus lining from breaking
down in order to keep the uterus lining
ready for implantation of the embryo
- it thickens the uterus lining even more
- it inhibits the production of FSH so it inhibits
the development of a new follicle
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Hormone
FSH
LH
Oestrogen
Progesterone
a fall in progesterone levels on days 23-28
removes the maintaining effect and causes
the uterus lining to break down
break down of the uterus lining leads to
menstruation
if fertilization occurs progesterone levels
stays high during pregnancy and no
menstruation occurs
Role during pregnancy
-remains low during pregnancy to prevent
ovulation
-stimulates the corpus luteum to produce and
secrete progesterone
- prepares the uterus for labour by increasing the
sensitivity of the uterus muscles so it can contract
during labour
- causes the lining of the vagina to become
thicker, tougher and have a better blood supply to
allow an easy passage during birth
- makes the placenta grow and function properly
- it triggers the development of the organs and
bone density of the developing foetus
- promotes growth of maternal breast tissue to
prepare the pregnant woman for lactation
- high levels of progesterone ensures that
pregnancy is maintained as it keeps the uterus
lining thickened
- high levels will inhibit contractions of the uterus
walls to prevent abortion of the foetus
- prevents menstruation during pregnancy
- helps to suppress immune responses of the
mother to the foetus antigens in order to prevent
rejection of the foetus
- prevents any further ovulation during pregnancy
by inhibiting the release of FSH
- stimulates the development of secretory tissues
in the mammary glands to prepare the breasts for
milk production and lactation
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4.3.2. Sexual reproduction in humans
Specific objective: identify from diagrams the parts of the male and female
reproductive system and state the function of these parts
Female reproductive system
Front view
side view
Functions of the parts of the female reproductive system
Ovaries
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a pair of tiny glands situated in the pelvic cavity, one on either side of the uterus
womb
ovaries are held in position by ligaments and are connected to the uterus by the
ovarian ligaments
Functions:
produce, store, ripen and release egg cells (ova)
produce and secrete the female sex hormones oestrogen and progesterone
Oviduct (fallopian tube)
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narrow tube that extends from each ovary to the uterus through which the ovum
travels
walls of the oviduct is lined with ciliated epithelium
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Functions:
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where the ovum is released into and travels down towards the uterus with the
help of the flicking movement of the cilia
where fertilisation occurs, the developing zygote and embryo are moved by
the cilia and peristaltic movements towards the uterus
Uterus (womb)
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a muscular organ about 7cm long and 5 cm wide
Functions:
provides a place where the embryo develops into a foetus
it is within the uterus that the foetus develops during pregnancy before birth
during birth the baby is pushed out by the walls of the uterus through a series of
contractions
Cervix
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lower part of the uterus that leads to the vagina
it is the neck of the uterus
Functions:
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it dilates (widens) during labour and birth for the baby to pass out to the outside
passage through which sperms travels through to fertilise an egg cell after sexual
intercourse
a mucus plug forms in the cervix to close off the uterus during pregnancy
Vagina
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muscular tube lined with a thick mucous membrane
it opens to the exterior through the vulva
Functions
receives the penis during sexual intercourse (copulation), semen is released into
the top part of the vagina
baby passes through the vagina during natural birth
has a membrane called the hymen that partially closes the vagina of a virgin
Vulva
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the outer part of the female genital, includes the opening of the vagina, the labia
and clitoris
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Male reproductive system
Front view
Side view
Functions of the parts of the male reproductive system
Scrotal sac (scrotum)
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consist of a sack of skin and smooth muscle and is located under the penis
Functions:
contains and encloses the testes and the epididymis and part of the vas deferens
regulates the temperature of the testes by maintaining it at 2 degrees below the
normal body temperature
this ensures an optimum temperature for sperm production
Testes (testicles)
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usually two testes located descended from the pelvis
the testis is surrounded by skin called the scrotum which hangs outside the body
between the legs
Functions
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where the production of sperm (male sex cells) takes place (in the seminiferous
tubules in the testes)
has cells that are responsible for the production of testosterone
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Epididymis
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a six meter long tube that lies on top of the testis
the tube that connects a testis to a vas deference
Functions:
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where sperm undergo maturation processes to make them functional and
mobile
where sperms produced in the testis enter to be stored
Vas deferens (sperm duct)
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tiny muscular tube that carries sperm from the epididymis to the ejaculatory duct
Functions:
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transports sperm by peristaltic movements towards different glands where
seminal fluid is added to the sperm
during ejaculation, the smooth muscle in the walls of the vas deferens contracts
reflexively to propel the sperm forward into the urethra
Seminal vesicle
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two small glands that lie directly above the prostate gland, attached to the vas
deferens
Functions:
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secrete the bulk of the seminal fluid, a combination of sperm and seminal fluid is
called semen
the fluid is rich in fructose an important source of energy for sperm to swim
the fluid also prevents sperm from drying out
Cowper’s gland
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secretes a sticky fluid before actual ejaculation of sperms and semen
Functions:
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the fluids are alkaline to protect sperms from the acidic environment of the male
urethra
the fluids helps in penetration by providing lubrication
the fluids make the semen alkaline, to increase survival in the acidic conditions of
the vagina
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Prostate gland
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located between the bladder and the penis just in front of the rectum
the urethra runs through the centre of the prostate from the bladder to the penis
Functions:
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secretes prostate fluid, one of the components of semen which energises the
sperm
secretes the alkaline fluid found in semen and makes the vaginal canal less
acidic
it squeezes fluid into the urethra as sperm move through during ejaculation
Urethra
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tube connected to the bladder
it travels through the penis and thus carries both urine and semen to the exterior
Functions:
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carries semen from the sperm duct to the penis during ejaculation
urine also passes through the urethra from the bladder during urination
Penis
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external male organ that transfers semen from the male to the female and also
serves as the urinary duct
Functions:
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it is the organ which is inserted into the vagina during sexual intercourse
contains erectile tissue (a collection of blood vessels separated by sheets of
connective tissue)
during an erection the blood vessels fill up with blood and causes the penis to
become erect
Specific objective: describe fertilisation as:
The fusion of the nuclei from a male gamete and a female gamete (egg cell/ovum) to
form a zygote
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During fertilization:
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the sperms propel themselves using its tail and swim through the cervix up to the
uterus and into the oviduct
egg cell released during ovulation has to be inside the oviduct for fertilization to
occur
when sperms reach the egg cell, only one sperm cell enters head first and loses
its tail in the process
the membrane around the egg cell becomes impermeable to other sperms that is
why only one sperm is capable of entering the egg cell
the haploid nucleus of the sperm cell fuses with the haploid nucleus of the egg
cell in the oviduct and thus fertilization occurs
the fertilized egg becomes a diploid zygote
Specific objective: describe the function of the placenta and umbilical cord in relation
to exchange of gases dissolved nutrients and excretory products
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after fertilization the fertilized egg (zygote) divides by mitosis and develops into
an embryo
the developing zygote and embryo are moved by the cilia in the oviduct towards
the uterus for implantation
implantation is the attachment of the developing embryo in the thickened uterine
walls
this thickened uterine walls is called the placenta
the placenta has two arteries and one vein (making up the umbilical cord)
connecting it to the developing foetus
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Functions of the placenta
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acts as an endocrine gland to release hormones (oestrogen and progesterone)
during pregnancy
it is used for the exchange of substances between the mother and the foetus
oxygen and nutrients like amino acids, glucose, water, vitamins) diffuse across
the placenta from the mother’s blood to the foetal blood
placenta allows the passage of antibodies from the mother to the foetus
waste excretory products like carbon dioxide and urea diffuse through the
placenta from the foetal blood to the mother’s blood
placenta forms a barrier between the blood systems, preventing maternal and
foetal blood systems from mixing
placenta prevents pathogens like viruses from passing through to the foetal
blood, but a damaged placenta can fail to do that
Functions of the umbilical cord
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connects the placenta to the navel of the foetus
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consist of two arteries and one vein which forms the only connection between
the foetus and the placenta
One vein inside the cord takes substances to the foetus from the placenta
Two arteries in the cord take substances from the foetus to the placenta
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The uterus, amniotic sac and amniotic fluid
Uterus
- organ which holds the
foetus during pregnancy
- muscular organ that
increases its size and
thickness during
pregnancy
Amniotic sac (amnion)
- strong membrane that
surrounds the foetus during
pregnancy and keeps
amniotic fluid in
Amniotic fluid
- fluid that surrounds the
foetus during pregnancy
- allows foetus to float and
move in the fluid
- produces and secretes
amniotic fluid
- keeps the foetus in a space
free from the effects of gravity
and pressure
- prevents entry of
pathogens from the vagina
- helps minimise temperature
changes
- collects foetal urine
- prevents drying out of the
foetus
- provides a sterile
environment for the foetus
Specific objective: outline the growth and development of the foetus in terms of
increasing complexity in the early stages and increasing size towards the end of
pregnancy
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Growth refers to the permanent irreversible increase in size and number of cells
of an organism
Growth is a result of cell division (mitosis)
Development refers to the stage of growth which includes changes in structure
and appearance of cells (differentiation and cell specialization)
The new cells forms new tissues and organs and the body becomes more
complicated and increases in complexity
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Stage in pregnancy
Week 1 to week 4
Week 5 to week 8
Month 3
Month 4
Month 5
Developmental changes
- amniotic bag and sac forms around embryo
- placenta and umbilical cord develops
- foetus forms a primitive face with large dark circles for
eyes
- mouth, lower jaw and throat start developing
- embryo size by week 4 is about 4mm in length
- other facial features including arms, legs and fingers
start forming
- the brain, spinal cord and other central nervous tissues
forms
- bone start to replace cartilage
- head enlarges in proportion to the body
- digestive tract and sensory organs begin to develop
- at 6 weeks a heartbeat can be detected
- embryo size by week 8 is about 2.5cm in length
- embryo is now called a foetus with fully formed limbs
- foetus can open and close its fists and mouth
- finger and toe nails began to develop
- beginnings of teeth start forming as well as reproductive
organs
- organs are present and foetus is fully formed and will
continue to mature in order to become functional
- foetus is about 9cm in length
- heart beat is audible
- limbs and nails and facial features are formed
- teeth and bones become denser
- nervous system start to function, foetus can stretch,
suck thumb or yawn
- reproductive organs are fully developed and can be
seen on ultra sound
- foetus is about 17cm in length
- foetal movement can be felt by mother as muscles are
developing
- hair grows on head
- baby skin is covered with a protective coating that is
shed before birth
- foetus is about 25cm in length
279
Month 6
Month 7
Month 8
Month 9
- foetal skin is reddish, wrinkled and veins can be seen
- finger and toe prints become visible
- foetus responds to sound by moving or increasing pulse
rate
- foetus born premature after week 23 can survive with
intensive care
- foetus is about 36cm in length
- foetus continues to mature and develops reserves of
body fat
- foetus changes position frequently and responds to
stimuli like sound, pain and light
- fluid begins to diminish
- foetus is about 38cm in length
- as foetus continue to mature, it starts kicking more
- brain develops rapidly and foetus can see and hear
- most internal systems are well developed, the lungs
may still be immature
- foetus is about 43cm in length
- growth and maturity continues, with lungs nearly fully
developed
- reflexes are coordinated (blinking, closing eyes, turning
head and responding to touch n sound)
- foetus moves less due to tight space, position is
changed with head facing down
- foetus drops in the pelvis area, skin is smooth
- foetus is about 49cm in length and ready to be born
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Specific objective: outline the processes involved in labour and birth
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At about forty weeks, the foetus is about to be born
Labour approaches as the mother ‘s body undergoes changes to prepare for the
birth of the baby
Labour is the time period that precedes birth
Oxytocin a hormone that stimulates contractions is secreted by the pituitary gland
Stage one: labour
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Contractions of the uterus occur (can be gentle at first with hourly intervals
between contractions)
As contractions become regular, the foetus’s head moves down into the
pelvis and cervix start to get thin and dilate
Mucus plug is pushed out, mucus near the cervix protects baby from bacteria
Next water will break (the amniotic sac ruptures and releases fluid as
contractions continue
The dilation of the cervix and position of the baby will be monitored
periodically
Contraction become longer, stronger and more frequent and the cervix
stretches the last few centimetres (cervix becomes fully dilated at ten
centimetres wide)
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Stage two: birth or delivery of the baby
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
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
Involves pushing the baby out through the vaginal canal
Pushing during contractions pushes the head of the foetus out of the
cervix into the vagina
Small cuts can be made on the vagina by the doctor to enlarge the vaginal
opening
Once the baby’s head fully appears, one more push is required to push
the rest of the baby’s body out
Immediately after delivery, the baby is expected to cry within one minute
Crying is important as it allows the baby to draw breath into its lungs and
start breathing on its own
After the baby starts to breath, the umbilical cord is clamped and cut
Stage three: delivery of after birth




Involves delivery of the placenta and umbilical cord (afterbirth)
Contractions will begin five minutes after birth signaling the time to deliver
the placenta
Placenta is expelled through the vagina
Any cuts and tears will be repaired by the doctor
Specific objective: discuss the advantages and disadvantages of breast-feeding
compared to bottle-feeding
Advantages of breast-feeding or
breast milk
Disadvantages of bottle-feeding or
formula milk
- always the correct temperature
- needs to be heated or cooled to correct
temperature
- contains no antibodies
- sometimes not available in shops
- contains protective antibodies
- more convenient and is always
available
- free of charge
- digested more easily
- free of harmful pathogens
- suckling from breast helps baby to
feel safe and develop a sense of
security
- can protect mother against ovarian
cancer
- very expensive
- more difficult to digest
- can contain harmful pathogens
- suckling from a bottle lets baby feel less
safe and does not help to develop a
sense of security
- no protection against ovarian cancer
282
Advantages of bottle feeding or
formula milk
- no pain because of cracked nipples,
bite marks or mastitis
- one can comfortable feed a baby
with a bottle any place
- entire family can share in the
experience of feeding the baby
- mothers are not required to change
their diet and can eat whatever they
want
- the mother knows exactly how much
milk the baby has consumed
Disadvantages of breast feeding or
breast milk
-breast-feeding comes with pain because
of cracked nipples, bit marks or mastitis
- feeding in public can be uncomfortable
- only the mother can feed the baby
- mothers are required to change their
diet and cannot eat whatever they want
- the mother does not know the exact
quantity the baby has had resulting in
over or under feeding
4.3.3 Methods of birth control and increasing fertility
Specific objective: outline the following methods of birth control: natural methods:
limited to abstinence, monitoring body temperature and cervical mucus; chemical
methods: IUD, contraceptive pill, implant and injection; barrier methods: limited to
condom, femidom and diaphragm; surgical methods: vasectomy and tubal ligation
Birth control (contraceptive)
The use of medications, practices or devices to prevent pregnancy
Birth control methods:
-
Natural methods
Chemical methods
Barrier methods
Surgical methods
283

-

-
Natural methods
Birth control that rely on observations about the woman’s body and menstrual
cycle
Examples include:
Abstinence
No sexual contact of any kind involved
It is 100% effective in preventing pregnancy and sexually transmitted infections.
Monitoring body temperature
A woman’s temperature rises slightly around the time of ovulation and this can
alert a woman when she is most fertile and likely to fall pregnant
A sustained increase in temperature that lasts for three days indicates that
ovulation has taken place
A woman can therefore identify when ovulation has passed and the fertile stage
of her menstrual cycle has finished
Cervical mucus
Cervical mucus changes in thickness throughout a woman’s menstrual cycle
As egg develops in the ovary, oestrogen is released making the mucus thinner
and slippery (sperms can then swim easily to the egg)
After the egg is released, an increase in progesterone causes mucus to thicken
and act as a barrier to other sperm
A woman can track the state of their mucus to give them insight into their fertility
Chemical methods
Involves the use of chemical hormones to control the menstrual cycle or
ovulation
Examples include:
IUD (intrauterine device)
A small, T-shaped device which is placed through the vagina directly in the
uterus by a doctor
The device contains hormones which prevents implantation
Some IUD releases copper ions which immobilise the sperm and makes it hard
for sperm to swim through the uterus
Once inserted in the uterus, it can stay in place for 5-10 years
After it is removed, the contraceptive wears off and one can become pregnant
fast
A cost effective option that is effective against pregnancy but not against STIs
Contraceptive pill
-
The 21 day pill is a birth control method designed to be taken orally every day, at
the same time of the day in order to prevent pregnancy
284
-
It contains a combination of oestrogen and progesterone hormones
The pill prevents the body from ovulating so no egg is available for fertilization to
occur
It prevents ovulation by inhibiting the production of FSH
The mini pill (28 day pill) contains progesterone only
Contains 28 pills which are all active
The progesterone thickens the mucus at the cervix, making it hard for sperm to
swim to an egg
This prevents the entry of sperm through the cervix into the uterus
The morning after pill is an emergency contraception used after unprotected
sexual intercourse has taken place
The morning after pill delays ovulation as well as prevents implantation or
increase the cervical mucus to prevent sperm from accessing the egg cell
Implant
-

-
Is a flexible plastic rod that is implanted under the skin (upper arm) by a health
care professional
The implant prevents pregnancy by releasing low but steady dose of
progesterone hormone to thicken the cervical mucus
The implant suppresses ovulation
It offers effective long term contraception against pregnancy but not against STIs
Injection (the shot)
A three-monthly contraceptive injection containing synthetic progesterone
hormone
Patients returns every twelve weeks for an injection by a doctor
The injection releases progesterone in to the blood stream which prevents
ovulation
It also thickens the cervical mucus to make it difficult for sperm to swim through
the cervix and thins the uterus lining to prevent implantation
Barrier methods
Involves physical barriers designed to stop sperm from entering a woman’s
uterus
Examples include:
Male condom
Tube of thin latex rubber material that is rolled onto an erect penis before sexual
intercourse
Semen is captured at the tip of the condom so that no sperm cells reaches the
ovum therefore preventing fertilisation
Condoms are very effective against pregnancy and STIs if used correctly
285
Female condom (femidom)
-
A long pouch with flexible rings that is inserted into the vagina before intercourse
The femidon covers the cervix, vaginal canal and semen is trapped inside the
femidom and thus sperms cannot reach the ovum and no fertilization can occur
They are very reliable if used correctly
Diaphragm
-
-
A soft, silicon rubber dome shaped device folded in half and inserted vaginally to
cover the cervix
It is inserted no more than 3 hours before intercourse and should be removed 824hours after intercourse
Cervix is covered to prevent sperm from reaching the uterus and oviduct to avoid
fertilisation
It works best with a spermicide (cream that kills sperm) and thus make them very
effective and reliable
It gives no protection no protection against STIs

Surgical methods
-
Involves carrying out surgical procedures on the reproductive organs
Examples include:
Vasectomy (male sterilisation)
-
A surgical procedure for male sterilisation
The vas deferens are cut and tied to prevent sperm from entering into the urethra
The testis will still produce sperms but will be broken down and absorbed by the
body so there will be no sperm in semen
It is considered a permanent method of contraception
It does not protect against STIs
Tubal ligation (female sterilisation)
-
A surgical procedure for sterilisation in which the woman’s fallopian tubes are
clamped and blocked or cut and sealed
This prevents sperms from reaching the eggs
Tubal ligation is considered a permanent birth control method
It is effective at preventing pregnancy but not STIs
286
Specific objective: outline the use of hormones in contraception and fertility treatments
-
Fertility is the natural capability of humans to produce offspring
Infertility is the failure to achieve pregnancy after twelve months or more of
regular unprotected sexual intercourse
Reasons why a couple may have difficulty conceiving a baby
In females
- Blocked oviducts
- No egg production
- An irregular menstrual cycle
- Hormonal insufficiency that
inhibits ovulation
- Unsuitable uterine environment
for implantation
- No proper development of the
uterus lining
- Post menopause
In males
- No sperm production
- Produces defective sperm cells
that cannot swim
- Low sperm count
- Blocked sperm ducts
- Reduced testosterone levels
- Due to STIs like gonorrhoea
Use of hormones in fertility treatments
-






This involves the use of using substances like fertility drugs to treat women who
experience difficulty in becoming pregnant
Fertility drugs causes the body to release hormones that trigger or regulate
ovulation
Fertility treatments are given to women because they don’t produce enough FSH
that allows their eggs to mature
The fertility drugs contain FSH and LH
A woman is injected with this drugs during the early stages of the menstrual cycle
FSH treatment will stimulate production, development and maturation of egg cells
LH stimulates ovulation
More eggs are then released which increases the chance of fertilization
287
Specific objective: discuss the social aspects of contraception and fertility treatments
Positive social implications and aspects of using fertility drugs


Enables infertile couples to have children
Can help increase populations in countries with falling birth rates
Negative social implications and aspects of using fertility drugs










Being infertile and undergoing fertility treatments can be very stressful for
couples wanting to conceive
Fertility treatments are expensive especially when it is repeated before success
Sometimes fertility drugs do not work and so infertility will not be treated
Fertility drugs can result in multiple births while couples only wanted one child
Multiple births are often risky and babies can be born prematurely and this can
bring financial hardships on parents
Fertility treatments can increase the human population which is already over
populated
There are many religious objections to the use of fertility drugs
Unused embryo may die or used in research and this may be considered as
murder
Social problems may arise from the donation of embryo from one woman to the
other
Elderly parents may raise concern about the child’s welfare in childhood and
beyond if the elderly parents die
4.3.4. Sexual transmitted infections (STIs)
Specific objective: define sexually transmitted infections as:
An infection that is transmitted via body fluids through sexual contact
Specific objective: describe the methods of transmission of the human
immunodeficiency virus (HIV) and the ways in which it can be prevented from
spreading
Methods of transmission of HIV


During unprotected sexual intercourse with an infected person as virus is
found in semen and vaginal secretions of infected persons
Drug users sharing contaminated hypodermic needles and syringes infected
with blood
288





Health workers getting stuck with an HIV contaminated needle or sharp object
During blood transfusions if given untested blood
An infected mother can pass it to the foetus during pregnancy if the placenta
is damaged
An infected mother can pass it on to the baby during labour and birth because
of blood contact
An infected mother can pass the virus to the baby during breast feeding
Methods of preventing HIV from spreading







Through abstinence
Using barrier methods of contraception like condoms and femidom
Drug users to use sterilised needles and not share needles
Ensure blood is screened and tested before given to anyone else
Promote health education to create awareness of the dangers of HIV
The use of Antiretroviral (ARV) drugs to prevent the spread of the virus in the
body
People to know their status by being tested
Specific objective: discuss the increased vulnerability of Namibians to other
illness due to the increased prevalence of HIV and AIDS






HIV and AIDS lead to a weakened immune system and reduced immunity to
diseases because white blood cells are destroyed
HIV and AIDS reduces the action of phagocytes and lymphocytes
This results in less antibody production and less phagocytosis
B-lymphocyte response is low, so TB and pneumonia pathogens can multiply
faster and are not destroyed before it causes diseases
The immune system is unable to fight TB and other opportunistic diseases
AIDS is a syndrome and not a disease, it is opportunistic as it results in an
opportunity for other infections because of a weakened immune system
289
Specific objective: discuss the socio-economic consequences of the HIV and
AIDS pandemic for Namibia







Loss of young skilled workers affect the country’s economy negatively due
to reduced labour force
Lower productivity as people are too sick to work and thus absent from
work
Cost pressure for companies increases as they have to provide benefit
payments for sick workers
Increases costs on medical care, drugs and funeral expenses
Healthy family members may miss work or school to take care of AIDS
sufferers
Cost incurred to take care of orphans
Indirect costs incurred to train and recruit new members to replace absent
workers
4.3.5 Growth and development
Specific objective: define growth
Growth is the permanent irreversible increase in size and dry mass of an
organism.
Specific objective: define development
Development is an increase in complexity of an organism as it grows.
Specific objective: describe the environmental conditions that affect
germination
Germination: the process by which an embryo plant grows from a seed after a
period of dormancy.
Environmental conditions needed for seeds to germinate:

-
Water
Water starts the germination process by breaking the dormancy of the seed
Water softens the seed coat and causes the seed to rupture for the embryo plant
parts to emerge
Water enters the seed through the micropyle, and activate enzymes to break up
stored insoluble nutrients into smaller soluble parts
Water acts as a solvent to transport soluble nutrients to the growing embryo
(radicle and plumule)
290
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
-
Suitable temperature
Suitable temperature provides an optimum temperature for enzymes to work
faster
A suitable temperature increases the metabolic rate in the seed to digest and
break down stored nutrients
Oxygen
Oxygen is needed for respiration of the embryo
Stored nutrients like glucose, fats and oils are used to release energy for the
growth of the embryo plant
That is why there is a decrease in dry mass during germination as some nutrients
are being used
Specific objective: investigate the environmental conditions necessary for seed
germination
Practical activity
Materials:
-
Five small paper cups
One black piece of felt
Pens
Plastic wrap
- fifteen viable bean seeds
- refrigerator or cooler
- cotton balls
- rubber bands
- potting soil
- masking tape
- petroleum jelly
Procedure:
1. Add potting soil to the cups so that each is about half full
2. Place 3 bean seeds in each cup at a depth equal to the length of the seeds.
Space seeds around the edge of the cup.
3. Label the cups 1 to 5:
Cup1: seeds deprived of water at room temperature
Cup 2: seeds watered at room temperature
291
Cup 3: seeds watered at cold temperature
Cup 4: seeds watered at room temperature with no light
Cup 5: seeds watered at room temperature with no oxygen
4. Add the following:
Cup 1: don’t add anything to this cup during the experiment other than the seeds
Cup 2: add water until the soil is damp. Do not soak the soil, water this cup each
day for the next 2 days
Cup 3: add water until the soil is damp. Water this cup each day for the next 2
days. Place this cup in a refrigerator or in a cooler for the rest of the experiment
Cup 4: add water until the soil is damp. Water this cup each day for the next two
days. Place a piece of black felt over this cup except for when you are watering
it.
Cup 5: add water until the soil is damp. Place cotton balls that have been
covered with petroleum jelly so they completely cover the soil surface. Smear a
layer of petroleum jelly around the top rim of the cup. Place a piece of plastic
wrap on top of the petroleum jelly so that it completely covers the opening of the
cup. Use a rubber band to firmly hold the seal in place and keep out oxygen.
5. Place the cups in a location where they will not be disturbed for the next 3
days
6. Each day for the next 2-3 days, observe what is occurring in each cup and
record it in a data table
Conditions for germination
Number of seeds that
germinated
After 24 hours
After 48 hours
After 72 hours
Cup A
Cup B
292
Cup C
Cup D
Cup E
7. Conclusion made from the data recorded.
Possible questions needing answers in the conclusion:
-
In which cups did germination occur and why; is light required for germination;
Is oxygen required for germination? Explain
Is water required for germination? Explain
Can seeds germination under cold conditions? Explain
List the conditions necessary for the germination of a seed
At what point in their development do you think plants must receive light to
survive and why?
How could you manipulate the variables to design a new experimental trial
Specific objective: compare the percentage germination of seeds stored for
different times or under different conditions
Percentage germination: the overall percentage of seeds that germinate to produce
a live plant.
-
Every plant species is adapted to certain conditions, thus its seeds are also
adapted to different storage conditions
Storage conditions can affect its germination percentage
Practical activity to compare the germination percentage of seeds stored under
different conditions
1. Take 20 quality seeds of the same species (maize, peas, beans) and store ten
seeds at 40C in a refrigerator and ten seeds at room temperature for the same
amount of time
2. Wash the seeds and soak in distilled water for 2-3 hours, rinse and remove
excess water with tissue paper. Allow seeds to dry
3. Sow the seeds kept in the refrigerator in a seed tray or pot labelled ‘’4 0C’’; and
saw the ones kept at room temperature in a seed tray or pot labelled ‘’room temp’’.
Use the same soil to plant the seeds in
4. Thoroughly water the seed trays or pots and place in the same conditions to
germinate. Supply the same amount of water regularly for 15 days
5. Find the rate of germination after 15 days. Calculate the percentage for both
treatments and compare them
293
Specific objective: describe methods of measuring growth in a herbaceous plant
and a mammal
Measuring growth in a herbaceous plant (measuring the height)


Using a ruler for smaller plants or measuring tapes for taller plants
Place the ruler at the base of the plant to its highest point and record the
measurement
Repeat measurements every two to three days
Calculate the average growth rate using the formula:
S2 – S1 ÷ T where S1 is the first measurement, S2 the second measurement
and T is the number of days between each
Measuring growth in a herbaceous plant (measuring dry weight)
Dry mass is the mass of an organism (plant) excluding the fluctuating water
concentrations. It is a more reliable measure than fresh mass
Procedure:









Plant a large number of seeds of the same plant in identical conditions in a pot or
large tray
Conditions like temperature, water supply, humidity and nutrients must be the
same for all seedlings
Every week randomly remove a sub sample of about 5 seedlings from the pot
Wash and remove all soil from the roots
Heat the seedlings in an oven at a suitable temperature of about 60-1100C to
remove all water from the plants
Cool the plants in a dry environment
Then weigh the dry plants (the plants now contain no water but dry mass only)
Every week record the dry weight of the five plants and calculate the average
Compare mass for the different days, an increase in dry mass show growth
Measuring growth in a mammal (measuring height, length, width and girth)
Growth of animals can be determined by measuring:
-
Its height
Its length
Its width
Its girth (measurement around something like a chest or waist)
294



Various measuring instruments like rulers, measuring tapes can be used
When determining height, ensure the animal is standing straight on levelled
ground
Measurements can be taken weekly or monthly to determine growth
Specific objective: measure and record the growth of an annual herbaceous plant
from sowing to maturity, and use these results to construct and explain a growth
curve
Practical activity to measure growth in a mammal using live weight



Weight of a chosen animal can be taken daily, weekly or monthly using a
measuring scale
The animal can be weighed before the morning feeding
Record the weight in a table and draw a line graph of the weight increase
Practical activity to measure growth of an annual herbaceous plant



An annual herbaceous plant dies completely at the end of the growing season or
after it has flowered and produced seeds or fruits
Growth can be measured using the dry mass procedure or height of the plant
starting from sowing stage until maturity
Draw a growth curve of the weekly measurements
Specific objective: discuss the advantages of different methods of measuring growth
Advantages of measuring height,
length, width and girth in organisms
-
Disadvantages of measuring height,
length, width and girth in organisms
Quick and easy
Does not kill or damage the
organism
Can be done anywhere suitable
-
295
Roots and shoots can vary in length
without an increase in the mass
Only measures one direction of
growth in plants
Two leaves of the same mass can
differ in length
Difficult to keep animals still to get
accurate measurements
Advantages of measuring dry mass of
organisms
- Gives accurate measurement
- Not influenced by water content
- Reflects permanent increase in
mass
- Not influenced by variation in
growth patterns
Advantages of measuring wet mass of
organisms
- quick process
- does not kill the organism
- more accurate than height and
length
Disadvantages of measuring dry mass
of organisms
- Time consuming process
- It kills the organism
- Two leaves of the same mass can
differ in length
Disadvantages of measuring wet mass
of organisms
- plants need to be uprooted to get
measurement
- difficult to measure large organisms
- water content in organism’s body is
variable
4.4 Inheritance
Specific objective: define inheritance as:
The transmission of genetic information from generation.
Specific objective: define chromosomes as:
Thread-like structure of DNA, carrying genetic information in the form of the
genes
Specific objective: define homologous chromosomes as:
Chromosomes having the same genes at the same loci but not identical with
different alleles.
296
Specific objective: define the terms haploid nucleus and diploid nucleus:
Haploid nucleus in a nucleus containing a single set of unpaired chromosomes
(e.g. in gametes)
Diploid nucleus is a nucleus containing two sets of chromosomes (e.g. in body
cells)
Specific objective: define gene as:
A length of DNA that codes for a protein
Specific objective: allele as:
A version of the same gene
Specific objective: explain the concept of genetic code with reference to the
sequence of bases in a gene to form amino acids in a correct order for a specific
protein



It is an order or sequence of DNA bases which determines the sequence of
the amino acids in protein.
There are three bases code for one amino acid.
The code defines how sequences of DNA bases called codons specify which
amino acid will be added next during protein synthesis.
Specific objective: explain how DNA controls cell function by controlling the
production of proteins, antibodies and receptors for neurotransmitters

-
By controlling the production of proteins
The nucleotide sequence that make-up DNA are a “code” for cell to make
hundreds of different type of proteins.
It is these proteins that function to control, regulate cell growth, division,
communication with other cells and other cellular functions.
297

-
DNA in cells produce antibodies.
An antibody is a protein used by immune system to identify and neutralize
pathogens like bacteria and viruses.
These proteins (antibodies) are made in plasma cells due to the genetic code on
the DNA which carries the code for its production.

-
By controlling the production of antibodies
By controlling the production of receptors for neurone transmitters
A gene or length of DNA does not produce a behavior, emotion or even a fleeting
thought.
DNA sequence codes for a specific protein and some of these proteins have a lot
to do with the behaviors, feelings and thoughts.
These proteins include hormones (which carry messages between cells) and
neurotransmitters (which carry messages between nerve cells).
They also include receptors that receive hormone and neurotransmitter
messages and the enzyme that synthesize and degrade those messages.
4.4.2 Monohybrid inheritance
Specific objective: define the terms:
Genotype as the genetic make-up of an organism in terms of the alleles present
Phenotype as the observable features of an organism.
Homozygous as having two identical alleles of a particular gene.
Heterozygous as having two different alleles of a particular gene.
Dominant as an allele that is expressed if it is present.
Recessive as an allele that is only expressed when there is no dominant allele of
the gene present.
298
Specific objective: Calculate and predict the results of monohybrid crosses
involving 1:1 and 3:1 ratios (use a genetic diagram)
Monohybrid crosses involving 1:1
299
Monohybrid cross involving 3:1
Specific objective: describe the inheritance of sex in humans with reference to XX
and XY chromosomes.

Sex (gender) is determined genetically.

There are two kinds of chromosomes: the autosomes and the sex chromosomes.

There are 23 pairs of chromosomes in humans: 22 autosomal pairs and one pair
of sex chromosomes.

The autosomal pairs are numbered 1 to 22 according to their size, while the sex
chromosomes are called X and Y.

Females are determined by two identical sex chromosomes, having a genotype
XX.

Males are determined by using X in combination with the second kind of sex
chromosome Y. Their genotype is XY.
300
The human karyotype
301
Specific objective: Describe how to use a test cross to identify an unknown
genotype.

A test cross is a way to determine the genotype of an organism.

A test cross is a cross of an individual of an unknown genotype with an
individual of a known genotype (homozygous recessive).

The pattern of observed phenotypes in the offspring can determine the unknown
genotype of the parent.
E.g. in humans, brown eyes (B) are dominant over blue eyes (b).
- When you look at someone who has a dominant trait, you cannot tell if they are
homozygous or heterozygous for that trait.
- For example, if someone has brown eyes and brown is the dominant trait, they could
be BB (homozygous) or Bb (heterozygous).
How can one use test cross to determine if the trait is BB or Bb?

Always start the test cross with a purebred homozygous (BB or bb) in the P1
generation.

Take a recessive (in this case blue eyes, or bb) and cross it with brown eyes
showing the dominant trait.

If in the F1 generation of offspring are all dominant, then the dominant parent is
probably a BB. If there are any recessives in the F1, then the dominant parent
has to be a Bb.
-
If a known parent is homozygous recessive (bb) and an unknown parent is
homozygous dominant for a trait then all offspring will be heterozygous and show
the dominant trait.
If a known parent is homozygous recessive (bb) and an unknown parent is
heterozygous dominant for a trait then 50% of the offspring will be dominant and
50% recessive.
-
302
Specific objective: Predict the results of monohybrid crosses involving codominance or sex- linkage and calculate phenotypic ratios (use genetic diagrams).
Co- dominance is when the two alleles of a gene both have an effect when they are
present together (equally dominant).

Neither of the two alleles for a particular trait is more dominant over the other
(equally dominant).

Example: Dark skin is dominant in one parent. Light skin is dominant in the
other parent. Breeding results in the skin colour not dark or light, but in between”.

In some breed of cattle, two of the alleles for coat colour are co-dominant (red
coat and white coat give a roan coat). Roan is a colour produced by a mixture of
red and white hair.

The letter C is used to indicate co-dominance.

Human blood groups A and B are co-dominant, that is why a child with alleles A
and B will have a blood group called AB

Blood group alleles are written as superscripts on the letter I e.g. IAIB

Sex-linked characteristics are those characteristics in which the gene responsible
is located on the sex chromosome

Examples of a sex-linked characteristic include a genetic disorder called
Haemophilia and colour blindness

Sex linked alleles are indicated as superscripts of the sex chromosomes X or Y
e.g. XHXH
303
304
Specific objective: Investigate inheritance of one or more characteristics using for
example tomatoes or genetic maize to stimulate genetic crosses.
-
A homozygous maize plant with black kernels is crossed with a homozygous
maize plant with yellow kernels.
305
4.4.3. Variation
Specific objective: define variation as:
The differences between individuals of the same species
There are two types of variation:


Continuous variation
Discontinuous variation
Specific objective: state that phenotypic variation is caused by both genetic and
environmental factors



Phenotypes are the traits or characteristics of organisms that are observable
Some phenotypes such as blood type cannot be seen on the outside
Other examples of phenotypes include size, colour, shape, gender etc.
What causes the different phenotypes?



Genes inherited from parents
Environmental factors
A combination of genes and environmental factors
Phenotypic variation


The variation that exists in a population
Examples of phenotypic variation include eye colour, tongue rolling ability, hair
colour, weight, leaf shape etc.
Specific objective: describe continuous variation as influenced by the environment
and genes as illustrated by height in humans
Continuous variation
A smooth gradation between individuals


Continuous variation is caused by genes and environment factors
Histograms are used to show continuous variation
Examples of continuous variation
Height, hair colour, nose shape, fingerprints, skin colour, weight, intelligence etc.
306
Example of a histogram to present continuous variation
Specific objective: record and present the results of investigations into
continuous and discontinuous variation with reference to height in humans,
length of leaves and mass of seeds
Activity on recording and presenting results of continuous variation
a) Measure and record the lengths of the leaves below along the lines indicated.
Question: does the result show continuous or discontinuous variation?
307
b) Use a measuring tape to measure the heights of 10 randomly selected girls in
the class and record the results a table
Question: do the results show continuous or discontinuous variation of height in
the 10 girls selected?
Specific objective: state that discontinuous variation is mostly caused by genes
alone e.g. A, B, AB and O blood groups
Discontinuous variation
Shows that there is no gradual change between two extremes




there are distinct categories so there is no in between category
discontinuous variation is caused by genes only
a bar chart is used to show discontinuous variation
human blood groups (A, B, AB and O), tongue rolling, ear lobes attached or free
are all examples of discontinuous variation
Example of a bar chart to present discontinuous variation
308
Activity on recording and presenting results of discontinuous variation


count the number of people in the class that can roll their tongues or whose ear
lobes are attached and loose
present the results on an appropriately labelled graph
Loose ear lobe
attached ear lobe
Specific objective: define gene mutation
Gene mutation is a change in the base sequence of DNA




gene mutation occurs when the sequence of basis (nucleotides) is not copied
precisely in replicating a strand of DNA
this results in a change in the instructions for the formation of proteins
this results in the formation of a faulty protein or even no protein at all
once this occurred, it is replicated in the formation of further strands of DNA
Examples of gene mutation

albinism (inheriting altered genes that do not make usual amounts of the skin
pigment melanin)
Specific objective: outline the effects of radiation and chemicals on the rate of
mutation

mutation is a spontaneous or sudden change in the genetic structure of an
organism but there are factors that can increase the chances of mutation to occur
Factors (mutagens) that increases the chances of mutation


radiation such as UV rays, X rays, gamma rays, alpha and beta particles
chemicals such as tar in cigarette smoke
309
Specific objective: describe mutation as a source of variation, as shown by Down’s
syndrome


the spontaneous unpredictable changes in the genetic structure of an organism
can be a source of variation between species
down’s syndrome is a result of chromosome mutation
a down’s syndrome person receives three copies of chromosome 21 (trisomy)
instead of two copies only
this chromosome mutation (95%) originates during meiosis to produce egg cells
in the mother
a zygote will have 47 chromosomes after fertilization instead of 46




Individuals with Down’s syndrome
often have learning difficulties
have thick set bodies
have flattened nose and face
have short fifth finger which curve inwards



Karyotype showing trisomy
Specific objective: describe the symptoms of sickle-cell anaemia
Sickle cell anaemia



is an inherited blood disorder where there is an abnormality in the oxygen
carrying protein haemoglobin (haemoglobin S) in red blood cells
the haemoglobin S changes the shape of red blood cells to form an abnormal
sickle (crescent) shape
red blood cells are normally bi-concave disc shaped
310
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



sickle cell anaemia is caused by a recessive allele represented as (Hbs)
normal haemoglobin allele is HbA
an individual with HbA HbA (homozygous dominant) have normal haemoglobin
an individual with HbA Hbs (heterozygous) is a carrier of the abnormal
haemoglobin but the individual does not have sickle cell anaemia
an individual with Hbs Hbs (homozygous recessive) have sickle cell anaemia
Symptoms of sickle cell anaemia








red blood cells become sickle-shaped in areas of low oxygen concentration in
tissues
person has fewer red blood cells
the sickle shape red blood cells have a reduction in surface area and volume so
they have less haemoglobin
less haemoglobin results in less efficient at transporting oxygen and delivers less
oxygen to the body’s tissues
less respiration occurs because of less haemoglobin to transport sufficient
oxygen
less respiration means less energy is released and the person experiences
fatigue, breathlessness and exhaustion
sickle shaped red blood cells may clump together and get stuck in capillaries
causing blockages
this increases the chances of thrombosis (clotting of the blood in the circulatory
system)
Specific objective: explain the distribution of the sickle-cell allele in human
populations with reference to the distribution of malaria



sickle cell anaemia disease is more common in people of African, Mediterranean,
Middle East, South and Central America descent
these are areas where malaria is common
it is an advantage to be a carrier of sickle cell anaemia in a malaria prone area
311





an HbA HbA person does not have sickle cell anaemia but will be susceptible to
malaria
a carrier HbA Hbs does not have sickle cell but is less susceptible to suffer from
malaria
an Hbs Hbs person have sickle cell anaemia but will not suffer from malaria
hence, it is a selective advantage to be heterozygous in a malaria area
this is because, the malaria parasite will not be able to survive and multiply on
sickle shaped red blood cells carrying less oxygen
4.4.4 Adaptive features, selection and evolution
Specific objective: define adaptive features as:
The inherited functional features of an organism that increase its fitness
Specific objective: define adaptation as:
A process, resulting from natural selection, by which populations become more
suited to their environment over generations
Specific objective: describe the adaptive features of hydrophytes and xerophytes
to their environment
Hydrophytes (aquatic plants)
This are plants adapted to growing and
living partially or completely in water

Xerophytes
This are plants adapted for growing in dry
environments

some species have leaves totally
submerged in water while other
species have leaves that float on
the water
312
can survive extremely dry
conditions and high temperatures
Examples of hydrophytes
Examples of xerophytes
Adaptations:
- large air spaces in the leaves to keep
them close to the water surface for light
absorption
- small roots to extract nutrients from the
surrounding water through tissues
- stomata mainly found on the upper
epidermis and are usually open all the
time for easy gaseous exchange
- lower leaves lack stomata to prevent
water entering the air spaces
Adaptations:
- thick waxy cuticle to reduce water loss
and the shiny layer to reflect heat and
lower temperature around the plant
- sunken stomata in pits in the epidermis
so that moist air is trapped and reduce
evaporation
- rolled leaves with stomata inside to
reduce surface area for transpiration
- leaves covered with hairs to trap moist
air and reduce water loss
- small leaves or leaves reduced to thorns
to reduce transpiration
- thick leaves or stems which stores water
Specific objective: define natural selection as:
The selection of only the best adapted organisms for survival and reproduction
313
Specific objective: describe natural selection with reference to: production of many
offspring, variation within populations, struggle for survival, competition for resources
and reproduction by individuals that are better adapted to the environment and
passing their genes to the next generation



variation among individuals within a population of animals give some individuals
a better chance of survival in any environment
in environments with limited resources, individual organisms best suited to such
conditions are the ones able to survive and reproduce
examples of such natural selection adaptations include:
 production of many offspring
 variation within populations
 struggle for survival
 competition for resources
 reproduction
Specific objective: define evolution as:
The process of change over a period of time
Specific objective: describe evolution as the change in adaptive features of a
population over time as the result of natural selection




over successive generations, populations are modified in response to changes in
the environment
individuals with different characteristics survive and reproduce to give rise to a
new set of alleles that are passed on to their offspring
over time, this will bring about a change in the characteristics of the species and
cause evolution
such characteristics enable individuals to become better suited or adapted to
their environment
Specific objective: outline the importance of natural selection as the mechanism for
evolution


natural selection is a main mechanism of evolution of species
mutations lead to genetic variations that improves the traits of individual
organisms
314


organisms that are able to survive and adapt to environmental changes,
reproduce and pass on their adaptive genes to the next generation
hence, new organisms carry inheritable adaptive traits different from their
descents and thus evolve
Specific objective: describe the development of strains of antibiotic resistant
bacteria as an example of evolution by natural selection




Antibiotics are drugs used to destroy disease causing bacteria
Examples of antibiotics include penicillin
Antibiotics cause the cell walls of a bacterium to become weaker during its
production
The bacteria will then not be able to resist turgor and it bursts and dies
How bacteria become resistant to antibiotics (an example of evolution by
natural selection)




Bacteria becomes resistant when the antibiotic loses its ability to effectively
control or kill the bacteria
This bacteria that could not be killed by the antibiotic becomes resistant and
continue to multiply in the presence of the antibiotic
Antibiotics are widely over used and also incorrectly prescribed
Patients most often do not complete the whole course of antibiotics as prescribed
and this results in bacteria becoming resistant
How natural selection occurs





The bacteria show genetic variation in resistance to antibiotics
Bacteria with the resistant gene is more likely to survive when given the
antibiotic, while the non-resistant bacteria are killed
The resistant bacteria divide and multiply and passes on the resistant allele to
the new strains of bacteria
This increases the antibiotic resistant population of bacteria
This is an example of natural selection
315
Specific objective: define artificial selection as:
The modification of species by selective breeding




Humans deliberately breed organisms with desired characteristics which may
not occur easily with natural selection
Bred animals are called breeds and bred plants are called varieties
When two animals are crossed together, they are referred to as cross breeds
When two plants are crossed, they are referred to as hybrids
Specific objective: describe the role of artificial selection in the production of
varieties of animals and plants with increased economic importance



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

Results in plant and animal varieties that are resistant to diseases
Cows which produces the best yield and quality of milk are selected to improve
yield and quality of milk
Breeding a good male horse with excellent racing qualities with a healthy horse
results in offspring with the desired traits of racing
Selecting animals to produce the best quality meat
Crops that are climate adapted can be produced
Crops with better storage qualities can be bred to reduce product loss in
storage
Crops with larger seeds and fruits can be bred compared to existing varieties
High yielding crops can be produced to increase food production
TOPIC 5
5. Relationships of organisms with one another and with their environment
5.1. Energy flow, food chains and food webs
Specific objective: Describe the flow of energy transferred as heat or light through
living organisms and its transfer to the environment


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

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Principal source of energy is the sun in the ecosystem.
During photosynthesis, chlorophyll in chloroplasts traps light energy which is
converted into chemical energy in plants.
Chemical energy is stored in organic molecules e.g. starch, proteins and lipids.
Animals to obtain organic molecules by eating during nutrition.
Energy released from organic molecule is used in all living cells during respiration.
Some of the energy is lost to the environment as heat.
316
Specific objective: Define the following terms

Producer as an organism that makes its own organic nutrients usually using
energy from the sunlight, through photosynthesis

Consumer as an organism that gets its energy by feeding on other organisms

A food chain as showing the transfer of energy from one organism to the next,
beginning with a producer

Food web as a network of interconnected food chains

Herbivore as an organism that gets its energy by eating plants

Carnivore as an animal that gets its energy by eating other animals

Decomposer as an organism that gets its energy from dead or waste organic
material

Trophic level as the position of an organism in a food chain, food web, pyramids
of numbers or pyramid of biomass.
317
Specific objective: construct simple food chains




Food chains always start with producers (green plants)
Consumers get their energy and biomass by eating other organisms
Flow of energy in a food chain is shown by arrows
Arrows always go in the direction in which the energy flows
Examples of food chains:
Grass  grasshopper  scorpion  baboon  leopard
Seed plant  antelope  lion  vulture
Specific objective: Identify producers, primary, secondary, tertiary and quaternary
consumers as the trophic levels in food chains, food webs, pyramids of numbers and
pyramids of biomass.
Producers


Producers are responsible for the input of energy into an ecosystem.
They occupy the first trophic level.
Examples of producers: green plants, some protoctista, phytoplankton and some
bacteria.
318
Examples of producers
Consumers
Types of consumers
(a) Primary consumers



Feed directly on producers.
are plant -eaters ( herbivores)
They occupy the second trophic level.
Examples: cattle, antelope, kudu, springbok, rhino, seed-eating birds
Examples of primary consumers
(b) Secondary consumers:


Are animals that feed on primary consumers.
they occupy the third trophic level
Examples: cheetahs, lions, leopards and eagles.
319
(c) Tertiary consumers


Feed on secondary consumers.
They occupy the fourth trophic level
Examples: hawks, owls, tick birds, hyenas and vultures
(d) Quaternary consumers
-
feed on meat and flesh of tertiary consumers
they occupy the fifth trophic level
Examples: sharks eat carnivorous fish and hawks that eat owls.
Definition of omnivore
An omnivore is an animal that gets its energy by eating other animals and plants.


Omnivores are secondary, tertiary or quaternary consumers.
These are animals that feed on a mixed diet of animals and plant material.
320
Decomposers
Examples of decomposers: fungi and putrifying (decomposition) bacteria.
Food web is a network interconnected food chains




The food web below shows seven food chains interlinked together.
The food web shows the feeding relationships between these living organisms.
This food web shows that most animals can have more than one food source,
e.g. the owls feed on frogs and mice, and mice feed on grasshoppers and plants.
A food web also shows the flow of energy in an ecosystem from the producers to
the consumers.
Diagram shows how food chains can be interconnected in a complex way.
321
Specific objective: Describe how energy is transferred between trophic levels




Energy is transferred from one trophic level to the next, in the form of chemical
energy in food and lost from the food chain as heat.
The amount of available energy decreases from one trophic level to the next.
Only around 10% of energy is passed on to the next trophic level.
The rest of the energy lost in form of heat, life processes, faeces and remains
passed to decomposers.
Flow of energy in a food chain
Specific objective: explain why food chains have usually fewer than five trophic levels




Less energy is transferred from one trophic level to the next (only 10%)
The biomass gets smaller
The longer the food chain, the more energy is lost, the less energy will be
available for the organisms at the highest trophic level.
Due to a decrease of energy and biomass from the lower to higher trophic
levels, so the length of food chains is limited.
322
Specific objective: describe and interpret pyramids of biomass

Biomass is the total weight of organisms in an area (ecosystem).
A pyramid of biomass
 Shows the dry mass of organisms at each trophic level in a food chain.
 Measures the total mass (weight) of biological material at each trophic level.
Calculation of biomass
 Number of individual organisms X mass of each organism.
Interpretation of pyramids of biomass
 The length of the bars represents the relative biomass of organisms
 The larger the bar, the greater the mass
 Bottom bar show producers followed by consumers
 Biomass in each trophic level is always less than the trophic level below
323
Examples of pyramids of biomass
Specific objective: Explain why there is an increased efficiency in supplying green
plants as human food compared to the relative inefficiency in feeding crop plants to
livestock that can be used as food





90% of energy is lost during transfer from one trophic level to the next.
More energy is available at the second trophic level than at the third trophic level,
More humans can be supported and fed.
Less land is needed for growing crops and there are fewer expenses involved.
More land must therefore be used as arable land to produce food.
324
5.2. Biochemical cycling
What is cycle?



A Cycle is when a substance is removed, used and released to be reused
again.
Biochemical is a chemical element used to make biomolecules such nitrogen
and carbon.
Biochemical cycle is the cyclic movement of biochemicals between the
environment and living organisms.
Specific objective: describe the nitrogen cycle
Nitrogen cycle: is the continuous movement of nitrogen from the atmosphere through
living organisms and returning it back to the atmosphere.
Importance of nitrogen
 For synthesis of amino acids, proteins and DNA.
 Used for plant growth in form of nitrates.
 For synthesis of chlorophyll in plants used during photosynthesis
-
Nitrogen is changed into more reactive forms such as ammonia or
nitrates during nitrogen fixation process, because it is unreactive by
nature.
325
Nitrogen cycle diagrams
326
Nitrogen cycle process occurs in the following ways
 Nitrogen fixation
 Nitrification
 Denitrification
 Decomposition (Ammonification)
1. Nitrogen fixation
A. By nitrogen fixing bacteria (rhizobium)
 Bacteria live in the soil and in root nodules of leguminous plants like
peas, beans, Lucerne, clover, soya beans and vachellia sp.
 Bacteria converts nitrogen (N2) into ammonia and ammonium
compounds into amino acids.
 Some biological nitrogen-fixing bacteria in the soil convert nitrogen into
ammonia and ammonium compounds or converted into nitrates.
B. Atmospheric nitrogen fixation (lightning)



Lightning breaks up nitrogen molecules for atoms to combine with
oxygen in the air forming nitrogen oxides.
Nitrogen oxides dissolves in rain, forming nitric and nitrous acids
that combine with other salts to form nitrates.
Oxides dissolve in rain and are washed into the soil in the form of
nitrates and absorbed by plant roots.
C. Industrial nitrogen fixation
During the harbour process at 600o C atmospheric nitrogen and

hydrogen are combined to form ammonia (NH3)
 Ammonia is used directly as fertilizer (NPK Fertilizers) that contain
nitrogen which forms nitrates when added to the soil and absorbed
by plants
 Ammonia is further processed into urea and ammonium nitrate.
2. Ammonification (decomposition)
 plants and animal remains are decomposed by bacteria and fungi
by removing an amino group and hydrogen to make ammonia
during deamination
 putrifying bacteria converts nitrogen containing compounds into
ammonia (NH3) and ammonium ions (NH4+)
327
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
Nitrogenous waste like urea ,uric acid (excretory products of
animals ) are decomposed to form ammonia and ammonium
compounds
Faeces and animal manure are decomposed into ammonia and
ammonium compounds
3. Nitrification
 Nitrifying bacteria oxidise ammonia and ammonium compounds
into nitrites and then into nitrates
4. Denitrification
 Is a process that occurs in anaerobic conditions where denitrifying
bacteria converts nitrates into nitrogen gas.
 Denitrifying bacteria obtain their energy by breaking down nitrates
into nitrogen gas.
Specific objective: state the roles of microorganisms in the nitrogen cycle limited to
decomposition, nitrification, nitrogen fixation and denitrification
 purifying bacteria
Converts nitrogen containing compounds into ammonia (NH3) and
ammonium ions (NH4+)
 Denitrifying bacteria
Convert nitrates into nitrogen gas
 Nitrifying bacteria
Oxidise ammonia and ammonium compounds into nitrites and then into
nitrates
 Nitrogen fixing bacteria
Converts nitrogen (N2) into ammonia and ammonium compounds.
Some biological nitrogen-fixing bacteria in the soil convert nitrogen into
ammonia and ammonium compounds then into nitrates.
328
Specific objective: Describe the carbon cycle limited to photosynthesis, respiration,
feeding, decomposition, fossilisation and combustion
Photosynthesis

Green plants remove carbon dioxide from the air and trap it in organic
compounds (starch and glucose)
Respiration
 Plants and animals break down glucose in the presence of oxygen to release
water and carbon dioxide in the air.
Feeding
 Animals obtain carbon during nutrition of carbohydrates, fats, proteins
Decomposition
 When animal and plants remains are broken down by decomposers and purifying
bacteria to release carbon dioxide in the air
Fossilization
 When coals and oil from dead plants and animal remains are subjected to high
pressure and does not decompose completely, traps in the carbon
Combustion
 Release of carbon dioxide in air when burning fossil fuels
329
Specific objective: discuss the effects of the combustion of fossil fuels and the cutting
down of forests on the balance between oxygen and carbon dioxide
Effects of combustion of fossil fuels on balance between oxygen and carbon dioxide


Burning of fossil fuels uses more oxygen and release more carbon dioxide in the
atmosphere than which is removed
This leads to an increase of carbon dioxide concentration in the atmosphere.
Effects of cutting down of forests on the balance between oxygen and carbon dioxide




The fewer the forests, the less carbon dioxide is removed from the atmosphere
by photosynthesis
This Leads to an increase in carbon dioxide levels
This leads to a reduction in oxygen levels
Resulting in desertification and climate changes
Specific objective: discuss the possible effects of a long-term increase in the
percentage of carbon dioxide in the atmosphere and on the carbon cycle
An increased carbon dioxide in atmosphere causes global warming leading to:
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

Climate change causing unexpected floods
Heat waves lead to heat-stroke
Violent storms increases lightning strikes
Polar ice-caps melting raises the sea level
Change in sea temperature, sea animal and plants may become extinct
Changes in world weather patterns cause droughts
More evaporation of water from fertile areas results in deserts
An increase in carbon dioxide causes greenhouse effect that leads to:
greenhouse gases absorbing heat energy and cannot be released
330
5. 3. Population
Specific objective: Define population as
A group of organisms of one species that live in the same area, at the same time
Examples of populations:

A group of camel thorn trees.

All black rhinos in Etosha National park.

All cheetahs (Acinonyx jubatus) belong to the same species.
Specific objective: state the factors affecting the rate of population growth for a
range of living organisms

food supply

availability of water

predators.

diseases.

size of the ecosystem and population

weather and climate

natural disaster/drought/flood

fertility rate

mortality (death rate) and natality rate (birth rate)

immigration and emigration

those factors above are limiting factors of the population because they control the
size of the population; either speed up or slow down or reduce the size of the
population
Specific objective: Define community and ecosystem as follows:
Community as all of the populations of different species in an ecosystem
Ecosystem as a unit containing the community of organisms and their environment,
interacting together (e.g. decomposing log and lake)
331
Specific objective: identify the phases of a sigmoid population growth curve for a
population growing in an environment with limited resources and explain the role of
limiting factors to each phase
Sigmoid curve





is used when an organism is in an environment where there are no limiting
factors
the curve will follow a characteristic S-shape known as a sigmoid curve.
When a limiting factor influences population growth a sigmoid curve is formed.
even when environmental pressures slow the rate of growth.
Sigmoid curve is a graph that has a lag phase, exponential phase, and stable
phase.
S-shaped population growth form (sigmoid growth curve).
332
Description of three phases of the sigmoid growth curve
Lag phase:

Population just introduced to the environment.

Slow growing population

Individual adapt and acclimatize

Shortage of reproducing individuals

Breeding partners do not meet

Individual seek out a new breeding partner

Population starts with very low number.
Log (exponential) phase

Rapid/increase/fast increase population

Due to availability of food/plenty of food

All conditions are favourable

There is little or no predator/few parasites

Little or no competition

Plenty of space

Birth rate exceed death , the faster the population growth
333
Stationary/stabilization phase
 Growth slow down/ is constant/stable

Zero population growth / reached its carrying capacity/maximum population the
environment can support

Increase in competition for food and space

Increase of predators/parasites

Increase in diseases/disease spreads faster

Birth rate is equal to death rate

Resources become scare
Specific objective: Describe the increase in human population size in the absence of
limiting factors and the social implications of current human survival rate on the
environment
1. Reduction of infectious diseases causes the population to increase due to the
following reasons:

Improvement in water supply.

Sewage treatment decrease diseases like dysentery, cholera, diarrhea etc.

Hygienic handling of food – fewer cases of food poisoning

general standard in cleanliness- eliminate diseases like typhoid

Immunisation against infectious diseases such as polio, meningitis, tetanus, and
chicken pox.

Antibiotics are used to treat diseases caused by bacteria such as syphilis,
gonorrhea and TB.
2. Increase in food production and supply causes and increase in human population
due to the following reasons:




More land brought under cultivation because of modern machinery and
equipment.
More efficient agriculture due to modern technology.
Fertilisers increase crop yield.
Pesticides control disease and pests and fewer crop are destroyed.
3. Natality greater than mortality
4. When natality and mortality are equal; zero population growth (ZPG)
5. Immigration
334
Social implications of current human survival rate on the environment








Famine, war and diseases may increase the death rate
More overcrowding and pollution will occur.
More unemployment.
Less land will be available for agriculture.
Growth in the population has effect on global climate change (global warming).
Increase in global demand for water for irrigation, household, and industrial use
is higher than supply.
An increased demand for oil, electricity, gas and fuel will result
An increased demand for raw materials and minerals will result
Specific objective: Interpret graphs and diagrams of human population growth
Data on human population growth is presented in different ways:
- Graphs
- pyramids.
Population pyramid
- Is a graphical representation and give information of the age and
gender for a specific country.
Shape of the pyramid

Shows the population’s potential for future growth

Bars represent number of males and females in each group

Males are on the left and females on right side
Population pyramids can have:
Classic pyramid shape:

more young people (broader base at bottom)

narrow at top (few old people)

Natality is high than mortality

Mostly in developing countries
335
Columnar shape:
Even spread of ages, mostly for developed countries.
Stable shape:

zero growth or negative population growth

fewer young people than old people

low rate of immigration

same mortality and natality rate

Very good medical care

High life expectancy.
The shape of the population pyramid convey more information about a country as it is
influenced by:

Birth and death rates and life expectancy.

Medical standard of the country.

Rate of immigration and emigration.

Educational status.
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5.4 Human influences on the ecosystem
5.4.1 Food supply, habitat destruction and pollution


Habitat is an environment or area that is inhabited by a particular species of
animal, plant or another type of organism.
Habitat destruction is the destruction of natural habitats in which living organisms
live.
Specific objective: describe the negative impacts to an ecosystem of large-scale
monocultures of crop plants and of intensive livestock production
Definitions
Monoculture is the practice of the production of a single crop in a given area year
after year.
Polyculture is where more than one crop is grown in the same space at the same
time, it is the alternative to monoculture.
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Negative impacts of large-scale monoculture of crop plants
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Monoculture increases the difficulty in nutrient recycling. All the plants are
removing the same minerals from the soil and because the crop is removed long
before the plants die and decompose, virtually nothing is returned to the soil.
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Monoculture decreases crop yields, because plants use the same minerals from
the soil each year and the soil becomes depleted of those minerals.
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It increases the use of artificial fertilisers to be added to restore soil fertility.
- But it is very expensive and increases the overhead costs for farmers in the longrun.
- Artificial fertiliser takes the place of compost, creating an increased risk for
environmental pollution.
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Planting a single type of crop gives pests a chance to destroy the whole crop.
- Monoculture leads to the quicker build-up of pests and diseases.
- If different crops are planted, these pests often die out or move away.
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Monoculture increases pesticide use.
- In monoculture, the farmer is completely dependent on chemical pesticides to
control insects and diseases.
- Pesticides make their way into ground, water or become airborne, creating
pollution.
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Farmers who practice monoculture are more vulnerable to natural disasters and
failed crops.
- A farmer who plants only maize may not be able to overcome a year of drought,
- But a farmer who plants maize, tomatoes, green beans and other vegetables may
be able to save one crop from failing.
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The lack of biodiversity of monoculture makes plants less adaptable to changes in
the environment, so a single threat can destroy an entire crop.
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Specific objective: Discuss the social, environmental and economic implications of
providing sufficient food for an increasing human global population
1. Rapid population growth leads to environmental change.
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Rapid population growth has swelled the ranks of unemployed men and women at
an alarming rate.
Due to this, a large number of people are being pushed into ecologically sensitive
areas such as hillsides and tropical forests.
This leads to the cutting of forests for cultivation, causing several environmental
changes.
The increasing population growth leads to the migration of large numbers to urban
areas with industrialisation.
This results in polluted air, water, noise and large populations in big cities and
towns.
2. Increased population means more mouths to feed which, in turn, puts pressure on
available stock of food.
 This is the reason why under-developed countries, with rapidly growing
populations, are generally faced with a problem of food shortage.
 Despite all their efforts for raising agricultural production, they are not able to feed
their growing population.
3. Food scarcity affects the economic development of a country in two ways.
 Firstly, the inadequate supply of food leads to undernourishment of the people
which lowers their productivity.
 It further reduces the production capacity of the workers.
 Secondly, the deficiency of food compels the country to import food grains, placing
unnecessary strain on their foreign exchange resources.
Specific objective: state the reasons that lead to habitat destruction
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Clearing land for agriculture
Irrigation
Mining
Urban development and infrastructure development
Natural disasters
Fire
Global warming
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Specific objective: explain the undesirable effects of deforestation on the environment
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More soil erosion, mudslides and silting up of rivers due to a shortage of tree
roots to stabilise the soil.
 Increased risk of flooding due to a shortage of trees and bushes to slow down
water flow.
 Leaching of the soil increases, because minerals are washed down and cause
soils to become infertile.
 Less decomposing of organic material also leads to infertile soils.
 Less photosynthesis occurs, so less carbon dioxide is removed from the air
which results in global warming.
 Less photosynthesis also causes a drop in oxygen levels in the atmosphere.
 Fewer trees cause less transpiration which can result in a reduction in rainfall.
 Reduction and destruction of habitats cause a decrease in animals and plant
numbers and biodiversity.
- Nesting sites for many species are reduced.
- Species are forced to settle in other species which cause food shortages.
 These decreases can cause the extinction of species and a loss of biodiversity.
 Extinction of species means the loss of genes.
 Extinction of species means the loss of sources for medicine and other
substances.
 Disruption of food chains and food webs is a direct result of deforestation.
 More pollution can occur because of burning and smoke (combustion).
 A loss of income can result because fewer tourists will visit the area.
 Removal of rainforests leads to loss of certain cultures, because people who live
there eventually disappear.
Specific objective: Explain the problems that contribute to famine including unequal
distribution of food, drought and flooding, increasing population and poverty
Famine is a widespread scarcity of food, caused by several factors.
Poverty is the scarcity or the lack of a certain amount of material possessions or
money.
Extreme poverty refers to the complete
 Unequal distributions of food
 Increasing population
 Poverty
 Drought and flooding
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Specific objective: state the sources and effects of pollution of water (rivers, lakes and
the sea) by chemical waste, discarded rubbish, untreated sewage and fertilisers
Pollution is the contamination of air, water or soil by substances that are harmful to
living things.
Main sources water pollution
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Polluted water from kitchens, bathrooms and toilets which contains human
faeces, detergents, bacteria, urine, parasites and other disease-causing
organisms
 Raw sewage can be discharged into lakes and rivers or allowed to soak into the
groundwater, which can cause this and drinking water to become contaminated
threaten the health of humans, wildlife
Specific objective: describe the undesirable effects of water pollution by sewage and
chemical waste (pesticides and herbicides), with reference to eutrophication
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Sewage can be washed away into streams, lakes, rivers and dams.
Large amounts of this organic matter, e.g. faeces are decomposed by aerobic
bacteria.
Bacteria respire and reduced the amount of oxygen in the water
Other aquatic organisms die or move away, which reduces the biodiversity.
A high content of nitrates, phosphates and ammonia, which are very soluble, are
also released into the water.
These nitrates, phosphates and ammonia in sewage then cause algae bloom
when algae grow very fast.
These algae block or reduced light, preventing photosynthesis in submerged
aquatic plants.
The aquatic plants start to die.
Decomposers, e.g. bacteria, now increase rapidly to use the dead plants as food
during the process of decomposition.
The bacteria start using up the available oxygen during respiration.
Other aquatic organisms like fish and snails start to die or start to migrate.
This process is known as eutrophication.
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Specific objective: explore how recycling can reduce pollution and improve healthy
living conditions
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Make a compost pile in your backyard.
- Adding scraps of fruits and vegetables to the heap all year round eventually
turns it into nutritious soil.
Place water bottles and cool drink cans into a recycling bin instead of the
rubbish bin.
Call your local collection centre to begin a recycling pick- up service.
Check if your town requires pre-sorted recycling.
Know the types of paper and cardboard that you can recycle.
- Almost all paper, as long as it is not food-soiled, can be recycled.
- This includes mail, newspapers, magazines, old books, egg cartoons, cereal
boxes and gift-wrap.
Know the types of plastic that can be recycled.
- If the plastic shows the famous recycling triangle it can be recycled.
Keep objects intact when recycling them.
- Shredded paper is less useful when recycled, as most of it becomes waste.
- With aluminium cans, crushing them flat can sometimes prevent them from
being recycled at all.
Collect and take your recycling bin to the curb the night before pick up.
Save returnable cans and bottles to make some money.
- In Namibia, you can return aluminium cans and glass bottles for a few cents at
your local recycling centre.
Know the things that you cannot recycle.
Such as: Pyrex glass, drinking glasses, foil-lined bags (such as chips, candy
and coffee packages), diapers, light bulbs, batteries and electronics, paint and
oils.
Specific objective: Describe the overuse and dangers of fertilisers on the land (e.g.
nitrates)
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Nitrates and phosphates in fertilisers are very soluble.
Any nitrates and phosphates not used by plants can be washed away (leaching)
into streams, lakes, rivers and dams.
These nitrates and phosphates then cause algae bloom when algae in water
grow very fast
The abundant algae can block out and reduce light, preventing photosynthesis
in submerged aquatic plants.
The aquatic plants start to die.
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Decomposers, e.g. bacteria, now increase rapidly to use the dead plants as
food during the decomposition process.
The bacteria start using up the available oxygen during respiration
This process is known as eutrophication.
Other aquatic organisms like fish and snails start to die or to start migrate.
Specific objective: Discuss alternatives to the use of large amounts of industrially
produced fertilisers.
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Use nitrogen-fixing crops (legumes like beans, peas, clover and Lucerne) to
add nitrates to the soil.
Use crop rotation by planting different crops, which need different nutrients,
each year.
Crop rotation will improve yield, soil structure and means fewer diseases and
pests of different plants and animals each year.
Fewer diseases mean we have to use fewer pesticides.
Nitrogen-fixing bacteria in root nodules of legumes put nitrates directly into the
soil which means no leaching, because the nitrates are directly available to the
plants.
The use of organic fertilisers like manure, fishmeal, bone meal, seaweed and
compost means less use of artificial inorganic fertilisers.
This is less expensive and one uses materials already available.
The use of green manure can also be a solution. Legumes are planted and the
entire plant is worked into the soil.
Green manure also improves the soil structure.
Responsible use of fertilisers is very important. Apply only the correct amount,
never apply when rain is forecasted, never spray near water sources and
never apply fertilisers to empty fields
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Specific objective: state the causes and effects of acid rain on the environment and
the measures that might be taken to reduce its incidence
Causes of acid rain
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Caused by emissions of Sulfur dioxide (SO2) and nitrogen oxide (NO2 / NO),
which react with the water molecules in the atmosphere to produce acids.
Most pollution by Sulphur dioxide is caused by large industries that burn coal,
such as power stations.
The major cause of nitrogen oxides is exhaust fumes from vehicles such as
cars and trucks.
Sources of acid rain gases (vehicles and industries)
Effects of acid rain on humans and human structures
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Acid rain is corrosive, so marble and limestone buildings, statues, monuments
and gravestones lose their shapes and features and even crumble and break.
It corrodes metals and causes rusting.
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Effects of acid rain on soil and soil organisms
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Soils become acidic which can cause leaching of minerals like magnesium and
calcium out of the soil. It, therefore, decreases soil fertility.
Acidic soils affect microorganisms, which cause less decomposition.
Effects of acid rain on water
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Acid rain dissolves and washes out poisonous aluminium and mercury from the
soil. It can end up in water sources and poison aquatic organisms.
Lakes and streams have a neutral pH level between 6 and 8. If the pH becomes
too acidic, it kills organisms like plankton, plant life, invertebrates and frogs,
which affect the entire food chain.
Damage is caused to groundwater and drinking water.
Effects of acid rain on plants
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Acid rain dissolves nutrients like magnesium, calcium, and potassium from
leaves which cause the leaves to turn brown and fall off.
Aluminium also damages fine root hairs and they are not able to absorb water
and nutrients.
This reduces the productivity of farm and forest land, because crops do not
grow well.
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Acid rain impact on trees by causing leaves to fall, damaging bark, stunting
growth and increasing the possibility of diseases.
It destroys the waxy cuticle on the leaves of plants, which means more water
loss from leaves.
Measures to reduce acid rain
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Emission (waste gases) from electricity-generating plants and power stations can
be passed through scrubbers to absorb and neutralise Sulfur dioxide.
Emissions from vehicle exhaust systems can be cleaned up by catalytic
converters, which convert nitrogen oxides into nitrogen gas, carbon dioxide and
water.
Lime or limestone can be added to neutralise acidic soils
Limestone or calcium carbonate can be added to paint on walls to neutralise
acids.
The use of fossil fuels must be limited, e.g. by using alternative fuel in cars or
electric cars.
Coal can be crushed and washed before it is burned. This can reduce Sulfur by
10%.
Using low Sulfur petrol and low Sulfur coal must be a priority. Oil can be treated
to remove Sulfur.
Specific objective: Discuss the effects of non-biodegradable plastics in the
environment, in both aquatic and terrestrial ecosystems
Biodegradable waste, e.g. food scraps and paper, will eventually become part of
the earth and soil, because bacteria and other microorganisms will break it down.
Biodegradable substances cause only short-term pollution problems.
Non-biodegradable waste, e.g. plastics, metal, glass, nylon, fibre glass, plastic
grocery bags and polystyrene cannot be broken down by bacteria and other
microorganisms and remain in the environment virtually forever.
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Disposal of non-biodegradable plastics is a problem.
It can be burned, but will remain in the soil for years.
It can be burned, but then it will release toxic gases.
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Plastic bags pose a serious danger to marine birds, turtles and mammals that
often mistake them for food.
Marine organisms like sea turtles, seabirds, shark and dolphins can get strangled
in nylon fishing lines and die.
Plastic bottles and containers are afloat all over the oceans of the world.
Effects of non-biodegradable plastics in terrestrials environments
 Herbivores like cows sometimes eat plastic, which results in the blockage of
their guts or them choking to death.
 Plastic bags clog waterways, spoil the landscape, block drains and end up in
landfills,
 The bags become wrapped in branches of trees and in fences, spoiling the
landscape.
 Producing plastic bags requires millions of litres of petroleum, which is a fossil
fuel. Manufacturing plastic bags, therefore, contribute to global warming, which is
the release of polluting gases into the atmosphere.
 Inadequate levels of recycling aggravate the waste problem.
Plastics pose many dangers to living organisms
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5.4.2 Conservation
Specific objective: Define the following terms:
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conservation as maintaining the environment and natural resources in a state
that maintains biodiversity
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Non-renewable resources are resources which cannot be used again and
again such as oil, coal and gas (fossil fuels)
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Sewage is waste matter such as faeces or dirty water from homes and
factories, which is carried away in sewers or drains for dumping or conversion
into a form that is not toxic.
Specific objective: Give reasons for the need to conserve non-renewable
resources with reference to fossil fuels
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It is important to conserve non-renewable sources of energy such as coal and oil
because they are running out.
The vast majority of resources are exhaustible and non-renewable, which
means they have a finite quantity and can be depleted if not managed properly.
If we burn all the oil to generate electricity, there will be none left for other things
we use oil for.
Although these things are naturally-occurring resources, they are in limited
supply.
Conservation efforts ensure these precious resources are not wasted and are
used more efficiently.
Fossil fuels (such as coal, petroleum and natural gas) and minerals are nonrenewable and cannot be replaced.
The over-use of natural resources has caused a serious imbalance in Nature.
Depletion of oil and gas all contribute to environmental issues such as global
warming and environmental pollution.
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Specific objective: define sustainable development as
Development, providing for the needs of increasing human population while
preserving the environment
Specific objective: state how forests and fish stocks can be sustained using
education, legal quotas and restocking
A forest is a large area of land dominated by trees.
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Forest areas need to be turned into large reserved or protected forests, which
refer to certain forests being granted protection by governments.
Ecotourism must be implemented in these reserves. An international ban on
trade in endangered forest species is another way of how forests can be
maintained.
There must be quotas and international regulations on the use of wood from
rainforests.
Education of consumers is important so that they can choose not to buy wood
from forests as well as goods manufactured from forest wood.
International aid needs to be provided to countries with tropical rainforests.
Ways must be found on how to make sustainable use of forests.
Only selective felling of trees must be allowed.
There must be a limit on the quantity of timber felled per time period.
How fish stocks can be sustained
Fish stocks are the numbers of fish in an area.
- As the human population has increased, so has the demand for fish.
- If fish are caught at a faster rate than the remaining fish can reproduce, the
numbers of fish (the fish stock) will decline
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Prevention of the disappearance of certain fish species in some areas is
by:
- Ensuring that fish populations remain at a sustainable level by
doing sustainable fishing.
- Setting fishing quotas can help to sustain fish stocks
- There must be regulations to conserve fish stocks.
- Setting fishing quotas for countries, individuals fishing vessels and
individuals fishermen limits the amount for fish which may be
caught, resulting in fewer fish being caught.
- These regulations and limits include the size of the fish one can
keep, total number of fish one can keep and time period when it is
legal to fish certain species.
- Education can help to sustain fish stocks
(Better information for fishermen and education about fishing
regulations is of utmost importance in sustaining fish numbers)
- Person, who catches fish for recreational purposes in Namibian
waters, must be in possession of a fishing permit.
- Fishing methods and fishing equipment must also not harm fish
populations.
- Legislation to limit the size of nets and increasing the size of the
holes in nets results that only mature, full sized fish can be caught
- Re-stocking of fish can help to sustain fish stocks (having
hatcheries to keep fish species)
Specific objective: outline how sewage is treated to make water safe for return to
the environment for human use
Sewage water treatment procedures
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Raw sewage and water go into the sewage collecting and pumping area.
The sewage inlet takes it to the screening area so that large, solid rubbish and
objects like tyres, wood, stones and tins/cans can be removed.
The rest of the sewage goes to the grit removal area where the grit is
removed and used as landfill.
Sewage then goes to the primary sedimentation tanks where solid matter
forms sludge, which is allowed to settle and then separated from the liquid
effluent.
The sludge is digested by anaerobic bacteria in the anaerobic digester. This
produces methane, which can be burned to power the sewage plant. The solid
matter can be dried and used as garden fertiliser.
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The liquid effluent is sprayed onto stones where aerobic bacteria and
Protoctista feed on sewage and remove harmful substances. This aerobic
stage kills anaerobic bacteria.
Clear effluent is produced which can be returned to the environment. e.g. by
pumping it into a river, or it can be purified and disinfected by chlorination to
be used again as drinking water
Sewage recycling plant
Reasons for treating sewage
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Sewage is treated to prevent the spreading of diseases as it removes harmful
bacteria and other pathogens.
Treatment helps to avoid pollution and remove harmful substances.
Sewage does not smell bad after treatment.
The water is recycled.
Eutrophication is prevented.
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Specific objective: explain why organisms become endangered or extinct, limited
to climate change, habitat destruction, pollution and hunting with reference to
poaching
a. Climate change endangers species
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Climate is usually defined as the average weather in a place. It includes patterns
of temperature, precipitation (rain or snow), humidity, wind and seasons.
Climate change refers to a change in average weather conditions.
Animals, cannot adapt very well to climate changes.
When the rainforest dries up or cool mountains in tropical zones heat up- animals
may simply become extinct.
Global warming could cause the extinction of plants and animals. All plants and
animals have a temperature range within which they thrive and outside of which
they suffer
Climate change has also been found to be a factor in habitat loss and
desertification.
Sea levels are rising and oceans are becoming warmer. Rising ocean
temperatures because of human-induced climate change could cause extinction
of oxygen – producing ocean phytoplankton, which likely would result in the mass
death of ocean animals.
Longer, more intense droughts threaten crops, wildlife and freshwater supplies.
b. Habitat destruction endangers species
- Every living organism needs a place to live, but a habitat is not just a residence, it
is also where an animal finds food, raises its young and allows the next generation
to take over.
Different ways in which humans destroy animal habitats:
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Building houses, industrial parks and shopping malls
Clearing forests to get timber
Clearing forests to plant crops and keep animals
Draining rivers to bring water to those crops
Paving over meadows to make streets and parking lots.
Pollution destroys natural animals habitats by making the food that animals eat
and the water that they drink unhealthy
Habitat destruction is the number one reason for animal endangerment and
extinction
Deforestation has killed off more species than we can identify
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C. Pollution endangers species
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Ships pollute saltwater by dumping wastes.
Oil spills kill large number of animals.
Many smaller spills and leaks go relatively undetected, but their cumulative
effects over the years also can injure wildlife.
Water is not the only element that suffers from pollution.
Factories and cars release chemicals into the air.
The chemicals are deposited on land by rain, causing pollution, including what is
known as acid rain.
Pesticides are another source of pollution.
Insects also carry pesticides. Pesticides remain in crops and in wild plants eaten
by herbivores (plant eaters).
Animals that eat herbivores get high concentrations of these chemicals in their
systems.
d. Illegal hunting endangers species
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When hunters ignore rules that regulate the number of animals that may be
hunted, they can reduce populations to the point that species become
endangered.
This is a practice known as poaching, done by illegal hunters. .
Specific objective: Describe how endangered species can be conserved limited to
monitoring and protecting species, habitats, education and captive breeding
programs and seed banks
How endangered species can be conserved
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Monitoring and protecting species
Africa has over 400 known species of endangered animals and monitoring them
is an essential and critical step in their conservation.
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Wildlife monitoring is essential for keeping track of animal
movement patterns, habitat utilisation, population demographics,
snaring and poaching incidents and breakouts.
Monitoring and protecting habitats
Habitat conservation is a management practice that seeks to conserve, protect
and restore habitat areas for wild plants and animals and prevent their extinction,
fragmentation or reduction.
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Education
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Many people are unaware of species extinction and do not realise
how their actions might be harming others.
Captive breeding programs
- Captive breeding is the process of breeding animals in controlled
environments within well-defined settings, such as wildlife reserves,
zoos and other commercial and non-commercial conservation
facilities.
Seed banks
- Seed banks serve a similar purpose for farmers and scientists who
work to conserve rare plants.
- Seed banks are where dried seeds are preserved by storing them
at very a low temperature for a long time.
- A seed bank stores seeds as sources for planting in case seed
reserves elsewhere are destroyed.
- Many botanical gardens use seed banks as a method of plant
conservation.
Specific objective: Discuss the advantages and disadvantages of tourism for
conservation
Advantages of tourism for conservation
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For many governments and nations, it is the revenue (taxes and visa fees)
generated through tourism that continues to support conservation and its
accompanying activities.
In many developing countries, protected areas like national parks rely heavily
on tourism fees (foreign exchange).Most of their income comes from tourism
revenue such as entrance fees, restaurants and accommodation fees.
Local inhabitants too have benefited from tourism. It has increased employment
opportunities (tour guides, hotel, lodge and restaurant workers) and in some
cases empower the entire community.
Tourism also drives changing attitudes towards conservation. Travelers
themselves might take on volunteer conservation tourism, communities can
help with reduction in poaching and governments can change legislation to
support sustainable tourism.
Tourism can help to maintain and increase wildlife in Namibia and can help to
prevent extinction of species and protect biodiversity.
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Tourism can raise awareness of poaching and even help in prevention of
poaching.
Disadvantages of tourism for conservation
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Globally, tourism currently contributes about 5% of greenhouse gas emissions,
primarily via airplanes or car. The remainder is derived from on-site impacts
associated with accommodation and leisure activities.
Damage to species habitats by vehicles, roads and camping sites can be a result
of tourists not obeying local laws.
Wildlife can be disturbed by human noise and movement of tourists.
Pollution by non-biodegradable materials like plastics, cans, bottles and
chemicals can negatively harm wildlife.
Conservation can cause animals to interfere with human activities, e.g. animals
can damage crops, homesteads and water supplies and injure and kill livestock.
Specific objective: explain the reasons for having conservation programs
Reason for conservation programs
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Conservation prevents the likelihood of extinction of species.
Food chains, food webs and entire ecosystems are preserved.
Conservation maintains the biodiversity (number and variety of organisms found
within an area).
All organisms remain interdependent.
Conservation has an aesthetic advantage for future generations and tourism.
Various species provide sources for medicine, drugs and oils and also serve as
food sources.
Conservation of species enable controlled breeding.
It also enables transfer of animals to repopulate new areas where they have
been depleted or exterminated.
Specific objective: investigate the impact of Rhinoceros poaching on the tourism
sector in Namibia
Investigate the impact of rhinoceros poaching on the tourism sector in Namibia
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Tourism forms an integral part of Namibia’s economy, contributing not
only significantly to the gross domestic product (GDP) of the country,
but also providing a means of improving livelihoods in remote areas
where there are few economic opportunities.
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The loss of our country’s iconic wildlife directly reduces the number of
wildlife tourists visiting the region and reflects significant potential losses
to the GPD.
Namibia has prioritised tourism as a major sector for driving economic
growth, employment creation and poverty reduction.
Therefore, the poaching of rhinoceros robs the country and communities
of their natural assets.
It undermines sustainable economic development and has serious
economic and social consequences, threatening the livelihoods of
communities that are dependent on wildlife tourism and natural
resources
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A sample of exam based questions
1. a) Test-tubes D, E and F, each containing a different concentration of reducing sugar,
were tested for reducing sugar.
i)
Describe the test for reducing sugar
(4)
ii)
Table 1.1 shows the result of this investigation.
Table 1.1
Solution
D
E
F
Observation
Yellow/orange
Green
Red
Determine the order of concentration of reducing sugar in the three solutions and
complete Table 1.2
Table 1.2
Concentration
High
Medium
Low
D,E or F
(2)
(b) A learner carried out an investigation to determine the relative concentrations of
reducing sugar in potato and onion tissue.
The two tissues were tested for reducing sugar and it was found that the onion tissue
contained a higher concentration of reducing sugar than the potato.
Explain how you would make sure that the two tests were carried out to allow a fair
comparison
(4)
[10]
2. Fig. 2.1 shows the experiment set up by a student to investigate the effect of osmosis.
Fig. 2.1
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The Visking tubing was filled with a strong sugar solution and its outer surface was rinsed
with distilled water. The Visking tubing was then dried and weighed. Then the Visking
tubing, containing the strong sugar solution, was placed in a test-tube containing distilled
water and left for two hours.
After the two hours, the Visking tubing was removed, dried and weighed again.
The following results were obtained.
- Mass before placing in distilled water: 34 g
- Mass after two hours in distilled water: 39.7 g
a) i) Suggest why Visking tubing allows water molecules to pass through more easily than
sugar molecules.
(1)
ii) Identify the parts of plant and animal cells that have similar properties to Visking tubing.
(1)
iii) Explain why the Visking tubing was rinsed with distilled water after it was filled with the
strong sugar solution.
(1)
iv) Explain why it was essential to dry the Visking tubing before it was weighed.
(1)
b) Calculate
i)
the exact change in mass.
ii)
the percentage change in mass. Show your working.
(1)
(2)
c) i) Explain why the mass changed the way it did.
(2)
ii)
Suggest what other change you might observe, in the Visking tubing, apart from
the change in mass.
(1)
[10]
1. Fig. 2.1 shows the appearance of a plant cell before and after it was immersed in
distilled water for 30 minutes.
Fig 2.1
358
(a) Give three reasons, evident from Fig. 2.1, that demonstrate that the cell was placed
in distilled water.
(3)
(b) Identify the parts labelled E, F and G.
(3)
(c) Explain the process that has taken place in the plant cell when placed in the distilled
water.
(3)
(d) The plant cell labelled before in Fig. 2.1 was then placed in a 20% sucrose solution.
In the space below, draw a diagram to show the appearance of this plant cell after 30
minutes in the sucrose solution.
(4)
[13]
3. Fig. 3.1 is a photomicrograph of a slide of the cellular and tissue structure of a
dicotyledonous leaf.
Fig 3.1
(a) Make a high power drawing of the whole cells of the part of the leaf shown in the
box in Fig. 3.1.
Label a palisade mesophyll cell and a cell wall.
(5)
b) Describe how you would make temporary slides of the upper or lower epidermis of a
leaf.
(4)
c) Fig. 3.2 and Fig. 3.3 show pollen grains of insect and wind pollinated flowers as seen
with the aid of an electron microscope.
Fig. 3.2 and Fig. 3.3 were magnified 300 times.
359
Fig 3.2
i)
Fig. 3.3
Calculate the actual size of grain X in Fig. 3.3.
Show your working.
[12]
(3)
4. Fig. 4 .1 shows the appearance of a cell from the epidermis of a leaf.
Three similar pieces of epidermis have been placed in different solutions and left
submerged for 30 minutes. One solution was pure water, another contained 1.5% sugar
solution and the third contained 5% sugar solution.
Figs. 4.2, 4.3 and 4.4 show a cell from each of these three pieces of epidermis.
Fig. 4.2
Fig. 4.3
Fig. 4.4
360
a) Measure the width of the cell contents along the lines drawn across each cell.
(2)
b) Suggest in which solution each of the cells, in Figs. 5.2, 5.3 and 5.4 was placed
and explain your choice.
(6)
[8]
1. Fig. 1.1 shows the apparatus used to determine the rate of photosynthesis of water
weed at different light intensities.
The rate of photosynthesis was measured by counting the number of bubbles produced
per minute. The only source of light in the room was the lamp.
The light intensity was decreased by moving the lamp away from the plant.
Fig. 1.2 shows the results of this experiment.
361
a) Describe the relationship shown by the graph.
(1)
b) Using the graph, state how many bubbles/min you would expect to be produced with
the lamp at a distance of 50 cm from the plant.
(1)
c) Sodium hydrogen carbonate was added to the water before the experiment began.
Explain why this was added.
(1)
d) Name the gas produced by the plant.
(1)
e) Describe the test for the gas named in d).
(1)
f) Describe and explain the results you would expect if the experiment was repeated
with the temperature of the water 10°C higher.
(2)
[7]
2. A potted plant with variegated leaves was destarched. Some of the leaves were
partially covered with tin foil and the whole plant was then left in the sunlight.
a) Describe how the plant could have been destarched.
(1)
b) Fig. 2.1 shows one of the leaves at the beginning of the experiment.
After a few hours in the sunlight, the leaf was tested for starch with iodine solution.
Complete Table 2.1.
362
(3)
c). When testing a leaf for starch, certain instructions need to be followed. The
instructions below are in the wrong order.
A Soften the leaf by dipping it into hot water for about 30 seconds.
B Place the leaf in alcohol and boil the tube in a beaker of hot water for about 5
minutes.
C Remove the leaf from the plant and dip it into boiling water for about one minute.
D Cover the leaf with iodine solution.
E Remove the leaf when it has lost its green colour.
F Observe and record the colour of the leaf.
a) Write the letters of the instructions in the correct order.
b) Explain why procedure B is necessary.
c) Explain why procedure B uses a beaker of hot water for boiling.
d) Explain why procedure C is necessary.
[10]
363
(2)
(2)
(1)
(1)
3. The apparatus shown in Fig. 3.1 was set up under bright light for a period of five
hours. At the start, the apparatus was completely full of water. During this time, a gas
collected at the top of the graduated tube.
Fig 3.1
a) How would you show this gas was oxygen?
b) Name the process within the plant responsible for the production of oxygen.
(1)
(1)
c) Determine the volume of gas collected in five hours and the rate of gas production
per hour.
(2)
d) How would you use this apparatus to obtain reliable results to show the effect of
differing light intensities on the production of oxygen?
(2)
[6]
364
1. Fig. 1.1 shows an enzyme controlled reaction at 37°C.
Fig 1.1
(a) Define the term enzyme.
(2)
(b) Identify the components labelled B and D.
(2)
(c) Describe and explain what happens to A and the effect that this has on the reaction
shown in Fig. 1.1 when the human body temperature rises above 40°C.
(3)
(d) Seeds contain stored food reserves.
With reference to named enzymes in seeds, describe the role they play in germination.
(4)
[11]
1. Fig. 1.1 shows an incomplete diagram of a vein transporting blood to the heart in
longitudinal section.
Fig 1.1
(a) (i) Complete Fig. 1.1 by drawing onto the diagram the valves. Show the direction of
blood flow through the vein using an arrow.
(2)
365
(ii) Describe the functions of the veins in the body and explain how the valves ensure
that the blood travels through them in one direction only.
(2)
(b) Blood vessels transport blood around the body.
Table 1.1 shows some features of what the blood transports around the body.
Complete Table 1.1.
Table 1.1
(4)
(c) Blood clotting is an important process that prevents excessive bleeding when a
blood vessel is damaged.
(i) State the component of blood and the plasma protein involved in blood clotting. (2)
(ii) Describe the sequence of events involved in blood clotting.
(4)
[14]
366
1. Fig. 1.1 shows the pulse rates of two students, A and B, before, during and after a
short period of vigorous exercise.
a) How long was the period of exercise?
(1)
b) (i) State three differences between the curve for student A and the curve for student
B.
(3)
(ii) Suggest two reasons why the results were different.
(2)
c) State three ways in which the increase in pulse rate during and after exercise would
affect the metabolism of student A.
(3)
d) (i) Suggest a suitable form of exercise for a class of students who were doing this
experiment.
(1)
(ii) Describe how you would measure your pulse rate.
(1)
[11]
367
2. Match the columns by writing the number of the term in Column A next to the correct
description.
(10)
[10]
1. Fig. 1.1 shows three stages in the germination of a bean seed.
Fig 1.1
(a) Study Fig. 1.1 and state two visible features that enable you to identify the bean
plant as a dicotyledonous plant.
(2)
(b) Name two conditions that are always necessary for the successful germination of a
seed, other than the presence of water.
(1)
(c) Describe an investigation that you could carry out to show the need in seed
germination for one of the conditions you named in (b).
(3)
368
(d) Fig. 1.3 shows seedlings that has been grown outside a cupboard and seedlings
that has been grown inside a cupboard.
Fig. 1.3
Record three visible differences between the seedlings grown outside and the
seedlings grown inside a cupboard.
(3)
[9]
369
1. Table 1.1 show the mass of some of the different food substances present in 100 g
of different cereals.
Table 1.1
(i) On the grid, draw a bar graph of the amount of fibre in the different cereals.
(4)
(ii) 1 g of fat gives 37 kJ of energy. Use the data from Table 3.1 to find the energy in fat
in 100 g of maize. Show your working.
(1)
(iii) Describe the importance of fibre in the diet.
(2)
[7]
370
1 (a) Fig. 1.1 shows the apparatus that can be used to show the loss of water from a
plant.
(i) Explain why oil is placed on top of the water in the flask.
(2)
(ii) Explain why the plant is added before the oil is carefully poured on top of the water.
(2)
(iii) Explain how you would use this apparatus to measure the rate of transpiration,
over seven days.
(4)
(iv) Outline the results you would expect if the experiment was repeated when the
temperature is 10°C warmer.
(1)
(v) Using the same apparatus, describe how you could show that there are more
stomata for water loss on the lower epidermis of the leaves than on the upper
epidermis.
(5)
[14]
371
1. Humans release carbon dioxide during respiration. This carbon dioxide enters the
atmosphere as part of the carbon cycle.
(a) (i) State one human activity that increases the amount of carbon dioxide in the
atmosphere.
(1)
(ii) During the carbon cycle, photosynthesis removes CO2 from the atmosphere, to
produce carbohydrates.
Name the form in which these carbohydrates are stored in a plant.
(1)
(iii) Explain what happens to the carbon containing compounds in plants and animals
when they die.
(2)
(b) Deforestation is one of the human activities that disrupt the process of
photosynthesis.
State three other undesirable effects of deforestation on the environment.
[7]
(3)
1. An ecosystem is made up of organisms that interact with each other and the
environment in which they live.
(a) (i) Name the primary source of energy in an ecosystem.
(1)
(b) Explain why it is more efficient for humans to eat plants instead of feeding plants to
cattle, goats or sheep.
(3)
(c) To survive, human beings need access to resources. Such resources need to be
conserved.
(i) Define the term conservation.
(2)
(ii) Describe four ways in which water can be conserved.
(4)
(iii) Discuss two advantages of tourism for conservation in Namibia.
(2)
(iv) Discuss two disadvantages of tourism for conservation in Namibia.
[14]
(2)
372