WORKBOOK ANSWERS
OCR A-level Biology A
Communication, homeostasis and
energy (Topic 8)
Genetics, evolution and ecosystems
This Answers document provides suggestions for some of the possible answers that might be
given for the questions asked in the workbook. They are not exhaustive and other answers
may be acceptable, but they are intended as a guide to give teachers and students feedback.
Module 5
Communication,
homeostasis and energy
Topic 8 Respiration
Respiration is important
1 It traps very small amounts of energy; it traps energy quickly and readily and releases it
equally readily; by conversion of ADP + Pi to ATP when energy is available and ATP to
ADP + Pi when the stored energy is needed; described as the energy currency of the
cells; ATP cannot be transported between cells — it must be released within each cell as it
is needed; as small amounts of energy are involved there is little energy loss during the
conversion.
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2
3 Photosynthesis is how sunlight energy is trapped and used as chemical energy. This
chemical energy is used by plants and other organisms for energy release during
respiration.
4 In microorganisms energy is needed for protein synthesis, cell division, growth and to drive
chemical processes.
5
Name of the process
Function
Active transport
Pump mechanisms, such as the movement of
molecules through cell membranes against
their concentration gradients:

uptake of many food molecules in the
gut

uptake of some ions in the kidney

Na+/K+ pump mechanism in nerve cells
to establish a resting or action potential
Activation
Energy is required for the activation of
molecules
Exocytosis
When large molecules are secreted or
released out of a cell
Endocytosis
When large molecules are taken into a cell
Protein synthesis
Building up polypeptides and proteins from
amino acids
DNA synthesis, cell division
Replicating DNA for cell division
The process of cellular respiration
1 It does not use or need any oxygen and occurs in both aerobic and anaerobic conditions.
It occurs in the cytosol (so all the enzymes necessary to convert glucose into pyruvate are
found there).
2 Both occur in the matrix of the mitochondria, where pyruvate is converted to acetate.
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3 They are highly folded so a large surface area is provided. They carry all the enzymes and
carrier molecules needed for the electron transport. These are embedded within the
membrane, along with ATP synthase. ATP synthase is where ATP is synthesised through
the phosphorylation of ADP.
4
a
Phosphorylation
Main reactions: one ATP molecule is hydrolysed for each step:
1
glucose to glucose 6-phosphate
2
fructose 6-phosphate to fructose 1,6-bisphosphate
So two ATPs are used up
What happens: glucose is activated and energy levels raised by adding phosphate
to form glucose 6-phosphate and later a second phosphate is added to fructose 6phosphate to form fructose 1,6-bisphosphate.
Energy and Pi are released for each of these steps.
b
Isomerisation
Main reaction: fructose 6-phosphate is formed.
What happens: glucose 6-phosphate is converted to fructose 6-phosphate.
c
Splitting the 6-carbon molecule
Main reaction: two molecules of 3-carbon triose phosphate (TP) are formed.
What happens: fructose 1,6-bisphosphate is split into two molecules, each one of
three carbons and with a single phosphate attached.
d
Oxidation
Main reactions: triose phosphate is oxidised and reduced NAD is formed.
What happens: two hydrogen atoms are removed to oxidise triose phosphate.
Dehydrogenase enzymes are used with the coenzyme NAD acting as the hydrogen
acceptor.
e
Substrate-level phosphorylation
Main reactions: two molecules of reduced NAD are formed and enough energy to
reform two ATP molecules.
What happens: enough energy is released to allow an ATP molecule to form
directly, when a substrate molecule (in this case triose phosphate) is modified by an
enzyme-controlled reaction.
f
Formation of pyruvate
Main reactions: a TP molecule is converted to pyruvate.
Two more ATP molecules form.
What happens: each triose phosphate molecule is converted to 3C pyruvate by
enzyme-controlled reactions.
Phosphorylation of two more ADP molecules forms two more ATP molecules by
substrate-level phosphorylation.
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5
6 Two molecules of reduced NAD. (The NAD coenzymes have been reduced by the
acceptance of hydrogen atoms.)
Two ATP molecules as a net gain. (Four are actually made but two are used up during
phosphorylation of glucose.)
Two molecules of pyruvate.
7 In the absence of oxygen pyruvate molecules are used in fermentation — either lactate
fermentation in animal cells or ethanol fermentation in yeast and some bacteria.
8 Decarboxylation: a carboxyl group is removed from pyruvate using pyruvate
decarboxylase. The carboxyl group forms carbon dioxide.
Dehydrogenation: the enzyme pyruvate dehydrogenase removes hydrogen atoms —
accepted by (coenzyme) NAD to form reduced NAD.
9
a
The products of the two molecules of pyruvate entering the link reaction: two
reduced NAD, two carbon dioxide molecules and two acetyl coenzyme A molecules.
b
Acetyl (a 2-carbon molecule) is the final product. Accepted by coenzyme A to form
acetyl coenzyme A. The 2-carbon acetyl group is carried as acetyl coenzyme A into
the Krebs cycle.
10 Release of coenzyme A: acetate drives the cycle with a series of enzyme-controlled
reactions, and releases coenzyme A to join oxaloacetic acid and form a 6C molecule.
Decarboxylation: citrate is oxidised to release two molecules of carbon dioxide.
Dehydrogenation: the citric acid molecule is oxidised by releasing eight hydrogen atoms.
Six of the hydrogen atoms are accepted by NAD and three molecules of reduced NAD are
formed. Two of the hydrogen atoms are accepted by FAD to form a single molecule of
reduced FAD.
Substrate-level phosphorylation: enough energy is released to form one molecule of
ATP directly.
Reformation of oxaloacetate: at the end of these steps, a 4C molecule is
dehydrogenated to reform oxaloacetate and continue the cycle again.
11 The products of two turns of Krebs cycle are: six reduced NAD, two reduced FAD, four
carbon dioxide molecules and two ATP molecules produced directly.
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12
Coenzymes
13 Coenzymes are non-protein organic molecules. They bind to the enzyme’s active site.
Some of them help in enzymatic oxidation and reduction reactions because the enzyme
needs a hydrogen acceptor or donor for the hydrogen atoms that it has removed.
The coenzyme becomes reduced when a pair of hydrogen atoms (which will later be
divided into protons and electrons) are accepted, and becomes oxidised when the
hydrogen is passed on.
14
a
NAD (nicotinamide adenine dinucleotide): a hydrogen acceptor molecule;
manufactured in the body from vitamin B3; nicotinamide accepts a pair of hydrogen
atoms and so becomes reduced; it is oxidised when it loses the hydrogen atoms; an
important molecule involved in all stages of respiration.
b
FAD (flavine adenine dinucleotide) (similar to NAD, but): contains the vitamin B2
riboflavin with adenosine and two phosphate groups; acts as a hydrogen acceptor
but it is tightly bound to the dehydrogenase enzyme embedded in the inner
mitochondrial membrane; therefore it does not pump the hydrogen atoms into the
intermembranal space; returns them to the matrix instead.
c
Coenzyme A (CoA): organic molecule made from vitamin B complex (pantothenic
acid), adenosine and three phosphate groups and the amino acid cysteine; carries
acetate made from pyruvate in the link reaction into the Krebs cycle; also carries
acetate groups made from fatty acids and some from amino acids to the Krebs
cycle; therefore is an important entry point for other respiratory substrates.
15 Phosphorylation is the addition of phosphate to a molecule. (For example, in the process
of ATP formation, using small amounts of energy to add inorganic phosphate to ADP and
form ATP.) In oxidative phosphorylation, the final acceptor is oxygen. When the final
electron acceptor is oxygen, the proton (hydrogen ion) rejoins the electron, and reduces
the oxygen to form water. Substrate-level phosphorylation and photophosphorylation are
ATP formation but without oxygen as the final acceptor. Photophosphorylation involves
trapping light energy.
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Chemiosmosis
16 Electrons flow along the carrier molecules and release energy; (the reduced coenzymes
are oxidised); some of the energy is used to pump the protons into the intermembranal
space; protons build up here and a proton and electrochemical gradient forms; no ATP is
used for this pump — energy comes from the electron flow or electromotive force; the
inner membrane is impermeable to ions (including protons); so ions cannot move across
the membrane but flow through ion channels (within the membrane) down the proton
gradient; the ion channels are part of ATP synthase; protons flow down the ion channel of
ATP synthase and the rotating part of the ATP synthase enzyme turns, with energy
released; used to build one ATP molecule from ADP and Pi.
17
Stage
Location
Substrate
molecule
Product
molecules
Number
of ATP
molecules
Type of
respiration
Glycolysis
Cytosol
6C
glucose
2 × 3C
pyruvate
2 × ATP
2 × reduced
NAD
2 directly
Aerobic and
anaerobic
Link reaction
Matrix of
mitochondria
Pyruvate
Acetate
Reduced
NAD and
CO2
Krebs cycle
Matrix of
mitochondria
Acetate
and CoA
Oxaloacetic
acid
recycled
2 × CO2
1 × ATP
6 × reduced
NAD
2 × reduced
FAD
2 directly
(for two
turns of
cycle)
Aerobic
Oxidative
phosphorylation
Cristae of
mitochondria
Reduced
NAD and
ADP
H2O and
26 x ATP
26–28
Aerobic
Total
Aerobic
30–32
Mitochondria
18
a
Outer membrane
Feature: protein channels and carriers
Function: passage of molecules (pyruvate and reduced NAD from glycolysis) into
the mitochondrion
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b
Matrix
Feature: enzymes in matrix; 4C oxaloacetate; reduced coenzyme NAD;
mitochondrial DNA and ribosomes (70s)
Function: catalyse the link and Krebs reactions; accepts acetate from the link
reaction to form a 6C molecule; carries H2 to the electron transport chain;
synthesises the proteins and enzymes needed
c
Inner membrane — cristae
Feature: a lipid inner membrane that is different from the outer membrane; highly
folded structure; electron carrier complexes; coenzyme FAD; cofactors; large
molecules of ATP synthase enzyme protruding into the matrix
Function: membrane is impermeable to ions, including H+ ions, so these are
prevented from passing through; increases surface area for more electron carriers
and enzymes, such as ATP synthase; each carrier is an enzyme and a cofactor,
arranged in a chain to allow the electron transport chain reaction to proceed; bound
to dehydrogenase enzyme within the cristae and reduced by accepting H atoms
(these H are pumped out to the matrix – not to the intermembranal space); haem- or
iron-containing groups (Fe2+) become reduced (to Fe3+) when accepting an electron
and oxidised (to Fe2+) when the electron is donated to the next carrier (an example
of a redox reaction)
Protons pass through down proton gradient from intermembranal space to matrix —
this rotates ATP synthase and allows ATP to be formed from ADP and Pi using
energy from the process called chemiosmosis
d
Intermembranal space
Feature: small space between inner and outer membranes
Function: protons are actively pumped into this space — they build up in this space
as they can enter through the outer membrane but cannot pass on through the inner
membrane; creates a proton gradient as the source of potential energy
Anaerobic respiration
1 Lactate fermentation (e.g. in the muscle cells of animals) where pyruvate acts as the
hydrogen acceptor for the hydrogen from reduced NAD.
Ethanol or alcohol fermentation (e.g. in fungi such as yeast, and in plants and some
bacteria) where ethanal acts as a hydrogen acceptor to remove the hydrogen atoms from
reduced NAD. Ethanol is first produced from pyruvate.
2
Molecules and
substances involved
in fermentation
Lactate fermentation
Ethanol fermentation
Pyruvate
Remains unchanged until it
accepts H
Loses a molecule of carbon
dioxide and becomes
ethanal
Intermediate molecule
None
Ethanal reduced to ethanol
H
Pyruvate
Ethanal
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Molecules and
substances involved
in fermentation
Lactate fermentation
Ethanol fermentation
Enzyme
Lactate dehydrogenase
Pyruvate decarboxylase
Ethanol dehydrogenase
NAD
Oxidised
Oxidised
End product
Lactic acid (lactate)
Ethanol (alcohol)
Number of molecules
reduced NAD
2
2
Number of CO2
molecules
0
2
Net gain of ATP
molecules
2
2
Respiratory substrates
3
a
Glucose is most commonly respired in most human tissues.
b
Red blood cells cannot respire any other substrate because they have no
mitochondria. Brain cells only respire glucose as only glucose can cross the blood/
brain barrier.
c
Fat and proteins/amino acids and other molecules can also be used as respiratory
substrates. Most animals eat different food, so their tissues use a variety of
substrates.
The respiratory quotient
4
a
RQ = 5.0/6.0 = 0.83
b
This value suggests a mixture of respiratory substrates, but with more fat than
carbohydrate or protein because the value is lower than 0.9.
RQ protein = 0.9, RQ carbohydrates = 1.0, RQ fats = 0.7
5
a
The bubble can move in either direction so that if there is more or less CO2 released
than O2 uptake the difference can be calculated.
b
From 0 to 4 minutes RQ = 1. At 6 minutes and 8 minutes RQ = 0.9. At 10 minutes
RQ = 0.95. (With sodium hydroxide the CO2 produced is removed, so the figures in
this column represent O2 uptake.)
c
0.33 mm−1 g−1 min–1 or 0.005 mm−1 g−1 s–1
d
An accurate value could be obtained by calculating the volume of gas in the
capillary tube.
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Exam-style questions
1 For active transport, for the uptake of mineral ions against their concentration gradient
from the soil; for the secondary pump required in the transport of food molecules by
translocation; for the light-independent stage of photosynthesis (although the ATP for
this is derived from the light-dependent stage).
2 B
3
a
i
Oxidative phosphorylation: the cristae/inner membrane
ii
The Krebs cycle: the matrix
iii
Decarboxylation: the matrix
iv
Electron transport chain: the cristae/inner membrane
b
Codes for proteins/enzymes required in respiration.
c
NAD or FAD
d
The protons are actively pumped into the intermembranal space; using energy from
some of the oxidised coenzymes; no ATP used at this step; proton gradient built up;
electrochemical gradient also built up; protons flow down this gradient via ion
channels; ion channels associated with ATP synthase; as the protons flow through
the ion channel of ATP synthase the enzyme is rotated; this releases energy; used
to build an ATP molecule.
e
When electron flow stops chemiosmosis stops; no ATP produced; lack of ATP
means muscles cannot contract; actin and myosin cannot slide past each other/no
active movement of the myosin head; muscles fail to contract; breathing movements
stop/intercostal muscles/diaphragm muscles stop breathing movements.
a
Glycolysis
b
In the cytosol
c
Pyruvic acid/pyruvate
d
4 ATP, but since two are used up early in the process there is a net gain of only 2
ATP.
e
Lactate production occurs in actively respiring cells, e.g. muscles; glycolysis results
in ATP, reduced NAD and pyruvate; the pyruvate acts as the H+ acceptor to recycle
the NAD; lactate dehydrogenase catalyses the reduction of pyruvate to lactate and
the oxidation of NAD.
4
Whereas, in yeast, alcohol fermentation; glycolysis results in ATP, reduced NAD
and pyruvate as for lactate fermentation; pyruvate loses CO2; catalysed by enzyme
pyruvate decarboxylase and coenzyme; forms ethanal; ethanal acts as hydrogen
acceptor from reduced NAD; now NAD oxidised and ethanal reduced to ethanol;
catalysed by enzyme ethanol dehydrogenase.
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5
a
A: Glycolysis
B: Lactate fermentation/anaerobic respiration
C: Aerobic respiration/Krebs cycle and oxidative phosphorylation
b
Pathway C
c
Pathway A
d
Alcohol and CO2
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Module 6 Genetics,
evolution and
ecosystems
Topic 1 Cellular control
1 Genetic variation may give rise to a new allele of a gene that gives a selective advantage
to the organisms and allows a species to adapt to a changing environment.
2 Alleles of genes that result in undesirable features will be selected against and either fall to
a very low frequency in, or be eliminated from, the population.
Types of gene mutation
3 A change in the DNA base sequence of a gene.
4
Type of gene mutation
What is it?
What is the impact?
Substitution
A nucleotide with a different
base from the original is
incorporated into a
developing strand of DNA
Causes a change in only one
of the base triplets, which
might not change the coded
amino acid and so have an
effect on the coded
polypeptide
Insertion
One or more nucleotides are
added to the DNA
Causes a change to all the
base triplets that follow the
insertion, i.e., causes a
frameshift
Deletion
Loss of one or more
nucleotides from the DNA
Causes a change to all the
base triplets that follow the
insertion, i.e. a frameshift
mutation
Stutter
The same nucleotide triplet is
repeated, usually many times
Causes the organism to
malfunction, e.g.
Huntington’s disease
5
a
Because most amino acids are encoded by more than one base triplet.
b
Because the code is degenerate, a change in one base may result in a triplet that
still codes for the same amino acid.
6 A genetic code based on just pairs of letters would give 16 possibilities (42). This is
insufficient for 20 amino acids.
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7 The new allele may be recessive and masked by a dominant allele. The gene may be only
one of a number involved in the expression of the characteristic and so the impact of the
new allele is reduced. Another gene may actively suppress the expression of this gene —
an effect called epistasis.
The effect of mutations on protein production and function
Beneficial mutations
8 The intensity of sunlight is too low in northern Europe to penetrate the skin, reducing the
ability of humans to synthesise vitamin D. A mutation that lowered the concentration of
melanin in the skin would enable humans to produce more vitamin D and so avoid rickets.
Neutral mutations
9 A mutation causing a change in eye colour is neutral because a change in eye colour is
unlikely to result in a selective advantage or selective disadvantage.
Regulatory mechanisms that control gene expression
10
a
One of the DNA strands is used as a template for the production of mRNA.
b
Pre-mRNA is edited by removal of the introns and splicing the exons back together.
c
A newly formed polypeptide chain is processed in the Golgi apparatus.
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Transcriptional level
11
Method
Process
Methylation
Methyl groups are added to
DNA nucleotides with an
enzyme to catalyse the
reaction
Acetylation
Transfer of acetyl group to
histone proteins; causes
chromosome to change
shape
Regulatory role
Stabilises cells so they can
no longer act as stem cells
Creates epigenetic effect
May prevent expression of
retroviral DNA, so helps to
control disease
When the chromosome
changes shape, the gene
becomes more or less
accessible
The accessibility determines
the expression of that gene
12 It is a short sequence of nucleotides close to, and upstream from, the gene. When one or
more transcription factors bind to the promoter region, it allows RNA polymerase to bind to
the gene and begin transcription.
13 Oestrogen binds to a receptor on the surface of target cells and, being lipid soluble,
diffuses through the cell surface membrane. It diffuses to the nucleus where it binds to an
oestrogen receptor within a protein complex. This receptor is a transcription factor that
binds to the gene promoter regions for lots of different genes and initiates transcription.
Once it is attached the oestrogen changes the receptor shape and leaves the proteincomplex, binds to the promoter region for one of the target genes and attracts cofactors.
The oestrogen receptor and cofactors allow RNA polymerase to bind and transcription of
that gene begins.
The lac operon
14 The lac operon is a functioning unit of the DNA containing a group of genes all involved (in
some way) in the digestion of lactose. They all are under the control of one promoter.
15
a
Region where the protein CRP (or cyclic AMP repressor protein) binds. Helps bind
the enzyme RNA polymerase to the promoters to allow transcription of the genes. If
cyclic AMP (cAMP) is present the CRP attaches to the DNA and allows RNA
polymerase to bind.
b
RNA polymerase attaches here if CRP–cAMP join with the DNA at the CRP site.
RNA polymerase cannot attach if this site is covered by the repressor and so DNA
transcription stops and no mRNA is produced for protein synthesis.
c
Located next to the structural gene. This is where the repressor binds. The region
acts as a switch, allows transcription and translation of structural genes. This gene
codes for the three enzymes if there is no inhibitor. When the repressor binds to the
operator site transcription and translation are stopped.
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d
In the lac operon there are three structural genes, alongside each other. These code
for β-galactosidase, lactose permease and another enzyme. All three can be
transcribed to mRNA together when the promoter and operator are switched on.
Post-transcriptional level
Editing of primary mRNA and removal of introns
16 Each gene of a eukaryote contains sequences of non-coding DNA, called introns. During
translation, these introns are copied to molecules of pre-mRNA; the editing removes these
intron regions from the mRNA and splices together the coding regions (exons).
17 Both are nucleotide sequences within a molecule of DNA/pre-mRNA (comparison for 1
mark).
Exons code for amino acid sequences but introns do not (contrast for 1 mark).
18 There are some introns that code for specific proteins themselves or can be further
processed to create non-coding RNA molecules that may have a gene regulatory function.
Post-translational level
Activation by cAMP
19 It is a derivative of ATP, formed by the action of an enzyme, adenylyl cyclase. It acts as a
secondary messenger for many processes such as hormone activation, cell transport, and
activation and regulation of ion channels.
Genetic control of the development of body plans
20 The sequence is determined by transcription factors (sequence specific DNA-binding
factors). These are proteins that bind to specific DNA sequences and control the rate of
transcription for those gene sequences into mRNA.
Homeobox gene sequences in plants, animals and fungi
21
a
Homeobox genes code for transcription factors. They are a sequence of 180 bases
(60 amino acids of a protein). They regulate transcription by the protein binding to
the DNA at one point and regulating transcription of genes further along the DNA.
Homeobox sequences are the same whatever organism they are found in, because
they all have the same function of coding for transcription factors. Homeobox genes
act as master genes controlling other genes and when they will function at different
stages
b
The homeobox controlling early development in animals, plants and fungi, by
turning specific genes on and off in the correct order.
Homeobox sequences give the basic body pattern by controlling segmentation
pattern (of insects and mammals) and the identity for the sections (what each will
become and which organs will be present). They determine the polarity of the
organism (i.e. the head and tail) and development of wings and limbs.
22 There are nine homeobox genes in the fruit fly, Drosophila melanogaster, which are
involved in control of development in the embryo in the correct sequence: head, thorax
and abdomen. The same genes are found in four Hox clusters in humans.
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Role of Hox genes in controlling body plan development
23
a
They are in organised groups known as Hox clusters.
b
Some organisms only have one or two Hox clusters. For example, roundworms
have just one Hox cluster whereas vertebrates have four clusters, each with 9–11
genes, found on different chromosomes.
24 The hox clusters contain genes in a linear order that is directly related to the order of the
regions they affect and when they affect the region. (For example, head–tail axis develops
first followed by segmentation in the embryo; when and where the legs, antennae and
wings will develop.)
If only a single mutation occurs, this will cause changes in the gene cluster, which in turn
will cause similar changes in all the regions under its control. Some Hox genes will
activate genes initiating apoptosis and control development by causing cell death in some
areas. (In Drosophila there is a Hox gene that activates a gene, called Rpr (meaning
reaper), that causes cell death in areas of the head lobes separating the maxillary and
mandibular head lobes.) The same idea is seen in finger and toe development.
The importance of the mechanisms of mitosis and apoptosis
25 Breakdown of the cell cytoskeleton; DNA becoming denser and more tightly
packed/condensation of the chromatin; breakdown of the nuclear envelope; biochemical
changes in the cytosol; formation of vesicles containing hydrolytic enzymes; finally,
phagocytosis by phagocytes.
26 Destroying the harmful self-targeting T lymphocytes during development of the immune
system, to prevent attack on the self body cells. Has a protective mechanism against
cancer — under the influence of the tumour-suppressor genes apoptosis destroys and
removes any genetically damaged cells (with damaged DNA) as they could give rise to
cancer. Role in embryonic development—by destoying unwanted cells during
development to ensure the final structure is correct. Role in maintaining healthy adult
tissues by destroying damaged or genetically deformed cells.
Response of genes to internal and external stimuli
27
Process
Factors in process
Outcome
Internal cell signalling
Cytokines, hormones, growth
factors, nitric oxide
Controls cell division and
apoptosis. Without control
apoptosis is reduced and cell
division is uncontrolled so
tumours may arise. Or if too
much apoptosis occurs
degeneration of tissues or
organs results.
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Process
Factors in process
Outcome
External factors (e.g.
pathogens or pollutants)
A protein —cyclin D—is
produced by cells when they
are affected by external
factors.
An upset in the
mitosis/apoptosis balance.
Cyclin D sets off a chain
effect—an enzyme cascade.
Genes are activated that
produce cyclins A and B.
For the cell cycle to proceed
cyclins A, B and D must be
produced.
External factors (e.g. those
increasing stress)
The mechanism of cell
signalling is disrupted.
The microbiome of the body
is disrupted.
The cell signalling no longer
effectively controls the
mitosis/apoptosis balance.
An imbalance of the
microbiome has been linked
to a wide variety of illnesses
in humans, such as
inflammatory bowel disease,
Crohn’s disease, some of the
autoimmune diseases and
anxiety
Epigenetic effects
Changes in gene expression
(epigenetics)
The DNA code itself is not
affected, although acetylation
of methylation will mean it is
presenting differently.
Exam-style questions
1 Level 3 (5–6 marks) Provides a comprehensive discussion of the mechanisms, with an
understanding of both. Well-developed line of reasoning that reads well is clear and logical
and uses scientific terminology where appropriate. All information is relevant and in
continuous prose. Reference to controlled by cyclins, cyclin-dependent kinases or CDKs
and other control measures, e.g. protooncogenes genes, which stimulate cell division, and
tumour suppressor genes which reduce cell division. Cyclins are the regulators and the
CDKs act as the catalysts once they are activated by the cyclins. When cyclins activate
the CDKs they catalyse phosphorylation of certain target proteins, which has the effect of
activating or inactivating them. This allows the cell cycle to move from one phase to the
next.
Apoptosis is programmed cell death. Also developmental changes in the embryo are
orchestrated so that cells are produced (by mitosis) where they are needed but later are
destroyed (by apoptosis) in areas where they are not needed. So here destruction is part
of the developmental process in organisms—for example—fingers and toes initially are
formed as a whole but then partially destroyed to allow them to separate.
Level 2 (3–4 marks) Describes some (a few) of the mechanisms, with some indication of
understanding both. A line of reasoning that is presented with some structure and use of
scientific terminology. Mostly relevant information.
Level 1 (1–2 marks) Describes some aspects of the mechanisms. The information has
only a little structure and lacks scientific terminology.
2 It is a section of DNA that carries genes coding for those enzymes needed to hydrolyse
lactose, which are only produced when lactose is needed and is available for use.
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There are two mechanisms that both ensure glucose is used, if available, before any
hydrolysis of lactose.
In the normal situation for the operon the enzymes needed for lactose hydrolysis are
prevented from being transcribed by the lac repressor. If there is no lactose this lac
repressor stops transcription of the genes, in the lac operon group, coding for the enzymes
that hydrolyse lactose.
If glucose is present (even when lactose is also in the medium) the activator protein,
needed for the enzymes’ production, stays inactive and the third enzyme lactose
permease is also not produced to stop lactose transport into the cells.
3 A role in polypeptide manufacture: when bound to cAMP receptor protein, it causes the
binding of RNA polymerase to gene promoter regions, activating transcription.
The activation of protein kinases: in eukaryotic cells, it works by activating protein kinase A
(PKA), an inactive or precursor enzyme. Once activated it is the enzyme phosphorylase
kinase. This enzyme is responsible for activating a number of proteins.
cAMP activates some proteins directly, e.g. exchange proteins are activated by cAMP with
a similar mechanism to the PKA mechanism.
Another mechanism involves control of translation involving siRNA (small interfering RNA).
These are short lengths of double-stranded RNA attached to protein complexes and the
mRNA targeted for breakdown. Therefore the siRNA controls whether certain proteins are
synthesised and so controls the gene expression.
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Topic 2 Patterns of inheritance
Phenotypic variation
1 Phenotypic variation refers to differences in the observable or measurable characteristics
of an organism. Genotypic variation refers to the combination of the different alleles of
each gene that an organism possesses.
2
Feature
Continuous variation
Discontinuous variation
Number of genes involved
Polygenic—many genes,
each with two or more alleles,
are involved in the
inheritance
Few—usually one or two,
each with two or more alleles
Effect on phenotype
The features show a gradual
change as there are many
intermediates
The features are in discrete
categories—either present or
absent
Effect on genotype
Environmental factors have a
significant effect
Environment has little
influence
Environmental factors
3 Poor diet may limit growth due to a lack of essential amino acids/protein so full potential of
height, for example, may not be realised unless the diet is well balanced and offers all the
essential nutrients and amino acids/protein.
4 Plants kept in the dark will become yellow and etiolated (over-long and weakly growing)
due to the action of the plant growth substance auxin. They will be unable to produce
chlorophyll if the chloroplasts are not subjected to light. However, once in the light
chlorophyll is produced and etiolation is slowly reversed and normal growth resumes.
Although the genes are present in the plants they cannot be expressed due to the
environmental conditions.
Genetic factors
5 The albino plant has a mutant gene that prevents production of chlorophyll. This is very
different from the environmental effect seen in question 4 because the plant cannot
produce any chlorophyll at all due to the lack of the necessary gene. The gene for
chlorophyll synthesis is expressed only in the light.
Genetic variation
6 Variation is increased by sexual reproduction because parents produce genetically
different gametes and fertilisation of these gametes occurs at random.
7
a
Changes the combination of the alleles of the genes on each chromosome.
b
The two gametes carry chromosomes with different alleles. Two different sources of
genetic information are brought together, so new combinations are possible.
Fertilisation is a random process with many sperm being possible fertilisers.
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c
Introduces new variants or changes in the genetic material, which may or may not be
harmful, e.g. non-disjunction, occurring during meiosis.
8 The chiasmata are points at which chromatids from homologous chromosomes become
entangled, break and rejoin. They help variation because parts of chromatids are
exchanged, leading to new combinations of alleles of the genes on each chromosome.
Genetic diagrams showing patterns of development
Monogenic inheritance
9 A (monohybrid) monogenic cross is the inheritance of a single pair of alleles of a single
gene. If one allele is dominant and the other is recessive, the expected phenotypic ratio is
3 : 1.
10
Parents:
GG
Gametes:
×
(green)
G
F1 :
gg
(yellow)
g
All Gg (green)
F1 Gametes:
G
or g
×
G or g
Punnett square showing genotypes of F2
F2 phenotype
G
G
GG
g
Gg
g
Gg
gg
3 green : 1 yellow
Codominance
11 The resulting phenotypic ratio is 1 : 2 : 1 ratio.
Explanation: When two alleles are codominant neither is dominant and so both are
expressed in the phenotype. When both alleles appear together the phenotype is a
mixture of both. This may be clarified with an example, e.g. red and white flowers
producing a pink flower if it is heterozygous.
12
a
There are two identical alleles for a gene coding for a characteristic.
b
The alleles for a gene coding for a characteristic are different.
c
Two alleles of a gene are neither dominant nor recessive so when they appear
together the phenotype will show the effects of both alleles.
d
A normal chromosome other than a sex chromosome.
e
Two or more genes are found on the same chromosome.
f
An allele that is shown in the phenotype only if it is present in the homozygous
condition.
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g
An allele always expressed in the phenotype even when another allele is also
present.
Dihybrid inheritance
13 Dihybrid; linked; crossing over; alleles; variation
14 Use T = tall, t = dwarf, C = cut-leaves, c = potato-leaves.
The F1 plants must have a heterozygous genotype because they have a range of different
phenotypes in their offspring in the F2 . So the F1 genotype = TtCc
The F1 cross is therefore =
The gametes for both are:
TtCc
TC
× TtCc
Tc
tC
tc
TC
Tc
tC
tc
TTCC
TTCc
TtCC
TtCc
Tc
TTCc
TTcc
TtCc
Ttcc
tC
TtCC
TtCc
ttCC
ttCc
tc
TtCc
Ttcc
ttCc
ttcc
TC
So the expected ratio is 9 : 3 : 3 : 1
Observed = 944 + 278 + 284 + 102 = 1608
1608/16 = 100.5
Therefore, expected = 904.5, 301.5, 301.5, 100.5, which is close to 9 : 3 : 3 : 1 ratio (need
a χ2 test to confirm)
Multiple alleles
15 Antigens; ABO; monohybrid; multiple allelic; two.
16
Blood Groups of the ABO system
Possible genotypes
A
IAIA or IAIO
B
IBIB or IBIO
AB
IAIB
O
IO IO
Use of phenotypic ratios and linkage
Autosomal linkage
17
a
A normal body chromosome—not a sex chromosome.
b
3 : 1 (or 1 : 2 : 1) — the same ratio as for monogenic inheritance.
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c
Crossing over at meiosis I can give rise to new combinations of alleles, so in a
population a small number will show a new linked combination.
Sex linkage
18 A carrier female has one X chromosome with the recessive allele for colour blindness and
one X chromosome with the dominant allele for colour vision. A male has a 1 in 2 chance
of inheriting the X chromosome carrying the allele for colour blindness so you might expect
half the woman’s sons to be colour blind. Whichever of her X chromosomes they inherit,
all the woman’s daughters will inherit their father’s X chromosome carrying the allele for
colour vision. Consequently, none of them will be colour blind.
19
a
XB Xb
b
As males only have one X chromosome they can only have a genotype of XB Y or
Xb Y, so cannot be tortoiseshell.
c
Parental phenotypes: Black female
Parental genotypes:
Parental gametes:
×
Xb Xb
Xb or
Ginger male
XB Y
Xb
XB or Y
Punnet Square:
XB
Y
Xb
XB Xb
Xb Y
Xb
XB Xb
Xb Y
F1 phenotype: (% of tortoiseshell cats): 50%
Epistasis
20 When two or more genes interact they all influence the expression of the same
characteristic, because the genes are involved in an enzyme pathway. In this case each
gene usually codes for (a polypeptide) an enzyme. If one of the genes has an allele coding
for a non-functioning enzyme the metabolic pathway stops and the characteristic cannot
be expressed.
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21
a
The ratio is close to a 9 : 4 : 3 ratio and both parental genotypes are PpRr.
b
Since the F1 ratio is close to a 9:3:3:1 ratio, the genes must be unlinked. Of the 16
possible genotypes:

9 will contain at least one dominant allele of both genes, so will have purple leaves

3 will have at least one dominant allele of gene A but will be homozygous
recessive for gene B and so will have red leaves

4 will be homozygous recessive for gene A and so will have white leaves.
Use these in the following Punnett square for a student check.
PR
Pr
pR
pr
PR
PPRR
PPRr
PpRR
PpRr
Pr
PPRr
PPrr
PpRr
Pprr
pR
PpRR
PpRr
ppRR
ppRr
pr
PpRr
Pprr
ppRr
pprr
Chi squared test
22 It is used on categorical data/data in categories to see if there is a significant difference
between the expected data and the observed data.
23
a
The null hypothesis was ‘there is no difference between the observed and the
expected data’.
b
Category
O
E
O–E
(O – E)2
(O – E)2 / E
Red uneven fruit
97
109
12
144
1.32
Red round fruit
48
36
12
144
4.00
Orange uneven
fruit
45
36
9
81
2.25
Orange round
fruit
4
12
8
64
5.33
χ2 = 12.90
c
4–1=3
d
p < 0.01
e
The probability that the difference between the observed results and the expected
results occurred by chance is less than 0.01. Since this is less than the accepted
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0.05 probability level, we can conclude that the difference was not due to chance
and can reject the null hypothesis.
Factors affecting evolution of a species
24
a
Maximum population size that a specific habitat is able to support indefinitely.
b
A combination of all the biotic and abiotic factors that limit any increases in
population size.
c
Any factor that affects the ability of some members of the population to survive and
reproduce.
25
Factor
Effect on population
Example
Stabilising selection
A population remains stable
for a characteristic if selection
removes the extremes.
The optimum (mode) is
maintained.
Stabilisation of human birth
mass around the optimum.
Directional selection
The characteristic will slowly
change in one direction. If an
environment changes any
individuals with features
better suited to the new
environment will be selected
and survive to breed.
Ground finches on the
Galapagos with larger beaks
survived at the expense of
those with smaller bills when
the available seeds became
larger.
Genetic drift
A random change in allele
frequencies that occurs in
small populations.
Red poppies over white
spotted marmots
Genetic bottleneck
A change in allele
frequencies that follows a
drastic reduction in population
size.
Hawaiian goose (ne ne),
northern elephant seal, giant
panda
Founder effect
The population shows low
genetic diversity because the
founding population
contained only a small
sample of the gene pool of
the population from which it
came.
Tibial muscular dystrophy—
an inherited mutant allele
(TTN) in Finland.
A Pacific island with high
numbers of an inherited eye
defect present as a recessive
mutant allele in just one of
the original colonisers.
Hardy-Weinberg equation for allele frequencies
26 The sum of all the alleles for a specific gene in the population.
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27
a
p + q = 1 refers to the relative frequencies of the alleles for a specific gene in the
population. p is the frequency of the dominant allele and q is the frequency of the
recessive allele for the gene. 1 represents the whole population.
b
p2 + 2pq + q2 =1 refers to the genotypic frequencies for the alleles for a characteristic.
Here p2 represents the frequency of the homozygous dominant genotype, q2 the frequency
of the homozygous recessive genotype and 2pq the frequency of the heterozygous
genotype.
28

There are no genetic changes in the population.

There is no migration (either emigration or immigration).

There is no selection.

There is no gene mutation.

The population is large and can breed randomly and freely within the population with
no limit on success.

There is no genetic drift.
Isolating mechanisms in evolution of new species
29
a
Populations become separated by a geographical barrier and so become
reproductively isolated. In the absence of gene flow, selection can result in these
isolated populations accumulating genetic differences that establish them as
separate species.
b
Sub-populations are not separated by a geographical barrier but by
spatial/temporal/behavioural differences. In the absence of gene flow, selection can
result in these isolated populations accumulating genetic differences that establish
them as separate species.
The principles of artificial selection
30 Maintaining a resource of organisms; providing a source of wild type which show more
genetic variation; increased milk yield in dairy cattle; variety and range of dog breeds
useful to man, all bred from grey wolves; improvement in bread wheat; larger-sized fruit
e.g. tomatoes, and other plants used for human consumption.
Ethical issues with artificial selection
31 Level 3 (5–6 marks) Provides a comprehensive discussion of the ethical issues with an
understanding of both sides of the argument. Well-developed line of reasoning that reads
well, is clear and logical and uses scientific terminology where appropriate. All information
is relevant and in continuous prose.
Level 2 (3–4 marks) Describes some (a few) of the issues with some indication of both
sides of the argument being at least considered. A line of reasoning that is presented with
some structure and use of scientific terminology. Mostly relevant information.
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Level 1 (1–2 marks) Describes some aspects of the issues. The information has only a
little structure and lacks scientific terminology.
Exam-style questions
1 Child 1 could have either Mr X or Mr Y as the male parent since the M could come from
the mother and N from either of the male fathers. Mr Y may have a recessive IO.
Child 2 must have Mr X as their father since the father must contribute an M antigen as
child 2 has two M alleles (Mr Y has no M to contribute) and for the ABO system the allele
IO will come from both parents to give an O group.
Child 3 cannot be the child of Mr X since neither Mr X nor Mrs X has antigen A to
contribute. This must have come from another father—likely to be Mr Y.
2 B = black, b = brown, S = short hair, s = long hair
F1 = BbSs × BbSs so F1 gametes =
BS
Bs
BS
Bs
bS
bs
bS
bs
BS
BBSS
BBSs
BbSS
BbSs
Bs
BBSs
BBss
BbSs
Bbss
bS
BbSS
BbSs
bbSS
bbSs
bs
BbSs
Bbss
bbSs
bbss
So 9 black, short hair (B_S_) : 3 black, long hair (B_ss): 3 brown, short hair (bbS_): 1
brown, long hair (bbss)
3
a
= bA
F1 = BbAa × bbAA, so F1 gametes for black agouti = Ba
bA and for brown agouti
bA
Ba
BbAa
bA
bbAA
So the F2 ratio is 1 : 1
b
The two genes are linked so there is no independent assortment of the two alleles for
the two genes.
c
This is a lethal gene, which halts development or causes death. Epistasis is where
one gene or allele masks the phenotypic expression of other genes or alleles in the
interaction. (needs the idea of masks)
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4
Albinism can only occur by having two recessive alleles for albinism (aa). Therefore, the
frequency of the homozygous recessive (q²) must be 0.0001 (since 1 in 10 000 people is
affected). So:
q=
0.0001 = 0.01
Using p + q = 1: p = 1 – 0.01 = 0.99
Now use 2pq (from the second equation p2 + 2pq + q2 = 1):
2 × 0.99 × 0.01 = 0.02 (to 2 d.p.). So the frequency of carriers of albinism is 0.02.
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Topic 3 Manipulating genomes
The uniqueness of the genome
1 Its genome
2 E. coli has a single circular loop of DNA, with relatively few protein coding genes (4288)
mostly organised into operons and in smaller loops called plasmids. There are also some
ribosomal RNA and transfer RNA genes. Homo sapiens has rod-shaped chromosomes
(23 pairs), each carrying 20 000 to 25 000 genes, and loops of DNA in the mitochondrial
matrix.
3 The study and application of genetic and molecular biology techniques to gene mapping
and gene sequencing (on the chromosomes or for an entire organism).
4
a
Any three from: structural genes; non-structural; ribosomal RNA and transfer RNA
genes; regulatory genes; promoters; hypervariable sequences; junk DNA. (Allow
introns and exons.)
b
Mitochondrial DNA (mtDNA) within the mitochondria or chloroplast DNA within
chloroplasts (ctDNA).
The principles of DNA analysis
5 Any two from: PCR (polymerase chain reaction); gel electrophoresis; DNA sequencing;
DNA profiling.
Polymerase chain reaction (PCR)
6 The sequence of bases in the primer must match (by complementary base pairing) the
DNA sequence just beyond the region of DNA being copied/amplified. Different DNA has
different base sequences.
Gel electrophoresis
7 The DNA sample is treated with restriction enzymes to ‘cut’ it into small sections or
fragments are amplified using PCR; a dye or a radioactive label is added and the
fragments are loaded into the wells in the gel.
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8
DNA sequencing and new sequencing techniques
The principles of DNA profiling
9 Basic principle—DNA contains a variable number of short tandem repeats; they are
unique to each individual.
Method—amplify fragments containing short variable repeats, separate by electrophoresis,
match bands between samples of DNA.
10 Minisatellites (variable number tandem repeats or VNTRs) are small sections (10–100
bps) of DNA in structural genes or in the non-structural DNA and are inherited from both
parents. These are repeated at various points along the chromosome and can be cut by
restriction enzymes.
Microsatellites (STRs) are much shorter sequences of nucleotides (2–5 nucleotides long
but repeated 10–30 times) which may be in the non-coding DNA between the structural
genes. These are easily identified and more unique as there is more variation between
individuals and they are less likely to degrade. Also, only a short length is needed, so DNA
profiling works well on a small sample.
11 The greater the number of STRs in the gene for the protein huntingtin, the greater the risk
of developing the symptoms of Huntington’s disease in later life.
Use in detecting disease
12 It is a variation in one nucleotide. A change in a single base will not affect the gene
reading frame so protein synthesis can still occur and a protein can still be produced. It
may be the same protein if the nucleotide change still codes for the same amino acid or it
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may be a different variation of a protein. Role: act as biological markers of disease; form a
large proportion of genetic variation in humans; occur more frequently than mutations.
Bioinformatics
13 The genes involved in development in Drosophila are similar in some ways to those found
in other animals including humans. This can be given as a percentage figure of degrees of
similarity, which is useful in uncovering evolutionary relationships and ancestry.
The principles and techniques of genetic engineering
14
a
An enzyme that forms phosphodiester bonds. Can be used to join a gene to a
plasmid.
b
A structure used to deliver a gene into the host organism as part of genetic
engineering.
c
A bacterial artificial chromosome.
d
A vesicle surrounded by plasma membrane.
e
recombinant DNA, formed when DNA from one organism is combined with the DNA
of another.
15
Applications
Role in genetic engineering
Restriction enzymes
Cuts DNA at specific base sequences
Ligase enzymes
Anneals or seals the DNA cut ends and forms a
phosphodiester bond between them
Reverse transcriptase enzymes
Makes cDNA from RNA
Plasmids
Is used to transfer DNA into the bacterial cells
Viruses/liposomes
May be used to transfer genes into human cells
16 It cuts across the DNA at a specific recognition sequence, comprising a short palindromic
sequence of bases.
17 This is an unpaired ‘overhanging’ sequence of bases on one of the polynucleotide chains.
It allows the annealing of another ‘overhanging’ sequence of complementary bases.
Humulin
18 The differences, although in only a few nucleotide base pairs, create a slightly different
protein. This difference triggers the human immune system and an immune response.
1. The mRNA produced from the gene for insulin is extracted from β cells in the
pancreas and converted to DNA using reverse transcriptase.
2. Sticky ends are added to the cDNA.
3. A plasmid from E. coli is extracted and cut using a restriction enzyme.
4. The cut plasmids are mixed with the cDNA with complementary sticky ends.
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5. A ligase enzyme anneals the cDNA and cut plasmid.
6. Plasmids are added to new bacteria treated with calcium followed by a cold shock.
7. Transformed bacteria are identified and pure continuous cultures of them are set up.
8. Insulin, produced in large quantities by the transformed bacteria, is harvested.
GM crops and ethics
20 Any two valid answers, for example golden rice that provides increased levels of vitamin A
precursors or GM bananas that act as a vaccine for hepatitis B.
21 Advantages: resistance to disease pests, drought and herbicides improves crop
production and reduces losses; an increase in nutritional value; production of
medicines/vaccines; allows production of crops that allow for climate change faster than
traditional selection processes; allows large-scale food production to keep pace with the
increase in human population growth.
Objections: may be too expensive for farmers in developing countries; possible transfer of
antibiotic-resistant genes or herbicide-resistant genes into other organisms; could lead to
antibiotic-resistant pathogens or weeds super resistant to herbicides; the genomes of crop
plants may be altered by introduction of foreign genes; mutations in foreign genes could
cause unknown effects in crop plants; possible creation of new pathogenic viruses; loss of
genetic diversity; no economic advantage to farmers as they still need to use herbicides
and pesticides and cannot keep seed for replanting as the crops do not breed true.
GM animals and ethics
22 Unspecified public concerns regarding use of GM animals; full effect of GM on the whole
genome of the animals is unknown; effect of GM animals in the food chain untested; lack
of public trust of scientists.
GM microorganisms and pathogens and ethics
23
GM organism
Use in genetic engineering
Application
Mouse
Production of monoclonal
antibodies
Detection and directed
cancer treatment
Virus
Modified to carry and deliver
genes
Gene therapy
The principles of gene therapy
24
a
Inserting genes into cells of an organism to correct or repair a genetic problem.
Usually a dominant allele is added into the cells where an individual has two
recessive alleles for the gene.
b
Somatic gene therapy: a gene is inserted into body cells of an organism so only
those cells benefit. Usually this must be repeated at regular intervals as the cells will
die and not pass on the gene. If inserted into a stem cell it may offer longer-term
treatment.
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Germ-line germ therapy: a gene is inserted into reproductive cells of an organism. In
this case the gene will be passed on to the gametes and so to any offspring.
Exam-style questions
1 Level 3 (5–6 marks) provides a comprehensive discussion of the issues, with an
understanding of both benefits and drawbacks of the argument. Well-developed line of
reasoning that reads well, is clear and logical and uses scientific terminology where
appropriate. All information is relevant and in continuous prose.
Level 2 (3–4 marks) Describes some (a few) of the benefits with some indication of
drawbacks being at least considered. A line of reasoning that is presented with some
structure and use of scientific terminology. Mostly relevant information.
Level 1 (1–2 marks) Describes some aspects of the issues and not well balanced at all.
The information has little structure and lacks scientific terminology).
2
a
i
Fluorescent markers, which can be detected by laser scanners.
ii
A type of gel electrophoresis is used. The DNA migrates through this gel to
the anode (in a similar way to the gel electrophoresis technique.
b
Their phosphate groups give DNA fragments a negative charge, so DNA fragments
migrate to the anode of an electric field. Pores in the agarose gel allow fragments to
migrate, with the smaller fragments moving faster, and so further, than the larger
fragments.
c
The DNA fragments are detected as they migrate through agarose gel in capillary
tubes, so there is no need to wait until the end of electrophoresis to begin analysis.
There are many capillary tubes—96 is the number frequently used—so more ‘runs’
can be performed at the same time. Analysis is by direct feed into a computer, so
the output is faster.
a
A retrovirus contains a core of RNA instead of DNA and an enzyme called reverse
transcriptase.
b
The virus infects the organism’s cells. Its RNA and enzyme enter the nucleus where
DNA is manufactured from viral RNA suing the host cell’s mechanism and the
enzyme. The resulting DNA is incorporated into the host DNA where it can lie
dormant for many years before becoming activated.
3
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Topic 4 Cloning and biotechnology
Cell division
1
Mitosis
Meiosis
Number of nuclear divisions
1
2 (meiosis I and meiosis II)
Number of daughter cells
produced
2
4
Chromosome sets in each
daughter cell
Same as parent cell
Half the number as the
parent cell
May involve crossing over
No
Yes
Results in genetic variation
No
Yes
Type of cell division
Feature
2
a
Cells that can divide to renew themselves and to produce a small variety of other
cells.
b
Cells that can divide and differentiate into any type of body cell and into extra
embryonic cells.
c
Cells that can divide and differentiate into any type of body cell but not into extra
embryonic cells.
3 Meristematic tissue — found at the tips of shoots or roots (apical meristems), in
axillary/lateral buds and between the xylem and phloem within vascular bundles (enabling
secondary thickening).
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Cloning in plants
Natural cloning
4
Vegetative reproduction
Type of tissue
Reproductive features
Runners
Horizontal stems
Stem grows over the soil
surface. New roots and
leaves and, later, new plants
form at the nodes.
Stolons
Thin, underground stems
Potato tubers are stolons that
swell with stored starch at the
tips. These eventually
produce shoots from the
lateral buds (‘eyes’ of the
potato) and new plants.
Bulbs
Thick, short stems; often
disc-like.
The leaf bases are swollen
with stored food, e.g. onion.
New plants grow in the early
spring.
Root tubers
Swollen root
The root is swollen with food.
New stems and new plants
grow from the tuber in the
spring.
Artificial cloning
5 Advantages: rapid growth, easy to establish; all plants will be of same reliable quality
because they are genetically identical; so identical crops such as cereals—rice, maize,
wheat and various fruits such as apples, pears and raspberries will all be of the same
quality; plants will breed true, which does not always happen from seed; plants may be
produced that are not weather dependent.
Disadvantages: all plants are genetically identical, so in the case of disease or pests all
will be equally susceptible; loss of variation and diversity; build-up of pests and diseases,
especially viruses.
6 Cloning in plants involves using tissue from the original parent such as a vegetative
structure, storage organ, cuttings, tissue cultures. As a result all the clones are genetically
identical. Formation of dog breeds is not cloning as they are not genetically identical. They
result from selective breeding over successive generations. So it is also a slow process.
7 The most commonly used plant growth substance is auxin (IAA, indoleacetic acid); at
appropriate concentrations it stimulates mitosis and the formation of adventitious roots.
8
a
Concentration of abscisic acid; concentration of gibberellin; temperature at which
seeds germinated; time for germination.
b
Age/variety/source of seeds used; pH of solutions; volumes of all liquids;
concentration of oxygen in Petri dishes; space between seeds.
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c
Temperature affects enzyme reaction rate and 25oC is optimum / suitable
temperature for seed germination OR to remove temperature as a confounding
variable;
d
Gibberellins stimulate germination; because in the absence of abscisic acid (water
only) there is a strong correlation between concentration of gibberellin and
percentage germination/reference to data; abscisic acid inhibits germination/has an
antagonistic effect on gibberellin; because in the presence of abscisic acid there is no
germination (whatever the concentration of gibberellin).
e
Any two of: encouraging root growth of cuttings; encouraging growth in length of
stem; discouraging growth in length of stem; discouraging growth of lateral buds;
inducing flowering; inducing fruit formation; stimulating fruit ripening; delaying fruit
ripening.
9 Callus: explants are removed by cutting leaves, stems or roots (all are differentiated cells);
stimulated to divide by mitosis into an undifferentiated mass—callus; callus is subdivided
to increase number of plantlets; cultured on a medium with plant hormones added; equal
concentrations of auxin and cytokinins give best growth; some cells isolated from the
callus may be cultured in a suspension.
Meristem: meristems are cut from axillary buds or apex of the stem; then cultured in same
way as for callus growth to promote mitosis.
Commercial value: numerous new plants can be cultivated each year from one original
callus or meristem, e.g. chrysanthemum apical meristem may produce 1 million new plants
in a year; all genetically identical so quality and desirability are known and reliable;
importance in fruit crops as rapid and reliable growth with reliable resulting fruit crop; also
in cereal crops; named examples such as apples, pears, rice, wheat; resistance to known
pathogens, diseases and pests is assured.
Commercially good as cost is lower; idea of less care needed when compared to seeds
(easier).
Disadvantages: mainly loss of variation and loss of genetic diversity; risk of loss of
adaptability in the case of environmental changes or new pathogens; build-up of pests and
diseases possible in cloning, especially viruses.
Cloning in animals
Natural Cloning
10 Identical twins are genetically identical and have same epigenetic imprinting since
produced from the same fertilised egg. Therefore scientific studies can focus on the
environmental effects as genetics is known. Epigenetics may be a result of environmental
changes. Non- identical twins are NOT genetically identical only as similar as any other
sibling.
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Artificial Cloning
11
12 In embryo cloning a single fertilised egg is cloned into two embryos—artificial identical
twins. So cloning involves the germ cell. The embryos are then implanted into surrogate
mothers.
In cloning of adult cells (somatic cell nuclear transfer) the nucleus is removed from the
adult somatic cell. The nucleus is also removed from an egg (cell) and is replaced by the
nucleus from the adult somatic cell, e.g skin or gut lining. The technique may involve
injecting the nucleus, using an electric current to fuse the somatic cell and the enucleated
egg cell. Cloning of the embryos may then also occur before implantation into a surrogate
mother. This may involve a surrogate of the same species OR of another, similar species
in the case of rare animals.
13 Level 3 (5–6 marks) Provides a comprehensive discussion of the ethical issues relating to
artificial cloning, with an understanding of both sides of the argument. Well-developed line
of reasoning that reads well, is clear and logical, and uses scientific terminology where
appropriate. All information is relevant and in continuous prose.
Level 2 (3–4 marks) Describes some (a few) of the issues, with some indication of both
sides of the argument being at least considered. A line of reasoning that is presented with
some structure and use of scientific terminology. Mostly relevant information.
Level 1 (1–2 marks) Describes some aspects of the issues. The information has little
structure and lacks scientific terminology.
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Biotechnology and microorganisms
14 Use of living organisms to produce useful products, e.g. drugs or food, or to carry out a
useful service, such as sewage disposal.
15 Primary metabolites are any compounds produced by the metabolic activity of an
organism during normal growth. Secondary metabolites are any compounds produced by
an organism after it has stopped growing.
Use in human food production
16
a
It is an enzyme used to tenderise the meat.
b
Papain is a protease that digests/hydrolyses protein fibres and any fibrous tissue.
c
Re-used for more brewing; used as yeast extract; used in cattle feed.
d
pH 3.5–4.5
Use in drug production
17 With batch-fed fermentation, food such as maize supplements the glucose and lactose
nutrients as they are being used up, and this stimulates the production of the antibiotic that
would not be produced or produced in small amounts with only glucose supplied. With
batch fermentation the only nutrients added are those at the start of the reaction, so the
quantity of product reduces as time progresses.
18
Use in bioremediation
19 The bacteria metabolise the hydrocarbons in the oil. The products are carbon dioxide and
water as well as biomass, which are harmless to the environment.
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Culturing microorganisms and aseptic techniques
20 Sources of carbon, energy, nitrogen, mineral salts, water and growth factors. (All prepared
under sterile conditions.)
21 Disinfection of surrounding areas before and after work to remove bacteria from work
surfaces; the agar/nutrient culture medium must be autoclaved to kill any bacteria it
contains; using protective clothing and gloves to reduce risk of transmission of bacteria
between self and culture; all apparatus to be sterilised in an autoclave or similar at over
120oC for 15–20 minutes to kill all bacteria on and in it; medium to be sterilised and poured
under sterile conditions; transfer of microbes to be with a flamed loop or pipette to kill
bacteria in/on it; flame loop after inoculation/place wet pipette into disinfectant after
inoculation since it is contaminated by the inoculum; lift lid of Petri dish as little as possible
to reduce risk of airborne bacteria falling on agar surface; have lighted Bunsen burner on
bench to create upward current away from culture.
22 Batch: production continues until substrate is used up; waste products accumulate; the
product must then be separated from enzyme and waste products; limited operation time;
reactor must be cleaned and sterilised before starting up again.
Fed-batch: operation time continued for longer as substrate is fed into fermenter at regular
intervals; production stops when waste products build up; temperature and pH constantly
monitored; product must be separated from enzyme and waste products.
Continuous: product is continuously collected from fermenter; pH and temperature
conditions are constantly monitored and substrate is added to enzyme continuously;
production can continue for long periods until reaction rate slows due to enzyme
damage/denaturing.
23
Condition
How the condition is
controlled
Reason for control
Temperature
Using a thermostatically
controlled heater
Enzymes work best at an
optimum temperature
pH
Use buffers to maintain
optimum pH
Enzymes have an optimum
pH
Carbon source
Add a supply of carbohydrate,
e.g.
monosaccharide/disaccharide
or polysaccharide; CO2 for
autotrophs
To enable microorganisms
to produce organic
molecules
Nitrogen source
Add amino acids, peptides
nitrates in autotrophic
organisms
To enable microorganisms
to produce amino acids
/polypeptides /proteins/
nucleotides
Using immobilised enzymes
24 An enzyme that is entrapped in an inert matrix or compound but remains active.
25 No need to keep adding enzyme because it can be re-used; no cost of separating enzyme
from final products because the enzyme remains in the fermenter.
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Exam-style questions
1 Packed-bed continuous flow is the best for industry because a stream of substrate is
passed over it and the product can be continually collected. There is no ‘down time’.
Immobilised enzymes will not form part of the product and so there is no costly separation
needed after the reaction. As they are trapped they will remain in the reactor for reuse
unlike soluble enzymes. Immobilised enzymes have greater thermostability as they remain
78% active at 70oC and 55% active at 80oC, unlike soluble enzymes. A thermostable
enzyme can be used at higher temperatures giving a faster rate of reaction; products will
be formed quicker and the yield per unit time is greater.
2 The smooth surfaces are easier to clean; reduces contamination; non-corrosive and inert,
so not affecting flavour; strong, so resists pressure build up; temperature control is easier;
can be made into large containers more easily than other materials.
3 Level 3 (5–6 marks) Provides a comprehensive discussion of the advantages AND
disadvantages of using microorganisms, with an understanding of both sides of the
argument. Well-developed line of reasoning that reads well, is clear and logical, and uses
scientific terminology where appropriate. All information is relevant and in continuous
prose.
Level 2 (3–4 marks) Describes some (a few) of the issues with some indication of both
sides of the argument being at least considered. Possibly not well balanced, with one side
better argued. A line of reasoning that is presented with some structure and use of
scientific terminology. Mostly relevant information.
Level 1 (1–2 marks) Describes some aspects of the issues. Possibly only one side
considered. The information has only a little structure and lacks scientific terminology.
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Topic 5 Ecosystems
1
a
The variety of life. This can be considered by the number of different ecosystems or
by the number of different species within them or by the genetic variation within each
of the species.
b
A measure of how close the numbers of organisms of each species are to each other
in a particular environment.
c
The number of different species present in an area.
2 Species richness only considers the number of different species present whilst species
diversity considers both the number of different species present and the relative
abundance of each of them.
3 Photoautotrophic organisms (such as green plants) synthesise complex organic
compounds using light energy. Heterotrophic organisms (such as animals) obtain organic
compounds from other organisms.
Sampling techniques for studying distribution and abundance
4 An abundance scale is a method of estimating the relative abundances of different species
against a known but subjective scale. Examples: DAFOR (Dominant, Abundant, Frequent,
Occasional and Rare) or ACFOR (Abundant, Common, Frequent, Occasional and Rare) or
the Braun-Blanquet scale (+ for less than 1% to 4 for between 51 and 75% coverage)
5 Longworth mammal traps/pitfall traps/light traps to trap animals; mark, release, recapture;
cameras for catching sight of wandering large animal; sweep nets for assessing
pond/stream animals.
6
a
Set out a suitable grid; generate random numbers for coordinates; devise method
for locating coordinates e.g paired working procedure; OR a line transect and
method of position quadrats along the line at suitable intervals; quadrat size
selected; identification of species; method of recording percentage cover for each
species; method of replication suggested; sampling for each slope.
b
For the south-facing slope N=46. For the north-facing slope N=41.
Species
South-facing
slope
(n/N)2
North-facing slope
(n/N)2
Calluna vulgaris
(heather)
14
0.092
9
0.048
Deschampsia
caespitosa (grass)
12
0.067
9
0.048
Trifolium repens
(clover)
9
0.038
6
0.021
Holcus lanatus
(grass)
6
0.017
3
0.005
Juncus effusus
(rush)
5
0.012
1
0.001
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Species
South-facing
slope
(n/N)2
North-facing slope
(n/N)2
Achillea millefolium
(yarrow)
0
0
6
0.021
Anthriscus
sylvestris (cow
parsley)
0
0
4
0.010
Epilobium hirsutum
(great willowherb)
0
0
3
0.005
N=
46
41
∑(n/N)2 = 0.226
c
∑(n/N)2 = 0.159
Carry out replicates in each site / repeat the investigation at another time.
Biotic factors
7
Biotic factor
Example
Relative benefits to populations involved
Intraspecific
competition
An example of plant or
animal species
Successful organisms gain water, nutrients, space,
shelter, light and a suitable mate
Intraspecific
cooperation
Social species, e.g.
termites, ants or honey
bees
Work together for benefit of whole group
Interspecific
cooperation
Mutualism, e.g. corals
and zooxanthellae,
lichen, mycorrhiza
Algae are protected from consumers or gain nitrogen
from nitrogenous waste or gain carbon dioxide from
coral respiration; coral gains carbohydrates
In lichen, fungi gain sugars and nutrients. Algae gain
a foothold and are protected from water loss
In mycorrhiza, fungi gain sugars, nutrients and
protection; plant gains ions
Predation
Carnivores
Carnivores gain food prey species suffer loss of
population members
Diseases
Parasites, e.g.
tapeworms, flukes,
plasmodium,
ichneumon wasp
Only the parasite benefits, with the host suffering to a
varying degree
Infectious diseases,
e.g. influenza
Causes harm to host and spreads to other individuals
Abiotic factors
8
a
Any abiotic factor related to the climate, e.g. wind exposure, temperature range,
rainfall/ precipitation.
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b
Any abiotic factor related to the soil, e.g. air (oxygen concentration), pH, soil texture,
soil type, soil depth, humus content, mineral ion concentration, water content/ water
table level.
c
Any physical factor such as aspect, gradient, drainage, altitude, topography, soil
erosion.
9 Soil type or crumb type/texture, moisture content, mineral ion content, humus content, pH,
compactness, temperature, soil air/oxygen.
Energy and biomass transfer
10 A measure of the organic and inorganic matter, excluding water, present in an organism or
in a trophic level.
11 Photoautotrophic and chemoautotrophic organisms. Accept green plants and
chemoautrophs,
12 Photoautotrophs fix carbon dioxide using light energy. Chemoautrophs fix carbon dioxide
using energy derived from the oxidation of inorganic substances.
13 A photoheterotrophic organism uses light energy but obtains carbon as complex carbon
compounds. A chemoheterotrophic organism obtains energy from complex carbon
compounds already built up by organisms in a lower trophic level.
14
Efficiency of energy transfer
15 Efficiency of energy transfer =
net productivity of trophic level 2
´100%
net productivity of trophic level 1
a
(44090/7.1 × 106) × 100 = 0.62%
b
(8760/44090) × 100 = 19.87%
c
3960 – 2300 = 1660
(1660/8760) × 100 = 18.94%
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Recycling in the ecosystem
Recycling of carbon
16 They have acted as a sink for carbon, which is now being released when fossil fuels are
combusted.
17 Level 3 (5–6 marks) Provides a comprehensive discussion of the number of different
producers. An understanding of the producer feeding relationship in the ecosystem /
trophic level. An understanding of the process of carbon fixation. Well-developed line of
reasoning that reads well is clear and logical and uses scientific terminology where
appropriate. All information is relevant and in continuous prose.
Level 2 (3–4 marks) Describes some (a few) of the issues with some indication that both
the producer role within the ecosystem and carbon fixation as a process are being at least
considered. A line of reasoning that is presented with some structure and use of scientific
terminology. Mostly relevant information.
Level 1 (1–2 marks) Describes some aspects of the producer issue or of the carbon
fixation process. Unlikely to be both but if it is then very patchy and superficial for both.
The information has little structure and lacks scientific terminology.
Recycling of nitrogen
18 Nitrogen is fixed in the form of nitrates (nitrification). Nitrate ions are taken up by
autotrophs, to be used to make amino acids. Amino acids are used to synthesise proteins
or used as amino groups for synthesis of purines and pyrimidines. Primary consumers eat
the autotrophs, hydrolyse the proteins to amino acids that are absorbed and reused in
protein synthesis. In turn, these new proteins are consumed by secondary consumers and
decomposers. Consumers may metabolise some amino acids into the Krebs cycle or
break down unwanted amino acids for conversion to glucose or glycogen.
Nitrogen fixation
19 Nitrosomonas—converts ammonia (NH3) or ammonium ions (NH4+) to nitrite ions (NO2–);
Nitrobacter—converts nitrite ions (NO2–) to nitrate ions (NO3–); Azotobacter—nitrogen
fixation; Rhizobium—nitrogen fixation.
20 Ammonia (NH3) or ammonium ions (NH4+); dinitrogen/nitrogen gas (N2); nitrite ions (NO2-)
nitrate ions (NO3-)
Succession
21 Ecological succession describes the series of changes (called seral stages) occurring over
time, in which each community changes the abiotic conditions, enabling other communities
to develop. Features common to all types of succession include:

loss of members of earlier communities through competition with populations in the
later communities

eventual development of a relatively stable climax community

progression from pioneer species that colonise the environment to the climax
community, associated with an increasing number of populations, an increase in the
complexity of the food webs, an increase in total biomass and an increase in the
stability of the community
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Primary succession
22 Primary succession begins in a barren environment, such as new land exposed by
volcanic activity, water collecting in a hanging lake or land exposed by glacial retreat.
Secondary succession begins in an environment that was previously colonised but has
become bare due to damage, e.g. a forest fire or a dried up pond. In this case the
environment might not be completely free of organisms, for example seeds or resistant
spores from the previous community might remain.
23 Marram grass and sedge are the pioneer species—only they can tolerate the harsh
conditions of the bare sand. Their death and decomposition adds humus to the sand,
changing its humus content and pH. This allows heather to colonise the developing soil.
The death and decay of the dominant plants in each community lowers the pH still further
and adds yet more humus to the soil, changing its colour. Eventually a climax community,
dominated by oak, develops.
Deflected succession
24 Chalk grassland, heathland, lawns and playing fields: grazing or human intervention such
as mowing keeps the seral stage an early stage—called plagioclimax. Agriculture:
ploughing, crop growth and harvesting cause deflected succession. In this case no
plagioclimax forms as the harvesting prevents this.
Exam-style questions
1
Abiotic factor
Effect on plant distribution
Temperature
Temperature determines species e.g. cool temperatures:
temperate plants; hotter climes: tropical plants. In any
ecosystem an increase in temperature causes a rise in
metabolism of organisms, e.g. an increase in the rate of
photosynthesis.
Light intensity
An increase in light intensity increases the rate of
photosynthesis in some plants but is harmful to those that
are adapted to shaded environments.
Oxygen concentration
Low oxygen concentration is rare in ecosystems in which
plants are able to photosynthesise rapidly. In aquatic
environments that are polluted by organic waste, bacterial
respiration might reduce the oxygen to such a low
concentration that plants will not survive.
pH
pH is a critical factor in determining the niche of a species.
Some need a neutral pH and others survive best in alkaline
or in acidic conditions
Wind speed
High wind speeds increase the rate of transpiration from
plant leaves as well as reducing the local temperature.
Woody xerophytes, such as pine, survive well in these
conditions but more fragile, herbaceous plants seldom do.
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2
a
Dry mass, wet mass and carbon mass. (For dry mass, the biological material is
weighed, dried at a temperature high enough to evaporate the water but not so high
that it will burn the organic matter, and reweighed until its mass is constant.)
b
Dry mass is an easy way to measure biomass, but the organisms must be
destroyed. Wet mass is easy to measure but includes water that is variable in
quantity and is not a component of assimilation. Carbon mass is a good comparator
as carbon is a key component of all assimilated organic compounds, but it is very
difficult to determine/does not measure nitrogen content does not measure
inorganic/mineral content.
c
Harvest the production from a known area. Burn the harvested material in such a
way that it heats a known volume of water/in a bomb calorimeter. Use the value of
the specific heat of water to calculate the energy needed to cause the measured
rise in water temperature. The units will be a unit of energy and a unit of area, e.g.
kJ m–2. To calculate productivity over a period, the units would include time, e.g. kJ
m–2 year–1.
3 Level 3 (5–6 marks) Provides a comprehensive discussion of the relationship between
energy and the ecosystem. Well-developed line of reasoning that reads well, is clear and
logical, and uses scientific terminology where appropriate. All information is relevant and in
continuous prose.
Level 2 (3–4 marks) Describes some (a few) of the issues with some indication of both
energy and ecosystem being at least considered. A line of reasoning that is presented with
some structure and use of scientific terminology. Mostly relevant information.
Level 1 (1–2 marks) Describes some aspects of the issues. The information has little
structure and lacks scientific terminology.
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Topic 6 Populations and sustainability
Factors determining population size
1
a
All the individuals of the same species living in the same area at the same time.
b
All the organisms of all the populations living in the same area at the same time.
2 Limiting factors are those factors that are closest to the minimum required and so will limit
the growth (in this case) of the population. There are a number of different abiotic and
biotic factors that affect population growth.
3 Members of the same species occupy the same niche and so will be competing for the
same resources, whilst those of different species have a different niche and are unlikely to
have identical requirements to support life.
4 Lag Description: the growth is very slow and shows little increase as the organisms
become accustomed to the environment. Explanation: In some cases specific enzymes
must be manufactured in order to make use of the available nutrients. Reproduction in the
population is slow because the initial numbers are low.
Log Description: this is the stage of exponential growth, so the curve is very steep.
Explanation: reproduction and growth are rapid and exceed death rate, resources are
freely available and not limiting growth.
Stationary Description: the curve levels off and flattens. Explanation: reproduction has
slowed and now is equal to death rate. Resources are limiting, so only a set number of
organisms can be supported (environmental resistance). The fermenter has reached
carrying capacity.
5 The curve will drop and eventually may become zero. This is because resources have
been used up and/or toxic waste products have accumulated.
6
a
The maximum population size / density that an ecosystem can support.
b
A precise description of the range of abiotic and biotic factors in which organisms of a
single population can survive.
c
When different species avoid direct competition by occupying different niches within
the same habitat.
d
When two species have the same niche but the more competitively successful
species eliminates the less successful species in competition.
Interaction between populations
Predator–prey relationships
7
a
Between day 4 and 14 the two populations fluctuate, with the predator cycle lagging
a little behind that of the prey. As the predator populations rises in number the prey
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population falls to a low point (of 26 at day 10). In turn this causes the predator
number to fall (26 at day 12). Prey population now increases as there are few
limiting factors once the food increases. This rise provides the predator with more
food so their number rises.
b
At A the prey numbers reach a peak due to few limiting factors. Predator numbers
reach a peak 2 days later. The predator numbers increase above the prey numbers
briefly.
c
Paramecium
Day 1: 162/15 cm 3 of medium
Day 16: 162/15 cm 3 of medium
Change = 0
Saccharomyces
Day 1: 75/0.1 cm 3 of medium
Day 16: 75/0.1 cm 3 of medium
Change = 0
8 The predator population is also likely fall, a little after the fall in the prey population.
However, the fall is not large because alternative food sources will also provide for
immediate food needs. Some organisms do not exist on one source of food.
9 Interspecific competition is between members of different species. Intraspecific
competition is between members of the same species.
10
a
For example, between grey (or black) and red squirrels or American bluebells and
English bluebells, or any suitable suggestion.
b
For example, between male robins or stags for a mate or territory, or any suitable
suggestion.
Conservation of biological resources
11 Conservation is the management of biodiversity and sustainability of habitats and species
whilst at the same time allowing managed use by humans. Preservation is protection of
ecosystems, habitats and species without allowing any human use.
12
Biological resource
Human use
Reason
Sustainability
Timber
Use as construction
material for
buildings/homes and
for paper
Economic value
Managing both natural
and plantation forests,
e.g. coppicing in
woodlands
Fish
As a protein source
(food) and a vitamin
source
Economic value
Managing fish stocks in
oceans by controlling
fishing methods,
volumes and areas
13 Banning fishing in certain areas (exclusion zones can be set up), limiting the number of
fishing boats; limiting the size of catch for each boat or the number of certain species;
banning fishing during the spawning season; regulation of the type of fishing gear, e.g. the
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size of fishing net, to allow young fish to escape and breed and restocking. Vital for
inspections to ensure all these points are enforced.
14 Protected by the Antarctic Treaty, which protects the area for science and conserves
marine life. This is a pioneering ecological approach, which means looking at population
dynamics but also at reducing the chances of harm, e.g. by reducing visitors, preventing
introduction of alien plant and animal species, long-term monitoring of species present
(e.g. Adele penguin) and especially reducing catches of krill. Setting up complete
exclusion zones is part of this. Standard methods do not monitor or aim to control
ecologically but use mainly population dynamics and monitoring.
15
Management of
timber
Method of removal of timber
Effect of management
Clear felling
Removal of trees of same age over
an area and so removing the
canopy.
Economic advantage but is
destructive because
biodiversity is affected and soil
is exposed and likely to erode.
Selective felling
Removal of mature, diseased and
unwanted trees. Canopy is mostly
preserved.
Economically less
advantageous but less
destructive to ecosystem.
Trees left in place allow
continued biodiversity and
natural regeneration.
Strip felling
Removal of small areas or strips.
Strips are replanted and adjacent
strips are only felled once these
have grown to a reasonable size.
Less disruptive with reduced
damage to biodiversity. Little or
no soil erosion.
Coppicing
Cutting trees to ground level but
leaving the stump to re-grow, with
several stems developing from
lateral buds. These grow very
rapidly because roots are well
established. Can be carried out in
rotation.
Wood is important for charcoal,
fencing poles and posts. Can
be used for paper or burning in
power stations. Can be
repeated indefinitely and a
range of trees are suitable.
The effects of humans on environment
16 For example:
Ecosystem
Specific interest
Possible human damage
Masai Mara in Kenya
Large mammal populations
of the savanna
Tourism; increase in local
human population; big game
hunters illegally killing
animals enticed out of the
reserves
Terai region in Nepal
Bengal tiger, one-horned
rhino and Indian elephant
Reducing numbers to critical
levels through illegal hunting
for ivory or tusks or other
body parts for Chinese
medicine; poaching by game
hunters; habitat reduction
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Ecosystem
Specific interest
Possible human damage
Peat bogs
Acidic soil caused by
constant leaching out of
bases increases abundance
of moss (Sphagnum sp.),
heather, deergrass, bog
myrtle, so it is unique;
provides a habitat for red
deer and rare birds
Measures to ‘improve’ the
soil using drainage or
afforestation; use of conifers
that are not native; reducing
the natural habitat
Snowdonia National Park
Scenic areas with mountain
and local species
Invasion by alien plants;
tourism including climbers
Antarctica
Penguins and marine life
Invasion by alien plants and
animals; overfishing of krill
Galapagos islands
Many endemic species
Alien species and tourism
17 Invading the areas and showing strong and vigorous growth that out-competes the natural
vegetation. In some areas river and stream banks have become overcrowded and choked.
Habitat change occurs and river flow is obstructed – leading to changes in the freshwater
species.
18 These are established and protected links that join up separated areas such as parks and
game reserves. They allow many animals to move from one to the other in a protected
way and take advantage of much larger spaces they need to search for food, mates and
places to live and survive. It is important they are maintained for these reasons.
19 Krill populations are important as a food source for a large variety of marine life such as
birds, seals, some whale species and many species of fish. Krill are at the primary
consumer level (trophic level) of feeding relationships with an important link to other
consumers. When numbers decline, other organisms have too little food and numbers of
all marine life decline. Krill feed on phytoplankton. With low numbers of primary
consumers, phytoplankton numbers will not be controlled so they will increase in numbers
and upset the balance of gases—oxygen and carbon dioxide—in the seas, eventually
causing large-scale death and decline. This will impact on the carbon levels in the
atmosphere as phytoplankton in the seas are the largest carbon sink on earth.
20 Measures to ‘improve’ the soil led local authorities to drain soil and plant trees (e.g. nonnative conifers) in an afforestation program funded by grants. This has reduced the natural
habitat including natural vegetation such as moss (Sphagnum sp.) heather, deergrass,
bog myrtle and cottongrass, which die as the soil is dried out. Animal species, such as red
deer and rare birds (e.g. golden plovers, golden eagles, greenshanks and dunlins) are
declining. It is a unique habitat.
Efforts to conserve the areas include reducing the use of non-native conifers and
replanting with native species, reducing the soil drainage program and removal of the tax
breaks that encouraged the process. Other actions include protecting the areas (e.g. the
RSPB has taken steps to prevent further development). Blocking drains and removing
non-native trees helps to restore the areas. Protection also includes setting up nature
reserves( e.g. at Forsinard Flows) and protecting areas by citing them as Sites of Special
Scientific Interest (SSSIs).
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21
a
Bengal tiger, one-horned rhino and Indian elephant.
b
Rhinos and elephants illegally hunted for horns and ivory, and tigers for body parts
for Chinese jewellery/artefacts and medicine. Poaching by game hunters. All cause
reduction in numbers to critical levels. Habitat reduction causes a reduction in other
species as well.
c
National parks, conservation units such as the Tiger Conservation Units and
policing/patrolling of these areas and game reserves units provided by the
government and the army.
Exam-style questions
1 Level 3 (5–6 marks) Provides a comprehensive discussion of the conflict issues, with an
understanding of both sides of the argument. Well-developed line of reasoning that reads
well, is clear and logical, and uses scientific terminology where appropriate. All information
is relevant and in continuous prose.
Level 2 (3–4 marks) Describes some (a few) of the issues with some indication of both
sides of the argument being at least considered. A line of reasoning that is presented with
some structure and use of scientific terminology. Mostly relevant information.
Level 1 (1–2 marks) Describes some aspects of the issues. The information has only a
little structure and lacks scientific terminology.
2
a
A non-living environmental factor that affects the ecosystem and distribution of
organisms.
b
Biotic = any other stream-living organism, e.g. shrimps, sticklebacks, minnows.
Abiotic = pH of water, flow rate of water, temperature of water, salinity etc.
c
Oxygen concentration is much lower than at the other sites; allow paired and
comparative data quotes that illustrate this point because high BOD of organic
pollutants or bacterial action decomposing pollutants; high numbers of tube worms at
site B and tube worms survive in/are indicator species of low oxygen concentration;
absence of species such as mayflies or stoneflies at B and these can only survive
in/are indicator species of unpolluted water.
d
Presence of limiting factors; competition between individuals within the same
species (intraspecific); the carrying capacity has been reached for that ecosystem.
e
Runoff of the manure fertiliser into stream at B; leads to eutrophication/or an
explanation of this; decomposers (aerobic bacteria) increase in number; as a result
use the oxygen; reduction in certain freshwater/aquatic species.
f
Protection of biodiversity by the conservation area; compromised by the farm
activity; educational facilities of trust affected; any research may be affected; effect
on fund raising; visual appearance may be affected; farmer interested in increasing
milk yield; and financial impact of moves/directives for change or alternative
wording; possibility of enforcement or protection legislation/a named reserve
designation.
© Jenny Wakefield-Warren 2017
Hodder Education
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