GENETICS: THE CODE BROKEN?
KEY WORDS AND TERMS USED IN THIS TOPIC
As you study this topic you should write the definitions for the following syllabus terms.
Term
polypeptide
gene expression
multiple alleles
ABO blood groups
the Rhesus factor
polygenic inheritance
DNA fingerprinting
diploid
haploid
somatic cells
gametic cells
dihybrid crosses
linked genes
chromosome mapping
Human Genome Project
recombinant DNA
gene therapy
trisomy
polyploidy
mutation
base substitution mutations
frameshift mutations
transposable genetic elements
germ line mutations
somatic mutations
gene cloning
whole organism cloning
selective breeding
gene cascades
gene homologues
Definition
GENETICS: THE CODE BROKEN?
SUMMARY OF THIS TOPIC
In this option, polypeptide synthesis is revised, as
is the structure of the DNA molecule. Polypeptide
synthesis is also referred to as ‘gene expression’,
because this is when the information in the original
DNA molecule actually expresses itself. Gene
expression is regulated by the action of other
‘regulatory’ genes, which produce proteins that
can control the transcription stage and other
aspects of protein synthesis.
Characteristics can be determined by more than
one pair of alleles (‘multiple alleles’) within a gene
pair; examples of this include the inheritance of
blood groups and Rh antigens on the red blood
cells. This situation differs from ‘polygenic
inheritance’, which occurs when many alleles,
often located on different chromosomes, control
the inheritance of a particular characteristic.
Examples of this include the inheritance of human
height and skin colour. The greatest degree of
genetic variation between two organisms occurs in
the non-coding regions of the DNA molecule
known as ‘introns’, and these regions are
consequently used to compare two individuals in
DNA fingerprinting techniques.
Somatic (body) cells contain the full complement
of chromosomes, known as the ‘diploid’ number,
whereas gametes (sex cells) contain only half this
number (i.e. the ‘haploid number’). Genes can be
inherited on different chromosomes or, in the case
of ‘linked’ genes, on the same chromosome.
Typical Mendelian ratios do not occur when
inheritance patterns of linked genes are traced;
these patterns can in fact be used to construct gene
linkage maps to determine the relative position of
linked genes along a chromosome.
An even more detailed study of gene locations is
being carried out by the Human Genome Project.
This task involves mapping all the genes in the
human genome, working out the DNA sequence of
each gene, and identifying disease-causing genes.
Recombinant DNA technology can also be used to
locate the positions of genes on chromosomes and
the actual nucleotide sequence on the DNA
molecule.
Modern genetic techniques have also enabled the
development of ‘gene therapy’, a process which
involves altering the genetic makeup of an
individual. Although mostly still in the trial stage,
treatments for diseases such as cystic fibrosis,
cancers and genetic diseases are showing promise.
The genetic makeup of an individual can change
through various types of mutations, often
producing harmful effects on human health.
Examples of such harmful mutations include
Down’s syndrome, muscular dystrophy and sicklecell anaemia. Despite this, genes damaged during
DNA replication are continually being repaired by
special DNA repair genes, which act to either
restore damaged bases or remove and replace
them. Genes may also be changed when sections
of DNA (‘transposable elements’) move from one
part of the genome to another.
Humans, through selective breeding and cloning,
can also alter the genomes of plants and animals
by favouring particular genes over others. Gene
cloning, which uses genetic engineering techniques
to produce multiple copies of a desired gene, can
also be used to produce transgenic organisms with
completely different genomes to the original ones.
The stages of embryonic development are
controlled by different sets of genes; regulatory
genes appear to control the expression of other
genes, with the result that not all genes are
expressed during each stage of an organism’s
development. Studies of regulatory genes such as
‘HOX’ genes and other genes have also helped
scientists to determine evolutionary relationships
between organisms.
GENETICS: THE CODE BROKEN?
MAJOR OBJECTIVES OF THIS TOPIC
As indicated in the HSC Biology syllabus, the
major outcomes of this topic include the ability to:
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describe the main stages of polypeptide
formation, and the roles of DNA and RNA
in this process
construct a model of DNA
explain current ideas about gene
expression
describe the inheritance of a trait
controlled by multiple alleles in an
organism other than humans
solve problems relating to the inheritance
of ABO blood groups and the Rhesus
factor in humans
describe an example of polygenic
inheritance
explain how variable genes are used in
DNA fingerprinting to discern genetic
differences between organisms
explain the terms ‘haploid’, ‘diploid’,
‘somatic’ and ‘gametic’ cells
solve problems involving dihybrid crosses
of independently inherited and linked
genes
outline the use of gene linkage maps and
explain how they can be made using the
results of experimental crosses
discuss the main objectives and limitations
of the Human Genome project
explain how recombinant DNA is
produced and outline how this technology
is used to locate genes on chromosomes
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describe the use of gene therapy in the
treatment of a genetic disease, a named
form of cancer, or AIDS
describe the effects of mutations on human
health, and describe the following types of
mutations:
rearrangements
changes in chromosome number
base substitution
frameshift
outline the role of DNA repair genes
explain how transposable genetic elements
can alter the genome
describe the difference between germ line
and somatic mutations in terms of their
effect on the genome
describe how selective breeding, cloning
and gene cloning can alter the genetic
makeup of a particular species
trace the history of the selective breeding
of a particular agricultural species
describe the process involved in animal
cloning
explain how embryonic development is
controlled by genes, including the role of
gene cascades in limb formation
describe how the occurrence of gene
homologues in different species can be
used to trace evolutionary relationships
1) The structure of a gene
provides the code for a
polypeptide
Eventually, a chain of amino acids is formed. This
process is known as translation. Fig. 6-1 shows the
main steps involved in protein synthesis, and Table
6-1, below, explains some of the terms commonly
used.
The role of DNA in polypeptide
synthesis
transcription
The steps involved in polypeptide synthesis are
outlined in chapter 2. Recall that the stages
involved in this process are as follows:
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DNA in the nucleus ‘unzips’, exposing
unpaired nitrogen bases.
Messenger RNA copies the code on a single
stranded DNA molecule in the nucleus. This
process is known as transcription. RNA is
single stranded and contains a ribose instead
of a deoxyribose sugar.
The messenger RNA moves from the nucleus
and attaches itself to the small subunit of a
ribosome in the cytoplasm.
Transfer RNA molecules carry a ‘triplet’ of
bases at one end. Each triplet codes for a
particular amino acid, and these are picked up
by the tRNA molecules in the cytoplasm.
The transfer RNA molecules match up with
their complementary base triplets on the
messenger RNA.
Further amino acids, carried by their transfer
RNAs, become attached to the messenger NA
and are joined to each other by peptide bonds.
translation
messenger RNA
transfer RNA
code
codon
anticodon
process where the
information on a single
DNA strand is copied
by a mRNA molecule
process in which
polypeptides are
assembled on the
ribosome using the
information in the
mRNA molecule
a type of RNA that
carries the genetic code
from DNA in the
nucleus to the ribosome
in the cytoplasm
a type of RNA that
carries amino acids to
the ribosome
the sequence of bases in
a DNA molecule
a set of three bases on
DNA or mRNA
a set of three bases that
complements a codon;
found on tRNA
molecules
Table 6-1 Terms used in polypeptide synthesis
ii)
aa
i)
mRNA
iii)
DNA
aa
Fig. 6-1 Polypeptide synthesis. In i), the DNA code is being copied by mRNA; in ii), mRNA moves to a
ribosome in the cytoplasm; in iii), tRNA carries amino acids to the ribosome where the base triplets match uphere, the amino acids will link together to form a polypeptide
The current understanding of gene
expression
THINK!!! In a segment of one strand
of a DNA molecule, the base sequence is CAA
CTA GAA. What would be the sequence of bases
in a mRNA molecule that has been transcribed
from the DNA? (Remember that in RNA the base
uracil replaces thymine).

As a requirement of this topic, you need
to perform a first-hand investigation to
construct a model of DNA.
James Watson, Frances Crick and Maurice Wilkins
received the Nobel Prize in 1962 for working out
the structure of the DNA molecule. Their model
revealed DNA as a double stranded helix,
consisting of alternating sugar and phosphate
groups which are linked by pairs of nitrogen bases.
Each strand of the molecule is complementary to
the other one; that is, the base sequence on one
strand is complementary to the base sequence on
the other. Guanine (G) always pairs with cytosine
(C), and adenine (A) always pairs with thymine
(T). Fig. 6-2, below, shows how a model of a DNA
can be made using pegs of four different colours,
wool or string, and shapes to represent sugar or
phosphate groups.
complementary
base pair
When the information in a gene is actually used to
manufacture a particular polypeptide, we say it is
being expressed. During the life of an organism,
many genes are only expressed at certain times;
during adolescence, for example, the genes
responsible for the production of hormones will
become ‘switched on' to a greater degree. Once
gene expression commences, transcription of the
DNA code onto a messenger RNA and translation
of this code into a series of amino acids on the
ribosome will occur. Current knowledge of this
process recognises the fact that gene expression is
regulated by the action of other genes. These
regulatory genes produce proteins that bind to
‘control element’ segments of the gene in question
and either activate it or suppress its expression.
Most gene regulation occurs during the
transcription step of polypeptide synthesis, and
involves ‘transcription factor’ genes. The
regulatory proteins produced by these genes can
control the number of RNA transcripts produced,
the rate of transcription and which section of the
DNA molecule is copied. By recognising the first
and last codons in the gene, the regulatory proteins
can also turn the transcription process on or off,
thus controlling the length of the messenger RNA
formed. Gene regulation also controls other
aspects of gene expression such as the splicing of
the initial large mRNA to remove the coding
sequences (exons) from the non-coding sequences
(introns). This process can affect the type of
protein that is ultimately produced. Other points of
control include the selection of mRNA molecules
that undergo translation on the ribosome and the
activation or inactivation of the actual proteins that
have been made.
sugar
group
phosphate
group
Fig. 6-2 Making a model of DNA
Within the control element section of
a gene, there are regulatory sequences and
promoter sequences. The regulatory sequence is
the region regulatory proteins bind to when
controlling the expression of a gene. The promoter
sequence is a region at the start of the gene that
makes sure DNA polymerase transcribes the DNA
in the correct direction during the transcription
process.
2) Multiple alleles and polygenic
inheritance provide further
variability within a trait
Group
Antigen
present
Antibody
present
A
B
AB
A
B
A, B
B
A
none
O
none
A,B
Multiple alleles: ABO blood groups
and the Rhesus factor
Multiple alleles occur when a characteristic is
determined by more than one pair of alleles within
a gene pair. In humans, for example, blood groups
are determined by three different alleles; A, B and
O. The combination of these alleles can result in
four different blood groups- A, B, AB or O. Group
A has the genotype IAi or IAIA, group B can be IBi
or IBIB, and group O is ii. Group AB has the
genotype IAIB and is an example of co-dominance
of both genes. Blood in each group except for
group O contains antigens on the surface of the red
blood cells; group A contains A antigens, group B
contains B antigens and group AB contains both A
and B antigens. Group O blood contains Antibody
A and Antibody B, group A contains Antibody B,
group B contains Antibody A, while group AB
contains no antibodies. As a result, care must be
taken when administering blood transfusions that
blood types are not incompatible (see table 6-2). A
person with group A blood, for instance, cannot
receive blood from group B or group AB because
Antibody B in the recipient’s blood will react with
the B antigen present in the donated blood and
cause the blood to agglutinate.
Another set of proteins found on the red blood
cells is determined by at least eight alleles. These
are called the Rh antigens, and the lack of antigen
D (the ‘rhesus factor’) in a mother’s blood can
result in a condition known as haemolytic anaemia
in newborn babies. A person without antigen D is
referred to as being Rh negative (genotype rr),
while those possessing the antigen are Rh positive
(genotypes RR or Rr). If an Rh negative mother is
carrying an Rh positive embryo, she may develop
antibodies against antigen D during childbirth,
which can be harmful to subsequent Rh positive
babies she may carry. Rh positive babies can only
arise if the father has the genotype RR or Rr.
A method now used in hospitals
involves injecting an Rh negative mother with Rh
antibodies so she does not need to make her own.
This means that her antibodies are not present in
subsequent pregnancies.
Can
donate
blood
to:
A, AB
B, AB
AB
A, B,
AB, O
Can
receive
blood
from:
A, O
B, O
AB, O,
A, B
O
Table 6-2 The compatibility of various blood
group combinations
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As a requirement of this topic, you need
to be able to solve problems to predict
the inheritance patterns of ABO blood
groups and the Rhesus factor.
Sample problems
i)
Is it possible for parents with blood types A
and B to produce a child with blood group O?
Explain your answer, showing working.
Answer
Yes; the cross involved is IAi x IBi
↓
IAIB, IAi, IBi, ii,
and ii is the genotype of group O.
ii) The possible genotypesof an Rh+ person arte
RR and Rr, while the genotype of an Rhperson is rr. What are the chances of an Rhwoman producing an Rh+ baby if her husband
is Rh+ and heterozygous for the Rhesus factor?
Show working.
Answer
The cross involved is Rr x rr
↓
Rr, Rr, rr, rr,
so 50% of the offspring must be Rh+ (Rr)
Multiple alleles in other organisms
Traits such as coat colour in animals can be
controlled by multiple alleles. The colour of mice,
for example, is controlled by three alleles,
producing four possible phenotypes as shown in
table 6-3.
Genotype
Aya
AyA
AA or Aa
aa
Coat colour
yellow
yellow
Agouti ( a light brown
colour)
black
Table 6-3 Multiple alleles in mice
Upon inspection of this table the alleles can be
listed in order of decreasing dominance as follows;
AY> A > a. Some sample problems associated
with these alleles are outlined below.
phenotypic expression of that characteristic will
not be represented by discrete alternatives, but will
instead show continuous variation. Although all
genes involved are inherited according to Mendel’s
laws, there are often so many of them that the
contribution of each one to the final phenotype will
be minimal. Moreover, the environment also
usually has a large effect on the phenotype in these
cases. Examples of features controlled by
polygenic inheritance include human height and
skin colour, leaf length, milk production in cattle
and egg weight in poultry. In each of these
situations a range of characteristics occurs, with
the measurable ones generally conforming to a bell
curve, as shown in Fig. 6-3.
Sample problems
i)
Predict the phenotypic ratio of the offspring
produced when a black mouse is crossed with
a heterozygous agouti mouse.
Answer
The cross involved is aa x Aa
↓
Aa, aa, Aa, aa ,
so 50% of the offspring are agouti and 50% are
black
Fig. 6-3 The height distribution of male spectators
at a football final. Human height is an example of
polygenic inheritance, which results in continuous
variation
ii) A yellow mouse was crossed with a black
mouse. The phenotypes of the resulting
offspring were in the ratio 50% yellow: 50%
black. What was the genotype of the yellow
parent? Show working.
Answer
AYa x aa
↓
AYa, AYa, aa, aa
= 50% yellow, 50% black, so the yellow mouse
must have had the phenotype AYa.
Polygenic inheritance
Polygenic inheritance occurs when multiple
alleles, often located on different chromosomes,
control the inheritance of a particular
characteristic. The result of this is that the
THINK!!! The colour of wheat
kernels is also determined by polygenic
inheritance. If a dark red kernel has the genotype
R1R1R2R2 and a white kernel has the genotype
r1r1r2r2, match the genotypes in column A of the
table below with their phenotypes in column B.
A
r1r1R2r2
R1r1R2R2
R1R1r2r2
R1r1R2r2
B
Medium-dark red
Medium red
Medium red
Light red
The use of highly variable genes for
DNA fingerprinting
DNA fingerprinting is a technique used to compare
the DNA from two sources to determine whether
they are identical or related. Its applications
include criminal investigations, paternity testing,
constructing animal pedigrees and the
development of desirable traits in livestock and
disease resistance in crops.
i) Criminal investigations - In criminal
investigations, the DNA of a suspect can be
compared with the DNA found at a crime scene.
DNA is taken from tissue samples such as blood,
skin, sperm and hair. The DNA is then cut into
fragments using restriction enzymes. The segments
used for DNA fingerprinting are usually the noncoding regions, known as introns, as these tend to
vary greatly among individuals. The DNA
fragments are then separated according to their
weight using electrophoresis, a process in which
the fragments move up an electrically charged gel.
After being transferred to a nylon sheet, the
location of these fragments is indicated by using
radioactive or fluorescent probes, which have a
complementary base sequence to the fragments.
The fragments are then transferred to photographic
film using X-rays. The resulting band patterns can
then be compared to determine whether the tissue
of a suspect matches that found at the crime scene.
DNA fingerprinting can be used to determine a
crime suspect’s innocence, but cannot be used to
conclusively prove their guilt.
In criminal investigations, the actual intron
sequences chosen for testing are called VNTR
markers (abbreviated from ‘variable number
tandem repeats’). Within a particular population, a
combination of four different VNTR markers has
only a 1 in 100,000 chance of occurring. However,
within a particular ethnic group, these odds may be
as low as 1 in 1000. This is one reason why DNA
evidence cannot be used to conclusively prove a
person’s guilt. As a consequence, although the
blood of a suspected murderer may match that
found at the scene of the crime, his defence
lawyers could argue that this is not conclusive
evidence. On the other hand, lawyers for the
prosecution can make sure that as many markers as
possible in each DNA sample are tested to rule out
virtually all other suspects .They could also send
the DNA to two or more different laboratories to
help eliminate any experimental error.
ii) Animal breeding – The fingerprinting methods
described above can be used by animal breeders as
early as the embryonic stage to determine whether
the genes for a particular trait are dominant or
recessive. In the area of dog breeding, genes
responsible for over 350 inherited diseases will
soon be able to be located using this technique, as
well as ‘trigger’ genes that are often associated
with polygenic disorders such as epilepsy. In
agriculture, work is currently underway in
Australia to develop a DNA fingerprinting
procedure that determines the paternity and
maternity of Merino lambs.
THINK!!! The diagram below shows
the band patterns produced when DNA fragments
from different crime suspects (lanes 1-4) were
subjected to electrophoresis and compared with
DNA from the crime scene (lane 5). Which of the
suspects is guilty?
___
___
___
___
___
___
1
___
___
___
___
___
___
___
___
___
___
___
___
___
___
___
___
___
___
___
___
___
___
___
___
2
3
4
5
3) Studies of offspring reflect the
inheritance of genes on different
chromosomes and genes on the
same chromosomes
Chromosome numbers in somatic
cells and gametes
Normal body cells are known as somatic cells,
whereas those produced by the gonads in meiosis
are called gametes. Somatic cells contain the full
complement, or diploid number of chromosomes,
whereas gametes contain only half this number
(the haploid number of chromosomes). In humans,
the diploid number is 46, consisting of 23 pairs of
chromosomes. During meiosis, which is a
reduction division, the resulting gametes contain
23 single chromosomes. The stages of meiosis are
described in table 6-4, below, and illustrated in
Fig. 6-3.
1
2
Chromosomes appear as pairs of chromatids,
joined at the centromere
Homologous chromosomes pair up with each other
and chromosomes ‘cross over’ by exchanging their
genetic material in some places
3
Members of each homologous pair separate from
4
each other (chromatids
still remain
together)
Chromosomes
become bound
by a nuclear
Dihybrid crosses involving
independently assorted genes
In some of his breeding experiments, Mendel
experimented with two traits at a time to find out
whether or not they are inherited independently of
each other. In one experiment he crossed round,
yellow-seeded plants with wrinkled, green seeded
plants. The round and yellow traits were dominant
and the wrinkled and green traits were recessive.
All the seeds produced in the F1 generation were
therefore round and yellow, as shown below.
membrane and two new cells are formed
5
RRYY x rryy
↓
all RrYy
Chromosomes, no longer in pairs, consisting of two
chromatids line up along the centre of the cell
spindle
6
Chromatid pairs separate and one of each pair
moves to opposite ends of the cell
7
The nuclear membrane reforms, forming four new
cell with only half the number of chromosomes as
the original cell
Table 6-4 The stages of meiosis
When these F1 plants were crossed he found that
totally new combinations of plants appeared in the
offspring, namely in a ratio of 9 yellow round: 3
yellow wrinkled: 3 green round: 1 green wrinkled.
This could only have occurred if the gametes were
allowed to combine at random; in other words,
either allele of a gene pair can combine with either
allele of another gene pair. This is known as
Mendel’s 2nd Law or the Law of Independent
Assortment. The Punnett square below shows how
the alleles sort independently to produce the
9:3:3:1 ratio of phenotypes in the offspring.
RrYy x RrYy
Gametes
RY
Ry
rY
ry

RY
RRYY
RRYy
RrYY
RrYy
Ry
RRYy
RRyy
RrYy
Rryy
rY
RrYY
RrYy
rrYY
rrYy
ry
RrYy
Rryy
rrYy
rryy
As a requirement of this topic, you need
to be able to analyse the outcome of
dihybrid crosses when both traits are
inherited independently.
Sample problems
Fig. 6-3 The stages of meiosis
i) Black hair colour in guinea pigs (B) is
dominant to white hair colour (b) and short
hair (S) is dominant to long hair (s).When a
heterozygous black, short haired guinea pig
(BbSs) was crossed with a white, long haired
guinea pig (bbss), 80 offspring were obtained.
a) What are the four possible phenotypes
produced?
b) Use a Punnett square to predict how many of
the 80 are likely to be of each type.
Answer
a) i) black, short; ii) black, long; iii) white, short;
iv) white, long
b) There should be 20 of each type;
Gametes
bs
bs
bs
bs
BS
BbSs
BbSs
BbSs
BbSs
Bs
Bbss
Bbss
Bbss
Bbss
bS
bbSs
bbSs
bbSs
bbSs
bs
bbss
bbss
bbss
bbss
ii) In Drosophila (fruit fly), the gene for grey body
colour(G) is dominant to black colour(g) , and the
gene for long wings(L) is dominant to short
wings(l) . Describe a test cross you could carry out
to determine whether a grey, long-winged fruit fly
was pure breeding ( homozygous) for each trait.
Draw a Punnett square to explain your answer.
Answer
The fly should be crossed with a pure breeding,
recessive fly (i.e.black, short-winged). If no
recessive offspring are obtained, the fly is probably
pure breeding (i.e. GGLL).
Gametes
gl
gl
gl
gl
GL
GgLl
GgLl
GgLl
GgLl
GL
GgLl
GgLl
GgLl
GgLl
GL
GgLl
GgLl
GgLl
GgLl
GL
GgLl
GgLl
GgLl
GgLl
Gl
Ggll
Ggll
Ggll
Ggll
a) Independently assorted genes
B= black hair; b= white hair; L = short hair; l=
long hair
Parents:
BBLL x bbll
Gametes: all BL
all bl
F1:
all BbLl (black, short haired)
F1 cross: BbLl x BbLl
Gametes:
BL, Bl, bL, bl
If, however, the gene for hair colour is located on
the same chromosome as the gene for hair length,
the following results would be expected:
b) Linked genes
Parents: BL/BL x bl/bl
Gametes: all BL
all bl
F1: all BL/bl (black,
short haired)
(/ represents linkage)
GL
GgLl
GgLl
GgLl
GgLl
All offspring here will be grey, long-winged flies.
If the original grey, long-winged parent was
heterozygous for either or both traits, some of the
offspring would exhibit recessive traits, as shown
below for the cross between GGLl and ggll flies.
Gametes
gl
gl
gl
gl
Genes that are located on the same chromosome
are said to be linked. As a result, they do not
undergo independent assortment during meiosis,
and only becoming separated from each other
when crossing over of chromatids occurs during
metaphase 1. The inheritance of two pairs of
linked genes thus produces different results to
Mendel’s dihybrid ratios involving independently
assorted genes. If the genes involved in a dihybrid
cross are assorted independently, we would expect
the following results when a black, short haired
animal is crossed with a white, long haired animal:
GL
GgLl
GgLl
GgLl
GgLl
Gl
Ggll
Ggll
Ggll
Ggll
Half the offspring here will be grey, long-winged,
and half will be grey, short-winged
Dihybrid crosses involving linked
genes
F1 cross: BL/bl x BL/bl
Gametes: BL or bl only
The F1 cross in b) is shown in the Punnet square
below.
Gametes
BL
bl
BL
BBLL
BbLl
bl
BbLl
bbll
The phenotypic ratio of the F2 offspring is thus 3
black, short haired: 1 white, long haired, assuming
no crossing over of the linked genes has occurred.
This obviously differs from the 9:3:3:1 ratio
expected from the F1 cross in a), where the genes
were independently assorted.

As a requirement of this topic, you need
to be able to analyse the outcome of
dihybrid crosses when both traits are
linked.
Sample problems
i)
Predict the phenotypic ratio of the offspring
produced when a heterozygous black, short
haired guinea pig (BbLl) is crossed with a
white, long haired guinea pig (bbll) if the
genes for each characteristic are linked
together on the same chromosome, and no
crossing over has occurred.
Answer
BL
BbLl
BbLl
bl
bbll
bbll
Phenotypic ratio of offspring;- 1 black, short: 1
white, long
ii) Parents with the genotypes GgDd and ggdd
were crossed. The chromosomes involved are
shown below.
Parent 1
D
Gametes
gd
gd
GD
GgDd
GgDd
gd
ggdd
ggdd
The only two possible genotypes of the
offspring are thus GgDd and ggdd.
c) The other possible genotypes among the
offspring if crossing over had occurred would
be: Ggdd, and ggDd. These would not occur in
a predictable ratio as crossing over occurs
randomly.
Gene linkage maps
Gametes
bl
bl
G
b)
Parent 2
g
d
x
g
g
d
d
a) Are the genes involved in this question linked?
Explain.
b) What are the only two possible genotypes of the
resulting offspring if crossing over does not occur?
Show your working.
c) What other genotypes are possible in the
offspring if crossing over does occur? Would you
expect these to occur in a predictable Mendelian
ratio? Explain.
Answer
a) Yes; G and D alleles are located on the same
chromosomes.
If, in a dihybrid cross, parents with the genotypes
GgDd and ggdd are mated, we would expect
offspring in the ratio of 1:1:1:1 (i.e. 1GgDd:1
Ggdd: 1 ggDd: 1 ggdd) when the genes are
assorted independently. If, however, the two pairs
of genes are linked, we would expect offspring in
the ratio of 1:1 (i.e. 1 GgDd: 1 ggdd). The working
involved in each case is shown in the punnet
squares below.
a) Independently assorted genes
Gametes
gd
gd
gd
gd
GD
GgDd
GgDd
GgDd
GgDd
Gd
Ggdd
Ggdd
Ggdd
Ggdd
gD
ggDd
ggDd
ggDd
ggDd
gd
ggdd
ggdd
ggdd
ggdd
b) Linked genes
Gametes
gd
gd
GD
GgDd
GgDd
gd
ggdd
ggdd
In reality, the phenotypic ratio of the offspring
produced in b) would not be 1: 1 because crossing
over of linked genes often occurs during meiosis.
Obviously, the closer together two genes are, the
less likely it will be that they will be involved in
crossing over , while the further apart they are on a
chromosome the more likely it is that this will
happen. The relative distance of one linked gene
from another, measured in ‘map units’, can be
estimated using the following equation:
No. of recombinant offspring x 100 = % recombinant
total number of offspring
offspring (map units)
Using this method, breeding experiments can be
conducted on genes that are known to be linked
and gene maps can be created to show the relative
positions of genes along a particular chromosome.
The gene linkage map, below, shows the relative
positions of the genes B, F and J
(‘m.u.’ = map units)
c) 1 grey, curved: 1 black, normal
d) % recombinant offspring = (43+ 45) x 100
806
= 10.9 %
= 10.9 map units
e) 1 grey, curved: 1 black, normal: 1 grey,
normal: 1 black, curved
B 3m.u. F
12 m.u
J
From this, it could be expected that genes B and J
are separated the most from each other during
crossing over, and that gene B crosses over with
gene F 3 % of the time, and with gene J 15 % of
the time. Gene F would be expected to cross over
with gene J 12% of the time.
Sample problem
A heterozygous grey bodied, normal winged fruit
fly (BbCc) is test-crossed with a black bodied,
curved winged fruit fly (bbcc). The 806 offspring
produced were as follows:
Bbcc
bbCc
BbCc
bbcc
560
558
43
45
a) What are the four different phenotypes
produced in the offspring?
b) Explain why these results indicate the two
genes involved are linked.
c) What would the expected phenotypic ratio of
the offspring be if no crossing over is
involved?
d) Use the formula in the passage above to
calculate the percentage of recombinant
offspring and hence the distance between the
two genes in map units.
e) What would be the expected phenotypic ratio
of the offspring in this cross if the genes were
independently assorted?
Gene linkage maps can be compared
to assess similarities between different species.
One comparison has found that a section of 17
linked genes on chromosome 3 of chickens has
remained intact during evolution and is almost
identical to the same section on a human
chromosome. Other studies have revealed humans
and mice, humans and pigs, and chickens and
cattle to share sections of linked genes. This
suggests the members of each pair may have
evolved from a common ancestor.

As a requirement of this topic, you need
to perform a first-hand investigation to
model gene linkage.
A chromosome pair (‘parent 1’) containing the
linked genes G and D can be constructed using
pipe cleaners and paper labels, as shown in Fig. 64.
G
g
D
d
4cm
Fig. 6-4 Modelling gene linkage
Answer
a) grey, curved; black, normal; grey, normal;
black, curved
b) The phenotypes of the offspring do not follow
Mendelian ratios.
Another chromosome pair (‘parent 2’) with the
genotype ggdd can be constructed, with each
linked gene positioned 4cm apart as before.
Assuming 1cm = 1 map unit = 1% recombinant
offspring produced, the percentage of recombinant
offspring produced when parents 1 and 2 are
crossed must be 4%. If 1000 offspring were
produced, a table of predicted offspring can be
completed, as shown below.
Genotype of offspring
Expected:
GgDd
ggdd
Recombinant: Ggdd
ggDd
Number of
offspring produced
475
485
19
21
Table 6-5 Modelling gene linkage
Each pair of linked genes can now be moved to
positions that correspond with, say, 68
recombinant offspring out of 1000 (i.e. 6.8%)
being produced.
THINK!! On the gene linkage map
shown below, what percentage of the offspring
would be recombinant if breeding experiments
tracing the genes C and E were carried out?
A 2m.u. E 1m.u. D 4m.u.
B
4m.u.
costing in excess of 6 billion dollars. The number
of genes in the human genome is roughly 100,000,
and 3 billion nucleotide pairs are involved. Despite
these seemingly impossible numbers, a draft
human genome map was completed in 2000, and
the sequencing of chromosomes 22, 21, 20 and 14
has been completed. Benefits arising from the
human genome project include a better
understanding of human evolutionary relationships
and the ability to detect, and even cure, up to 4000
genetic disorders. Moreover, a knowledge of the
genes causing hereditary diseases means that a
whole new range of drugs can now be developed
which target these diseases at the molecular level.
Information from the Human Genome
Project is already being applied in medicine.
Hundreds of tests have been developed to screen
people for mutations associated with genetic
disorders such as muscular dystrophy, sickle cell
anaemia, cystic fibrosis and Huntington’s disease.
In addition, tests have recently been developed to
detect mutations for breast, ovarian and colon
cancers. Knowledge of an individual’s risk levels
for these diseases can aid in their prevention –
those at a higher risk of breast cancer, for instance,
will have the option of undergoing more regular
mammograms.
C

4) The Human Genome project
is attempting to identify the
position of genes on
chromosomes through whole
genome sequencing
The Human Genome Project
A genome is defined as the full set of genetic
information on an organism’s chromosomes. The
main purposes of the human genome project are a)
to map all the genes in the human genome; b) to
deduce the DNA sequence of each gene and c) to
identify disease- causing genes. This undertaking
involves scientists from all over the world and is
As a requirement of this topic, you need
to use secondary sources to assess
reasons why the Human Genome
Project could not be achieved by
studying linkage maps.
Information for the human genome project has
mostly been achieved using recombinant DNA
technology, rather than by studying linkage maps,
because the latter method is too simplistic. One of
the reasons for this is that some segments of DNA
overlap and repeat themselves; in addition, linkage
maps only provide information about known
genes, which may comprise less than 50% of the
total genes on a chromosome. Gene linkage maps
may, however, be used to indicate the general area
in which a gene is located, and this section of the
chromosome can be cloned and studied later in
greater detail using recombinant sequencing
techniques.
Recombinant DNA technology
Recombinant DNA is DNA from a particular
organism that has been genetically modified to
contain sections of DNA from another organism.
Areas in which recombinant DNA has been useful
include the alteration of plants and livestock to
produce genetically modified, or ‘transgenic’
species, which possess beneficial features such as
disease resistance and improved yields. The
insertion of genes into bacteria has also enabled
the rapid production of chemicals such as insulin
and human growth hormone, which can be used to
treat people who need them. Another use of
recombinant DNA is in gene therapy, a process
being developed in which the DNA in body cells
can be altered, thus curing genetic diseases such as
diabetes, and even, in some cases, providing
immunisation against diseases such as Hepatitis.
Recombinant DNA is also used to test the effects
of certain human diseases on animals.
The stages involved in the production of
recombinant DNA are described below:
The required section of DNA is cut from a donor
cell’s DNA with restriction enzymes. At the same
time a bacterial plasmid is split with restriction
enzymes.
↓
Care must be taken that the ‘promoter sequence’
code is still attached to the donated DNA segment,
or the introduced gene will not be able to be
switched on or off.
↓
The ‘sticky’ ends of the two pieces of DNA join
together, with the help of DNA ligase, to form a
recombinant bacterial plasmid (see Fig. 6-5)
↓
The recombinant bacterial plasmid is then
introduced into host cells using microinjection
techniques or particle guns.
↓
The DNA may also be inserted into bacterial cells
via plasmids or bacteriophages (viruses that can
multiply inside bacteria), and the bacterial cells
can be cultured to produce useful products such as
insulin.
enzymes
cut desired
section of
DNA
Donor
cell
gene is
inserted
into
bacterial
DNA
Bacterial DNA
is introduced
into host cells
or is cultured
itself to make
useful products.
Fig. 6-5 The production of recombinant DNA
Restriction enzymes are obtained
from bacterial cells. They are special ‘cutting’
enzymes, and each one cuts the nucleotide
sequence of the DNA molecule at a specific point.
The enzyme EcoRI, for instance, makes a cut
between adjacent guanine and adenine nucleotides
when it recognises a GAATTC sequence. The cut
results in an overhanging section of single stranded
DNA, referred to as a ‘sticky’ end, which can be
attached to a similar sticky end from another piece
of DNA.Other restriction enzymes include HindIII
and BamHI.
Using recombinant DNA technology
to locate genes
The location of genes along a chromosome is
determined using restriction maps and DNA
sequencing techniques.
a) Restriction mapping – This technique allows
scientists to locate the positions of DNA fragments
along a chromosome. The steps involved are
described below:
Sections of DNA are cloned, using DNA
polymerase, and then cut using restriction
enzymes.
↓
The sections are cut into smaller fragments using
two different restriction enzymes; some are cut
using one restriction enzyme only and some are cut
using both of them. Each restriction enzyme
cleaves the DNA molecule at a specific site.
↓
Three types of fragments are produced, and they
are separated by gel electrophoresis according to
their length (i.e. the number of base pairs present).
↓
One end of each section is labelled with a
radioactive phosphorous marker.
↓
The phosphorous marker and the known length of
each fragment are used to determine the position of
each fragment on the original piece of DNA
↓
The resulting map will reveal the positions on the
chromosome of the fragments that have been cut
by particular restriction enzymes.
Genetic probes can then be used to locate
particular genes on a chromosome. This involves
again separating DNA into fragments using gel
electrophoresis, and treating the resulting bands
with chemicals or heat. This separates the DNA
strands in each fragment. Identified genes or pieces
of DNA with known sequences are then tagged
with radioactive or fluorescent markers and
hybridised with the unknown DNA fragments.
Using this information and knowledge of the
fragment’s position on a DNA restriction map, a
particular gene can be located on a chromosome.
b) DNA sequencing – A process known as the
Maxim-Gilbert process, can be used to determine
the actual sequence of nucleotides on a DNA
molecule. The steps involved are described below:
The DNA is cut with restriction enzymes as
before, and specific bases (e.g. Adenine) are
chemically removed from different parts of each
fragment. Each fragment is labelled at one end
with radioactive phosphorous.
↓
The fragments are again subjected to gel
electrophoresis and they migrate to different areas
along the gel representing different nucleotide
lengths.
↓
The length of the fragments which contain
radioactive phosphorous can be determined, and
the position of the missing bases will obviously be
one nucleotide length longer than these. If a
particular fragment, for instance, is found to be 6
nucleotides long, the missing base must have been
at position 7 along this strand (the phosphorous
marker is at the other end).
↓
This process is repeated for all the nitrogen bases
until their positions along the DNA strand are
deduced.
(Another similar sequencing method,
known as the Sanger Method, modifies each
fragment so that it terminates at one of the four
bases.)
Fig. 6-6 shows an autoradiograph of an
electrophoresis gel. The wide bands represent
DNA fragments from an original, larger fragment,
that have been labelled with radioactive
phosphorous (on the left hand end of each
fragment). Each of the fragments has been
chemically treated so that a guanine base has been
removed from the non-labelled end.
-------------========
---------------------------------------========
--------------------------========
--------------
10
9
8
7
6
5
4
3
2
1
Fig. 6-6 Electrophoresis bands produced from the
Maxim Gilbert method of DNA sequencing
Assuming that the fragments have been separated
according to length, with the numbers at the side
representing the number of nucleotides present, we
can deduce that the guanines were in 3rd, 6th and
10th positions along each fragment, respectively
(i.e. moving from the bottom to the top).
5) Gene therapy is possible once
the genes responsible for
harmful conditions are identified
Gene therapy
Gene therapy involves altering the genetic makeup of an individual in an attempt to cure diseases
caused by faulty genes. In most cases, ‘normal’
genes are inserted into the liver or lung cells to
replace the function of faulty genes. These normal
genes then produce their particular proteins, which
are released to the rest of the body. In some cases,
the abnormal gene may actually be replaced by the
normal gene, or it may be repaired through
selective reverse mutation. Most genes are
introduced into the body cells via recombinant
retroviruses, which are capable of introducing their
genes to human cells. Other types of recombinant
viruses used for this purpose include adenoviruses
and Herpes simplex viruses. In some cases,
recombinant genes may also be introduced via
plasmids or yeast cells.
Gene therapy is still only in the experimental
stage, with the FDA (Food and Drug
Administration) not yet approving any gene
therapy products for sale. Some experimental
successes have, however, included the treatment of
children with X-SCID (severe combined
immunodeficiency).Children with this disease are
unable to produce the enzyme adenosine
deaminase (ADA). Retroviruses containing the
normal gene for ADA production are inserted into
a culture of the child’s stem cells, and these cells
are then injected back into the body. A temporary
halt was called on these trials in 2003, however,
when a child treated for X-SCID developed
leukaemia. Sickle cell anaemia is another disease
currently undergoing experimental gene therapy;
in this treatment, a drug is introduced which
switches a faulty gene back on so that the patient
can manufacture normal haemoglobin. Other
research is being carried out to investigate methods
of inserting genes into cancer cells which can be
identified by the immune system and attacked.
Gene therapy involving the repair of messenger
RNA from defective genes is also being tested on
diseases such as thalassaemia, cystic fibrosis and
some cancers. Trials in which interferon genes are
administered to asthma patients are another
example of gene therapy that shows future
promise. Various vaccines, in which the gene
coding for the surface of a particular antigen is
incorporated into the recipient’s cells, are also
proving successful in trials. In this situation, the
body cells actually begin to manufacture the
antigen, therefore triggering an immune response.
A recombinant influenza vaccine is among several
vaccines showing promise in this area.
Problems associated with gene therapy – These
include the following:
i) Unless the introduced gene is incorporated into
gametes before fertilisation, it is impossible to alter
the whole genome of a person to produce
permanent genetic changes.
ii) The introduced gene may be regarded as a
foreign body by the patient’s immune system, thus
stimulating the production of antibodies and
making it more difficult to carry out subsequent
gene therapy procedures.
iii) Viral vectors may become virulent again once
inside the patient’s cells, or they could stimulate
the patient’s immune system.
Besides viruses and plasmids, other
gene therapy vectors may soon include liposomes,
which are lipid spheres containing the required
DNA. These liposomes are capable of penetrating
cell membranes. In the USA some researchers
have even managed to transfer genes into the brain
(which is inaccessible to viral vectors) in this
manner, a technique that may be useful in treating
Parkinson’s disease.
6) Mechanisms of genetic change
Mutations of chromosomes
There are a number of ways chromosomes can
become changed, or ‘mutated’. These include:
i) Rearrangements – These mutations include
inversions, duplications, amplifications and
translocations. Inversions occur when a whole base
triplet within the DNA molecule is back to front.
Duplications involve an extra copy of a particular
nucleotide sequence occurring along the length of
a chromosome. Amplification mutations are
similar, but involve multiple copies of a particular
base sequence on the same chromosome.
Translocations are mutations in which a base
sequence from one chromosome joins onto another
one.
ii) Gene mutations – These involve changes in the
actual DNA sequence and include the substitution
of one base for another or the insertion or deletion
of a base. Insertions and deletions result in the
production of a completely different amino acid
sequence from the point of the mutation onwards,
a phenomenon known as a ‘frameshift mutation’.
iii) Trisomy – An example of this is Down’s
syndrome, in which a child inherits three copies of
chromosome 21 instead of two.
iv) Polyploidy – This occurs when chromosomes
fail to separate during mitosis, resulting in an
organism inheriting multiple sets of chromosomes.
Polyploid plants, often superior to diploid ones,
can in fact be deliberately produced by the use of
the chemical colchicine, which prevents the
separation of chromosomes during mitosis.
examples of commercially produced polyploid
plants include wheat, cotton, tobacco, bananas,
potatoes, pansies and daylilies.
THINK!!! Can you name the type of
gene mutation occurring in a) and b) below?
Which one of these will result in a frameshift
mutation?
a) original DNA C A G T A G G T C
mutated DNA C A G A AG G T C
b) original DNA
mutated DNA

CAGTCGGTA
CGTCGGTA
As a requirement of this topic, you need
to describe the effects of one named and
described genetic mutation on human
health.
Some diseases caused by genetic mutations are
described below:
i) Down’s syndrome – This is an example of
trisomy, a mutation in which three chromosomes
are inherited instead of two. In the case of Down’s
syndrome, chromosome pair 21 fails to separate
during meiosis, a situation that is referred to as
non-disjunction. Down’s syndrome can also result
from translocation mutations, with the result that a
large portion of an extra chromosome 21 becomes
attached to chromosome 15. Symptoms of this
disease include mental deficiency, stocky body
type and sensitive skin.
ii) Muscular dystrophy – this is caused by a
mutation of the dystrophin gene, resulting in the
body’s inability to produce the proteins required
for muscle strength.
iii) Wilm’s tumour – This cancer occurring in
young children is. Symptoms of the disease
include cancer of the kidney caused by the deletion
of a base on chromosome 11. Other cancers have
been found to be related to gene translocations.
iv) Sickle cell anaemia – This is caused by the
substitution of adenine for thymine in a particular
base triplet of the sickle cell gene. As a result, the
amino acid valine replaces glutamic acid when the
gene expresses itself during protein formation.
Symptoms include the formation of sickle shaped
red blood cells, which cannot carry oxygen.
v) Thalassemia – This is also caused by a
substitution mutation. In this case, uracil replaces
cytosine in the haemoglobin gene. This results in
symptoms of anaemia.
DNA repair genes
The DNA molecule is continually being damaged
as it undergoes replication during the ‘S’
(synthesis) stage of the cell cycle, and without the
proteins coded for by the DNA repair genes,
serious diseases can result. An example of this can
be seen in the disease xeroderma pigmentosa, in
which a set of DNA repair genes are faulty.
Symptoms of this disease include vulnerability to
UV light and increased susceptibility to skin
cancers. Another gene, known as ‘p53’, is capable
of producing proteins that can stop the cell cycle
during the first (G1) growth stage to allow for the
repair of damaged DNA by other proteins. If the
p53 gene itself becomes mutated (mutagens
include some viruses, nicotine and certain fungal
toxins), uncontrolled division of damaged DNA
can occur, thus causing tumours.
The DNA repair genes operate in two main ways:
a) Chemical reversal of the damaged gene - This
type of repair occurs most often when a cytosine
base(C) has been chemically altered to thymine (T)
by the addition of a methyl group (-CH3-).
Enzymes called glycosylases are produced by
DNA repair genes in response to this type of
mutation, and act to restore the original cytosine.
b) Excision repair
i) Base excision repair- Here, a range of DNA
repair genes produce DNA glycosylases, which
remove damaged bases and their corresponding
phosphate group. Other repair genes then produce
DNA polymerases, which insert the correct
nucleotide back into the DNA. DNA ligases
complete the repair by joining the break in the
DNA strand. These enzymes are also used to repair
general breaks in DNA strands.
Somatic and germ line mutations
and the effect of mutations on species
ii) Nucleotide excision repair- Although similar
to the above process, this repair mechanism
involves the removal of a whole group of
nucleotides around the damaged area. The
enzymes involved include transcription factor
enzymes, DNA polymerases and DNA ligases.
Approximately 15-18 enzymes are required for
cutting out the nucleotide sequence, and more than
12 are needed to repair this section of the DNA.
Somatic cell mutations are mutations that occur in
the body cells, and are therefore not passed on to
offspring. They can, however, cause diseases such
as cancer. Germ line mutations, on the other hand,
are mutations that occur in the gametes, and can
therefore be inherited. A favourable germ line
mutation can result in a species becoming better
adapted to its environment, so affecting its
evolution. An unfavourable germ line mutation,
however, can cause diseases such as sickle cell
anaemia and haemophilia which will adversely
affect the survival of a species.
iii) Mismatch repair- Mistakes in normal base
pairing can be corrected by some of the excision
repair enzymes mentioned above, and also by
cutting enzymes (encoded by the MLH1 gene) and
DNA polymerases. The initial mismatch is
recognised by proteins produced by the MSH2
gene. Damage to either the MLH1 gene or the
MSH2 gene can result in colon cancer.
7) Selective breeding is different
to gene cloning but both
processes may change the
genetic nature of species
Transposable genetic elements
Selective breeding
Some segments of DNA are transposable, which
means they can move from one part of the genome
to another. Some of the effects of this include the
attachment of new genes onto unrelated
chromosomes and mutations where the broken
section of DNA has been re-sealed. Transposable
elements can also alter parts of the DNA molecule
in their vicinity by causing deletion or addition
mutations. The impact of these elements on the
genome includes their ability to alter gene
expression, thus affecting the evolution of a
particular species. This has also led some scientists
to believe that they are involved in the ability of
bacteria to rapidly evolve resistance to antibiotics.
Haemophilia and leukaemia are among some
genetic diseases thought to be due to the action of
transposable genetic elements.
Transposable genetic elements, or
‘jumping genes’, were discovered by Barbara
Mc Clintock in the 1940s. She noticed that
mutations affecting pigmentation in maize kernels
were caused by genes that moved from one
position to another. This phenomenon usually
involved two genes, with one gene controlling the
action of the other one. Her work was not
recognised until similar elements were discovered
in bacteria more than twenty years later.
Selective breeding is practised by plant breeders
because they can select for more desirable genes
such as disease resistance, higher crop yields,
longer flowering periods and drought tolerance.
Animal breeders can select for features such as
increased protein content and flavour in meat and
improved shell hardness in eggs. Examples of the
use of selective breeding include the development
of hybrid corn species which have increased kernel
size and nutritional content. Fig. 6-7 shows the
method used to produce such hybrid plants. Inbred
strains of corn are crossed with other, genetically
different inbred strains. The hybrid plants
produced from two such crosses are then crossed
with each other to produce seed.
Inbred A x inbred B
AB
seeds
Inbred C x inbred D
CD
Fig. 6-7 The production of hybrid corn
The worldwide need for food has also been
addressed recently by the selective breeding of
wheat and rice to produce higher yields. Wheat has
also been hybridised to produce increased rust
resistance and drought tolerance. Cattle and pigs
have been bred selectively for 8000 years to
produce high yields of food for human
consumption. Table 6-6 describes some selective
breeding techniques currently in use.
Hybridisation
Artificial pollination
Artificial insemination
Cloning
A process in which a
breeder combines
desirable traits from
two or more varieties
to produce a plant that
is often superior to its
parents.
A technique in which
the pollen from
selected plants is
dusted over fertile
stigmas. A method of
controlling the genetic
composition of
offspring plants.
Results in greater
variation and better
seed production.
Involves inserting
semen from selected
male livestock into a
female animal.
A method which
produces genetically
identical offspring to
the parent; in plant
breeding this is
achieved through tissue
culture techniques.
more widely used as an edible oil after the 1940s.
The first commercial canola crop was grown in
Australia in 1969. These initial crops, however,
had relatively high levels of erucic acid and
glucosinolates. Moreover, the two major varieties
of Canola were susceptible to ‘blackleg’ disease,
and by the 1970s, many losses were occurring as a
result of this. Victoria established a breeding
program in 1970 to address these problems, with
Western Australia and NSW following in 1973.
The initial varieties produced by these breeding
programs included ‘Wesreo’ and ‘Wesway’ in the
late 1970s. These plants had lower erucic acid
levels and greater blackleg resistance. In 1980,
another variety, known as ‘Marnoo’ was
developed in Victoria. This plant had lower
glucosinolate levels, and was much higher yielding
than the other varieties. Its limited resistance to
blackleg disease, however, resulted in NSW
growers preferring another variety with greater
disease tolerance called ‘Jumbuck’. The first
hybrid canola, Hyola 30, was developed in 1988.
By 1987, two Canola varieties (‘Maluka’ and
‘Shiralee’) were finally produced that combined
improved yields and disease resistance with low
erucic acid and glucosinolate levels. In addition,
new varieties resistant to the herbicide Triazine
were developed in 1993, which meant they could
now successfully compete with weeds. This
characteristic was later combined with the ability
to mature earlier or later in the year, resulting in
the more widespread cultivation of Canola across
the country. Although the Triazine tolerant
varieties produce slightly less oil and have a lower
yield than other varieties, they are still favoured by
many growers, especially in Western Australia.
Table 6-6 Current artificial selection techniques

As a requirement of this topic, you need
to trace the history of the selective
breeding of one species for agricultural
purposes and describe the changes that
have occurred in this species as a result
of this selective breeding.
Canola breeding in Australia
Canola, originally called rapeseed, is a plant that is
grown for its oil production. Although it was first
used in Canada as a lubricant in ships, it became
THINK!!! Explain why NSW Canola
growers preferred to use ‘Jumbuck’ Canola instead
of the ‘Marnoo’ variety in the early 1980s.
Gene cloning and its uses
Gene cloning refers to the various processes used
to produce multiple copies of a gene or a section of
DNA. Once a bacterial cell has had a recombinant
plasmid inserted into it, it can divide to produce
many new recombinant bacterial cells. In this way,
genes for hormones such as human growth
hormone can be reproduced rapidly; up to one
billion copies can be made in 24 hours.
Polymerase chain reactions can also be used to
copy large numbers of DNA sequences or genes in
a short space of time. In this procedure, the
enzyme DNA polymerase is mixed with the DNA
to be copied and nucleotide sequences called
‘primers’, and heated to separate the DNA strands.
After cooling, the primers bind to the DNA strands
and the DNA polymerase begins to make
complementary strands. Special replicator
machines can now use this process to produce 100
billion copies of a DNA section in the space of a
few hours. Uses for gene cloning include the large
scale production of chemicals such as synthetic
human interferon and human hormones such as
insulin and growth hormone. Gene cloning is also
used to manufacture genes for research purposes
and for the production of genes for use in
transgenic species.
The difference between gene cloning
and whole organism cloning is that all cells of a
cloned organism are genetically identical to the
parent, whereas in gene cloning only selected
genes or DNA segments of an organism are
cloned. DNA hybridisation and DNA
fingerprinting techniques can be used by scientists
to verify that one organism is the clone of another.
Cloning animals and plants
Cloning is a process in which offspring are
produced which are genetically identical to the
parent. In plants, this can be achieved through
traditional methods, such as taking cuttings, bulb
division and grafting. A more recent and effective
technique involves the use of tissue culture, a
process in which the cells of a parent plant are
cultured on a nutrient medium and used to produce
identical offspring.
Animal cloning is still in its developmental stages,
but success has been achieved using the nuclear
transfer method to produce frogs (shown in Fig 68), mice and more recently, a cloned sheep (see
chapter 2) In this procedure, adult body cells are
taken from the parent and inserted into an egg cell
that has had its nucleus removed. In mammals, the
egg is then implanted in a female and develops
into an embryo which is a clone of the original
parent. The nuclear transfer method has so far
proved less successful using the nuclei from
embryos. Another method currently being
developed, however, involves removing an actual
cell from an embryo and culturing this cell to
produce numerous embryos.
a)
b)
c)
d)
Fig. 6-8 The stages involved in the nuclear transfer
method of cloning;
a) nucleus is taken from the intestines of a tadpole;
b) UV radiation destroys nuclear material in egg
cell; c) nucleus is inserted into egg cell using a
micropipette; d) adult frog
Cloning provides a fast and efficient method of
producing plants or animals with desired
characteristics. Tissue culture has proved
extremely useful in the production of crop and
nursery plants, and is also used to provide multiple
copies of transgenic plants such as insect-resistant
Bt cotton and tomatoes that have been genetically
modified to increase their shelf life. In the future,
animal cloning could prove to be equally effective,
producing animals on a large scale with favourable
qualities. Examples include high milk production
in cows and the ability to secrete antitrypsin (used
in the treatment of lung conditions) in the milk of
transgenic sheep.
8) The timing of gene expression
is important in the
developmental process
The role of genes in embryonic
development
Not all genes are expressed during each stage of an
organism’s development. In the case of vertebrate
embryos, ‘structural’ genes responsible for the
development of the organism are controlled by
regulatory genes that control their expression. In
the case of limb development, special regulatory
genes called HOX genes can ‘switch’ the structural
genes on or off during the transcription stage of
polypeptide synthesis. As with other regulatory
genes (mentioned earlier in this chapter), they do
this by binding to areas on the DNA of the
structural genes called ‘control elements’
.Complicated gene cascades can result, in which
the HOX genes decide which genes are expressed,
and the proteins produced by these genes in turn
trigger the production of proteins by other genes.
As a result of this process, the order in which
limbs develop is as follows:

and the nucleotide sequence of MSX genes, which
code for a specific protein, is identical in most
animals. These examples provide further evidence
for evolution from common ancestors.
A study of mutations in gene homologues is also
providing scientists with useful information about
the mechanisms of evolution. Mutations in some
homologous genes, for instance, have resulted in
characteristics that are more primitive than those
occurring when the genes are expressed normally.
In addition, the fact that these mutations can occur
rapidly, creating obvious structural differences,
adds further support to the punctuated equilibrium
theory of natural selection.
limb buds appear on the upper body
↓
limb buds appear on the lower body
↓
fingers form
↓
toes form
Useful websites to refer to in this
topic:
As a requirement of this topic, you need
to assess the evidence that analysis of
genes provides for evolutionary
relationships
ii) Cystic fibrosis gene therapy:
http://www.personalmd.com/news/a1999031902.s
html
Genes and evolution
The analysis of genes can help to determine
evolutionary relationships between organisms. One
example of this is the occurrence of gene
homologues -identical gene clusters- in groups of
seemingly unrelated organisms. The HOX genes,
for instance, which code for skeletal and
neurological development in vertebrates, are also
found in lower animal classes such as molluscs and
worms. It has been shown, using transgenic
techniques, that a HOX gene inserted into a
vertebrate from a lower animal can perform the
same regulatory functions as the vertebrate’s own
HOX genes. This, and the fact that HOX genes are
located in clusters in similar positions along the
chromosomes in many species, suggests that the
many animal types in existence evolved from a
common ancestor. The genes themselves are
thought to have evolved through cluster
duplication. Other gene homologues include the
Pax-6 gene, which is present both in vertebrates
and lower animal groups, and performs a similar
role (eye development) in each. Many DNA repair
genes are also similar in a large number of species,
i) Restriction enzymes:
http://users.rcn.com/jkimball.ma.ultranet/BiologyP
ages/R/RestrictionEnzymes.html
iii) Asthma gene therapy:
http://www.usforacle.com/vnews/display.v/ART/2
003/11/07/3faba0c3cbff8
iv) Selective breeding:
http://www.nd.edu/~chem191/al.html
v) Canola breeding:
www.biotechnology.gov.au/.../food/herbicide_tolera
nt_canola/
better_canola_farm/solution_to_weeds.htm
vi) Transposable genetic elements:
http://biocrs.biomed.brown.edu/Books/Essays/Jum
pingGenes.html
REVIEW QUESTIONS
MULTIPLE CHOICE
1. The process in which polypeptides are
assembled on the ribosome using the
information in the mRNA molecule is known
as :
a) transcription
b) gene linkage
c) translation
d) transposition
2. In RNA,as opposed to DNA, the base uracil is
present instead of:
a) thymine
b) adenine
c) cytosine
d) guanine
3. The function of regulatory genes is to:
a) produce DNA ligases
b) control the action of restriction enzymes
c) control the expression of other genes by
switching them on or off
d) repair mutations in genes
4. The sequence of bases transcribed onto a
mRNA molecule from a DNA strand with the
sequence GAA CGT ATG would be:
a) CTT GCA TAC
b) AGG TAC GCA
c) TCC ATG CGT
d) CUU GCA UAC
5. Which of the following is not an example of
the action of DNA repair genes?
a) Glycosylases are produced by repair genes
to convert thymine to cytosine.
b) The repair genes bind to the control
element sections of the damaged DNA
c) Damaged bases are removed and are
replaced with the help of DNA
polymerases produced by the repair genes.
d) Excision repair enzymes, cutting enzymes
and DNA polymerases are produced by the
repair genes to replace mis-matched bases
with the correct ones.
6. Which of the following lists human
chromosomes that have been completely
sequenced by the Human Genome Project?
a) Chromosomes 23, 14, 21 and 8
b) Chromosomes 22, 21, 20 and 14
c) Chromosomes 5, 21, 20 and 7
d) Chromosomes 22, 20, 14 and 11
7. In fruit flies, the gene for grey body colour (G)
is dominant to black colour (g), and the gene
for long wings (L) is dominant to short wings
(l). Assuming genes are sorted independently,
the phenotypic ratios of the offspring when a
heterozygous grey, long winged fly (GgLl) is
crossed with a heterozygous grey, short
winged fly (Ggll) would be:
a) 8 grey, long winged: 8 grey, short-winged
b) 9 grey, long winged: 3 grey, short winged;
3 black, long winged: 1 black, short
winged
c) 6 grey, short winged: 6 grey, long winged:
2 black, long winged: 2 black, short
winged
d) 8 black, long winged: 8 grey, short winged
8. In a certain animal, the gene for hair colour is
located on the same chromosome as the gene
for hair length. Black hair, B, is dominant to
white hair, b, and long hair, L, is dominant to
short hair, l. When two heterozygous black,
short haired animals (both BL/bl) are crossed,
the phenotypic ratio of the offspring would be:
a) 3 black, short haired: 1 white, long haired
b) 9 black, short haired: 3 black, long haired:
3 white, short haired: 1 white, long haired
c) 3 black, long haired; 1 white, short haired
d) 9 black, long haired: 3 white, long haired:
3 black, long haired: 1 black, short haired
9. The gene linkage map, below, shows the
relative positions of the genes B, F and J
(‘m.u’ = map units).
B 4m.u. F 6 m.u. G
J
10 m.u
The gene pair that would be separated the least
during crossing over would be:
a) B and J
b) B and F
c) F and J
d) B and G
J
10. Which of the following can all be used in the
production of recombinant DNA?
a) biotransformation, micro-injection,
polymerase chain reactions
b) fermentation, electrophoresis, DNA ligases
c) bacteriophages, plasmids, DNA ligases
d) electrophoresis, fermentation,
biotransformation
b) SHORT ANSWER AND
LONGER RESPONSE QUESTIONS
11. a) Explain what is meant by ‘multiple alleles’.
b) The possible genotypes of an Rh+ person
are RR and Rr, while the genotype of an
Rh- person is rr. What are the chances of a
heterozygous Rh+ woman producing an
Rh- baby if her husband is Rh-? Show
working.
c) Explain, showing working, why it is
possible for parents with blood types A
and B to produce a child with blood group
O.
12. a) Briefly explain what occurs in each of the
following chromosomal mutations:
i) trisomy; ii) polyploidy; iii) inversions;
iv) duplications; v) translocations; vi)
amplifications.
b) Name the type of gene mutation occurring
in the diagram below:
CAGCAGGTC
CAGCTAGGTC
c) List two uses for gene cloning.
16. a) Briefly describe the main purposes of the
human genome project.
b) The human genome project has been
largely realised through the use of
recombinant DNA technology. Briefly
describe what is involved in each of the
following techniques:
a) Electrophoresis; b) The use of genetic
probes; c) Polymerase chain reactions; d)
restriction mapping; e) DNA sequencing.
17. a) Briefly outline the role of regulatory genes
such as HOX genes in embryonic
development.
b) Describe how the occurrence of gene
homologues in different species has
indicated evolutionary relationships.
18. a) Explain what is meant by ‘polygenic
inheritance’.
b) Human height is controlled by polygenic
inheritance. The table below shows the
height distribution of male spectators at a
football final. Plot them to form a line
graph, and use the resulting curve to
comment on whether the expression of this
trait follows predictable Mendelian ratios.
Height (cm)
150
160
170
180
190
Frequency (%)
1
4
25
10
3
13. Name two human diseases caused by mutations
and briefly describe the type of mutation
involved.
14. a) Name three organisms that have been
subjected to selective breeding programs ,
and describe one favourable characteristic
each has been selected for.
b) Briefly describe the processes involved in
the following selective breeding
techniques:
i) hybridisation; ii) cloning; iii) artificial
pollination; iv) artificial insemination.
15. a) Explain what is meant by ‘gene cloning’.
How does it differ from whole organism
cloning?
b) Describe one method used to clone genes.
19. Outline the main stages involved in the
production of recombinant DNA. Include the
words listed in the box below in your answer.
plasmids, promoter sequence, DNA ligase,
restriction enzymes, bacteria, bacteriophages,
microinjection
20. a) Most animal cloning experiments involve
the nuclear transfer method, shown below.
Explain what is happening at parts a) to e)
in the diagram.
Gl
Gl
gl
gl
GGLl
GGLl
GgLl
GgLl
GGll
GGll
Ggll
Ggll
GgLl
GgLl
ggLl
ggLl
Ggll
Ggll
ggll
ggll
Genotypes = 6 grey, long; 6 grey, short; 2 black,
long; 2 black, short
8. c);
Gametes
bl
bl
d)
BL
BbLl
BbLl
bl
bbll
bbll
Phenotypic ratio of offspring; 1 black,long: 1
white, short
c)
a)
9. b); These genes are the closest together and so
would be separated the least during
crossing over.
e)
b)
10. c)
b) Describe one modern and one traditional
method used to clone plants.
c) Outline one reason why cloning animals and
plants can be of benefit to humans.
11. a) Multiple alleles occur when a characteristic
is determined by more than one pair of
alleles.
b) Rr x rr
↓
Rr, Rr, rr, rr
so the chance of
producing Rh- offspring is 50%
ANSWERS
a) MULTIPLE CHOICE
1.
c)
2. a)
3. c); Examples of regulatory genes are HOX
genes, which bind to the control element
sections of the DNA of other genes,
switching them on or off.
4. d); uracil replaces thymine in RNA
5. b); This is how regulatory genes operate, not
DNA repair genes.
6. b)
7.
c);
Gametes
GL
Gl
b) SHORT ANSWER AND
LONGER RESPONSE
QUESTIONS
gL
gl
c) The parents would have to have the
genotypes IAi and IBi. The cross involved
is
IAi x IBi
↓
IAIB, IAi, IBi, ii
and ii is the genotype of group O
12. a) i) Trisomy occurs when offspring inherit
three copies of a particular chromosome
instead of two; an example is Down
Syndrome.
ii) Polyploidy is a condition where an
organism has multiple sets of
chromosomes, usually resulting from the
failure of chromosomes to separate
during mitosis.
iii)
Inversions are mutations in which a
particular nucleotide sequence is
reversed.
iv) Duplications are mutations in which
extra copies of a particular nucleotide
sequence are made.
v) Translocations are mutations in which a
base sequence from one chromosome
joins onto another one.
vi) Amplifications are mutations in which
many additional copies of a particular
base sequence occur on the same
chromosome.
b) This is an insertion mutation (T has been
inserted).
13. i)
ii)
14. a)
b)
Down syndrome( trisomy 21) is caused
by the failure of the chromosome 21 pair
to separate during meiosis. This situation
is known as non-disjunction. Symptoms
include mental deficiency, stocky body
type and other disorders.
Sickle cell anaemia is caused by a
substitution mutation; symptoms include
poor oxygen carrying capacity in the
blood, due to the formation of sickleshaped red blood cells.
Corn- selected for increased kernel; size
and nutritional content; wheat- selected
for increased yield and rust resistance;
cattle- selected for high food yield.
i) A process in which desirable traits
from two or more varieties are combined
to produce a superior plant; ii) Produces
genetically identical offspring to the
parent. Involves tissue culture
techniques, grafting, cuttings etc. in
plants; iii) The pollen from selected
plants is dusted over fertile stigmas; iv)
Involves inserting semen from selected
male livestock into a female animal.
cloned using the polymerase chain
reaction in replicator machines.
c) i) The large scale production of
chemicals such as interferon and
hormones such as insulin; ii) The
production of genes for use in transgenic
species.
16. a) i) To map all the genes in the human
genome; ii) to determine the DNA
sequence of each gene and iii) to identify
disease-causing genes.
b) Electrophoresis: A procedure which
separates DNA fragments by passing
them through and electrically charged
gel; Genetic probes: These are
radioactive or fluorescent
complementary segments of DNA which
locate the DNA fragments that have
been separated by electrophoresis;
Polymerase chain reactions: These use
DNA polymerase to make copies of
genes; Restriction mapping: A process
that uses restriction enzymes to cut DNA
at particular locations in order to
determine the position of particular base
sequences along the DNA strand; DNA
sequencing: A procedure used to
determine the sequence of nucleotides
on a DNA molecule- involves cutting
with restriction enzymes, removing a
particular base and separating the
fragments using electrophoresis.
17. a)
b)
15. a)
b)
Gene cloning involves the various
processes used to produce multiple
copies of a gene or a section of DNA. It
differs from whole organism cloning
because this latter technique involves
cloning all the cells of an organism.
One method used to clone genes involves
the formation of recombinant bacterial
plasmids which are inserted into
bacterial cells. The bacteria then
multiply, forming multiple copies of the
introduced gene. Genes can also be
HOX genes are responsible for
regulating the expression of the genes
involved in embryonic limb
development by switching them on or
off. These genes produce regulatory
proteins that bind on to the control
element sections of other genes, thus
switching them on or off.
The fact that these gene homologues have
been found to occur in many species,
and seem to perform similar roles in
each, suggests that the many animal
types in existence evolved from a
common ancestor. Examples include
HOX genes, the Pax-6 gene and MSX
genes.
18. a)
Polygenic inheritance occurs when
multiple alleles, often located on
different chromosomes, control the
inheritance of a particular characteristic.
b)
c)
The resulting curve is bell-shaped,
indicating the inheritance of height shows
continuous variation, unlike Mendelian
ratios, which would show discontinuous
variation (i.e. not a bell shape).
19.
The required section of DNA is cut from a
donor cell’s DNA with restriction
enzymes. At the same time a bacterial
plasmid is split with restriction enzymes.
↓
Care must be taken that the ‘promoter
sequence’ code is still attached to the
donated DNA segment, or the introduced
gene will not be able to be switched on or
off.
↓
The ‘sticky’ ends of the two pieces of
DNA join together, with the help of DNA
ligase, to form a recombinant bacterial
plasmid.
↓
The recombinant bacterial plasmid is then
introduced into host cells using
microinjection techniques or particle guns.
↓
The DNA may also be inserted into
bacterial cells via plasmids or
bacteriophages (viruses that can multiply
inside bacteria), and the bacterial cells can
be cultured to produce useful products
such as insulin.
20. a) a) nucleus is taken from an adult ewe udder
cell; b) UV radiation destroys nuclear
material in egg; c) nucleus is inserted into
egg cell using a micropipette; d) egg is
implanted into a surrogate mother; e) adult
sheep
b) Modern: tissue culture techniques, in
which the cells of a parent plant are
cultured on a nutrient medium;
Traditional: taking cuttings, bulb division
and grafting.
Cloning provides a fast and efficient
method of producing plants and animals
with desired characteristics; e.g. the
production of nursery plants. Animal
cloning may in the future produce
animals on a large scale with favourable
qualities, such as high milk production
in cows.