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BIO4: Genetics
1.
The structure of a gene provides the code for a polypeptide
DESCRIBE THE PROCESSES INVOLVED IN THE TRANSFER OF INFORMATION FROM DNA THROUGH RNA TO THE
PRODUCTION OF A SEQUENCE OF AMINO ACIDS IN A POLYPEPTIDE
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Each gene is a certain sequence of bases along a DNA strand, and is the information required to make a
particular polypeptide.
o Each triplet of bases, called a codon, codes for one
amino acid
o 61 different codons specify 20 different amino acids
Messenger RNA (mRNA): a ribonucleic acid which carries
information from DNA in nucleus to ribosomes in cytoplasm
Transfer RNA (tRNA): a ribonucleic acid consisting of a single
strand, which brings amino acids to the ribosome to be linked
together
o A different type of tRNA for each amino acid
 Each tRNA contains an anticodon which is
complementary to a codon on mRNA
Ribosomes: Made up of 2 subunits, and act as site for protein
synthesis in cytoplasm
o 3 binding sites to hold mRNA strand and 2 tRNA
molecules together temporarily
Transcription: process by which information on DNA is copied onto an
RNA molecule
1) DNA strand in nucleus unwinds in area of required gene
2) RNA polymerase moves along the strand linking complementary RNA nucleotides
to form the pre-mRNA strand
o Start (AUG) and stop (UAA, UAG, UGA) codons (on RNA) control length of
mRNA strand
o Uracil (U) replaces thymine (T) as the complementary RNA nucleotide of
Adenine (A)
3) After the whole gene is copied, introns (non-coding regions) are spliced from the
pre-mRNA and exons (coding regions) are joined together to form mRNA
4) A poly-A tail, and a methylguanosine chemical cap, are added. The chemical cap
assists in binding the mRNA to ribosomes.
5) Modified mRNA moves from nucleus into cytoplasm
Activation of Amino Acids:
 In cytoplasm, the enzyme amino acyl tRNA synthetase attaches amino acids to tRNA molecules with the correct
anticodon
Translation: process by which information on RNA molecule is used to make a new protein
1) mRNA and 2 tRNA molecules attach to ribosome
o The end of mRNA with start codon AUG binds on to ribosome
o tRNA carrying amino acid methionine (AUG) and anticodon UAC binds to start codon on mRNA within
ribosome
o Second tRNA binds to the next mRNA codon
2) The 2 amino acids link by a peptide bond
3) Previous tRNA is released from ribosome, ribosome moves along mRNA strand one codon at a time, another
tRNA binds within the ribosome and its amino acid links up to the previous. This step is repeated…
4) When a ‘stop’ codon is reached the polypeptide chain is released into the cytoplasm
5) Further processing is necessary before final protein is formed, as polypeptide chain is only primary structure
PROCESS INFORMATION FROM SECONDARY DATA TO OUTLINE THE CURRENT UNDERSTANDING OF GENE
EXPRESSION
Gene expression is the process by which genes are used to synthesise gene products, hence controlling phenotype.
 Information from a gene is used to synthesise a polypeptide. The sequence of bases on the gene determines the
sequence of amino acids in the polypeptide, through transcription and translation
 Polypeptides join to become proteins.
 Proteins control all chemical reactions in the body, and hence phenotype
 Therefore, genes determine the characteristics of an organism.
The Need to Regulate Gene Expression:
 Some genes are always active in every cell, making mRNA and proteins
o e.g. cellular respiration genes
 In multicellular organisms, each differentiated cells only has some of its genes activated, corresponding to its
function. Therefore, some tissue-specific genes may be expressed only in certain cells
o E.g. muscle cells have genes that control muscle factors turned on
o E.g. genes that control pigment production are expressed in melanocyte cells which are present around
growing hairs, in skin, and in the iris
Regulation of Gene Expression: Gene expression is controlled by various regulatory DNA sequences. Control points
include DNA unpacking, transcription, splicing, moving mRNA into the cytoplasm, translation, and activation/inactivation
of proteins.
 DNA is unpacked by removing methyl groups from DNA, and by histone acetylation
 Most gene regulation occurs in the transcription step. ‘Transcription factor’ genes produce proteins that bind to
‘control element’ segments of the targeted gene to activate/inactivate its expression. They can regulate which
sections of DNA are copied, the number of mRNA transcripts produced, and the rate of transcription
 Introns are cut out of the mRNA strand. Introns may have a role in regulating the genome, or in controlling
developmental processes
 Not all mRNA molecules are selected to move out of the nucleus into the chromosome. Transport out of the
nucleus through nuclear pores is controlled by exportin proteins
 In travelling from the nucleus to translation site, mRNA will be degraded unless it is specifically protected from
degradation by modifications such as the methylguanosine cap and poly-A tail
 Gene regulation can also be effected by controlling which mRNA molecules undergo translation
 After they are made, some proteins may be labelled for proteasomal degradation by the addition of ubiquitin
monomers
CHOOSE EQUIPMENT OR RESOURCES TO PERFORM A FIRST-HAND INVESTIGATION TO CONSTRUCT A MODEL OF DNA
2.
Multiple alleles and polygenic inheritance provide further variability within a
trait
GIVE EXAMPLES OF CHARACTERISTICS DETERMINED BY MULTIPLE ALLELES IN AN ORGANISM OTHER THAN HUMANS
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In the case of multiple alleles, there are more than 2 possible alleles for a gene in the population. Any 2 of these
can occur in one individual.
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𝑁𝑜. 𝑜𝑓 𝑔𝑒𝑛𝑜𝑡𝑦𝑝𝑒𝑠 =
o
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𝑛
(𝑛
2
+ 1) … where n = No. of alleles
The greater the number of alleles for a given gene, the greater the number of possible
genotypes/phenotypes that can exist in the population.
White spotting in dogs
o s = white spots absent
o si = Irish spotting
o sp = piebald spotting
o se = extreme spotting
Eye colour in Drosophila fruit fly [HIPA]
o w+ = red
o w = white
o wh = honey
o wi = ivory
o wp = pearl
o wa = apricot
COMPARE THE INHERITANCE OF THE ABO AND RHESUS BLOOD GROUPS
and
SOLVE PROBLEMS TO PREDICT THE INHERITANCE PATTERNS OF ABO BLOOD GROUPS AND THE RHESUS FACTOR
ABO Blood Groups:
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ABO blood groups are an example of multiple alleles
There are 3 alleles: IA, IB, i
o IA  produce antigen A
o IB  produce antigen B
o i  produce neither antigen
IA and IB are co-dominant, and both are dominant over i
There are 4 possible blood groups: A, B, AB, O
o A: presence of antigen A
o B: presence of antigen B
o AB: presence of both antigens A and B
o O: absence of both antigens
Genotype Antigens
Present
IA IA
A
A
I i
A
B
B
I I
B
IB i
B
A
B
I I
A and B
ii
neither
Blood
Group
A
A
B
B
AB
O
Frequency in
Population
38%
10%
3%
49%
Rhesus Factor:
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Rhesus factors are an example of Mendelian dominant inheritance
There are 2 alleles: D, d
D is dominant over d
There are 2 possible phenotypes: Rh+, Rho Rhesus positive (Rh+): antigen D is present
o Rhesus negative (Rh-): antigen D is absent
Comparison of Inheritance Patterns:
 ABO blood group has 3 alleles, 6 genotypes, 4 phenotypes
 Rhesus factor has 2 alleles, 3 genotypes, 2 phenotypes
Example Punnet Squares:
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IB i
IB
DD x
i dd
o 75% B+, 25% O+
Genotype
DD
Dd
dd
Phenotype
Rh+
Rh+
Rh-
IB D
iD
IB d
IB IB Dd
IB i Dd
id
IB i Dd
i i Dd
DEFINE WHAT IS MEANT BY POLYGENIC INHERITANCE AND DESCRIBE ONE EXAMPLE OF POLYGENIC INHERITANCE IN
HUMANS OR ANOTHER ORGANISM.
A polygenic trait is defined as being controlled by more than one pair of genes.
Simple Model of Polygenic Inheritance:
 Several loci are involved in expression of the trait
 Genes at each locus behave as if they follow codominance
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The loci act in an additive fashion, each adding or detracting a small amount from the phenotype
Environment interacts with genotype to produce final phenotype
Usually forms a graduated series of continued phenotypic variation rather than one or two discrete phenotypes
The individual genes of polygenic traits usually show conventional (Mendelian) patterns of inheritance, and the
apparent phenotypic ‘blending’ effect is a result of codominance of multiple genes for the same trait.
Regression towards the mean: offspring of parents who have ‘extreme’ phenotypes of a polygenic trait will tend to
have offspring closer to the population average.
PROCESS INFORMATION FROM SECONDARY SOURCES TO IDENTIFY AND DESCRIBE ONE EXAMPLE OF POLYGENIC
INHERITANCE
Skin Colour:
The inheritance of skin colour is best demonstrated with a
simplified model.
Assumptions:
3 genes control the trait. Each gene has two alleles. A capital
letter denotes the allele that adds dark pigment (melanin). A
lowercase letter denotes the allele that adds light pigment, or does not contribute to the trait. Alleles that contribute
dark pigment act in a cumulative way and are codominant. Therefore, the number of capital letters in the genotype
corresponds to darkness of skin colour, e.g. AaBbCc (3) and AABbcc (3) both have intermediate skin colour
Results of Crossing 2 Intermediate Genotypes (AaBbCc x AaBbCc):
Each parent produces 8 different types of gametes (2 x 2 x 2). These gametes combine in 64 different ways (8 x 8),
resulting in seven skin colours. The histogram curve with skin colour on the horizontal axis and frequency on vertical axis
approximates a bell-shaped curve called a normal distribution.
The Reality of Skin Colour:
There is further variation due to environmental factors such as
exposure to UV radiation. These variations smooth out the
histogram into a smooth curve. As a general pattern, people with
ancestors from tropical regions and higher altitudes (greater UV
exposure) will have darker skin than people with ancestors from
subtropical regions.
Skin colour is controlled by at least 4 polygenes, contributing in
complex, additive and non-additive combinations. One gene, SLC24A5, has just 2 variations. Nearly all Europeans have
the version with amino acid threonine; nearly everyone else has the other, alanine. Researchers found that people with
predominantly the threonine variant were lightest and people with predominantly the alanine variant were darkest, and
subjects with both versions of the gene had a variety of intermediate skin colours.
OUTLINE THE USE OF HIGHLY VARIABLE GENES FOR DNA FINGERPRINTING OF FORENSIC SAMPLES, FOR PATERNITY
TESTING AND FOR DETERMINING THE PEDIGREE OF ANIMALS
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Hypervariable genes
o DNA in some regions of the human chromosome consists of specific non-coding sequences that are
repeated in tandem. The number of repeats of a given sequence varies from person to person.
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e.g. a person may have 4 repeats (CATCATCATCAT) and 6 repeats (CATCATCATCATCATCAT) on
his homologous pair of number-7 chromosomes
o These variable regions are inherited as codominant multiple alleles. Monozygous
identical twins have the exact same DNA
o Include VNTRs and STRs
How DNA fingerprinting works
1) DNA is extracted from the sample by treating
with chemicals
2) In RFLP, the DNA is cut into fragments using
restriction enzymes. The sizes of these
fragments vary for different people.
 In more modern methods which can
use much smaller DNA samples,
primers specific to the flanking regions
of the sequence may be used to make
many copies of the relevant
hypervariable genes using PCR
3) Gel electrophoresis: an electric field is used to
separate the fragments by their size. Smaller
fragments travel further through the gel.
4) Southern blotting: The gel is placed in an alkaline solution. Nitrocellulose paper is placed on top of the
gel and weighed down with paper towels. The alkaline solution is pulled upward by capillary action,
transferring the DNA fragments from the gel to the paper while retaining their relative positions.
5) The nitrocellulose paper is placed in a tray, and bathed in a solution containing DNA probes. A DNA
probe is a short piece of DNA containing a radioactive isotope, which attaches to its complementary
sequence in the sample DNA. Excess probe is washed off.
6) Autoradiography: The nitrocellulose paper is placed in contact with photographic film. Radioactivity
from the probes forms dark bands on the film. This is a ‘DNA fingerprint’, unique to the length of each
DNA sequence being tested in each individual.
Variable number tandem repeats (VNTRs) or minisatelites:
o Alex Jeffreys identified repeats of one particular sequence of about 13 bases in many VNTRs. He used a
restriction enzyme specific to this base sequence to cut out these VNTRs, from various chromosomes in
a sample of a person’s DNA.
o He used a radioactively labelled ‘multi-locus probe’ which was also specific to this base sequence, to
separate the VNTR fragments by length and make them visible as a pattern of bands.
o Alternatively, a single-locus probe can be used, which binds to DNA from just one locus. A single-locus
probe is specific to the gene locus, and separates a person’s alleles for that gene by their length.
 A homozygous individual will have 1 band, and a heterozygous individual will have 2 bands.
 Using more single-locus probes in combination reduces the chance that 2 unrelated individuals
will have a matching DNA profile
Short tandem repeats (STRs) or microsatellites:
o This new technique uses 2 or 4-base repeats. Each locus has 5-20 different alleles, and there is a 5-20%
chance that 2 individuals have the same allele, so using multiple single-locus probes drastically
decreases the chances of accidental misidentification.
DNA Fingerprinting of Forensic Samples:
 Forensic samples are obtained, e.g. hairs from a crime scene, or vaginal swabs from a rape victim
 Samples are delivered to a forensic laboratory, with utmost care taken to avoid contamination
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DNA fingerprinting involves comparing the DNA profiles from different individuals, e.g. to determine if a hair
from the crime scene matches the DNA of a suspect.
Identical patterns of fragment sizes (i.e. matching bands when the fragments are separated by length) suggest
identical sources of DNA. As more identical DNA sequences or matching bands are found, the probability of
identification increases.
DNA fingerprinting can contribute to conviction (e.g. proving that a suspect’s blood is mixed with that of a
murder victim) or acquittal (e.g. proving that semen in a rape victim does not belong to the suspect)
Paternity Testing:
 Half of a child’s DNA comes from its mother and the other half from its father. When comparing profiles of
hypervariable regions of DNA, half of a child’s bands come from its mother and the other half from its father.
Each band in a child must come from one or the other of its biological parents.
 Therefore, if the child has a lot of fragments that are neither present in the mother nor the father, it is highly
likely that one of them is not the child’s biological parent.
Determining the Pedigree of Animals:
 In breeding pedigreed animals, knowledge of an animal’s sire (father) and dam (mother) must be certain.
 Mistakes in assigning parents can occur due to semen/embryo mix-ups during artificial insemination, mistakes
in record keeping, or when accidental matings occur.
 The use of DNA profiling to definitively identify an animal’s biological parents allows breeders to be certain that
their animals have the ancestry which they claim, and also prevents breeders from faking pedigrees.
DNA fingerprinting is more accurate than blood typing because there is a greater chance that two individuals will have
identical blood/tissue types. Also, only tiny amounts of tissue are needed to obtain DNA fingerprinting results, as
samples can be multiplied using PCR
3.
Studies of offspring reflect the inheritance of genes on different chromosomes
and genes on the same chromosomes
USE THE TERMS ‘DIPLOID’ AND ‘HAPLOID’ TO DESCRIBE SOMATIC AND GAMETIC CELLS
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Body/somatic cells in eukaryotic cells contain the diploid number (2n) of chromosomes.
o These cells may be called diploid cells
Gametes such as sperm and ova contain only one chromosome of each homologous pair due to meiosis. They
have half the diploid number of chromosomes, i.e. the haploid number (n).
o These cells may be called haploid cells
On fertilisation, two haploid cells (sperm and ovum) unite to form a zygote with the diploid number of
chromosomes
In humans, the diploid number is 46, and the haploid number is 23. Gametes have 23 chromosomes. Somatic
cells have 23 homologous pairs, or 46 chromosomes
DESCRIBE OUTCOMES OF DIHYBRID CROSSES INVOLVING SIMPLE DOMINANCE USING MENDEL’S EXPLANATIONS
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Tall stem (T) is dominant over short stem (t), and round seed (R) is dominant over wrinkled seed (r)
o Two plants are crossed, each heterozygous for both stem height and seed shape
o Parents: TtRr x TtRr
o Possible gametes: TR, Tr, tR, tr and TR, Tr, tR, tr
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To find the probability of any phenotype combination, we can treat this dihybrid cross as a combination of two
monohybrid crosses, then multiply the probabilities together. This is because chromosomes segregate
independently and genes are inherited independently, so the probabilities are independent of each other.
o Tt x Tt  1/4 TT : 1/2 Tt : 1/4 tt  3/4 tall : 1/4 short
o Rr x Rr  1/4 RR : 1/2 Rr : 1/4 rr  3/4 round : 1/4 wrinkled
gametes TR
Tr
tR
tr
o Multiplying probabilities gives: 9/16 tall-round : 3/16 tallTTRR TTRr TtRR TtRr
TR
wrinkled : 3/16 short-round : 1/16 short-wrinkled
Alternatively, we can use a Punnet square. Because chromosomes
TTRr TTrr TtRr Ttrr
Tr
segregate independently, there is an equal chance of the 4 difference
TtRR TtRr ttRR ttRr
tR
types of possible gametes (TR, Tr, tR, tr) being produced.
o Phenotypic ratios: 9 tall-round : 3 tall-wrinkled : 3 shortTtRr Ttrr ttRr ttrr
tr
round : 1 short-wrinkled
PREDICT THE DIFFERENCE IN INHERITANCE PATTERNS IF TWO GENES ARE LINKED
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Mendel’s model of inheritance assumed that genes showed independent
assortment. In actuality, it is the chromosomes that segregate
independently.
If the loci of 2 genes are on the same chromosome and are physically
close, they are said to be linked. We cannot consider the segregation of
linked alleles as independent events.
E.g. A/a and B/b are 2 linked genes. A person has AB on one
chromosome, and ab on its homologue. Mendelian inheritance would
predict that there is a 25% chance of gametes AB, Ab, aB and ab due to
independent assortment. However, if crossing over does not occur, the
only possible gametes will be AB and ab.
Crossing over occurs in meiosis when homologous chromosomes
exchange segments of DNA. This can separate linked alleles.
Recombinant gametes carry new combinations of alleles due to crossing
over. Parental gametes carry the original combination of alleles. Note
that every gamete will have some crossing over; we say gametes are
‘recombinant’ or ‘parental’ only with respect to the linked genes in question.
If 2 linked genes are further apart on a chromosome, the chance that they will be separated by crossing over is
greater. In general, the relative distance between two loci in map units corresponds to the probability of
recombinant gametes/offspring being produced.
o Continuing our prior example, if alleles A-B are 10 map units apart and a-b are 10 map units apart on
the homologous chromosome, there is a 10% chance of recombinant gametes. The gametes will be 45%
AB, 45% ab, 5% Ab, and 5% aB
o *Note: a ‘map unit’ is equivalent to a centrimorgan (cM), and corresponds to about 1 million bases. A
map unit/centrimorgan corresponds to the occurrence of 1% recombinant gametes from the relevant
test cross
PROCESS INFORMATION FROM SECONDARY SOURCES TO ANALYSE THE OUTCOME OF DIHYBRID CROSSES WHEN
BOTH TRAITS ARE INHERITED INDEPENDENTLY AND WHEN THEY ARE LINKED
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A double heterozygous male is crossed with a double homozygous
recessive female in a test cross: AaBb x aabb
If these traits were inherited independently, male gametes would be AB,
Ab, aB, ab. Female gametes would all be ab
o Phenotype: 25% AB, 25% Ab, 25% aB, 25% ab
AB
Ab
aB
ab
ab AaBb Aabb aaBb aabb
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If these traits were linked by 10 map units, male
45% AB
5% Ab
5% aB
45% ab
gametes would be 45% AB, 5% Ab, 5% aB, 45%
100% ab 45% AaBb 5% Aabb 5% aaBb 45% aabb
ab. Female gametes would be all ab.
o Phenotype: 45% AB, 5% Ab, 5% aB, 45% ab
In linked genes, offspring are more likely to have parental combinations of alleles. The total percentage of
recombinant offspring in the test cross corresponds to the separation of the 2 gene loci in map units.
EXPLAIN HOW CROSS-BREEDING EXPERIMENTS CAN IDENTIFY THE RELATIVE POSITION OF LINKED GENES
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The above example used a test cross between a double heterozygous individual and a double homozygous
recessive individual. From the percentage of offspring that are recombinant, we can identify the relative
position of linked genes.
In the above example, 10% of offspring/gametes were recombinant. Therefore, the linked genes are about 10
map units apart
Note*: As the distance between two loci increases, so does the probability of a double crossover which results
in linked genes being inherited together. The maximum percentage of ‘recombinant offspring’ cannot
theoretically exceed 50%. Our method of using cross-breeding experiments to identify the relative position of
linked genes is only reliable up to about 40 map units. If about 50% of offspring are recombinant, we can only
say that the genes are more than 50 map units apart. This is because if two genes are more than 50 map units
apart, the probabilities of double cross-overs are such that there is an equal chance the genes will be inherited
together or separately.
Genes Involved
A and B
A and C
B and C
% Recombinant Offspring
15
5
10
Separation of Loci (cM)
15
5
10
Example:
 In the table, the cross of A and B
is a test cross, i.e. AaBb x aabb
 If AB are linked and ab are
linked, then 15% recombinant offspring means that 42.5% are AB, 42.5% ab, 7.5% Ab, and 7.5% aB. This tells us
that the loci are separated by 15 cM.
 From these results, we can draw a simple chromosome map denoting the relative positions of the 3 genes
o A—(5)—B—(10)—C
 Chromosome maps based on crossover frequencies can only be used to identify the relative position of genes,
not their absolute loci
DISCUSS THE ROLE OF CHROMOSOME MAPPING IN IDENTIFYING RELATIONSHIPS BETWEEN SPECIES
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More closely related species tend to show fewer differences in their chromosomes. We can use chromosome
maps to establish degrees of genetic similarity between species.
Chromosome number
o Chromosome mapping shows us the diploid number of a species. Species that have a recent common
ancestor will tend to have similar or identical diploid numbers.
 E.g. Eucalyptus have 22, Felidae (cats, lions) have 38, Hominidae (great apes) have 48, and
humans have 46
Banding patterns
o When chromosome maps or karyotypes are made, we can see visible banding patterns on
chromosomes. Closely related species will tend to have similarities in their banding patterns
Gene linkage groups
o Since crossing over causes the exchange of homologous regions of chromosomes, the gene linkage
group is maintained even though linked alleles are separated.
o
o
Gene loci which are close together can only be ‘unlinked’ by significant changes in chromosome
structure, which move one gene locus to a different section of the same chromosome, or a different
chromosome altogether. When such a drastic change in DNA occurs in a germline cell, the organism is
unlikely to survive, and the genetic change is unlikely to become prevalent in the animal population.
Rats and mice have many linkage groups in common. About 50 genes on the rat’s number-3
chromosome also appear on the mouse’s number-2 chromosomes. Because they are closely related,
rats and mice share many gene linkage groups, as certain genetic loci remain close together over
generations.
PERFORM A FIRST-HAND INVESTIGATION TO MODEL LINKAGE
4.
The Human Genome Project is attempting to identify the position of genes on
chromosomes through whole genome sequencing
DISCUSS THE BENEFITS OF THE HUMAN GENOME PROJECT
DESCRIBE AND EXPLAIN THE LIMITATIONS OF DATA OBTAINED FROM THE HUMAN GENOME PROJECT
OUTLINE THE PROCEDURE TO PRODUCE RECOMBINANT DNA
EXPLAIN HOW THE USE OF RECOMBINANT DNA TECHNOLOGY CAN IDENTIFY THE POSITION OF A GENE ON A
CHROMOSOME
PROCESS INFORMATION FROM SECONDARY SOURCES TO ASSESS THE REASONS WHY THE HUMAN GENOME PROJECT
COULD NOT BE ACHIEVED BY STUDYING LINKAGE MAPS
5.
Gene therapy is possible once the genes responsible for harmful conditions
are identified
DESCRIBE CURRENT USE OF GENE THERAPY FOR AN IDENTIFIED DISEASE
and
PROCESS AND ANALYSE INFORMATION FROM SECONDARY SOURCES TO IDENTIFY A CURRENT USE OF GENE THERAPY
TO MANAGE A GENETIC DISEASE, A NAMED FORM OF CANCER OR AIDS
𝑪𝒚𝒔𝒕𝒊𝒄 𝑭𝒊𝒃𝒓𝒐𝒔𝒊𝒔 (𝑪𝑭)
General Information on Gene Therapy:
Gene therapy is the insertion of genes into an
individual’s cells to treat a disease. This is
done by cutting genes from DNA of healthy cells,
and inserting them into DNA of defective
cells/tissues, hence allowing them to produce the
functional protein that is missing in the diseased
state.
Transfer of large genetic sequences into cells is
difficult because DNA is a negatively charged molecule
and will not easily cross cell membranes.
3 ways to deliver genes to target cells are:
 Viral vectors which normally infect specific tissues
 DNA combined with lipids to form a liposome which can cross cell membranes
 Direct injection of ‘naked’ DNA into cells.
Ex vivo approach (outside body): Cells taken from a patient are cultured outside the body, the required gene added,
and then modified cells returned to the patient. Can only be used when modified cells can be easily returned to patient
to repopulate the affected tissue. (shown in Figure: Cell-based Delivery)
In vivo approach (within body): Copies of required gene are introduced directly into patient’s cells, e.g. using an aerosol
containing the gene for lung cells. (see Figure: Direct Delivery)
Cystic Fibrosis, the Disease:
Cystic fibrosis (CF) is a Mendelian recessive condition caused by a mutation in the cystic fibrosis transmembrane
conductance regulator (CFTR) gene. In most people with CF, their defective gene on chromosome 7 is missing three
adjacent base pairs at the 508th triplet, and defective protein missing just one amino acid of 1480. The recessive gene
does not allow sufferers to produce an ion channel protein that controls movement of chloride ions across cell
membranes. This leads to an osmotic imbalance, and water from tissue fluid enters the cells. This results in secretion of
thick mucus, which blocks airways and pancreatic ducts.
Why is cystic fibrosis well suited for gene therapy?
It is a single gene defect. Most severely affected organ (the lung) is relatively accessible for treatment. Studies have
shown that only 5-10% of normal CFTR gene expression is needed to prevent lung manifestations of cystic fibrosis. CF
patients have virtually normal lungs at birth, allowing treatment to be started before significant lung damage occurs.
Problems with Using Gene Therapy to treat CF:
1) Immune response to vector may: cause inflammation, prevent virus from accessing target cells, and reduce
effectiveness of viral vector in repeat dosings *Note that viral vectors are more likely to cause an immune response.
Non-viral vectors mainly have to contend with gene delivery and gene expression issues.
2) Carrying capacity required by a vector to transport CFTR cDNA and promoter sequence
3) Gene delivery: Efficiency of viral vectors in infecting lung cells, or ability of non-viral vector to pass the lipid bilayer
4) Effectiveness of gene expression: Efficient promoter
sequences which work for a long time are required.
Human promoters such as ‘ubiquitin’ which can drive
gene expression for three months are currently being
tested.
5) Duration for which the treatment is effective, and ability
to be safely re-administered: Normal gene does not
usually become incorporated into chromosomal DNA, and
bronchial epithelial cells have a lifetime of 120 days, so
gene therapy must usually be repeated.
Gene Therapy Using Viral Vectors:
How it works: Normal CFTR gene is cloned in bacteria, then
inserted into a virus. Virus solution can be dripped into the
lung via a thin tube, or delivered via nasal cavities using nebulisers (aerosol delivery).
Problems specific to viral vectors:
 Viruses may infect cells other than the targeted diseased ones
 New gene might be inserted in the wrong location in DNA causing mutations
 Changes may be passed on to children if virus infects reproductive cell
 Transferred genes could be overexpressed producing harmful levels of the missing protein
 There may be an immune reaction to the viral vector
Clinical Trials with Adenovirus and Adeno-associated Virus:
Early clinical trials with adenovirus were undergone in 1993: These trials failed because the adenovirus caused
inflammation of the recipient’s lungs, and the immune response removed the viruses before they could deliver the gene
to lung cells. Jesse Gelsinger’s death after participating in a clinical trial for OTC deficiency using an adenovirus vector
severely curtailed research into adenovirus vectors.
Clinical trials with adeno-associated virus (AAV): Schaffer induced mutations in AAV, cultured these variants, took
those with improved infection rate, and repeated the process. The final AAV strain was able to bind to a more plentiful
receptor on lung cells, and had enhanced ability to slip past the cell membrane. Other serotypes of AAV have proved
more effective when administered to the lungs, and are currently being trialled as a promising treatment.
Pros and Cons of adeno-associated viruses (AAV):
 AAV causes a milder immune response than adenoviruses
 AAV can infect non-dividing cells as well as dividing cells
 AAV has the ability to stably integrate into host cell genome in chromosome 19. Note that recombinant AAV
does not integrate into the genome, but rather fuses at its ends into a functional circular form
 However, AAV has smaller packaging capacity than adenovirus, which gives little space to include both the CFTR
cDNA and a promoter.
Other Vectors Are Being Developed:
Pseudotyped lentiviral vectors are under development: These provide sustained gene expression as they insert their
genome into target cell’s DNA. There are no suitable receptors on the surface of airway epithelial cells, so lentiviruses
with other surface proteins have been developed (pseudotyping).
Parainfluenza virus vectors are under development. They are more effective
than adenovirus or adeno-associated virus at delivering the CFTR gene to
airway epithelial cells.
Liposomes are also being researched and have undergone some clinical
trials. Cationic liposomes are mixed with plasmid DNA to form a
complex, which enhances transport of DNA past the lipid bilayer of cell
membranes. There is no immune response; however, the amount of
gene transfer achieved is too small, and gene expression is ineffective.
Summary of Current Use:
Various vectors such AAV and liposomes are in clinical trial. Methods for increasing the efficiency of these are being
researched. Other vectors are being developed and tested, e.g. pseudotyped lentivirus. Due to various difficulties, gene
therapy is not yet used in the mainstream to manage cystic fibrosis. However, it offers a promising treatment for the
future. Further research and clinical trials are necessary before gene therapy can be effectively used to manage CF.
6.
Mechanisms of genetic change
DISTINGUISH BETWEEN MUTATIONS OF CHROMOSOMES, INCLUDING
REARRANGEMENTS
CHANGES IN CHROMOSOME NUMBER, INCLUDING TRISOMY, AND POLYPLOIDY AND MUTATIONS OF GENES
INCLUDING
BASE SUBSTITUTION
FRAMESHIFT
OUTLINE THE ABILITY OF DNA TO REPAIR ITSELF
DESCRIBE THE WAY IN WHICH TRANSPOSABLE GENETIC ELEMENTS OPERATE AND DISCUSS THEIR IMPACT ON THE
GENOME
DISTINGUISH BETWEEN GERM LINE AND SOMATIC MUTATIONS IN TERMS OF THEIR EFFECT ON SPECIES
PROCESS AND ANALYSE INFORMATION FROM SECONDARY SOURCES TO DESCRIBE THE EFFECT OF ONE NAMED AND
DESCRIBED GENETIC MUTATION ON HUMAN HEALTH
7.
Selective breeding is different to gene cloning but both processes may change
the genetic nature of species
EXPLAIN, USING AN APPROPRIATE EXAMPLE FROM AGRICULTURE, WHY SELECTIVE BREEDING HAS BEEN PRACTISED
DESCRIBE WHAT IS MEANT BY ‘GENE CLONING’ AND GIVE EXAMPLES OF THE USES OF GENE CLONING
DISTINGUISH BETWEEN GENE CLONING AND WHOLE ORGANISM CLONING IN TERMS OF THE PROCESSES AND
PRODUCTS
DISCUSS A USE OF CLONING IN ANIMALS OR PLANTS THAT HAS POSSIBLE BENEFITS TO HUMANS
ANALYSE AND PRESENT INFORMATION FROM SECONDARY SOURCES TO TRACE THE HISTORY OF THE SELECTIVE
BREEDING OF ONE SPECIES FOR AGRICULTURAL PURPOSES AND USE AVAILABLE EVIDENCE TO DESCRIBE THE SERIES
OF CHANGES THAT HAVE OCCURRED IN THE SPECIES AS A RESULT OF THIS SELECTIVE BREEDING
IDENTIFY DATA SOURCES, CHOOSE EQUIPMENT OR RESOURCES, GATHER, PROCESS AND ANALYSE INFORMATION
FROM SECONDARY SOURCES TO DESCRIBE THE PROCESSES USED IN THE CLONING OF AN ANIMAL AND ANALYSE THE
METHODOLOGY TO IDENTIFY WAYS IN WHICH SCIENTISTS COULD VERIFY THAT THE ANIMAL PRODUCED WAS A
CLONE
8.
The timing of gene expression is important in the developmental process
IDENTIFY THE ROLE OF GENES IN EMBRYONIC DEVELOPMENT
SUMMARISE THE ROLE OF GENE CASCADES DETERMINING LIMB FORMATION IN BIRDS AND MAMMALS
DESCRIBE THE EVIDENCE WHICH INDICATES THE PRESENCE OF ANCESTRAL VERTEBRATE GENE HOMOLOGUES IN
LOWER ANIMAL CLASSES
DISCUSS THE EVIDENCE AVAILABLE FROM CURRENT RESEARCH ABOUT THE EVOLUTION OF GENES AND THEIR
ACTIONS
IDENTIFY DATA SOURCES, GATHER, PROCESS AND ANALYSE INFORMATION FROM SECONDARY SOURCES AND USE
AVAILABLE EVIDENCE TO ASSESS THE EVIDENCE THAT ANALYSIS OF GENES PROVIDES FOR EVOLUTIONARY
RELATIONSHIPS