Comprehensive course notes with examples of high-scoring exam responses written in direct response to study design dot points for maximum relevancy. Written by a student with 99+ ATAR and a raw 45 in Biology.
Biology Study Design Checklist Name: _______________________
Unit 3 How do cells maintain life?
AOS 1 - What is the role of nucleic acids and proteins in maintaining life?
In this area of study students explore the expression of the information encoded in a sequence of DNA to form a
protein and outline the nature of the genetic code and the proteome. They apply their knowledge to the structure and
function of the DNA molecule to examine how molecular tools and techniques can be used to manipulate the
molecule for a particular purpose. Students compare gene technologies used to address human and agricultural issues
and consider the ethical implications of their use.
On completion of this unit the student should be able to analyse the relationship between nucleic acids and proteins,
and evaluate how tools and techniques can be used and applied in the manipulation of DNA.
The relationship between nucleic acids and proteins
Nucleic acids as information molecules that encode instructions for the synthesis of
1 proteins: the structure of DNA, the three main forms of RNA (mRNA, rRNA and tRNA) and
a comparison of their respective nucleotides
DNA:
** 3’ and 5’ orientations stand for the direction of the deoxyribose sugar in the strand. At
the 3’ end, the third carbon on the sugar faces outwards while on the 5’ end, the fifth
carbon (phosphate end) faces outwards. DNA is antiparallel – one strand has its 5’ end one
way while the complimentary strand has the 5’ end on the other side.
**RNA polymerase reads DNA 3’ to 5’; RNA is thus synthesized in a 5’ to 3’ direction and
the ribosome reads the mRNA in a 5’ to 3’ direction because that is how it has been
synthesized.
mRNA tRNA
Differences Found in the nucleus and the Only found in cytosol
cytosol Double stranded
Single stranded Purpose: to hold specific
Purpose: to act as a copy of amino acids, and bring those
DNA, holding the instructions amino acids to the codons
to synthesize specific that code for them through
proteins. the tRNA’s complementary
anticodons.
Similarities:
Nucleotide structure: pentose sugar, ribose sugar; uracil not thymine
Usually more unstable than DNA
USUALLY single stranded (bar part of tRNA)
, Biology Study Design Checklist Name: _______________________
The genetic code as a universal triplet code that is degenerate and the steps in gene
2 expression, including transcription, RNA processing in eukaryotic cells and translation by
ribosomes
Triplet code – universal
Nucleotides are grouped in threes, termed triplets in DNA and codons in RNA. The triplet
code is universal – all organisms have the same genetic material with nucleotides grouped
into threes.
Degeneracy in the triplet code:
Different arrangements of the nucleotides allow for 64 possible triplets/codons that code
for a total 20 different amino acids. Therefore, multiple triplets code for the same amino
acid, demonstrating degeneracy within the triplet code.
(Degeneracy is an important protection against mutations causing dysfunctional proteins –
if the last nucleotide in most triplets is changed, the amino acid it codes for is likely to
remain the same, therefore protecting against mutations in the last nucleotide of most
triplets.)
Transcription:
Creates a copy of the genetic material to synthesize proteins as DNA is too big to leave the
nucleus.
1. Initiation – RNA primers attach to the promoter region of the target gene on the
template DNA strand. RNA polymerase binds to the promoter region of the
template strand.
2. Elongation – the RNA polymerase reads the template strand in a 3’ to 5’ direction.
It covalently bonds free RNA nucleotides via condensation polymerisation
reactions to form a complimentary RNA strand to the template. The RNA
nucleotides are bound together by phosphodiester bonds. The mRNA is
antiparallel to the DNA – runs 5’ to 3’.
3. Termination – once the RNA polymerase hits the termination sequence of the
target gene, it detaches from the template strand, allowing the DNA to go back to
its double helix formation. The pre-mRNA is free in the nucleus.
Post transcriptional modifications
Are completed in order to:
a. Stabilise the mRNA
b. Prevent other free nucleotides from joining
c. Allow the mRNA to easily bind to ribosomes
Types include:
1. Capping and tailing – the addition of methyl caps to the 5’ end and poly-A tails to the 3’
end of the mRNA strand.
2. alternative splicing – the removal of all transcribed introns (non-protein-coding regions)
and possible removal of exons and rearrangement of exons (protein-coding regions) from
the mRNA. The possible removal of exons allows one gene to code for multiple proteins,
depending upon which exons have been spliced. Spliceosomes do the splicing.
*Alternative splicing occurs before capping and tailing
, Biology Study Design Checklist Name: _______________________
Exam eg. Explain the process of alternative splicing and its benefits
Introns are removed from pre-mRNA. Exons can be either removed or rearranged
during this process.
This allows multiple proteins to be synthesized using the same mRNA from the same
gene.
Removing introns stabilises the mRNA and allows for a functional protein to be
produced as only coding regions (exons) are left.
Translation:
1. The mature mRNA leaves the nucleus via nuclear pore and either becomes
associated with a free ribosome, or a ribosome itself associated with the rough ER.
2. The ribosome reads the mRNA in a 5’ to 3’ direction, starting at the start (AUG)
codon.
3. It reads in codons (groups of three nucleotides), as each codons code for a specific
amino acid.
4. As it reads, tRNA with complimentary anticodons carrying specific amino acids
attach to mRNA codons via hydrogen bonding. When 2 tRNA molecules are next to
each other, the ribosome will facilitate that binding of the two amino acids
together via condensation polymerisation reaction, leaving the two amino acids
bound together via peptide bond.
5. As it reads the mRNA, the ribosome joins more and more amino acids together via
peptide bond, until eventually it finishes reading and a full polypeptide chain is
created.
6. Also remember polysomes – many ribosomes working off of the same mRNA.
Exam eg. Identify and explain how mRNA becomes a polypeptide chain
Translation occurs
The mRNA associates with a ribosome which reads it three nucleotides – one
codon – at a time.
tRNA with a complimentary anticodon and a specific amino acid will attach to
each codon.
Amino acids coded for adjacently will be bound together via peptide bond
through a condensation polymerisation reaction.
The structure of genes: exons, introns, and promoter and operator regions
3 All eukaryotic genes are structured as such:
Promoter, Operator, Introns and Exons, Terminator.
The promoter region is where RNA polymerase and RNA primers bind to catalyse
transcription.
, Biology Study Design Checklist Name: _______________________
The operator region is where transcription factors bind to during gene regulation.
Exons are the regions of the gene that code for the proteins and are expressed during
protein synthesis.
The introns are the non-coding regions of the gene involved in regulating exon and whole
gene expression.
The terminator region has codons signalling the end of the gene, and tells RNA polymerase
where to stop transcribing the gene.
The basic elements of gene regulation (eukaryotes/prokaryotes): prokaryotic trp operon
4 as a simplified example of a regulatory process
Gene regulation
Is necessary because:
a. (Largely prokaryotes): Cells need to be energy efficient, which means that they
need to be able to stop producing specific proteins if there is already enough of
that substance wherever in the body it is needed, in order to save energy.
b. (Largely eukaryotes): Different cells in organisms have different roles in the
organism, and therefore must synthesize different proteins. All cells have the
entirety of an organism’s genome within them, so they must discriminate between
which proteins they produce to fulfil their role within the organism (eg. You don’t
want toes salivating)
Eukaryotes:
1. Pre-transcription regulation (also in prokaryotes):
The existence of transcription factors – repressors and activators – that either repress or
activate gene expression. These are proteins/RNA transcribed from genes. Repressors bind
to the operator of a gene in order to stop RNA Polymerase alongside other activating
transcription factors from being able to bind to the promoter/operator. Activators bind to
the regulatory genes of DNA to induce bending in the DNA which allows RNA Polymerase to
reach and bind to the promoter of genes to express them. Occurs before the initiation
stage of transcription.
2. Post-transcriptional regulation:
Splicing, capping and tailing. Already discussed.
3. Translational regulation:
The amount of time for which mRNA can survive; the availability of translation factors such
as ribosomes, tRNA, amino acids.
4. Epigenetic regulation:
Either the addition of DNA tags or the coiling of the histone proteins themselves.
a. DNA tags are chemicals which attach themselves to the DNA and prevent
enzymes from attaching to repress gene expression.
b. Histone modifications, the addition of methyl groups or histone
acetylation encourages the nucleosomes to coil around each other very
tightly and therefore inhibits gene expression, or results in loose packing
of nucleosomes promoting gene expression, respectively.
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