Nucleic Acids

Question 1

Background

Answer 1

• 1953 – James Watson and Francis Crick introduced elegant double-helical model for DNA
• DNA can direct its own replication from monomers
• DNA contains hereditary information that controls your biochemical, anatomical, physiological and
behavioural traits

Question 2

Evidence that DNA can transform bacteria

Answer 2

• 1928 – Frederick Griffith tired to develop vaccine
against pneumonia -> streptococcus
• He used two strains:
o Virulent (capsule) smooth-strain ->
pathogenic
o Non-virulent (a-virulent), rough strain ->
harmless

• Griffith wanted to know whether injections of
heat-killed virulent pneumococci could be used
to immunise against pneumonia
• He was surprised to find that when heat killed
pathogenic bacteria was mixed to living,
harmless bacteria, some of the living bacteria
was transformed to pathogenic bacteria
• This meant some chemical substance (DNA)
caused this heritable change
• This also meant that proteins couldn’t be the
factor that converted them, as proteins would
have been denatured when heat was applied
• This phenomenon is known as transformation – a
change in the genotype and phenotype due to
the assimilation of external DNA by a cell.

Question 3

Only discovered in 1943 to be the DNA by Avery and his co-workers

Answer 3

• They created two cultures
o One where DNA was broken down leaving proteins
o And one where protein was broken done leaving DNA
o When the heat-killed pathogenic bacteria were mixed with harmless bacteria, the
conversion to pathogenic bacteria only occurred when the tube containing DNA and
broken down protein was used
o See next page for experiment example of this method

Question 4

Further evidence that DNA is the genetic material

Answer 4

• Erwin Chargaff analysed the base composition of
DNA from a number of organisms
• In 1950 he reported that the base composition of
DNA varies from species to species
• He also noticed that the regularity of the ratios of
nucleotide bases
o In DNA -> the no. of adenine equalled the
thymines and the guanines equalled the
cytosines

• He developed the Chargaff rules:
1. Base composition varies between species
2. Within a species A and T bases are equal, and
G and C bases are equal

Question 5

How is the genetic info contained in the DNA?

Answer 5

To fulfil its biological role, the following four must be
met by the genetic material
1. It must carry the genetic information from
parent cell to daughter cell and from
generation to generation
2. It must contain information for producing a
copy of itself
3. It must be chemically stable
4. However, it must be capable of mutation

Question 6

Structure of DNA

Answer 6

• DNA are polynucleotides (polymers) = monomers of
nucleotides
• Nucleotide -> a nitrogenous base + phosphate group +
pentose (five-carbon) sugar
• Nucleoside -> above without phosphate group
• Adenine and guanine are large two fused carbon rings
-> purines
• Uracil, thymine and cytosine are smaller with a smaller
single carbon ring -> pyrimidines
• One purine always attaches to one pyrimidine ->
creates uniform length diameter
• Adenine bonds to thymine with two hydrogen bonds
• Guanine bonds to cytosine with three hydrogen bonds
• Deoxyribose (sugar) has one OH on its 5 carbon ring
• Ribose has two OH’s on its 5 carbon ring
• Sugar-phosphate forms the backbone of a helix strand of DNA
• Adjacent nucleotides are connected by phosphate group.
• The phosphate of one nucleotide is attached to the sugar of the next nucleotide
• 5 prime (5’) end of DNA molecule is where phosphate molecule is sticking out
• 3 prime (3’) end of DNA molecule is where sugar is sticking out
• The subunits of the two strands of the molecule run in opposite directions. This is
known as anti-parallel nature of DNA

Question 7

DNA double helix

Answer 7

• Molecular architecture of DNA enables replication of genes
• Although the complementary base-pairing rules dictate the combinations of
nitrogenous bases that form the rungs of the double helix, they do not restrict the
sequence of nucleotides along each DNA strand
• 10 base pairs present in one full turn of the helix (3.4nm)

Question 8

DNA Replication: The Basic Principle: Base Pairing to a Template Strand

Answer 8

• First hypothesized by Francis Crick and James Watson
• When a double helix replicates, each of the two daughter
molecules will have one old strand, from the parental
molecule, and one newly made strand -> semiconservative
model (see end of section for exp. evidence)

Question 9

DNA Replication: A Detailed look

Answer 9

• E coli has a single chromosome (5 million base pairs) à 60 minutes for the cell to replicate
• Human cell (6 billion base pairs) à few hours to replicate
• Enormous amount of replication with very few errors occurs (1 per 10 billion nucleotides)
• High accuracy and speed requires over a dozen enzymes and other proteins

Question 10

RNA Replication: Initiation

Answer 10

• The replication of a chromosome begins at particular sites called origins of replication à these are short
stretches of DNA having a specific sequence of nucleotides
• Bacteria have circular chromosomes with a single origin
• Proteins that initiate DNA replication recognize this sequence and attach to the DNA, separating the two
strands and opening up a replication ‘bubble’
• Replication of DNA then proceeds in both directions until the entire molecule is copied
• A eukaryotic cell may have hundreds or thousands of origins à multiple bubbles form and eventually fuse,
speeding up the process of replication in very long strands
• At the end of each replication bubble is a replication fork, is a Y-shaped region where parental strands of
DNA are being unwound

Question 11

Enzymes and Proteins in DNA replication

Answer 11

• Helicase is an enzyme that untwists the double helix at the replication forks, separating the parental strands
and making them available as template strands
• After the parental strands are separated, single-strand binding proteins bind to the unpaired DNA strands,
stabilizing and keeping them from re-pairing
• The unwinding/untwisting of the DNA strand causes tension and strain ahead of the replication fork à
topoisomerase helps relieve the strain by breaking, swivelling and re-joining the DNA strands
• The unwound sections of parental DNA strands are now available to serve as templates for the synthesis of
new complementary DNA strands
• However, the enzymes that synthesize DNA cannot initiate the synthesis of a polynucleotide à they can only
add DNA nucleotides to the end of an already existing strand that is base paired with a template strand
• The initial nucleotide chain that is produced during DNA synthesis is actually a short stretch of RNA
o This RNA strand is called a primer and is synthesized by primase
o Primase starts a complementary RNA chain from a single RNA nucleotide, using the DNA strand as a
template
o The completed primer (5-10 nucleotides long) is base paired to the template strand and the new
DNA strand will start from the 3’ end of the primer

Question 12

Synthesizing a New DNA Strand: Elongation

Answer 12

• Enzymes called DNA polymerases catalyse the synthesis of new DNA by adding nucleotides to a pre-
existing chain

• Two of these enzymes play a major role in replication -> DNA polymerase I and III
• Most of the DNA polymerases require a primer and template strand -> along which complementary
nucleotides are lined up
• In bacteria DNA pol. III adds a DNA nucleotide to the RNA primer and then continues to add nucleotides to
the growing end of the new DNA strand (500 per second in bacteria, 50 in humans)

Question 13

Antiparallel Elongation

Answer 13

• As noted the two ends of a DNA strand are different, giving each strand directionality, like a one-way street
• In addition, the two strands of DNA in a double helix are antiparallel, meaning they are orientated in opposite
directions of each other
• Therefore, the two new strands formed during DNA replication must also be antiparallel to their template
strands
• Because of the structures of DNA polymerases, they can only add nucleotides to the 3’ prime end of a
primer or growing DNA strand, never to the 5’ end
• Thus a new DNA strand can elongate only in the 5’->3’ direction

Question 14

Leading strand

Answer 14

• Along one template strand, DNA pol III can
synthesize a complementary strand
continuously by elongating the new DNA in the
mandatory 5’à3’ direction
• DNA poll III remains in the replication fork on that
template strand and continuously adds
nucleotides to the new complementary strand
as the fork progresses
• Strand made by this mechanism is known as
the leading strand
• Only one primer is required for DNA poll III to
synthesize an entire leading strand

Question 15

Lagging strand

Answer 15

• To elongate the other new strand of DNA in the
mandatory 5’à3’ direction, DNA poll III must
work along the other template strand in the
direction away from the replication fork
• The DNA strand elongating in this direction
called the lagging strand
• Even though synthesis of both strands occur
simultaneously and at the same rate, it is known
as the lagging strand because synthesis is
delayed slightly relative to the synthesis of the
leading strand à enough template strand
needs to be exposed first before elongation can
begin
• Unlike leading strand (continuous), the lagging
strand is synthesized discontinuously, as a
series of segments
• These segments on the lagging strand are
called Okozaki fragments
• Whereas one primer is needed for the
leading strand, each Okozaki fragment
on the lagging strand must be primed
separately
• After DNA pol III forms an Okozaki
fragment, DNA pol I replaces the RNA
nucleotides of the adjacent primer with
DNA nucleotides
• But DNA pol I cannot join the final
nucleotide of this replacement DNA
segment to the first DNA nucleotide of
the adjacent Okozaki fragment
• Another enzyme, DNA ligase, is able to
complete this task and joins all Okozaki
fragments into a continuous DNA strand
-> acting like a glue

Question 16

RNA & Protein Synthesis: The Flow of Genetic Information

Answer 16

• The information content of genes is in the form of specific sequences of nucleotides along strands of DNA,
the genetic material
• The DNA of an organism leads to specific traits by dictating the synthesis of proteins -> proteins are the link
between genotype and phenotype
• Gene expression is the process by which DNA directs the synthesis of proteins (via translation and
transcription)
• This is also known as The Central Dogma of Biology, which describes and provides an explanation to the
basic framework of how genetic information flows from a DNA sequence to a RNA sequence to a
synthesized protein product inside cells. The central dogma also suggests that DNA contains the information
needed to make all of our proteins, and that RNA is a messenger that carries this information to the
ribosomes, where they are synthesized. DNA -> RNA -> protein

Question 17

RNA & Protein Synthesis: Basic principles of Transcription and Translation

Answer 17

• Genes provide the instructions for making a protein but a gene does not build a protein directly
• The bridge between DNA and protein synthesis is RNA
• RNA is similar to DNA except that:
o It contains ribose instead of deoxyribose as its sugar
o It contains the nitrogenous base uracil rather than thymine
o It is a single strand
• Transcription is the synthesis of RNA using information in the DNA
o The information is simply transcribed from one DNA to RNA in the form of triplet codons
o DNA serves as the template strand to assemble the complementary RNA strand
o The RNA is thus a transcript of the instructions that code for a specific polypeptide (mRNA)
• Translation is the synthesis of a polypeptide using information in the mRNA
o There is a change in language -> the cell must translate the nucleotide sequence of an mRNA
molecule into the amino acid sequence of a polypeptide -> codons match with anticodons
o The sites of translation are ribosomes, molecular complexes that facilitate the orderly linking of amino
acids

Question 18

RNA & Protein Synthesis: Transcription and Translation in bacteria and eukaryotes

Answer 18

• Major difference in translation and transcription between eukaryotes
and bacteria is the fact that in bacteria there is no nuclear envelope
that separates DNA from protein-synthesizing equipment
• Therefore, in bacteria, the lack of compartmentalization allows for
translation and transcription to occur simultaneously. In eukaryotes,
however, the nuclear envelope prevents this from happening

• The transcription of a protein-coding eukaryotic gene results in pre-
mRNA and further processing yields functional mRNA, that travels

into the cytoplasm -> the initial RNA transcript is more generally
termed a primary transcript

Question 19

Protein Synthesis: A Detailed View:

Molecular components of Transcription

Answer 19

• mRNA is transcribed from the template strand of a
gene
• An enzyme called RNA polymerase pries the two
strands of DNA apart and joins together RNA
nucleotides complementary to the DNA template
strand, thus elongating the RNA polynucleotide
• Like DNA polymerases that function in DNA
replication, RNA polymerases can assemble a
polynucleotide only in its 5’->3’ direction
• Unlike DNA polymerases, RNA polymerases are able
to start a chain from scratch, they don’t need a primer
• The specific sequences of nucleotides along the DNA
strand mark where transcription ends and begins
• These are known as the promotor and terminator
sequences
• The stretch in between these two sequences is
known as the transcription unit
• Bacteria have a single type of RNA polymerase whilst
eukaryotes have at least three (the one used to make
pre-mRNA is RNA pol II)
• There are three major stages of transcription are:
initiation, elongation and termination

Question 20

Protein Synthesis: Initiation

Answer 20

• The sequence of DNA where RNA polymerase II attaches and
initiates transcription is known as the promotor
• Promotor (small string of DNA) contains within itself the start point -> place where transcription actually starts eg. TATA
Box
• In eukaryotes transcription factors (proteins) mediate the
binding of RNA polymerase to the promotor and the initiation
of transcription
• In bacteria the RNA polymerase binds directly to the promotor
• Only after transcription factors are attached to promoter does
RNA polymerase II bind to it
• The whole complex of transcription factors and RNA
polymerase II bound to the promoter is called a transcription
initiation complex
• Once the appropriate transcription factors are firmly attached
to the promotor DNA and the polymerase is bound in in the
correct orientation, does the enzyme unwind the two DNA
strands and begins transcribing the template strand at the
start point

Question 21

Protein Synthesis: Elongation

Answer 21

• As RNA polymerase moves along DNA, it untwists 10-20
DNA nucleotides at a time for pairing with RNA nucleotides,
thereby, elongating the RNA in the 5’->3’ direction
• The enzyme adds nucleotides to the 3’ end of a growing RNA
molecule as it continues along the double helix
• In the wake of the advancing wave of RNA synthesis, the new
RNA molecule peels away from its DNA template, and the
DNA double helix reforms
• A gene can be transcribed simultaneously by several
molecules of RNA polymerase -> this helps cell make the
protein in large amounts

Question 22

Protein Synthesis: Termination

Answer 22

• In bacteria transcription proceeds through a terminator sequence in the DNA. The transcribed
terminator (an RNA sequence) functions as the termination signal causing, polymerase to detach from
DNA and release the transcript, which requires no modification before translation
• In eukaryotes, RNA polymerase II transcribes a sequence on the DNA called the polyadenylation signal
sequence, which specifies a polyadenylation signal (AAUAAA) in the pre-mRNA. This is called a signal
because once this stretch of 6 RNA nucleotides appears, it is immediately bound by certain proteins in
the nucleus. The proteins then cut it free from the polymerase, releasing the pre-mRNA. The pre-mRNA
then undergoes processing. However, the RNA polymerase continues to transcribe. Since the new 5’
end isn’t protected by a cap, enzymes degrade it. The polymerase continues to transcribe until the
enzymes catch up to dissociate it from the DNA.

Question 23

Modification of pre-mRNA: in Eukaryotic cells

Answer 23

• Before being sent to cytoplasm the transcript is modified and undergoes RNA processing by enzymes in the
nucleus
• During this processing, both ends of the RNA transcript are altered
• Also some interior sections of the RNA molecule are cut out and the remaining parts spliced together.

Question 24

Alteration of mRNA ends

Answer 24

• The 5’ end is synthesized first and receives a 5’
cap (modified form of guanine)
• The 3’ end is modified before mRNA exits the
nucleus, an enzyme adds 50-250 more
adenine nucleotides, forming a poly-A-tail
• Both of these serve important functions:
1. Facilitate the export of the mature mRNA
from the nucleus
2. They help protect the mRNA from
degradation by hydrolytic enzymes
3. They help ribosomes attach to the 5’ end of
the mRNA once it reaches the cytoplasm
• The ends aren’t translated and neither are UTR
(untranslated regions) -> they assists in
ribosome binding

Question 25

Split genes and RNA splicing

Answer 25

• Large portions of the RNA molecule that is
initially synthesized is removed through a
process called splicing
• RNA is 27k nucleotides long, average size
protein of 400 amino acids requires only 1200
nucleotides
• Therefore, there are large sections of noncoding
RNA in the transcript
• Theses noncoding parts are interspersed
amongst the coding parts -> the DNA
nucleotides that code for a eukaryotic
polypeptide is not continuous
• Non coding segments are introns
• Coding segments are called exons -> they are
eventually expressed (however UTR’s within
exons don’t code for amino acids)
• The introns are cut out and the exons are
spliced together forming a continuous coding
sequence
• The splicing of pre-mRNA and the removal of
introns involves the use of spliceosomes (a
complex of proteins and small RNAs)
• The complex binds to the intron, releasing it
(where it rapidly degrades), and then joins
together the exons
• The small RNAs assist in splice site recognition
as well as catalysing the splicing reaction

Question 26

Molecular components of Translation

Answer 26

• In the process of translation, a cell reads a genetic
message and builds a polypeptide accordingly
• Message is a series of codons along mRNA and
the translator is tRNA (transfer)
• Function of tRNA is to transfer amino acids from
cytoplasmic pool of amino acids to a growing
polypeptide in a ribosome
• Cell always keeps stock of the 20 amino acids,
either by synthesis or uptake from surrounding
solution
• Ribosome is a structure made of proteins and
RNAs that adds each amino acid brought to it by
tRNA to the growing end of a polypeptide chain

Question 27

Structure and Function of tRNA

Answer 27

• Each tRNA translates a codon into an amino acid
• tRNA bears a specific amino acid at one end whilst
at the other end is a nucleotide triplet that can base
pair with a complementary codon on mRNA
(anticodon)
• tRNA has parts that fold back due to
complementary base pairs, thereby forming a clover
leaf and a complex 3D shape
• Transcribed from DNA template in nucleus before
moving to cytoplasm
• Each tRNA molecule is used repeatedly, picking up
its amino acid, depositing it and then returning to do
the same thing again
• An accurate translation of a genetic message
requires two instances of molecular recognition:
o First a tRNA that binds to an mRNA codon
specifying a particular amino acid must carry
that amino acid and no other to the ribosome
(correct matching up of tRNA and amino acid is
called aminoacyl-tRNA synthetases)
o 20 types of synthetases for 20 amino acids
o Synthetase catalyses attachment of amino acid
to its tRNA in a process driven by hydrolysis of
ATP
o The aminoacyl is then released and allowed to
deliver the amino acid to the growing chain
o The second instance is the pairing of anticodon
to mRNA codon
o Some tRNAs however can pair up with more
than one kind of codon
o This is because the third nucleotide base of a
codon and the corresponding base on the
tRNA are relaxed compared to the other two
bases
o For example U at the 5’ end of a tRNA
anticodon can bond to either an A or a G in the
third position on an mRNA codon -> known as
wobble

Question 28

Ribosomes

Answer 28

• Facilitates the specific coupling of tRNA anticodons
with mRNA codons during protein synthesis
• Consists of a large and small subunit, each made up
of proteins and one or more rRNAs
• Made in nucleus from transcribed rRNA and imported
amino acids
• The completed subunits are then exported to
cytoplasm
• Subunits join to form functional ribosome only when
when attached to an mRNA molecule
• Thousands of ribosomes = rRNA (2/3 of ribosome
mass) is the most abundant type of cellular RNA
• Eukaryotic ribosomes are larger and have different
molecular composition -> antibiotics only affect
bacterial ribosomes
• In addition to a binding site, each ribosome has three
binding sites for tRNA
o P site -> peptidyl-tRNA, holds tRNA carrying
growing polypeptide chain
o A site -> aminoacyl-tRNA, holds tRNA
carrying next amino acid to be added
o E site -> exit, discharged tRNA exits through
here

• Holds mRNA and tRNA in close proximity to allow
amino acids to be added to polypeptide
• Also catalyses the peptide bond between amino
acids
• When polypeptide is long enough it passes through
exit tunnel in the large subunit where it is released

Question 29

Translation: Initiation

Answer 29

• The initiation stage brings together mRNA, a tRNA bearing the first amino acid of the polypeptide and two
subunits of a ribosome
• First, the small subunit binds to both mRNA and a specific initiator tRNA (carries amino acid methionine)
• In bacteria the small subunit binds to the mRNA just upstream of the start codon (AUG), thereafter the initiator
tRNA binds to start codon
• In eukaryotes, the small subunit, with initiator tRNA already bound, binds to the 5’ cap of mRNA. It then
moves downstream until it reaches the start codon where the initiator tRNA then binds to it
• The union of the small subunit, tRNA and mRNA is followed by the attachment of the large subunit
completing the translation initiation complex
• Proteins called initiation factors are required to bring all these components together
• GTP (guanosine triphosphate) is hydrolysis to expand energy needed to form this complex
• After the initiation process, the initiator tRNA sits in the P-site and the vacant A-site is available for next
aminoacyl-tRNA

Question 30

Translation: Elongation

Answer 30

• In the elongation process amino acids are added one by one to the the previous amino acid in the growing
chain
• Each addition involves several proteins called elongation factors and occurs in a three step cycle (see figure
below)
• Energy expenditure occurs in the first and third steps
o Codon recognition requires hydrolysis of GTP to increase accuracy and efficiency of this step
o One more GTP is required to provide energy for translocation step
• The mRNA moves through the ribosome in the 5’->3’ direction
• The ribosome and mRNA move relative to each other, unidirectionally, codon by codon

• The three step cycle is repeated until the chain is complete, thereafter, the tRNAs are released form the E-
site where they will be reused

Question 31

Translation: Termination

Answer 31

• Elongation continues until a stop codon on the mRNA reaches the A-site
• The base triplets UAG, UAA and UGA do not code for amino acids -> act as stop signals
• A release factor, a protein shaped like an aminoacyl-tRNA, binds directly to the stop codon in the A-site
• The release factor causes the addition of a water molecule instead of an amino acid to the polypeptide chain
• This reaction breaks (hydrolyses) the bond between the completed polypeptide and the tRNA in the P-site,
releasing the polypeptide through the exit tunnel of the large subunit
• The remainder of the translation assembly comes apart aided by protein factors and the hydrolysis of two or
more GTP molecules

Question 32

Completing and Targeting the Functional Protein

Answer 32

• Process of translation is not sufficient to make functional proteins; they have to undergo the following
processes:
1. Protein Folding and Post-Translational Modifications
2. Targeting Polypeptides to Specific Sites

Question 33

Protein Folding and Post-Translational Modifications

Answer 33

• During synthesis a polypeptide begins to coil and fold spontaneously as a consequence of its amino acid
sequence (primary structure), forming a protein with a specific shape: a three-dimensional molecule with
secondary and tertiary structure
• The gene determines the primary structure, and the primary structure in turn determines shape
• In many cases a chaperone protein (chaperonin) helps the polypeptide fold correctly
• Additional steps (post-translation modification) may be required before protein can fulfil its role
o This could include chemical modification by adding sugars, lipids, phosphate groups etc.
o Enzymes may remove amino acids from the leading (amino) end of the polypeptide or from middle
sections (in some cases separating the chain into two parts which come together later)

Question 34

Targeting Polypeptides to Specific Sites

Answer 34

• In eukaryotes there are two types of ribosomes: bound and free
o Free ribosomes are suspended in cytosol and synthesize proteins that stay and function in the
cytosol
o Bound ribosomes are attached to the cytosolic side of the Rough ER or on the nuclear envelope.
They synthesize proteins of the endomembrane system as well as secretory proteins (eg. Insulin)

• It is important to note that bound and free ribosomes are identical and can change roles
• Polypeptide synthesis always begins in a free ribosome in the cytosol
• If the polypeptide is bound for the endomembrane system or export it will be marked with a signal peptide,
which targets the protein to the ER

• This signal peptide is recognized as it emerges from the ribosome by protein-RNA complex called a signal-
recognition particle (SRP) where it follows the process illustrated in the figure below

Question 35

Mutations that affect protein structure and function

Answer 35

• Changes in genetic information of cells are called mutations
• They are responsible for diversity amongst organisms because they are the source of new genes
• Mutations that affect one (point mutations) or a few pairs nucleotides are called small scale
mutations
• Large scale mutations are caused by chromosomal rearrangements that affect long segments of
DNA
• If point mutations occur in gametes or cells that give rise to gametes they may be transferred onto
future generations, which may result in adverse side effects to the phenotype of organisms resulting
in a genetic disorder
• A change of a single nucleotide in the DNA’s template strand could result in the production of an
abnormal protein

Question 36

Types of Small-Scale Mutations

Answer 36

1. Single nucleotide-pair Substitutions

2. Nucleotide-pair Insertions or Deletions

Question 37

Single nucleotide-pair Substitutions

Answer 37

• Substitution is the replacement of one
nucleotide and its partner with another pair of
nucleotides
• Some substitutions have no effect on encoded
protein (due to redundancy of genetic code)
• A substitution that results in a different codon
being translated into the same amino acid is
called a silent mutation
• Substitutions that change one amino acid into
another are called missense mutations
• Such a mutation may have little effect on the
protein because:
o New amino acid may have similar
properties to original amino acid
o Location of new amino acid is in region
of protein where it isn’t essential to
function of protein

• However, a substitution in a crucial area may
significantly alter protein activity
• Occasionally this mutation leads to an improved
protein but more often it is neutral or detrimental
(leads to a useless or less active protein)
• Substitutions are usually missense mutations ->
the altered codon still codes for a amino acid
which makes sense, although it may not be the
right sense
• A point mutation can also change a codon for
an amino acid to a stop codon -> this is called
a nonsense mutation, which causes translation
to end prematurely, resulting in a shorter
polypeptide

• Most nonsense mutations result in non-
functional proteins

Question 38

Nucleotide-pair Insertions or Deletions

Answer 38

• Insertions and deletions are additions or losses
of nucleotide pairs in a gene
• These may affect the reading frame (the triplet
grouping of nucleotides on the mRNA) of the
genetic message
• Such a mutation is called a frameshift mutation
and occurs whenever the number of nucleotides
inserted or deleted is not a multiple of three
• All nucleotides downstream of the insertion or
deletion will be improperly grouped leading to
extensive missense and resulting sooner or later
in nonsense and premature termination
• Unless the frameshift is very near the end of the

gene, the protein is almost certain to be non-
functional

Question 39

Mutagens and New Mutations

Answer 39

• Substitutions, insertions and deletions can arise
in number of ways.
• Errors during DNA replication can lead to these
mutations or large scale ones
• If a mutation occurs, usually it is corrected by
DNA proofreading and repair systems
• If it is missed it will result in the mutation serving
as a template in the next round of replication ->
this is called spontaneous mutations
• Mutagens are agents that interact with DNA and
cause mutations
o Mutagens include physical (UV radiation,
X rays) as well as chemical agents