TMA02 groups of three nucleotides called codons,

TMA02                                                                                           Linsey Duffin



          Smooth Red Petals = Rs/Rs

          Serrated White Petals = rS/rS













Offspring Genotype = Rs/rS

Offspring Phenotype = Serrated Red Petals

know the phenotype of the F1 Offspring will be serrated red petals, as the
dominant genotypes from the parent plants are Red (as shown by the R) and
Serrated (as shown by the S) so from the genotype of Rs/rS, we take the
dominant genes in order to establish the phenotype of the F1 Offspring.


b-i)     Two phenotypes which have arisen due to
meiotic recombination are:

                    Red Serrated petals (Genotype

                    White Smooth Petals
(Genotype rs/rs)


know this because when the F1 Offspring (Rs/rS = Red Serrated Petals) are
crossed with Smooth White petals (rs/rs) they should produce the following












Offspring = Rs/rs Red Smooth

                    = rs/rS White Serrated


from Table 1 we can see that there are four different F2 progeny produced,
suggesting that meiotic recombination is responsible for the two unexpected


The recombination frequency is 30%.

of plants from the unexpected progeny divided by the total number of plants
produced, multiplied by 100 to give the answer as a percent.

Serrated = 769 plants                      White
Smooth = 811 plants

plants produced = 5267


          769 + 811 = 1580                                 

= 0.30

           0.30 x 100 = 30


frequency = 30%


2) DNA


a)     mRNA transcribed = 5′


b)    Using the genetic code Book 1
Ch. 5 Fig. 1.3, the sequence of the polypeptide is:






c)      If a single Guanine nucleotide
was inserted into the template DNA where indicated in the question, this could
create an mRNA Transcription which is incorrect and does not stop where
expected. As DNA is read in groups of three nucleotides called codons, the
added guanine will cause a shift within the codons. As the order of nucleotides
within the codon is what dictates its corresponding amino acid, the amino acids
present will be different to those expected. Previously, there was a codon
which issued a STOP signal at the 8th codon. However, once the
guanine is added this STOP codon moved to the 11th codon. With the
shift in position of the STOP codon, the chain will feature amino acids which
would not have been present otherwise. Consequently, the chain does not have
the expected pattern of amino acids, which causes the polypeptide to be
non-functional. This occurs because the amino acid chain is longer and
different to that expected and therefore will not fold correctly due to
different bonding characteristics. The expected polypeptide is –
Met-Glu-Tyr-Pro-Arg-Thr-Val-Asn – STOP, however with the added guanine, the
resultant polypeptide is Met-Glu-Tyr-Pro-Glu-Asn-Ser-Glu-Leu-Lys-STOP.



As seen in Figure 2, the level of reporter gene expression differs within the
cell lysates containing various plasmids. The expression increases to a maximum
of approximately 16000 at pLuc-1430 before trending to a low of approximately
500 at pLuc-42. It is also noteworthy that pLuc-441 and pLuc-649 are showing
the same expression with a value of approximately 8000 light units.


The region of DNA between -6270 and -2480 contain a binding site for a
transcriptional repressor. We know this as Figure 2 shows the light units
emitted by the luciferase drops dramatically from 15000 units for pLuc -2840 to
approximately 8000 units for pLuc -6270. The presence of a transcriptional
repressor would explain why the light units emitted reduces so greatly as it
would reduce the speed of which the DNA translation takes place, reducing the
total expression.


The region of DNA presented between -1430 and -649 contain a binding site for a
transcriptional activator. As figure 2 shows the light units emitted by the
luciferase increases from approximately 8000 units for pLuc -649 to
approximately 16000 units for pLuc -1430. The presence of a transcriptional
activator would explain the increase of light unit emitted as it would increase
the speed at which the DNA translation takes place, increasing the total


b) An
experiment with no drug added was used as the control.

Hedamycin reduces the light emitted by the survivin gene expression. It appears
to act as a transcriptional repressor as the experiment with No Drug added see’s
8000 light units expressed compared with less than 2000 light units expressed
when 10nm of hedamycin is introduced. As seen in figure 3, there is a further
reduction between 10nm and 25nm of hedamycin, however between 25nm and 100nm of
hedamycin there is almost no change, with 25nm, 50nm and 100nm of hedamycin all
staying around 500 light units expressed.

During S294, we have come across Immunolabelling which which could also be used
to measure the effect of hedamycin on the expression of the survivin protein as
we know it allows for the tagging of proteins, by using an enzyme which
produces light.

iv) A
77.5% reduction in light emitted by the survivin gene expression is observed
between the control (No Drug (8000 light units)) and 10nm of hedamycin (1800 light

of hedamycin percentage = (1800/8000) x 100 = 22.5%.

100% –
22.5% = 77.5%

the percentage change in the survivin gene expression is 77.5%.



c)    Hedmycin inhibits the growth of
the HeLa cells, as indicated by Figure 4, the culture dishes were incubated for
a period of 48 hours, after which time, there was an abundance of cells visible
on the 0µM culture dish. There are remarkably less HeLa cells on the 0.05µM Culture
Dish and less again on the 0.5µM Culture Dish.


d)    Figure 3 shows a reduction in
expression of the survivin gene when hedamycin is introduced to the experiment.
Figure 4 shows far less HeLa cells on culture dishes which include hedamycin.
This indicates hedamycin reduces the rate at which HeLa cells can grow by
reducing the expression of the survivin gene.





Histones are the proteins in eukaryotic cells which associate with DNA to form

Chromatin gains the ‘beads on a string’ appearance due to the DNA wrapping
around each individual histone, but never breaking its line, therefore appearing
as string connecting the histone beads.

c) It
is suggested that chromatin fibre (compacted chromatin polymer) is around
5-100nm in diameter which causes it to be below the resolution limit of light
microscopy. Chromatin is not electron dense, and therefore cannot be picked up
in electron microscopy.

According to the article, two main methodological advances represented by the
ChromEMT technique used by OU et al. allowing them to visualise chromatin
fibres within the nucleus are the use of fluorescent dye which binds to DNA,
which can be photo converted to produce electron dense precipitate and
Tomography, slicing cells and tilting them in an electron beam, which allows
them to be imaged from many angles.


The classic model of chromatin organisation comes from biomechanical in-vitro
studies which suggest that DNA forms higher order structures in the shape of
chromatin fibres. Ou et al. have shown that there was little evidence for
higher order structures as the classic model theory would suggest. It is stated
that Ou et al. found the majority of the chromatin in the nucleus to be organised
as a disorganised polymer of 5-24nm. They suggest that past the primary
structure of beads on a string, there is minimal organisation.

f) According
to the article there is little difference in the higher order organisation of
chromatin fibres, regardless of whether it is euchromatin or heterochromatin.
This is important as it is not what would be expected. Heterochromatin was
expected to be tightly packed as it is DNA that is stored and unable to be
transcribed, instead it simply appears follow the ‘beads on a string’
configuration, as shown by the observation of the 5nm-24nm diameter fibres. The
article also states that there is a ‘striking’ similarity between interphase
and mitotic chromosomes, which would never be expected due to the morphological
changes that chromosomes experience during the chromatin compaction of mitosis.
The conclusion that can be drawn from these observations is that regardless of
whether the DNA was mitotic, in interphase, euchromatin or heterochromatin, all
that was seen was the 5nm-24nm fibres with no sign of the higher structure
which was expected. Where a higher structure would be expected, all that was
found was a high density of ‘bead on a string’ fibres.


g) The
authors of the article cite:

          S. S. Teves et al., eLife 5, e22280 (2016)

     A paper called ‘A Dynamic Mode of Mitotic
Bookmarking by transcription factors’.

The article outlines the concern that ‘ChromEMT relies on chemical fixation of
samples, which could introduce artifacts.’ However, it is countered that it is
most likely unlikely that the structures observed by the ChromEMT are due to
fixation artifacts as the ChromEMT process uses cross-linking fixation (a
method by which the cross linking fixatives create a covalent chemical bond
between proteins in a tissue) which would only produce higher order structures
rather than the small fibres which were observed.





5) Essay

The structural Hierarchy of Proteins

          Introduction:  What are proteins?

Body:              Primary Structure
– Chemical/Structural features

Structure –Chemical/Structural Features

Structure – Chemical/Structural Features

                              Diagram – Disulphide Bonds

Quaternary Structure – Chemical/Structural Features

                              Diagram – The Four Levels of Protein

Structure Study technique

                      What are
homomeric and heteromeric Proteins



Structural Hierarchy of Proteins

are essential to all living organisms. They are made from long chains of amino
acids and can have many different functions. These functions are decided based
on the structure of the protein, and the structure is dependent on the amino
acids involved. There is a four level hierarchy to the structure of the protein
– primary, secondary, tertiary and quaternary and these will all be discussed
within the body of this essay. (The Open University, 2014)

primary structure of a protein consists of a unique chain of amino acids. These
amino acids are joined by peptide bonds, which are created when enzymes
catalyse a condensation reaction. Similarly, peptide bonds can be broken down
by enzyme catalysed hydrolysis reactions.

secondary structure of a protein can be identified by polypeptide chains being
folded or pleated into different shapes. Common shapes to identify are Alpha
Helices and Beta Pleated Sheets. Alpha helices are identifiable by their
right-handed spiral structure and Beta Pleated Sheets are identifiable by their
beta strands which are connected by two or three hydrogen bonds. These can
exist as both parallel and anti-parallel sheets. In both these shapes, the
structure is held together due to strong hydrogen bonds which give the shapes
their stability. These bonds are found between carbonyl and amino groups. In
alpha helices, this hydrogen bond actually helps in creating its spiral shape
as the carbonyl amino acid 1 (C=O) bonds to the hydrogen of amino acid 5 (N-H)
which encourages the curl of the structure in order to keep these bonds
together. (Khan Academy, 2016)

Structures are the final 3 Dimensional structure of a protein before it
connects to other polypeptide chains. These 3D shapes take two forms – Globular
– ball like structures which are common in metabolic roles such as enzymes, and
Fibrous – Long fibres which are insoluble in water and tend to take on
structural roles such as collagen and keratin. The tertiary structure is mainly
caused by interactions of the ‘R’ group of the amino acids. There are four
types of bonds possible in a tertiary structure – Disulphide, Ionic, Hydrogen
and Hydrophobic and Hydrophilic Interactions. Disulphide bonds form when there
are two cysteine monomers found in the amino chain as shown in Figure 1.

two sulphur molecules within the cysteine monomers form a strong covalent bond
and these bonds assist in giving the tertiary structure its 3D shape.

As Figure 1 shows, the cysteine
monomers do not have to be directly beside each other, just close enough for
the covalent bond to form. This is why the bond can assist in providing the 3D
shape to tertiary structures, as the amino chains will bend to allow for the
covalent bond.


bonds form within the tertiary structure when two oppositely charged ‘R’ groups
come into close contact and bond together. This occurs as many ‘R’ groups
become slightly positively or negatively charged, and they will always look to
balance out their charges by forming ionic bonds. Hydrogen bonds work exactly
the same within tertiary structures as they did in secondary structures, once
again influencing the shape of the structure by the strength of their bonds.
Hydrophobic and Hydrophilic Interactions within tertiary structures occur
within water based environments. Here the amino acids will arrange themselves
so that the hydrophobic amino acid ‘R’ groups are turned to the inside of the
structure while the Hydrophilic amino acids face outwards towards the water
present. All tertiary bonds can be broken by the application of heat, so much
that even once cool again the protein cannot reform its original complex shape.
A protein which has lost its shape in this way is said to be denatured.

Structures are seen when two or more polypeptide groups join together. Examples
of these are haemoglobin and collagen. Haemoglobin is a water soluble globular
protein found mainly in blood. It contains two alpha helices and two beta
pleating sheets. Collagen is a fibrous protein. It contains three alpha helices
which are coiled around themselves. Hydrogen bonds strengthen collagen further
by bonding not only within its own alpha helix, but with the alpha helices
which are coiled beside them. Collagen has the ability to form Covalent Cross
Links which means that the collagen molecules can form larger chains with other
collagen molecules and continue to extend hydrogen bonds between these chains.
This adds to the strength and stability of the molecule, assisting in its
structural functionality within bones, cartilage and tendons among others. (A
Level Notes, n.d)

is important to note that not all proteins will form quaternary structures.

2 below shows clear examples of protein in its primary amino acid chain,
secondary alpha helix structure, tertiary globular structure and quaternary
structure of four connected polypeptide groups.

The shape of the protein after
either the tertiary or quaternary level is very important as this will lead to
the shape of the proteins binding site. Without the correct shape at its
binding site, the protein will be unable to bond with the relevant substrate
and perform the task it was created to do.


Crystallography is one technique which can be used for studying the structure
of proteins. By using this method it is possible to determine the 3 dimensional
structure of protein molecules (among others) at atomic resolution. Electron
density can be reconstructed using this technique; by collecting the data of
diffracting x-rays from a single crystal, however this crystal must have an
ordered and regularly repeating arrangement of atoms. It is important to note
that not all proteins can be studied under x-ray crystallography. (T. M.
Picknett and S. Brenner, 2001)

can be described as either homomeric or heteromeric. Where the subunits of the
quaternary structure are all the same the protein is described as being
homomeric. An example of a homomeric protein is Beta Galactosidase. Beta
Galactosidase consists of four identical polypeptides (Oxford Reference, n.d).
Most Proteins are homomeric. When the subunits are different, the protein is
described as being heteromeric (or heteromultimeric). Once such example is the
voltage-gated potassium channel in the plasma membrane of a neuron, as these
are made up from four different subunits (Wikipedia, 2017)

described above, each level of the proteins’ structure is key to its function
as its shape will dictate what it can interact with. Every bond within the
polypeptide plays a part in twisting and intertwining the chains so they can
fulfil this function. By combining simple building blocks and putting together
multiple pieces, incredibly complex structures can be made from simple amino
acids. Part of the key to understanding how a protein functions is to
understand its shape and that is why technique’s like X-ray Crystalography is
so important in protein research.



Level Notes (n.d) Protein Structure Online
Available at:
(Last Accessed: 29.1.18)

Academy (2016) Order of Protein Structure
Online Available at:
(Last Accessed: 29.1.18)

Reference (n.d) Homomeric Protein
Online Available at:
(Last Accessed: 29.1.18)

T. M.
Picknett and S. Brenner (2001) X-Ray
Crystallography Online Encyclopaedia of Genetics, Available at:
(Last Accessed 29.1.18)

Open University (2014) Book 2: Working
Cells, 3rd Edition, Milton Keynes, The Open University.

(2017) Protein Complex, Homomultimeric
and Heteromultimeric Proteins Online Available at:
(Last accessed 29.1.18)