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TOPIC 1: CELLULAR FUNCTIONS – Part 10

LEARNING OUTCOME

j)Investigate and explain the effects of temperature, pH, enzyme concentration and substrate concentration on the rate of enzyme catalysed reactions, and explain these effects.

ESSAY ANSWER

Temperature
The rate of reaction increases with temperature until the optimum temperature is reached. Optimum temperature is the temperature at which the enzyme is functioning at its maximum rate. The effects of temperature on the rate of reaction is expressed as the temperature coefficient (Q10 value) and it is calculated as follows:

Q10 = rate of reaction at (10+x) deg C/ rate of reaction at x deg C

For most enzymes, the Q10 value is approximately 2. This means that the rate of reaction doubles for every increase of 10ºC. This is only applicable between an approximate temperature range of 15°C to 25°C.
Certain organisms have been able to adapt to high temperatures. Enzymes in
these heat-tolerant organisms have a higher optimum temperature than most
normal organisms, eg. thermophilic bacteria. Despite this, the rate of reaction takes on the same trend (i.e. decreases rapidly) as temperature increases beyond its optimum temperature.

pH
Optimum pH is pH is at which the rate of an enzyme-catalysed reaction is
at its maximum. Usually matches the usual pH environment which the enzyme
is found in. At this pH, the intramolecular bonds which maintain the
secondary and tertiary structures of the enzyme are intact. Conformation of active sites is most ideal for binding of substrate and the frequency of successful collisions between enzyme and substrate molecules is the highest. The number of enzyme-substrate complexes formed per unit time is at its highest and the rate of reaction is at its maximum.

At pH lower or higher than the optimum, the concentration of hydrogen
ions (H+) would have changed and this would alter the charges on the R
groups of the amino acid residues of the enzyme molecule. The ionic bonds
and hydrogen bonds that help maintain the conformation of the enzyme
molecule would be disrupted and the binding of the substrate would be
affected. Enzyme work within a narrow range of pH. If the pH is altered by a small extent from the optimum pH, the effects are normally reversible. If the pH is restored to the optimum pH, the maximum activity of the enzyme will be restored. However, if the pH is altered by a large extent, the conformation of the enzyme molecule would be severely affected and denaturation of the enzyme might be irreversible.

Enzyme concentration
An increase in the enzyme concentration leads to an increase in the rate of reaction. The rate of reaction is directly proportional to the enzyme
concentration. The more enzyme molecules present, the more active sites
available for the substrate to be catalysed, hence the higher the rate of
reaction. The following conditions must be present for the rate of reaction to increase linearly in proportion to the enzyme concentration:
– Substrate concentration must be kept at high level and must not be a
limiting factor.
– Concentration of the enzyme is much less than that of the substrate.
– Factors such as pH and temperature must be constant.

At low enzyme concentration, adding more enzymes increases the rate of
reaction. As enzyme concentration increases, the frequency of successful
collisions between the enzyme and substrate molecule increases. With more
enzyme presents, it is more likely that a substrate will bind to an empty active site on an enzyme. More enzyme-substrate complexes are formed. Rate of reaction increases linearly with increasing enzyme concentration.

At very high enzyme concentration, the substrate concentration is the
limiting factor. An increase in the enzyme concentration would not result in any further increase in the rate of reaction.

Lastly, for reactions with different enzyme concentrations, if the amount of substrate were the same for all reactions, the total amount of product will eventually be the same.

Substrate Concentration
At fixed enzyme concentration, reaction rate increases as substrate
concentration increases; i.e. the rate of reaction is directly proportional to the substrate concentration.

At low substrate concentration; as substrate concentration increases, the
frequency of successful collisions between the enzyme and substrate
molecules increases. More enzyme-substrate complexes are formed and rate
of reaction increases. Active sites of the enzyme molecules are not all
occupied at this point. Rate of reaction is limited by substrate
concentration.

At high substrate concentration; the rate of reaction will continue to increase with increases in substrate concentration, until a point where further increase in substrate concentration will no longer produce a significant change in the rate of reaction. This is because the active sites of all enzyme molecules are saturated with substrate molecules. Any extra substrate molecule has to wait until the enzyme-substrate complex has released the products before it can bind to the active site. Enzyme concentration is now the limiting factor.

The maximum rate for a particular enzyme reaction is denoted as Vmax. The
Michaelis constant (KM) is the concentration of substrate required to make
the reaction proceed at half the maximum rate. It is the same for a particular enzyme but varies from one enzyme to another. It is a measure of the affinity of the enzyme for its substrate. A low KM value indicates a high affinity of the enzyme for the substrate, and vice versa.

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TOPIC 1: CELLULAR FUNCTIONS – Part 11

LEARNING OUTCOME

(k)Explain the effects of competitive and non-competitive inhibitors (including allosteric inhibitors) on the rate of enzyme activity.

An enzyme inhibitor is a substance that prevents an enzyme from
catalysing the reaction. Thus, it decreases the rate of an enzyme-catalysed reaction. It does so by combining with the enzyme (either at the active site or elsewhere) to form an enzyme-inhibitor complex, which in turn restricts the enzyme from combining with its substrate. Their binding to the enzyme may be reversible or irreversible.

Competitive inhibition
The inhibitor is structurally similar to that of substrate. It competes with the substrate for the active site of enzyme. The inhibitor associates with active site but is unable to react with it. This prevents access of any molecules of the true substrate and decreases the affinity of active site for substrate (Km increases). If substrate concentration increases, rate of reaction increases since the degree of inhibition is lessened. This kind of inhibition can be overcome by increasing the concentration of substrate because this increases the likelihood of an enzyme-substrate complex being formed (as opposed to an enzyme-inhibitor complex). Therefore, at very high substrate concentration, the rate of reaction can reach its maximum value (Vmax does not change).

Non-Competitive Inhibitors
The non-competitive inhibitor does not resemble the substrate, i.e. has no
structural similarity to the substrate. The inhibitor does not directly
compete with the substrate for the active site and binds to a part of the
enzyme away from the active site. This interaction causes the enzyme
molecule to change its shape. The active site of enzyme is altered, resulting in:
i. substrate can no longer bind to active site (or)
ii. the enzyme becomes less effective at catalysing the reaction. Substrate
can still bind to the altered active site of enzyme but active site functions less effectively Therefore, increasing the substrate concentration has no effect on the inhibition since the inhibitor and substrate do not compete for the same site on the enzyme. Vmax cannot be reached and Km remains unchanged as the inhibitors render a portion of enzymes non-functional, effectively “lowering”
the concentration of enzymes present.

Reversible Non-Competitive Inhibitors – Allosteric Inhibitors
Allosteric inhibitors bind to a regulatory site known as allosteric site. They Binds to the allosteric site via noncovalent interactions, which are weak and hence inhibition is reversible. This type of inhibitors dissociate from the enzyme at low concentrations. Binding of an allosteric inhibitor results in a conformation change in active site and stablises the inactive form of the enzyme, thus, inhibiting the activity of the enzyme.

Irreversible Non-Competitive Inhibitors
Toxins and poisons are often irreversible enzyme inhibitors. For non-competitive irreversible inhibition, the inhibitor is strongly attached to the enzyme by covalent bonds with R-groups. For example: the enzyme precipitates when its sulfhydryl groups combine permanently with the inhibitor. Some chemicals are known to cause irreversible inhibition of enzymes, e.g. mercury, silver, arsenic, nerve gas DFP.

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TOPIC 1: CELLULAR FUNCTIONS – Part 12

LEARNING OUTCOME

(l)Explain the importance of mitosis in growth, repair and asexual reproduction.

ESSAY ANSWER

Mitosis being a type of nuclear division aids in the formation of daughter cells that are exact copies of the parents. This maintenance of identical chromosomal numbers, exact genetic information, and same ploidy as the parental cells is crucial in that new resulting cells must have the same type and number of chromosomes as the cells which they are replacing in the following 3 processes:

1. Growth
It is an increase in number of cells within the organism. For tissue
growth, it is important new cells are identical to existing cells so that they carry out the same function.

2. Repair
It is the regeneration of cells and tissues lost in normal processes of
wear and tear, aging, damage, and disease. It is important that damaged
cells are replaced with exact copies of the original cells in order for the tissue to function properly.

3. Asexual Reproduction
It is the reproduction of a unicellular organism without production of
gametes. For example; binary fission (in bacteria, Amoeba, and
Paramecium), budding (in Hydra and yeast), and vegetative propagation
in potato, onion, and ginger. It is important to ensure that offspring are genetically identical to parent.

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TOPIC 1: CELLULAR FUNCTIONS – Part 13

LEARNING OUTCOME

(m)Explain the need for the production of genetically identical cells and fine control of replication.

ESSAY ANSWER

The production of genetically identical cells is needed for growth, repair and asexual reproduction [refers to LO (l)].
Fine control of replication is needed for the following reasons:

1. Ensure that all the genetic information is retained during the doubling of the DNA, i.e. the integrity of the genetic information is maintained. Genetic stability for growth, repair, and asexual reproduction results from the semi-conservative replication of DNA during the S phase.

2. Ensure that no genetic variation (not including spontaneous mutations) occurs in mitosis so that the chromosomes in each daughter cell carry the same hereditary content as the parental cells in terms of number and type.

3. Prevent the formation of cancerous cells. Cell division in normal cells is regulated by a cell cycle control system. Cancerous cells have
mutations in genes that regulate cell growth and division protooncogenes
and tumor-supressor genes), leading to changes in regulation of the cell cycle control, thus dividing excessively to form a mass of cells – tumor.

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TOPIC 1: CELLULAR FUNCTIONS – Part 14

LEARNING OUTCOME

(n)Explain how uncontrolled cell division can result in cancer, and identify causative factors (e.g. genetic, chemical carcinogens, radiation, loss of immunity) which can increase the chances of cancerous growth. (Knowledge that dysregulation of checkpoints of cell division can result in uncontrolled cell division and cancer is required, but detail of the mechanism is not required.)

KEY WORDS
• Gain of function mutation
• Loss of function mutation
• Accumulation of mutations in a single cell
• Uncontrolled cell division, will eventually contribute to cancer

ESSAY ANSWER

Causative factors

Genetic factors, e.g. inherited oncogenes or defective tumour suppressor genes
Chemical carcinogens, e.g. ethidium bromide
Tar in cigarette smoke
Ultraviolet radiation
Ionising radiation, e.g. X-ray or gamma ray
Loss of immunity due to increasing age

Explanation
(Named factor from above) causes mutation in genes controlling normal cell division, i.e. proto-oncogenes and tumour suppressor genes.

When proto-oncogenes mutate to form oncogenes, this is known as gain of
function mutation. When both copies of a tumour suppressor are mutated, there is a loss of function mutation.

With accumulation of these mutations and other mutations to genes which
encode for enzymes like telomerase in a single aberrant cell, this abnormal cell divides uncontrollably and more rapidly than normal by mitosis. This leads to formation of tumours, i.e. groups of cells which are abnormal and unspecialised. When cells of a tumour invade surrounding tissue, and generally metastasize to other sites, the tumour is considered to be malignant and is defined as a cancer

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TOPIC 1: CELLULAR FUNCTIONS – Part 15

LEARNING OUTCOME

(o) Describe with the aid of diagrams, the behaviour of chromosomes during the mitotic cell cycle and the associated behaviour of the nuclear envelope, cell membrane and centrioles. (Names of the main stages are expected)

ESSAY ANSWER

Prophase
At the start of prophase, chromatin fibres become more tightly coiled,
condensing into discrete chromosomes. The duplicated chromatin fibres from S phase become visible as duplicated chromosomes. Each duplicated
chromosome has a pair of identical sister chromatids held together by a
centromere. The two pairs of centrioles move to opposite poles of cell. The mitotic spindle (spindle fibres) begins to form. The mitotic spindle is made up of microtubules and proteins. The nuclear envelope disintegrates and the nucleolus gradually disappears.

Metaphase
Chromosomes are aligned at the metaphase plate / equator of spindle. The two pairs of centrioles have reached the opposite poles of the cell. Kinetochore microtubules attach to the kinetochores at the centromere of each chromosome. The nuclear envelope and the nucleolus are absent.
Anaphase During anaphase, the centromere divides into two, releasing each sister chromatid. The sister chromatids from each chromosome separate and move to opposite poles of the spindle, centromeres first. Each separated sister chromatid is now known as a chromosome. The kinetochore microtubules shorten, helping sister chromatids to move to opposite poles of the cell. The non-kinetochore microtubules lengthen, causing the cell to elongate. The nuclear envelope and the nucleolus are absent.

Telophase
Individual chromosomes arrive at opposite poles of the cell. Chromosomes
decondense to become chromatin fibers. Spindle fibres disintegrate. The
nuclear envelope reforms, forming two daughter nuclei. The nucleoli
reappear. There is a centriole associated with each nucleus.

Telophase often overlaps with cytokinesis. In animal cells, cytokinesis occurs through a process known as cleavage, where the centre of the parent cell is constricted from the outside inwards. A cleavage furrow forms near the old metaphase plate. On the cytoplasmic side of the furrow is a contractile ring of microfilaments. As the dividing cell’s ring of microfilaments contracts, the cleavage furrow deepens until the parent cell pinches into two daughter cells. In plant cells, after nuclear division, Golgi vesicles derived from the Golgi
apparatus line up in the middle of the parent cell. The Golgi vesicles fuse to form a cell plate that extends across the equator of the parent plant cell. The contents of the Golgi vesicles contribute to the new cell walls (middle lamella and cell wall matrix) of the two daughter cells, whilst their membranes form the new plasma membranes of the daughter cells. The cell plate eventually fuses with the parent cell wall and cell membrane, separating the two daughter cells.

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TOPIC 1: CELLULAR FUNCTIONS – Part 16

LEARNING OUTCOME

(p)Explain what is meant by homologous pairs of chromosomes.

KEY WORDS

• Identical order of gene loci
• Centromeres in the same position
• Arms of the same length

ESSAY ANSWER

Homologous chromosomes are two chromosomes where one derived from
each parent and both determine the same characteristics
Homologous chromosomes pair with each other during prophase I of
meiosis, have identical order of gene loci and have centromeres in the
same position and have arms of the same length as each other.

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TOPIC 1: CELLULAR FUNCTIONS – Part 17

LEARNING OUTCOME

(q)Explain the need for reduction division (meiosis) prior to fertilisation in sexual reproduction.

KEY WORDS

Conservation of diploid number

ESSAY ANSWER

Meiosis halves the number of chromosomes going into gametes, so that the
diploid number in the zygote can be conserved when the gametes fused
together during fertilisation.

During fertilisation, the number of chromosomes doubles in the zygote. In order to conserve the diploid number in the organism, there has to be an equivalent reduction in the chromosome number in the gametes that fused together.

This reduction is made possible by meiosis, which is also known as reduction division.

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TOPIC 1: CELLULAR FUNCTIONS – Part 18

LEARNING OUTCOME

(r)Explain how meiosis and random fertilisation can lead to variation

KEY WORDS

• Chiasmata formation and crossing over during prophase I
• Independent assortment of chromosomes at metaphase I
• Independent assortment of chromatids at metaphase II
• Each gamete has a unique combination of genes

ESSAY ANSWER

Meiosis introduces genetic variation through producing new combination of alleles in gametes.

Chiasmata formation and crossing over during prophase I During prophase I, synapsis occurs and homologous chromosomes pair up to form bivalents.

As a result of chiasmata formation, crossing over between non-sister
chromatids of homologous chromosomes where breakage and rejoining
between the 2 non-sister chromatids resulting in the exchange of alleles. This leads to recombinant chromatids being formed. This results in new combinations of alleles on chromosomes of the gametes, causing genetic variation.

Independent assortment of homologous chromosomes at metaphase I
During metaphase I, each pair of homologous chromosomes aligns in 2 rows at the equator.

The arrangement of each pair of homologous chromosome is completely
independent of the other pairs of homologous chromosomes.

Subsequent separation of the homologous chromosomes during anaphase I will produce different combinations of chromosomes in the daughter cells at the end of meiosis I.

Independent assortment of chromatids at metaphase II
During metaphase II, there is independent assortment of non-identical sister chromatids. Subsequent separation of chromatids during anaphase II will further contribute to different combinations of chromosomes in the four daughter cells produced at the end of meiosis II.

Random fertilisation of an ovum by a sperm
Two gametes fuse to form a zygote.
Each gamete has a unique combination of genes.
Any male gamete can fertilise any female gamete, thus making each zygote
unique which contributes to genetic variation

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TOPIC 1: CELLULAR FUNCTIONS – Part 19

LEARNING OUTCOME

(s)Describe, with the aid of diagrams, the behaviour of chromosomes during meiosis, and the associated behaviour of the nuclear envelope, cell membrane and centrioles. (Names of the main stages are expected, but not the subdivisions of prophase)

ESSAY ANSWER

Prophase I
Condensation of the chromatin occurs until the chromosomes become visible. Synapsis occurs, where homologous chromosomes pair
up, forming bivalents. Crossing-over occurs between non-sister chromatids of homologous chromosomes, at chiasmata. Crossing over enables exchange of genetic material between homologous chromosomes
and results in new combination of alleles on chromosomes of gametes. Nucleoli disintegrate and nuclear envelope disappears. Centrioles
moves to opp. poles

Metaphase I
2 rows of chromosomes are now aligned along the equator of the spindles, each chromosome is still paired with its homologous partner. The two hromosomes of each homologous pair faces opposite poles. If crossing over and exchange of genetic material occurred in Prophase I, non-sister chromatids will carry genetic material from the other chromosome. Independent assortment of chromosomes also occurs at this stage.
Kinetochore microtubules attach to the kinetochores of chromosomes

Metaphase I
2 rows of chromosomes are now aligned along the equator of the spindles, each chromosome is still paired with its homologous partner. The two chromosomes of each homologous pair faces opposite poles. If crossing over and exchange of genetic material occurred in Prophase I, non-sister chromatids will carry genetic material from the other chromosome. Independent assortment of chromosomes also occurs at this stage.
Kinetochore microtubules attach to the kinetochores of chromosomes

Anaphase I
The homologous chromosomes are separated. One homologous chromosome
is pulled to one pole, and the other to the opposite pole. Centromeres remained intact and sister chromatids are not separated at this stage. Kinetochore microtubules shorten to pull the chromosomes to their respective poles.

Telophase I & Cytokinesis I
Chromosomes reach opposite spindle poles. Separation of homologous chromosomes implies that chromosome number is halved after cytokinesis. Each nucleus has a haploid (n) chromosome set. Each chromosome exist
as two sister chromatids.

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TOPIC 2: DNA AND GENOMICS – Part 1

LEARNING OUTCOME

(a)Describe the structure and roles of DNA and RNA (tRNA, rRNA and mRNA).(Mitochondrial DNA is not required.)

ESSAY ANSWER
STRUCTURE OF DNA

The DNA molecule consists of two polynucleotide chains twisted around each other to form a double helix. Each nucleotide is made up of deoxyribose sugar, phosphate group and the nitrogenous bases can be adenine, thymine, cytosine or guanine.

In each polynucleotide chain, the alternating pentose sugars and phosphate groups form the sugar phosphate backbone. The nitrogenous bases project into the helix. The bases of both chains are flat structures and can therefore be stacked. The bases lie perpendicular to the central axis. The nitrogenous bases are 0.34 nm apart.

The two polynucleotide chains (or strands) are antiparallel. The two polynucleotide strands are held together by hydrogen bonds that form between the nitrogenous bases of opposite strands. Complementary base-pairing rule govern the formation of hydrogen bonds. Adenine pairs with thymine and forms two hydrogen bonds, while cytosine pairs with guanine and forms three hydrogen bonds.

Hydrogen bonds between bases of two polynucleotides, together with hydrophobic interactions between the stacked bases, stabilises the structure of the double helix. The strong covalent phosphodiester bonds between adjacent nucleotides maintain the integrity of the DNA base sequence.

The width between the two sugar-phosphate backbones of a DNA molecule is constant at 2 nm. This is equal to the width of a purine and a pyrimidine. Each complete turn of the helix is 3.4 nm; there are 10 bases per turn in each chain.

ROLE OF DNA
DNA functions in DNA replication. DNA replication ensures that the amount of DNA in the daughter cells is the same as that in the parent cells at the end of mitosis. It also ensures that genetic information is passed down accurately from parent cell to daughter cells during mitosis.

STRUCTURE OF mRNA
mRNA consists of 1 polynucleotide chain. It is made up of nucleotide monomers. Each nucleotide monomer consists of a phosphate group, ribose sugar and the nitrogenous bases adenine, uracil, cytosine or guanine.
ROLE OF mRNA The role of mRNA is to serve as a template for protein synthesis in translation.

STRUCTURE OF tRNA
tRNA consists of 1 polynucleotide chain. It is made up of nucleotide monomers. Each nucleotide monomer consists of a phosphate group, ribose sugar and the nitrogenous bases adenine, uracil, cytosine or guanine.

tRNA has a clover leaf structure that results from the hydrogen bonds between complementary base pairs at certain regions. The 5’ end of the tRNA always ends in guanine. The 3’ end is the amino acid attachment site. It binds to an amino acid specific to the tRNA molecule. The anticodon is made up of a specific triplet base sequence. The triplet base sequence at the anticodon is complementary to the codon
found on the mRNA.

ROLE OF tRNA
The role of tRNA is in matching the correct amino acid to its codon in mRNA, in the process of translation.

STRUCTURE OF rRNA
rRNA consists of 1 polynucleotide chain. It is made up of nucleotide monomers. Each nucleotide monomer consists of a phosphate group, ribose sugar and the nitrogenous bases adenine, uracil, cytosine or guanine.

ROLE OF rRNA
The role of rRNA is in ribosome synthesis. rRNA is synthesized in the nucleolus. The nucleolus combines proteins imported from the cytoplasm with rRNA to form the large and small subunits of the ribosomes.

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TOPIC 2: DNA AND GENOMICS – Part 2

LEARNING OUTCOME

(b)Describe the process of DNA replication and the experimental evidence for semi-conservative replication.

ESSAY ANSWER

PROCESS OF DNA REPLICATION

Before DNA replication occurs, free deoxyribonucleotides are synthesized in the cytoplasm and transported into the nucleoplasm via pores in the nuclear envelope. DNA Replication starts at specific sites known as origins of replication. The enzyme helicase causes DNA to unwind at the origin of replication through catalysing the breakage of hydrogen bonds between the complementary bases of the two polynucleotide chains.

This causes the two DNA strands to separate. The separated strands of DNA interact with single stranded binding protein (SSBP). The role of the single stranded binding protein is to stabilize the separated single-stranded so that the unwound region can serve as a template. Each separated strand of DNA can now act as a template for DNA synthesis. The template strand is also known as the parental strand.

At each DNA template, DNA synthesis is initiated by the synthesis of an RNA primer. An RNA primer is a short RNA chain consisting of about ten ribonucleotides. Its sequence is complementary to its corresponding DNA template strand. This RNA primer is needed at the beginning of all new DNA chains because the enzyme responsible for the synthesis and elongation of DNA, DNA Polymerase III, needs a free 3’ OH group to add the new complementary nucleotide. The RNA Primer is synthesized by the enzyme primase. After the RNA primer has been synthesized, DNA
Polymerase III catalyses DNA synthesis and elongation by catalysing the formation of a phosphodiester bond between new deoxyribonucleotides and the last ribonucleotide of the RNA primer.

Elongation of new DNA strands occurs in the 5’ to 3’ direction. Nucleotides are added one by one to the growing end of the new strand by complementary base pairing to the bases of the ‘old’ strand. Complementary base pairing involves the formation of hydrogen bonds between complementary bases (A to T; C to G). Nucleotides are joined
via the hydrolysis of two phosphates from the NTPs. This releases energy to drive the polymerization of the nucleotides. DNA polymerase III selects free deoxyribonucleotides that are complementary to those on the parental strand.

The leading daughter strand is synthesized continuously from the parental strand. The lagging daughter strand is synthesized discontinuously from the parental strand, forming of Okazaki fragments. As DNA Polymerase requires a 3’ OH end to elongate DNA, every Okazaki fragment must be “primed”. Hence, the lagging strand has more
than 1 primer. The primers are later excised, replaced with DNA. This is aided by DNA polymerase I. DNA Ligase catalyses the formation of a phosphodiester bond between two Okazaki fragments. DNA Polymerase I also completes a proof reading function of the newly synthesized strand of DNA.

The two separate strands of DNA winds back into a double helix after DNA
Polymerase I has completed its action.

EXPERIMENTAL EVIDENCE FOR SEMI-CONSERVATIVE REPLICATION

Meselson and Stahl used two different isotopes for nitrogen in their experiments: 14N, the ordinary isotope of nitrogen and15N, the heavy isotope of nitrogen.

In this experiment, Escherichia coli was cultured in a medium that contained nucleotide precursors labeled with heavy isotope 15N, for several generations. Nitrogen (N) is incorporated in the nitrogenous bases of these nucleotides. This causes all DNA in the bacteria to be labeled with 15N. This same group of bacteria was then transferred to medium with only 14N, a lighter isotope.

Two DNA samples were taken from this flask, one at 20 minutes and one at 40 minutes, after the 1st and 2nd replications, respectively Meselson and Stahl could distinguish DNA by their weight/ by centrifuging DNA extracted from the bacteria. DNA containing only 15N is denser (heavier) than DNA containing only 14N. Due to the weight difference, 15N would form a band below 14N in caesium chloride solution. Hybrid DNA (14N-5N),
consisting of one 14N strand and one 15N strand, would form a band halfway between the 14N DNA band and 15N DNA band.

The results from the “First Replication” were a single DNA band was precisely halfway between those of 14N DNA and 15N DNA. Hence, only14N15N DNA was found in these first generation cells. The results from the “Second Replication” were two bands of DNA, one above the other This indicated that half of the DNA molecules were of hybrid type
(14N-15N) and the other half were pure 14N DNA.

Meselson and Stahl compared their results to those predicted by each of the three models in the first replication in the 14N medium produced 1 band of hybrid (14N-15N)DNA. This result eliminated the conservative model. The second replication produced 1 light and 1 hybrid DNA bands, a result that eliminated the dispersive model and supported the semi-conservative model. From these results, Meselson and Stahl concluded that DNA replication is semi-conservative.

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TOPIC 2: DNA AND GENOMICS – Part 3

LEARNING OUTCOME

c)Describe how the information on DNA is used to synthesise polypeptides in prokaryotes and eukaryotes. (Description of the processes of transcription, formation of mRNA from premRNA and translation is required.)

ESSAY ANSWER

TRANSCRIPTION
Transcription is the process by which RNA is synthesized from DNA within the nucleus. The synthesis of RNA begins in the nucleus of a living cell that is not undergoing cell and nuclear division. Within the nucleus, messenger RNA (mRNA) is transcribed from the template DNA strand of a gene in three main phases: Initiation, Elongation and Termination.

INITIATION:
In prokaryotes, transcription is initiated when RNA polymerase binds to the pribnow box in the promoter. In eukaryotes, transcription is initiated when general transcription factors bind to the TATA box of the promoter. This facilitates the binding of RNA polymerase to the promoter. In eukaryotes, RNA Polymerase I transcribes rRNA, RNA polymerase II transcribes mRNA and RNA Polymerase III transcribes
tRNA. In both prokaryotes and eukaryotes, RNA polymerase unwinds both strands of DNA apart for elongation.

ELONGATION:
RNA polymerase adds ribonucleotides through complementary base-pairing rules: A – U; C- G. RNA Polymerase also catalyse the formation of phosphodiester bonds between ribonucleotides. mRNA is synthesized in the 5’
3’ direction. 5’
3’elongation of the mRNA strand occurs until the transcription stop site is reached.

TERMINATION:
In prokaryotes, transcription proceeds through a terminator sequence in the DNA.The transcribed termination (an RNA sequence) functions as the termination signal,causing the polymerase to detach from the DNA and release the transcript, which is available for immediate use as mRNA. In eukaryotes, RNA polymerase II transcribes a sequence on the DNA called the polyadenylation signal (AAUAAA) in the pre-mRNA. Then, at a point about 10 to 35 nucleotides downstream from the AAUAAA signal,proteins associated with the growing RNA transcript cut it free from the polymerase,releasing the pre-mRNA.

POST-TRANSCRITIONALMODIFICATION of pre-mRNA
In eukaryotes, mRNA undergoes post-transcriptional modification. A 5’ cap consisting of 7-methyguanosine is added to the 5’ end of pre-mRNA. This is followed by RNA splicing. Finally, a poly-A tail is added to the pre-mRNA. RNA splicing is carried out by spliceosomes, which is a complex of small ribonucleoproteins (snRNPs), at the specific splice sites. In this process, the noncoding introns are removed and exons are spliced together. The addition of poly-A tail is catalysed by poly-A polymerases. The 5’ cap and poly-A tail protects the mRNA from degradation by nucleases. Also, both are needed for ribosomal binding. The poly-A tail also directs the mRNA from the nucleus to the cytoplasm for translation.

TRANSLATION
Translation is the process where mRNA is used as a template for the synthesis of a polypeptide.

ACTIVATION OF AMINO ACIDS:
Aminoacyl-tRNA synthetase attaches the correct amino acid to its tRNA, based on its codon sequence.

INITIATION:
The small subunit of the ribosome recognizes and binds to the 5’ end of the mRNA. This is followed by the binding of initiator tRNA with amino acid methionine to the start codon, The initiator tRNA has the anticodon sequence UAC, complementary to that of the start codon, AUG. Finally, the arrival of the large subunit completes the translation initiation complex. GTP provides energy for the formation of this complex.

ELONGATION:
The initiator tRNA carrying methionine is at the P site. The next tRNA with an anticodon sequence complementary to that of the codon at the A site, binds to the Asite.GTP is required for codon-recognition. Peptidyl transferase catalyzes the transfer of amino acid on the P-site to the amino acid on the A-site, forming a peptide bond between them. This is followed by the shift of the ribosome by one codon downstream. The initiator tRNA is now at the E site, and the second tRNA is now at the
P site. GTP drives the translocation process. The tRNA on the E-site is now free to exit the ribosome. The A-site is now available for the next tRNA.

TERMINATION:
Elongation of the polypeptide continues until the ribosome reaches a termination codon (UAA, UAG or UGA). The A-site accepts a protein called a release factor instead of a tRNA at the stop codon. The release factor causes the addition of water instead of an amino acid to the polypeptide chain. This causes the hydrolysis of the polypeptide chain from the tRNA at the P-site, with the provision of energy by GTP.

Polyribosomes may trail along the same mRNA.

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TOPIC 2: DNA AND GENOMICS – Part 4

LEARNING OUTCOME

(d) Explain how a change in the sequence of the DNA nucleotide (gene mutation)
may affect the amino acid sequence in a protein, and hence the phenotype of the
organism e.g. sickle cell anaemia and cystic fibrosis. (Knowledge of substitution, addition, deletion and frameshift mutation may be required.)

ESSAY ANSWER

Sickle cell anaemia Mutation and effect on amino acid sequence and phenotype

Base substitution, where thymine is replaced with adenine in the beta haemoglobin gene at chromosome 11, where CTT which codes for polar
glutamic acid is substituted to CAT which codes for non-polar valine.

Glutamic acid and valine are amino acids with very different properties.
Glutamate is hydrophilic while valine is hydrophobic;

At low oxygen concentration, hydrophobic areas of different molecules would
stick together, resulting in HbS molecules polymerizing/ precipitating into
fibres. This will cause conformation changes to the tertiary structure of HbS
and affect the quarternary structure of the haemoglobin.

These long fibres distort the membrane of the red blood cell giving it its
distinct sickle shape.

Effect of sickle shape red blood cells

Sickle shaped cells blocked capillaries which will result in organ damage and
have a shorter lifespan due to increase in wear and tear.

Cystic fibrosis
Mutation and effect on amino acid sequence and phenotype

Cystic fibrosis is a genetic disorder affecting chromosome number 7, which is
caused by inheriting two recessive mutated alleles for Cystic Fibrosis
Transmembrane Conductance Regulator (CFTR).

CFTR is a transmembrane protein (located in the cell surface membrane) needed
for the regulation of diffusion of Cl− ions into and out of the epithelial cells.

Most cases of CF is caused by a mutation that results in the deletion of 3 base
pairs from the CFTR gene – codon number 508 in the mRNA sequence (F508).

As a result, the amino acid phenylalanine is missing at position 508 in the
protein. This in turn changes the 3D configuration of CFTR protein,
preventing the normal efflux of Cl- out from the cells, thus, Cl- ions accumulate in the cells.

Being negatively charged, the built-up of Cl- ions in the cells causes the influx of
Na+ ions into the cells to balance the negative charges
High concentration of ions inside the cell in turn prevents water from leaving the cells causing copious mucus production and cysts to develop in the lungs,
pancreas and liver

Effect of thick mucus

The thick mucus produced in CF patients clogs up the airways of the lungs, the
branches of the pancreatic duct and the bile duct from the liver into the gut.
This results in:

Repeated lung infections as mucus interferes with gaseous exchange in the
lungs.

Poor release of pancreatic enzymes as thick much clogs up the branches of
pancreatic duct.

Poor absorption of digested food as thick mucus clogs up the branches of bile
duct and pancreatic duct preventing the emulsification of fats and digestion of
carbohydrates.

Affected males are almost always infertile and females are frequently infertile.

CF sufferers have shorter than normal life expectancy. The shorter life
expectancy is due to their susceptibility to:

Bacteria lung infection especially pneumonia Development of enlarged liver / spleen. An inability to digest or absorb digested food products efficiently
The development of diabetes

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