B. Proteins Flashcards

1
Q

What is the RNA interference Pathway? (3)

A
  • RNAi is a biological process and is a sequence-specific (homology-dependent) gene silencing pathway.
  • It is initiated by double-stranded RNA (dsRNA).
  • RNAi is a method of gene regulation that works by silencing a gene
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2
Q
  • RNAi has 3 pathways:
A

1) microRNA (miRNA) pathway – This is the only important one as it is the only one found in humans.
2) Piwi-interacting RNA pathway
3) Endogenous small interfering RNA pathway.

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3
Q

Where is RNAi present?

A
  • RNAi is present in most multicellular eukaryotes, some unicellular eukaryotes but not in prokaryotes.
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4
Q

What is gene expression responsible for?

A
  • Gene expression is responsible for every aspect of a cell’s life (growth, division, death etc.) and since RNAi is a gene regulator it regulates all these processes. Examples include haematopoiesis (formation of blood cellular components), differentiation, tumour suppression, and developmental timing.
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5
Q
  • Antiviral defence and _____ silencing are also regulated by RNAi but are regulated by the Piwi and Endogenous pathways and are thus not present in humans.
  • RNAi was discovered in _____ by Andrew Fire and Craig C. Mello.
  • microRNA (miRNA) was discovered in _____.
A

transposon
1998
2001

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6
Q

What is the miRNA Pathway? (4)

A
  • miRNA is first transcribed (from RNA Polymerase II promoters) like any other gene and is first expressed as large RNA called pri-miRNA (primary microRNA).
  • Since mRNA (messenger RNA) is also transcribed from RNA Polymerase II, miRNA looks very similar to mRNA. It has a 5’ cap and a 3’ polyadenylated tail. They are both large sequences.
  • These large structures fold up on themselves to form hairpin loops/stem loops. This is because at physiological temperatures and physiological pH single-stranded RNA’s will fold up into secondary structures. tRNA is an example of this – it folds into the cloverleaf structure. The same is true of any RNA. They all fold into a secondary structure.
  • Pri-miRNA folds into hairpin loops (stem loops) and can have a single (monocistronic) or multiple (polycistronic) hairpin loops.
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7
Q

How is miRNA produced? (9)

A
  1. miRNAs are first transcribed as primary miRNA (pri-miRNA), which are usually produced from RNA polymerase II promoters. Pri-miRNAs have highly structured RNA sequences that fold upon themselves to form hairpin loops.
  2. Within the nucleus, the pri-miRNAs are processed by an enzyme complex called the microprocessor complex, which consists of Drosha and its dsRNA-binding partner DGCR8.
  3. This processing step releases hairpin loops or pre-miRNAs, which are approximately 70-80 nucleotides long.
  4. The pre-miRNAs are then exported to the cytoplasm through the action of exportin-5 (Exp-5 or XPO5)
  5. In the cytoplasm, the pre-miRNAs undergo a second processing step mediated by an enzyme called Dicer, along with its dsRNA-binding partner TRBP. Dicer processes the pre-miRNAs into miRNA duplexes.
  6. The miRNA duplexes then associate with the RNA-induced silencing complex (RISC), which contains Ago-2 (Argonaute-2).
  7. Within the RISC, one strand of the miRNA duplex, called the passenger strand, is removed.
  8. The remaining strand, known as the mature miRNA or guide strand, guides the RISC complex to complementary mRNA molecules.
  9. The guide strand recognizes specific target mRNA sequences through base pairing. The interaction between the guide strand and the mRNA can lead to the degradation of the mRNA if there is a perfect match or inhibit translation if there is an imperfect match.
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8
Q

Some important points
- We know that all mRNA is _______ from DNA.
- DNA has coding portions and non-coding portions.
- mRNA is a protein coding RNA, and thus is the only RNA transcribed from the coding portion of
DNA. All other RNAs (including miRNA) are transcribed from the ____-_____ portions of DNA.
- Drosha and Dicer both cleave the RNA and are called ______ III enzymes. When they cleave the RNA,
they both end up leaving 2 nucleotide long 3’ overhangs.

A

transcribed
non-coding
RNase

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9
Q

miRNA Dysregulation:

The miRNA pathway is part of normal human funtioning and can be very beneficial. When it is _________, however, it can cause severe problems such as cancer. An example of this is Chronic Lymphocytic Leukemia. When a 30 kb (kilobases) DNA deletion occurs on Chromosome ___, miR-15 and miR-16 (microRNA 15 and 16) are affected and this causes the cancer. This is because once the DNA portion that codes for these miRNA’s are deleted, they can no longer silence their targets. If their targets are not silenced, we get _____-_____ of cell proliferating and cell division markers, thus leading to more cell proliferation and growth, ultimately leading to cancer. Numerous cancers work in a similar way. Even though miRNA may not cause the cancer, it is still very important in the development of the cancer. We will focus specifically on cell _______ and cell division.

A

dysregulated
13
over-regulated
proliferation

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10
Q

The relationship between miRNA and cancer
There are two ways through which cancer can be caused due to miRNA dysregulation: (2)

A

1) Downregulation of miRNA.
A decrease in the expression of a miRNA targeting an oncogene leads to increased expression of the oncogene.

2) Upregulation of miRNA.
An increase in the expression of a miRNA targeting a tumour suppressor leads to decreased expression of the tumour suppressor.

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11
Q

1) Downregulation of miRNA.
A decrease in the expression of a miRNA targeting an oncogene leads to increased expression of the oncogene.

How does this process occur? (7)

A
  • Through mutation (a miRNA’s promoter might be mutated), less miRNA can be produced or though deletion of DNA as we saw in Chronic Lymphocitic Leukemia less or no miRNA may be produced. For whatever reason, less pri-miRNA is produced.
  • We now have less pri-miRNA produced, which leads to less pre-mRNA produced, which leads to less miRNA duplex produced, which leads to less entry into RISC and fewer guides being produced.
  • This means that there will be less targeting of the gene that is the target of the miRNA.
  • If that gene happens to be an oncogene (genes which stimulate cell growth and division) we will
    have less silencing, which means more expression of that gene. If the miRNA is downregulated, its
    target (such as an oncogene) is upregulated.
  • Since the oncogene is not silenced, we have increased cell growth and proliferation which leads to
    cancer.
  • miR-15 and miR-16 are frequently downregulated in B-cell Chronic Lymphocytic Leukemia.
  • If the downregulation of a miRNA causes cancer, its target must be an oncogene.
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12
Q

2) Upregulation of miRNA.
An increase in the expression of a miRNA targeting a tumour suppressor leads to decreased expression of the tumour suppressor.

How does this process occur? (7)

A
  • Through mutation (a miRNA’s promoter might be mutated) or duplication of DNA, more miRNA can be produced. For whatever reason, we have more pri-miRNA produced.
  • More pri-miRNA = more pre-miRNA = more miRNA duplex = more acivation of RISC = more targeting.
  • If the target happens to be a tumour supressor (suppress tumours by inhibiting growth or cell division), the tumour suppressor will be silenced. We say the tumour supressor is downregulated.
  • If the tumor suppressor is silenced, it can no longer downregulate cell growth and cell division.
  • Since the tumour suppressor is silenced, we have increased cell growth and proliferation which
    leads to cancer.
  • miR-17~92 is frequenlty upregulated in B cell lymphomas.
  • If upregulation of a miRNA causes the cancer, its target must be a tumour suppressor.
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13
Q

Viruses have been shown to encode their own miRNA (miRNA found in their own genome) in humans and exploit the RNAi and miRNA pathway. Viruses are very good at exploting the cell machinery of their host. They can exploit the transcription and translation of a cell they infect. Viruses can hijack the miRNA pathway in 2 ways:

A

1) They target the host’s genes.
2) They target their own genes.

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14
Q

Viruses and miRNA
Viruses have been shown to encode their own miRNA (miRNA found in their own genome) in humans and exploit the RNAi and miRNA pathway. Viruses are very good at exploting the cell machinery of their host. They can exploit the transcription and translation of a cell they infect. Viruses can hijack the miRNA pathway in 2 ways:
1) They target the host’s genes.

How does this work? (2)

A
  • Viruses have been shown to inhibit cellular factors involved in innate or adaptive immunity. They can encode miRNA’s that downregulate the natural killer cell ligand MICB (involved in the immune response). Viruses that do this include Human Cytomeglovirus, Kaposi’s Sarcoma-associated Herpesvirus, and Epstein-Barr Virus.
  • Epstein-Barr virus encoded miRNA also downregulates the pro-apoptotic (apoptosis is programmed cell death – cells are programmed to die when they are infected) protein PUMA.
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15
Q

Viruses and miRNA
Viruses have been shown to encode their own miRNA (miRNA found in their own genome) in humans and exploit the RNAi and miRNA pathway. Viruses are very good at exploting the cell machinery of their host. They can exploit the transcription and translation of a cell they infect. Viruses can hijack the miRNA pathway in 2 ways:
2) They target their own genes.

How does this work? (2)

A
  • Viruses can encode miRNA that downregulates the expression of viral proteins. What this means is that viruses such as Herpes Simplex Virus 1 encodes miRNA that target early transactivators such as ICP0 and ICP4, which are needed for the virus to begin reproduction. This causes the virus to stay latent and not replicate. In this latent state, it is hidden and the immune system cannot recognise and eliminate it.
  • When you are healthy and your immune system is strong, the virus stays hidden. When you become sick or stressed and your immune system is compromised, the virus turns off its miRNA and begins reproduction to infect you.
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16
Q

Exploting the RNAi pathway to our advantage
Exploting the pathway can be very useful to silence any gene you want silenced. In theory, you would produce an miRNA complementary to the gene you want silenced. There are 2 systems to do this:

1) Synthetic systems
2) Expressed systems

A
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17
Q

Why is exploiting the RNAi pathway useful?

A

Exploiting the RNAi pathway is used in functional genomics where a gene is silenced and the effect is observed. This tells us what that gene is resposible for. It is also used in gene therapy where nucleic acid sequences like these mimics are produced to bring about a therpeutic effect, such as a guide against a gene that belongs to a virus to eliminate the virus or a gene involved in cancer development to suppress the cancer.

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18
Q

What is hybridisation? (2)

A

Hybridisation is simply when the 2 complementary sequences of DNA come together – they hybridise and form Hydrogen bonds. When DNA is heated, it will denature and separate. When it is cooled, it will join, and we call this hybridisation. The interaction is maintained through 3 Hydrogen bonds between G and C and 2 Hydrogen bonds between A and T/U.

Hybridisation is the establishment of a sequence-specific interaction between two complementary nucleic acid sequence

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19
Q

First technique: FISH – Fluorescence in situ Hybridisation

  • Probes are simply pieces (fragments) of DNA or RNA manufactured in a lab. They can be labelled fluorescently or radiocatively. Fluorescent _____ have a bright colour attached to them so that we can keep track of the probe when using it.
  • ____-_______ probes bind to a particular chromosomal region. This is useful when trying to determine on which chromosome a particular gene is located.
  • As the F in FISH suggests, FISH makes use of fluorescent probes.
  • In Situ means in its original place.
A

probes
Locus-specific

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20
Q

What is FISH? (2)

A
  • FISH is a molecular cytogenic (at the chromosome level) technique that uses fluorescent signals
    and nucleic hybridisation to detect a sequence of interest.
  • Essentially, FISH is a technique where we make a specific fragment of DNA in a lab, and then insert
    it into a DNA sample to see if it binds. If it binds to the DNA sequence, we know that a specific DNA
    sequence is present.
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21
Q

How can FISH be used? (3)

A
  • FISH can be used for the identification of presence and location of nucleic acids within metaphase
    chromosomes, interphase nuclei, fixed tissues and cells in culture.
  • FISH is used in conjunction with flourescence microscopy (to see the results). If the fluorescent
    probe hybridises to the DNA sample, the fluorescent probe-DNA sequence can be studied under a fluorescent microscope. Note that it will only bind if the probe and sample sequence have high sequence complementarity.
  • There are different types of samples: formalin-fixed parrafin embedded tissues or fixed cell suspension.
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22
Q

What is the process of FISH? (5)

A

a) A probe and sample are prepared.
b) The probe is labelled (very important step)
c) The probe and sample are denatured (to make them single-stranded so they can bind).
d) The probe is hybridised to the sample.
e) The sample is imaged.

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23
Q

What is the process of FISH? (5)

A

a) A probe and sample are prepared.
b) The probe is labelled (very important step)
c) The probe and sample are denatured (to make them single-stranded so they can bind).
d) The probe is hybridised to the sample.
e) The sample is imaged.

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24
Q

FISH
- A probe is designed against a a DNA sequence of interest.
- It is then labelled through: (2)

A

1) Direct labelling using flourophores, which are fluorescent molecules

2) Indirect labelling using haptens, which are small molecules against which a fluorescently labelled antibody can be raised. The probe is essentially flourescenlty labelled, but with extra steps.

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25
Q

FISH:

  • Samples are prepared on glass slides.
  • The sample is imaged by _______ microscopy.
  • If a signal is observed, the probe is present and so is the sequence of interest.
  • If there is no signal observed, it means the probe did not ______, which means the probe is
    absent and so is the sequence of interest.
  • The position (on which chromosome) can be determined through FISH, as the name suggests (in
    situ).
A

fluorescence
hybridise

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26
Q

We can perform FISH when the sample is in interphase or metaphase.
- Interphase:

A
  • Interphase:
    DNA is less condensed and not duplicated.
    The results of FISH perfomed during interphase appear as depicted →
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27
Q

We can perform FISH when the sample is in interphase or metaphase.
- Metaphase:

A

Metaphase is induced.
DNA is duplicated in metaphase and exists as sister chromatids.
The DNA is highly condensed in metaphase.
The results of FISH performed during metaphase appear as depicted →

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28
Q

1) Preparing a Probe
There are many ways to prepare a probe, but we focus on Nick translation, which is where we make nicks or cuts in DNA. You then use nuceotides that are labelled to fill in the gaps and thus your probe ends up being labelled.

What is the process is preparing a probe? (4)

A
  • Begin with a double stranded probe.
  • Introduce an enzyme called DNase I which
    will randomly cut the DNA (make nicks).
  • Introduce DNA Polymerase to fill in the
    nicks with modified dNTPs (deoxynucleotide triphosphates). DNA Polymerase extends chains (5’ to 3’) from nicks with modified dNTPs.
  • When the probe gets filled, it incoporates these modified nucleotides, and the probe thus gets fluorescently labelled.
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29
Q

1) Preparing a Probe
There are many ways to prepare a probe, but we focus on Nick translation, which is where we make nicks or cuts in DNA. You then use nuceotides that are labelled to fill in the gaps and thus your probe ends up being labelled.

What is the process is preparing a probe? (4)

A
  • Begin with a double stranded probe.
  • Introduce an enzyme called DNase I which
    will randomly cut the DNA (make nicks).
  • Introduce DNA Polymerase to fill in the
    nicks with modified dNTPs (deoxynucleotide triphosphates). DNA Polymerase extends chains (5’ to 3’) from nicks with modified dNTPs.
  • When the probe gets filled, it incoporates these modified nucleotides, and the probe thus gets fluorescently labelled.
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30
Q

2) Denature Probe and Sample
Once the probe and sample have been prepared, they need to both be heated to denature them.
Chemicals can also be used to denature them. This is to make them both _____ _______.

A

single stranded

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31
Q

3) Hybridisation:

The single-stranded DNA sample and fluorescently labelled probe must now be hybridised together.
- The _______ probe is applied to the slide/tissue.
- The slide is incubated to allow hybridisation.
- A probe that has not hybridised must be _____ off.

A

denatured
washed

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32
Q

4) Image sample
The sample must now be imaged by fluorescence microscopy.
- A laser is used to excite the_______ (a particle which has the ability to absorb light at a certain wavelength and then emit that light at another wavelength), which then emits the fluorescent signal.
- Fluorescent hybridisation signals are then detected.

A

flourophores

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33
Q

What are the applications of FISH? (4)

A
  • Identification of chromosomal abnormalities.
  • Helps with gene mapping (localisation of gene in the genome – chromosome location), analysis of
    chromosome structural aberrations (things away from the norm), toxicological studies, and ploidy
    determination.
  • Cancer diagnostics – deletion of tumour suppressors, duplication of an oncogene, and translocation
    of two genes.
  • Genetic abnormality testing, such as monosomy (being one chromosome short) or trisomy (havine
    once extra chromosome).
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34
Q

Examples:

You have performed FISH using a red and a green fluorophore to identify 2 sets of genes very close to each other on the chromosome.

  • Seeing a yellow signal is normal because the genes are so close to each other. The red and green signal are super-imposed on each other forming a yellow signal. Two yellow signals will be shown because humans are normally _______
  • ______ is having more than 2 sets of chromosomes, such as being _____ (having 69 chromosomes).
  • Aneuploidy is having an extra chromosome or have a chromosome less. Down’s Syndrome is when a person has 2 of each chromosome (as is normal), except chromosome ____ , of which they have 3.
A

Diploid
Polyploid
Triploid
21

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35
Q

Second technique: PCR – Polymerase Chain Reaction PCR is divided in 3 steps:

A

1) Denaturation (950C)
2) Annealing/ Hybridisation (~55oC)
3) Exension/ Elongation (68-72oC)

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36
Q

PCR
1) Denaturation (950C):

A
  • Double-stranded DNA is heated to 95 degrees to disrupt hydrogen bonds between complementary bases to form single-stranded DNA.
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37
Q

PCR
2) Annealing/ Hybridisation (~55oC): (3)

A
  • Primer (short fragments of DNA complementary to target sequence) anneals the single-stranded DNA.
  • The difference between this and FISH is that here, 2 primers on each side of the region you are trying to amplify are needed, and they are not fluorescently labelled.
  • See the diagram alongside. One primer is needed to anneal to the one DNA strand upstream and one primer is needed to anneal to the other strand downstream.
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38
Q

PCR
3) Exension/ Elongation (68-72oC):

A
  • Polymerase (as stated by P in PCR) extends primers by adding dNTPs in a template dependent fashion, forming new strands of DNA in both directions.
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39
Q

PCR
These steps are repeated ___ to ___ times to get an exponential amplification of the products. It is hence a chain reaction, as stated in the CR of PCR.

A

30
50

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39
Q

PCR
These steps are repeated ___ to ___ times to get an exponential amplification of the products. It is hence a chain reaction, as stated in the CR of PCR.

A

30
50

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40
Q

What are the components of a PCR reaction? (5)

A
  • Enzyme – Thermostable DNA Polymerase
  • Buffers and MgCl2 – Make the environment condusive for the PCR to occur by providing the correct ions to sustain the reaction and for the enzyme to work optimally.
  • Primers – 2 primers that hybridise at each end of the fragment to be amplified.
  • dNTPs (deoxynucleotide triphosphates) – These are the bases that DNA Polymerase will add to the
    growing strand of DNA.
  • Template DNA.
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41
Q

What is PCR Automation? (3)

A
  • PCR is now fully automated. It has thermal cyclers that do the temperature changes automatically.
  • Most enzymes (such as DNA Polymerase) denature at high temperatures. Therefore after each
    cycle, new enzyme must be added and this causes problems.
  • A major breakthrough for PCR was the discovery of Thermostable DNA Polymerases. These are DNA
    Polymerases found in bacteria called Extremeophiles that are found in places that have high temperatures, such as thermal vents on the ocean floor or hot springs. The Polymerase is thus used to hot temperatures and does not denature at high temperatures, making it useful to PCR.
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41
Q

What is PCR Automation? (3)

A
  • PCR is now fully automated. It has thermal cyclers that do the temperature changes automatically.
  • Most enzymes (such as DNA Polymerase) denature at high temperatures. Therefore after each
    cycle, new enzyme must be added and this causes problems.
  • A major breakthrough for PCR was the discovery of Thermostable DNA Polymerases. These are DNA
    Polymerases found in bacteria called Extremeophiles that are found in places that have high temperatures, such as thermal vents on the ocean floor or hot springs. The Polymerase is thus used to hot temperatures and does not denature at high temperatures, making it useful to PCR.
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42
Q

What is the PCR mechanism?

A

PCR yields a great amount of product.
- If you know the location of the primers, you can determine the size of the PCR product since it will be from the beginning of one primer till the end of the other primer.

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43
Q

What is the PCR mechanism?

A

PCR yields a great amount of product.
- If you know the location of the primers, you can determine the size of the PCR product since it will be from the beginning of one primer till the end of the other primer.

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44
Q
  • If you start with one double-stranded DNA sequence at the beginning of the reaction, you have 2 dsDNA at the end of the first cycle. Remember that this is repeated ___-____ times.
  • In the second cycle, you are starting with __ dsDNA, and you will yield __ dsDNA at the end of the cycle.
  • This continues for 30 times or more, doubling at each cycle.
  • This is what is meant by exponential _______. At the 30th cycle, we have over 1 billion copies.
  • This is very useful for _____ scenes, where small amounts of DNA can be amplified.
  • Covid-19 tests are PCR tests.
A

30-50
2
4
Amplification
Crime

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45
Q

A formula can be derived to calculate number of copies produced.

𝑦 = 𝑎 × 2^𝑥 where:

A
  • 𝑦 is the copy number produced.
  • 𝑎 is the starting copy number.
  • 𝑥 is the number of cycles completed.
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46
Q

Detecting PCR Products:

DNA is ultimately the end product of PCR, but it is not visible to the naked eye. A process called Gel electrophoresis is used. This is the movement of charged particles in a fluid under the influence of an electric field.

How are PCR products detected? (3)

A
  • PCR products are placed in a buffer in a gel.
  • An electric current is passed through, and cations migrate toward the cathode and anions toward
    the anode.
  • DNA is negatively charged due to its Phosphate group, and will migrate from the negative to the
    positive electrode.
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47
Q

What is Agarose Gel Electrophoresis? (3)

A
  • Not all DNA migrates at the same rate.
  • Larger DNA molecules do not go through the matrix of the gel as quickly as smaller DNA fragments.
  • Thus, when we load pieces of DNA on an
    Agarose gel, we can effectively separate DNA based on size, with the smallest being furthest away from the wells where you loaded them, and the largest being closer.
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48
Q

How do we get different sizes of DNA in a PCR reaction? (4)

A
  • Where primers bind determine the size of the PCR product.
  • Lets say the first primer binds to bases 100
    to 119 on the one strand and the other
    primer binds to bases 481 to 500 on the other.
  • Effectively, the product will be from 100 to 500, which is 400 bps long. This Polymerase, however, tends to add one extra base, and thus 401. This extra base is usually an A added at the tail – recall the Polyadenylated tail added by RNA Polymerase II.
  • If you want to check that your PCR went well, run a sample of the completed product on an Agarose gel under electrophoretic conditions, and with the help of a special dye that stains DNA, you can visualise your PCR products under UV light and make sure that the size is correct.
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49
Q

What are the PCR products?

A

On the left we have a DNA ladder consisting of various fragments of DNA length. This helps determine size of PCR prodcuts.

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50
Q

What are the applications of PCR? (6)

It is argued that PCR is the most versatile molecular tool to date.

A
  • Detection of heriditary diseases.
  • Forensic DNA detection.
  • Identification of genetic fingerprints (paternity testing, forensic testing).
  • Cloning of genes.
  • Identification of transgenic plants.
  • Sequencing.
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51
Q

What is the process of RT-PCR? (6)

A
  • This uses RNA as a starting template instead of DNA, as is done with PCR.
  • Thus, an extra step is required to convert the RNA to cDNA (complementary DNA) and this is done
    with the help of an enzyme called Reverse Transcriptase.
  • The product is double-stranded cDNA.
  • This is useful when you need to determine the abundence of specific RNA transcripts present in a
    cell or tissue or to measure gene expression.
  • Note that the presence of a gene does not mean that it is being transcribed, and so measuring the
    amount of specific RNA is a better indicator of whether the gene is expressed.
  • There is another PCR method called Real-Time PCR, and it is different to RT-PCR. The abbreviation for Real-Time PCR is Q-PCR. Q-PCR uses flourescence to monitor the amplification reaction in real
    time.
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52
Q

What are Restriction Enzymes?

A

Restriction enzymes cut DNA at specific sequences and are called Restriction Endonucleases.
- Restriction Enzymes/ Restriction Endonucleases act as molecular scissors.
- They each recognise a different sequence and only cut the DNA where that sequence is present.

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53
Q

Restriction Enzymes

An example is EcoRI (the I is a Roman
numeral for 1):
- EcoRI has cut the DNA at the ______ sequence shown.
- Results in two fragments of DNA with _____ ends.
- Look at how EcoRI has cut. It has not done so cleanly between the six bases. EcoRI left a _____ tail on the left and a complementary _____ tail on the right, and these are known as sticky ends.
- Not every enzyme cuts with sticky ends. Some cut and leave no tail.
- There are many Restriction Enzymes – some leave sticky ends and some leave _____ ends.

A

recognition
sticky
TTAA
AATT
blunt

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54
Q

Structure of an Amino Acid: (5)

A
  • Amino acids are organic molecule.
  • It consists of a Carbon attached to an Amino group (H3N+), a Hydrogen (H), a Carboxyl group (COO), and a variable side chain (R).
  • The variable side chain is what gives an amino acid its specificity because each amino acid has a different variable side chain.
  • Glycine is the simplest amino acid because its variable side chain is just one Hydrogen.
  • The way in which variable sidechains interact with each other determines how a protein folds, and
    how a protein folds determines its function.
55
Q

What are Hydrophobic (water-hating) Amino Acids? (4)

A
  • Hydrophobic amino acids are non-polar.
  • Hydrophobic Amino Acids hate water and thus tend to be found away from the outside of the protein that encounters water/liquids.
  • Hydrophobic amino acids are the largest amino acids.
  • There are two types of Hydrophobic Amino Acids:
56
Q

1) Aliphatic Hydrophobic Amino Acids:

A

These are called Aliphatic because they consist of a variable side chain consisting of Carbon and Hydrogen.
You do not need to know the structure of each amino acid →

57
Q

2) Aromatic Hydrophobic Amino Acids:

A

These are called Aromatic because they have a benzene ring in their variable side chains.
You do not need to know the structure of each amino acid →

58
Q

2) Aromatic Hydrophobic Amino Acids:

A

These are called Aromatic because they have a benzene ring in their variable side chains.
You do not need to know the structure of each amino acid →

59
Q

What are hydrophilic (water-loving) Amino Acids? (2)

A
  • Hydrophilic amino acids are charged.
  • Hydrophilic amino acids tend to be found on the surface of proteins as they can interact with water or fluid.
60
Q
  • There are three types of Hydrophilic Amino Acids. Two of these are charged: (3)
A

1) Basic Hydrophilic Amino Acids
2) Acidic Hydrophilic Amino Acids
3) Polar Hydrophilic Amino Acids

61
Q

1) Basic Hydrophilic Amino Acids:

A

These have a positively charged variable side chain.
An NH2 is found in their side chain.
You do not need to know the structure of each amino acid →

62
Q

2) Acidic Hydrophilic Amino Acids:

A

These have a negatively charged variable side chain.
A COO is found in their side chain.
You do not need to know the structure of each amino acid →

63
Q

2) Acidic Hydrophilic Amino Acids:

A

These have a negatively charged variable side chain.
A COO is found in their side chain.
You do not need to know the structure of each amino acid →

64
Q
  • The last type of Hydrophilic Amino Acids is Polar:

3) Polar Hydrophilic Amino Acids:

A

These amino acids are uncharged.
There are two types of Polar Hydrophilic Amino Acids:
1) Polar Hydrophilic Amino Acids containing an Amide group. H2N – C = O
2) Polar Hydrophilic Amino Acids containing a Hydroxyl Group. OH

65
Q
  • The last type of Hydrophilic Amino Acids is Polar:

3) Polar Hydrophilic Amino Acids:

A

These amino acids are uncharged.
There are two types of Polar Hydrophilic Amino Acids:
1) Polar Hydrophilic Amino Acids containing an Amide group. H2N – C = O
2) Polar Hydrophilic Amino Acids containing a Hydroxyl Group. OH

66
Q

Other Amino Acids:
Sulphur containing R groups
- These amino acids have Sulphur in their ______ side chains.

A

variable

67
Q

What is proline? (2)

A
  • This amino acid has a cyclic sidechain, but its not a Benzene ring so it is not an Aromatic Amino Acid.
  • This amino acid’s side chain is made of only Carbons and Hydrogens, which normally would make it an Aliphatic Amino Acid, but it is in a ring and thus cannot be Aliphatic.
68
Q

How are peptide bonds formed? (3)

A
  • Two amino acids are linked together through peptide bond formation.
  • The peptide bond is formed through Dehydration Synthesis, and we have loss of a water molecule.
  • TakenoteoftheC=OandN–H;itwillbeusefullater.
68
Q

How are peptide bonds formed? (3)

A
  • Two amino acids are linked together through peptide bond formation.
  • The peptide bond is formed through Dehydration Synthesis, and we have loss of a water molecule.
  • TakenoteoftheC=OandN–H;itwillbeusefullater.
69
Q

A sequence of amino acids:
- Note all the different variable side chains. It is the interaction of these side chains with each other which determines how a protein folds, and how a protein folds determines its function.
- Proteins are synthesized from the __-____ to the C-terminal (carboxy terminal).
- Note that Lysine (Lys) is _____ charged, and Glutamic acid (Glu) is _____ charged. They will
attract each other, and this affects structure.

A

N-terminal
Positively
Negatively

70
Q

What is the genetic code?

A
  • The genetic code is the specific sequence of nucleotides in mRNA that determines what amino acid is required through complementary base pairing with the anticodons on tRNA’s. Hence, the genetic code determines polypeptide amino acid sequence.
71
Q
  • The nucleotides of mRNA are organized into code words called _____.
  • There are 20 different amino acids in the human body, and if each mRNA codes for a specific amino acid, there must be at least 20 mRNA codons as well.
  • mRNA is made up of 4 different _____ (A, U, G, and C).
A

codons
nucleotides

72
Q

What is each codon made of?

A
  • Each codon is made of 3 nucleotides and is called a triplet code (such as AUG), and thus means that there are 64 possible mRNAs. Think about it. If codons were made of 2 nucleotides, we would only have 16 possible combinations, but there are 20 amino acids, and so we need more than 16 codons.
73
Q

What is each codon made of?

A
  • Each codon is made of 3 nucleotides and is called a triplet code (such as AUG), and thus means that there are 64 possible mRNAs. Think about it. If codons were made of 2 nucleotides, we would only have 16 possible combinations, but there are 20 amino acids, and so we need more than 16 codons.
74
Q
  • UAA, UGA and UAG do not code for anything. Recall that they are ____ codons (chain terminators).
  • AUG codes for Methionine and recall that it is a ____ codon.
A

stop
start

75
Q
  • There are thus 61 codons that code for 20 amino acids, thus making the genetic code _____ (each amino acid is coded for by 2 or more different codons).
  • Codons are ______– the same codon will always code for the same amino acid.
  • Codons are non-overlapping, non-punctuated, and _____.
A

degenerate
unambiguous
universal

76
Q

What is translation? (3)

A
  • Translation is protein synthesis mediated by the interplay of mRNA, tRNA, activating enzymes, protein factors and ribosomes attached to endoplasmic reticulum (ER) or free in the cytoplasm.
  • Translation begins with the addition of COOH (carboxyl) group of an amino acid to the 3’ end of a specific tRNA.
  • The amino acid is added to the tRNA by an enzyme called Aminoacyl tRNA Synthetase. There are 20 amino acids, and each one has its own Aminoacyl tRNA Synthetase. There are thus 20 Aminoacyl tRNA Synthetases. Each one of these enzymes recognise all tRNA’s for its specific amino acid.
77
Q

Ribosomes
- Recall that a ribosome has a large and small subunit. When these are together, we get three sites formed, namely and _ site, a _ site, and an _ site. See below:

A

A
P
E

78
Q

The process of Translation:
1) Initiation =(6)

A
  • The 5’ end of mRNA binds to the small ribosomal subunit.
  • The ribosome moves along the mRNA until the initiator codon (AUG) is in the P site.
  • A tRNA anticodon complementary to AUG binds (this is UAC).
  • Methionine is the amino acid coded for by the AUG codon or UAC anticodon.
  • Methionine is the starting amino acid of all proteins but is sometimes removed after translation.
  • Eukaryotic initiation factors (eIFs) regulate the
    binding of the large ribosomal subunit to the small ribosomal subunit.
78
Q

The process of Translation:
1) Initiation =(6)

A
  • The 5’ end of mRNA binds to the small ribosomal subunit.
  • The ribosome moves along the mRNA until the initiator codon (AUG) is in the P site.
  • A tRNA anticodon complementary to AUG binds (this is UAC).
  • Methionine is the amino acid coded for by the AUG codon or UAC anticodon.
  • Methionine is the starting amino acid of all proteins but is sometimes removed after translation.
  • Eukaryotic initiation factors (eIFs) regulate the
    binding of the large ribosomal subunit to the small ribosomal subunit.
79
Q

The process of Translation:
- Elongation Step 1 = (3)

A
  • After initiation, the A site of the ribosome was empty.
  • Now, a tRNA with an anticodon complementary to the mRNA codon (now in the A site) binds.
  • This requires GTP (Guanine Triphosphate).
80
Q

The process of Translation:
- Elongation Step 2 = (3)

A
  • Elongation is the process by which amino acids are added one after the other to from the chain of amino acids or polypeptide of protein.
  • A ribozyme called Peptidyl Transferase forms a peptide bond between the amino acid in the A site (which was Methionine – the initiatal amino acid) and the amino acid in the P site. These 2 amino acids (dipeptide) now bound to each other are on the tRNA in the A site and the tRNA in the P site that had the Methionine has no amino acid and is thus uncharged.
  • This process is a dehydration reaction and results in the release of water.
81
Q

The process of Translation:
- Elongation Step 3 = (5)

A
  • The large ribosomal subunit translocates (shifts) to the right, and then the small ribosomal subunit translocates to the right. This requires GTP.
  • The dipeptide is in the P site.
  • The tRNA without an amino acid is in the E site (Exit site) and
    is released.
  • The A site is empty.
  • Eukaryotic elongation factors (eEFs) regulate this process.
82
Q

Elongation:
- The process continues.
- A tRNA complementary to the mRNA will bind in the __ site.
- A peptide bond forms between the new amino acid in the A site and the and the existing ______ in the P site, forming a _______ in the A site.
- Ribosome shifts to the right, and the tripeptide is in the P site and the uncharged tRNA with no
amino acid is in the __ site and is released.
- A site is empty (has no tRNA) and a new corresponding tRNA will bind.
- This continues over and over.

A

A
dipeptide
tripeptide
E

83
Q

The process of translation:
Termination = (7)

A
  • Termination occurs when a stop codon (UAG, UGA, UAA) appears in the A site.
  • Stop codons do not code for an amino acid because no tRNAs recognise stop codons. tRNA does not
    bind to stop codons.
  • The stop codon is recognised by a release factor.
  • The release factor binds to the stop codon and adds H2O (water) to the end of the polypeptide
    chain.
  • The polypeptide is released from the P site, and ribosome subunits dissociate from each other.
  • Remember that only uncharged tRNA’s with no amino acid exit through the E site.
  • Eukaryotic release factors (eRFs) regulate this process.
84
Q

What are polyribosomes? (4)

A
  • Multiple ribosomes read the same mRNA strand at the same to produce many proteins at once.
  • These polyribosomes (multiple ribosomes) can
    be present in the cytoplasm or on the
    endoplasmic reticulum (ER).
  • If a protein is needed in the cytoplasm, it will be made in the cytoplasm.
  • If a protein is required to be secreted or act as a membrane receptor or is present in the lysosome, it will be made directly in the ER.
85
Q

A Summary of Transcription and Translation:

A
86
Q

How are proteins formed in the endoplasmic reticulum? (8)

A
  • A short signal sequence on the newly formed peptide (the short sequence of amino acids still growing) directs the mRNA and attached ribosome to ER.
  • The ribosome remains in the cytoplasm on the ER and produces the growing polypeptide into the ER (the growing polypeptide passes into the cisterna of the ER).
  • The signal sequence initially remains attached to the receptor but is soon clipped off by an enzyme.
  • As the protein synthesis continues, sugar or carbohydrate groups may be added to the protein.
  • Once the protein is completed, it is released from the ribosome into inside the ER and folds into its 3-D conformation, and this is aided by molecular chaperones.
  • Once in its mature form, vesicles containing the protein will bud off from the endoplasmic
    reticulum. These are known as transport vesicles.
  • Transmembrane proteins (proteins that are found on the membranes of cells) remain embedded in the membrane – they are not released into the ER but remain on the ER membrane. This portion of
    the ER membrane may be pinched off into a transport vesicle as well.
  • The transport vesicles travel to the Golgi apparatus, where further processing and sorting of the proteins occur.
87
Q

Processing of Proteins in the Golgi Apparatus
The Golgi Apparatus consists of stacked and flattened membranous sacs shaped like hollow dinner plates called cisternae, associated with many tiny membranous vesicles.

  • The transport ______ travels from the ER through the cytoplasm to the Golgi apparatus and fuses with the Golgi apparatus at its cis face.
  • The proteins from the transport vesicles can now be found in the ______ of the Golgi apparatus.
  • Newly formed vesicles become transport vesicles and they either become secretory vesicles, membrane vesicles, or ______ vesicles.
  • The major function of the Golgi apparatus is to modify, concentrate, and package the proteins
    made at the rough ER.
A

vesicle
lumens
lysosomal

88
Q

There are thousands of proteins in our bodies. They may be sorted into:

A
89
Q

What are proteins?

A

Proteins are the most abundant organic molecule in living systems. Recall that there are 20 different amino acids and the only difference between each of them is their variable side chain.

90
Q
  • The beginning of a protein is called the amino end/______.
  • The end of a protein is called the ______ end/terminus.
  • Proteins are produced as linear molecules.
    When they fold, the hydrophobic (non-
    polar) side chains move to the centre where
    they will be away from water and the hydrophilic (polar) side chains move the outside where they will encounter ______ solutions.
  • Peptides have __ to 30 amino acids.
  • Proteins are made of __ to 1000 amino acids.
A

terminus
carboxyl
aqueous
2
50

91
Q

There are 4 levels of protein structure:

A

1) Primary structure
2) Secondary structure
3) Tertiary structure
4) Quaternary structure

92
Q

There are 4 levels of protein structure:
1) Primary structure: (2)

A
  • The sequence of amino acids in the polypeptide chain as determined by mRNA codons. Contains peptide bonds.
  • Ultimately, the unique sequence of any protein is determined by the gene that encodes the protein. Any change in the gene sequence may lead to a different amino acid being added to the polypeptide chain, causing a change in protein structure and function.
93
Q

2) Secondary structure
Recall that when amino acids bind, we get a __-___ and C=O.

  • Secondary structure results from the hydrogen-bonding between the N-H and C=O groups in the ________ backbone.
  • This causes ______ and there are two regular folding patterns found as dictated by amino acids in primary sequence – secondary structure.
A

N-H
Polypeptide
Folding

94
Q

1) Secondary structure: α helix (2)

A
  • In α helix, a single polypeptide chain twists to form a rigid cylinder.
  • Hydrogen bonds within the polypeptide chain are spaced four amino acids apart.
95
Q

2) Secondary structure: β pleated sheet

A
  • Rigid planar structures with hydrogen bonds between amino acids in adjacent strands form.
  • β sheets can form either from neighbouring polypeptide chains that run in the same orientation (parallel chains) or from a polypeptide chain that folds back and forth upon itself, with each section of the chain running in the opposite direction of its immediate neighbours (antiparallel chains).
96
Q

There are 4 levels of protein structure:
3) Tertiary structure = (5)

A
  • Tertiary structure refers to the unique 3-D structure of a polypeptide.
  • The structure is caused by the chemical interactions between amino acid side chains.
  • These interactions include disulphide bonds, hydrogen bonds, ionic bonds, hydrophobic interactions, and disulphide bonds.
  • Forms by secondary structures folding with each other.
  • Tertiary structure is the highest level of organisation in monomeric proteins (monomeric means molecules that can combine with other to form polymers).
97
Q

There are 4 levels of protein structure:
4) Quaternary structure = (3)

A
  • Quaternary structure describes the number and relative positions of subunits in multimeric proteins (proteins composed of several subunits).
  • The image showing a quaternary protein has 4 subunits and is called a tetramer:
  • If the quaternary protein was composed of three subunits, it would be called a trimer.
98
Q

Protein Structure Summary:

A
99
Q

What are protein domains? (3)

A
  • A protein domain is a portion of a protein that is composed of 40-350 amino acids and has tertiary structure of its own (meaning it is biologically active) and is independently folded.
  • It is the modular unit from which many larger proteins are constructed.
  • Different domains are often associated with different
    functions.
100
Q
  • Small proteins have a single domain, and larger
    proteins can contain as many as several dozen domains, usually connected to each other by short, relatively unstructured lengths of polypeptide chain.

Provide an example.

A
  • An example is the SH3 (SRC Homology-3) domain. SH3 domains are small domains of around 50 amino acids residues that are involved in protein-protein interactions. SH3 domains have a characteristic 3D structure, as seen in the image alongside→
    SH3 domains occur in a diverse range of proteins with
    different functions, including adaptor proteins, phospholipases and myosins. MCK is a cytoplasmic adaptor protein that contains multiple SH3 domains. It is involved in transducing signals from growth factor receptor Tyrosine Kinase (recall from the first page that tyrosine kinase is involved in cell signalling) to downstream signal recipients.
  • Multiple domains can occur in a single protein, and these can be the same or different.
101
Q

What is Post Translational Modification (PTM)?

A
  • PTM refers to the modification of proteins after they have been synthesized resulting in changes in cellular functions, structure and can impact the interactions of proteins. We can call proteins dynamic since they can be modified to elicit changes in cellular functions.
102
Q
  • The Human Proteome (which refers to the proteins that can be expressed by a cell, tissue or organism) is more complex than the Human Genome (which refers to the complete set of genes of an organism). This is because single genes can code for multiple proteins, and hence there are more proteins expressed than genes. PTMs further increase this ______.
  • PTMs _____ localisation, activity and interactions with other cellular molecules.
A

diversity
regulate

103
Q
  • PTMs can occur at any time during the life cycle or biosynthesis of a protein: (2)
A

1) PTMs can occur after translation to mediate proper folding of the protein. This can increase stability of the protein or help localise the protein to a distinct cellular compartment.

2) PTMs can occur after protein folding and localisation. This can alter the biological activity of the protein.

104
Q
  • PTMs can occur at any time during the life cycle or biosynthesis of a protein: (2)
A

1) PTMs can occur after translation to mediate proper folding of the protein. This can increase stability of the protein or help localise the protein to a distinct cellular compartment.

2) PTMs can occur after protein folding and localisation. This can alter the biological activity of the protein.

105
Q
  • PTM can occur through the addition of chemical groups. To be modified, proteins may undergo phosphorylation, acetylation, hydroxylation and ______.
A

methylation

106
Q

What is Phosphorylation? (3)

A
  • Phosphorylation involves the addition of a phosphate group on serine, threonine or tyrosine residues and is one of the most important and extensively studied PTM in both prokaryotes and eukaryotes.
  • The addition of a phosphate group can convert a previously uncharged pocket of protein into a negatively charged and hydrophilic protein, thereby inducing conformational change in the protein.
  • The transfer of the phosphate is facilitated by enzymes known as protein kinases, and enzymes that
    help to remove the phosphate group are called Phosphorylases. Since phosphate groups can be
    added or removed, we say that the proteins have reversable modification.
107
Q

What is an example of phosphorylation?

A
  • An example is P53, a tumour suppressor protein used in cancer therapy that is activated by
    phosphorylation of its N terminal (beginning terminal) by several kinases.
108
Q

What is an example of phosphorylation?

A
  • An example is P53, a tumour suppressor protein used in cancer therapy that is activated by
    phosphorylation of its N terminal (beginning terminal) by several kinases.
109
Q

What is methylation? (2)

A
  • Methylation refers to the addition of a methyl group to Lysine or Argenine residues of a protein.
  • Methylation involves the transfer of C (methyl groups) to Nitrogen or Oxygen to amino acid side
    chains to make the protein more Hydrophobic. Methylation can neutralise a negative amino acid
    charge when bound to Carboxylic Acids.
110
Q

Methylation:

  • _______can be methylated once or twice, while Lysine can be methylated once, twice, or thrice.
  • Methylation is achieved by enzymes called methyl transferases.
  • Methylation has been widely studied in histones, wherein histone methylation (on the N-terminal)
    can lead to gene activation or repression, based on the _____ that is methylated.
  • P53, the tumour suppressor mentioned above can be methylated to regulate gene_______.
  • Histone methylation of Lysine and Argenine residues in the N-terminal tail of histone H3 can
    influence ______ organisation.
  • When such methylation reactions occur within gene _______ regions, they can activate or repress
    gene transcription, depending on their location.
A

Argenine
Residue
Transcription
Chromatin
Promotor

111
Q

What is Acetylation? (2)

A
  • Acetylation regulates many diverse functions, including DNA recognition, protein-protein interactions, and protein stability.
  • Protein acetylation plays a particularly important role in chromatin remodelling and is associated with the activation of gene transcription.
112
Q

What is Glycosylation? (3)

A
  • Glycosylation refers to the addition of saccharides (sugars) to proteins to produce proteoglycans in the ER (endoplasmic reticulum), Golgi, or cytoplasm.
  • They have critical roles in protein sorting, immune recognition, receptor binding, inflammation, and pathogenicity (the property of causing disease).
  • Glycosylation plays an important role in protein folding, conformation, distribution, stability, and activity.
113
Q

What are the examples of Glycosylation?

A
  • Examples of Proteoglycans include the ABO blood group proteins on red blood cells and gp120 (glycoprotein 120) of HIV.
114
Q

What is autophagy? (3)

A
  • Autophagy is the destruction of “large” cytoplasmic contents.
  • During the process of autophagy, the cell produces a membrane which surrounds the proteins or
    organelles that are going to be destroyed. This forms an Autophagosome.
  • The Autophagosome combines with a Lysosome and the contents of the Autophagosome are
    digested.
115
Q
  • This process is not very discriminating as autophagy can: (3)
A
  • Destroy organelles no longer required by the cell.
  • Destroy large protein aggregates – this is when there has been oxidative damage to the cell and many proteins have been damaged.
  • Activated in starvation to produce nutrients for the cell. This why when you diet, you breakdown muscle to feed your body.
116
Q

We need a more specific and targeted process to destroy specific proteins: (2)

A
  • This can include abnormal proteins, normal proteins no longer required by the cell, or foreign proteins such as those produced by viral infections within in the cell.
  • Abnormal proteins can include mutant proteins, oxidised/denatured proteins or incorrectly folded proteins.
117
Q

What is protein folding?

A
  • Protein folding is a complex process, and every cell has a process for correcting abnormal folding, as it is very dangerous to have abnormally folded proteins in the cell. This is because you could have hydrophobic proteins on the outside which can then interact with various membranes within the cell and cause damage to the cell.
  • You do not need to know the following diagram, just read it. Note that this process requires energy in the form of ATP.
118
Q

Normal Protein Levels in the Cell:

  • There must be a _____ between protein production and loss or destruction of proteins in a cell.
  • Normal proteins must sometimes be destroyed as proteins within a cell may need to increase or
    decrease depending on the _______ or response of a cell.
  • Both the _______ and timing of the presence of proteins in a cell is important for cell function.
  • Some proteins are short-lived, whist others’ life spans can vary.
A

balance
requirements
concentration

119
Q
  • Proteins whose concentration needs to be controlled may include: (5)
A
  • Transcription factors – these need to be regulated so that they are present when the cell wishes to transcribe genes and absent at other times so that the gene is kept silent.
  • Cell cycle control proteins – these need to control the phases of the cell cycle.
  • Inducible proteins (like some enzymes) – the more substrate there is, the more enzyme need to be produced, and we need to control for that.
  • DNA repair proteins – these should only be present when there is DNA breakage or a problem with the DNA needs to be sorted out. Extra proteins should not be kept in the cell if they are not required.
  • NFkB signalling – will be discussed in immunology, but it is a signalling pathway that is critically dependent on the breakdown of certain proteins.
120
Q

What is the proteasome?

A

The name Proteasome is derived from Protease, an enzyme which breaks down proteins.
- The Proteosome is found in both the nucleus and the cytoplasm, and proteins can thus be broken
down on either of the places.

121
Q
  • The components of the Proteasome include: (2)
A
  • Core particle – contain catalytic (proteolytic) sites which do the actual breaking down of the protein.
  • Regulatory caps – found on each end and each made of a base and a lid. Regulatory caps protect proteins that should not be destroyed from getting into the core particle. They act as “stoppers” on each end which protect and limit the number of proteins that can get inside the catalytic sites.
122
Q

How does the Proteasome know which proteins to destroy?

A
  • Answer: The Ubiquitin protein – derived from the word “ubiquitous” meaning “found everywhere” or “widely present”.
  • A chain of ubiquitin molecules (polyubiquitin) is attached to the targeted protein as a signal for destruction.
  • Ubiquitin is always present in a cell and then can be used to attach (as a chain of ubiquitin molecules) to a protein to allow the Proteasome to recognise the protein for destruction.
  • The ubiquitins bind to the Regulatory cap of the Proteasome and are released by the cap. It is not broken down and can attach itself to another protein.
  • The protein is unfolded and passes into the cylinder of the proteasome, where the catalytic sites will digest the protein into small peptides or even individual amino acids.
123
Q

What is Polyubiquitin Tagging?

A
  • This is the process through which ubiquitin is added to a protein.
124
Q

What are the 3 enzymes involved in Polyubiquitin Tagging?

A
  • E1, which is the Ubiquitin-Activating Enzyme.
  • E2.
  • E3.
125
Q

What is the process of Polyubiquitin Tagging? (8)

A
  • The first step is with the Ubiquitin-Activating Enzyme (E1), which activates the ubiquitin. This uses ATP but is not a Phosphorylation reaction. It uses the energy of ATP to change the energy state of ubiquitin, and the ubiquitin is then bound to E1.
  • Once the ubiquitin is bound to E1, the E1 associates with an E2 and E3 and the E1 transfers the activated ubiquitin onto E2. The E1 then leaves the complex.
  • The E2 (with ubiquitin bound) and E3 are now ready to receive the protein that needs to be tagged.
  • The target protein is now bound to this ubiquitin ligase.
  • The particular shape of the E3 will recognise the particular shape of the protein, as E3 provides a
    specific binding site for the targeted protein. This holds the protein for this process to occur.
  • The E3 ligases catalyse the transfer of Ubiquitin from E2 directly to a Lysine in the substrate protein.
  • The ubiquitin is transferred from the E2 directly to the protein.
  • This whole process is repeated several times until we get a polyubiquitin chain. The initial ubiquitin
    is added directly to the protein. Subsequent ubiquitin molecules are added to the previous ubiquitin molecule.
126
Q

Specificity of the Tagging Process

  • The specificity of the Polyubiquitin Tagging process is achieved through having multiple versions of the 3 enzymes required (E1, E2, and E3).
  • In humans there are: (3)
A
  • 2 different versions of E1.
  • +/- 60 versions of E2.
  • +/- 600 versions of E3
126
Q

Specificity of the Tagging Process

  • The specificity of the Polyubiquitin Tagging process is achieved through having multiple versions of the 3 enzymes required (E1, E2, and E3).
  • In humans there are: (3)
A
  • 2 different versions of E1.
  • +/- 60 versions of E2.
  • +/- 600 versions of E3
127
Q
  • If you combine a specific ___, a specific E2, and a specific E3, you get a pathway that is very specific for a particular protein. Once the E2 and E3 are brought together, it will only recognise 1 particular protein and no others.
A

E1

128
Q

Activation of this Pathway needs to be Specific and Controlled:

  • As stated previously, we have to control the concentrations and the timing of certain proteins in the cell with absolute ___-____ timing. If not, control of the cell may be lost and this is a very dangerous process.
  • There are different ways in which you can change either the protein or the ___ to recognise each other.
  • If you change the protein, then an E3 which is already there can recognise it.
  • If you change the E3 by _______ it in some way, it will be able to recognise the protein.
A

split-timing
E3
modifying

129
Q

How can the Protein be changed? (3)

A
  • The first way is through Phosphorylation. What the protein looks like changes when it has a phosphate group added to it. Phosphorylation is an important way of changing proteins.
  • The second way is removing an extra protein subunit or associated protein that is masking the binding site for the E3 ligase.
  • The third way is by cleaving off a part of the protein that is hiding the binding site.
130
Q

How can the Ubiquitin Ligase be activated/change its recognition site for the protein? (3)

A
  • The first way is through Phosphorylation. Phosphorylation is an important way of changing proteins (like E3) and allows them to change their function.
  • The second way is through adding a ligand. This brings about a change in the conformation of the E3.
  • The third way is through adding a protein, which will change the conformation of the E3 and allow it to recognise the protein that it needs to bind to.
131
Q

What is the role of the proteasome in the immune system? (2)

A
  • Short pieces of foreign peptides (from viruses and bacteria) are displayed on the surface of cells as part of the immune process.
  • The proteasome is specially modified (by the use of a different cap and proteolytic enzymes), to produce that specific length of peptide required.
132
Q

Malfunctioning of The Proteasome Pathway: (3)

A
  • Dysregulation of the proteasome can cause neurodegenerative diseases such as Alzheimer’s disease and Parkinson’s Disease.
  • In Parkinson’s, PARK2 (which encodes the E3 ligase parkin) mutations lead to familial Parkinson’s disease.
  • In both Alzheimer’s and Parkinson’s disease, there is an intracellular or intraneuronal accumulation of misfolded proteins which would normally be broken down by the proteasome system.
133
Q

Malfunctioning of The Proteasome Pathway: (3)

A
  • Dysregulation of the proteasome can cause neurodegenerative diseases such as Alzheimer’s disease and Parkinson’s Disease.
  • In Parkinson’s, PARK2 (which encodes the E3 ligase parkin) mutations lead to familial Parkinson’s disease.
  • In both Alzheimer’s and Parkinson’s disease, there is an intracellular or intraneuronal accumulation of misfolded proteins which would normally be broken down by the proteasome system.
134
Q

Manipulation of The Proteasome Pathway
- Bortezomib is an anti-cancer drug. It blocks the proteasome and appears to cause increase apoptosis (programmed cell death) and decreased proliferation of some types of cancer cells. It blocks NFkB signalling (recall that this signalling relies on the breakdown of proteins, and if the proteosome is blocked, proteins will not be broken down).
- Bortezomib combats these cancers in two ways: (2)

A

1) Myelomas (cancer of plasma cells) and Pancreatic Carcinomas (tumour of the Pancreas) are
treated with Bortezomib. This is because these cancers rely on the NFkB pathway for signalling and proliferation of the cells. Plasma cells normally use the NFkB pathway, and Myeloma which is cancer of the plasma cells thus also uses the NFkB pathway. Inhibiting the NFkB pathway therefore inhibits the proliferation of the Myeloma cells.

2) Because Myelomas come from plasma cells (which produce antibodies), they are also cells which produce large numbers of proteins. In any cell that produces proteins, there is going to be cases of proteins which have not folded correctly. If these misfolded proteins are allowed to accumulate, it will cause damage to the cell and the cell will die. Since Bortezomib blocks the proteasome, these misfolded antibody proteins will accumulate and damage the cancer cells. Pancreatic Carcinomas are another type of cancer in which large amounts of protein are produced, and Bortezomib will allow the accumulation of misfolded proteins which will damage the cancer cells and allow them to die.