Målsætninger i molekylær biologi B
forår 2003

skrevet af Ana-Marija Hristovska

PROTEIN-DNA INTERAKTIONER

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1. Beskrive de vigtigeste principper for DNA-protein interaktion og angive hvilken betydning DNA-major groove har. Beskrive helix-turn-helix motiver og zink-finger. Definere leucin-zippers og beskrive hvorledes de er involveret i protein-protein interaktion med DNA

Principper for DNA-protein interaktion
Devlin, s. 387
Devlin 390-391, fig. 9.27
Stryer, s.795, fig. 28.20
Stryer, s. 867

The entire idea with DNA-protein interaction is the need of controlling the gene expression. Two things have to be considered:

• a regulatory sequence on the DNA, ca. 10 nucleotides long

• a protein that binds to the regulatory sequence

There are ca. 30000 genes in the human genome, and each of them has a regulatory sequence called the transcription factor binding element. This element can be:

- a promoter - located upstream and close to the target gene
- an enhancer - located thousands of nucleotides away from the target gene

The DNA-binding proteins, known as transcription factors or regulatory proteins recognize this regulatory sequence due to non-covalent interactions between the protein and DNA, and bind to it.

The interplay between the proteins DNA-binding site and the DNA-sequence controls the gene expression at transcription level. The transcription factor usually binds the RNA polymerase that transcribes the gene.

fx.
DNA-binding proteins utilize a variety of strategies for interaction with DNA. I will take the TATA-TBP complex as an example.

The TATA box is the promoter sequence in DNA found in eukaryotes, while TBP is the TATA-box binding protein.

TBP is a saddle-shaped protein consisting of two similar domains, each composed of a curved antiparallel B-sheet with a concave surface. The TATA box of DNA binds to the concave surface of TBP. The double helix is substantially unwound to widen its minor groove, enabling it to make very extensive contact with the antiparallel B-strands on the concave side of TBP.
Immediately, outside the TATA box, classical B-DNA resumes.

TBP bound to TATA box is the heart of the initiation complex. The surface of TBP saddle provides docking sites for the binding of other components, among which RNA polymerase II.

Betydning af DNA-major groove
Devlin, s. 40
Devlin, s. 38-39 - fig.2.15; 2.16
Stryer, s. 748, fig. 27.8

The major groove of DNA is the binding site for the different transcription factors. Its larger size makes it more accessible for interactions with proteins that recognize specific DNA sequences. 

Because each of the four bases has its own orientation with respect to the rest of the helix, each base always displays the same atoms into the grooves. These atoms then constitute an important means of sequence-specific recognition of DNA by other molecules such as proteins.

fx.
the N7 of purines is always displayed in the major groove and can serve as hydrogen-bond acceptor in interactions with donor groups of proteins

Helix-turn-helix (HTH)
Devlin, s.354-355, fig. 8.24
Devlin, s. 387,  fig. 9.20
Stryer, s. 874

This is the most common structural motif found in many DNA-binding proteins. It is made up of a pair of alfa-helices and a tight turn:

- the first alfa-helix -  made up of 7 amino acids, participates primarily in interactions with the DNA-backbone, as well as hydrophobic interactions with the second helix, stabilizing the structure.

- the second alfa-helix - also called  the recognition helix is made up of 9 amino acids and is positioned across the major groove, where side chain residues of the helix form specific non-covalent interactions with the base sequence of the target DNA.

Most HTH-proteins bind as dimers, each monomer domain binding at 2 adjacent turns of the major groove of DNA on the same side of the molecule. The distance between the monomers is 34Å, which corresponds in one turn of the DNA-molecule. The dimer dual-binding interaction makes a much stronger and more specific protein-DNA interaction than possible for a single monomer protein-DNA interaction. The interaction induces a distorsion in the DNA structure so that the protein is easily and better accomodated.

Zink-finger
Devlin, s. 389, fig. 9.22
Stryer, s. 880

The name of the zink-finger domain comes from the importance of the Zn-jons it contains. The Zn-jons stabilize the structure of the domain; without them the domains unfold.

The primary structure for the motif contains 2 close cysteins separated by 12 a. a. from a second pair of Zn-liganding amino acids (most often to histidines, but sometimes they can be substituted by cysteins). The two cysteins in the first pair are separated by only two amino acids, while the histidines/cysteins in the second pair can be separated by three amino acids.

The motif contains an alfa-helix segment that can bind within the major groove at its target site in DNA and makes specific interactions with the nucleotide sequence. The alfa-helix segment is located at the end of the Zn-finger domain.

Leucine zippers
Devlin, s. 356 - fig. 8.26
Devlin, s. 389 - fig. 9.24

Leucine zippers, also called basic-region leucine zippers, are formed from a region of alfa-helix that contains at least 4 leucines, each leucine separated by six amino acids from one another. With 3.6 residues per turn of the alfa-helix, the leucines align on the edge of the helix, with a leucine at every second turn of the helix.

Protein-protein interaction: The leucine rich helix forms a hydrophobic interaction with a second leucine helix on another protein to zipper the two together to form a dimer. The formation of the zipper is necessary in order to form a strong bond with the DNA.

The leucine zipper motif does not interact directly with DNA.

The DNA-binding region is adjacent to the zipper motif. It is a basic region, defined by the presence of arginine and lysine (that is why the protein is called basic-region leucine zippers). This basic region assumes the conformation of two alfa-helices with a small break that allows the helices to follow the major groove of the DNA.

The basic amino acids serve to stabilize the DNA-protein association through electrostatic interactions between the positively charged amino acids and the negatively charged DNA backbone.

Many regulatory proteins containing a leucine zipper are oncogenes. If the regulatory proteins are mutated or produced in a non-regulated fashion, the cell can be transformed into a cancer cell. 

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2. Beskrive p53-DNA komplexet som et eksempel på protein-DNA interaktion. Angive at p53 inaktiverende muationer kan oftest lokaliseres i det DNA integrerende domæne
Devlin, s. 391-2, fig. 9.28.a, c.
Devlin, s.217, clinical correlation 5.3

The p53 protein is also called a tumor suppressor protein. Its absence allows mutated DNA in the cell to accumulate, leading to its transformation to a cancer cell. 

Its main 3 functions are:

- the p53 protein is a transcription factor that controls the checkpoint between G1 and S phases of the cell cycle, and on sensing damaged DNA upregulates the expression of genes that inhibit cell division, to give the cell the time to repair the damaged DNA.

- it can also promote transcription of the DNA-repair genes.

- alternatively, it can instruct the cell to undergo apoptosis, if the DNA-damage is too extensive to repair.

All of these actions would counteract the neoplasmatic transformation of the cell.

The DNA-binding region of p53 consist of a central fold (similar to an immunoglobulin fold), made up of 2 beta-sheets with antiparallel strands.

This central fold provides the scaffolding (platform) for:

•  loop-sheet-helix motif containing an alfa hellix (H2)

•  2 loops: loop1 (L1) and loop3 (L3)

These motifs interact with DNA, in the following way:

• H2 and L1 fit into the major groove

• L3 interacts with the adjacent minor groove

p53 binds DNA as a tetramer. Each monomer of the tetramer binds to a discrete consensus target DNA. 

Mutant or interactive p53 forms are found in majority of human cells. Somatic mutations can be identified in about half of all human cancers. Mutations represent a loss of function, either affecting the stability of p53 or its DNA-binding ability.

Mutations found in p53 from tumors affect the DNA-binding domain of the protein. fx. nearly 20% of all mutated residues involve mutation at two positions of p53. Both mutated amino acids are arginines that form under normal circumstances H-bonds with DNA. This decreases the ability of p53 to regulate transcription.

fx.
Arginine in position 248 forms H-bonds with the minor groove of the DNA-helix with a tymine oxigen and a ring nitrogen of adenine. Mutation disrupts this H-bonded network and therefore the ability of p53 to regulate transcription.

NUKLEINSYREKEMI

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1. Beskrive og tegne hvordan et nukleotid i en nukleinsyre er sammensat af komponenterne: base (purin/pyrimidin), sukker, fosfat
Devlin, s. 29-31;, fig. 2.5
Devlin, s.33 - fig. 2.8
Stryer, s.118-120, fig. 5.3

Nucleic acids are linear polymers consisting of repeating nucleotide units.

Each nucleotide consists of three components:
 

·                    a pentose sugar - D-ribose (in RNA) and 2-deoxy-D-ribose (in DNA).

The deoxy prefix in deoxyribose in DNA indicates that the 2´ carbon atom of the sugar lacks the oxygen atom that is linked to the 2` carbon atom in ribose in RNA
 

·                    a phosphate group - monofosfat, difosfat or trifosfat.

The sugars in nucleic acids are linked to one another by phosphodiester bridges of a monophosphat group.  Specifically, the 3`-OH group of the sugar moiety of one nucleotide is esterified to a phosphate group, which is, in turn, joined to the 5`-hydroxyl group of the adjacent sugar.

DNA and RNA consist of polymers joined by 3`- 5` phosphodiester bonds.

·                    heterocyclic base - purines and pyrimidines.

The major purine derivates are guanine (G) and adenine (A), which are found in  both DNA and RNA and are attached to the sugar at N-9.

The 2 major pyrimidine derivates are cytosine (C), uracil (U) and thymidine (T). Cytosine is present in both DNA and RNA. Uracil is found generally in RNA, while thymidine in DNA. They are all linked to the sugar through N-1 position.

The bond between the base and the sugar is a beta-glycosidic linkage.

A difference should be made between:

• nucleoside - a base glycosylated with either pentose sugar

• nucleotide -  phosphate esters of nucleosides, contains: base, sugar and a phosphate ester

• oligonucleotide - linear sequence of under 50 nucleotides

• polynucleotide - linear sequence of over 50 nucleotides

• backbone - the chain of sugars linked by phosphodiester bridges

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2. Beskrive hvordan nukleosidanaloger kan bruges kemoterapeutisk
Devlin, s. 854
Devlin, s.166

Nucleoside - a base glycosylated with either pentose sugar. The four nucleoside units in RNA are adenosine, guanosine, cytidine and uridine, while in DNA deoxyadenosin, deoxyguanosine, deoxycytidine and thymidine.

Analogs to these nucleosides can be used in chemotherapy.

De novo synthesis of purine and pyrimidine nucleotides is critical for normal cell replication, maintenance and function. When a normal cell transforms into a cancer cell, our ultimate goal is to interrupt the intercellular processes to that degree that the cell eventually dies. One way of doing it, is to prevent it from replicating, by preventing its synthesis of the DNA and RNA building blocks.

For that purpose, nucleotide analogs can be used. Despite very high accuracy of polymerases, they can incorporate nucleotide analogs. Nucleoside analogs are often used in chemotherapy to kill rapidly growing cancer cells or viruses.

Analogs that are phosphorylated to nucleotides can be incorporated into DNA, where they can inhibit further synthesis or lead to high level of mutation.

Many compounds, both synthetic and natural products of plants, bacteria, fungi, are structural analogs of the bases or nucleotides used in metabolic reactions.  

Differences in the ability of viral or bacterial polymerases to incorporate nucleotide analogs can provide a therapeutic window, allowing physicians to target the infected cells.

DNA STRUKTUR

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1. Tegne et A-T og G-C basepar
Devlin, s.39- fig.2.16
Stryer, s.122 - fig.5.12

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2. Beskrive de kræfter (Watson-Crick baseparring og "base-stacking") som holder en DNA dobbelt helix sammen

2 types of forces are used to stabilize the DNA-double helix:

Watson-Crick base pairing:
Devlin, s.39, fig.2.16,17
Stryer, s.10

The relationship between bases in the double helix is described as complementarity.

Bases are complementary because every base of one strand is matched by a complementary hydrogen bonding to a base on the other strand. These complementary base pairs are called Watson-Crick base pairs.

Fx. for each adenine projecting towards the common axis of the double helix, a thymine must be projected from the opposite chain so as to fill exactly the space between strands by hydrogen bonding. No other base would fit.

• Adenosine and thymine are linked by two hydrogen bonds, while cytosine and guanine with three H-bonds.
   

• -NH and NH₂ groups are proton donors (have partial positive charge, because H has a small electronegativity - a less strong attraction to the shared electrons then N, that has a higher electro negativity), while -O=  and =N- groups are proton acceptors (have partially negative charge because they have a stronger attraction to shared electrons and have a free electron pair) for the hydrogen bond
   

• The hydrogen bond is not very strong bond in itself, but because of the double helix confirmation, as well as the huge lenght of the DNA molecule, it becomes strong enough to keep it all together
   

• Uracil (RNA) has the same donor and acceptor groups as thymidine, meaning an RNA molecule can make base pairing with a DNA molecule
   

• The complementary bases have a space confirmation that favors base stacking.

Base stacking:
Devlin, s. 34
Stryer, s.123-4

The faces of the base rings tend to avoid contact with water, unlike their edges that contain polar groups that interact with other polar groups (Watson-Crick base pairing) or surrounding water molecules.

Therefore, the faces of the rings interact with one another, in order to produce a stacked confirmation - base stacking. Base stacking introduces 2 types of bonds:

• hydrophobic bonds between the non-polar rings of the bases. These bonds reduce the hydrophobic surface area of the DNA molecule by exposing the more polar surfaces to the surrounding water.  This also causes release of water into the bulk solvent, which is entropically favorable.
   

•  Van der Waals forces or London dispersion forces - these forces are produced between hydrophobic molecules, when the electrons moving around the nucleus of one atom can cause a dipole in the electron density of the neighbor atom when the electron orbitals of the two atoms approach to a close distance . These induced dipoles are very temporary, but they still contribute to the stabilization on the bond. Because the strength of these electronic interactions is very dependent upon distance, no empty space remains between the stacked bases.

Polynucleotides adopt confirmations that maximize the favorable stacking interactions between neighboring bases.

When the two strands of DNA come close to each other, it is also because of base stacking that it is impossible for two purines or pyrimidines to make a hydrogen bonds together. Two purines fill up too much, while two pyrimidines too little. That way, gaps or bulges are created in the DNA which are unfavorable to base stacking.

On top of everything, the phosphate groups in the DNA backbone do not want to be close to eachother since they all have negative charges, which has a destabilizing effect on the molecule. This problem is solved by introducing catjones which bind with the negative backbone and stabilize the entire strucutre.

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3. Beskrive polariteten af nukleinsyrer (5'-3' - ender) og angive at strengene i en DNA dobbelthelix er antiparallele
Stryer, s. 120
Devlin, s.39

A DNA chain has a polarity. One end of the chain has a free 5`-OH group (or a 5`-OH group attached to a phosphate), whereas the other end has a 3`-OH group, neither of which is attached to another nucleotide.

By convention, the base sequence is written in the 5`- to - 3` direction. This also underlines the fact that the DNA-molecule consists of polymers that are joined by  3`- 5` phosphodiester bonds.

The double helix DNA molecule contains 2 strands that are antiparallel to each other. This basically means that the two strands are aligned in opposite direction. This supports the complementarity principle.

fx.
if two adjacent bases in the same strand, thymine (T) and cytosine (C), are connected in 5`-3` direction, their complementary bases adenine (A) and guanine (G) will be linked in 3`-5` direction.

This antiparallel alignment produces a stable association between strands to the exclusion of the alternate parallel arrangement. The 2 strands have a very high affinity for each other.

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4. Definere major og minor groove i en DNA dobbelthelix
Stryer, s.748, fig.27.7; fig. 27.8
Devlin, s. 40
 

Double helical nucleic acid molecules contain two grooves between the sugar-phosphate backbones of the two helices, called the major and the minor groove. They are produced by the interwinding of the two strands.

The grooves arise because the glycosidic bonds between the bases and the backbone pentose are not arranged diametrically opposite to each other, but are displaced towards the minor groove.

Because of the non-simetrical glycosidic bonds, the nucleotide sequence of DNA can be detected by looking inside the grooves, without being necessary to dissociate the double helix.

Each base always displays the same atoms in the groove. This is very important, because a lot of transcription factors can recognize their DNA binding site by being able to scan the DNA sequence and recognize the specific atoms. It is usually the major groove that serves as a site for DNA-protein interaction.

In B-DNA, the major groove is wider (12Å vs. 6Å) and deeper (8.5Å vs. 7.5Å) in comparison to the minor groove.

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5. Angive de omtrentlige dimensioner af DNA.
Stryer, s. 122, fig 5.11
Stryer, s. 750, tab. 27.1
Devlin, s.47, tab.2.3

• the bases are nearly perpendicular to the helix axis - 1˚ tilted, and adjacent bases are separated by 3,4 Å.
   

• there are 10,4 bases per turn of helix - the helical structure repeats every 34Å (3,4nm)
   

• the diameter of the helix is ca 23,7 Å (2.4 nm)
   

• the length of the DNA molecule depends on the species. In a bacteriofag, there are 5000 base pairs per DNA molecule. In one human DNA molecule there are over 10⁸ base pairs.
   

• major groove - 12Å wide, 8.5Å deep
   

• minor groove - 6Å wide, 7.5Å deep

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6. Angive andre konformationer i DNA end Watson-Crick dobbelthelix, fx. cruciform DNA, tre-strenget helix, fire-strenget DNA og slipped/mispaired DNA
Devlin, s.44-48, fig. 2.33
Stryer, s.747-750, fig. 27.4, 27.10
Stryer, s.750 table 27.1

There are several confirmations of double helical DNA. Depending on conditions and base sequence, the double helix can acquire various forms of distinct geometrics. 11 forms have been described. Thou, there are 3 main conformations:

B-DNA (the Watson-Crick DNA)

·                     base pairs are relatively perpendicular to the helix axis
·                     has a major and minor groove
·                     right-handed helix
·                     appears at high humidity and low salt concentration
·                     app. 10 base pairs per helical turn

 Most DNA is in B-from under physiological conditions.

A-DNA 

·                     the base pairs are tilted from the plane perpendicular 19˚ to the helix axis
·                     major and minor groove
·                     right handed helix
·                     appears at low humidity and high salt concentration
·                     shorter and thicker then B-DNA
·                     11 base pairs per helical turn  

Double stranded regions of RNA and some RNA-DNA hybrids adopt the form very similar to A-DNA.

Z-DNA

·                     left-handed helix
·                     zigzag backbone
·                     only one groove

Triple stranded DNA
Devlin, s.50-53, fig. 2.29; 2.30

Triple-stranded DNA can be formed by antiparallel binding of an oligonucleotide to the homopurine strand of a Watson-Crick double helix. Purines have two groups that can participate in additional base pairing:

• in adenine - N-7 and and the second H in (NH₂)-6, since the first H is already used in Watson-Crick base pairing
   

• in guanine - N-7 and O-6

Selective triplets can be formed between A and A-T pairs as well as G and G-C pairs. The new DNA molecule has 3 strands.  (fig.2.29)

Base stacking plays an important role in stabilizing the seqeunce, but because there are three negatively charged phosphate-sugar backbones, the repulsion between them is bigger leading to a less stable stucture. The structure can be stabilized by adding Mg⁺⁺ - jons, that shields the negative phosphate charges.

Potential triple helices regions are especially common near sequences involving gene regulation.

Four-stranded DNA
Devlin, s. 54-56, fig. 2.32; 2.33
 

Four-stranded DNA is made up by stacking of the so-called G-quartets.

G-quartet - tetrameric structures containing guanine nucleotides and highly G-rich polynucleotides. They can stack upon eachother and form a multylayered structure.

G-quartets are formed when 4 coplanar guanines of different G-rich polynucleotides form a tetrameric structure making Hoogstein hydrogen bonds. The cavity in the center of the quartet can accomodate Na⁺ og K⁺ jons, which interact with guanine oxigens (O-6) and stabilize the hole. 

The exsistance of four-stranded DNA in vivo has never been proved. A possible function can have something to do with telomeres that contain repetitive G-sequences - TTAGGG, fx. initiation and termination of replication and recombination.

In vitro experiments indicated that oligonucleotides with the appropriate seqeunce can form G-quartets in telomeres.

Cruciform DNA
Devlin, s.49-50, fig. 2.28
 

Cruciform DNA is created as a result of inverted repeats. Inverted repeats are wide spread in the human genome. It has been speculated that they may function as molecular switches for replication and transcription.

An inverted repeat is a DNA sequence on one strand that has its complementary sequence on the same strand, just a little bit more down the sequence. If the complementary sequence is read in 5'-3' direction, it is exactly the same as the template.

Imagine a nucleotide sequence (which i will call the "template" sequence in the lack of a better word) that appears on one DNA strand and pairs with its complementary (progeny) sequence on the antiparallel strand. The same template sequence is found few nucleotides down on the antiparallel strand, just in the opposite direction and this one pairs with the complementary sequence on the first DNA strand. This means that we have the template and complementary sequence on the same strand, just divided by some nucleotides. Both the template and the complementary seqeunce lieing on the same strand have the same seqeunce of nucleotides if read in 5'-3' direction.

The reason why no symmetric inverted repeat is formed, is that there is fx. a uneven number of bases in the sequence, so there can be no symmetry, plus the seqeunce lies further down the molecule.

Disruption of hydrogen bonds between the complementary sequences and formation of intra-strand hydrogen bonds (between the template and complementary sequence on the same strand of DNA) on both helices creates a cruciform DNA.

The loops generated by cruciform formation require unpairing of 3-4 bases at the end of the hairpin. These were actually the bases that were in between the template and complementary sequence on the same strand.

The presence itself of invert repeats is favorable, but not enough to cause formation of cruciform DNA. The double helix DNA finds in thermodynamically unfavorable to create cruciform DNA. The DNA needs to be unwinded if the cruciform structures is to be formed. Unwinding favors the intra-strand hydrogen bonding.

No function of the cruciform DNA has been established. 

Slipped/mispaired DNA 
Devlin, s. 56-8, fig 2.35; fig. 2.36
 

It appears in DNA regions with direct repeat symmetry - presence of two adjacent tandem repeats. This means that the same template DNA sequences are found on one DNA strand right after each other.

 The formation of slipped DNA involves the unwinding of the double helix and realignment of the strands in that way that the first DNA template sequence pairs with the complementary sequence which was previously paired with the second template sequence or vice versa. Therefore, 2 isomeric forms can be made. 

This realignment generates two single-stranded loops, the second template strand and the first complementary strand in the example I gave. It is vice versa with the isomeric for.

Slipped DNA is involved in spontaneous frame shift mutagenesis, when 1 base is slipped. Depending if the base has been slipped in the complementary or template strand, it can manifest as addition or deletition of the base.

A deletition of addition of a longer segment during DNA replication can be seen due to larger sequences of slipped DNA. (fig. 2.36)

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7. Definere begreberne DNA denaturering (melting), renaturering (re-annealing) og hybridisering af prober
Stryer, s.124-5
Devlin, s.42 - s.44, fig.2.20

DNA denaturizing (melting) - disruption of the hydrogen bonds between the bases in the DNA sequence. The secondary structure - the DNA double helix is lost.  

During DNA replication and other processes, the two strands of the double helix must be separated from one another, at least in a local region. This can also be accomplished in laboratory conditions by heating up the DNA. The heating disrupts the hydrogen bonds between the base pairs, and thereby causes the strands to separate.
Inside living cells, the double helix is not melted by addition of heat. Instead, proteins called helicases use chemical energy to disrupt the structure of the double helix.

Besides the temperature, other factors have also influence on this DNA-melting:

• the pH value - at pH of 11.3, the NH- groups of the basic rings become unprotonated (the H-atoms disassociate), thereby loosing their ability to make Watson-Crick bonding

• the salt-concentration - a high concentration of catjons, which bind to the negative phosphate backbone, stabilize the DNA structure and decrease the repulsive forces between the two strands. So, lower salt-concentration, bigger denaturizing.  

DNA-renaturizing (re-annealing) - if appropriately treated, the separated strands of DNA can re-associate to form a double helix (fx. if the temperature is lowered).

Annealing is possible even after complementary strands have been completely separated. One DNA strand needs to find its complementary strand among a lot of other DNA molecules. In order to find the fitting sequence, the bases form a lot of hydrogen bonds among each other thereby testing if they have found the right strand. The "wrongly" paired strands have a short life, because the bases that lie around the complementary sequence can not pair. Once the correct bases begin to pair by chance, the double helix is rapidly reformed over the entire DNA. The formation of the first base pair is always slow. The second base pair makes hydrogen bonds faster, the third even faster... and so on as the sequence expands. 

Hybridization of probes

Hybridization - association between two polynucleotide chains, which may be of same or different origin or length, provided that a base complementarity exists between these chains.

Probe - short single stranded RNA and DNA oligonucleotides (app. 15-20 nucleotides) that are complementary to a specific sequence of interest of the genomic DNA.

The following hybrids can be formed:

DNA-DNA
DNA-RNA
RNA-RNA

The hybridization of probes can be used for several purposes:

1. location of any given DNA sequence of interest by annealing with a probe, that is appropriately tagged for easy detection of the hybrid (location of genes in DNA that correspond to a particular RNA, by using the RNA as a probe)

2. DNA molecules from 2 different species can be melted and allowed to hybridize in the presence of each other. The degree of hybridization is an indication of the relatedness of the genomes and hence the organisms.

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8. Beskrive hypokromisk og hyperkromisk effekt og middeltemperatur (Tm)
Devlin, s.43 - fig. 2.20; fig.2.22

The denaturizing of nucleic acids can be easily followed by monitoring their absorption of UV-light at 260 nm. There are 2 possibilities:

hypochromicity - absorption of less UV-light due to stacking. The total absorbance of stacked bases may reduce the UV absorption by as much as 40% compared to an unstacked state.

hyperchromicity -  absorption of more UV-light due to denaturizing of the DNA and unstacking of bases. When DNA is heated, its absorbance of UV-light at 260 nm increases with rising temperature, because more and more bases are unstacked. When the DNA is completely denaturized, the absorption is 100%.

Tm - at this temperature, half of the DNA is single-stranded (denaturized), and half of it is found in its double stranded helix form. This basically means that a half of the maximum optical absorbtion is reached.

The Tm gets higher, when the DNA has more cytosine (C) - guanine (G) content, because more heat is needed to break the 3 hydrogen bonds that link these 2 bases. The Tm is, as expected, lower if there are more A=T base pairs.

In general nucleosides absrob the most UV-light followed by nucleotides, single stranded DNA absorbes even less and double stranded DNA absorbs the least. The reason is, of course, that stacking is smalllest in nucleosids and biggest in double stranded DNA.

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9. Beskrive forekomsten og funktionen af supercoiling af DNA: positive og negative supercoils (superhelix, supertwist), samt funktionen af topoisomeraser (type I og II). Beskrive L = T + W.
Devlin, s.60- s.62
Devlin, s.64 - fig.2.43
Devlin, s.65 - c.c. 2.7
Stryer, s.125, fig.5.18
Stryer, s.754-7
 

The DNA in our cells can be found in the following forms:

relaxed DNA - no manipulation with the DNA molecule. No super helical turns. This form of DNA has a greatly reduced activity in number of crucial biological processes, fx. replication, translation and recombination.

supercoiling - coiling of the axis of the double helix. Can be created by either unwinding or overwinding the DNA molecule. Supercoling can hinder or favor the capacity of the double helix to unwind and thereby affect the interactions between DNA and other molecules.

• negative supercoling - a right-handed coil is assigned a negative number. It is the biologically active form of DNA. Most naturally occurring DNA molecules are negatively supercoiled. In essence, negative supercoiling prepares DNA for processes requiring separation of the DNA strands, such as replication or transcription.
  The negatively supercoiled DNA is more unstable because it is unwinded (twisted one time less), and the hydrogen bonds between the base pairs are more easily broken.
  - On the site of replication, DNA is unwinded.
  - It also appears when the DNA wraps itself around the histones.

• positive supercoiling - a left-handed coil is assigned a positive number. It makes strand separation more difficult and isn’t biologically active, because the DNA is overwinded (twisted one time extra) and is extremely stable. During replication, overwinding appears after the replication site as a result of unwinding of the helix on the replication site. These positive supercoils must be removed if DNA replication is to continue.
  - before and after the replication fork, the DNA is positively supercoiled

L = T + W

L - linking number - number of times that a strand of DNA winds in the right-handed direction around the helix axis, when the axis is constrained to lie in plane.   

(antallet af gange en DNA streng snor sig i højre handed retning rundt om den anden DNA-streng)

T -twisting number - measure of helical winding of the DNA strands around each other

(antallet af dobbelt helix snoringer)

W - writing number - measure of the coiling of the axis of the double helix

(antallet af supercoils)

Topoisomers - DNA molecules differing only in linking number. Topoisomers of DNA can be interconverted only by cutting one or both DNA strands and then rejoining them.

Topoisomerases - enzymes that catalyze the interconversion of topoisomers of DNA. These enzymes alter the linking number of DNA by catalyzing a 3-step process:

1. The cleavage of one or both strands of DNA. Topoisomerases form a transient break in the DNA backbone and then resealing it. The break is not formed by hydrolysis of the sugar-phosphate backbone, but by a transesterification reaction that creates a phosphate-enzyme bond as a transient intermediate. 

2. The passage of a segment of DNA through the break

3. The releasing of the DNA. Rejoining the backbone phosphodiester displaces the enzyme. 

There is a difference between topoisomerase in prokaryotes and eukaryotes, which is EXTREMLY confusing and difficult to remember.

Topoisomerase I

• catalyze the relaxation of supercoiled DNA, (in prokaryotes removes negative supercoils, in human cells removes both positive and negative supercoils)

• cleaves only one strand of DNA

• does not use ATP, because this is a thermodynamically favorable process

Topoisomerase II

• adds negative supercoils to relaxed or supercoiled DNA

• cleaves both strands of DNA

• uses ATP

• found only in prokaryotes, the human topoisomerase has a different function

The degree of supercoiling of DNA is determined by the opposing actions of the two enzymes. Negative supercoils are introduced by topoisomerase II and relaxed by topoisomerase I. The amounts of these enzymes and their activities are regulated to maintain an appropriate degree of negative supercoling.

In treatment of cancer, both topoisomerase I and II can be targeted, by interfering with the enzyme-catalyzed rejoining of the two DNA strands. This may result in degradation of DNA, introduction of mutations or inhibition of translation or replication.

Man kan måske finde en bedre forklaring på human topoisomerase-funktionen på:
http://www.smi.stanford.edu/projects/helix/geis/documentaries/topo.html

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10. Definere begrebet palindromer (symmetrisk inverterede repeats) i DNA,
Devlin, s.72

Palindrome - symmetrical inverted repeats found in DNA. This means, that two complementary sequences on the DNA helix, one on each strand, lying opposite each other have the same order of the bases when they are read in 5`-3` direction.

Palindromes are used and recognized by specific restriction enzymes that cleave both complementary DNA molecules at a specific symmetrical place.

The term "inverted repeats" is also used in cruciform DNA, but the difference is that the inverted repeats in cruciform DNA are not symmetrical. This basically means that, the template sequence on the first DNA strand has its inversed repeat on the second DNA strand couple of nucleotides down the sequence, and is not directly complementary with it because there is an uneven number of nucleotides in the template sequence (as there is in symmetrical inverted repeats)

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11. Beskrive DNA bindingsområder for proteiner og hvordan proteiner genkender disse bindingsområder

The protein binding regions of DNA are called transcription factor binding element. As their name suggest, they serve as binding sites for proteins that control the transcription. They are usually no longer then 10 nucleotides.  

The protein recognizes the binding sequence of the DNA by forming non-covalent bonds. This usually occurs in the major groove, because the base pairs that project into this groove are more accessible for scanning and binding.

Se også:

- målsætning (1)
- målsætning (8)
- målsætning (16) - sidste afsnit

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12. Beskrive kortfattet hvordan DNA er organiseret i nukleosomer og kromatosomer ved hjælp af histon proteiner.
Stryer, s.875-7, fig. 31.16
Devlin, s.68-9, fig. 2.47
Devlin, s.348-9, fig. 8.18

Histones - small basic proteins that are used for packing DNA. They have a high content of the basic amino acids lysine and arginine, and are bound to the negatively charged DNA backbone. A histone is built up of:

• central non-polar region - forms a globular structure

• a -N and a -C terminal regions - contain most of the basic groups

Five major histones are present in DNA: H1, H2A, H2B, H3 and H4.
- H1 is larger then the other histones, has more basic amino acids and is species and tissue specific.
- an additiional histone, called H5 can also be used for DNA-packing and has a similar function to H1.
- H3 and H4 are very similar in stricture and are the most susteined ones in evolution.  

Nucleosome - disk-shaped structure consisting of:

• a core of eight proteins, octameric histone cluster:

- 2 x H3
- 2 x H4
- 2 x (H2A-H2B)

•  negatively supercoiled DNA segment, wrapped around the histone octamer 1.75 times

Because the DNA makes a negative supercoil around the histones, when the DNA is straightened out, it will be unwinded. This unwinding is exactly what is needed to separate the two DNA strands during replication and transcription. 

Nucleosomes are not radnomly (periodically) created in the chromatine, it all depends on the DNAs willingness to bend. Long A-tracts and regions rish in C-G resist bending, while short A-tracts are willing to bend.

Chromatosome

• a core of eight proteins - octameric histone cluster:

- 2 x H3
- 2 x H4
- 2 x (H2A-H2B)

• negatively supercoiled DNA segment, wrapped around the histone octamer 2 times.
   

• protein H1, which has a different structure from the other histones and is species specific, seals off the nucleosome at the location at which a linker DNA enters and leaves the nucleosome.

The linker DNA joins the nucleosomes into a polynucleosome.

Some interesting numbers:

• About 200 bp of DNA make up a single nucleosomal unit, with about 130-160 bp in direct contact with the octamer core (average 146bp)

• Some of the remaining DNA binds histones H1 and H5 and the rest is linker DNA between nucleosomes. Linker-DNA is 20-90 bp long.   

In order for the DNA to interact with transcription factors, two conditions need to be met:

• The DNAs major groove needs to be facing away from the histone. The major groove is the one contacted by the transcription factors. A pattern has been established that in A-T rich regions the minor groove is in the inside of the nucleosome core, while the major groove is left free for protein interactions (and v.v. for C-G rich region). However, there is still no clear way to predict which groove will face the outside of the nucleosome.
   

• The interaction between the DNA and the histone proteins need to be week. Acetylation of ξ – amino group near the N-terminus of the histones reduces the positive charge carried by the histone and thus weakens the electrostatic interaction between the histones and the DNA. In general, the acetylation (euchromatin) leads to activation of gene expression, while deacetylation (heterochromatin) reverses the effect.
  (Devlin, s.348 - fig.8.18)

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13. Definere begreberne kodende og "junk" områder i DNA sekvenser
Devlin, s.74
http://www.psrast.org/junkdna.htm

The human genome codes for ca. 35000 genes.

Only 2-4% of DNA in mammalian cells codes for structural genes.
Some of the remaining DNA (2-3%) is found in telomeres, centromeres and DNA sequences that code for RNA.  

Over 95% of DNA has largely unknown function. It was called "junk-DNA" because molecular biologist weren’t able to ascribe any function to it. It was assumed to be molecular garbage. The idea that a major part of our DNA is "garbage" ignored the fact that a key feature of biologic organisms is optimal energy expenditure. To carry enormous amounts of unnecessary molecules is contrary to this fundamental energy saving feature of biological organisms.

If the non-coding DNA is mostly molecular garbage, then it should have had a completely random bas sequences. This is not the case.

Increasing evidence is now indicating many important features of junk DNA:

• junk DNA plays a vital role in the regulation of gene expression during embrionic development

• junk DNA enhances the transcription of proximal genes (genes close to the respective DNA)

• junk DNA acts as silencer for suppression of transcription of proximal genes

• junk DNA regulates translation of proteins 

Still, there is a little knowledge about the relationship between junk DNA and coding DNA. This adds to other factors making it impossible to foresee and control the effect of artificial insertion of foreign genes, fx. gene therapy.

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14. Beskrive organiseringen af mitokondrielt DNA og angive i hvilke organeller/organismer DNA er linært eller cirkulært.
Devlin, s.59-60, fig.2.37
Stryer, s.125 - fig.5.18

Genomic DNA can be linear or circular.  

Linear genome is found in eukaryotes, while circular mainly in prokaryotes.

Circular DNA results from the formation of phosphodiester bonds between the 3`- and 5`- termini of linear polynucleotides.

• Many viruses have circular DNA at some point in their life cycle.

• Most (if not all) DNA in bacteria is circular. This includes the bacterial chromosomes as well as smaller extra-chromosomal DNA called plasmids.

• Mitochondria have also circular DNA.

mtDNA
Stryer, s.493
Devlin, s.587-8, fig. 13.61

 0.3% of human DNA is found in mitochondria.

Mitochondria are semiautonomous organelles that live in an endosymbiotic relation with the host cell. Mitochondria generate most of the ATP required by aerobic cells. Ca. 100 mitochondria are found in a metabolitic active cell.

These organelles contain their own genome, a circular double-stranded DNA (mtDNA). The mitochondrial DNA comprises 16,569 bp and encodes:

• 13 respiratory chain proteins

• 2 mitochondrion-specific rRNAs, a small and a large one and

• 22 tRNA species, which are required for protein synthesis in mitochondria. 

All the genes lie on the external strand of the circular DNA and are very closely packed, that sometimes one sequence is used by two adjacent genes.

Mitochondria have the ability to transcribe and their own DNA, but they are not self-replicating organelles. Over 90% of all mitochondrial proteins are encoded in nuclear DNA and imported in the cytosol. There are ca. 100 proteins all in all.

Mitochondria’s genetic code is slightly different:

• UGA codes for tryptophan instead of being a stop codon

• AUA codes for methionine and not isoleucine, as it does in the cytosolic DNA

• AAG and AGA are stop codons, instead of coding for arginine

Mitochondria are though to be descendants of aerobic prokaryotes that invaded and set up a symbiotic relationship with eukaryotic cells. That is why they have some prokaryotic characteristics, so antibiotics can slightly effect them too.

DNA REPLIKATION OG REPAIR

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1. Forklare at DNA replikation foregår semikonservativ, og at princippet i DNA replikation er den samme i prokaryote og eukaryote organismer. Replikation foregår samtidig på begge strenge og er derfor bidirektionel.
Stryer, s.123
Devlin, s.162-3

The double-helical model of DNA and the presence of specific bas pairs immediately suggested how the genetical material might replicate.

The sequence of bases on one strand of the double helix precisely determines the sequence of the other strand. Thus, separation of the double helix into two component chains would provide two single stranded templates, onto which two double helices could be constructed. 

This is called semi conservative replication, because the new DNA helix will contain one parent strand and a newly synthesized complementary strand (a copy of the complementary parental strand).

The basic principle of replication can be seen both in prokaryotes and eukaryotes. There are, of course, some differences. Fx. there are different enzymes involved in the process.

The replication begins at specific DNA sequences called organs of replication, and proceeds bidirectionally and at the same times on both strands. One strand is made continuously while the other is synthesized in fragments.

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2. Angive at et prokaryot kromosom har et enkelt replikation-origin (OriC) mens eukaryote kromosomer indeholder mange replikation-origins.
Devlin, s.177-178, fig.4.14
Stryer, s.760, s.764 

Origins of replication (OriC) are specific sites on the DNA molecules from where the replicating begins and proceeds in both directions (bidirectional).

Known origins of replication contain:

• multiple, short, repeated sequences that bind specific proteins

• A=T rich regions on which the initial separation of parental strands occurs

 A prokaryotic DNA-helix has only one origin of replication (fx. E.Coli, 245bp-region acts as OriC. It contains four repeats of a sequence that together act as a binding site for an initiation protein called DnaA. It also contains a tandem array of three nearly identical 13-nucleotide sequences rich in A=T)

In eukaryotic cells, there are thousands of origins. These OriC are tandemly arrayed along the chromosome, placed ca. 50.000-100.000 bp apart. In human beings, replication requires about 30.000 origins of replication, with each chromosome containing several hundred.

Adjacent OriC tend to function together.  The replication proceeds bidirectionally from an OriC - there are two replication forks created, two DNA polymerases III move in opposite directions,, each synthetising one leading and one lagging strand. This basically means that each new DNA-strand is synthesized partially as leading and partially as a lagging strand.

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3. Beskrive hvordan polymeriseringsprocessen (DNA-synthese) foregår fra 5' imod 3'
Stryer, s.760-1, fig.27.27
Stryer, s.763-4, fig. 27.33
Devlin, s.166-7, fig.4.5

Both strands of parental DNA serve as templates for synthesis of new DNA. The site of DNA synthesis is called the replication fork. There are two replication forks created from each origin of replication, going in two oppsite directions.  

Replication fork - complex formed by the newly synthesized daugther strands arising from the parental duplex. DNA polymerase III moves in the direction of the fork opening. Therefor there are also two DNA polymerases III that go in oppsite directions from one origin of replication.

The two strands are antiparallel, meaning they run in opposite direction. One of the daughter strands has its 3`-end towards the replication fork, which is fine because the DNA polymerase III adds nucleotides to the free 3´-OH group of the previous nucleotide in the sequence.

DNA polymerase III only adds nucleotides in the 5´-3´direction.

The polymerisation reaction - The phosphodiester bond of the dNTP connecting the first ( α ) phosphate attached to 5`-carbon of the deoxyribose and the two other ( β, γ ) phosphates undergoes a nucleophillic attack from the free 3`-OH group of the previous nucleotide attached to the DNA. That way, a new bond is created between the a  phosphate of dNTP and the  3`-OH group of deoxyribose.The terminal two phosphates are released as inorganic phosphate and hydrolyzed in the cells. This makes the reaction irreversible.

This process is, as already mentioned, catalysed by DNA polymerase III.  

However, the strand that has its 5´-end towards the replication and this presents a problem. The mode of synthesis of the lagging strand is therefore more complex.

A looping of the template for the lagging strand places it in position for 5`-3` polymerization. The looped lagging strand template passes through the polymerase site in one subunit of a dimeric DNA polymerase III, while the leading strand passes through the other subunit.

DNA polymerase lets go of the lagging strand after adding about 1000 nucleotides. The synthsiezed fragment is called an Okazaki fragment. A new loop is then formed and primase again synthesizes a short RNA primer to initiate for another Okazaki fragment.

This way all synthesis goes on in the 5`-3` direction.

The gaps between these fragments are filled up by DNA polymerase I. This enzyme also has a 5´-3´ endonuclease activity to remove the RNA primer. In the end, DNA ligase connects the fragments.

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4. Beskrive hvordan følgende elementer er involveret i DNA replikation: template, primer, substrater (deoxynucleosid triphosphater, dNTPs), primase, DNA polymerase I og III, DNA ligase, leading og lagging strands (Okazaki fragmenter) og diskontinuerlig syntese,
Devlin, s.163-169, fig.4.2
Stryer, s.750
Stryer, s.761

Template - A sequence of DNA or RNA that directs the synthesis of the complementary sequence.  

Primer - short stretches of RNA (ca. 8-10 nucleotides long) that are already base-paired to the template and provide the free 3`-OH to which the nucleotides are added by DNA polymerases. The primers are synthesized by primase. They are found in between the Okazaki fragments of the lagging strand and on the beginning on both daughter strands.  

Primers are necessary because the DNA polymerase does not join the first two nucleotides to start a strand. The requirement for a primer is not due to the chemistry of phosphodiester bonds, since RNA polymerases can start a strand on their own.

The likely reason is low accuracy of proofreading preformed by DNA polymerase on the first few nucleotides in the sequence. The possibility to identify and remove the primer later, as well as fill the gap more accurately contributes to less errors in DNA synthesis.

 The primer can be removed by DNA polymerase because of its 5`-3`activity. This enzyme can also fill in the gap by adding nucleotides, while DNA ligase seals the nick between the newly added DNA sequence and the DNA that already was there.

Primase - an RNA-polymerase that synthesizes the primer sequence using ATP, UTP, GTP, CTP.

Substrates
Devlin, s.164 - fig.4.2

Substrates - the building blocks of DNA are deoxy - nucleoside triphospahts (5`-dNTP) that lie around in the nucleus of the cell: dATP, dTTP, dGTP, dCTP. They are the precursors of the DNA nucleotides, because the addition of a mononucleotide to a growing chain is not a spontaneous process.  

This entire DNA-synthesis process is catalyzed by DNA polymerases. When a deoxy-nucleoside triphosphate incomes, it first forms a hydrogen bond with the complementary base and only then the DNA polymerase links the incoming base with the predecessor in the chain by a nucleophilic substitution reaction.

The phosphodiester bond of the dNTP connecting the first (α) phosphate attached to 5`-carbon of the deoxyribose and the two other ( β,γ ) phosphates undergoes a nucleophillic attack from the free 3`-OH group of the previous nucleotide attached to the DNA. That way, a new bond is created between the a  phosphate of dNTP and the  3`-OH group of deoxyribose.

The terminal two phosphates are released as inorganic phosphate and hydrolyzed in the cells. This makes the reaction irreversible.

 DNA polymerase I:

1. polymerase activity - fills the gaps left after fx. the RNA primer is removed. A gap means that at least one nucleotide is missing.

2. 5´- 3´ exonuclease activity - it can remove the RNA primers of the lagging strand and act as a repair system

3. 3´- 5´exonuclease activity - proof reading, removes the wrongly incorporated nucleotide at the 3`-OH end

4. high processivity - lets go of the DNA molecule after adding 20 nucleotides

5. low speed - adds 10 nucleotides/s

DNA polymerase III:
Stryer, s.763-764, fig. 27, 31;27.33

1. The replication unit of DNA polymerase III is a asymmetric dimer. It is structured as a dimer in order to be able to replicate both parental DNA strands in the same place at the same time. It is symmetric because the leading and the lagging strands are synthesized differently.

2. Polymerase activity - synthesizes DNA by adding dNTP in the 5´-3´direction, meaning adds nucleotides to a free 3´- OH group. Is not used to fill in gaps.

3. no 5´-3´ exonuclease activity - meaning it can not remove the RNA primer and it can not act as a repair system

4. 3´-5´exonuclease activity - proof reading, removes the wrongly incorporated nucleotide at the 3`-OH end

5. extremly high processivity - the processivity unit of DNA polymerase III is formed as a star-shaped ring. In the center of the ring there is a hole that can readily accommodate a duplex DNA molecule, yet leaving enough space between the DNA and the enzyme to allow rapid sliding and turning during replication. The DNA polymerase does not let go of the DNA until the entire replication is finished

6. high speed - adds 1000 nucleotides/s

    

DNA ligase
Stryer, s.761

DNA ligase - after the RNA primer is removed and the gape is filled by DNA polymerase I, there is still a nick left which is sealed of by DNA ligase. A nick is an interruption in the phosphodiester backbone with no missing nucleotides.

DNA ligaze catalyses the formation of a phosphodiester bond between the 3`-OH group at the end of one DNA molecule and the 5´-phosphate group at the end of the other. DNA ligaze can not add nucleotides to a single stranded DNA, it needs a double helix in order to be operational.

The energy source in the eukaryotes is ATP, which is cleaved to AMP and PPi; in prokaryotes NAD⁺ which breaks to AMP and NMN.

Leading strand - the daughter strand of DNA that is replicated continuously from the origin of replication in the direction 5´-3´. It has its 3´-end towards the replication fork and the DNA polymerase III has not problems adding nucleotides on the free 3´-OH group - continuous replication.
 

Lagging strand - the strand which has its 5´-end towards the replication fork. The DNA polymerase adds nucleotides only in 5´-3´direction, while the lagging strand has the 3´-5´ direction, meaning has a free 5`-OH group.

That is why this strand is synthesized in small segments, called Okazaki fragments, after the template DNA for the lagging strand makes a loop, which allows the DNA polymerase to add nucleotides in the 5´-3´direction. The gaps left after the removal of the primers are filled by DNA polymerase I.

The synthesis of the lagging strand is therefore said to be discontinuous.

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5. Beskrive hvordan DnaA, DnaB+C, ssDNA og topoisomeraser er involveret i DNA replikationen. Devlin, s.169
Devlin, s.173 – fig.4.11
Devlin, s.171 - fig.4.8

DnaA, DnaB and DnaC have an important role in initiating the replication process. (fx E.Coli) 

DnaA – also called initiator protein. It binds to the origin of replication, where it recognizes 4 sequences of 9 nucleotides that are methylated. 4 DnaA molecules bind as a tetramer on one 9-nucleotide sequence.

DnaC – binds after DnaA is bonded and acts as a “matchmaker”, because by associating to OriC it allows DnaB to bind.

DnaB – DnaB is a helicase. Helicases are enzymes that separate the parental strands locally before the replication. Helicases bind to single stranded DNA (doesn't matter which one) and move along it in fixed direction, with each step requiring hydrolysis of ATP. This pushes the parental strand apart. The replication fork is created by DnaB.  

ssDNA binding protein – after the two complementary strands of DNA have been separated by helicases, there is a big chance that they are going to form hydrogen bonds between each other, since they are complementary. A special protein called singe-stranded DNA binding protein, that has a big affinity for single stranded DNA, binds to the separated strands. It has the following functions:

• prevents the DNA from reannealing

• prevents the DNA from making hairpins

• aligns the template strands for rapid DNA synthesis

• falls off when DNA polymerase has passed over

These proteins are also important in recombination and repair.   

Topoisomerase – Replication requires removal of positive supercoils created by overwinding of the helix immediately after the unwinding site.

In E.Coli:

• Topoisomerase I removes negative supercoils, not sufficient for replication.

• Topoisomerase II (gyrase) removes positive supercoils by inducing negative ones.

In human cells:

• Topoisomerase I is capable of removing both negative and positive supercoils

• Topoisomerase II is not a gyrase, but is essential for the termination step, segregation of chromosomes and appears to attach DNA to special sites to special site during interphase.

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6. Definere begreberne proofereading og processivitet og angive i hvilen grad de forskellige DNA-polymeraser har disse egenskaber.

Proofreading
Stryer, s.752-3
Devlin, s. 164-5
Stryer, s.750 - fig. 27.11

DNA polymerases help ensure the accuracy of the replication process in two ways:

Initial selection - based on the fit of the incoming nucleotide into the enzymes active site. Only an incoming nucleotide that makes the correct hydrogen bonds with the nucleotide on the template strand can be added to the growing chain. The error rate is 10^-4

Proofreading - the proofreading mechanism relies on the increased probability that the end of the growing strand with an incorrectly incorporated nucleotide will leave the polymerase site and transiently move to the exonuclease site. This site has a 3´-5´endonucleatic activity. This means that it removes the last few incorporated bases on the 3´-end, including the wrongly incorporated base.

In other words, a primer with a properly base paired nucleotide is a good substrate for addition of the next nucleotide, and a poor substrate for the 3´-5´ endonuclease. On the other hand, a primer with a mismatched terminal nucleotide is a bad substrate for addition of the following nucleotides, and a good substrate for the 3´-5´endonuclease.

The result of proofreading is removal of virtually all mismatched nucleotides before the next nucleotide is added. It also leads to removal of some properly incorporated nucleotides.

DNA polymerases did not evolve ever-increasing accuracy of proofreading, probably because of the following reasons:

• extremely accurate proofreading is energetically costly

• it the process, a lot of correctly incorporated bases will be removed

• DNA repair systems found in the cell deal with residual errors

• a certain degree of mutations produces a population with some degree of genetical variation, which are better able to survive in changing conditions 

Processivity
Devlin, s.172, fig.4.10
Stryer, s.762

Processivity - the ability of an enzyme to catalyze many consecutive reactions without releasing its substrate.

The processivity unit of DNA polymerase III is a protein, called sliding clamp. They hold the DNA polymerase III in contact with the growing DNA chain. It is formed as a star-shaped ring. In the center of the ring there is a hole that can readily accommodate a duplex DNA molecule, yet leaving enough space between the DNA and the enzyme to allow rapid sliding and turning during replication.

A clamp-loading protein binds to DNA and assembles the sliding clamp. DNA polymerase binds in the place of the clamp loader and becomes processive.

A high processivity means also a fast addition of nucleotides, since the DNA is not released and caught again a lot of times.

DNA polymerase I has a endonuclease 3´-5´proofreading system, a relatively high processivity, (releases the DNA after adding 20 nucleotides) and a relatively fast synthesis (10 nucleotides/s).  

DNA polymerase III has also a endonuclease 3´-5´proofreading system, even higher processivity (doesn’t let go of DNA until the entire replication is finished) and an extremely fast synthesis (1000 nucleotides/s).

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7. Beskrive hvordan DNA syntesen termineres.
Devlin, s.181, fig 4.17
Stryer, s.765

The termination of the replication is different in eukaryotes and prokaryotes.

In prokaryotes (that have a circular genome), the termination generally occurs 180 degrees away from the OriC. In some small viruses, the termination occurs wherever the two replication forks meet.

On the other hand, in E.Coli, there are special termination sequences that constrain termination within a region, by preventing forks from proceeding past the region.

In eukaryotes (having a linear genome) the free ends of the linear DNA molecules introduce several problems.

The continuous, leading strand can be theoretically synthesized all the way to the end of its template, by the 5´-3´ polymerase activity of  DNA polymerase III.

The discontinuous, lagging strand can not be synthesized all the way to the end of the template strand, because there is no place to synthesize a primer to which the nucleotides opposite the end of the template can be added. If the last nucleotides should be added, then the primer should be placed just outside the template strand, and that is not possible.

Even if a primer is created complementary to the very end of the template strand, after the removal of the primer, it will leave a gap and an incomplete 5´end of the lagging strand is created, since no polymerase adds nucleotides to the 5´-end of the strand.

This may not be a problem in a single generation, but after many cycles of replication, chromosome ends would be shortened to that degree that essential genes become lost and the cell dies.

In order to avoid this problem, the ends of the eukaryotic linear chromosomes are special structures called telomeres, which contain many repeats of a six-nucleotide, G-rich repeat sequence.

Human telomeres contain hundreds of tandem repeats of the sequence TTAGGG. The 3´-end of one strand of the chromosome extends about 18 nucleotides beyond the complementary strand with 5´-end, leaving three repeats as an overhang.

The overhanging 3´-end fold back on itself and makes G-G hydrogen bonds. It also binds extra proteins that define the chromosomes length and protect it from recombination. (since the end of the DNA molecule tend to trigger recombination)

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8. Definere mutagener og carcinogener og forklare hvordan mutationer kan induceres af kemikalier, bestråling, polymerase felj og interkalering i DNA helix.
Devlin, s.192-194
Stryer, s. 768-770

Mutagen – An agent, such as a chemical, ultraviolet light, or a radioactive element, that can induce or increase the frequency of mutation in an organism.

Carcinogen – substance that raises the incidence of cancer. They can be direct carcinogens or procarcinogens. Procarcinogens in their native form do not damage DNA, but they can be activated by metabolitic processes into carcinogens and damage DNA.  
All direct carcinogen substances are also mutagens, but all mutagens are not carcinogens.  

Mutations are changes in the DNA sequence. They can result from:

1. replication error

2. damage to the DNA

3. errors during repair of the damage

Point mutations – change in the single base pair.

There are several types of mutations:

• Substitution – one base pair is substituted with another. It can be a transition, in which one purine is substituted for another or one pyrimidine is substituted for another. It can also be a transversion, where a purine is substituted for a pyrimidine or vice versa.

• Deletition – of one or more base pairs

• Insertion – of one or more base pairs.

• Triplet expansion – a great increase in the number of repeating triplets. It number tends to become bigger from one generation to another. Can be seen in Huntington disease.

Chemical agents can induce mutations by chemically modifying the DNA.

Fx. HNO₂ reacts with bases containing an amino group. Treatment of DNA with HNO₂ results in the conversion of adenine to hypoxanthine. Hypoxanthine pairs with cytosine, inducing a transition from A-T to C-G after the replication.

Ultraviolet light can also act as a mutagen. The ultraviolet component of sunlight is a ubiquitous DNA-damaging agent. Its major effect is to covalently link adjacent pyrimidine residues along a DNA-strand. Such a pyrimidine dimer can not fit into a double helix and so replication and gene expression are blocked until the lesion is removed.  

X-rays are also powerful mutagens. They cause jonisation of the base pairs, so that:

• Instead of A-T, we have A* - G – transversion. (T, pyrimidine is replaced with G, purine)

• Instead of T-A, we have T* - G – transition. (A, purine is replaced with G, purine)

• No mutation, in the case of C* - G.

X-rays produce also double stranded breaks at any stage of the cell cycle in a dose-dependant linear fashion. 
 

Mutations can also be induced by substances that can intercalate in the DNA-helix. They are flat aromatic molecules that slip in between adjacent base pairs in the DNA double helix. That way DNA polymerase can introduce extra base pairs (which make hydrogen bonds with the intercalated molecules) on the daughter strand. They shift the reading frame in translation.

DNA polymerase can also make mistakes, even though it has proofreading and initial selection activity. Because of these two properties, the error rate is reduced to 10 ^{– 9}. So, there are still mutations. That is why repair systems are needed.

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9. Gøre rede for vigtigheden af DNA repair. Beskrive princippet i DNA repair ved base excision og nukleotid excision i prokaryote og eukaryote celler
Devlin, s.191
Devlin, s.194-6, fig.4.28, fig. 4.29
Stryer, s.770-3

The accuracy of DNA replication is very big, but not 100%.

DNA polymerase increases the replication accuracy by 2 processes. The initial selection brings the error rate down to 10^{- 4} and the proofreading down to 10 ^{- 9}.

Still, there is handful of errors introduced during each round of replication.   

The maintenance of the integrity of the genetic message is key to life. Consequently, all cells possess mechanisms to repair damaged DNA.

A difference should be made whether the base itself is damaged or if the wrong base is incorporated!!!

If the base itself is damagd, there are three pathways of repair:

• Direct pathway - fx. an enzyme binds to the distorted region (pyrine dimer) and uses light energy to cleave it into the original state

• Base-excision pathway

• Nucleotide excision pathway

The basic principle of excision-repair pathway is the following:

• Removal of the damaged nucleotide

• Leaving a gap in the DNA

• Re-synthesis using the genetic information on the opposite strand

• Ligation to restore continuity of the DNA

Base-excision repair

Used in removal of damaged bases (methylated, delaminated, oxidized). I will explain it stepwise:

1. Detection: The damaged DNA is detected by a DNA glycosylase. Binding of the glycosylase flips the mismatched base into the active site of the enzyme.
    

2. Excision: removal of the single base from the DNA backbone by a glycosylase that cuts the N-glycosidic bond between the sugar and the base. The sugar-phosphate backbone is not broken. An AP (abasic) site is created, that must be removed. The sugar-phosphate backbone is cleaved at 5` to the abasic site by an AP endonuclease, but the sugar is still attached with its 3` end to the next nucleotide. AP lyase cuts this bond and completely removes the sugar-phosphate component that was linked to the damaged base.
    

3. Re-synthesis: The gap is filled by DNA polymerase I (prokaryotes) or DNA polymerase b (eukaryotes)
    

4. The remaining nick is sealed by DNA-ligase.

Nucleotide excision repair

Acts on a variety of damages, typically involving large adducts or distortion of the double helical structure of the DNA.

Stepwise:

1. Detection: the damaged base is recognized by a DNA repair complex
    

2. Excision: the segment around the damage is excised by an enzyme complex that makes two nicks (breaks the sugar-phosphate backbone) in the same strand, one on each side of the damage. An oligonucleotide (27-29nt in humans, 12-13 in E.Coli) is released.
    

3. Re-synthesis: The gap is filled by DNA polymerase I (prokaryotes) or DNA polymerase b (eukaryotes). The 3`- end of the nicked strand is the primer, and the intact complementary strand is the template.
    

4. The remaining nick is sealed by DNA-ligase.
   
    

Mismatch Repair
 

Mismatches are not DNA-damage, there are no damaged or modified bases present, just the wrong one of the four bases is incorporated. Thus recognition of mismatches relies on the distortion of the double helical structure.

When a mismatch is recognized, the cell has a problem because both bases are normal; it does not know which one of them should be excised.  The base lying on the newly synthesised strand must be the one which is wrongly incorporated, therefore the repair system must figure out which one of the two strands is the daughter strand.

The brief time during which newly synthesized DNA is not methylated provides time within which the strand can be recognized. DNA in most organisms is methylated at specific positions. When the repair complex recognizes these non-methylated positions, it can remove the wrongly incorporated base by excision repair.

Fx.
Hereditary nonpolyposis colorectal cancer (HNPCC or Lynch syndrome) results from defective DNA mismatch repair. This is not a rare disease, ca. 1 in 200 people will develop this form of cancer.
Mutations in two genes, called hMSH2 and hMLH1, account for most cases of the hereditary predisposition to cancer. These two genes incode proteins that obviously participate in recognizing, binding and cleaving of the mismatched basepairs.
This leads to accumulation of mutations thoriught the genome. In time, genes important in cell proliferation become altered, resulting in onset of cancer

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10. Beskrive telomere samt hvordan og hvorfor de bliver repareret. Angive at de fleste humane kræftceller indeholder aktiv telomerase mens normale somatiske celler (undtagen stamceller) ikke gør.
Devlin, s.181, fig 4.17
Stryer, s. 765-6, fig.27-35
http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/T/Telomeres.html

Telomeres are special structures on the end of eukaryotic linear chromosomes, which contain many repeats of a six-nucleotide, G-rich repeat sequence.

Human telomeres contain hundreds of tandem repeats of the sequence TTAGGG. The 3´-end of one strand of the chromosome extends about 18 nucleotides beyond the strand with 5´-end, leaving three repeats as an overhang. 

The overhanging 3´-end folds back on itself and makes G-G hydrogen bonds. It also binds extra proteins that define the chromosomes length and protect it from recombination (since the ends of the DNA molecule tend to trigger recombination). 
 

Telomeres are replicated and maintained by telomerase, a ribonucleoprotein complex that adds six-nucleotide repeats to the 3´-end of a telomere. Telomerases are specialized polymerases that carry their own RNA template.

The telomerase binds complementary to the template strand with the 3`-OH ending, binding a part of its RNA sequence with hydrogen bonds to the last few nucleotides of the template strand.
 

• Telomerase RNA then acts as template for the reaction

• The protein part of the enzyme acts as a reverse transcriptase, which synthesizes DNA using RNA as a template.

After 6 nucleotides are added, the telomerase can disassociate and bind again, adding 6 extra nucleotides for each time it binds.  

Telomeres do not have to remain exactly the same length. Shortening is not a problem as long as the lost sequences do not encode proteins.

So, the cells need to maintain and repair their telomeres, in order not to lose gene coding sequences localized towards the end of the chromosome.

Cells that have differentiated and no longer divide, or will divide only a limited number of times, do not express telomerase; it is turned off during embryonic development. Somatic cells in the human body do not have active telomerase and the chromosomes shorten 50 base pairs for every cell division.

If cells from a new born baby are cultivated in the lab and left to divide, they would divide around 100 times and then die. A 70-year old men cell would divide only dozen times before they die. In the last case, the telomeres are already extensively shortened and it does not take long before gene-coding sequences are lost.

That is why telomeres are associated with aging.

Telomerase is expressed in:

• Embryonic stem cells - these cells are rapidly dividing and differentiate. Telomerase therapy can be used in exhausted stem cells; it can renew their ability to divide. 
   

• Cancer cells - these cells produce abnormally high levels of telomerase. The enzyme is found freely flowing in the nucleus. Thus, cancer cells can multiply at will and divide a lot of times without losing coding sequences. That is why telomerase in these cells is an attractive target for chemotherapy. Still, the effect would be delayed by many cell cycles.

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11. Beskrive specifik og homolog rekombination, herunder hvornår, hvordan og hvorfor det sker
Stryer, s.766-8
Devlin, s.186-8, fig. 4.23

Recombination – the exchange of genetic material.

Homologous recombination – exchange of genetic material between identical or nearly identical sequences.

Enzymes called recombinases catalyze the exchange of genetic material during recombination. A key intermediate in the process is a cross-like structure, called Holiday junction, formed by the four polynucleotide chains of the 2 chromosomes.

I will take is stepwise:

1. The process begins with recombinase binding to the DNA substrates and aligns them properly. Single strand breaks are made at homologous positions in strands of same polarity. (fx. both have 5`-3`).  

2. Strands partly unwind and one strand from DNA helix 1 reciprocally base pairs with the opposite homologous DNA molecule from DNA helix 2, forming a structure in which two intact strands are joined by two crossed strands. Regions in which one strand was from DNA molecule 1 and the other one is the homologous strand from DNA molecule 2 are called heteroduplexes. In the human genome, where two individuals differ by one base pair in 1000, there is at least one mismatched base pair per heteroduplex. It can undergo mismatch repair, converting one allele into another, by gene conversion.
 

3. Then the position of the cross-over moves by Branch migration – there is simultaneous unwinding of the DNA molecule 1 and 2, followed by rewinding of the two heteroduplexes, but the total number of hydrogen bonds remains unchanged. This process creates larger heteroduplexes.
 

4. DNA ligase seals the original nicks of the DNA and all four crossover strands are held on one crossover point.

5. Then the DNA molecules rotate, so that they form the holiday junction, which is a cross like structure held by recombinases. There are two resolutions to the holiday junction, where molecules are separated by a symmetrical cut in either horizontal or vertical direction.

A horizontal cut will cause no recombination, while a vertical one would cause recombination (if the gene does not lie in the heteroduplex region)

Homologous recombination occurs in mejosis, where two DNA molecules recombine with each other to form new DNA molecules that have segments from both parental strands. This process shuffles the combination of genes before they are passed to the next generation, generating genetical diversity.

Specific recombination

I was not able to find the therm *specific recombination* in the book. As I understand it, (which is not necesserily correct), specific recombination is the same as non-homologous recombination. So, the sequences exchanged between the chromosomes are not homologue.

There are 3 types on non-homologous recombination:

1. Site-specific recombination – a non-homologous recombination, in which specific enzymes catalyse the integration of a sequence into particular sites of DNA.

Fx.
Integration of bacteriophage λ in the bacterial genome of E. Coli. The λ-integrase catalyses specific nicking of λ-DNA and of a special sequence of E.Coli chromosome and the resealing involved. The process is reversable.

2. Transposition

- see målsætning (30)

3. Illegitimate recombination

No homology at all needed. No special seqeunces.
When genes are introduces in the human genome (with gene therapy) they randomly insert into the chromosomes probably using the enzymes involved in double strand repair/recombination. The DNA fragment can insert in a middle of a gene seqeunce causing muations. This is a major limitation of gene therapy.

(30)
12. Beskrive transposoner og deres forbindelse til antibiotika resistens
Devlin, s.190-1
Devlin, s.345-6, cc. 8.1

Transposition – movement of specific pieces of DNA in the genome.

Transposons – some pieces of DNA, that have two key features that enable them to move nearly anywhere into a target chromosome:

• They encode transposase enzymes

• They have insertion sequences that are recognised by the transposase, usually short inverted terminal repeat sequences, that are also used to define the transposon

Transposase catalyses the process of excision from one place in the genome and insertion into another; also replicative insertion into a new place without loosing the original piece of DNA.

Transposons can integrate into other chromosomes and move to new places within a chromosome. A significant fraction of the human genome has resulted from accumulation of transposons and insertion sequences.  

Transposons vary tremendously in length; they can consist of few thousand base pairs and contain two or three genes or many thousand base pairs and several genes.

Besides the transposase enzymes, transposons can also code for:

• Toxins

• Proteins that hydrolyse antibiotics. Transposons contain antibiotic resistance sequences and therefore have a key role in antibiotic resistance.

Plasmids - independent DNA molecules found in bacteria, contain genes that facilitate their transfer from one bacterium to another. Plasmids can contain transposons incorporated in their DNA.

As the plasmids transfer, e.g. between different infecting bacterial strains, the transposons containing antibiotic resistance genes are moved into the new bacterial strain. Once inside a new bacterium, the transposon can duplicate onto the chromosome and become permanently established in the cell’s lineage. The result is that more and more pathogenic bacterial strains have become resistant to increasing number of antibiotics.

Many cases have been documented in which a bacterial strain in a patient being treated with one antibiotic suddenly becomes resistant to that antibiotic, and simultaneously, to several other antibiotics, even though the bacterial strain had never previously been exposed to these other antibiotics

RNA STRUKTUR OG TRANSKRIPTION

(31)
1. Forklare forskelle mellem RNA og DNA
Stryer, s.26-28
Devlin, s.78

There are 3 important differences between DNA and RNA: 

• RNA’s sugar component is ribose that has a 2`-OH group, while DNA has deoxyribose that has only –H atom in position 2`. The 2`-OH group makes RNA unstable because it becomes more susceptible to a base-catalysed hydrolysis. The removal of the 2`-OH group from ribose and replacement with 2`-H atom decreases the rate of hydrolysis by approximately 100-fold under neutral conditions and perhaps even more under extreme conditions.
   

• RNA contains uracil, while DNA contains thymine, which is a methylated uracil. The methylation facilitates the repair of damaged DNA.

(these two differences make DNA more suitable as genetic material)

• RNA molecules are usually one-stranded molecules, while DNA is double stranded. Thus, molar ratios (A+U) and (G+C) are not equal. In some viruses double stranded RNA can also be found.
   

• RNA has enzymatic activity and DNA doesn't
   

• DNA has permanent mutations, while if a mutation fx. happens in an mRNA molecule, only that mRNA and the protein that is translated from it are affected.

(32)
2. Beskrive hvordan nukleotider er opbygget og forbundet i et RNA molekyle
Devlin, s.78

RNA is an unbranched linear polymer of ribonucleoside monophospahtes – ribonucleotides, joined together by a 3´-5´- phosphodiester bond.

Each ribonucleotide consists of the following components: 

• Ribose sugar – contains 2`-OH group which makes RNA more susceptible to hydrolysis.
    

• Phosphate-groups: monophosphat, diphsophat and triphosphat.

The RNA molecule is built in the way that there is phosphodiester bond of a monophosphat is formed between the 3´-OH end of one ribose to the 5`-OH group to the next ribose. The standard direction of RNA, as in DNA is 5´-3´, because the first base in the linear polynucleotide has a 5`-free end and it’s 3`-OH group is linked with a phosphodiester bond to the 5`-OH of the next nucleotide, which is what happens in the last nucleotide.

• Heterocyclic bases: purines – adenine and guanine linked to the sugar by at glycosidic bond though N-9 and pyrimidines – cytosine and uracil linked to the sugar thru N-1. These bases extend away from the axis of the backbone and can pair with complementary bases by intramolecular base pairing to form double-helical base paired regions called hairpins. They are the secondary structure of RNA.

(33)
3. Angive de tre hovedformer af RNA (mRNA, tRNA og rRNA) samt beskrive deres funktioner
Stryer, s. 129
Devlin, s. 82-85

mRNA – the template for protein synthesis or translation. It can be:

• Polycistronic – One mRNA is produced for a group of genes in prokaryotes.

• Monocistronic -- In eukaryotes a distinct mRNA is produced for each gene. (rarely also seen in prokaryotes)

In the cytoplasm, mRNAs have a relatively short lifespan. Some mRNAs are synthesised and stored in an inactive or dormant state in the cytoplasm, ready for a quick response when protein synthesis is required (fx. Fertilized egg synthesizes proteins without transcription).

mRNA is transcribed by RNA polymerase II. After the transcription, the mRNA is called the primary transcript or the pre-mRNA. It has to go through several modification: capping, slicing and polyadenylation, before it becomes mature mRNA.

mRNA comprises for only 5% of the entire amount of RNA-molecules.

tRNA – carries amino acids in an activated form to the ribosome for peptide bond formation, in a sequence dictated by the RNA template. There is at least one kind of tRNA for each amino acid.

tRNAs have a cloverleaf secondary structure, with three loops and a L-form tertiary structure. They contain an anticodon sequence on the anticodon loop, which is used for base pairing with the complementary sequence of the mRNA.

The acceptor stem - the CCA terminal is used for bringing the right amino acid, which is going to be incorporated in the growing nucleotide chain.

tRNA is synthesized by RNA polymerase III.

The average length of tRNA is ca. 75 nucleotides. It accounts for 15% of all RNAs.

rRNA – major component of ribosomes, constitutes nearly 2/3 of the entire ribosome masse. It plays both a catalytic and structural role in protein synthesis. It is the most abundant of the RNA molecules. (80% in E.Coli)

Different RNAs are formed from cleavage of one big primary transcript into smaller components. All rRNAs are synthesized by RNA polymerase I in the nucleolus, except 5S rRNA which is synthesized by RNA polymerase III in the nucleoplasma.

A short overview over which RNAs are synthesized by which polymerases:

mRNA - RNA pol II
tRNA - RNA pol III
rRNA - RNA pol I
5sRNA - RNA pol III

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4. Beskrive (kortfattet) funktionen af en RNA polymerase under transkriptionen
Devlin, s.208-210
Stryer, s.130-131
Stryer, s.782

Transcription or RNA-synthesis is the process by which RNA chains are made of DNA template. DNA nucleotide sequence information is transcribed into RNA sequence information. The enzymes that catalyse the transcription process are called RNA polymerases.

The transcription process takes place in three stages: initiation, elongation and termination. RNA polymerase performs multiple functions in this process:

1. It searches DNA for initiation sites, also called promoter sites (promoters). These DNA-sequences are on the same strand as the gene itself.
    

2. It unwinds a short stretch of double-helical DNA to produce a single-stranded DNA template from which it takes instructions.
    

3. It selects the correct ribonucleoside triphosphate and catalyses the formation of a phosphodiester bond. The 3´-OH group at the terminus of the growing chain makes a nucleophilic attack on the innermost α - phosphate of the incoming nucleoside triphosphate with the release of a pyrophosphate.
    

4. It detects termination signals that specify where a transcript ends.
    

5. It interacts with transcription factors - activator and repressor proteins that modulate the rate of transcription initiation by binding to specific sequences in the DNA. 

RNA polymerase requires the following components for transcription:

1. A template - the preferred template is a double stranded DNA molecule.

2. Activated precursors – all four ribonucleosides are required: ATP, GTP, UTP and CTP.

3. A divalent metal jon, fx. Mn²⁺

The synthesis of RNA is like that of DNA in several aspects:

• The direction of the synthesis is 5´-3´on the new synthesized RNA-strand.

• The mechanism of elongation is similar: the 3´-OH group at the terminus of the growing chain makes a nucleophilic attack on the innermost phosphate of the incoming nucleoside triphosphate.

There are though two major differences in contrast with DNA polymerase:

• RNA polymerase lacks the activity of proofreading and excising mismatched nucleotides.

• RNA polymerase doesn’t require a primer in order to start the transcription.

All cellular RNA is made by RNA polymerase. The chemistry of RNA-synthesis is identical for all RNAs, but the location of synthesis and the RNA polymerase type are different.
 

Angive at transkription i prokaryote celler kun involverer en RNA polymerase, mens tre forskellige RNA polymeraser er nødvendige for eukaryot transkription
Stryer, s. 783
Stryer, s.793, table 28.2
Devlin, s.210

There is a difference between the transcription processes in prokaryotes and eukaryotes when it comes to the number of RNA polymerases involved.  

In prokaryotic cells there is only one RNA polymerase that catalyses the transcription process. I will take the RNA polymerase in E.Coli as an example.

RNA polymerase from E.Coli is a very large (ca. 400 kDa) and complex enzyme consisting of 5 subunits (a₂, b, b`, s). The enzyme can be divided in the following manner:

• The σ - subunit - that helps find the promoter site which directs the RNA polymerase to the initiation site. A large fragment of the σ-factor is an α-helix that recognises the -10 sequence (the TATAAT sequence) on the DNA template. Thus, it participates in the initiation of RNA synthesis and then disassociates from the rest of the enzyme, when the newly synthesized RNA strand has reached 9 or 10 nucleotides in length. The σ subunit plays a key role in determining where RNA polymerase initiates transcription. 
  The σ-factor is also involved in transcription regulation; if it doesn’t bind, transcription doesn’t occur.
  The σ-factor decreases RNAs affinity for random DNA for 10^-4 times. It walks on the DNA until it finds the compatible seqeuence.    

• The core enzyme – the two α subunits, the β and β` subunits that contain the catalytic site. The catalytic site resembles the DNA polymerase. The core enzyme can not start the transcription on the promoter by itself. After the s factor is lost, the core enzyme makes a tighter bond to the DNA template.  

The core enzyme and the σ - subunit form the holoenzyme, which is capable of specific RNA synthesis in vivo and in vitro.       

 In eukaryotic cells, there are three different RNA-polymerases involved.

• RNA polymerase I – located in the nucleolus, where it transcribes the tandem array of genes for 18S, 5.8S and 28 S rRNA.
   

• RNA polymerase II – located in the nucleoplasm, synthesizes the precursors of mRNA as well as several small RNA molecules, such as those of the splicing apparatus.
   

• RNA polymerase III – located also in the nucleoplasm, where it synthesizes all tRNAs and the 5.S rRNA

(35)
5. Beskrive virkningen af speciffike inhibitorer på transkriptionsprocesserne
Stryer, s.791, 793-4
Devlin, s.210-1, cc. 5.1

RNA polymerase is an essential enzyme for life since transcription is the first step to gene expression. No RNA polymerase means no other enzymes.

α-amantin – binds very tightly to RNA polymerase II and thereby blocks the elongation phase of RNA synthesis. (RNA polymerase III gets affected at higher levels of α-amantin)

α-amantin is produced by the poisonous mushroom Amanita Phalloides; the poisoning starts with relatively mild gastrointestinal symptoms, followed 48 hours later by massive liver failure as essential mRNAs and their proteins are degraded but not replaced by newly synthesized molecules. The only therapy is supportive including liver transplantation.  More then 100 deaths result worldwide per year from the indigestion of this mushroom.

Rifampicin – an antibiotic that inhibits the transcription in prokaryotic cells, fx. in bacteria. Rifampicin specifically inhibits the initiation of RNA synthesis. Rifampicin does not block the binding of RNA polymerase to the DNA template; rather, it interferes with the formation of the first few phosphodiester bonds on the RNA chain.

Rifampicin blocks the channel in RNA polymerase through which the DNA-RNA hybrid just generated by the enzyme must pass for the first time.

Rifampicin does not hinder chain elongation one initiated, because the DNA-RNA hybrid has already passed though the channel and prevents the antibiotic from binding.

Rifampicin is used in the treatment of tuberculosis.  

Since mammalian RNA polymerase differs from the prokaryotic, inhibition of RNA polymerase in prokaryotes by Rifampicin is possible without great toxicity to the host. Rifampicin is 1000 times more toxic on bacteria then on human cells. This implies a good therapeutic index for the drug, that is, the ability to treat a disease without causing undue harm to the patient.

(36)
6. Beskrive hvordan transkriptionsprocessen (bakterier og højere organismer) involverer: RNA polymerase, promotorsekvens i DNA, DNA unwinding (negativ supercoiling), DNA template strand, nukleotid triphosphat (NTP) substrater, initiering, elongering og terminering af transkription
Stryer, s. 783-789, fig. 28, 7; fig. 28, 8; fig. 28-10; fig. 28,12
Devlin, s.208-214, fig. 5.2; fig. 5.3; fig. 5.4

RNA – polymerase:

RNA polymerase performs multiple functions in the transcription process:

1. It searches DNA for initiation sites, also called promoter sites (promoters). These DNA-sequences are on the same strand as the gene itself.
    

2. It unwinds a short stretch of double-helical DNA to produce a single-stranded DNA template from which it takes instructions.
    

3. It selects the correct ribonucleoside triphosphate and catalyses the formation of a phosphodiester bond.
    

4. It detects termination signals that specify where a transcript ends.
    

5. It interacts with transcription factors - activator and repressor proteins that modulate the rate of transcription initiation by binding to specific sequences in the DNA. 

It is important to mention that RNA polymerase has a very high processivity, meaning once it is bound to the template DNA, it won’t let go until the entire gene is transcribed.

Another characteristic of RNA polymerase is that it has no proofreading ability. This leads to higher errors rates, ca. 10⁵ times higher then DNA polymerase, but this is still acceptable since the errors are not given transmitted to progeny.

Promoter sequence in DNA - a specific sequence on the template strand of DNA that directs the RNA polymerase to the initiations site. In prokaryotes there are two promoter sequences, located 10 and 35 nucleotides upstream the start site of transcription. They have a consensus sequence, which has been depict by comparing the sequences of many prokaryotes.

Prokaryotic transcription begins with binding RNA polymerase recognising and binding its  σ -factor to a gene’s promoter sequence. The core enzyme is unable to start transcription at the promoter site.

The promoter sequence also regulates the transcription, depending on if it is a weak or strong promoter. A strong promoter has a sequence that corresponds in high degree to the consensus sequence.

In eukaryotes, there are three important consensus sequences: the TATA box, the CAAT box and the GC box. They all lie upstream from the start site, the TATA box lying closest. These promoter sequences do not bind the RNA polymerase directly as in prokaryotes; rather they bind to a transcription factor, which binds to other transcription factors and RNA polymerase II making a huge enzyme complex.

DNA-unwinding: Although RNA polymerase can search for promoter sequences when bound to double helical DNA, a region of the duplex helix must be unwinded before synthesis can begin. A region of duplex DNA must be unpaired so that nucleotides on one of its strands become accessible for base-pairing with incoming ribonucleoside triphosphates.

It is the RNA polymerase that unwinds the DNA helix. It has been estimated that each bound RNA polymerase molecule unwinds a 17-bp segment of DNA, which corresponds to 1.6 turns of B-DNA helix.

 The state in which DNA is double-helical is called a closed promoter complex, while the single stranded DNA is the open promoter complex.   

If the RNA chain would be copied without opening of the DNA double helix, the transcription complex and the newly synthesized RNA chain would have to wind around the double helix as they travel from the begging towards the end of the gene.

DNA template: A DNA template is, of course, necessary for the transcription process. The DNA strand which actually contains the gene sequence is called the coding (sense) strand. The template strand is complementary to both the coding DNA strand and the newly synthesized RNA strand, and is called the anti-coding strand.
This basically means that RNA polymerase uses the anticoding strand as a template.

This also means that the newly synthesized RNA and the coding DNA have the same sequence; just that RNA has U instead of T.

Nucleoside triphosphate substrates: All four nucleoside triphosphates: ATP, GTP, UTP, and CTP are necessary for the RNA polymerase. The principle of linking of their linking is the same as in the replication process. The 3´-OH group at the terminus of the growing RNA chain makes a nucleophilic attack on the innermost phosphate of the incoming nucleoside triphosphate with the release of a pyrophosphate – PPi.

Initiation: The initiation process requires that RNA polymerase has recognized the promoter sequence and bound its σ-factor to the promoter sequence. All this happens in a closed promoter complex, meaning the DNA helix is still intact. RNA polymerase then binds more tightly to the promoter sequence and the DNA helix is unwinded, creating the open promoter complex.

The DNA helix is unwinded in the -10 nucleotide promoter sequence, because the consensus sequence is rich in T and A, that make double H-bonds to the complementary strand, which are more easily broken.

The σ-factor is at this point still bonded to RNA polymerase. 

The unwinded DNA-strand binds the first incoming, meaning the initiating ribonucleotide and RNA polymerase then forms the first phosphodiester bond. This marks the beginning of the elongation.

Elongation: The elongation process begins with the formation of the first phosphodiester bond. After the eighth or ninth phosphodiester bond, the s factor disassociates from the RNA polymerase, which promotes a tighter bond between the RNA polymerase and the DNA template. 

As RNA polymerase continues down the double helix, it continues to separate the two strands of the DNA template. The region containing RNA polymerase, DNA template and the nascent RNA is called the transcription bubble. The newly synthesized RNA forms a hybrid complex, with the template DNA-strand. This RNA-DNA helix is about 8 bp long.

The lengths of the RNA-DNA hybrid and of the unwinded DNA region remain constant during transcription. This means that DNA is rewinded at the back at the same rate as it is unwinded in the front. It also means that the RNA-DNA hybrid is separated in the back as a new ribonucleotide is added in the front. This is all being catalysed by RNA polymerase.

The process of unwinding and restoring the DNA double helix is aided by DNA topoisomerase I and II, which are components of the transcription complex.

Topoisomerase I relaxes the positive supercoils introduced by the local unwinding of the helix, because it removes the negative supercoils introduced by the RNA polymerase on the site of transcription. This means that topoisomerase I goes in front and behind the transcription bubble.

Topoisomerase II introduces negative supercoils in DNA and can increase the efficiency of promoters on distant sites. 

During the elongation process, it is only one RNA polymerase that transcribes the entire gene, meaning RNA polymerase has a very high processivity and it won’t let go until the entire gene is transcribed.

Termniation:

The termination of the transcription can be ρ -dependant or ρ - independent.

ρ -independent termination:

The template DNA (the anticoding strand) contains stop signals. These contain a GC-rich palindromic sequence followed by an AT-rich region.

The GC-palindromic region transcribed into RNA is self-complimentary, so it makes a very stable hairpin, since the C and G form three hydrogen bonds. RNA polymerase seems to release the DNA-RNA hybrid as soon it has encountered a hairpin.

The AT-rich sequence on the DNA template is transcribed into a U-rich sequence on the RNA strand. The bonds between the RNA and DNA hybrid are now the ones between rU – dA. They are the weakest of all, which leads to disassociation of the RNA from DNA and release of the RNA polymerase.

After that, the template DNA pairs with the coding DNA and the transcription bubble closes.  

ρ - dependent termination:
Stryer, s. 789 - fig.28.12

The ρ-factor is a hexameric protein that has an essential RNA-dependant ATP-ase activity. It hydrolyses ATP only in the presence of a single stranded RNA, but not in the presence of DNA or duplex RNA. The ρ - factor is brought into action by sequences located on the nascent RNA rich in C and poor in G. 

The ρ - factor binds RNA in that way that the RNA molecule passes through the center of the protein structure.  The ATP-ase activity of the ρ - factor enables the protein to pull the nascent RNA while pursuing the RNA polymerase. When ρ catches RNA polymerase at the transcription bubble, it breaks the RNA-DNA hybrid by functioning as a helicase. 

There are 2 major differences in the transcription process in eukaryotes and prokaryotes:

• In eukaryotes, transcription takes place in the nucleus, while translation happens in the cytoplasm a little bit later. In prokaryotes, which have no nucleus, translation of the RNA begins while the transcription still goes on.

• Eukaryotes very extensively process nascent mRNA, and prokaryotes do not. Nearly all mRNA precursors in eukaryotes are spliced.

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7. Beskrive hvordan transkriptionsfaktorer og enchancere er involveret i regulæring af eukaryot transkription
Stryer, s. 794
Stryer, s. 878
Devlin, s.214-216

Transcription factors – protein molecules that bind to the DNA sequences of cis-acting elements. The cis-acting elements can be promoters or enhancers; the name is due to the fact that these sequences are on the same molecule as the gene which transcription is regulated.

The transcription factors in eukaryotes work in a different way then the σ-factor of prokaryotes. The protein binding factors bind to the specific sequence on the DNA (enhancer or promoter) and then bind the RNA polymerase. This mechanism is called recruitment.

Enhancer – DNA sequences, 20-30 nucleotides long,  that can increase the effectiveness of the promoter enormously. The position of the enhancers relative to the promoters is not fixed; it can vary substantially. Enhancers can exert their stimulatory activity over distances over several thousand base pairs. They can be located upstream, downstream and also in the middle of the transcribed gene. Enhancers can also be present on the coding or on the anti-coding DNA strand.

Enhancers function by serving as binding sites for specific regulatory proteins, which influence transcription by perturbing the local chromatin structure to expose a gene or its promoter sequence so it is more accessible to RNA polymerase II (rather then binding to RNA polymerase II themselves)

They are found in eukaryotes and their viruses.

An enhancer is only affected in specific cells – in which the appropriate regulatory proteins are expressed.  
fx.
Immunoglobulin enhancers function only in B-lymfocyttes. 

Transcription factors and other proteins that bind to regulatory sites on the DNA can be regarded as passwords that cooperatively open multiple locks, giving RNA polymerase access to specific genes. This discovery makes it possible to understand how genes are selectively expressed in eukaryotes.

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8. Beskrive de grundlæggende træk for typiske prokaryote og eukaryote promotorer
Stryer, s. 784-785
Stryer, s. 794-795
Devlin, s.216-217

Promoter – specific sequence on the DNA template that direct the RNA polymerase to the proper initiation site for transcription.
 

Prokaryotic promoter

A striking pattern is evident when the sequences of many prokaryotic promoters are compared. Two common motifs, each 6 nucleotides long, are present on the upstream side of the start site.

• The first one is the -10 sequence, called also the “Pribnow box”, which is located 10 nucleotides upstream from the start site and has the consensus (average) sequence of TATAAT. This sequence is recognised by the s factor in prokaryotes.
   

• The second one is the -35 sequence, TTGACA, located 35 nucleotides upstream from the start site. It is also recognised by the σ-factor, which is a part of the RNA polymerase.

Therefore, it is the RNA polymerase itself that recognizes the promoter sequences in the prokaryotes, in comparison to eukaryotes, where a special protein, called the transcription factor recognizes a specific sequence and binds the RNA polymerase afterwards.
 

Promoters can be strong and weak.

Strong promoters have sequences that correspond closely to the consensus sequence. Genes with strong promoters are transcribed very often because the s-factor recognizes the promoter sequence more often (fx. Every 2 sec. in E.Coli)
Weak promoters have multiple substitutions in comparison to the consensus sequence. Genes with weak promoters are transcribed not that frequently. (fx. Every 10 min. in E.Coli)

Thus, the strength of a promoter sequence serves to regulate transcription. Regulatory proteins that bind to specific sequences near promoter sites and interact with the RNA polymerase also markedly influence the transcription of many genes.    

Eukaryotic promoter:

In eukaryotes, each of the three types of the RNA polymerases has its distinct promoter.

RNA polymerase I – transcribes from a single type of promoter, present only in rRNA genes, that encompasses the initiation site.

RNA polymerase III - responds to promoters that lie upstream or downstream the initiation site.

RNA polymerase II has a more complex mechanism. As already mentioned, RNA polymerase transcribes mRNA.  

Promoters for RNA polymerase II are also located upstream from the start site. The three most important promoter consensus sequences are:

• The TATA-box, called the proximal promoter, usually located between -30 and -100 (devlin, -25) nucleotides from the start site. The mutation of a single base of the TATA-box markedly impairs promoter activity.
   

• The CAAT and the GC box, the distal promoters, which are located further upstream, between -40 and -150 nucleotides away from the start site.

The positions of these upstream sequences can vary in comparison to prokaryotic promoters that have quite constant locations at -10 and -35.

The promoter sequences do not contact the RNA polymerase II directly. Rather, they require the binding of a transcription factor, a protein, in order to function. These protein factors do not bind only to their promoter, but also to another transcription factors and to RNA polymerase creating a large enzyme complex.

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9. Forklare kort hvordan translationel kontrol af genekspression kan reguleres af RNA bindende proteiner, der påvirker messanger stabilitet og, eller translationel initiering
Stryer, s.887-890; fig. 31.38; fig. 31.39
Devlin, s. 1059-1060; fig. 24.3; fig. 24.4.; fig. 24.5

Modulation of the rate of transcriptional initiation is the most common mechanism of gene regulation. However, the gene expression can also be modulated at posttranscriptional level.

A typical example is the iron metabolism in humans. Iron is an essential nutrient required for the synthesis of haemoglobin and many other proteins. However, excess iron can be harmful because it can initiate a range of free-radical reactions that damage proteins, lipids and nucleic acids.

There are sophisticated systems in the cell for the accumulation of iron in times of scarcity and for the safe storage of excess iron for later use.

There are three proteins that should be considered:

• Transferrin – transport protein that carries iron absorbed in the intestines through the serum into the liver
   

• Transferrin receptor – membrane protein that binds the iron loaded transferrin and initiates its entry into the cell
   

• Apoferritin – an iron-storing protein found in the liver and in the kidney. 24 polypeptide chains of apoferritin store 2400 iron atoms; one iron atom per aminoacid, given that a polypeptide chain consists of 100 a.a.

I will consider apoferritin first. The apoferritin mRNA has a stem-loop structure, termed the iron-responsive element in its 5’-untranslated region.

1. Under low iron concentrations, (when the iron should not be stored for later) the iron-responsive element is bounded by a specific protein, called aconitase, that blocks the translation of the mRNA. This means less apoferritin, and more free iron in the cells.

2. When the level of iron in the cell is increased, the iron binds to aconitase, which no longer can bind to the iron-responsive element on the mRNA. This means that mRNA can be translated, and new apoferritin is created that stores the excess iron. 

Now to the transferrin receptor. The mRNA for the transferrin receptors has also iron-responsive elements, but they are located towards the 3’ untranslated region of the mRNA.

1. Under low iron concentration, aconitase binds to these iron-responsive regions, so that exonucleases can not eat the mRNA from its 3’-end before enough transferrin receptor molecule have been synthesized in order to bind more iron-loaded transferrin.

2. Under high iron concentration, iron binds to aconitase, so the 3’-end of mRNA is no longer protected, so that exonucleases can eat the mRNA before a lot of transferrin receptor molecules are made, so less iron is absorbed in the cell. 

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10. Angive visse transkriptions faktorers rolle i carcinogenese: proto-oncogener og tumor-supressor gener
Devlin, s.217, cc. 5.3
Blicher Pedersen, Kræftens biolog; s.170

Oncogenes – genes whose mutated or over-expressed products contribute to carcinogenesis.

Proto-oncogenes – normal, non-mutated cellular analogs of the oncogenes.

• Their protein products are transcription factors that regulate cell growth and differentiation of a normal cell.

• Some proto-oncogene products cell membrane receptors that are involved in transduction of hormonal signals or recognition of cellular growth factors

- se også målsætning (102) 

Tummor-supressor gene - codes for proteins, that can stop the cell's growth and division. If a tumor-supressor gene mutates, it is uable to code for a protein at all or codes for a non-fucntional protein.

A tumor-supressor gene is completely inactive when both copies of the gene are mutated at the same time.
Thou, a single copy of the mutant gene causes Li-Fraumeni syndrome, an inherited condition predisposing to carcinomas of the breast and adrenal cortex, sarcomas, leukaemia and brain tumors.

p53 - a tumor – suppressor protein, product of a dominant proto-oncogene. It has 3 essential functions:

1.       p53 is a transcription factor that, on sensing damaged DNA, promotes transcription of genes that inhibit the cell division, giving  the cell time to repair the damage.

2.       It can alternatively promote transcription of DNA-repair genes.

3.       Under specific circumstances, it can promote programmed cell death – apoptosis.

Thus, the biochemical actions of p53 serve to keep cell growth regulated, maintain the information content of the genome and finally, eliminate damaged cells. All of these functions would counteract neoplastic transformation of a cell.    

If the the tumor-supressor gene that codes for p53 mutates, the DNA-binding power of p53 is affected. That way, it can not control the transcription of the genes needed to inhibit cell division, repair DNA or cause apoptosis.

Somatic mutation of p53 is found in about half of all human cancers.

Growth-stimulation of a cell is under control of proto-oncogenes, while growth-inhibition is controled by tumor-suppresor genes.

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11. Angive at de primære transkripter fra tRNA, rRNA og mRNA bliver processeret
Devlin, s.220-3, fig.5.10; fig. 5.11
Stryer, s. 797-8, fig. 28.23

All RNA copies of DNA sequences must be modified to mature, functional molecules in eukaryotes. In prokaryotes, tRNA and rRNA are modified, but not mRNA.

 The reactions of RNA processing can include:

• Removal of extra nucleotides

• Addition of nucleotides

• Base modification

• Separation of different RNA sequences by the action of specific nucleases

tRNA precursors are converted into mature tRNAs by a serious of alterations:

1. cleavage of a 5`-leader sequence (by RNase P)

2. splicing to remove an intron (exonuclease and ligase)

3. replacement of the 3´-terminal UU by CCA (with a free 3`-OH end)

4. modification of several bases

rRNA primary product of transcription is a long rRNA, termed 45S RNA, which contains the sequences of 38S, 5.8S and 18S rRNAs. Processing of 45S rRNAs goes on in the nucleolus. The processing is carried out by large ribonucleotide protein assemblies and follows a sequential order.

mRNA

- se målsætning (42)

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12. Beskrive (kortfattet) følgende for mRNA: splicing (fjernelse af introns), forekomst af alternativ og ukkorekt splicing (og hvordan alternativ splicing udvider repertoiret af genprodukter, mens inkorrekt splicing kan medføre sygdom), 5'-capping og 3'-polyadenylering. Angive kodende og ikke kodende sekvenser

Perhaps the most extensively modified transcription product in eukaryotes is the one of RNA polymerase II – mRNA. The immediate product of the RNA polymerase II is referred to as primary transcript or pre-mRNA.

 Pre-mRNA is modified before it leaves the nucleus. The enzymes that participate in this modification are recruited by RNA polymerase during the transcription process. RNA polymerase II contains a C-terminal domain that is extensively phosphorylated when the nascent RNA is about 25 nucleotides long. The phosphorylated C-domain recruits the capping enzyme, splicing and polyadenylation complexes to the transcript. This means that all the modifications are made during transcription. mRNA gets a cap on the 5’-end, a tail of ca. 250 adenylate residues on the 3’-OH and all the introns are spliced out.

5´-Capping
Stryer, s. 798; fig. 28.24
Devlin, s.225

 
As the transcription complex moves along the DNA, the capping enzyme complex modifies the 5`-end of the nascent RNA. The 5`-end of the DNA has a triphosphate-end which releases a phosphate by hydrolysis. The diphosphate-end then attacks the a phosphate atom of a GTP to form a very unusual 5´-5´triphospahte linkage. This distinctive terminus is called a cap. 

The N-7 terminus of the guanine is additionally methylated (cap 0). The adjacent riboses can also be methylated (cap 1 and 2). The latter methylation occurs by adding -CH₃ on the 2'-OH group of ribose.

Function of caps:

• Caps contribute to stability of mRNAs by protecting their 5`-ends from phosphatases and nucleases.  

• They also enhance the translation of mRNA.

Splicing
Stryer, s. 799-801, fig. 28.29
Devlin, s.226-5, fig. 5.14

Splicing – the process in which introns (intervening sequences) in the pre-mRNA are excised and the exons (expressing sequences) are linked together. Spicing must be exquisitely sensitive; a one nucleotide slippage point in a splice site would shift the entire reading frame and gives an entirely different and therefore non-functional protein.

Three sequences are necessary for the splicing process:

• The intron start-sequence, which is always GU. This is the 5’-splice point.
   

• The intron end sequence, which is invariant AG and represent the 3’-splice point.
   

• The branch site – which is an internal intron sequence located 20 to 50 nucleotides upstream the 3’-splice point. It contains an adenine which 2’-OH group of the ribose is used in the splicing.

The splicing principle is the following:

1. The 2’-OH group of the A-residue of the branch site attacks the 5’-splice point on the beginning of the intron, and cleaves the phosphodiester bond between the upstream exon and the GU sequence of the intron.
    

2. A 2’-5’ phosphodiester bond is formed between the A-residue and the GU of the intron by transesterification, meaning that the phosphodiester bond that used to join the end of the exon and the beginning of the intron now joins the beginning of the intron and its branch site.
    

3. This leaves the 3’-OH group of the upstream exon free. This group attacks the phosphodiester bond between the intron (the AG-sequence) and the first nucleotide of the downstream exon (3’-splice point).
    

4. The two exons are joined by a transesterification reaction, while the intron is released with its 5’-splicing point bound to the branch site and a free 3’-splice point.

Thus the splicing process consists of two transesterifications and uses no energy since the number of phosphodiester bonds stays the same.

Incorrect splicing
Devlin, s.227, fig. 5.16
Devlin, s.223; s.224 - cc.5.4

Mutations in the splicing points cause human diseases. A mutation in the invariant sequences on the beginning and in the end of an intron means that the correct splice junction can not be recognised. This leads to shifting of the reading frame, and an incorrect and non-functional protein.

Fx. if there is a mutation in the intron-start sequence, GU, then the splicing won’t happen because the exon-intron junction can not be recognized. The upstream exon is going to be longer, meaning the protein is going to have extra amino acids and probably be non-functional.

If there is a mutation in the intron stop sequence, AG, the splicing won’t happen here, so the downstream exon is going to be removed until a accidental AG sequence, where the upstream exons is going to be connected. The mRNA is shorter; the protein is missing amino acids.

It is also possible that a stop codon in the mRNA is encountered and the protein is terminated sooner.

Mutations that interfere with introns removal are a major cause of human diseases, fx. b-talasemia.

Alternative splicing
Stryer, s.803, fig. 28.33
Devlin, s.227, fig. 5.17

Alternative splicing – inclusion of different exons in the mature mRNA. There is only one pre-mRNA, but after the alternative splicing and the translation of the mature mRNA, distinct forms of proteins for specific tissues or developmental stages can be produced.

It is a widespread mechanism for generating protein diversity.

fx.
A typical example is tropomyosin proteins which are found in muscles. Each type of muscle has its own tropomyosin protein, even though a single gene is transcribed into primary transcript in all cells. Each cell type processes the primary transcript in a characteristic fashion.

 Recent evidence suggests that 30% of human genes are alternatively spliced.  

Polyadenylation
Stryer, s. 798
Devlin, s. 226, fig.5.15

Pre-mRNA is also modified at the 3-end. Most eukaryotic mRNAs contain a polyadenylate - poly (A) tail at the end, added after transcription has finished. 

The enzyme that creates this tail is called poly (A) polymerase. This is the principle:

• A cleavage sequence, AAUAAA on the pre-mRNA is recognised by specific endonucleases that cleave the primary transcript about 20 nucleotides downstream from the sequence.

• Poly (A) polymerase then adds about 250 adenylate residues to the 3’end of the mRNA.

The role of polyadenylation:

1. Increases the stability of mRNA

2. Enhances translation efficiency

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13. Beskrive strukturen og funktionen af et spliceosom
Stryer, s. 800-803, tab. 28.3, fig. 28-31

Spliceosome – a large, dynamic assembly composed of:

• snRNPs – small nuclear ribonucleoprotein particles, also called *snurps*. They are made of snRNA (small nuclear RNAs) associated with specific proteins. The snRNAs is transcribed by RNA polymerase II. The snRNAs involved in the spliceosome are: U1, U2, U4, U5, and U6.
   

• proteins called splicing factors
   

• mRNA precursor being processed  

The spliceosome has a vital role in splicing process.

U1 binds to the 5’-splice site and U2 binds to the branch point. A preformed U4-U5-U6 complex then joins the assembly to form the complete spliceosome.

U2 and U6 form the catalytic center of the spliceosome. 

The spliceosome uses ATP energy to carry out the accurate removal of the intron.

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14. Beskrive hvilke interaktioner der danner de sekundære og tertiære struktur af stabile RNA'er (tRNA og rRNA), herunder tRNAs hårnålestruktur, kløverbladsstruktur og L-form
Devlin, s.79-82; fig. 2.56; 2.58; 2.59; 2.60
Stryer, s.815-817; fig. 29.4; 29.6

RNA’s secondary structure

RNA molecules are single stranded, and their secondary structure results from relatively short regions of intramolecular base pairing, that is, base pairs formed from complementary sequences in the same molecule. The base pairing is created of double or triple hydrogen bonds between the complementary bases.

Even non-paired sequences of single stranded RNAs contain considerable helical structure, due to strong base stacking. The base stacking phenomenon is due to Van der Walls and hydrophobic bonds between the heterocyclic rings of the bases.

There are 11 bp. per turn; RNA resembles the A-DNA form.
 

Hairpins - representing RNA secondary structure, are double helical stem-loop regions in RNA-molecules. They can vary in connection with the length of the base paired regions and the size and number of loops.

RNA’s tertiary structure – results from base-stacking and hydrogen bonds between different parts lying far away from eachother in the molecule. 

tRNAs are excellent examples for both base-stacking, hydrogen bonding in a single RNA strand, creating a secondary and tertiary structure of the molecule.

tRNA Hairpins – double helical stem-loop regions in the tRNA. There are all in all four hairpins in tRNA, which I will describe in the cloverleaf structure.

Cloverleaf structure of tRNA

 tRNA is a single chain containing between 73 and 93 ribonucleotides. About 60% of bases are paired in four double-helical stems. Five regions are not paired: 

• the 3’-CCA terminal region

• TψC loop

• The extra arm, which contains variable number of residues

• DHU loop

• the anticodon loop, made up of seven nucleotides. This loop is present near the center of the entire tRNA sequence. The partial helix caused by base stacking in this loop, binds, by base-pairing to a complementary codon in the mRNA so translation can occur. 

These unpaired regions have the capability of forming base-pairs with bases in the same or other looped regions, contributing in this way to the tertiary structure of the tRNA.
 

tRNA L-form – a tertiary structure of tRNA. It is a very compact folded structure, stabilized by:  

1. hydrogen bonds (Watson-Crick base pairing) between bases laying in distant parts of the tRNA

2. base interaction involving more then two nucleotides. The bases can donate hydrogen atoms to the phosphate backbone. The 2’-OH of ribose can be an important acceptor or donor of hydrogen. (fig.2.60)

All these interaction contribute to the folding of the tRNA.

In the L-form of the tRNA, the CCA terminus extends from one end of the L, while the anticodon site is located on the other end of the L. (fig. 2.59)

Interactions found in tRNA are also seen in rRNA.

rRNAs are folded into definite structures with many short duplex regions and a lot of loops.

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15. Beskrive tRNAs adapterfunktion (anticodon) og aminoacyl 3'-ende (aktiveret aminosyre)
Stryer, s.815-8
Devlin, s. 82

Transfer RNA serves as the adapter molecule that binds to a specific codon of the mRNA and brings with it an amino acid for incorporation into the polypeptide chain.

Both functions are reflected in the fact that tRNA has two essential parts:

 The 3’-OH terminal CCA sequence – to which specific amino acids are attached enzymatically. The CCA sequence is located on the amino acid acceptor stem, which is 7 nucleotides long. The CCA sequence is not base paired. The amino acid attachment site is the 3’-OH group of the adenosine residue of the CCA sequence.

The anticodon triplet, which recognizes the complementary mRNA sequence. The anticodon triplet is found on the anticodon loop - a sequence 7 nucleotides long located almost in the middle of the tRNA molecule.

Three nucleotides in the mRNA make a codon. The codon and anticodon triplets form base pairs during peptide-bond formation. The base pairing is possible due to the base-stacking of the anticodon triplet making a partial helical structure that is capable of binding to it complementary region on the mRNA.

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16. Angive fordelingen af rRNA i ribosomerne
Stryer, s.824-6, fig. 29.17; 29.16
Devlin, s.83, s.243-tab.6.7

rRNA constitutes for 2/3 of the entire mass of the ribosome (1/3 made of protein). rRNAs are folded into defined structures with many short duplex regions. The key sites of ribosomes are made of RNA, while the proteins serve as structural scaffolding.  

Each ribosome is made up of a big and a small subunit, containing different rRNAs. The rRNA molecules in the ribosomes are named after their sedimentation constant.

There is a difference in the rRNA content in prokaryotes and eukaryotes:

• Eukaryotes – one ribosome (80S) is made up of 4 rRNAs and 82 proteins. The small subunit (40S) is made up of 18S RNA and 33 proteins. The large subunit (60S) is made up of 28S, 5.8 and 5S rRNA and 49 proteins.
   

• Prokaryotes - one ribosome (70S) is made up of 3 rRNAs and 55 proteins. The small subunit (30S) contains 16S rRNA and 21 proteins, while the large subunit (50S) has 34 proteins and 5S and 23S rRNA.

In eukaryotes, all rRNAs are transcribed by RNA polymerase I in the nucleolus, except 5S rRNA which is synthesized by RNA polymerase III in the nucleoplasm.

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17. Beskrive hvad antibiotiske inhibitorer af translationen er, og hvordan antibiotisk resistens kan opstå
Devlin, s.254-5
Devlin, s.347
Stryer, s. 838-4

Protein synthesis is central to the continuing of life and reproduction of cells. An organism can gain a biological advantage by interfering with the ability of its competitors to synthesize proteins, and many antibiotics and toxins function this way.

Many antibiotics work by inhibiting proteins synthesis. They exploit the difference between prokaryotic and eukaryotic ribosomes.  

Several examples:

• Streptomycin – binds to the 30S subunit of prokaryotic ribosomes, interferes with the initiation process and causes misreading of mRNA (initiation)

• Tetracycline – Binds to the 30S subunit and inhibits binding of aminoacyl tRNA. (initiation)

• Puromycin - resembles an aminoacyl t-RNA and binds to the A site, acting as an acceptor in the peptidyltransferase reaction. However, it can not serve as a donor, so it terminates the translation prematurely, leading to the release of an incomplete and non-functional protein. Has effect on both pro- and eukaryotes. Not too good for our cells.  (elongation)

• Chloramphenicol – binds to the peptidyltransferase center in the 50S subunit and inhibits forming of a peptide bond. (elongation)
   

Antibiotic resistance can occur with the help of transposons – small DNA sequences that have the ability to jump between chromosomes. Transposon sequences contain genes that facilitate their “jumping”, and often contain antibiotic resistance sequences, toxin production sequences etc.

The antibiotic resistance sequence can code for a protein that hydrolyses the antibiotic.

Plasmids, independent DNA molecules in bacteria, contain genes that facilitate their transfer from one bacterium to another. Plasmids can contain transposons.

As the plasmids transfer, e.g. between different infecting bacterial strains, the transposons containing antibiotic resistance genes are moved into the new bacterial strain. Once inside a new bacterium, the transposon can duplicate onto the chromosome and become permanently established in the cell’s lineage. The result is that more and more pathogenic bacterial strains have become resistant to increasing number of antibiotics.

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18. Angive at RNA kan have katalytisk aktivitet: selv-splicing og hydrolytisk aktivitet (hammerheads, Rnase P, groupe I og groupe II introns)
Devlin, s.86-88, fig. 2.64, fig. 2.65
Stryer, s.804-5

RNA can be catalytic. Enzymes whose RNA subunits carry out catalytic reactions are called ribozymes. There are five types of ribozymes. The first three carry out self-processing reaction, while the last two act as true catalysts on separate substrates:

1. Group I self-splicing introns – the self-splicing reaction requires an added guanosine nucleotide, that is necessary as a cofactor. The chemistry of the self-splicing is two transesterification reactions. This RNA has been found in bacteria and bacteriophage.
    

2. Group II self-splicing introns – the attacking moiety here is the 2’-OH group of an adenylate residue in the intron. It is found in the mtRNA precursors in prokaryotes.
    

3. Hammer head ribozymes are self-cleaving viral RNAs. These RNAs self-cleave, during generation of single genomic RNA from large multimeric precursors. The catalytic activity of this ribozymes is due to the fact that it can adopt complex structures, and the substrate RNA molecule adapts to the structure by making base pair. One part of the substrate can not make base pairs and becomes “cracked” and is cleaved in the process.  
    

4. Ribonuclease P – contains both a protein and a RNA component. It acts as a true enzyme in cells, cleaving tRNA precursors.  Ribonuclease P recognizes a constant structure in the tRNA, fx. its 3’-CCA end.
    

5. The ribosome itself in the translation process! The active site of the ribosome is on the larger subunit. There are no protein functional groups close enough to catalyse peptide bond formation, leading to the conclusion that some part of the RNA chain serves as the catalyst for the peptide bond formation.

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19. Forklare hvad der menes med det evolutionære begreb *The RNA World*
Devlin, s.87
Stryer, s.23

RNA molecules can serve both as catalysts and carriers of genetic information. This has raised the possibility that the earliest living organisms were based entirely on RNA and that DNA and proteins evolved later. This model is sometimes referred to as “the RNA world”.

In the RNA world, RNA played all major roles, including those important for heredity, the storage of information, and promotion of specific reactions, that is, biosynthesis and energy metabolism. 

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20. Beskrive RNA turnover (endo- og exonucleases) og forklare hvorfor RNA molekylers levetid er kortere end DNA molekyler
Devlin, s.229-30
 

The different roles of DNA and RNA in genetic expression are reflected in their metabolic rates.

DNA is a very stable molecule and has a longer life-time because of the following reasons:

• DNA is the genetic information in our cells that need to be given from generation to generation.
   

• It also serves as a template for replication and transcription during the entire life. The DNA molecule must therefore be preserved and various DNA-repair systems are found in the cell.
   

• When the DNA is not replicating or being transcribed, it is a metabolically inert molecule.
   

• It doesn’t have a 2’OH group as RNA, so it isn’t that prone to hydrolysis
   

• It is wrapped around the basic histones that protect the molecule from interaction with other basic molecules.

RNA isn’t that stable and has therefore a shorter life. There are several reasons for this:

• RNA, unlike DNA, has a 2’-OH group attached to its ribose, which makes it more susceptible to hydrolysis.
   

• RNA is a dispensable molecule and can be synthesized over and over from the DNA template. Therefore there is no RNA repair systems found in the cell which makes RNA more prone to mutation. This decreases the stability and life time of a RNA molecule. If there is a mutation in the RNA molecules, fx. mRNA, it will only affect the protein product of the respective mRNA and nothing else. The cell would waste too much energy and time to repair the mutation in mRNA which doesn’t have that bad affect overall.

The defective RNAs are degraded into nucleotides, which are reused for synthesizing new molecules.

So, all in all, RNA molecules are unstable. However, even the stable ones are degraded (turn over) pretty fast. Fx. tRNA in the liver has a half-life of 5 days, while mRNA has a half life of 30 hours.

Removal of RNAs from the cytoplasm is accomplished by cellular ribonucleases. There are two types:

• exonucleases – that degrade RNA from the 5’-3’ end

• endonucleases – that cleave phosphodiester bond within the RNA-molecule
   

Turnover number – number of substrate molecules converted into product by an enzyme molecule in a unit of time when the enzyme is fully saturated with substrate. This definition can be applied for RNA molecules, if we think of RNA as substrate molecules, while the exo- and endonucleases are the enzymes that degrade the mRNA.

So, the RNAs turnover number must be number of mRNA molecules that are converted by the endo- and exonucelases in a unit of rime, when the nucleases are maximaly saturated with RNA.

RIBOSOMER OG TRANSLATION

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1. Forklare ribosomets funktion. Kort beskrive dets struktur som store rRNA-protein komplekser opbygget af to subunits i alle organismer
Devlin, s.242-245
Stryer, s. 823

Function - Ribosomes are workbenches for protein synthesis. They coordinate the interplay of charged tRNAs, mRNA and proteins, which leads to protein synthesis.

Structure - Ribosomes are complex ribonucleoprotein particles made up of 2 dissimilar subunits - a big and a small one. Each subunit includes an RNA core, folded into a specific 3-dimensional structure, upon which proteins are positioned through protein-RNA and protein-protein interactions. The RNA makes up a total of 2/3 of the ribosome, while the protein part 1/3. The key sites in a ribosomes are composed almost entirely of RNA.

Polysome - several ribosomes can translate a single mRNA molecule simultaneously.

Ribosomes occur either free in cytosol or are bound to rER membranes.

The free ribosomes synthesize proteins that remain within the cytosol or become targeted to the nucleus, mitochondria or some other organelle.

Membrane-bound ribosomes synthesizes proteins that will be secreted from the cell or make up and function in other cellular membranes and vesicles

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2. Angive ribosomets tre sites for tRNA (A, amino; P, peptide, og E, exit site). Definere funktionen af disse tre sites
Stryer, s. 828-829, fig. 29.22; fig. 29.23; fig. 29.24
Devlin, s 250

Ribosomes have three tRNA-binding sites that bridge the small and large subunit.

The binding sites for the amino acids carried by tRNA, (the amino acids of the growing polypeptide chain) are located on the large subunit, while the binding site for the codon-anticodon interactions is located on the small subunit. This is possible since tRNA has a L-from tertiary structure, which allows that the acceptor stem is in one side of the L, while the anticodon on the other.

Thus, each tRNA molecule contacts both the large and the small subunit and both subunits have an A, P and E site.

A-site - for aminoacyl. This is also the so-called acceptor-site.
It is called like that because the aminoacyl-tRNA specified by the next codon in the mRNA message is bound to it. After translocation it is empty.  
In the A-site of the small subunit tRNAs anticodon is bound to the mRNA codon. 

P-site - for peptidyl, also called the donor site. At the beginning of the translation, the initiating methionyl-tRNA is placed into this position so that its methionyl residue may be transferred to the free α-amino group of the incoming aminoacyl-tRNA. As the translation goes on, new tRNA molecules carrying the adjacent a. a. occupy this site.
The P-site of the small unit contains the anticodon of the donor-tRNA before the translocation and afterwards, it contains the anticodon of the ex-acceptor tRNA.

 
E-site - for exit, the binding place for tRNA after it has completed its role in translation. When bound to this site, tRNA is freed from the ribosome.
 

The mechanism of protein synthesis is the following: The cycle begins with a peptidyl tRNA in the P site. An aminoacyl tRNA binds to the A site. When both sites are occupied, a new peptide bond is formed. The tRNAs and the mRNA are translocated through the action of an elongation factor, which moves the deacylated tRNA to the E site. Once there, it is free to disassociate and complete the cycle.

The ex-aminoacyl tRNA, which is now the peptidyl-tRNA occupies the P site, while a new aminoacyl t-RNA comes into the A site and the cycle begins again.   

The other end of each tRNA molecule interacts with the large subunit, where the amino acids are bound in the growing polypeptide chain. The acceptor stems of the tRNA molecules occupying the A and the P site converge at a site where a peptide bond is formed. The site is connected to the back of the ribosome by a channel, through which the nascent polypeptide chain passes during synthesis.

(53)
3. Beskrive at rRNA'er udgør fundamentale bestandele af ribosomet, samt angive deres distribuering
Devlin, s.243,fig.6.7

rRNA is the fundamental component of a ribosome. It was though that the catalytic part of the ribosome is a protein, since they are enzymes. The active site of the ribosome is located on the large subunit, and there are no protein functional groups close enough to catalyse the peptide bond formation, so it must be a part of the RNA chain that serves as a catalyst.

Therefore, the key sites in ribosomes are composed almost entirely of RNA.
The protein component serves as a structural scaffolding and reassembly.

Ribosomes are made of two subunits, a large and a small one. Each subunit includes an RNA core, folded into a specific 3-dimensional structure, upon which proteins are positioned through protein-RNA and protein-protein interactions.  

There is a difference in the structure of ribosomes in prokaryotes and eukaryotes.
 

Prokaryote ribosomes (70S):  

The large subunit is called 50S and is made up of:

• 23S rRNA

• 5S rRNA and

• 34 proteins

The small subunit is called 30S and contains:

• 16S rRNA

• 21 proteins
   

Eukaryote ribosomes (80S) are bigger and more components:

The large subunit is called 60S and contains:

• 28S rRNA

• 5S rRNA

• 5.8S rRNA

• 49 proteins

The small subunit, 40S:

• 18s rRNA

• ca. 33 proteins

(54)
4. Angive at det genetiske kode betsår af 64 tri-nukleotid kodons (61 svarende til aminosyrer samt 3 stop-kodons). Angive at der er undtagelser fra denne universelle kode i mitokondrier. Genkende start og stop kodons
Devlin, s. 235-6, tab. 6.1
Stryer, s.133-137

Genetic code – the relation between the sequence of bases in DNA (or its RNA transcripts) and the sequence of amino acids in proteins.

Codon – a group of three bases that encodes an amino acid. There are all in all 64 codons, since there are 4 bases and a codon consists of 3 nucleotides. (4³ = 64) 

The characteristics of the genetic code:

1. The genetic code is degenerated. There are all in all 64 codons and only 20 amino acids. This means that multiple codons encode for one amino acid. The degeneracy minimises the deleterious effects of mutations. 61 of the codons encode for an amino acid and three of them are stop codons.
    

2. The genetic code is non-overlapping.
    

3. The genetic code has no punctuations and no commas. The sequence of bases is read sequentially from the initiation codon to the stop codon.
    

4. The genetic code is nearly universal. A specific codon is translated into a specific amino acid in both prokaryotes and eukaryotes. Fx. mRNAs can be correctly translated by the protein-synthesizing machinery of different species. That way, we can express human genes in bacteria. However, the mitochondrial genetic code of different species differs slightly from the standard genetic code (fx. UGA is not a stop codon, but a Trp codon). The reason for the mitochondrial deviation from the rest of the cell can be because mitochondrial DNA encodes for distinctive set of tRNAs. The codon is not the only thing that can deviate. Protein-synthesis systems can also read the codon differently (fx. in ciliates). Both mitochondria and ciliates are organisms that branched off very early in eukaryotic evolution.

Start signal – or the translation initiation codon is AUG. It is usually the first AUG sequence encountered by the ribosome in the 5’-3’ direction of the mRNA. This codon specifies methionine, which also is the initiating amino-terminal residue in the polypeptide chain. It the AUG sequence is found in the middle of mRNA, then it codes for a normal methionine incorporated in the polypeptide chain.

Stop signals – or the termination codons are: UAG, UAA and UGA. They specify no amino acid. They are therefore not read by tRNA, but rather by specific releasing factors. When these factors bind to the ribosome, the newly synthesized polypeptide chain is released.

These signals are, of course, located on mRNA.

(55)
5. Definere begrebet 'reading frame' for mRNA. Angive at mRNA altid læses fra 5' imod 3'
Stryer, s.135; s.827
Devlin, s.236, s.245

Reading frame – groups of three non-overlapping nucleotides on the mRNA. The reading frame is established after the start codon AUG is located.
 

mRNA is always read in the 5’-3’ direction. This has important consequences.
The transcription also has the direction 5’-3’. If the direction of translation would be opposite that of transcription, only fully synthesized mRNAs could be translated. In contrast, because the directions are the same, mRNA can be translated while it is being synthesized. This actually occurs in prokaryotes, where almost no time is lost between transcription and translation; they are closely coupled in space and time.
On the other hand, in eukaryotes, the newly synthesized mRNA must first exit the nucleus and go to the free ribosomes or to the rER before it can be translated.

As already said, mRNA is always written, transcribed and read in the 5’-3’ direction.
Amino acids are both written and synthesized from the amino-terminal residue to the carboxy-terminus. Comparisons of mRNA sequences with sequences of the proteins they encode reveal a perfect, collinear, non-overlapping and gap-free correspondence. Deduced sequences can however differ from the mRNA sequences due to posttranscriptional modifications.

(56)
6. Angive at kodons i mRNA er komplementære til og læses af anticodons i tRNA
Devlin, s.236, fig. 6.1
Devlin, s.79
Stryer, s.831-832

Translation of the codons in mRNA involves direct interaction between the codon sequences of mRNA with complementary anticodon sequences in tRNA. Each of the bases of the codon forms a Watson-Crick type of base-pair with a complementary base on the anti-codon. The codon and the anticodon would then be lined up in an antiparallel fashion.

The anti-codon region in tRNA is an unpaired, base stacked loop of seven nucleotides. The partial helix cause by base stacking in this loop binds, by base pairing, to the complementary codon in mRNA. 

Each tRNA species carries a unique amino acid, and each has a specific three-base anticodon sequence. When the codon-anticodon interaction is made, the amino-acid is incorporated into the growing polypeptide chain.

The amino-acid in aminoacyl-tRNA does not play a role in selecting a codon. The codon is strictly selected by complementary codon-anticodon base pairing.

(57)
7. Definere begrebet "wobble hypothese" (kodons degenereret i tredje position). Beskrive hvorfor dette er særligt vigtigt i mitokondrier
Devlin, s.236-7, tab. 6.1
Stryer, s.832-3, tab. 29.3

Due to the degeneration of the genetic code, several tRNAs can code for one amino acid. Also one tRNA molecule can recognize more then one codon.

This can be explained by the wobble hypothesis, which permits less stringent base pairing between the first position of the anticodon and the last, degenerate position of the codon.

• If U is the first base in the anticodon, it can bind to both A and G in the codon.

• If G is the first base, it can bind to both U and C, which take the 3^rd position in the codon.

• If inosine, which is an unusual nucleotide in tRNA, occupies the first position in the anticodon, it can bind to three bases in the last position of the mRNA codon: U, C and A. This means that the same tRNA molecule comes for different 3rd positions of the codons.  

Two generalisations of the codon-anticodon interaction can be made:

1. The first two bases of the codon pair in a standard way. Recognition is precise. Hence, codons that differ in either of their first two bases must be recognized by different tRNA.
    

2. The first base of the anticodon determines whether a particular tRNA molecule reads one, two or three codons: C and A (one codon), U and G (two codons) or I (three codons). Thus, part of the degeneracy of the code comes from the imprecision (wobble) in the pairing of the third base of the codon with the first base of the anticodon. 

The reason for the steric freedom in the binding of the third base of the codon is found in the small subunit of the ribosome. It has two adenine bases (A1492; A 1493) that form hydrogen bonds on the minor groove of the codon-anticodon duplex and check if Watson-Crick base pairing is present in the first two positions, but not in the third.

There are 61 codons, but cells have no more then 51 tRNAs. Some codons are read more efficiently by one anticodon then another.  

The wobble hypothesis is very important in mitochondria, whose genetic code is different from the one of the rest of the cell. Mitochondria have only 22tRNAs, 61 codons and they need to code for 20 amino acids.  The low number of tRNAs and high number of codons means that the wobble hypothesis is highly applied and therefore extremly important in mitochondria.

(58)
8. Beskrive monocystron (eukaryout) og polycystron (prokaryot) mRNA
Devlin, s.235

I am also going to describe the general differences between eukaryotic and prokaryotic mRNA.  

Eukaryotic mRNA has the following characteristics:

• it is monocistronic – one mRNA molecule encodes a single polypeptide

• has a cap of 7-methylguanosine at the 5’ end

• has a 5’ non-translated region, which can be short or up to a few hundred nucleotides

• has a translation initiation signal, which is very often the first encountered AUG sequence when the mRNA is read in the 5’-3’ direction

• has a translation termination signal

• has a 3’ non-translated sequence, usually up to 100 nucleotides

• has a poly (A) tail of up to 250 A-residues

Prokaryotic mRNA:

• most mRNA is polycistronic, meaning it encodes for two or more polypeptide chains, and it also includes two or more AUG initiation sequences. The first AUG sequence is found about 25 nucleotides from the 5’-end of the mRNA

• a ribosome positioning sequence, called Shine-Delgarno sequence, is found 10 nucleotides upstream from the AUG initiation sequence

• it has a 3’ non-translated sequence

• no cap, no poly (A)-tail

(59)
9. Beskrive hvordan tRNA aminoacyleres (charges) i 3'-enden med en aminosyre, herunder hvordan følgende elementer indgår i processen: korrekt (cognate) tRNA, aminosyre, aminoacyl-tRNA sythetase og ATP
Devlin, s.239-242
Stryer, s.817-822, fig. 29.7

In order to be incorporated in proteins, amino acids must first be activated by linkage to their appropriate tRNA carriers. This two-step process is catalysed by a family of aminoacyl-tRNA synthetases, called activating enzymes. 

I will describe the catalysed process stepwise: 

• The amino acid reacts with ATP. The carboxyl group of the amino acid gets linked to the α-phosphate group of AMP – aminoacyl - AMP. 

• The aminoacyl-group of aminoacyl – AMP is then transferred to a particular tRNA, forming an aminoacyl – tRNA in which the a. a carboxyl group is linked to the OH-group of adenylate ribose.

 Two molecules of ATP are consumed in the process; one in forming the ester linkage of aminoacyl t-RNA; the second one in driving the reaction forward.

The linkage of an amino acid to a tRNA is crucial for 2 reasons:

1. An attachment of an amino acid to a particular tRNA establishes the genetic code. When an amino acid has been linked to a tRNA, it will be incorporated into the growing polypeptide chain at a position dictated by the anticodon of tRNA.
    

2. The formation of a peptide bond between free amino acids is not thermodynamically favourable. The amino acid must first be activated for protein synthesis to proceed. The activated intermediates in protein synthesis are amino acid esters in which the carboxyl group of an amino acid is linked to either the 2’- or 3’-OH group of the ribose of the adenylate residue of the acceptor stem CCA.

Aminoacyl t-RNA – an amino acid ester of tRNA. Only one amino acid is attached by an ester bond to an OH-group of the adenylate ribose.

Polypeptidyl t-RNA – polypeptide ester of tRNA. An entire polypeptide chain is attached to the ribose.

The correct selection of both tRNA and the amino acid by synthetase is central to fidelity in protein synthesis.  It has been established that the aminoacyl-tRNA synthetases have separate structural domains that are involved in the catalytic process, tRNA and amino acid recognition.

1. Selection of the correct amino acid – some amino acids are easily recognized by their bulk, lack of bulk or by positive or negative charges on their side chains.  Other amino acids, which differ only in an OH- or CH₃ -group are more difficult to discriminate. If the amino acids differ in their chemical properties but are very similar structurally, the enzyme contains a jon in its binding site that can interact with fx. the OH-group but not with the methyl group. (threonine and valine)

If the two amino acids have similar chemical properties and almost same structure, an additional proof-reading step is required because the synthetase can not discriminate between the two.

 The error rate of synthetase is ca. 1 in 10 ^{4, 5} after the proofreading.  

2. Selection of the correct tRNA – tRNA molecules are more complex then amino acids, so it should be easier to recognise them, even though they have the same basic cloverleaf structure. The right amino acid can be recognized by:

·         their anticodon loop, when base pairs in the loop form hydrogen bonds with specific part of the enzyme

·         base pairs in their acceptor stem (several nucleotides besides the CCA sequence)

·         parts of the variable loop

·         micro-helix, containing only 1/3 of the native tRNA can sometimes be enough

(60)
10. Definere begrebet proofreading samt beskrive denne process på syntetase enzymet
Devlin, s. 241
Stryer. s. 820, fig. 29.9; fig. 29.10

Proofreading by aminoacyl-tRNA synthetases increases the fidelity of protein synthesis. Proofreading is used when tRNA is supposed to see the difference between two amino acids whose chemical properties are almost the same, so they can’t be separated by a specific jon that binds to only one of them in the catalytic site, but are structurally very similar, fx. one has an extra CH₃ – group. 

For that reason, most aminoacyl-tRNA synthetases contain an extra editing site in addition to the amino-acylation site. These two sites act as a double sieve to ensure very high fidelity. The flexible CCA arm of the aminoacyl-tRNA can move the amino acid between the activation site (a. acyl-site) and the editing site.  

Two amino acids that are chemically identical but structurally different, must differ in size. Thus, an incorrect amino acid can either be bigger or smaller then the correct one. 

The acylation site is hydrophobe, and rejects amino acids bigger then the correct one because there is insufficient room for it. So, if the amino acid does not fit in the activation site, it is automatically rejected, due to hydrophobic steric forces.

If the amino acid fits well into the editing site, the amino acid is removed by hydrolysis. The editing site is hydrolytic, and it cleaves amino acids that are smaller then the correct one.

The correct aminoacid would fit perfectly in the acylation site and won't at all fit in the editing site, because it would be too big.

Just to sum up, the aminoacyl t-RNA synthetase has several structural domains with different functions:

• the activation site, (amino-acyl site) that catalyses the esterification of the amino acid to the acceptor stem and rejects amino acids bigger then the correct one due to hydrophobic steric forces

• the editing site that is hydrolytic and cleaves amino acids that fit into it, because they are smaller then the correct one

• a site used for recognition of the right tRNA

(61)
11. Kort beskrive de tre faser af translationen: initiering, elongering og terminering, samt angive at der under alle disse faser forbruges energi i form af GTP

Initierings fasen – start-kodonen findes af det lille subunit på ribosomet. Det store subunit kommer til og det første aminoacyl-tRNA indsættes i P-stedet (intieringskomplex)

Elongation – ny indkommende aminoacyl t-RNA bindes til A-sitet i ribosomet. Peptidet overrækes til denne tRNA. Denne tRNA sættes nu ind i P-sitet. tRNAet uden peptid sidder nu i E-sitet, hvorfra den forlader ribosomet.

Terminering – stop-kodon i A-sitet (UAA, UAG, UGA). Termineringsfaktorer bindes og proteinet frigives.

Under alle disse faser forbruges energi i form af  GTP. Der bruges alt i alt 4 molekyler GTP pr. cyklus.

(62) 
12. Beskrive kort initiering af translation ud fra følgende:
Stryer, s. 827-9
Stryer, s.833, fig. 29.27
Stryer, s.837
Devlin, s.245-7

Prokaryote celler:

Ribosom positioneringssekvens (Shine-Delgarno)

The Shine-Delgarno sequence is a purine-rich sequence (GAGGGG) found in the initiator region of prokaryotic mRNA. It is located about 10 nucleotides upstream the initiator codon AUG.

The Shine-Delgarno sequence binds to a complementary region near the 3’-end of the 16S rRNA (small subunit of the ribosome). The number of base pairs between these mRNA and rRNA ranges from three to nine.

This region therefore has influence on where the translation process starts. Protein synthesis actually begins with the interaction of the Shine-Delgarno sequence of the mRNA with the rRNA of the ribosome.

AUG startkodon i mRNA

Another region, which of course has influence on where the translation starts is the initiation sequence AUG. Prokaryotic mRNA is polycistronic, meaning it encodes for several polypeptide chains, so it has several AUG starting sequences. The very first AUG sequence is always located more then 25 nucleotides from the 5’-end of the mRNA.

When the complex between the Shine-Delgarno sequence and the rRNA of the small subunit is formed, the initiator fMet-tRNA (charged with formylmethionine) binds to the initiator AUG sequence through Watson-Crick complementary base pairing.

Important: Shine-Delgarno of mRNA binds to rRNA, while AUG of mRNA binds to tRNA.

30S ribosomal subunit
 

The 30S ribosomal subunit, the small subunit, contains a 16S rRNA that has a pyrimidine rich sequence towards its 3’-end, which binds complementary to the purine-rich Shine-Delgarno sequence and helps position the mRNA in the right way. So, the codon-anticodon interaction between mRNA and tRNA happens in the small subunit.

The 30S ribosomal unit forms a complex with IF1 and IF3, which prevents premature joining between the large and small subunit, creating a dead-end ribosome.

The 30S subunit has an A, P and E site. It is in the P-site that the tRNA and mRNA make the codon-anticodon interaction. The A site is free for the next tRNA complementary to the second codon in the sequence.

 
Initieringsfaktorer (tre proteiner)
 

mRNA and fMet-tRNA must be brought to the ribosome for protein synthesis to begin.

There are three proteins, called initiation factors, IF1, IF2 and IF3 that are essential for the translation.  

IF1 and IF3 bind to the small subunit of the ribosome and prevent it from premature rejoining with the large subunit.

IF2 binds GTP, and the conformational change introduced in IF2 enables it to bind with fMet-tRNA.

The IF2-fMet-tRNA-GTP complex can then bind to mRNA, correctly positioned due to the Shine-Delgarno sequence with IF1-IF3-16srRNA, and the 30S initiation complex is formed.
Stryer, s.833 - fig. 29.27

fMet-tRNA (i P-sitet)
Stryer, s.828 - fig.29.21

Prokaryotic translation is initiated by a formyl-methionine tRNA. The initiator tRNA is first charged with methionine by aminoacyl-tRNA synthetase. The methionyl-tRNA is then formylated by transformylase.

 The fMet-tRNA is recognized by IF2, which in turn has to be bound to GTP in order to bind the small subunit of the ribosome. The tRNA is located in the P-site, since it is the donor tRNA, and it will give its amino acid to the growing chain.   

 The anticodon loop of tRNA complementary binds to the codon sequence of mRNA, after mRNA has been positioned rightly by the Shine-Delgado sequence and 16SrRNA.

Other methionines that have to be incorporated in the growing polypeptide chain are not formylated.

30S initiation complex:

1. mRNAs Shine-Delgado sequence binds complementary to the 16S rRNA of the 30S subunit

2. IF1 and IF3 bind to the 30S subunit and prevent it from prematurely joining the 50S subunit

3. IF2 binds GTP and after that binds the fMet-tRNA

4. fMet-tRNAs anticodon sequence recognises AUG sequence in the mRNA

Eukaryote celler:

Første AUG fra 5´-CAP enden af mRNA er startkodon

Eukaryotic mRNA is monocistronic, meaning it only encodes for one polypeptide chain. That means that there is only one initiation sequence, which here also is AUG.

It is usually the first AUG encountered in mRNA from its 5’-end that is the initiation sequence, but sometimes the surrounding nucleotide sequence or secondary structure is not appropriate for initiation and a later AUG is then selected.

The AUG codon is recognized by the anticodon sequence of the initiating met-tRNA.

40S ribosomal subunit
Stryer, s. 837 - fig.29.33
 

The 40S unit is contacted by an initiation factor, bound to GTP and Met-tRNA _i. The initiation factor prevents the 40S ribosomal subunit from binding with the large subunit.

This complex then attaches to the cap of the 5’-end of mRNA and searches for the AUG initiation sequence, so there is no need for a Shine-Delgarno sequence. The scanning process of mRNA utilises ATP.  

When the anticodon of Met-tRNA is paired with the AUG codon, the target has been found. 

The 40S subunit has an A, P and E site. It is in the P-site that the tRNA and mRNA make the codon-anticodon interaction. The A site is free for the next tRNA complementary to the second codon in the sequence.
 

 
Initierings faktorer (mange proteiner)
 

Eukaryotes initialise many more initiation factors then prokaryotes and their interplay are more complicated. The proteins are called eIF: eIF1, eIF2 anf eIF3.   

Met-tRNA (i P-sitet)
 

The initiation tRNA in eukaryotes is also Met-tRNA _i, but it isn’t formylated. No other tRNA can replace it in the initiation complex, not even the Met-tRNA_e, which carries a methionine which is to be incorporated during elongation of the polypeptide chain.

The initiation tRNA is bounded by an initiation factor-GTP complex, and the associated with the 40S subunit which then scans the mRNA.   

As usually, the anticodon of tRNA recognizes the AUG codon of mRNA.

Initiation in eukaryotes:

1. an initiation factor binds to ATP, binds to Met-tRNA _i

2. they bind to the 40S subunit

3. the complex scans for the AUG

4. complementary pairing between tRNA anticodon and mRNA codon

The basic difference in the initiation between prokaryotes and eukaryotes is that eukaryotes do not have a Shine-Delgarno sequence, so the ribosome has to scan the mRNA instead of using ribosome positioning. Prokaryotes have polycistronic mRNA, meaning more Shine-Delgarno sequences, more AUG start codons and can not be scanned.
 

(63)
13. Kort beskrive elongeringsfasen ud fra følgende:
Stryer, s.829-835
Devlin, s.245-251, fig. 6.8
 

Prokaryote celler:

50S ribosomal subunit samles med 30S (= 70S ribosom)

When the fMet-tRNA recognises the AUG of mRNA and binds to it, the 50S ribosomal subunit contacts the 30S subunit.

The 50S ribosomal subunit has three sites: A, P and E site.

When the first methionine-tRNA comes in, it binds to the P site. An aminoacyl (charged) tRNA, whose anticodon is complementary to the codon following the initiation AUG, takes the A site. The stage is set for formation of a peptide bond: the formylmethionine linked to the initiator tRNA will be transferred to the amino group of the amino-acid in the A-site. This transfer takes place at a special site of the peptidyltransferase enzyme.

After the formation of the peptide-bond, the uncharged CCA-region of the methionine tRNA is in the E site, while the CCA-region of the tRNA containing the growing polypeptide chain is located in the P-site, while their anti-codons are still in, respectively, the P and A-site of the small subunit.  

The polypeptide chain remains in the P-site of the large subunit, presumably growing in the channel that opens on the other side of the ribosome.
 

 
Elongeringsfaktorer
Stryer, s.834 - fig.29.28
Stryer, s.835 - fig.20.30
 

The process of elongation begins when the aminoacyl-tRNA complementary to the second codon is inserted into the A-site. The aminoacyl t-RNA doesn’t simply leave the aminoacyl t-RNA synthetase and diffuse to the A-site, rather it is delivered to the ribosome by an elongation factor – EF-Tu.

EF-Tu binds to the aminoacyl-tRNA only when bound to GTP. It protects the fragile ester bond between the adenylate-ribose and the carboxyl group of the amino acid. When the appropriate interaction is formed between the aminoacyl-tRNA anti-codon and the mRNA codon, the GTP is hydrolysed and EF-Tu releases tRNA. If the perfect bond is not formed, there is no hydrolysis and the aminoacyl tRNA is not transferred to the ribosome. 

A second factor, EF-Ts helps EF-Tu release the GDP and bind new GTP so it is capable of binding and delivering a new aminoacyl tRNA.  

EF-G is the prokaryotic enzyme that catalyses translocation of the uncharged tRNA from the P site to the E site of the smaller subunit, and the translocation of the polypeptidyl tRNA from the A to the P site again on the smaller subunit. A protein domain mimics the anticodon stem of tRNA.

GTP is used in the process, and the A-site is left completely empty.
 

Korrekt (cognate) ladet tRNA (i A-sitet)

Two enzymes play a crucial role:

Aminoacyl t-RNA synthetase is the enzyme that has to incorporate the correct amino acid to the CCA – acceptor stem of the tRNA. It has several structural domains that are used to find the right amino acid to the right tRNA.

It can determine if the right amino acid has been found by its bulk or lack of it, the positive or negative side chains. In the case when two amino acids are structurally similar but chemically different, it uses its catalytic site, which contains a jon which binds to one group but not to the other. If the amino acids are chemically similar, but have different structure, then a proofreading is required. For that purpose, synthetase has an extra editing site that is hydrolytic and cleaves amino acids that are smaller then the correct one, while the catalytic site is hydrophobe and cleaves amino acids that are too big.

The synthetase also checks if it has the right tRNA, by checking its anticodon sequence, its acceptor stem, the variable region and sometimes, the so-called micro-helices, which contain about 1/3 of all the sequences of an tRNA molecule. 

EF-Tu – the enzyme that brings the aminoacyl t-RNA to the ribosome after it has been released from the synthetase. It is a G-protein that binds the tRNA in its GTP form and disassociates from it if the anticodon triplet forms a complementary bond to the mRNA codon, since only then the GTP disassociates to GDP and Pi. If the codon-anticodon interactions are not complementary, then hydrolysis of GTP does not occur and the aminoacyl-tRNA is not released in the A-site.
 

Eukaryote celler:
Devlin, s.250
Stryer, s.837-838
 

60S ribosomal subunit samles med 40S (=80S ribosom)

When the AUG codon is located by the scanning complex of 40S subunit, Met-tRNAi and GTP, the large subunit is bound with the help of many initiation factors. GTP is then released from the complex.

The principle of moving the tRNAs from A to P and E site of the big subunit in eukaryotes is basically the same as what goes on in prokaryotes.
 

 
Elongeringsfaktorer (EF-1alfa og EF-2)

EF - 1α in eukaryotes has the same function as EF-Tu in prokaryotes.

EF - 1βγ is the equivalent of EF – Ts.

EF - 2 is the eukaryotic enzyme that catalyses translocation and it pretty much has the same function as EF-G in prokaryotes. (not sure if it also mimics the anticodon stem as EF-G does)

EF - 1α and EF-2 are G-proteins, bound to GTP they have a high affinity for aminoacyl t-RNA and ribosome, while bound to GDP they disassociate from them more easily. Same goes for the prokaryotic equivalents.

Korrekt (cognate) ladet tRNA (i A-sitet)

Again, completely the same principle as in prokaryote, just that EF-Tu is EF-1α.

Peptidyl transferase aktivitet (peptid binding)

Peptydiltransferase is an enzyme that transfers the growing peptide (or in the first step the initiating methionine residue) from its carrier tRNA to the -amino group of the amino-acid residue of the aminoacyl t-RNA specified by the next codon. 

Whatever the length of the growing chain, peptide bond formation always occurs through attack of the α-amino group of the incoming aminoacyl-tRNA on the carbonyl group of the ester linkage of the peptidyl t-RNA to form a tetrahedral intermediate that collapses to form the peptide bond and the deacylated tRNA.

The transfer occurs in a ribosome site called the peptidyltransferase center.  

Peptydiltransferase has never been dissociated from the large subunit or characterised as a specific ribosomal protein. It is thought that the large subunit is a complex ribozymes in which peptide bond formation is a RNA-catalysed reaction.
 

Translokation
 

After the formation of the peptide-bond, the uncharged CCA-region of the methionine tRNA is in the E site, while the CCA-region of the tRNA containing the growing polypeptide chain is located in the P-site, while their anticodons are still in, respectively, the P and A-site of the small subunit. 

The mRNA must move by a distance of three nucleotides and the dipeptidyl-tRNA must be repositioned to permit another elongation cycle to begin. The repositioning happens on the small subunit because the respective areas of the tRNA on the large subunit have already been moved by peptidyltransferase.

This process is called translocation and is catalysed by EF2 in eukaryotes and EF-G in prokaryotes, enzymes also called translocases.  

A protein domain of EF-G mimics the anticodon stem of tRNA! In the process, the uncharged tRNA is moved out of the P-site of the small subunit into the E-site and completely disengages from the mRNA. (The entire uncharged tRNA is now moved to the E-site)

The dipeptidyl (later polypeptidyl) tRNA is moved from the A to the P site. (The entire polypeptidyl tRNA is now moved to the P-site)

 The A-site is completely vacant (both in the large and small subunit) and a new cycle can begin. 

During translocation, GTP is used and splits to GDP and Pi.
 

Frigivelse af brugte (deacetylerede) tRNA’er gennem E-sitet

The uncharged tRNAs are released from the mRNA in the process of translocation by translocases.

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14. Beskrive termineringsfasen (stopkodon i A-sitet + releasefaktorer)
Devlin, s.251
Stryer, s. 835-6, fig. 29.31
 

The final phase of translation is termination, when a stop codon of mRNA occupies the A-site.

Aminoacyl tRNA does not normally bind to the A site of a ribosome if the codon Is UUA, UAG or UGA, because normal cells do not contain tRNAs with anticodons complementary to these trop signals. Instead, the stop codons are recognized by release factors.
 

The release factors are G-proteins that mimic the tRNA structure. They use ATP to transfer a water molecule to the growing polypeptide chain (the reaction of transfering a normal amino acid to the growing polypeptide does not require ATP since the energy has already been stored when the amino acid is acitvated).

The realease factors contain a special tRNA fold that contains a water molecule that can be brought in the peptidyltransferase center and hydrolyse the ester bond between the carboxy-group of the amino acid (peptide chain) and the OH-group of the adenylate ribose. The water molecule is basically transferred to the growing polypeptide chain and forms its –COOH end.

The polypeptide chain is then freed. GTP of the release factor is hydrolyzed GDP and Pi, presumably altering the confirmation of the release factor. The entire complex then disassociates.

So, the release factors free the polypeptide chain and disassemble the ribosome

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15. Beskrive hvordan mutationer kan resultere i ukorrekte proteinstrukturer
Devlin, s.238-9

Mutation – a change in a gene. Mutations happen on DNA level, and can be caused by a replication error, damage to the DNA or errors during repair of the damage. The DNA is then transcribed to mRNA and in the end result in a wrong amino acid sequence in proteins.

Point mutations are changes in a single base pair in DNA, thus in mRNA.

Silent mutations – a change in a single base pair occurs in the third position of a degenerated codon. The resulting protein is completely normal, since the same specific amino acid is incorporated in the growing polypeptide chain.  

Missence mutations – a change in a single base results in incorporation of a wrong amino acid. This mutation can also be in the third position of the codon, but the codon in that case is no degenerated. The created protein is not functioning completely, but normal amounts of the protein are still found in the cell.

Nonsense mutation - a type of a missence mutation, in which the point mutation transforms an amino acid codon into a stop codon or v.v. The result is, respectively a shorter or a longer protein. The translation of the longer protein goes on until a new termination codon is reached. The finished protein is completely not functioning.

Frameshift mutation – an insertion or deletion of a single base pair (due to intercalated molecules or inverted repeats in the DNA) changes completely the reading frame resulting in a completely wrong protein, often prematurely terminated because a stop codon is reached.

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16. Beskrive hvordan translationen kan hæmmes i bakterier af antibiotika og i eukaryote celler af toxiner
Devlin, s.254-5
Stryer, s.838-9

Protein synthesis is central to the continuing of life and reproduction of cells. An organism can gain a biological advantage by interfering with the ability of its competitors to synthesize proteins, and many antibiotics and toxins function this way.  

In bacteria:

Many antibiotics work by inhibiting proteins synthesis. They exploit the difference between prokaryotic and eukaryotic ribosomes.

Several examples:

• Streptomycin – binds to the 30S subunit of prokaryotic ribosomes, interferes with the initiation process and causes misreading of mRNA (initiation)

• Tetracycline – Binds to the 30S subunit and inhibits binding of aminoacyl tRNA. (initiation)

• Puromycin - resembles an aminoacyl t-RNA and binds to the A site, acting as an acceptor in the peptidyltransferase reaction. However, it can not serve as a donor, so it terminates the translation prematurely, leading to the release of an incomplete and non-functional protein. Has effect on both pro- and eukaryotes. Not too good for our cells.  (elongation)

• Chloramphenicol – binds to the peptidyltransferase center in the 50S subunit and inhibits forming of a peptide bond. (elongation)

In eukaryotic cells:
 

Eukaryotic translation is inhibited by diphtheria toxin, a product of Corynebacterium diphtharieae, a bacterium that grows in the upper respiratory tract of the infected person. The toxin binds at the cell membrane and a subunit enter the cytoplasm, where it inactivates the EF2-elongerings factor which catalyses the translocation. The tRNA is not moved to the adjacent site so elongation is inhibited.

Ricin – N-glycosidase that cleaves a single adenine from the 60S subunit of the eukaryotic ribosome, which becomes completely deactivated by this minor damage. The ribosome can not assemble for translation.

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17. Forklare hvorfor proteinsynthesen foregår lidt anderledes i mitochondrier
Devlin, s.251-254

Mitochondria are cell organelles that contain their own circular DNA molecule. They encode for 13 respiratory chain proteins and RNA molecules which are mitochondrion specific: 22 tRNA, which are enough to translate all amino acids and 2 rRNAs, a small and a large one.

Their translocation is independent of the rest of the cell, and they have besides their own tRNAs and rRNA - an RNA polymerase, aminoacyl tRNA synthetase and ribosomes. All these are unique for the mitochondrion, but are all synthesized in the cytosol and have to be imported. That is why mitochondria are not self-replicating organelles; they depend on the nuclear DNA. Around 90% of all proteins required by mitochondria are synthesized in the cytosol.

Few differences between the mitochondrion and cytosolic translation:   

• The number of tRNAs is smaller in mitochondria, only 22

• The genetic code is slightly different. It is mainly because of the fact that mitochondrias genetic code is not universal. Fx. UGA, the termination sequence in eukaryote cells codes for tryptophan in mitochondria. The stop codons in mitochondria are AGG and AGA

• Mitochondrial ribosomes are smaller, the rRNAs shorter

• The initiation tRNA is fMet-tRNAi, like in prokaryotes, but the formyl-modification process is different

• The aminoacyl tRNA synthetase is different

Mitochondria have a high degree of prokaryotic characteristics, since they are though to be descendants from aerobic prokaryotes that invaded and set up a symbiotic relationship with a eukaryotic cell. That is why antibiotics that have effect on bacteria and other prokaryotes can have slight influence on our mitochondria (but still 1000 times less influence on mitochondria then on bacteria).

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18. Angive at proteiner sythetiseres i retningen N mod C og at de kan modificeres og udskilles posttranskriptionelt
Devlin, s.255-271
Color atlas of biochemistry, s. 210

Proteins are synthesized from their NH₂ –terminal towards their COOH terminal. The initiation amino acid, methionine is bound with its –COOH-terminal to the OH-group of the acceptor stem’s ribose (on addenine) by an ester bond.  This ester bond is broken by the large subunits peptidyltransferase activity and replaced with a peptide bond with the following amino acid encoded by the mRNA.

So, the first amino acid in a polypeptide chain has a free –NH₂ terminal and its –COOH terminal makes a peptide bond with the following amino acid.

The last amino acid in the polypeptide chain must therefore have a free COOH-terminal, created by the hydrolysis of the ester bond between the amino acid and the tRNA. This reaction is catalysed by a termination factor that mimics the adjacent tRNA and provides the water molecule.  

Some