Principles of replication, transcription, translation and their regulation

Question 1

Life can look many different ways, what is the main structure that is the foundation for life? Why?

Answer 1

The plasma membrane. It consists of lipids and proteins that provide a sturdy yet pliable, hydrophobic barrier to the outside world. The plasma membrane helps with cellular integrity throughout the whole life cycle and blocks charged/polar molecules from entering. It is selective in what it takes in.

Cells also have the cytosol on the inside, an aqueous solution of soluble molecules (like RNA, enzymes, cofactors) and ions. Most cells also have a compartment for their genome; nucleoid/nucleus.

Question 2

Explain the difference between aerobic and anaerobic conditions.

Answer 2

aerobic - with oxygen.
anaerobic - without oxygen.

Organisms living in aerobic conditions obtain energy from transferring electrons from fuel molecules to oxygen, while anaerobic organisms transfer electrons to other electron acceptors, like nitrate (forming N2), sulfate (forming H2S), or CO2 (forming CH4).

Question 3

Organisms can be classified by their energy source. Explain the difference between phototrophs and chemotrophs.

Answer 3

Phototrophs trap and use sunlight to obtain energy and carbon, and chemotrophs derive their energy from oxidation of a chemical fuel.

Question 4

Describe the biggest differences between animal and bacterial cells.

Answer 4

Bacteria have no mitochondria and no nucleus, a circular genome and have different membrane structure than animal cells, like containing a peptidoglycan. Animal cells are generally much bigger and have extensive compartmentalization and membrane enclosed organelles, with a less rigid phospholipid bilayer plasma membrane.

Plant cells are very similar to animal cells, but additionally have a cell wall and chloroplasts.

Question 5

Biochemistry is both a biological but also a chemical science. What does all four of the “molecules of life” have in common?

Answer 5

That they are polymers of smaller monomers:

• Proteins are polymers of amino acids
• Nucleic acids are polymers of nucleotides (nitrogenous bases + sugars)
• Carbohydrates are polymers of sugars
  ( -Lipids are polymers of hydrocarbons)

The structural hierarchy: the cell and its organells > supramolecular complexes (held together by non-covalent interactions) > macromolecules (held together by covalent bonds) > monomeric units.

Question 6

What is important to keep in mind when studying biomolecules in vitro?

Answer 6

In vitro studies have of course revealed a lot of important biochemical information, but isolated components can (and will) behave a lot different in vitro than when it can interact with the rest of the cell components in vivo.

Question 7

What four elements most abundant in living cells? Why?

Answer 7

The four most abundant elements in living organisms, in terms of percentage of total number of atoms, are hydrogen, oxygen, nitrogen, and carbon, which together make up more than 99% of the mass of most cells.

They are the lightest elements capable of efficiently forming one, two, three, and four bonds, respectively; in general, the lightest elements form the strongest bonds.

Question 8

Name five important functional groups occurring in biomolecules, and how they are structured.

Answer 8

• Carbonyls: aldehyde R-C(=O)-H, carbon with a double bound oxygen or ketone R-C(=O)-R.
• Amino group R-N+H3
• Disulfide R-S-S-R
• Thioester R-C(=O)-S-R
• Phosphoanhydride R-O-P(-O neg, =O)-O-P(-O neg, =O)-R. (breaking this bond generate a lot of energy).
• Mixed anhydrides (very reactive) R-C(=O)-O-P(-O neg, =O)-OH

Question 9

All biochemistry follows the first law of thermodynamics: energy can not be created nor destroyed. Explain how energy from sunlight can be used to maintain DNA integrity.

Answer 9

The energy from sunlight is used to reduce carbon dioxide, and the energy released from this reaction is stored in covalent bonds in glucose. Grucose can then be oxidized in glycolysis (and in the cirtic acid cycle) and the potential energy in the bonds is used to phosphorylate ATP, or transfer electrons to electron carriers which through the ETC drives ATP synthase. The energy stored in the bonds of ATP can then be coupled to energetically unfavorable reactions to drive them, such as maintaining DNA integrity.

The energy could be converted to several types of work, like chemical transformations, heat, metabolism (producing compounds simpler than original –> higher entropy), all of which result in decreased entropy by polymerization, or information storage.

Question 10

How does cells perform reactions with a positive delta G (endergonic reactions)?

Answer 10

By coupling them to exergonic reactions (with a larger negative delta G), so that the overall reaction becomes exergonic which provides enough energy that can be used to carry out the endergonic reaction.

Example, the phosphoanhydride bonds in ATP breaking is an exergonic reaction, that is often coupled to endergonic reactions such as DNA synthesis to supply enough free energy to make the endergonic reaction favorable.

Question 11

Explain in short how enzymes work.

Answer 11

Enzymes are biocatalysts that lover the free-energy required to get over the activation barrier for specific reactions, without being consumed in the process.

They do this by providing a more “comfortable” fit for the transition state, a surface that complements the transition state in stereochemistry, polarity, and charge. The binding of enzyme to the transition state is exergonic, and the energy released by this binding reduces the activation energy for the reaction and greatly increases the reaction rate.

A further contribution to catalysis occurs when two or more reactants bind to the enzyme’s surface close to each other and with stereospecific orientations that favor their reaction. This increases the probability of productive collisions between reactants and leads to the reaction being a looot faster than the uncatalyzed reaction.

Question 12

Life evolved in water. What makes water a good solvent?

Answer 12

Water molecules are polar and with its almost tetrahedral structure (the two hydrogens and the two free electron pairs) it has the ability to form four hydrogen bonds. This is what makes liquid water to be stable over a large temperature span and make it a powerful solvent as it can stabilize ions, and polar biomolecules.

Question 13

Even though non-covalent interactions are weaker than covalent ones, The non-covalent binding of a substrate to an enzyme can be very stable, why?

Answer 13

Enzymes and their substrates interact by more than one non-covalent interaction, for example, it could interact by several hydrogen bonds and one ionic interaction with additional interactions through hydrophobic effect and van der waals interactions. Even though these are all individually weak and only last for a very short period of time, all of these would need to dissociate at the same time for the substrate to get unbound, which is highly unlikely. So the cumulative stability of non-covalent interaction is much higher than their individual strength summarized!

The energy released when an enzyme binds noncovalently to its substrate (through many weak interactions) is the main source of the enzyme’s catalytic power.

Question 14

Explain in short what acids and bases are (Bronsted-lowry).

Answer 14

Acids = proton donors and bases = proton acceptors.

Question 15

What does the pKa value say about an acid?

Answer 15

Low pKa = strong acid, high tendency to donate its proton(s).

High pKa = weak acid, low tendency to donate proton.

Question 16

What does the “p” in pH stand for?

Answer 16

“p”=the negative logarithm of. So pH means the negative logarithm of [H+] (in M), at neutral pH, the concentration of H+ = 10^-7 = pH 7 because log (1/10^-7)=7.

Question 17

How does buffer systems work?

Answer 17

Buffer systems consists of a conjugate acid-base pair that can “neutralize” small changes in pH (small additions of weak base or acid). They do this by reacting with the proton released from the acid or with the base. A good example is HAc (CH3COOH), which can ionize to form H+ and Ac- (acetate: CH3COO-), with it’s own equilibrium constant. When H+ is added the protons can be taken up by Ac- to reform HAc, and when OH- is added, an H+ can be donated from HAc to form acetate and H2O. This results in very minor changes in overall pH.

Conjugate acid base pairs have a buffering capacity +- 1 around their pKa, so HAc for example have a pKa of ~4.7, which means it has buffering capacity between pH 3.7 and 5.7.

Question 18

Give one example of a buffering system of cells.

Answer 18

One example of a buffering system is the side chain of histidine residues of proteins, which are weak acids with a pKa of 6. In natural pH, they are deprotonated and below 6 they are protonated, so they are an effective buffering system at natural pH.

If a weak acid is added and the pH falls, the side chain can take up the H+ and neutralize the pH. Fine tuning of pH is very important, as enzymes in different compartments have different optimal pH.

Some other examples: organic acids buffer the vacuoles of plant cells, ammonia buffers urine, bicarbonate buffering in the blood which regulate the rate of respiration (s. 326) and the phosphate buffer system, which acts in the cytoplasm of all cells, consists of H2PO−4 as proton
donor and HPO2−4 as proton acceptor:

H2PO(^−)4 ⇌ H+ + HPO(^2−)4

Question 19

Since DNA is complementary, one strand is in 5’->3’ direction while the other is in 3’->5’ direction. What consequence does this have for replication?

Answer 19

Since the strands are anti-parallell and replication only occurs in a 5’->3’ direction, one strand is replicated continuously (leading strand) and one is replicated discontinuously (lagging strand) in the opposite direction as the replication fork. The fragments are called Okazaki fragments (that are then ligated together by DNA ligase)

Question 20

DNA replication is semi-conservative, what does this mean?

Answer 20

That the two daughter DNA molecules contain one parent stand and one daughter stand, each parent strand is the template for the daughter strand.

Question 21

Can DNA replication start anywhere in the genome?

Answer 21

No! Replication always start at an Ori (origin of replication), the number of Ori in a genome differs between species.

Question 22

What are the two types of nucleases called and what do they do?

Answer 22

Endo- and exonucleases. Endonucleases cleave and degrade nucleic acids (DNAses if specific to DNA) in the middle of the DNA molecule, while exonucleases degrade DNA from either of the ends of the DNA molecule, either in a 3’->5’ direction or 5’->3’ direction.

Restriction endonucleases selectively cleave at specific sequences (some only where modification has happened, like methylation), very important in molecular biology and biochemistry.

Question 23

What enzyme catalyzes DNA replication? Explain the reaction in short.

Answer 23

DNA polymerase.

The reaction carried out by DNA polymerase is a phosphoryl (PO4^-) transfer, where the OH group of the 3’ end of the growing strand performs a nucleophilic attack on the alpha-phosphorous in the deoxynucleoside-triphosphate (A, G, C or T), forming a covalent bond between the 3’O and the nucleoside, elongating the strand and releasing an inorganic pyrophosphate (PPi). All DNA polymerases requre 2 mg2+ ions as cofactors. The deprotonation of OH -> O- is mediated by one of the mg^2+ the other mg2+ binds to the incoming dNTP and facilitates the departure of PPi, which in turn are stabilized by asp residues.

Each nucleotide added cost the equivalent of 1 ATP (GTP, CTP, TTP), so this is a very energy demanding process!

Question 24

Besides cofactors and substrate, DNA polymerases need something else to function, what?

Answer 24

They need a free 3’-OH group to start the catalysis, so a primer is needed. The primer is usually an complementary RNA molecule that is synthesized by specialized enzymes (eg. primase, a type of RNA polymerase) that synthesize primers when and where they are required.

Question 25

Why are deoxynucleotides (dNTPs) used in DNA synthesis instead of NTPs?

Answer 25

dNTPs are used in DNA replication instead of NTPs because dNTPs have a deoxyribose sugar, lacking a 2' hydroxyl group, which provides stability to the DNA structure. DNA polymerases specifically recognize and incorporate dNTPs, ensuring high fidelity in replication. Incorporating NTPs, which have a 2' hydroxyl group, would disrupt the DNA double-helix structure and decrease replication accuracy.

Question 26

What is meant by "processivity" of an polymerization enzyme?

Answer 26

Processivity describes how many monomers (nucleotides in the case of DNA Pol) on average are incorporated in the growing chain before the enzyme dissociates from/falls off the template, and lets you compare the rate of polymerizing enzymes. Different DNA polymerases vary greatly in processivity.

Question 27

Explain in short how DNA polymerases are structured.

Answer 27

Most DNA polymerases are shaped kind of like a human hand gripping the DNA, with a thumb and fingers. The palm of the "hand" contains the active site which in turn consists of two parts, the insertion site; where a new dNTP is added, and a postinsertion site. When the phosphate bond brakes, the enzyme moves forward, so that the inserted dNTP is moved into the postinsertion site.

Question 28

DNA polymerases have extremely high fidelity (accuracy). In E. Coli a mistake is only made about every 10^9-10^10 nucleotide, and considering that the whole genome is in the 10^6 order, this means it has extreme fidelity! What three things makes this possible?

Answer 28

- The optimal hydrogen bonding of A=T and G≡C only accounts for a part of this high fidelity. - Geometry in the active site also contributes. The shape of the active site is constructed so that only the correct pairings fit well, so if for example A is in the template, a G or C doesn't fit quite right sterically and therefore is not incorporated. An incorrect nucleotide may be able to hydrogen-bond with a base in the template, but it generally will not fit into the active site. - The biggest contributor to the high fidelity is that DNA polymerase also have "proof reading" in the form of 3'-5' exonuclease activity in a position close to the active site. So after each base is added, the enzyme repositions so that the base is in the exonuclease site, which is highly specific to incorrect matches. If the pairing is incorrect, the enzyme wont be able to position back and the phosphodiester bond is hydrolyze the the mispaired nucleotide and then repositions back to add the correct one. - Besides all this, there are also many DNA repair enzymes that go through the DNA to find and correct mistakes. Note: Some DNA pol doesn't have proof reading activity, eg Taq polymerase. Wonder how Thermus aquaticus survives without high fidelity?? Question for another day.

Question 29

Compare DNA pol I, II and III in terms of processivity and polymerization rate.

Answer 29

DNA pol III, responsible for replication, has a processivity of over 500 000 and a polymerisation rate of about 250-1000 nts per second! It has two beta-clamps that grip both strands of DNA, with lagging strand looped, which increases the processivity by several orders. The others are mainly for DNA repair and have much lower polymerization rate (10-40) and fidelity (3-1500). DNA pol I in E.coli also have 5'-3' exonuclease activity, which enables it to produce a nick where it finds an error, break down part of the strand and synthesize a new one to repair the error in a process called "nick translation" (also inportant to remove primers)

Question 30

DNA polymerase alone is not enough to efficiently and accurately replicate DNA, the replisome of E. Coli consists of DNA pol III + 20 other proteins/cofactors. Name three of these and explain what they do.

Answer 30

- Helicase: rips open the dsDNA to make the template strand accessible. - Topoisomerase: Relieves the strain on the surrounding double helix caused by the unwinding around the replication fork. - Primase: Synthesizes the RNA primer needed for the polymerization to be started. - DNA ligase: Covalently closes/seals nicks between the Okazaki fragments in the discontinuous strand. - DNA-binding proteins stabilize the separated strands.

Question 31

Describe the initiation of replication in E. Coli in detail.

Answer 31

Initiation: - 8 DnaA proteins binds to the consensus sequences in OriC in a helical formation which puts strain on the helix, which in turn denatures the double helix in the A=T rich DUE sequence (ATP dependent). - DnaB (hexamer) binds to both strands (loaded by DnaC) and migrates along the single-stranded DNA in the 5′ → 3′ direction, unwinding the DNA as it travels, creating the two replication forks. - SSBs (single stranded binding proteins) bind to and stabilize the separated strands, and DNA gyrase (DNA topoisomerase II) relieves the topological stress induced ahead of the fork by the unwinding reaction. - The initiation of replication is regulated by methylation. The oriC DNA is methylated by the Dam methylase that binds to a palindromic sequence GATC and methylates the adenine. When replication has happened, the OriC is hemimethylated, and DnaA can only bind to fully methylated OriC to initiate replication again. So Dam methylase needs to fully methylate OriC after replication, which regulates it.

Question 32

Name two important elements/consensus sequences of OriC.

Answer 32

The consensus sequences are short repetitive sequences which are the DnaA binding sites, five repeats of a 9 bp sequence and the DNA unwinding element (DUE), a region rich in A═T base pairs.

Question 33

Describe elongation of replication in E.coli in detail.

Answer 33

Elongation: - For the leading strand (5'-3' direction) primase synthesizes a 10-60 NT long primer and DNA pol III extends it until it gets back to the start, very straightforward. - For the lagging strand it's a few more steps: - Primase synthesizes primers for each Okazaki fragment and Pol III extends. - Pol I removes the primers and fills in the gap, DNA ligases then seals the nicks between each fragment (ATP of NAD+ dependent). - The coordination for synthesizing both strands at the same time is the complex part, The clamps need to be "juggled" with each Okazaki fragments, releasing an reattaching from the clamp holder in a ATP dependent reaction.

Question 34

Explain termination of replication in E. Coli in detail.

Answer 34

Termination occur at a given region "Ter", a region containing multiple copies of a 20 bp sequence in opposite directions, which acts as binding sites for the protein Tus (terminus utilization substance). - Tus binds and halts the replication fork. (The Tus-Ter complex can arrest a replication fork from only one direction, so only one Tus-Ter complex functions per replication cycle) which result in catenated DNA molecules (rings connected) - The DNA molecules are then nicked, separated and resealed by topoisomerase IV. The complex first encountered by either replication fork and halts it, which may prevent overreplication by one fork in the event that the other is delayed or halted by an encounter with DNA damage or some other obstacle. Remember that replication in eukaryotes are more complex! Eg waaay more Ori!

Question 35

Why does DNA contain thymidine instead of uracil?

Answer 35

Thymidine is like a tagged U, So that the maintenance enzymes recognize that it's DNA (selectivity) and can also be selective in finding errors.

Question 36

What is transcription and why is it such a highly regulated process?

Answer 36

Transcription is the process by which a cell makes an RNA copy of a piece of DNA. It's an energetically expensive process, each nucleotide added onto an RNA costs one ATP. Important no to transcribe things that are not needed.

Question 37

Describe the major similarities and differences between replication and transcription.

Answer 37

The direction of synthesis, the template, the chemical reaction are the same in replication and transcription. The major differences are that transcription don't require primers and instead begins at promoters, only regard a part of the DNA molecule as template and only one of the strands act as template, the 3'->5' strand. The other strand is called the coding/nontemplate strand and has the same sequence as the resulting RNA. (RNA also uses regular ribose's and U instead of T)

Question 38

What enzyme catalyze transcription and what does it need to function?

Answer 38

RNA polymerase catalyze transcription and requires a DNA template, ribonucleoside 5′-triphosphates (ATP, GTP, UTP, and CTP, note, U instead of T), Mg2+ (as DNA pol), the reaction is the same as for DNA pol.

Question 39

Which of the DNA strands is the coding strand vs the template strand? Why?

Answer 39

The coding strand is the non template stand, in the 5'-3' direction, while the template strand is the antiparallell 3'-5' direction strand. This because the RNA synthesis follows the rules for 5'->3' synthesis, and therefore is complementary to the 3'-5' strand. This means that the RNA will have the same sequence as the coding strand (with U instead of T). Note: The coding strand for a particular gene may be located in either strand of a given chromosome By convention, the regulatory sequences that control transcription are designated by the sequences in the coding strand.

Question 40

Transcription doesn't require topoisomerases or helicases to denature the DNA strands and aid in strain from opening up the transcription bubble, how?

Answer 40

Because RNA pol does all this itself!

Question 41

Transcription doesn't require a primer, but is instead initiated by the promoter. What about the promoter is important?

Answer 41

The promoter contains highly conserved sequences, but it is the topographic structure of the sequence in DNA that is most important, so some nucleotides are not as well conserved because they don't change the structure so much, while others are more important. Note, the first nucleotide added after the promoter is not cleaved, and this is what is recognized by the capping machinery in eukaryotes.

Question 42

Describe how RNA polymerase in E. Coli is structured.

Answer 42

RNA pol in E. Coli consists of five subunits, 2 alpha units in the back, two beta units that make up the "groove" which all bind other protein involved in transcription and one gamma in the lower back that together has a molecular weight of ~400 kDa. It also requires a sigma factor (of varying types but σ 70 is most common) that directs to which promoter it should bind and together tall these six units constitute the RNA polymerase holoenzyme.

Question 43

How does the fidelity of RNA pol compare to that of DNA pol? Why?

Answer 43

The fidelity of RNA pol is much lower than DNA pol, mainly because RNA pol lacks proof reading activity. Also, it is not nearly as important that an RNA is correct, as it is less short lived, so it makes sense.

Question 44

It would be extremely wasteful to start transcription at random sites in DNA, instead transcription is directed by promoters. Describe the important parts of a bacterial promoter in short.

Answer 44

Promoters are ~100 bp long sequences located about -70 to +30 upstream of the transcription start site. The specificity of promoters comes from their consensus sequences, which are optimized for binding RNA polymerase. The consensus sequences that the σ 70 factor bind to are the −10 region is (5')TATAAT(3′) and the −35 region (5')TTGACA(3′). Between the consensus sequences there are spacers, with highly conserved lengths but not sequence. For some promoters for highly expressed genes, there is a third one; the UP (upstream promoter) element (-60 to -40) rich in A=T.

Question 45

Why are the length of spacers in between consensus sequences in the promoters conserved but not the sequence itself?

Answer 45

It is the outside/topographical cylinder structure of the promoters that matter for the recruitment of RNA oplymerase, not the sequence itself. Therefore, the length of the spacers is more important than the sequence itself, while the consensus sequences are important for binding.

Question 46

The promoter consensus sequences could be regarded as the first layer of regulation, how?

Answer 46

Deviations from the consensus sequences make the binding sites less optimal for the RNA pol holoenzyme, which can be utilized to regulate transcription: By having the promoters for the most important/frequently used genes close to the consensus sequence and other promoters for less important/frequently used genes deviate more from the consensus sequences you ensure that the more important genes are transcribed more efficiently. Basically, the closer to the consensus sequences you get, the more transcripts you get.

Question 47

Explain bacterial (E. Coli as model) transcription in short.

Answer 47

1. Initiation: - RNA pol associates with the sigma factor and binds to the promoter to form a closed complex. - The transcription bubble is formed and transcription is initiated after promoter clearance. 2. Elongation: - NTPs are added according to the template - sigma factor disassociates - RNA pol is processive (does not leave the template until termination occurs) 3. Termination, there are two types of termination in E.Coli: - rho dependent termination: A helicase recognize and bind to a rut ( (rho utilization) sequence and migrates along the transcript until it reaches the RNA pol and physically separate the mRNA from the DNA template (ATP dependent) - rho independent termination: a terminator sequence of self-complimentary sequences is transcribed, forming a hairpin structure that put so much strain on the RNA-DNA hybrid in the RNA pol that it releases the mRNA.

Question 48

Eukaryotic transcription is waaay more complex than prokaryotic. What are the three different RNA polymerases, I, II and III responsible for?

Answer 48

RNA pol I: rRNA RNA pol II: mRNA RNA pol III: tRNA and other sRNAs.

Question 49

Compare eukaryotic and prokaryotic transcription and point out the biggest differences.

Answer 49

- The transcription machinery in eukaryotes is highly dependent on transcription factors, and each RNA pol has it's own setup of required TFs. Different TFs are active in different tissues, which result in cell differentiation. The comparable thing in prokaryotes is the sigma factor. - RNA pol II has a CTD (carboxyl terminal domain which is subject to regulation) - The transcripts of both bacteria and eukaryotes are processed to reach mature RNA, but the transcripts of eukaryotes are processed to a much larger extent, splicing, capping, polyadenylation, cleavage, pseudouridination etc.

Question 50

Transcription is not only regulated by the consensus sequences, but mainly with repressors and activators. Give one example of each and what they do.

Answer 50

Activator: CRP (cAMP receptor protein) is activated by cAMP (higher affinity to DNA) when glucose levels are low to activate transcription of genes for metabolism of other sugars. When glucose is high, the cAMP levels sink which basically shuts off all other sugar metabolism pathways as glucose is the "best" source of energy. An example of a repressor is the LacI repressor, Which when bound to the promoter block transcription of the genes for the enzymes of lactose metabolism when lactose is unavailable. When lactose is available, lactose binds to the repressor which blocks the repressor from binding to the promoter, and ensures that genes for metabolizing lactose are transcribed.

Question 51

Explain the three major modifications that are done to mRNA to reach maturity in short.

Answer 51

1. 5'-capping: A 7-Methylguanosine (m7G) residue is joined to the 5′ end of the mRNA in an unusual 5′,5′-triphosphate linkage. The reaction has four steps, each carried out by a different transferase enzyme. The gamma phosphoryl in the mRNA 5' end is removed, the beta and gamma phosphoryl are removed from GTP, the two are linked together and is finally methylated. The 5'-cap acts as protection from exonucleases and as recognition structure. The cap is added after initiation of transcription, when transcription is still ongoing. 2. Splicing: Introns are removed and exons ligated together. There are several types of splicing, group I and II are self splicing and but these are not common in eukaryotes and are only present in special cases of bacteria. - Group I: a free guanosine in the 3'OH end of the intron perform a nucleophilic attack on the phosphodiester bond between the next exon and the intron, cleaves off itself, and the 3'OH of the next exon becomes a nucleophile that attacks the 5'P end of the first exon, which ligates them together. - Group II: splicing via lariat formation: a specific adenosine (in the branch site) attacks the phosphodiester bond between intron and the next exon, and forms a lariat, while the 3'OH of the exon attack the phosphodiester bond between the intron and the first exon, cleaving away the lariat intron. - in Eukaryotes: spliceosome: complex of several snRNA (U1, U2 etc) that cleaves off intron in a reaction very similar to group two but more complex, and requires ATP. 3. Polyadenylation: The addition of a poly(A) tail in the end of the mRNA transcript. - an endonuclease recognize the signal sequence AAUAAA in the end of the transcript and cleave it, producing a free 3' end with an end that can be recognized. - polyadenylate polymerase recognizes the end of the transcript, binds and adds roughly 200 A nucleotides. Poly A binding proteins bind and this structure functions as protection but mainly recognition and can also be used for mRNA regulation.

Question 52

What is the hypothesized main function of splicing?

Answer 52

Alternative splicing, which can result in a loooot more protein diversity from the same template. Common with alt splicing in diff cell types. Can for example be different poly a sites, different subunits of a protein etc, or can lead to unfunctional protein if for example involved in sex differentiation.

Question 53

tRNA is also processed, both in bacteria and eukaryotes to ensure it takes the form of a upside down L. Explain how.

Answer 53

tRNA is subject to: - 5' & 3' end cleavage, RNase D cuts off chunks. - base modification, pseudouridination: backbone sugar moved to another site than in regular U. - addition of CCA in the 3' end.

Question 54

There are seven different points of regulation of gene expression, which?

Answer 54

- Transcription initiation - Posttranscriptional processing - RNA stability - Translational regulation - Protein modification - Protein transport - Protein degradation

Question 55

Transcription can be under negative or positive regulation. Explain the difference with two examples of each.

Answer 55

Negative regulation: When repressor binds=transcription is turned off. Can be either: - repressor being bound, inhibiting transcription which can be removed by a molecular signal that cause it to fall off and enable transcription. - repressor being unbound and transcription being turned on, and molecular signal causing repressor to bind and turn off transcription. Positive regulation: When activator binds=transcription turned on. Can be either: - activator being bound, enabling transcription which can be removed by a molecular signal that cause it to fall off and disable transcription. - Activator being unbound and transcription being turned off, and molecular signal causing repressor to bind and turn on transcription.

Question 56

Where does activators and repressors generally bind?

Answer 56

At regulatory sequences around the promoter.

Question 57

There are three major classes of DNA binding proteins, which? Explain their structure in short.

Answer 57

- Helix-turn-Helix proteins: In a DNA helix there are a minor and a major groove, and a alfa-helix fits perfectly into the major groove of DNA. One alfa helix turn is roughly 3.6 residues, and the side chains of these interact with the Backbone of DNA in the major groove to bind. The helix-turn-helix is usually just a smaller motif on bigger proteins and the two alpha helixes are kind of like "feet" that anchor the protein to the DNA, eg the LacI repressor have this to bind to the promoter. - Zinc finger proteins (eukaryotes mainly): the motif is structured with a helix, loop, beta sheet, loop and a beta sheet in the other direction. The helix fit in the groove and the loop residues interact with DNA. - Basic leucine zipper proteins: Structured with two longer alpha helices that fit in the major groove with a leucine residue sticking out (every 7th residue) stabilizing the protein.

Question 58

DNA binding proteins can be unspecific or specific, explain.

Answer 58

- Unspecific: interacting with the phosphate backbone (positive residues) of DNA - Specific: interacting with exposed bases in the grooves, often dependent of hydrogen bonding, eg T=A interacting with glutamate or asparagine or GC interacting with Arginine.

Question 59

Explain how the Lac operon is regulated.

Answer 59

The Lac operon is under both negative and positive regulation. - Negative: LacI bound = no transcription. When lactose is present, lactose binds to the repressor causing it to fall off (helix-turn-helix motif changes so it cant hold on to the DNA anymore) which enables transcription. - Positive: The lac promoter is quite weak, meaning it has low homology to the consensus sequences and therefore not optimized for RNA pol binding, so even when LacI is not bound = low transcription. BUT! CRP (cAMP receptor protein) is activated by cAMP (higher affinity to DNA) when glucose levels are low to activate transcription of genes for metabolism of other sugars, like lactose. The CRP has a helix turn-helix-motif that binds at a regulatory region upstream of the promoter when cAMP is bound, and upon binding, it bends the DNA which enables RNA pol to bind better than just without the repressor. (When glucose is high, the cAMP levels sink which basically shuts off all other sugar metabolism pathways as glucose is the "best" source of energy.)

Question 60

Explain how the Trp operon is regulated.

Answer 60

The Trp operon codes for genes that synthesize Tryptophan, which is enegetically expensive to make, so it's under tight regulation and it has two layers of negative regulation: - The trp repressor binds when trp is abundant in the cell, inhibiting transcription - attenuation with 2 trp codons, if trp is abundant this goes fast and a terminator structure forms. If trp is low, the translation of these goes slow and an antiterminator structure forms, allowing transcription.

Question 61

What are the major differences between eukaryotic and prokaryotic regulation of transcription?

Answer 61

- In prokaryotes, transcription and translation is coupled, which make processes like attenuation possible. - no operons in eukaryotes, more fine tuned and waaay more complex regulation. - DNA is tightly packed in euk, so transcription depends heavily on remodeling enzymes and transcription factors.

Question 62

What is a codon?

Answer 62

Three nucleotides in a row that encodes for a specific amino acid. The corresponding anti-codon on tRNA is complementary to the codon.

Question 63

What is meant by reading frame?

Answer 63

Depending on the starting point, the entire protein encoding sequence can be different as one codon includes three nucleotides. So if an insertion or deletion happen in the DNA template of RNA, this can disrupt a whole protein encoded by the gene.

Question 64

What is the definition of an open reading frame (ORF)?

Answer 64

In general, a reading frame without a termination codon among 50 or more consecutive codons is referred to as an open reading frame (ORF). Long ORFs usually correspond to genes that encode proteins.

Question 65

Explain the concept "wobble base"

Answer 65

The third base in a codon is also referred to as the wobble base, amino acids that are encoded by several codons usually differ in the third position of the codon, which binds to the base in the first position in the anticodon, the wobble base, and this is often an inosine in tRNAs with several codons. Inosine can base pair with A, U and C. This allows the cell to have less tRNAs than there are codons, saving energy AND the weaker interactions between the wobble base and codon balances the requirements for accuracy and speed.

Question 66

The genetic code is said to be resistant, what does this mean?

Answer 66

The code is structured in a way that purine to purine/pyrimidine to pyrimidine mutations lead to an amino acid with similar properties being encoded instead, minimizing the effect of the mutation on the resulting protein.

Question 67

To understand how tRNAs can have the pivotal roles as adaptors, translating the language of the genetic code into proteins. We need to understand their structure. Describe the most important parts of a tRNA.

Answer 67

Mature tRNA is structured king of like a upside down L, with an amino acid arm to the upper right, where the specific amino acids are loaded by being esterified by its carboxyl group to the 2'- or 3'-hydroxyl group of the A residue at the 3' end of the tRNA, an anticodon arm downwards which contain the anticodon, a D arm and a TψC. The D and TψC arms contribute important interactions for the overall folding of tRNA molecules, and the TψC arm interacts with the large-subunit rRNA.

Question 68

Explain the reaction mechanism for the aminoacyl loading of tRNA.

Answer 68

The enzyme catalyzing the activation of tRNA is aminoacyl-tRNA synthetases. They recognize the tRNAs using specific conserved base pairs, eg tRNA-ala is recognized by a single G-U base pair in the amino acid arm. Step 1: The carboxyl group of the amino acid attacks the alpha phosphorous of ATP, which results in the formation of an aminoacyl adenylate, which remains bound to the active site. (one phosphate bond broken) Step 2. the aminoacyl group is then transferred to the tRNA. The mechanism of this step is somewhat different for the two classes of aminoacyl-tRNA synthetases. - For class I enzymes, the aminoacyl group is transferred first to the 2'-hydroxyl group of the 3'-terminal A residue, then to the 3'-hydroxyl group by a transesterification reaction. - For class II enzymes, the aminoacyl group is transferred directly to the 3'-hydroxyl group of the terminal adenylate on the tRNA. During the reaction, the leaving pyrophosphate (PPi) is hydrolyzed to drive step 2. Thus, two high-energy phosphate bonds are ultimately expended for each amino acid molecule activated, rendering the overall reaction for amino acid activation essentially irreversible.

Question 69

A specific amino acid initiates protein synthesis, which?

Answer 69

N-formylmethionine. All organisms have two tRNAs for methionine, while only one codon (AUG) encodes for Met. One tRNA is used exclusively when (5')AUG is the initiation codon for protein synthesis. The other is used to code for a Met residue in an internal position in a polypeptide.

Question 70

How is N-formylmethionine synthesized in E. Coli in short?

Answer 70

The synthesis happens via two successive reactions. - First, methionine is attached to tRNA-fMet by the Met-tRNA synthetase (which in E. coli aminoacylates both tRNA fMet and tRNA Met) - Next, a transformylase transfers a formyl group (-CH=O) from N 10-formyltetrahydrofolate to the amino group of the Met residue. The transformylase is more selective than the Met-tRNA synthetase; it is specific for Met residues attached to tRNA-fMet, presumably recognizing some unique structural feature of that tRNA. Addition of the N-formyl group to the amino group of methionine by the transformylase prevents fMet from entering interior positions in a polypeptide while also allowing fMet-tRNA fMet to be bound at a specific ribosomal initiation site that accepts neither Met-tRNA Met nor any other aminoacyl-tRNA.

Question 71

What is the Shine-Dalgarno sequence?

Answer 71

The SD sequence is a well conserved consensus sequence of purines (A and G) located 8 bp upstream of the start codon in bacteria, that base pairs to a part (16S rRNA) of the small ribosomal subunit (30S) which guides the start codon to the correct position in the ribosome.

Question 72

How is translation initiated in bacteria?

Answer 72

Bacterial ribosomes have three sites that bind tRNAs, the aminoacyl (A) site, the peptidyl (P) site, and the exit (E) site. The A and P sites bind aminoacyl-tRNAs, whereas the E site binds only uncharged tRNAs that have completed their task on the ribosome. 1. Factor IF1 binds at the A site of the 30S subunit which prevents tRNA binding at this site during initiation and factor IF-3 binds to the E site. The complex then binds to mRNA. (The initiating (5')AUG is positioned at the P site, the only site to which fMet-tRNA fMet can bind, all other aminoacyl-tRNAs bind to the A site first.) 2. IF-2-GTP binds to the 30S subunit and this rectruits the fMet activated tRNA, which base pairs with the start codon. 3. The 50S subunit is recruited, IF-2 hydrolyses GTP and all the initiation factor disassociate and the initiation complex have been formed.

Question 73

Explain the elongation step of translation in bacteria in detail.

Answer 73

The elongation stage is a three step reaction that is repeated until a stop codon is reached. 1. An incoming aminoacyl-tRNA binds the elongation factor Tu that is bound to GTP. The whole complex then binds to the A site. The GTP is hydrolyzed and the EF-Tu–GDP complex is released from the ribosome. The EF-Tu is then recycled and charged with a new GTP to bind to the next incoming aminoacyl tRNA. 2. The α-amino group of the amino acid in the A site acts as a nucleophile, attacking the fMet in the P site, and displaces the tRNA in the P site to form a peptide bond. 3. EF-G bound to GTP binds to the A site, and hydrolysis of the GTP to GDP provides energy to move the ribosome one codon towards the 3' end of the mRNA, which causes translocation of the tRNA in the P site to the E-site. The cycle can now repeat. In summary: For each amino acid residue correctly added to the growing polypeptide, two GTPs are hydrolyzed to GDP and Pi as the ribosome moves from codon to codon along the mRNA toward the 3' end, an energetically expensive process! During the EF-tu and Ef-G complexes are present, the ribosome does proofreading, which improves fidelity.

Question 74

Explain how translation is terminated in bacteria.

Answer 74

When a stop codon enters the A site, a release factor (RF) binds at the termination codon and induces peptidyl transferase to transfer the growing polypeptide to a water molecule rather than to another amino acid. This leads to hydrolysis of the ester linkage between the nascent polypeptide and the tRNA in the P site and release of the completed polypeptide. Finally, the mRNA, deacylated tRNA, and release factor leave the ribosome, which dissociates into its 30S and 50S subunits and a new cycle can begin.

Question 75

In total, what is the summarized energetic cost of formation of a peptide bond between to specific amino acids?

Answer 75

On average, the cost is 4 ATP equivalent. - Formation of each aminoacyl-tRNA uses one high-energy phosphate groups, and since two needed, it takes 2 ATP equivalents. (An additional ATP is consumed each time an incorrectly activated amino acid is hydrolyzed by the deacylation activity of an aminoacyl-tRNA synthetase as part of its proofreading activity. ) - A GTP is cleaved to GDP and Pi during the first elongation step, and another during the translocation step. 2 ATP equivalents. Thus, on average, the energy derived from the hydrolysis of more than four NTPs to NDPs is required for the formation of each peptide bond of a polypeptide.