DNA & Proteins

Question 1

The 4 Hypotheses on the Origin of Life on Earth

Answer 1

1. Organic chemical synthesis in a reducing atmosphere.
2. Carriage by meteorites.
3. Organic chemical synthesis in deep ocean vents.
4. RNA world.

Question 2

1. Organic Chemical Synthesis in a Reducing Atmosphere

Answer 2

It was thought that early Earth had a reducing atmosphere, rich in hydrogen and methane. The methane, ammonia and hydrogen mix to electrical discharges (lightning) in the presence of water. This resulted in a prebiotic soup (amino acids and nucleotides). There is no data on how soup forms organic networks (lipids and macromolecules) encompassed by a membrane. The question is whether the atmosphere at the time reducing and the current consensus says no.

Question 3

2. Carriage by Meteorites/Comets

Answer 3

Panspermia is the attractive theory due to the sudden appearance of life on Earth and its amazing uniformity (no actual data). Organic compounds are common in space e.g. amino acid (glycine) found on a comet 2009. Based on studies complex oranic chemicals could arise of Titan (moon of Saturn). This theory only moves the question how did life originate in space. The blast impact of meteors and comets also make it very difficult for organic matter to survive so that is also disputed.

Question 4

3. Synthesis on Metal Sulphides in Deep Sea Vents

Answer 4

Vents are sites of abundant biological activity which is independent of solar energy. Chemical energy sources lead to another prebiotic soup theory. Prebiotic soup self-organises into life-supporting networks on metal sulphide surfaces. Networks must incorporate membranes however there is no data on this.

Question 5

4. RNA World

Answer 5

The question as to whether the first self-replicating entity simpler than a cell. Short RNA molecules were discovered that can store information and catalyse chemical reactions (ribozymes). RNA molecules have been synthesised that are capable of self-replication. The question of how lipid membranes formed is still in question and unexplained.

Question 6

Representations of DNA

Answer 6

Chromatin, chromosomes, double-helix, uncondensed.

Question 7

Discovering DNA Structure

Answer 7

This was done by Watson and Crick 1953. This discovery was facilitated by many other people. Watson sees X-ray diffraction image of DNA and began to work with Crick who was working on helical diffraction in proteins in the same lab. The idea was then made to use X-ray diffraction to find the structure of DNA. This was assisted by Franklin who developed a better X-ray image of DNA. The first model by Watson and Crick produced a 3-stranded DNA model which was inconsistent with the images that Franklin had gathered. Chargaff found that there was a unity of the A/T and G/C ratios which pointed more inconsistencies in Watson and Cricks model. With Franklins work they discovered that DNA had a double helix structure.

Question 8

Helical Diffraction

Answer 8

A technique by which to identify the structure and then assume and identify the function of proteins.

Question 9

DNA

Answer 9

This is the genetic material. It is base-paired, anti-parallel, right-handed double helix. The code is cracked (triplets of A,T,C and G) code for individual amino acids. Amino acids are the building blocks of proteins. Gene to protein relationships established, the control of gene expression partly elucidated and large scale sequencing of genomes is now common (cost and speed).

Question 10

How Genetic Code Works

Answer 10

Peptides (met-enkephalin) are present in humans made up of an amino acid sequence (Met, Tyr, Gly, Gly, Phe, Met) which has a DNA code (ATG, TAT, GGT, GGT, TTT, ATG). The 3 bases code for a single amino acid e.g. TTT = Phe. There are 64 possible combinations of genetic code that are possible which codes for 22 amino acids and ‘stop’ codes. The difference in the number of codes compared to amino acids means that some amino acids can be coded by multiple different base codes.

Question 11

Parts of an Amino Acid Sequence

Answer 11

The start of this is known as an ‘N terminus’ while the end is known as a ‘C terminus’. In between these are where the different amino acids are placed. The ‘N terminus’ is adjacent to the 5’ end of DNA while the ‘C terminus is adjacent to the 3’ end.

Question 12

DNA Code in Context

Answer 12

Control regions on DNA have specific sequences of G, A, T and C but not organised in triplets. These control regions also have certain characteristics as the upstream (before) controls has identifiers for the RNA to replicate and downstream (after) controls has stop codons. These also control the time and expression ability of the code for regulation. Code is composed of triplets of any of the 4 bases with each triplet being used for a particular specification. This is comparable to the 1/0 in binary code.

Question 13

The Central Dogma

Answer 13

DNA self replicates -> The DNA is transcribed by RNA -> the RNA is translated into proteins. In RNA there can be reverse transcription where it becomes double stranded and can enter your DNA and replicate. This is done by certain viruses e.g. HIV and COVID19.

Question 14

Nucleotides

Answer 14

The basic structural units of both RNA and DNA. This consists of a sugar, a nitrogenous base and a phosphate group.

Question 15

Nucleosides

Answer 15

These are also structural units of both RNA and DNA. This consists of a sugar and a nitrogenous base only.

Question 16

Naming

Answer 16

If a sugar is deoxyribose the prefix for the name should be ‘deoxy’ e.g. deoxyadenosine, deoxyadenosine monophosphate (dAMP). If the sugar is ribose the prefix should be ‘ribo’ e.g. riboadenosine triphosphate (rATP). DNA (deoxyribonucleic acid) contain deoxyribose sugar whereas RNA (ribonucleic acid) contains ribose sugar.

Question 17

How the DNA/RNA Chain is Linked

Answer 17

The 5’ phosphate can form a phosphodiester bond by reacting with the 3’ hydroxyl group of another nucleotide.

Question 18

Base Pairing

Answer 18

Between nitrogenous bases there is hydrogen bonding. A only combines with T (vis versa) in DNA whereas G only combines with C (vis versa) which is known as complementary bonding. Between A and T groups there is only 2 hydrogen bonds whereas with C and G groups there are 3 hydrogen bonds meaning C and G bonds are harder to break.

Question 19

Antiparallel

Answer 19

The strands of DNA are in opposite directions e.g. if 1 strand is 5’-3’ then the opposite stand is 3’-5’.

Question 20

Information Encoding

Answer 20

information is encoded by the order of bases 5’-3’. One strand is the coding strand while the other is the non-coding strand. The bases are read in triplets known as codons with 1 codon coding for 1 amino acid.

Question 21

Right-Handed Double Helix

Answer 21

Looking down the double helix follows a clockwise pattern. The DNA forms major and minor grooves due to the right-handed double helix shapes. The major grooves help proteins to interact with the DNA molecule to control the expression of certain genes.

Question 22

Modern Day DNA

Answer 22

Transgenics and gene knock out/in occurs in order to identify the purposes of genes to test organisms (typically bacteria) with lost/inactive genes (knock-out mutants) and those with additional genes (knock-in mutants). Genetic screening is used for personalised medicine. Viruses and living cells are created from synthetic DNA constructs. ‘Bioinformatics’ is a new scientific field now available.

Question 23

Modern DNA Research

Answer 23

‘Junk DNA’ is not junk trying to find other purposes. DNA can change to other forms (A, Z, G). Chromosomal position and movement within the nucleus is preserved across species and affects gene expression.

Question 24

General Features of Chromosome Replication

Answer 24

The complementary base-airing enables semiconservative DNA replication, DNA synthesis initiates at origins, synthesis usually moves bidirectionally away from an origin via 2 replication forks which creates a replication bubble, this moves in a 5’-3’ direction and the synthesis of new DNA always requires a primer.

Question 25

Complementary Base Pairing

Answer 25

Each strand of a double stranded DNA molecules serves as a template for synthesis of a new complementary strand. A binds only with T while G only binds with C. The 5’ strand is called the S strand while the 3’ strand is called the S’ strand. The reason that DNA replication is accurate and identical is because it uses each of the old strands as a template in order to gain new bases and form 2 identical strands.

Question 26

Semiconservative DNA Replication

Answer 26

Each strand of a DNA molecule serves as a template for synthesis of a new complementary strand. Each daughter molecule has parental strand plus a new strand. The accuracy and speed of replication is 1000 nucleotides per second without error. The group of proteins meet to operate as a protein machine moving along a replication fork. DNA polymerase adds nucleotides to the 3’ end of the new strand. DNA polymerase has proofreading property to reduce the error rate.

Question 27

Origins

Answer 27

Double stranded DNA are pried apart at this point by helicase at an identified position with a particular DNA sequence. Shorter strands will only have one of these but larger DNA strands will have multiple. Strands with multiple of these will combine in order to synthesise even more DNA.

Question 28

Bidirectional

Answer 28

From the origins the DNA will replicate in both directions.

Question 29

Replication Forks

Answer 29

These structures are the points at which the DNA is not yet separated and seems to converge. There are 2 of these at every origin. All of the new DNA is synthesised at these points in a 5’-3’ direction. Due to the directionality (3’-5’) of one of the lagging strands Okazaki fragments are formed wheras on the leading strand the polymerisation process occurs continuously.

Question 30

Okazaki Fragments

Answer 30

The the discontinued fragments of DNA added to the lagging strand which it is forced to do so as the direction of polymerisation is 5’-3’. Because one strand isn’t moving in this direction polymerisation must occur in small segments as the DNA continues to be separated.

Question 31

Replication Bubble

Answer 31

This is created by the separation of the 2 strands of DNA during replication which leaves a space in the middle which looks similar to a bubble.

Question 32

DNA Polymerase

Answer 32

This enzyme adds each deoxyribonucleotide to the 3’ end of a primer strand attached to the template strand.

Question 33

DNA Primase

Answer 33

This is an enzyme that synthesises a short strand of RNA on a DNA template. On the leading strand in creates 1 RNA primer in order to synthesise the entire strand. On the lagging strand each Okazaki fragment requires an RNA primer meaning it makes jumps to place primers further down the strand to ensure there is a place for DNA polymerase to attach and continue polymerisation.

Question 34

DNA Ligase

Answer 34

This enzyme joins the Okazaki fragments by their sugar-phosphate backbones and removes the primer once the polymerisation process is finished.

Question 35

Single-Strand DNA Binding Proteins

Answer 35

These help to stabilise single stranded DNA and aid the helicase.

Question 36

Helicase

Answer 36

This enzyme pries apart (unwinds) the double helix in order to form 2 single stranded DNA molecules for replication.

Question 37

Sliding Clamp

Answer 37

This protein maintains the attachment of DNA polymerase to DNA strand and pushes it along making it more efficient at polymerisation.

Question 38

Clamp Loader

Answer 38

This protein assembles the full clamp on the DNA is association with the sliding clamp and DNA polymerase using ATP energy. It also can assist in the efficiency of polymerisation.

Question 39

DNA Polymerase I (E. Coli)

Answer 39

This form of DNA Polymerase is seen in numbers of 400 per cell where it functions in DNA repair as well as the maturation of Okazaki fragments (removes RNA primer and fills in gaps), It is slow moving adding 20 nucleotides per second and has a low affinity for nucleotides.

Question 40

DNA Polymerase II (E. Coli)

Answer 40

This form of DNA Polymerase is seen in number of 100 per cell where it functions in DNA repair it is the slowest form adding 5 nucleotides per second and has a low affinity for nucleotides.

Question 41

DNA Polymerase III (E. Coli)

Answer 41

This form of DNA Polymerase is seen in numbers of 10-20 per cell with it being the main DNA replication enzyme (for both strands), it is the fastest form adding 1000 nucleotides per second and has a high affinity for the nucleotides.

Question 42

DNA Polymerase in Mammals

Answer 42

There are 5 types of these in mammals consisting of alpha, beta, gamma, delta and epsilon. Alpha is found in the nucleus and is involved with primase and elongates the primer with a short length of DNA. Beta is found in the nucleus and is involved in DNA repair. Gamma is found in the mitochondria and involved in the replication of mitochondrial DNA. Both delta and epsilon are found in the nucleus and their functions are highly debated and controversial however they both seem to be involved in the synthesis of the lagging and leading strands respectively.

Question 43

Mutation

Answer 43

Any permanent and heritable change in the DNA sequence of an organism. Damaged DNA will cause problems with DNA replication which can potentially be lethal. Restoration of the correct nucleotide sequence can result in an incorrect nitrogenous base being incorporated. To overcome these issues all living cells have mechanisms for DNA repair.

Question 44

How Changes in DNA Sequence Occur

Answer 44

Replication errors are very rare as DNA has a high fidelity rate. Incorrect copying occurs at a rate of 1/1,000,000,000.

Question 45

High Fidelity Rate

Answer 45

The ability of DNA to have a high fidelity is due to the base paired structure of DNA, the primer requirement of DNA polymerase and the ‘proof reading’ that DNA polymerase does.

Question 46

Environmental Factors for Mutations

Answer 46

These consist of chemicals (mutagenic chemicals and nucleotide instability), radiation (UV and gamma) and mobile DNA (infectious agents and transposable elements).

Question 47

Nucleotide Instability

Answer 47

This can occur through depurination where the base is lost from the DNA backbone. The other method is through deamination where the bases lose the alkaline groups.

Question 48

Deamination

Answer 48

When this happens to a nitrogenous base the base pairing will no longer occur. When replicated there will be one new strand of DNA which is the same as before. The second replicated strand will have a different base pair in that place as it will select the nitrogenous base that will pair with the changed base. This means that a mutated strand becomes present. This may change the protein which may impair a bodily function.

Question 49

Depurination

Answer 49

When this happens one nitrogenous base is deleted. When replicated one strand will be the same as before any change. The second replicated strand however will be missing one base pair. Because the DNA is read in triplets the entire code will be changed. This will most likely cause massive change.

Question 50

Mutagenic Chemicals

Answer 50

These chemicals can either cause alkylation or intercalation. Alkylation is when electrophiles add alkyl groups to nitrogenous bases and stalls replication e.g. carcinogens. Intercalation is when a compound inserts into the double stranded helix leading to distortion however it doesn’t change the bases e.g. ethidium bromide.

Question 51

Mutations by UV Light

Answer 51

This can cause the formation of thymine dimers. Where 2 Ts next to one another will attach to each other and distort the DNA.

Question 52

Mutation by Radiation

Answer 52

Gamma and X-rays attack DNA bonds as they directly produce free electrons which attack the backbone of DNA or indirectly by generating hydroxide free radicals. Both of these result in single and double stranded breaks in the DNA backbone.

Question 53

Mutation by Infectious Agents

Answer 53

This occurs through mobile DNA which are viruses and bacteriophages or transposons (pieces of DNA which jump into cells) which have the ability to insert or recombine into a target DNA molecule.

Question 54

Recombination

Answer 54

This is the breaking and rejoining of DNA molecules to form a different combination. This can be non-homologous which means there is no similarity between DNA molecules is required which is site-specific and catalysed by enzymes called integrases and transposases. They can also be homologous which means both the donor and acceptor DNA molecules have similarities in DNA sequence.

Question 55

Retroviruses

Answer 55

An example is HIV and their equivalent in bacteria which can integrate into host DNA. They act as parasites and utilise the host cells replication machinery. They can be lytic (enter and lyse (replicate and kills) the host) or lysogenic (enter and integrate into host chromosome life cycles). Insertion of foreign DNA can physically disrupts a coding region e.g. HIV as transcriptionally active genes are more likely to be replicated. This isn’t a common feature for all viruses.

Question 56

Transposons

Answer 56

These are linear DNA molecules that move within and between chromosomes and inset into many different DNA sequences. This insertion will physically disrupt a gene, an excision of this can result in a small duplication of DNA which will again disrupt a gene, this is common in bacterial DNA rearrangement (bacteria changing their DNA for better fitness). These target direct repeats of bases

Question 57

Non-Homologous Recombination

Answer 57

This is also known as transposition. This can occur in either a cut-and-paste non-replicative transposition and replicative transposition. In non-replicative the transposon in the DNA will encode for the transposases which will then cut it out of the DNA and then place the transposon into a new DNA strand. In replicative the transposon will be copied and then placed into a new DNA strand meaning that the original and new DNA strands have the transposon in them.

Question 58

DNA Repair

Answer 58

Mechanisms for repair result in the restoration of the original sequence the 2 types are mis-match repair system and homologous recombination.

Question 59

Mismatch Repair

Answer 59

DNA replication without this mechanism the error rate is 1/10^7 whereas with this feature it is 1/10^9 meaning it 100x the accuracy. This focuses on mis-paired nucleotides and repair proteins recognise and excise the strand of DNA containing the mismatch using 4 steps. The repair proteins patrol the DNA and bind to a mismatched sequence (look for distortion). The mismatched region is excised by nucleases creating a single stranded DNA patch. A second strand is synthesised by repair DNA polymerase using the free 3’ hydroxyl group as a primer. Finally ligation of the DNA backbone is done by DNA ligase.

Question 60

Homologous Recombination Repair

Answer 60

This occurs when regions of very similar sequences align. The double strands are broken and a cross over occurs. Repair then occurs to generate new combinations of DNA. A double stranded break is typically introduced to a chromosome. Exonuclease will remove nucleotides from the 5’-3’. The single stranded 3’ overhand migrates into the recipient chromosome where the sequences are homologous. DNA polymerase synthesises a new complementary strands. Rotations of crossed strands will allow a section of one strand to be joined to the section of another (DNA is exchanged). Nucleotide sequence at the site of exchange is unaltered (no addition or subtraction of bases).

Question 61

The Central Dogma Reexplained

Answer 61

Portions of DNA sequence (genes and intergenic spaces) are copied into RNA. The process by which eukaryotic and prokaryotic RNA is processed differs. Not all RNA is translated only the messenger RNA (mRNA) relates directly to protein synthesis. DNA is double-stranded and one strand moves 5’-3’ with the other going from 3’-5’ and has the bases A,T,C and G. RNA is single-stranded which moves 5’-3’ and has the bases A,U,C and G. Due to the single-stranded nature of RNA it is more unstable and able to make complex structures.

Question 62

RNA Structure

Answer 62

This has ribose sugar, contains a different base to DNA (A,U,C,G), it is single-stranded compared to DNA being double stranded. The reason for uracil being in the strand is hypothesised to be due to its ease of production or as it is the deaminated version of cytosine which spread around naturally.

Question 63

Transcription

Answer 63

This is the production of RNA from a DNA template. The DNA sequences are written in the 5’-3’ direction. The ‘coding’ or ‘sense’ DNA strand is written from 5’-3’. RNA polymerase makes a complementary copy of the anti-parallel strand (template) in the 3’-5’ direction from the promoter (P) to the terminal (T). Transcribed mRNA is written in the 5’-3’ direction and is the same sequence as the coding strand in the DNA except that U is substituted for T. This starts with a ribosome bonding site (RBS), start codon, rest of the code and the stop codon.

Question 64

RNA Synthesis

Answer 64

This process requires 4 ribonucleoside 5’ triphosphates (5’ATP, 5’GTP, 5’CTP, 5’UTP), Mg2+ ions, DNA template with no primer requires and RNA polymerase (RNAP). The mechanisms of RNA synthesis is initiation, elongation and termination. This process is similar to DNA replication but only one strand is copied.

Question 65

RNA Polymerase

Answer 65

This is similar to DNA polymerase as it forms phosphodiester bonds between the ribonucleosides and it uses energy stored in the ribonucleosides triphosphates for polymerisation. Unlike DNA polymerase however it can start without a primer, it has a higher error rate 1/10,000 < 1/10,000,000 and the unwinding of the DNA template doesn’t require helicase or ATP (it rewinds after copying and only opens a small part of the DNA).

Question 66

E. Coli RNA Polymerase

Answer 66

This form has 5 subunits which are 2x alpha, 1x beta, 1x beta’ and 1x sigma which make a complex called the core enzyme. The sigma subunit is responsible for promoter recognition (pyridine-rich DNA sequence of 10 or more bases) and associates with it to form the holoenzyme. Once correct initiation has been achieved the sigma subunit dissociates from the holoenzyme and the core enzyme continues elongation of an RNA chain.

Question 66

Upstream Initiation Signals (Prokaryote Transcription)

Answer 66

This occurs first through initiation signals. The transcription start point which is referred to as ‘+1 base’ which will continue to count down as transcription occurs e.g. +1,0,-1 etc. This start point will be found by an area between -5 and -10 called the pribnow box.

Question 67

Downstream Termination Signals (Prokaryote Transcription)

Answer 67

The coding strand can end anywhere between -10 and -35. There are signals which are achieved either by an inverted repeat followed by a stretch of T bases (hairpin of DNA) or by a protein factor being bound (Rho) which will stop coding.

Question 68

Prokaryote Transcription

Answer 68

In the holoenzyme the promotor region of DNA is found by the sigma subunit. The beta and beta’ subunits are the RNA polymerase and attach to the template strand and pries open and begins to replicate the section of the strand. At this point the sigma subunit leaves to attach to another core enzyme. A new messenger RNA is synthesised until the hair-pin loop or protein factor is met after which the new mRNA produced is release and the RNA polymerase is released to be used again.

Question 69

Prokaryote Messenger RNA

Answer 69

These mRNA molecules are a copy of the coding strand of DNA sequences (amino acid sequence of all proteins). A DNA segment corresponding to 1 protein chain is called a cistron. A single mRNA encoding a single polypeptide is called a monocistronic mRNA. In prokaryotes polycistronic mRNAs are quite common and code for several different polypeptide chains. mRNA’s have 5’ leaders, 3’termini and for polycistronic mRNAs have intercistronic regions called spacers. These mRNA have short half lives (few minutes only) meaning they produce proteins very quickly. The primary transcript isn’t processed.

Question 69

Eukaryote Messenger RNA

Answer 69

These have genes containing coding and non-coding regions (exons and introns). When RNA is processed RNA capping (G with methyl group is added to the 5’ end), polyadenylation (a series of ‘A’ is added to the 3’ end) and then the RNA molecule is identified as mRNA. The introns are then removed by splicesomes which recognises the boundaries between introns and exons with the exons being stitched back together and eventually translated. The mRNA molecules eventually degrade. Protein synthesis occurs in the cytoplasm not in the nucleus like in prokaryotes.

Question 70

Classes of RNA

Answer 70

Messenger RNAs (mRNA) which code for proteins, ribosomal RNA (rRNA) which is part of ribosomes ( protein synthesis), transfer RNA (tRNA) (adaptors between mRNA and amino acids) and small RNAs which are the pre-spliced mRNA which are involved in the transport of proteins to ER and other cellular processes.

Question 70

DNA to Proteins

Answer 70

The production of a protein by a eukaryotic cell. The final level of each protein in a cell depends on the efficiency of each step. All genes are on the 5’-3’ strand meaning the 3’-5’ strand is what is used as the template. Within the nucleus there is also many steps involved which allow for the final transcription product (capping, elongation, splicing and polyadenylation termination). This RNA will then leave the cell and any corrupted RNA will slowly degrade. A ribosome will then find the mRNA and from the start codon will begin to synthesise the protein. After this protein alteration and folding occurs and any errors in this process causes the protein to degrade or create adverse effects in the cell.

Question 70

3 Major RNA Molecules

Answer 70

Messenger RNA (mRNA) is a copy of DNA which codes for the protein. Transfer RNA (tRNA) is a small adaptor molecules which align specific amino acids opposite their triplet codon in the mRNA molecule during translation. Ribosomal RNA (rRNA) is an integral part of ribosomal structures and is an important part of protein synthesising machinery. These 3 interact to enable the translation of mRNA into proteins.

Question 70

Eukaryotic vs Prokaryotic Protein Synthesis

Answer 70

In prokaryotes both the transcription and translation occurs together and occurs simultaneously for speed and efficiency. In eukaryotes the transcription and translation processes occur separately. These activities are compartmentalised which enables greater control.

Question 70

Eukaryotic vs Prokaryotic RNA

Answer 70

On eukaryotic cells there is a 5’ cap and a 3’ end polyadenylation which assist the export of mRNA molecules from the nucleus, it protects mRNA from degradation and increases its half life and enables the binding of ribosomes and promotes translation. In bacteria there is no modification of the mRNA. The cap is made up of methylguanosine. The Poly-A tail which is a long string of adenosines linked to the 3’ end which is not incorporated in the DNA sequence and is the binding site of several proteins as it varies from 150-250 adenosine molecules.

Question 70

Transfer RNA

Answer 70

Amino acids don’t directly recognise the mRNA codon and therefore requires an adaptor. These molecules are the adaptors which are 80 nucleotides long. They align specific amino acids opposite their triplet codon in the mRNA molecule during translation. Bases other than A,G,C and U are present in this as a result of post-transcriptional modification e.g. inosine (modified adenine), pseudouridine, dihydrouridine.

Question 70

Transfer RNA Structure

Answer 70

The internal complementary base pairing of this RNA gives a cloverleaf structure. It contains anticodons (triplets of bases) which determines the mRNA codon it binds to. The amino acid is attached at the 3’ end. It has modified bases such as dihydrouracil and pseudouridine. The 3D shape determines the attachment of the correct amino acid (matching the codon/anticodon pair) by aminoacyl-tRNA synthetases.

Question 70

Adaptors For Translation

Answer 70

The first one of these is aminoacyl-tRNA synthetase couples a particular amino acid with its corresponding tRNA. The second one is tRNA which has an anticodon that forms base pairs with a specific codon of mRNA.

Question 70

Aminoacyl-tRNA Synthetase

Answer 70

For every amino acid there is one of these molecules that exists meaning that it links a particular amino acid to one of the several acceptable tRNA molecules. These enzymes are responsible for the attachment of the correct amino acid to the tRNA meaning the amino acid specified and the amino acid attachment sequence of the 3’ end. When the correct amino acid is attached to its tRNA it is said to be charged or acylated and called an amino acyl tRNA. This combination requires an ATP hydrolysis reaction. An incorrect amino acid attachment to a tRNA is known as mischarging.

Question 70

Ribosomal RNA

Answer 70

The mRNA message is decoded on ribosomes. The ribosome is a very large multi-domain complex rRNA and protein macromolecule. They scan along the mRNA and captures the correct aminoacyl tRNA. Correct complementary (base-pairing) results in the covalent linking of an amino acid onto the growing polypeptide chain.

Question 71

Prokaryotic Ribosome Structure

Answer 71

rRNAs are an integral part of the structure of this molecule and are important parts of the protein synthesising machinery. Prokaryotes has 5S, 23S rRNA in their large ribosomal subunits (50S) and a 16S rRNA in their small ribosomal subunits (30S).

Question 72

Eukaryotic Ribosome Structure

Answer 72

The rRNAs have 5S, 5.8S and 28S rRNA in the large ribosomal subunits (60S) and 18S rRNA in their small ribosomal subunits (40S). Despite differences in protein and rRNA components both prokaryotic and eukaryotic ribosomes have nearly the same structure and function.

Question 73

Ribosome Binding Sites

Answer 73

The small subunits match the tRNA anticodon to the codon on the mRNA. The large subunit catalyses the formation of the peptide bonds that link the amino acids together into a polypeptide. Each ribosome has one of these spots for mRNA and 3 of these for tRNA. The sites for tRNA are the A (aminoacyl-tRNA - new tRNA enters the ribosomal complex), P (peptidyl-tRNA - tRNA is attached to the polypeptide chain) and E (exit - empty tRNA exits the ribosomal complex).

Question 74

Elongation of Polypeptides

Answer 74

1. Aminoacyl-tRNA binds to the A-site and spent tRNA molecules leave from the E-site.
2. New peptide bonds are formed between the amino acids and the polypeptide chain.
3. The large subunit moves to the next codon.
4. The small subunit moves to the next codon, the A-site is now empty and ready for the next aminoacyl-tRNA.
5. Steps 1-4 Repeat themselves over and over again until a protein is formed.

Question 75

Single Prokaryotic mRNA Codes for Several Proteins

Answer 75

Unlike with eukaryotic ribosomes which recognise a 5’ cap, prokaryotic ribosomes initiate translation at ribosome binding sites, which can be located in the interior of an mRNA molecule. This feature means prokaryotes can synthesise more than one type of protein from a single polycistronic mRNA molecule.

Question 76

Polyribosomes

Answer 76

Many ribosomes can simultaneously translate the same mRNA molecule. This increases the production capacity of a particular protein. These are very large assemblies of several ribosomes as close as 80 nucleotides apart of a single mRNA molecule.

Question 77

The ‘Universal’ Genetic Code

Answer 77

For most organisms the bases in DNA are A, C, T and G whereas in RNA it is A, C, U and G. In transcription there is a similar language with 4 bases matching 4 bases (copy). During translation this is different as 4 bases 20 amino acids translate. The DNA/RNA still have 4 bases however the maximum number of codons is calculated as 4**3 = 64 possible codons. Due to most amino acids being coded for multiple codons (except methionine and tryptophan) the code is called redundant or degenerate.

Question 78

Unique Codons

Answer 78

AUG only codes for methionine. This amino acid is found at the start of every open reading frame and is the initiation of translation. In prokaryotes it is formyl-methionine whereas in eukaryotes it is simply methionine. The 3 codons that aren’t recognised by tRNA are the stop codons found at the end of every translated open reading frame which terminates the signal for translation.

Question 79

Open Reading Frame

Answer 79

Because of the large size of DNA consisting of millions of bases. A lot of space on that DNA doesn’t code for genes however the parts that do a represented by the sites where there is a large number of base pairs without a successive stop codon close by.

Question 80

Cracking the Genetic Code 1

Answer 80

In 1961 Nirenberg and Matthaei produced an artificial RNA composed entirely or uracil, poly (U) and were able to synthesise a protein (poly-phenylalanine). They used a cell-free system. To do this they took E.coli and did a cytosol fraction where the membrane was removed and only cytosol was kept, the mRNA within the cell was destroyed and the radiolabeled amino acids were added. This experiment was repeated 20 times with different combination of mRNA to see what amino acids were created.

Question 81

Cracking the Genetic Code 2

Answer 81

After this organic chemist Gobind Khorana developed a method of making specific RNA strands of repeating di/tri/tetra nucleotide sequences through the transcription of DNA into RNA with RNA polymerase. It was found that different patterns of genetic code would sometimes produce the same amino acid.

Question 82

Cracking the Genetic Code 3

Answer 82

Nirenberg and Leder realised that they could trap trinucleotides (codon sized) with the ribosome and this would also trap the corresponding charged aminoacyl-tRNA which was radiolabeled. In this way they were able to decipher al the remaining codon sequences e.g. what amino acid each codon codes for. In the final analysis they found there are 61 codons for 20 amino acids this would in turn mean that 61 anticodons of tRNA were requires however only 50 of these existed.

Question 83

Wobble Hypothesis

Answer 83

A theory developed by Crick which stated that due to the degeneracy of the third base pair it was loose and so could interact with other bases to a certain degree however not completely however due to the full interaction between the first 2 bases in tRNA the connection is still made. Some tRNA’s can bind at more than one codon. This ability was referred to as ‘wobble’. If inosine (modified A) is in the first position in the anticodon (adjacent to the third position in the codon) it can bind to A, U or C in the third position of the codon. This means that that tRNA can recognise 3 different codons.

Question 84

Conventions

Answer 84

You can read DNA in the 5’-3’ direction and using teh genetic code indicates what amino acids should be encoded. The DNA is typically directly translated to amino acids as writing out the RNA as well can be tedious. Upstream from the gene will be written above or in front of the 5’ end whereas downstream from the gene will be written below or behind the 3’ end.

Question 85

Protein Synthesis

Answer 85

This process is divided into 3 stages.
1. Initiation - mRNA, ribosomes and initiating tRNA all combine to begin the process.
2. Elongation - peptide bonds form while the ribosome moves along the mRNA.
3. Termination - the ribosomes are dissociated and the peptides are freed.

Question 86

Genetic Code Reading

Answer 86

Since there are 3 nucleotides for each codon (functional unit) there are 3 potential translational reading frames (frame 1, frame 2 and frame 3) for each strand however only 1 frame is used for translation.

Question 87

Translocation

Answer 87

The direction of movement undertaken by the ribosome during translation when reading the DNA. This moves in the 5’-3’ direction.

Question 88

Translation Initiation Signals

Answer 88

An ‘AUG’ is a translation initiation signal however methionine also occurs in other places in the code. In order to choose correct spot to start translation other initiation signals have to be available. Sometimes this is a ribosome binding site motif in the mRNA which helps the ribosome detect the correct starting codon, some ribosomes have a scanning ability in order to find the correct site while in some bacteria the small ribosomes binds to the ‘Shine Dalgarno sequence (5’AGGAGGU3’)’ which is located upstream of the start ‘AUG’ codon.

Question 89

Initiation of Translation in Eukaryotes

Answer 89

1. The initiation tRNA (Met-tRNAi) is first loaded into the ‘P’ site of the small ribosomal subunit with initiation factors (eukaryotic Initiation Factor (eIF2-GTP)).
2. The loaded small ribosomal subunit attaches to the 5’ end of the mRNA (assisted by the 5’ end cap).
3. The ribosome scans itself along the mRNA (5’-3’ direction) until it identifies the first AUG codon surrounded by a consensus sequence (Kozac sequence). If this Kozac sequence is degenerate (there are multiple of these) and the ribosome may initiate translation from multiple start codons which produces truncated polypeptides (different functions for different areas).
4. The initiation factors detach and the large subunit can bind on to complete the ribosomal complex. The first amino acid is in the ‘P’ site ready for a chain elongation.

Question 90

Initiation of Translation in Bacteria

Answer 90

1. The 30S ribosome interacts with an initiation factor (IF) and this complex binds the Shine Dalgarno sequence upstream of the ‘AUG’ start codon.
2. The initiator tRNA (fMet-tRNAi) aligns with the start codon ‘AUG’ in the mRNA.
3. The 50S portion of the ribosome associates with the 30S one which releases the IF and forms the 70S complex. tRNA occupies the peptidyl ‘P’ site of the 50S subunit (aminoacyl (A) site is empty).

Question 91

Elongation tRNA-Met vs Initiation tRNA-Met

Answer 91

This methionine in tRNA in bacteria has a formylated methionine (N-formyl-methionine tRNA (fMet-tRNAi)) whereas in eukaryotes this form is not formylated (N-methionine tRNA (Met-tRNAi)). The second form of methionine is the same for both eukaryotes and prokaryotes and is different to the first type (methionine tRNA (Met-tRNA)) as it has a different stem loop structure that preferentially binds to elongation co-factors.

Question 92

Elongation

Answer 92

This is a 3-step cycle which is repeated over and over during the synthesis of a protein chain.

1. an aminoacyl-tRNA molecule binds to the empty ‘A’ site on the ribosome.
2. A new peptide bond is formed between the amino acid attached to the tRNA in the ‘A’ site and the previous amino acid attached to the tRNA in the ‘P’ site (directly after initiation this is ‘AUG’ (methionine). This combination creates a tRNA-peptide in the ‘A’ site.
3. The mRNA moves a distance of 3 nucleotides through the small subunit (uses energy from ATP). The de-acylated tRNA is ejected from the ‘E’ site, the ‘A’ site tRNA peptide moves into the ‘P’ site.

The ‘A’ site is now empty and located over the downstream codon and is ready to receive another aminoacyl-tRNA molecule which base pairs with this codon. The position at which the growing peptide chain is attached to a tRNA doesn’t change during this cycle (it is always linked to the tRNA present at the ‘P’ site of the large subunit.

Question 93

Elongation Reactions

Answer 93

The ribosomes are made up of rRNA (this forms a highly structured pocket of hydrogen networks and has catalytic properties) and polypeptides (structural). This means ribosomes are ribozymes as the rRNA performs the condensation of the carboxy-terminus of the amino acid on the tRNA in the ‘P’ site with the amino-terminus of the amino acid in the tRNA in the ‘A’ site. The peptide-tRNA in the ‘A’ site. The peptide-tRNA in the ‘A’ site moves to the ‘P’ site as the ribosome moves down the mRNA in the 3’ direction.

Question 94

Termination

Answer 94

This is processed initiated by 1 of 3 stop codons which are ‘UAA’, ‘UAG’ and ‘UGA’ all of which don’t have a matching tRNA anticodon. The releasing factors bind to the stop codons instead when it reaches the ‘A’ site of the ribosome. This process alters the activity of the peptidyl transferase (large ribosome) and adds a water molecule to the peptide and releases it. The ribosome then releases the mRNA and then the 2 subunits dissociate and can bind to the same or different strands of the 5’ end of mRNA.

Question 95

Antibiotics

Answer 95

The antibacterial action of some antibiotics is based on inhibiting translation e.g. streptomycin/neomycin/kanamycin bind the 30S (small) subunit prevents the transition from initiation to elongation. Tetracyclines/puromycin mimic the structure of charged tRNA’s and block the ‘A’ site of the prokaryotic ribosome. Chloramphenicol blocks the peptidyl transferase reaction in prokaryotes. Erythromycin blocks the translocation (movements of the ribosome) reaction.

Question 96

Eukaryote vs Prokaryote Translation Machinery

Answer 96

The subtle differences between prokaryotic (bacterial) and eukaryotic ribosomes is enough to create high affinity interaction between antibiotics and the prokaryotic ribosome. Eukaryotic ribosomes are not usually inhibited (some ribosomes in the mitochondria may be as they are similar to prokaryotic ones e.g. chloramphenicol). These differences are the differences in the size of both large and small subunits as well as the amounts of rRNA and proteins that make them up. There is also differences in the binding sites for ribosomes and the type of start codon that is present.