Nucleotides and Nucleic Acids

Question 1

Which nucleotides are purines, and what are their structures?

Answer 1

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Question 2

Which nucleotides are pyrimidines, and what are their structures?

Answer 2

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Question 3

Why is DNA a double helix rather than a single helix?

Answer 3

A double helix is more energetically stable than a single helix.

The bases are protected within the sugar-phosphate backbone, so the genetic code cannot be easily damaged.

Question 4

How does DNA fit inside eukaryotic cells?

Answer 4

DNA folds up by wrapping around positively charged histone protein complexes, so the protein will bind tightly to the negatively charged phosphates.

The complex consists of 2 of each of H2A, H2B, H3 and H4. so these form a histone octamer. The octamer with the DNA wrapped around it is called a nucleosome.

Each nucleosome is separated by about 50bp of linker DNA, so it all looks like ‘beads on a string’. H1 binds to the linker DNA and coils the DNA into chromatin fibres called solenoids.

The chromatin fibres then form loops with the help of nonhistones, and this is called supercoiling.

Question 5

How does DNA fit inside prokaryotic cells?

Answer 5

There is no nuclear envelope within prokaryotic cells, so the condensed chromosome with its associated proteins lie free in the cytoplasm.

The structure it forms is called a nucleoid.

Question 6

How is the phosphodiester bond formed in DNA?

Answer 6

1 → The 5’ group of a nucleotide is held close to the free 3’ hydroxyl group of a nucleotide chain.

2 → The 3’ hydroxyl group forms a bond to the alpha P (via nucleophilic attack), and the bond between the alpha P and the adjacent O atoms breaks.

3 → A phosphodiester bond now joins the two nucleotides together.

4 → A pyrophosphate group has been liberated.

5 → The pyrophosphate group is hydrolysed forming two phosphate ions. This releases energy and drives the reaction forward to completion.

Question 7

What 5 things does replication require to begin?

Answer 7

1. An origin
2. A single-stranded region
3. A primer (can be pre-existing DNA or RNA)
4. Primase (binds primer to DNA)
5. DNA polymerases

Question 8

How is DNA synthesis initiated?

Answer 8

1. DnaA (associated with ATP) recognises a sequence of 9 nucleotides repeated 4 times (called 9-mers), binds to this region and causes the DNA to curl around DnaA.
2. The DNA helix opens at a nearby region of 13-mers (13 NTPs repeated 3 times), forming an open complex.
3. DnaB is a helicase and DnaC is a DnaB load helper. Two molecules of DnaB (one for each strand) are escorted by DnaC. One molecule attaches to the template strand and moved 5’ → 3’ while the other molecule attaches to the coding strand and moves 3’ → 5’.
4. Single-strand binding proteins then bind to the DNA. This makes the DNA rigid without bends or kinks, so it is a good template for DNA synthesis.

Question 9

How is DNA replicated on each strand?

Answer 9

Replication on the leading strand is continuous in the 5’ → 3’ direction.

Replication on the lagging strand is discontinuous. This is because DNA polymerase 3 cannot orientate itself in the other direction.

Instead, DNA polymerase 3 still synthesises in the 5’ → 3’ direction, but in short fragments. When the helix has unwound further, polymerase loops back on itself to synthesise another Okazaki fragment.

DNA ligase then joins these fragments together.

Question 10

What do topoisomerases do?

Answer 10

Topoisomerases release tension that is created when the double helix is unwound during replication.

Type 1 cuts and unwinds a single strand of DNA, so the double helix has one less twist. It does this by cutting one strand so that the other can pass through.

Type 2 cuts and unwinds both strands of the double helix when it exists as a helix with another double strand.

Topoisomerases are important in:

Growing fork movement

Untangling finished chromosomes after DNA replication

Initiating DNA replication

Question 11

How does DNA polymerase ‘proof-read’ DNA?

Answer 11

DNA polymerase senses the distortion of the double helix from the insertion of the incorrect base.

It then ‘close’ the fingers of the hand, and moves the DNA fom the polymerase domain to the exonuclease domain.

The incorrect base is removed by 3’ → 5’ exonuclease activity, and then DNA is moved back to the polymerase domain.

Question 12

What other method is there of repairing incorrect base pairing?

Answer 12

MutHLS system is present in E. coli and a similar one is present in humans.

MutHLS binds to the incorrect sequence and removes it.

The mutation itself, as well as the bases adjacent to it, are removed and replaced.

DNA polymerase 3 then replaces it with the correct sequence.

Question 13

How can DNA be damaged and what effect does this have?

Answer 13

Both physical and chemical agents can cause DNA damage.

Physical agents: UV and ionising radiation

Chemical agents: All are mutagens because they alter one or more nucleotides

These can cause:

Altered bases (bulky adducts)

Lost bases

Dimerisation of adjacent bases

Breakage of phosphodiester bonds

Covalent linking of strands

Depurination (loss of a purine due to being hydrolysed)

Deamination (producing U from C by removing an amine group)

Question 14

What are thymine dimers and how are they repaired?

Answer 14

Thymine dimers are formed on expose of DNA to UV light; this is why bacteria are killed by UV light.

The bases become covalently linked between C₅ + C₅ and C₆ + C₆. This means the dimer cannot fit properly into the double helix; instead it bulges out and blocks both transcription and replication.

This is fixed by excision repair:

1 → thymine dimer and 30 surrounding dNTPs are excised from DNA

2 → The exposed, undamaged DNA must be protected from nuclease attack, so it is protected by various proteins

3 → DNA polymerase 1 (prokaryotes) / Pol β (eukaryotes) and DNA ligase repair the damaged strand by putting in the correct bases

Question 15

How does DNA sequencing (Sanger/dideoxy/chain termination) work?

Answer 15

DNA replication is performed in the presence of dideoxynucleotide triphosphates.

ddNTPs prevent the daughter strand from increasing in length because they do not have any OH groups on the ribose sugar, so they cannot perform a nucleophilic attack on the next dNTP in the sequence.

The ddNTPs are labelled with different fluorochromes and are present in a concentration that allows each position to have a ddNTP there.

The DNA is then ordered by size by electrophoresis and the order is read using UV.

Question 16

How does PCR work?

Answer 16

1 → The double helix is heated to 90ºC to separate the two strands.

2 → The mixture is cooled to 60ºC and incubated with 2 DNA primers, which are each complementary to one of the strands. The primers anneal to the DNA.

3 → Heated to 72ºC and incubated with thermostable DNA polymerase. The primers direct polymerase to copy each strand, producing 2 times the number of original templates.

Question 17

What is base-excision repair and how does it work?

Answer 17

Base-excision repair is the method used to repair deamination^1a and depurination^1b.

1a → U does not form part of undamaged DNA, so it is recognised by uracil-DNA glycosidase. This leaves a gap in the DNA, and because no enzyme can reattach a C, apyrimidinic endonuclease (AP) recognises the gap and removes the ribose sugar by breaking the phosphodiester bonds on either side.

1b → When DNA has been damaged by depurination, apurinic endonuclease (AP) also removes the sugar by breaking the phosphodiester bonds.

2 → The repair process for inserting a purine or pyrimidine is the same: DNA polymerase 1 replaces the correct dNTP, and DNA ligase seals the strand.

Question 18

What is the general structure for the promotor region in prokaryotes?

Answer 18

Transcription starts at the +1 site.

There are two recognition sequences upstream of the +1 site. These are at -10bp and -35bp, and are 16-18bp apart.

RNA polymerase binds to the promoter region, and this will initiate transcription.

Question 19

What are transcription factors and what are their structure?

Answer 19

They are proteins that bind to DNA and control the rate of transcription.

Transcription factors contain one or more DNA binding domains; these can be activators or repressors, among other things.

Examples: helix-turn-helix motif, helix-loop-helix motif

Both of these bind to the major groove. The former includes lac and trp repressors, CPR protein, homeodomain proteins, etc

zinc finger motif

leucine zipper motif → this ‘unzips’ as it binds to DNA

Question 20

How does transcription stop in prokaryotes?

Answer 20

Transcription stops at the termination site.

These often have repeated regions of Gs and Cs that form a secondary structure of a hairpin loop in the mRNA, followed by a stretch of A/U.

This causes the transcription bubble to shrink, and this cannot be maintained due to weak A-U bonds.

The two strands reconnect, RNA polymerase is released, transcription stops and the double helix reforms.

This is known as Rho-independent termination.

In Rho-dependent termination, the protein Rho is an ATP-dependent helicase that dislodges the 3’ end of the mRNA from the active site of RNA polymerase.

Question 21

What are the different components of the lac operon?

Answer 21

lac Z → codes for beta-galactosidase (cleaves lactose into glucose and galactose, converts lactose into allolactose)

lac Y → codes for galactosidase permease (transports lactose (and other substances) into the cell)

lac A → detoxifies other products brought into the cell by galactoside permease that aren’t useful

lac I → transcribes lac I mRNA, which codes for the repressor protein

Question 22

How does the lac operon work?

Answer 22

Without lactose present:

Repressor protein binds to the operator region

RNA polymerase cannot bind to the promoter region

Structural genes not transcribed

With lactose present (no glucose):

Allolactose binds to the repressor protein, changing its shape and preventing it from binding to the operator region

RNA polymerase binds to promoter region

Genes are transcribed

With lactose present (low glucose %):

Low transcription of lac Z and lac Y

In response to glucose:

Levels of cAMP are inversely proportional to that of glucose. At low glucose%, cAMP binds to CAP (catabolite activator protein), which in turn binds to the CAP binding site, upstream of the lac promoter.

This causes a 90º bend in the DNA, allowing RNA polymerase to bind and transcribe the lac genes.

Question 23

How is transcription regulated in eukaryotes?

Answer 23

Eukaryotes do not have operons in their genome.

They are monocistronic, that is, each gene has its own promoter.

Eukaryotic genomes contain trans-regulatory elements and cis-regulatory elements.

Trans elements code for a regulatory protein, whereas cis elements act as binding sites for the regulatory protein eg. the TATA box.

Factors specific to the cell ensure that genes are expressed in the correct time and space.

The correct spatial expression is important for body formation.

Question 24

What are introns and exons, and what is their purpose?

Answer 24

Introns → spliced out of pre-mRNA as they do not code for proteins

Exons → not spliced out of pre-mRNA as they do code for proteins

Different exons can be spliced out, forming different proteins from the same genetic code

However, disease can form from inaccurate splicing

Question 25

What are general transcription factors, and what do they do?

Answer 25

They help start transcription by aiding binding of RNA polymerase to the transcription start site by forming the transcription preinitiation complex.

General transcription factors are present in different specialised cells; proteins are expressed at different levels in different cells due to the presence of specific regulatory transcription factors, rather than due to the presence of GTFs.

Question 26

How does the transcription preinitiation complex work?

Answer 26

1 → TATA binding protein (a subunit of TFIID) binds to the TATA box

2 → TFIIA and TFIIB are recruited to the promoter

3 → RNA polymerase and TFIIF are recruited to the promoter

4 → TFIIE and TFIIH are recruited to the promoter

5 → TFIIH has ATPase and helicase activity, so creates negative supercoil tension, thereby creating the transcription bubble

6 → the tail of RNA polymerase is phosphorylated; RNA polymerase leaves the promoter region and transcription begins

Question 27

What are eukaryotic enhancers and how do they work?

Answer 27

Enhancers are short regions of DNA that can be many kbp away from the promoter region. They increase the rate of transcription of the associated gene.

Activator proteins bind to enhancer region; this causes the DNA to bend to bring the enhancers closer to the core promoter.

Co-activators link the enhanceosome to the GTF TFIID, which helps to position TFIID on the promoter region.

Other GTFs bind as well as RNA polymerase, and transcription begins.

Question 28

How many different codons are there, and what different things can they specify?

Answer 28

64 codons (4 x 4 x 4)

61 codons code for amino acids

This includes 1 start codon = methionine

This is in both prokaryotes and eukaryotes, but in prokaryotes methionine is f-met, ie. an aldehyde group has been added to methionine to act as an initiator

3 stop codons → UAA, UAG, UGA

There are 21 amino acids, but 61 codons for amino acids; this means the genetic code is degenerate as more than one codon can specify a particular amino acid

Question 29

How can different polypeptides be made from the same mRNA sequence?

Answer 29

There are 3 possible ways mRNA can be read:

Frame 1 = start on base 1 (usually frame 1 is used)

Frame 2 = start on base 2

Frame 3 = start on base 3

However, most mRNAs only give one reasonable frame because the other 2 frames introduce stop codons that mean translation is stopped before a functional protein is formed.

Question 30

What types of mutation can occur during translation?

Answer 30

Frameshift → one base is deleted. thereby affecting the rest of the sequence

Missense → One base is substituted for another

Nonsense → one base is substituted, incorporating a stop codon into the sequence

The latter two are called point mutations

Question 31

What molecules are involved in translation?

Answer 31

mRNA

rRNA

tRNA

aminoacyl-tRNA synthetases → these are enzymes

that load each tRNA with the corresponding amino acid.

→ there is one enzyme for each amino acid

Question 32

What is the structure of ribosomes?

Answer 32

Ribosomes are ribonucleoproteins that synthesise proteins.

they are made up of 2 subunits that consist of several different rRNAs and more than 50 proteins.

Their size is measured in Svedberg units.

Prokaryotes, mitochondria, chloroplasts:

50S + 30S = 70S

Eukaryotes:

60S + 40S = 80S

Question 33

What is the function of tRNAs and aminoacyl-tRNA synthetases?

Answer 33

All tRNAs have two functions: to covalently link with an amino acid, and base pair with a codon in mRNA

Each of the aminoacyl-tRNA synthetases recognises one amino acid and all of the compatible tRNAs.

When the amino acid links with the tRNA it forms a high-energy ester bond via:

ATP → AMP + PPi

Question 34

How can some cells have fewer than 61 tRNAs?

Answer 34

A single tRNA can recognise more than one codon; this is called wobble.

Wobble occurs in the 3rd base position of the mRNA codon, which is the 1st base position of the anticodon on the tRNA.

Inosine is the deaminated derivative of adenosine and only occurs on tRNA anticodons.

I can pair with C, U and A, therefore the tRNA can pair with 3 different codons.

U can also pair with A and G

Question 35

How do ribosomes bind to mRNA in prokaryotes?

Answer 35

The 16S rRNA recognises an 8bp sequence called the Shine-Dalgarno sequence, which lies upstream of the initiator methionine.

The small ribosomal subunit is recruited and binds to the mRNA, then the large subunit is recruited and binds.

Question 36

How do ribosomes bind to mRNA in eukaryotes?

Answer 36

Both subunits bind to the 5’ capped end and scan for the Kozak sequence, which has the initiator methionine embedded within it.

Translation then begins.

Question 37

What are the different sites within the ribosome, and what do they do?

Answer 37

Initiator methionine binds to the P site

Aminoacyl site: tRNA carrying its amino acid enters the ribosome here, and the codon binds to the tRNA anticodon

Peptidyl site: a peptide bond forms between amino acids; this is catalysed by peptidyl transferase. 3 proteins called elongation factors help to elongate the polypeptide chain.

Exit site: mRNA exits the ribosome here

Translation then stops when a stop codon is reached because these do not code for an amino acid, so there is no associated tRNA molecule

Question 38

How does translation stop?

Answer 38

Termination release factors recognise different stop codons.

They promote cleavage of the peptide and tRNA so that the peptide chain is freed from the ribosome.

It does this by binding to the stop codon in the aminoacyl site.

Prokaryotes:

RF1 → recognises UAG and UAA

RF2 → recognises UGA and UAA

RF3 → dissociates RF1+2 from ribosome

Eukaryotes:

eRF1 → recognises all stop codons

eRF3 → releases eRF1 from ribosome

All the translational components are then recycled.

Question 39

How do antibiotics interfere with translation?

Answer 39

Different antibiotics interfere at different points:

Chloramphenicol: interferes with A binding site so proteins cannot form → bacteria dies

Diptheria: inhibits A-P translocation

Question 40

What is Western blotting and how does it work?

Answer 40

It is a method to separate proteins by size and detect specific ones.

1 → Add 2-mereaptoethanol to the protein sample to break any disulfide bonds

2 → Add SDS, an anionic detergent to the sample to coat all the proteins with a negative charge.

3 → Place sample in a polyacrylamide gel (to provide a matrix) and separate using electrophoresis. Proteins will separate by size and move towards the positive electrode due to SDS.

4 → Proteins are stained with a blue dye, or alternatively a silver stain for more sensitive detection.

5 → A nylon membrane is placed against the gel so that the pattern of proteins is preserved on the nylon membrane.

6 → The membrane is incubated with an antibody specific to the desired protein. The antibody will then bind to the protein.

7 → A 2nd antibody is added that is specific to the first antibody. This is attached to an enzyme and its substrate that make a coloured product, which can then be detected.

Question 41

What are the steps involved in pre-mRNA processing in eukaryotes?

Answer 41

1 → the 5’ terminal triphosphate group has a G added via a 5’ - 5’ phosphodiester link. The G is then methylated forming a 7-methyl G cap.

2 → The poly-A site is cleaved at the 3’ end.

3 → The poly-A tail is added by poly-A polymerase without using a template strand; the tail can be between 100 - 250 bp in length.

4 → pre-mRNA may be spliced ie. introns are cut out by spliceosomes

These steps (excluding number 4) increase the lifetime of the RNA molecule by protecting it from digestion by nucleases.

Question 42

What is reverse transcription?

Answer 42

Reverse transcriptase is an enzyme used to generate cDNA from mRNA. So, this process is called reverse transcription.

These are both used to describe a sequence that translates into a protein.

cDNA is more robust and easier to work with.

Question 43

What are microarrays?

Answer 43

Aka gene chips, are small glass wafers with single-stranded cDNA attached.

mRNA is isolated from a cell, tagged with a fluorescent dye and will then hybridise to the cDNA.

Computer algorithms then analyse the patterns made for a particular microarray.

Question 44

How does the trp operon work?

Answer 44

The trp operon has 5 structural genes that code for enzymes that synthesise tryptophan. The operon is only transcribed if the % of tryptophan is low.

trpR codes for an inactive repressor protein called an aporepressor. When the tryptophan % is high it binds to the repressor protein to form an active repressor complex, called a corepressor.

The corepressor then binds to the operator region, preventing transcription of the operon.