MCB 3: DNA Mechanisms - Replication and Repair

Question 1

Definition of Replication

Answer 1

• the process by which a copy of a DNA molecule is made

Question 2

Briefly describe the cell cycle

Answer 2

Question 3

What is semi-conservative DNA replication?

Answer 3

• in each round of DNA replication, each of the two strands of DNA is used as a template for the formation of a new complementary strand
• it is semi-conservative because each daughter DNA double helix is composed of one conserved strand and one new strand

Question 4

What enzyme is responsible for DNA strand separation?

Briefly describe what it does

Answer 4

• DNA helicase
• it uses ATP as an energy source and breaks hydrogen bonds between base pairs

Question 5

Describe how structure of DNA helicase facilitates DNA strand separation

Answer 5

• DNA helicases have a hexagonal arrangement of six identical subunits, forming a ring
• the ring is not 6-fold symmetric but slightly squished
• two of the opposing bind sites bind to ATP tightly, two others are more likely to bond to ADP and phosphate and 2 are empty
• as ATP is hydrolysed, their binding sites inter-convert, creating a ripple effect
• there are loops, proposed to bind to DNA, that extend into the centre hole of the ring
• they oscillate up and down, and these oscillating loops pull a DNA strand through a central hole, thus unwinding the double helix

Question 6

What is chain elongation?

Answer 6

• the process of adding nucleotides to the 3’ end of a growing chain to synthesise new DNA (using the enzymes DNA polymerase)

Question 7

What do DNA polymerases require to allow chain elongation?

Answer 7

• a template strand
• an oligonucleotide primer (they cannot start a new strand from scratch)
• a supply of deoxynucleotide triphosphates (dNTPs)

Question 8

Explain how chain elongation works

Answer 8

• the deoxyribonucleoside triphosphate hydrogen bonds to the template strand
• the innermost phosphorus atom of the incoming deoxyribonucleoside triphosphate undergoes nucleophilic attack by the 3’-hydroxyl group
• this forms a phosphodiester bridge, and release pyrophosphate

Question 9

What is the replication fork?

Answer 9

• a Y-shaped junction that is the point at which DNA replication takes place

Question 10

Why is the replication fork asymmetrical?

Answer 10

• as chain elongation only proceeds in a 5’ to 3’ direction, this creates a problem due to the anti-parallel nature of DNA
• the replication fork is asymmetrical with the leading stranding being synthesised continuously, and the lagging strand synthesised in short section called Okazaki fragments

Question 11

Explain how the leading strand is primed to allow DNA synthesis

Answer 11

• an RNA primer is needed to start replication at a replication origin
• on the RNA primer, there is a free 3’ hydroxyl group to allow for chain elongation to occur
• DNA polymerase copies the DNA sequence of the template strand in a continuous manner

Question 12

Explain how the lagging strand is primed for DNA synthesis

Answer 12

• DNA primase makes a new RNA primer
• DNA polymerase adds to the RNA primer from the exposed hydroxyl on the 3’ end of the primer group, synthesising the new Okazaki fragment
• the old RNA primer is erased by ribonuclease
• the gap is filled by repair DNA polymerase
• DNA ligase joined the new Okazaki fragment to the growing chain

Question 13

Why would the lagging strand stop short and not be able to replicate the end? And how is this resolved?

Answer 13

• this is called the end replication problem
• on the lagging strand, RNA primers provide 3’ hydroxyl groups at regular intervals along the template
• even if a final RNA primer were built at the very end of the chromosome, the lagging strand stops short because the primers need to be removed and a 5’ section of DNA is exposed instead
• there is no 3’-hydroxyl group available at the end of the chromosome to prime DNA synthesis
• at the end of chromosomes, there are a section of G-rich series of repeats, called a telomere
• telomerase recognises this sequence and using an RNA template within the enzyme, it elongates the parental strand in the 5’ to 3’ direction
• the lagging strand is then completed by DNA polymerase alpha, which carried DNA primase as one of it subunits
• in this way, the end is completely copied in new DNA

Question 14

What is the overall error rate in DNA replication?

Answer 14

• difficult to estimate
• may be 1 - 2 bases per genome
• may sound low but humans have approx 6 million base pairs per diploid cell

Question 15

Why can the difference between a DNA mutation and a variant (polymophism) be subtle?

Answer 15

• All changes to the DNA sequence from the original template are considered to be mutations. However, as DNA base changes occur very frequently in a general population, and that genes spread throughout the population, we may never know the original true sequence - and such differences seen in a population are referred as variants. All variants arise from DNA mutation.
• A mutation can be regarded as being a change in the DNA sequence away from that observed in the general population.

Question 16

What are the five types of DNA repair?

Answer 16

• direct repair
• excision repair
• mismatch repair
• non-homologous end-joining
• homologous-directed repair

Question 17

Explain direct repair

Answer 17

• This is the simplest repair process, acting on damaged nucleotides to restore their original structure.
• For example: Guanine may become alkylated leading to it binding to Thymine, rather than to Cytosine, as it can form only 2 hydrogen bonds.
• This can be corrected by enzymes with the ability to transfer the alkyl group to themselves such as the MGMT enzyme in humans or the Ada enzyme in E. coli.
• By this transfer, the original structure and function of the enzyme can be restored.
• Due to its simplicity, only a few mutations can be repaired in this way

Question 18

Explain excision repair

Answer 18

• This is a more complicated, multi-step process which involves the complete removal of the damaged nucleotide before correctly filling the gap
• A type of DNA glycosylase is used to perform the excision of the damaged nucleotide by flipping it out of the helix and then cutting the b-N glycosidic bond between the damaged base and the sugar in the backbone.
• This baseless site is known as the AP site and the 5’ and 3’ phosphates are cut by the action of an AP endonuclease or Phosphodiesterase (for the 3’ side).
• The gap can then be filled with the correct nucleotide by DNA polymerase and sealed by DNA ligase.

Question 19

Explain mismatch repair

Answer 19

• Direct and Excision repair work by recognising abnormal chemical structures in DNA but cannot correct a mismatch caused by a DNA replication error.
• As the mismatched nucleotide is not abnormal, the mismatch repair mechanism recognises improper base pairing between the parent and daughter strands.
• This mechanism is not yet well understood in eukaryotes, however in E. colithe two strands are distinguished by DNA methylation on the parent strand at the GATC, CCAGG, and CCTGG sequences while the daughter strand is not yet methylated.
• MutS binds to the site of the mismatched nucleotide while MutH binds to unmethylated GATC sequences on the daughter strand and cuts it at the methylation site.
• DNA helicase is then used to detach a segment of the daughter strand from the cut site, allowing an exonuclease to degrade the strand from the cut site and beyond the site of the mismatch.
• The gap can then be correctly filled by DNA polymerase Iand joined by DNA ligase.

Question 20

Explain non-homologous end-joining

Answer 20

• In the event of a double-strand break, Ku proteins bind to the broken ends of DNA
• The Ku proteins have an affinity for one another so bring themselves along with the ends of the DNA close enough for DNA ligase to re- join the ends.
• This is an error-prone mechanism as the nucleotides flanking the break are often lost, or hybrid structures form if 2 chromosomes happen to be broken.

Question 21

Explain homology-directed repair

Answer 21

• This is a mechanism by which a double-strand break is repaired using a homologous template, often a sister chromatid.
• The 5’ ends of the broken DNA strand are first degraded by endonucleases to leave long 3’ tails.
• With the help of recombination proteins, a 3’ tail is able to invade the homologous strand to create a ‘D loop’ (displacement loop) and undergo DNA synthesis using the homologue as a template until it is complementary to the 3’ tail on the other side of the break.
• Once reattached, the remaining gaps are then filled and sealed by DNA polymeraseand DNA ligase.
• As it uses a homologous template, it is more precise than NHEJ.

Question 22

How can a specific DNA fragment be amplified by PCR?

Answer 22

1. 95°C:
   - DNA strands denature and separate from one another
2. 55°C:
   - 2 specific oligonucleotide primers which are complementary to short stretches on either side of the desired fragment hybridise to the DNA
3. 72°C:
   - DNA polymerases and nucleotide triphosphates are added to extend the primers and synthesises 2 molecules of dsDNA.
   - The cycle is repeated multiple times and by the completion of the 30thcycle, the product consists almost entirely of the target sequence.

Question 23

What analysis could be used for high throughput diagnosis?

Used to examine many different mutations across the gene?

Answer 23

• Microarrays:
• use multiple probes to test for the presence/absence of multiple different mutations
• Multiplex PCR:
• using different colour fluorophores to label the different primers, or simply using the difference in size of the products, we can perform PCR assays for different targets at the same time.

Question 24

What method do we use to detect DNA variants if we do not know the exact target sequence?

Describe how the method works

Answer 24

• whole genome sequencing (Sanger sequencing)
• this uses a mixture of dNTPs and ddNTPs (didioxyneucleotides), which cause the chain to terminate when it is reached since ddNTPs lack 3’-OH groups and cannot form phosphodiester bonds.
• This can be repeated using ddNTPs for each base to generate DNA sequences for each and putting them together like a jigsaw.
• If electrophoresis is used, the smallest fragments run down the gel fastest and the sequence can be determined by reading it from smallest to largest DNA fragment.
• With scientific advancements, the ddNTPs can now be radioactively/fluorescently labelled and run through automated sequencing machines.