MCG - Maintenance and use of Genetic Information

Question 1

Give a general overview of replication.

Answer 1

The DNA unwinds to expose the bases.

The daughter strands of DNA are synthesised, using one parent strand as template (complementary base pairing).

This occurs through semi conservative replication
(each new DNA molecule contains one parent and one daughter strand).

Question 2

Describe the process of initiation of replication.

Answer 2

The base pairs of the double helix must be broken (the strands must unwind). The DNA is unwound by helicase (so bases are now accessible).

The immediate re-formation of the double helix is prevented by single strand binding proteins.

Pulling strands apart increases their winding about each other further down the molecule, introducing positive supercoiling. The helix would eventually snap (double strand breaks are the most toxic type of DNA damage).

Topoisomerase breaks a phosphodiester bond in one of the parental strands ahead of the replication fork, providing a degree of freedom around which the remainder of the helix can unwind.

DNA polymerase now has access to the bases and can synthesize new strands using the parental ones as a template.

Two “replication forks” are formed.

Question 3

Describe the process of replication.

Answer 3

DNA polymerase synthesizes new strands using the old ones as templates.

DNA polymerase only synthesizes DNA in the 5’ to 3’ direction.

This results in a different mode of replication for the two parent strands.

One strand is synthesized continuously (the leading strand).
The other must be synthesized discontinuously (the lagging strand).

Question 4

How is DNA polymerase prepared?

Answer 4

DNA polymerase cannot initiate DNA synthesis on its own, as it can only add nucleotides to pre-existing chains.

RNA polymerase can initiate RNA synthesis, so DNA synthesis begins with synthesis of a short RNA primer.

The RNA primer, 8-10 nucleotides long, is synthesized by primase (an RNA polymerase).

DNA polymerase then takes over, extending the 3’ end of the RNA primer.

Question 5

How is the lagging strand different to the leading strand?

Answer 5

The lagging strand will consist of a series of Okazaki fragments which must be joined up.

Question 6

Describe the events that join up the adjacent Okazaki fragments.

Answer 6

DNA polymerase extends DNA
5’ in the 5’ to 3’ direction until it reaches the next RNA primer.

The primer is degraded by an exonuclease, leaving a gap.

DNA polymerase continues synthesis across the gap.

Two Okazaki fragments are now next to each other, and the missing phosphodiester bond is put in place by DNA ligase.

Question 7

How is leading and lagging strand synthesis coordinated by DNA polymerase?

Answer 7

The DNA polymerase folds the lagging strand so that it faces backwards, and thus both can be added to in the ‘same direction’ as the DNA polymerase moves across them.

Question 8

Quickfire question:

What is the function of helicase?

Answer 8

It unwinds DNA double-strands.

Question 9

Quickfire question:

What is the function of topoisomerase?

Answer 9

It releases supercoils in DNA.

Question 10

Quickfire question:

What is the function of single strand binding protein?

Answer 10

It stablises single stranded DNA.

Question 11

Quickfire question:

What is the function of primase?

Answer 11

It makes RNA primer so that DNA synthesis can begin.

Question 12

Quickfire question:

What is the function of DNA polymerase?

Answer 12

It synthesises DNA.

Question 13

Quickfire question:

What is the function of exonuclease?

Answer 13

It removes the RNA primer.

Question 14

Quickfire question:

What is the function of DNA ligase?

Answer 14

It links the adjacent Okazaki fragments.

Question 15

What will happen in the DNA polymerase adds an incorrect nucleotide?

Answer 15

The DNA polymerases also posses 3-5’ exonuclases. The nuclease will remove the wrong nucleotide, and the DNA polymerase will add the correct one.

This shows the DNA polymerase proof-reading ability.

Question 16

What results in the end replication paradox?

Answer 16

We have multiple replication origins per eukaryotic chromosome.

The DNA polymerase only extends 3’ OH ends.

This results in the end replication paradox i.e. a small amount of DNA is lost from each end of a linear chromosome after each round of DNA replication.

This is due to the fact that the primer needs to sit onto a bit of DNA, but is later on removed, so that bit of DNA is lost in translation.

Question 17

How is this loss of information prevented in DNA, and how is the structure formed?

Answer 17

The ends of chromosomes shorten with every round of replication.

A loss of information is prevented by capping the DNA using telomeres.

They are composed of hundreds of copies of 5’ TTAGGG 3’.

The formation of these telomeres is by the enzyme telomerase.

Telomerase extends DNA (without using a chromosomal template), since the enzyme carries its own template (a short stretch of RNA).

Question 18

When does telomerase get switched off, and what happens as result?

Answer 18

In somatic cells, telomerase is switched off. So with every division, telomeres get shorter.

Eventually, the telomeric sequences are lost, and useful DNA will start being lost due to the end replication defect.

This is thought to make a significant contribution to ageing.

Question 19

What is the Hayflick limit?

Answer 19

There is a limited repetitive cell lifespan for cells known as the Hayflick limit.

You can increase that limit by extending the lifespan of DNA.

Question 20

What are the different types of DNA mutations?

Answer 20

• point mutations
• small scale insertions and deletions
• changes in gross morphology of chromosomes

Question 21

Describe point mutations.

Answer 21

It is when a single base is changed in the DNA sequence.

Some consequences of point mutations:

• silent mutations (the amino acid made is the same as before)
• missense mutation (non-synonymous - another amino acid is made)
• nonsense mutation (a stop codon is generated)

Question 22

Give some examples of missense mutations.

Answer 22

• haemoglobin in sickel cell anaemia (Glu -> Val)
• N-Ras (oncognene is activated in rhabdosarcoma) (Gln -> His)

The N-Ras activates the Ras enzyme constantly, telling cells to divide all the time, which is a contributory mutation to that lead to rhabdosarcomas.

Question 23

Describe small scale insertions and deletions.

Answer 23

It is when two or more bases are changed in the DNA sequence.

If the change is in a multiple of three, then there is maintenance of the reading frame, but if it is not, then the reading frame is lost.

Some small scale mutations might not have serious consequences to human health if occuring in a non-coding region (which is the main reason why microsatellites are variable).

However, some could have serious consequences to human health if in a coding region, such as Huntington’s Disease.

The Htt (HD) locus encodes huntingtin, which features a trinucleotide CAD 10-34 times.

The increase in CAG causes the Huntingtin to denature and aggregate with itself. It comes out of solution, and when it precipitates in cells, it leads to neuronal decay.

Question 24

Describe changes in gross morphology of chromosomes.

Answer 24

These are large-scale changes in the sections of the chromosomes.

These changes could be:

• inversions of part of the chromosome
• deletions
• duplications
• translocations

These normally originate from a double stranded break in the DNA, which leads to the DNA chunk having the ability to change as such.

Two double strand breaks can lead to translocations.

Question 25

Give some examples of clinical conditions due to large-scale chromosomal changed.

Answer 25

DELETION - Cri du chat syndrome

Deletion of the end of chromosome 5 (leading to limited physical and mental development). It is genetic.

TRANSLOCATION - Chronic myelogenous leukaemia (CML)

Pieces of chromosome 9 and 22 swap (the new small chromosome 2 is called Philadelphia chromosome). 25% of adult leukaemia is CML. This can occur during one’s lifetime.

Question 26

How do these mutations appear in our genomes?

Answer 26

They can be:

• spontaneous mutations
• induced mutations

Question 27

Describe the different ways in which spontaneous mutations can occur.

Answer 27

1. Errors in DNA replication.
   These escape the proofreading and repair mechanisms.
2. Replication slippage (which is more likely at repetitive regions).
3. Deamination.
   - C turns to U (now pairs with A) (this happens 100 times /cell/day)
   - A to Hypoxanthine (now pairs with C)

Question 28

Describe the different ways in which induced mutations can occur.

Answer 28

They can be physical:

• ionizing radiation - causes single or double strand breaks
• ultraviolet light - thymine dimers (UV light causes a covalent association between two thymine on the same strand, leading to thymine dimers)

They can also be chemical, occurring through agents that react with bases:

• nitrous acid: cytosine to uracil, CG → TA
• alkylating agents: guanine modification, GC → AT
• free radicals: strand breaks and base modification

Question 29

Describe the damaged caused by UV irridation.

Answer 29

UV-C (180-290 nm) - not found in daylight as it is absorbed by the ozone layer - the most energetic and lethal (used as a sterilization agent)

UV-B (290-320 nm)- major mutagenic fraction of sunlight

Question 30

What are the different ways in which damage is repaired?

Answer 30

• direct repair:
  (for e.g. dealkylation by enzyme)
• removal of the damaged region followed by re-synthesis (Nucleotide excision repair (NER))
• repair of double strand breaks (homologous recombination)

Question 31

Describe DNA repair of double strand breaks (homologous recombination).

Answer 31

1. DNA damage occurs (double strand break).
2. An exonuclease removes a few bases leaving protruding 3’ single- strand ends end (resection).
3. Free strand invades homologous intact duplex, promoted by recA (prokaryotes) Rad51 (homologue in eukaryotes).
4. Donor strand used as template for repair.
5. Resolution at junctions to give two duplexes.

Question 32

Describe how the mutant form of BRCA2 leads to cancer.

Answer 32

BRCA2 mutations confer a 66% lifetime risk of developing breast cancer (12% risk for ovarian cancer).

1. BRCA2 binds to single stranded DNA at a double strand break.
2. It enables orderly assembly of RAD51 onto single stranded DNA, forming a nucleoprotein filament.
3. Rad51 is inefficiently recruited to double strand breaks in cells with mutated BRCA2 (inefficient homologous DNA repair).

In BRCA2 mutant cells, there is a high degree of chromosome instability, including chromosome breaks and radial chromosomes (where mutations accumulate). Cells where this is happening are going to be accumulating mutations faster than normal cells, accelerating your path to developing tumours.

Question 33

Describe a situation in the body where constant cell division is a feature.

Answer 33

Epithelial cell populations are in constant flux; each minute approximately 5 million cells in the colon epithelium die and equal number of new cells replace them.

This layer is the site of most of the pathological changes associated with intestinal carcinomas (caused by accumulation of mutations over time).

Question 34

What are the different ways in which chromatin can be modified to enable access to DNA?

Answer 34

• nucleosome sliding

- DNA pulled away from nucleosome

Question 35

What are the different ways in which gene expression is tightly regulated?

Answer 35

1. Tissue specific expression.
2. External signals lead to changes in the gene expression profile.
3. Regulation in time and space.

Question 36

Describe how gene expression is regulated through tissue specific expression.

Answer 36

All somatic cells contain the same DNA, yet different cell types vary dramatically in both structure and function.
This is largely due to different genes being expressed in different tissues.

Question 37

Describe how gene expression is regulated through external signals that lead to changes in the gene expression profile.

Answer 37

For example, during intense exercise or starvation, levels of glucose in the blood supply will be low and must be raised. Glucocorticoid hormones are released.

These reach the liver via the blood supply, and signal the liver to increase the expression of genes that catalyze production of glucose from amino acids

When the hormone is no longer present, production of these enzymes drop to their normal levels.

Question 38

Describe how gene expression is regulated through regulation in time and space.

Answer 38

During development, certain genes must be switched on or off at the right time in the right place.

Question 39

How can gene expression be disturbed to lead to disease?

Answer 39

• activation of genes that promote cell division
• decreasing transcription of genes that inhibit cell growth
• increasing transcription of genes that promote angiogenesis (proliferation of blood vessels so that tumours can receive oxygen and nutrients)