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FBS1- CGCC > Maintenance And Use Of Genetic Information > Flashcards

Maintenance And Use Of Genetic Information Flashcards

1
Q

What are the 4 parts of the cell cycle

A

G1
S
G2
M

(G0)

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2
Q

What is G1

A

The growth or gap phase- the amount in time between the M and S stages

The cell is reacting to the environment by ensuring it has the nutrient sufficient for cell division and in particular DNA replication and also ensuring that there are enough growth factors to promote it to S phase

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3
Q

What happens during G2

A

Preparation for mitosis-

The chromosomes are segregated accurately
The spindles start to form

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4
Q

Summarise replication

A

DNA is unwound- this exposes the bases

Daughter strands of DNA synthesised, using parent strand as a template
Semi Conservative Replication

Each new DNA molecule will contain:

  • one original parent strand
  • one daughter strand
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5
Q

How is DNA replication initiated

A

Not initiated at random point- originates with proteins interacting with DNA sequences at “origin of replication”

Base pairs of the double helix must be broken to allow the DNA molecule to unwind, exposing the bases. Allows for DNA polymerase to access the bases- synthesising the new strands using original ones as a template

there are multiple pints of origin or replication along its length

Replication forks are formed

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6
Q

Describe what is done to ensure that base pairs do not reform

A

The most energetically favourable state of the DNA is when the two bases are paired up- energy is required to break H bonds.

  • DNA Helicase unwinds the DNA to expose the bases
  • The bases pairs will want to reform hence the single stranded binding proteins will attach into each strand, preventing the intermediate reforming
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7
Q

Describe Problem 1

A

POSITIVE SUPERCOILING - by pulling the strands apart, it increases their winding about each other along the molecule - can lead to a double strand break

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8
Q

How is positive supercoiling prevented

A

A topoisomerase breaks the phosphodiester bond in one of the the parental strands- this provides a degree of freedom around which remained of helix can unwind.

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9
Q

Comment on the direction for which DNA polymerase works in

A

DNA polymerase can only synthesise in the 5’ —> 3’ direction ( the new strand )

This results in a different mode of replication for the two parents strands- ie DNA polymerase running in topologically opposite directions

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10
Q

Leading strands

A

The strand that is synthesised continuously

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11
Q

Lagging strand

A

The strand that is synthesises discontinuously- this is because DNA polymerase synthesising in 5’—>3’ - the synthesis on this strand must occur again at the beginning of every fork

NB OKAZAKI FRAGMENT

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12
Q

Describe problem 2

A

DNA cannot initiate the DNA Synthesising, it can only add the nucleotides to pre existing chains

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13
Q

How is Problem 2 solved

A

RNA polymerase initiates RNA synthesis so DNA synthesis can start on the short RNA primer

The RNA primer, 8-10 nucleotides long, leaves a free 3’ end. Allows for DNA polymerase to take off extending the 3’ end of the RNA primer

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14
Q

Describe how the Okazaki fragments can be joined up

A

DNA polymerase extend DNA 5’-3’ direction until it reaches the next RNA primer

Exonuclease degrades primer leaving a gap

DNA polymerase continuous the extension of the strand across the gap.

The Okazaki fragments are now next to each other. A 5’-3’ phosphodiester bond is now put ion place by DNA ligase. This joins the fragments together

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15
Q

How are the leading and lagging strand synthesis coordinated?

A

Despite the fact that the two strands are running in topologically opposite directions, a lagging strand loop helps with this

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16
Q

How can it be assured that the newly synthesised DNA is a perfect copy

A

There can be a situation where wrong nucleotide is added by DNA polymerase

DNA polymerase also possesses a 3’-5’ exonuclease - removing that base

DNA polymerase then adds the correct nucleotide

17
Q

Describe Problem 3

A

Small amount of DNA is lost from each end of the linear chromosome after each round of the DNA replication.

The loss of information is prevented by telomeres which are formed of hundreds of 5’ TTAGGG 3’
These are catalysed by telomerase which extends the DNA without using a template since the enzyme carries its own template

18
Q

What is the exception to the problem 3’s solution

A

In somatic telomerase is switched off- eventually the telomere come sequences will be lost- useful DNA lost due to the end replication defect.

19
Q

What is a point mutation

A

A single base is changed

20
Q

What are the types of consequences resulting from the point mutation

A

Silent mutations
Missence mutation- eg GAG—> VAL (sickle cell Anaemia)
Nonsense mutation- amino acid codon- stop codon

21
Q

What are Small Scale Insertions and Small Scale Deletions called ?

A

Indels—>

Multiple of 3- maintains the reading frame ( loss of 508th CFTR)

No multiple of 3- reading frame is lost (loss of 32 BP in CCR5- individuals homozgous for this are resistant to HIV-1 )

22
Q

what are the clinical considerations with indels

A

Might now have serious consequences to human health if in a non coding regions

But could if it is in a coding region

Eg Huntington disease

23
Q

examples of changes in gross morphology of chromosomes

A

Inversions
Deletions- Cri du chat (limited physical and mental development)
Translocations- CML
Duplications

24
Q

How do mutations appear in the genome

A

Spontaneous mutations

Induced mutations

25
Q

Describe spontaneous mutations

A

Errors in DNA replication- if it has been missed by proofreading and repair mechanisms

Replication slippage-
Gain of repeats- reverse slippage
Loss of repeats- forward slippage

Deamination- C—>U ( now pairs with A)
A—> Hypoxanthine (pairs with C)

26
Q

Describe induced mutations

A

Physical-

Ionising radiation- causes single/double strand breaks
UV light- eg. UV b - major mutagenic fraction of sunlight- induces chemical bonds between adjacent thymine distorting DNA- causing problems during DNA replication

Chemical-

Agents that react with bases:

Nitrous Acid- cytosine —> uracil
Alkylating agent- guanine modification
Free Radicals- strands break/ modification of bases

27
Q

how is DNA damage repaired

A

Removal of damaged region followed by resynthesis

1) Distorted helix
2) Region around the damage removed by nuclease and helicase
3) Re-synthesis

Direct Repair- O^6 methylguanine dealkylated by MTase to guanine

28
Q

How is information in DNA accessed?

A

Genes are transcribed by RNA polymerase which binds to the promoter regulated by TF

Chromatin are highly condensed- Tight association of the DNA with nucleosomes inhibits access of the DNA to proteins that involved in transcription

Regions that bear actively transcribed genes must be decondensed- DNA can then be accessible to RNA polymerase , TF

29
Q

How can chromatin be modified?

A

Nucleosome sliding

DNA being pulled away from nucleosome

30
Q

How is Gene expression regulated

A

Tissue Specific expression :
All somatic cells contain the same DNA yet different cell types vary dramatically- STRUCTURE—> FUNCTION
Largely due to different genes being expressed in different tissues

External Signals leads to changes in gene expression profile:

  • During intense exercise/starvation, glucose concentration in blood will be low, glucocorticoid hormones released
  • when these reach the liver via the blood supply and signal to increases the expression of genes that catalyse production of glucose from amino acid
  • hormone is no longer present- production of these enzyme drop to normal

Regulation in time and space- during development , certain genes must be switched on or off at the right time in the right place

31
Q

How can disease be caused by disturbance of gene expression?

A

Change in gene expression makes a significant contribution to tumour development
Eg.

Activation of genes that promote cell division
Decrease transcription of genes that inhibits cell growth

Increasing transcription of genes that promote angiogenesis

Change in gene expression contributes to metasis

FBS1- CGCC (7 decks)

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