Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

Genes and Genomes > Lectures 1-18 > Flashcards

Lectures 1-18 Flashcards

1
Q

Polymerase I is used mainly to

A

replace RNA primers in DNA replication, due to the presence of a 5’ to 3’ exonuclease

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Polyerase II is used mainly for

A

DNA repair

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

Polymerase III is used mainly for

A

General DNA synthesis, has a 3’ to 5’ exonuclease that can “proofread” for errors

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

DNA ligase links

A

3’ hydroxyl to a 5’ phosphate

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

DNA replication in prokaryotes is triggered by

A

accumulation of dnaA protein which binds to recognition sequences at the ori and promoting strand separation

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

FACT protein

A

binds to proliferation cell nuclear antigen (PCNA) and destabilises nucleosomes to allow replication to occur

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

Linear chromosomes present an issues because of

A

telomeres; there is perpetual degradation of the DNA at the end of the linear chromosome. This is overcome by telomerase, which acts on the highly repetitive nature of telomeres to add bases and protect the coding DNA

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Mutation is

A

a heritable change in the DNA sequence (not the same as damage to DNA)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Transitions occur when

A

a purine is swapped for another purine (G and A) or a pyrimadine is swapped for another pyrimadine (C and T)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

Transversions occur when

A

a purine becomes a pyrimadine

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

Streisinger model of slippage

A

during DNA replication, where strands don’t reanneal with the correct base esp. in sequences of many same-base repeats

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Depurination occurs when

A

a base is lost, sugar phosphate backbone remains intact but a nucleotide is cleaved off

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

Deamination occurs when

A

Deamination of 5 methile C bases results in T production for which there are no mechanisms to fix it, and this results in an increase in spontaneous mutation, specifically, a transion

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Oxidation occurs when

A

uncontrolled free radicals cause damage to the DNA such that G pairs with A and causes a transversion

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

Mutagens are

A

any agent that causes an increase in mutation rate above the spontaneous level

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

base analogues

A

Similar enough to a normal base to be incorporated into DNA and undergo tautomeric shift at increased frequency. - Example: 5-Bomouracil which causes transitions

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

Mis-pairings can be caused by

A 
Different mutations within this class cause different mutation through the same mechanism 
Hydroxlyamine which causes C to pair with A, alkylating agents which are very potent mutagens causing G to pair with T, intercollating agents which cause additions and deletions of single base pairs - possibly by stabiliging loops made in the Streizinger model
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

DNA strand breakage

A

Strand breaks can occur spontaneously but is increased by high energy ionizing radiation such as gamma and X-rays. In general these issues can be repaired without loss of information however if the breaks are double stranded this can be lethal to the cell. Incorrect re-joining can result in chromosomal rearrangements such as inversions, deletions, duplications and translocations

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

The Ames test looks at

A

reversion rates in salmonella typhi to show how commonly mutation is being caused

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

photoreactivation acts to

A

reverse pyrimidine dimers - requires light to function

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

Nucleotide excision repair

A

uvrA, B and C, all needed for successful repair system. When it finds a bulky lesion (pyrimadine dimers and others) it has the capacity to cleave DNA phosphodiester bonds and cut the mutation out, so that uvrD (helicase) can release the fragment and DNA PolI can fill the gap while DNA ligase seals this

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

Mismatch repair

A
  • MutS binds to the mismatch, MutH recognises the helimethlyated GATC and complexes with MutS and MutL which requires DNA looping
  • MutH cuts the unmethylated strand and creates a SS region that can be corrected by an exonuclease, PolIII and Ligase
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

SOS Bypass Replication (Translesion Synthesis)

A
  • In E.coli:
    ○ SOS genes UmuC and UmuD can form DNA Pol V (bypass polymerase) does not have ability to proofread and have lower specificity - meaning they can randomly pair up bases in areas that are damaged and are skipped by PNA Pol III. In the context of a highly damaged cell this is ideal over having highly damaged DNA even thought it results in mutation
    ○ recA mutant is more UV sensitive due to the accumulation of mutations that the cell cant deal with due to the lack of SOS genes being expressed
  • In Eukaryotes:
    ○ Bypass polymerase is expressed constitutively but remains inactive until ubiquitin and Rad6 bind to it
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

Non-homologous end joining (NHEJ)

A

Ku70/Ku80 heterodimer detects and binds the two strands of broken DNA together with a protein kinase and ligase.Ends are joined back together without the need for homology - this can explain translocations

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
25
Q

Homologous end joining (SDSA)

A

Requires homology through the active generation of complementary DNA bases through end trimming - Rad51 checks for homology. Can result in re-annealing of the original strands resulting in no crossing over or DNA synthesis and second end capture resulting in tangling which much be resolved and this can result in a cross over event or may not this mechanism can be used to explain recombination events.

26
Q

Cis factors (used in transcription) are

A

factors that present in the actual DNA sequence that facilitate encoding i.e. a promoter sequence

27
Q

Trans factors (used in translation) are

A

factors encoded for by a gene that facilitate encoding i.e. ribosomes and polymerases

28
Q

Sequences required for gene transcription include

A
  • promoter
  • terminus
  • -35 box
  • -10 (Pribnow) box which has low GC content
29
Q

Sigma-70 is used for

A

transcription in prokaryotes, has high association with promoter sequences, can be reused

30
Q

Intrinsic termination is

A

Rho independent, uses sequences transcribed into mRNA which have dyad symmetry. This can form a hairpin molecule, unlikely to denature. Formation of stem loop causes RNA polymerase to stall and causes the end of transcription, while the U rich allows for destabilisation of the polymerase from the RNA

31
Q

Extrinsic termination

A

requires rho factor, which has helicase activity and binds to rut sequences, causing polymerase to stall. Rho then catches up and unwinds RNA from DNA

32
Q

Differences in the Fate of Transcripts in E.coli vs. Eukaryotes

A

In E.coli the transcripts can be immediately translated which cannot happen in eukaryotes because of the location in the nucleus. Because of this, pre mRNA must be modified before leaving the nucleus which occurs while the transcript is being produced:

  • Addition of a 5’ cap
  • Addition of a polyA tail
  • Removal of introns
33
Q

The main promoter in eukaryotes is the

A

TATA box

34
Q

Mis-splicing can affect

A

the translation reading frame, result in normally coding areas being spliced out, or non-coding regions being left in

35
Q

the three sites of an RNA are

A
  • the A-site: aminoacyl
  • the P-site: peptidyl
  • the E-site: exit
36
Q

in prokaryotes, transcription initiation requires

A

the shine-delgarno sequence

37
Q

in eukaryotes, transcription initiation requires

A

the 5’ cap, and the Kozak sequence which contains the start codon (AUG)

38
Q

types of mutations in coding regions

A
  • silent
  • missense
  • nonsense
  • indels
39
Q

mutations in introns generally

A

have no effect unless they alter the consensus sequences required for splicing i.e. GU-AG rule

40
Q

Mutations may:

A
  • reduce functionality
  • alter the level or function of the gene product
  • have no effect on gene function
  • produce conditional mutants
  • produce gain of function mutants
41
Q

a forward genetics approach

A

starts by looking at the phenotype

42
Q

reversion is when

A

the original mutation is reverted back to the WT phenotype

43
Q

intragenic suppression is when

A

a second mutation is in the same gene and counteracts the effects first mutation

44
Q

extragenic suppression is when

A

a second mutation is in a different gene than the first and counteracts the effects of the first mutation

45
Q

segregation analysis involves

A
  • Cross back to WT
  • All progeny of this cross will be WT in the case of a reversion
  • When the suppressor is intragenic all parental offspring will be WT and the very rare recombinants will be mutant
  • In the case of extragenic suppression the frequency of the mutants depends on whether the genes are linked or not i.e. 1:1:1:1 or recombinant classes are smaller
46
Q

C-value is

A

the amount of DNA per haploid cell or number of kilobases per haploid cell

47
Q

Features of a prokaryotic genome

A
  • Supercoiled
  • Base of loops tethered to a protein core
  • Most non-coding regions are associated with gene regulation
  • Very dense genomes
  • Low level repetitive DNA made up of insertion sequences/transposons
  • Lack histones
  • May include plasmids
48
Q

Features of a eukaryotic genome

A
  • Includes organelle genomes
  • Multipartite, linear chromosomes
  • Binds to histone proteins to create chromatin
49
Q

The C-Value Paradox

A
  • In eukaryotes, the amount of DNA in the haploid cell of an organism is not absolutely correlated to its complexity
  • Related organisms have genomes of different sizes
  • Genome size does not scale precisely with gene number
  • Suggests variable gene density in eukaryotes - other elements in their genome
  • Differences in amount of intergenic DNA may account for C-value paradox
50
Q

Melt curves

A
  • Heat breaks weak hydrogen bonds and renders single stranded
  • Transition from DS to SS is measured with UV light
  • SS DNA absorbs more light
  • Tm is the temperature at which half of the DNA is rendered SS
  • Tm depends on the CG content of the genome i.e. higher CG = large Tm value
  • Can be used to distinguish genomes of different species
51
Q

Density Gradient Centrifugation

A
  • Creation of concentration gradient in a tube
  • Denser items move to the bottom at high speeds
  • Solution is a heavy salt (Caesium Chloride)
  • DNA will come to occupy positions matching its own density
  • Can distinguish between DNA conformation and/or base composition
  • CG content and DNA shape contribute to where it will sit in the tube
  • Bands at top represent satellite DNA
52
Q

Re-association Analysis (C0t curves)

A
  • Analysing the rate at which denatured strands of DNA reanneal/reassociate/hybridise
  • Random collisions between two complementary strands of DNA
  • Temp, DNA and salt concentration all influence the rate of reassociation
  • Larger C0t values means that that DNA is reannealing slower and vice versa
  • Size of genome influences C0t - more collisions are needed by chance for complementary stands find each other
  • Renaturation rate is inversely proportional to the size of the genome
53
Q

Transposable elements are

A

segments of DNA that can move from one position of the genome to another

54
Q

Two types of transposition

A

replicative (copy) and conservative (cut)

55
Q

Types of transposons in prokaryotes

A
  • IS (insertion-sequence elements): low copy number, simple, TIR, uses transposase for cut and paste mechanism
  • Simple transposons: small flanking inverted repeats, TIR, may contain other function genes
  • Complex transposons: two separate IS elements with genes unrequired for transposition which move as a single unit
56
Q

What is needed for copy and paste transposition mechanism

A

Resolvase

57
Q

Types of transposition in eukaryotes

A
  • Class II TEs: cut and paste, can be autonomous or affected by mutations that immobilise them
  • Class I TEs (LTR retrotransposons): long TIR, encode reverse transcriptase, suggests transcription occurs via RNA intermediate
58
Q

Characteristics of protein-coding genes

A
  • Open reading frame
  • Presence of promoter/consensus sequences (Pribnow Box/Shine-Delgarno Sequence in prokaryotes, TATA Box and Kozak sequence in eukaryotes)
  • Codon bias
59
Q

Paralogs are

A

derived from the same ancestral gene by gene duplication which often evolve to have different functions

60
Q

Orthologs are

A

derived from the same ancestral gene by speciation

61
Q

Duplications within the genome arise from

A

unequal crossovers between repetitive elements that results in tandem duplications

62
Q

Whole genome duplications arise from

A

Errors during meiosis leads to formation of diploid gametes through non-disjunction

Genes and Genomes (1 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions