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Genes and Environment > W10 > Flashcards

W10 Flashcards

1
Q

what is the structural difference between DNA and RNA?

A

similarities:

  1. same nucleotide structure consisting of a phosphate group, sugar and nitrogenous base
  2. both are polynucleotides
  3. 2 pyrimidines and 2 purines

differences:

  1. RNA has many copies per cell whereas DNA only has one copy unless it is about to divide
  2. RNA has ribose sugar rather than DNA which has deoxyribose
  3. RNA has uracil instead of thymine
  4. RNA is mostly single stranded, but DNA is in double helix and only separates during synthesis
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2
Q

what is mRNA used for

A
  1. formed by transcription in nucleus from DNA
  2. complementary to DNA template: beware! uracil instead of thymine
  3. template used during protein synthesis
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3
Q

what are the roles of non-translated RNAs

A

miRNA:

    1. endogenous type of RNA interference
  • 21-25 nucleotides
    1. transcribed from DNA
  • 2.requires processing to become single stranded
  • 3.can bind imperfectly to mRNA also transcribed from dna
    1. blocks translation or makes mRNA for degradation

siRNA:

  • 20-25 nucleotides
    1. synthetic and 2. double stranded
  • requires unwinding
  • 3.perfectly binds mRNA transcribed from dna
    1. mRNA complexes are degraded
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4
Q

how does interference RNA work

A
  1. the XIST gene is on the X chromosome
  2. transcription of the XIST gene makes an interference RNA
  3. RNA binds to the X chromosome from which it is transcribed
  4. methylation and histone deacetylation attract chromosomal proteins that form heterochromatin, inactivating the chromosome
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5
Q

what are the building blocks of proteins

A

amino acids in the primary structure
- amino acids have the same backbone, but with different side chains
- there are 20 different amino acids with different chemical properties
(amino and carboxyl group)

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6
Q

how are these protein building blocks arranged

A

primary: amino acid sequence
secondary(folding): loops, turns, alpha helices and beta pleated sheets
- carboxyl and amino groups can form bonds to give rise to alpha helices and beta sheets
tertiary(packing): 3D conformation of polypeptide
- the chemical properties of amino acids can lead to interactions
- tertiary packaging often involves interactions of more distant amino acids
quaternary(interaction): proteins with multiple peptide subunits
- individual components encoded by different genes

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7
Q

outline transcription initiation

A

both PRO/EUKAR:

  • RNA pol. unwinds dna to form transcription bubble
  • sigma factor and general transcription factors must dissociate for RNA pol to be able to move along DNA
  • the transcription machinery is RNA polymerase: 1 in prokaryotes and 3 in eukaryotes (RNA pol. 2 does mRNA)

function of RNA pol.?

  • needs to identify gene to transcribe in that cell
  • needs to identify start: consensus sequences (promotor and upstream) which have similar sequence and location across genes, but different in prokaryotes and eukaryotes
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8
Q

outline transcription elongation

A
  • elongation has DIRECTIONALITY and the two DNA STRANDS have different nomenclature

DNA coding strand=non-template
- STRAND SIMILAR TO mRNA EXCEPT FOR u instead of T
DNA template strand
- strand used to generate the complementary mRNA sequence
- template strand is read 3’ to 5’, but RNA strand is produced from 5’ to 3’

  1. using unwound DNA TEMPLATE strand to add mRNA building blocks one at a time
  2. transcription bubble exists inly within RNA polymerase (unwinds at front and rewinds at back)
  3. inside the bubble, individual RNA nucleoside triphosphate (NTP) are added
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9
Q

outline transcription termination

A

in both: there are specific termination sequences that are transcribed as part of the mRNA

prokaryote:
- Hairpin is formed- signals end and dissociation of mRNA and RNA pol.

eukaryote:
- special signal recruits 4 cleavage factors that bind to make a complex, cleave the mRNA and dissociate mRNA and RNA pol.

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10
Q

contrast transcription initiation in prokaryotes and eukaryotes

A

prokaryotes:
- whole holoenzymes (RNA pol+sigma) binds, sigma factor recognises consenses sequence in promotor

eukaryotes:

  1. general transcription factors bind to promotor sequence sequentially and recruit RNA pol.
  2. specific transcription factors bind upstream and control rate
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11
Q

describe how new cells can arise

A

EACH cell contains the ENTIRE human genome (all genes), but differential gene expression allows for different cells to make different proteins

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12
Q

outline gene structure and punctuation

A

by consensus, genes are read and transcribed from 5’ to 3’

  1. regulatory sequence; controls differential gene expression
  2. promotor (start of transcription)
  3. coding sequence (template for elongation of transcription)
  4. terminator: stop of transcription
  5. untranslated sequence (start and stop of translation)
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13
Q

outline transcription overall

A

transcription machinery is RNA polymerase

  1. initiation (start where RNA pol. binds)
  2. elongation (generation of mRNA (RNA pol moves along)
  3. termination (stop of transcription (RNA pol dissociates)
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14
Q

outline prokaryotic transcription initiation in terms of RNA pol. and its subunits, functioning of RNA pol. and process of action

A
  • only one RNA polymerase composed of 5 subunits; 2 alpha, 1 beta, 1 beta’, 1 omega(core enzyme) and a detachable sigma factor
    core enzyme+sigma factor=holoenzyme
    SIGMA FACTOR: determines the start by binding promotor and then dissociates ie. not required for elongation or termination of transcription
    RNA pol:
  • looks for consensus sequences in the promotor region

binding process:

  1. sigma factor(as part of holoenzyme) binds promotor=closed complex
  2. RNA pol. switches to open complex and melts the double strands of DNA to form the transcription bubble
  3. sigma factor then dissociates, which is required for RNA pol. to escape promotor
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15
Q

what are consensus regions

A

highly conserved in different genes. differ between prokaryotes and eukaryotes in terms of sequence and location and there are some within and upstream of the core promotor region

  • do not have to match 100%, pattern is recognised
  • notation: +1 refers to the first b.p sequenced
  • whilst both are upstream, pro. sequence are separate whereas eukaryote. sequences are close together
  • the sequence itself is also diff.
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16
Q

outline eukaryotic promotor binding initiation process

A
  1. mediated by general transcription factors that:
  2. recognise consensus regions
  3. bind to the DNA in a particular sequence
  4. recruit RNA polymerase
  5. pre initiation complex formed
  6. unwind DNA (helices activity)
  7. specific transcription factors (proximal and distal) are required for binding important for differential gene expression
    - has activators and repressors that can bind to specific regions, this leads to a higher rate of action
17
Q

what is the transcription bubble?

A

since transcription requires access to the DNA

  • RNA pol. needs to keep unwinding the DNA
  • transcription bubble describes the unwound DNA base pairs DNA that has already been transcribed rewinds behind the bubble
18
Q

multiple RNAs can be transcribed (…) ie. (…) genes

A

multiple RNAs can be transcribed simultaneously ie. rRNA genes

19
Q

outline prokaryotic transcription termination

A
  • recognition of gene end and dissociation of RNA polymerase
  1. RNA polymerase encounters chain termination sequence
  2. self complementary RNA forms hairpin (stem and loop structure)
  3. RNA and RNA polymerase dissociate from DNA
20
Q

outline eukaryotic transcription termination

A
  • more complicated and different for each RNA polymerase

FOR RNA pol2:

  1. polymerases passes termination sequence
  2. RNA is cleaned by cleavage complex

A. cleave and polyadentation specificity factor
B. cleave stimulating factor
C. cleavage factor protein CF1
D. cleavage factor protein CF2

21
Q

how is RNA processed after transcription

A
  • occurs after transcription termination
  • occurs in nucleus of eukaryotes
  • only occurs in eukaryotes and generates Mature mRNA from pre-mRNA
    1. 5’ capping: addition of 7 methyl guanine cap at 5’ end
    2. 3’ polyadenylation: addition of Poly A tail to 3’ end
    3. RNA splicing: removal of introns and joining of exons by spliceosomes
    4. alternative splicing: results in inclusion and exclusion of different axons and can be used to make tissues specific versions of a protein
22
Q

what are the role of these processing steps

A
  1. 5’ capping whereby there is addition of the 7 methyl guanosine cap protects RNA from degradation by nucleases and is recognised by translation machineru
  2. 3’ polyadenylation: up to 200 are added. it Is added by polymerase and enhances RNA stability and regulates nucleus to cytoplasm transport
23
Q

what is the genetic code

A

3 bases=1 amino acid=4^3 COMBINATIONS+64 combinations

KEY TRAITS:

  1. triplet code: 3 RNA base pairs=codon, encode one amino acid
  2. the code is degenerate: more than one codon per amino acids. codons that encode the same amino acids are synonymous codons
  3. wobble of the codon describes codons where the last base is irrelevant
  4. only one start codon, AUG
  5. 3 stop codons
24
Q

which are the start and stop codons

A

AUG: start

stop: UAA, UAG, UGA. not encode aa, just a signal

25
Q

what does ‘wobble’ refer to

A

where the last base is irrelevant, this occurs in all Amino acid boxes with more than > 2 combinations for that particular amino acid ie. only those with 3 or more synonymous codons

26
Q

what does synonymous refer to

A

degenerate=synonomous

- codons that encode the same amino acid since more than one codon per amino acid

27
Q

what are the three key stages of translation

A

translation machinery is the ribosome (rRNA+protein)

  1. initiation: ribosomes finds and binds to start codon on mRNA
  2. elongation: polypeptide chain grows by addition of individual amino acids carried to the ribosome by tRNA
  3. termination: translation finds stop codon and dissociates
28
Q

outline both translation initiation

A

INITIATION: PROKARYOTES

  1. recognition of consensus ribosome binding site: Shine-dalgarno sequence located in the 5’UTR
  2. binding of small ribosome unit and initiation factors
  3. special initiator tRNA in prokaryotes: formyl methionine (fMet-tRNA) binds
  4. initiation factors are released and the large ribosome unit binds forming the initiation complex

INITIATION: EUKARYOTES

  1. small ribosome unit binds initiator tRNA and eukaryotic initiation factors form the pre-initiation complex
  2. the pre-initiation complex finds and binds the 5’ cap
  3. the pre-initiation complex moves along the mRNA scanning for the KOZAK consensus sequence, which indicates the start site(containing AUG)
  4. initiation factors dissociate and large ribosome unit binds forming the initiation complex
29
Q

what is the process of RNA splicing

A
  1. snRNP (small nuclear RNA-sRNA+protein) finds and binds splice sites
  2. snRNP and other proteins associate to form the spliceosome
  3. the spliceosome cuts at the exon-intron boundary and the intron forms a loop
  4. the two ends are joined and the intron is degraded
30
Q

outline alternative splicing and the effects of it

A
  • can make different mRNA from the same gene
  • allows slightly different proteins to be made from the same gene, which may be more suited to different cells/tissues/organs: advantage
  1. exon skipping- leave out an exon
  2. mutually exclusive exon- multiple slight different exon options
  3. alternative 5’ spice donor sites- multiple options of splice start
  4. alternative 3’ spice donor sites- multiple options of splice end
31
Q

outline the process of drosophila sex determination

A
  1. protein ratio as determinant of TF
  2. only females make enough X product to activate transcription of sex lethal
  3. sex Lethal (females only) leads to alternative splicing to make bypass a stop codon and make TRA protein
  4. TRA protein leads to alternative splicing two make DSX-F females vs. no TRA protein makes DSX-M males
32
Q

the genetic code is (…) and starts at a (…)

A

the genetic code is non overlapping and starts at a fixed place

33
Q

what are the components involved in transcription

A

key components

  1. mRNA is the transcribed, processed gene message template
  2. rRNAs are key components of the small and large ribosome unfits carrying out the actual translation process
  3. tRNA bind one specific amino acid depending on the amino-acyl transferase and deliver the amino acid to the ribosome
34
Q

contrast eukaryotic and prokaryotic translation

A
  1. the ribosome binding site is the SHINE DALGARNO SEQUENCE in the 5’ UTR in prokaryotes whereas eukaryotic initiation begins with the pre-intiation complex formed finding and binding the 5’ cap. this later binds the Kodak consensus sequence
  2. whilst initiation factors are released in prokaryotes before the large ribosome unit binds forming the initiation complex, initiation factors dissociate and large ribosome unit binds forming the iniation complex
35
Q

outline translation elongation

A

ELONGATION
- three tRNA binding sites within the ribosome
A site: codon recognition between anticodon (aminoacyl-tRNA) and codon (mRNA)- mismatch leaves, match will bind
P site: peptide bond formation and amino acid translocation
E site: exit of uncharged tRNA AND translation machinery progresses
- progression of tRNA through ribosome sites A-P-E are due to ribosome moving along

36
Q

outline translation termination

A

TERMINATION

  • when STOP codon is encountered
    1. release factor recognises the STOP codon and enters the A site
    2. Stop codons to not encode amino acids and the release factors do not carry any amino acids
    3. release factors separates the last tRNA from the last AA

Genes and Environment (12 decks)

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