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BMSC320 > Information Transfer > Flashcards

Information Transfer Flashcards

1
Q

How big is the human genome in diploid cells?

A

6x10^9 bp

6x10^9 bp * 0.34nm/bp = 2.04x10^9nm = 2m/cell

37x10^12 cells in human body = 74x10^9 km of DNA in body

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2
Q

What is the structure of chromosomes?

A
  • Single DNA molecule wrapped around scaffold of proteins
  • highest condensed structure
  • made up of condensed chromatin fibers
  • average human mitotic chromosomes is 5μm in length and contains 5cm of DNA
  • unwound within the nucleus, only appear during mitosis
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3
Q

What is the structure of chromatin?

A
  • first step in the folding events to condense DNA 10,000 fold from it’s extended form into compact mitotic chromosome
  • made up of DNA, histone protein, and RNA
  • made up of nucleosomes
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4
Q

what is a karyotype?

A

full set of chromosomes of an individual
humans have 22 pairs of autosomes and 1 pair of sex chromosomes

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5
Q

How is there only one genome?

A

Every cell has the same genome, but the information differs from person to person
- epigenetics is the reason the body can make different cell types and organs
- some genes are modified to either be expressed or not expressed
- whole different transcriptome can be expressed in different cells
- not always considered a change in sequence, but cancer cells may have mutations related to epigenetics

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6
Q

What are the two levels of epigenetic modifications?

A

DNA modifications
Histone modifications

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7
Q

How are histones involved?

A
  • 150bp wrapped twice around histone core protein
  • each histone has N-terminal tail that’s unstructured and chemically modified by many types of enzymes
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8
Q

what are the three tail residues and how can they be modified?

A

K (Lysine): acetylated, methylated, ubiquitinated
T/S (Threonine/serine): phosphorylated

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9
Q

What are histones?

A
  • key DNA protectors of information storage
  • main packaging proteins
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10
Q

What are the 5 classes of histones?

A

H1: linker histone (not as conserved between species)
H2A, H2B, H3, H4: core histones
- 2 copies of each make up nucleosome octamer which forms the “beads on a string” structure

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11
Q

what is an important characteristic of histones?

A
  • very basic, rich in lysine and arginine
  • arginine can be methylated
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12
Q

what are introns and exons?

A

intron: intervening sequences
exon: expressed sequences

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13
Q

What are 3 things found in the promoter region (where transcription factors bind to recruit RNA pol)?

A
  1. Trans-acting elements (DNA)
  2. Cis-acting elements (proteins) that bind to trans elements
  3. TATA box: recruits binding factors that initiate transcription
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14
Q

where does transcription start and stop?

A

start codon: ATG
stop codon: AUG

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15
Q

What is gene expression?

A
  • series of events where info in DNA sequence is converted to RNA product and then to protein, which performs biochemical/biological function in the cell or organism
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16
Q

what is the central dogma of gene expression?

A

Gene (DNA) –––(transcription, processing)––→RNA–––(translation)––→ Protein

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17
Q

what are 3 things that gene expression helps us understand?

A
  1. why different types of cells have different functions (1 genotype, many phenotypes)
  2. why organisms differ from one another
  3. how cells adapt and respond to different signals and changes in their environment
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18
Q

what is a gene?

A

the entire DNA sequence that is necessary for the synthesis of a protein or RNA molecule
- includes the promoter, translational coding region, untranslated regions, introns, transcription termination signal, and regulatory sites

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19
Q

what 5 things determines the amount of each protein in a cell?

A

abundance of mRNA is determined by:
1. the rate of its synthesis (transcription of gene)
2. the rate of mRNA degradation (stability)
3. the availability of the RNA molecule (mRNA sequestration)

  1. efficiency of translation
  2. processing and stability of the protein
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20
Q

purpose of prokaryotic gene control

A

allows cells to optimize growth and division in response to changing environments

21
Q

purpose of eukaryotic gene control

A

regulate a genetic program that underlies embryonic development and tissue differentiation

22
Q

mRNA processing

A

-RNA synthesized by RNA pol II is termed pre-transcript/pre-mRNA and is extensively modified before transported to cytosol for translation

23
Q

4 reasons for processing mRNA

A
  1. identifies it as mRNA
  2. allows transport to the cytosol
  3. makes it competent for translation
  4. stabilizes and protects in from degradation (RNA is inherently unstable)
24
Q

Overview of eukaryotic mRNA processing

A

(DNA)
- transcription by RNA pol II
- addition of 5’ cap when 20-30 nt of pre-mRNA made
- addition of 3’ polyA tail (longer tail = more stable the pre-mRNA)

(pre-mRNA)
- RNA splicing: introns removed

(mRNA)
- export out of nucleus
- translation

(Polypeptide)

25
Q

architecture of final mRNA product

A
  • 5’ cap is guanosine added to 5’ end of pre-mRNA backwards (via 3 phosphates instead of by sugar) and methyl group is added to it to protect mRNA from degradation and determine translation efficiency
  • open reading frame (ORF) determines amino acid sequence of protein
  • polyadenylation signal ( in the 3’ UTR) determines stability
26
Q

2 main parts of mRNA processing

A
  1. capping
  2. splicing
27
Q

describe capping

A

first 2 ribose sugars are methylated (guanine cap and first nt after it), first methylation of G cap is Cap0 and nt methylation is Cap1
- functions to regulate mRNA transport out of nucleus, prevent 5’ exonuclease degradation, promotes translation via binding to ribosomes, and promotes 5’ proximal exon excision

28
Q

what are the 4 steps of 5’ capping?

A
  1. RTPase removes a phosphate from pre-mRNA
  2. GTase pulls a phosphate off of GTP and adds it back to 5’ end of pre-mRNA backwards which regenerates triphosphate
  3. N7MTase (methyltransferase) pulls methyl group from S-adenosyl methionine (SAM) and adds it to G cap (Cap0)
    - SAM becomes SAH (S-adenosyl homocysteine)
    - if not enough SAM, can’t properly methylate cap = problems
  4. repeat reaction to methylate Cap1 (catalyzed by 2’OMTase)
29
Q

How is capping catalyzed?

A

-guanylyltransferase (GTase) is recruited by the CTD
- GTase is associated with CTD to ensure that each mRNA is capped as its transcribed and in position to interact with the newly synthesized mRNA 5’ end
- GTase only dissociates once pre-mRNA is capped, then the cap-binding complex (CBC) binds with CTD and G cap so the 5’ end is kept protected

30
Q

describe splicing together of exons and removal of introns

A
  • exons: expressed sequences
  • introns: intervening sequences
  • coding sequence starts with exon and ends with exon
  • eukaryotic genes can have dozens of introns
  • exons are usually short (50-250 bp) and introns range from short to long (50-20,000 bp)
31
Q

why did exon/intron structure evolve in eukaryotic genes?

A
  • may be result of organisms attempting to increase # of genes
  • enhances rate of expression of genes
  • differential splicing provides a way to obtain different proteins from a single gene
  • makes it hard for toxic agents to damage coding sequence when low percent of gene is actually expressed
32
Q

What is the structure of pre-mRNA

A

5’exon-GU———branch point A—–(pyr rich region)—-AG-3’exon

33
Q

what are the splice sites of introns?

A
  • always has 5’ and 3’ splice sites
  • 5’ splice site has a conserved GU
  • 3’ splice site has conserved AG
  • branch point A is always within 15-45 bases from the 3’ splice site
34
Q

what is the structure of U#snRPs in splicing

A
  • U1/U# is a complex of RNA and protein
  • snRP = small nuclear RNA protein
  • RNAs are always highly structured and form stem loops when homology exists which form binding sites for protein
  • explains why all pre-mRNAs begin with an exon: splicing apparatus (U1snRP) needs 5’ intron GU site (which base pairs with U1’snRP’s 5’ RNA end) to initiate cleavage and is just downstream of first exon
  • if started with intron, the first intron wouldn’t be spliced out because GU isn’t there
35
Q

9 steps of splicing cycle

A
  1. U1 (5’ end) binds 5’ splice site
  2. U2 (5’ end) binds internal branch point A (but has no U to pair with branch point A so that is free to attack the GU at the 5’ end)
  3. U4/5/6 recruited and bring together U1/2
  4. U1 is ejected from complex which allows U6 to bind 5’ splice site
  5. U4 is ejected and NTC enters which activates the complex (U4 is only to stabilize complex before activation)
  6. 5’ splice site is brought to branch point A
  7. 5’ splice site is joined to branch point A (branch point A attacks GU, 5’ exon is cleaved but not released by the complex)
  8. the 5’ and 3’ exons are brought together and joined
  9. intron lariat (discarded product) is detached and entire complex is recycled for next intron
36
Q

Describe polyadenylation

A
  • most eukaryotic mRNAs have 50-200 A residues at the 3’ end which are not coded for by the gene
  • this poly A tail is added on by an enzyme (poly A polymerase: PAP) following the cleavage of mRNA at specific site near the 3’ end
  • mRNAs have polyadenylation signal AAUAAA near the 3’ ends, coded for by the last exon in the gene which dictates where the polyA tail should be added
  • polyA tail shortens as mRNA ages
37
Q

How does polyadenylation happen?

A
  1. polyadenylation signal triggers cleavage 10-35 nts downstream of poly A signal on pre-mRNA
  2. PAP polyadenylates the tail
38
Q

what are 3 main roles of polyadenylation

A
  1. protect from degradation
  2. signal for mRNA transport out of nucleus
  3. required for efficient translation
39
Q

what allows assembly of pre-cleavage complex?

A
  • CPSF-73 (processing nuclease) performs the cleavage event including bending the RNA to expose the cleavage site
40
Q

steps of how RNA pol II terminates transcription

A
  1. RNA pol II transcribes exons and introns
  2. splicing occurs and introns are removed
  3. RNA pol II reaches polyA signal
  4. mRNA gets cleaved and polyadenylation occurs
  5. RNA pol II continues even though the mRNA is complete and reaches the cotranscription cleavage site (CoTC) and transcribes RNA that isn’t needed
  6. exonuclease Xm2 recognizes the cleaved site on the useless RNA and begins chewing it back
  7. Xm2 reaches RNA pol II and it is released from the template, resulting in termination
41
Q

what are 6 additional ways of transcript processing?

A
  1. RNA editing
  2. selection of different promoters
  3. Selection of different AUG translation start sites
  4. selection of alternative polyadenylation sites
  5. Alternative splicing
  6. mRNA stability
42
Q

describe RNA editing

A

C to U occurs by deamination and can result in premature stop codon at truncated protein

  • mechanism: bisulfite deaminates C to U - used to determine amount of C methylation in a sequence since methylated Cs are protected from bisulfite conversion
  • C residue is selected; mRNA forms hairpin loop; ACF is recruited and clamps on to RNA; ACF recruits APOBEC1 (deaminating enzyme)
  • if APOBEC1 is overexpressed there are too many conversions which can lead to cancer
  • mRNA melts so that APOBEC 1 has a C in it’s grip and converts it to U
  • complex dissolves and hairpin RNA reforms but with C as U
43
Q

describe selection of different promoters

A
  • ex: PPAR𝜸 gene has many promoters and 2 start sites that can be used
  • transcription can start 1 of 3 promoters and translation can start from one of the 2 AUG start codons in the transcribed RNA
  • if starting from 𝜸1, transcription of A1 and A2 but start site B is cut out
  • if starting from 𝜸2 or 𝜸3, only get transcription of B or A2, respectively
  • use of 𝜸1 and 𝜸3 results in different mRNAs but the same protein because they use the same start codon
  • differential selection of transcription start site is often coupled with differential splicing
44
Q

describe selection of different AUG translation start sites

A
  • if there are three potential start sites you can have three different proteins
  • in most cases the AUGs are all in frame, so everything downstream of the start (whether it’s AUG1, 2, or 3) will be all in frame and the triplet codes all match us so amino acid sequence will be the same
45
Q

Describe selection of alternative polyadenylation sites

A
  • 70% of human gene transcripts undergo polyadenylation
  • can have impact on 3’ UTR region and regulatory sequences therein such as miRNA binding sites
  • selection of polyA site is linked to rate of transcription and loading of polyA complex onto the site
  • weaker polyA sites will be passed over if pol II is moving quickly
  • RNA binding protein regulators affect interactions between core machinery and poly A sites
  • strongest poly A site is usually most distal, and weakest most proximal
46
Q

describe alternative splicing

A
  • ex: calcitonin gene can be expressed into 2 different gene products (calcitonin and CGRP which have different biological activities)
  • involves alternative splicing and alternative selection of polyA sites
  • some exons may be spliced out
  • decision of which exon is removed is determined by RNA sequence elements and protein regulators, and is associated with rates of transcriptional elongation
47
Q

what are the 4 categories of sequence elements in alternative splicing

A
  1. ESEs (exonic splicing enhancers)
  2. ESSs (exonic slicing silencers)
  3. ISEs (intronic splicing enhancers)
  4. ISSs (intronic splicing silencers)
    - all are cis-elements that are recruited to DNA
  • enhancers in exons/introns facilitate splice-site recognition and exon/intron definition and are important when they’re weak non-consensus sites
  • they slow down elongation of RNA pol II which may facilitate recognition of weak splice sites
  • speeding up rate may facilitate non-recognition
  • there are specific proteins that bing to these mRNA sequence elements
  • if proteins bind enhancer sequence they facilitate splicing/recruit spliceosome
  • of proteins bind silencer sequence, they silence splicing and don’t recruit spliceosome
  • splice site recognition can be inhibited which allows some exons to be excluded (some sequence elements inhibit site recognition by binding proteins that in turn block access of SNRPs or enhancer binding proteins so splicing of an exon can’t occur)
48
Q

describe mRNA stability

A
  • half lives of eukaryotic mRNAs vary significantly, so mRNA from housekeeping genes generally have long half-lives and mRNA genes that need to be dynamically regulated and expressed in short bursts have shorter half lives

BMSC320 (9 decks)

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