Chem 153B: Weeks 8-10

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What do class Ⅱ genes produce?

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1

What do class Ⅱ genes produce?

Transcribes mRNA-encoding genes which promote initiation to produce proteins and some small RNAs

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2

What kind of genes are excluded from class Ⅱ?

Genes that trascribe gene encoding the 45S rRNA precursor and genes that transcribe for tRNAs, 5S RNAs and other small RNA

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3

What is the basic layout of a class Ⅱ gene?

TATA box in promoter, cis-regulatory module (CRM)/enhancers are upstream/downstream of promoter and exons after promoter

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4

What are general factors?

analogous to sigma factors in prokaryotes. They direct core polymerases to promoters and multiple are required to do so. Highly conserved

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5

What prokaryotic protein are general factors analogous to?

sigma factors

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6

What do general factors recognize?

the TATA box and hydroxyls such as serine, in the Pol Ⅱ C terminal domain and act as a kinase to phosphorylate it

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7

What is the mechanism for TFⅡD and and why is it unusual?

TFⅡD has phenylalanines which will wedge themselves into AT base pairs to have hydrophobic, base-stacking interactions which is unusual because it happens in the minor groove which is weird because it has sequence-specific interactions

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8

What is the relationship between the prokaryotic Pribnow and the TATA box?

It is a homologous conserved feature of prokaryotes and eukaryotes

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9

What function do cis-regulatory elements serve?

They are a collection of binding sites for several regulatory factors, either enhancing or inhibiting

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10

Know the steps of initiation at a class Ⅱ promoter

  1. IID binds and recruits IIA and IIB

  2. Pol II and IIF join

  3. IIE and IIH join onto Pol II. This forms the closed complex, which is incredibly unstable.

  4. IIH acts as a helicase to unwind DNA and promoter melts into open complex. This step uses ATP.

  5. Promoter clearance and IIH acts as a kinase to phosphorylate the CTD.

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11

What is analogy between prokaryotic transcription initiation and class Ⅱ eukaryotic initiation?

IIH is acting as a helicase and the formation of the closed and open complex

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12

What factors recognize the TATA box?

IID, IIA, IIB, IIF

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13

What is the RNAP Ⅱ CTD?

C terminal domain of the RNAP. This gets phosphorylated by IIH. The sequence (YSPTSPS)n is how it is recognized

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14

When is the CTD phosphorylated?

when promoter clearance occurs, this serves as a signal for the transition from initiation to the elongation phase of transcription.

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15

What is the general structure of a eukaryotic transcription factor?

polypeptide with a DNA binding domain that has specific binding to the promoter. It has an activation and repression domain that stabilizes or destabilizes. Both the DNA binding domain and the activation/repression domain are connected with a flexible linker.

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16

How is a general transcription factor a modulator structure?

the set up is always the same and can be divided into strongly interacting parts that are relatively autonomous with respect to each other

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17

Connect structure of general TFs to eukaryotic DNA binding motifs.

Activation or repression domain is used to recruit or destabilize RNAP. Eukaryotes also have an extra layer of regulation: chromatin modification and remodeling. Zinc fingers and leucine zipper are both present in the DNA binding domain.

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18

Why is a mediator of transcription necessary in eukaryotes?

It is an integrator of regulatory signals. TFs interact with the different subunits of the mediator, translating their signals (activation, repression, etc) to interactions with RNAP, which the mediator is bound to. This is important to Eukaryotes because the regulatory network is large and complex. RNAP alone cannot handle integration of these signals without a secondary co-factor.

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19

What generally is a transcriptional co-regulator and what do they do?

They are proteins convert the wide variety of environmental signals into a finite set of outputs. Depending on the environmental signals, they can repress or activate RNAP binding. and modulate elongation by RNAP. They can also modulate chromatin structure

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20

What is the most important co-regulator?

the mediator of transcription

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21

What is a suppressor mutation?

suppresses the effect of an earlier mutation at a different site.

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22

How did a suppressor mutation allow us to better understand the function of the mediator?

The cold sensitive phenotype in yeast proved that the interaction between CTD and the mediator creates transcriptional activity for RNAP. Several suppressor mutations got rid of the cold sensitive phenotype.

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23

What is the general scheme for interaction between RNAP Ⅱ, cis-regulatory elements (like activators) and the mediator of transcription?

The holoenzyme consists of polymerase and the mediator. The tail of the mediator most often contacts activators while the middle and head modules contacts core polymerase. CTD makes extensive contact with the head module and adds greatly to the stability of the complex. It runs through the cleft between the head and middle nodules and interacts with RNAP. Stabilizing contact between activator and mediator help recruit Pol II to promoter.

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24

Epigenic regulation

Alteration of phenotype through non-genetic variables

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25

What is chromatin?

Chromatin is a plastic (alterable) structure of eukaryotic DNA wrapped around histones that can be compacted into dense inaccessible units or unwound to create accessibility, which is accomplished through post translational modifications of histones. It helps package a meter of DNA into nucleus

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26

What is the function of nucleosomes?

Nucleosome is a histone octamer wrapped with DNA. It helps to regulate transcription by unwinding chromatin or keeping chromatin wound.

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27

histone octamer

two each of H2A, H2B, H3, and H4 form the protein complex which DNA wraps around. Alpha helices cross over and stack on top of each other to hold itself together

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28

What kind of amino acids populate histone tails?

positively charged AAs because the H2A/H2B dimer is negatively charged

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29

How do histone tails create tight associations between nucleosomes?

Positively charged histone tail has electrostatic interactions with nucleosomes that are negatively charged to create tight associations

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30

how does chromatin condensation /de-condensation alter accessibility?

When chromatin is condensed, it is difficult to access the DNA so that it can be transcribed. Lysine is not acetylated at this time. Chromatin decondensing is achieved through the acetylation of Lys, which breaks the interaction between the tail and the nucleosome. This allows for RNAP to bind and for transcription to begin.

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31

How could the use of DNAse I measure accessibility?

condensed, less accessible. decondensed, more accessible

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32

lysine acetylation

Co-activators use HAT to acetylate the Lysine on the nucleosome tail to weaken the interaction between nucleosomes by removing the positive charge on lysine. This causes chromatin to be decondensed.

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33

Lysine Deacetylation

Co-repressors use HDAC to restore the positive charge on Lysine found on the nucleosome tail. This causes the nucleosome tail to interact with the negatively charged surface on the H2A/H2B dimer, and thus chromatin is condensed.

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34

Lysine methylation

Methylated lysines can either lead to activation or repression, depending on their lcoation on the tail.

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35

What is epigenetic inheritance? How does it (likely) work?

Both acetylation and methylation are conserved in mitosis, which allows gene activity state to be inherited. This works because when parent cell divides, half of histones in one daughter cell and other half of histones in other daughter cell, but when DNA replicated, will have twice the amount of histones

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36

What do bromodomains bind to?

acetylated lysines

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37

How does the recruitment of bromodomain containing co-factors impact chromatin condensation?

Bromodomain-containing factors include histone remodeling complexes, which mobilize the histone octamers and thus push them out of the way of transcriptional machinery. Chromatin remodeling complexes also couple ATP hydrolysis to movement along chromatin, allowing them to push nucleosomes out of the way. Bromodomain also recruits other cofactors to unfold chromatin which can initiate transcription.

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38

How would mRNA levels vary over time given a two-state chromatin model?

The open/closed dynamic will create a bursty expression as mRNA expression has high period and low periods with drastic switches because mRNA can be degraded quickly

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39

Burst size

how many transcripts are transcribed

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40

Burst frequency

how frequently on states appear

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41

How do mRNA levels relate to protein levels (for a particular gene?)

Protein level tracks roughly with transcript number

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42

How can protein-level variability lead to divergent cell outcomes?

Some cells have high protein levels and others have low protein levels depending on bursting which will lead to potentially different cell fates. There is a randomness with chromatin opening since there is the same evel of chromatin markers near promoter. Slight memory is causesd by epigenetic marks which may affect protein expression.

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43

How does information from the environment influence gene regulation?

environmental signals to transcription factors to genes

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44

What is a transcription network?

System of circuitry illustrating the biological information which mediates the flow of information from the environment and computes a response through transcription factors. Genes that encode a transcription factor are centers and other genes are edges.

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45

What are transcription motifs?

A motif is a conserved pattern appearing more frequently than chance would expect. Network motifs are common patterns of gene interactions (as mediated through transcription factors).

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46

Why do transcription motifs exist?

They exist because they are evolutionary useful as repetition of certain motif implies important functionality

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47

How could you recognize a transcription motif?

It can be recognized by the amount of self arrows as there are more self arrows (more autoregulation) in a real network than a randomized network

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48

Understand how you would determine the rate of change of protein concentration (as well as its steady state) given simple regulation of a protein?

the change in protein concentration is the difference between the production rate and the degradation rate. Xst is the steady state concentration of the protein, when dX/dt = 0.

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49

beta in change of protein conc

protein production rate

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50

alpha in change of protein conc

protein degradation rate

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51

What is autoregulation?

Regulation of a gene by its own gene product (a transcription factor modifies its own transcription)

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52

How do we know autoregulation is an important network motif?

Important network motif because significant more self-edges than expected (40 vs 1 for E. coli)

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53

Negative autoregulation

repression of a gene by its own gene product

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54

How does negative autoregulation change protein kinetics?

Given strong autorepression, the self-repression threshold dominates steady state because as the protein concentration exceeds the self-repression threshold, it inhibits itself whereas if if the protein concentration was under the threshold, it would allow the protein concentration to increase. The response rate (T ½) is completely defined by the repression threshold and production rate

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55

How can negative autoregulation create a rapid response?

Negative autoregulatory systems can use a strong promoter (consensus sequence) to gain X quickly and use their autorepression to stop at the desired steady-state concentration

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56

How must a gene be designed to accomplish this goal of rapid response?

The promoter must have the consensus sequence (or something similar) so that RNAP can bind quickly and thus the protein concentration can reach steady state faster.

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57

How does the steady-state concentration of a negative autoregulatory system create robustness?

Negative autoregulation is defined by the the repression interaction, so the system is not easily influenced as opposed to simple gene regulation. Production fluctuations do not influence the system.

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58

Why is robustnes in a negative autoregulatory system desirable?

It is not easily influenced by production fluctuations and this creates a buffer when it comes to changing environmental factors.

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59

How does positive autoregulation change protein kinetics?

the increasing production rate pushes steady state higher until saturation which creates a delay in reaching steady state.

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60

What is bistability?

There are two states for X: when there is no external input, there is no X because the process has yet to be initiated. When there is external input, X increases because the process has been initiated by the external signal. X amplifies the signal, so it can continue to turn itself on. The system is stable whether there is an external signal or not.

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61

How is bistability accomplished in positive autoregulation?

Once the initial activator kicks in, the newly synthesized protein will take over the activation of its own production and help the system reach the steady state (even though the threshold continues to get pushed back). In short, the whole mechanism is either turned on by the initial activator then reached the steady state, or remained turned off when the activator is absent.

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62

How does positive autoregulation create memory?

Feedback keeps system going, making it a memory circuit because its activity continues after the signal is gone (remembers it)

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63

Where does prokaryotic translation occur compared to eukaryotic translation?

Prokaryotic translation occurs in ribosomes whereas eukaryotic translation occurs in the nucleus and travels to the cytosol

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64

What are the consequences of the difference between location of prokaryotic translation and eukaryotic translation?

Prokaryotic transcripts can be fed directly into ribosomes, so the transcripts are functional immediately. Eukaryotic transcripts aren't funcitonal immediately after transcription, since they need to be transported to the cytosol. They need to go through processing so that they can be functional.

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65

What purpose(s) does the 5' cap serve? (some of this information is in our section on translation)

Ribosomes recognize an mRNA by the 5' cap, since methylated guanosine is not usually present in mRNA. Without it the ribosome would not attach to the mRNA and translation would not occur. Additionally, it allows for the transcript to be protected from 5' exonucleases and marks RNAP II's switch from initiation to elongation

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66

When is the 5' cap added? How is this timing regulated?

Once the transcript is 30 nt long, the cap is added. Capping machinery acts concurrently with transcription.

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67

Know the mechanistic steps of the capping process

  1. Removal of terminal phosphate from 5' triphosphate by phosphorylase

  2. Guanylation of mRNA capping enzyme (using GTP) by guanylyl transferase. Pyrophosphate leaves

  3. Methylation of guanine by guanine-7-methyltransferase. Methyl is added to oxygen.

  4. Can also methylate N1 and N2 by O-2' methyl transferase

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68

How is the 5' guanine added?

The terminal phosphate leaves the NTP. The O on the second phosphate then attacks the phosphorous on the innermost phosphate on GTP, causing pyrophosphate to leave. This creates the 5' to 5' triphosphate bridge, and thus 7-methyl-guanine is added.

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69

How is RNAP's CTD involved in the process of adding the 5' cap?

Capping enzyme binds to phosphorylated CTD, so capping is concurrent with RNAP II's switch from initiation to elongation. CTD recruits capping enzymes when it is phosphorylated.

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70

What nucleotides end up methylated?

guanosine and N1/N2 are optional. Methyl on O2' varies depending on the species

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71

What purpose(s) does the poly-A tail serve?

Both the 5' cap and the poly A tail stabilize the RNA transcript by preventing 5' and 3' exonuclease activity. Transcription termination in eukaryotes is imprecise, so adding the poly A tail defines the 3' end.

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72

How does the addition of the poly-A tail "standardize" eukaryotic transcripts?

Defines the 3' end because length of transcript is heterogeneous and the poly-A tail will help normalize the transcript

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73

What role does CPSF play in initiating the creation of the poly-A tail? What transcript sequence does it recognize? Where does the cleavage occur?

The cleavage and poly-adenylation specificity factor recognizes and binds to the AAUAAA sequence and phosphorylated CTD to activate the poly(A) polymerase (PAP). PAP uses ATP to generate poly A tail. CPSF and CStF work together to recruit elements that cut at a specific location (the CF, which is 15-25 nts past AAUAAA) and promote poly adenylation by attracting PAP.

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74

What enzyme does CPSF recruit to synthesize the poly-A tail?

PAP

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75

How is RNAP's CTD involved in this process?

CTD recruits CPSF and regulates CPSF's activity

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76

Why are there inconsistencies between gene sizes and protein lengths?

RNA splicing- removes introns (long "junk" DNA) and alternative splicing

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77

Exon shuffling is a process in which:

Recombination event leads to illegitimate recombination which is useful for bringing new protein domains together and allowing them to have new functions

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78

Propagation of junk DNA

The equivalent to a DNA parasite. The junk DNA can inject itself into a protein coding region of DNA and this can propagate generation after generation

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79

Invariant BP for 5' slice site

GU

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80

Invariant BP for 3' slice site

AG

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81

Invariant BP for branch point

A makes the nucleophilic attack

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82

Which nucleotide forms a nucleophilic attack and what nucleotide receives this attack?

The A on the branch point makes the nucleophilic attack on the G of the 5' splice site

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83

What comprises the spliceosome?

Comprised of snRNAs, proteins (which form snRNPs), Mg ions, and the intron itself

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84

What does the spliceosome do?

Recognizes intron/exon boundaries and branch point and cuts out the intron. It also prevents the 5' exon from floating away

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85

Why does the spliceosome require ATP?

ATP ensures that the attack happens at the right spot. This also allows conformational changes to occur and snRNAs to be recruited.

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86

U1 snRNA

recognizes and binds to 5' splice junction to hold it in place with its complementary sequence

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87

U2 snRNA

recognizes and binds to sequence to form branch point by pushing out a nucleotide, typically adenosine (sometimes G), so it can attack

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88

U4, U5, and U6 snRNA

form conglomerate that form a bend that bring U1 and U2 together, also bringing the 5' splice site and branch point closer together to encourage nucleophilic attack

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89

How does the U2 snRNA establish the branchpoint mRNA structure?

Forms bulge at 2nd nucleotide on branch point (usually with A or sometimes with G)

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90

How would you disrupt the branchpoint?

Non-canonical bases through mutations (non-A/G)

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91

What do metal ions accomplish in the spliceosome?

active site magnesium stabilizes the oxyanions in the transition state

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92

How can alternative splicing create tissue specificity?

Specificity for different splice junctions will lead to different exon combinations which will lead to different types of proteins that are translated across different tissues. Combinatorial control is a lot of complexity from a small set of data.

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93

How is sex determined in Drosophila?

Chromosome number has different steady state protein concentrations. XX(females)has nearly a double steady state protein concentration than XY(males) Different protein concentration leads to differential splicing

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94

How do female flies ensure a functional concentration of Sxl?

Female flies have 2 X chromosomes and each X chromosome has a copy of Sxl. That means female flies have roughly 2 times the concentration of Sxl and the steady state concentration is high enough for Sxl to bind and block U2AF, leading to a different splice site, which promotes female development

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95

How does Sxl alter splicing specificity of tra? What are the two possible gene products?

There is more Sxl in the nucleus of females to bind to the preferred site of the U2AF regulatory splicing factor. U2AF cannot bind to this area bc Sxl is there, so it binds at its second preferred site. This creates the two possible gene products depending on where the U2AF binds and determines if the blue region of exon 2 is included in the final protein or not. If the blue region is included, then it will lead to Male development. If the blue region isn't included in the final protein, then it will lead to female development.

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96

What has to be present to allow for self splicing?

free guanosome

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97

What forms the active site of the tetrahymena pre rRNA self splicing reaction?

RNA sequence creates own active site. It interacts w itself. Folding of the ribosome results in the formation of an active site. Metal ions stabilize the transition state

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98

Why is the Kcat of the Tetrahymena self-splicing reaction lower than the Kc?

It is a ribozyme, not an enzyme and it is not regenerative so there is not constant rate of movement. Kcat is the rate of product release but Kc is the concentration at equilibrium

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99

Which codons don't have associated tRNAs?

Stop codons (Ter)

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100

How is it possible for one tRNA to recognize multiple codons?

The code is degenerate meaning that multiple codons can code for the same amino acid

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