Exam 3

Question 1

U2 snRNA

Answer 1

Bind to the branch point of the splicesome

Question 2

Differences found in RNA but not DNA

Answer 2

Ribose sugar (less stable), uracil (additional pairing and unusual bases), and is single stranded (can form different shapes)

Question 3

Base pairs in RNA (in order of decreasing strength)

Answer 3

G-C, A-U, and G-U

Question 4

Base pairs in RNA (in order of decreasing strength)

Answer 4

G-C, A-U, and G-U

Question 5

The function of uncommon bases in RNA

Answer 5

Usually found in non-coding tRNA (usually pseudouridine) and they are believed to stabilize RNA structure. They can modify base pairing if found in mRNA.

Question 6

Name the non-watson-crick base pairings

Answer 6

C-G-C base triplet, T-A-A base triplet, T-A-T base triplet, and C-G-G base triplet

Question 7

Coding RNAs and Non-coding RNAS: which feature is most important?

Answer 7

The sequence is most important for coding RNAs and the structure is most important for non-coding RNAs.

Question 8

How are secondary RNA structures formed? What are their names?

Answer 8

By canonical base-pairing (non-watson and crick). Hairpin loop (@ the end), internal loop, stem, bulge, and external base (just one base sticks out)

Question 9

Secondary structure vs tertiary structures

Answer 9

Typically easier to predict, more stable, and more evolutionarily conserved than tertiary structures.

Question 10

Function of different RNAses

Answer 10

RNAse V1- cleaves dsRNA
RNAse 1 - cleaves ssRNA, 3’ end with a C or U
RNAse T1 - ssRNA, 3’ end with a G

Question 11

Use of RNAse digestion assay

Answer 11

Can’t be used in live cells, but useful for 1 by 1 secondary structure interpretation. Very slow method for large RNA amounts.

Question 12

Use of DMS seq

Answer 12

Assay for RNA secondary structure which can assess all cellular RNAs at once in-vivo (live cells). It modifies the As and Cs in ssRNA with DMS to show what the secondary structure looks like upon cDNA sequencing

Question 13

Use of SHAPE-seq

Answer 13

Modifies the 2’-OH group of all ssRNA bases. Like DMS seq, it can be performed in vivo or in vitro and is genome-wide. Unlike DMS seq, you can identify the base pairing in hairpin structures and signal resolution is increased based on the burying of the base within the folded structure.

Question 14

Tertiary structures of RNA

Answer 14

A form helix is a right-handed 11 bp/turn
B form helix is a right-handed 10 bp/turn

Duplex, triplex, G-quadruplex, pseudoknot, and kissing loops

Question 15

What percentage of RNA is translated into protein?

Answer 15

<5%

Question 16

Name 6 types of RNA based on their size/function/expression level

Answer 16

RNA genomes (ex: RNA viruses), protein-coding RNAs (mRNA), non-coding RNAs for translation (tRNA, rRNA), non-coding RNAs for RNA processing (gRNAs, snRNAs), non-coding RNAs for DNA replication (telomerase RNA), non-coding RNA for protein targeting (7SL RNA, 4.5S RNA)

Question 17

Signal Recognition Particle (SRP)

Answer 17

Protein sorting RNA which directs membrane destined proteins for nuclear/rough er localization.

The SRP in eukaryotes is comprised of 7SL RNA

Question 18

pRNA

Answer 18

Packing RNA is the most powerful molecular motor that packages RNA into viral capsids.

Question 19

Ribozymes

Answer 19

RNAs that act as enzymes (natural or artificial). The sequences are able to form multiple secondary structures which can allow them to act in different ways.

Question 20

Riboswitch

Answer 20

RNAs which can sense environmental changes and their functions can change as a result. Ex: bacterial RNA thermometer- only when the environment is above 42 degrees can the ribosome can bind to the RNA to create protein

Question 21

Aptamer

Answer 21

RNA molecules that bind specific ligands and are usually less active than ribozymes. Selected through SELEX process.

Question 22

What is the case for RNA being the first molecule to create life?

Answer 22

RNA is the only macromolecule that can store genetic information AND catalyze reactions. There were also once self-replicating RNAs that are no longer found in nature.

Question 23

What are the alternatives that were proposed instead of RNA backbones in the first living organisms?

Answer 23

Treose (TNA), peptide (PNA), Glycerol (GDNA), and Pyranosyl RNA

Question 24

RNA Transcription

Answer 24

Synthesis is 5’-3’, not restricted to S-phase, more error prone than DNA synthesis.

Question 25

Operons

Answer 25

Very efficient bacterial DNA regions of related genes which can be turned on or off concertedly. multiple proteins with 1 RNA molecule

Question 26

Bacterial RNA polymerase

Answer 26

Alpha, beta, beta', and sigma are the main catalytic subunits. Sigma serves as a promoter recognition and is necessary for circular DNA there are several sigma-factors. Alpha causes chain initiation. Beta leads to chain initiation and elongation. Beta' causes DNA binding. and

Question 27

RNA polymerization: Initiation

Answer 27

1) The sigma-factor recognizes and binds the promoter to form a closed complex. 2) The RNA polymerase then unwinds the DNA to form an open complex transcription bubble (no helicase needed). 3) The first phosphodiester bond is formed between rNTPs to form an unstable ternary complex. 4) The sigma-factor is then released after ~10 nt to form a stable ternary complex.

Question 28

Sigma promoter sequences

Answer 28

Usually a -35 sequence and -10 sequence (Pribnow box/TATA box). The strongest promoter sequence match the consensus. Some genes also have Upstream Promoters (UP) or enhancers that increase polymerase affinity.

Question 29

DNA footprinting

Answer 29

Identifies RNA polymerase binding sites on DNA.

Question 30

What does ChIP-seq do?

Answer 30

DNA binding proteins are introduced to a sample which bind to target DNA sequences. The DNA is then chopped into fragments. Antibodies for the proteins are then immunoprecipitated and isolated. DNA is released from the proteins and then is sequenced to identify the regions which bind to the proteins. Useful for identifying DNA modifications such as methylation/acetylation.

Question 31

Abortive initiation

Answer 31

The first few rounds of polymerization are non-productive because sigma blocks the RNA exit channel. Several short strands of RNA are formed until sigma dissociates to form a stable ternary complex and productive transcription starts.

Question 32

Termination sequence of bacterial transcription

Answer 32

GC rich inverted repeats cause the formation of a hairpin loop. The sequence then is followed by poly-A, which causes weak binding and then dissociation.

Question 33

Rho-dependent termination of transcription

Answer 33

Rho is an ATP-dependent RNA helicase that unwinds the DNA:RNA hybrids. This is required for sequences without the terminator sequence (GC + A). Not all RNAs will terminate at the same base with RHO, causing different-sized transcripts.

Question 34

Post-transcriptional modifications in bacteria

Answer 34

An addition of a poly-A tail causes RNA degradation (opposite of eukaryotes) because RNAses like the ssRNA. The hairpin 2 structure at the ends prevents degradation, but still incredibly unstable (which makes cotranslation a benefit)

Question 35

Use of (q)RT-PCR

Answer 35

Common assay of RNA abundance on a limited number of RNAs (very low throughput, must be 1 RNA at a time). cDNA formation through PCR and then quantitatively monitor DNA amplification with intercalating dyes. Lower Ct value = higher abundance bc it crosses threshold faster.

Question 36

Use of RNA-seq

Answer 36

Assay for RELATIVE abundance of RNA in the entire genome. From this method you can compare the cDNA sequence to the DNA sequence to identify the number of exons/introns.

Question 37

RNA polymerase I

Answer 37

Nuclear eukaryotic RNA polymerase which transcribes rRNA only (80% of total RNA)

Question 38

RNA polymerase II

Answer 38

Nuclear eukaryotic RNA polymerase which mostly transcribes mRNA (only 1-5% of total RNA). Structurally homologous to bacterial polymerase

Question 39

RNA polymerase III

Answer 39

Nuclear eukaryotic RNA polymerase which transcribes tRNA and some small RNA sequences.

Question 40

RNA Pol II (eukaryotic)

Answer 40

TFIID - Recognizes the TATA box and other sequences near the transcription starting point. ~12 subunits to regulate the start of transcription TFIIH - unwinds DNA at the transcription start point. Phosphorylates Ser5 at the RNA polymerase CTD and releases RNA polymerase from the promoter.

Question 41

Eukaryotic promoters

Answer 41

Difficult to predict due to wide variability. There is typically a core promoter ~4bp from TSS, proximal promoter ~200-40 bp from TSS, and an enhancer that is many kbs away from the TSS.

Question 42

Eukaryotic enhancer

Answer 42

Activators bind to this sequence which is kb away from the TSS. A mediator is a large multiprotein complex that brings activators and basal transcription factors together to increase transcription. Insulators establish chromatin boundaries to help maintain the specificity of interactions despite the great distance between promoters.

Question 43

Eukaryotic transcription initiation by poll II

Answer 43

TFIID (through Tata binding protein) binds the TATA box. TFIIB is then recruited to the TBP-TATA box complex to recruit RNA polymerase to the promoter. TFIIH serves as the helicase and phosphorylates the CTD to stimulate the release of the mediator. This causes the beginning of transcription.

Question 44

Eukaryotic RNA elongation

Answer 44

Elongation factors help. The 7- methylG cap is added on the 5' end to prevent degradation as it emerges. Splice junctions are marked by the binding of splice factors and then splicing occurs co-transcriptionally.

Question 45

What does GRO-seq do?

Answer 45

Assay for active transcription by RNA pol. UTP is replaced with BrUTP which has specific antibodies for it. This can help identify alternative splicing mechanisms and led to the discovery that transcription can go in both directions.

Question 46

Co-transcriptional modification of eukaryotic mRNA

Answer 46

Multiple phosphorylations of the CTD: unphosphorylated form is for activation, serine 5P causes early termination, & serine 2P causes late torpedo termination through polyadenylation.

Question 47

Nucleosomes in eukaryotic rna transcription

Answer 47

Chaperone proteins can displace the octamer behind the polymerase, the polymerase can traverse without displacement, or the cotranscriptional histone modification can cause displacement.

Question 48

Allosteric termination in eukaryotic RNA transcription

Answer 48

The binding of 3' processing factors lead to rearrangement of the elongation complex and then termination

Question 49

Torpedo termination in eukaryotic RNA transcription

Answer 49

A nuclease degrades the nascent RNA from the 5' end, catching up with the elongation complex and then finally displaces poll II from DNA.

Question 50

CTD/CTD cycle

Answer 50

The CTD of rpb1 contains many heptapeptide repeats: YSPTSPS. The CTD cycle is the patterned phosphorylation of the series at different stages of transcription.

Question 51

Splicesome

Answer 51

5 subunit which removes the majority of introns. The major spliceosome cuts at 5' GU and AG 3'. The minor spliceosome cuts at 5' AU and AC 3'. THE BRANCH POINT IS ALWAYS AN A, forming a lariat structure. Cut by U6 RNA ribozyme and exon junction complexes mark the spliced sites.

Question 52

Group II self-splicing

Answer 52

The 2' OH group of an adenine attacks the 3' splice site and forms a lariat. No spliceosome mechanism.

Question 53

Group I self-splicing

Answer 53

The 3' OH group of a guanine residue attacks the 5' splice site and does not form a lariat.

Question 54

tRNA splicing

Answer 54

An endonuclease cleaves the 3' splice site and cleaves again at the ACC 5' site. The ACC is retained and ligated. Maturation of tRNA is caused by the RNAse P.

Question 55

T or F: Post-transcriptional base editings can happen in eukaryotes

Answer 55

T, it is most common in tRNA.

Question 56

A-to-I editing

Answer 56

Commonly used in the synaptic transmission of neurons as a normal part of embryonic neuronal development.

Question 57

C-to-U editing

Answer 57

ApoB in the liver normally is 4563 AA long. When C is changed to U via oxidative deamination in the intestine, the ApoB is 2153 AA long.

Question 58

mRNA methylation in eukaryotes

Answer 58

Transient modification, most common is the m6A methylation. This usually is used to alter stability, translation, and modulate splice activity. Readers recognize & bind to the modified base, writers modify the base, erasers reverse the modification.

Question 59

How are mature eukaryotic mRNAs exported from the nucleus?

Answer 59

Via the nuclear pore.

Question 60

nonsense-mediated mRNA decay

Answer 60

Premature stop codons typically arise due to preserved introns. Exon junction complexes stay at the splicing sites on introns which is recognized as abnormal and targeted for degradation.

Question 61

Non-stop decay

Answer 61

processes mRNA transcripts without a stop codon. These are typically caused by abortive transcription where the polymerase is removed before the full sequence is transcribed. The mRNA is stuck to the ribosome until proteins remove it and then endonucleases degrade it at the 3' end.

Question 62

No-go decay

Answer 62

The ribosome stalls due to strong secondary structure formation by intron retentions. Endonucleases cut the RNA in the middle and both parts get degraded since both ends don't have the post-transcript capping.

Question 63

Housekeeping RNA

Answer 63

rRNA, tRNA, snRNA, snoRNA, RNAse P, and 7SL RNA. Non-coding RNAs which are essential for cell survival

Question 64

miRNA

Answer 64

type of ncRNA which is part of the RISC (RNA-induced Silencing Complex). It binds target mRNAs at the 3' UTR and suppresses translation. Single active/leader strand imperfectly base pairs with the structure.

Question 65

siRNA

Answer 65

miRNAs which signal the degradation of mRNAs. Perfect pairing with the target mRNA through the RISC pathway.

Question 66

lncRNA

Answer 66

long-non-coding RNAs that are typically found within the introns or can form from bidirectional promoters of coding RNAs. Most have unknown functions but those characterized are implicated primarily in gene expression/chromatin structure. Often have very long half-lives and stay close to sites of transcription.

Question 67

3 types of gene expression regulation in bacteria

Answer 67

Constitutive expression (always on), repressible (normally on, turned off by a repressor), and inducible (normally off, can be turned on by an inducer)

Question 68

Lac Operon

Answer 68

Inducible operon in bacteria. Lac I creates repressor which bins to the operator. In the absence of lactose, the repressor protein is made and binds to the operator. This prevents polymerase binding. In the presence of lactose, the repressor binds to allolactose which allows polymerase binding.

Question 69

Functions of the Lac genes

Answer 69

Lac I creates repressor which bins to the operator. Lac Z codes for b-galactosidase to form allolactose from lactose. Lac Y codes for the permease which brings lactose into the cell.

Question 70

CAP binding

Answer 70

Necessary for the efficient transcription of lac genes. CAP requires cAMP to bind the binding site, which increases when glucose is low.

Question 71

Trp operon

Answer 71

Repressible operon with two levels of control: repression and attenuation. Codes for the enzymes necessary to synthesize trp.

Question 72

Trp operon: Repression

Answer 72

When Trp is low, the repressor can not bind and transcription proceeds. When Trp is high, the repressor forms a complex with the Trp and it can bind the operator to stop transcription.

Question 73

Trp operon: Attenuation

Answer 73

When Trp mRNA is low, ribosome stalls and it favors the alternative loop so that transcription proceeds. When Trp mRNA is high, the sequence transcribes quickly and forms a terminator loop so transcription stops.

Question 74

Chromatin and eukaryotic gene expression

Answer 74

Heterochromatin, nucleosome positions, and histone methylation can decrease transcription

Question 75

Transcriptional co-regulators

Answer 75

Co-transcribed sequences on eukaryotic genes that can either activate or repress transcription by binding cooperatively to DNA.

Question 76

Retinoid X Receptor (RXR)

Answer 76

3 closely related genes with multiple isoforms due to alternative splicing. A heterodimeric partner for other nuclear receptors to regulate cell differentiation, lipid and glucose metabolism, development, and immune response. Recruits co-regulators to activate or repress transcription.