Transciption

Question 1

In prokaryotes, the coding region of a gene is often a

Answer 1

Single, continuous unit (CO-LINEAR)

Question 2

Only the untranscribed part is called the

Answer 2

Shine-Dalgarno sequence

Question 3

Do transcription and translation occur at the same time in prokaryotes

Answer 3

yes :)

Question 4

What is a protein coding segment

Answer 4

Exon

Question 5

What segments do not code for proteins

Answer 5

Introns

Question 6

Do transcription and translation occur at the same time in eukaryotes

Answer 6

No :(

Question 7

Types of RNA

Answer 7

Messenger RNAs (mRNA) - intermediates that carry genetic info from DNA to ribosomes
Transfer RNAs (tRNA) - adaptors between AA and codons in mRNA
Ribosomal RNAs (rRNA) - structural and catalytic components of ribosomes
Small nuclear RNA (snRNA & snoRNA) - spliceosomes, rRNA, tRNA modification, respectively
MicroRNAs (miRNAs, siRNA & Crispr RNA) - short RNAs that block expression of complementary mRNAs
Long noncoding RNAs - long RNAs that regulate gene transcription

Question 8

What is the RNA synth to the DNA template strand

Answer 8

Complementary and antiparallel

Question 9

RNA synthesized in _’ to _’ direction while DNA is the opposite

Answer 9

5,3

Question 10

Transcription requires a _’ to _’ template and 4 ribonucleoside triphosphates and a DNA-dependent RNA polymerase.

Answer 10

3,5, rNTPs

Question 11

Initiation in Prokaryotes cells

Answer 11

RNA polymerase binds to promoter, unwinds it and joins the first 2 nucleotides
Initiation of RNA synthesis DOES NOT require a primer
Recognition of the gene promoter region requires the intact RNA polymerase holoenzyme
σ factor recognizes and binds to the -35 element in promoter region properly position the RNA polymerase to begin transcription
2 important promoter sequence elements are the -35 element to which σ factor binds and -10 element which is prone to unwinding due to A/T rich content (fewer H bonds)
-35 sequence: 5’ TTGACA 3’, separated from -10 by ~16-19 bp
-10 sequence: 5’ TATAAT 3’ (aka “Pribnow” box)
Transcription initiates about 5-9 bp downstream from the end of the -10 sequence by looking for a purine (A or G) on the 5’ to 3’ strand, finding opposing pyrimidine (3’ to 5’ strand - template), and attaching the first purine onto template strand (synthesize in 5’ to 3’ direction)

Question 12

Elongation in prokaryote cells

Answer 12

complementary nucleotides continue to be added during elongation process
Elongation occurs when σ factor is released and RNA polymerase begins to move along 3’ to 5’ DNA template strand
Localized DNA unwinding ahead of RNA polymerase generates “transcription bubble” (18 bp)
Transcription bubble moves with the RNA polymerase and the unwound DNA rewinds behind it
Positive supercoils formed in dsDNA ahead of advancing RNA polymerase are removed by topoisomerases (gyrase)
RNA polymerase has both helix UNWINDING & REWINDING activities

Question 13

Termination in prokaryote cells

Answer 13

complementary nucleotides continue to be added during elongation process
Elongation occurs when σ factor is released and RNA polymerase begins to move along 3’ to 5’ DNA template strand
Localized DNA unwinding ahead of RNA polymerase generates “transcription bubble” (18 bp)
Transcription bubble moves with the RNA polymerase and the unwound DNA rewinds behind it
Positive supercoils formed in dsDNA ahead of advancing RNA polymerase are removed by topoisomerases (gyrase)
RNA polymerase has both helix UNWINDING & REWINDING activities

Question 14

Transcription in eukaryotes, Five RNA polymerases

Answer 14

Refer to the notes for an answer its a big chart.

Question 15

In eukaryotes, ____ recognize each of these specific types of promoters (through interaction with their DNA sequences) and ____ the appropriate polymerase for transcription

Answer 15

accessory proteins, bind/recruit

Question 16

Initiation in eukaryotic cells

Answer 16

(for pol. II) - involves stepwise assembly of general Transcription Factors of pol. II (TFII A, B, D, E, F, and H)
TFIID assembles first at the TATA box followed by remaining general transcription factors (TFs) and finally, RNA polymerase II
This forms the preinitiation complex (PIC) which initiates transcription

Question 17

Elongation in eukaryotic cells

Answer 17

many general transcription factors remain at promoter region, making for quick re-initiation with a new Pol. II
An ~8 bp transcription bubble is generated by RNA:DNA binding
This with DNA unwinding ensures that free RNA 3’-OH terminus is free for new rNTP addition

Question 18

Termination in eukaryotic cells

Answer 18

involves cleavage of pre-mRNA and 5’ to 3’ degradation of remaining RNA by Rat1 exonuclease
RNA po. II transcribes well past the coding sequence of the gene
Transcription terminates when Rat1 catches up to pol. II (follows up behind it degrading excess mRNA as it goes)
Transcript is polyadenylated (adds a bunch of A bases on 3’ end)
An important inhibitor of transcription is α-amanitin (found in Amanita mushrooms) - inhibits RNA pol. II at both initiation & elongation states of transcription (genes are not expressed)

Question 19

The removal of introns from pre-mRNA are required to form

Answer 19

mature mRNA that is translated into a polypeptide

Question 20

Removal of introns

Answer 20

1. Addition of 7-Methyl Guanosine (7-MG) Cap - unique linkage to pre-mRNA by 5’ phosphate of 7-MG and 5’ phosphate of first ribonucleotide in the RNA (5’ to 5’ phosphate linkage)
   Addition of 7-MG cap occurs early in elongation process
2. Addition of PolyA tail - Eukaryotic pre-mRNA is cleaved 11-30 nt following 5’ AAUAAA 3’ sequence
   PolyA polymerase adds ~200 A residues at cleaved end
3. Removal of Introns - introns in pre-mRNA removed by specialized process called “RNA splicing”
   Removal of introns must be precise in order to properly fuse 3’ end of one exon to 5’ end of next
   Every intron has 2 conserved sequences that are required for its precise removal:
   5’ & 3’ splice sequences containing junction sequences “5’ GU & 3’ AG”, respectively
   Intron Branch point - conserved “A” residue
   Splicing is accomplished by RNA/protein complex known as the spliceosome
   Spliceosomes contain five snRNAs - U1, U2, U4, U5, and U6
   snRNAs associate with about 40 small proteins to form small nuclear ribonucleoproteins (snRNPs, pronounced “snurps” lol)
   snRNPs U1, U2, U4, U5, and U6 assemble to form a complete spliceosome
   snRNP U1 binds to 5’ splice site, snRNP U2 binds to branch site
   Complete spliceosome assembles and cleaves transcript at 5’ splice site
   5’ end of intron joined to A in branch site to form a lariat structure, U1 & U4 are released
   Lariat structure - linkage between 5’ phosphate of G and 2’ OH of A (fused loop)
   3’ splice site is cleaved and 5’ end of exon 2 is joined to 3’ end of exon 1, lariat-shaped intron is released along with U2, U5, and U6
   Summary - snRNP assembly → 5’ splice site cleaved → lariat loop forms → 3’ splice site cleaved → exons joined and intron removed

Question 21

Alternate splicing

Answer 21

Only one type of pre-mRNA made, can result in exons being removed along with introns, different mature mRNAS

Question 22

Multiple 3’ cleavage sites

Answer 22

One exon may have more than one 3’ cleavage cute which results in 2 different chunks of that exon

Question 23

Rna editing is carried out by

Answer 23

guide RNAs (gRNA)

Question 24

Cytidine deaminase converts _ to _ and in one specific case transforms a glutamine codon (CAA) into a ____ codon (UAA)

Answer 24

C,U,Termination

Question 25

Transfer RNAs

Answer 25

Transfer RNAs (tRNAs) - specialized RNAs that have a role as adaptor molecules in translation 
First postulated by Crick in 1956 as adaptors between AA and codons in mRNAs
At 3’ end of tRNA there is a tag: 5’ CCA 3’ that binds to AA covalently 
Specificity of tRNA to AA is dictated by the anticodon 
tRNAs have modified bases in some parts that assist with structure & function

Question 26

Ribosomal RNAs

Answer 26

rRNAs) - key component of ribosome
Ribosomes composed of large and small subunits that are assembled from many different proteins and rRNAs
Prokaryotes - 5S and 23S rRNA in 50S (large) subunit of prokaryotic ribosomes, 16S rRNA in 30S (small) subunit (ribosome is 70S, densities are NOT additive)
16S rRNA has a sequence complementary to shine-dalgarno sequence that is found at the 5’ end of prokaryotic mRNA
Small subunit is essential for initiation of translation in prokaryotes
rRNA synthesis and ribosome assembly occur in the cytoplasm
tRNA is encoded in rRNA gene and trimmed out in rRNA processing
Eukaryotes - 5S, 5.8S, and 28S rRNA associate with 60S (large) subunit, 18S rRNA in 40S (small) subunit (ribosome is 80S)
rRNAs are synthesized and ribosomes are assembled in nucleolus (eukaryotic)
rRNAs are processed through methylation, cleavage, and trimming before being put to use

Question 27

Small nuclear RNAs (snRNAs)

Answer 27

snRNAs act in complexes with proteins, play roles in post-transcriptional processing of RNA such as splicing (spliceosome assembly)

Question 28

Small nucleolar RNAs (snoRNAs)

Answer 28

also act in complexes with proteins, involved in chemical modifications of rRNAs, tRNAs, and snRNAs

Question 29

Small micro RNAs

Answer 29

(siRNA, miRNA, Crispr RNA) - act differently in eukaryotes than in prokaryotes
Eukaryotes - act as short (~22nt) ssRNAs that bind to complementary sequences in mRNA
Produced by cleavage of mRNAs, RNA transposons, and RNA viruses
Regulate and control gene expression in different ways
siRNA bind to mRNA and inhibit translation, miRNA bind to mRNA and degrade it
Prokaryotes - Crispr RNA is encoded by DNA in prokaryotes, works in association with Cas9 nuclease to cleave foreign DNA that enters host cell
Bacterial defense system that prevents incorporation of foreign DNA into host genome

Question 30

Long noncoding RNA

Answer 30

Comprise about 80% of mammalian genome, regulate and control gene expression