RNA and Proteins

Question 1

Gene

Answer 1

– a segment of DNA that contains all the information necessary for the synthesis of a function product (protein or RNA)
o Composed of:
 untranscribed promotor and enhancer regions that are necessary to regulate transcription
 transcribed coding regions (exons) that are translated into protein
 transcribed non-coding (introns) which are removed post transcriptionally
 transcribed untranslated regions (UTR) at the 5’ and 3’ ends that regulate translation and mRNA stability

Question 2

Transcription Initiation

Answer 2

o RNA polymerase does not need a primer to provide first 3’OH; it binds to DNA weakly and non-specifically
o 2 type of transcription factors that help RNA polymerase begin to transcribe in the proper location and at correct rate to make required amount of mRNA
 Basal/general transcription factors
 Transcriptional activators (transcription factors, TSFs)

Question 3

Basal/general transcription factors

Answer 3

• Bind as multi-protein complex to TATA box and interacts with RNA pol.
• Same complex for every gene
• Regulates low, basal level transcription
• Not regulated in response to environment
 TATA Binding Protein (TBP) – 1st transcription factor to bind to initiation complex
• Part of TFIID that binds directly to TATA
• Completion of initiator complex causes conformational change
• Kinks DNA to separate strands
 TFIIH
• Last basal transcription factor to bind to initiation complex
• Multi-subunit protein with helicase activity that uses ATP to unwind helix and separate DNA strands
• Kinase activity phosphorylates the C terminal domain of RNA polymerase to signal transcription to begin
• Also plays roll in transcription-coupled repair of DNA

Question 4

Transcriptional Activators

Answer 4

 Bind to enhancer elements in regulatory regions of genes and influence rate of transcription initiation
• Recruit histone acetyl transferases and/or chromatin remodeling complexes
o Acetylation leads to euchromatin formation
o Deacetylation leads to heterochromatin formation
• AND/OR have direct interactions with general transcription factors, or indirect interactions through mediator proteins
 Each factor binds to a certain nucleotide sequence in DNA (consensus site; enhancer element) via its DNA binding domain
• Consensus binding sequences – binding site of a given transcription factor can be slightly different on different genes – purines/pyrimidines may vary
o Varying binding affinity
• Each gene is bound by a different set of activators
 Regulate transcription in response to environmental signals
 Facilitate the formation of basal initiation complex

Question 5

Repressors

Answer 5

– can interfere with the activity of transcriptional activators in multiple ways
o Competition for same binding site as activator
o Binding to and obstructing the activation domain of activator
o Recruiting histone deacetylases or other proteins to cause heterochromatin formation

Question 6

Transcriptional Elongation

Answer 6

o One strand (coding strand) is held out of way while the other strand (template strand) is copied
o 1st nucleotide remains a triphosphate and is protected by adding a 5’ cap (= 7 methylguanosine)
o RNA sequence is complementary to the template strand except with U’s replacing T’s
o 3 major RNA polymerases
o DNA will re-anneal back together after transcription

Question 7

5’ Cap

Answer 7

– 7 methylguanylate – occurs very quickly
 Backward 5’ to 5’ linkage
 1st 1 or 2 ribonucleotides are also methylated
 Added by a capping enzyme (guanyl transferase) that associated with the phosphorylated CTD of RNA polymerase II
 Provides mRNA stability
 Aids in transport of mature mRNA from nucleus to cytoplasm
 Necessary for efficient translational initiation

Question 8

post-transcriptional processing of mRNA

Answer 8

 RNA polymerase II carries some pre-mRNA processing proteins on its tail (the phosphorylated C-terminal domain – CTD); the proteins are transferred to newly transcribed mRNA when necessary
• Some DNA repair enzymes also travel with RNA polymerase in this way

Question 9

splicing

Answer 9

 Introns are intervening sequences that do not encode proteins; they are removed in the nucleus and the exons are spliced together before mature mRNA is transported out
• Involves small nuclear RNAs (snRNAs) which are associated with proteins to form small ribonucleoprotein particles (snRNPs)
• Some snRNPs recognize consensus sequences at exon/intron and intron/exon junctions
 Alternative splicing – some mRNAs can be spliced in different ways to yield different proteins from the same pre-mRNA transcript; ~60% of genes
 ~15% of all genetic diseases result from mutations that affect mRNA splicing

Question 10

polyaadenylation

Answer 10

 mRNA is cleaved downstream of polyA signal (AAUAAA)
 polyA polymerase adds 100-200 A’s to the 3’ end; then polyA binding protein binds
 nuclease binds the 5’ end and degrades the trailing RNA; when it reaches the RNA polymerase, transcription is terminated
 poly A is added in the nucleus but sometimes lengthened in the cytosol
 provides mRNA stability
 involved in transport of mature mRNA from nucleus to cytosol
 Contributes to efficient translational initiation

Question 11

tRNA structure

Answer 11

– cloverleaf structure – 70-90 nt long – 31 kinds
o 3’ end – attaches amino acid
o Anticodon
o D and T loops

Question 12

tRNA processing

Answer 12

o 16 nucleotide sequence at 5’ end is cleaved by RNase P
o 14 nucleotide intron in the anticodon loop is removed
o Uracil residues at 3’ end are replaced by CCA sequence – found in all mature tRNAs
o Many bases are converted to characteristic modified bases

Question 13

ribosomal RNA

Answer 13

o Single gene encodes large 45S precursor
o 250-300 copies of the gene clustered together
o Some nucleotides of rRNA precursor are chemically modified
o Processed by cleavage rRNA is responsible for most of the catalytic activity of ribosome
o Ribosomal subunits are assembled in nucleolus by ribosomal proteins

Question 14

miRNA

Answer 14

– estimated to be ~1000
o may be involved in regulation of 2/3 of genes
o transcribed by RNA polymerase II
 some processed from introns or 3’UTR of pre-mRNA
 most processed from pri-miRNAs – transcripts of 100-1000 nucleotides
o miRNA acts as a “guide sequence” that brings nuclease into contact with targets mRNAs
o if sequence is complimentary to miRNA, argonaute slices the mRNA, removing the polyA tail and making the mRNA vulnerable to nucleases
o if sequence is not fully complimentary, no splicing occurs but translation is inhibited and the mRNA is destabilized by moving it to P-bodies (processing bodies) in the cytosol where it is eventually degraded

Question 15

processing of miRNA

Answer 15

o miRNA sequence folds to form hairpin of ~70 nucleotides with imperfect base pairing in stem
o an RNase (Drosha) cleaves hairpin to produce pre-miRNA
o pre-miRNA goes to cytosol
o dicer RNase process the pre-miRNA to produce a double stranded miRNA that’s 21-23 nt long
o one strand binds with an argonaute protein to form a mature RISC (RNA-induced silencing complex)

Question 16

Genetic Code

Answer 16

– 3 mRNA nucleotides form a codon which encodes an amino acid
o	Non-overlapping, no punctuation – starts at AUG (codes for methionine) until an in-frame stop codon is reached (UAA, UAG, UGG)
o	Specific – given code always encodes the same amino acid
o	Degenerate/redundant – given amino acid may be encoded by more than 1 codon
	Due to “wobble” in the binding between tRNA and mRNA codon
•	5’ base of tRNA anticodon undergoes more movement than the other 2
o	Amino (N) terminus of protein corresponds to the 5’ end of mRNA
o	Carboxy C terminus of protein corresponds to the 3’ end of mRNA

Question 17

Types of Point Mutations

Answer 17

o Frameshift – caused by insertion or deletion of any number of nucleotides not divisible by 3
o Missense – new codon encodes different amino acid
o Nonsense – new codon is a stop codon resulting in premature termination
o Silent – new codon encodes the same amino acid; change in DNA sequence but not protein

Question 18

Activation or Charging of tRNAs

Answer 18

– aminoacyl tRNA synthetases
o Family of 20 enzymes
o Each recognizes one amino acid
o 2 active sites – 1 for amino acid and 1 for tRNA
o “double sieve” proofreading – makes sure correct amino acid is added to tRNA
 Synthesis site - Rejects amino acids that are larger than correct one
 Editing Site – rejects amino acids that are smaller than the correct one

Question 19

Initiation of Protein Synthesis - characteristics

Answer 19

 Involves eukaryotic initiation factors (eIFs)
• eIF-2-GTP has to be recycled – regulation point
 involves special initiator tRNA charged with methionine
 involves recognition of 5’ cap and polyA tail by eIF’s
 use energy from ATP and GTP
 eIF-2-GDP must be reactivated to eIF-2-GTP before it can participate in more initiation
• allows for global regulation of translation under certain conditions

Question 20

initiation of protein synthesis mechanism

Answer 20

o initiation begins begins by attaching methionine amino acid to a specific initiator tRNA
o initiator met-tRNA forms comples with eIF-2-GTP
 other eIFs separate the ribosomal subunits and bind the small 40S subunit
o initiator tRNA complex binds to the small ribosomal unit to form an initiator complex
o mRNA is bound by initiation factors, including some that recognize the 5’ cap and polyA tail
o initiator complex binds to mRNA and uses energy from ATP to move along the mRNA until it find the AUG codon; complementary base pairs are formed between start codon and anticodon of tRNA; tRNA for methionine goes directly into P site
o 60S subunit is added in a reaction that hydrolyses GTP and releases initiation factors

Question 21

Elongation of Peptide Chain

Answer 21

The initiator met-tRNA complex occupies the P site (the “peptidyl” or donor site)
o A tRNA containing an anticodon complementary to the next codon and charged with the correct amino acid enters the A site (the “acceptor”, “aminoacyl” or entry site).
o If the hydrogen bonding between the codon and the anticodon is correct, GTP is hydrolyzed and the elongation factor is released along with GDP.
o Peptidyl transferase is an enzymatic activity associated with an rRNA in the large 60S ribosomal subunit. It transfers the peptide chain from the tRNA in the P site to the amino group of the amino acid attached to the tRNA at the A site to form a peptide bond. This is a condensation reaction
o Peptide bond formation is accompanied by conformational changes in the large ribosomal subunit that result in shifting the two tRNAs to the E and P sites.
o Another series of conformational changes moves the mRNA exactly three nucleotides through the ribosome and resets the ribosome so the A site is open
o Elongation proceeds as a repetition of these reactions, adding one amino acid at a time to the growing peptide chain.

Question 22

Termination of Protein Synthesis

Answer 22

o When a termination codon is in the A site, eRF-GTP (release factor) binds in the A site
o Protein is released from tRNA by peptidyl transferase in a reaction that involves addition of water to the COO- group of the last amino acid and the hydrolysis of GTP
o Uncharged tRNA dissociated from ribososme and the ribososme dissociates from mRNA

Question 23

polyribosomes

Answer 23

• More than one ribosome can translate the same mRNA at the same time (polyribososmes)
o eIF that interacts with both 5’ and 3’ end of mRNA transcript and forms a circle so that the ribosome can restart another translation as soon as it finishes a previous one