- Transcription
1. Define transcription.
a. Transcription is the process of using DNA as a template to synthesize a complementary messenger RNA, which can be used to translate protein. RNA is synthesized in the 5’ to 3’ direction, reading off the template DNA strand in the 3’ to 5’ direction, and is identical in sequence to the nontemplate strand (except for T and U difference).
2. Contrast the structure of RNA and DNA. a. Both are 5-carbon membered, furanose sugars. b. Deoxyribose lacks a hydroxyl group at the 2-carbon.
3. Describe the structural elements of a eukaryotic gene and connect each element to its role in transcription.
a. Eukaryotic genes that produce mRNA contain a basal promoter region. Transcription factors, which recruit RNA Polymerase, bind here. Promoters contain a number of conserved sequences.
i. A TATA (Hogness) Box, containing the consensus sequence TATATAA, is located about 25 bp upstream from the transcription start site.
ii. A CAAT box is frequently found about 70 bp upstream from the start site.
iii. GC-rich regions (GC boxes) often occur between -40 and -110 bp from the start site.
b. Enhancers are DNA sequences that function in the stimulation of the transcription rate.
i. Enhancers can be located thousands of bp upstream or downstream from the start site
ii. Other sequences called silencers function the inhibition of transcription.
c. Pre-mRNA (heterogeneous nuclear RNA, hnRNA) initially produced by RNA polymerase containing introns (sequences that are removed and are not contained in the mature RNA) and exons (coding regions). hnRNA modified in the nuclease before mRNA released into cytoplasm through nuclear pores.
d. The primary transcript (hnRNA) is capped by a 7-methylguanosine cap using a 5’ to 5’ triphosphate linkage during transcription.
e. A poly(A) tail, 20 to 200 nucleotides in length, is added to the 3’ end of the transcript. The sequence AAUAAA in hnRNA serves as a signal for cleavage of the hnRNA and addition of the poly(A) polymerase
f. Splicing rxns remove introns and connect the exons.
i. The splice point at the left flank of an intron usually has the sequence AG followed by an invariant GU. AT the right flank, an Invariant AG is frequently followed by GU
ii. Small nuclear RNAs complexed with protein (snRNPs) are involved in the cleavage and splicing process. A lariat structure is generated during the splicing rxn.
4. Understand the effect of DNA methylation on gene expression.
a. Cytosines can be methylated. Methylation occurs exclusively on CpG dinucleotides (CpG Islands when densely clustered).
b. Methylation is inheritable modification, so each new strand receives the methylation, but only one strand methylated à hemimethylated. DNA methyltransferase recognizes hemimethylated state
c. Methylation serves to repress gene expression, and some cancers are hypermethylated at CpG islands.
5. Discuss how chromatin packaging of DNA impacts transcription.
a. Nucleosomes, the basic unit of chromatin, may inhibit promoter region accessibility. Acetylation of histones by histone acetyl transferase complexes and the remodeling or movement of nucleosomes by chromatin remodeling complexes allow accessibility for transcription factors and RNA polymerase. A second remodeling protein might attach to nucleosome to allow for elongation to proceed unhindered.
6. Describe how transcription factors regulate gene expression and how this contributes to heterogeneity in gene expression across tissues and development.
a. Transcription is major mechanism by which protein levels are controlled.
b. Transcription factors can activate or repress transcription
c. Recruit proteins to modify histones
d. Recruit chromatin remodeling proteins
e. Directly recruit RNA polymerase and general transcription factors
f. The combinatorial action of transcription factors and histone modifications in promoter regions controls the level of transcription at any one promoter
g. Mediator is a 1.2 megadalton 26+ subunnit comples (huge)
h. Mediator significantly enhances pol II recruitment and stabilizes transcription complexes at the promoter
i. May act as a scaffold and integrator of long range information (such as interacting with distant promoter sequences)
7. Explain the general process of transcription, including the three major steps of initiation, elongation and termination.
a. Initiation
- i. TATA binding protein (TBP), which is part of TFIID, recognizes TATA box and binds to minor groove of DNA, bending it 800 to help recruit other transcription factors, TFIIB and TFIIA.
- ii. Recruitment of TFIIE and F and RNA polymerase II ensues
- iii. Followed by Recruitment of TFIIH (multisubunit)
- iv. Helicase activity (ATPase)
- v. Promotes RNA polymerase phosphorylation at the CTD (c-terminus)
- vi. Start site of initiation is dictated by the location of the TATA box and is a property of TFIIB
- vii. TFIID can remain to initiate add’l synthesis
b. Elongation
i. RNA polymerase attached to template strand of DNA proceeds in 3’à5’ direction on DNA template and adds RNA to developing hnRNA in 5’à3’.
c. Termination
i. Transcription termination in eukaryotes involved cleavage of the new transcript followed by template independent addition of As at its 3’ end, polyadenylation.
8. List the types of RNA polymerase found in the eukaryotic cell and the different RNA products to which they give rise.
a. RNA Polymerase I is found in the nucleolus and used for synthesis of 45S rRNA.
b. RNA Polymerase II is found in the nucleus and transcribes hnRNA (pre-mRNA), miRNA, and snRNA
c. RNA Polymerase III is found in the nucleus and is used to synthesize snRNA, tRNA, long ncRNA, 5S rRNA.
9. Summarize the ways in which hnRNA is modified on its way to becoming mature mRNA after transcription in eukaryotic cells.
a. Pre-mRNA (heterogeneous nuclear RNA, hnRNA) initially produced by RNA polymerase containing introns (sequences that are removed and are not contained in the mature RNA) and exons (coding regions). hnRNA modified in the nuclease before mRNA released into cytoplasm through nuclear pores.
b. The primary transcript (hnRNA) is capped by a 7-methylguanosine cap using a 5’ to 5’ triphosphate linkage during transcription by guanosyltransferase.
c. A poly(A) tail, 20 to 200 nucleotides in length, is added to the 3’ end of the transcript. The sequence AAUAAA in hnRNA serves as a signal for cleavage of the hnRNA and addition of the poly(A) polymerase.
d. Splicing rxns remove introns and connect the exons.
i. The splice point at the left flank of an intron usually has the sequence AG followed by an invariant GU. AT the right flank, an Invariant AG is frequently followed by GU
ii. Small nuclear RNAs complexed with protein (snRNPs) are involved in the cleavage and splicing process. A lariat structure is generated during the splicing rxn.
10. Describe the ways in which the primary transcript of a tRNA is modified to become a mature tRNA.
a. RNAase P trims the 5’ end.
b. 10% of bases are modified
i. 3’ UU converted to CCA
ii. Addition of methyl and isopentenyl to purines.
iii. Methylation of 2’-OH
iv. Conversion of Uridine.
11. Diagram the processing of eukaryotic pre-rRNA 45S and prokaryotic pre-rRNA 35S.
a. Eukaryotes
i. A 45S precursor is produced by RNA polymerase I from the rRNA genes located in the fibrous region of the nucleolus. Many copies of the genes are present, linked together by spacer regions.
ii. The 45S precursor is modified by methylation at certain 2’ hydroxyls and undergoes a number of cleavages that ultimately produce 18S and 28S rRNA; the latter is hydrogen-bonded to a 5.8S rRNA.
iii. 18S rRNA complexes with proteins and forms the 40S ribosomal subunit. iv. The 28S, 5.8S, and 5S (formed by RNAP III outside nucleolus) rRNAs -complex with proteins and for the 60S ribosomal subunit.
b. Prokaryotes
Small (30S) = 16S + proteins; large (50S) = 5S + 23S + proteins; total 70S
12. Comprehend how alternative RNA splicing allows an organism to produce more proteins than the number of genes it possesses.
a. Alternative RNA splicing allows for the selection of certain exons in hnRNA for inclusion in mature mRNA such that one gene may encode different proteins.
b. Enhancer elements in the particular type of cell help dictate which splicing elements are used to eventually comprise the final transcript.
13. Understand how RNA levels and ultimately gene expression can be modulated by mechanisms such as RNA turnover, RNA export, ribo-switches and non-sense-mediated decay.
a. RNA turnover – process by which mRNA in cytoplasm is degraded by removal of poly-a-tail, hydrolysis of the cap, and degradation by 5’ exonuclease. Helps control amount of protein produced, so that once mRNA is created, cell doesn’t just keep churning out protein.
b. RNA export – regulating the export of RNA into the cytoplasm can regulate how much protein is translated, as translation occurs exclusively in the cytoplasm.
c. Ribo-switches – when part of an mRNA molecule binds a small target molecule which regulates the gene’s activity. Helps control the rate at which additional mRNA are produced à regulate protein levels.
d. Non-sense-mediated decay – cellular mechanism of mRNA surveillance that serves to detenct nonsense mutations and prevent the expression of truncated/non-functional protein.
14. Describe the functional roles of non-protein coding RNAs such as miRNAs, snRNAs, and Xist in gene regulation.
a. miRNAs in conjunction with RNA induced silencing complex (RISC) serve to inhibit translation or cleave mRNA.
b. snRNAs (small nuclear) are transcribed by RNAP II or RNAP III, and serve to assist with hnRNA splicing.
c. Xist is RNA gene on X-chromosome that serves to inactivate one of the X-chromosomes in each cell. It is expressed on the inactive X, but not on the active one.
Translation
1. Define translation.
a. Protein synthesis is the process of anabolic metabolism (simple substances converted into more complex compounds) that forms new proteins from amino acids. Processed mRNA from the nucleus is used to synthesize these proteins.
2. Interpret the codon table.
a. Need three nucleotide codon because there are four nucleotides and 20 amino acids.
b.
3. Apply the genetic code to predict the amino acid sequence of a protein based on the mRNA sequence.
4. Explain the significance of the wobble base. Understand wobble base-pairing rules in the antiparallel binding of aminoacyl-tRNA to mRNA. Explain why the genetic code is considered to be degenerate but unambiguous.
a. WOBBLE hypothesis says that one tRNA can base pair with more than one codon. Attempts to synthesize the apparent discrepancy that there are 32 prokaryotic tRNAs, 20 amino acids, and 61 different codons. tRNA has an extra base – inosine that can pair with C, A, and U. tRNAs designed so that they only have to pair with the first two nucleotides in a codon, and the third one has some flexibility. Wobble position is 3rd on mRNA, 1st on tRNA.
b. The first base of the anticodon indicates whether the tRNA can read one, two, or three different codons:
i. Anticodons beginning with A or C à read only one codon
ii. Anticodons beginning with G or U à read two codons
iii. Anticodons beginning with I à read three codons
c. Degenerate – more than one codon can code for one amino acid, but unambiguous because a given codon only codes for one amino acid.
5. Utilize the codon table to predict the result that a mutation at the DNA level would have on the resulting protein. Distinguish between silent, missense and nonsense point mutations and frame-shift mutations due to insertions or deletions.
a. Silent mutations are changes in single amino acids that do not change the eventual amino acid encoded for.
b. Missense mutations are changes in amino acids that change the amino acid encoded for.
c. Nonsense mutations result in a change of an amino acid to a stop codon.
d. Insertion of one or more nucleotides causes a frameshift in the 3 nucleotide continuous coding, so that most (if not all) amino acids downstream will be affected.
e. Deletions of one or more nucleotides also cause a frameshift, leading to a change in the downstream amino acids.
6. Recognize the cellular and molecular components required for protein synthesis and recall their subcellular localization
a. mRNA – cytoplasm
b. tRNA – cytoplasm
c. ribosomes – free floating in cytoplasm, or attached to ER
d. activated amino acids – attached to tRNA
e. release factor – to terminate translation
7. Describe the roles of tRNA, rRNA and mRNA in translation.
a. mRNA provide the template from which the polypeptide can be formed. Contain three nucleotide codon regions.
b. tRNA serve to recruit the appropriate amino acids to the forming polypeptide. Contain anti-codon region that complement codon region.
c. rRNA are part of the ribosome and provide mechanism for decoding mRNA into appropriate amino acids. Also interact with tRNA during translation by providing peptidyl transferase activity.
8. Illustrate/represent the process by which a tRNA is joined to the appropriate amino acid to form an aminoacyl-tRNA.
a. Amino acids are activated and attached to their corresponding tRNAs by highly specific enzymes known as aminoacyl-tRNA synthetases
b. Each aminoacyl-tRNA synthetase recognizes a particular amino acid and the tRNAs specific for that amino acid.
c. Amino acid first activated by reacting its carboxyl group with ATP to form aminoacyl AMP (unstable, even in synthetase) and pyrophosphate.
d. Then carboxyl group of amino acid is transferred to sugar on 2’ or 3’ end of tRNA resulting in an ester linkage.
e. SOURCES OF SPECIFICITY for synthetase:
i. Selection of the correct amino acid
1. Active-site pocket performs amino acid recognition
2. Hydrolytic editing at the second pocket (following tRNA binding)
ii. Selection of the correct tRNA
1. Anticodon recognition
2. tRNA nucleotide recognition
9. Acknowledge the role of aminoacyl-tRNA synthetases in maintaining the fidelity of protein synthesis by selecting the right amino acid and tRNA to join in an aminoacyl-tRNA and proofreading the accuracy of the aminoacyl-tRNAs formed.
10. Describe the structure, function and subcellular localization of ribosome.
a.
b. Composed of over 50 ribosomal proteins and several rRNA molecules.
c. Eukaryotic ribosomal subunits are assembled in the nucleolus after import of ribosomal proteins into the nucleus. After assembly the subunits are exported to the cytoplasm, where translation occurs.
d. The ribosome facilitates the specific coupling of tRNA anticodons to the codons on mRNA such that amino acids are added in the correct order to form a functional polypeptide.
e. The small subunit provides a framework on which the tRNAs can be accurately matched to the codons of the mRNA, while the large subunit catalyzes the formation of the peptide bonds that link the amino acids together into a polypeptide chain.
11. Summarize the process of translation (including initiation, elongation and termination) specifically listing the enzymes, protein factors and energy sources that are needed in each step.
a. Initiation
i. Formation of Pre-initiation complex
1. Entry complex: small ribosomal subunit + charged met-tRNA in P site + GTP + elF2 (elongation factor)
2. Distringuish mRNA from other RNA types
a. elF-4E recognizes 5’ cap
b. elF-4G recognizes poly A tail
3. RNA is loaded into the 40 S subunit
4. Scanning to identify the start codon (sets the reading frame) à this step requires energy à ATP hydrolysis.
ii. Formation of Initiation complex
1. The 60S subunit joins the complex along with elF5 à this step requires GTP Hydrolysis
2. Initiation factors and GDP are released.
3. 80S ribosome formed
b. Elongation
i. The charged tRNA for the next aa comes into the A site. An elongation factor (EF1) helps the selection and positioning. This step requires GTP hydrolysis.
ii. Peptidyl transferase catalyzes the formation of a peptide bond between the amino acid.
iii. Translocation: Hydrolysis of another GTP provides the energy that allows the large and small subunit undergo a conformational change that catalyzes peptide bond formation and moves the tRNAs into the E and P sites.
iv. the tRNA that is moved to the E site dissociates and the process begins again.
v. Peptide Bond formation
1. Catalyzed by peptidyl Transferase (ribozyme (RNA enzyme) associated with the large ribosomal subunit).
2. Peptidyle transferase mechanism is catalyzed by the N3 of an Adenine from the 23S rRNA.
c. Termination
i. When a ribosome reaches a stop codon on MRNA, the A site of the ribosome accepts a protein called a release factor instead of tRNA.
ii. The release factor hydrolyzes the bond between the tRNA in the P site and the last amino acid in the polypeptide chain. à polypeptide is freed
iii. Two ribosomal subunits and other components of assembly disassociate.
iv. This process requires GTP hydrolysis.
v. Release factor is a protein that looks and acts like tRNA.
12. Recognize the different post-translational modifications of proteins (including both addition of chemical groups and proteolytic cleavage)
a. Before or after folding, the new polypeptide may undergo enzymatic processing, including:
i. Removal of one or more aa
ii. Addition of acetyl, phosphoryl, methyl, coboxyl, or tother groups to certain aa residues
iii. Proteolytic cleavage (like in collagen or insulin)
iv. Attachment of oligosaccharides or prosthetic groups.
13. Enumerate the differences between prokaryotic and eukaryotic protein translation and relate those differences to the mechanism of action of the various drugs that inhibit protein synthesis in bacteria.
a. Translation in prokaryotes occurs while transcription is still ongoing (as there is no nuclear envelope), which in eukaryotes mRNA is processed in nucleus and transported to cytoplasm before translation can occur. Also, the ribosomal subunits in prokaryotes and eukaryotes are different, so antibiotics may be specific for one type and not the other.
b. Streptomycin changes shape of 30S rRNA and causes mRNA to be read incorrectly
c. Chloramphenicol binds to 50S rRNA and inhibits formation of peptide bond
d. Erythromycin binds to 50S rRNA and prevents movement along mRNA
e. Prokaryotes
i. Tetracycline: binds the 30S ribosomal subunit and blocks binding of aminoacyl tRNA to the A site.
ii. Streptomycine: prevents the transition from translation initiation to chain elongation. Also causes misreading of the mRNA (miscoding)
iii. Chloramphenicol: blocks the peptidyl transferase reaction on ribosomes by binding to the 50S subunit. Same effect on mitochondria.
iv. Erythromycin: binds in the exit channel of ribosomes thereby blocking translocation and inhibits elongation of the peptide chain.
v. Linezolid: Blocks the formation of the 70S complex
f. Eukaryotes
i. Cycloheximide: blocks the translocation reaction on ribosomes by inhibiting eukaryotic peptidyl-tRNA on the 80S ribosomal subunit
ii. Ricin: Inactivates 28S ribosomal RNA
iii. Abrin: Inactivates the large ribosomal subunit.
g. Both
i. Puromycin: causes the premature release of nascent polypeptide chains by its addition to the growing chain end (mimicks the action of an aminoacyl-tRNA.