Structure & functions of RNA

1. Deepa John Harini Chandra Affiliations Structure & functions of RNA Ribonucleic acid (RNA) is a long polymer of nucleic acid monomers that is structurally similar to DNA but has a vast array of diverse functions, the most important being its central role in protein synthesis.

2. Master Layout (Part 1) 1 This animation consists of 3 parts: Part 1 – Structure of RNA Part 2 – Different classes of RNA Part 3 – Functions of different classes RNA in protein synthesis 2 Sugar phosphate backbone Nucleoside 3 Nitrogenous bases – Adenine/ Guanine/ Uracil/ Cytosine Base Ribose sugar 4 P Base Phosphate Ribose sugar Nucleotide 5

3. Definitions of the components:(Part 1 – Structure of RNA) 1 1. Nucleoside:A base bound to a sugar, either ribose or deoxyribose, by means of a b-glycosidic linkage. 2. Nucleotide: The subunit or chain link in DNA or RNA composed of a sugar, a base and at least one phosphate group. It is more specifically known as ribonucleotide in RNA. 3. Ribonucleic acid (RNA): A polymer composed of ribonucleotides, linked together by phosphodiester bonds. 4. Ribose sugar: A monosaccharide, aldopentose sugar that is abundant in nature in its D isomeric form and is the sugar component of RNA. 5. Pyrimidine: An organic compound similar to benzene and pyridine that is composed of a heterocyclic, aromatic six-member ring having nitrogen atoms at positions 1 and 3. The nitrogenous bases found in RNA, cytosine and uracil, are derivatives of pyrimidine. 6. Purine: These are the most abundant nitrogen-containing heterocyclic compounds in nature and are composed of a pyrimidine ring fused with an imidazole ring. Derivatives of this aromatic compound occur in both DNA & RNA in the form of adenine and guanine. 2 3 4 5

4. Definitions of the components:(Part 1 – Structure of RNA) 1 7. Adenine (A): This is a purine nucleobase found in both DNA and RNA that pairs with thymine (T) through two hydrogen bonds in the double stranded RNA (dsRNA) structure. In addition to being a component of genetic material, it is also essential for synthesis of various cofactors in the body. 8. Guanine (G):This is another purine nucleobase found in both DNA and RNA. This planar, bicyclic molecule base pairs with cytosine in dsRNA. 9. Cytosine (C): This pyrimidine derivative found to be a component of both DNA and RNA base pairs with guanine in the dsRNA structure. 10. Uracil: This pyrimidine derivative found to be a component of only RNA base pairs with adenine in the dsRNA structure. 2 3 4 5

5. Part 1,Step 1: 1 Phosphodiester bond Polynucleotide chain 2 Base Nucleoside Nucleotide 3 3’ 5’ P P P P P Ribose sugar Base Base Base Base 5’ 5’ 5’ 5’ 3’ 3’ 3’ 3’ 4 Action Description of the action Audio Narration First show the pink pentagon below with label, followed by blue ‘base’ being attached .The curly bracket must appear with the label ‘nucleoside’. Then the ‘pink circle as depicted in the animation must appear followed by the next curly bracket and label ‘nucleotide’. The thick black line must then appear with label followed immediately by the next unit and so on. As shown in animation. RNA is made up of three basic components – a sugar, a nitrogenous base and a phosphate group. The sugar and base are linked to form a nucleoside and attachment of the phosphate group results in a nucleotide. Many such nucleotide units are linked together by means of a covalent bond known as the phosphodiester bond. This is formed between the 3’ carbon of one sugar and 5' carbon of the next sugar via a phosphate group to give rise to a polynucleotide chain. 5

6. Part 1, Step 2: 1 Purines 2 3 Pyrimidines 4 Action Description of the action Audio Narration Show the structures above with their labels and numbering as depicted. (Please use black background & redraw all figures) Make the four structures appear one at a time as depicted with their labels and numbering around the structure. RNA is composed of four different nitrogenous bases that are derivatives of the heterocyclic, aromatic compounds, purines and pyrimidines. Adenine and guanine are purines while uracil and cytosine are the pyrimidines. 5 Source: www.mun.ca/biochem/courses/3107/.../bases_and_chains.html, www.bio.miami.edu

7. Part 1, Step 3: 1 5’ - U U G G U G G A G U C U G C A A C U G A C U C C A U U G C A - 3' Single stranded RNA A single stranded RNA molecule may fold back on itself to form a secondary structure (stem and loop) Single stranded RNA tends to assume right handed helical confirmation 2 3 Bases are shown in gray. Phosphate atoms in yellow and ribose in green. 4 RNA secondary structure Action Description of the action Audio Narration RNA exists mainly as a single-stranded molecule. The base stacking interactions often tend to make the RNA assume a right-handed helical conformation. Single stranded RNA also forms secondary structures by folding back on itself resulting in formation of loops and hairpins due to base pairing interactions. Functional RNA molecules often require a specific tertiary structure, the scaffold of which is provided by the secondary structure. These RNA due to their large negative charge are stabilized by metal ions. (Please use black background & redraw all figures) First show the linear sequence of alphabets on top. Next show the left arrow, the respective text followed by the winding of the sequence of top around a vertical axis to give the figure on the left bottom. Then show the right arrow and respective text followed by the bending of the sequence to give the right bottom figure with the alphabets in red next to each other as shown. Show all the forms of RNA 5 Source: Biochemistry by Lehninger, 4th edition (ebook), Biochemistry by Stryer,5th edition(ebook)

8. Master Layout (Part 2) 1 This animation consists of 3 parts: Part 1 – Structure of RNA Part 2 – Different classes of RNA Part 3 – Functions of different classes RNA in protein synthesis Acceptor stem Large subunit 2 TyC loop D loop Small subunit Ribosomal RNA (rRNA) Variable loop 3 Anticodon loop 5’ 3’ Transfer RNA (tRNA) Messenger RNA (mRNA) 4 Coding sequence Cap 5‘UTR Stop Poly A tail 3‘UTR Start 5

9. Definitions of the components:Part 2 – Different classes of RNA 1 1. mRNA: The messenger RNA is a long sequence of nucleotides that serves as a template for protein synthesis. It is transcribed from a DNA template by RNA Polymerase and gets translated into the amino acid sequence of the corresponding protein. Eukaryotic mRNA requires extensive processing to form the mature mRNA while prokaryotic mRNA does not. Typical mRNA structure is composed of the following regions: a) Cap: An altered nucleotide consisting of a methylated guanosine residue bound through 5’ - 5’ triphosphate linkage to the first transcribed nucleotide of the mRNA. Capping takes place only in eukaryotes and is a vital process to produce mature mRNA and provide a recognition site for binding of ribosomes. b) 5’ UTR: An untranslated region in mRNA that is present before the start codon and plays an important role in providing mRNA stability, facilitating mRNA localization and providing a ribosome binding site. c) Start site: It is the site from where translation is initiated. The codon at this site is usually AUG coding for methionine. d) Coding sequence: These are the regions in mRNA that encode specific amino acid sequences for the process of translation into a polypeptide chain. It begins with a start codon and ends with a stop codon. e) Stop site: The site where translation is terminated. The codon here is usually UAA, UAG and UGA. These do not code for any amino acids. 2 3 4 5

10. Definitions of the components:Part 2 – Different classes of RNA 1 f) 3’ UTR: An untranslated region in mRNA that is present after the stop codon. It has several roles including mRNA stability and localization. g) Poly A(poly adenylic acid) tail: A sequence of about two hundred adenine residues added to the end of eukaryotic mRNA that plays an important role in nuclear export, translation and providing stability to the mRNA. 2. tRNA- A relatively small RNA molecule involved in protein synthesis that binds an amino acid at one end and base pairs with an mRNA codon at the other, thus serving as an adaptor that translates an mRNA code into a sequence of amino acids. A tRNA molecule consists of the following components: a) Acceptor stem: This is a stem of around 7 base pairs that is formed by base pairing of the two ends of tRNA, the 5'-terminal nucleotide with the 3'-terminal nucleotide. b) TyC loop:Thisis a 5 base pair stem containing the sequence TΨC in which the Ψ stands for a modified nucleotide, pseudouridine. Pseudouridine is similar to normal uridine except that the base is linked to the ribose through 5th carbon of the base instead of the nitrogen-1. c) Variable loop: The region between the anticodon loop and T loop which is so called because it varies in length from 13 to 14 nucleotides. 2 3 4 5

11. Definitions of the components:Part 2 – Different classes of RNA 1 d) Anticodon loop: The loop, conventionally drawn at the bottom of tRNA molecule, that contains a 3 base sequence that base pairs with a specific codon of mRNA. e) D-loop: It is dihydrouracil loop made up of a 4 base pair stem. It is so named because of the modified uracil base that the region contains. 3. rRNA- rRNA forms the central component of ribosomes. It has both catalytic and structural roles in protein synthesis. The ribosome that houses this rRNA consists of a large subunit and a small subunit. 2 3 4 5

12. Part 2, Step 1 1 Mature mRNA structure 3‘UTR Poly A tail Start Coding sequence 5’ Cap 5‘UTR 3’ Stop 5’ 2 AUG UAG/UAA/UGA AAAAAAAAAAA 3 4 Action Description of the action Audio Narration Messenger RNA is formed from a DNA template by transcription. This mRNA is often referred to as the pre-mRNA in eukaryotes since it undergoes further processing to form a mature mRNA. A fully processed eukaryotic mRNA includes a 5’ cap, where the nucleotide at the 5’ end is modified by addition of 7-methyl guanosine and a poly A tail at the 3’ end which serves to protect the mRNA from degradation by exonucleases. The mRNA also contains 5’ and 3’ UTRs that contain signal sequences and serve as binding sites for various proteins. The coding sequence is flanked by start and stop codons that define the beginning and end of the gene to be transcribed. (Please use black background & redraw all figures) Show the pale green ‘mRNA’ structure which is present in the background. Next, each region must be highlighted sequentially with the label for that particular region appearing. In places indicated above, the details of the structure of the highlighted region must be shown in a zoom box as depicted in the animation. As shown in animation. 5

13. Part 2, Step 2 1 v Dihydrouridine tRNA processing & structure Modified nucleotide bases in tRNA Acceptor stem RNAses 2 TyC loop D loop 3 Variable loop Anticodon loop Pseudouridine Tertiary structure of tRNA Mature tRNA Secondary structure of tRNA Primary transcript 4 Action Description of the action Audio Narration First show image on left and yellow segments being removed followed by folding to form image in the centre & then right. Longer RNA precursors are modified by enzymatic removal of nucleotides from the 5’ and 3’ ends to form the tRNA structure. Additional processing of the tRNA such as attachment of the 3’ CCA trinucleotide unit and modification of certain bases takes place in ceratin bacteria and almost all eukaryotes. All tRNAs have a common secondary structure represented by a clover leaf having four base-paired stems. The anticodon loop recognizes the corresponding mRNA codon while the acceptor stem adds the suitable amino acid to the growing polypeptide chain. (Please use black background & redraw all figures) First show the figure on the left followed by the colored ‘RNAses’. These must cut out the yellow regions of the figure which must then disappear. The green portion at the bottom must join with the hanging end on the other side to form a loop as shown in the centre panel and a pink region on the right top of the figure must be added. Two of the zoomed in structures must then be shown and finally, this structure must be shown to fold on itself and the figure on right must appear. 5 Source: Biochemistry by Lehninger, 3rd edition (ebook); www.sparknotes.com,molecular biology by Rober Weaver.

14. Part 2, Step 3 1 Pre-rRNA transcript (30S) Prokaryotic ribosome 5S 16S tRNA 23S Large subunit - 50S Methylation 2 Cleavage by specific RNAses & nucleases 3 Methylated sites on rRNA Small subunit - 30S 16S tRNA 23S 5S 4 Mature rRNAs Action Description of the action Audio Narration (Please use black background & redraw all figures) First show the two units on the left and then the larger unit moving down and joining the smaller one. Next the small unit must be zoomed into and the figure on top right must appear with labels. The small orange circle must then appear and traverse the length of the rectangle up to the end of ‘23S’. As it moves, small green stars must appear as shown in the middle figure. Next the pie-shaped objects must appear at the sites indicated and must cut through the rectangle. These blue portions must then be removed and the bottom figure must appear. rRNA is the central component of the ribosome involved in protein synthesis in all living cells. Prokaryotic 70S ribosome is composed of 50S and 30S subunits where S is a measure of the rate of sedimentation of the respective components in a centrifuge. rRNAs are derived from longer precursors called pre-rRNA. A single 30S rRNA precursor is processed by several enzymes to give rise to 16S, 23S and 5S rRNAs in bacteria. As shown in animation. 5 Source: www.ncbi.nlm.nih.gov

15. Part 2, Step 4 1 Pre-rRNA transcript (45S) Eukaryotic ribosome 28S 18S 5.8S Large subunit - 60S Methylation 2 Cleavage by small nucleolar RNAs 3 Methylated sites on rRNA Small subunit - 40S 28S 18S 5.8S 4 Mature rRNAs Audio Narration Description of the action Action (Please use black background & redraw all figures) First show the two units on the left and then the larger unit moving down and joining the smaller one. Next the small unit must be zoomed into and the figure on top right must appear with labels. The small orange circle must then appear and traverse the length of the rectangle up to the end of ‘23S’. As it moves, small green stars must appear as shown in the middle figure. Next the pie-shaped objects must appear at the sites indicated and must cut through the rectangle. These blue portions must then be removed and the bottom figure must appear. Eukaryotic 80S ribosome is composed of 60S and 40S subunits where S is a measure of the rate of sedimentation of the respective components in a centrifuge. In eukaryotic vertebrates, a single 45S rRNA precursor is processed by several enzymes to give rise to 18S, 5.8S and 28S rRNAs. As shown in animation. 5 Source: www.ncbi.nlm.nih.gov

16. Master Layout (Part 3) 1 This animation consists of 3 parts: Part 1 – Structure of RNA Part 2 – Different classes of RNA Part 3 – Functions of different classes of RNA in protein synthesis 2 Growing polypeptide chain Amino acid Ribosome 3 Outgoing tRNAs Incoming aminoacyl-tRNAs 3’ 5’ Start codon Stop codon A site P site 4 mRNA Movement of ribosome 5

17. Definitions of the components:Part 3 – Functions of different classes of RNA in protein synthesis 1 1. Translation:A process by which the mRNA sequence is read in the form of three letter codes known as codons to incorporate the corresponding amino acids in the growing polypeptide chain with the active involvement of rRNA, tRNA and several other enzymes. 2. mRNA: The messenger RNA is the intermediate between DNA and the protein which encodes the required “blueprint” of the protein product. The RNA obtained from DNA immediately after transcription is known as the pre-mRNA and is made up of both coding (exons) and non-coding (introns) regions. This mRNA is further processed to give the mature mRNA which contains coding and other essential sequences for protein synthesis. 3. Ribosome: An RNA- protein particle that is involved in translation of the mRNA into protein. Prokaryotic 70S ribosome is composed of 30S and 50S subunits, while eukaryotic 80S ribosome is composed of 40S and 60S subunits, where S is a measure of the rate of sedimentation of the respective components in a centrifuge (Svedberg). 4. Incoming aminoacyl tRNAs: The tRNA carrying the amino acid specified by the next subsequent codon on the mRNA chain. 5. Outgoing tRNAs: The uncharged tRNA exiting from the ribosome after transferring the amino acid to the growing polypeptide chain attached to the tRNA in the P site. 2 3 4 5

18. Definitions of the components:Part 3 – Functions of different classes of RNA in protein synthesis 1 6. Growing polypeptide chain: A single protein chain that progressively increases in length during translation until it reaches the stop codon. It is composed of specific sequence of amino acids linked together by peptide bonds. 7. Start codon: The position at which initiation of translation takes place and most often contains the codon AUG that codes for methionine. 8. Stop codon: The site that codes for the termination of translation and consists of one of the three codons - UAA, UAG or UGA. 9. A site: The site on the ribosome to which the charged, incoming amino acyl-tRNA (except the first one) binds. 10. P site: The site on ribosome to which the peptidyl-tRNA is bound at the time that a new amino acyl-tRNA enters the ribosomal A site. 2 3 4 5

19. Part 3, Step 1 1 Initiation Large ribosomal subunit 2 UAC 3’ Stop codon AUG Start codon mRNA 5’ 3 Initiator fMet tRNA P site A site Small ribosomal subunit 4 Description of the action Action Audio Narration Initiation of protein synthesis is carried out by binding of the mRNA to the small ribosomal subunit such that its initiation codon, most often an AUG sequence, is aligned at the P site. The initiator tRNA that carried a modified methionine amino acid on its acceptor stem then binds to the ribosomal subunit by means of codon-anticodon interactions. The large subunit is then assembled on top of this to form the initiation complex. Other initiation factors are also involved which ensure correct positioning of all the components. (Please use black background )First show the small unit at the bottom followed by binding of the blue strand to this pink unit. The marked AUG sequence must fall in the ‘P site’. Next the blue figure on top which must move down and bind to the ‘AUG’ below. Then the large pink subunit must be assembled on top of these. As shown in animation. 5

20. Part 3, Step 2 1 Elongation Incoming aminoacyl tRNAs 2 Peptide bond formation 3’ UAC Stop codon AUG Start codon Large ribosomal subunit mRNA 5’ 3 Initiator fMet tRNA P site A site Small ribosomal subunit 4 Action Description of the action Audio Narration The next incoming aminoacyl tRNA carrying the amino acid corresponding to the next codon occupies the A site. A peptide bond is then formed between the amino acid in the A site and the P site with the P site amino acid beingtransferred to the A site. The unbound tRNA then leaves the P site and is moved to the exit or E site briefly before being removed. (Please use black background ) The colored figures on top must be shown to enter the frame and then the orange figure must move down and bind to the ‘A site’. Once this happens, the blue circle in the ‘P site’ must move and bind on top of the orange circle in the ‘A site’. Next the blue figure must be moved out of the P site as depicted in animation. As shown in animation. 5

21. Part 3, Step 3 1 Elongation Incoming aminoacyl tRNAs 2 Peptide bond formation 3’ UAC Stop codon AUG Start codon Large ribosomal subunit mRNA 5’ 3 Initiator fMet tRNA P site A site Small ribosomal subunit 4 Action Description of the action Audio Narration (Please use black background ) The pink subunits must move a little bit to the right such that the orange figure with blue circle on top now occupies the ‘P site’. Then the next green figure on top must enter the ‘A site’ and the same sequence of events mentioned in step 2 must be repeated such that the orange and blue circles are transferred on top of the green circle and the empty orange figure leaves the figure. Then again the pink subunits must move and so on until many colored circles have been added. Once the peptide bond has been formed, the ribosome moves one codon towards the 3’ end of the mRNA such that the tRNA in the A site now occupies the P site and the A site is again free for the next incoming aminoacyl tRNA. Multiple such rounds of elongation followed by translocation of the tRNAs are carried out to form the growing polypeptide chain. As shown in animation. 5

22. Part 3, Step 4 1 Termination Polypeptide chain released Dissociation of all components 2 3’ Stop codon AUG Start codon mRNA Large ribosomal subunit 5’ Release factor 3 P site A site Small ribosomal subunit 4 Action Description of the action Audio Narration When the ribosome encounters the termination sequence, typically UAA, UAG, UGA, a release factor binds to the vacant A site and the polypeptide chain is hydrolyzed and released. Other termination factors also aid this process. Once synthesis is complete, the ribosomal subunits dissociate from each other and all components are separated until commencement of the next round of translation. (Please use black background ) Once several circles have been added and the pink subunits reach the ‘stop’ signal on the blue chain, an oval shape must enter the ‘A site’ with the label ‘release factor’. Once this happens, the chain of circles must be dissociated from the last colored ‘tRNA’ figure (red rectangular figure above). After this, all the other components i.e. the pink units and the blue chain must also move apart to show that they are dissociating as depicted in the animation. As shown in animation. 5

23. Interactivity option 1:Step No:1 1 What inference can be drawn from this experiment? A) Subunit exchange between ribosomes does not occur after each round of translation. 2 B) Subunit exchange between ribosomes occurs after each round of translation. C) One of the two subunits had degraded 3 D) Only one of the two subunits dissociates. 4 Results Interacativity Type Options Boundary/limits (Please use black background ) User must be allowed to choose one of the four options after the experiment displayed in the next slide is shown. B is the correct answer,. If user chooses B, it must turn green with the remark ‘correct answer’. If user chooses any of the remaining options, it must turn red with the remark ‘incorrect answer’. Choose the correct option 5

24. Interactivity option 1:Step No:2 1 The following experiment was performed…. After 3.5 generations, isolate the riobosmes and analyze by sucrose density gradient centrifugation with 14C labeled riobosomes for comparison. 2 Place cells with labeled heavy ribosomes in medium with ordinary light isotopes of N, C and H. Label ribosome by growing E. coli in presence of heavy isotopes like N, C and H and make them radioactive by including some 3H. Heavy ribosomes radioactively labeled(indicated by asterisk sign) 3 Labeled whole ribosome had a hybrid sedimentation coefficient in between the light and heavy ribosome sedimentation coefficient 4 Light ribosome Labeled Whole Ribosome (hybrid sedimentation coefficient) Heavy ribosome subunit + 5

25. Questionnaire 1 1. The form of genetic information used directly in protein synthesis is Answers: a) DNA b) mRNA c) rRNA d)‏ ribosomes 2. The process in which ribosome get engaged Answers: a) transcription b) translation c) replication d) cell division‏ 3. The bases of RNA are the same as those of DNA with the exception that RNA contains Answers: a) cysteine instead of cytosine b) uracil instead of thyminec) cytosine instead of guanine d)‏ uracil instead of adenine 4. The nucleotide sequences on DNA that actually have information encoding a sequence of amino acids are Answers: a) introns b) exons c) UAA d) UGA‏ 5. Which one of the following is not a type of RNA? Answers: a) rRNA b)nRNA c)mRNA d)‏ tRNA 2 3 4 5

26. Links for further reading Books: Genetics by Peter.J.Russell, 5th edition Molecular biology by Robert Weaver, 4th edition Biochemistry by Lubert Stryer et al., 6th edition (ebook) Molecular Biology of the Gene by James Watson et al., 5th edition