1 Nucleic Acids Structures of Nucleic Acids DNA Replication RNA and Transcription

2 Nucleotides Nucleic acids consist of nucleotides that have a sugar, nitrogen base, and phosphate nucleoside Sugar Base PO 4

3 Nitrogen-Containing Bases

4 Sugars

5 Nucleosides in DNA BaseSugarNucleoside Adenine (A)DeoxyriboseAdenosine Guanine (G)DeoxyriboseGuanosine Cytosine (C)DeoxyriboseCytidine Thymine (T)DeoxyriboseThymidine

6 Nucleosides in RNA BaseSugarNucleoside Adenine (A)riboseAdenosine Guanine (G)riboseGuanosine Cytosine (C)riboseCytidine Uracil (U)riboseUridine

7 Example of a Nucleoside

8 Nucleotides in DNA and RNA DNA dAMPDeoxyadenosine monophosphate dGMPDeoxyguanosine monophosphate dCMPDeoxycytidine monophosphate dTMPDeoxythymidine monophosphate RNA AMPadenosine monophosphate GMPguanosine monophosphate CMPcytidine monophosphate UMPuridine monophosphate

9 Structure of Nucleic Acids Polymers of four nucleotides Linked by alternating sugar-phosphate bonds RNA: ribose and A, G, C, U DNA: deoxyribose and A,G,C,T nucleotide nucleotide nucleotide nucleotide P sugar base P sugar base P sugar base P sugar base

10 Nucleic Acid Structure 3,5-phosphodiester bond 3 5

11 Double Helix of DNA DNA contains two strands of nucleotides H bonds hold the two strands in a double- helix structure A helix structure is like a spiral stair case Bases are always paired as A–T and G-C Thus the bases along one strand complement the bases along the other

12 Complementary Base Pairs Two H bonds for A-T Three H bonds for G-C

13 Double Helix of DNA

14 Learning Check NA1 Write the complementary base sequence for the matching strand in the following DNA section: -A-G-T-C-C-A-A-T-G-C-

15 Solution NA1 Write the complementary base sequence for the matching strand in the following DNA section: -A-G-T-C-C-A-A-T-G-C- -T-C-A-G-G-T-T-A-C-G-

16 DNA Replication DNA in the chromosomes replicates itself every cell division Maintains correct genetic information Two strands of DNA unwind Each strand acts like a template New bases pair with their complementary base Two double helixes form that are copies of original DNA

17 DNA Unwinds G--C A--T C--G T--A G-C A-T C-G T-A

18 DNA Copied with Base Pairs Two copies of original DNA strandG-CA-TC-G T-AG-A

Nucleic Acid Chemistry Where the info is…interpreting the blueprint

Central Dogma DNA  RNA  protein Replicati on transcription translation

Central Dogma Replication –DNA making a copy of itself Making a replica Transcription –DNA being made into RNA Still in nucleotide language Translation –RNA being made into protein Change to amino acid language

Replication Remember that DNA is self complementary Replication is semiconservative –One strand goes to next generation –Other is new Each strand is a template for the other –If one strand is 5’ AGCT 3’ –Other is: 3’ TCGA 5’

Replica Write the strand complementary to: 3’ ACTAGCCTAAGTCG 5’ Answer

Replication is Semiconservative

Replication Roles of enzymes –Topoisomerases –Helicase –DNA polymerases –ligase DNA binding proteins –DNA synthesis Leading strand Lagging strand

Replication

Helix opens –Helicase Causes supercoiling upstream –Topoisomerases (gyrase) DNA Binding Proteins –Prevent reannealing

Replication

Leading strand –3’ end of template –As opens up, DNA polymerase binds –Makes new DNA 5’ -  3’ Same direction as opening of helix Made continuously

Replication

Lagging strand –5’ end of template Can’t be made continuously as direction is wrong –RNA primer –New DNA made 5’  3’ Opposite direction of replication Discontinuous –Okazaki fragments Ligase closes gaps

Transcription DNA template made into RNA copy –Uracil instead of Thymine One DNA strand is template –Sense strand Other is just for replication –Antisense (not to be confused with nonsense!) In nucleus –nucleoli

Transcription From following DNA strand, determine RNA sequence 3’ GCCTAAGCTCA 5’ Answer

Transcription

DNA opens up –Enzymes? RNA polymerase binds –Which strand? –Using DNA template, makes RNA 5’-  3’ Raw transcript called hnRNA

Transcription How does RNA polymerase know where to start? upstream promotor sequences Pribnow Box TATA box RNA polymerase starts transcription X nucleotides downstream of TATA box

Introns and Exons Introns –Intervening sequences –Not all DNA codes for protein –Regulatory info, “junk DNA” Exons –Code for protein

Processing of hnRNA into mRNA 3 steps –Introns removed Self splicing –5’ methyl guanosine cap added –Poly A tail added Moved to cytosol for translation

Processing of hnRNA into mRNA

Translation RNA --  Protein –Change from nucleotide language to amino acid language On ribosomes Vectorial nature preserved –5’ end of mRNA becomes amino terminus of protein –Translation depends on genetic code

Genetic Code Nucleotides read in triplet “codons” –5’ -  3’ Each codon translates to an amino acid 64 possible codons –3 positions and 4 possiblities (AGCU) makes 4 3 or 64 possibilities –Degeneracy or redundancy of code Only 20 amino acids Implications for mutations

Genetic Code

Not everything translated AUG is start codon –Find the start codon Also are stop codons To determine aa sequence –Find start codon –Read in threes –Continue to stop codon

Translation Steps: –Find start codon (AUG) –After start codon, read codons, in threes –Use genetic code to translate Translate the following: GCAGUCAUGGGUAGGGAGGCAACCUGAACCGA C Answer

Translation Process Requires Ribosomes, rRNA, tRNA and, of course, mRNA –Ribosome Made of protein and rRNA 2 subunits Has internal sites for 2 transfer RNA molecules

Ribosome Left is cartoon diagram Right is actual picture

Transfer RNA Mostly double stranded –Folds back on itself Several loops –Anticodon loop Has complementary nucleotides to codons 3’ end where aa attach

Transfer RNA

Translation Initiation –Ribosomal subunits assemble on mRNA –rRNA aids in binding of mRNA Elongation –tRNAs with appropriate anticodon loops bind to complex – have aa attached (done by other enzymes) –Amino acids transfer form tRNA 2 to tRNA 1 –Process repeats Termination –tRNA with stop codon binds into ribosome –No aa attached to tRNA –Complex falls apart

Translation

Happening of process (circa 1971) O0iCLwwhttp:// O0iCLww

Mutations Changes in nucleotide sequence Can cause changes in aa sequence –Degeneracy in genetic code can prevent Two types –Point mutations Single nucleotide changes –Frame shift Insertions or deletions

Point Mutations Single nucleotide changes Old sequence AUG GGU AGG GAG GCA ACC UGA ACC GAC aa: G R E A T New sequence AUG GGU AGU GAG GCA ACC UGA ACC GAC aa: G S EAT

Point mutations Depending on change, may not change aa sequence Old sequence AUG GGU AGG GAG GCA ACC UGA ACC GAC aa: G R E A T New sequence AUG GGU AGA GAG GCA ACC UGA ACC GAC aa: G R EAT

Point Mutations Change could make little difference –If valine changed to leucine, both nonpolar Change could be huge, –Could erase start codon Old sequence AUG GGU AGG GAG GCA ACC UGA ACC GAC aa: G R E A T New sequence AUU GGU AGA GAG GCA ACC UGA ACC GAC aa: no start codon…protein not made

Point Mutations Other possibilities, –Stop codon inserted Truncated protein –Stop codon changed Extra long protein Bottom line, –Depends on what change is

Frame Shift mutations Insertions or deletions –Change the reading frame Insertion example Old sequence AUG GGU AGG GAG GCA ACC UGA ACC GAC aa: G R E A T New sequence AUG GGU AGG AGA GGC AAC CUG AAC CGA C aa: G R RGN L N R

Frame Shift Mutations Deletion example Old sequence AUG GGU AGG GAG GCA ACC UGA ACC GAC aa: G R E A T New sequence Delete second A (Underlined above) AUG GGU GGG AGG CAA CCU GAA CCG AC aa: G G RQP G P

Complementary DNA Strand Template: 3’ ACTAGCCTAAGTCG 5’ 5’ TGATCGGATTCAGC 3’ Back

RNA Transcript DNA 3’ GCCTAAGCTCA 5’ RNA 5’ CGGAUUCGAGU 3’ Back

Translation Answer Find start codon GCAGUCAUGGGUAGGGAGGCAACCUGAACCGAC Read in threes after that: AUG GGU AGG GAG GCA ACC UGA ACC GAC Using Genetic code AUG GGU AGG GAG GCA ACC UGA ACC GAC G R E A T stop After stop codon…rest is garbage Back