Presentation on theme: "Welcome Back!! The SL material you learned last year is important to review before each unit this year… Today’s Opener: Draw and label a simple diagram."— Presentation transcript:
1Welcome Back!!The SL material you learned last year is important to review before each unit this year…Today’s Opener:Draw and label a simple diagram of the molecular structure of DNA. (4 marks)
3Happy Monday! 8/23/2010 The structure of the DNA double helix was described by Watson and Crick in Explain the structure of the DNA double helix, including its subunits and the way in which they are bonded together. (Total 8 marks)
4subunits are nucleotides; one base, one deoxyribose and one phosphate in each nucleotide;description / diagram showing base linked to deoxyribose C1 and phosphate to C5;four different bases – adenine, cytosine, guanine and thymine;nucleotides linked up with sugar-phosphate bonds;covalent / phosphodiester bonds;two strands (of nucleotides) linked together;base to base;A to T and G to C;hydrogen bonds between bases;antiparallel strands;double helix drawn or described;Accept any of the points above if clearly explained in a diagram.
77.1.1 Describe the structure of DNA, including the antiparallel strands3¢–5¢ linkagesH bonding between purines (A, G) and pyrimidines (C, T)A/T 2 H bondsG/C 3 H bondsMajor and minor grooves, direction of the “twist”, alternative B and Z forms, and details of the dimensions are not required.
87.1.2 Outline the structure of nucleosomes. Nucleosome = DNA wrapped around 8 histone proteins; held together by another histone protein
9A Human chromosome can be 4 cm long! 7.1.3 State that nucleosomes help to supercoil chromosomes and help to regulate transcription.A Human chromosome can be 4 cm long!DNA wraps twice around 8 core histones(DNA - , histone +)No transcription when packaged so tightly—regulation!
107.1.4 Distinguish between unique or single-copy genes & highly repetitive sequences in nuclear DNA. Highly repetitive sequences constitutes 5–45% of the genomesequences typically base pairs per repeat, and may be duplicated as many as 105 times per genome“satellite DNA” = clustered regions of repeatscentromeresMost is dispersed throughout genomeProbably don’t codeTOK: once classified as “junk DNA”, showing a degree of confidence that it had no role; research has been sparse. This addresses the question: To what extent do the labels and categories used in the pursuit of knowledge affect the knowledge we obtain?Transposable elements...can move around w/in the genome (McClintock, 1950)
117.1.5 State that eukaryotic genes can contain exons and introns. Less than 2% of human genome is genes that code for proteinsHGProject, mid-70s – 2001Introns, exons (expressed)
19Happy Thursday!!!! 9/15Living organisms use DNA as their genetic material. Explain how DNA is replicated within the cells of living organisms.(Total 8 marks)
20helix is unwound;two strands are separated;helicase (is the enzyme that unwinds the helix separating the two strands);by breaking hydrogen bonds between bases;new strands formed on each of the two single strands;nucleotides added to form new strands;complementary base pairing;A to T and G to C;DNA polymerase forms the new complementary strands;replication is semi-conservative;each of the DNA molecules formed has one old and one new strand;
227.2.1 State that DNA replication occurs in a 5¢ ® 3¢ direction. 5¢ end of free DNA nucleotide is added to the 3¢ end of chain of nucleotides that is already synthesized.
23Origin of replication (bubble; prok has one origin, euk has several) Explain the process of DNA replication in prokaryotes, including the role of enzymesOrigin of replication (bubble; prok has one origin, euk has several)Helicase:uncoils the double helixRNA primase:@ replication fork; short sequence of RNA (5-10 n’tides)DNA polymerase III:adds DNA n’tides in 5’ to 3’ directionN’tide is actually a deoxynucleoside triphosphate (dNTP)2 P groups lost during bonding energy for bondingDNA polymerase I:removes primer from 5’ end and replaces it with DNA n’tides
24ANTI-PARALLEL...OKAZAKI FRAGMENTS! Explain the process of DNA replication in prokaryotes, including the role of enzymesANTI-PARALLEL...OKAZAKI FRAGMENTS!Can only go 5’ to 3’ b/c DNA Pol IIILeading strandFast; toward rep fork in 5’ 3’Needs primase, primer, DNA pol III only onceLagging strandsSlower; Fragments away from rep fork, 5’ 3’Each fragment needs primase, primer, DNA pol IIIDNA ligase attaches S-P backbones of Okazaki fragments to make 1 strand
26IB Book, fig 7.7 page 200 Examiner’s Hint Draw the figure from memory Annotate what’s happening at specific locations
277.2.3 State that DNA replication is initiated at many points in eukaryotic chromosomes. 1. replication begins at origin, strands separate b/c helicase breaks H bondsReplication fork at each end of bubble (DBL strand opens to expose 2 template strands)Bubble enlarges in both directions (bidirectional) & eventually fuse togetherMultiple bubbles replication faster for the BIG eukaryotic genomes
38Happy ….Explain the process of transcription in eukaryotes.(8)
39RNA polymerase controls transcription / is the enzyme used in transcription; DNA is unwound by RNA polymerase;DNA is split into two strands;mRNA is made by transcription;promoter region (by start of gene) causes RNA polymerase to bind;anti-sense / template strand of DNA is transcribed;direction of transcription is ;free nucleotide triphosphates used;complementary base pairing between template strand and RNA nucleotides / bases;Accept this marking point if illustrated using a diagramRNA contains uracil instead of thymine;terminator (sequence) stops RNA polymerase / transcription;mRNA is released / RNA polymerase released; 8 max
40TUUUUUUUESDAY, 8/31!!! List three of the other molecules, apart from mRNA, required for transcription.3 mks
41DNA;RNA polymerase;(ribose) nucleotides / ribonucleotides / RNA nucleotides;transcription factors;nucleoside / ribonucleoside triphosphates; 3 maxAny two of the following: A / C / G / U;
437.3.1 State that transcription is carried out in a 5¢ ® 3¢ direction. SIMILAR to replicationNo helicase!RNA polymerase separates DNA strandsbinds to promoter region of DNA5’ 3’The 5¢ end of the free RNA nucleotide is added to the 3¢ end of the RNA molecule that is already synthesized.
447.3.2 Distinguish between the sense & antisense strands of DNA. sense strand = coding strand, carries the genetic codehas the same base sequence as mRNA with uracil instead of thymineantisense strand = template strandis transcribed, has the same base sequence as the tRNAPromoter region determines which is antisenseIs always the same strand for a particular geneCan differ for different genes
45Front of RNA polym unwinds the DNA helix Explain the process of transcription in prokaryotes, including the role of the promoter region, RNA polymerase, nucleoside triphosphates terminator.Front of RNA polym unwinds the DNA helixAdds nucleoside triphosphates to produce mRNA (2 P groups released), nucleotides added to 3’ end of growing strandAnti-sense is templateBack of RNA polym rewinds DNA strandsTerminator: DNA sequence, transcribed, causes RNA polym to detach & transcription stops
55THURSDAY, 9/2The information needed to make polypeptides is carried in the mRNA from the nucleus to the ribosomes of eukaryotic cells. This information is decoded during translation. The diagram below represents the process of translation.State the name of the next amino acid which will attach to the polypeptide. (1)Explain how the amino acid was attached to the tRNA. (3)
56Alanine / Ala 1an activating enzyme attaches amino acid to the tRNA;specific enzyme for specific tRNA;recognizes tRNA by its shape / chemical properties;energy (ATP) is needed;amino acid attached at end;amino acid attached at CCA; 3 max
57Happy Thursday! 9/2/2010 Explain the process of translation. (Total 9 marks)
58consists of initiation, elongation and termination; mRNA translated in a 5' to 3' direction;binding of ribosome to mRNA;small sub-unit then large;first / initiator tRNA binds to start codon / to small subunit of ribosome;AUG is the start codon;second tRNA binds to ribosome;large subunit moves down mRNA after a second tRNA binds;amino acid / polypeptide on first tRNA is transferred / bonded to amino acid on second tRNA;peptide bonds between amino acids / peptidyl transferase; requires GTP;movement of ribosome / small subunit of ribosome down the mRNA;loss of tRNA and new tRNA binds;reach a stop codon / termination;polypeptide released;tRNA activating enzymes link correct amino acid to each tRNA;(activated) tRNA has an anticodon and the corresponding amino acid attached;
60H bonds form in 4 areas, create “clover” Explain that each tRNA molecule is recognized by a tRNA-activating enzyme that binds a specific amino acid to the tRNA, using ATP for energy.3’ end is free, has CCASite of a.a. attachmentH bonds form in 4 areas, create “clover”1 loop of clover has anticodon(unique to each tRNA)Each a.a. has a specific tRNA-activating enzyme (aminoacyl-tRNA synthetase; 20 of them)Active site fits only 1 a.a. & its tRNA, rxn requires ATP“activated” amino acid, tRNA takes it to ribosomeThe shape of tRNA and CCA at the 3’ end should be included.
61mRNA binding site – in cavity b/w 2 subunits Outline the structure of ribosomes, including protein and RNA composition, large and small subunits, three tRNA binding sites, mRNA binding sites.Lg & sm subunitsrRNA (2/3 of its mass) and many RNA proteinsProkaryotic – smaller than eukaryotic3 tRNA binding sitesA site: holds tRNA carrying the next amino acid to be added to the polypeptide chainP site: holds the tRNA carrying the growing polypeptide chainE site: site from which tRNA that has lost its amino acid is dischargedmRNA binding site – in cavity b/w 2 subunits
637.4.3 State that translation consists of initiation, elongation, translocation, termination. Start codon AUG on 5’ end of all mRNAActivated amino acid (methionine + tRNA w/anticodon UAC) attaches to mRNA & small ribosomal subunitSmall subunit travels down mRNA to start codon (AUG)Starts translationH bonds b/w initiator tRNA & start codonLarge ribosomal subunit attachesElongationTranslocationTermination
647.4.3 State that translation consists of initiation, elongation, translocation, termination. tRNAs bring amino acids to mRNA-ribosomal complex in order specified by codons on mRNAElongation factors (ptns) bind tRNAs to exposed A sitetRNA P site nextRibosomes catalyze peptide bonds forming b/w amino acids (condensation)TranslocationTermination
657.4.3 State that translation consists of initiation, elongation, translocation, termination. During elongation phasetRNAs move down mRNABinds with A siteIts amino acid bonds to polypeptide (attached to A site)Moves to P siteTransfers polypeptide to new tRNA in A siteEmpty tRNA to E site, released5’ to 3’ direction (ribosome moves along mRNA TOWARD 3’ end—start codon was on 5’ end)Termination
667.4.3 State that translation consists of initiation, elongation, translocation, termination. 1 of 3 stop A siteRelease factor (protein) fills A site (doesn’t have amino acid)Catalyzes hydrolysis of bond linking P site’s tRNA to polypeptideAll released
677.4.4 State that translation occurs in a 5¢ ® 3¢ direction. During translation, the ribosome moves along the mRNA towards the 3¢ end. The start codon is nearer to the 5¢ end.
687.4.5 Draw and label the structure of a peptide bond between two amino acids.
69Explain the process of translation, including ribosomes, polysomes, start codons and stop codons.
70Polysomes: string of ribosomes all translating the same mRNA common
71State that free ribosomes synthesize proteins for use primarily within the cell, and that bound ribosomes synthesize proteins primarily for secretion or for lysosomes.“bound” to what???ER!