2 Topic Outline DNA Structure DNA Replication Transcription Translation ProteinsEnzymesHOME
3 Topic 6.1 - DNA Structure 6.1.1 Outline the structure of nucleosomes. A nucleosome is a basic unit of DNA packaging. It consists of 8 histones(histone is a kind of protein) with theDNA double helix wrapped around it.This "bead' is fastened onto the"string“of DNA by another histone.MAIN PAGE
4 6.1.2 State that only a small proportion of the DNA in the nucleus constitutes genesand that the majority of DNA consistsof repetitive sequences.Only a small proportion of the DNAin the nucleus constitutes genesand that the maority of DNA consists
5 6.1.3 Describe the structure of DNA including the antiparallel strands, 3'-5' linkages and hydrogen bonding between purines and pyrimidines.The sides of the ladder of DNA consist of alternating phosphate groups and deoxyribose (a sugar). The two sides are antiparrallel, meaning that the sugar and phosphates are running in opposite directions (one looks "upside down").
6 Each side has a 5' end and a 3' end. If a strand is structured from 3' to 5', thatmeans that the sugar-phosphate backboneruns from sugar to phosphate. Since thesides are antiparallel, one side goes in the3' to 5' direction, and the other goes in the5' to 3' direction. There are two types ofnucleotides, pyrimidines and purines.Pyrimidines hydrogen bond to purinesto create the rungs of the DNA ladder
7 Thymine and cytosine are pyrimidines, adenine and guanine are purines. Helpful hint:Thymine and cytosine are pyrimidines,adenine and guanine are purines.
8 Topic 6.2 - DNA Replication 6.2.1 State that DNA replcation occurs in a 5' to 3' direction.DNA replcation occurs in a 5' to 3' direction.MAIN PAGE
9 6.2.2 Explain the process of DNA replication in eukaryotes including the role of enzymes(helicase, DNA polymerase III, RNA primase,DNA polymerase I, and DNA ligase),Okazakifragments and deoxynucleoside triphosphates.The process of replication begins at specificnucleotide sequences called the originsof replication on the DNA strand.
10 It is at these points that helicase splits the DNA into its two antiparallel strands. Onthe strand running in the 5'--->3' direction,DNA polymerase III latches on at one endof the opening, called the replication bubble,and begins to continuously lay a new DNAstrand from free nucleotides in the nucleus.
11 As always, an exact copy of the now-detached strand is formed from this template due tobase-pairing rules. At the same time DNApolymerase III is laying new DNA, helicaseis continuing to split the strands, thus allowingreplication to continue uninterrupted. On theopposite strand running 3'--->5',replication is not so simple.
12 Because new strands have to be laid in the 5'---3' direction, DNA polymerase III cannotlay continuously as it can on the other strand.Instead, RNA primase lays short segments ofRNA primer nucleotides at many points alongthe strand. When one segment of primer comesin contact with another,DNA polymerase I attachesand replaces the primer with DNA.
13 These segments of DNA are called Okazaki fragments. Once these fragments have beenlaid, they are joined by yet another enzymeknown as DNA ligase, which attaches DNA intothe gaps between fragments and completesthe new strand. The 3'--->5' strand with Okazakifragments is called the lagging strand,while the leading strand is thecontinuously replicating one.
14 6.2.3 State that in eukaryotic chromosomes, replication is intiated at many points.In eukaryotic chromosomes, replicationis intiated at many points.
15 6.3.1 State that transcription is carried out in a 5' to 3' direction. Topic Transcription6.3.1 State that transcription is carried out in a 5' to 3' direction.Transcription is carried out in a 5' to 3' direction (from phosphate to sugar). The 5' end (phosphate) of the RNA nucleotide is added to the 3' end (sugar) of the part of the RNA molecule which has already been synthesized.MAIN PAGE
16 6.3.2 Outline the lac operon model as an example of the control of gene expression in prokaryote.E. Coli bacteria use three enzymes to break downlactose (milk sugar). These three genesare next to one operon (operon is the genes +operator + promoter). this operon is called thelac operon. A regulator gene, located outsideof the operon, codes for a protein that preventsRNA polymerase from binding to the promoterregion, thus preventing the synthesisthose three enzymes.
17 However, an isomer of lactose will bind to the repressor protein, and change its shape so itdoesn't fit with the operator. This allows forthe production of the three enzymes.Therefore, if there is no lactose (andtherefore no isomers of lactose), therepressor protein will prevent thetranscription of the enzymes.
18 If there is lactose, the repressor protein will be inactivated and the enzymes will be produced.Operons are founds only is prokaryotes.
19 6.3.3 Explain the process of transcription in eukaryotes including the role of thepromoter region, RNA polymerase,nucleoside triphosphates and the terminator.RNA polymerase separates the two strandsof the DNA and bonds the RNA nucleotideswhen they base-pair to the DNA template.RNA polymerase binds to parts of the DNAcalled promoters in order for separationof DNA strands to occur.
20 Transcription proceeds as nucleoside triphosphates (type of nucleotide) bind to the DNA templateand are joined by RNA polymerase in the 5' to 3‘direction. Transcription ends when RNA polymerasereaches a termination site on the DNA. Whenit reaches the terminator, the RNA polymerasereleases the RNA strand.
21 6.3.4 Distinguish between the sense and antisense strans of DNAThe sense strand is the coding strand and has thesame base sequence as the mRNA (with uracilinstead of thymine). The antisense strandis transcribed and has the samebase sequence as tRNA.
22 6.3.5 State that eukaryotic RNA needs the removal of introns to form mature mRNA.Eukaryotic RNA needs the removalof introns to form mature mRNA.
23 6.3.6 State that reverse transcriptase catalyses the production of DNA from RNA.Reverse transcriptase catalyses the production of DNAfrom RNA. This helps to show the aspects of the DNAviral life cycle to that of the AIDS virus (an RNA virus).
24 6.3.7 Explain how reverse transcriptase is used in molecular biology.This enzyme can make DNA from mature mRNA(eg insulin) which can then be spliced into host's(eg baceria)DNA without the introns. Then when thehost's DNA is transcripted, proteins like insulin aremade. It is important that the DNA created by reversetranscriptase have no introns, because the host doesnot have the genes (and therefore proteins) necessaryto remove the introns
25 Topic Translation6.4.1 Explain how the structure of tRNA allows recognition by a tRNA-activating enzyme that binds a specific amino acid to tRNA, using ATP for energy.Each amino acid has a specific tRNA-activating enzyme that help to tRNA to combine with its comlimentary mRNA codon.MAIN PAGE
26 The enzyme has a 3-pat active site that recognizes three things: a specific amino acid, ATP, anda specific tRNA. The enzyme attaches the aminoacid to the 3' end of the tRNA. The aminoacid attachment site is always the base triple CCA.
27 It is important to note that each tRNA molecule can attach to one specific amino acid, but anamino acid can have a few tRNA moleculeswith which is can combine.
28 6.4.2 Outline the structure of ribosomes including protein and RNA composition,large and small subunits, two tRNAbinding sites and mRNA binding sites.A ribosome consists of two subunits: small and large.These two subunits separate when they are notin use for protein synthesis.
29 In eukaryotes, the large subunit consists of three different molecules of rRNA (ribosomal RNA)and about 45 different protein molecules. A smallsubunit consists of one rRNA molecule and33 different protein molecules. A ribosome canproduce all kinds of proteins. It has a mRNAbinding site, and two tRNA binding siteswhere the tRNA are in contact with the mRNA.
30 6.4.3 State that translation consists of intitiation, elongation, and termination.Translation consists of initiation,elongation, and termination
31 6.4.4 State that translation occurs in a 5' to 3' direction.Translation occurs in a 5' to 3' direction, the ribosomemoves along the mRNA toward the 3' end.The start codon is nearer to the 5' endthan the stop codon.
32 6.4.5 Explain the process of translation including ribosomes, polysomes, startcodons, and stop codons.There are three stages in translation. The first is:1. Initiation Once the RNA reaches the cytoplasm, itattaches its 5' end to the small subunit of theribosome. AUG is called the start codon(remember: codon is a base triple onthe mRNA) because it initiatesthe translation process.
33 The anticodon on one end of a tRNA molecule is complimentary to a specific condon on the mRNA,meaning that the anitcodon and the codon bondby hydrogen bonds. The codon AUG hydrogenbonds with a tRNA molecule holding the aminoacid methionine (the initiator tRNA).
34 The large ribosomal subunit has two tRNA binding sites: the P site and the A site. tRNAmolecules in these sites attach to the mRNA"conveyor belt". tRNA molecules move fromthe A site to the P site (because the mRNAmoves that way), so when initiation is donethe initiator tRNA is in the P site.
35 2. Elongation Another tRNA (let's call it tRNA "X") carrying The next stage is:2. Elongation Another tRNA (let's call it tRNA "X") carryinga specific amino acid attaches itself to thenext codon at the A site of the larger subunit.These two amino acids (methionin and theamino acid on X) now make a peptide bondwith each other. The initiator tRNA breaksoff and tRNA X moves from theA site to the P site.
36 Then another tRNA molecule (tRNA "Y") attaches to the codon in the A site. Theamino acid that is attached to tRNA Ymakes a peptide bond with the aminoacid from tRNA X. The codon keepsmoving through the ribosome in a5' to 3' direction (from A to P). (Inactuality, it is the ribosome that ismoving across the mRNA chain.)This makes the A site vacant foranother tRNA to attach a newamino acid etc. etc.
37 The final step of translation is: 3. Termination The stop codon is one that does notcode for an amino acid and that terminatesthe translation process. The polypeptideis released and the mRNA fragments returnto the nucleus. These nucleotides are recycledand used for RNA and DNA synthesis. tRNAalso is returned to its free state and attachesto its specific amino acid so as to be readyfor the translation process when needed.
38 6.4.6 State that free ribosomes synthesize proteins for use primarily within the cell andthat bound ribosomes synthesize proteinsprimarily for secretion or for lysosomes.Free ribosomes synthesize proteins for useprimarily within the cell and bound ribosomessynthesize proteins primarily forsecretion or for lysosomes.
39 Topic Proteins6.5.1 Explain the four levels of protein structure, indicating each level's significance.The primary structure is the basic order of amino acids in the polypeptide protein chain, before any folding or bonding between amino acids has occured. Proteins are usually not functional on the primary level, and all proteins have a primary structure.MAIN PAGE
40 The secondary structure is the repeated, regular structure protein chains take dueto hydrogen bonding between amino acids.The secondary structure is usually in theform of an alpha helix (similar to a DNAchromosome) or a beta- pleated sheet(similar in form to the corrugations of cardboard).
41 The third structure is the tertiary structure. It is the complex, three-dimensional proteinshape resulting from the folding of thepolypeptide due to different types of bondsbetween the amino acids. These bondsinclude hydrogen bonds, disulphide linkagesand electrovalent bonds. A disulphidelinkage is where one of the aminoacids, cysteine, contains sulfur.
42 Two of these can have a bond between their sulfuratoms which results in a disulphide linkagewhich leads to a folding in the chain. Aneloctrovalent bond is between the negative andpositive molecules or amino acidgroups along the chain.
43 The quaternary structure is the most complicated form of a protein. It encompasses the primary,secondary and tertiary, and then adds anotherlevel: the quaternary is one or more peptidechains bonded together. This is the functionalform of many proteins, but again, just as notall proteins have secondary or
44 tertiary structure, not all proteins have a quaternary structure. For a commonexample of a quaternary-structuredprotein, see hemoglobin.
45 6.5.2 Outline the difference between fibrous and globular proteins, with referenceto two examples of each protein type.Fibrous proteins are in their secondary structure,which could be in the alpha helix or betapleated forms. They are made of a repeatedsequence of amino acids that can be coiledtightly around in a pattern that makesit a very strong structure.
46 Two examples are keratin (in hair and skin) and collagen (in tendons, cartilage, and bones).Glbular proteins are in their tertiary or quaternarystructure, which is folded, creating a globular,three-dimensional shape. Two examples are allenzymes and microtubules (form centrioles,cilia, flagella, and cytoskeleton).
47 6.5.3 Explain the significance of polar and non-polar amino acids.Non-polar amino acids have non-polar (neutrallycharged) R groups. Polar amino acids have Rchains with polar groups (charged eitherpositive or negative). Proteins with a lot ofpolar amino acids make the proteins hydrophyllicand therefore able to dissolve in water.
48 Proteins with many non-polar amino acids are more hydrophobic and are less soluble in water.With these abilities, proteins fold themselvesso that the hydrophilic ones are on the innerside and allows hydrophilic molecules and ionsto pass in and out of the cells through the channelsthey form. These channels are vital passagesfor many substances in and out of the cell.
49 6.5.4 State six functions of proteins, giving a named example of each.One function is transport. An example ishemoglobin, which transports oxygen aroundthe body within a blood cell. Another functionis the contraction of muscles. An exampleis actin and myosin which are involved inthe contraction of muscles. Enzymesare all proteins and catalyze reactions
50 Some examples are trypsin and amylase. Some hormones are proteins. In example is insulin,a growth hormone. Another function isantibodies that fight against disease are madeof proteins. The sixth function is membraneproteins. They function as receptors, channels,enzymes or antigens studded throughoutthe phospholipid bilayer.
51 Topic Enzymes6.6.1 State that metabolic pathways consist of chains and cycles of enzyme- catalysed reactions.Metabolic pathways consist of chains and cycles of enzyme catalysed reactions.MAIN PAGE
52 6.6.2 Describe the induced fit model. This is an extension of the lock-and-key model. Itaccounts for the broad specificity of someenzymes. The induced fit model of enzymesstates that the active site of an enzymedoes not fit perfectly with the substrate,as would a lock and key.
53 Instead, the fit is not quite perfect; thus when a substrate collides with an enzyme and entersthe active site, the substrate is stressed ina way that allows its bonds to break easier.When another collison occurs with thesubstrate in the active site, the bonds arethen broken and the substrate released.Enzymes can process reactions suchas this very rapidly.
54 6.6.3 Explain that enzymes lower the activation energy of the chemical reactions that they catalyse.All reactions, either with or without enzymes, needcollisons between molecules in order to occur.Many molecules have strong bonds holdingthem together, and as such require powerfulcollisions at high speed in order to break these bonds.
55 However, increasing the rate of collision to a rate at which these reactions would occur wouldrequire prohibitive amounts of energy, usuallyin the form of heat. Enzymes, by stressingsubstrate bonds in such a way that a weakercollision is required to break them, reduce theamount of energy needed to cause thesereactions to occur, or the activation energy.
56 6.6.4 Explain the difference between competitive and non-competitive imhibition, withreference to one example of each.Competitive inhibition occurs when an inhibitingmolecule structurally similar to the substratemolecule binds to the active site, preventingsubstrate binding. Examples are the inhibitionof butanedioic acid (succinate) dehydrogenaseby propanedioic acid (malonate) in the Krebscycle, and inhibition of folic acid synthesis inbacteria by the sulfonamide prontosil (an antibiotic).
57 Non-competitive are an inhibitor molecule binding to an enzyme (not to its active site) that causesa conformational change in its active site,resulting in a decrease in activity. Examplesinclude Hg, Ag, Cu and CN inhibition of manyenzymes (eg cytochrome oxidase) by bindingto SH groups, thereby breaking -S-S- linkages;and nerve gases like Sarin and DFP (diisopropylfluorophosphate) inhibiting ethanyl(acetyl) cholinesterase.
58 6.6.5 Explain the role of allostery in the control of metabolic pathways byend product inhibition.Allostery is a form of non-competitive inhibition.The shape of allosteric enzymes can be alteredby the binding of end products of an enzymeto an allosteric site (an area of the enzymeseperate from the active site), therebydecreasing its activity.
59 Metabolites, a type of allosteric inhibitor, can act as allosteric inhibitors of enzymes earlier ina metabolic pathway and regulate metabolismaccording to the requirements of organisms;they are a form of negative feedback. Examplesinclude ATP inhibition of phosphofructokinasein glycolysis and inhibition of aspartatecarbamoyltransferase (ATCase) whichcatalyses the first step in pyrimidine synthesis.MAIN PAGE