Molecular Biology: DNA, gene, chromosome and genome (Learning Objectives)

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1 Molecular Biology: DN, gene, chromosome and genome (Learning bjectives) Nucleic acid structure and composition ompare and contrast the structure of DN and RN: features they share and how do they differ? (number of stands, sugar, and nitrogen bases). Which nitrogen bases are found in DN or RN. Learn the base-pairing rule, its underlying reason and its effect on the DN structure. Distinguish between DN, gene, chromosome and genome. Explain the DN structure: number of strands, polarity (5-3 ), complementary strands and their anti-parallel nature. Distinguish between the strong covalent bonds that hold the sugar-phosphate backbone of nucleic acids and the weaker chemical bonds form between the nitrogen bases and hold the two strands of DN. Learn how DN is packaged as chromatin inside id the nucleus in association with histone proteins. DN replication Explain the steps for DN replication: strand separation, name of protein enzyme, pairing of complementary nucleotides, polymerization, termination. Distinguish between the DN template strand and new strand. In what direction is the new strand made? ompare and contrast t the leading and lagging newly synthesized DN strands: which h is continuously made and which is discontinuously made? Role of DN ligase.

2 Molecular Biology: DN, gene, chromosome and genome (cont d) ranscription Explain the purpose for this process and its sub-cellular compartment. Learn the three steps of transcription ti and the bio-molecules l involved. Name the stretch of nucleotides in the DN that binds RN polymerase to initiate transcription. Name the DN stretch that causes RN polymerase to come off DN terminating transcription. Explain the split nature of eukaryotic genes. Distinguish between exon, intron. Which contains information that will specify the amino acid sequence of the protein product? Explain RN processing of eukaryotic cells. enetic code Explain the language of nucleic acids: letters and words (nucleotides and codons), number of nucleotides that make up a codon, and total number of codons. Which h codon marks the initiation iti of translation, ti what amino acid does it specify? How many stop codons specify the termination of translation. ranslation or protein synthesis Where does it take place, what components are necessary, and what are the three steps. Which biomolecule can interpret the language of nucleic acids into the language of proteins Mutation Define the term mutation, how do they arise and list the types of mutations. What impact might a mutation have on the protein product? Use the genetic code table to translate a sequence of codons into a sequence of amino acids and the reverse as well as the sequence of normal and mutated proteins

3 HE FLW F ENEI INFRMIN he DN of the gene is transcribed into RN which is translated into the polypeptide (protein) DN ranscription RN Protein ranslation Figure 10.6

4 Nucleic cid hemical Structure DN and RN are polymers of nucleotides Sugar phosphate backbone Phosphate group Nitrogenous base Sugar Phosphate DN group nucleotide P Nitrogenous base (,,, or ) H 3 N H H H N 2 hymine () H H H H H Sugar (deoxyribose) DN nucleotide Figure 10.2 DN polynucleotide

5 DN has four kinds of nitrogenous bases: DN has four kinds of nitrogenous bases:,,, and H H N H H N N H N H H 3 H H N N H N N N N H N N N H N N H H H H H H H hymine () ytosine () denine () uanine () Purines Pyrimidines Figure 10.2B

6 RN is also a nucleic acid with a slightly different sugar U instead of Nitrogenous base (,,, or U) Key Phosphate Hydrogen atom group arbon atom H N H Nitrogen atom xygen atom P H 2 H N Phosphorus atom Uracil (U) H H H H H Figure 10.2, D Sugar (ribose)

7 ene: a linear stretch of nucleotides with information for one product (polypeptide or protein) hromosome: a very long stretch of DN carrying many genes. DN is always associated with protein as chromatin DN oiling into chromatin and condensed chromosomes dna coiling.htm enome: totality of DN in a cell

8 Erwin hargaff, 1947 Biochemical analysis of DN nucleotides from different species = & = Human DN = 30.9% = 29.4% = 19.9% = 19.8%

9 DN Shape DNis a double stranded helix James Watson and Francis rick worked out the three dimensional structure of DN, based on work by Rosalind Franklin Figure 10.3, B

10 hestructure of DN two polynucleotide strands wrapped around each other in a double helix Figure 10.3 wist

11 ovalent bonds hold the sugar phosphate backbone Hydrogen bonds between bases hold the two strands Each base pairs with a complementary partner with, and with Figure 10.3D Base pair H P Hydrogen bond H H 2 2 P H 2 P H 2 P H 2 H 2 P P H 2 H 2 P H H 2 P H Ribbon model Partial chemical structure omputer model

12 DN REPLIIN DN replication depends on specific base pairing Starts with the separation of DN strands protein enzyme uses each strand as a template to assemble new nucleotides into complementary strands Figure 10.4 Parental molecule of DN Nucleotides Both parental strands serve as templates wo identical daughter molecules of DN Builda DN Molecule

13 DN replication is a complex process the helical DN molecule must untwist or unwind several proteins are involved including DN Polymerase it start at specific DN sequences, origin of replication Figure 10.4B

14 Replication of long stretches of DN Begins at multiple specific sites on the double helix rigin of replication Parental strand Daughter strand Bubble Figure 10.5 wo daughter DN molecules

15 Each DN strand of the double helix is oriented in the opposite direction 5 end 3 end Figure 10.5B P H P 5 4 P P P P P H P 3 end 5 end

16 he enzyme DN polymerase uses a single strand and makes a new complementary strand in a 5 to 3 direction one daughter strand is made as a continuous piece the other strand is synthesized as a series of short piece which are then connected by the enzyme DN ligase 5 3 DN polymerase 3 molecule 5 Daughter strand Parental DN synthesized continuously Daughter 3 strand 5 synthesized in pieces 5 3 Figure 10.5 DN ligase verall direction of replication

17 ranscription Producing copies of the genetic messages in the form of RN ranscription of a gene 1. DN: double stranded 2. Enzyme: RN Polymerase 3. Monomers: RN nucleotides 4. Steps: Initiation Start at promoter Elongation ermination Stop at termination signal Figure 10.9B Promoter DN RN polymerase 1 Initiation 2 Elongation DN of gene 3 ermination ompleted RN rowing RN erminator DN RN polymerase

18 RN polymerase unwinds the two DN strands Uses one strand as a template strand Polymerizes RN nucleotides against the template strand followingthe base pairing rules Single stranded messenger RN (mrn) peels away from the template strand he DN strands rejoin RN polymerase RN nucleotides U Figure 10.9 Direction of transcription Newly made RN emplate Strand of DN

19 Split enes of Eukaryotes Eukaryotic RN is processed before leaving the nucleus Noncoding segments called introns are spliced out a cap and a tail are added to the ends Exon Intron Exon Intron Exon DN ap ranscription ddition of cap and tail RN Introns removed transcript ail with cap and tail Exons spliced together mrn oding sequence Nucleus Figure ytoplasm

20 HE FLW F ENEI INFRMIN FRM DN RN PREIN he information carried idby sequence of DN bases constitutes an organism s genotype he DN genotype is expressed as proteins, which provide the molecular basis for the phenotype

21 enetic information written in a code that is translated into amino acid sequences he words of the DN language are triplets of bases (3b bases long) called codons Each codons in a gene specify one amino acid sequence of the polypeptide

22 Protein synthesis ranslation is the RN directed synthesis of a polypeptide ranslation of the language of nucleic acids into thelanguage of proteins (amino acids) ne codon one amino acid

23 DN molecule ene 1 ene 2 ene 3 DN strand template ranscription RN ranslation U UU U U U U odon Polypeptide Figure 10.7 mino acid

24 In-class activity/enetic code Use the genetic code table to answer the following questions 1. How many codons are there for leu (leucine)? 2. How many codons are there for Met (Methionine)? 3. How many codons are there for Phe (phenylalanine)? Draw a conclusion about the number of codons for amino acids. 4. How many stop codons are there? nswer the following questions using this genetic code: 5 -UUUUUUU-3 5. How long is this message in nucleotides? 6. Is this the information present in DN or in mrn? Explain your answer. 7. Write down the sequence of amino acids coded for by the above stretch of nucleotides. how long is this polypeptide?.

25 hegenetic code is the Rosetta stone of life U UUU UU UU Second base U Phe Leu UU U U U Ser UU U yr U Stop U Stop UU U ys U Stop U rp U Nearly all organisms use exactly the same genetic code Figure 10.8 irst base F UU U U U UU U Leu Ile U U Met or start U U Pro hr U U His ln sn Lys U U UU U U U U sp Val la U lu U rg Ser rg ly U U U

26 n exercise in translating the genetic code Strand to be transcribed (template) DN ranscription RN U U U U U Start condon ranslation Stop condon Figure 10.8B Polypeptide Met Lys Phe

27 codon start codon within the mrn message marks the translation initiation and a stop codon marks the end of translation Start of genetic message End Figure 10.13

28 ranslation akes place in cytoplasm Involves: Ribosomes: wo subunits (each made of many proteins & r RN) mrn t RN (interpreter) 20 amino acids

29 ransfer RN molecules serve as interpreters during translation mino acid attachment site Hydrogen bond RN polynucleotide chain Figure nticodon

30 Each trn molecule is a folded molecule bearing a base triplet ti tcalled an anticodon on one end specific amino acid is attached to the other end mino acid attachment site nticodon Figure 10.11B

31 ribosome attaches to the mrn translates its message into a specific polypeptide aided by transfer RNs (trns) Protein Synthesis animation hill.com/sites/ /student_view0/chapter1 5/animations.html# ranslation Intro (Flash nimation)

32 1. Initiation Steps of protein Synthesis a. Binding of mrn to small ribosomal subunit b. Binding of Met trn to the initiation codon U on mrn c. Binding of large ribosomal subunit 2. Elongation 3. ermination

33 mrn, a specific trn, and the ribosome subunits assemble during initiation Initiator trn Met U U P site Start codon mrn Small ribosomal 1 subunit 2 Met U U Large ribosomal subunit site Figure 10.13B

34 1. Initiation Steps of protein Synthesis a. binding of mrn to small ribosomal subunit b. binding of Met trn to U on mrn c. Binding of large ribosomal subunit 2. Elongation a. binding of a second trn to next codon b. formation of peptide bond c. sliding of ribosome by one codon mino acids are added to the growing polypeptide chain until a stop codon is reached. 3. ermination

35 Elongation steps Polypeptide mrn P site odons site mino acid nticodon 1 odon recognition mrn movement Stop codon 2 Peptide bond formation New Peptide bond Figure ranslocation

36 1. Initiation Steps of protein Synthesis a. binding of mrn to small ribosomal subunit b. binding of Met trn to U on mrn c. Binding of large ribosomal subunit 2. Elongation a. binding of a second trn to next codon b. formation of peptide bond c. sliding of ribosome by one codon mino acids are added to the growing polypeptide chain until a stop codon is reached. 3. ermination Disassembly of the protein synthesis machinery

37 Review: he flow of genetic information in the cell is DN RN protein he sequence of codons in DN, via the sequence of codons in mrn spells out the primary structure t of a polypeptide

38 Mutations are changes in the DN base sequence aused by errors in DN replication or recombination, or by mutagens Normal hemoglobin DN Mutant hemoglobin DN mrn mrn U Figure Normal hemoglobin lu Sickle cell hemoglobin Val

39 Substituting, inserting, or deleting nucleotides alters a gene with varying effects on the organism. Normal gene mrn Protein U U U U Met Lys Phe ly la Base substitution U U U U Met Lys Phe Ser la Base deletion U Missing U U U U Figure 10.16B Met Lys Leu la His