Exam 2 Notes.docx

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Biological Sciences
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BIOSC 0160
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Bledsoe

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Exam 2 Notes BIO 0160 DNA AS GENETIC MATERIAL I. DNA (Deoxyribonucleic acid) a. Structure i. The double helix 1. Consists of two strands 2. Each strand has a “backbone” of alternating sugar and phosphate 3. Each sugar has a base attached to it –either: a. A, adenine (a purine) b. C, cytosine (a pyrimidine) c. G, guanine (a purine) d. T, Thymine (a pyrimidine) i. Purine – two nitrogenous rings ii. Pyrimidine – one nitrogenous ring ii. Complementary and antiparallel strands 1. Strands held together by hydrogen bonds between bases 2. Base-pairing rules a. A bonds with T (two hydrogen bonds) b. G bonds with C (three hydrogen bonds) 3. Antiparallel strands (Fig. 14.4B) a. Deals with strand directionality b. Each strand has two ends i. 5’ phosphate end 1. Also known as “free” phosphate a. Covalent bond to only one sugar…..that’s why it is free ii. 3’ hydroxyl end 1. A free OH at the 3’ position of the last sugar c. Strands are parallel but run in opposite directions – hence “antiparallel” d. Consider this drawing (lines are hydrogen bonds) 5’____________________3’ 3’____________________5’ b. Replication i. Copying DNA ii. Using enzymes called DNA polymerases iii. Begins at origin of replication 1. Many, many, all along the linear DNA 2. At each, strands separate to yield two replication forks, where new DNA is formed 3. Each replication bubble is associated with two replication forks moving in opposite directions Exam 2 Notes BIO 0160 iv. Complementarity of strands key to accurate replications 1. Semiconservative model a. Each original (parental) strand operates as a template for each new strand b. Meselson-Stahl experiment (Fig 14.5)*****LOOK IN BOOK 2. Leading and lagging strands a. Some first principles i. Nucleic acids are always synthesized in the 5’ to 3’ direction ii. Hence template is “read” in the 3’ to 5’ direction iii. DNA replication requires RNA primers 1. Primers inserted into single strand DNA so DNA polymerase can bind to the double strand and start b. Leading strand i. Grows continuously because replication fork leads (is in front of) the new DNA strand ii. Direction of DNA synthesis is the same as direction of movement as the replication fork c. Lagging strand i. Grows in short pieces (Okazaki fragments) because the fork lags behind (is in back of) the new DNA strand ii. Direction of DNA synthesis is the opposite of the movement of replication fork Exam 2 Notes BIO 0160 iii. Ligase must join Okazaki fragments 3. Proteins in DNA replications a. Helicase – first enzyme; untwists the DNA double helix, opening the replication fork b. Primase – synthesizes RNA primer on leading and lagging strands c. DNA polymerase III – synthesizes new DNA strands, beginning at RNA primers d. Nuclease – excises errors in replication, using excision repair e. DNA polymerase (I &II) – fill in gap created by nucleases during excision repair f. DNA polymerase I – removes and replaces RNA primer with DNA in preparation for ligase action g. DNA ligase – joins DNA fragments during replication and repair h. Topoisomerase – relieves twisting forces during replication i. Sliding clamp – holds DNA polymerase in place c. Repair i. Mismatch repair 1. DNA polymerase III proofreads as it goes 2. Checks each added nucleotide, fixes if needed (Fig 14.14 pg 271) ii. Excision repair (Fig 14.15 pg 272) 1. UV light can cause covalent bond to form between adjacent thymine’s forming “thymine dimer” 2. Nuclease spots problem, and excises DNA 3. Produces gap 4. DNA polymerase fills gap (polymerase I & II) 5. Ligase seals remaining nick (Fig 14.16 pg 273) d. The special case of telomeres i. Telomeres 1. Special, non-genic DNA at end of chromosomes 2. Short segment repeated 100-1000 times 3. Typical repeat: TTAGGG 4. Function to prevent the end-replication problem a. Gap left by primer removal unfilled! b. Gap at 5’ end is unfilled c. After many cell divisions, eaten-away portion can extend into genes! 5. Telomerase a. RNA-containing enzyme that can make telomeres b. Present in gametes and in cancerous cells II. The Elegant Experiments of the 1940’s and 1950’s a. What was known in 1928 (Griffith) i. Disease-causing but dead bacteria could transform harmless ones into disease-causing ones Exam 2 Notes BIO 0160 ii. Only purified DNA could cause such transformation from dead disease- causing bacteria to living harmless bacteria iii. But proteins more attractive as hereditary material b. Avery, McCarty, and MacLeod, 1944 i. Announced transforming agent was DNA c. Chargaff, 1947 i. Looked at base composition among life ii. Found great diversity in base composition iii. Yet oddly: 1. Amount of A always equaled amount of T 2. Amount of C always equaled amount of G 3. These are called “Chargaff’s rules” d. Hershey and Chase, 1952 i. Grow phages (virus that infects bacteria) in radioactive sulfur 1. Tags proteins with radioactive sulfur ( S) ii. Infect bacteria, then centrifuge cells iii. Result: little radioactive sulfur in cell pellet at bottom of test tube after centrifuge iv. Grow phages in radioactive phosphate 32 1. Tags DNA with radioactive phosphate ( P) v. Infect bacteria, then centrifuge vi. Result: LOTS of radioactive phosphate in cell pellet vii. Interpretation: DNA, not protein, injected into bacteria as phage hereditary material e. Franklin, early 1950’s i. Took crucial x-ray diffraction (crystalize something then see how x-ray diffract off crystalline structure) photo of DNA ii. Watson happened to see photo iii. It provided crucial information to Watson and Crick! f. Watson and Crick (ASSHOLES), 1953 i. Proposed the double-helix model of DNA ii. Did not escape their notice: a built-in copying mechanism 1. Called semi-conservative replication MOLECULAR GENETICS I. One gene-one polypeptide hypothesis a. Based on Beadle and Tatum’s original work i. Deduced gene-protein relationship from mutants in mold (Fig 15.2 pg 278) ii. “one gene-one protein” hypothesis b. Extension to polypeptides i. Many proteins contain several polypeptides encoded by different genes c. Exceptions and modifications i. Control sequences and introns 1. Non-coding sequences 2. Will to these later… ii. tRNA’s (transfer RNA’s) and rRNA’s (ribosomal RNA’s) 1. Non-proteins coded by genes Exam 2 Notes BIO 0160 2. Used in protein synthesis iii. Overlapping reading frames (boundaries of genes) 1. Genes don’t usually overlap with their reading frames but in some instances they do….makes more than one product 2. Some genes overlap with others in viruses d. Definition of “gene” i. A region of DNA whose final product is either a polypeptide or an RNA molecule II. Overview a. Central Dogma (pg 280) i. DNARNAproteins b. Revised Central Dogma (Bledsoe Version) i. DNA(transcription)RNA(translation)protein DNA (Reverse Transcription)RNA DNA Replication ---- DNAmore DNA c. Transcription i. Conversion of DNA information into an mRNA intermediate ii. M stands for messenger iii. Carries information to ribosomes d. Translation i. “Reading” mRNA information into amino acid sequences ii. Involves protein synthesis iii. Occurs on ribosomes iv. Information in language of nucleic acids in “translated” into language of proteins e. Posttranslational Modification i. After protein synthesis ii. Some proteins get modifies (e.g., by addition of a sugar group) III. The Genetic Code --The dictionary linking DNA/RNA information and amino acids a. Codons i. Sets of 3 RNA nucleotides (A,U,G,C,) ii. Listed in 5’ to 3’ direction iii. Total of 64 possible codons iv. They stand for: 1. 20 amino acids (fig 15.6 pg 284) 2. 1 start translation signal (AUG) 3. 3 stop translation signals (UAA, UAG, UGA) b. Reading Frame i. “Sending the right message” ii. Start signal establishes the reading frame (fig 15.7) iii. No overlap in codons iv. A frame shift drastically alters the resulting protein c. Redundancy i. 2 amino acids have only one codon 1. Tryptophan Exam 2 Notes BIO 0160 2. Methionine ii. Other 18 have several codons iii. Hence, the code is redundant d. Code is nearly universal – very few exceptions i. Code must have arisen early in life’s history ii. Exceptions; for example… 1. Certain single-celled eukaryotes have a few codons that differ 2. A few mitochondrial and chloroplast codons that don’t fit with the rest IV. Transcription (DNA to RNA) a. Basic process i. RNA polymerase unzips DNA strands and synthesizes RNA ii. RNA polymerase binds to the promoter upstream of sequence to be transcribed iii. Protein transcription factors also bind to yield transcription initiation complex (bind at and around promoter) iv. All of the above are the binding and initiation phases of transcription v. Then elongation phase: 1. RNA polymerase moves along DNA, making RNA 2. Many RNA polymerase molecules can do this at the same time on one piece of DNA (allows more rapidly expression of the gene) vi. Then termination 1. A special DNA sequence called a terminator is transcribed 2. The resultant RNA sequence functions as a termination signal b. RNA processing i. 5’ cap 1. A single modifies guanine 2. Added immediately after transcription begins 3. Prevents RNA degradation at the 5’ end 4. Is an attachment signal to ribosomes ii. 3’ poly-A tail 1. 30-200 adenines added to 3’ end 2. Prevents RNA degradation at the 3’ end of the transcript iii. RNA splicing 1. Eukaryotic genes have: a. Introns: intervening sequences i. Not translated: spliced out b. Exons: transcribed sequences i. Linked together: spliced (Fig16.6a pg 294) c. Occurs in spliceosome i. Uses snRNP’s 1. Small nuclear ribonucleoproteins d. RNA in snRNP is catalytic! V. Translation -Production of protein from mRNA information a. tRNA i. Transfer RNA ii. At 3’ end, has a specific amino acid Exam 2 Notes BIO 0160 iii. Hence each amino acid has its own tRNA(s) iv. Near 5’ end, it has an anticodon 1. Binds to complementary codon on mRNA v. Hence, amino acids are deposited in prescribed order (Fig 16.13 pg 299) vi. Wobble: relaxation of base-pairing rules vii. Amino-acyl tRNA synthetase 1. Loads an amino acid onto tRNA 2. One such synthetase for each of the 20 amino acids b. rRNA i. Ribosomal RNA ii. Forms ribosomes iii. Typically two subunits: large and small c. Basic process i. Binding and Initiation 1. mRNA and small subunit of rRNA bind 2. Large subunit then attaches 3. Initiation factors are required (Fig 16.15 pg 301) ii. Elongation 1. Codon recognition ushers tRNA into A site on ribosome 2. Then shifts (translocation) to P site after amino acid is added to growing polypeptide 3. Then shifts (translocation) to the E site where tRNA is released (Fig 16.16 pg 302) iii. Termination 1. Occurs at an mRNA stop codon 2. A release factor (protein) binds to the stop codon at the A site 3. Water added to polypeptide 4. This causes release from ribosome (Fig 16.17 pg 303) d. Signal peptides i. Short peptides that signal destination of protein (e.g. endoplasmic reticulum) **RNA is catalytic….Ribozyme  catalytic nucleic acids** VI. Mutations a. Alterations of DNA sequence b. Point Mutation - Change or alteration of a single nucleotide in a gene (Table 15.1) i. Silent Mutation 1. No change in amino acid sequence…code is redundant ii. Missense Mutation 1. Change in nucleotide that changes the amino acid iii. Nonsense Mutation 1. Mutation in DNA that causes a “STOP CODON” iv. Frameshift 1. Add or delete single nucleotide in gene altering the reading frame c. Chromosome-level Mutations i. Polyploidy – more than two of each basic chromosome Exam 2 Notes BIO 0160 ii. Aneuploidy – an extra/missing a chromosome (Trisomy 21) iii. Inversion – whole region of chromosome is flipped (i.e. inverted) iv. Translocation – region of one chromosome moves to a different chromosome CONTROL OF PROKARYOTIC GENE EXPRESSION I. Overview a. Binding to the promoter i. For transcription, RNA polymerase needs to bind efficiently to the promoter 1. Promoter  a short, protein binding DNA sequence just upstream (toward 3’ end of the gene) of the gene to be transcribed ii. Some proteins block access of the promoter (turns gene off) iii. Some proteins enhance the affinity of RNA polymerase for the promoter (turns genes on for longer time: UP-regulate gene expression) 1. Changing affinity of RNA polymerase alters how much transcription occurs iv. Most gene regulation involves promoters b. Prokaryotes i. Genes can be rapidly turned on, turned off, and turned on again 1. In association with rapid response to changing environmental conditions c. Eukaryotes i. Gene regulation is more complex ii. It occurs throughout development iii. Different tissues express different genes iv. In many tissues, genes are activated briefly and then shut down permanently v. In many tissues, many genes are never expressed II. Viral Cycles (Prokaryotes) a. Lytic cycle (Bacteriophages) i. Phage attaches to the host ii. Phage injects its genome iii. Phage causes host chromosome degradation iv. Phage uses host metabolism to make phage genomes and capsid protein (outer part of phage particle) v. Phage particles form vi. Host cell lyses (breaks open and dies – phage particles released and phage has reproduced) b. Lysogenic cycle i. Phage attaches to the host ii. Phage injects its genome iii. Phage genome inserts into host chromosomes, as a prophage (sets up bacteria recombination) iv. Phage spreads in host population v. Under environmental stress, prophage excises and enters lytic cycle III. Prokaryotic gene regulation uses operons a. Basic structure and function i. Operon make-up: 1. A cluster of metabolically-related genes Exam 2 Notes BIO 0160 2. A single promoter (control sequence that RNA polymerase) 3. An operator ii. Operon functions to: 1. Regulate synthesis of enzymes 2. In response to environmental conditions: a. Presence of a substance to be degraded (e.g. lactose) b. Absence of substance to be synthesized (e.g. tryptophan [trp]) b. Repressible operons i. Transcription inhibited by a molecule that would otherwise need to be synthesized ii. The molecule activates a repressor iii. Details – the trp operon 1. Regulatory gene makes an inactive repressor 2. If tryptophan (co-repressor) is present: a. It binds to and activates the repressor b. The repressor then binds to the operator c. Upstream, the promoter has RNA polymerase d. The activated repressor bound to the operator blocks the polymerase, thus blocking transcription 3. If tryptophan is absent: a. The inactive repressor can’t bind to the operator b. Upstream, the promoter has RNA polymerase c. The polymerase isn’t blocked, and hence transcribes genes for trp synthesis c. Inducible operons i. Transcription stimulated by a molecule that needs to be degraded ii. The molecule inactivates a repressor iii. Details – the lac operon 1. Regulatory gene encodes an active repressor 2. If lactose is absent: a. The repressor binds to the operator b. The active repressor on the operator now blocks RNA polymerase, thus blocking transcription 3. If lactose (the inducer) is present: a. It binds to and deactivates the repressor b. The inactive repressor falls off the operator c. Thus, the RNA polymerase isn’t blocked, and hence transcribes genes for lactose degradation Exam 2 Notes BIO 0160 d. Association between repressible and inducible operons i. Repressible operons are associated with anabolic (build up) pathways ii. Inducible operons are associated with catabolic (break down) pathways e. Positive control i. Example: when both lactose and glucose are present 1. Cells prefer to use glucose (more ATP out of glucose degradation) 2. Thus, if it’s present, transcription at the lac operon is suppressed, but not eliminated 3. Details: a. Cyclic AMP (cAMP) builds up when glucose is absent i. It binds to and activated a regulatory protein, cAMP receptor protein (CRP/CAP) ii. Active CRP binds near promoter and enhances RNA polymerase binding b. cAMP goes way down when glucose is present i. CRP is thus inactive ii. It fails to bind near the promoter iii. RNA polymerase binding is much less efficient iv. Hence, there is little transcription of genes for breaking down lactose IV. Prokaryote genetic recombination a. Transformation i. Uptake of naked DNA from other, dead bacteria ii. DNA becomes integrated into the bacterial chromosome b. Transduction i. Generalized Transduction (lytic cycle) 1. Phage-mediated genetic recombination a. Phage injected into host cell b. Phage particle contains parts of host cell DNA c. If next host cell fights off phage, then former host cell DNA gets put into new host cell DNA genome ii. Specialized Transduction (lysogenic cycle) 1. Virus attaches and inserts genome 2. Excises itself to enter lytic cycle 3. Takes some genic info from host 4. Attaches to different host and displays first host genes c. Conjugation i. DNA transfer between two living bacterial cells ii. One way transfer from donor cell to recipient cell V. Viral gene expression a. Viral genomes i. Genome can be either: 1. dsDNA [double stranded DNA](e.g. herpes simplex I) 2. ssDNA [single stranded DNA] (e.g. roseola: baby measles) 3. dsRNA (e.g. diarrhea) 4. ssRNA, for replication and protein synthesis (e.g. HIV) b. The interesting case of HIV i. Each HIV viral particle has two single stranded molecules of RNA Exam 2 Notes BIO 0160 ii. In addition, attached to each single strand is a molecule with an enzyme called reverse transcriptase iii. Once viral genome is in human cell, reverse transcription occurs (reverse transcriptase does this) iv. Produces RNA-DNA hybrid v. DNA compliment made by the single strand DNA vi. Double strand is then incorporated into host genome vii. Once gene is expressed, repli
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