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BIO 140 Notes 10/4/2016- 10/11/2016

by: Jay Ty

BIO 140 Notes 10/4/2016- 10/11/2016 BIO 140

Marketplace > James Madison University > Biology > BIO 140 > BIO 140 Notes 10 4 2016 10 11 2016
Jay Ty
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These are the notes from last week, and today! I will post the notes I typed from Thursday class on Friday <3 Good luck studying.!
Foundations of Biology I
Terrie Rife
Class Notes
Biology, life, DNA, RNA, Chemicals
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This 8 page Class Notes was uploaded by Jay Ty on Tuesday October 11, 2016. The Class Notes belongs to BIO 140 at James Madison University taught by Terrie Rife in Summer 2016. Since its upload, it has received 4 views. For similar materials see Foundations of Biology I in Biology at James Madison University.


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Date Created: 10/11/16
10/4/2016 BIO 140 Cresawn Notes I. Cell Organization (Second half of Chapter 5) I. Eukaryotic Cells vs. Prokaryotic Cells II. Did Group work III. Know the simplicity of Prokaryotic Cells IV. Know the Complexity of Eukaryotic Cells 10/6/2016 BIO 140 Cresawn Notes I. View Full PowerPoint II. Genetic Disorders in Wildlife a. Nucleic Acids and the Encoding of Biological Information i. Macromolecule Structure and Function ii. Genetic Information 1. DNA stores genetic information, which encodes for proteins that provide structure and do much of the workof the cell 2. Genetic Information is stored in genes,similar to how textual information is organized inthe form of words 3. Turning on a gene, or gene expression, ultimately causes and effect on the organism a. Whether that segment of DNA is transcribing to mRNA i. DNAtranscriptionpre-mRNAForm endsSplicemRNAtranslationproteinpo st-translational modificationsfunctional protein 4. Gene regulations controls when this process should occur a. How we control whether or not a particular gene is expressed 5. How we know DNA is the hereditary material? Evidence? a. Chromosomes are 50% proteinand 50% nucleic acid b. There are 20 different amino acids c. 4 different building blocks iii. Griffith’s Experiment 1. Was testing pneumonia inmice 2. Griffith injected virulent bacteria into the mice, and they did terrible 3. Griffith injected the nonvirulent bacteria into the mice, and they remained healthy 4. Killed virulent bacteria injected, mice remained healthy 5. Killed virulent and live nonvirulent bacteria caused the mice to die from pneumonia a. Control: Type of mice, un-manipulated scenarios (Griffith injected virulent bacteria into the mice, and they did terrible;Griffith injected the nonvirulent bacteria into the mice, and they remained healthy) b. Experimental treatment: c. Independent variable:The bacterial combination d. Dependent variable:Health of the mouse e. Hypothesis: f. Conclusion: 6. The untreated extract can transform nonvirulent cells into virulent cells. 7. DNA is extracted from heat-killed virulent cells,along with trace amounts of RNA and protein 8. By using enzymes that breakdownRNA, protein, and DNA, the phenomenon kept occurring until DNase was used to breakdown DNA. a. RNase breaks down just the RNA that’s present b. Protease breaks downproteins c. DNase breaks downDNA i. The breakdown of DNA ceasedthe transformation on nonvirulent cells into virulent cells ii. This proves that DNA was used as hereditary material 9. Only extracts treated withthe enzyme that destroys DNA were unable to transform nonvirulent bacteria iv. In 1953, James Watson and Francis Crick announced the three- dimensional structure of DNA 1. Rosalind Franklin figured this out 2. Maurice Wilkins stole her picture 3. Wilkins gave her photo to Francis Crickand James Watson 4. They changed their model and copiedher work, and denied Franklin credit v. DNA structure 1. Double helix 2. Moves 3. Not static 4. Nucleotide Structure a. Ionized hydroxyl groups b. Phosphate group c. Five carbonsugar (deoxyribose inDNA) d. Base (A, G, T, or C)  CalledNitrogenous Bases i. Have a lot of Nitrogen 1. Adenine (A) a. Purine 2. Guanine (G) a. Purine 3. To remember: Purines have tworings (Think PURE AG, as in Pure Silver) 4. Thymine (T) a. Pyrimidines 5. Cytosine (C) a. Pyrimidines 6. To remember: Pyrimidines have one ring 7. Thymine, Cytosine, and Pyrimidines all have the letter “y” e. Sugar + Base + Phosphate = Nucleotide i. Once a phosphate is added to a Nucleoside, it becomes a Nucleotide; the number of phosphates present dictate which nucleotide it is 1. Nucleotide monophosphate 2. Nucleoside diphosphate 3. Nucleoside triphosphate f. Sugar + Base = Nucleoside i. Nucleoside g. Phosphodiester Bonds: betweensugar and phosphate i. Polarirt: 5’-AGCT-3’ (from the top) 1. Goes from 5’ to 3’ ii. Polarity: 3’-TCGA-5’ (from the bottom) 1. Goes from 3’ to 5’ 2. The two strands are anti-parallel, because the polarities are going in opposite directions h. Hydrogen BONDING AND Base Stacking i. What are the properties of H-bonds? 1. Hydrogen bonds are weak 2. Individually, all the hydrogen bonds are weak, but there are strength in numbers so the double helix actually becomes quite stable ii. A and T are held together by two hydrogen bonds iii. C and G are held together by three hydrogen bonds 1. More hydrogen bonds, stronger, and harder to separate iv. Major groove: biggap in the helix 1. See the sequence of the nitrogenous bases 2. Where proteins recognize the sequences v. Minor groove: small gap in the helix vi. How do we know? 1. ErwinChargaff… Chargaff Rules! 2. DNA: though to be repeating polymer a. All four bases present inequal amounts b. A:G ratio would be? 3. Chargaff discovered:ratios varied!! a. Ratios for nitrogenous bases varied in organisms! 4. [A]= [T] and [C]= [G] 5. Also: [A + T ] ≠ (does not equal) [C + G] vi. Practice Questions 1. If the C content of a preparation of DNA is 15%, what is the A content? a. 35%, because C= G b. If C = 15%, then G = 15% c. 100- 30 = 70% for A and T d. A=T so 70/2= 35% e. A = 35% i. According toChargaff’s Rules, which of the following statements would be true? (More than can be) 1. [A] + [T] = [G] + [C] : False 2. [A]/[T]=1 : True 35/35 = 1 3. [G] = [C] : True 15 = 15 4. [A] + [G] = [T] + [C] : True 35+15= 35+15 ii. In DNA of certain bacterial cells, 16% of the nucleotides are adenine. What are the percentages of the other nucleotides in the bacterial DNA? 1. Answer: 16% thymine, 34% guanine, 34% cytosine f. Purine Pyrimidine base pairing will give an evenly spaced double helix! g. Imagine that you are an interior architecture/biology double-major. While at home during break, one of your friends asks you todesign and help build a spiral staircase that is modelled aftera DNA molecule. The staircase must have 12 steps,and eachstepneeds to be 6 feetwide. What list of materials do you needto build the staircase,necessaryto make a staircase based on the molecular structure of a DNA molecule? i. C: 2 railings and 24 boards (12are 4 feet,12are 2 feetlong) because purines are wider nucleotides 10/11/2016 BIO 140 Cresawn Notes III. Eukaryotic DNA is organized into Chromosomes i. Human Genome: 3 *10^9 nucleotides distributed over 24 linear chromosomes 1. Bacterial chromosomes are circular ii. Eukaryotic chromosomes: long, linear,molecules of DNA 1. Associated withProteins 2. Folded and packed into compact structure iii. DNA packaging 1. Human genome: 6 billionbase pairs a. 2 meters of DNA i. Nucleus is only 10 micrometers in diameter ii. How does it all fit and still function? b. 10 nm nucleosome willhold 70 nm of DNA i. Packaging ratio of 7:1 iv. Histones and Nucleosomes 1. Basic amino acids are positivelycharged 2. The charge on the DNA backbone is negatively charged 3. Ionic bond forms 4. Histones a. Rich inpositivelycharged acids,basic proteins- lysine and arginine (positivelycharged) b. Enable them to form ionic bonds withthe negatively charged sugar phosphate backbone of the DNA c. Five types 5. DNA and histones= nucleosomes 6. 146bp DNA around histone octamer (8 histones) 7. Getting the idea of the structure: a. The amino acids on the tails of the proteins willchange how tightly the DNA is packaged b. The tighter it’s packaged, the harder it is to express the genes. Too tightly can leadto disease i. Not a change in the DNA, but a change inthe DNA packaging c. The easier it’s packaged, the easier it is to express the genes and continue v. 30 nm fibers 1. Coil the histones on top of one another to make a slightly thicker fiber and package the DNA more tightly a. 30 nm fibers are packaged ina ‘zig-zag’ model 2. Info from Chromosome Coiling: Packed into chromosomes (when cells are dividing) 3. At the end of cell division,DNA becomes less organized vi. Supercoiling 1. Enzymes prevent supercoiling IV. Chromosome Structure a. Important parts of chromosomes i. Have at leastone replicationorigin ii. Need to have one centromere 1. Centromere: allowsone copy of each duplicated chromosome to go to eachcell in meiosis 2. Does not have to be inthe center of the chromosome 3. DNA is always tightly packed 4. Spindle grabs on to the chromosomes at the centromere and pulls the chromosomes apart during celldivision iii. 2 telomeres, one on eachend 1. Repeated sequences at tips 2. Humans: TTAGGG 3. Repeated 500- 5,000 times 4. Telomeres are highly conserved a. Enzyme that adds the repeats is called telomerase 5. Important inreplicating the ends of the chromosomes, and keeps chromosomes from sticking to one another 6. Telomeres are requiredfor complete replicationof chromosomes 7. Protective caps: a. Interaction betweenchromosomes and nuclear envelope b. Prevent chromosome fusion 8. They are non-coding 9. Prevent chromosome fusion iv. Telomeres in Aging 1. Cells in culture- don’t divide indefinitely a. Dramatic decrease in telomere length 2. Same in elderlyadults- no telomerase a. Telomere shortening- ‘to crisis’ 3. Add telomerase: extend lifespan a. Increased telomerase in mice extended life b. Study from May 2012 4. But telomere shortening; protects from cancer a. 90% of cancers have an active telomerase enzyme b. Report January 2013 v. Telomere lengths also affected by stress 1. Lower ranking hyenas 2. Increased stress- have to work harder for food a. Results inmore stress hormones, peroxides,and reactive O2 3. Results inshorter telomeres a. Leads to cell death V. DNA vs. RNA a. DNA is larger than RNA b. DNA is always a double helix, while RNA has more complex and variable structures c. RNA is ribose:has a hydroxyl (-OH) group where deoxyribose (DNA) has a hydrogen (-H) VI. DNA Structure a. Purines: A, G b. Pyrimidines: T, C, U (Uracil) c. DNA Structure Bases: i. A, G, C, T VII. RNA Structure a. Question: which of the followingstatements about RNA is correct? (Look at Slide [her powerpoint] for Reference) i. Answer: All of the answers are correct b. RNA Structure Bases i. A, G, C, U ii. Uracil has a hydrogen (-H) where thymine has a methyl (-CH3) group c. In RNA, hydrogen bonding occurs between nts on the same strand, resulting in tertiary structure d. Functions of RNA i. All RNAs are transcribed, not all are used to make proteins ii. Discovery for ribozymes providedstrong evidence for the RNA World Hypothesis 1. Altman and Cech wonNobel Prize in Chemistry in 1989 VIII. DNA Replication a. Structure of DNA replicationmade easy b. DNA acts as a template for its ownduplication c. Two strands separate, new bases can be added in d. Three possiblemodels of DNA replication i. Which one is accurate based on figures we have used? 1. Semiconservative 2. Dispersive 3. Conservative 4. Answer: Semiconservative e. Parental strands serve as the templates for the daughter strands i. Original strands remain intact ii. Each new daughter DNA is one oldstrand and one new strand iii. DNA begins at replicationorigins 1. Replicationorigins are species specific 2. Bacterial: AT-richso easy to open. Has one originof 100bp 3. Human genome: 10,000 origins a. What is the benefit to so many origins? i. It’s more efficient, can replicate multiple parts of the DNA all at once b. Initiator protein attracted to origin which then attracts other components: Function as a proteinmachine 4. Replicationbubble starts at the origin a. Two forks form ineach bubble 5. Synthesis is biodirectional and rapid a. 1000 nucleotide pairs/secin bacteria b. Only 100 nucleotide pairs/secin humans c. Why so much lower? f. DNA Synthesis i. Enzyme: DNA polymerase 1. Adds nucleotides added to the 3’ end (-OH group) 2. Synthesis from 5’  3’ 3. DNA polymerase remains attached


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