BIOL230W Week 2

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Pennsylvania State University *

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230W

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Biology

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Feb 20, 2024

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: BIOL230W Week 2 - Inside the Nucleus During Week 1 we explored the symptoms of cystic fibrosis in the context of transport across membranes, the endon}cmhmne system, and the Central Dogma. This was an introduction to some of the major processes we will study in more depth this semester. Our first topic is the structure of DNA and its role in information storage. Day 1 Learning Objectives 1. Describe the evidence supporting RNA as the early information storing molecule. 2 Descn_be why DNA is better as an information storage molecule compared to RNA. 3. Explain the benefits of a nucleus. 4. Identify and describe the major features of the monomeric and polymeric forms of DNA (ex.backbone, base, sugar, etc). 5. Describe how monomers form the DNA polymers including the rule of base pairing (A:T and G:C). 6. Describe the role of hydrogen bonding between DNA strands and how it affects DNA strand separation. 6. Evaluate how changes to the structure of DNA may impact function. 7. Describe the activity of major DNA modifying enzymes. Small Group Activity: Recap of the Central Dogma and the importance of a nucleus and DNA 1. This week’s Exploration Homework explored the RNA world hypothesis. Please list the evidence provided in the article that supports the hypothesis that RNA acted as our hereditary molecule before the evolution of DNA? woh ¢ opiluty v 2. Why is it important that RNA can fold into unique motifs (conformations)? — . . ) Adla. 4‘,/‘ 2 LSE ’ oL 78 o UNR'g : NarL Yy MO LCA LS 3. In eukaryotic cells, the nuclear membrane separates the DNA from the cytoplasm. What was the original benefit of the evolution of simple cellular compartments in the RNA world? ¢ prans plAnas W and "i“‘n| wanslAtion - occny ; \ ) Sir AEX (20jon S winih inccesdes yegulrbon aned proce ) - J 4. What are the benefits of DNA vs. RNA as an information storage molecule? FNA Yos on exva rydx) ayou) WY Ch Odcveadel 118 STAVIMTH- A Byntlwsi 2iney e ‘,/i.ui." YO ESTEY
BIOL230W Week 2 — Inside the Nucleus Worksheet - DNA: From Monomer to Polymer A nucleotide is the monomeric building block of the polymer of DNA. To better understand the characteristics ul.l)T\A, 1Lis important to understand how the monomers join to form the polymer. Use the following diagram of DNA (and the diagram of a nucleotide in the PowerPoint presentation) to label the listed features: 3" and 5 ends, nucleotide, phosphate group, sugar group, phosphodiester bond, nitrogenous base (pyrimidine and purine), and hyvdrogen bonds. 5. Looking at the polymer, why is a DNA considered antiparallel? \ J v S AL\ v < )\ 6. Why are the ends of DNA strands named 3" or 5°? Vs A 1% s ey i ¥ *(nuf“) Wk ar A OW geoup oF Ha . 1 b o TP rnirel nypor 1 4 Sugar Illuj [ J ' = W evgl o e ;“}sm‘.fil Yhat hes e fith carben in Hu 7. How do A:T interactions differ compared to G:C interactions? How does this affect stability of the bonded nucleotide pairs? L«‘.L'nyffigk\Dfl; A E‘HOWSPV Aue 4+ an ;f(‘(:;."-l<HUM( H-bondl A 5!?“‘7(& re £ N mor e WwiHh mave G C Pare rq"') NW"A be S‘f"‘a/gnf Y one with - AT ":\1 rs: 8. During gel electrophoresis, to which electrode (+ or -) does DNA migrate? Why? The DA 1§ vl W QoSitive Decaube of e elefvical current thet 3§ t’!«.:;zcm\::) “'va:i)"? e (Jé“ trom - Hh T. 5
BIOL230W — Week 2, The Nucleus and DNA Day 2 Learning Objectives 1. Use a temperature vs. UV absorbance chart to determine characteristics about the DNA. 2. Describe the relationship between DNA, chromosome, gene and allele using cystic fibrosis as an example. 3. Explain the overall structure of a nucleosome. 4. Compare and contrast the structure and function of euchromatin vs. heterochromatin. Small Group Activity: Recap of DNA Structure 1. How does the structure of DNA change if phosphodiester bonds are targeted for hydrolysis? N 2. How does the structure of a DNA molecule change if all hydrogen bonds between nucleotides are broken? 3. Does enzymatically removing the phosphate group from a 5’ end of one strand of DNA change the directionality of the DNA strand? Why or why not? /{0 ’ ~ . Airechoredity , PAT dAre 7o d v B 0 ) s / A melting, is the process by which double- stranded DNA separates into single- stranded DNA via heat or other denaturation agents. The process can be observed by using absorption of UV light: double-stranded DNA has a relative absorbance of UV light at 1; the relative absorbance increases as the double- stranded DNA becomes single-stranded. Denaturing DNA can tell scientists information about the specific DNA being examined. Tm = temperature at which half of the DNA sample is single-stranded. Use the diagram below to answer the J questions. 2 2 DNA denaturation, also called DNA + Temperature (°C) 4. What is the x-axis measuring? What is the y-axis r&a&r&gl PR ~ mperature of \51’1 DA (- E‘ki'w’ B M omgurk of UVl o s been)&” J aysovioed by the DNA
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BIOL230W — Week 2, The Nucleus and DNA 5. What is the relationship between temperature, UV absorbance and DNA denaturation? 10 | { {5; 7. T 4 1 fo A A vavialzles 6. What bonds are broken in the DNA molecule due to the heating processes? 7. A DNA sample shows a curve shifted to the right so that that the Tm is higher (i.e., 80 degrees C). What does that tell you about the A:T, G:C content of the DNA sample compared to the known sample measured above? e res, [More erexveny s . - S SRLNAS Veécaude ol +Han odari 8. The length of all DNA in a human nucleus (if you added all 46 chromosomes together) is about six feet in total. This is analogous to fitting a spool of thread into a Tylenol pill capsule. How would you do that while protecting the thread (DNA) from knotting, tangling or tearing? i / 2l vV iTivie, ns VI y rovngl Nt/ weve, floing ovned 1 \ / '] - 2 — (2 | l,') tiyn, MY pe -?r_"i‘r.'szlf ‘J’):,W”/aix Review Activity: Chromosome vs. Gene. vs. Allele 9. In the image below distinguish in some way a chromosome, a gene and an allele so that the difference makes sense to you. e A= Norme [ ,,.« CHROMOSOMES ( o -mutent cneo” et lfl gumcf \ 7@ A baby receives ONE copy i 2 _1‘ ene- (Y/O"‘d_) l n | of the gane from EACH parent. I m. 3 U } \, (B maton | L—) wWith 1 ¢ snc_ allede The baby wil have ONE of the four following combinations of genes: / SHE R B . ) m\ s £ alleles exgress e B0 %% oo x‘(l;,v\ or fl or hJ o [{ ™ i { L > U U U g e " Non-carter Carvier Camer Aflectsd WthCF 3 7 L 75% chance wnaftectsd healthy) ——— L— 25% chan la allete CNOWASSAAL

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