Structure. Section 6.5 Questions, Page 294 1. the Following Structures Are

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Structure. Section 6.5 Questions, Page 294 1. the Following Structures Are Section 6.5: DNA Organization in Eukaryotes and Prokaryotes Mini Investigation: Packing DNA, page 292 A. Answers may vary. Sample answer: No, I was not successful in fitting the thread into the capsule. I wasn’t able to use an efficient packing system, so the thread may not have been packed to take up the least possible space. Also, the thickness of the thread takes up space, so I couldn’'t fit in as much as I thought I might be able to. B. This activity mimics DNA packing in the nucleus because DNA molecules are very long and can move and tangle similar to the string. The DNA is required to have efficient packing structures to fit inside the cell nucleus, like the efficient packing structures required to fit the string into the capsule. C. Answers may vary. Sample answer: This activity differs from the packing of DNA into a nucleus in that the latter utilizes a variety of associated protein complexes to add packing efficiency. D. DNA packed in chromatin has the advantage of being protected and has a reduced volume so that it fits more easily into the nucleus. E. If the thread was the true thickness of DNA it would have easily fit into the relatively large capsule, but still would have become tangled if it did not have an efficient packing structure. Read pg. 291-294 Do Q#1-6, 8 and 9 pg. 294 Read pgs. 295-298 Do Q#1-7 pg. 298 Section 6.5 Questions, page 294 1. The following structures are in order of size from smallest to largest: histone, nucleosome, solenoid, and chromosome. –5 2. Given: length of a nucleus, LN = 10 m –8 length of a solenoid, LS = 3 × 10 m Required: number of solenoids that can fit sided by side in the nucleus, NS Solution: Step 1. Divide the length of the nucleus by the length of a solenoid. NS = LN ÷ LS –5 –8 = (10 m) ÷ (3 × 10 m) 2 NS = 3.3 × 10 Statement: This nucleus can fit 300 solenoid structures side by side within it. 3. DNA wraps tightly around the histone complex because of polar and ionic interactions. The histones are positively charged and the DNA is negatively charged. 4. Answers may vary. Sample answer: DNA-packing Strategies Prokaryotes Eukaryotes - contain circular - contains linear chromosomes chromosomes - uses supercoiling to - use chromatin and 5. Answers may vary.increase Sample packing answer: Due tohistones the volume to increase of DNA that must be replicated in eukaryotic cells, efficiencyas compared to bacteria, packingthe process efficiency would simply take too long in eukaryotic cells -if therefree floatingwere only DNA one replication- contains site. telomeres An alternative to hypothesis is that the small size of the circular chromosomes ofprotect bacteria the prevent ends ofs replication from occurring in more than one spot simply due to the lack chromosomesof space; multiple sites cannot be unwound simultaneously in bacteria. - DNA contained in a 6. (a) Methylation occurs when a methyl groupnucleus is added to a molecule. Some arginine and lysine side chain residues undergo post-translational methylation. (b)5. TheAnswers histone may “tails vary.” refer Sample to regions answer: of D theue tohistone the volume that modulate of DNA DNA that must accessibility be replicated withinin eukaryotic a nucleosome. cells, as The compared tails contain to bacteria arginine, the and process lysine would residues simply that aretake methylated too long in Copyright © 2012 Nelson Education Ltd. Chapter 6: DNA: Hereditary Molecules of Life 6.5-1 andeukaryotic demethylated cells ifto there modulate were structureonly one and/orreplication interactions site. An of alternative the core histone hypothesis tails ,i sor that to servethe small as binding size of sites the forcircular ancillary chromosomes proteins or of enzymes. bacteria prevent s replication from occurring 7.in Answers more than may one vary. spot Sample simply aduenswer: to the Immunodeficiency, lack of space; multiple centromeric sites cannot instability, be unwound and facialsimultaneously dysmorphism in bacteria. (ICF) syndrome is a very rare autosomal disorder caused by a mutation6. (a) Methylation to the gene occurs (DNMT3B when) encodinga methyl grDNAoup methyltransferaseis added to a molecule.-3B resulting Some arginine in and hypomethylatiolysine side chainn of residues histone underproteins.go post ICF- translationalis a very serious methylation. disease characterized by facial dysmorphism,(b) The histone recurrent “tails” referand prolonged to regions respiratory of the histone infections, that modulate infections DNA of accessibilitythe skin and digestivewithin a system,nucleosome. and variable The tails immune contain and arginine immunoglobulin and lysine residues deficiencies. that are Since methylated histones areand involved demethylated in the mostto modulate basic structure structure of and/or DNA coilinginteractions, I would of the expect core thathistone histone tails , or to defectsserve ascould binding have sites serious for ancillaryconsequences. proteins or enzymes. 8.7. T Answershe benefit may to having vary. Sample large non answer:-coding Immunodeficiency, repeating sections centromericin the telomeres instability, of DNA and is thatfacial these dysmorphism protect the coding(ICF) syndrome regions from is a beinveryg rare lost autosomalduring replication. disorder Somecaused DNA by a at the endmutation of the strandto the geneis removed (DNMT3B and )not encoding replaced DNA with methyltransferase every replication.- 3BThe resulting presence in of non- codinghypomethylatio regions protectsn of histone the important proteins. coding ICF is regionsa very serious in the centredisease of characterized the DNA strand. by facial 9.dysmorphism, During the organisms recurrent’ firstand prolongedreplications respiratory there will infections,be little effect infections of losing of the nonskin- and codingdigestive regions. system, The and larger variable the regions immune are, and presumably immunoglobulin the longer deficiencies. the organisms Since can histones reproduceare involved without in the suffering most basic any structure effects. The of DNA real problemcoiling, Istarts would as expectthe organism that histone undergoesdefects could continued have serious reproduction. consequences. Eventually there will be no more non-coding regions left8. Tinhe the benefit telomeres to having and the large coding non -regionscoding repeatingof the DNA se ctionswill be in affected. the telomeres Losing of telomere DNA is regionsthat these probably protect limits the coding the age regions a cell might from beinachieve,g lost therefore during replication. limiting lifespan. Some DNA at the end of the strand is removed and not replaced with every replication. The presence of non- coding regions protects the important coding regions in the centre of the DNA strand. 9. During the organisms’ first replications there will be little effect of losing the non- coding regions. The larger the regions are, presumably the longer the organisms can reproduce without suffering any effects. The real problem starts as the organism undergoes continued reproduction. Eventually there will be no more non-coding regions left in the telomeres and the coding regions of the DNA will be affected. Losing telomere regions probably limits the age a cell might achieve, therefore limiting lifespan. Copyright © 2012 Nelson Education Ltd. Chapter 6: DNA: Hereditary Molecules of Life 6.5-2 Copyright © 2012 Nelson Education Ltd. Chapter 6: DNA: Hereditary Molecules of Life 6.5-2 .
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