Chapter 11: Problem 7
Why might we predict that the organization of eukaryotic genetic material will be more complex than that of viruses or bacteria?
Chapter 11: Problem 7
Why might we predict that the organization of eukaryotic genetic material will be more complex than that of viruses or bacteria?
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Get started for freeDescribe the transitions that occur as nucleosomes are coiled and folded, ultimately forming a chromatid.
In this chapter, we focused on how DNA is organized at the chromosomal level. Along the way, we found many opportunities to consider the methods and reasoning by which much of this information was acquired. From the explanations given in the chapter, propose answers to the following fundamental questions: (a) How do we know that viral and bacterial chromosomes most often consist of circular DNA molecules devoid of protein? (b) What is the experimental basis for concluding that puffs in polytene chromosomes and loops in lampbrush chromosomes are areas of intense transcription of RNA? (c) How did we learn that eukaryotic chromatin exists in the form of repeating nucleosomes, each consisting of about 200 base pairs and an octamer of histones? (d) How do we know that satellite DNA consists of repetitive sequences and has been derived from regions of the centromere?
What are the molecular composition and arrangement of the components in the nucleosome?
Mammals contain a diploid genome consisting of at least \(10^{9}\) bp. If this amount of DNA is present as chromatin fibers, where each group of 200 bp of DNA is combined with nine histones into a nucleosome and each group of six nucleosomes is combined into a solenoid, achieving a final packing ratio of \(50,\) determine: (a) the total number of nucleosomes in all fibers. (b) the total number of histone molecules combined with DNA in the diploid genome. (c) the combined length of all fibers.
The human genome contains approximately \(10^{6}\) copies of an \(A l u\) sequence, one of the best-studied classes of short interspersed elements (SINEs), per haploid genome. Individual Alus share a 282 -nucleotide consensus sequence followed by a 3 '-adeninerich tail region. Given that there are approximately \(3 \times 10^{9}\) bp per human haploid genome, about how many base pairs are spaced between each Alu sequence?
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