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Ch. 12 - DNA Organization in Chromosomes

Chapter 12, Problem 27

Spermatogenesis in mammals results in sperm that have a nucleus that is 40 times smaller than an average somatic cell. Thus, the sperm haploid genome must be packaged very tightly, yet in a way that is reversible after fertilization. This sperm-specific DNA compaction is due to a nucleosome-to-nucleoprotamine transition, where the histone-based nucleosomes are removed and replaced with arginine-rich protamine proteins that facilitate a tighter packaging of DNA. In 2013 Montellier et al. showed that replacement of the H2B protein in the nucleosomes with a testis-specific variant of H2B called TSH2B is a critical step prior to the nucleosome-to-nucleoprotamine transition. Mice lacking TSH2B retain H2B and their sperm arrest late in spermatogenesis with reduced DNA compaction. Based on these findings, would you expect that TSH2B-containing nucleosomes are more or less stable than H2B-containing nucleosomes? Explain your reasoning.

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Hey everyone. Let's take a look at this question together to fit the DNA inside the nucleus. The DNA is wrapped around the his stones forming the structure called nuclear zones. What makes the D. N. A. Strongly attached to the his stones. So let's recall what we've learned about the makeup of DNA. To try to figure out why it's strongly attached to his stones. So we know that one key characteristic about DNA is that it is negatively charged. And this is due to the phosphate groups found within D. N. A. And this is important because when we're talking about its strong attachment to his stones, the reason for that being his stones are actually positively charged. And we know that when we have one thing being negative and one thing being positive we have that strong attachment between the two. And so the reason why D. N. A. Is strongly attached to the his stones is due to their charge differences or answer choice. A The correct answer because we know that D. N. Is entirely negatively charged because of its phosphate groups and his stones are positively charged. So we get this attraction between the two because of those opposite charges. And that allows D. N. A. To be strongly attached to the his stones due to their charge differences or answer choice. A. I hope you found this video to be helpful. Thank you and goodbye
Related Practice
Textbook Question
While much remains to be learned about the role of nucleosomes and chromatin structure and function, recent research indicates that in vivo chemical modification of histones is associated with changes in gene activity. One study determined that acetylation of H3 and H4 is associated with 21.1 percent and 13.8 percent increases in yeast gene activity, respectively, and that histones associated with yeast heterochromatin are hypomethylated relative to the genome average [Bernstein et al. (2000)]. Speculate on the significance of these findings in terms of nucleosome–DNA interactions and gene activity.
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Textbook Question
An article entitled 'Nucleosome Positioning at the Replication Fork' states: 'both the 'old' randomly segregated nucleosomes as well as the 'new' assembled histone octamers rapidly position themselves (within seconds) on the newly replicated DNA strands' [Lucchini et al. (2002)]. Given this statement, how would one compare the distribution of nucleosomes and DNA in newly replicated chromatin? How could one experimentally test the distribution of nucleosomes on newly replicated chromosomes?
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Following is a diagram of the general structure of the bacteriophage chromosome. Speculate on the mechanism by which it forms a closed ring upon infection of the host cell.
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Because of its rapid turnaround time, fluorescent in situ hybridization (FISH) is commonly used in hospitals and laboratories as an aneuploid screen of cells retrieved from amniocentesis and chorionic villus sampling (CVS). Chromosomes 13, 18, 21, X, and Y (see Chapter 8) are typically screened for aneuploidy in this way. Explain how FISH might be accomplished using amniotic or CVS samples and why the above chromosomes have been chosen for screening.
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Textbook Question
To gauge the fidelity of DNA synthesis, Arthur Kornberg and colleagues devised a technique called nearest-neighbor analysis, which determines the frequency with which any two bases occur adjacent to each other along the polynucleotide chain (J. Biol. Chem. 236: 864–875). This test relies on the enzyme spleen phosphodiesterase (see the previous problem). As we saw in Figure 11-8, DNA is synthesized by polymerization of 5'-nucleotides—that is, each nucleotide is added with the phosphate on the of deoxyribose. However, as shown in the accompanying figure, the phosphodiesterase enzyme cleaves DNA between the phosphate and the C-5' atom, thereby producing 3'-nucleotides. In this test, the phosphates on only one of the four nucleotide precursors of DNA (cytidylic acid, for example) are made radioactive with ³²P and DNA is synthesized. Then the DNA is subjected to enzymatic cleavage, in which the radioactive phosphate is transferred to the base that is the 'nearest neighbor' on the 5' side of all cytidylic acid nucleotides. Following four separate experiments, in each of which a different one of the four nucleotide types is radioactive, the frequency of all 16 possible nearest neighbors can be calculated. When Kornberg applied the nearest-neighbor frequency test to the DNA template and resultant product from a variety of experiments, he found general agreement between the nearest-neighbor frequencies of the two. Analysis of nearest-neighbor data led Kornberg to conclude that the two strands of the double helix are in opposite polarity to one another. Demonstrate this approach by determining the outcome of such an analysis if the strands of DNA shown here are (a) antiparallel versus (b) parallel:
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