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.

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?

Experimental evidence demonstrates that the nucleosomes present in a cell after the completion of S phase are composed of some 'old' histone dimers and some newly synthesized histone dimers. Describe the general design for an experiment that uses a protein label such as ³⁵S to show that nucleosomes are often a mixture of old and new histone dimers following DNA replication.

DNase I cuts DNA that is not protected by bound proteins but is unable to cut DNA that is complexed with proteins. Human DNA is isolated, stripped of its nonhistone proteins, and mixed with DNase I. Samples are removed after 30 minutes, 1 hour, and 4 hours and run separately in gel electrophoresis. The resulting gel is stained to make all DNA fragments in it visible, and the results are shown in the figure. DNA fragment sizes in base pairs (bp) are estimated by the scale to the left of the gel. Draw a conclusion about the organization of chromatin in the human genome from this gel.

DNase I cuts DNA that is not protected by bound proteins but is unable to cut DNA that is complexed with proteins. Human DNA is isolated, stripped of its nonhistone proteins, and mixed with DNase I. Samples are removed after 30 minutes, 1 hour, and 4 hours and run separately in gel electrophoresis. The resulting gel is stained to make all DNA fragments in it visible, and the results are shown in the figure. DNA fragment sizes in base pairs (bp) are estimated by the scale to the left of the gel. Examine the gel results and speculate why longer DNase I treatment produces different results.

Genomic DNA from the nematode worm Caenorhabditis elegans is organized by nucleosomes in the manner typical of eukaryotic genomes, with 145 bp encircling each nucleosome and approximately 55 bp in linker DNA. When C. elegans chromatin is carefully isolated, stripped of nonhistone proteins, and placed in an appropriate buffer, the chromatin decondenses to the 10-nm fiber structure. Suppose researchers mix a sample of 10-nm–fiber chromatin with a large amount of the enzyme DNase I that randomly cleaves DNA in regions not protected by bound protein. Next, they remove the nucleosomes, separate the DNA fragments by gel electrophoresis, and stain all the DNA fragments in the gel.

Approximately what range of DNA fragment sizes do you expect to see in the stained electrophoresis gel? How many bands will be visible on the gel?

Human chromosome 5 and the corresponding chromosomes from chimpanzee, gorilla, and orangutan are shown here. Describe any structural differences you see in the other primate chromosomes in relation to the human chromosome.

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 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.

Give descriptions for the following terms:

Histone proteins