Answer2 - MCB 161 Lecture 2 Structure of DNA chromosomes...

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Unformatted text preview: MCB 161 Lecture 2: Structure of DNA, chromosomes, and genomes MBOC problems 4-7, 4-11, 4-12, 4-17, 4-29, 4-44, 4-56 MBOC 4-7 The complementary strand has the sequence 5’ GTGCACCAT 3’ MBOC 4-11 Study Problems In all samples of double-stranded DNA, the mole percents of A and T are equal as are the mole percents of G and C. Results such as the one presented in the problem stood out as odd in the days before the structure of DNA was known. Now it is clear that, while all cellular DNA is double stranded, certain viruses contain single-stranded DNA. The genomic DNA of the M13 virus is single stranded. In single-stranded DNA, A is not paired with T, nor G with C, and so the A = T and C = G rules do not apply. MBOC 4-12 The 5’ end of this DNA molecule is at the top, while the 3’ end is at the bottom. Remember that the carbons in the ribose sugar ring are numbered clockwise around the ring, starting with C1’, the carbon to which the base is attached, and ending with C5’, the carbon that lies outside the ribose ring. MBOC 4-17 This experiment implicates DNA as the genetic material. Clearly, by giving rise to progeny, bacteriophage T4 must possess some form of genetic material. Since T4 contains only protein and DNA, there are only two choices for the genetic material. By separating the bacteria from the bacteriophage after infection and showing that the bacteria contained only the 32P-label (DNA), Hershey and Chase were able to demonstrate a clean separation of DNA from protein. The ability of these infected cells (in the absence of T4 proteins) to generate progeny virus shows that DNA must be the bacteriophage’s genetic material. There are caveats associated with these experiments, as the authors clearly recognized. For example, an absolute separation of bacteriophage from bacteria could not be accomplished: there was 1% 35S associated with the infected cells. Also, not all proteins can be labeled with 35S-methionine (some don’t have methionine in their amino acid sequence). Thus, these experiments don’t absolutely rule out protein as the genetic material. Nevertheless, the weight of the argument falls on the side of DNA. These experiments, coupled with other experiments such as the demonstration by Avery, MacLeod, and McCarty that DNA was the transforming principle of Streptococcus pneumoniae, convinced scientists of the role of DNA as the genetic material. Reference: Hershey AD & Chase M (1952) Independent functions of viral protein and nucleic acid in growth of bacteriophage. J. Gen. Physiol. 36, 39–56. MBOC 4-29 False. In living cells nucleosomes are packed upon one another to generate regular arrays in which the DNA is more highly condensed, usually in the form of a 30-nm fiber. The beads-on-a-string form of chromatin is usually observed only after the 30-nm fiber has been experimentally treated to unpack it. MBOC 4-44 The total length of DNA in chromosome 1 is 9.5 x 107 nm [(2.8 x 108 bp) x (0.34 nm/bp)], which is 9.5 x 104 mm. In mitosis the chromosome measures 10 mm. Therefore the DNA molecule in chromosome 1 is compacted 9500fold (9.5 x 104 mm/10 mm) at mitosis. MBOC 4-56 The presence of a nuclease-resistant fraction in chromatin—but not in naked DNA—suggests that the Martian DNA is associated with a nucleosomelike structure that protects it from micrococcal nuclease. Since extensive digestion produced a limit product of about 300 nucleotides, the nucleosomelike structure must protect about this length of DNA. The smear of digestion products indicates that the nucleosomelike structures are not regularly spaced along the DNA as Earthly nucleosomes are. If they were regularly spaced, they would have given a ladder of bands analogous to those seen in rat liver. Additional problems 1. Why does GC-rich DNA have a higher melting temperature than AT-rich DNA? GC basepairs are held together by three hydrogen bonds, whereas AT basepairs are held together by two hydrogen bonds. For a DNA duplex of a given length, greater thermal energy is required to disrupt one with predominantly GC pairing than one with mostly AT pairing. 2. Examine the structure of the nucleosome as shown in Figure 7.22. How many core histone proteins are there? How many times is DNA wrapped around these core histones? Look at the histone tails in this structure. What functions do these tails have? What effect does acetylation of these tails have on gene expression? Why does acetylation affect gene expression? There are 4 types of core histone proteins; they form an octomer shaped a bit like a hockey puck around which DNA wraps 1.65 times. Histones have unstructured tails that extend out past the wrapped DNA. When unacetylated, histone tails help to promote binding of histones to DNA (they are positively charged and the sugar backbone of DNA is negatively charged). They also participate in the formation of the 30-nm fiber form of DNA by interacting with core histones on adjacent nucleosomes. In particular, it has been shown that lysine 16 of histone H4 (abbreviated H4 K16) binds to an acidic patch on the histone H2A of an adjacent nucleosome. Acetylation of histone tails interferes with both these functions, resulting in less tightly packed DNA. In fact, acetylation of H4 K16 inhibits formation of the 30 nm fiber in vitro. Thus histone acetylation can make DNA more accessible to transcription factors and increase gene expression by inhibiting chromatin compaction. ...
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