LEHNINGER PRINCIPLES OF BIOCHEMISTRY - VOL. 3. INFORMATION PATHWAYS - 2017

PART III. INFORMATION PATHWAYS

24. GENES AND CHROMOSOMES

Questions and Problems

1. DNA Packaging in Viruses.

The Molecular Weight of bacteriophage T2 DNA is 120 • 10⁶; this DNA molecule is packaged within a bacteriophage HEAD 210 nm long. Calculate the length of the DNA (determine the molecular weight knowing that The Genome contains 650 nucleotide pairs) and compare it with the size of the T2 head.

2. M13 Phage DNA.

The Nucleotide Composition of M13 phage DNA is: 23% A, 36% T, 21% G, 20% C. What can be deduced about the M13 phage DNA from these data?

3. Mycoplasma Genome.

The complete genome of the smallest bacterium, Mycoplasma genitalium, consists of 580,070 bp and is organized as a circular DNA molecule. Determine the molecular weight and length (in the relaxed state) of this molecule. What is the $L_r^0$ of the Mycoplasma chromosome? If $\sigma = -0.06$, what is the $Lk$?

4. Size of Eukaryotic Genes.

An enzyme isolated from rat Liver contains 120 amino acid residues and is encoded by a Gene 1,440 bp in size. Explain the relationship between the number of ami

no acid residues in the enzyme and the number of nucleotide pairs in its gene.

5. Linking Number.

A covalently closed circular DNA molecule in a relaxed form has $Lk = 500$. Estimate the approximate number of nucleotide pairs in this DNA. How does the linking number change (increases, decreases, remains unchanged, cannot be determined) if: (a) a protein complex is attached to a nucleosome; (b) One DNA strand is broken; (c) DNA gyrase and ATP are added to the DNA solution; (d) The Double Helix is denatured by heating?

6. DNA Topology.

In the presence of eukaryotic condensin and type II topoisomerase, the $L_r$ of a relaxed covalently closed circular DNA does not change. However, numerous nodes appear within it.

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Node formation requires DNA breakage, the passage of a DNA segment through the break, and ligation mediated by a topoisomerase. Given that the topoisomerase reaction must be accompanied by A change in the linking number, explain the conservation of the $Lk$ value.

7. Supercoiled DNA Density.

Bacteriophage $\lambda$ infects E. coli by integrating its DNA into the bacterial chromosome. The success of this recombination depends on the topology of the E. coli DNA. If the superhelix density of E. coli DNA is $\sigma > -0.045$, the probability of integration is less than 20%; if $\sigma < -0.06$, the probability exceeds 70%. It has been established that the length of plasmid DNA isolated from an E. coli culture is 13,800 bp, and $L_r = 1,222$. Calculate $\sigma$ for this DNA and predict the probability that bacteriophage $\lambda$ will be able to infect these Cells.

8. Alteration of the Linking Number.

(a) What is the $L_r$ of a circular double-stranded DNA molecule of 5,000 bp with a single-strand break (nick)? (b) What is the $L_r$ of the molecule from part (a) after ligation of the break (relaxation)? (c) How would the $Lk$ of the molecule from part (b) change As a result of the action of a single E. coli topoisomerase I molecule? (d) What is the $L_r$ of the molecule from part (b) after eight catalytic cycles of a single DNA gyrase enzyme molecule in the presence of ATP? (e) What is the $Lk$ of the molecule from part (g) after four catalytic cycles of a single bacterial topoisomerase I molecule? (f) What is the $L_r$ of the molecule from part (g) after binding of a single nucleosome?

9. Chromatin.

Experimental evidence that helped determine The Structure of nucleosomes was obtained by agarose gel Electrophoresis (electrophoretogram shown below), in which the bands correspond to DNA molecules. The samples for electrophoresis were obtained by mild Treatment of chromatin with a DNA-cleaving enzyme, followed by the removal of all Proteins. The positions to which linear DNA fragments of corresponding sizes migrate are indicated on the side. What can be concluded about Chromatin Structure from these data? Why are the DNA bands broad and diffuse rather than narrow and concentrated?

10. Introduction/20.html">DNA Structure.

Explain how partial unwinding of B-DNA can facilitate or stabilize The formation of Z-DNA.

11. Maintenance of DNA Structure.

(a) Name two structural properties required to maintain negative supercoiling in a DNA molecule. (b) Name three structural changes that occur more readily upon negative DNA Supercoiling. (c) Which enzyme, with the participation of ATP, can introduce negative supercoils into DNA? (d) Describe the physical MECHANISM OF ACTION of this enzyme.

12. Yeast Artificial Chromosomes (YACs).

YACs are used to clone large DNA fragments in yeast cells. What Three types of DNA sequences are necessary for the correct Replication and propagation of a YAC within a yeast Cell?

13. STRUCTURE OF THE Bacterial Nucleoid.

In Bacteria, the METABOLISM/31.html">Transcription of gene clusters depends on DNA topology: relaxation of DNA may enhance expression, but more frequently diminishes it. When the bacterial chromosome is cleaved by a specific restriction enzyme (which recognizes and cleaves long and consequently rare sequences), expression is enhanced or diminished only for nearby genes (within 10,000 bp). The transcription of other chromosomal genes remains unaffected. Explain this observation. (Hint: See Fig. 24-37).

14. DNA Topoisomers.

During electrophoretic Separation of DNA in an agarose gel, short fragments migrate faster than long ones. Circular DNA molecules of the same size but with different linking numbers can also be resolved in an agarose gel: highly twisted and therefore more compact topoisomers migrate faster (from top to bottom in the gels on the right). The dye chloroquine was added to the gels; it intercalates between Base Pairs and stabilizes the least twisted DNA molecules. When the dye binds to the relaxed form of circular DNA, it unwinds the DNA in the regions where the dye molecules have bound, while compensatory positive supercoils are formed elsewhere. In the experiment shown here, topoisomerases were used to prepare samples of the same circular DNA with varying superhelix densities (σ). Fully relaxed DNA migrated to position N (nicked), whereas highly twisted DNA (beyond the limit where individual topoisomers can be resolved) migrated to position X.

a) Why does gel A show multiple bands in the lane with σ = 0 (i.e., in the DNA sample where the average σ = 0)?

b) Does the DNA in gel B from the sample with σ = 0 carry negative or positive supercoils in the presence of the intercalating dye?

c) In both gels, the lane with σ = -0.115 exhibits two bands: one corresponding to supercoiled DNA and the other to relaxed DNA. Explain the presence of relaxed DNA in this and other samples.

d) The native DNA sample (far-left lane in each gel) represents the same circular DNA isolated from bacterial cells without topoisomerase treatment. What is the approximate superhelix density of this native DNA?

Analysis of Experimental Data

15. Functional Elements of Yeast Chromosomes.

Figure 24-9 illustrates the main Structural elements of chromosomes from baker's yeast (Saccharomyces cerevisiae). Hieter, Mann, Snyder, and Davis (1985) determined The properties of several of these elements. Their experiment was based on the observation that during mitosis in yeast cells, Plasmids (containing genes and an origin of replication) behave differently from chromosomes (which contain the same elements as well as centromeres and telomeres). Plasmids are not controlled by the mitotic apparatus and are distributed randomly between daughter cells. In the absence of a selective marker forcing the host cell to retain the plasmids (see Fig. 9-4, Vol. 1), they are rapidly lost. In contrast, chromosomes are controlled by the mitotic apparatus and, even without selective markers, are lost extremely rarely (at a frequency of about 10^-5 per Cell Division).

Hieter and co-workers set out to identify the most critical elements of yeast chromosomes. To do this, they constructed plasmids containing various segments of chromosomes and monitored whether these "synthetic chromosomes" segregated properly during mitosis. Determining the number of abnormally distributed chromosomes required a rapid assay method capable of counting the number of copies of synthetic chromosomes in individual cells. They chose a method based on the fact that wild-type yeast colonies are white on nutrient medium, whereas colonies of certain adenine-requiring (ade^-) mutants are red. Specifically, ade2^- cells lack functional AIR carboxylase (The enzyme catalyzing step 6a in Fig. 22-33), causing AIR (5-aminoimidazole ribonucleotide) to accumulate in their Cytoplasm. Excess AIR is converted into a red pigment. In addition, the study utilized the SUP11 gene, which encodes an ochre suppressor (a type of nonsense suppressor; see Box 27-4) that suppresses the phenotype of certain ade2 mutants. Hieter et al. began by studying a diploid yeast strain homozygous for the ade2^- mutation; these cells have a red color. If these mutant cells carry a single copy of SUP11, the metabolic defect is partially suppressed, and the cells turn pink. If a cell contains two or more copies of SUP11, the defect is fully suppressed, and the cells appear white.

The researchers inserted a copy of SUP11 into synthetic chromosomes carrying various elements deemed essential for chromosomal function, and then monitored how successfully the chromosomes were transmitted from one generation to the next. Pink cells were plated onto non-selective media, and The behavior of the synthetic chromosomes was observed. In particular, colonies were identified in which the synthetic chromosomes missegregated at the First Division after plating, producing colonies where half of the genes belonged to one type and half to another. Because yeast cells are stationary, the colonies grew as sectors: half one color, half the other.

a) One mechanism that accounts for mitotic nondisjunction is that chromosomes replicate, but sister chromatids fail to separate, so that both copies of the chromosome end up in the same daughter cell. Explain how the nondisjunction of synthetic chromosomes could lead to the formation of two-colored red-and-white colonies.

(b) Another type of mitotic defect is chromosome loss, where the chromosome fails to enter the Nucleus of the daughter cell or fails to replicate. Explain how the loss of a synthetic chromosome could lead to the formation of two-colored red-and-pink colonies.

Counting Different types of colonies allowed Hieter and coworkers to estimate the frequency of mitotic defects for various types of synthetic chromosomes. First, they sought to determine the required size of the centromeric fragment using synthetic chromosomes with inserts of DNA of varying lengths containing a known centromere. The experimental results are presented below.

c) Based on the provided data, draw a Conclusion about the size of the centromeric fragment required for normal chromosome segregation. Explain your reasoning.

d) Interestingly, all the synthetic chromosomes generated in these experiments were circular and lacked telomeres. Explain how they could replicate more or less successfully.

Next, Hayter and colleagues constructed a set of linear synthetic chromosomes containing functional centromeric sequences and telomeres, and determined the total rate of mitotic errors (% chromosome loss + % nondisjunction) as a function of chromosome size. The experimental results are presented below.

e) Based on these data, draw a conclusion about the chromosome size required for normal segregation. Explain your reasoning.

f) Normal yeast chromosomes are linear, ranging from 250 to 2000 kb in length, with a mitotic error rate of no more than 10^-5 per cell division. Using this information, answer the following question: did the centromeric and telomeric sequences used in the experiments ensure the mitotic stability typical of normal yeast chromosomes, or are other elements required for this process? Explain your reasoning. (Hint: Consider the plot of the logarithm of the error rate versus chromosome length.)