The length of chromosome 2 would be 0.34 nm/bp × 243 million bp = 82.62 million nm = 82,620 μm, or about 13,770 times longer than the diameter of the nucleus.

Lk₀ = (4,200 bp) / (10.5 bp/turn) = 400. From Equation 9-1, ΔLk = Lk − Lk₀ = 374 − 400 = −26. Substituting the values for ΔLk and Lk₀ into Equation 9-2: σ = ΔLk/Lk₀ = −26/400 = −0.065. The superhelical density is negative, so the DNA molecule is negatively supercoiled. When the same molecule has an Lk of 412, ΔLk = 412 − 400 = 12, and σ = 12/400 = 0.03. The superhelical density is positive, so the molecule is positively supercoiled.

(a) The DNA has ∼171,000 bp; at 0.34 nm/bp, the DNA length is 58,140 nm. The DNA is almost 600 times longer than the JS98 head. (b) 170,523 bp.

The content of A does not equal the content of T. The simplest explanation is that the DNA is single-stranded.

The DNA has a molecular weight of 580,070 bp × 650/bp = 377,045,500. The contour length is 197,224 nm. Lk₀ = (580,070 bp) / (10.5 bp/turn) = 55,245. If σ = −0.06, Lk = 55,245 − (55,246 × 0.06) = 51,930.

The DNA has ∼5,250 bp. (a) In the absence of strand breakage and resealing, Lk is unchanged; a positive supercoil must form elsewhere in the DNA to compensate. (b) Lk is undefined. (c) Lk decreases. (d) No change.

Lk remains unchanged because the topoisomerase introduces the same number of positive and negative supercoils.

Lk₀ = (13,800 bp) / (10.5 bp/turn) = 1,314. σ = (Lk − Lk₀)/Lk₀ = 292/1,314 = 20.07. Superhelical density is the same for the cellular chromosome and the plasmid, so the probability of infection is >70%.

(a) Lk undefined. (b) Lk = 500. (c) No effect. (d) Lk = 484. (e) Lk = 488. (f) Lk = 484.

Z-DNA is a left-handed double helix. Underwinding of the right-handed B-form helix will make a left-handed helix easier to form.

(a) The DNA must be unbroken and topologically constrained so that Lk < Lk₀. (b) Strand separation, formation of hairpins and cruciforms, and formation of Z-DNA are all more favorable in negatively supercoiled DNA. (c) DNA gyrase introduces negative supercoils into DNA, with the aid of ATP. (d) The mechanism involves creation of a double-strand break, passage of an unbroken DNA segment through the break, followed by strand resealing. Transient phosphotyrosyl-DNA intermediates form, and the conformational changes are coupled to hydrolysis of ATP.

S-10

The DNA must include origins of replication, required for DNA replication; a centromere, for proper segregation of the chromosome at cell division; and telomeres, to protect the chromosomal ends.

(a) The lower, faster-migrating band is negatively supercoiled plasmid DNA. The upper band is nicked, relaxed DNA. (b) DNA topoisomerase I relaxes the supercoiled DNA. The lower band will disappear, and all of the DNA will converge on the upper band. (c) DNA ligase produces little change in the pattern. Some minor additional bands may appear near the upper band, due to the trapping of topoisomers not quite perfectly relaxed by the ligation reaction. (d) The upper band will disappear, and all of the DNA will be in the lower band. The supercoiled DNA in the lower band may become even more supercoiled and migrate somewhat faster.

(a) When DNA ends are sealed to create a relaxed, closed circle, some DNA species are completely relaxed but others are trapped in slightly under- or overwound states. This gives rise to a distribution of topoisomers centered on the most relaxed species. (b) Positively supercoiled. (c) The DNA that is relaxed despite the addition of dye is DNA with one or both strands broken. DNA isolation procedures inevitably introduce small numbers of strand breaks in some of the closed-circular molecules. (d) σ ≈ 20.05. This is determined by comparing native DNA with samples of known σ. In both gels, the native DNA migrates most closely with the sample of σ = −0.049.

Form I DNA was negatively supercoiled. When spread on an electron microscope grid, the DNA would tend to fold onto itself, creating DNA crossings or nodes. In form II DNA, the circles are relaxed.

The pattern in lane 2 is produced by DNA gyrase; that in lane 3 by DNA topoisomerase III (a type I topoisomerase).

The superhelical density would increase. The new mutations occur in the two genes encoding subunits of DNA polymerase II (gyrase), such that the activity of gyrase declines.

(a) 25 nodes. (b) Removal of 25 of 667 DNA turns would correspond to a σ of −0.037. (c) ΔLk = 25/(−0.89) = −28; thus, σ = −0.042. (d) No.