Bio-366-Key-Test2-Fall2009

Bio-366-Key-Test2-Fall2009 - Bio-366 Test-2 September 17,...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Bio-366 Test-2 September 17, 2009 Name: UT EID: Please answer briefly and to the point. I A . Could you suggest mechanisms by which DNA within a cell might acquire negative supercoiling? During DNA replication unwinding of DNA strands behind the replication fork generates negative supercoiling, and compensatory positive supercoiling is generated in front of the fork. The positive supercoiling is released by a topoisomerase (gyrase, for example during E. coli DNA replication), resulting in a buildup of net negative supercoiling. By a similar mechanism negative supercoiling can be generated during transcription as well . B . What is the magnitude of each node formed by the crossing of one B form double helical segment over another in a covalently closed circular DNA? If this crossing results in a right handed plectonemic helix, what is the sign of the node? Is the DNA positively or negatively supercoiled? The magnitude of the node (a superhelical node) is 1. If the plectonemic superhelix is right handed, the node sign is negative. The DNA is negatively supercoiled. Negative node formed by the crossing of the DNA axis: C . Imagine a circular DNA molecule of 10,000 base pairs (B form right handed double helix) in the relaxed state. Imagine that an enzyme cut one strand, removed 50 right handed turns and joined the strand again. Assume that there are 10 base pairs per turn in the relaxed state. What is the Lk0 (relaxed state linking number) for this molecule? What is the linking number Lk after the action of the enzyme? What is the superhelical density of this DNA molecule? Is the DNA positively or negatively supercoiled? For 1,000 bp DNA circl e,
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Lk = Tw +Wr (Lk is the linking number, tw is the twist and Wr is the writhe). In the example, Lk0, the linking number in the relaxed state = Tw0 +Wr0 = 100 (1000/100); In the relaxed state, Wr is zero, so Lk0 = Tw0 = 100 Since 50 turns have been removed, the Lk of the molecule is = 100-50 = 50 Superhelical density, s = [Lk-Lk0] / Lk0 = [50 - 100] /100 = -0.5 The DNA is underwound, negatively supercoiled. For 10,000 bp circular DNA, Lk = Tw +Wr (Lk is the linking number, tw is the twist and Wr is the writhe). In the example, Lk0, the linking number in the relaxed state = Tw0 +Wr0 = 1000 (10000/100); In the relaxed state, Wr is zero, so Lk0 = Tw0 = 1000 Since 50 turns have been removed, the Lk of the molecule is = 1000-50 = 950 Superhelical density, σ = [Lk-Lk0] / Lk0 = [950 - 1000] /100 = -0.05 C . Imagine one conformation of the above molecule in which 50% of the unwinding was partitioned into writhe and 50% into twist. What is the average number of base pairs per turn in this topoisomer? You may give the answer as the numerator divided by the denominator. You
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 10

Bio-366-Key-Test2-Fall2009 - Bio-366 Test-2 September 17,...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online