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Chapter 29: Problem 13

(Integrates with Chapter 11 .) The SWI/SNF chromatin-remodeling complex peels about 50 bp from the nucleosome. Assuming B-form DNA, how long is this DNA segment? In forming nucleosomes, DNA is wrapped in turns about the histone core octamer. What fraction of a DNA turn around the core octamer does 50 bp of DNA comprise? How does 50 bp of DNA compare to the typical size of eukaryotic promoter modules and response elements?

Short Answer

Expert verified
The DNA segment is about 17 nm long. The 50 base pairs of DNA comprises about 3.03 turns around the core octamer. And, 50 base pair of DNA is larger than the typical size of eukaryotic promoter modules and response elements.

Step by step solution

01

Size of B-DNA

A B-DNA helix has a periodicity of about 10.5 base pairs per turn of the helix. Thus, for 50 base pairs of DNA, it will form about 4.76 turns of the helix. Each base pair is 3.4 Å apart, so 50 base pairs will measure approximately \( 50 \times 3.4 = 170 \) Å, or 17 nm.
02

Fraction of a turn

Now, the DNA that is wrapped around the histone core makes about 1.65 turns, which is about 16.5 base pairs per turn. So, if the one turn of the DNA helix contains 16.5 base pairs, the fraction of a DNA turn around the histone core for 50 base pairs would be approximately 3.03 turns.
03

Comparing with promoter modules

Typical eukaryotic promoter modules and response elements can vary in size, and usually range from about 5 to 20 base pairs. Hence, a segment of 50 base pairs is significantly larger in comparison.

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Most popular questions from this chapter

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