I. What is the most common form of DNA in a cell? _____________________ II. Each

Question

I. What is the most common form of DNA in a cell? _____________________
II. Each nucleotide pair of DNA double helix weighs about 1x 10-21g. If the human body contains approx. 0.5g of DNA, how many nucleotide pairs of DNA are there in the human body? ______________________________________________________

III a. If you assume that all the DNA is in the B-DNA form (and not condensed into
chromosomes), how far would the above 0.5g of DNA reach if stretch end to end? (Show your calculations clearly).
III b.. How many helical turns would there be in total? _______________________

Explanation / Answer

I. most common form of DNA in a cell is B-DNA form

II. Weight of each Nucleotide pair of DNA double helix = 1 x 10^-21 g

Mass of DNA is human body = 0.5 g

Number of nucleotide pairs of DNA in human body = 0.5 / 1 x 10^-21 = 5 x 10^20 nucleotide pairs / human

III.(a) Average distance between each nucleotide pair in B-form of DNA = 3.4 nm

Number of nucleotides per human = 5 x 10^20

Distance stretched by DNA from end to end = 5 x 10^20 * 3.4 = 1.7 x 10^21 nm = 1.7 x 10^12 m

III.(b) In B-DNA there are 10 nucleotides per turn

Number of helical turns = 5 x 10^20 / 10 = 5 x 10^19 helical turns

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