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Chapter_9_trouble_problems

# Chapter_9_trouble_problems - histones 1.65 x 10...

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Chapter 9 trouble problems 9.10 A) There are 2 nucleotides per base pair so 2x 200,000 = 400,000 B) 200,000 (base pairs) x .34 nm (Average distance base pairs are stacked apart)= 68,000 nm Tall 68,000 nm/ 3.4 nm (distance per spiral) = 20,000 spirals C) 400,000 (One phosphorus atom per nucleotide) 9.17 The solution to this problem is on page 215, to answer it you would just compare and contrast the facts listed in the text. ** Dr. Larkin said you will not see a problem like this in the future** 9.22 Note: The information you need to solve this problem is found on page 225. The nucleosome core is made up of 146 nucleotide pairs and each linker consists of 60 pairs of nucleotides. So together they consist of 206 nucleotide pairs. A diploid Drosophila melanogaster nucleus contains 3.4 x 10 8 nucleotide pairs So 3.4 x 10 8 (nucleotide pairs) / 206 (nucleotide pairs per nucleosome/linker unit) = 1.65 x 10 6 nucleosomes 146 nucleotide pairs of DNA wrap as 1 ¾ turns around an octamer (8 proteins) of

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Unformatted text preview: histones 1.65 x 10 6 (nucleosomes) x 8 histones = 1.32 x 10 7 / 4 (4 histone proteins involved ) = 3.3 x 10 6 molecules of each histone (H2a, H2b, H3 and H4). 9.34 3 x 10 9 nucleotide pairs per genome (bp/genome) Average nucleotide pair MW = 660 g/mol So MW of haploid human genome = 3 x 10 9 x 660 g/mol = 1.98 x 10 12 AVOGANDRO’S #: Note: one Mole of any substance contains 6.02 x 10 23 molecules/mole 6.02 x 10 23 (genome copies or ”molecules”) / 1.98 x 10 12 (grams of genome) = 3.04 x 10 11 (genome copies per gram) So since 10 6 μg = 1g there are 3.04 x 10 5 copies of genome per μg Or since 10 3 mg = 1g there are 3.04 x 10 8 copies of genome per mg ** The question asks for copies per mg, so I am not sure why the solutions manual gives the answer in copies per μg. I figured this might be confusing so I included the answer in both forms....
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Chapter_9_trouble_problems - histones 1.65 x 10...

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