Unformatted text preview: as G and C, but since G must equal C, we know that G = C = 30 %. 30 % of 8.4 million is 2.52 million G’s in one copy of the chromosome. 2. If there are 8.4 million deoxynucleotides in one copy of the chromosome (see part 1 above), there must also be 8.4 million deoxyriboses, as there is one deoxyribose in each deoxynucleotide. 3. There is, on average, a distance of 3.4 Å (= 0.34 nm = 3.4 E-10 meters) between adjacent base pairs. So the total length of one chromosome will be (3.4 E-10 meters/bp) (4.2 E6 bp) = 1.428 E-3 meters =1.428 mm. If you wanted to be more precise you could say that for 4.2 million bp there are really only 4,199, 999 3.4 Å spaces between bp and so the actual distance is 3.4 Å less than 1.428 mm, but 1.428 mm is plenty good enough for these purposes!...
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- Spring '06
- DNA, base pairs, nucleotide, Base pair