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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 Gs 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 (= 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 spaces between bp and so the actual distance is 3.4 less than 1.428 mm, but 1.428 mm is plenty good enough for these purposes!...
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- Spring '06