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Welcome to Biology.SE! I think you've come to a reasonable conclusion, but not for exactly the right reasons. The genomes of several strains of Escherichia coli have been sequenced (e.g. https://www.ncbi.nlm.nih.gov/nuccore/U00096) so you could predict the sizes of the fragments you would expect from your digest. The genome size of E. coli is about 4.6 ...


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The individual values in the OP's onion label are not out of line with the U.S. Department of Agriculture's reference data. There are two of the latter, one for "Onions, raw" and one for "Onions, sweet, raw". Their figures would lead to expecting carbohydrates of 13.8g or 11.2g, respectively, for a 148g sample. The negative difference can be explained by ...


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The values on the nutrition label from the question (11 g total carbs, 9 g sugars, 3 g fiber) could be actually explained by rounding: Total carbohydrates: 11.45 (rounded to 11) Sugars: 8.55 g (rounded to 9 g) Fiber: 2.55 g (rounded to 3 g) ...which would mean there are additional 0.35 grams of non-sugar, non-fiber carbohydrates... ...BUT... ...using ...


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the answer will be mature DNA. We know that except for nucleus other organelle such as mitochondria also have ability to replicate an as mentioned in the question the enucleated ovum (ovum lacking its nucleus) do have the other cell organelles, whereas for the mature erythrocytes, they lack nucleus as well as other cell organelles such as mitochondria and ...


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