5'+to+3'+explained

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Explain the 5' - 3' 5' + 3' directions (DNA)?



First, you should understand what 5' and 3' mean. You'll understand when you look at how DNA is formed.

As you can see, each nucleotide comprises: one ribose (a 5 carbon sugar ring - in the box on the diagram - where each carbon is numbered 1 to 5); one phosphate group; and one nitrogenous base (A, T, C or G). The phosphate group is attached to C5 (the 5th carbon) of ribose, which sticks out from the ring. The nitrogenous base (purine or pyrimidine) is attached to C1 of the ribose. Notice that there's a hydroxyl group (-OH) attached to C3 of the ribose. This is important to DNA formation. DNA strands are formed when the phosphate of one nucleotide attacks and replaces the -OH on C3 of another nucleotide. The result, as shown by the two joined nucleotides at the top of this diagram, is a "phosphodiester bond": two nucleotides held together by a phosphate. Notice that the first nucleotide (the one with "T" attached to it) still has its C5 phosphate, but as lost its C3 -OH. The second nucleotide (the one with "C" attached) has its C5 phosphate now bound to C3 of the first nucleotide. This phosphate to -OH attack is repeated many times as the nucleotides string together to form a long DNA strand. When the strand has been completed and no more attacks take place, you've got a strand with two ends that look very different. At one end, you've got the very first nucleotide that was attacked. It's got its C5 phosphate group sticking out into space. This is what we call the "5' end." At the other end, you've got the very last nucleotide added to the strand. It's the only one that still has its C3 -OH. This is what we call the "3' end." DNA, or more specifically, the sequence of bases (A, T, C, G) attached to the nucleotide chain, is "read" from the 5' end to the 3' end. This is what we mean when we say "5' to 3' direction."