Draw the structure of a four-residue segment of DNA with the f...

Question

Draw the structure of a four-residue segment of DNA with the following sequence. (3′end) G-T-A-C (5′ end)

Accepted Answer

All right. Hi, everyone. So this question says to write the appropriate structure of a DNA segment that contains four nucleotides with the sequence shown below. Now here the sequence is T AC G going from the three prime end to the five prime end. Now recall that the nucleotides of DNA and RN A are composed of three main things, right? That all have to combine together. The first of which is your sugar oops. Now recall that in DNA, the sugar backbone is two or D two deoxy rivals, which is a five member ring, right? And a hydroxy group on carbon three. Now, on carbon one is the connection between the sugar backbone and a nitrogen space. Whereas on carbon four prime of the ring, our positions here are described as prime, carbon four prime connects to carbon five prime outside of the ring, which is subsequently connected to our third and final component, which is a phosphate group. Now, the structure shown in the question here gives us four different nucleotides each with a different base. However, what I want to point out here is that when drawing the nucleotide sequences of DNA, you should always do. So from the five prime end to the three prime end. So the order written in the question is actually backwards, right? We should be drawing our nucleotides in the opposite order starting with guanine, followed by cytosine, adenine and thymine. So the first thing I'm going to do when it comes to my structure is just go ahead and draw my sugar backbones connected together. He's my first one and here is my first phosphate group, making sure to add my hydrogens as well. So when it comes to linking our sugar molecules together, the three prime hydroxy group of the first sugar is going to connect with the phosphate group of the second and so on. Right. So the second phosphate group is only going to have one negative charge because one oxygen is connected to our first ring here. So with that, I draw my second sugar with all of my hydrogens. And then later on, I'm going to attach each each base. But for now, I just want to make sure I have the foundation right now, I'm going to connect the three prime hydroxyl group of my second sugar to the phosphate group of the next nucleotide. And with that, I connect the third sugar. And later on, I will go ahead and attach that base. But for now, right, the last step that I need to do is connect the three prime hydroxy group of my third sugar to the fourth and final phosphate group here of my last nucleotide. And with that, I draw my last sugar now because this fourth and final sugar is not going to connect to any other phosphate groups. Considering our sequence is only four nucleotides long, I'm going to go ahead and put a free hydroxy group on its carbon three prime. So now let's go ahead and scroll upwards a bit to talk about our basis. Now, as discussed previously, right nucleotides are to be drawn from the five prime end to the three prime end. So the order is going to be guanine followed by cytosine, adenine and thymine. Now recall that adenine and guanine are the purine bases of DNA and Arne. Whereas cytosine and thymine are referred to as the pyrimidines. Now there is one more pyrimidine il but that one is only present in D in R and excuse me. So the first space that we're going to draw here is going to be Guan oops. But before I do that, let me actually go ahead and move this first. So now, right, guanine is going to be the base associated to the first nucleotide that I have here. So the connection between the the nitrogen is space and the sugar backbone is going to occur on carbon one prime. So here I'm going to draw a line upwards from carbon one prime and that is where I connect make one. So from this nitrogen that's connecting to carbon one prime I go ahead and I fill in the rest of the structure. Right. Now, the purines look very similar to each other in that they both have this fused two ring structure. However, one primary difference between them, right is it Guan in has a car bono group in addition to a free amino group outside of the rink. So now this is Guan in. And so next on the list is going to be cytosine. So cytosine, I'm going to go ahead and draw connected to carbon one prime of my second sugar backbone. And so here right, cytosine is a pyrimidine, which means that it is going to have only one ring in its structure. But in this case, it's also going to have a car bal and a free amino group. So now this is cytosine and up next is Adoni. Now adenine is going to have a very similar structure to guanine on carbon one prime of my third sugar. And the reason why I say that it's going to have a similar structure is because Adenine is going to have a very similar looking two ring structure with nitrogen insect, right? But the difference is that Adam does not have a car bono group, the way that Guan Yin does. So here is adenine and last but not least is thyme and thymine being another pyrimidine is going to have a similar structure as cytosine in that it also has a one ring structure. Now thymine is different from cytosine is that it actually has two car bono groups and one methyl group. So there you have it. This last base I do here is timing and so that completes our nucleotide structure. And here what I also want to bring to your attention is that when we go from the five prime end to the three prime end, we're going from the phosphate group on the outside to this inner hydroxy group on carbon three prime. So I know there was a lot, right? But if you stuck around to the end, thank you so very much for watching. And I hope you found this helpful.

Draw the structure of a four-residue segment of DNA with the following sequence. (3′end) G-T-A-C (5′ end)

Key Concepts

DNA Structure

Base Pairing

Directionality of DNA Strands

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