If you haven't yet answered question 10, pause the video now and try it. If you have ribose or deoxyribose for that matter and it's linked to a base like adenine, for example, what you have is a nucleoside. Because the side ending as opposed to the tied ending means that there's no phosphate group here. So looking at our molecule that we are given which is again ribose attached to adenine, we would call that molecule adenosine. That's the name of that particular nucleoside. It would not be a deoxyribonucleoside because we're using ribose not deoxyribose. While adenine is appearing, it's not a nucleotide because there's no phosphate group present. And likewise, 'c' is wrong because it's not a nucleotide and also adenine is not even a pyrimidine so it's like double wrong. Adenosine monophosphate implies that there's a phosphate group attached, which there's not. So that's out too, leaving us with 'e'.
Looking at question 11, the double helix form of DNA is stabilized by 'd', nonspecific base stacking interaction between 2 adjacent bases in the same strand. So if we look at a strand of DNA, that's saying that the 2 bases along one strand have an interaction between them. And here are 2 bases on the other strand. So if we zoom in on this image, basically our bases are kind of stacked on top of each other. And they have these stacking forces that are holding them together. These are in fact hydrophobic forces. But it's worth noting that water interacts with the outside of DNA. I'm drawing little water molecules here. And you might remember that the outside of DNA has all those phosphate groups on it which are negatively charged, meaning water is going to play nice with them. So those are actually hydrophilic forces that help stabilize the structure as well just on the outside as opposed to the inside. And, just looking at the wrong answer choices real quick. Covalent bonds between the 3 prime end of one strand and the 5 prime of the other. Nope. There's a phosphate group in the middle there. Right? We have that phosphodiester bond. And, looking at 'b', hydrogen bond between the phosphate groups, no. We have hydrogen bonding between the bases. Hydrogen bonds between the ribose. Again, no. Hydrogen bonds are between the bases. And ribose interactions with the planar base pairs, while they're covalently bonded, that's about it. So, and and again, all those don't help stabilize the structure. So, 'd' is the correct answer.
Looking at question 12, double stranded RNA takes an A form and A forms and B forms are right handed always. Those are right handed helices. Z is a left handed helix. So double stranded RNA takes the A form. B form is what we find most DNA in. We see the z form around regulatory sequences.
Looking at 13, what is the complementary strand to the one given? Well, I've given you a little hint by including the 5 prime and 3 prime up there, because remember, DNA is always written 5 prime to 3 prime. So while 'a' might look like a good answer, it is actually a trap. And that's because here is our 5 prime end and here is our 3 prime end. 'B' is the correct answer. 'B' is the correct answer and that is because if we write it backwards we have our 3 prime end t aacgtaand 5 prime. And you can see that that is complementary to the strand given there. Now, it's worth noting that this question could have been fair game without the 3 prime and 5 prime indicated. You're supposed to know that stuff.
Alright. When DNA is heated to about 95 degrees Celsius, which change does not occur. And while UV light absorbance does increase, right, this is called the hyperchromic shift. Do not break. Those will stay together. Those are covalent bonds. They're really strong. They're not going to be broken down by temperatures of, temperatures that are that low. The helix will unwind. Right? The DNA is going to denature and, the strands are going to separate which means that the hydrogen bonds between the bases are going to break. And additionally, the viscosity of the solution will decrease. That's not really a fact you need to worry about too much. The facts in 'a', 'c', and 'd' are much more important for you to understand.
So looking at question 15, what is the role of dideoxy CTP in DNA sequencing by the classical Sanger method? The answer is 'd'. The role of dideoxy CTP is to occasionally terminate formation of the complementary strand. All the dideoxynucleotides are used in order to terminate formation of the complementary strand early so that we develop fragments of different lengths. 'A' is wrong because we wouldn't include the same dideoxynucleotide in all four reaction mixtures. We would only include it in, we would include 1 in each, right? So we'd have one that's a, one that is t, one that is C, and one that is G. And it doesn't necessarily have to be, you know, that particular order but they're done separately from each other. Specific enzymes are used to cut the product into smaller pieces. No. The pieces are made because of early termination of synthesis. You know, we use a much smaller concentration. We use like millimolar, concentration of dNTPs and like micromolar concentration of ddNTPs. And so every now and then, one of these, dideoxynucleotides is added and because it lacks a 3 prime hydroxyl group, synthesis can't continue on that strand so you get, early termination. So you wind up with fragments of the whole strand that you're trying to replicate. Dideoxynucleotides are present at low levels, right? Micromolar compared to millimolar, much, much smaller like over a 1000 fold difference. The template strand is radioactive so that it can cleave the product into small pieces. This is nonsense. We use radioactively labeled dideoxy nucleotides so that we can, figure out, you know, like which strands they're attached to and and, you know, use it to read them on gels. But the radioactivity is just in order to identify the molecules, not to cut anything. It's not really how radioactivity works anyways. With that, let's flip the page.
