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Labeling Reading FramesFor questions 12, writc thc first thrcc DNA nucleotides of cach of the six reading frames of the DNA duplex in the image below. Be sure to la...

Question

Labeling Reading FramesFor questions 12, writc thc first thrcc DNA nucleotides of cach of the six reading frames of the DNA duplex in the image below. Be sure to labcl each~AGCACCAGAC - ~TCGTGGTCTG -Question (1 point)The first three DNA nucleotides of the +1 reading frame areQuestion (1 point)The first three DNA nucleotides of the +2 reading frame areQuestion (1 point)The first three DNA nucleotides of the +3 reading frame are 3'Question (1 point)The first three DNA nucleotides of the readi

Labeling Reading Frames For questions 12, writc thc first thrcc DNA nucleotides of cach of the six reading frames of the DNA duplex in the image below. Be sure to labcl each ~AGCACCAGAC - ~TCGTGGTCTG - Question (1 point) The first three DNA nucleotides of the +1 reading frame are Question (1 point) The first three DNA nucleotides of the +2 reading frame are Question (1 point) The first three DNA nucleotides of the +3 reading frame are 3' Question (1 point) The first three DNA nucleotides of the reading frame are Question 10 (1 point) The first threc DNA nucleotides of the reading frame are Question 11 (1 point) The first thrcc DNA nucleotides of the reading frame are Question 12 (2 points) Write the first three RNA nucleotides that would be produced the +3 reading frame was transcribed.



Answers

How is the reading frame of a nucleotide sequence set?

Hello everyone. Welcome to chapter 10 for the question number nine. Here we have two questions A. And B. First of all let's look at part one. The D. N. A sequence from its fine end to three and is right starting from the bottom of the jail Where the smallest d. a fragments mega rate each band results from the incorporation of the appropriate eight the D. U. S. Nuclear tides try false fate. And as expected there were no two bands that have the same mobility. This allows one to determine the D. N. A. Sequence by reading of the bands in strict order, proceeding upward from the bottom of the jail and assigning the correct nuclear tied according to which lane the band is in the nucleotide sequence of the top. Dead. Okay. Was behind the directly from the data of figure in question # nine and the bottoms dried was deduced from the complementary base pairing roots. So much for the question eight let's come to be the D. N. A. Sequence can then be translated into an immuno acid sequence using the genetic code. However there were two strengths of the A. That could be transcribed into R. N. A. And three possible reading frames for each spread. Thus there are six you know essayed sequences that can be in principle be encoded by this stretch of the A. Of the three reading frames possible from the top. Instead only one is not interrupted by a stop Carden From the bottoms. Just two of the three reading frames also have stop cordons. The third frame gives the following sequence. Now look at this sequence. It is not possible from the information given to tell which of the two open reading frames can response to the actual pertain encoded by the stretch of the age. What additional experiment could distinguish between these two possibilities. So much for this. Thank you for listening.

Hi welcome to another numerator video. The purpose of this video is to kind of give a visual representation of the processes of transcription and translation. Kind of help understand it a little bit better. It may sound like I'm kind of rambling but I am actually trying to answer some questions that come at the end of Chapter seven in the book. Biology concepts and investigations third edition. Um So that there there is uh in order to it if it hopefully it will it will appear that way transcription is the process of taking the D. N. A. A. Strand of DNA. And using it to build a strand of RNA trans translation. Then is the process of taking that RNA and using it to build a polyp peptide. And so in this picture the blue material is the DNA and the green material is the messenger RNA that we're that we're building from the D. N. A. Now i it's getting kind of messy so I won't draw it in. Didn't mean to do that. Um Yeah it's getting a little messy so I won't draw it in. But there would be a a molecule of RNA preliminary race involved in this. Now If on the DNA over to the left you see that's the three prime end and the right is a five prime end. So on the messenger RNA it's said to be anti parallel. So the right is going to be the three prime ends of the messenger RNA. And the The far left would be the five from in. And so it's the three prime end where the transcription is happening. And so this end down here is where the preliminary star in a preliminary race would be preliminary race. Anytime you hear that a sc ending, you know, you're talking about an enzyme polymer is as a molecule made of multiple parts, multiple sub units bonded together. And so RNA polymerase is an enzyme that is bonding the subunits of RNA together. It's building the RNA. So the art april and race would be working on that end. Now there are a couple of things we can tell from looking at this picture. The first is that transcription is happening along the DNA strand. So the DNA strand is opened up and it is create is using that to build a messenger RNA strand. And we could actually tell which bases are going to be um added next if we follow this sequence. So here we have guangming and the D. N. A. And that and that is going to pair with a side of scene. So on the RNA strand you're going to put aside a scene right here. Next is a dinning. And what would bind with at a mean is thing I mean, except I mean is not found in messenger RNA. RNA instead it's going to be you're a cell. And so then another year or so would go here then this thing I mean is going to pair up with anatomy. This side is seen is going to pair up with the guarani. This I've seen with the guarani. This thing I mean with an ataman adami this quantity with cida scene. So you can just go down the line here and know which bases are gonna come next and this stranded messenger already. So that is the idea of transcription that you've taken DNA and built RNA from it. Now if this were a eukaryotic cell, this strand of messenger RNA would have to leave the nucleus in order to go out and join with the river is home and to start translating that code into actual protein or poly peptide. This appears to be a pro carry attic sell because the translation process is happening basically at the same time as transcription. So transcription you were taking DNA and making messenger RNA translation. You're using the messenger RNA to build a polyp peptide in this picture there to polyp peptides being built And that's what's happening here with the red. The red represents transfer RNA. So the transfer RNA is going to bring a um an amino acid and hook up with the code on on the messenger RNA strand. And then the next piece of transfer RNA will hook up with the next coat on and as they do that will bring the amino acids close together and then they will form a bond. Obviously there are other enzymes involved with that. But then those so those amino acids will be bonded together, eventually forming a polyp peptide. So we see that this is happening simultaneously. We're wielding to polly peptides here, polly peptide B. And polly peptide A. So we could ask ourselves, you know, what's going to be the next amino acid that comes into the into the polyp peptide B. So we we see that glue scene has been added. Um The transfer RNA has what's called an anti code on that matches with a coat on Which is a three bay sequence on the messenger RNA. So if we look at the very next code on and our messenger RNA, it's A A. G. And that's going to require an anti code on uh uracil here is still side A scene. Or you you see and so we can ask what's going to be the next thing that bonds with this. You you see and I have a chart in the book where we can look that up. So we can look up at the code on A A. G. Which corresponds to the anti code on you. You see codes for the amino acid lysine. So the next amino acid here in this polytech Diaby is going to be lysine. Now, another thing that we can tell from this picture is that we're not seeing the entire sequence that this polyp peptide is going to form. Um if if we keep looking at these code ons here, G U C C U A G and so forth. We don't find the stop code on. Um the stop code ons are you A U A G and U G A. We're not finding that stop code on. And so we assume that this is not the complete um sequence of amino acids in this polyp peptide that it's going to continue building until it gets to a stop code on. Once it does get to a stop code on and the probably peptides are going to be released and they will probably then fold into a different configuration depending upon how long this polyp peptide is. It might be a complete protein by itself or it might then join with other polyp peptides to create a protein and then that protein will then carry out its function. So you can see so much going on in a picture like this. But just just review um transcription is taking the DNA code which we see in the blue down here in the bottom and using that to build a molecule of messenger RNA. Then translation is taking those code ons that are on the messenger RNA. And transfer RNA comes in the anti code on on transfer RNA binds to the code on on the messenger RNA, bringing in place an amino acid. Those amino acids are then put together to form a polyp peptide. So that's several things you can see that are going on with transcription and translation Again. I encourage you to read about it in the book and then um and then try and reproduce some of this by yourself. You'll learn better that way. Hope this helps have a great day.

In here us that the DNA is derived from a middle of a city and a loan of a 1,000,000 protein. So it asked us if we can determine that amino acid sequence of this portion of protein. So essentially it is not going to be possible due to the fact that from the information that we're given, we cannot determine the amino acid sequence here because of the three reading frames, all of which do not contain a start code on here. And that's the reason why we cannot tell anything about this particular sequence. Um, and that's going to be for part B. For part eight. Ask what's the sequence of DNA? And that's used in the sequencing reaction shown here. So we have four legs that show the products of sequencing reactions that contained the DDG D D A and DDT, which are video nucleotides, which are used in singer sequencing on Essentially in orderto determined the sequence itself. We look at what DeOssie died, the oxy I'm nucleotide that to be used for example, G g a. T versi here, and essentially whatever is gonna end up at the last year is going to be the sequence. So we go from bottom top in this particular case,

So this question gives us, um, six different sequence reads, and so we're going to be using these six sequence reads to pray a sequence content of this part of the Gino. So you're just going to align the sequences to determine their overlap? Um, and once you overlap them according to where they would match up with each other, you create the, uh, kontic of, um um, then, since you do not know if the DNA sequence represents a temple Ishan for the Amorin A or is a complementary strength and there are two possible, um, transcripts. So you either turn all the cheese into use or, um, you will turn off the tees into a zone all the seas into G's and vice versa. So then that question be asks us to transit the sequence contact each of all possible reading frames. So the translation of the first possible transcripts starting at the first letter reads, uh, C l P Okay, translation starting at the second letter. So sissy. First, this is the second translation, starting at the third reads for the second part, services first and now for the second possible. This transcript, starting at the first letter. It will just read Stop for the second possible transcripts, starting at the second letter and then for the third. And now we have, ah, questions. See, that asks us to go to the blast page of the, um and C b i website and see if you can identify the gene. So using the nucleotide sequence of the con take and performing the blast, um, or the possible translation products and performing t blast in You will discover that the sequence in the translation product with above licit first match perfectly with the region of Exxon 19 of, um human. See if tr gene


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