Transcription and Translation Practice Problems 

13:3 what is the middle sequence ma

THIS! First video actually someone saying if its 5 to 3 or 3 to 5 on DNA or mRNA, because it's opposite so it's really hard without being told exactly the way it is ;P

Thanks, glad it worked for you! I'll be adding more videos this fall so let me know what bio topics would be most useful and I'll see what I can do...

@@professorisin Awesome! Transcription, translation, Prokaryotic cells, Eukaryotic cells, reproduction cycle, genetics, punnett squares, and hardy weinberg. Any or all of these would be AMAZING!

@@here2learn847 I second this pleassssse

samee.... are u from MIT,Manipal?!

@@vittalprabhu379 Ayo wtf YES. What section dawg . I'm from CC

You're welcome, glad it helped! Thanks for subscribing to my channel!

When reading the mRNA strand from 5' to 3', you go along one base at a time until you encounter AUG, the start codon. You should ignore any stop codons that might appear before then. So if you have a stretch like 5' - CCUGAUGGGA... - 3', then you'd start 1 base at a time: CCU (nope), CUG (nope), UGA (stop codon, ignore and nope), AUG (that's it, the start codon!) From AUG, you'd translate 3 at a time from that point: AUG, GGA, and so forth. You'll do this until you encounter the first stop codon. At the first stop codon, a release factor enters the ribosome and the complex falls apart -- the protein has been made. So any other stop codons will not be seen by the ribosome.

In this video, we start with the template strand in the 5' to 3' direction because it's convention to write a segment of DNA in the forward direction (5' to 3'). It is true that the template strand is always read in the 3' to 5' direction, because the new DNA or RNA is synthesized from 5' to 3', in the opposite direction. Other videos may have already done the work for you and started with the sequence in the 3' to 5' direction, while in this video, you need to rewrite the piece of DNA in the 3' to 5' direction before you begin replicating or transcribing it. Or alternatively, you can just rewrite the resulting RNA in the 5' to 3' direction before you translate it, as is done here. This video starts with a 5' to 3' sequence so we can learn to become more comfortable checking the ends to make sure the RNA strand is in the 5' to 3' configuration before translation.

Great,I love this explanation.

There's a lot to consider in your question! First, if you're given one strand of *DNA* and no additional information, then you'd likely have to assume this is the template strand you should use to *transcribe* the mRNA sequence. And if the DNA sequence is written with the 3' end on the left and the 5' end on the right side, then it's all set up to be transcribed into an mRNA sequence that would have the 5' end on the left and 3' end on the right. Because *translation* into a protein occurs in the 5' to 3' direction, the resulting mRNA is ready to be translated into a protein. Start at the 5' end of the mRNA and walk one base at a time until you find AUG, the start codon. From there, you proceed three bases at a time (the three bases are called a "codon"). IRL you should encounter a stop codon at some point to indicate the end of the protein. Remember there are three stop codon possibilities: UAA, UAG, UGA.

Check the sequence you wrote again because it's easy to get shifted if you're writing out the nucleotides. I double-checked and there are no extra nucleotides before the UGA stop codon. Because you're right -- once you find the start codon, AUG, then the RNA is read in 3-nucleotides chunks (called "codons") until a stop codon is encountered.

Sure! Here we are working with trying to become more comfortable with DNA and RNA sequences of nucleotides. DNA is double stranded, and there can be genes on both stands. In our case, the transcribed gene is from the top strand of DNA. So when working with this sequence, you either have to reverse the initial DNA template strand so it's 3' to 5', or you have to reverse the resulting RNA transcript afterwards so it's 5' to 3' before you translate it -- and that's what we did here. (On exams or practice problems, professors often will give you the template strand in the 3' to 5' direction already to make it easier for you to translate the result immediately. However, when you're looking at a gene in real life the sequence of DNA may be in the 5' to 3' direction because it's convention to write DNA in this direction. This video is a little more realistic about the strands in order to remind us to watch the ends and make sure the RNA sequence is 5' to 3' before we translate it to the protein. In the cell, RNA polymerase and the ribosome will always bind on the correct ends because there are factors and proteins that assemble there to recruit them.) Does all that make any sense? :)

5’ AUG UGA CGC GAU ACG CUU UAU AGC UUC AGU GGG CCG AUC UGA 3’

@@rjaguirre3637 It might be worthwhile to ask your professor if they intended this. You're right that the ribosome would stop at the first stop codon, leaving a single methionine in the protein (which is just an amino acid, not a functional protein). Or maybe that was the point of this question? Tricky!

Uracil is necessary to start translation as part of AUG, the start codon. If there's no U in the mRNA, then there will be no protein produced. (If you have a strand without any U, double check to make sure it's not DNA. If it's DNA, then you have to transcribe it into mRNA before translating the new sequence into a protein.)

There should always be a stop codon if you started at a start codon and went in three-letter chunks (codons). If you're doing a practice question and don't find a stop codon, you may want to check that A) the mRNA is in the 5' to 3' direction, and B) you went 1 nucleotide at a time until you got to the start codon, AUG, and then went codon by codon. Remember, too, that there are 3 stop codon options: UAA, UAG, and UGA!

First, make sure it's the mRNA and not DNA. Next, go one nucleotide at a time, 5' to 3', until you find "AUG." That's where the protein will start being translated and where you will start reading 3 nucleotides each time, using the "genetic code" key, until you reach a stop codon.

The mRNA is translated (read by the ribosome) in the 5' to 3' direction. When you finish transcribing it from the DNA, it's in the 3' to 5' direction, from left to right. You don't have to flip it, but if you didn't you'd have to read the sequence from right to left. So I always find it easier to set it up 5' to 3', left to right, before I translate it.

If it's mRNA and there's no AUG, then translation won't start. But remember that the ribosome moves 1 nucleotide at a time until AUG, and *then* it reads in 3 nucleotide codons until a stop codon is reached.

It's actually thymine (T) that is replaced by uracil (U) in RNA. And that means that adenine (A) will base pair with uracil (U). Check out this video for more explanation of how DNA and RNA are made: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-ORPtbsJD6kU.htmlfeature=shared&t=236 (The RNA section starts at 8:21 into the video)

"5 prime" and "3 prime" indicate the ends of a strand of DNA. Check out this video for why we call the ends this: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-8l00SCMn2nM.htmlfeature=shared&t=145

If you're talking about the translation of mRNA transcripts from eukaryotic genomic DNA, then the ribosome wouldn't be able to translate this into a protein without the AUG start codon. There are a few exceptions to this rule in prokaryotes and mitochondria, however. Otherwise, if this question appeared on a test about eukaryotic translation, it could be a trick question (no protein would be made) or it could just be an exercise to see if you can use the genetic code (but not necessarily how the ribosome would).

That's right! (Except there's thymine in DNA and uracil in RNA.) You can compare the coding strand of DNA and the mRNA transcript at ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-8xamWBYhv_w.html.

Cool idea! Hmm, I wonder how one becomes a voice actor -- research time!🙂

Yes, RNA is synthesized by RNA polymerase and translated by ribosomes in the 5' to 3' direction. This video is more about becoming comfortable working with the sequences of DNA and RNA. DNA is double stranded, and you might be given a DNA strand written in the 5' to 3' direction (like here), so it's good to know how to transcribe and translate that sequence. In the laboratory and in scientific publications, DNA is usually written in the 5' to 3' direction (check out ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-bmsvXbOhhNI.html)

We know that this is a DNA strand because it has thymine (T) and not uracil (U) in the sequence. If this sequence is used as the template to make a piece of mRNA, then you'd use the complementary base in the mRNA. We know that thymine (T) basepairs with adenine (A), and cytosine (C) basepairs with guanine (G). So if the DNA strand starts with TAC, as in your example, then the corresponding mRNA would start with ATG. And then we'd continue down the sequence using the basepairing rules. (On a side note, we have to make some basic assumptions when answering this question. For instance, we have to assume this is the template strand of DNA, not the coding strand, because the template strand is used as the "template" to make the mRNA -- hence it's name! We also don't know the direction of the DNA, like which is the 5' or 3' end. That doesn't change the mRNA sequence, just how the RNA polymerase would create the strand because it makes the new strand in the 5' to 3' direction.)

@@professorisin dont we know its 3 5 direction since it has the TAC sequence?

How so? Remember that our starting piece of DNA here is the template strand for the mRNA, not the coding strand of DNA 🧬

No, in this example the top strand of DNA is being used as the template strand to transcribe mRNA. The sequence you list would be the answer if the question asked to transcribe the bottom DNA strand. It's good practice (although the resulting transcript will have only the start codon and not the stop codon).