Question

I need help in understanding transcription/translation with regards to Question 3d. found on the problem set http://ocw.mit.edu/courses/biology/7-01sc-fundamentals-of-biology-fall-2011/molecular-biology/transcription-translation/MIT7_01SCF11_2.3sol.pdf

Solution 3d. says the first 5 amino acids translated are N Met-Leu-Tyr-Pro-Ala C ---- how is this possible looking at the strands and also answer to 3c. ? I am really confused as 3c. answer says the first 15 nucelotides are 5’ CUAAUAUUGUGAGAU 3’, would this not encode an entirely different set of amino acids? Am i missing something? HELP!

Answer 1

for 3c,
the question says 'transcription begins with and includes the red and underlined C/G (top strand/bottom strand) base pair and RNA polymerase proceeds from left to right along the DNA.'

Like DNA, RNA strand is made from 5' to 3'. If you are not sure which strand RNA polymerase will use as template, the question gives you a hint: RNA polymerase proceeds from left to right along the DNA. So you pick the bottom strands because the polymerase will run from left to right to make new rna from 5' to 3' direction. So your newly made mrna will be CUAAUAUUGUGAGAU.

for 3d,
now, here comes the translation part. There are 15 nucleotides on mrna, this means you have 5 codon because one codon has three base. Codon base pair with anticodon which is found on trna. If your first codon is CUA, the anticodon will be GAU. The first amino acid codes by this codon CUA is leucine but the answer showed is methionine. which means the mrna is read from 3' to 5'. But as far as i know, ribosome read mrna in 5' to 3' direction since the polypeptide chain is made from N terminus to C terminus and amino acid is attached at the 3' end of trna (and you read trna from 3' to 5' direction).

Answer 2

Yukitou -Thanks for your reply. I got the answer to 3c.
Now for 3d. Ok so what you are trying to say is that the newly made mRNA - CUAAUAUUGUGAGAU the anti codons will be
GAUUAUAACACUCUA. So working on this antcodon, if I start at 3` to 5`, but read the condons from 5` to 3`, i get
CUA ACU ACA UAU GAU --- Is this what you meant? I still don't get where Met comes in... Could you please clarify?

Answer 3

yes, you're right on reading codon.

I think i type it wrongly for mentioning the methionine. I think, although i read mrna in 5' to 3' or 3' to 5' direction, there is no genetic code that codes for methionine. Methionine is coded by AUG.

Answer 4

So if this is correct:
CUA ACU ACA UAU GAU then this should code for amino acids Leu-Thr-Thr-Tyr-Asp. If i consider that there is no genetic codon for Met but it codes for AUG, where is AUG? Or do we just insert it by default? Am more confused than ever.

Answer 5

yes, you're right for thee amino acids.
i also doubt if we can get the protein translated out if there is no start codon.
Don't worry, you're right all the way.

Answer 6

Hi Yukitou... All I want to know how it came out as N Met-Leu-Tyr-Pro-Ala C
I have this CUA ACU ACA UAU GAU and its anti codon GAU UAU AAC ACU CUA. Neither of them are coding for that set amino acids! Which one do i use and how do i read it?

Answer 7

i always use codon to read out.

Maybe the answer is wrong. Or maybe i'm the wrong one.

Answer 8

Yes i have a feeling the answer maybe wrong. I have posted this on another forum as well. If i get some answers will post back here. Will try some more problems sets from other links to see if this kind of thing repeats. But thank you very much for taking the time to reply.
If anybody else can shed light to this problem I would be grateful.

Answer 9

Sure. this is a study process and I also need some 'light' too :D

Answer 10

Hi Yukitou - I have solved this issue for 3d.. Here is the solution:

1. If you have single strand from 5' to 3' - first write down complementary strand. In this case complementary strand is already given so we are ok here.
2. Look for ORF (starting with ATG start sequence) reading only from 5' to 3' on both directions. In our case it starts at the 21st position.
3. Then encode starting with the ATG (MET) with the genetic code look up table.
4. So first 15 nucelotides we will have the codons--->... ATG TTA TAT CCC GCC ...
5. Corresponding mRNA (convert all T to U) ---> ... AUG UUA UAU CCC GCC ...
6. The corresponding amino acids are ----> N Met Leu Tyr Pro Ala C

---Phew!

Answer 11

oh, i get it, the question 3c ask us to list the first 15 nucleotides on the mrna made, and the actual mrna is way longer. To find the methionine, we need to contiunue reading the mrna.
Ok, i fell into the trap already :P

Thanks :D

Answer 12

Yes I hope this is the right way to get at the answers. Will try more problem sets using the same technique

Answer 13

When transcribing we start at the promoter sequence. When translating start at the START codon AUG. I think if we remember this we can solve any sequence problem :)
Cheers!