Section 3: Transcript

Image of DNA transcription to RNA and then to translation to protein

Lecture 1: Transcription  


Transcription and mRNA processing | Biomolecules | MCAT | Khan Academy

Lecture 2: Translation


Translation (mRNA to protein) | Biomolecules | MCAT | Khan Academy

In section 1: Genes, you learned how to translate DNA to mRNA and finally to tRNA through traditional methods. The processes of transcribing DNA to RNA and producing a protein by translation provide several opportunities for applying creative thinking to problem-solving.

     To illustrate, short polypeptides composed of three or more amino acids can be generated such that one letter taken from each three-letter amino acid abbreviation will form a word.  Many different words can be formed using letters from the amino acid abbreviations.  Mental flexibility/creativity is used to determine which letter of each amino acid is to be used to reveal the hidden word.

     Several possibilities exist for making words from the amino acid abbreviations.  One is using only the first letter of each amino acid: the tetrapeptide (protein having four amino acids)  Leu-Ala-Ser-Trp produces the word LAST.  A second is using only the second letters of the amino acids: the tetrapeptide gLu-hIs-tRp-vAl produces the word LIRA.  A third is using only the third letter of each amino acid:  the tetrapeptide asP-alA-hiS-meT produces the word PAST.

One skill in solving this type of exercise/puzzle is to identify which letters from a mixture (letters in the tetrapeptide) are needed for the solution, e.g. LAST above.  The same type of skill would be used in finding the word LAST from an alphabetical list of the same amino acid letters e.g.  a,a,e,e,l,l,p,r,r,s,t,u.  However, solving a polypeptide for a word also involves learning about genetics.

Specific Examples

Identifying which letters in a group (of letters) are needed for a solution involves mental flexibility to determine which letters are to be used. For Example: Find the word hidden in the tetapeptide leu-ala-ser-trp. 

A solution might be:

  1. Take the letter “u” of leu as the first letter of the hidden word,
  2. Then take the “a” from ala as the next letter of the word (ua could be the first two letters of a word).
  3. Next, take a letter from ser. Taking “e” from ser would make the first letters of the potential solution uae. However, these three letters would make the first three letters are not likely to make a word by adding one more letter.
  4. Go back to step 2 and try a different letter from ser as the third letter of the hidden word. Taking “s” from ser would make the partial solution uas which could make a word… or
  5. Go back to step 2 and try a different letter from ala as the second letter of the hidden word… or
  6. Go back to step and try a different letter from leu as the first letter of the hidden word.

This method will eventually show that beginning with “u” will not produce a word so the next step would be to try the second letter of the first amino acid and then try combinations of letters from the remaining amino acids.  This process continues until the solution is reached.

 As more and more pieces are put together correctly, the final solution might be revealed before all of the pieces have been assembled.  Seeing the final solution from a partial solution involves creativity.  For example, putting together the l from leu, the a from ala and the s from ser forms the first three letters of a partial solution las.  From las the words lash or last could be made.   The last piece of information needed is that trp, the fourth amino acid, has no “h”, but it does have a “t” so the solution must be the word LAST.

Using creativity to find words hidden in letters of amino acids in a peptide can also be combined with the language of transcription and translation to create more creative thinking exercises.

To illustrate: a correctly completed set will show all bases in DNA, mRNA, tRNA and amino acids would reveal word SIGMA:


    Original TEMPLATE backbone      = A G A T A A G T C T A C C T A
 Original NON-TEMPLATE backbone     = T C T A T T C A G A T G G A T
 mRNA from TEMPLATE DNA backbone    = U C U A U U C A G A U G G A U
                       tRNA's       = A G A U A A G U C U A C C U A
                   amino acids      = ser   ile   gln   met   asp   = SIGMA

Letters in amino acid abbreviations, DNA and/or RNA can be replaced with blanks, necessitating creative thinking in arriving at the solution.  For example :

     Template DNA               =  C _ C _ T A A _ C C A A
     Messenger RNA (codons)     =  _ _ _ _ _ _ _ _ _ _ _ _
     Transfer RNA (anticodons) = _ U C G _ _ _ C _ C _ _
     Amino acids (peptide)      =  ---  ---   ---   ---

Using the relationships between DNA and mRNA and mRNA and amino acids, the blanks can be filled in: C in DNA would produce a G in mRNA, T in DNA would produce A in mRNA,etc.  While a G in a mRNA codon would produce a C in the corresponding tRNA anticodon, A in mRNA would produce U intRNA, etc. If you need a quick review of the relationships, go back and review “What is DNA”

Using these relationships, the blanks in the above example would be completed to give:

     Template DNA               =  C T C G T A A C C C A A
     Messenger RNA (codons)     =  G A G C A U U G G G U U
     Transfer RNA (anticodons)  =  C U C G U A A C C C A A
     Amino acids (peptide)      =  ---   ---  ---   ---
 

Now the amino acids and mRNA codons can be filled in as the codon GAG will produce “glu” in the protein, the codon CAU will produce “his” in the protein.

     Template DNA               =  C T C G T A A C C C A A
     Messenger RNA (codons)     =  G A G C A U U G G G U U
     Transfer RNA (anticodons)  =  C U C G U A A C C C A A
     Amino acids (peptide)      =  glu   his   trp   val

Finally deciding which letters to use from each three-letter amino acid abbreviation will produce    

  • Using the first letters of each amino acid abbreviation    -> ghtv
  • Using the second letter of each amino acid abbreviation ->  lira  
  • Using the third letter of each amino acid abbreviation    ->  uspl.

     Using logic and mental flexibility it is possible to complete these types of exercises, while using the concepts of genetics – strengthening knowledge of genetics and enhancing logic/creative thinking skills.

     Deductive reasoning can also be strengthened using exercises that vary the way the amino acid abbreviations are presented.

One method is to let a single lowercase letter represent each amino acid, i.e., deleting two of the three letters.  For example s, e, or r could be used to represent ser, similarly h, i or s  could represent his, etc.  The letter g could represent arg, gln, glu, or gly, etc.

An example of such an exercise could be:

     Template DNA               =  C _ C _ _ _ A _ C _ _ A
     Messenger RNA (codons)     =  _ _ _ _ _ _ _ G _ _ _ _
     Transfer RNA (anticodons)  =  _ _ _ G _ _ _ _ _ _ _ _
     Amino acids (peptide)      =  g     i     t     v

The logic (steps) used to complete this set could be :

  1.  only one amino acid has v in its abbreviation (val) 
    1. thus, the last amino acid must be val, 
    2. its codon (mRNA) is G U U,
    3. its DNA triplet is C A A,
    4.  and its anticodon (on tRNA) is C A A.
  2.  the amino acid designated by t could be either met, thr, trp, or tyr.
    1. Filling in the blanks above, the codon for t is UGG. 
    2. Thus, the amino designated by t (and whose codon is UGG) is trp.
  3. the amino acid designated by g could be either arg, gln, glu or gly. 
    1. Filling in the blanks above, the codon for the amino acid represented by  gis G_G. 
    2. The codon for arg is CGU, the codon for gln is CAG,
    3. the codon for glu is GAG
    4. and the codon for gly is GGU.
    5. Thus, the amino acid designated by g must be glu
      1. and its codon is GAG,
      2.  its anticodon is CUC,
      3. and its DNA triplet is CTC.
  4. the amino acid designated by i could be his or ile. 
    1. The codon for this amino acid is C_ _. 
    2. Only his has a codon beginning with C (CAU).
    3. Ile has a codon of AUU and cannot be the designated amino acid.

Thus the correct solution for this example is

     Template DNA               =  C T C G T A A C C C A A
     Messenger RNA (codons)     =  G A G C A U U G G G U U
     Transfer RNA (anticodons)  =  C U C G U A A C C C A A
     Amino acids (peptide)     =  glu   his   trp   val

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