AP Biology · Lesson 2

Transcription, RNA processing and translation

Know Know it

Transcription copies a gene into RNA. RNA polymerase binds at the promoter, unwinds the DNA and reads the template strand 3' to 5', building RNA 5' to 3'. It needs no primer. The RNA has the same sequence as the other DNA strand, the coding strand, except that it has U where the coding strand has T. Besides mRNA, genes also make tRNA and the rRNA of ribosomes.

In eukaryotes the first transcript, pre-mRNA, is processed in the nucleus. A 5' cap is added to the front and a poly-A tail to the end; both protect the mRNA and help it get translated. Introns are cut out and exons joined. With alternative splicing, different cells keep different exons, so one gene can give several related proteins. Bacteria do not process their mRNA this way, and their ribosomes can start translating while the mRNA is still being made.

Translation builds a polypeptide from mRNA. A ribosome reads the mRNA in three-base codons, starting at AUG, which codes for methionine. Each tRNA carries one amino acid and has an anticodon that pairs with the codon. The ribosome links each new amino acid to the chain by a peptide bond. At a stop codon (UAA, UAG or UGA) no tRNA fits, and the chain is released.

The genetic code is redundant: most amino acids have more than one codon, often differing only in the third base. It is also nearly universal: the same codons stand for the same amino acids in almost all living things, which is evidence that they share common ancestry. It is also why a bacterium given the coding sequence of a human gene, with the introns already removed, can make the human protein.

Trap The mRNA matches the coding strand, not the template

RNA polymerase reads the template strand, so the mRNA is complementary to the template and has the same sequence as the coding strand, with U for T. If a question gives you the template, pair the bases; if it gives you the coding strand, just swap T for U.

Worked example Easy

A gene's template strand reads TACCCG, read from its 3' end to its 5' end. Which mRNA is made from it, read from its 5' end to its 3' end?

  1. ATGGGC
  2. UACCCG
  3. AUGGGCAnswer
  4. GCCCAU

How to solve it

  1. Pair each template base with an RNA base: T with A, A with U, C with G, G with C.
  2. TACCCG pairs with AUGGGC.
  3. The template ran 3' to 5', so the mRNA runs 5' to 3': AUGGGC.

Why each choice is right or wrong

  • A. Incorrect. ATGGGC has thymine, which is a DNA base; RNA polymerase puts uracil in its place.
  • B. Incorrect. UACCCG copies the template letter for letter, but the mRNA must pair with the template.
  • C. Correct. Pairing T-A, A-U, C-G, C-G, C-G and G-C gives AUGGGC, which starts with the start codon AUG.
  • D. Incorrect. GCCCAU does not pair with the template base by base; it is the template's letters with U for T, written in reverse order.

Explain Explain it

Task verbs: determine, explain

To determine a protein from an mRNA, find the first AUG, split the message into codons from there, and look up each codon until you reach a stop codon. Count the amino acids you added, not the codons, since the stop codon adds none. To explain an effect, say which step, transcription, processing or translation, is changed and what it does to the protein.

Worked example Medium

Part of the genetic code (mRNA codons)

CodonAmino acid
AUGMethionine (start)
AAALysine
AAGLysine
GGAGlycine
GGCGlycine
UUCPhenylalanine
UGGTryptophan
CCAProline
UAAStop
UAGStop
UGAStop

An mRNA reads AUG UUC GGA AAG UGG UAA, from its 5' end. Using the table, which polypeptide does it code for?

  1. Methionine, phenylalanine, glycine, lysine, tryptophanAnswer
  2. Methionine, phenylalanine, glycine, lysine, tryptophan, stop amino acid
  3. Phenylalanine, glycine, lysine, tryptophan
  4. Methionine, phenylalanine, glycine, proline, tryptophan

How to solve it

  1. Start at AUG: methionine.
  2. Read on in threes: UUC is phenylalanine, GGA is glycine, AAG is lysine, UGG is tryptophan.
  3. UAA is a stop codon, so the chain ends with five amino acids.

Why each choice is right or wrong

  • A. Correct. AUG, UUC, GGA, AAG and UGG give methionine, phenylalanine, glycine, lysine and tryptophan, and UAA ends the chain.
  • B. Incorrect. A stop codon has no tRNA and adds no amino acid, so there is no sixth amino acid called stop.
  • C. Incorrect. The start codon AUG is itself translated, so methionine is the first amino acid, not left out.
  • D. Incorrect. The fourth codon is AAG, which the table gives as lysine; proline would need CCA.

Determine Your turn

A gene's coding strand reads 5'-ATGTTCAAA-3'. First, determine the sequence of the mRNA transcribed from this gene, written from its 5' end to its 3' end. Then explain why the mRNA has the same sequence as the coding strand, except for one kind of base, even though RNA polymerase reads the other DNA strand.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Gives the mRNA as 5'-AUGUUCAAA-3'
  • Explains that RNA polymerase builds the mRNA by pairing bases with the template strand, so the mRNA is complementary to the template
  • Explains that the coding strand is also complementary to the template, so the mRNA and the coding strand match, with U in the mRNA where the coding strand has T

Model answer

The mRNA is 5'-AUGUUCAAA-3'. RNA polymerase reads the template strand and builds the mRNA by pairing an RNA base with each template base, so the mRNA is complementary to the template. The coding strand is also complementary to the template, since the two DNA strands pair with each other. Two sequences that both pair with the same template must match each other, so the mRNA has the coding strand's sequence. The one difference is that RNA uses uracil in place of thymine, so each T in the coding strand is a U in the mRNA.

Apply Apply it

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