Free AP Biology Gene Expression and Regulation

AP Biology · Gene Expression and Regulation · free lesson

Translation

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

From message to protein

Think of a cell as a kitchen, the DNA as a recipe book kept safe on a shelf, and the mRNA as a photocopy of one recipe carried to the counter. Translation is the cooking: the cell uses the mRNA as a template to build a polypeptide, converting information written in nucleotides into a chain of amino acids. The order of those amino acids sets the protein's shape and job, and the proteins of an organism help produce its traits. That chain of cause is how a genotype determines a phenotype: the DNA sequence is copied to mRNA, the mRNA sequence is read into an amino acid sequence, and the protein does the work that we observe.

Where translation happens

The workbench of translation is the ribosome, a complex of rRNA and many proteins. Ribosomes are present in the cytoplasm of both prokaryotic and eukaryotic cells. In eukaryotic cells, ribosomes are also found on the cytoplasmic surface of the rough endoplasmic reticulum, which is why that membrane looks studded or "rough".

A common mix-up is to place translation in the nucleus because the nucleus holds the DNA. In a eukaryote, the nucleus is where the mRNA is made; the mRNA then reaches ribosomes in the cytoplasm, where the protein is built.

Bacteria have no nucleus, so nothing separates the DNA from the ribosomes. In bacteria, translation of an mRNA begins while the mRNA is still being transcribed. In the picture, the RNA polymerases move along the DNA from left to right. The mRNA that started earliest is the longest, and ribosomes are already reading its older end while its other end is still being made. A eukaryotic cell does not work this way: transcription happens in the nucleus and translation in the cytoplasm, so a ribosome cannot start on an mRNA while that mRNA is still being transcribed.

Reading the message: codons and the genetic code

The mRNA has four letters (A, U, G and C) and proteins are built from 20 kinds of amino acids, so one letter cannot name one amino acid. The sequence of nucleotides on the mRNA is read in triplets called codons. Pairs of letters would give just 16 combinations, too few for 20 amino acids; triplets give 64, which is plenty. Each codon encodes a specific amino acid, which can be found with a genetic code chart.

The codons are not all different in meaning. Of the 64 codons, 61 specify an amino acid and three are stop codons (UAA, UAG and UGA) that specify no amino acid at all. That leaves more codons than amino acids, so many amino acids are encoded by more than one codon. Here they are for three amino acids.

Leucine has $6$ codons ( $\text{UUA, UUG, CUU, CUC, CUA, CUG}$ ), while methionine has $1$ ( $\text{AUG}$ ) and tryptophan has $1$ ( $\text{UGG}$ ). Many codons for the same amino acid are different versions of one meaning, so the code is like a dictionary in which several words can mean the same thing. It is not a dictionary in which one word has several meanings: a given codon calls for at most one amino acid (one amino acid, or a stop signal).

Definition.

Codon: a group of three mRNA nucleotides that specifies one amino acid or a stop signal. Start codon: AUG, which codes for the amino acid methionine and is where translation begins.

Nearly all living organisms use the same genetic code. A codon that means leucine in a bacterium also means leucine in a human, a tulip or a mushroom. This is evidence for the common ancestry of all living organisms: it would be a remarkable coincidence for lines of life that arose separately to settle on one code.

The three steps: initiation, elongation, termination

Translation involves many sequential steps, including initiation, elongation and termination.

Initiation. Translation is initiated when the rRNA in the ribosome interacts with the mRNA at the start codon, AUG, which codes for methionine. The ribosome therefore starts at the start codon and not at the very first nucleotide of the mRNA. The mRNA ahead of the start codon is not translated, and the start codon sets the grouping of the triplets that follow, called the reading frame.
Elongation. A tRNA is a small RNA molecule with a three-nucleotide anticodon at one end and an amino acid attached at the other. The tRNA brings the correct amino acid to the place specified by the codon on the mRNA: its anticodon pairs by complementary bases with the codon. The amino acid is then transferred to the growing polypeptide chain, and the process continues along the mRNA, codon by codon.
Termination. The process continues along the mRNA until a stop codon is reached. A stop codon has no amino acid to bring, so translation terminates with the release of the newly synthesized protein.

Here is the middle of elongation, translating the mRNA GCAUGUUUGGCUAAGG. The two leading bases are skipped, because the start codon is the first AUG. The ribosome has read AUG and UUU, so the chain already holds methionine and phenylalanine; the tRNA at the next codon, GGC, is bringing glycine. The anticodons are written 3' to 5' because they pair in opposite directions to the codons.

Finishing the job, the next codon is UAA, a stop codon, so the chain $\text{Met-Phe-Gly}$ is released: $3$ amino acids, one for each codon before the stop codon.

Who does what

Molecule	Role in translation
mRNA	the template, read in codons
tRNA	brings the amino acid that matches a codon, by its anticodon
rRNA	part of the ribosome; it interacts with the mRNA at the start codon
ribosome	the site where the polypeptide is built

The key idea is the match between a codon and an anticodon. If a codon reads 5'-CUA-3', the tRNA that binds it has the complementary anticodon 3'-GAU-5' and carries leucine. The tRNA is the adaptor: the codon is a fact about the mRNA, the amino acid is a fact about the protein, and the tRNA connects them.

A special case: retroviruses

The usual flow of information in a cell is DNA to RNA to protein. Genetic information in retroviruses is a special case with an alternate flow of information, from RNA to DNA. A retrovirus carries an RNA genome. Inside the host cell, an enzyme called reverse transcriptase copies the viral RNA genome into DNA. This DNA integrates into the host genome. The host's machinery then transcribes and translates it for the assembly of new viral progeny.

Warning.

Four mix-ups to leave behind. First, translation does not take place in the nucleus of a eukaryote; ribosomes are in the cytoplasm and on the rough ER. Second, the stop codon does not add an amino acid; it ends the chain. Third, the amino acids are not specified one nucleotide at a time: each codon is three nucleotides. Fourth, one amino acid can have several codons, but a single codon does not call for several amino acids.

Adapted from Biology by OpenStax (Rice University), licensed under CC BY 4.0. Access for free at https://openstax.org/details/books/biology

Worked examples

Each example is the kind of question the AP exam asks on this topic, worked step by step.

Example.

1. Reading an mRNA. The mRNA segment 5'-GCAUGUUUGGCAAAUAAGG-3' is translated by a ribosome. Which polypeptide is made, and how many amino acids does it have?

Translation is initiated at the start codon, AUG, so the ribosome starts at the first AUG and not at the first base. From there the mRNA is read in triplets until a stop codon is reached.

\begin{aligned} t &= \operatorname{translation}(m) \\ &= \operatorname{translation}(\text{GCAUGUUUGGCAAAUAAGG}) \\ &= \left\{ \begin{aligned} \text{codons} &= \left[ \text{``AUG''},\ \text{``UUU''},\ \text{``GGC''},\ \text{``AAA''},\ \text{``UAA''} \right] \\ \text{protein} &= \text{Met-Phe-Gly-Lys} \\ \text{one\_letter} &= \text{MFGK} \\ \text{length} &= 4 \\ \text{start} &= 3 \\ \text{stop} &= \text{UAA} \end{aligned} \right. \\ n &= \text{length of }t \\ &= 4 \end{aligned}

The codons from the start codon to the stop codon are $\text{AUG, UUU, GGC, AAA, UAA}$ . The first, AUG, is methionine. The last, $\text{UAA}$ , is a stop codon, so it adds no amino acid and ends the chain. The polypeptide is $\text{Met-Phe-Gly-Lys}$ , with $4$ amino acids. The leading GC and the trailing GG are not translated.

Result.

$\text{Met-Phe-Gly-Lys}$ : $4$ amino acids. A stop codon is read, but it does not add an amino acid.

Example.

2. From the DNA to the protein. The template strand of a gene reads 3'-TACAAACCGATT-5'. Which polypeptide does the gene specify?

The mRNA is made by pairing bases with the template strand, with uracil in place of thymine, so it is complementary to the template.
The mRNA is then read from its start codon in triplets, each matched to an amino acid by the genetic code.

\begin{aligned} rna &= \operatorname{transcribe}(dna, \text{template}) \\ &= \text{AUGUUUGGCUAA} \\ t &= \operatorname{translation}(rna) \\ &= \operatorname{translation}(\text{AUGUUUGGCUAA}) \\ &= \left\{ \begin{aligned} \text{codons} &= \left[ \text{``AUG''},\ \text{``UUU''},\ \text{``GGC''},\ \text{``UAA''} \right] \\ \text{protein} &= \text{Met-Phe-Gly} \\ \text{one\_letter} &= \text{MFG} \\ \text{length} &= 3 \\ \text{start} &= 1 \\ \text{stop} &= \text{UAA} \end{aligned} \right. \end{aligned}

The mRNA, written 5' to 3', is AUGUUUGGCUAA. Its codons are $\text{AUG, UUU, GGC, UAA}$ . The first codon, AUG, is the start codon and codes for methionine, and the last, $\text{UAA}$ , is a stop codon. The polypeptide is $\text{Met-Phe-Gly}$ .

Result.

mRNA 5'-AUGUUUGGCUAA-3', polypeptide $\text{Met-Phe-Gly}$ . Information flows from the template strand to the mRNA to the amino acid sequence.

Example.

3. The tRNA for a codon. A ribosome is reading the codon 5'-CUA-3' of an mRNA. Which anticodon is on the tRNA that binds it, and which amino acid does that tRNA carry?

The anticodon pairs with the codon by complementary bases: A with U and G with C. The two pair in opposite directions, so the anticodon is written 3' to 5' under the codon.

Pairing C with G, U with A and A with U, the bases of CUA give the anticodon 3'-GAU-5'. The codon CUA specifies leucine, so the tRNA with that anticodon carries leucine. The tRNA brings the amino acid to the place that the codon specifies, and the amino acid is then transferred to the growing polypeptide chain.

Result.

Anticodon 3'-GAU-5', carrying leucine.

Example.

4. Counting nucleotides. A protein is made of 120 amino acids. How many nucleotides are in the mRNA from the start codon through the stop codon?

Each codon is three nucleotides, and each amino acid in the chain is specified by one codon of this mRNA. The stop codon is one more codon, with no amino acid.

\begin{aligned} codons &= aas + 1 \\ &= 120 + 1 \\ &= 121 \\ nt &= 3codons \\ &= 3 \times 121 \\ &= 363 \end{aligned}

There are $120$ codons for the amino acids and 1 stop codon, so $121$ codons, and 3 times $121$ = $363$ nucleotides. Counting just the amino acids would give 3 times $120$ = $360$ , which leaves out the stop codon.

Result.

$363$ nucleotides.

Example.

5. Explaining observations. (a) In a bacterium, a student sees ribosomes attached to an mRNA whose other end is still attached to the DNA. Explain. (b) A virus has an RNA genome, yet a DNA copy of it is found in the DNA of infected human cells. Explain.

(a) A bacterium has no nucleus, so the mRNA is available to ribosomes while it is still being transcribed. In prokaryotic organisms, translation of the mRNA molecule occurs while it is being transcribed, which is what the student sees.
(b) This is a retrovirus. Its enzyme, reverse transcriptase, copies the viral RNA genome into DNA, and that DNA integrates into the host genome. The host then transcribes and translates it to make the parts of new virus particles.

Result.

(a) Transcription and translation overlap in bacteria. (b) The flow of information is RNA to DNA in a retrovirus, made possible by reverse transcriptase.

Practice

Answer AP-style questions on translation. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise translation with new numbers