Translation: Study Guide (Biology 6.4)

From genotype to phenotype

Your genotype is the set of genes you carry. Your phenotype is what you can observe: eye colour, enzyme activity, whether you can digest lactose. The link between them is protein. A gene’s base sequence sets the amino acid sequence of a protein; the amino acid sequence sets how the protein folds; the shape sets what the protein does. Translation is the step that turns a base sequence into an amino acid sequence.

Where translation happens

Translation takes place on ribosomes. Ribosomes are found:

• free in the cytoplasm of both prokaryotic and eukaryotic cells;
• attached to the cytoplasmic surface of the rough endoplasmic reticulum (rough ER) in eukaryotic cells.

In eukaryotes, the mature mRNA must first leave the nucleus. In prokaryotes there is no nucleus, so transcription and translation happen in the same space. Ribosomes attach to the 5′ end of an mRNA and start translating while RNA polymerase is still making the rest of it. This coupling lets bacteria respond very quickly when they need a new protein.

The genetic code

An mRNA has four bases; proteins are made from 20 amino acids. One base cannot specify 20 amino acids, and nor can pairs (only 4 × 4 = 16 combinations). Triplets give 4 × 4 × 4 = 64 combinations, which is enough. So the mRNA is read in codons: groups of three bases, read one after another without overlapping.

Key features of the code:

• Start codon: AUG. It codes for methionine (Met) and marks where translation begins. This is the one codon you must know.
• Stop codons: three codons (UAA, UAG, UGA) code for no amino acid. They mark where translation ends. You do not need to memorise them; read them from the table.
• Redundancy: 61 codons code for amino acids, but there are only 20 amino acids, so many amino acids have more than one codon. Leucine has six; methionine and tryptophan have only one each.
• No ambiguity: each codon codes for only one amino acid.
• Near-universal: almost all living organisms use the same codon assignments, from bacteria to plants to humans. A few small variations exist, for example in mitochondria.

Table 1. The genetic code (mRNA codons). Find the first base in the left column, the second base along the top, then pick the row with the right third base.

First base	Second base U	Second base C	Second base A	Second base G
U	UUU Phe	UCU Ser	UAU Tyr	UGU Cys
U	UUC Phe	UCC Ser	UAC Tyr	UGC Cys
U	UUA Leu	UCA Ser	UAA Stop	UGA Stop
U	UUG Leu	UCG Ser	UAG Stop	UGG Trp
C	CUU Leu	CCU Pro	CAU His	CGU Arg
C	CUC Leu	CCC Pro	CAC His	CGC Arg
C	CUA Leu	CCA Pro	CAA Gln	CGA Arg
C	CUG Leu	CCG Pro	CAG Gln	CGG Arg
A	AUU Ile	ACU Thr	AAU Asn	AGU Ser
A	AUC Ile	ACC Thr	AAC Asn	AGC Ser
A	AUA Ile	ACA Thr	AAA Lys	AGA Arg
A	AUG Met (start)	ACG Thr	AAG Lys	AGG Arg
G	GUU Val	GCU Ala	GAU Asp	GGU Gly
G	GUC Val	GCC Ala	GAC Asp	GGC Gly
G	GUA Val	GCA Ala	GAA Glu	GGA Gly
G	GUG Val	GCG Ala	GAG Glu	GGG Gly

Why a shared code points to common ancestry

If each group of organisms had evolved its own code separately, we would expect many different codes. Instead, nearly every organism reads the same codons as the same amino acids. The simplest explanation is that the code was already in place in a common ancestor of all life and has been passed down ever since. A practical result: a gene from one species can be read correctly by the cells of a very different species. Bacteria given a human gene (in a form without introns) can make the human protein.

The players

• mRNA: the message, read 5′ → 3′, codon by codon.
• tRNA: the adaptor. Each tRNA carries one specific amino acid and has an anticodon, three bases that pair with a codon. A tRNA with the anticodon 3′-AAA-5′ pairs with the codon 5′-UUU-3′ and carries phenylalanine.
• Ribosome: two subunits, made of rRNA and proteins. The ribosome holds the mRNA and tRNAs in position and joins each new amino acid to the growing chain.

Three stages of translation

1. Initiation. The small ribosomal subunit binds the mRNA. Its rRNA lines the mRNA up so that translation begins at the start codon, AUG. A tRNA carrying methionine pairs with AUG, and the large subunit joins to complete the ribosome. Because the reading begins at AUG, the start codon also fixes the reading frame: how the rest of the mRNA is divided into triplets.

2. Elongation. This cycle repeats for each codon:

1. A tRNA whose anticodon matches the next codon enters the ribosome and pairs with it, bringing the correct amino acid to the place the codon specifies.
2. The growing chain is transferred onto the new amino acid. A peptide bond joins them, so the chain is now one amino acid longer.
3. The ribosome moves along the mRNA by one codon (three bases) towards the 3′ end. The empty tRNA leaves and can pick up another amino acid of its kind.

3. Termination. When a stop codon reaches the ribosome, no tRNA pairs with it. Instead, the finished polypeptide is released, and the ribosome comes apart from the mRNA. The polypeptide then folds, sometimes with help, into its working shape.

The many proteins and enzymes that help each stage are not required for the course. Concentrate on the sequence of events and on what codons, anticodons and rRNA do.

A special case: retroviruses

Usually information flows DNA → RNA → protein. Retroviruses, such as HIV, have genomes made of RNA and reverse the first step:

1. The virus enters a host cell and releases its RNA genome and an enzyme, reverse transcriptase.
2. Reverse transcriptase uses the viral RNA as a template to make a DNA copy (RNA → DNA).
3. The viral DNA integrates into the host cell’s own DNA.
4. The host cell’s RNA polymerase transcribes the viral genes, and the host’s ribosomes translate the viral mRNA into viral proteins.
5. New viral RNA genomes and proteins are assembled into new viruses.

Because the viral DNA is now part of the host’s genome, it is copied whenever the cell replicates its DNA. Human cells do not use reverse transcriptase to express their own genes, which is one reason it makes a useful drug target.

Worked example 1: translating an mRNA

Question. Part of a mature mRNA is 5′-CCGAUGUUUGGACAUCGUUAGGCA-3′. Using Table 1: (a) give the amino acid sequence of the polypeptide made; (b) give the anticodon of each tRNA used, written 3′ → 5′.

(a) Step by step.

1. Read from the 5′ end and find the first AUG. Here it begins at nucleotide 4 (the bases CCG before it are not translated).
2. From AUG, split the mRNA into triplets: AUG | UUU | GGA | CAU | CGU | UAG | …
3. Look up each codon in Table 1: AUG = Met, UUU = Phe, GGA = Gly, CAU = His, CGU = Arg, UAG = Stop.
4. Stop at the stop codon. The bases after it (GCA) are not translated.

Polypeptide: Met–Phe–Gly–His–Arg (5 amino acids).

(b) Each anticodon is complementary and antiparallel to its codon:

Codon (5′ → 3′)	AUG	UUU	GGA	CAU	CGU
Anticodon (3′ → 5′)	UAC	AAA	CCU	GUA	GCA
Amino acid carried	Met	Phe	Gly	His	Arg

Check. Five codons were translated, so five tRNAs were used. No tRNA is used for the stop codon. Compare the first four codons with Figure 1, which shows a moment part-way through this example.

Worked example 2: counting codons and predicting a change

Question. A fictional mature mRNA is 1200 nucleotides (nt) long, not counting its poly-A tail. Its start codon begins at nucleotide 61, and its stop codon (UGA) ends at nucleotide 1011. (a) How many amino acids are in the polypeptide? (b) How many nucleotides of the mRNA are not translated? (c) Predict the effect if an error in the mRNA turned codon 120 into UAA. (d) Predict the effect if codon 50 changed from GGU to GGC.

(a)

1. The translated stretch, including the stop codon, runs from nucleotide 61 to 1011: 1011 − 61 + 1 = 951 nt.
2. Codons: 951 ÷ 3 = 317.
3. The last codon is the stop codon, which adds no amino acid: 317 − 1 = 316 amino acids (counting the first methionine).

(b) Before the start codon: 60 nt. After the stop codon: 1200 − 1011 = 189 nt. Total untranslated: 249 nt.

(c) UAA is a stop codon. Translation would end at codon 120, so the polypeptide would have only 119 amino acids. It would probably be unable to fold into the normal shape, so the protein would likely not work.

(d) GGU and GGC both code for glycine (Table 1). The amino acid sequence is unchanged, so the protein should work normally. This is the redundancy of the code in action.

Common misconceptions

• “The first three bases of an mRNA are the first codon.” Translation begins at the start codon AUG, which is usually some way along the mRNA.
• “The tRNA anticodon is the same as the codon.” It is complementary and antiparallel to the codon.
• “Stop codons code for a ‘stop’ amino acid.” They code for no amino acid; they trigger release of the polypeptide.
• “Each amino acid has one codon.” Most have two or more; only Met and Trp have just one.
• “Every organism has its own genetic code.” Nearly all use the same code, which is evidence of common ancestry.
• “All ribosomes are on the rough ER.” Many ribosomes are free in the cytoplasm, and prokaryotes have no ER at all.
• “Prokaryotes finish transcription before translation begins.” Without a nucleus, they translate mRNA while it is still being made.
• “Retroviruses break the rules completely.” They add an extra step (RNA → DNA), but their genes are then transcribed and translated normally by the host cell.

Where this leads

Every cell in your body has the same genes, yet a liver cell and a nerve cell make very different proteins. In Topic 6.5, Regulation of Gene Expression, you will see how cells control which genes are transcribed and when. To review how the mRNA was made and processed, revisit Topic 6.3, Transcription and RNA Processing. Test yourself with the practice questions, then use the revision notes and the checklist to consolidate.