Educerie

Educerie · AP · Biology

AP Biology 6.4 Translation

In one breath

  • Translation builds a polypeptide on ribosomes, in the cytoplasm and on the rough ER, and in prokaryotes it begins while the mRNA is still being transcribed.
  • The ribosome's rRNA binds the mRNA at the start codon AUG (methionine), and the mRNA is read 5′ to 3′ in codons.
  • For each codon a tRNA with the complementary anticodon brings its amino acid, which is joined to the chain by a peptide bond, until a stop codon releases the polypeptide.
  • The chart gives codons: there are 64, most amino acids have several, and nearly all life uses the same code, which points to a common ancestor.
  • Retroviruses copy their RNA into DNA with reverse transcriptase; that DNA integrates into the host genome and is transcribed and translated into new viruses.

What you must be able to do

  • Explain how the genotype decides the phenotype through translation: where it happens, initiation at AUG, codons read in threes, tRNA bringing amino acids, termination at a stop codonTranslate an mRNA using a code chart; count amino acids; predict the effect of blocking a ribosome; explain why the near-universal code points to common ancestry
  • Explain the special flow of information in retroviruses: RNA to DNA by reverse transcriptase, then integration, transcription and translationOrder the steps of a retrovirus life cycle; predict the effect of a reverse transcriptase inhibitor

Once you have read the note, tick off each one you can do in Check your understanding.

In the exam. Section I multiple choice gives you an mRNA (or a DNA strand) and a genetic code chart and asks for the amino acids, the number of amino acids, or the tRNA anticodon; it also asks where translation happens, and what follows when a step is blocked, including in a retrovirus. In Section II, Analyze model or visual representation questions use a drawn ribosome with tRNAs, Conceptual analysis questions ask why a shared genetic code is evidence of common ancestry, and Interpreting and evaluating experimental results questions can involve a drug that blocks a ribosome or reverse transcriptase.

1Where translation happens

Ribosomes are found in the cytoplasm of both prokaryotic and eukaryotic cells, and in eukaryotes also on the cytoplasmic surface of the rough ER. Ribosomes free in the cytoplasm tend to make proteins that stay in the cell's fluid; ribosomes on the rough ER make proteins that go into membranes or are secreted.

In prokaryotes there is no nucleus, so nothing separates the DNA from the ribosomes. Ribosomes attach to an mRNA and start translating it while it is still being transcribed. In eukaryotes transcription is in the nucleus and translation in the cytoplasm, so the mRNA must be processed (topic 6.3) and leave the nucleus first. That separation is what gives eukaryotes their extra steps for controlling genes.

2The three stages

Figure 1 shows a ribosome part-way through translating an mRNA.

Translation at the ribosome A U G U U C C A U G C A U A A G U A His C G U Ala Phe Met 5′ 3′ mRNA peptide bond codon anticodon tRNA amino acid growing polypeptide large subunit small subunit
Figure 1 · Translation: tRNAs whose anticodons pair with the mRNA codons bring amino acids to the ribosome, where each is added to the growing polypeptide

Initiation. The ribosome's small subunit attaches to the mRNA; the rRNA in the ribosome interacts with the mRNA at the start codon, AUG. A tRNA carrying methionine pairs with AUG, the large subunit joins, and the ribosome is ready. AUG sets the reading frame: from here on, the bases are read in non-overlapping threes.

Elongation.

  1. The next codon sits in the ribosome. The tRNA whose anticodon is complementary to it binds there, bringing its amino acid. (The anticodon pairs antiparallel with the codon, A with U and G with C.)
  2. The amino acid is transferred to the growing polypeptide: the ribosome joins it to the chain by a peptide bond.
  3. The ribosome moves along the mRNA by one codon, 5′ to 3′. The empty tRNA leaves, to be loaded with a new amino acid.

Termination. When a stop codon (UAA, UAG or UGA) reaches the ribosome, no tRNA matches it. The polypeptide is released, and the ribosome comes apart from the mRNA.

The CED does not examine the names of the protein factors involved in these steps, or the names of the ribosome's binding sites. Know the order of events and what each part does.

3Reading the genetic code

The genetic code is the set of rules linking each codon to an amino acid. With four bases read three at a time there are 4 × 4 × 4 = 64 codons: 61 code for amino acids and 3 are stop codons. Since there are only 20 amino acids, many amino acids have more than one codon (leucine has six). That redundancy matters in topic 6.7: a change in the third base of a codon often leaves the amino acid the same.

You do not have to memorise the code, except that AUG is the start codon and codes for methionine. The exam gives you a chart. Figure 2 is one: find the first base on the left, the second base along the top, and the third base on the right.

The genetic code as a table of mRNA codons second base first base third base U C A G U UUU Phe UUC Phe UUA Leu UUG Leu UCU Ser UCC Ser UCA Ser UCG Ser UAU Tyr UAC Tyr UAA Stop UAG Stop UGU Cys UGC Cys UGA Stop UGG Trp U C A G C CUU Leu CUC Leu CUA Leu CUG Leu CCU Pro CCC Pro CCA Pro CCG Pro CAU His CAC His CAA Gln CAG Gln CGU Arg CGC Arg CGA Arg CGG Arg U C A G A AUU Ile AUC Ile AUA Ile AUG Met ACU Thr ACC Thr ACA Thr ACG Thr AAU Asn AAC Asn AAA Lys AAG Lys AGU Ser AGC Ser AGA Arg AGG Arg U C A G G GUU Val GUC Val GUA Val GUG Val GCU Ala GCC Ala GCA Ala GCG Ala GAU Asp GAC Asp GAA Glu GAG Glu GGU Gly GGC Gly GGA Gly GGG Gly U C A G AUG = start (Met) Stop = stop codon
Figure 2 · The genetic code: the amino acid for each of the 64 mRNA codons

Worked example. Translate the mRNA 5′-GCAUGGCUAAAGGUUGGUAGCC-3′.

  1. Find the first AUG, reading 5′ to 3′: GCAUG…. The two bases before it are not translated.
  2. Split the rest into codons from AUG: AUG · GCU · AAA · GGU · UGG · UAG · CC.
  3. Look each one up: Met · Ala · Lys · Gly · Trp · stop.
  4. The polypeptide is Met-Ala-Lys-Gly-Trp, five amino acids. The stop codon adds nothing, and the bases after it are not read.

Counting amino acids. From AUG to the stop codon inclusive, divide the number of bases by three and subtract one for the stop. A coding region of 1,503 bases is 501 codons, so 500 amino acids. (Before you use a protein, cells often trim it; the exam will say so if that matters.)

Working back to the tRNA. A tRNA's anticodon is complementary and antiparallel to the codon. The codon 5′-GGC-3′ (glycine) is read by the anticodon 3′-CCG-5′. Always convert an anticodon back to its codon before you use the chart: the chart lists codons, not anticodons.

The mRNA is read 5′ to 3′ in codons from the start codon AUG; each codon is matched by a tRNA anticodon, and the chart gives codons, never anticodons.

4One code for nearly all life

The same codons mean the same amino acids in bacteria, fungi, plants and animals, with a few small exceptions (some mitochondria and a few microbes reassign one or two codons). The CED makes this a piece of evidence: such an arbitrary, detailed set of rules is far more likely to be shared because all living things inherited it from one common ancestor than to have arisen many times independently.

It is also why genetic engineering works. A human gene put into a bacterium is translated into the same amino acid sequence, because the bacterium reads the codons the same way (topic 6.8).

5Retroviruses: information flowing backwards

In cells, information flows DNA → RNA → protein. Retroviruses, such as HIV, are the CED's special case. Their genome is RNA, and they carry an enzyme called reverse transcriptase, which copies RNA into DNA.

  1. The virus enters a host cell and releases its RNA and reverse transcriptase.
  2. Reverse transcriptase copies the viral RNA into DNA.
  3. That viral DNA integrates into the host cell's genome, becoming part of a host chromosome.
  4. The host's own 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 assemble into new viruses, which leave the cell.

So the flow is RNA → DNA → RNA → protein. A drug that blocks reverse transcriptase stops step 2, so no viral DNA can integrate and no new viruses are made, even though the viral RNA got into the cell.

Practice questions

8 questions on this note, from easy to hard. Each one you get right lights a star of this topic's constellation.