mindstair

Home / Courses / Genetics / Unit 7

Genetics Unit 7: Gene Expression: Transcription and Translation

Genes act by being transcribed into RNA and translated into protein. This unit covers each step, from promoter to polypeptide, and the rules of the genetic code.

Covers: Central dogma and one gene–one enzyme, Transcription in bacteria and eukaryotes, Template and coding strands, RNA processing and splicing, The genetic code and reading frames, tRNA, wobble and aminoacyl-tRNA synthetases, Ribosomes and the steps of translation

Study guide

The central dogma describes the usual flow of information: DNA is transcribed to RNA, which is translated to protein. Beadle and Tatum's Neurospora experiments linked genes to enzymes: each nutritional mutant was blocked at one step of a pathway, and a mutant grows if given any compound after its block. This led to the one gene–one enzyme idea, later refined to one gene–one polypeptide, since some genes encode RNA products and some proteins have several different chains.

Transcription is carried out by RNA polymerase, which reads the template strand 3′→5′ and builds RNA 5′→3′ without a primer. The RNA matches the coding strand except that U replaces T. In bacteria, the sigma factor directs the polymerase to promoter sequences around −35 and −10, and transcription ends either at a GC-rich hairpin followed by U's (intrinsic termination) or with the Rho protein. In eukaryotes, RNA polymerase II transcribes mRNA genes; general transcription factors bind promoter elements such as the TATA box and recruit the polymerase.

Eukaryotic pre-mRNA is processed in the nucleus. A 7-methylguanosine cap is added to the 5′ end, and the 3′ end is cleaved and given a poly-A tail; both help export, stability and translation. Introns are removed by the spliceosome, which recognizes GU at the 5′ end and AG at the 3′ end of each intron and joins exons. Alternative splicing lets one gene make several mRNAs by including or skipping exons. In bacteria, there is no nucleus, so translation can begin while transcription is still underway.

The genetic code is read in non-overlapping triplets called codons. With four bases, 4³ = 64 codons exist: 61 specify amino acids and 3 (UAA, UAG, UGA) are stops. AUG codes for methionine and usually starts translation. The code is degenerate (several codons per amino acid) but unambiguous, and nearly universal across life. The reading frame is set by the start codon; inserting or deleting bases not in multiples of three shifts the frame. tRNAs carry amino acids, and their anticodons pair antiparallel with codons. Wobble at the third codon position lets one tRNA read several codons.

Translation takes place on ribosomes. In bacteria, the small subunit binds the Shine-Dalgarno sequence near AUG; in eukaryotes, it binds the cap and scans to the first good AUG. The initiator tRNA sits in the P site. In elongation, a charged tRNA enters the A site, the large subunit's rRNA (peptidyl transferase) forms a peptide bond, and translocation shifts tRNAs to the P and E sites, exposing the next codon. A stop codon in the A site binds a release factor, freeing the chain. Accuracy depends heavily on aminoacyl-tRNA synthetases, because the ribosome checks only codon-anticodon pairing.

Key ideas

Formulas and rules

Codon combinations = 4³ = 64 (61 sense + 3 stop)
Coding nucleotides = 3 × (amino acids) + 3 for the stop codon
Peptide bonds = number of amino acids − 1
Anticodon (5′→3′) = reverse complement of codon
Ribosome sites in order of tRNA movement: A → P → E
Nutritional mutant grows on any intermediate after its block

Common mistakes

Before the test, make sure you can

Words to know

Central dogma
Information flows DNA → RNA → protein.
Template strand
The DNA strand read by RNA polymerase.
Coding strand
The DNA strand with the same sequence as the mRNA (T for U).
Promoter
A DNA sequence where RNA polymerase binds to start transcription.
Sigma factor
Bacterial protein that lets RNA polymerase recognize promoters.
TATA box
A eukaryotic promoter element bound by a general transcription factor.
5′ cap
A modified guanine added to the 5′ end of eukaryotic mRNA.
Poly-A tail
A string of adenines added to the 3′ end of eukaryotic mRNA.
Intron
A noncoding sequence removed from pre-mRNA by splicing.
Exon
A sequence kept in the mature RNA.
Spliceosome
The snRNP-protein complex that removes introns.
Alternative splicing
Different exon combinations from one gene, making different mRNAs.
Codon
A three-nucleotide mRNA unit that specifies an amino acid or stop.
Anticodon
A three-base tRNA sequence that pairs with a codon.
Wobble
Flexible pairing at the third codon position.
Aminoacyl-tRNA synthetase
Enzyme that attaches the correct amino acid to its tRNA.
Shine-Dalgarno sequence
Bacterial mRNA sequence that positions the ribosome at the start codon.
Release factor
Protein that recognizes a stop codon and releases the polypeptide.

Sample practice questions

30 more questions like these, from level 1 to 5, are in the practice set.

Question 1 (level 1)

Which strand of DNA has the same sequence as the mRNA (with T in place of U)?

  1. Either strand, chosen at random for each base
  2. The template (antisense) strand
  3. Neither strand, since mRNA is edited after transcription
  4. The coding (nontemplate) strand
Show the answer and steps

Answer: D) The coding (nontemplate) strand

  1. RNA polymerase reads the template strand and makes RNA complementary to it.
  2. The RNA is therefore identical in sequence to the other strand, except U replaces T.
  3. That other strand is called the coding or sense strand.
Question 2 (level 1)

Which codon is the usual start codon, and what amino acid does it specify?

  1. AUG, methionine
  2. UAA, methionine
  3. AUG, tryptophan
  4. UGA, methionine
Show the answer and steps

Answer: A) AUG, methionine

  1. Translation almost always begins at AUG.
  2. AUG codes for methionine (formyl-methionine in bacteria).
  3. UAA and UGA are stop codons.
Question 3 (level 3)

An mRNA reads 5′-GCAUGUUUGGCUAAGC-3′. Starting from the first AUG, how many amino acids are in the resulting peptide?

  1. 5
  2. 4
  3. 2
  4. 3
Show the answer and steps

Answer: D) 3

  1. The first AUG begins at the third base: AUG UUU GGC UAA.
  2. AUG = Met, UUU = Phe, GGC = Gly, UAA = stop.
  3. The peptide is Met-Phe-Gly: 3 amino acids; the stop codon adds none.