The shape of the lesson
By the end of the lesson, learners can
- Transcription: a gene copied into mRNA: Press Transcribe and watch RNA polymerase (green) move along the gene, opening the DNA just around itself.
- Translation: codons meet anticodons: Press Translate.
- Introns out, exons joined: Press Remove the introns.
- mRNA = complement of the template strand = coding strand with U for T.
- Anticodon on tRNA is complementary to the codon on mRNA.
- Start AUG (Met); stop codons have no tRNA.
- Introns removed, exons joined, before mRNA leaves the nucleus.
One picture every idea returns to
A polypeptide is coded for by a gene, a sequence of nucleotides forming part of a DNA molecule. The genetic code is a triplet code: each group of three bases codes for one amino acid, or acts as a start or stop signal, and it is universal (the same in almost all organisms). In transcription, in the nucleus, RNA polymerase separates the strands of the gene and builds mRNA complementary to the template (transcribed) strand; the other is the non-transcribed strand. In eukaryotes the primary transcript is modified: introns (non-coding) are removed and exons (coding) are joined. In translation, at a ribosome in the cytoplasm, each tRNA carries a specific amino acid and has an anticodon complementary to an mRNA codon. Starting at AUG, amino acids are joined by peptide bonds until a stop codon is reached.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, From gene to polypeptide.. A gene is a stretch of DNA that codes for one polypeptide. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Transcription: a gene copied into mRNA | Press Transcribe and watch RNA polymerase (green) move along the gene, opening the DNA just around itself. Project tab 4 of the lab (Nucleic acids lab, transcription) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Translation: codons meet anticodons | Press Translate. Project tab 5 of the lab (Nucleic acids lab, translation) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Introns out, exons joined | Press Remove the introns. Project tab 6 of the lab (Nucleic acids lab, introns and exons) and let a student drive it while the class predicts. | Think-Pair-Share |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 18 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Introns out, exons joined, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Think-Pair-Share
Pose the question, give thirty seconds of silent thinking, then a minute to compare with a partner before a few pairs share with the class. The silent minute is the part that cannot be cut.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| mRNA is complementary to the template strand, not to the coding strand. | So the mRNA reads the same as the coding (non-transcribed) strand, with U instead of T. If you are given the template strand, pair every base and swap T for U. |
| Remember which is which: ex ons are ex pressed (they stay in the mRNA); in trons are in the way (they are removed). | It is the RNA that is edited, not the DNA: the introns stay in the gene. |
Transcription and translation: worksheet
6 questions, 18 marks. Written in the style of the papers; no past paper question is reproduced.
State what is meant by a gene, and name the DNA strand that is used in transcription.
Describe how an mRNA molecule is made from a gene in the nucleus of a eukaryotic cell.
Part of a template DNA strand reads 3′ TAC AAA CCG ATT 5′. Using the information that AUG = Met, UUU = Phe, GGC = Gly and UAA = stop, give the mRNA sequence, the tRNA anticodons and the amino acid sequence.
Explain how the sequence of codons in mRNA determines the sequence of amino acids in a polypeptide.
A eukaryotic gene contains 3 exons of 420, 270 and 231 nucleotides and 2 introns. The mRNA includes a start codon and a stop codon and no other non-coding bases. Calculate the number of amino acids in the polypeptide.
Suggest why a human gene for insulin can be translated into human insulin inside a bacterium, but the bacterium could not use the gene directly as it is found in human chromosomes.
Answer key and marking notes
- A gene is a sequence of nucleotides, forming part of a DNA molecule, that codes for a polypeptide.
- The template strand (also called the transcribed strand); the other strand is the non-transcribed strand.
- RNA polymerase binds to the gene, breaks hydrogen bonds and separates the two DNA strands.
- Free RNA nucleotides pair with exposed bases on the template strand by complementary base pairing (A-U, T-A, C-G, G-C).
- RNA polymerase joins the nucleotides with phosphodiester bonds to form the primary transcript.
- Introns are removed and exons are joined together to form mRNA, which leaves through a nuclear pore.
- mRNA: 5′ AUG UUU GGC UAA 3′.
- Anticodons: UAC, AAA, CCG (none for the stop codon).
- Amino acids: Met-Phe-Gly, then the polypeptide is released at UAA.
- mRNA binds to a ribosome; translation starts at the start codon AUG.
- Each tRNA carries a specific amino acid and has an anticodon.
- The anticodon pairs with the complementary codon on the mRNA by hydrogen bonds, so the right amino acid is brought for each codon.
- A peptide bond forms between adjacent amino acids; the ribosome moves along one codon at a time.
- At a stop codon no tRNA binds and the polypeptide is released; so the codon order fixes the amino acid order.
- Coding mRNA = 420 + 270 + 231 = 921 nucleotides = 307 codons.
- The stop codon codes for no amino acid, so 306 amino acids.
- The genetic code is universal, so the bacterial ribosomes and tRNAs read the same codons as the same amino acids.
- The human gene contains introns, and bacteria cannot remove introns, so a version without introns (made from mRNA) must be used.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
In a eukaryotic cell, where do transcription and translation take place?
- Atranscription in the nucleus; translation at ribosomes in the cytoplasm
- Btranscription at ribosomes; translation in the nucleus
- Cboth in the nucleus, before the polypeptide leaves through a pore
- Dtranscription in the cytoplasm; translation on the Golgi body
The non-transcribed strand of part of a gene reads 5′ ATG CTA GGA 3′. What is the sequence of the mRNA made from this gene?
- A5′ AUG CUA GGA 3′
- B5′ UAC GAU CCU 3′
- C5′ ATG CTA GGA 3′
- D5′ TAC GAT CCT 3′
What is an anticodon?
- Athree bases on a tRNA molecule complementary to a codon on mRNA
- Bthree bases on the template DNA strand complementary to the coding strand
- Ca stop codon that ends translation of the mRNA
- Dthree bases on mRNA that code for one amino acid
Use the table to find the polypeptide made from the mRNA 5′ AUG AAA UUU GAA UGA UGG 3′.
- AMet-Lys-Phe-Glu
- BMet-Lys-Phe-Glu-Trp
- CLys-Phe-Glu-Trp
- DMet-Lys-Phe-Glu-Stop-Trp
Which statement about tRNA is correct?
- Aeach tRNA carries one specific amino acid and has one anticodon
- Beach tRNA carries one amino acid chosen at random at the ribosome
- CtRNA is made by DNA polymerase in the cytoplasm
- DtRNA attaches to the stop codon to release the polypeptide
A primary transcript is 3600 nucleotides long. After the introns are removed, the mRNA is 1350 nucleotides long. What percentage of the primary transcript was intron?
- A62.5%
- B37.5%
- C266.7%
- D2250%
Proteins are built from 20 types of amino acid. Why can the genetic code not be a doublet code using two bases per codon?
- A42 = 16 combinations, too few for 20 amino acids
- B4 × 2 = 8 combinations, too few for 20 amino acids
- C24 = 16 combinations, but no stop codons would remain
- Da doublet code cannot be read by a ribosome in a cell
An mRNA codon is 5′ GCA 3′. What is the matching triplet on the template strand of the DNA?
- A3′ CGT 5′
- B5′ CGT 3′
- C3′ CGU 5′
- D5′ GCA 3′
Quiz key
The DNA stays in the nucleus, so transcription happens there; mRNA leaves through a nuclear pore to ribosomes in the cytoplasm (free or on rough ER) for translation. Ribosomes do not transcribe DNA. Polypeptides are not made inside the nucleus. The Golgi body modifies proteins after translation; it does not translate.
The mRNA is complementary to the template strand, so it has the same sequence as the non-transcribed strand with U in place of T: AUG CUA GGA. UAC GAU CCU is complementary to the strand given, which would only be right if it were the template. ATG CTA GGA still contains T, which RNA lacks. TAC GAT CCT is the template DNA strand itself.
An anticodon is a triplet of unpaired bases on tRNA that pairs with the complementary codon on mRNA, so each tRNA delivers the right amino acid. A triplet on the template DNA strand is not called an anticodon. A stop codon is an mRNA codon with no matching tRNA. Three mRNA bases coding for one amino acid is the definition of a codon.
Reading AUG AAA UUU GAA gives Met-Lys-Phe-Glu; UGA is a stop codon, so translation ends and UGG is never read. Adding Trp ignores the stop codon. Leaving out Met forgets that the start codon also codes for methionine. Stop is not an amino acid, so it cannot be part of the chain.
Each tRNA has a particular anticodon and carries the amino acid that matches it, which is how the codon sequence fixes the amino acid sequence. Amino acids are not attached at random. tRNA is RNA, transcribed in the nucleus by RNA polymerase. No tRNA pairs with a stop codon; that is why the polypeptide is released there.
Introns = 3600 − 1350 = 2250 nucleotides; 2250 ÷ 3600 × 100 = 62.5%. 37.5% is the exon (mRNA) percentage. 266.7% divides the transcript by the mRNA. 2250% forgets to divide by the transcript length.
Each of the two positions can be any of 4 bases, giving 4 × 4 = 16 doublets, fewer than the 20 amino acids. A triplet code gives 43 = 64, enough for every amino acid plus start and stop signals, with spare codons (degeneracy). 4 × 2 adds rather than multiplies the choices. 24 has the base and power swapped. The ribosome argument is not a reason based on the numbers.
Pair each base (G-C, C-G, A-T) and keep the strands antiparallel: the template reads 3′ CGT 5′. Writing CGT 5′ to 3′ reverses the direction. CGU contains uracil, which is not in DNA. 5′ GCA 3′ is the non-transcribed (coding) strand, which has the same sequence as the mRNA.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.