DSE Biology | Gene Expression and Protein Synthesis Lesson-Prep Guide (for tutors) | transcription, codon arithmetic, four common errors

2026-11-05

Tutor lesson-prep guides · Secondary Biology

Students can chant DNA then RNA then protein, but ask how many amino acids a 900-base mRNA carries and the division goes wrong, and they treat every mutation as an amino acid swap. This lesson transcribes and splices one real sequence, then counts codons, amino acids and peptide bonds.

Objectives — Students can: (1) write the mRNA from a DNA template strand (A pairs with U, T pairs with A) and state that transcription happens in the nucleus under RNA polymerase; (2) take a 1200-base pre-mRNA, remove 300 bases of introns and convert 900 bases into 300 codons, 299 amino acids and 298 peptide bonds, explaining why the stop codon is deducted; (3) separate silent, missense, nonsense and frameshift mutations and say which one almost always destroys the protein.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
DNA structure and base pairing	How many hydrogen bonds join A to T, and G to C	Two for A-T, three for G-C
Central dogma	Which molecule carries information to which	DNA to RNA to protein, in that order
Organelles	Where are ribosomes and what are they made of	Free or on rough ER, made of rRNA and protein

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-6 min	Prior check	Hydrogen bond counts, the order of the dogma, ribosome site	Answer orally, errors go in the board corner
6-22 min	Act 1 transcription	Build the mRNA from the template strand, U instead of T	Fill each base in turn and compare with a partner
22-36 min	Act 2 processing and translation	Splice out 300 bases, then read codon by codon	Count codons, amino acids and peptide bonds
36-48 min	Act 3 mutation	Try a substitution, a deletion and an insertion	State the effect of each on the protein
48-56 min	Question drill	One item from each of the four types, timed	Work alone, then swap and mark
56-60 min	Exit check	Collect three items and mark on the spot	Books closed

03 Teaching script (read it out)

Scene 1: Act 1: transcription copies DNA into RNA, it does not copy DNA

Write the template strand TAC GGA TTC and ask for the mRNA. Eight out of ten write ATG CCT AAG, because they reuse the DNA replication rule. Correct it at once: RNA uses uracil instead of thymine, so A pairs with U and T pairs with A.
The second point is the word template. Only one strand is read; the other is the coding strand, whose sequence matches the mRNA apart from T becoming U. Run both strands once: read the wrong one and every codon downstream is wrong.
The third point is location and enzyme: transcription happens in the nucleus, RNA polymerase unwinds the helix and adds complementary ribonucleotides at about 40 nucleotides per second, so a 900-base mRNA takes roughly 22.5 seconds. Misconception box: transcription does not copy the whole chromosome, only the gene being expressed.

Students restate: RNA uses U not T, only the template strand is read, transcription runs in the nucleus.

Scene 2: Act 2: splice first, then read three bases at a time

Write a pre-mRNA of 1200 bases with 300 of them introns. The first reflex is to divide 1200 by three. Make them splice first: 1200 minus 300 leaves 900 bases of mature mRNA, and introns never reach the ribosome. The splicing happens in the nucleus before the mRNA leaves through a pore.
Then divide: 900 over three gives 300 codons. The step students skip is the stop codon, which codes for no amino acid and only tells the ribosome to stop, so the amino acid count is 300 minus one, which is 299, and the peptide bonds number 298. Write all three numbers on the board so the class sees that bases, codons, amino acids and bonds are four different quantities.
Then the scale of the code: 4 bases in triplets give 64 codons, 3 of them stop signals, about 4.69 percent, against only 20 amino acids, which is why several codons share one amino acid, the degeneracy of the code. AUG is both the start codon and the codon for methionine, a double role that gets examined.
Misconception box: codon and anticodon are not interchangeable. The codon sits on mRNA, the anticodon on tRNA, and the two are complementary.

Students restate: splice the introns, three bases per codon, deduct the stop codon before counting amino acids.

Scene 3: Act 3: a mutation need not change an amino acid, but a frameshift almost always ruins the protein

Build one table with four rows. Silent: the base changes but the amino acid does not, thanks to degeneracy. Missense: one amino acid is swapped and the protein may partly work, as in the sickle-cell haemoglobin case. Nonsense: an amino acid codon becomes a stop codon and the protein is truncated. Frameshift: an insertion or deletion that is not a multiple of three shifts every codon downstream.
Work a concrete case: the codon GGU for glycine becomes GAU for aspartate through a substitution, a missense change; insert one A instead and everything from that point is regrouped, which is the frameshift. Stress why frameshifts are worst: not one letter changed but the whole reading frame, so a stop codon usually appears soon and the protein is short and useless.
Close with one question: why is a silent mutation still a mutation? Because the DNA sequence really changed even though the phenotype does not, and it is still inherited. That question separates memorisers from understanders.

Students restate: silent keeps the amino acid, missense swaps one, nonsense stops early, frameshift breaks everything after the point.

04 Question types and worked steps

Type	Example	Steps	Answer
Transcribe a sequence	Template strand TAC GGA TTC, write the mRNA	1 pair each base 2 A with U and T with A 3 write in 5 prime to 3 prime order	AUG CCU AAG
Base arithmetic	1200 bases with 300 intron bases, find the amino acid count	1 splice 2 divide by three 3 deduct the stop codon	900 bases, 300 codons, 299 amino acids
Classify a mutation	GGU becomes GAU, which type	1 compare the two codons 2 check whether the amino acid changes 3 name the type	Missense, one amino acid replaced
Frameshift estimate	Insert one A after the third base	1 regroup from the insertion 2 translate each new codon 3 find the first stop	Every codon after the point changes and the chain ends early

Sanity check (teach this): Check plausibility: amino acids must be fewer than codons (299 against 300) because the stop codon is deducted, and peptide bonds fewer than amino acids (298 against 299) because n amino acids make n minus one bonds. A student reporting more amino acids than codons forgot to divide by three; equal bonds and amino acids means minus one was skipped. A scale check: 64 codons against 20 amino acids gives about three codons per amino acid, so an answer of 64 amino acid types confuses codons with amino acids.

05 Independent practice (3 questions)

Question	Answer	What to watch
A pre-mRNA has 1200 bases and 300 of them are introns. Find the mature base count, codon count and amino acid count.	900 bases, 300 codons, 299 amino acids	Introns spliced first, stop codon deducted
The template strand is TAC GGG CCA. Write the mRNA and name the three amino acids, given GGG is glycine and CCA is proline.	AUG CCC GGU, methionine then proline then glycine	A correctly paired with U
If base six changes from G to A so that CCC becomes CCA, does the amino acid change?	No, both code for proline, so it is a silent mutation	Degeneracy used as the reason, not a guess

06 Common mistakes and what to say

Mistake	Student writes	Say this
Writing T inside RNA	mRNA written as ATG CCT AAG	RNA uses U, not T. Circle any T the moment you see it.
Forgetting to splice	Dividing 1200 by three to get 400 codons	Introns never leave the nucleus, so splice first and use the mature count.
Amino acids equal to codons	Answering 300 amino acids	The last triplet is a stop codon and codes for nothing, so deduct one.
Treating every mutation as harmful	Answering that any mutation disables the protein	Silent changes nothing, missense sometimes keeps partial function, so state the degree.

07 Homework (paper, section and question numbers)

Redo this chapter — Redo practice items 1 to 3 (questions 1-4), writing bases, codons and amino acids for each
Biology chapter index — Pick the genetics and molecular biology chapter and do question 3, drawing the transcription and translation flow
Build a codon chart — Mark the 3 stop codons and AUG among the 64 codons and state what each does

08 Exit check (last 5 minutes)

Question	Answer	Decision
Which RNA base replaces DNA thymine	Uracil	Answering T means replay act one
How many amino acids come from 300 codons	299	Answering 300 means replay act two
Which mutation comes from an insertion that is not a multiple of three	A frameshift	A wrong answer means replay act three

09 Tutor FAQ

Why distinguish pre-mRNA from mature mRNA?

Eukaryotic genes contain introns, so the transcript must be spliced before it leaves the nucleus and is translated.

How many codons can one amino acid have?

Up to six, and that degeneracy is why some base changes leave the protein untouched.

Do prokaryotes splice?

Generally no introns, and translation starts while transcription is still running, so their mRNA is short-lived.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.