Answer key · Unit 5

DNA and proteins.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 4

Lesson 1 of 4 · About 20 minutes

The double helix

How DNA is built: two strands, four bases and one pairing rule, the hydrogen bonds that hold the ladder together, Chargaff's rule, and how one molecule becomes two.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Two strands and a ladder
Practice · Choose
Question
Each rung of the ladder is a pair of bases, one from each strand. Which base always sits opposite G?

Correct answer
- Correct C
Setup, choices and feedback
The Scene at this step
G–C pairs
50 %
Highlight
All
Double helix / Copied
Double helix
Choices
Shown in a random order.
- Correct C
- Wrong T
  Misconception: Thinks any two bases can pair, or that like pairs with like (A with A, or A with G).
  Feedback from the catalogue
  The pairing is fixed: A always faces T, and G always faces C. The reason is shape: A and G are large bases with two rings, and T and C are small, with one. Every pair is large against small, so the gap between the strands is the same all along the helix, and only these pairs form hydrogen bonds that line up with each other.
- Wrong A
  Misconception: Thinks any two bases can pair, or that like pairs with like (A with A, or A with G).
  Feedback from the catalogue
  The pairing is fixed: A always faces T, and G always faces C. The reason is shape: A and G are large bases with two rings, and T and C are small, with one. Every pair is large against small, so the gap between the strands is the same all along the helix, and only these pairs form hydrogen bonds that line up with each other.
- Wrong G, another G
  Misconception: Thinks any two bases can pair, or that like pairs with like (A with A, or A with G).
  Feedback from the catalogue
  The pairing is fixed: A always faces T, and G always faces C. The reason is shape: A and G are large bases with two rings, and T and C are small, with one. Every pair is large against small, so the gap between the strands is the same all along the helix, and only these pairs form hydrogen bonds that line up with each other.
Feedback for a right answer
G faces C, and A faces T, in every pair of the helix. So reading one strand is enough to know the other.

Hint
Set Highlight to G–C and see what sits opposite each G.

Worked solution
1. The helix has four bases: A, T, G and C.
2. The pairing rule: A pairs only with T, and G only with C.
3. So C always sits opposite G.
If you chose A, you paired two large, two-ringed bases: a pair like that is too wide for the gap between the strands.
Step 2 of 5
Open this step
Hydrogen bonds
Practice · Calculate
Question
The helix in the Scene has $20$ ⁠ base pairs, and $8$ ⁠ of them, the highlighted ones, are G–C pairs. The rest are A–T. How many hydrogen bonds hold the two strands together?

Correct answer
48 ± 0.1
Setup, choices and feedback
The Scene at this step
G–C pairs
50 %fixed
Highlight
G–C
Double helix / Copied
Double helixfixed
Accepted within
± 0.1
Computed from the Scene at this step’s setting
Hydrogen bonds holding the two strands together: two per A–T pair, three per G–C pair/nodes/dna/outputs/hydrogenBonds
Wrong answers it recognises
- 40 ± 0.1
  Misconception: Gives every base pair the same number of hydrogen bonds.
  Feedback from the catalogue
  Not every pair is held as tightly. An A–T pair is held by two hydrogen bonds, and a G–C pair by three. So count them separately: two times the A–T pairs, plus three times the G–C pairs. That is also what makes GC-rich DNA more stable.
- 60 ± 0.1
  Misconception: Gives every base pair the same number of hydrogen bonds.
  Feedback from the catalogue
  Not every pair is held as tightly. An A–T pair is held by two hydrogen bonds, and a G–C pair by three. So count them separately: two times the A–T pairs, plus three times the G–C pairs. That is also what makes GC-rich DNA more stable.
- 52 ± 0.1
  Misconception: Gives every base pair the same number of hydrogen bonds.
  Feedback
  You swapped them: an A–T pair has two hydrogen bonds, and a G–C pair three. The right sum: $2 \times 12 + 3 \times 8$ ⁠.
Feedback for a right answer
$12$ ⁠ A–T pairs with two bonds each, and $8$ ⁠ G–C pairs with three: $24 + 24 = 48$ ⁠ hydrogen bonds.

Hint
How many A–T pairs are there? How many bonds in each kind of pair?

Worked solution
1. A–T pairs: $20 - 8 = 12$ ⁠.
2. Bonds in the A–T pairs: $12 \times 2 = 24$ ⁠.
3. Bonds in the G–C pairs: $8 \times 3 = 24$ ⁠.
4. Together: $24 + 24 = 48$ ⁠ hydrogen bonds.
If you got $40$ ⁠ or $60$ ⁠, you gave every pair the same number of bonds. If you got $52$ ⁠, you swapped two and three.
Step 3 of 5
Open this step
Chargaff's rule
Practice · Calculate
Question
Now only $4$ ⁠ of the $20$ ⁠ pairs are G–C, that is $20\%$ ⁠ of the pairs. What percentage of all the bases in the helix, on both strands together, are A?

Correct answer
40 ± 0.5
Setup, choices and feedback
The Scene at this step
G–C pairs
30 %fixed
Highlight
G–C
Double helix / Copied
Double helixfixed
Accepted within
± 0.5
Computed from the Scene at this step’s setting
Share of all the bases that are A (as many as T), 0..0.5/nodes/dna/outputs/adenineShare
Wrong answers it recognises
- 80 ± 0.5
  Misconception: Gives the share of base pairs as if it were the share of bases, forgetting that each pair holds two different bases.
  Feedback from the catalogue
  Every pair holds two different bases. If $20\%$ ⁠ of the pairs are G–C, then G is only half of that share of all the bases, $10\%$ ⁠, and C another $10\%$ ⁠. Likewise for A–T: $80\%$ ⁠ of the pairs give $40\%$ ⁠ A and $40\%$ ⁠ T. That is Chargaff’s rule: exactly as much A as T, and as much G as C.
- 10 ± 0.5
  Misconception: Gives the share of base pairs as if it were the share of bases, forgetting that each pair holds two different bases.
  Feedback
  $10\%$ ⁠ is the share of G. The question is about A, which sits in the A–T pairs.
Feedback for a right answer
$16$ ⁠ A–T pairs give $16$ ⁠ A bases out of $40$ ⁠ bases: $40\%$ ⁠. In the same way there is $40\%$ ⁠ T, $10\%$ ⁠ G and $10\%$ ⁠ C.

Hint
How many bases are there in all? Each A–T pair has one A and one T.

Worked solution
1. Bases in all: $20$ ⁠ pairs times $2$ ⁠ = $40$ ⁠.
2. A–T pairs: $20 - 4 = 16$ ⁠, each with one A: $16$ ⁠ A bases.
3. The share: $\frac{16}{40} = 40\%$ ⁠.
4. Check with Chargaff’s rule: A and T $40\%$ ⁠ each, G and C $10\%$ ⁠ each, and together $100\%$ ⁠.
If you answered $80\%$ ⁠, you gave the share of A–T pairs among all the pairs. But only half the bases in each of those pairs are A.
Step 4 of 5
Open this step
A sturdier helix
Practice · Reach the target
Question
A helix of $20$ ⁠ pairs is sturdier the more hydrogen bonds it has. Change G–C pairs until the helix is held by exactly $56$ ⁠ hydrogen bonds, then press Check.

Goal
- 56 ± 0.1
  At the start of the step: 48
Setup, choices and feedback
The Scene at this step
G–C pairs
50 %
Highlight
G–C
Double helix / Copied
Double helixfixed
Read from the Scene at
Hydrogen bonds holding the two strands together: two per A–T pair, three per G–C pair/nodes/dna/outputs/hydrogenBonds
One setting that reaches it
G–C pairs
75 %
Highlight
All
Feedback for a right answer
$16$ ⁠ G–C pairs and $4$ ⁠ A–T pairs: $3 \times 16 + 2 \times 4 = 56$ ⁠. GC-rich DNA is harder to pull apart, so bacteria living in hot springs tend to be rich in G–C.

Feedback for a wrong answer
Not yet. Count: each G–C pair gives $3$ ⁠ bonds, and each A–T pair $2$ ⁠.

Hint
If all $20$ ⁠ pairs were A–T, there would be $40$ ⁠ bonds. Each G–C pair adds one bond.

Worked solution
1. With $x$ ⁠ G–C pairs and $20 - x$ ⁠ A–T pairs there are $3x + 2(20 - x) = 40 + x$ ⁠ bonds.
2. You want $40 + x = 56$ ⁠, so $x = 16$ ⁠.
3. Move G–C pairs until $16$ ⁠ pairs are highlighted as G–C. On this helix that happens around $75\%$ ⁠.
The percentage on the slider does not always come out exactly: the helix picks its bases one by one, so count the highlighted pairs.
Step 5 of 5
Open this step
One molecule becomes two
Practice · Predict, then look
Question
Before a cell divides it copies its DNA. When the copying in the Scene is done there will be two helices. What will each of them hold?

Correct answer
- Correct One old strand and one new strand
Setup, choices and feedback
The Scene at this step
G–C pairs
50 %
Highlight
All
Double helix / Copied
Double helixfixed
Choices
- Correct One old strand and one new strand
- Wrong One wholly old helix, and beside it a wholly new one
  Misconception: Thinks copying DNA leaves the old helix whole and builds a completely new one beside it.
  Feedback from the catalogue
  DNA is not photocopied. The two strands come apart, and each old strand is the template for a new strand built against it. So each of the two helices at the end holds one old strand and one new. This is called semi-conservative replication, and Meselson and Stahl showed it by experiment in $1958$ ⁠.
- Wrong Old and new pieces alternating along each strand
  Feedback
  That was one of the ideas before the Meselson and Stahl experiment, but each old strand stays whole and is the template for a whole new strand.
Once the prediction is locked in, the Scene changes
Double helix / Copied
Copiedfixed
The Scene then shows: 2 · Double helices: 1, or 2 once copying is complete/nodes/dna/outputs/helices
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Each old strand is a template, and the pairing rule decides what is built against it. So the two helices match each other and the original, and each holds an old strand and a new one.

Hint
The strands open like a zip. What is each new strand built on?

Worked solution
1. The strands separate: each of them carries the whole message.
2. A new strand is built against each old one, A opposite T and G opposite C.
3. So each helix at the end has one old strand and one new: semi-conservative replication.
In the Scene the new strands are drawn in a paler tone, and each helix keeps one old strand.

Lesson 2 of 4 · About 25 minutes

From gene to protein

How a gene becomes a protein: transcription into messenger RNA, translation in triplets of letters, start and stop codons, and why the genetic code uses three letters.

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The 6 steps and their answers

Step 1 of 6
Open this step
Genes make proteins
No practice question in this step.
The Scene at this step
One-letter change
Nonedefaulthidden
Gene / Folded protein
Gene
Step 2 of 6
Open this step
Transcription
Practice · Choose
Question
The gene’s coding strand begins like this, after its first three letters: ATG AAG GAG. What does the messenger RNA say at the same place?

Correct answer
- Correct AUG AAG GAG
Setup, choices and feedback
The Scene at this step
One-letter change
Nonedefaulthidden
Gene / Folded protein
0.34 (Gene → Folded protein)
Choices
Shown in a random order.
- Correct AUG AAG GAG
- Wrong UAC UUC CUC
  Misconception: Writes the mRNA as the complement of the coding strand, mixing up the coding strand with the template strand.
  Feedback from the catalogue
  The RNA is built against the template strand, so it is the complement of that strand and matches the coding strand exactly, except for U in place of T. A gene is written on its coding strand: if that reads ATG AAG, the RNA reads AUG AAG. If you complemented each letter (A to U, T to A and so on), you wrote the complement of the RNA, not the RNA.
- Wrong ATG AAG GAG
  Misconception: Keeps T in the RNA copy.
  Feedback from the catalogue
  RNA has no T. In its place it has U, uracil, which pairs with A just as T does. So the copy of ATG is AUG: the same letters, with U wherever the gene has T.
- Wrong TAC TTC CTC
  Feedback
  That is the DNA’s other strand, the template strand. The RNA is built against it, so it is its complement, and comes out like the coding strand with U in place of T.
Feedback for a right answer
The same letters, with U in place of T: AUG AAG GAG. The AUG is the start codon: this is where the ribosome will begin to read.

Hint
The RNA has the same letters as the coding strand, with one swap.

Worked solution
1. RNA polymerase builds the RNA against the template strand, so the RNA matches the coding strand.
2. The only difference: RNA has U in place of T.
3. ATG AAG GAG becomes AUG AAG GAG.
If you complemented each letter, you wrote the complement of the RNA. If you kept T, you forgot RNA has no T.
Step 3 of 6
Open this step
Reading in threes
Practice · Calculate
Question
The Scene’s gene has $51$ ⁠ letters. The start codon AUG takes letters $4$ ⁠ to $6$ ⁠, and the stop codon UAA letters $46$ ⁠ to $48$ ⁠. How many codons does the ribosome read, the stop codon included?

Correct answer
15 ± 0.1
Setup, choices and feedback
The Scene at this step
One-letter change
Nonedefaulthidden
Gene / Folded protein
0.5 (Gene → Folded protein)
Accepted within
± 0.1
Computed from the Scene at this step’s setting
Codons the ribosome reads, from AUG to the stop codon inclusive/nodes/gene/outputs/codons
Wrong answers it recognises
- 17 ± 0.1
  Misconception: Starts reading codons at the first letter of the gene instead of at the start codon AUG.
  Feedback from the catalogue
  The ribosome does not start at the first letter. It slides along the RNA to the first AUG, and only from there reads triplets. The letters before the AUG, and those after the stop codon, are not translated. In the Scene’s gene the AUG starts at letter $4$ ⁠, so the first three letters, GCC, are not part of the protein.
- 45 ± 0.1
  Misconception: Matches one letter with one amino acid, instead of three letters, a codon, with one.
  Feedback from the catalogue
  One amino acid is coded by three letters, a codon. Four letters alone could code only four amino acids, and there are twenty. So a stretch of $6$ ⁠ letters is $2$ ⁠ codons, and $2$ ⁠ amino acids, not $6$ ⁠.
Feedback for a right answer
From letter $4$ ⁠ to letter $48$ ⁠ there are $45$ ⁠ letters, that is $15$ ⁠ codons: $14$ ⁠ that bring an amino acid, and the stop.

Hint
How many letters are there from letter $4$ ⁠ to letter $48$ ⁠, both included?

Worked solution
1. Reading starts at the AUG, at letter $4$ ⁠, and ends at the end of the stop codon, letter $48$ ⁠.
2. Letters read: $48 - 4 + 1 = 45$ ⁠.
3. Codons: $\frac{45}{3} = 15$ ⁠.
If you got $17$ ⁠, you divided all $51$ ⁠ letters by three, as if reading started at the first letter and ran to the end.
Step 4 of 6
Open this step
The chain
Practice · Calculate
Question
How many amino acids are in the chain the ribosome builds from this gene?

Correct answer
14 ± 0.1
Setup, choices and feedback
The Scene at this step
One-letter change
Nonedefaulthidden
Gene / Folded protein
0.74 (Gene → Folded protein)
Accepted within
± 0.1
Computed from the Scene at this step’s setting
Amino acids in the protein (0 when there is no start codon)/nodes/gene/outputs/aminoAcids
Wrong answers it recognises
- 15 ± 0.1
  Misconception: Counts the stop codon as an amino acid.
  Feedback from the catalogue
  A stop codon brings no amino acid. No tRNA fits UAA, UAG or UGA, so when the ribosome reaches one it lets the chain go. The number of amino acids is one less than the codons read, because the last of them is the stop.
- 17 ± 0.1
  Misconception: Starts reading codons at the first letter of the gene instead of at the start codon AUG.
  Feedback from the catalogue
  The ribosome does not start at the first letter. It slides along the RNA to the first AUG, and only from there reads triplets. The letters before the AUG, and those after the stop codon, are not translated. In the Scene’s gene the AUG starts at letter $4$ ⁠, so the first three letters, GCC, are not part of the protein.
- 51 ± 0.1
  Misconception: Matches one letter with one amino acid, instead of three letters, a codon, with one.
  Feedback from the catalogue
  One amino acid is coded by three letters, a codon. Four letters alone could code only four amino acids, and there are twenty. So a stretch of $6$ ⁠ letters is $2$ ⁠ codons, and $2$ ⁠ amino acids, not $6$ ⁠.
Feedback for a right answer
$14$ ⁠: one amino acid per codon, except the stop codon, which brings nothing and releases the chain.

Hint
$15$ ⁠ codons were read. What does the last of them bring?

Worked solution
1. The ribosome reads $15$ ⁠ codons, from AUG to UAA.
2. Every codon but the last brings one amino acid.
3. The stop codon codes for no amino acid.
4. In the chain: $15 - 1 = 14$ ⁠ amino acids.
If you answered $15$ ⁠, you counted the stop codon too. If you answered $51$ ⁠, you gave each letter an amino acid.
Step 5 of 6
Open this step
Where the chain ends
Practice · Choose
Question
What does the ribosome do when it reaches the codon UAA?

Correct answer
- Correct It releases the chain: no tRNA fits this codon
Setup, choices and feedback
The Scene at this step
One-letter change
Nonedefaulthidden
Gene / Folded protein
0.74 (Gene → Folded protein)
Choices
Shown in a random order.
- Correct It releases the chain: no tRNA fits this codon
- Wrong It adds one last amino acid, then stops
  Misconception: Counts the stop codon as an amino acid.
  Feedback from the catalogue
  A stop codon brings no amino acid. No tRNA fits UAA, UAG or UGA, so when the ribosome reaches one it lets the chain go. The number of amino acids is one less than the codons read, because the last of them is the stop.
- Wrong It skips it and goes on to the next codon
  Feedback
  If the ribosome skipped it, the chain would run on to the end of the RNA. A stop codon is exactly what decides where the protein ends.
Feedback for a right answer
UAA, UAG and UGA are stop codons. They have no tRNA, so a special protein, a release factor, takes the place instead and the chain comes free.

Hint
Count again: $15$ ⁠ codons, $14$ ⁠ amino acids.

Worked solution
1. Every ordinary codon fits a tRNA carrying an amino acid.
2. The three stop codons have no tRNA.
3. So when the ribosome reaches UAA it releases the chain, and the RNA and ribosome part.
In the Scene: $15$ ⁠ codons were read, and the chain has $14$ ⁠ amino acids. The difference is the stop codon.
Step 6 of 6
Open this step
The genetic code
Practice · Calculate
Question
A codon is three letters, and each of them can be any of four: A, C, G or U. How many different codons can be written?

Correct answer
64 ± 0.1
Setup, choices and feedback
The Scene at this step
One-letter change
Nonedefaulthidden
Gene / Folded protein
Folded protein
Accepted within
± 0.1
A fixed value in the question
Wrong answers it recognises
- 12 ± 0.1
  Misconception: Matches one letter with one amino acid, instead of three letters, a codon, with one.
  Feedback
  $4 \times 3 = 12$ ⁠ counts letters, not combinations. Each of the four choices for the first letter has four for the second, and each of those four for the third.
- 20 ± 0.1
  Misconception: Thinks every change of a letter changes the protein.
  Feedback
  $20$ ⁠ is the number of amino acids. There are more codons than that, so most amino acids have several codons.
Feedback for a right answer
$4 \times 4 \times 4 = 64$ ⁠ codons: $61$ ⁠ code for amino acids and $3$ ⁠ are stops. With two letters there would be only $16$ ⁠, fewer than the twenty amino acids.

Hint
Count as in a tree: how many choices for the first letter, how many for the second, how many for the third?

Worked solution
1. Choices for the first letter: $4$ ⁠.
2. For each of them, $4$ ⁠ for the second: $4 \times 4 = 16$ ⁠.
3. For each of those, $4$ ⁠ for the third: $16 \times 4 = 64$ ⁠.
4. $61$ ⁠ codons code for $20$ ⁠ amino acids, and $3$ ⁠ are stops.
That is why the code uses triplets: pairs of letters give only $4^2 = 16$ ⁠ combinations, too few for twenty amino acids.

Lesson 3 of 4 · About 25 minutes

Mutations

What a change of one letter in a gene does: silent, missense and nonsense mutations and frameshifts, and why losing three letters does less harm than losing one.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
The same meaning
Practice · Predict, then look
Question
Letter $15$ ⁠ of the gene turns from G to A, so the codon CTG becomes CTA. In the code table both mean leucine. What will happen to the protein?

Correct answer
- Correct Nothing: exactly the same protein
Setup, choices and feedback
The Scene at this step
One-letter change
Nonefixed
Gene / Folded protein
Folded protein
Choices
- Correct Nothing: exactly the same protein
- Wrong One amino acid will be swapped for another
  Misconception: Thinks every change of a letter changes the protein.
  Feedback from the catalogue
  Most amino acids have several codons, often differing only in the third letter: CTG and CTA both mean leucine. A mutation that turns one into the other is silent: the gene changed, the protein did not. Besides, most of a cell’s DNA does not code for protein at all.
- Wrong The chain will stop early
  Misconception: Thinks every change of a letter changes the protein.
  Feedback from the catalogue
  Most amino acids have several codons, often differing only in the third letter: CTG and CTA both mean leucine. A mutation that turns one into the other is silent: the gene changed, the protein did not. Besides, most of a cell’s DNA does not code for protein at all.
- Wrong Every amino acid from here on will change
  Misconception: Thinks every change of a letter changes the protein.
  Feedback from the catalogue
  Most amino acids have several codons, often differing only in the third letter: CTG and CTA both mean leucine. A mutation that turns one into the other is silent: the gene changed, the protein did not. Besides, most of a cell’s DNA does not code for protein at all.
Once the prediction is locked in, the Scene changes
One-letter change
Letter 15: G to Afixed
Gene / Folded protein
Folded protein
The Scene then shows: 0 · Position in the chain of the first amino acid that differs, counting from 1 (0 when none does)/nodes/gene/outputs/changedAt
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The gene changed, but the protein did not: leucine stays leucine, and the fold stays the same fold. This is a silent mutation.

Hint
The ribosome only sees which amino acid the codon stands for.

Worked solution
1. Before: CTG, leucine. After: CTA, also leucine.
2. The ribosome brings the same amino acid, so the chain is identical.
3. An identical chain folds to the same shape.
After the reveal, the original and the new protein are the same, and no amino acid changed first: the number the Scene shows is $0$ ⁠.
Step 2 of 5
Open this step
A different amino acid
Practice · Calculate
Question
Now letter $20$ ⁠ turns from T to A, so the codon GTG, letters $19$ ⁠ to $21$ ⁠, becomes GAG. Which amino acid in the chain, counting from the first, is swapped?

Correct answer
6 ± 0.1
Setup, choices and feedback
The Scene at this step
One-letter change
Letter 20: T to Afixed
Gene / Folded protein
Folded protein
Accepted within
± 0.1
Computed from the Scene at this step’s setting
Position in the chain of the first amino acid that differs, counting from 1 (0 when none does)/nodes/gene/outputs/changedAt
Wrong answers it recognises
- 7 ± 0.1
  Misconception: Starts reading codons at the first letter of the gene instead of at the start codon AUG.
  Feedback from the catalogue
  The ribosome does not start at the first letter. It slides along the RNA to the first AUG, and only from there reads triplets. The letters before the AUG, and those after the stop codon, are not translated. In the Scene’s gene the AUG starts at letter $4$ ⁠, so the first three letters, GCC, are not part of the protein.
- 20 ± 0.1
  Misconception: Matches one letter with one amino acid, instead of three letters, a codon, with one.
  Feedback from the catalogue
  One amino acid is coded by three letters, a codon. Four letters alone could code only four amino acids, and there are twenty. So a stretch of $6$ ⁠ letters is $2$ ⁠ codons, and $2$ ⁠ amino acids, not $6$ ⁠.
Feedback for a right answer
The sixth. From letter $4$ ⁠ the codons start at letters $4, 7, 10, 13, 16, 19$ ⁠, so GTG is the sixth codon: valine, which shuns water, is swapped for glutamate, which likes it, and the fold changes. This is a missense mutation.

Hint
Reading starts at the AUG, at letter $4$ ⁠. What number is the codon that starts at letter $19$ ⁠?

Worked solution
1. The first codon starts at letter $4$ ⁠, and each codon is three letters after the last.
2. The codon starting at letter $19$ ⁠: $\frac{19 - 4}{3} + 1 = 6$ ⁠.
3. The chain is M K E L S V…, and the sixth, V (valine), becomes E (glutamate).
If you answered $7$ ⁠, you counted triplets from the gene’s first letter, but the three letters before the AUG are not read. If you answered $20$ ⁠, you gave the letter’s number, not the amino acid’s.
Step 3 of 5
Open this step
An early stop
Practice · Calculate
Question
Now letter $22$ ⁠ turns from C to T, so the seventh codon, CAG, becomes TAG. How many amino acids will the chain have?

Correct answer
6 ± 0.1
Setup, choices and feedback
The Scene at this step
One-letter change
Letter 22: C to Tfixed
Gene / Folded protein
Folded protein
Accepted within
± 0.1
Computed from the Scene at this step’s setting
Amino acids in the protein (0 when there is no start codon)/nodes/gene/outputs/aminoAcids
Wrong answers it recognises
- 14 ± 0.1
  Misconception: Thinks a change of one letter can change at most one amino acid.
  Feedback from the catalogue
  A change of one letter can do much more. If it makes a stop codon, a nonsense mutation, the chain is cut short and every amino acid after it is lost. And if a letter is lost or added, a frameshift, the ribosome still reads in triplets, so every codon after the change is grouped differently, until a stop codon turns up by chance.
- 7 ± 0.1
  Misconception: Counts the stop codon as an amino acid.
  Feedback from the catalogue
  A stop codon brings no amino acid. No tRNA fits UAA, UAG or UGA, so when the ribosome reaches one it lets the chain go. The number of amino acids is one less than the codons read, because the last of them is the stop.
Feedback for a right answer
$6$ ⁠: UAG is a stop codon, so the ribosome releases the chain after six amino acids instead of fourteen. This is a nonsense mutation, and a chain cut this short usually cannot work.

Hint
In RNA, TAG is UAG. What does it mean in the code?

Worked solution
1. The seventh codon becomes UAG, a stop codon.
2. The ribosome adds amino acids only for the six codons before it.
3. In the chain: $6$ ⁠ amino acids.
If you answered $14$ ⁠, you thought one letter changes only one amino acid. If you answered $7$ ⁠, you counted the stop codon.
Step 4 of 5
Open this step
A lost letter
Practice · Predict, then look
Question
Now one letter, letter $13$ ⁠, is lost from the gene. What will happen to the protein?

Correct answer
- Correct Every amino acid from that place on will be garbled
Setup, choices and feedback
The Scene at this step
One-letter change
Nonefixed
Gene / Folded protein
Folded protein
Choices
- Correct Every amino acid from that place on will be garbled
- Wrong Only one amino acid will change
  Misconception: Thinks a change of one letter can change at most one amino acid.
  Feedback from the catalogue
  A change of one letter can do much more. If it makes a stop codon, a nonsense mutation, the chain is cut short and every amino acid after it is lost. And if a letter is lost or added, a frameshift, the ribosome still reads in triplets, so every codon after the change is grouped differently, until a stop codon turns up by chance.
- Wrong The chain will simply lack one amino acid
  Feedback
  That happens when three letters are lost together, a whole codon. One letter shifts every triplet after it.
- Wrong Nothing
  Misconception: Thinks every change of a letter changes the protein.
  Feedback from the catalogue
  Most amino acids have several codons, often differing only in the third letter: CTG and CTA both mean leucine. A mutation that turns one into the other is silent: the gene changed, the protein did not. Besides, most of a cell’s DNA does not code for protein at all.
Once the prediction is locked in, the Scene changes
One-letter change
Letter 13 lostfixed
Gene / Folded protein
Folded protein
The Scene then shows: 8 · Amino acids in the protein (0 when there is no start codon)/nodes/gene/outputs/aminoAcids
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
A frameshift: the first three amino acids stay, and from there every codon is grouped differently. A stop codon turns up by chance, and the chain ends after $8$ ⁠ amino acids, most of them wrong.

Hint
The ribosome always reads triplets from the AUG. What happens to the triplets when a letter is missing?

Worked solution
1. Letter $13$ ⁠ is the first of the fourth codon, CTG.
2. Without it, every letter after it moves back one place, and the triplets are read differently.
3. From the fourth amino acid on the chain differs, until a stop codon turns up by chance.
4. The Scene shows a chain of $8$ ⁠ amino acids.
Losing one letter is almost always worse than losing three: three letters take out a whole codon and leave the frame in place.
Step 5 of 5
Open this step
Three letters
Practice · Calculate
Question
Suppose that instead of one letter, the three letters CTG are lost, the whole fourth codon. How many amino acids will the chain have?

Correct answer
13 ± 0.1
Setup, choices and feedback
The Scene at this step
One-letter change
Letter 13 lost
Gene / Folded protein
Folded protein
Accepted within
± 0.1
A fixed value in the question
Wrong answers it recognises
- 11 ± 0.1
  Misconception: Matches one letter with one amino acid, instead of three letters, a codon, with one.
  Feedback from the catalogue
  One amino acid is coded by three letters, a codon. Four letters alone could code only four amino acids, and there are twenty. So a stretch of $6$ ⁠ letters is $2$ ⁠ codons, and $2$ ⁠ amino acids, not $6$ ⁠.
- 8 ± 0.1
  Misconception: Thinks a change of one letter can change at most one amino acid.
  Feedback
  $8$ ⁠ was the chain’s length after losing one letter. Three letters are a whole codon, so the frame does not move.
Feedback for a right answer
$13$ ⁠: only the leucine is missing, and everything else reads as before. The commonest mutation behind cystic fibrosis looks like this: three letters lost, one amino acid gone.

Hint
Three letters are one codon. Do the triplets after it move?

Worked solution
1. Three letters in a row starting at the start of a codon are a whole codon.
2. Once they are gone, the triplets after them start in the right place again: the frame is kept.
3. Of $14$ ⁠ amino acids one is missing: $14 - 1 = 13$ ⁠.
If you answered $11$ ⁠, you took away one amino acid for each letter lost.

Lesson 4 of 4 · About 25 minutes

Editing genes with CRISPR

How Cas9 finds one site in a genome, why a wrong letter near the PAM stops it and one at the far end does not, and how the cell's repair knocks a gene out or writes new letters into it.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Searching for one site
Practice · Calculate
Question
Cas9 does not read all the DNA. It slides along it and stops only at a PAM, any letter followed by GG, and checks its guide RNA there. Slide Run the edit slowly. At how many places along the stretch in the Scene can Cas9 stop?

Correct answer
3 ± 0.1
Setup, choices and feedback
The Scene at this step
Guide RNA
Matches the gene
Repair
Join the ends
Run the edit
Cas9 arrives
Accepted within
± 0.1
Computed from the Scene at this step’s setting
PAMs (NGG) on the stretch of DNA, where Cas9 stops to check/nodes/edit/outputs/pams
Wrong answers it recognises
- 1 ± 0.1
  Misconception: Thinks Cas9 cuts only where all twenty letters of the guide match.
  Feedback
  Cas9 does not know in advance where its target is. It stops at every PAM and checks, and this stretch has two more PAMs besides the target.
Feedback for a right answer
Three PAMs: at the first and the third the guide does not match, and the DNA closes again. Only at the middle one do all twenty letters pair.

Hint
When Cas9 stops, the DNA opens a little beside it. Count the stops.

Worked solution
1. Cas9 lands and slides along the DNA.
2. At each PAM, a letter followed by GG, it opens the DNA a little and checks the guide RNA.
3. The stretch has three PAMs, so three stops: two fail, and at the target it locks on.
Searching by PAM is what lets Cas9 find one site in a genome of billions of letters without checking every one.
Step 2 of 6
Open this step
The far end
Practice · Predict, then look
Question
Now the guide RNA has one wrong letter, at the end far from the PAM. $19$ ⁠ of its $20$ ⁠ letters match the target. Will Cas9 cut the gene?

Correct answer
- Correct Yes, it locks on and cuts
Setup, choices and feedback
The Scene at this step
Guide RNA
Matches the genefixed
Repair
Join the ends
Run the edit
Cas9 arrivesfixed
Choices
- Correct Yes, it locks on and cuts
- Wrong No, any wrong letter stops it
  Misconception: Thinks Cas9 cuts only where all twenty letters of the guide match.
  Feedback from the catalogue
  Cas9 checks the guide RNA from the PAM onwards, letter by letter. A wrong letter near the PAM stops it, but one wrong letter at the far end is tolerated, and Cas9 locks on and cuts anyway. That is one reason CRISPR sometimes cuts where nobody meant it to, so guides are designed to have no close match elsewhere in the genome.
Once the prediction is locked in, the Scene changes
Guide RNA
One letter off, far endfixed
Run the edit
0.6 (Cas9 arrives → Gene repaired)fixed
The Scene then shows: 1 · 1 if Cas9 cuts the target, else 0/nodes/edit/outputs/cuts
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Cas9 opens from the PAM, and the first $19$ ⁠ letters it checks pair. One wrong letter at the far end is tolerated, and it cuts. That also means cuts where none were meant: off-target cutting.

Hint
Cas9 opens the DNA from the PAM onwards. When does it meet the wrong letter?

Worked solution
1. Cas9 checks the guide letter by letter, starting at the end beside the PAM.
2. The first $19$ ⁠ letters pair, and the opening is almost complete.
3. One wrong letter at the far end is not enough to stop it, so it locks on and cuts: the Scene shows $1$ ⁠, a cut.
So researchers check the genome has no other site within a letter or two of the guide.
Step 3 of 6
Open this step
Near the PAM
Practice · Predict, then look
Question
And now the wrong letter in the guide RNA sits near the PAM, three letters from it. Will Cas9 cut?

Correct answer
- Correct No: the DNA closes again and Cas9 moves on
Setup, choices and feedback
The Scene at this step
Guide RNA
Matches the genefixed
Repair
Join the ends
Run the edit
Cas9 arrivesfixed
Choices
- Correct No: the DNA closes again and Cas9 moves on
- Wrong Yes: here too $19$ ⁠ of $20$ ⁠ match
  Feedback
  The count is not enough: where the wrong letter sits matters. Cas9 opens from the PAM, and meets it after only three letters, before it can lock on.
Once the prediction is locked in, the Scene changes
Guide RNA
One letter off, near the PAMfixed
Run the edit
Gene repairedfixed
The Scene then shows: 3 · Letters the guide unzips at the target PAM before it locks on or gives up/nodes/edit/outputs/opened
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Cas9 opens only $3$ ⁠ letters, meets the wrong one, and the DNA closes. The region beside the PAM, the seed, has to match exactly.

Hint
From which side does Cas9 start checking?

Worked solution
1. Cas9 checks from the end beside the PAM.
2. The wrong letter is the third it meets.
3. A wrong letter in the seed stops the opening: the DNA closes, and there is no cut.
4. The Scene shows only $3$ ⁠ letters opened.
Compare the last question: the same number of wrong letters, one, but in another place, and the opposite result.
Step 4 of 6
Open this step
Joining the ends
Practice · Calculate
Question
The guide matches, Cas9 cuts, and the cell joins the ends and slips in one extra letter, a T. Before the edit the stretch reads as $25$ ⁠ codons with no stop. The cut falls after codon $16$ ⁠, and on the codon strip the stop codon is the fourth after the break. How many amino acids are coded now before it stops?

Correct answer
19 ± 0.1
Setup, choices and feedback
The Scene at this step
Guide RNA
Matches the genefixed
Repair
Join the endsfixed
Run the edit
Gene repaired
Accepted within
± 0.1
Computed from the Scene at this step’s setting
Codons read before a stop in the stretch shown, after the edit/nodes/edit/outputs/aminoAcids
Wrong answers it recognises
- 25 ± 0.1
  Misconception: Thinks a change of one letter can change at most one amino acid.
  Feedback from the catalogue
  A change of one letter can do much more. If it makes a stop codon, a nonsense mutation, the chain is cut short and every amino acid after it is lost. And if a letter is lost or added, a frameshift, the ribosome still reads in triplets, so every codon after the change is grouped differently, until a stop codon turns up by chance.
- 20 ± 0.1
  Misconception: Counts the stop codon as an amino acid.
  Feedback from the catalogue
  A stop codon brings no amino acid. No tRNA fits UAA, UAG or UGA, so when the ribosome reaches one it lets the chain go. The number of amino acids is one less than the codons read, because the last of them is the stop.
Feedback for a right answer
$16 + 3 = 19$ ⁠: the extra letter shifts the frame, three codons are misread, and then TGA turns up. The protein is cut short and the gene knocked out. That is how CRISPR is used to find out what a gene does: switch it off and see what changes.

Hint
$16$ ⁠ codons before the break, and after it three garbled codons, then a stop.

Worked solution
1. Before the break: $16$ ⁠ codons as before.
2. The extra T shifts the frame, so the codons after it are read differently.
3. Three garbled codons, and the fourth is a stop codon, which codes no amino acid.
4. Together: $16 + 3 = 19$ ⁠ amino acids.
If you answered $25$ ⁠, you thought one letter cannot change more than one codon. If you answered $20$ ⁠, you counted the stop codon.
Step 5 of 6
Open this step
Copying a template
Practice · Calculate
Question
This time the researchers give the cell a template, and it copies six new letters from it into the break: CAC CAT. How many codons are read along the stretch now, with no stop codon?

Correct answer
27 ± 0.1
Setup, choices and feedback
The Scene at this step
Guide RNA
Matches the genefixed
Repair
Copy a templatefixed
Run the edit
Gene repaired
Accepted within
± 0.1
Computed from the Scene at this step’s setting
Codons read before a stop in the stretch shown, after the edit/nodes/edit/outputs/aminoAcids
Wrong answers it recognises
- 31 ± 0.1
  Misconception: Matches one letter with one amino acid, instead of three letters, a codon, with one.
  Feedback from the catalogue
  One amino acid is coded by three letters, a codon. Four letters alone could code only four amino acids, and there are twenty. So a stretch of $6$ ⁠ letters is $2$ ⁠ codons, and $2$ ⁠ amino acids, not $6$ ⁠.
- 19 ± 0.1
  Misconception: Thinks a change of one letter can change at most one amino acid.
  Feedback
  $19$ ⁠ was the count after joining the ends. Six letters are two whole codons, so the frame does not shift and there is no early stop.
Feedback for a right answer
$25 + 2 = 27$ ⁠: six letters are two new codons, histidine and histidine, and the reading does not shift. That is how a chosen change is written into a gene.

Hint
Six letters make how many codons? Does the frame shift?

Worked solution
1. The template writes six letters, that is $\frac{6}{3} = 2$ ⁠ whole codons.
2. Six is a multiple of three, so the codons after the break start in the right place.
3. The stretch now reads as $25 + 2 = 27$ ⁠ codons, all sound.
If you answered $31$ ⁠, you gave each new letter a codon.
Step 6 of 6
Open this step
Who repairs
Practice · Choose
Question
Who writes the new letters into the gene?

Correct answer
- Correct The cell, when it repairs the break
Setup, choices and feedback
The Scene at this step
Guide RNA
Matches the gene
Repair
Copy a template
Run the edit
0.6 (Cas9 arrives → Gene repaired)
Choices
Shown in a random order.
- Correct The cell, when it repairs the break
- Wrong Cas9, after it cuts
  Misconception: Thinks Cas9 itself writes the new letters into the gene.
  Feedback from the catalogue
  Cas9 only cuts: it finds the place and breaks both strands. The cell repairs the break, with tools it always had. Joining the ends in a hurry often slips in an extra letter and knocks the gene out. When the cell has a template the researchers supplied, it copies from it, and that is how new letters are written in.
- Wrong The guide RNA, which stays inside the DNA
  Misconception: Thinks Cas9 itself writes the new letters into the gene.
  Feedback from the catalogue
  Cas9 only cuts: it finds the place and breaks both strands. The cell repairs the break, with tools it always had. Joining the ends in a hurry often slips in an extra letter and knocks the gene out. When the cell has a template the researchers supplied, it copies from it, and that is how new letters are written in.
Feedback for a right answer
Cas9 lets go right after the cut. The cell repairs the break with its own tools: a quick joining of the ends, or copying from a template when there is one.

Hint
Watch what happens to Cas9 after the cut.

Worked solution
1. Cas9 finds the site with its guide RNA and cuts both strands.
2. Then it lets go, and the break remains.
3. The cell repairs it: joining the ends often slips in an extra letter, and copying from a template writes in what the template says.
So to write a chosen change you must give the cell a template too, not only Cas9.

Across the unit · 13 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

The double helix

Two strands and a ladder

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Hydrogen bonds

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Chargaff's rule

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

A sturdier helix

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

One molecule becomes two

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Hint

Worked solution

From gene to protein

Genes make proteins

Transcription

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Reading in threes

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The chain

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Where the chain ends

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

The genetic code

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Mutations

The same meaning

Question

Correct answer

The Scene at this step

Choices