Translation and Protein Synthesis

Central Dogma

Part 1 of 7 — Central Dogma

Your DNA holds instructions, but DNA itself does not build anything. Proteins do the work of the cell: enzymes, channels, receptors, structural fibers and signals. The central dogma describes how information stored in DNA becomes a working protein:

$\text{DNA} \xrightarrow{\text{transcription}} \text{RNA} \xrightarrow{\text{translation}} \text{protein}$

The key idea is sequence → sequence: the order of nucleotides in DNA determines the order of bases in mRNA, which determines the order of amino acids in the protein. Amino acid order determines how the protein folds, and its shape determines its function.

🔑 Key Concept: A gene is a stretch of DNA whose sequence is transcribed into a functional RNA. For most genes, that RNA is an mRNA that is then translated into a polypeptide.

Three kinds of RNA, three jobs

The CED stresses that an RNA's base sequence and its structure together determine its function.

RNA versus DNA, in brief

Because RNA is single-stranded, it can fold into specific three-dimensional shapes, which is why tRNA and rRNA can do jobs that DNA cannot.

Check Your Understanding 🎯

Where it happens: prokaryotes versus eukaryotes

Because eukaryotes separate the two steps with the nuclear envelope, they gain extra control points: the RNA can be processed, checked and exported only when it is ready. Prokaryotes gain speed instead.

Exceptions to the usual flow

• Retroviruses such as HIV carry an RNA genome. Their enzyme reverse transcriptase copies RNA into DNA, which is then inserted into the host's chromosome and transcribed by the host. Information flows RNA → DNA, the reverse of transcription.
• Many other RNA viruses copy RNA directly into more RNA.

📝 AP Connection: The CED names retroviruses as an exception to the usual flow of information. Expect questions asking what reverse transcriptase does and why drugs that block it slow HIV.

One code for (nearly) all life

mRNA is read in groups of three bases called codons. There are $4^3 = 64$ possible codons:

• 61 codons specify the 20 amino acids, so most amino acids have more than one codon (the code is redundant).
• 3 codons are stop signals. (A codon chart identifies them; you do not need to memorize which ones they are.)
• AUG codes for methionine and also serves as the usual start codon. AUG is the one codon the AP exam expects you to know without a chart.

Nearly every organism, from bacteria to plants to people, uses the same codon assignments. A human gene placed in a bacterium can be translated into the correct human protein, which is how bacteria are engineered to make human insulin.

Why it matters: a shared genetic code is strong evidence that all living organisms descend from a common ancestor. The few known variations (for example, in mitochondria) are small changes to that shared code.

💡 Misconception alert: "Redundant" does not mean "ambiguous." Several codons can mean the same amino acid, but each codon means only one amino acid (or stop).

Apply It 🔬

Roadmap for this topic

🎯 Takeaway: DNA sequence → mRNA sequence → amino acid sequence → protein shape → function. Every later part fills in one arrow of that chain.

Match the Molecule to Its Role 🔍

Transcription

Part 2 of 7 — Transcription

Transcription is the synthesis of an RNA molecule complementary to one strand of a gene. The enzyme that does it is RNA polymerase. Three rules from the CED drive almost every transcription question:

1. RNA polymerase uses one DNA strand, the template strand, to direct which base is added next.
2. RNA is built in the 5′ → 3′ direction, so the template is read 3′ → 5′.
3. Which strand serves as the template depends on the gene: different genes on the same chromosome can use opposite strands.

Naming the two strands

🔑 Key Concept: Because the mRNA is complementary to the template, and the coding strand is also complementary to the template, the mRNA and the coding strand match each other (except U for T).

How RNA polymerase works

RNA polymerase is a single machine that does several jobs at once:

• Opens the double helix over a short stretch, exposing the template bases. (Unlike DNA replication, it needs no separate helicase.)
• Needs no primer: it can start a new chain from scratch.
• Adds ribonucleotides (A, U, G, C) that pair with the template: A on the template pairs with U in RNA, T pairs with A, G with C and C with G.
• Links each new nucleotide to the 3′ end of the growing chain, which is why RNA grows 5′ → 3′.
• Lets the DNA rewind behind it, releasing the RNA as a single strand.

Three stages

Several RNA polymerases can transcribe the same gene one after another, so one gene can produce many mRNA copies quickly. This is the first point at which gene expression is amplified.

Check Your Understanding 🎯

Worked example: writing the mRNA

Problem. Part of a gene's template strand reads:

3′ – T A C G G A T T C – 5′

Write the mRNA and the coding strand, with ends labeled.

Step 1. Pair each template base. T → A, A → U, C → G, G → C.

Step 2. Label the ends. The mRNA is antiparallel to the template, so it starts with its 5′ end opposite the template's 3′ end:

mRNA: 5′ – A U G C C U A A G – 3′

Step 3. Write the coding strand. It has the mRNA's sequence with T for U:

Coding: 5′ – A T G C C T A A G – 3′

Check: the coding strand and mRNA read the same in the same direction, and both are complementary to the template. The mRNA begins with AUG, the start codon you will meet in Part 4.

💡 Misconception alert: Students often write the mRNA as the coding strand's complement. That gives the template sequence (with U), which is wrong. Always pair the mRNA against the template.

Write the Sequences 🧮

A different gene's template strand reads 3′ – T A C C T T G A C – 5′.

1) Write the mRNA, 5′ → 3′, as one string of letters (no spaces). 2) Write the coding strand, 5′ → 3′, as one string of letters. 3) How many uracils does the mRNA contain?

Which strand is the template?

Along one chromosome, gene 1 may be transcribed using the "top" strand as template and gene 2, a few thousand bases away, using the "bottom" strand. The promoter's orientation decides: RNA polymerase binds facing one way and can only move in the direction that lets it read its template 3′ → 5′.

Transcription versus DNA replication

📝 AP Connection: Free-response questions often give a short DNA sequence and ask for the mRNA or the amino acids. Points are lost for missing 5′/3′ labels and for using T in RNA. Label both ends every time.

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RNA Processing

Part 3 of 7 — RNA Processing

In a eukaryotic cell, the RNA that comes off RNA polymerase is not ready to be translated. This first product is the pre-mRNA (primary transcript). Before it leaves the nucleus, enzymes modify it in three ways, which the CED lists explicitly:

The result is the mature mRNA, which is exported through a nuclear pore to the cytoplasm.

🔑 Key Concept: Exons are expressed: they remain in the mature mRNA. Introns are intervening sequences: they are cut out and broken down in the nucleus.

The cap and tail: protection and a passport

Enzymes in the cytoplasm called nucleases attack RNA from its ends. The cap and the tail act like protective end-pieces:

• Experiments that inject mRNA into cells show that mRNA without a poly-A tail is degraded much faster than the same mRNA with one.
• An mRNA without a 5′ cap is poorly translated, because the small ribosomal subunit normally recognizes the cap when it binds.

Because cap and tail both affect how long an mRNA survives and how often it is translated, they also influence how much protein a gene produces.

Splicing: cutting out introns

Most eukaryotic genes are split: coding exons are separated by noncoding introns, which are often much longer than the exons.

• Introns begin and end with short signal sequences that mark the cut sites.
• A complex called the spliceosome, made of small RNAs and proteins, recognizes those signals, cuts the pre-mRNA at both ends of an intron and joins the neighboring exons.
• In the spliceosome, RNA helps catalyze the reaction. RNA molecules that act as catalysts are called ribozymes, another example of RNA's structure determining its function.

💡 Misconception alert: Introns are removed from the RNA, not from the DNA. The gene in the chromosome still contains every intron.

Check Your Understanding 🎯

Alternative splicing: one gene, several proteins

Exons are not always joined the same way. In alternative splicing, different combinations of exons are kept in the mature mRNA, so one gene can produce several different polypeptides.

Example: the calcitonin gene. In thyroid cells, its pre-mRNA is spliced to make the hormone calcitonin. In certain neurons, the same pre-mRNA is spliced differently to make a different signaling peptide (CGRP).

This helps explain how humans make far more kinds of proteins than they have protein-coding genes (roughly 20,000).

Worked example: counting splice variants

Problem. A gene has five exons (E1–E5). E1 and E5 are always kept. E2 and E4 can each be kept or skipped independently. E3 is always kept. How many different mature mRNAs can be made? Give the exon order of the shortest one.

Step 1. Identify the choices. Only E2 and E4 are optional, and each has 2 options (in or out).

Step 2. Multiply. $2 \times 2 = 4$ different mRNAs.

Step 3. List them to check:

Answer. 4 mRNAs; the shortest is E1–E3–E5.

Note that exon order is preserved: splicing removes pieces but does not shuffle the exons into a new order.

Processing Arithmetic 🧮

A pre-mRNA is 6,400 nucleotides long (not counting the cap and tail). It has four exons of 300, 450, 250 and 200 nucleotides; the rest is introns.

1) How many nucleotides of the pre-mRNA are intron sequence? 2) How many nucleotides are in the spliced mRNA (exons only)? 3) If exons 2 and 3 can each be independently included or skipped, while exons 1 and 4 are always kept, how many different mature mRNAs are possible?

Prokaryotes skip this step

Bacterial genes rarely have introns, and bacterial mRNA has no 5′ cap or poly-A tail of the eukaryotic kind. That difference has a practical consequence:

• A bacterium given a human gene with its introns cannot splice them out, so it produces a faulty protein.
• Biotechnologists therefore start from the human mature mRNA and use reverse transcriptase to make an intron-free DNA copy (cDNA), which bacteria can express correctly.

Evidence that introns exist

When a mature mRNA is allowed to base-pair with its single-stranded gene and viewed with an electron microscope, the DNA shows loops that have no partner in the mRNA. Each loop is an intron, a stretch of the gene with no matching sequence in the processed message.

📝 AP Connection: Expect to explain why a mutation in a splice site can change a protein drastically: the spliceosome may skip an exon or keep an intron, changing a large block of the amino acid sequence or creating an early stop codon. You will practice this in Part 6.

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Match the Modification to Its Role 🔍

Translation

Part 4 of 7 — Translation

Translation converts the base sequence of an mRNA into the amino acid sequence of a polypeptide. It happens on ribosomes in the cytoplasm, and it needs three kinds of RNA working together.

The ribosome has a small and a large subunit, and three tRNA-binding sites:

• A site: accepts the incoming tRNA carrying the next amino acid.
• P site: holds the tRNA attached to the growing polypeptide.
• E site: the exit, where the empty tRNA leaves.

Three stages of translation

1. Initiation

• The small ribosomal subunit binds the mRNA (in eukaryotes, at the 5′ cap) and moves along it to the start codon, AUG. The CED describes this as rRNA in the ribosome interacting with the mRNA at the start codon.
• An initiator tRNA carrying methionine pairs with AUG.
• The large subunit joins, completing the ribosome with the initiator tRNA in the P site.

2. Elongation (repeats once for each amino acid)

1. Codon recognition: a tRNA whose anticodon matches the codon in the A site binds there.
2. Peptide bond formation: the rRNA catalyzes a bond that transfers the growing polypeptide from the tRNA in the P site onto the amino acid in the A site.
3. Translocation: the ribosome moves one codon toward the mRNA's 3′ end. The empty tRNA shifts to the E site and leaves, and the A site is open for the next codon.

3. Termination

• A stop codon enters the A site. No tRNA matches it, so no amino acid is added.
• The finished polypeptide is released, and the ribosome comes apart.

(The names and details of the other proteins and factors that assist each stage of translation are beyond the scope of the AP exam.)

The polypeptide grows from its amino end to its carboxyl end, matching the mRNA read 5′ → 3′. Several ribosomes can translate one mRNA at the same time, forming a polyribosome, so one mRNA can yield many protein copies quickly.

Check Your Understanding 🎯

Reading the genetic code

On the AP exam you are given a codon chart when you need one; you are not expected to memorize it. Here are the codons used in this topic's examples:

Worked example: translating an mRNA

Problem. Translate: 5′ – G C A U G C C U A A G U G G G A A U A A G C – 3′

Step 1. Find the start codon. Scan from the 5′ end for the first AUG. It begins at the 4th base: G C A U G ...

Step 2. Divide into codons from AUG. AUG | CCU | AAG | UGG | GAA | UAA | GC

Step 3. Translate each codon until a stop.

Answer. Met – Pro – Lys – Trp – Glu (5 amino acids). The bases before AUG and after UAA are not translated.

Anticodons. The anticodon is complementary and antiparallel to the codon. For the codon 5′–AUG–3′, the anticodon is 3′–UAC–5′, which is also written 5′–CAU–3′. Always check which direction a problem uses.

Translation Practice 🧮

Use the codon table above.

mRNA: 5′ – A U G U U U G G C U A C U G A – 3′

1) How many amino acids are in the polypeptide? 2) What is the third amino acid? (Give the three-letter abbreviation.) 3) A different mRNA has 450 nucleotides from the first base of AUG through the last base of the stop codon. How many amino acids does its polypeptide contain?

Coupling in prokaryotes

In bacteria, ribosomes attach to an mRNA while RNA polymerase is still making it. Electron micrographs show a DNA strand with RNA strands growing from it, each RNA studded with ribosomes. This coupled transcription and translation lets bacteria respond to change within minutes.

Translation is a target for antibiotics

Bacterial ribosomes differ in structure from the ribosomes in a eukaryote's cytoplasm. Several antibiotics bind bacterial ribosomes and block translation, while leaving the host's cytoplasmic ribosomes largely unaffected.

📝 AP Connection: Because nearly all organisms share the same genetic code and the same basic translation steps, these processes are evidence of common ancestry. The structural differences in ribosomes are what make selective antibiotics possible.

Apply It 🔬

Order the Events of Elongation 🔍

Choose the step number for each event in one elongation cycle.

Post-Translational Modification

Part 5 of 7 — Post-Translational Modification

A polypeptide leaving the ribosome is a chain of amino acids, not yet a working protein. Before it can do its job it must fold, often be chemically modified, and be delivered to the right place in or out of the cell. These steps connect gene expression (Unit 6) to protein structure (Unit 1) and the endomembrane system (Unit 2).

Folding: sequence decides shape

The amino acid sequence (primary structure) contains the information for folding. As the chain emerges, hydrophobic side chains cluster away from water, hydrogen bonds form helices and sheets, and ionic bonds and disulfide bridges stabilize the final shape.

• Chaperone proteins help many polypeptides fold correctly by shielding them from improper interactions while they fold. They do not supply the shape; they prevent mistakes.
• Misfolded proteins are usually tagged and destroyed. In the most common cystic fibrosis allele, a deletion of one amino acid causes the CFTR chloride channel to misfold; the cell degrades most of it before it reaches the plasma membrane, so chloride transport fails.

💡 Misconception alert: A mutation does not have to change the "active site" to ruin a protein. Changing a single amino acid anywhere can alter folding, and a misfolded protein may never reach its destination.

Targeting: where will the protein go?

Every polypeptide begins on a free ribosome in the cytosol. What happens next depends on whether its first amino acids form an ER signal sequence.

Check Your Understanding 🎯

Chemical modification

Worked example: from gene to cell-surface receptor

Follow a membrane receptor protein from its gene to its working position on the cell surface. Each step depends on the one before it.

Predicting disruptions. A change at any step changes what the cell can do:

• No signal sequence (step 2): the receptor is finished in the cytosol and never reaches the membrane.
• A mutation that prevents folding (step 4): the receptor is degraded, so the cell cannot respond to the receptor's signal.
• Vesicles that cannot leave the Golgi (step 6): the receptor accumulates in the Golgi instead of reaching the surface.

Turnover: proteins have lifespans

Proteins are not permanent. Cells tag many proteins for destruction and break them down, some within minutes and others over days. How long a protein lasts is one more way a cell controls how much of it is present, and so the cell's phenotype.

Reading data: a pulse-chase experiment

Pancreatic cells are given radioactive amino acids for 3 minutes (the pulse), then normal amino acids (the chase). The location of radioactivity is followed over time.

Interpretation. Newly made proteins move in order: rough ER → Golgi → vesicles → secretion. Only proteins made during the pulse are labeled, so the experiment tracks one batch of proteins over time.

📝 AP Connection: Questions may ask where a secreted protein is glycosylated, what happens when one step of the pathway is blocked, or what a pulse-chase table shows. Tie each answer to a specific structure or step.

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Where Does It Happen? 🔍

Transcription & Translation: Problem-Solving Workshop

Part 6 of 7 — Problem-Solving Workshop

This workshop combines Parts 2–5 into the multi-step problems the AP exam favors: go from DNA to protein, predict what a mutation does, and interpret experiments that block one step of gene expression.

The DNA → protein routine

1. Identify the strand. Is the given DNA the template or the coding strand? Check the 5′/3′ labels.
2. Write the mRNA 5′ → 3′ (complement of the template, or coding strand with U for T).
3. Find the first AUG from the 5′ end; that sets the reading frame.
4. Read codons in triplets with the codon chart until a stop codon.
5. Count and check: number of amino acids = number of codons before the stop.

Worked problem 1: template to polypeptide

Problem. A gene's template strand reads 3′ – TAC AAA CTT CCG ATT – 5′ (spaces mark triplets). Give the mRNA and the polypeptide.

Step 1–2. Complement each base of the template:

mRNA: 5′ – AUG UUU GAA GGC UAA – 3′

Step 3–4. AUG starts the frame.

Answer. Met – Phe – Glu – Gly (4 amino acids).

We will use this "wild-type" sequence to see what different mutations do.

Worked problem 2: classifying mutations

Each row changes one thing in the wild-type mRNA AUG UUU GAA GGC UAA.

What to notice.

• A substitution affects one codon. Its effect ranges from none (silent) to severe (nonsense).
• An insertion or deletion of 1 or 2 bases shifts the reading frame, scrambling every codon downstream and often creating an early stop. Inserting or deleting 3 bases adds or removes one amino acid but keeps the frame.
• Effects on the phenotype depend on where the change falls: a missense change in a crucial region can be as harmful as a nonsense mutation, while one in a flexible surface loop may have no effect.

Real case: sickle cell disease. In the β-globin gene, codon 6 changes from GAG to GUG, a missense change from glutamic acid (charged, hydrophilic) to valine (nonpolar). The hydrophobic valine makes hemoglobin molecules stick together when oxygen is low, forming fibers that distort red blood cells into a sickle shape.

Check Your Reasoning: Mutations 🎯

Worked problem 3: locating a block in gene expression

Cells are treated with one of two drugs, then given a labeled precursor. New mRNA and new protein are measured after 30 minutes (100 = untreated level).

Reasoning.

• Drug B: mRNA is made normally, but almost no protein. The block is in translation.
• Drug A: almost no new mRNA. The block is in transcription. Protein synthesis is only partly reduced after 30 minutes because mRNAs made before the drug was added are still being translated. Over a longer time, protein synthesis would also fall as those mRNAs are degraded.

📝 AP Connection: The trap is to say Drug A "blocks both steps." The evidence shows the drug acts on transcription; the drop in protein is a downstream consequence that lags because existing mRNAs persist.

Worked problem 4: mutations outside the coding sequence

Check Your Reasoning: Experiments 🔬

FRQ practice: a model response

Prompt (AP style). A mutation deletes a single nucleotide from the tenth codon of a gene that encodes a 300-amino-acid enzyme. (a) Describe the effect on the mRNA. (b) Predict the effect on the enzyme. (c) Justify your prediction.

Model response.

(a) The mRNA will be one nucleotide shorter, missing one base in codon 10.

(b) The enzyme will be nonfunctional.

(c) Deleting one base shifts the reading frame from codon 10 onward, so every later codon is read incorrectly. This changes the amino acid sequence after position 9 and is likely to create a premature stop codon, producing a short polypeptide that cannot fold into the enzyme's active shape.

Classify the Mutation 🔍

Wild-type mRNA: AUG AAA UGG GGC UAA (Met–Lys–Trp–Gly–Stop).

Codon chart entries you need: AAA and AAG = Lys; UGG = Trp; GGC = Gly; GCC = Ala; UAA and UGA = Stop.

Transcription & Translation: Synthesis & AP Review

Part 7 of 7 — Synthesis & AP Review

This part ties the whole pathway together, from a gene in the nucleus to a working protein at its destination.

The pathway on one page

Prokaryote versus eukaryote, one more time

The seven traps that cost the most points

1. Pairing mRNA with the coding strand. mRNA is complementary to the template and matches the coding strand.
2. Using T in RNA. RNA has U.
3. Dropping the 5′/3′ labels. mRNA is built 5′ → 3′; the template is read 3′ → 5′.
4. Starting at the first base. Translation starts at the first AUG, not at the first nucleotide.
5. Counting the stop codon as an amino acid. Stop codons add nothing.
6. Thinking introns are removed from DNA. They are removed from the pre-mRNA; the gene keeps them.
7. Assuming every mutation changes the protein. Silent mutations and most intron mutations do not.

Quick reference: mutation outcomes

Mixed Review, Set 1 🎯

FRQ practice: tying the steps together

Prompt (AP style). A researcher compares two alleles of a gene. Allele 1 produces a normal-length protein. Allele 2 has a single base substitution, and cells with allele 2 produce normal amounts of mRNA but a protein that is less than half the normal length. (a) Identify the type of mutation most likely present in allele 2. (b) Explain why mRNA levels are unaffected. (c) Propose a measurement that would test your answer to (a).

Model response.

(a) A nonsense mutation: the substitution converted a codon into a stop codon about midway through the coding sequence.

(b) The substitution lies in the coding region, not the promoter, so RNA polymerase still binds and transcribes the gene at the normal rate; the change only matters once the ribosome reads the mRNA.

(c) Sequence the mRNA (or the gene) of allele 2 and compare it with allele 1: a codon that the codon chart lists as a stop codon, at the position where translation ends, would confirm a nonsense mutation.

📝 Scoring notes: (b) earns the point only by linking the mutation's location to the step it affects. Saying "the mutation does not affect transcription" without a reason is not enough.

Mixed Review, Set 2 🔬

Last-mile checklist

• I can write an mRNA from either DNA strand, with 5′ and 3′ ends labeled.
• I can translate an mRNA with a codon chart, starting at AUG and stopping at a stop codon.
• I can name the three mRNA processing steps and what each one does.
• I can count splice variants and calculate exon and intron lengths.
• I can describe initiation, elongation and termination, including the A, P and E sites.
• I can classify a mutation as silent, missense, nonsense or frameshift and predict its effect.
• I can interpret an experiment that blocks transcription or translation.
• I can explain why the universal code supports common ancestry, and name the retrovirus exception to the usual flow.

🎯 Final takeaway: Sequence determines sequence, and sequence determines shape. Locate any change on the DNA → RNA → protein path and predict forward from there.

Final Match: Which Step Is Affected? 🔍