DP Biology HL D1.2 Protein Synthesis

DP Biology HL D1.2 Protein Synthesis - Full notes

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DP Biology HL | D1.2 Protein Synthesis

DP BIOLOGY · THEME D: CONTINUITY AND CHANGE

D1.2

Protein Synthesis

Full notes for HL · IB-style

Contents

1	From gene to protein
2	Transcription
3	The genetic code
4	Translation
5	Mutations and the code: sickle cell anaemia
6	tRNA activation and the ribosome in detail (HL only)
7	RNA processing and post-translational modification (HL only)
8	From gene to protein
9	Transcription in detail
10	Template and coding strands
11	The genetic code
12	Translation: initiation
13	Translation: elongation and termination
14	tRNA structure and activation
15	Polysomes
16	Using the codon table
17	Mutations and protein synthesis
18	Post-transcriptional modification (HL)
19	Post-translational modification (HL)
20	Mixed review: worked problems
21	Additional exam-style practice
22	Antibiotics that target translation
23	Gene expression and cell differentiation
24	Practical: modelling the genetic code
25	RNA types and roles
26	Common misconceptions and links
27	Exam-style question with markscheme
28	Check your understanding
29	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
outline the flow of information from DNA to RNA to protein
explain transcription by RNA polymerase
explain the genetic code: codons, degeneracy and universality
use a codon table to deduce amino acid sequences
explain translation, including the roles of mRNA, tRNA and ribosomes
explain how a base substitution causes sickle cell anaemia
explain tRNA activation and the structure of the ribosome
explain splicing of pre-mRNA and alternative splicing
explain post-translational modification, including insulin

Sections marked HL only (Sections 6 and 7) are Additional Higher Level content. Every other section is common to SL and HL.

Guiding questions

How is the information in DNA used to make proteins? How can a change of one base change a whole protein?

How this topic is examined

D1.2 is examined through diagrams of transcription and translation, codon table exercises, and the sickle cell example. HL questions include the roles of the ribosome sites, splicing and modification.

Distinguish codons (mRNA), anticodons (tRNA) and base triplets (DNA) carefully.

Command term	What the examiner wants
State / Identify	a brief answer with no explanation needed
Outline	a brief account or summary
Describe	a detailed account, including trends or features, without reasons
Explain	a detailed account including causes, reasons or mechanisms
Distinguish / Compare and contrast	give the differences (and similarities), point by point
Calculate	a numerical answer with working and units
Deduce / Suggest	reach a conclusion from data; propose a reasonable explanation
Evaluate / Discuss	weigh strengths and limitations and reach a judgement

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section; the answer comes in the section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the biology shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 From gene to protein

DNA makes RNA makes protein.

Think about it

Every cell in your body has the same DNA, yet a pancreas cell makes insulin and a red blood cell precursor makes haemoglobin. How does a cell turn a gene into a protein?

A gene is a sequence of DNA bases that codes for a polypeptide (or a functional RNA). The information flows in two steps. In transcription, the base sequence of a gene is copied into a molecule of messenger RNA (mRNA). In translation, the base sequence of the mRNA is used to assemble a specific sequence of amino acids, forming a polypeptide, on a ribosome.

This one-way flow of information (DNA → RNA → protein) is sometimes called the 'central dogma' of molecular biology. In eukaryotes, transcription takes place in the nucleus and translation in the cytoplasm.

2 Transcription

Copying a gene into mRNA.

Transcription is the synthesis of RNA using DNA as a template. The enzyme RNA polymerase binds to the DNA at the start of a gene (the promoter). It separates the two DNA strands by breaking hydrogen bonds over a short section.

Only one of the two strands, the template strand, is transcribed. RNA polymerase moves along it, adding free RNA nucleotides that are complementary to the template bases: A pairs with U (uracil, in RNA), T with A, C with G, G with C. RNA polymerase joins the RNA nucleotides with covalent bonds, forming the mRNA strand. Behind the polymerase, the DNA strands re-form the double helix.

When RNA polymerase reaches the end of the gene (the terminator), it releases the mRNA. The mRNA has the same base sequence as the coding strand (non-template strand) of DNA, except that it has U instead of T.

Figure 1. Transcription.

WORKED EXAMPLE 1 Transcription

A template strand reads 3' TACGGCATT 5'. Write the mRNA produced.

Solution

Complementary bases, with U instead of TM1

mRNA: 5' AUGCCGUAA 3'A1

Common mistake

Pairing A with T in mRNA. RNA contains uracil, so adenine on the template pairs with uracil in mRNA.

Quick check

1. Which enzyme carries out transcription?

2. Which base pairs with adenine during transcription?

Answers: 1. RNA polymerase 2. uracil

3 The genetic code

Three bases specify one amino acid.

The genetic code is read in groups of three bases, called codons, on the mRNA. Each codon codes for one amino acid. There are 4³ = 64 possible codons but only 20 amino acids, so most amino acids are coded by more than one codon: the code is degenerate. Three codons (UAA, UAG, UGA) are stop codons, which end translation. AUG is the start codon, coding for methionine; translation always begins at AUG.

The code is universal: almost all organisms use the same codons for the same amino acids, which is strong evidence for common ancestry (A4.1) and allows genes to be transferred between species (for example, human insulin genes expressed in bacteria).

Codon	Amino acid	Codon	Amino acid
AUG	methionine (start)	GGC	glycine
CCG	proline	UUU	phenylalanine
GAG	glutamic acid	GUG	valine
UAA, UAG, UGA	stop	CAU	histidine

WORKED EXAMPLE 2 Using the code

Using the table, give the amino acid sequence coded by the mRNA AUG CCG GGC UUU UAA.

Solution

AUG = methionine; CCG = proline; GGC = glycine; UUU = phenylalanine; UAA = stopM1

Met–Pro–Gly–PheA1

Quick check

1. Why is the code described as degenerate?

2. What is the start codon?

Answers: 1. most amino acids have more than one codon 2. AUG

4 Translation

Ribosomes read mRNA and link amino acids.

Translation is the synthesis of a polypeptide using the base sequence of mRNA. It takes place on ribosomes in the cytoplasm (free or on rough ER). Three types of RNA are involved: mRNA, carrying the code; transfer RNA (tRNA), carrying amino acids; and ribosomal RNA (rRNA), which, with proteins, forms the ribosome.

Each tRNA molecule has an anticodon, three bases complementary to a codon on the mRNA, and carries the specific amino acid that corresponds to that codon.

Steps: (1) the mRNA binds to the small subunit of a ribosome; (2) the ribosome moves to the start codon (AUG), and a tRNA with the complementary anticodon (UAC) carrying methionine binds; (3) a second tRNA, whose anticodon matches the next codon, binds alongside; (4) the ribosome catalyses the formation of a peptide bond between the two amino acids (a condensation reaction); (5) the ribosome moves along the mRNA by one codon; the first tRNA, now without its amino acid, leaves to collect another; (6) the process repeats, adding one amino acid at a time; (7) when a stop codon is reached, the polypeptide is released.

Many ribosomes can translate the same mRNA at once (a polysome), producing many copies of the polypeptide quickly.

Figure 2. Translation.

	Transcription	Translation
Location (eukaryotes)	nucleus	cytoplasm (ribosomes)
Template	DNA template strand	mRNA
Product	mRNA	polypeptide
Enzyme / structure	RNA polymerase	ribosome
Bonds formed	between RNA nucleotides	peptide bonds

Quick check

1. What is an anticodon?

2. What bond forms between amino acids?

3. What ends translation?

Answers: 1. three tRNA bases complementary to an mRNA codon 2. peptide bond 3. a stop codon

5 Mutations and the code: sickle cell anaemia

One base change, one amino acid change, major consequences.

A change in the base sequence of a gene (a mutation, D1.3) can change the amino acid sequence of the polypeptide. Sickle cell anaemia is caused by a base substitution in the gene for the beta chain of haemoglobin. In the DNA, the triplet CTC becomes CAC; in the mRNA, codon GAG becomes GUG; so the sixth amino acid changes from glutamic acid to valine.

Valine is hydrophobic, while glutamic acid is charged. The altered haemoglobin (HbS) molecules stick together in low oxygen conditions, forming long fibres that distort red blood cells into a sickle shape. Sickle cells carry less oxygen, block capillaries and are destroyed quickly, causing anaemia, pain and organ damage.

People with one copy of the allele (heterozygotes) have sickle cell trait, which gives some protection against malaria. This explains why the allele is common in regions where malaria is widespread (D4.1).

	Normal (HbA)	Sickle cell (HbS)
DNA triplet	CTC	CAC
mRNA codon	GAG	GUG
Amino acid 6	glutamic acid	valine
Red blood cells	biconcave discs	sickle-shaped at low O₂

Quick check

1. What base change causes sickle cell anaemia?

2. Why do sickle cells cause problems?

Answers: 1. A substituted for T (CTC → CAC); GAG → GUG 2. carry less O₂, block capillaries, destroyed quickly

6 tRNA activation and the ribosome in detail (HL only)

Each tRNA must be loaded with the right amino acid.

HL only

This section is Additional Higher Level content. SL students do not need it.

Each tRNA has a characteristic clover-leaf shape with an anticodon loop and an attachment site (CCA at the 3' end) for its amino acid. tRNA-activating enzymes (aminoacyl-tRNA synthetases), one for each of the 20 amino acids, attach the correct amino acid to the correct tRNA, using energy from ATP. The enzyme recognizes both the amino acid and the shape of the tRNA. This step ensures the genetic code is translated correctly, and stores energy that is later used to form the peptide bond.

The ribosome consists of a small subunit (which binds mRNA) and a large subunit. The large subunit has three tRNA binding sites: the A site (aminoacyl: where an incoming tRNA binds), the P site (peptidyl: holds the tRNA carrying the growing polypeptide) and the E site (exit: from which the empty tRNA leaves). The peptide bond is formed by the rRNA of the large subunit, which acts as a ribozyme. Translation proceeds in the 5' to 3' direction along the mRNA.

Ribosome site	Role
A site	incoming tRNA with amino acid binds
P site	tRNA holding the growing polypeptide
E site	empty tRNA exits

7 RNA processing and post-translational modification (HL only)

Editing the message and finishing the protein.

HL only

This section is Additional Higher Level content. SL students do not need it.

In eukaryotes, the RNA first made by transcription is pre-mRNA. It contains introns (non-coding sequences) and exons (coding sequences). Before leaving the nucleus, the introns are removed and the exons joined together by splicing. A cap is added at the 5' end and a poly-A tail at the 3' end, which protect the mRNA and help it to be translated.

In alternative splicing, different combinations of exons are joined, so one gene can produce several different mRNAs and therefore several different proteins. This greatly increases the number of proteins (the proteome) that can be made from a limited number of genes.

Post-translational modification: many polypeptides are modified after translation before they become functional. Insulin is first made as a single polypeptide, preproinsulin; a signal sequence is removed to give proinsulin, which folds and forms disulfide bonds; a section (the C-peptide) is then cut out, leaving two chains (A and B) joined by disulfide bonds: active insulin. Other modifications include adding carbohydrate chains (glycosylation) in the Golgi apparatus. Proteasomes break down proteins that are damaged or no longer needed, controlling protein levels.

Figure 3. Post-translational modification of insulin.

Quick check

1. What are introns?

2. Why does alternative splicing increase protein diversity?

3. Outline the modification of insulin.

Answers: 1. non-coding sequences removed from pre-mRNA 2. different exon combinations give different proteins from one gene 3. preproinsulin → proinsulin → C-peptide removed → insulin

8 From gene to protein

The central dogma.

Genetic information flows from DNA to RNA (transcription) to protein (translation). A gene is a sequence of DNA that codes for a polypeptide (or functional RNA). The base sequence determines the amino acid sequence.

WORKED EXAMPLE 3 Central dogma (basic)

State the two stages of protein synthesis and where they occur in eukaryotes.

Solution

Transcription in the nucleusA1

Translation at ribosomes in the cytoplasmA1

9 Transcription in detail

Making mRNA.

RNA polymerase binds to the promoter, separates the DNA strands, and adds RNA nucleotides complementary to the template (antisense) strand, joining them 5′→3′. Uracil pairs with adenine. Transcription stops at a terminator sequence and the mRNA is released.

WORKED EXAMPLE 4 Complementary mRNA (standard)

A template strand reads 3′-TACGGATTC-5′. Write the mRNA.

Solution

5′-AUGCCUAAG-3′A2

Quick check

1. Write the mRNA transcribed from template 3′-AAGTCC-5′.

2. What is the coding (sense) strand?

Answers: 1. 5′-UUCAGG-3′ 2. the DNA strand with the same sequence as the mRNA (T instead of U)

10 Template and coding strands

Which strand is read?

The template (antisense) strand is read by RNA polymerase. The coding (sense) strand has the same sequence as the mRNA, except T replaces U. Different genes may use different strands as template.

WORKED EXAMPLE 5 Coding strand (standard)

Given mRNA 5′-AUGGCA-3′, write the coding strand.

Solution

5′-ATGGCA-3′A1

11 The genetic code

Codons and amino acids.

The genetic code is a triplet code: each codon of three mRNA bases specifies one amino acid. With 64 codons and 20 amino acids, the code is degenerate (several codons per amino acid). AUG is the start codon (methionine); UAA, UAG and UGA are stop codons. The code is universal.

Codon	Amino acid
AUG	Met (start)
UUU, UUC	Phe
GCU, GCC, GCA, GCG	Ala
UAA, UAG, UGA	stop

WORKED EXAMPLE 6 Degeneracy (standard)

Explain what is meant by the degenerate genetic code.

Solution

More than one codon can code for the same amino acidR1

12 Translation: initiation

Assembling the ribosome.

The small ribosomal subunit binds to the mRNA and moves to the start codon AUG. A tRNA carrying methionine with anticodon UAC binds the start codon. The large subunit joins, forming the complete ribosome with three tRNA sites (A, P, E).

Quick check

1. What is the anticodon of the tRNA that binds AUG?

Answers: 1. UAC

13 Translation: elongation and termination

Building the polypeptide.

A tRNA with an anticodon complementary to the next codon enters the A site. A peptide bond forms between the amino acids (catalysed by rRNA). The ribosome moves along by one codon; the empty tRNA leaves from the E site. At a stop codon, a release factor frees the polypeptide.

WORKED EXAMPLE 7 Sequence (exam level)

Outline the events of elongation.

Solution

tRNA with complementary anticodon binds the codon in the A siteR1

Peptide bond formed between adjacent amino acidsR1

Ribosome moves one codon along (translocation)R1

14 tRNA structure and activation

Adaptor molecules.

Each tRNA folds into a clover-leaf shape with an anticodon loop and an amino acid attachment site (CCA at the 3′ end). Specific enzymes (aminoacyl-tRNA synthetases) attach the correct amino acid using ATP.

Quick check

1. Why are there specific tRNA-activating enzymes for each amino acid?

Answers: 1. to ensure each tRNA carries the correct amino acid

15 Polysomes

Many ribosomes on one mRNA.

Several ribosomes can translate the same mRNA simultaneously, forming a polysome. This produces many copies of a polypeptide quickly. In prokaryotes, translation can begin before transcription has finished.

WORKED EXAMPLE 8 Polysomes (standard)

State the advantage of polysomes.

Solution

Many polypeptides made from one mRNA at the same timeA1

16 Using the codon table

Decoding sequences.

To find the amino acid sequence from DNA: write the mRNA, divide into codons from the start codon, and read the amino acids from the codon table until a stop codon.

WORKED EXAMPLE 9 Decoding (exam level)

mRNA: 5′-AUG UUU GCA UAA-3′. Give the polypeptide.

Solution

Met–Phe–AlaA1

UAA is a stop codonA1

Quick check

1. How many amino acids are coded by an mRNA of 300 bases, including a stop codon?

Answers: 1. 99 (100 codons, one is stop)

17 Mutations and protein synthesis

Changing the sequence.

A base substitution may change one amino acid (missense), create a stop codon (nonsense) or have no effect (silent, due to degeneracy). Insertions and deletions shift the reading frame (frameshift), usually changing every amino acid after the mutation.

WORKED EXAMPLE 10 Effects (standard)

Explain why a base substitution may have no effect on the protein.

Solution

The new codon may code for the same amino acidR1

The genetic code is degenerateR1

18 Post-transcriptional modification (HL)

Splicing and processing.

In eukaryotes, pre-mRNA is modified: a 5′ cap and a 3′ poly-A tail are added; introns are removed and exons spliced together. Alternative splicing allows one gene to produce several different polypeptides.

WORKED EXAMPLE 11 Alternative splicing (exam level)

Explain how one gene can code for several proteins.

Solution

Different combinations of exons are joinedR1

Different mRNAs translated into different polypeptidesR1

19 Post-translational modification (HL)

After the ribosome.

Polypeptides may be modified after translation: folding, cleavage (proinsulin is cut to form active insulin), addition of carbohydrates (glycoproteins) or phosphate groups, and combination with other polypeptides.

Quick check

1. How is insulin produced from proinsulin?

Answers: 1. proinsulin is cleaved, removing the C-peptide

20 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 12 Mixed problem (exam level)

Compare transcription and translation.

Solution

Transcription: DNA → mRNA, RNA polymerase, nucleusR1

Translation: mRNA → polypeptide, ribosomes and tRNA, cytoplasmR1

Both involve complementary base pairingR1

21 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 13 Paper 1 question (standard)

Which molecule carries amino acids to the ribosome? A. mRNA B. rRNA C. tRNA D. DNA

Solution

CA1

WORKED EXAMPLE 14 Paper 2 question (exam level)

Explain how the base sequence of a gene determines the structure of a protein.

Solution

Gene transcribed into mRNA codonsR1

Codons specify amino acid sequence via tRNA anticodonsR1

Sequence determines folding and 3D structureR1

22 Antibiotics that target translation

Prokaryotic ribosomes.

Some antibiotics (streptomycin, tetracycline) bind to 70S ribosomes and block translation in bacteria without affecting the 80S ribosomes in human cytoplasm.

Quick check

1. Why do these antibiotics not harm human cells much?

Answers: 1. human cytoplasmic ribosomes are 80S, with a different structure

23 Gene expression and cell differentiation

Selective transcription.

All body cells have the same genes, but only some are transcribed in each cell type. Transcription factors control which genes are expressed, giving different proteins and specialized cells.

Quick check

1. Why do red blood cell precursors make haemoglobin but skin cells do not?

Answers: 1. haemoglobin genes are transcribed only in red blood cell precursors

24 Practical: modelling the genetic code

Decoding exercises.

Decoding exercises using the codon table and simple models with cards (mRNA, tRNA, amino acids) reinforce how translation works and the effects of mutations.

WORKED EXAMPLE 15 Model (standard)

In a card model, what do the tRNA cards represent?

Solution

Adaptor molecules linking codons to specific amino acidsR1

Try this

Write a 'secret message' as DNA, transcribe it, and decode it with the codon table using one-letter amino acid codes.

25 RNA types and roles

mRNA, tRNA, rRNA.

Type	Function
mRNA	carries the genetic code from DNA to ribosomes
tRNA	brings specific amino acids; anticodon pairs with codon
rRNA	structural and catalytic component of ribosomes
regulatory RNAs (miRNA)	control gene expression

26 Common misconceptions and links

Students often think…	Actually…
the codon is on tRNA	codons are on mRNA; anticodons on tRNA
both DNA strands are transcribed	only the template strand is transcribed
each amino acid has one codon	the code is degenerate
introns code for protein	introns are removed; exons code

How D1.2 connects to the rest of the course

Topic	Connection
A1.2 Nucleic acids	DNA and RNA structure
B1.2 Proteins	peptide bonds
B2.2 Organelles	nucleus and ribosomes
D1.3 Mutations	base substitutions
D2.2 Gene expression (HL)	control of transcription

27 Exam-style question with markscheme

Try this under exam conditions (about 15 minutes) before reading the markscheme.

Text A. In the haemoglobin beta-chain gene, the DNA triplet CTC is changed to CAC in people with sickle cell anaemia.

(a) Deduce the mRNA codon for the sixth amino acid of normal haemoglobin from the DNA triplet CTC. [1]

(b) Explain how the base substitution changes the protein. [3]

(d) Explain the role of tRNA in translation. [3]

(e) Explain how one gene can code for several proteins. [2]

Markscheme

Part	Answer	Marks
(a)	GAG	A1
(b)	CTC → CAC; codon GAG → GUG	R1
	glutamic acid replaced by valine	R1
	haemoglobin forms fibres; sickle cells	R1
(c)	transcription in nucleus, translation in cytoplasm	A1
	DNA template vs mRNA template	A1
	mRNA product vs polypeptide product	A1
(d)	carries a specific amino acid	R1
	anticodon pairs with complementary codon	R1
	brings amino acids in the correct order	R1
(e)	alternative splicing of pre-mRNA	R1
	different combinations of exons joined	R1

Examiner's comments. In (a), many answers wrote CTC or CUC. Transcription gives the complementary base sequence with U.

28 Check your understanding

Section A: multiple choice

1. Transcription produces

A. mRNA B. protein C. DNA D. tRNA only

2. The enzyme that transcribes DNA is

A. DNA polymerase B. helicase C. RNA polymerase D. ligase

3. How many bases form a codon?

A. 1 B. 3 C. 2 D. 4

4. The start codon is

A. UAA B. UGA C. GUG D. AUG

5. Peptide bonds are formed in

A. the nucleus B. ribosomes C. the Golgi D. lysosomes

6. In sickle cell anaemia, glutamic acid is replaced by

A. valine B. glycine C. proline D. methionine

7. tRNA-activating enzymes

A. form peptide bonds B. remove introns C. transcribe DNA D. attach the correct amino acid to each tRNA

8. The empty tRNA leaves from the

A. A site B. P site C. E site D. small subunit

Section B: structured

9. Explain transcription. [4]

10. Explain translation. [5]

11. Explain why the genetic code is described as degenerate and universal. [3]

12. Explain splicing and alternative splicing. [3]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. RNA polymerase binds promoter, separates strands, adds complementary RNA nucleotides (U for T) to template strand, releases mRNA.

10. mRNA on ribosome; start codon; tRNAs with anticodons; peptide bonds; ribosome moves; stop codon.

11. Degenerate: several codons per amino acid; universal: same in almost all organisms.

12. Introns removed, exons joined; different exon combinations give different mRNAs and proteins.

29 Summary and topic map

Idea	Remember
Flow	DNA → mRNA → polypeptide
Transcription	RNA polymerase; template strand; nucleus
Genetic code	codons; degenerate; universal; start AUG; stop codons
Translation	ribosome, mRNA, tRNA anticodons, peptide bonds
Sickle cell	GAG → GUG; glutamic acid → valine
HL	tRNA activation; A, P, E sites; splicing; insulin modification

Before the exam, make sure you…

•write mRNA with U, not T
•use codon tables carefully
•distinguish codon and anticodon
•explain sickle cell from base to phenotype

Protein Synthesis on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 4. Topic map for Protein Synthesis.

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