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IB Biology D1.2: Protein synthesis worked solutions

Worked solutions to every practice question in D1.2: Protein synthesis of Photon Biology: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

[1 mark]
Which process makes mRNA, and where does it happen in a eukaryotic cell? (A) Translation, cytoplasm (B) Transcription, nucleus (C) Transcription, ribosomes (D) Replication, nucleus
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B. Transcription copies a gene into mRNA in the nucleus. Translation happens on ribosomes in the cytoplasm, and replication makes DNA.

Final answer

B

Question A2

[1 mark]
A section of template strand reads TAGGCA. Which is the mRNA copied from it? (A) ATCCGT (B) UAGGCU (C) AUCCGU (D) TAGGCA
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C. Pair each base: T with A, A with U, G with C, G with C, C with G, A with U, which gives AUCCGU. A and D contain T, so they are DNA, and B is not complementary.

Final answer

C

Question A3

[1 mark]
The mRNA codon is CAG. What is the anticodon of the tRNA that pairs with it? (A) GUC (B) GTC (C) CAG (D) CUG
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A. C pairs with G, A with U and G with C, so the anticodon is GUC. An anticodon is RNA, so B, which contains T, is wrong. C is the codon itself.

Final answer

A

Question A4

[1 mark]
The mRNA sequence is AUG UUU GGC. Use the genetic code table to find the amino acid sequence. (A) Met-Leu-Gly (B) Met-Phe-Ala (C) Phe-Met-Gly (D) Met-Phe-Gly
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D. AUG is methionine, UUU is phenylalanine and GGC is glycine. B has alanine, which is GCN, and C has the first two in the wrong order.

Final answer

D

Question A5

[1 mark]
A polypeptide has 50 amino acids. How many mRNA bases code for these amino acids, not counting the stop codon? (A) 50 (B) 150 (C) 300 (D) 17
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B. Each amino acid needs one codon of three bases, so 50\times3=150. A uses one base per amino acid and C counts both strands.

Final answer

B

Question A6

[1 mark]
Which feature do replication and transcription share? (A) A ribosome reads it (B) A polypeptide is made (C) Pairing with a DNA template (D) Both strands, whole genome
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C. Both use a DNA template and pair bases by complementary base pairing. Only translation involves a ribosome or makes a polypeptide, and only replication copies both strands of the whole genome.

Final answer

C

Question A7

[1 mark]
What does it mean that the genetic code is degenerate? (A) One codon, many amino acids (B) It differs between species (C) Some codons are stop codons (D) Many codons, one amino acid
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D. Degenerate means that most amino acids are coded for by more than one codon. A reverses this, B describes a code that is not universal, and C describes stop codons.

Final answer

D

Question A8

HL[1 mark]
RNA polymerase is making an mRNA. At which end of the growing mRNA is the next nucleotide added? (A) 3′ end of the mRNA (B) 5′ end of the mRNA (C) 3′ end of the template (D) 5′ end of the template
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A. RNA polymerase joins each new nucleotide to the free 3′ end, so mRNA is built from 5′ to 3′. The template is not extended.

Final answer

A

Question A9

HL[1 mark]
Which of these DNA regions are non-coding: a telomere, a promoter and an exon? (A) Telomere only (B) Telomere and promoter (C) Promoter and exon (D) All three
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B. A telomere and a promoter are never translated into a polypeptide. An exon codes for part of a polypeptide.

Final answer

B

Question A10

HL[1 mark]
Which change happens to a eukaryotic pre-mRNA before translation? (A) Exons out, introns joined (B) Cap added at the 3′ end (C) Introns out, poly-A tail added (D) Anticodons added
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C. Introns are removed and a poly-A tail is added. Exons are kept, the cap is added at the 5′ end, and anticodons belong to tRNA.

Final answer

C

Tutorial practice

Question B1

[3 marks]
Outline the roles of mRNA and tRNA in translation.
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mRNA carries the base sequence (codons) copied from the gene to the ribosome (1). Each tRNA carries one particular amino acid (1). The anticodon of the tRNA pairs with the codon by complementary base pairing, so the amino acids are placed in the order the mRNA specifies (1).

Final answer

mRNA carries the codons copied from the gene to the ribosome; tRNA brings one specific amino acid; its anticodon pairs with a codon so the amino acids join in the coded order.

Question B2

[4 marks]
A ribosome and a tRNA molecule each have a structure that fits their job. (a) State the two kinds of molecule that make up a ribosome. [1] (b) Explain how the structure of a tRNA lets it bring the correct amino acid to the ribosome. [3]
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(a) Ribosomes contain rRNA and protein (1). (b) One end of the tRNA has an anticodon of three bases (1) that pairs by complementary base pairing with the codon of the mRNA (1). The other end binds one specific amino acid, which fits the anticodon, so the correct amino acid is brought for each codon (1).

Final answer

(a) rRNA and protein. (b) The anticodon pairs with the codon; the amino acid attachment end carries only the matching amino acid; so amino acids are placed in the correct order.

Question B3

[5 marks]
Outline the process of transcription.
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RNA polymerase attaches to the DNA at the start of the gene (1). Hydrogen bonds between the base pairs break and the two strands separate (1). One strand, the template, is copied (1). Free RNA nucleotides pair with the template by complementary base pairing, with A pairing with U (1). RNA polymerase joins the nucleotides into a strand (1). The mRNA is released and the DNA strands come together again (1). Any five points.

Final answer

RNA polymerase binds at the gene and separates the strands; one strand is the template; RNA nucleotides pair with it, A with U; RNA polymerase joins them; mRNA released and DNA closes.

Question B4

[3 marks]
Heart muscle cells do not divide after they mature, yet each one makes mRNA from the same genes for decades. (a) Explain why the base sequence of its DNA must stay unchanged. [1] (b) Explain why transcription does not use up or change the DNA template. [2]
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(a) The base sequence is the code for the proteins the cell needs, and the cell cannot replace its DNA by division, so a change would be passed to every mRNA made from it (1). (b) Transcription makes a complementary RNA copy and leaves the DNA template unchanged (1); the strands close again, so the same gene can be transcribed many times (1).

Final answer

(a) A change would alter the mRNA and the protein made, harming the cell. (b) Transcription copies the sequence into RNA; the template is not altered, so it can be read again and again.

Question B5

[6 marks]
A section of the template strand of a gene is TAC CCG TGC AAT ATC. Use the genetic code table. (a) Deduce the base sequence of the mRNA. [1] (b) Deduce the amino acid sequence of the polypeptide. [2] (c) State the number of amino acids in the polypeptide. [1] (d) The fourth template triplet changes from AAT to ATT. Deduce the effect on the polypeptide. [2]
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(a) Pair each base: AUG GGC ACG UUA UAG (1). (b) Met-Gly-Thr-Leu (1), with translation stopping at UAG (1). (c) Four amino acids; the stop codon does not code for one (1). (d) AAT changes to ATT, so UUA changes to UAA (1), a stop codon, so the polypeptide ends early with three amino acids, Met-Gly-Thr (1).

Final answer

(a) AUG GGC ACG UUA UAG. (b) Met-Gly-Thr-Leu. (c) 4. (d) AAT to ATT gives UAA, a stop codon: polypeptide Met-Gly-Thr, 3 amino acids.

Question B6

[4 marks]
The table describes four ribosomes on one mRNA. Distance is measured from the left end of the mRNA. Ribosome 1 is 90 bases from the left end with a chain of 30 amino acids; ribosome 2, 270 bases, 90; ribosome 3, 450 bases, 150; ribosome 4, 630 bases, 210. (a) Deduce, with a reason, which end of the mRNA, left or right, is the 5′ end. [2] (b) Explain the advantage to the cell of having several ribosomes on one mRNA. [2]
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(a) The ribosome with the longest chain, ribosome 4, started first (1) and has moved furthest from the left end, so the ribosomes move rightwards and the left end is the 5′ end (1). (b) Several ribosomes translate the same mRNA at the same time (1), so many copies of the polypeptide are made quickly from one mRNA (1).

Final answer

(a) Ribosome 4 started first; the right end is the 3′ end, so the left end is 5′. (b) Many copies of the same polypeptide are made at once, so protein is made faster.

Question B7

[5 marks]
Cells were given a hormone at time 0. The amounts of mRNA and protein from one gene were measured (arbitrary units). Time 0, 1, 2, 4 hours. mRNA: 5, 40, 120, 120 units. Protein: 0, 0, 10, 60 units. (a) Calculate the percentage increase in mRNA between 0 and 2 hours. [2] (b) Compare the timing of the changes in mRNA and in protein. [2] (c) Suggest why the protein lags behind the mRNA. [1]
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(a) (120-5)\div5\times100=2300\% (1) with the working shown (1). (b) The mRNA amount rises at 1 hour, before any protein is detected (1); protein appears only by 2 hours and keeps rising to 4 hours while the mRNA stays level (1). (c) Time is needed to translate the mRNA and fold the polypeptide, so protein lags behind the mRNA (1).

Final answer

(a) 2300%. (b) mRNA rises first (by 1 h) and protein only later (from 2 h). (c) Translation and folding take time after the mRNA is made.

Question B8

HL[3 marks]
A mutation deletes the promoter of gene Y in one cell line. (a) State what a promoter is. [1] (b) Explain the effect of the deletion on the expression of gene Y. [2]
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(a) A non-coding stretch of DNA at the start of a gene, to which RNA polymerase and transcription factors bind (1). (b) Without the promoter RNA polymerase cannot bind and start transcription (1), so no mRNA and no protein Y are made (1).

Final answer

(a) A non-coding DNA sequence at the start of a gene where RNA polymerase binds. (b) RNA polymerase cannot bind, so gene Y is not transcribed and its protein is not made.

Question B9

HL[4 marks]
Gene W has five exons. In cell type 1 the mature mRNA contains exons 1, 2, 3, 4 and 5. In cell type 2 it contains exons 1, 2, 4 and 5. (a) Name the process that produces the two mRNAs. [1] (b) Explain how one gene gives two different polypeptides. [2] (c) State the number of introns in the pre-mRNA. [1]
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(a) Alternative splicing (1). (b) Different combinations of exons are joined (1), so the mature mRNAs have different codon sequences and are translated into different polypeptides (1). (c) The pre-mRNA has four introns between five exons (1).

Final answer

(a) Alternative splicing. (b) Different exon sets give different mRNA codon sequences, so different polypeptides. (c) 4.

Question B10

HL[4 marks]
Each ribosome has A, P and E sites for tRNA. (a) State the site that holds the tRNA carrying the growing polypeptide. [1] (b) A drug blocks the A site. Explain the effect on translation. [2] (c) State the site from which an empty tRNA leaves. [1]
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(a) The P site (1). (b) A drug that blocks the A site stops an incoming tRNA binding the next codon (1), so no peptide bond forms and the polypeptide cannot grow (1). (c) Empty tRNA leaves from the E site (1).

Final answer

(a) P site. (b) No new tRNA can bind, so no peptide bond forms and translation stops. (c) E site.

Advanced practice

Question C1

HL[7 marks]
Explain how a polypeptide is made on a ribosome after a mature mRNA reaches the cytoplasm.
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The mRNA binds to the small ribosomal subunit (1). An initiator tRNA carrying methionine pairs with the start codon AUG (1) and the large subunit joins (1). A tRNA whose anticodon is complementary to the next codon enters the ribosome (1). A peptide bond forms between the amino acids (1). The ribosome moves along the mRNA one codon at a time, 5′ to 3′, and the cycle repeats (1). At a stop codon there is no tRNA, so the polypeptide is released (1). Any seven points.

Final answer

mRNA binds the small subunit; initiator tRNA with methionine pairs with AUG; large subunit joins; tRNAs enter by codon-anticodon pairing; peptide bonds form; the ribosome moves codon by codon; stop codon releases the polypeptide.

Question C2

HL[6 marks]
Insulin is made as pre-proinsulin of 110 amino acids. A signal peptide of 24 amino acids is removed to give proinsulin. Insulin has 51 amino acids. (a) Calculate the number of amino acids in proinsulin. [1] (b) Calculate the number of amino acids in the part removed to turn proinsulin into insulin. [1] (c) Suggest an advantage of making insulin as an inactive precursor. [2] (d) Outline how proinsulin becomes insulin. [2]
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(a) 110-24=86 (1). (b) 86-51=35 amino acids removed: the 31-amino-acid C-peptide plus 4 linking residues (1). (c) The inactive proinsulin can be stored without acting on cells (1) and converted to the active hormone quickly when it is needed (1). (d) Disulfide bonds form between the chains (1) and the C-peptide is cut out, leaving the A and B chains linked (1).

Final answer

(a) 86. (b) 35 (the 31-amino-acid C-peptide plus 4 linking residues). (c) Inactive precursor can be stored safely and made active quickly when needed. (d) Disulfide bonds form between the chains and the C-peptide is cut out.

Question C3

HL[6 marks]
Protein P was incubated with proteasomes in a test tube. The table shows the percentage of protein P broken down after 30 minutes (illustrative data). Proteasome without ATP, 8 percent; with ATP, 52 percent; with ATP and an inhibitor, 6 percent. (a) Calculate the increase in percentage broken down when ATP is added to normal proteasomes. [1] (b) Deduce what the inhibitor result shows. [2] (c) Explain the value to the cell of breaking proteins down to amino acids. [2] (d) State where in the cell proteasomes are found. [1]
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(a) 52-8=44 percentage points (1). (b) With the inhibitor present, breakdown falls to 6%, no higher than without ATP (1), which shows the proteasome is responsible for breaking down protein P (1). (c) Proteasomes break proteins to amino acids (1) which are reused to make new polypeptides, so damaged or unwanted proteins are removed and the proteome stays functional (1). (d) Proteasomes are found in the nucleus and in the cytoplasm (1).

Final answer

(a) 44 percentage points. (b) The inhibitor lowers breakdown to about the level without ATP, so the proteasome is doing the breaking down. (c) Amino acids are recycled to make new polypeptides, so the cell conserves resources and keeps its proteome working.

Question C4

HL[7 marks]
A gene is transcribed into a pre-mRNA of 3100 bases. The exons together contain 1005 bases, including the stop codon. Ignore the cap and the poly-A tail. (a) Calculate the number of bases in the introns. [1] (b) Calculate the number of amino acids in the polypeptide. [2] (c) Calculate the percentage of the pre-mRNA removed by splicing. [1] (d) The same gene also gives a mature mRNA of 822 bases. Deduce how this arises and the effect on the polypeptide. [3]
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(a) 3100-1005=2095 bases (1). (b) 1005\div3=335 codons, and the stop codon is not an amino acid, so 334 amino acids (2). (c) 2095\div3100\times100=67.6\% (1). (d) The two mRNAs of 1005 and 822 bases differ by 1005-822=183 bases (1), which is an exon left out by alternative splicing (1); 183\div3=61, so the shorter polypeptide has 61 fewer amino acids (1).

Final answer

(a) 2095. (b) 334. (c) 67.6%. (d) A second mRNA of 822 bases lacks an exon of 183 bases, which is alternative splicing, and the polypeptide is 61 amino acids shorter.

Question C5

HL[6 marks]
An antibiotic binds to the A site of ribosomes. In treated bacteria the amount of mRNA is unchanged but protein synthesis falls by 95%. (a) State why the mRNA amount is unchanged. [1] (b) Explain how blocking the A site stops protein synthesis. [2] (c) Explain why measuring mRNA alone would not show the effect of the drug. [2] (d) Predict the length of the chains made by treated ribosomes. [1]
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(a) The drug blocks translation, which is after transcription, so the mRNA is still made (1). (b) An incoming tRNA cannot bind to the A site (1), so no peptide bond forms and the chain cannot be extended (1). (c) The amount of mRNA shows how much of the gene is transcribed, not how much protein is made (1), because translation, folding and breakdown follow (1). (d) Mostly very short chains, only the start of the polypeptide (1).

Final answer

(a) The drug acts on translation, not transcription. (b) The tRNA cannot enter, so no peptide bond forms. (c) mRNA is made before translation, so mRNA does not report protein made. (d) Short chains of one or two amino acids.

Connected questions

Question X1

Connected questions[6 marks]
D1.2 + D1.3 + B1.2. The sixth codon of the \beta-globin mRNA is GAG. A point mutation changes it to GUG, and the haemoglobin made by the mutant allele forms fibres when oxygen is low. (a) State the template triplet before and after the mutation. [2] (b) Name the type of mutation. [1] (c) Explain how the mutation changes the properties of haemoglobin. [3]
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(a) The template triplet CTC becomes CAC (1) and the mRNA codon is GUG (1). (b) A substitution, a point mutation (1). (c) The codon GUG codes for valine instead of glutamic acid (1); valine has a different R group, so the interactions that hold the folded protein change (1) and the haemoglobin molecules stick together into fibres (1).

Final answer

(a) Template CTC to CAC; mRNA GAG to GUG. (b) Substitution. (c) Glutamic acid becomes valine, a different R group; the tertiary and quaternary structure changes so haemoglobin sticks together.

Question X2

Connected questions[6 marks]
D1.2 + D1.1 + D2.2. Liver cells and neurons from one person have the same DNA but different proteins. (a) Compare transcription with DNA replication. [3] (b) Explain how the two cell types can have different proteins. [3]
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(a) Replication copies both strands (1) of the whole genome, using DNA polymerase, and makes DNA (1); transcription copies one strand of one gene, using RNA polymerase, and makes RNA with U instead of T (1). Any three. (b) Only some genes are transcribed in each cell (1), because transcription of each gene is switched on or off (1); mRNA is made only from the genes that are on, so the proteins differ even though the DNA is the same (1).

Final answer

(a) Replication copies both strands with DNA polymerase to make DNA; transcription copies one strand of one gene with RNA polymerase to make RNA with U. (b) Different genes are transcribed in each cell type; transcription is switched on or off.

Question X3

Connected questions[6 marks]
D1.2 + B1.2 + C1.1. A mutation changes the mRNA codon of one amino acid in an enzyme, far from its active site. In another mutant, a codon becomes UAA. (a) Suggest why the enzyme may still work. [2] (b) Explain the effect of the UAA codon. [2] (c) Explain why the amino acid sequence determines how the enzyme works. [2]
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(a) The new codon may code for the same amino acid because the code is degenerate (1), or for one that does not change folding, so the active site keeps its shape (1). (b) UAA is a stop codon with no tRNA (1), so the polypeptide ends early and the enzyme is not formed (1). (c) The amino acid sequence decides which R groups interact when the polypeptide folds (1), and so the shape of the active site that the substrate fits (1).

Final answer

(a) The new codon may code for the same amino acid (degenerate) or one that does not alter the shape, so the enzyme works. (b) UAA is a stop codon: the polypeptide is cut short and the enzyme is not made. (c) The sequence decides the interactions between R groups that fold the chain into the shape of the active site.

From the book

Photon Biology: Study Guide for the IB Diploma (SL & HL)

Protein synthesis from the core ideas to full exam questions, with question types, worked examples and graded practice.

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