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IB Biology B1.2: Proteins worked solutions

Worked solutions to every practice question in B1.2: Proteins 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 part of an amino acid is different in each of the twenty amino acids used to make proteins? (A) the amine group (B) the carboxyl group (C) the R group (D) the central hydrogen
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The amine group, the carboxyl group and the hydrogen atom are the same in every amino acid. Only the R group (side chain) differs (C).

Final answer

C

Question A2

[1 mark]
The diagram shows part of a polypeptide chain. Which numbered bond is a peptide bond? [Graph: A polypeptide fragment drawn with four numbered bonds: 1 is the alpha carbon to carbonyl carbon bond, 2 is the carbonyl carbon to nitrogen bond, 3 is the nitrogen to alpha carbon bond and 4 is the C=O double bond.] (A) 1 (B) 2 (C) 3 (D) 4
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A peptide bond joins the carbon of a carboxyl group to the nitrogen of the next amino acid, so it is the C to N bond marked 2 (B). Bond 1 and bond 3 are inside amino acids, and bond 4 is the C=O double bond.

Final answer

B

Question A3

[1 mark]
Which row describes the formation of a peptide bond between two amino acids? (A) hydrolysis; water used (B) hydrolysis; water released (C) condensation; water used (D) condensation; water released
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Joining two amino acids is a condensation reaction and releases one water molecule (D). Hydrolysis is the reverse reaction and uses water.

Final answer

D

Question A4

[1 mark]
A polypeptide of 25 amino acids in one unbranched chain is hydrolysed completely. How many water molecules are used? (A) 24 (B) 23 (C) 25 (D) 26
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A chain of 25 amino acids has 25-1=24 peptide bonds, and each bond broken uses one water molecule, so 24 (A).

Final answer

A

Question A5

[1 mark]
What is a non-essential amino acid? (A) one that the body does not need (B) one found only in plant foods (C) one that the body can make from other amino acids (D) one that must be eaten every day
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A non-essential amino acid can be made by the body from other amino acids (C). It is still needed, so A is wrong. The word essential refers to the diet, not to plant or animal origin (B, D).

Final answer

C

Question A6

[1 mark]
An enzyme solution is heated from 25\,{}^\circ\mathrm{C} to 90\,{}^\circ\mathrm{C} and loses all of its activity. Which change explains this? (A) The peptide bonds are hydrolysed. (B) The amino acid sequence changes. (C) The enzyme is digested to amino acids. (D) R-group interactions break and the shape changes.
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Heat breaks the weak interactions between R groups, so the shape changes and the active site is lost (D). The peptide bonds and the amino acid sequence are not changed (A, B, C).

Final answer

D

Question A7

HL[1 mark]
Which bonds hold the turns of an alpha helix together? (A) hydrogen bonds between backbone groups (B) disulfide bonds between cysteines (C) ionic bonds between R groups (D) peptide bonds between turns
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The helix is held by hydrogen bonds between backbone C=O and N–H groups (A). Disulfide bonds and ionic bonds are between R groups and belong to tertiary structure, and peptide bonds run along the chain.

Final answer

A

Question A8

HL[1 mark]
A globular protein dissolves in water. Which statement describes its R groups? (A) All of the R groups are hydrophobic. (B) Hydrophobic groups face the water. (C) Hydrophilic groups outside, hydrophobic inside. (D) Hydrophilic groups are mainly inside.
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Hydrophilic R groups are mainly on the surface, where they bond to water, and hydrophobic R groups cluster in the core (C).

Final answer

C

Question A9

HL[1 mark]
Which of these proteins is conjugated? (A) insulin (B) haemoglobin (C) collagen (D) no prosthetic group
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Haemoglobin carries a haem prosthetic group, so it is conjugated (B). Insulin and collagen are quaternary but have no prosthetic group, and a single chain without one is not conjugated.

Final answer

B

Tutorial practice

Question B1

[6 marks]
Amino acids join to form polypeptides. (a) Draw the generalised structure of an amino acid and label the R group. [2] (b) Draw the formation of a dipeptide, with the peptide bond and the other product, and name the reaction. [4]
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(a) The alpha carbon is drawn at the centre (1) with an amine group, a carboxyl group, a hydrogen atom and an R group each bonded to it (1). (b) The two amino acids are drawn with the carboxyl group of one next to the amine group of the other (1); the dipeptide has a C–N peptide bond with the C=O and N–H kept (1); water is shown as the other product (1); the reaction is named as condensation (1).

Final answer

(a) A central carbon with H, NH_2, COOH and R attached. (b) Two amino acids with OH leaving the carboxyl group of the first and H leaving the amine group of the second; a C–N peptide bond drawn; water shown as a product; the reaction is condensation.

Question B2

[6 marks]
Wheat and soya were compared with a reference pattern of essential amino acids (illustrative data, mg per g of protein).\begin{array}{lrrr}\hline \text{}&\text{Lysine}&\text{Methionine + cysteine}&\text{Tryptophan}\\ \hline \text{Reference pattern}&50&25&7\\ \hline \text{Wheat protein}&25&38&11\\ \hline \text{Soya protein}&62&26&13\\ \hline\end{array} (a) Outline what is meant by an essential amino acid. [2] (b) Identify the amino acid limiting wheat and calculate its value as a percentage of the reference. [2] (c) Evaluate soya protein as the only source of these three amino acids. [2]
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(a) Essential amino acids cannot be synthesised by the body (1) and must be obtained from the diet (1). (b) Lysine, because the wheat value of 25 is below the reference of 50 (1); \dfrac{25}{50}\times100=50\% (1). (c) Soya is at or above the reference for all three amino acids shown (62, 26, 13 against 50, 25, 7), so it could be the only source of these (1); the table shows only three of the essential amino acids and says nothing about digestibility or how much is eaten, so adequacy is not proved (1).

Final answer

(a) An amino acid that cannot be made by the body, so it must come from the diet. (b) Lysine; 25/50\times100=50\%. (c) Soya meets the reference for all three amino acids (62 against 50, 26 against 25, 13 against 7), so it supplies them; but the table covers only three amino acids and does not show digestibility or the amount eaten.

Question B3

[6 marks]
Relative activity (%) of enzyme X, from a human, and enzyme Y, from a hot-spring bacterium, at different temperatures.\begin{array}{lrrrrrr}\hline \text{Temperature / }^\circ\text{C}&\text{20}&\text{40}&\text{50}&\text{60}&\text{70}&\text{80}\\ \hline \text{Enzyme X}&\text{40}&\text{100}&\text{25}&\text{0}&\text{0}&\text{0}\\ \hline \text{Enzyme Y}&\text{5}&\text{25}&\text{50}&\text{80}&\text{100}&\text{30}\\ \hline\end{array} (a) State the optimum temperature of enzyme Y, from the table. [1] (b) Describe the change in activity of enzyme X between 40 and 60\,{}^\circC. [2] (c) Explain the change in activity of enzyme X in terms of protein structure. [3]
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(a) 70\,{}^\circC, the temperature with the highest activity (1). (b) Activity falls from 100\% at 40\,{}^\circC to 0\% at 60\,{}^\circC (1), by 75 units between 40 and 50\,{}^\circC but only 25 units between 50 and 60\,{}^\circC (1). (c) Higher temperature makes the molecule vibrate more and breaks the weak interactions between R groups (1); the shape changes, so the active site no longer fits the substrate (1); the protein is denatured, though the peptide bonds and sequence are unchanged (1).

Final answer

(a) 70\,{}^\circC. (b) From 100\% at 40\,{}^\circC to 25\% at 50\,{}^\circC and 0\% at 60\,{}^\circC, fastest between 40 and 50\,{}^\circC. (c) Heat breaks interactions between R groups, the shape changes, the active site no longer fits the substrate, and the protein is denatured; the amino acid sequence stays the same.

Question B4

[5 marks]
A peptide hormone is a single unbranched chain of 9 amino acids. (a) State the number of peptide bonds, and of water molecules released, when the hormone was made, and calculate the number of different sequences of 9 amino acids. [2] (b) Explain how only 20 types of amino acid can give a vast variety of polypeptides. [3]
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(a) 9-1=8 peptide bonds, each made with one water molecule released, so 8 of each (1); 20^9=5.12\times10^{11} (1). (b) There are twenty types of amino acid (1) and any of them can be placed at any position, in any order (1); chains can also have any length, from a few to thousands, so the number of possible polypeptides is extremely large and each has its own sequence of R groups (1).

Final answer

(a) 8 peptide bonds and 8 water molecules; 20^9=5.12\times10^{11}. (b) Any order of the 20 amino acids is possible at each position, chains can be any length, so the number of sequences is very large; each sequence places different R groups in different positions.

Question B5

HL[5 marks]
Part of a protein forms a coil in which hydrogen bonds link each turn to the next. (a) Name this secondary structure and state the two groups that form the hydrogen bonds. [2] (b) Distinguish secondary structure from tertiary structure. [3]
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(a) Alpha helix (1); between backbone C=O and N–H groups (1). (b) Secondary structure is a regular, local pattern such as the helix or sheet (1) held by hydrogen bonds between backbone groups (1); tertiary structure is the overall three-dimensional fold of one polypeptide (1) held by interactions between R groups (maximum 3).

Final answer

(a) Alpha helix; the C=O group of one peptide link and the N–H group of another, in the backbone. (b) Secondary structure is a regular local pattern held by hydrogen bonds between backbone groups; tertiary structure is the fold of the whole chain held by interactions between R groups (hydrogen, ionic, disulfide, hydrophobic).

Question B6

HL[5 marks]
A mutation replaces core valine 54 of a soluble enzyme with aspartate, whose R group is negatively charged. Activity falls. (a) State what is meant by primary structure. [1] (b) Explain how the change at position 54 could alter the shape of the enzyme, and suggest why the activity falls. [4]
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(a) The order and number of amino acids, held by peptide bonds (1). (b) The charged, hydrophilic R group lies in a core made of hydrophobic groups (1) so it cannot form the hydrophobic interactions that held the fold (1) and may attract water or opposite charges, changing the tertiary structure (1). The shape of the active site changes, so the substrate fits less well and activity falls (1).

Final answer

(a) The sequence and exact position of the amino acids. (b) A charged hydrophilic group sits in the non-polar core, away from water; it cannot form hydrophobic interactions and may attract water or oppositely charged groups, so the fold changes; the active site changes shape, so activity falls.

Question B7

HL[5 marks]
Insulin has two polypeptide chains joined by two disulfide bonds and no non-protein part. (a) State the level of structure that joins the chains and deduce whether insulin is conjugated, with a reason. [2] (b) Compare the structure of insulin with that of haemoglobin. [3]
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(a) Quaternary structure (1); not conjugated, because there is no prosthetic group (1). (b) Both have more than one polypeptide chain, so both have quaternary structure (1); haemoglobin has four chains and four haem prosthetic groups, so it is conjugated (1); insulin has two chains and no prosthetic group, so it is not (1).

Final answer

(a) Quaternary; not conjugated, because it has no prosthetic group. (b) Both are quaternary proteins. Haemoglobin has four chains and four haem groups, so is conjugated; insulin has two chains and no prosthetic group, so is not.

Question B8

HL[4 marks]
Collagen is a fibrous protein and insulin is a globular protein. (a) Outline how the structure of collagen makes it strong. [2] (b) Explain why insulin is soluble but collagen is not. [2]
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(a) Collagen has three polypeptide chains wound into a triple helix (1) and many molecules lie side by side in fibres, so it is strong and resists stretching (1). (b) Insulin is compact with hydrophilic R groups on its surface, so it is soluble and can be carried in the blood (1); collagen is long and rope-like and insoluble in water, which suits a structural role (1).

Final answer

(a) Three chains wound into a triple helix, with many molecules side by side in fibres, resist pulling. (b) Insulin is compact with hydrophilic R groups on the outside, so it dissolves in blood; collagen is long and its fibres do not dissolve.

Advanced practice

Question C1

[8 marks]
Young rats ate diets with the same total protein for three weeks. Diet P had all the essential amino acids, diet Q the same with lysine halved, and diet R was Q with lysine added back. Mean mass gain \pm standard deviation, n=10 per group.\begin{array}{lrr}\hline \text{Diet}&\text{Mass gain / g}&\text{Standard deviation}\\ \hline \text{P}&62&5\\ \hline \text{Q}&31&6\\ \hline \text{R}&60&5\\ \hline\end{array} (a) Calculate the percentage reduction in mass gain on diet Q compared with diet P. [2] (b) Compare the mass gain on diets Q and R. [2] (c) Explain the result for diet Q and the result for diet R. [2] (d) Evaluate the claim that vegans should plan their diet to include lysine-rich foods. [2]
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(a) \dfrac{62-31}{62}\times100=50\% (1 for the working, 1 for the value). (b) Rats on Q gained 60-31=29 g less than those on R (1); the ranges do not overlap (25 to 37 and 55 to 65), so the difference is likely to be real (1). (c) Lysine is an essential amino acid and cannot be made by the body, so with half the lysine protein synthesis and growth are limited (1); adding it back removed the shortage and restored growth to about the level of diet P (1). (d) The data support the claim that a shortage of one essential amino acid limits growth (1); but the study used rats for three weeks, changed only lysine and used small groups, so it cannot justify firm advice for people (1).

Final answer

(a) 50\%. (b) Q gained 29 g less than R (31 against 60), and the ranges do not overlap. (c) Lysine is essential, so growth needs it from the diet; half the lysine limited protein synthesis, and adding it back restored growth. (d) Supports the idea that a shortage of one essential amino acid limits growth, but rats, three weeks, one amino acid and small groups do not justify a firm recommendation for people.

Question C2

[7 marks]
A polypeptide of 60 amino acids in one chain is digested by a protease into amino acids. The mass of amino acids released in 10 min was measured in buffers of different pH (illustrative data).\begin{array}{lrrrr}\hline \text{pH}&\text{2}&\text{5}&\text{7}&\text{10}\\ \hline \text{Amino acids released / mg}&0.4&4.8&9.6&1.2\\ \hline\end{array} (a) State the number of peptide bonds, and of water molecules needed to hydrolyse them all. [1] (b) Identify the optimum pH and calculate the percentage decrease in activity at pH 10 from it. [2] (c) Explain the low activity of the protease at pH 2. [2] (d) Outline how water breaks a peptide bond. [2]
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(a) 60-1=59 peptide bonds, and one water molecule for each, so 59 (1 for both values). (b) pH 7, the highest value (1); \dfrac{9.6-1.2}{9.6}\times100=88\% decrease (1). (c) At pH 2 the R groups gain or lose H^+ so their charges change (1); the interactions holding the shape break, the active site changes and the substrate no longer fits (1). (d) A water molecule adds OH to the carbonyl carbon and H to the nitrogen (1), which breaks the C–N peptide bond and releases two amino acids (1).

Final answer

(a) 59 peptide bonds, so 59 water molecules. (b) pH 7; (9.6-1.2)/9.6\times100=88\% decrease. (c) At pH 2 the charges on R groups change, interactions break, the shape of the enzyme changes and the active site no longer fits. (d) Water adds OH to the carbonyl carbon and H to the nitrogen, which breaks the C–N bond.

Question C3

HL[8 marks]
In a soluble enzyme, an ionic bond between a lysine side chain (-\mathrm{NH_3^+}) and a glutamate side chain (-\mathrm{COO^-}) helps to hold two parts of the chain together. (a) Write the change in the lysine side chain when the pH rises and it loses a proton. [1] (b) Explain why this removes the ionic bond. [2] (c) Explain why enzyme activity falls. [2] (d) Explain why the sequence is unchanged although the shape changes, and whether this is reversible. [3]
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(a) -\mathrm{NH_3^+}\rightarrow-\mathrm{NH_2}+\mathrm{H^+}, a balanced change of atoms and charge (1). (b) The side chain loses its positive charge (1), so there is no electrostatic attraction to the negative glutamate side chain and the ionic bond is lost (1). (c) The loss of ionic bonds changes the three-dimensional fold, the tertiary structure (1); the shape of the active site changes, so the substrate no longer fits and activity falls (1). (d) The sequence is unchanged because the covalent peptide bonds are not broken (1); only weak interactions between R groups are lost, so the shape changes but the order of amino acids does not (1). If the pH returns to its earlier value the lysine regains H^+ and the bond can re-form, so a mild change may be reversible, but an extreme change denatures the enzyme for good (1).

Final answer

(a) -\mathrm{NH_3^+}\rightarrow-\mathrm{NH_2}+\mathrm{H^+}. (b) The lysine side chain has no charge, so it no longer attracts the negatively charged glutamate side chain. (c) Loss of this bond (and others) changes the tertiary structure, so the active site changes shape and activity falls. (d) The amino acid sequence is unchanged because peptide bonds are not broken; only the R-group interactions are, so the shape changes.

Question C4

HL[7 marks]
In an exercise with invented data, cryogenic electron microscopy images of protein P (large outline) mixed with protein T (small triangular outline) were classified, 200 particle images per condition. At a low concentration of T, 40 images showed P with T attached; at a higher concentration, 160 did. (a) Calculate the percentage of images that show the complex at each concentration of T. [2] (b) State the conclusion that the counts support. [2] (c) State one thing the images cannot show about the interaction. [1] (d) Another protein imaged in this way has two polypeptide chains and one haem group. Deduce its level of structure and whether it is conjugated. [2]
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(a) \dfrac{40}{200}\times100=20\% and \dfrac{160}{200}\times100=80\% (1 each). (b) The fraction of images with the complex rises from 20\% to 80\%, four times as common (80/20=4), at the higher concentration of T (1), so more T gives more association of P and T under matched classification (1). (c) The images show the outlines in contact, but outline contact does not identify an ionic bond, hydrogen bond or any other bond (1). (d) A protein with two polypeptide chains has quaternary structure (1), and the haem group is a non-protein prosthetic group, so it is conjugated (1).

Final answer

(a) 20\% and 80\%. (b) More T makes the complex four times as common among the images. (c) The images show where P and T touch; they do not show which kind of bond holds them. (d) Quaternary structure; conjugated, because haem is a prosthetic group.

Question C5

HL[7 marks]
Explain how the primary, secondary and tertiary structures of an enzyme are related to the way it works.
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One mark for each point, maximum 7: the primary structure is the exact sequence of amino acids joined by peptide bonds (1); each position has an R group with particular properties (1); the sequence therefore decides which interactions can form (1); the secondary structure is helices and sheets held by hydrogen bonds between backbone groups (1); the tertiary structure is the fold of the whole chain held by hydrogen bonds, ionic bonds, hydrophobic interactions and disulfide bonds between R groups (1); hydrophobic R groups cluster in the core and hydrophilic ones face the water, so the enzyme is soluble (1); the fold gives an active site of a particular shape that fits only its substrate, so function depends on structure (1); a change in the sequence can change the fold and the active site (1).

Final answer

Seven points from: primary structure is the sequence of amino acids joined by peptide bonds; the sequence puts particular R groups in particular places; the secondary structure forms helices and sheets by backbone hydrogen bonds; tertiary folding by R-group interactions (hydrogen, ionic, disulfide, hydrophobic); hydrophobic groups cluster inside a soluble enzyme and hydrophilic groups face the water; the fold gives a specific shaped active site; substrate fits, so it is specific; a change of sequence can alter the fold and the function.

Connected questions

Question X1

Connected questions[6 marks]
B1.2 + B1.1. Proteins and starch are both digested by hydrolysis. (a) Name the bond that joins amino acids and the bond that joins glucose units in starch. [2] (b) State what is added to each bond when it is broken. [1] (c) Calculate the water molecules needed to break a polypeptide of 41 amino acids and an unbranched starch chain of 100 glucose units completely into monomers. [2] (d) Explain why a protease does not digest starch. [1]
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(a) The amino acids are joined by peptide bonds (1) and the glucose units by glycosidic bonds (1). (b) Each bond is broken by the addition of a water molecule (1). (c) A chain of 41 amino acids has 41-1=40 peptide bonds, so 40 waters (1); an unbranched chain of 100 glucose units has 100-1=99 glycosidic bonds, so 99 waters (1). (d) The active site of a protease has a shape that fits polypeptides and not starch, so the enzyme is specific (1).

Final answer

(a) Peptide bond; glycosidic bond. (b) Water. (c) 41-1=40; 100-1=99. (d) The active site of the protease is shaped to fit the polypeptide substrate and not starch.

Question X2

Connected questions[6 marks]
B1.2 + C1.1. An enzyme has an active site made of the R groups of several amino acids. (a) Explain how the amino acids in the active site make the enzyme specific. [2] (b) A mutation changes one amino acid far from the active site and the enzyme works less well. Suggest why. [2] (c) Explain why high temperature permanently reduces activity but low temperature does not. [2]
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(a) The folded chain brings particular R groups together in the active site (1); their shape and charge fit the substrate and not other molecules, so the enzyme is specific (1). (b) A mutation far from the active site can change an R-group interaction (1) and so the tertiary structure, which alters the shape of the active site (1). (c) High temperature breaks interactions between R groups and changes the shape for good (1); low temperature slows molecules down but does not change the shape, so activity returns on warming (1).

Final answer

(a) The R groups in the active site have particular shapes and charges, so only a substrate of matching shape and charge binds. (b) The mutation changes the tertiary structure, so the active site changes shape. (c) High temperature breaks the interactions that hold the shape, so the denaturation is permanent; low temperature only slows the molecules and the shape is not changed.

Question X3

Connected questions · HL[6 marks]
B1.2 + D1.3. In sickle-cell haemoglobin one codon change replaces a glutamate on the outside of a chain with valine. (a) State the level of protein structure that is changed first. [1] (b) Explain how the change can make haemoglobin molecules stick together. [2] (c) State the prosthetic group of haemoglobin. [1] (d) Predict the effect of the change on the solubility of haemoglobin and the consequence in the red cell. [2]
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(a) The primary structure, the order of amino acids, is changed first (1). (b) Glutamate has a charged, hydrophilic R group that faces the water (1), while valine is hydrophobic, so the surface patch is attracted to a hydrophobic site on another haemoglobin molecule and the molecules stick together (1). (c) Haem is the prosthetic group, so haemoglobin is a conjugated protein (1). (d) The sticking reduces the solubility of the haemoglobin (1), so long fibres can form in the cell (1).

Final answer

(a) Primary structure. (b) Glutamate is charged and hydrophilic; valine is hydrophobic. A hydrophobic patch on the surface is attracted to hydrophobic patches on other molecules, so they stick together. (c) Haem. (d) Lower solubility in the red cell.

From the book

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

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