Biological molecules

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1
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Describe how a molecule of beta glucose is different from a molecule of alpha glucose. (2 marks)

  • Hydroxyl group on carbon 1 is in a different position

  • In alpha glucose it is below the ring, in beta glucose it is above the ring

2
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Glycogen and cellulose are both carbohydrates. Describe two differences between the structure of a cellulose molecule and a glycogen molecule. (2 marks)

  • Cellulose is made up of β-glucose monomers and glycogen is made up of α-glucose

  • Cellulose molecule has straight chain and glycogen is branched

  • Glycogen has 1,4 and 1,6 glycosidic bonds and cellulose has only 1,4 glycosidic bonds

3
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A covering of lignin protects cellulose from enzyme attack . Use your knowledge of the way in which enzymes work to explain why cellulose-digesting enzymes do not digest lignin. (2 marks)

  • Enzymes are specific

  • Shape of lignin molecules will not fit active site (of enzyme)

4
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Describe the structure of a cellulose molecule and explain how cellulose is adapted for its function in cells. (6 marks)

  • made from β-glucose

  • joined by condensation, glycosidic bond

  • hydrogen bonds linking long straight chains

  • cellulose fibres makes cell walls strong

  • can resist osmotic pressure

  • bond difficult to break; resists digestion / enzymes

5
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Explain how cellulose molecules are adapted for their function in plant cells (3 marks)

  • Long and straight chains

  • Become linked together by many hydrogen bonds to form microfibrils

  • Provide strength to cell wall

6
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<p>Use the information in the figure to explain two ways in which fatty acids are important in the formation of new cells. (4 marks)</p>

Use the information in the figure to explain two ways in which fatty acids are important in the formation of new cells. (4 marks)

1. Fatty acids used to make phospholipids; Phospholipids in membranes; More phospholipids more membranes made;
2. Fatty acids respired to release energy; More triglycerides more energy released; Energy used for cell production

7
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Draw a diagram to show the structure of a triglyceride molecule. (2 marks)

knowt flashcard image
8
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Define the quaternary structure of a protein. (1 mark)

Multiple polypeptide chains held together by hydrogen bonds and ionic bonds

9
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Explain how two enzymes with different amino acid sequences can catalyse the same reaction. (2 marks)

Both active sites have similar tertiary structures so can form enzyme-substrate complexes with the same substrates

10
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Explain how a change in the primary structure of a globular protein may result in a different three-dimensional structure. (3 marks)

  • Sequence of amino acids change

  • Tertiary structure folds in a different way

  • As bonds form in different places

11
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Explain how the structure of fibrous proteins is related to their functions.

  • Long chains of amino acids;

  • Folding of chain into a coil /pleated sheet;

  • Association of several polypeptide chains together

  • H bonds / Disulphide bonding (In context);

  • Fibres provide strength (and flexibility);

  • Sheets provide flexibility;

  • Example e.g. keratin in hair, collagen in bone; (MUST be in context)

  • Insoluble because external R-groups are non-polar

12
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Explain why initial rate of this reaction was faster at 65 °C than it was at 55 °C (3 marks)

  • Particles have more KE at higher temperatures; move faster

  • Greater chance of collision

  • More enzyme-substrate complexes form

13
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Raffinose is an example of a trisaccharide made up of glucose, fructose and galactose. The chemical formulae of these monosaccharides are listed below: 

Glucose = C6H12O6

Fructose = C6H12O6

Galactose = C6H12O6

Determine the number of carbon, hydrogen, and oxygen atoms in a single molecule of raffinose. (1 mark)

18 carbon atoms, 32 hydrogen atoms, 16 oxygen atoms (two molecules of H2O are lost when forming the glycosidic bonds between the three monosaccharides)

14
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Describe the test for starch.

  • Add iodine solution to the sample.

  • If the solution goes from orange-brown to blue-black starch is present.

15
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Describe the test for lipids.

  • Emulsion test - add ethanol to the sample and shake the test tube then add water to it.

  • Lipids are present if a white emulsion appears.

16
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Describe the test for reducing sugars.

  • Add Benedict's reagent to the sample then heat the solution gently.

  • If it changes from light blue to a brick-red precipitate, indicates that reducing sugars are present.

17
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Describe the test for non-reducing sugars.

  • Add HCl to the sample and heat gently, neutralise the sample with NaHCO3 solution.

  • Add Benedict's reagent to the sample and heat in water bath

  • If change from light blue to brick-red precipitate, then result is positive.

18
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Describe the test for proteins.

  • Add Biuret solution

  • If proteins present, solution turns from blue to purple

19
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When the reaction with catalase is carried out in a test tube, the test tube feels warm at the end of the reaction. Explain why. (2 marks)

  • Energy in products less than energy in substrate

  • Energy is released as heat

20
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A student carried out the Benedict’s test. Suggest a method, other than using a colorimeter, that this student could use to measure the quantity of reducing sugar in a solution. (2 marks)

  • Filter and dry the sample

  • Find the mass

21
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<p><span>In an investigation, a student wanted to identify the solutions in two beakers, A and B. She knew one beaker contained maltose solution and the other beaker contained glucose solution. Both solutions had the same concentration.</span><br><span>She did two separate biochemical tests on a sample from each beaker.</span><br><span>Test 1 - used Benedict's solution to test for reducing sugar.</span><br><span>Test 2 - added the enzyme maltase, heated the mixture at 30 °C for 5 minutes, and then used Benedict's solution to test for reducing sugar.</span><br></p><p>Explain the results for beakers A and B in the table. (2 marks)</p>

In an investigation, a student wanted to identify the solutions in two beakers, A and B. She knew one beaker contained maltose solution and the other beaker contained glucose solution. Both solutions had the same concentration.
She did two separate biochemical tests on a sample from each beaker.
Test 1 - used Benedict's solution to test for reducing sugar.
Test 2 - added the enzyme maltase, heated the mixture at 30 °C for 5 minutes, and then used Benedict's solution to test for reducing sugar.

Explain the results for beakers A and B in the table. (2 marks)

  • A = glucose and B = maltose

  • Because more sugar after hydrolysis

22
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Use of a colorimeter in this investigation would improve the repeatability of the student’s results. Give one reason why. (1 mark)

  • Gives quantitative results

  • So standardises the method

23
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<p>Name the type of peptidase which will hydrolyse the bond labelled G in the diagram. (1 mark)</p>

Name the type of peptidase which will hydrolyse the bond labelled G in the diagram. (1 mark)

Endo(peptidase)

24
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Describe the induced-fit model of enzyme action and how an enzyme acts as a catalyst. (3 marks)

  • Substrate binds to the active site

  • Active site changes shape (slightly) so it is complementary to substrate

  • Reduces activation energy

25
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When bread becomes stale, the structure of some of the starch is changed. This changed starch is called retrograded starch. Scientists have suggested retrograded starch is a competitive inhibitor of amylase in the small intestine.

Assuming the scientists are correct, suggest how eating stale bread could help to reduce weight gain. (3 marks)

  • Less hydrolysis of starch

  • To maltose

  • So less absorption of glucose

26
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Describe how the structure of a protein depends on the amino acids it contains. (5 marks)

  • Structure is determined by (relative) position of amino acids

  • Primary structure is sequence/order of amino acids

  • Secondary structure formed by hydrogen bonding

  • Tertiary structure formed by interactions (between R groups)

  • Creates active site in enzymes

27
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Explain how the active site of an enzyme causes a high rate of reaction. (3 marks)

  • Lowers activation energy

  • Induced fit causes active site (of enzyme) to change shape

  • So enzyme-substrate complex causes bonds to form/break

28
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<p>A solution contained a mixture of three different amino acids. A scientist passed an electric current through the solution to separate the amino acids. She placed a drop of the mixture at one end of a piece of filter paper, attached an electrode to each end of the paper and switched on the current, then off the current after 20 minutes and stained the paper to show spots of the amino acids at new positions.</p><p>Her results are shown in the diagram.</p><p>Explain what the positions of the spots in the diagram show about these amino acids. (3 marks)</p>

A solution contained a mixture of three different amino acids. A scientist passed an electric current through the solution to separate the amino acids. She placed a drop of the mixture at one end of a piece of filter paper, attached an electrode to each end of the paper and switched on the current, then off the current after 20 minutes and stained the paper to show spots of the amino acids at new positions.

Her results are shown in the diagram.

Explain what the positions of the spots in the diagram show about these amino acids. (3 marks)

  • Moved to negative electrode because positively charged

  • Spots move different distances because (amino acids) different charge/mass

  • Two spots (not three) because these two amino acids same charge/mass

29
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The secondary structure of a polypeptide is produced by bonds between amino acids. Describe how. (2 marks)

  • Hydrogen bonds

  • Between NH (group of one amino acid) and C=O

    (OR Forming β pleated sheets / α helix)

30
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Formation of an enzyme-substrate complex increases the rate of reaction. Explain how. (2 marks)

  • Reduces activation energy

  • Due to bending bonds

31
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<p>Lyxose binds to the enzyme. </p><p>Suggest a reason for the difference in the results shown in the graph with (<strong>L</strong>) and without lyxose (<strong>C</strong>). (3 marks)</p>

Lyxose binds to the enzyme.

Suggest a reason for the difference in the results shown in the graph with (L) and without lyxose (C). (3 marks)

  • Binding alters the tertiary structure of the enzyme

  • This causes active site to change shape

  • So more successful E-S complexes form

32
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A change from Glu to Lys at amino acid 300 had no effect on the rate of reaction catalysed by the enzyme. The same change at amino acid 279 significantly reduced the rate of reaction catalysed by the enzyme.

Suggest reasons for the differences between the effects of these two changes.

  • Change at amino acid 300 does not change the shape of the active site

  • Amino acid 279 may have been involved in a (ionic, disulfide or hydrogen) bond and so the shape of the active site changes

33
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Suggest two variables the biochemist controlled when investigating the effect of temperature on the rate of breakdown of a protein by an enzyme. (2 marks)

  • Enzyme concentration

  • pH

34
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Use your knowledge of protein structure to explain why enzymes are specific and may be affected by non-competitive inhibitors. (6 marks)

  • Each enzyme has specific primary structure / amino acid sequence

  • Folds in a particular way / has particular tertiary structure giving an active site with a unique structure

  • Shape of active site complementary to substrate

  • Inhibitor fits at site on the enzyme other than active site

  • Distorts active site

  • So substrate will no longer fit to make enzyme-substrate complex

35
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Describe how a phosphodiester bond is formed between two nucleotides within a DNA molecule. (2 marks)

  • Condensation reaction

  • Between phosphate and deoxyribose

  • Catalysed by DNA polymerase

36
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Describe two similarities and two differences between the structure of a DNA molecule and the structure of a molecule of ATP. (4 marks)

Similarities:

  • Both contain adenine

  • Both contain a phosphate group

Differences:

  • ATP has 3 phosphates, DNA nucleotide has 1

  • ATP contains ribose, DNA contains deoxyribose

37
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Describe the structure of DNA. (5 marks)

  • Polymer of nucleotides

  • Each nucleotide formed from deoxyribose, a phosphate and a nitrogenous base

  • Phosphodiester bonds between nucleotides

  • Double helix held by hydrogen bonds

  • Hydrogen bonds between adenine, thymine and cytosine, guanine

38
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Use your knowledge of semi-conservative replication of DNA to suggest:

1) the role of the single-stranded DNA fragments

2) the role of the DNA nucleotides (3 marks)

Role of single-stranded DNA fragments:

  • Template

  • Determines order of nucleotides/bases

Role of DNA nucleotides:

  • Forms complementary pairs (A – T, G - C)

39
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Describe the role of two named enzymes in the process of semiconservative replication of DNA. (3 marks)

  • DNA helicase causes breaking of hydrogen bonds between DNA strands

  • DNA polymerase joins the DNA nucleotides

  • Forming phosphodiester bonds

40
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<p>Scientists investigated the function of a eukaryotic cell protein called cyclin A which is thought to be involved with the binding of one of the enzymes required at the start of DNA replication.</p><p>Treated cultures of cells in the following ways:</p><p>C – Control cells, untreated</p><p>D – Added antibody that binds specifically to cyclin A</p><p>E – Added RNA that prevents translation of cyclin A</p><p>F – Added RNA that prevents translation of cyclin A and added cyclin A protein</p><p>They then determined the percentage of cells in each culture in which DNA was replicating. Their results are shown in the table.</p><p>Suggest explanations for the results in the table. (3 marks)</p>

Scientists investigated the function of a eukaryotic cell protein called cyclin A which is thought to be involved with the binding of one of the enzymes required at the start of DNA replication.

Treated cultures of cells in the following ways:

C – Control cells, untreated

D – Added antibody that binds specifically to cyclin A

E – Added RNA that prevents translation of cyclin A

F – Added RNA that prevents translation of cyclin A and added cyclin A protein

They then determined the percentage of cells in each culture in which DNA was replicating. Their results are shown in the table.

Suggest explanations for the results in the table. (3 marks)

  • Treatment D - Antibody binds to cyclin A so it cannot bind to DNA and initiate DNA replication

  • Treatment E - RNA interferes with polypeptide formation (so cyclin A not made)

  • In Treatment F added cyclin A can bind to enzyme to initiate DNA replication

41
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<p>The arrows in Figure 2 show the directions in which each new DNA strand is being produced.</p><p>Use Figure 1, Figure 2 and your knowledge of enzyme action to explain why the arrows point in opposite directions.</p>

The arrows in Figure 2 show the directions in which each new DNA strand is being produced.

Use Figure 1, Figure 2 and your knowledge of enzyme action to explain why the arrows point in opposite directions.

  • Figure 1 shows that DNA has antiparallel strands

  • Also shows shape of the nucleotides is different

  • Enzymes have active sites with specific shape

  • Only substrates with complementary shape can bind with active site of enzyme / DNA polymerase.

42
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Write a simple equation to show how ATP is synthesised from ADP. (1 mark)

ADP + Pi ATP

43
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ATP is useful in many biological processes. Explain why. (4 marks)

  • Releases energy in small, manageable amounts

  • Broken down in one step

  • Immediate energy compound

  • Can be made again

44
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Describe how an ATP molecule is formed from its component molecules. (4 marks)

  • Adenine, ribose, three phosphates

  • Joined in a condensation reaction

  • Catalysed by ATP synthase

45
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<p>The diagram shows the structure of molecules found in organisms</p><p>Complete the table by putting the correct letter, A, B, C or D, in the box next to each statement. Each letter may be used once, more than once, or not at all. (4 marks)</p>

The diagram shows the structure of molecules found in organisms

Complete the table by putting the correct letter, A, B, C or D, in the box next to each statement. Each letter may be used once, more than once, or not at all. (4 marks)

B

D

C

B

46
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State and explain the property of water that can help buffer changes in temperature. (2 marks)

  • Water has a relatively high specific heat capacity

  • Can gain / lose a lot of heat without changing temperature

47
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Explain five properties that make water important for organisms. (5 marks)

  • A metabolite in photosynthesis/respiration

  • A solvent so allowing transport of substances

  • High specific heat capacity so buffers changes in temperature

  • Large latent heat of vaporisation so provides a cooling effect through evaporation

  • Cohesion between water molecules so supports columns of water (in plants)

48
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Give two properties of water that are important in the cytoplasm of cells. For each property of water, explain its importance in the cytoplasm.

  • Universal solvent

    • Metabolic reactions occur faster in solution

  • Reactive

    • Takes place in hydrolysis / condensation

49
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A high concentration of sodium in the blood can affect blood volume and cause hypertension.

Use your knowledge of water potential to suggest how high sodium concentrations in the medicines taken could affect blood volume. (3 marks)

  • Sodium ions lower the water potential of blood

  • Water would move into the blood by osmosis

  • Increasing the blood volume

50
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Describe the roles of iron ions, sodium ions, and phosphate ions in cells. (5 marks)

Iron ions:

  • Haemoglobin binds/associates with oxygen

Sodium ions:

  • Co-transport of glucose/amino acids (into cells)

  • Affects osmosis/water potential

Phosphate ions:

  • Phosphorylates other compounds making them more reactive

  • Hydrophilic part of phospholipid bilayer