Topic 1: Biological Molecules Exam Questions Flashcards

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1
Q

Explain why water is considered so important for life to occur. [5]

A
  1. A metabolite in condensation/hydrolysis/photosynthesis/respiration;
  2. A solvent so (metabolic) reactions can occur
    OR
    A solvent so allowing transport of substances;
  3. High (specific) heat capacity so buffers changes in temperature;
  4. Large latent heat of vaporisation so provides a cooling effect (through evaporation);
  5. Cohesion (between water molecules) so supports columns of water (in plants);
  6. Cohesion (between water molecules) so produces surface tension supporting (small) organisms.
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2
Q

State and explain the property of water that helps to prevent temperature increase in a cell. [2]

A
  1. High (specific) heat capacity;
  2. Buffers changes in temperature.
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3
Q

State and explain the property of water that can help to buffer changes in temperature. [2]

A
  1. (water has a relatively) high (specific) heat capacity;
  2. Can gain / lose a lot of heat / energy without changing temperature;
    OR
    Takes a lot of heat / energy to change temperature.
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4
Q

Give two properties of water that are important in the cytoplasm of cells.
For each property of water, explain its importance in the cytoplasm. [4]

A
  1. Polar molecule;
  2. Acts as a (universal) solvent;
    OR
  3. (Universal) solvent;
  4. (Metabolic) reactions occur faster in solution;
    OR
  5. Reactive;
  6. Takes place in hydrolysis / condensation reaction.
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5
Q

Describe how you could confirm the presence of non-reducing sugar in a solution. [3]

A
  1. Do Benedict’s test and stays blue/negative;
  2. Boil with acid then neutralise with alkali;
  3. Heat with Benedict’s solution/reagent and becomes orange/red (precipitate)
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6
Q

What is the test for starch? [2]

A
  1. Add iodine in potassium iodide solution;
  2. Positive result: goes blue-black.
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7
Q

Starch is a carbohydrate often stored in plant cells.

Describe and explain two features of starch that make it a good storage molecule. [2]

A

Any two from:
1. Branched / coiled / helical so makes molecule compact;
2. Polymer of alpha glucose so proves glucose for respiration;
3. Insoluble (in water) so doesn’t affect osmosis;
4. Branched/more ends, for fast breakdown/enzyme action so glucose readily released;
5. Large (molecule), so can’t cross the cell surface membrane.

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8
Q

Describe four ways in which the structure of starch is different from the structure of cellulose. [4]

A

Starch: Cellulose:
1. Contains alpha glucose Contains beta glucose;
2. Helical/coiled Straight;
3. Branched Unbranched;
4. No microfibrils Can form microfibrils;
5. All glucose / monomers Alternate glucose / monomers
same way up upside down.

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9
Q

Describe three ways in which the structure of cellulose is similar to the structure of starch. [3]

A
  1. Both polysaccharides / glucose polymers / made of glucose monomers;
  2. Both contain glycosidic bonds (between monomers);
  3. Both contain carbon, hydrogen and oxygen / C, H and O.
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10
Q

The structure of cellulose is related to its role in plant cell walls. Explain how. [3]

A
  1. Long / straight / unbranched chains (of glucose);
  2. (joined to each other by) hydrogen bonds;
  3. Form microfibrils;
  4. Provide rigidity / strength /support.
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11
Q

Compare and contrast the structure of cellulose and the structure of glycogen. [6]

A
  1. Both polysaccharides / glucose polymers / made of glucose monomers;
  2. Both contain glycosidic bonds;
  3. Both contain carbon, hydrogen and oxygen/C, H and O;
  4. Glycogen has alpha-glucose and cellulose has beta-glucose;
  5. Glycogen is helical and cellulose is straight;
  6. Glycogen is branched and cellulose is straight;
  7. Cellulose can form microfibrils and glycogen does not;
  8. Glycogen has glucose (monomers) the same way up and cellulose has alternate glucose (monomers) flipped upside down.
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12
Q

Describe the chemical reactions involved in the conversion of polymers to monomers and monomers to polymers.

Give two named examples of polymers and their associated monomers to illustrate your answer. [5]

A
  1. A condensation reaction joins monomers together and forms a bond and releases water;
  2. A hydrolysis reaction breaks a bond between monomers and uses water;
    3/4. Amino acid and polypeptide/protein/enzyme
    OR Nucleotide and polynucleotide
    OR Alpha glucose and starch/glycogen
    OR Beta glucose and cellulose;
  3. Reference to a correct bond within a named polymer
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13
Q

What is a monomer? [1]

A

A (smaller, repeating) unit/molecule from which larger molecules/polymers are made.

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14
Q

Explain one way in which starch molecules are adapted for their function in plant cells. [2]

A
  1. Insoluble;
  2. Doesn’t affect water potential;
    OR
  3. Helical;
  4. Compact;
    OR
  5. Large molecule;
  6. Cannot leave cell.
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15
Q

Explain how cellulose molecules are adapted for their function in plant cells. [3]

A
  1. Long and straight chains;
  2. Become linked together by many hydrogen bonds to form microfibrils;
  3. Provide strength (to cell wall).
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16
Q

Describe how a triglyceride molecule is formed. [3]

A
  1. One glycerol and three fatty acids;
  2. Condensation (reactions) and removal of three molecules of water;
  3. Ester bonds (formed)
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17
Q

Here is the structure of a molecule of glycerol and a molecule of fatty acid.
Draw a diagram to show the structure of a triglyceride molecule. [2]
H
|
H — C — OH
|
H — C — OH R — COOH
|
H — C — OH
|
H

Glycerol Fatty acid

A

3 fatty acids attached;
Ester bonds correct

     H                  
     | H — C — OOCR
     | H — C — OOCR
     | H — C — OOCR
     |
     H
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18
Q

Organisms that move, e.g. animals, and parts of organisms that move, e.g. some plant seeds, use lipids rather than carbohydrates as an energy store. Suggest one reason why this is so. [3]

A
  1. Lipids provide more energy per gram than carbohydrates when aerobically respired;
  2. Therefore a lower mass of lipid needs to be stored relative to carbohydrate;
  3. Lipid is therefore a lighter energy storage product - major advantage if an organism (or see) needs to move around.
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19
Q

Describe how a saturated fatty acid is different from an unsaturated fatty acid. [1]

A

Saturated single/no double bonds (between carbons)

OR

Unsaturated has (at leat one) double bond (between carbons)

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20
Q

Explain the arrangement of phospholipids in a cell-surface membrane. [2]

A
  1. Bilayer;
  2. Hydrophobic (fatty acid) tails point away/are repelled from water
    OR
    Hydrophilic (phosphate) heads point to/are attracted to water.
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21
Q

Compare and contrast the structure and properties of triglycerides and phospholipids. [5]

A
  1. Both contain ester bonds;
  2. Both contain glycerol;
  3. Fatty acids on both may be saturated or unsaturated;
  4. Both are insoluble in water;
    5.Both contain C, H and O but phospholipids also contain P;
  5. Triglycerides have three fatty acids and phospholipids have two fatty acids plus a phosphate group;
  6. Triglycerides are hydrophobic/non-polar and phospholipids have hydrophobic and hydrophilic regions;
  7. Phospholipids form bilayer (in water) but triglycerides don’t.
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22
Q

The general structure of a fatty acid is RCOOH.

Name the group represented by COOH. [1]

A

Carboxyl

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23
Q

Describe how you would test for the presence of a lipid in a liquid sample of food. [2]

A
  1. Add ethanol then add water and shake;
  2. White/milky (emulsion)
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24
Q

Describe how an ester bond is formed in a phospholipid molecule. [2]

A
  1. Condensation (reaction);
  2. Between glycerol and fatty acid.
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25
Q

Some seeds contain lipids. Describe how you could use the emulsion test to show that a seed contains lipids. [3]

A
  1. Crush/grind;
  2. With ethanol;
  3. Then add water;
  4. Forms white/milky emulsion.
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26
Q

What is an unsaturated fatty acid? [2]

A
  1. Double bond(s);
  2. Between carbon
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27
Q

Why are hydrogen bonds important in cellulose molecules? [2]

A
  1. Holds chains/cellulose molecules together
    OR Forms microfibrils
  2. Hydrogen bonds strong in large numbers.
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28
Q

A starch molecule has a spiral shape. Explain why this shape is important to its function in cells. [1]

A

Compact / occupies small space / tightly packed

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29
Q

Describe how a peptide bond is formed between two amino acids to form a dipeptide. [2]

A
  1. Condensation (reaction)/loss of water;
  2. Between amine and carboxyl.
30
Q

Describe how monomers join to form the primary structure of a protein. [3]

A
  1. Condensation reaction between amino acids;
  2. (Forming) peptide bonds;
  3. Creating (specific) sequence/order of amino acids.
31
Q

The secondary structure of a polypeptide is produced by bonds between amino acids. Describe how. [2]

A
  1. Hydrogen bonds;
  2. Between NH (group of one amino acid) and C=O (group).
32
Q

It is possible for two proteins to have the same number and type of amino acids, but different tertiary structures. Explain why. [2]

A
  1. Different primary structure/sequence of amino acids;
  2. Forms ionic / hydrogen / disulfide bonds in different places.
33
Q

Describe how the structure of a protein depends on the amino acids it contains. [5]

A
  1. Structure is determined by (relative) position of amino acids/R groups/interactions;
  2. Primary structure is sequence/order of amino acids;
  3. Secondary structure formed by hydrogen bonding (between amino acids);
  4. Tertiary structure formed by interactions (between R groups);
  5. Creates active site in enzymes
    OR Creates complementary/specific shapes in antibodies/carrier proteins/receptor (molecules);
  6. Quaternary structure contains >1 polypeptide chain
    OR Quaternary structure formed by interactions/bonds between polypeptides.
34
Q

Describe a biochemical test to confirm the presence of protein in a solution. [2]

A
  1. Add biuret (reagent);
  2. (Positive result) purple/lilac
35
Q

A dipeptide consists of two amino acids joined by a peptide bond. Dipeptides may differ in the type of amino acids they contain.

Describe two other ways in which all dipeptides are similar and one way in which they might differ. [3]

A

Similarities:
1. Amine (group at end);
2. Carboxyl (group at end);
3. Two R groups;
4. All contain C and H and N and O;

Difference:
5. Variable/different R group(s).

36
Q

Describe how you would test a piece of food for the presence of lipid. [2]

A
  1. Dissolve in ethanol, then add water and shake;
  2. White emulsion shows presence of lipid
37
Q

An electric current is passed through a solution containing a mixture of 3 different amino acids to separate the amino acids.

A drop of the mixture is placed at one end of a piece of filter paper, attached to an electrode at each end and the current is switched on. The current is switched off after 20 minutes and the spots of the amino acids at new positions is stained.

There are two dots and both have moved towards the negative electrode. Explain what the positions of the spots show about these amino acids. [3]

A
  1. Moved to negative (electrode) because positive(ly charged);
  2. (Spots move) different distances/rates because (amino acids) different charge/mass;
  3. Two spots (not three) because (amino acids) same charge/mass
    OR
    One spot has 2 amino acids because (amino acids) same charge/mass.
38
Q

Describe the induced-fit model of enzyme action. [2]

A
  1. (before reaction) active site not complementary;
  2. Shape of active site changes as substrate binds/as enzyme-substrate complex forms;
  3. Stressing/distorting/bending bonds (in substrate leading to reaction).
39
Q

Formation of an enzyme-substrate complex increases the rate of reaction.

Explain how. [2]

A
  1. Reduces activation energy;
  2. Due to bending bonds
    OR
    Without enzyme, very few substrates have sufficient energy for reaction.
40
Q

Explain what is meant by a polymer. [1]

A

(Molecule) made up of many identical/similar molecules / monomers / subunits

41
Q

Give one example of a biologically important polymer other than starch or protein. [1]

A

Cellulose / glycogen / nucleus acid / DNA / RNA

42
Q

In an investigation, the enzyme amylase was mixed in a test tube with a buffer solution and a suspension of starch. The amylase broke down the starch to maltose. When all the starch had been broken down, a sample was removed from the test tube and tested with biuret reagent.

(i) Explain why a buffer solution was added to the amylase-starch mixture. [2]
(ii) What colour would you expect the sample to go when tested with biuret reagent? [1]
(iii) Give an explanation for your answer to part (ii). [2]

A

(i) 1. To keep pH constant
2. A change in pH will slow the rate of reaction / denature the amylase.

(ii) Purple/lilac.

(iii) 1. Protein present / the enzyme/amylase is a protein;
2. Not used up in the reaction / still present at the end of the reaction.

43
Q

An enzyme has an optimum pH of 6.

Explain why the rate of reaction of the enzyme is lower at pH 5. [3]

A
  1. (change in pH) leads to breaking of bonds holding tertiary structure / changes charge on amino acids;
  2. Enzyme/protein/active site loses shape/denatured;
  3. Substrate will not bind with active site / fewer/no enzyme-substrate complexes formed.
44
Q

Explain why an enzyme that works over a larger range of pH would be better for using in washing powders. [1]

A

More resistant to changes in pH and washing conditions variable / works in alkaline pH and washing powders alkaline

45
Q

A competitive inhibitor decreases the rate of enzyme-controlled reaction.

Explain how. [3]

A
  1. Inhibitor similar shape to substrate;
  2. Fits/binds to active site;
  3. Prevents/reduces enzyme-substrate complex forming.
46
Q

Describe how a non-competitive inhibitor can reduce the rate of an enzyme-controlled reaction. [3]

A
  1. Attaches to the enzyme at a site other than the active site;
  2. Changes (shape of) the active site;
  3. (So active site and substrate) no longer complementary so less/no enzyme-substrate complexes form
47
Q

Give four differences between competitive and non-competitive inhibitors. [4]

A

Competitive inhibitors: Non-competitive inhibitors:
1. Similar shape to substrate Different shape to substrate;
2. Binds to active site Binds to allosteric site;
3. Doesn’t change the shape of Changes the shape of the
the active site active site;
4. Increasing substrate concentration Increasing substrate concentration
can overcome inhibition cannot overcome inhibition
(Vmax unchanged) (lower Vmax)

48
Q

Describe the induced-fit model of enzyme action and how an enzyme acts as a catalyst. [3]

A
  1. Substrate binds to the active site
    OR Enzyme-substrate complex forms;
  2. Active site changes shape (slightly) so it is complementary to substrate OR Active site changes shape (slightly) so distorting/breaking/forming bonds in the substrate;
  3. Reduces activation energy.
48
Q

Explain how the active site of an enzyme causes a high rate of reaction. [3]

A
  1. Lowers activation energy;
  2. Induced fit causes active site to change shape;
  3. (So) enzyme-substrate complex causes bonds to form/break
49
Q

Lactose is a disaccharide sugar which can be broken down by the enzyme lactase into two monosaccharides, glucose and galactose. What is the formula for lactose? [2]

A

C12H22O11

50
Q

Glucose and what other monosaccharide form sucrose? [1]

A

Fructose

51
Q

The enzyme sucrase catalyses the breakdown of sucrose into monosaccharides. What type of reaction is this breakdown? [1]

A

Hydrolysis

52
Q

Name the monomers that form a cellulose molecules. [1]

A

Beta glucose

53
Q

Starch molecules and cellulose molecules have different functions in plant cells. Each molecule is adapted for its function. Describe how the structure of starch and cellulose molecules are related to their function. [5]

A

Starch (max 3):
1. Helical so compact;
2. Large (molecule)/insoluble so osmotically inactive;
3. Branched so glucose is (easily) released for respiration;
4. Large (molecule) so cannot leave cell/cross cell-surface membrane;

Cellulose (max 3):
5. Long, straight/unbranched chains of beta glucose;
6. Joined by hydrogen bonding;
7. To form microfibrils;
8. Provides rigidity/strength.

54
Q

Describe the structure of glycogen. [2]

A
  1. Polysaccharide/polymer of alpha glucose;
  2. Joined by glycosidic bonds
    OR Branched structure.
55
Q

Suggest how glycogen acts as a source of energy. [2]

A
  1. Hydrolysed (to glucose);
  2. Glucose used in respiration.
56
Q

Suggest one advantage of using a pH meter rather than a pH indicator. [1]

A

Numerical readings / not subjective / colour change subjective / gives quantitative data / not qualitative / gives continuous data / greater accuracy.

57
Q

Explain why the pH decreases when lipase is added to milk. [1]

A

Fatty acids produced

58
Q

Some seeds contain lipids. Describe how you could use the emulsion test to show that a seed contains lipids. [3]

A
  1. Crush/grind;
  2. With ethanol;
  3. Then add water;
  4. Forms emulsion / goes white/milky.
59
Q

The structure of a phospholipid molecule is different from that of a triglyceride. Describe how a phospholipid is different. [2]

A
  1. Phosphate group;
  2. Instead of one of the fatty acids / and two fatty acids.
60
Q

What is meant by an unsaturated fatty acid? [2]

A
  1. Double bond(s) between carbon atoms;
  2. In fatty acid.
61
Q

What is the meaning of hydroxylating? [1]

A

Addition of hydroxyl/OH group/

62
Q

What two substances are formed when two amino acid molecules join together? [1]

A

Dipeptide and water.

63
Q

Explain how a change in the primary structure of a globular protein may result in a different three-dimensional structure. [3]

A
  1. Sequence of amino acids changes;
  2. Tertiary structure changes/folds in a different way;
  3. Bonds form in different places.
64
Q

What is an enzyme? [2]

A
  1. Protein;
  2. Catalyst;
  3. (For reaction involving a) specific substrate;
  4. Lowers activation energy.
65
Q

Explain why maltase:
- only breaks down maltose
- allows this reaction to take place at normal body temperature. [5]

A
  1. Tertiary structure of enzyme (means);
  2. Active site complementary to maltose/substrate / maltose fits into active site;
  3. As substrate binds, active site changes shape to become fully complementary;
  4. Enzyme is a catalyst/lowers activation energy;
  5. By forming enzyme-substrate complex
    Accept idea that binding stresses the bonds so more easily broken
66
Q

Describe the induce-fit model of enzyme action. [2]

A
  1. (Before reaction) active site not complementary to substrate;
  2. Shape of active site changes as substrate binds/as enzyme-substrate complex forms;
  3. Stressing/distorting/bending bonds (in substrate leading to reaction.
67
Q

Give a suitable control for an experiment looking into the effect of an enzyme. [1]

A

Use boiled enzyme.

68
Q

Describe how a non-competitive inhibitor can reduce the rate of an enzyme-controlled reaction. [3]

A
  1. Attaches to the enzyme at a site other than the active site/allosteric site;
  2. Changes (shape of) the active site
    OR Changes tertiary structure (of enzyme);
  3. (So active site and substrate) no longer complementary so less/no substrate can fit/bind.
69
Q

The formation of an enzyme-substrate complex increases the rate of reaction. Explain how. [2]

A
  1. Reduces activation energy;
  2. Due to bending bonds
    OR
    Without enzyme, very few substrates have sufficient energy for reaction.
70
Q

Galactose has a similar structure to part of the lactose molecule. Explain how galactose inhibits lactase. [2]

A
  1. Galactose is a competitive inhibitor/attaches to the active site (of lactase);
  2. Fewer enzyme-substrate complexes formed.