Exam Questions by Topic Chapter 1 Biology Flashcards
(53 cards)
Describe the chemical reactions involved in the conversion of polymers to monomers and monomers to polymers. (2)
1) A condensation reaction joins two monomers together, forms a chemical bond and releases water.
2) A hydrolysis reaction splits two monomers apart by breaking the chemical bond and uses water.
What is a monomer? (1)
1) A repeating unit that forms polymers.
A biochemical test for reducing sugar produces a negative result with raffinose solution.
Describe a biochemical test to show that raffinose solution contains a non-reducing sugar. (3)
1) Heat with acid and neutralise.
2) Heat with Benedict’s solution.
3) Red precipitate forms.
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)
1) Filter and dry the precipitate.
2) Find the mass.
Use of a colorimeter in this investigation would improve the repeatability of the student’s results. Give one reason why
1) Colorimeter is quantitative.
Describe three differences between the structure of a cellulose molecule and a glycogen molecule.
1) Cellulose is made up of beta glucose, glycogen is made up of alpha glucose.
2) Cellulose is a straight chain, glucose is branched.
3) Glycogen has 1,4 and 1,6 glycosidic bonds, cellulose has only 1,4 glycosidic bonds.
Describe and explain five features of starch that make it a good storage molecule.
1) Insoluble so it doesn’t affect the water potential.
2) Branched so it is more compact.
3) Polymer of alpha glucose so it provides glucose for respiration.
4) Branched so more ends for enzyme action.
5) Large so can’t cross the cell membrane.
Describe the structure of glycogen. (3)
1) Polysaccharide of alpha glucose.
2) Joined by glycosidic bonds.
3) Branched.
Suggest how glycogen acts as a source of energy. (2)
1) Hydrolysed into glucose.
2) Glucose is used in respiration.
Explain the difference in the structure of the starch molecule and the cellulose molecule. (2)
1) Cellulose is made up of beta glucose, starch is made up of alpha glucose.
2) Position of the hydrogen and hydroxyl group are inverted on carbon 1.
Explain how cellulose molecules are adapted for their function in plant cells. (3)
1) Long and straight chains.
2) Joined by hydrogen bonds to form fibrils.
3) Provides strength to cell wall.
Describe how a triglyceride molecule is formed. (3)
1) One glycerol and three fatty acids.
2) Condensation reaction and removal of 3 water molecules.
3) Ester bonds formed.
Describe how an ester bond is formed in a phospholipid molecule. (2)
1) Condensation reaction
2) Between glycerol and fatty acid.
Describe the induced-fit model of enzyme action and how an enzyme acts as a catalyst. (3)
1) Substrate binds to active site.
2) Active site changes shape so it is complementary to the substrate.
3) Reducing the activation energy.
A competitive inhibitor decreases the rate of an enzyme-controlled
reaction. Explain how. (3)
1) Inhibitor is similar in shape to the substrate.
2) Binds to active site.
3) Prevents enzyme-substrate complex from forming.
Describe how the structure of a protein depends on the amino acids it contains. (5)
1) Structure is determined by R groups.
2) Primary structure is the sequence of amino acids.
3) Secondary structure formed by hydrogen bonding.
4) Tertiary structure formed by ionic bonds, hydrogen bonds and disulfide bridges.
5) Quaternary structure contains more than one polypeptide.
Explain how the active site of an enzyme causes a high rate of reaction. (3)
1) Lowers activation energy.
2) Induced fit causes active site of enzymes to change shape.
3) So enzyme-substrate complex causes bonds to form.
Describe how amino acids join to form a polypeptide so there is always NH2 at one end and COOH at the other end. (2)
1) One amine group joins to a carboxyl group to form a peptide bond.
2) So in the chain there is a free amine group on one end and a free carboxyl group on the other end.
Describe two other ways in which all dipeptides are similar and one way in which they might differ. (3)
Similarities:
1) Amine group at one end.
2) Carboxyl group at the other end.
Difference:
1) Different R group.
Describe how a non-competitive inhibitor can reduce the rate of an
enzyme-controlled reaction. (3)
1) Attaches to the allosteric site.
2) Changes shape of active site.
3) Substrate is no longer complementary so no substrate can bind.
Describe how a peptide bond is formed between two amino acids to form a dipeptide. (2)
1) Condensation reaction
2) Between amine group and carboxyl group.
The secondary structure of a polypeptide is produced by bonds between amino acids. (2)
1) Hydrogen bonds.
2) Forming alpha helix or beta pleated sheets.
Two proteins have the same number and type of amino acids but different tertiary structures. (2)
1) Different sequence of amino acids.
2) Forms ionic, hydrogen and disulfide bonds in different places.
Formation of an enzyme-substrate complex increases the rate of reaction. Explain how. (2)
1) Reduces activation energy.
2) Due to bending bonds.