mark scheme standard Flashcards
Suggest one way the structure of the chromosome could differ along its length to result in the stain binding more in some areas (1 mark)
Different base sequence
What is a homologous pair of chromosomes
2 chromosomes that carry the same genes
Give 2 ways in which the arrangement of prokaryotic DNA is different from the arrangement of the human DNA (2 marks)
Prokaryotic DNA is :
- Circular ( as opposed to linear)
- Not associated with proteins
- Only one piece of DNA
Describe how 1 amino acid is added to a polypeptide that is being formed at a ribosome during translation (3 marks)
- tRNA brings specific amino acid to ribosome
- Anticodon on tRNA binds to codon on mRNA
- Amino acids join by condensation reaction
Explain how the formation of an enzyme-substrate complex increase the rate of reaction
- Reduces activation energy
- Due to bending bonds / without enzyme, very few substrates have sufficient energy for reaction
Suggest why iron-deficient plants have a reduced growth rate
- Less thylakoid membrane
- Smaller SA / less chlorophyll
- So reduced light absorption
- So slower rate of photosynthesis
What is a monomers
Molecule from which larger molecules are made
Describe a biochemical test to show a solution contains a non-reducing sugar (3 marks)
- Heat with acid and neutralise
- Heat with Benedict’s solution
- Red precipitate
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 precipitate);
Accept: correct reference to evaporation after filtration
2. Find mass/weight;
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 (monomers); - Cellulose molecule has straight chain and glycogen is
branched; - Cellulose molecule has straight chain and glycogen is coiled;
- glycogen has 1,4- and 1,6- glycosidic bonds and cellulose has
only 1,4- glycosidic bonds;
Ignore ref. to H bonds / microfibrils
Describe and explain two features of starch that make it a good storage molecule. (2 marks)
- Insoluble (in water), so doesn’t affect water potential;
- Branched / coiled / (α-)helix, so makes molecule compact;
OR
Branched / coiled / (α-)helix so can fit many (molecules) in
small area; - Polymer of (α-)glucose so provides glucose for respiration;
- Branched / more ends for fast breakdown / enzyme action;
- Large (molecule), so can’t cross the cell membrane
Require feature and explanation for 1 mark
- Accept Ψ or WP
- Accept Insoluble so doesn’t affect osmosis
- Do not allow ref to ‘doesn’t affect water leaving cells
- Ignore ‘surface area’
- Accept ‘branched so glucose readily released’
Describe the structure of glycogen
- Polysaccharide of α-glucose;
OR
polymer of α-glucose; - (Joined by) glycosidic bonds
OR
Branched structure
During early pregnancy, the glycogen in the cells lining the uterus is an important energy source for the embryo.
Suggest how glycogen acts as a source of energy.
Do not include transport across membranes in your answer.
- Hydrolysed (to glucose);
- Glucose used in respiration;
1. Ignore ‘Broken down’
2. ‘Energy produced’ disqualifies mp2
Name the monomers from which a maltose molecule is made
Glucose (and glucose);
Name the type of chemical bond that joins the two monomers to form maltose.
(α1,4) Glycosidic;
Explain the difference in the structure of the starch molecule and the
cellulose molecule (2 marks)
- Starch formed from α-glucose but cellulose formed from β-glucose;
- Position of hydrogen and hydroxyl groups on carbon atom 1
inverted.
Explain one way in which starch molecules are adapted for their function in plant cells. (2 marks)
- Insoluble;
- Don’t affect water potential;
OR - Helical;
Accept form spirals - Compact;
OR - Large molecule;
- Cannot leave cell.
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
fibrils; - Provide strength (to cell wall).
Name the group represented by COOH
Carboxyl;
Describe how you would test for the presence of a lipid in a liquid sample of food.
- Add ethanol/alcohol then add water and shake/mix
OR
Add ethanol/alcohol and shake/mix then pour into/add water;
Reject heating the emulsion test
Accept ‘Add Sudan III and mix’
Ignore a second shake - White/milky (emulsion)
OR
(emulsion) test turns white/milky;
Ignore cloudy
Reject precipitate
Accept (for Sudan III) top (layer) red
Describe how a triglyceride molecule is formed
- One glycerol and three fatty acids;
- Condensation (reactions) and removal of three molecules of water;
- Ester bond(s) (formed);
Describe how an ester bond is formed in a phospholipid molecule.
- Condensation (reaction)
OR
Loss of water; - Between of glycerol and fatty acid;
Describe the induced-fit model of enzyme action and how an enzyme acts as a catalyst (3 marks)
- Substrate binds to the active site/enzyme
OR
Enzyme-substrate complex forms;
Accept for ‘binds’, fits - 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; - Reduces activation energy;
A competitive inhibitor decreases the rate of an enzyme-controlled
reaction.
Explain how.
(3 marks)
- Inhibitor similar shape to substrate;
Reject same shape
Accept ‘complementary to active site’ - Fits/binds to active site;
- Prevents/reduces enzyme-substrate complex forming
Describe how the structure of a protein depends on the amino acids it contains. (5 marks)
- Structure is determined by (relative) position of amino acid/R
group/interactions;
Accept for ‘interactions’, hydrogen bonds / disulfide
bridges / ionic bonds / hydrophobic hydrophilic
interactions - Primary structure is sequence/order of amino acids;
- Secondary structure formed by hydrogen bonding (between amino
acids);
Accept alpha helix/β-pleated sheet for ‘secondary
structure’ - Tertiary structure formed by interactions (between R groups);
Accept for ‘interactions’, hydrogen bonds / disulfide
bridges / ionic bonds / hydrophobic hydrophilic
interactions - Creates active site in enzymes
OR
Creates complementary/specific shapes in antibodies/carrier
proteins/receptor (molecules); - Quaternary structure contains >1 polypeptide chain
OR
Quaternary structure formed by interactions/bonds between
polypeptides;
Accept for ‘intereactions’, hydrogen bonds/ disulfide
bridges/ionic bonds/hydrophobichydrophilic
interactions
Accept prosthetic (group)
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;
Accept: description, of induced fit
Accept: enzyme-substrate complex causes stress/strain
on bonds.
Describe a biochemical test to confirm the presence of protein in a
solution. (2 marks)
- Add biuret (reagent);
Accept sodium hydroxide (solution) and copper sulphate
(solution)
Reject addition of other incorrect chemicals - (Positive result) purple/lilac/violet /mauve;
Reject other colours
Ignore references to heating
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 marks)1
Similarities
1. Amine/NH2 (group at end);
Accept amino/NH3+
2. Carboxyl/COOH (group at end);
Accept carboxylic / COO−
3. Two R groups;
4. All contain C and H and N and O;
Differences
Variable/different R group(s);
Describe how a non-competitive inhibitor can reduce the rate of an
enzyme-controlled reaction. (3 marks)
- Attaches to the enzyme at a site other than the active site;
Accept ‘attaches to allosteric/inhibitor site’ - Changes (shape of) the active site
OR
Changes tertiary structure (of enzyme); - (So active site and substrate) no longer complementary so
less/no substrate can fit/bind;
Describe how a peptide bond is formed between two amino acids to form a dipeptide. (2 marks)
- Condensation (reaction) / loss of water;
Accept each marking point if shown clearly in diagram. - Between amine / NH2 and carboxyl / COOH;
The secondary structure of a polypeptide is produced by bonds between amino acids.
Describe how.
(2 marks)
- Hydrogen bonds;
Accept as a diagram
Reject N - - - C / ionic / disulfide bridge / peptide bond - Between NH (group of one amino acid) and C=O (group);
OR
Forming β pleated sheets / α helix;
Two proteins have the same number and type of amino acids but different tertiary structures.
Explain why
(2 marks)
- Different sequence of amino acids
OR
Different primary structure; - Forms ionic / hydrogen / disulfide bonds in different places;
Formation of an enzyme-substrate complex increases the rate of reaction.
Explain how.
- Reduces activation energy;
Accept ‘reduces Ea’. - Due to bending bonds
OR
Without enzyme, very few substrates have sufficient energy for
reaction;
Describe the structure of DNA (5 marks)
- Polymer of nucleotides;
Accept ‘Polynucleotide’
Accept for ‘phosphate’. phosphoric acid - Each nucleotide formed from deoxyribose, a phosphate (group) and an organic/nitrogenous base;
- Phosphodiester bonds (between nucleotides);
- Double helix/2 strands held by hydrogen bonds;
- (Hydrogen bonds/pairing) between adenine, thymine and
cytosine, guanine;
Describe how a phosphodiester bond is formed between two nucleotides within a DNA molecule.
- Condensation (reaction)/loss of water;
- (Between) phosphate and deoxyribose;
- (Catalysed by) DNA polymerase;
In the process of semi-conservative DNA replication, the two strands within a DNA molecule are separated. Each then acts as a template for the formation of a new complementary strand.
Describe how the separation of strands occurs. (2 marks)
- DNA helicase;
- Breaks hydrogen bonds between base pairs/ AT and GC/complementary bases
OR
Breaks hydrogen bonds between polynucleotide strands;
Reject hydrolysis of hydrogen bonds
Describe the role of DNA polymerase in the semi-conservative replication of DNA. (2 marks)
- Joins (adjacent DNA) nucleotides;
Reject suggestions that it forms hydrogen bonds or
joins complementary bases.
Reject ‘nucleotide bases’. - (Catalyses) condensation (reactions);
- (Catalyses formation of) phosphodiester bonds (between adjacent
nucleotides)
Name the two scientists who proposed models of the chemical structure of DNA and of DNA replication.
Watson and Crick
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
1. Template;
2. Determines order of nucleotides/bases;
Role of DNA nucleotides
3. Forms complementary pairs / A – T, G - C
OR
Forms complementary (DNA) strand;
Ignore forms complementary bases
Accept sequence/ chain for strand
Give two features of DNA and explain how each one is important in the semi-conservative replication of DNA (2 marks)
- Weak / easily broken hydrogen bonds between bases allow two
strands to separate / unzip;
may appear in the same feature - Two strands, so both can act as templates;
may appear in the same feature - Complementary base pairing allows accurate replication;
Allow description of complementary base pairing
and accurate replication.
Describe the role of two named enzymes in the process of semiconservative replication of DNA. (3 marks)
- (DNA) helicase causes breaking of hydrogen/H bonds (between DNA strands);
Reject ‘helicase hydrolyses hydrogen bonds’. - DNA polymerase joins the (DNA) nucleotides;
Reject if suggestion that DNA polymerase joins the
complementary nucleotides or forms H bonds.
Reject if joining RNA nucleotides or forming RNA. - Forming phosphodiester bonds;
The enzymes DNA helicase and DNA polymerase are involved in DNA
replication.
Describe the function of each of these enzymes.
1. DNA helicase
2. DNA polymerase
(2 marks)
- DNA helicase – (unwinding DNA and) breaking hydrogen bonds / bonds between chains / bases / strands;
- DNA polymerase – joins (adjacent) nucleotides
OR
forms phosphodiester bond / sugar-phosphate backbone
Adenosine triphosphate (ATP) is a nucleotide derivative.
Contrast the structures of ATP and a nucleotide found in DNA to give two differences.
- ATP has ribose and DNA nucleotide has
deoxyribose; - ATP has 3 phosphate (groups) and DNA
nucleotide has 1 phosphate (group); - ATP – base always adenine and in DNA
nucleotide base can be different / varies;
Describe how an ATP molecule is formed from its component molecules (4 marks)
- and 2. Accept for 2 marks correct names of three components adenine, ribose/pentose, three phosphates;;
Accept for 1 mark, correct name of two components
Accept for 1 mark, ADP and phosphate/Pi
Ignore adenosine
Accept suitably labelled diagram - Condensation (reaction);
Ignore phosphodiester - ATP synthase;
Water is used to hydrolyse ATP.
Name the two products of ATP hydrolysis.
Adenosine diphosphate and (inorganic) phosphate;
ATP is an energy source used in many cell processes. Give two ways in which ATP is a suitable energy source for cells to use. (2 marks)
- Releases relatively small amount of energy / little energy lost as
heat;
Key concept is that little danger of thermal death of cells - Releases energy instantaneously;
Key concept is that energy is readily available - Phosphorylates other compounds, making them more reactive;
- Can be rapidly re-synthesised;
- Is not lost from / does not leave cells