IFYBI003 Biology THE NCUK INTERNATIONAL FOUNDATION YEAR IFYBI003 Biology EXEMPLAR EXAMINATION MARK SCHEME Exemplar © NCUK Ltd. 2021 Page 1 of 15 IFYBI003 Biology This mark scheme should be used in conjunction with the Centre Marking and Recording Results section of the NCUK Academic Handbook. Contact your Principal/ Academic Manager if you do not have access to this. NB – Credit should be awarded for any pertinent answers not included in the mark scheme, not exceeding the total mark allowed for the question. Notice to markers. If a student has answered more than the required number of questions, credit should only be given for the first n answers, in the order that they are written in the student’s answer booklet (n being the number of questions required for the examination). Markers should not select answers based on the combination that will give the student the highest mark. If a student has crossed out an answer, it should be disregarded. Copyright: Please note that, unless otherwise stated, all diagrams have been produced from DK Publishing (Dorling Kindersley); Har/Com/Ps edition (30 Mar. 2009) Exemplar © NCUK Ltd. 2021 Page 2 of 15 IFYBI003 Biology Questions 1-9 Answer ALL questions. These questions carry 40 marks in total. Question Answer 1 2 3 4 5 6(a) 6(b) 6(c) 6(d) 6(e) B C D A D C C C D A Assessment Objective AO1 AO1 AO1 AO1 AO1 AO2 AO2 AO2 AO2 AO2 Question 7 (a) Mitochondrion/mitochondria; [1] (b) P = outer membrane; Q = cristae; R = matrix. Award 1 mark for each of the following points [to Max 2 marks]: [3] (c) (d) [2] muscle requires large amounts of energy; mitochondrion/mitochondria site of respiration; respiration releases energy Examples may include: [1] liver cell or sperm cell. (e) Note: accept any cell that is very active. Chloroplast diagram showing correct locations of: stroma ; thylakoid membranes / granum ; inner and outer membrane ; [3] AO1 - 7 marks AO2 – 3 marks Exemplar © NCUK Ltd. 2021 Page 3 of 15 IFYBI003 Biology Question 8 (a) (b) (c) (d) Award 1 mark for each correct answer. [4] W = ATP ; X = reduced NAD/NADH ; Y = reduced FAD/FADH2 ; Z = CO2/carbon dioxide ; matrix of mitochondrion/mitochondria ; Note: Answer must include matrix pyruvate ; production of reduced NAD ; release of CO2 ; any two comparative descriptions from: aerobic uses oxygen produces H2O + CO2 more energy / ATP [1] [3] [2] anaerobic no oxygen produces lactic acid less energy / ATP AO1 - 8 marks AO2 – 2 marks Question 9 (a) [2] to identify anomalous results; to find an average value at each measurement/check accuracy of the measurements. (b) Graph should look like the one below: [4] 70 60 50 40 30 20 10 0 0 5 10 15 20 25 30 Time in minutes The y axis units should be distance glucose solution moved (in mm) Exemplar © NCUK Ltd. 2021 Page 4 of 15 IFYBI003 Biology Award 1 mark for each of the following points: (c) Correctly labelled x and y axes with units; Suitable scale used on BOTH axes; Correct plotting of data; Line of best fit drawn. Award 1 mark for each of the following points: [3] glucose lowers water potential (makes it more negative); water moves in from beaker/from higher water potential; by osmosis. (d) suitable line drawn above the on graph; [1] AO1 - 0 marks AO2 – 10 marks Exemplar © NCUK Ltd. 2021 Page 5 of 15 IFYBI003 Biology Questions 10-14 Answer 3 questions. These questions carry 60 marks in total. Question 10 (a) Award 1 mark for each of the following points [to Max 10 marks]: (b) DNA is a polymer of nucleotides that forms a double helix; The four nucleotides are adenine/A, Thymine/T, Guanine/G and Cytosine/C; Nucleotides show complimentary base paring, A-T C-G; With/using hydrogen bonding; Topoisomerase breaks, swivels and rejoins the DNA molecule ahead of the replication fork; This relieves the strain on the molecule during unwinding; ”unwindase”/helicase unwinds the DNA double helix by breaking hydrogen/H bonds forming a replication fork; This gives two single stranded molecules/forms a leading strand and a lagging strand; These single stranded molecules act as templates for the synthesis of new DNA molecules; An RNA primer binds to the leading strand; DNA polymerase adds nucleotides to the RNA primer; Nucleotides are added continuously in a complementary manner to the original leading strand template; In 5’ to 3’ direction forming a continuous double strand/new DNA; The lagging strand elongates/adds nucleotides discontinuously; Multiple primers are added to the lagging strand; Nucleotides are added and short DNA fragments formed between primers; These short segments of DNA are called Okazaki fragments; DNA ligase then joins the Okazaki fragments into one continuous DNA molecule; Result is two new DNA molecules are synthesised, with each new double helix molecule having one strand running in the 5’ to 3’ direction and one in the 3’ to 5’ direction. Award 1 mark for each of the following points [to Max 10 marks]: Exemplar [ 10 ] [ 10 ] They grew/cultured bacteria; in a medium containing heavy nitrogen/15N; 15 N was incorporated into the DNA of the growing cells; They switched/transferred some of the cells to a medium containing naturally occurring (light) nitrogen/14N; These cells were allowed to replicate/grow; DNA was extracted from cells after the first replication and another DNA sample was taken/extracted after the second replication; Bacterial DNA ‘weighed’/centrifuged after each replication; 100% heavy became 100% intermediate/the first replication in the 14 N medium produced a band of hybrid /14N-15N DNA; Then 50% intermediate and 50% light/the second replication produced both light and hybrid DNA; This experiment allowed Meselson and Stahl to prove that © NCUK Ltd. 2021 Page 6 of 15 IFYBI003 Biology replication of DNA was semi conservative/replicated by the semi conservative method. AO1 - 10 marks AO2 – 10 marks Exemplar © NCUK Ltd. 2021 Page 7 of 15 IFYBI003 Biology Question 11 (a) Award 1 mark for each of the following points [to Max 10 marks]: (b) [ 10 ] Gene is isolated from DNA in cell of interest/isolation of gene from (usually) eukaryotic/ “foreign” DNA; Using restriction (endonuclease) enzymes; These enzymes cut the DNA at specific points/restriction sites; Plasmid is removed from a bacterial cell/plasmid removed to act as a cloning vector; The plasmid is cut/treated using the same restriction enzyme (used to cut the eukaryotic (“foreign”) DNA at similar restriction sites; Complementary sticky ends are formed in both plasmid and eukaryotic (“foreign”) DNA/complementary plasmid to sticky ends of isolated gene; The plasmid DNA and eukaryotic (“foreign”) DNA are then mixed together (under suitable conditions)/cut plasmid is added to isolated gene; Base pairing of plasmid and eukaryotic (“foreign”) DNA produces recombinant DNA/binding of eukaryotic gene to plasmid By the formation of hydrogen/H-bonding between complementary bases/base pairs; The recombinant plasmid placed/inserted into suitable bacterium; The transformed bacterium/recombinant bacterium is placed into suitable growth medium; The bacterium reproduces/grows and in doing so multiplies the recombinant DNA (containing the eukaryotic (“foreign”) gene/DNA; Reference to selection of transformed bacteria. Award 1 mark for each of the following points [to Max 5 marks]: [5] Medical: Used to produce safer drugs; Used to repair defective genes; Examples may include discussion/explanation of one or more of the following: Anti-thrombin (prevents clottng in surgery); Factor VIII (haemophiliac treatment); HGH (stimulates growth); Penicillin (antibiotic); Gene therapy; Insulin production; Industrial/agricultural: (c) BST (increased yield in cattle); Rennin production (cheese making); Pesticide production (crop yield); Delayed ripening. Award 1 mark for each of the following points [to Max 5 marks]: Exemplar [5] Caused by a recessive allele; Loci on the X-chromosome; © NCUK Ltd. 2021 Page 8 of 15 IFYBI003 Biology More common in males due to presence of Y chromosome / lack of another X chromosome; Reference to females haemophiliacs being homozygous recessive; Results in lack of clotting factor VIII or IX; Blood unable to clot effectively; Leads to internal bleeding or bleeding out after trauma/surgery. Note: Allow marks for any other correct points made if another example of a disease is given in the student’s answer. AO1 - 10 marks AO2 – 10 marks Exemplar © NCUK Ltd. 2021 Page 9 of 15 IFYBI003 Biology Question 12 (a) Award 1 mark for each of the following points [to Max 10 marks]: [ 10 ] Formation of tertiary structure: Proteins are formed from amino acids; Amino acids have specific R-groups; Amino acids are joined by peptide bonds; Description of primary structure as a specific sequence of amino acids; Description of secondary structure as alpha helix or beta pleated sheet folding; Tertiary structure involves 3D folding of polypeptide chain; Tertiary structure involves interactions between the side chains/R groups of the various amino acids; Tertiary structure is maintained by: Hydrogen bonding; Ionic bonding; Covalent bonding; Description of cysteine residues covalent bonding to form disulphide bridges; Hydrophobic/hydrophilic interactions; Van der Walls forces. Importance/Significance of tertiary structure: (b) Allows protein stability; Description of effect of temperature change and idea of optimum temperatures; Description of effect of pH range and idea of optimum pH values; Allows active site formation for enzymes; Binding site for hormones/ligands etc; Reference to specificity. Award 1 mark for each of the following points [to Max 10 marks]: [ 10 ] Proteins can be: Exemplar Structural/supporting, for example keratin, horn, feathers; Collagen and elastin proteins provide a framework around animal/human connective tissues/cytoskeleton; May function in movement; for example myosin and actin used in the contraction of muscle tissue; Proteins function in membrane transport processes, for example in ion channels/pumps; Specific proteins play a role in active transport/facilitated diffusion; Transport for oxygen, for example haemoglobin carries oxygen; Act as storage molecules, for example ovalbumin of egg white, © NCUK Ltd. 2021 Page 10 of 15 IFYBI003 Biology used as an amino acid source for the developing embryo; Proteins in DNA folding/formation of chromosomes; Proteins act as hormones in the co-ordination of metabolic activities of an organism, for example insulin; Function to protect against disease/defensive proteins, for example antibodies; Act as receptors, for example signal transduction in membranes of nerve cells; Cell-cell recognition. AO1 - 10 marks AO2 – 10 marks Exemplar © NCUK Ltd. 2021 Page 11 of 15 IFYBI003 Biology Question 13 (a) Award 1 mark for each of the following points [to Max 10 marks]: [ 10 ] Interphase: Alternates with the mitotic phase of the cell cycle; Consists of a number of different phases including the G1, S and G2 phases;Interphase accounts for about 90% of the cell cycle; G1/first gap Phase: Intense metabolic activity occurs in the cell during this phase/ normal cellular activity occurs; Cellular components such as mitochondria. Endoplasmic reticulum/ER and other organelles are duplicated during this phase/cellullar components are duplicated; S phase or synthesis phase: DNA replication/duplication of the chromosomes occurs entirely in this phase; G2/second gap phase: Cell growth continues/fair amount of metabolic activity takes place; Prepares for mitosis/DNA checking before the M phase of the cycle; M phase/mitotic phase: Mitosis takes place during this phase; Consists of prophase, metaphase , anaphase, telophase, and cytokinesis; Cleavage furrow/cell plate forms/pinches cell into two; Importance of regulation by cell cycle: (b) Prevents uncontrolled cell division; Prevents cancer/tumour formation. Differences: [5] mitosis 1 cell division 2 daughter cells Diploid cells formed Genetically identical daughter cells No crossing over Forms somatic cells Exemplar meiosis 2 cell divisions 4 daughter cells Haploid (Genetic) variation Crossing over Forms gametes © NCUK Ltd. 2021 Page 12 of 15 IFYBI003 Biology (c) [5] spermatogenesis Forms spermatocytes Occurs in testes Cells divide equally Four gametes produced Continuous process Starts at puberty Continues throughout life oogenesis Forms ova ovaries Unequal cell division One gamete/ovum and 2-3 polar bodies produced Interrupted process Starts in foetus Stops at menopause AO1 - 10 marks AO2 – 10 marks Exemplar © NCUK Ltd. 2021 Page 13 of 15 IFYBI003 Biology Question 14 (a) Award 1 mark for each of the following points [to Max 15 marks]: Mutations generally affect one or a few nucleotide pairs; Mutations may be referred to as point mutations; Two categories of point mutations, which are single nucleotide substitutions and nucleotide pair insertions or deletions; Insertions and deletions can involve more than one nucleotide pair; [ 15 ] Substitutions: One base pair is replaced by another/a different base pair; Some substitutions have no effect on the reading frame/no effect on the encoded protein; This is because of the redundancy of the genetic code; Silent mutation (a substitution mutation): A change in a nucleotide pair may transform/alter one codon for another that translates the same amino acid; This would have no effect on the phenotype of the organism; Mis-sense mutation (a substitution mutation): May be similar to a silent mutation in that such a mutation may have little effect on the protein function; Idea of degenerate code; Different codons code for same amino acid; Non-sense mutation (a substitution mutation): This is where there is the insertion of a stop codon; Prematurely shortens protein during synthesis/causes translation to be terminated prematurely/early; May have a marked effect on the function of a protein/nearly all nonsense mutations lead to non-functional proteins; Nucleotide pair insertions and deletions: These are where there are additions or losses of nucleotide pairs (in a gene/protein);These mutations tend to have a marked/disastrous effect on the resulting gene/protein (compared to substitution mutations); Deletion mutation: Removes a base pair/base pairs from a DNA/gene sequence; Causes a frame shift to occur in triplet codons/alters the normal DNA nucleotide sequence; Leads to abnormal translation of the protein/gene; (Generally leading) to a huge effect on protein structure; Generally occurs if deletion of base pairs is not in a multiple of three bases; Same/similar effects are seen with insertion mutations; Note: Allow similar points if made for insertion mutations, but only allow a mark once for a similar or the same mark point. Suitable examples of gene mutation: Exemplar © NCUK Ltd. 2021 Page 14 of 15 IFYBI003 Biology e.g. sickle cell anaemia, haemophilia, any other correct example. Note: Allow 1 mark for one or two examples of human disease caused by mutations. (b) Award 1 mark for each of the following points [to Max 5 marks]: [5] Trisomy is a chromosome aberration in which the chromosomes fail to separate / nondisjunction; This occurs during meiosis; The non-disjunction can take place either in the formation of the egg or the sperm; The result is an extra chromosome present in all of the body cells / has a total of 47 chromosomes instead of 46 in each cell; Trisomy often leads to miscarriage (eg trisomy 16); An example of trisomy is Down/Edward/Patau syndrome; Down syndrome occurs due to trisomy 21 (Edwards trisomy 18 and Patau trisomy 13); Downs syndrome is associated with physical growth disruption and learning difficulties; AO1 - 10 marks AO2 – 10 marks Exemplar © NCUK Ltd. 2021 Page 15 of 15
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