Level 3 Biology (90715) 2011 Assessment Schedule

NCEA Level 3 Biology (90715) 2011 — page 1 of 1

Assessment Schedule – 2011

Biology: Describe the role of DNA in relation to gene expression (90715)

Assessment Criteria

QUESTION ONE

Achievement / Achievement with Merit / Achievement with Excellence
Any TWO of:
•Describes the purpose of DNA replication.
Eg: Produces an exact copy of the original DNA before cell division.
•Enzymes are named.
A = helicase
B = DNA polymerase (iii)
C = DNA ligase.
•Describes role of enzymes.
Eg: Helicase unwinds the DNA helix
DNA polymerase iii joins new nucleotides to DNA strand
DNA ligase joins short strands / Okazaki fragments of DNA together (at the backbone). / Any TWO of:
•Explains the purpose of DNA replication.
Eg: Produces a copy of the original DNA before cell division / new cells formed (from mitosis) have an exact copy, with same instructions (information) / structure / function.
•Explains role of three enzymes in replication.
Eg: Helicase unwinds the DNA helix into single (template) strands by breaking the H bonds between base pairs.
DNA polymerase iii joins new nucleotides onto the 3’ end of the newly forming replicated DNA strand / DNA polymerase can only synthesize in the 5’ to 3’ direction,
DNA ligase joins short strands / Okazaki fragments of DNA together / forming phosphodiester bonds / sugar phosphate backbone.
•Explains consequences ofany enzyme failure in replication.
Eg: If Helicase fails to unwind the DNA, the bases will not be exposed for base pairing, so replication will not happen. / Discusses role of three enzymes.
Eg: Helicase unwinds the DNA helix into single strands by breaking the H bonds between base pairs.
DNA polymerase iii joins new nucleotides onto the 3’ end of the newly forming replicated DNA strand / DNA polymerase can only synthesize in the 5’ to 3’ direction.
DNA ligase joins Okazaki fragments of DNA together / forming phosphodiester bonds / sugar phosphate backbone.
AND
Discusses consequence of enzyme failures in replication.
Eg: DNA polymerase iii fails to attach to the DNA or attaches an incorrect matching base (A-C instead of A-T). This results in a mutation.

QUESTION TWO

Achievement / Achievement with Merit / Achievement with Excellence
Any TWO of:
•Transcription is described.
Eg: Production of mRNA copy of DNA / genetic material
•Translation is described.
E.g. information from mRNA is used to join amino acids into a polypeptide.
•A role of RNA is described.
Eg: mRNA carries the code from a gene to / site of synthesis / ribosome.
Eg: tRNA carries an amino acid which connects with the codon of mRNA at the ribosome. / Any TWO of:
•Explains the roles of transcription.
Eg: Transcription is the process where mRNA is manufactured by attaching free nucleotides to complementary DNA strand.
•Explains the role of translation.
Eg: Translation is the process happening at the ribosome, where RNA codons are matched with tRNA anti codons, resulting in the joining of amino acids.
Each amino acid is specific to an anticodon / codon.
•Explains a link between amino acids and protein folding
Eg: The interactions between amino acids sequence can affect the final shape / folding of the protein. / Discusses how DNA sequence results in the formation of a protein.
Eg: During transcription, mRNA is produced through complementary base pairing with exposed bases on DNA. mRNA is read via translation at the site of the ribosome (rRNA), where codon sequences are translated via tRNA into a chain of amino acids. tRNA attach to specific amino acids and contain an anti codon, complementary to the codons of the mRNA.
AND
The final protein / order in which the amino acids are joined is a result of the DNA sequence / genetic code.
OR
The order of amino acids affects the folding of the protein structure due to the interactions between them. Sulfur bridges / bonds between cysteines / hydrophobic / hydrophilic interactions / will lead to folding of the polypeptide / protein chain.

QUESTION THREE

Achievement / Achievement with Merit / Achievement with Excellence
Any TWO of:
•Describes linkage
Linked genes are carried on the same chromosome.
•Describes expected or observed outcome.
Eg: Expected 9:3:3:1, but in the observed there were extra parental phenotypes / purple long /
OR
Eg: 12:1:1:2 observed results produced extra parental phenotypes / purple long than expected.
•Describes linkage cross
Crossing over / recombination is the source of Purple round / red long offspring
OR
expected outcome from cross is 3 Purple long: 1 red round. / Any TWO of:
•Explains by comparing observed outcome with expected outcome.
Eg: If genes are not linked, will get 9:3:3:1, but observed outcome is different (12:1:1:2). Observed creates extra parental phenotypes (purple-long and red-round).
•Explains linkage cross
Eg Genes P and L / p and l are on the same chromosome so are inherited together so we get a 3: 1 ratio.
•Explains appearance of recombinants.
Crossing over separates P and L (purple and long) and p and l (red and round) in a few cases. This makes the other 2 phenotypes possible. / Discusses the linkage cross in terms of the observed results,
Eg:Linked genes are on the same chromosome, therefore they will not sort independently (separate) during meiosis and will be inherited together (giving a 3 (purple long):1(red round) ratio).
AND
The observed results show a result like this, but there are small numbers of recombinants (Pl 5% and pL 5%). These outcomes are due to (recombination) crossing over.
Evidence can be taken from a diagram. Eg:

(It is the new Pl and pL chromosomes that give rise to the recombinants.)