NCEA Level 2 Biology 91159 Gene expression: past papers, practice and marking
- NCEA Level 2
- Biology
- AS91159
- 4 credits
- External exam
About 91159
AS91159, Demonstrate understanding of gene expression. DNA and RNA structure, antiparallel strands and complementary base pairing, coding and template strands, triplets, codons and anticodons, degeneracy of the code, reading a codon table, transcription and translation, mRNA and tRNA, proteins as the products of genes, mutagens and the effect of substitution, insertion and deletion mutations on a protein, metabolic pathways with one gene for each enzyme, and how the environment changes phenotype without changing genotype: the 4-credit external, in the six question types it keeps reusing.
What 91159 covers
- DNA and RNANucleotides of phosphate, sugar and base, the sugar-phosphate backbone, antiparallel strands, A with T or U and C with G, double-ring purines with single-ring pyrimidines, two and three hydrogen bonds, DNA compared with mRNA, coding and template strands
- Genetic codeTriplet on DNA, codon on mRNA, anticodon on tRNA, start and stop codons, template strand to mRNA to amino acids with a codon table, tRNA anticodons, degeneracy (64 codons for 20 amino acids) and why it can make a mutation silent
- Protein synthesisTranscription in the nucleus from promoter to terminator, translation at the ribosome codon by codon, mRNA and tRNA structure, function and stability, why DNA is not translated directly, proteins as the products of genes with structural and catalytic roles, amino acid sequence deciding folding, shape and function
- MutationsA mutation as a permanent change in the DNA base sequence, errors in DNA replication and named mutagens, substitution (silent, missense, nonsense), insertion and deletion with a frameshift, the effect on amino acid sequence, polypeptide length, folding, shape, function and phenotype, comparing severity
- Metabolic pathwaysA series of enzyme-controlled reactions where the product of one is the substrate of the next, precursor, intermediate and final product, one gene for one enzyme, a mutation in gene 1, gene 2 or both, build-up of a substrate and a missing product, phenotype from the presence, absence or amount of a product
- EnvironmentGenotype plus environment gives phenotype, a temperature-sensitive enzyme, light, nutrients and UV changing phenotype without changing genotype, every cell having the same genotype, telling a non-mutagenic environmental factor from a mutagen
91159 past papers
Open a 91159 past paper beside your own Ako page, then get it marked, part by part.
Work through a past paper All NCEA Level 2 Biology past papers
Practice questions
Question 1: DNA and RNA
A 12-base-pair section of DNA contains 7 A–T pairs and 5 C–G pairs. The four DNA bases are adenine (A), guanine (G), cytosine (C) and thymine (T). Adenine and guanine are purines, with a double ring; cytosine and thymine are pyrimidines, with a single ring.
- State the complementary base-pairing rule for DNA, and for DNA pairing with mRNA.(up to Achieved)
- Calculate the number of hydrogen bonds holding this section together.(up to Achieved)
Check your answer
Part a: what gets the marks
- In DNA, adenine pairs with thymine (A–T) and cytosine pairs with guanine (C–G).
- When a DNA template pairs with mRNA, the bases pair as A–U, T–A, C–G and G–C, with uracil in the RNA in place of thymine.
A full answer: In DNA, A pairs with T and C pairs with G. When mRNA is made against a DNA template, A on DNA pairs with U on mRNA, T with A, C with G and G with C. RNA has uracil (U) in place of thymine.
Part b: what gets the marks
- Each A–T pair is held by 2 hydrogen bonds and each C–G pair by 3 hydrogen bonds.
- Total: 7 × 2 + 5 × 3 = 14 + 15 = 29 hydrogen bonds
A full answer: Each A–T pair has 2 hydrogen bonds and each C–G pair has 3: 7 × 2 + 5 × 3 = 29 hydrogen bonds.
What gets each grade
Question 2: DNA and RNA
Each mānuka cell keeps its genes as double-stranded DNA, even though only one strand of a gene is read to make mRNA.
- Describe the structure of ONE DNA nucleotide.(up to Achieved)
- Explain why it is an advantage for the cell that DNA has two strands, when only the template strand is read.(up to Merit)
Check your answer
Part a: what gets the marks
- A DNA nucleotide is made of a phosphate group, the sugar deoxyribose and one nitrogenous base (A, T, G or C).
A full answer: A DNA nucleotide is a phosphate joined to the sugar deoxyribose, which is joined to one base: A, T, G or C.
Part b: what gets the marks
- Because the coding strand is complementary to the template strand, a damaged base on the template can be repaired by pairing against the coding strand, so the gene's code is not lost.
- In a double strand the bases are paired on the inside of the helix, with the sugar-phosphate backbones on the outside, so the bases that carry the code are shielded from damage.
A full answer: The two strands are complementary, so the coding strand acts as a back-up copy: if a base on the template strand is damaged, it can be repaired by pairing against the coding strand. The paired bases also sit inside the double helix, with the sugar-phosphate backbones outside, so the bases that hold the code are protected.
What gets each grade
Question 3: DNA and RNA
In both DNA and RNA in kākāpō cells, the bases pair in a fixed way when two strands line up.
- Describe TWO similarities between DNA and mRNA.(up to Achieved)
- Explain why adenine pairs only with thymine (or uracil), and cytosine pairs only with guanine.(up to Merit)
Check your answer
Part a: what gets the marks
- Both DNA and mRNA are made of nucleotides, each with a phosphate, a sugar and a base.
- Both DNA and mRNA contain the bases adenine (A), guanine (G) and cytosine (C).
A full answer: Both are made of nucleotides, each made of a phosphate, a sugar and a base. Both contain the bases adenine, guanine and cytosine.
Part b: what gets the marks
- Adenine pairs with thymine in DNA (or uracil in RNA), and cytosine pairs with guanine.
- Adenine and guanine are large two-ring purines and thymine, uracil and cytosine are small one-ring pyrimidines, so each pair is one large plus one small base, which keeps the two strands the same distance apart; two large bases would not fit and two small bases would leave a gap.
- A and T (or U) form two hydrogen bonds while C and G form three, so A cannot bond properly with C, and G cannot bond properly with T.
A full answer: A pairs with T (or U) and C pairs with G. A and G are large purines with two rings, and T, U and C are small pyrimidines with one ring, so every pair is a purine with a pyrimidine and fits the same width between the strands. The pairs are also held by hydrogen bonds: two between A and T, three between C and G. A and C cannot line up their bonds, so they do not pair.