NCEA Level 2 Biology 91157 Genetic variation and change: past papers, practice and marking
- NCEA Level 2
- Biology
- AS91157
- 4 credits
- External exam
About 91157
AS91157, Demonstrate understanding of genetic variation and change. Dihybrid crosses from P to F2 and the 9:3:3:1 ratio, linked and unlinked genes, meiosis with crossing over, segregation and independent assortment, complete, incomplete and co-dominance, multiple alleles, lethal alleles, test crosses and sex linkage, mutation as the source of new alleles, and how natural selection, migration, genetic drift, founder effects and bottlenecks change allele frequencies in New Zealand populations: the 4-credit external, in the seven question types it keeps reusing.
What 91157 covers
- Dihybrid crossesP, F1 and F2 genotypes, gametes, the 4 × 4 Punnett square, the 9:3:3:1 ratio matched to the right phenotypes, ratios as fractions, percentages and decimals, expected numbers, working back from offspring to parents, why the F1 makes four kinds of gamete
- LinkageLinked genes close together on one chromosome, unlinked genes on different chromosomes or far apart on the same one, crossing over between non-sister chromatids of homologous chromosomes, segregation, independent assortment, parental and recombinant offspring, reading a ratio to decide whether genes are linked
- Monohybrid crossesComplete dominance and 3:1, incomplete dominance and 1:2:1, co-dominance, multiple alleles with an order of dominance, lethal alleles and the 2:1 ratio, a dominant lethal allele that persists, the test cross and why one litter cannot prove a genotype
- Sex linkageXX female and XY male, genes on the X chromosome, carrier females, why males cannot be carriers and show recessive X-linked conditions more often, reciprocal crosses
- MutationA permanent change in the DNA base sequence, the only source of new alleles, gametic mutations passed on and somatic mutations not, beneficial, harmful and neutral alleles and what natural selection does to each
- Natural selectionVariation, a selection pressure, individuals with a favourable phenotype surviving and reproducing more and passing on their alleles, allele frequency calculated from genotype counts, sexual selection, immigration and emigration as gene flow, telling selection from drift
- Genetic driftRandom change in allele frequency, the larger effect in small populations, loss (0 %) and fixation (100 %), the founder effect, the bottleneck effect, low genetic diversity, inbreeding and its costs, and translocation and breeding programmes for New Zealand species
91157 past papers
Open a 91157 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: Dihybrid crosses
In guinea pigs, black fur (B) is completely dominant to white fur (b), and rough fur (R) is completely dominant to smooth fur (r). The genes are not linked. A breeder crosses many pairs of guinea pigs that are both BbRr. Over two years, 320 young are born.
- Define the terms genotype and heterozygous.(up to Achieved)
- Calculate the expected number of young in each phenotype.(up to Achieved)
Check your answer
Part a: what gets the marks
- A genotype is the combination of alleles an individual has for a gene, for example Bb.
- Heterozygous means having two different alleles of a gene, for example Bb.
A full answer: Genotype: the combination of alleles an individual has for a gene (for example Bb). Heterozygous: having two different alleles of a gene (for example Bb or Rr).
Part b: what gets the marks
- The expected phenotype ratio is 9 : 3 : 3 : 1, a total of 16 parts.
- Each of the 16 parts of the ratio is 320 ÷ 16 = 20 young.
- Numbers: black rough 9 × 20 = 180; black smooth 3 × 20 = 60; white rough 3 × 20 = 60; white smooth 1 × 20 = 20
A full answer: BbRr × BbRr gives 9 black rough : 3 black smooth : 3 white rough : 1 white smooth. 320 ÷ 16 = 20 young per part, so 180 black rough, 60 black smooth, 60 white rough and 20 white smooth.
What gets each grade
Question 2: Dihybrid crosses
A class is given an ear of maize (corn) from an F2 generation. The kernels show two genes: purple kernel colour (P) is completely dominant to yellow (p), and smooth kernels (S) are completely dominant to wrinkled kernels (s). The class counts 392 kernels: 220 purple smooth, 79 purple wrinkled, 68 yellow smooth and 25 yellow wrinkled.
- State the genotype of the F1 plants that produced this F2, and the genotypes possible for a purple wrinkled kernel.(up to Achieved)
- Calculate the expected numbers of each phenotype if the ratio is 9:3:3:1.(up to Achieved)
Check your answer
Part a: what gets the marks
- The F1 plants are PpSs.
- A purple wrinkled kernel is PPss or Ppss.
A full answer: The F1 plants were PpSs. A purple wrinkled kernel is PPss or Ppss.
Part b: what gets the marks
- One part: 392 ÷ 16 = 24.5
- Expected: 9 × 24.5 = 220.5; 3 × 24.5 = 73.5; 3 × 24.5 = 73.5; 1 × 24.5 = 24.5
A full answer: 392 ÷ 16 = 24.5 kernels per part. Expected: 220.5 purple smooth, 73.5 purple wrinkled, 73.5 yellow smooth, 24.5 yellow wrinkled.
What gets each grade
Question 3: Dihybrid crosses
A pea plant has 7 pairs of homologous chromosomes. A student asks why a pure-breeding round yellow parent (RRYY) can only make one kind of gamete for these two genes, but its F1 offspring (RrYy) can make four.
- Define homologous chromosomes.(up to Achieved)
- Explain why an RRYY plant makes only one kind of gamete.(up to Merit)
Check your answer
Part a: what gets the marks
- Homologous chromosomes are a matching pair, one from each parent, of the same size and with the same genes at the same loci, possibly with different alleles.
A full answer: Homologous chromosomes are a matching pair of chromosomes, one inherited from each parent, that carry the same genes at the same positions (loci), though they may carry different alleles of those genes.
Part b: what gets the marks
- In segregation each gamete receives one allele of each gene.
- The RRYY plant is homozygous: both alleles of each gene are the same, so every gamete it makes is RY, however the chromosomes line up.
A full answer: In meiosis, segregation puts one allele of each gene into each gamete. The plant is homozygous for both genes: both homologous chromosomes carry R, and both carry Y. Whichever chromosome of each pair a gamete receives, it gets R and Y, so every gamete is RY.