U3 Topic 1
U3 Topic 2
Unit 4 Topic 1
Unit 4 Topic 2
Random
100
Identify the correct hierarchical sequence in the Linnaean classification system.
Genus --> Kingdom → Phylum → Class → Order → Family → Genus → Species.
100
Explain why energy decreases as it moves between trophic levels.
Energy is lost through respiration, heat, movement, waste and other metabolic processes, so only a proportion is transferred to the next trophic level.
100
Identify structure C, D and E
C - Hydrogen Bond, D - Deoxyribose Sugar, E - Phosphate
100
Compare stabilising, directional and disruptive selection.
Stabilising favours intermediate phenotypes; directional favours one extreme; disruptive favours both extremes over intermediate phenotypes.
100
Define genetic drift
Genetic drift is a change in the traits of a group over time caused purely by random chance rather than survival of the fittest.
200
Explain one limitation of the biological species concept.
It cannot be reliably applied to asexual organisms, fossils or populations that do not naturally interbreed despite being capable of doing so.
200
Compare gross primary productivity and net primary productivity
GPP is the total energy/biomass fixed by producers; NPP is GPP minus energy used in producer respiration.
200
Describe the roles of helicase and DNA polymerase during DNA replication
Helicase separates the DNA strands; DNA polymerase adds complementary nucleotides to the growing strands.
200
Calculate the frequency of allele A if a population contains 70 AA, 20 Aa and 10 aa individuals.
Total alleles = 200. A alleles = 140 + 20 = 160. Frequency of A = 0.80 or 80%.
200
Explain how crossing over contributes to genetic variation.
Homologous chromosomes exchange sections of DNA during meiosis, creating new combinations of alleles.
300
Explain why a quadrat would be more appropriate than capture–recapture when investigating the abundance of grasses.
Grasses are stationary, so individuals can be sampled within defined areas; capture–recapture requires mobile organisms that can be captured, marked and recaptured.
300
Predict two consequences of removing a keystone predator from a food web.
Its prey may increase, potentially causing overconsumption of lower trophic levels and altering community structure and biodiversity.
300
Describe the roles of helicase and DNA polymerase during DNA replication
Helicase separates the DNA strands; DNA polymerase adds complementary nucleotides to the growing strands.
300
Compare allopatric, sympatric and parapatric speciation
Allopatric occurs through geographic isolation; sympatric occurs without geographic separation; parapatric occurs between neighbouring populations with limited gene flow.
300
A species of flowering plant has two alleles for flower colour. The allele for purple flowers (P) is dominant over the allele for white flowers (p). Two heterozygous purple-flowered plants are crossed.
Determine the expected phenotypic ratio of the offspring and predict how many purple- and white-flowered plants would be expected from a sample of 80 offspring.
For 80 offspring:
- Purple = 80×(3/4)= 60
- White = 80×(1/4)=20
400
Compare (3) the reproductive strategies of r-selected and K-selected species.
r-selected species generally mature early, reproduce rapidly and produce many offspring with low parental investment; K-selected species mature later, produce fewer offspring and invest more parental care.
400
Explain why pioneer species are effective colonisers.
They can tolerate harsh conditions, reproduce rapidly, disperse effectively and modify the environment, making it more suitable for later species.
400
Explain how PCR and gel electrophoresis can be combined to produce a DNA profile.
.PCR amplifies selected DNA regions; gel electrophoresis separates DNA fragments according to size, producing a characteristic banding pattern.
400
Analyse why two species with similar morphology may not be closely related.
Similar morphology may result from convergent evolution, where unrelated species independently evolve similar traits due to similar selection pressures.
400
Determine the mRNA sequence transcribed from the DNA template strand TAC–GGA–CTT–ACT.
AUG–CCU–GAA–UGA
500
Infer why a species may have a high percentage frequency but a low percentage cover in a grassland
It occurs in many sampling sites but occupies relatively little area at each site, such as small individual plants dispersed throughout the ecosystem.
500
Nitrogen Cycle Diagram
I: Ammonium, II: Nitrites: III: Nitrates
500
Compare spermatogenesis and oogenesis.
Spermatogenesis produces four functional sperm from a primary spermatocyte and occurs continuously after puberty; oogenesis produces one functional ovum plus polar bodies and follows a cyclic pattern.
500
Explain how temporal isolation can lead to speciation
Temporal isolation occurs when populations differ in their periods of reproductive activity, so genes are no longer exchanged. This disruption to gene flow can result in two populations evolving independently to the point that they are no longer able to breed.
500
Explain the process of translation, leading to polypeptide formation.
- Initiation: The mRNA attaches to a ribosome. The ribosome recognises the start codon (AUG), and a complementary tRNA carrying the amino acid methionine binds to it.
- Elongation: tRNA molecules bring specific amino acids to the ribosome. Their anticodons bind to complementary mRNA codons. The amino acids are joined by peptide bonds, forming a growing polypeptide chain.
- Termination: The ribosome reaches a stop codon. No tRNA binds to the stop codon, and the completed polypeptide is released.
- The polypeptide then folds into its specific three-dimensional structure, allowing it to function as a protein.