School / IB / BIOLOGY HL / Theme D: Continuity and change (HL) Exam-style + marking analysis
Theme D: Continuity and change (HL) HL-only depth on DNA replication, gene expression, mutation, meiosis, inheritance, population genetics and hormonal homeostasis.
46 activities ≈ 110 minutes
Biology HL Theme D: Continuity and change (HL)
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1 Outline the role of the enzyme DNA helicase and the enzyme DNA polymerase during DNA replication. Easy 3 marks No calculator + 2 Marking analysis: A learner attempts the following task: “Outline the role of the enzyme DNA helicase and the enzyme DNA polymerase during DNA replication.” Their response addresses only this point: “States that DNA helicase unwinds the double helix and separates the two strands by breaking the hydrogen bonds between complementary base pairs.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks No calculator + 3 Explain the role of DNA ligase in joining Okazaki fragments on the lagging strand during replication. Easy 2 marks No calculator + 4 Marking analysis: A learner attempts the following task: “Explain the role of DNA ligase in joining Okazaki fragments on the lagging strand during replication.” Their response addresses only this point: “States that the lagging strand is synthesized discontinuously as a series of short Okazaki fragments, each begun from a separate primer.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 2 marks No calculator + 5 Outline the process of post-transcriptional modification of pre-mRNA in eukaryotic cells, including splicing. Easy 3 marks No calculator + 6 Marking analysis: A learner attempts the following task: “Outline the process of post-transcriptional modification of pre-mRNA in eukaryotic cells, including splicing.” Their response addresses only this point: “States that the primary transcript (pre-mRNA) contains both exons (coding sequences) and introns (non-coding sequences).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks No calculator + 7 Explain how alternative splicing of the same pre-mRNA transcript can produce different protein products. Easy 3 marks No calculator + 8 Marking analysis: A learner attempts the following task: “Explain how alternative splicing of the same pre-mRNA transcript can produce different protein products.” Their response addresses only this point: “States that alternative splicing means different combinations of exons can be joined together (or some exons excluded) from the same pre-mRNA transcript.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks No calculator + 9 Outline the role of transcription factors in controlling gene expression in eukaryotic cells. Easy 3 marks No calculator + 10 Marking analysis: A learner attempts the following task: “Outline the role of transcription factors in controlling gene expression in eukaryotic cells.” Their response addresses only this point: “States that transcription factors are proteins that bind to specific regulatory DNA sequences (e.g. promoters or enhancers) near a gene.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks No calculator + 11 Outline how DNA replication is described as a semi-conservative and bidirectional process at a replication origin. Easy 2 marks No calculator + 12 Marking analysis: A learner attempts the following task: “Outline how DNA replication is described as a semi-conservative and bidirectional process at a replication origin.” Their response addresses only this point: “States that replication begins at a specific site (origin of replication) where the double helix unwinds, forming a replication bubble/fork.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 2 marks No calculator + 13 Explain why a mutation in a non-coding region of DNA (e.g. an intron) is less likely to affect the phenotype than a mutation in an exon. Easy 3 marks No calculator + 14 Marking analysis: A learner attempts the following task: “Explain why a mutation in a non-coding region of DNA (e.g. an intron) is less likely to affect the phenotype than a mutation in an exon.” Their response addresses only this point: “States that introns are removed from the pre-mRNA during splicing and are not present in the mature mRNA that is translated.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks No calculator + 15 Outline the role of proofreading by DNA polymerase in maintaining the accuracy of DNA replication. Easy 2 marks No calculator + 16 Marking analysis: A learner attempts the following task: “Outline the role of proofreading by DNA polymerase in maintaining the accuracy of DNA replication.” Their response addresses only this point: “States that DNA polymerase can detect when an incorrectly paired (mismatched) nucleotide has been added to the new strand.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 2 marks No calculator + 17 Outline how the enzyme reverse transcriptase is used in genetic engineering to produce complementary DNA (cDNA) from an mRNA template. Easy 3 marks No calculator + 18 Marking analysis: A learner attempts the following task: “Outline how the enzyme reverse transcriptase is used in genetic engineering to produce complementary DNA (cDNA) from an mRNA template.” Their response addresses only this point: “States that reverse transcriptase is an enzyme that synthesizes a strand of DNA using an RNA molecule (e.g. mRNA) as its template.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks No calculator + 19 Distinguish between the structure and role of stamens and carpels in a flowering plant. Medium 4 marks No calculator + 20 Marking analysis: A learner attempts the following task: “Distinguish between the structure and role of stamens and carpels in a flowering plant.” Their response addresses only this point: “States that a stamen consists of an anther (which produces pollen containing the male gametes) supported on a filament.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Medium 4 marks No calculator + 21 Explain two ways in which meiosis generates genetic variation among gametes. Medium 4 marks No calculator + 22 Marking analysis: A learner attempts the following task: “Explain two ways in which meiosis generates genetic variation among gametes.” Their response addresses only this point: “States that crossing over occurs between non-sister chromatids of homologous chromosomes during prophase I, exchanging alleles and producing new combinations on each chromatid.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Medium 4 marks No calculator + 23 In sweet pea plants, flower colour (purple P dominant to red p) and pollen shape (long L dominant to round l) are controlled by genes on the same chromosome, 10 map units apart. A plant heterozygous for both genes (PpLl, in coupling: PL/pl) is test-crossed. State the expected phenotype ratio of the offspring, and explain why it does not match the standard 9:3:3:1 dihybrid ratio. Medium 5 marks Calculator + 24 Marking analysis: A learner attempts the following task: “In sweet pea plants, flower colour (purple P dominant to red p) and pollen shape (long L dominant to round l) are controlled by genes on the same chromosome, 10 map units apart. A plant heterozygous for both genes (PpLl, in coupling: PL/pl) is test-crossed. State the expected phenotype ratio of the offspring, and explain why it does not match the standard 9:3:3:1 dihybrid ratio.” Their response addresses only this point: “States that the two genes are linked (on the same chromosome), so they do not assort independently as required for the standard dihybrid ratio.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Medium 5 marks Calculator + 25 Distinguish between polygenic inheritance and epistasis, each with a brief example. Medium 4 marks No calculator + 26 Marking analysis: A learner attempts the following task: “Distinguish between polygenic inheritance and epistasis, each with a brief example.” Their response addresses only this point: “States that polygenic inheritance is when a single phenotypic trait is controlled by two or more genes at different loci, each contributing additively to the phenotype.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Medium 4 marks No calculator + 27 A population of 500 beetles is in Hardy-Weinberg equilibrium for a gene with two alleles, A (dominant) and a (recessive). 180 beetles show the recessive phenotype. Calculate the allele frequencies of A and a, and the number of heterozygous beetles. Medium 5 marks Calculator + 28 Marking analysis: A learner attempts the following task: “A population of 500 beetles is in Hardy-Weinberg equilibrium for a gene with two alleles, A (dominant) and a (recessive). 180 beetles show the recessive phenotype. Calculate the allele frequencies of A and a, and the number of heterozygous beetles.” Their response addresses only this point: “Calculates q² = 180/500 = 0.36.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Medium 5 marks Calculator + 29 In fruit flies, the genes for body colour and wing length are found to be linked, with a recombination frequency of 18%. Explain what this recombination frequency indicates about the relative positions of the two genes. Easy 3 marks No calculator + 30 Marking analysis: A learner attempts the following task: “In fruit flies, the genes for body colour and wing length are found to be linked, with a recombination frequency of 18%. Explain what this recombination frequency indicates about the relative positions of the two genes.” Their response addresses only this point: “States that a recombination frequency of 18% corresponds to a map distance of 18 map units between the two genes.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks No calculator + 31 Explain the role of non-disjunction during meiosis I in causing Down syndrome (trisomy 21). Medium 4 marks No calculator + 32 Marking analysis: A learner attempts the following task: “Explain the role of non-disjunction during meiosis I in causing Down syndrome (trisomy 21).” Their response addresses only this point: “States that non-disjunction is the failure of homologous chromosomes (chromosome 21) to separate properly during anaphase I.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Medium 4 marks No calculator + 33 Explain how directional selection can change the allele frequencies of a population over time, using antibiotic resistance in bacteria as an example. Medium 4 marks No calculator + 34 Marking analysis: A learner attempts the following task: “Explain how directional selection can change the allele frequencies of a population over time, using antibiotic resistance in bacteria as an example.” Their response addresses only this point: “States that within a bacterial population there is already variation, with a small number of individuals carrying a resistance allele (e.g. due to a pre-existing mutation).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Medium 4 marks No calculator + 35 Distinguish between a gene pool and a gene, and explain how genetic drift can reduce genetic diversity in a small population. Medium 4 marks No calculator + 36 Marking analysis: A learner attempts the following task: “Distinguish between a gene pool and a gene, and explain how genetic drift can reduce genetic diversity in a small population.” Their response addresses only this point: “States that a gene is a heritable factor at a specific locus that controls a characteristic, while a gene pool is the complete set of all alleles of all genes present in a population.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Medium 4 marks No calculator + 37 Explain how the loop of Henle establishes a concentration gradient in the medulla that enables the kidney to produce concentrated urine. Medium 5 marks No calculator + 38 Marking analysis: A learner attempts the following task: “Explain how the loop of Henle establishes a concentration gradient in the medulla that enables the kidney to produce concentrated urine.” Their response addresses only this point: “States that the ascending limb of the loop of Henle is impermeable to water but actively transports Na+ and Cl− ions out into the surrounding medullary tissue.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Medium 5 marks No calculator + 39 Explain the role of antidiuretic hormone (ADH) in regulating the water permeability of the collecting duct, and predict the effect of dehydration on ADH secretion. Medium 4 marks No calculator + 40 Marking analysis: A learner attempts the following task: “Explain the role of antidiuretic hormone (ADH) in regulating the water permeability of the collecting duct, and predict the effect of dehydration on ADH secretion.” Their response addresses only this point: “States that ADH is released from the posterior pituitary gland and travels in the blood to the kidney.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Medium 4 marks No calculator + 41 Outline the roles of follicle-stimulating hormone (FSH) and luteinising hormone (LH) in controlling the menstrual cycle. Medium 4 marks No calculator + 42 Marking analysis: A learner attempts the following task: “Outline the roles of follicle-stimulating hormone (FSH) and luteinising hormone (LH) in controlling the menstrual cycle.” Their response addresses only this point: “States that FSH, secreted by the pituitary gland, stimulates the growth and development of ovarian follicles.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Medium 4 marks No calculator + 43 Explain the hormonal changes that occur if fertilisation does not take place, leading to menstruation. Easy 3 marks No calculator + 44 Marking analysis: A learner attempts the following task: “Explain the hormonal changes that occur if fertilisation does not take place, leading to menstruation.” Their response addresses only this point: “States that if fertilisation does not occur, the corpus luteum degenerates towards the end of the cycle.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks No calculator + 45 A woman takes a combined oral contraceptive pill containing synthetic oestrogen and progesterone. Explain how this prevents pregnancy. Easy 3 marks No calculator + 46 Marking analysis: A learner attempts the following task: “A woman takes a combined oral contraceptive pill containing synthetic oestrogen and progesterone. Explain how this prevents pregnancy.” Their response addresses only this point: “States that the constant, artificially elevated levels of oestrogen and progesterone provide negative feedback to the pituitary gland.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks. Marking analysis Easy 3 marks No calculator + Self-assessed 0 / 0
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