IB DP Biology: Diversity of organisms: species, naming, karyotypes and genomes (DP Biology) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover sub-topic A3.1, Diversity of organisms, from the IB Biology guide, first assessment 2025. Questions 1 to 9 cover A3.1.1 to A3.1.11 (SL and HL). Questions 10 to 12 cover A3.1.12 to A3.1.15 and are HL only. Teaching is in the study guide and the revision notes. Course hub: IB DP Biology. Checklist: IB DP Biology checklist. Also see the Theme A practice questions and the free diagnostics.

Questions

1. The African elephant is named Loxodonta africana.

(a) Identify the genus and the species name. [2]

(b) State how the name should be written by hand. [1]

2. State the biological species concept and outline one problem with applying it to fossils. [3]

3. Two populations of a beetle live on opposite sides of a mountain range. After many generations they differ slightly in colour and in the length of their antennae. Males and females from the two populations mate in the laboratory and produce some fertile offspring and some infertile offspring.

Explain why it is difficult to say whether they are one species or two. [4]

4. A micrograph of a human chromosome is magnified ×2500. On the image the chromosome is 30 mm long and the short arm is 9 mm long.

(a) Calculate the actual length of the chromosome in µm. [2]

(b) Calculate the short arm as a fraction of the total length. [1]

(c) Describe two other features, besides centromere position, used to classify chromosomes in a karyogram. [2]

5. A human cell has 46 chromosomes, whereas a chimpanzee cell has 48. One hypothesis states that human chromosome 2 formed by fusion of two chromosomes in a shared ancestor. A second statement says that the fusion was part of a plan to make humans more intelligent.

(a) Outline two lines of evidence for the fusion hypothesis. [2]

(b) Distinguish between the two statements in terms of testability. [2]

(c) State the chromosome number of an ancestor that matched the chimpanzee, and explain how one fusion leads to the human number. [2]

6. A plant species has a diploid number of 38.

(a) State the haploid number and the number of homologous pairs. [2]

(b) Explain why a diploid number is normally even. [1]

(c) Predict the diploid number of a lineage in which two chromosomes of each haploid set fuse into one. [1]

7. The table shows genome sizes from a database (data supplied).

Organism	Group	Genome size (Mb)
Nematode worm	Invertebrate	100
Pufferfish	Vertebrate	390
Salamander	Vertebrate	32 000

(a) Calculate how many times larger the salamander genome is than the pufferfish genome. [2]

(b) Compare genome size with the complexity of the organisms using these data. [2]

(c) Suggest what the data indicate about whether genome size can predict complexity. [2]

8. Assume a human haploid genome of 3200 Mb. Two unrelated people differ at 0.08% of positions. A human and a chimpanzee differ at 1.2% of positions (data supplied).

(a) Calculate the number of differing bases between the two people. [2]

(b) Calculate how many times greater the difference is between a human and a chimpanzee. [1]

(c) Explain what single-nucleotide polymorphisms are and how they show both unity and diversity within a species. [2]

9. The first human genome sequence cost about 2.7 × 10⁹ US dollars. A genome can now be sequenced for about 1000 US dollars (data supplied).

(a) Calculate the factor by which the cost has fallen. [1]

(b) State one current use of whole genome sequencing. [1]

(c) Discuss one potential future use of whole genome sequencing and one concern that it raises. [2]

10. (HL only) A mare (2n = 64) is crossed with a male donkey (2n = 62) and the offspring is a mule.

(a) Deduce the number of chromosomes in a mule. [2]

(b) Explain why the mule is usually sterile. [2]

(c) Explain why the biological species concept is difficult to apply to bacteria. [2]

11. (HL only) A student collects five leaf-litter animals: a woodlouse (seven pairs of legs), a centipede (one pair of legs per body segment, many segments), a spider (eight legs, two body regions), a ground beetle (six legs, hard wing cases) and an ant (six legs, no wing cases, narrow waist). Construct a dichotomous key that identifies all five. [5]

12. (HL only) Water samples from three sites on a stream were analysed for DNA of an amphibian. Each sample was tested in three repeats. Site 1: DNA found in 3 of 3 repeats. Site 2: 1 of 3. Site 3: 0 of 3. A control sample of pure water gave 0 of 3.

(a) Outline the steps from collecting the water to identifying the species. [4]

(b) Explain why the result at Site 3 does not prove that the amphibian is absent. [2]

(c) Suggest one advantage and one limitation of using environmental DNA to survey the stream. [2]

Answers

1. (a) Genus: Loxodonta [1]; species name: africana [1]. (b) Underlined, because italics cannot be handwritten, with the genus capitalised and the species name lower case [1]. [3] Examiner insight: Answers gain credit when the genus and species name are separated correctly; calling the whole binomial “the species” is a common way to lose the second mark.

2. A species consists of organisms able to breed together [1] and produce fertile offspring [1]. Problem: fossils cannot be tested for breeding, so the concept cannot be applied to them [1]. [3] Examiner insight: The word fertile is what usually separates a full definition from a partial one, and the problem needs a link to fossils rather than a general comment.

3. The populations have started to diverge in traits [1] and gene flow is limited by the mountain range [1]. Speciation is gradual, so an intermediate stage is expected [1]. The partial interbreeding (some fertile, some infertile offspring) means the biological species concept gives no clear answer, so the decision is arbitrary [1]. [4] Examiner insight: Stronger answers say speciation is gradual and link the laboratory result to an intermediate stage, instead of simply choosing “one species” or “two”.

4. (a) Actual length = 30 ÷ 2500 [1] = 0.012 mm = 12 µm [1]. (b) 9 ÷ 30 = 0.30 [1]. (c) Length [1] and banding pattern [1]. [5] Examiner insight: Show the conversion from mm to µm; a correct number with a missing or wrong unit tends to lose credit.

5. (a) Any two of: banding of chromosome 2 matches two ape chromosomes placed end to end; telomere-type sequences lie in the middle of chromosome 2; a remnant of a second centromere lies on the long arm [2]. (b) The fusion hypothesis is testable because it predicts evidence (such as telomere sequence in the middle) that could turn out to be absent [1]; the plan statement is not testable because no observation could show it to be wrong [1]. (c) The ancestor had 48 [1]; fusing two chromosomes into one removes one, giving 46 [1]. [6] Examiner insight: For the testable claim, state a prediction that evidence could contradict; saying only “it can be tested” does not explain why.

6. (a) n = 19 [1]; 19 homologous pairs [1]. (b) Chromosomes occur in homologous pairs, so the diploid total must be two times the pairs [1]. (c) Each haploid set falls by one, so the diploid number falls by two, to 36 [1]. [4] Examiner insight: In part (c), state the effect on the haploid number first; the diploid number then follows and stays even.

7. (a) 32 000 ÷ 390 [1] = 82 times (82.05) [1]. (b) The salamander has by far the largest genome [1] yet is a vertebrate like the pufferfish, with no greater complexity [1]. (c) Genome size does not reliably predict complexity [1] because much DNA does not code for proteins or is repeated (accept an equivalent reason) [1]. [6] Examiner insight: Quote values from the table in a comparison, and state a conclusion that matches only the data given.

8. (a) 3200 × 10⁶ × 0.0008 [1] = 2.56 × 10⁶ bases [1]. (b) 1.2 ÷ 0.08 = 15 times [1]. (c) An SNP is a single base position where individuals have different nucleotides [1]; most of the genome is shared (unity) while SNPs give small differences between individuals (diversity) [1]. [5] Examiner insight: Convert the percentage to a decimal before multiplying, and keep the genome size in bases rather than Mb.

9. (a) 2.7 × 10⁹ ÷ 1000 = 2.7 × 10⁶-fold [1]. (b) Research into evolutionary relationships [1]. (c) Personalized medicine, where treatment is chosen using the patient’s genome [1]; concern, such as privacy of genetic information [1]. [4] Examiner insight: A “discuss” question needs a benefit and a limitation of the same use; giving only one side tends to earn half the marks.

10. (a) Mare gamete 64 ÷ 2 = 32 [1]; donkey gamete 62 ÷ 2 = 31, so the mule has 32 + 31 = 63 [1]. (b) The odd number means the chromosomes cannot all pair [1], so meiosis cannot produce balanced gametes [1]. (c) Bacteria reproduce asexually, so there is no breeding to test [1]; horizontal gene transfer moves genes between species, blurring boundaries [1]. [6] Examiner insight: Part (c) needs both asexual reproduction and gene transfer; one reason alone usually earns one mark.

11. Any valid key. Example: 1a. More than fourteen legs: centipede. 1b. Fourteen legs or fewer, go to 2. 2a. Eight legs: spider. 2b. Six or fourteen legs, go to 3. 3a. Seven pairs of legs: woodlouse. 3b. Six legs, go to 4. 4a. Hard wing cases: ground beetle. 4b. No wing cases: ant. Marks: each step is a pair of contrasting statements [1]; features are observable and constant, not size or colour alone [1]; four pairs for five organisms [1]; every ending names one animal [1]; the key gives the correct name for each of the five when tested [1]. [5] Examiner insight: A list of descriptions is not a key. Answers gain credit when each step splits the remaining set in two with contrasting statements.

12. (a) Collect a water sample [1]; extract and filter the DNA [1]; amplify a barcode region by PCR and sequence it [1]; match the sequence to a reference database to name the species [1]. (b) The DNA may be at a level too low to detect, or degraded [1]; the amphibian could be present but undetected, since eDNA shows only what has been shed into the water [1]. (c) Advantage: rapid and non-invasive, without capturing animals [1]; limitation: needs a reference database, which may be incomplete [1]. [8] Examiner insight: Use the data: 3 of 3 at Site 1 against 1 of 3 at Site 2 shows a weaker signal there, and the clean control supports the reliability of the results.

Where marks are usually lost

• Writing both parts of a binomial with capital letters.
• Defining a species without fertile offspring.
• Omitting the unit or the mm to µm conversion in magnification questions.
• Confusing the diploid number with the number of pairs.
• Treating a larger genome as proof of more complexity.
• Writing eDNA steps in the wrong order, or leaving out the reference database.
• Giving a key as a paragraph instead of paired statements.

Next steps

• Revision notes
• Study guide
• IB DP Biology course hub
• IB DP Biology checklist
• All free 10-minute diagnostics: diagnostics
• Book a free trial class: trial class

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, published by the International Baccalaureate Organization. Sub-topic A3.1, Diversity of organisms.