AQA A-level Biology: Species, Taxonomy and Investigating Diversity (7402) – 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 – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow AQA A-level Biology (7402), sections 3.4.5, Species and taxonomy, and 3.4.7, Investigating diversity, in the AQA AS and A-level Biology specification for first teaching 2015. Explanations are in the study guide and revision notes. Course pages: AQA A-level Biology hub and the printable checklist. The data are invented, and the marks are not an official tariff.

Questions

1. Two populations of a grass are crossed. The hybrid plants grow well and flower, but they produce no viable seeds. Explain whether the two populations belong to the same species. [3]

2. Two cricket species live in one meadow. Males of each produce a different chirp pattern, and a female approaches only the pattern of her own species. (a) Explain how this courtship behaviour helps a female. [2] (b) Females of species 1 are played both songs. They approach one recording and ignore the other. Explain what this shows about courtship. [2]

3. The table shows part of the classification of three mammals.

Taxon	Red fox	Grey wolf	Eurasian otter
Order	Carnivora	Carnivora	Carnivora
Family	Canidae	Canidae	Mustelidae
Genus	Vulpes	Canis	Lutra

(a) Which two species are most closely related? Explain your answer. [2] (b) Name the lowest taxon shared by all three species in the table. [1] (c) Write the binomial for the grey wolf (species name lupus) in the correct form for handwriting. [1] (d) A new species is placed in the genus Vulpes. State the taxa in the table it must share with the red fox. [2]

4. A student writes the name of the cabbage as: Brassica Oleracea. (a) Identify two errors in how the name is written. [2] (b) Give one advantage of using binomials rather than common names. [1]

5. An antibody was made against serum albumin from species P. Samples of albumin from four species were mixed with the antibody. The relative amount of precipitate formed was: P 100, Q 94, R 58, S 22 (arbitrary units). (a) Explain how the results show that R is more closely related to P than S is. [2] (b) Place Q, R and S in order, closest relative of P first. [1] (c) Explain why the antibody reacts less with the albumin of a distant relative. [2]

6. Describe how investigating genetic diversity has changed, and give two advantages of the newer method. [4]

7. A fragment of 40 aligned bases from the same gene was compared in four species. The numbers of bases that differ are shown.

	W	X	Y	Z
W	–	2	6	14
X		–	7	15
Y			–	13

(a) Calculate the percentage difference between W and Z. [1] (b) State which pair is most similar and give the percentage difference. [1] (c) Suggest how Y and Z are related to W and X. [2] (d) Give two limitations of this conclusion. [2]

8. The template strand of DNA for individual 1 reads TAC TTT GGG CCA ACT. Codon data: AUG Met, AAA Lys, AAG Lys, CCC Pro, UCC Ser, GGU Gly, UGA stop. (a) Give the mRNA sequence. [1] (b) Give the amino acid sequence. [1] (c) Individual 2 has TTC in place of TTT. Explain the effect on the amino acid sequence. [2] (d) Individual 3 has AGG in place of GGG. Give the amino acid sequence. [1] (e) Compared with individual 1, state the number of differences in the template DNA and in the amino acid sequence for individuals 2 and 3. [2]

9. In random samples of 120 plants from two meadows, meadow A had 36 pink-flowered plants and meadow B had 78. (a) Calculate the percentage with pink flowers in each meadow. [2] (b) Suggest what this shows about the two populations. [1] (c) Give two reasons why this evidence is limited. [2]

10. A privet hedge is 2 m tall. Describe how to take random samples of 30 leaves from the top 50 cm and the bottom 50 cm, to compare leaf length. [5]

11. The tail length of 40 lizards was measured on each of three islands.

Island	Mean / mm	Standard deviation / mm
1	62.0	3.1
2	65.5	3.4
3	62.4	8.2

(a) State which island shows most variation and explain how you know. [2] (b) Calculate the range from one standard deviation below to one above the mean for islands 1 and 2. [2] (c) Comment on whether the difference between the means of islands 1 and 2 is likely to be real. [2]

12. The first 12 amino acids of a protein were compared (one letter per amino acid).

W  GKVDLRSTAPEM
X  GKVDLRSTGPEM
Y  GKADLRNTGPEM
Z  GRADIRNSGQEL

(a) Calculate the percentage difference between X and Z, and between Y and Z. [2] (b) Use the sequences to suggest the order of relatedness of the four species, with reasons. [3] (c) The DNA coding for W and X differs at 4 of 36 bases. Explain why the DNA percentage difference is greater than the amino acid percentage difference. [2] (d) Give two reasons why conclusions from these amino acid data are tentative. [2]

Answers

1. A species is a group able to produce fertile offspring [1]. The hybrids make no viable seeds, so they are not fertile [1]. The populations therefore fail the test and are not the same species [1]. [3] Examiner insight: the word “fertile” links the hybrid evidence to the definition.

2. (a) She recognises males of her own species by the chirp pattern [1], so she does not waste gametes on a mating that gives no offspring or sterile hybrids [1]. [2] (b) Females respond only to their own species’ song [1], so courtship is species specific and allows species recognition before mating [1]. [2] Examiner insight: name “species recognition” and give its consequence for mating.

3. (a) Red fox and grey wolf [1], because they share the family Canidae whereas the otter is in a different family [1]. [2] (b) Order Carnivora [1] [1] (c) Canis lupus, with each word underlined separately, a capital C and a lower case l [1] [1] (d) It must share the genus Vulpes [1] and every higher taxon in the table: the family Canidae and the order Carnivora [1]. [2] Examiner insight: name the shared taxon when explaining why species are related.

4. (a) The species word oleracea should not have a capital letter [1]. The name should be in italics or underlined [1]. [2] (b) It is universal, so scientists worldwide use the same name for the same species [1]. [1] Examiner insight: common names vary by language and place, so say “the same name worldwide”.

5. (a) The antibody binds proteins similar to the one it was made against [1]. R gives more precipitate (58) than S (22), so R’s albumin is more similar to P’s [1]. [2] (b) Q, then R, then S [1] [1] (c) A distant relative’s albumin has more amino acid differences, so its shape differs from the antigen [1]. More differences have built up since the common ancestor [1]. [2] Examiner insight: quote the table, then state what it means for relatedness.

6. Before, differences in DNA were inferred from observable or measurable characteristics [1]. Now the base sequences of DNA (or of mRNA, or the amino acid sequence of proteins) are compared directly [1]. Any two advantages from: DNA is not changed by the environment, but characteristics can be; changes with no visible effect can be detected; many genes can be compared; the comparison is objective. [2] [4] Examiner insight: give the old and new methods as two separate points.

7. (a) 14 / 40 × 100 = 35.0 % [1] [1] (b) W and X, 2 / 40 × 100 = 5.0 % [1] [1] (c) Y is closer to W and X (15.0 % and 17.5 %) than Z is (35.0 % and 37.5 %) [1]. Z is the most distant relative, sharing an ancestor with the others longest ago [1]. [2] (d) The fragment is only 40 bases from one gene [1]. Counting differences is a simple measure, as a site may have changed more than once [1]. [2] Examiner insight: give the percentage for each pair and express each relationship in terms of common ancestry.

8. (a) AUG AAA CCC GGU UGA [1] [1] (b) Met-Lys-Pro-Gly [1], with the stop codon not coding for an amino acid [1] (c) The mRNA codon becomes AAG [1]. AAG and AAA both code for Lys, because the code is degenerate, so the amino acid sequence is unchanged [1]. [2] (d) AUG AAA UCC GGU UGA gives Met-Lys-Ser-Gly [1] [1] (e) Individual 2: one DNA difference, no amino acid difference [1]. Individual 3: one of each [1]. [2] Examiner insight: write out the mRNA codons so each change is visible.

9. (a) Meadow A: 36 / 120 × 100 = 30.0 % [1]. Meadow B: 78 / 120 × 100 = 65.0 % [1]. [2] (b) The populations may differ genetically, as pink is more frequent in B [1]. [1] (c) Flower colour can be influenced by the environment, such as soil [1]. One characteristic reflects only a small part of the genome [1]. [2] Examiner insight: use “may” or “suggests” for an inference from characteristics.

10. Number every leaf in the top 50 cm [1]. Generate random numbers [1]. Select the leaf with each number, replacing repeats until there are 30 [1]. Repeat for the bottom 50 cm [1] and measure all leaves the same way [1]. [5] Examiner insight: state how chance is used; “choose leaves randomly” repeats the question.

11. (a) Island 3 [1] has the largest standard deviation (8.2 mm), so its tail lengths are the most spread out [1]. [2] (b) Island 1: 62.0 − 3.1 = 58.9 to 62.0 + 3.1 = 65.1 mm [1]. Island 2: 65.5 − 3.4 = 62.1 to 65.5 + 3.4 = 68.9 mm [1]. [2] (c) The ranges overlap, from 62.1 to 65.1 mm, so the difference may be due to chance [1]. A statistical test is needed to decide [1]. [2] Examiner insight: refer to both the means and the spread.

12. (a) X and Z differ at 7 of 12 positions: 7 / 12 × 100 = 58.3 % [1]. Y and Z differ at 5 of 12: 5 / 12 × 100 = 41.7 % [1]. [2] (b) W and X are most similar, with 1 difference (8.3 %) [1]. Y is next, with 2 differences from X and 3 from W [1]. Z is most different from all, so it is the most distant relative [1]. [3] (c) 4 / 36 × 100 = 11.1 %, compared with 1 / 12 × 100 = 8.3 % for the protein [1]. The genetic code is degenerate, so some base changes do not change the amino acid and appear only in the DNA [1]. [2] (d) Only 12 amino acids are compared, so one change alters the percentage greatly [1]. A single protein reflects only a little of the genome [1]. [2] Examiner insight: give a number for each step and say what it means for common ancestry.

Where marks are usually lost

• Writing that species “can breed” instead of “produce fertile offspring”.
• Describing courtship without mentioning species recognition.
• Errors in writing binomials.
• Quoting data without saying what it means for relationships.
• Concluding relatedness from appearance alone.
• Saying “randomly” with no method that uses chance.
• Ignoring standard deviations when comparing means.

Next steps

• Review the revision notes.
• Tick off the course on the printable checklist and visit the AQA A-level Biology hub.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

These questions follow sections 3.4.5 and 3.4.7 of the AQA AS and A-level Biology specification (7401 and 7402) for first teaching 2015, published by AQA.