Modelled on a typical one-hour first round of a national biology olympiad for students in their last two years of school: 30 single-answer multiple-choice questions across the syllabus followed by two data-analysis questions; the medal bands are only a rough guide for a strong student.
60minutes50marksCalculator allowed
Answer every question. The paper has 50 marks: 30 in Section A and 20 in Section B.
Section A: choose one answer for each question. There is no penalty for a wrong answer.
Section B: write your answers in full sentences where an explanation is asked for, and show your working in calculations.
A calculator is allowed. Give numerical answers to a sensible number of significant figures, with units.
The clock starts when you press Start. Multiple-choice answers are marked for you; for written answers you get the mark scheme and give yourself the marks. Your answers stay on this device.
0:00
Feeder round mock paper 2 TheLucidSTEM · Biology olympiad feeder round mock · 60 minutes · 50 marks · Calculator allowed Name: ______________________ Date: ____________ · MARK SCHEME
Data sheet
Useful relationships: solute potential of a solution ψs = -iCRT, where i = 1 for sucrose, C is the concentration in mol m-3, R = 8.31 J mol-1 K-1 and T is the temperature in K (1 Pa = 1 J m-3). Spearman's rank correlation coefficient rs = 1 - 6Σd2 / (n(n2 - 1)), where d is the difference in rank for each pair and n the number of pairs. Percentage change = 100 × (final - initial) / initial. 1 dm3 = 1000 cm3; 1 cm3 = 1000 mm3.
Section A: multiple choice
Suggested time 35 minutes. One mark each, no penalty for wrong answers.
1[1]
Which polysaccharide is made of β-glucose units joined by 1,4 glycosidic bonds into long unbranched chains that lie side by side, held together by many hydrogen bonds?
Answer C. Cellulose is made of β-glucose; every second unit is turned over, so the chains are straight and form hydrogen bonds with neighbouring chains, giving strong microfibrils. Amylose is also unbranched and 1,4-linked, but it is made of α-glucose and coils into a helix; glycogen and amylopectin are branched α-glucose polymers.
2[1]
Fish living in cold water keep their cell membranes fluid. Which change in the membrane phospholipids would make a membrane MORE fluid at a low temperature?
Answer A. The C=C double bonds of unsaturated tails put kinks in them, so the phospholipids cannot pack closely and the bilayer stays fluid in the cold. Saturated and longer tails pack more tightly and make the membrane more rigid; adding channel proteins does not raise fluidity.
3[1]
An enzyme made of four polypeptide chains is heated to 80 °C and loses its activity for good. Which level of its structure is NOT changed by the heating?
Answer D. Heat breaks hydrogen bonds, ionic bonds and hydrophobic interactions, so the helices and sheets, the folding of each chain and the way the four chains fit together are all lost. The primary structure is the sequence of amino acids held by covalent peptide bonds, which heating to 80 °C does not break.
4[1]
During the aerobic respiration of glucose, in which stages is carbon dioxide released?
Answer B. Pyruvate is decarboxylated in the link reaction (1 CO2 per pyruvate) and two more decarboxylations happen in each turn of the Krebs cycle. Glycolysis splits glucose into two pyruvate without losing any carbon, and oxidative phosphorylation makes water, not CO2.
5[1]
A diploid animal cell in G1 of the cell cycle contains 6 pg of DNA in its nucleus. How much DNA is in each nucleus formed at the end of telophase II of meiosis?
Answer B. DNA is copied in S phase, so the cell enters meiosis I with 12 pg. Meiosis I halves this to 6 pg per nucleus, and meiosis II halves it again to 3 pg, half the G1 value, as expected for a haploid gamete. 6 pg is the content after meiosis I only.
6[1]
In mice, the allele for yellow coat (AY) is dominant to the allele for agouti coat (A), but embryos with the genotype AYAY die before birth. Two yellow mice are mated many times. What fraction of the offspring that are born are yellow?
Answer C. AYA × AYA gives 1 AYAY : 2 AYA : 1 AA. The AYAY quarter dies, so of the mice born, 2 out of 3 are yellow. 3/4 is the usual monohybrid answer, which forgets that one class never appears.
7[1]
Snails of one species live on a patchy shore of pale sand and dark rocks. Birds catch most of the snails with shells of intermediate shade, while very pale and very dark snails survive well. What kind of selection is this?
Answer A. Selection against the middle of the range and in favour of both extremes is disruptive selection, which can split a population into two forms. Stabilising selection is the opposite, favouring the intermediate; directional selection favours one extreme only.
8[1]
Which pair of structures are homologous?
Answer D. The whale flipper and the human arm have the same set of bones (humerus, radius, ulna, carpals, digits) inherited from a common tetrapod ancestor, even though they do different jobs. The other pairs do similar jobs but evolved separately (convergent evolution), so they are analogous. The shark fin has no such limb bones.
9[1]
In phloem tissue, which cell has a nucleus and many mitochondria and actively loads sucrose into the neighbouring conducting cell?
Answer B. The companion cell uses ATP to pump protons out and then takes up sucrose by co-transport, passing it into the sieve tube element through plasmodesmata. The mature sieve tube element has lost its nucleus and most organelles; tracheids belong to the xylem and fibres are dead support cells.
10[1]
After fertilisation in a flowering plant, which part of the ovule develops into the seed coat (testa)?
Answer C. The integuments, the outer layers of the ovule, harden into the testa. The ovary wall is tempting, but it becomes the fruit wall (pericarp), which surrounds the whole seed; the fused polar nuclei and second male nucleus form the endosperm.
11[1]
The tip of an oat coleoptile is cut off and left on an agar block for two hours. The block is then placed on the LEFT half of the cut stump, and the stump is kept in the dark. What happens to the stump?
Answer A. Auxin from the tip collected in the agar. It diffuses down the left side of the stump only, where it makes the cells elongate faster, so the left side grows longer and the stump bends to the right. Light is not needed for this response because the uneven auxin is supplied directly.
12[1]
A CAM plant such as a pineapple opens its stomata mainly at night and closes them by day. At which time is the malic acid content of its leaf cell vacuoles highest?
Answer D. Through the night, CO2 taken in through the open stomata is fixed by PEP carboxylase and stored as malic acid in the vacuoles, so the store is fullest at the end of the night. During the day the malate is broken down to release CO2 for the Calvin cycle, so it is lowest at dusk.
13[1]
A plant grown without magnesium shows yellowing (chlorosis) first in its OLDEST leaves, while the young leaves stay green for longer. Why?
Answer B. Magnesium is at the centre of every chlorophyll molecule and is mobile in the phloem, so the plant takes it from old leaves to supply new growth, and the old leaves turn yellow first. An immobile element such as iron or calcium gives the opposite pattern, with the youngest leaves affected first.
14[1]
Chief cells in the stomach wall secrete the inactive enzyme pepsinogen. What first converts it into the active enzyme pepsin?
Answer C. Hydrochloric acid from the parietal cells removes a short peptide from pepsinogen, and the pepsin formed then activates more pepsinogen. Enterokinase is the tempting answer, but it acts in the duodenum and activates trypsinogen, not pepsinogen.
15[1]
Which event directly triggers ovulation in the human menstrual cycle?
Answer A. Near mid-cycle the high oestrogen from the mature follicle switches to positive feedback on the hypothalamus and pituitary, giving the LH surge that makes the follicle rupture about a day later. The fall in progesterone at the end of the cycle leads to menstruation, not ovulation.
16[1]
Antibodies of class IgG pass from a mother across the placenta into her fetus. What kind of immunity does this give the baby?
Answer B. The baby receives ready-made antibodies (passive) by a normal body process (natural). It makes no memory cells of its own, so the protection fades within months. Artificial passive immunity is an injection of antibodies, such as antivenom.
17[1]
When a skeletal muscle fibre contracts, which part of each sarcomere stays the same length?
Answer D. The A band is the length of the myosin (thick) filaments, which do not shorten; the filaments slide past each other. The I band and H zone, where the filaments do not overlap, both get shorter, and so does the sarcomere.
18[1]
In a vertebrate embryo, which germ layer gives rise to the brain and spinal cord?
Answer C. The neural plate is a strip of ectoderm that rolls up into the neural tube, which becomes the brain and spinal cord. The notochord, made from mesoderm, signals to the ectoderm above it to form the plate but does not become nervous tissue itself.
19[1]
A garden snail pulls back its tentacles when a twig touches it. If the harmless touch is repeated every minute, the snail soon stops responding, but it still withdraws when touched by something new. What kind of behaviour is the fading response?
Answer A. Habituation is the simplest kind of learning: a response to a repeated stimulus that brings neither reward nor harm gradually wanes. Conditioning needs a link between a stimulus or action and a reward or punishment; imprinting is learning during a short sensitive period early in life.
20[1]
Crows break open whelk shells by flying up and dropping them onto rocks. The table shows the mean number of drops needed to break a whelk from different heights.
Drop height / m
1
3
5
7
10
Mean number of drops
40
11
5.0
4.0
3.4
If a crow should keep the total height it climbs per whelk as small as possible, which drop height is best?
Answer C. Total climb = height × drops: 40 m at 1 m, 33 m at 3 m, 25 m at 5 m, 28 m at 7 m and 34 m at 10 m. The minimum is at 5 m. 10 m is tempting because it needs the fewest drops, but each drop costs twice the climb.
21[1]
A rare plant survives only as one small, isolated population of 30 individuals. Apart from chance disasters, why is it at greater risk of extinction than a large population?
Answer B. In a small population drift removes alleles at random and relatives mate, so harmful recessive alleles become homozygous (inbreeding depression) and there is less variation for adapting to change. The mutation rate per individual does not depend on population size, and selection is in fact less effective, not stronger, when drift is strong.
22[1]
Which land biome has the highest mean net primary productivity per square metre?
Answer D. Tropical rainforest has high light, warmth and rainfall all year and a long growing season, giving a mean NPP of about 2000 g m-2 per year, the highest of these biomes. Boreal forest has a short cold growing season and deserts are limited by water.
23[1]
On a rocky shore, the barnacle Chthamalus lives only on the upper shore where another barnacle, Balanus, is present. When Balanus is removed, Chthamalus settles and survives much lower down. The upper zone that Chthamalus occupies when Balanus is present is its
Answer A. The fundamental niche is the full range a species could occupy (here the upper and lower shore, seen when Balanus is removed). Competition from Balanus shrinks this to the realised niche, the upper zone only. Competitive release is the spreading out that happens after the competitor is removed.
24[1]
Use this key to identify a woodlouse, which has seven pairs of jointed legs, two pairs of antennae and a body made of plates.
1
Jointed legs present
go to 2
No jointed legs
go to 4
2
Three pairs of legs
Group P
More than three pairs of legs
go to 3
3
Four pairs of legs, no antennae
Group Q
More than four pairs of legs, two pairs of antennae
Group R
4
Body divided into many ring-like segments
Group S
Body not divided into segments
Group T
Answer C. Jointed legs: go to 2. More than three pairs: go to 3. Seven pairs (more than four) and two pairs of antennae: Group R, the crustaceans. Group Q (four pairs of legs, no antennae) is the arachnids; Group S needs no jointed legs, so the plated body does not lead there.
25[1]
Which description fits a cnidarian such as a jellyfish or a sea anemone?
Answer B. Cnidarians have two cell layers (ectoderm and endoderm with jelly-like mesoglea between), radial symmetry and a gut with one opening that serves as mouth and anus, ringed by tentacles with stinging cells. No true tissues and a porous body describes sponges; segmentation with a coelom describes annelids.
26[1]
Which feature is found in flowering plants (angiosperms) but NOT in conifers (gymnosperms)?
Answer D. 'Angiosperm' means enclosed seed: the ovules sit inside an ovary, which becomes the fruit. Gymnosperm ovules lie exposed on cone scales. Both groups make seeds, both are vascular plants and both have a dominant sporophyte.
27[1]
Which of these organisms is a eukaryote that is neither a plant, an animal nor a fungus?
Answer A.Amoeba is a single-celled eukaryote with a nucleus and is classed among the protists. Staphylococcus is a bacterium, so it is not a eukaryote at all; yeast is a fungus, even though it is single-celled, and a liverwort is a plant.
28[1]
A student wants to know whether a fungicide reduces the germination of seeds. She soaks 50 seeds in a solution of the fungicide dissolved in ethanol and counts how many germinate. Which is the best control?
Answer D. The control must differ from the test in one thing only, the fungicide, so any effect of the ethanol solvent and of the soaking appears in both groups. Unsoaked seeds are the tempting answer, but then a lower germination rate could be caused by the ethanol rather than the fungicide.
29[1]
In a survey of 20 towns, the number of nesting storks is strongly and positively correlated with the number of babies born each year (r = 0.85). What is the best conclusion?
Answer B. A correlation, however strong, does not show that one variable causes the other. Larger towns have more roofs for nests and more people, so town size is a likely confounding variable. An r of 0.85 is a strong correlation, so it is not meaningless.
30[1]
A suspension of yeast cells is counted with a haemocytometer. Each large square is 0.2 mm × 0.2 mm and the chamber is 0.1 mm deep. A total of 100 cells is counted in 5 large squares. What is the concentration of cells in the suspension?
Answer A. Volume of one square = 0.2 × 0.2 × 0.1 = 0.004 mm3; five squares = 0.02 mm3. 100 / 0.02 = 5000 cells per mm3 = 5.0 × 106 per cm3. Using the volume of one square only gives 2.5 × 107; forgetting to convert mm3 to cm3 gives 5.0 × 103.
Section B: data analysis
Suggested time 25 minutes (about 12 minutes per question). 20 marks.
31Finding the water potential of a potato[10 marks]
A student cut cylinders of potato tuber with a cork borer, trimmed them to the same length, blotted them dry and weighed them. One cylinder was put into each of six sucrose solutions and left for 24 hours at 20 °C. The cylinders were then blotted and weighed again. The results are in the table.
Sucrose concentration / mol dm-3
Initial mass / g
Final mass / g
Percentage change in mass
0.00
2.50
2.90
+16.0
0.10
2.40
2.63
0.20
2.50
2.58
+3.2
0.30
2.60
2.52
0.40
2.50
2.29
-8.4
0.50
2.50
2.21
-11.6
(a) Calculate the two missing percentage changes in mass, for 0.10 and 0.30 mol dm-3.
0.30 mol dm-3: 100 × (2.52 - 2.60) / 2.60 = -3.1% (accept -3.08 or -3.0); the minus sign is needed [1]
Dividing by the final mass instead of the initial mass (giving 8.7% and 3.2%) scores 0 for that value.
(b) Explain why the student worked out the percentage change in mass rather than simply the change in mass.
[1]
Mark scheme
Answer: The cylinders did not all start with the same mass, so percentage change allows a fair comparison.
The initial masses were different (for example 2.40 g and 2.60 g), so the percentage change allows the cylinders to be compared fairly / it takes account of the different starting masses [1]
(c) Use the data to estimate the sucrose concentration at which the potato tissue would neither gain nor lose mass. Show how you obtained your answer.
[2]
Mark scheme
Answer: About 0.25 mol dm-3
The change in mass is zero between 0.20 (+3.2%) and 0.30 mol dm-3 (-3.1%); method shown, by reading the x-intercept of a graph of percentage change against concentration, or by interpolation: 0.20 + 0.10 × 3.2 / (3.2 + 3.1) [1]
= 0.25 mol dm-3 (accept 0.24 to 0.26) [1]
An answer without any method shown scores the second mark only.
(d) At the concentration found in (c), the water potential of the potato cells equals that of the solution. Calculate the water potential of the potato tissue in MPa. Use the data on the cover.
[2]
Mark scheme
Answer: About -0.61 MPa
C = 0.25 mol dm-3 = 250 mol m-3 and T = 293 K, so ψs = -1 × 250 × 8.31 × 293 [1]
= -6.1 × 105 Pa = -0.61 MPa (the solution has no pressure potential, so this is its water potential); the minus sign is needed (allow error carried forward from (c), for example 0.24 gives -0.58 MPa) [1]
Common errors: using 0.25 instead of 250 (gives -610 Pa), or using 20 instead of 293 K.
(e) Explain, in terms of water potential, why the cylinder in 0.50 mol dm-3 sucrose lost mass.
[2]
Mark scheme
Answer: The solution had a lower (more negative) water potential than the cells, so water left the cells by osmosis.
The 0.50 mol dm-3 solution has a lower (more negative) water potential than the potato cells [1]
Water moves out of the cells by osmosis, down the water potential gradient, through the partially permeable plasma membranes (the cells may become plasmolysed) [1]
Do not credit 'sucrose moves into the cells' or answers about water 'concentration' without the idea of water potential or osmosis.
(f) Suggest one way to make the estimate in (c) more reliable.
[1]
Mark scheme
Answer: Repeat with several cylinders at each concentration and use mean values (or add more concentrations between 0.20 and 0.30 mol dm-3).
Any one of: use three or more cylinders at each concentration and calculate a mean; use more concentrations between 0.20 and 0.30 mol dm-3; blot every cylinder in the same standard way before weighing; cut all cylinders from the same tuber [1]
32Walking back in time across a sand dune[10 marks]
Ecologists ran a belt transect inland from the strand line of a sandy coast. New dune ridges form at the seaward edge, so the further a ridge is from the sea, the older it is; the ages were found from old maps. At each of six sites they recorded the soil, the number of plant species in ten quadrats and the percentage of bare sand. Fresh beach sand contains many shell fragments made of calcium carbonate.
Site
Distance from strand line / m
Approximate age / years
Soil organic matter / % of dry mass
Soil pH
Number of plant species
Bare sand / %
A
20
5
0.2
8.6
3
85
B
80
30
0.8
8.2
7
50
C
150
80
1.9
7.5
14
15
D
250
150
3.6
6.8
22
4
E
400
250
6.0
5.9
19
2
F
600
400
9.5
4.8
11
0
Site A is dominated by marram grass, site D by a mixed dune grassland and site F by heather with young pine trees.
(a) Describe and explain how the soil organic matter and the soil pH change from site A to site F.
[3]
Mark scheme
Answer: Organic matter rises (0.2% to 9.5%) as dead plant material builds up as humus; pH falls (8.6 to 4.8) as rain leaches out the calcium carbonate and decaying humus releases acids.
Description: organic matter rises from 0.2% to 9.5% while pH falls from 8.6 to 4.8 (both trends with figures, or the idea that the soil becomes richer in organic matter and more acidic with age) [1]
Organic matter: pioneer plants such as marram grass, and later more plants, add dead leaves and roots, which are decomposed into humus and build up over time [1]
pH: rain leaches the calcium carbonate shell fragments out of the older sand, and the decomposition of humus (and heather litter) releases organic acids, so the soil becomes acidic [1]
(b) Calculate the mean rate at which soil organic matter built up between sites C and E, in % per year.
[2]
Mark scheme
Answer: About 0.024% per year
Change in organic matter = 6.0 - 1.9 = 4.1% over 250 - 80 = 170 years [1]
Rate = 4.1 / 170 = 0.024% per year (accept 0.0241) [1]
Using the distances (4.1 / 250 m) scores 0 for the second mark, because the rate is asked per year.
(c) A student calculates Spearman's rank correlation coefficient for soil organic matter against the number of plant species at the six sites. For six pairs, the critical value at p = 0.05 is 0.886. Calculate rs and state what it shows.
[3]
Mark scheme
Answer: rs = 0.60, below 0.886, so there is no significant correlation at p = 0.05.
Ranks for organic matter A to F: 1, 2, 3, 4, 5, 6; ranks for species: 1, 2, 4, 6, 5, 3, so d = 0, 0, -1, -2, 0, 3 and Σd2 = 14 [1]
0.60 is less than the critical value 0.886, so the correlation is not significant at p = 0.05 / the null hypothesis of no correlation is accepted (allow error carried forward from a wrong rs) [1]
Credit also, but not instead of the conclusion: species number rises and then falls, so the relationship is not monotonic and Spearman's test is not a good way to describe it.
(d) Suggest two reasons why the number of plant species is lower at site F than at site D.
[2]
Mark scheme
Answer: Dominant heather and pine shade out and outcompete smaller plants; the acidic soil (pH 4.8) is tolerated by fewer species.
Any two of, one mark each [2]:
a few large dominant species (heather, pine) shade out / outcompete the smaller plants for light, water or nutrients (competitive exclusion);
the soil is now acidic (pH 4.8) and few species can tolerate such acid soil;
acid soil and slow decomposition of heather litter leave few available nutrients, such as nitrate;
no bare sand is left for pioneer and early species to colonise.
Do not credit 'there is less organic matter' (the data show more) or 'the soil is too young'.
TheLucidSTEM practice paper. Questions are original and written in the style of a first national round; this is not an official olympiad paper.