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 1 TheLucidSTEM · Biology olympiad feeder round mock · 60 minutes · 50 marks · Calculator allowed Name: ______________________ Date: ____________ · MARK SCHEME
Data sheet
Useful relationships: Hardy-Weinberg p + q = 1 and p2 + 2pq + q2 = 1. Logistic growth dN/dt = rN(1 - N/K). Hamilton's rule: an altruistic act can spread when rB > C. Q10 = (rate at T + 10 °C) / (rate at T). 1 dm3 = 1000 cm3.
Section A: multiple choice
Suggested time 35 minutes. One mark each, no penalty for wrong answers.
1[1]
Ouabain blocks the Na+/K+-ATPase in the basal membrane of intestinal epithelial cells. Which process is the first to slow down as a direct result?
Answer C. The co-transporter is secondary active transport: it relies on the low Na+ concentration inside the cell that the pump keeps up. Stop the pump and the Na+ gradient runs down, so glucose uptake from the lumen slows. GLUT2 is facilitated diffusion down the glucose gradient, which only changes later.
2[1]
A hydrolytic enzyme is made on the rough endoplasmic reticulum and must end up inside lysosomes. Which tag, added in the Golgi apparatus, sends it there?
Answer A. Mannose-6-phosphate is recognised by receptors in the trans-Golgi that package the enzyme into vesicles for lysosomes. KDEL returns proteins to the ER lumen; ubiquitin marks proteins for the proteasome; a nuclear localisation signal acts on cytosolic proteins.
3[1]
In the coding strand of a gene, a single base substitution changes the mRNA codon UAC (tyrosine) to UAG. What is the most likely effect on the protein?
Answer D. UAG is a stop codon, so this is a nonsense mutation and the polypeptide is cut short. A substitution never shifts the reading frame; that needs an insertion or deletion that is not a multiple of three.
4[1]
A pure non-competitive inhibitor binds an enzyme at a site away from the active site and binds equally well whether or not substrate is bound. Compared with no inhibitor, what happens to Vmax and Km?
Answer B. The inhibitor in effect removes a fraction of active enzyme, so Vmax falls; the enzyme that remains free has its normal affinity, so Km is unchanged. Unchanged Vmax with higher Km is competitive inhibition; both falling together is uncompetitive inhibition.
5[1]
The coding region of an mRNA, from the first base of the start codon AUG to the last base of the stop codon, is 1503 nucleotides long. How many amino acids does the newly made polypeptide contain?
Answer A. 1503 / 3 = 501 codons. The last one is the stop codon, which codes for no amino acid, so the chain has 500 amino acids (the first is methionine from AUG).
6[1]
A PCR reaction starts with 5 double-stranded copies of a target sequence. If every cycle doubles the number of copies, how many copies are there after 8 cycles?
Answer C. Copies = 5 × 28 = 5 × 256 = 1280. 40 assumes linear growth (5 × 8); 390 625 is 58.
7[1]
Red-green colour blindness is X-linked recessive. A woman who is a carrier has children with a man who has normal colour vision. What is the probability that their next child is a colour-blind boy?
Answer D. The child must be a boy (1/2) and receive the mother's recessive X (1/2): 1/2 × 1/2 = 1/4. The tempting 1/2 is the probability that a SON is affected, not that a child is an affected boy.
8[1]
A recessive condition affects 1 in 2500 babies in a large, randomly mating population. About what fraction of the population are unaffected carriers?
Answer B. q2 = 1/2500, so q = 0.02 and p = 0.98. Carriers = 2pq = 2 × 0.98 × 0.02 = 0.0392, about 1 in 26. 1 in 50 is q itself, not 2pq.
9[1]
Two true-breeding white-flowered lines of sweet pea are crossed and all the F1 plants have purple flowers. Selfing the F1 gives 9 purple : 7 white in the F2. If every white F2 plant is selfed, what fraction of them will give only white offspring?
Answer D. A 9:7 ratio means purple needs a dominant allele at both genes (A_B_). Every white plant is aa or bb (or both). Selfing an aa plant gives only aa offspring, and selfing a bb plant gives only bb offspring, so every white plant breeds true for white. 3/7 counts only the plants homozygous at both genes, but that is not needed.
10[1]
A plant heterozygous at two linked genes (AaBb, with A and B on the same chromosome) is test-crossed with aabb. The offspring are 42 AB, 8 Ab, 9 aB and 41 ab. What is the distance between the two genes?
Answer A. The recombinants are the two small classes: (8 + 9)/100 = 17%, so 17 map units (cM). 83 is the parental fraction; 8.5 halves the recombinant fraction for no reason.
11[1]
Two populations of a tree frog breed in the same ponds. One population calls and mates in early spring, the other in early summer, so they never interbreed. What kind of reproductive isolation is this?
Answer C. Different breeding seasons in the same place is temporal (prezygotic) isolation. It is not geographical, because they share the ponds, and hybrid inviability acts only after a zygote has formed.
12[1]
A plant species with 2n = 14 is crossed with a related species with 2n = 18. The hybrid is sterile, but a doubling of its chromosomes gives a fertile new species. What is the chromosome number of this new species?
Answer B. Gametes carry 7 and 9 chromosomes, so the hybrid has 16. Doubling gives an allotetraploid with 32, which is fertile because every chromosome now has a partner at meiosis.
13[1]
A well-watered tree is transpiring rapidly on a sunny afternoon. Where is the water potential lowest (most negative)?
Answer D. Water moves down a water potential gradient from soil to root to leaf to air. Unsaturated air has a very negative water potential (often below -50 MPa), far lower than any part of the plant.
14[1]
Isolated chloroplasts are lit only with far-red light of wavelength above 690 nm, which photosystem I absorbs well but photosystem II hardly absorbs. What do the chloroplasts mainly produce?
Answer B. With PSII idle, water is not split, so little O2 is released and no electrons flow from water to NADP+. PSI can still drive cyclic electron flow, which pumps protons and makes ATP.
15[1]
When a barley grain absorbs water, amylase appears in the aleurone layer and starts to digest the starch in the endosperm. Which signal triggers this?
Answer A. The embryo releases gibberellin, which switches on transcription of the amylase gene in the aleurone cells. Abscisic acid works the other way, keeping the seed dormant.
16[1]
In the root endodermis, the Casparian strip blocks the apoplast pathway. To reach the xylem, water and mineral ions must therefore cross
Answer C. The waterproof suberin band stops movement in the cell walls, so water and ions must pass through a plasma membrane into the cytoplasm of the endodermal cells. This lets the plant control which ions enter the xylem.
17[1]
At rest a person has a heart rate of 70 beats per minute, an end-diastolic volume of 120 cm3 and an end-systolic volume of 50 cm3. What is the cardiac output?
Answer B. Stroke volume = 120 - 50 = 70 cm3. Cardiac output = 70 × 70 = 4900 cm3 per minute = 4.9 dm3 per minute. 8.4 uses the end-diastolic volume instead of the stroke volume.
18[1]
A toxin blocks the voltage-gated K+ channels of a neurone but leaves the Na+ channels working. What is the main effect on an action potential?
Answer D. Depolarisation depends on Na+ channels, which still work. Repolarisation normally depends on K+ leaving through the voltage-gated channels; with them blocked, the membrane returns to rest only slowly, through leak channels and Na+ channel inactivation.
19[1]
How does a cytotoxic (CD8) T cell recognise a body cell infected with a virus?
Answer A. Almost all nucleated cells display fragments of their own proteins, including viral ones, on MHC class I, which CD8 T cells inspect. MHC class II is found on antigen-presenting cells and is read by helper (CD4) T cells.
20[1]
A person breathes 12 times per minute with a tidal volume of 500 cm3. The anatomical dead space is 150 cm3. What is the alveolar ventilation?
Answer C. Each breath brings (500 - 150) = 350 cm3 of fresh air to the alveoli: 350 × 12 = 4200 cm3 per minute. 6000 is the total (minute) ventilation; 1800 is the dead-space ventilation.
21[1]
When acidic chyme enters the duodenum, cells in its wall release the hormone secretin into the blood. What is the main effect of secretin?
Answer D. Secretin makes the pancreatic duct cells secrete hydrogencarbonate, which neutralises the acid. Gall bladder contraction is mainly caused by cholecystokinin (CCK), and secretin reduces, not increases, stomach acid secretion.
22[1]
In untreated type 1 diabetes, glucose appears in the urine. Why?
Answer B. Blood glucose is so high that more glucose is filtered than the limited number of Na+-glucose co-transporters in the proximal convoluted tubule can take back (the transport maximum is exceeded). Reabsorption of glucose in the kidney does not need insulin.
23[1]
The net primary productivity of a grassland is 20 000 kJ m-2 per year. 10% of it is passed to herbivores, and 15% of the herbivores' energy is passed to carnivores. How much energy reaches the carnivores?
Answer D. 20 000 × 0.10 = 2000 kJ to herbivores; 2000 × 0.15 = 300 kJ m-2 per year to carnivores. 2000 stops one level too early; 3000 uses 15% of the producers.
24[1]
A population grows logistically with intrinsic rate of increase r = 0.4 per year and carrying capacity K = 1000. Using dN/dt = rN(1 - N/K), what is the largest possible growth rate of the population?
Answer A. Growth is fastest at N = K/2 = 500: dN/dt = 0.4 × 500 × (1 - 0.5) = 100 individuals per year. 400 is rK, which ignores the brake from crowding.
25[1]
A bird gives up raising one of its own offspring in order to help its full sibling (coefficient of relatedness 0.5) raise extra offspring. According to Hamilton's rule, this helping can spread only if the sibling raises at least how many extra offspring as a result?
Answer C. Hamilton's rule: rB > C. With C = 1 offspring and r = 0.5, B must be greater than 1/0.5 = 2. Counting genes in offspring gives the same result: the bird shares 0.5 with its own offspring and 0.25 with its nephews and nieces, so 0.25B > 0.5 × 1, again B > 2.
26[1]
Which process in the nitrogen cycle converts ammonium ions into nitrite ions?
Answer B.Nitrosomonas oxidises NH4+ to NO2-, the first step of nitrification; Nitrobacter then oxidises nitrite to nitrate. Denitrification turns nitrate into N2 gas.
27[1]
Which feature is present in every chordate at some stage of its life?
Answer A. All chordates, including lancelets and tunicates, have a notochord (with a dorsal hollow nerve cord, pharyngeal slits and a post-anal tail) at some stage. Only vertebrates have a vertebral column, and many chordates have no limbs.
28[1]
Which feature distinguishes archaea from bacteria?
Answer C. Archaeal membrane lipids are isoprenoid chains joined to glycerol by ether bonds, whereas bacteria (and eukaryotes) use ester-linked fatty acids. Bacteria, not archaea, have peptidoglycan walls; both are prokaryotes with 70S ribosomes.
29[1]
A student measures the heights of 30 plants grown with a new fertiliser and 30 grown without it. The heights in each group are normally distributed. Which statistical test should she use to see whether the mean heights differ significantly?
Answer B. Comparing the means of two independent samples of normally distributed, continuous data calls for a t-test. Chi-squared compares counts in categories; the two correlation tests look for a relationship between two variables measured on the same individuals.
30[1]
A bacterial culture is diluted by a factor of 106, and 0.1 cm3 of the diluted suspension is spread on an agar plate. After incubation the plate has 48 colonies. What was the concentration of living bacteria in the original culture?
Answer A. 48 colonies in 0.1 cm3 means 480 per cm3 of diluted suspension. Multiply by the dilution factor: 480 × 106 = 4.8 × 108 per cm3. Forgetting the 0.1 cm3 gives 4.8 × 107.
Section B: data analysis
Suggested time 25 minutes (about 12 minutes per question). 20 marks.
31Keeping warm: a mouse and a lizard[10 marks]
The rate of oxygen consumption of a mouse and of a lizard, each of mass 25 g, was measured at different air temperatures. Each animal was rested in a sealed chamber for one hour at each temperature before the reading. The results are in the table.
Air temperature / °C
Mouse: O2 consumption / cm3 g-1 h-1
Lizard: O2 consumption / cm3 g-1 h-1
5
7.2
0.05
10
6.0
0.08
15
4.8
0.12
20
3.6
0.18
25
2.4
0.27
30
2.4
0.40
35
2.7
0.60
Take the energy released per dm3 of oxygen consumed as 20 kJ. The body temperature of the mouse stays close to 36 °C throughout.
(a) State the range of air temperatures over which the mouse's metabolic rate is at its minimum. Explain why its oxygen consumption rises as the air temperature falls below this range.
[3]
Mark scheme
Answer: 25-30 °C (thermoneutral zone); below it the mouse loses more heat and must make more heat by respiration to keep body temperature constant.
Range 25 to 30 °C (accept 'between about 25 and 30 °C', or 'thermoneutral zone 25-30 °C') [1]
The mouse is an endotherm and keeps its body temperature (about 36 °C) constant; at lower air temperature the gradient from body to air is larger, so it loses heat faster [1]
It must produce extra heat to replace this loss, by a higher rate of respiration (for example shivering, brown fat / non-shivering thermogenesis), which uses more oxygen [1]
Do not credit 'the mouse is colder so it respires faster' without the idea of keeping body temperature constant.
(b) The temperature coefficient Q10 is the factor by which a rate increases for a 10 °C rise in temperature. Calculate Q10 for the lizard's oxygen consumption between 20 °C and 30 °C.
[2]
Mark scheme
Answer: Q10 = 0.40 / 0.18 = 2.2
Q10 = rate at 30 °C / rate at 20 °C = 0.40 / 0.18 [1]
= 2.2 (accept 2.22 or 2.2 to 2.3) [1]
An answer of 0.45 (ratio upside down) or a difference such as 0.22 scores 0.
(c) Calculate the energy the mouse would need per day to stay at an air temperature of 10 °C.
[2]
Mark scheme
Answer: 72 kJ per day
Oxygen per hour = 6.0 × 25 = 150 cm3 = 0.150 dm3, so energy per hour = 0.150 × 20 = 3.0 kJ [1]
Per day = 3.0 × 24 = 72 kJ [1]
Common errors: forgetting to multiply by the mass (2.9 kJ per day), or treating cm3 as dm3 (72 000 kJ). Allow error carried forward for the second mark.
(d) Between 5 °C and 25 °C the mouse's oxygen consumption falls in a straight line as air temperature rises. Extend this line to find the air temperature at which it would reach zero, and explain what this temperature means.
[2]
Mark scheme
Answer: About 35 °C, close to the mouse's body temperature: when air and body are at the same temperature there is no heat loss to make up.
Gradient = -1.2 per 5 °C = -0.24 cm3 g-1 h-1 per °C; zero at 5 + 7.2/0.24 = 35 °C (accept 34 to 36 °C) [1]
This is (about) the body temperature of the mouse; heat loss is proportional to the difference between body and air temperature, so at this air temperature no extra heat production would be needed [1]
(e) At 10 °C the mouse uses 75 times as much oxygen per gram as the lizard. Suggest why the lizard can survive on much less food than the mouse.
[1]
Mark scheme
Answer: The lizard is an ectotherm: it does not use respiration to keep its body warm, its body cools to air temperature and its metabolic rate falls.
Any one of: the lizard is an ectotherm and does not spend energy producing heat to keep body temperature above air temperature; its body temperature falls with the air temperature, so its metabolic rate is low (Q10 effect) [1]
32Mutants of the lac operon[10 marks]
In E. coli the enzyme β-galactosidase is coded by the gene lacZ in the lac operon. The operon has a promoter and an operator, and is controlled by a repressor protein coded by the separate gene lacI. Lactose (converted inside the cell to the inducer allolactose) can bind to the repressor.
The activity of β-galactosidase was measured in wild-type bacteria and in mutants P, Q, R and S, grown with and without lactose. Each mutant has one mutation in or near the lac operon. Partial diploids were then made by giving each mutant a plasmid carrying a complete wild-type copy of lacI and of the lac operon (promoter, operator, lacZ and the permease gene lacY). One working copy of lacZ gives 1 unit when fully repressed and 1000 units when fully induced.
Strain
No lactose / units
Lactose / units
Wild type
1
1000
P
1000
1000
Q
1
1
R
1000
1000
S
0
0
P + plasmid
2
2000
Q + plasmid
2
2
R + plasmid
1000
2000
S + plasmid
1
1000
(a) Explain the results for the wild type.
[2]
Mark scheme
Answer: Without lactose the repressor binds the operator and blocks transcription; the inducer binds the repressor, changes its shape and releases it, so lacZ is transcribed.
Without lactose, the repressor binds to the operator and stops RNA polymerase transcribing the operon (the 1 unit is a low basal level) [1]
With lactose, the inducer binds to the repressor and changes its shape so it can no longer bind the operator; the genes are transcribed and the enzyme made [1]
(b) Mutants P and R behave the same on their own. Use the partial diploid results to decide which one has a mutation in lacI and which one has a mutation in the operator. Explain your reasoning.
[4]
Mark scheme
Answer: P is the lacI mutant (repressor missing, made good in trans by the plasmid); R is the operator mutant (acts only in cis, so its own operon stays switched on).
P has the lacI mutation (no working repressor) [1]
In P + plasmid, the repressor made from the plasmid's lacI is a diffusible protein that binds the operators of BOTH operons, so both are switched off without lactose (2 units) and both are induced with lactose (2000 units): the mutation is recessive and can be complemented in trans [1]
R has the operator mutation (repressor can no longer bind the operator) [1]
In R + plasmid, the wild-type repressor is present but still cannot switch off the chromosomal operon, which keeps making 1000 units without lactose; only the plasmid operon is repressed. An operator affects only the genes on the same DNA molecule (acts in cis) [1]
Reversing P and R scores at most the two reasoning marks if the cis/trans idea is clearly correct.
(c) Suggest what is wrong with the repressor in mutant Q, and explain why the Q mutation is dominant in the partial diploid.
[2]
Mark scheme
Answer: Q makes a repressor that cannot bind the inducer (a super-repressor); it stays on both operators even when the wild-type repressor is released.
The Q repressor still binds the operator but cannot bind the inducer / allolactose (accept: the inducer-binding site is altered), so it is never released [1]
It binds to the operators of both the chromosomal and the plasmid operon, so even though the wild-type repressor is released by lactose, both operons stay blocked (2 units) [1]
Do not accept 'Q has no repressor' (that would give constitutive expression) or 'Q has a faulty lacZ' (the plasmid would then give 1000 units with lactose).
(d) Mutant S makes no β-galactosidase at all. Suggest two different places where its mutation could be, both consistent with the data, and explain why the data rule out a mutation in lacY alone.
[2]
Mark scheme
Answer: In lacZ (no working enzyme) or in the promoter of the chromosomal operon (no transcription). A lacY mutant would still have a working chromosomal lacZ, giving 2 and 2000 units with the plasmid.
Both of: in the coding sequence of lacZ (for example a nonsense or frameshift mutation, giving no active enzyme); in the promoter of the chromosomal operon so RNA polymerase cannot bind [1]
With the plasmid, S gives only 1 and 1000 units, the values for ONE working lacZ (the plasmid's), so the chromosomal lacZ is not expressed or not working; a lacY-only mutant has a working chromosomal lacZ and, with permease supplied by the plasmid, would give 2 and 2000 units (accept: on its own a lacY mutant would still show basal activity, not 0) [1]
TheLucidSTEM practice paper. Questions are original and written in the style of a first national round; this is not an official olympiad paper.