IB Biology Tools for Biology worked solutions

The dependent variable is the quantity that is measured to see the response: the number of bubbles per minute (C). Distance from the lamp is the independent variable, and water temperature and mass of weed are controlled variables.

0.01/2.50\times100=0.4\% (B).

Electrons have a much shorter wavelength than visible light, so the resolution is greater (A). Electron microscopes cannot image living cells (B), use magnetic lenses (C) and usually need a stain or coating (D).

Magnification = image size over actual size =45/0.15=\times300 (C).

Graphite does not dissolve in the solvent, so it does not move up the paper and add coloured spots (D). Ink dissolves and would run with the pigments.

DNA is negatively charged, so it moves towards the positive electrode, and the smallest fragments move furthest (B).

N=M\times n/R=24\times30/6 (1) =120 (1).

Mean =(4+6+6+9+15)/5=40/5=8.0 (1). Median is the middle value =6 (1).

Since 2.9>2.14 the null hypothesis is rejected and the difference between the means is significant at p<0.05 (A). The test does not prove the hypothesis (D).

(a) 30-40=-10 percentage points (1). (b) (30-40)/40\times100=-25\% (1).

The five species are separate categories, so a bar chart with gaps is correct (A). A histogram is for a continuous variable, and a scatter graph or line graph needs two continuous variables.

The logger takes readings automatically at set intervals while no one is present (1). Accept: frequent readings, catches rapid changes, avoids observer error.

(a) Independent variable: temperature of the water bath (1). Dependent variable: volume of carbon dioxide collected in 5 minutes (1). (b) Two from: mass or volume of yeast; concentration and volume of sugar solution; pH; type of sugar; time (1 each, max 2). (c) The tubes must reach the bath temperature before the reaction starts, otherwise the temperature during the run is not the one recorded (1).

(a) First: \pm1\,\mathrm{cm^3}, 1/20\times100=5.0\% (1). Second: \pm0.25\,\mathrm{cm^3}, 0.25/20\times100=1.25\approx1.3\% (1). (b) The 25\,\mathrm{cm^3} cylinder, because its percentage uncertainty is smaller (1). (c) 14.0/35=0.40\,\mathrm{cm^3\,s^{-1}} (1).

(a) 16\,\mathrm{mm}=16\,000\,\mu\mathrm{m} (1); M=16\,000/20=\times800 (1). (b) Actual width =56/800=0.070\,\mathrm{mm} (1) =70\,\mu\mathrm{m} (1). (c) Ribosomes are about 25\,\mathrm{nm}, far below the 0.2\,\mu\mathrm{m} resolution of a light microscope, so an electron microscope is needed (1).

(a) 6.0/7.5=0.80, 4.5/7.5=0.60, 1.5/7.5=0.20 (2 for three correct, 1 for two). (b) The spot at 4.5\,\mathrm{cm} has R_f=0.60 (1). (c) Pencil does not dissolve in the solvent, so it does not run with the pigments (1). The solvent front must be marked at once because it evaporates and its position is lost (1).

(a) Taq polymerase adds nucleotides to the primers to make new strands (1). It is heat-stable, so it survives the 95\,{}^\circ\mathrm{C} denaturing step in each cycle (1). (b) Suspect 2, whose bands are at the same positions as the scene (1). (c) The 300 bp fragment (1), because smaller fragments move more easily through the gel (1).

(a) Area sampled =20\times0.5=10\,\mathrm{m^2} (1); density =90/10=9.0\,\mathrm{m^{-2}} (1). (b) 9.0\times400=3600 (1). (c) Lay tapes along two edges as axes (1); choose pairs of random numbers as coordinates and place each quadrat there (1). (d) Meadow A has the higher index so it is more diverse (1), because its individuals are spread more evenly among more species.

(a) N=M\times n/R=50\times64/8 (1) =400 (1). (b) Any two: the marks are permanent and do not harm the snails; there is no migration, birth or death between samples; marked snails mix randomly; sampling is random (1 each, max 2).

(a) 51.0/6=8.5\,\mathrm{cm} (1). (b) Differences -0.4,\ 0.8,\ -0.9,\ 0.3,\ 0.5,\ -0.3 (1); squares sum to 2.04 (1); s=\sqrt{2.04/5}=0.64\,\mathrm{cm} (1). (c) \mathrm{SE}=0.639/\sqrt6=0.26\,\mathrm{cm} (1). (d) The SE is the SD divided by \sqrt n, so it is smaller than the SD for any sample of more than one value (1).

(a) The null hypothesis is that there is no difference between the mean masses (1). (b) 1.9<2.10, so the difference is not significant (1) and the null hypothesis is not rejected (1). (c) \chi^2=(52-50)^2/50+(48-50)^2/50=0.08+0.08=0.16 (1). (d) 0.16<3.84 (1), so the counts fit a 1:1 ratio.

(a) A line graph (1), with temperature, the independent variable, on the horizontal axis (1). (b) The rate rises from 0.8 to 5.0\,\mu\mathrm{mol\,min^{-1}} between 10 and 40\,{}^\circ\mathrm{C} (1), then falls to 0.3 at 60\,{}^\circ\mathrm{C} (1). (c) The optimum is about 40\,{}^\circ\mathrm{C}, the temperature with the highest rate (1); it could lie between 30 and 50. (d) r=+0.98 is very strong and positive (1). (e) A coefficient only shows that the variables change together (1); cause needs a controlled experiment in which only temperature changes.

(a) Expected =160\times9/16,\ 3/16,\ 3/16,\ 1/16=90,\ 30,\ 30,\ 10 (1). (b) \chi^2=8^2/90+3^2/30+2^2/30+3^2/10=0.71+0.30+0.13+0.90=2.04 (2). (c) Degrees of freedom =4-1=3, critical value 7.81 (1); 2.04<7.81, so the null hypothesis is not rejected and the data fit the ratio (1).

(a) P: \pi\times9^2=254\,\mathrm{mm^2} (1). Q: \pi\times7^2=154\,\mathrm{mm^2} (1). (b) Any two: same size of disc and volume of antibiotic; same bacteria and thickness of lawn; same incubation time and temperature (1 each, max 2). (c) A disc soaked in sterile water is a control: any zone around the others is caused by the antibiotic and not by the paper (1).

(a) A=62/25\,000=0.00248\,\mathrm{mm} (1) =2.48\,\mu\mathrm m\approx2.5\,\mu\mathrm m (1). (b) 25\,\mathrm{nm}\times25\,000=625\,000\,\mathrm{nm} (1) =0.63\,\mathrm{mm} (1). (c) A ribosome would look less than a millimetre across, too small to measure accurately with a ruler (1). (d) Print or view at a higher magnification (1); a scale bar would also let the size be checked without relying on the stated magnification (1).

(a) t=0.7/\sqrt{1.4^2/10+1.1^2/10}=0.7/\sqrt{0.317}=0.7/0.563=1.24 (2). 1.24<2.10, so red and blue do not differ significantly (1). (b) t=4.2/\sqrt{1.1^2/10+1.0^2/10}=4.2/\sqrt{0.221}=4.2/0.470=8.9 (2); 8.9>2.10, so red and green differ significantly (1). (c) Any two controls or improvements: the lamps give equal light intensity at the same distance; same compost, water and temperature; more seedlings or repeats (1 each, max 2, may be given with a reason).

(a) X is present in 40 and bracken in 26 quadrats. Expected both =40\times26/80=13; X only =40\times54/80=27; bracken only =40\times26/80=13; neither =40\times54/80=27 (2 for all four, 1 for two). (b) \chi^2=(12-13)^2/13+(28-27)^2/27+(14-13)^2/13+(26-27)^2/27=0.077+0.037+0.077+0.037=0.23 (2). (c) Degrees of freedom =1, critical value 3.84 (1); 0.23<3.84, so there is no significant association (1). (d) A larger number of quadrats or repeating in other areas (1 if included).

(a) The best-fit line through the origin has gradient 3.33/30=0.111 per \mathrm{mmol\,dm^{-3}} (1), so 0.28/0.111=2.5\,\mathrm{mmol\,dm^{-3}} (2; reading 2.5 from a drawn line also earns the marks). (b) A blank of reagent with no coloured product sets the zero, so only the coloured product is measured (1). (c) A reading above 0.44 would be beyond the standards, so the line may not continue straight there; dilute the sample and repeat (1). (d) Draw a best-fit line through the points (1).

(a) (8.0-5.0)/10=0.30\,\mathrm{mg\,dm^{-3}\,min^{-1}} (1). (b) (8.0-7.2)/2=0.40\,\mathrm{mg\,dm^{-3}\,min^{-1}} (1). (c) The rate falls over the run (1), probably because the falling oxygen concentration limits aerobic respiration (1). (d) The rate falls from 0.40 in the first interval to 0.20 in the last (1). (e) The logger gives frequent readings without an observer, so the change in rate is clear (1).

(a) The percentage identity falls from 99\% for the chimpanzee to 48\% for the zebrafish (1). (b) The more similar the sequence, the more recently the species shared an ancestor with humans (1), so the chimpanzee is the closest relative and the zebrafish the most distant of the four (1). (c) One protein may evolve at a different rate from the whole genome, so the conclusion should be checked with other genes (1). The student should record the database, the date and whether the entry was curated by experts (1).

(a) The sign changes between the two solutions. Zero change at 0.2+0.2\times6.0/(6.0+1.0) (1) =0.37\,\mathrm{mol\,dm^{-3}} (1). (b) Above the isotonic concentration the solution has a lower water potential than the cells (1), so water leaves the cells by osmosis and the strips lose mass (1). (c) \mathrm{SE}=1.1/\sqrt5=0.49 (2).

(a) Year 1: 50\times60/10=300 (1). Year 2: 50\times70/7=500 (1). (b) (500-300)/300\times100=+67\% (2). (c) Any two: more food, fewer predators, milder weather, immigration (1 each, max 2). (d) The marked beetles did not mix with the others, or marks harmed them (1).

(a) 12\times2^8=12\times256=3072 (2). (b) DNA polymerase can only add nucleotides to an existing 3^{\prime} end (1), so the primers give it a starting point on each template strand (1). (c) At 95\,{}^\circ\mathrm{C} the hydrogen bonds between the strands break (1).

(a) Volume =\pi r^2h=\pi\times1.0^2\times24 (1) =75\,\mathrm{mm^3} (1). (b) Rate per gram per minute =75.4/(2.0\times10) (1) =3.8\,\mathrm{mm^3\,g^{-1}\,min^{-1}} (1). (c) The potassium hydroxide removes the carbon dioxide produced, so the volume change is due to oxygen uptake alone (1). (d) A control tube with glass beads of the same volume shows changes in temperature or pressure that are not due to the seeds (1).