IB Biology The exam worked solutions

C. Convert first: 18\ \mathrm{mm}=18\,000\ \mu\mathrm{m}. Then magnification =\text{image}\div\text{actual}=18\,000\div3.0=6000. The value \times6 comes from dividing 18 by 3 without converting units; \times600 and \times60\,000 are powers-of-ten slips.

C. Percentage change =(52-40)\div40\times100=30\%. The value 12\% is the raw difference in milligrams; 23\% divides by the new value instead of the old one; 130\% is the new mass as a percentage of the old.

A. Distinguish needs a difference with both items stated. B gives a likeness, not a difference. C and D are true but each describes only one of the two vessels.

B. Heavy overlap of standard deviation bars means the data do not support a difference at that point. A and C are too strong: overlap does not prove equality or certainty about chance. D cannot be decided from the overlap alone.

C. q^2=0.04, so q=0.20 and p=1-0.20=0.80. Carriers are 2pq=2\times0.80\times0.20=0.32, which is 32\%. The value 16\% forgets the factor 2; 64\% is p^2 (homozygous dominant).

(a) Oxygen release rises from 2.1 to 4.0\ \mathrm{mm^3\,min^{-1}} between low and medium light [1], then levels off at about 4.4, a change within the error bars [1]. (b) (8.9-4.4)\div4.4\times100=102\% [2: 1 for the correct substitution, 1 for the answer]. (c) At low light the light-dependent reactions limit the rate; they supply too little ATP and reduced NADP to use extra carbon dioxide [2]. (d) Any two: temperature, shoot mass or leaf area, wavelength of light, distance of the lamp, time of exposure, species or source of the pondweed [1 each].

(a) ATP supplies the energy, when a phosphate group is transferred from it to the pump [1]. (b) Any three: the ion binds to the pump on the side of low concentration [1]; the phosphate from ATP changes the shape of the pump [1]; the ion is released on the side of higher concentration [1]; the pump returns to its original shape [1].

(a) N=\dfrac{M\times n}{m}=\dfrac{40\times50}{8}=250 snails [2: 1 for the substitution, 1 for the answer]. (b) Any two: no immigration, emigration, births or deaths between samples; marked snails mix randomly with the rest; marks are harmless and do not fall off; sampling is random [1 each]. (c) Too many marked snails appear in the second sample, so m is too large and the estimate is too low [1].

(a) Control: (38+40+39)\div3=39 beats per 15 s, so 39\times4=156 beats per minute [1]. Caffeine: (44+47+45)\div3=45.3, so 45.3\times4=181 beats per minute [1]. (b) Strengths, up to two: every caffeine animal beats faster than every control animal; the mean is about 25 beats per minute (16%) higher; a control in pond water was used [2]. Limitations, up to two: only three animals per group and no standard deviation or statistical test; one concentration and one source of animals; temperature and animal size not stated as controlled; a 15 s count multiplied by four magnifies counting error [2]. The data are consistent with the conclusion but too limited to prove it.

Any seven of: root hairs give a large surface area for uptake [1]; mineral ions are actively pumped in, lowering the water potential of the root cells [1]; water enters by osmosis from the soil [1]; water crosses the cortex to the xylem [1]; xylem vessels are hollow, with no end walls, forming a continuous column [1]; lignin in the walls prevents collapse under tension [1]; evaporation from the leaf mesophyll through the stomata (transpiration) creates tension that pulls water upwards [1]; hydrogen bonds give cohesion between water molecules, so the column is pulled as one [1]; adhesion to the xylem wall helps to hold the column [1].

(a) Any four: temperature; pH; substrate concentration; enzyme concentration; presence of an inhibitor [1 each]. (b) Any four: the inhibitor has a shape similar to the substrate [1]; it binds to the active site [1]; it blocks the substrate from binding, so fewer enzyme–substrate complexes form [1]; the rate falls [1]; the binding is reversible [1]; a higher substrate concentration can overcome the inhibition [1]. (c) Any seven: a pathway is a chain of enzyme-catalysed steps [1]; the end product binds to the first enzyme of the pathway [1]; it binds at a site other than the active site [1]; the active site changes shape and the substrate cannot bind [1]; the activity of the first enzyme falls and so does the whole pathway [1]; as the end product accumulates, inhibition grows (negative feedback) [1]; when the end product is used up it detaches and the enzyme works again [1]; this stops intermediates and the end product building up and saves substrate and energy [1]; the concentration of the product stays within narrow limits [1]. In a live paper one more mark is available for the construction of the whole answer.