Biology 0610 | Paper 4 Mock 01 | Theory (Extended)

This question uses trial data on herbicide-resistant GM soya to test the idea of genetic modification, competition, and a percentage-change calculation.

(a) A genetically modified organism is one whose genetic material (DNA) has been changed [1], usually by inserting a gene from another organism [1]. [2]

(b) Comparing the sprayed fields, spraying kills the weeds in the GM field (only 4 remain) [1] but does not damage the GM crop [1]; in contrast, spraying ordinary soya damages the crop and lowers its yield (from 2.1 to 1.4 t per ha) [1]. [3]

(c) The sprayed GM soya yields most because there are far fewer weeds, so there is less competition with the crop [1] for light, water, mineral ions and space [1]; and because the herbicide does not harm the resistant GM crop, more soya can grow [1]. [3]

(d) \[ \text{percentage decrease} = \frac{42-4}{42}\times100 = \frac{38}{42}\times100 \approx 90.5\% \] So the number of weeds falls by about 90.5% (accept 90-91%). [2, 1 method + 1 answer]

(e) Any two disadvantages [3, max 3]: weeds may themselves become resistant to the herbicide over time [1]; loss of biodiversity as fewer wild plants and the animals depending on them survive [1]; GM seed is expensive and farmers become dependent on the seed company [1].

(f) The enzyme used to cut the resistance gene out of another organism's DNA is a restriction enzyme. [1]

(a) The organ that monitors and controls blood glucose is the pancreas [1].

(b) When blood glucose rises, the hormone released is insulin [1].

(c) Insulin brings a raised glucose level down by making the liver and body cells take up glucose from the blood [1]; the liver converts the excess glucose into glycogen [1], which is stored, so the blood glucose concentration falls back to normal [1].

(d) When glucose falls too low, the pancreas releases the hormone glucagon [1]; glucagon makes the liver convert stored glycogen back into glucose [1]; the glucose is released into the blood, raising the concentration back to normal [1].

(e) A person with type 1 diabetes may need to inject insulin after a meal because they cannot make their own insulin [1]; so after a sugary meal their blood glucose would rise too high and stay high [1]; the injected insulin makes the liver and cells take up the glucose, lowering the level back to normal [1].

This question uses the lock-and-key idea to explain enzyme specificity and the effect of temperature and pH on amylase.

(a) Amylase breaks starch down into maltose (a sugar) [1].

(b) Amylase breaks down starch but not protein because the active site of amylase has a specific shape [1]; only starch (its complementary substrate) fits into that active site [1]; protein molecules have a different shape, so they cannot bind to the active site and are not broken down [1].

(c) As the temperature falls below 37 °C the enzyme and substrate molecules have less kinetic energy [1]; so they move more slowly and collide less often [1]; this means fewer enzyme-substrate complexes form per second, so the rate of reaction falls [1]. (Note that the enzyme is not denatured by cooling; activity returns if it is warmed again.)

(d) Amylase stops working in the stomach because the stomach is very acidic (a low pH) [1]; this is far below the optimum pH for amylase [1]; the low pH denatures the enzyme, changing the shape of its active site [1]; so the starch substrate can no longer bind to the active site and no reaction occurs [1].

(e) Any one variable that must be controlled when investigating temperature [1]: pH; enzyme (amylase) concentration; substrate / starch concentration; or the volume of the solutions used.

Exam tip: distinguish the two effects carefully: cold slows an enzyme by reducing collisions (reversible), while extreme pH or high heat denatures it by changing the active site (usually permanent).

This question tests reading catch data and explaining how a catch limit lets an overfished population recover.

(a) From year 1 to year 4 the mass caught fell sharply from year 1 to year 2 (85 to 40 thousand tonnes) [1], then rose again from year 2 to year 4 (40 up to 70) [1]. The initial fall reflects the limit imposed in year 2, and the later rise shows the stock recovering.

(b) A catch limit lets the population recover because fewer fish are removed / caught [1]; so more fish survive to reach breeding age and reproduce [1]; and the population can increase back towards its original size [1].

(c) Any two other conservation methods (1 each): control net mesh size so young fish can escape; set a minimum catchable size; create protected areas or closed seasons; or impose fishing quotas [2].

(a) The atmospheric gas cycled in the carbon cycle is carbon dioxide [1].

(b) The labelled processes are: A = photosynthesis (removes carbon dioxide from the air into plants) [1]; B = respiration (returns carbon dioxide to the air) [1].

(c) Decomposers return carbon to the air because decomposers (bacteria and fungi) feed on / break down dead organisms and waste [1]; as they do so they respire [1], which releases carbon dioxide back into the atmosphere [1].

(d) Burning fossil fuels can raise Earth's temperature because fossil fuels contain carbon (locked up from ancient organisms) [1]; combustion releases this as carbon dioxide [1]; carbon dioxide is a greenhouse gas that absorbs / traps heat re-radiated from the Earth's surface [1]; the resulting enhanced greenhouse effect raises the global temperature [1].

(e) Two ways, other than decomposition, that add carbon dioxide to the air: respiration (of plants and animals) [1]; combustion / burning of fuels [1].

(f) Two ways humans could reduce carbon dioxide released (1 each): burn less fossil fuel / use renewable energy sources [1]; plant more trees and reduce deforestation (so more CO2 is removed by photosynthesis) [1] (max 2).

This question is about genetic modification, using a Bt maize example and yield data.

(a) A genetically modified organism is one whose genes / DNA have been changed, for example by adding a gene from another organism [1].

(b) Mean yield of the GM variety across the three seasons:

\[ \frac{9.4 + 9.1 + 9.8}{3} = \frac{28.3}{3} = 9.43 \text{ t/ha} \]

So the mean yield is 9.43 t/ha (accept 9.4) [2; 1 for correct method, 1 for the answer].

(c) Any two differences from the table [1 each]: the GM variety has a higher yield in every season (about 9.1 to 9.8 t/ha compared with 6.8 to 7.5 t/ha); and the GM crop needs / uses only a small amount of insecticide (about 0.4 to 0.6 kg/ha).

(d) The bacterial gene increases yield because it makes the maize produce a (Bt) protein / toxin [1]; this toxin kills the insect pests that feed on the crop [1]; so less of the leaf and crop is eaten or damaged [1]; leaving more healthy leaf for photosynthesis and more grain surviving to harvest [1].

(e) Any two disadvantages, developed for the 4 marks [2 marks each, max 4]: loss of biodiversity / harm to non-pest (beneficial) insects such as pollinators [1][1]; pests may become resistant to the toxin over time, making it useless [1][1]; the GM seed is expensive and farmers become dependent on the company that sells it [1][1]; possible unknown effects on human health, or the escape / spread of the gene to wild plants [1][1].

(f) The process used to insert the bacterial gene into the maize cells is genetic engineering (genetic modification) [1].

Exam tip: when a calculated mean does not divide exactly, quote it to the same precision as the data (one decimal place here) and show the sum over the number of values for the method mark.