EXAM QUESTIONS

• FRESHWATER MARSHES HAVE ONE OF THE HIGHEST RATES OF GROSS PRIMARY PRODUCTION (GPP) AND NET PRIMARY PRODUCTION (NPP) OF ALL ECOSYSTEMS.

  CARBON EFFICIENCY (CUE) IS THE RATIO NPP:GPP FRESHWATER MARSHES HAVE A HIGH CUE.

  USE YOUR KNOWLEDGE OF NPP TO EXPLAIN WHY FRESHWATER MARSHES HAVE A HIGH CUE.

  1. EXPLANATION: low respiration.
  2. ADVANTAGE: more growth / biomass / colonisation
• FRESHWATER MARSH SOILS ARE NORMALLY WATERLOGGED. THIS CREATES ANAEROBIC CONDITIONS.

  USE THE NITROGEN CYCLE TO SUGGEST WHY THESE SOILS CONTAIN RELATIVELY HIGH CONCENTRATIONS OF AMMONIUM COMPOUNDS AND LOW CONCENTRATIONS OF NITRITE IONS AND NITRATE IONS.

  1. Less Nitrification / active nitrifying bacteria
  2. Less oxidation of ammonium ions to nitrite ions and to nitrate ions
  3. more denitrification / denitrifying bacteria
  4. so more nitrate ions converted to nitrogen gas
• (IMAGE) THE STUDENT USED THE EQUATION, HOWEVER SHE SUBSTITUTED HEIGHT FOR BIOMASS BECAUSE DIDNT WANT TO DESTROY THE PLANTS TO MEASURE THEIR BIOMASS.

  STATE THE ASSUMPTION SHE HAS MADE AND SUGGEST WHY THIS ASSUMPTION MIGHT NOT BE VALID.

  1. assumed height is directly proportional to biomass
  2. plants may put biomass into other aspects of growth other than height such as root growth.

  
• DESCRIBE THE ROLE OF SAPROBIONTS IN THE NITROGEN CYCLE

  1. They use enzymes to decompose proteins / DNA / RNA / urea
  2. releasing ammonium ions ( they perform Ammonification )
• ONE ENVIRONMENTAL ISSUE ARISING FROM THE USE OF FERTILISERS IS EUTROPHICATION. EUTROPHICATION CAN CAUSE WATER TO BECOME CLOUDY.

  YOU ARE GIVEN SAMPLES OF WATER FROM THREE DIFFERENT RIVERS.

  DESCRIBE HOW YOU WOULD OBTAIN QUANTITIVE MEASUREMENT OF THEIR CLOUDINESS.

  1. use of colorimeter
  2. measure absorbance / transmission of light
  3. standardisation of method - same volume of water, zeroing colorimeter, same wavelength of light, shaking the sample
• (IMAGE) SUGGEST WHY IN THIS INVESTIGATION THE SCIENTISTS USED THESE UNITS

  1. Because there’s very little ammonia in the soil
  2. because avoids use of many decimal places in the results
  3. g-1 to allow comparison between samples

  
• (IMAGE) THE SCIENTISTS CONCLUDED THAT THE SOIL MIXTURE EXPERIMENT SHOWED THERE WERE DIFFERENT COMMUNITIES OF BACTERIA IN SOILS A AND B.

  WHAT EVIDENCE FROM FIGURE 1 SUPPORTS THEIR CONCLUSION? GUVE REASONS.

  1. pH 4.3 / B has the fastest breakdown of ammonia
  2. A and B / mixture of pH 6.9 slowest
  3. suggests bacteria at B doesn’t work well at pH 6.9

  
• (IMAGE) IN WHICH SPECIES WAS THE NUMBER OF COPIES OF MRNA MORE AFFECTED BY CHANGES IN SOIL PH FROM 4.9 TO 7.5? USE A CALCULATION

  SPECIES S BECAUSE

  1. Species S change of 990,000 per gram of soil
  2. species T change of 9,900 per gram of soil

  
• THIS METHOD ALLOWED THE SCIENTISTS TO ESTIMATE THE EXPRESSION OF THE amoA GENE IN EACH CULTURE BUT NOT THE GROWTH OF THE BACTERIAL POPULATION IN EACH CULTURE. EXPLAIN WHY?

  1. They didn’t count bacteria
  2. number of mRNA related to amount of enzyme / amoA produced / translated
  3. don’t know how much mRNA / amoA produced by each cell
  4. don’t know if amoA (mRNA/enzyme) is linked to cell division / growth of population.

  
• THE SCIENTISTS SET UP THEIR CULTURES IN STERILE GLASS BOTTLES.

  SUGGEST ONE SUITABLE METHOD FOR STERILISING THE BOTTLES AND EXPLAIN WHY IT WAS NECESSARY TO STERILISE THEM.

  1. SUITABLE METHOD: boiling water, wash in disinfectant
  2. REASON: kill other bacteria

  
• FARMERS USE ARTIFICIAL FERTILISERS TO MAINTAIN OR INCREASE YIELD FROM GRAIN-PRODUCING CROP PLANTS SUCH AS WHEAT.

  ARTIFICIAL FERTILISER IS USED TO REPLACE MINERAL IONS REMOVED FROM THE LAND WHEN CROPS ARE HARVESTED. ONE OF THE MINERAL IONS IS NITRATE.

  GIVE TWO EXAMPLES OF BIOLOGICAL MOLECULES CONTAINING NITROGEN THAT WIULD BE REMOVED WHEN A CROP IS HARVESTED.

  1. amino acid / protein
  2. Nucleic acid
  3. DNA
  4. RNA
  5. ATP / DNA
  6. chlorophyll
• (IMAGE) USE THE DATA IN THE GRAPH TO EVALUATE THE USE OF ARTIFICIAL FERTILISERS ON GRAIN-PRODUCING CROPS IN INDIA

  1. using more but getting less response over time
  2. the graph shows correlation but doesn’t prove changes in yield due to fertiliser / but there could be other factors
  3. becomes less cost effective with time

  
• (IMAGE) EXPLAIN WHY THE SCIENTISTS STERILISED THE SURFACES OF THE SEEDS AND GREW THEM IN SOIL THAT HAD BEEN HEATED TO 85 DEGREES CELSIUS FOR 2 DAYS.

  1. to kill any fungus / bacteria on surface of seeds or in soil
  2. so only the added fungus has any effect

  
• (IMAGE) EXPLAIN WHY IT WAS IMPORTANT THAT THE SOIL CONTAINED NO MINERAL IONS USEFUL TO THE PLANTS

  so that only nitrate or ammonia / type of fertiliser affects growth

  
• (IMAGE) EXPLAIN HOW GROUPS A AND C ACT AS CONTROLS

  1. So that effects of nitrate or ammonium alone could be seen
  2. so that effects of fungus can be seen

  
• (IMAGE) SUGGEST WHAT THE SCIENTISTS SHOULD HAVE DONE DURING THE DRYING PROCESS TO BE SURE THAT ALL OF THE WATER HAD BEEN REMOVED FROM THE PLANT SAMPLES.

  1. weigh samples at intervals during drying
  2. to see if weighing became constant (by 3 days)

  
• (IMAGE) WHAT CONCLUSIONS CAN BE DRAWN FROM THE DATA IN THE TABLE ABOUT THE FOLLOWING?

  • THE EFFECTS OF FUNGUS ON GROWTH OF THE PEA PLANTS
  • THE EFFECTS OF NITRATE AND AMMONIUM FERTILISER ON GROWTH OF THE PEA PLANTS

  WITH LIVE FUNGUS - SHOWING EFFECTS OF THE FUNGUS

  1. Fungus increases growth f roots and shoots in both
  2. produces greater growth with nitrate

  WITH HEAT TREATED FUNGUS - SHOWING EFFECTS OF FERTILISER

  1. Similar dry masses for roots and shoots
  2. probably no significant difference because SDs overlap

  
• (IMAGE) EXPLAIN WHY DETERMINATION OF DRY MASSES WAS AN APPROPRIATE METHOD TO USE IN THIS INVESTIGATION

  1. Dry mass measured / determined increase in organic material
  2. water content varies

  
• (IMAGE) WHICH TREATMENT GAVE TE BEST RESULT IN COMMERCIAL TERMS? JUSTIFY YOUR ANSWER

  1. fungus with nitrate-containing fertiliser gave largest shoot : root ratio
  2. and largest dry mass of shoot
  3. 6.09:1 compared with ammonium-containing fertiger 4.18:1

  
• (IMAGE) APART FROM TEMPERATURE AND PH, GIVE TWO VARIABLES THE SCIENTIST WOULD HAVE CONTROLLED WHEN PREPARING THE LIQUID MEDIUM CULTURE

  1. volume of bacteria culture
  2. concentration of bacteria culture
  3. concentration of glucose
  4. volume of ammonium chloride
  5. time bacteria left to divide
  6. concentration of oxygen
  7. concentration of nitrogen

  
• (IMAGE) A STUDENT CONCLUDED THAT THIS INVESTIGATION SHOWED THAT AMMONIA INHIBITS NITROGENASE ACTIVITY NITROGEN-FIXING BACTERIA. USE ALL THE INFORMATION TO EVALUATE THE STUDENTS CONCLUSION.

  FOR

  1. nitrogenase activity decreases with increase in ammonium chloride concentration
  2. nitrogenase activity zero with high concentration ammonium chloride

  AGAINST

  1. only used one species
  2. inhibition / results may be due to chloride ions

  
• NITROGENASE CATALYSES THE REDUCTION OF NITROGEN DURING NITROGEN FIXATION. THE REACTION REWUIRES 16 MOLECULES OF ATP FOR EACH MOLECULE OF NITROGEN THAT IS REDUCED. WHEN AMMONIA INHIBITS NITROGENASE ACTIVITY, NITROGEN-FIXING BACTERIA MAY BENEFIT. EXPLAIN HOW.

  1. Less ATP required
  2. ATP can be used for growth

  
• (IMAGE) USING FIGURE 2, GIVE THREE CONCLUSIONS YOU CAN MAKE ABOUT THE DISTRIBUTION OF TESE LEMUR SPECIES

  1. E. Rufus in north (west)
  2. E. Rufifrons in south
  3. actual distribution similar to expected distribution

  
• (IMAGE) SUGGEST HOW SPECIATION HAPPENED TO PRODUCE TWO SPECIES OF LEMUR

  1. geographical isolation
  2. reproductive isolation
  3. different selection pressures
  4. variation due to mutations in different populations
  5. advantageous alleles passed on
  6. eventually different species can’t interbreed to produce fertile offspring

  
• (IMAGE) SUGGEST ONE PRECAUTION NEEDED WHEN MARKING THE LEMURS TO MAKE SURE THE ESTIMATE OF THE NUMBER OF LEMURS IS VALID

  marking doesn’t affect survival / predation / recapture

  
• IN PHOTOSYNTHESIS, WHICH CHEMICALS ARE NEEDED FOR THE LIGHT DEPENDENT REACTION?

  NADP, ADP, Pi and water

  
• DESCRIBE WHAT HAPPENS DURING PHOTOIONISATION IN THE LIGHT DEPENDENT REACTION

  1. Chlorophyll absorbs light causing the electrons to become excited
  2. electrons are then lost from the chlorophyll causing it to become positively charged
• (IMAGE) EXPLAIN WHY THE STUDENT MARKED THE ORIGIN USING A PENCIL RATHER THAN USING INK

  • the ink and pigments would mix
  • with the ink the line would move to a different position
  • with pencil the line is still in the same position
  • with pencil the lines is still visible

  
• (IMAGE) DESCRIBE THE METHOD THE STUDENT USED TO SEPARATE THE PIGMENTS AFTER THE SOLUTION OF PIGMENTS HAD BEEN APPLIED TO THE ORIGIN

  1. Level of solvent below the original position / line
  2. remove the chromatography paper before the solvent reaches the top / end

  
• (IMAGE) THE PIGMENTS IN LEAVES ARE DIFFERENT COLOURS. SUGGEST AND EXPLAIN THE ADVANTAGE OF HAVING DIFFERENT COLOURED PIGMENTS IN LEAVES.

  Each pigment absorbs different wavelengths of light for photosynthesis

  
• (IMAGE) HEAT STRESS DECREASES THE LIGHT DEPENDENT REACTION OF PHOTOSYNTHESIS. EXPLAIN WHY THIS LEADS TO A DECREASE IN THE LIGHT-INDEPENDENT REACTION.

  1. less ATP
  2. less reduced NADP

  
• (IMAGE) ANOTHER EFFECT OF HEAT STRESS IS A DECREASE IN THE ACTIVITY OF THE ENZYME RUBISCO. A DECREASE IN THE ACTIVITY OF AN ENZYME MEANS THAT THE RATE OF THE REACTION IT CATALYSES BECOMES SLOWER.

  A DECREASE IN THE ACTIVITY OF THE ENZYME RUBISCO WOULD LIMIT THE RATE OF PHOTOSYNTHESIS. EXPLAIN WHY.

  1. Less carbon dioxide reacts with RuBP
  2. less GP

  
• WHERE PRECISELY IS RUBISCO FOUND IN A CELL?

  in the stroma of a chloroplast
• (IMAGE) USE ALL THE INFORMATION TO EVALUATE THEOR CONCLUSION.

  1. Rubisco activity increases with temperature
  2. rubisco activase activity decreases at high temperatures above 25 degrees celsius
  3. the graph could possibly show that activase doesn’t affect activity of rubisco
  4. however these results are only for cotton so may be different in other types of plant and isolated enzymes
  5. results are for isolated enzymes
  6. no stats test

  
• (IMAGE) THE SOLUTION THAT THE STUDENT USED TO PRODUCE THE CHLOROPLAST SUSPENSION HAD THE SAME WATER POTENTIAL AS THE CHLOROPLASTS.

  EXPLAIN WHY IT WAS IMPORTANT THAT THESE WATER POTENTIALS WERE THE SAME.

  1. So that osmosis doesn’t occur
  2. so that the chloroplasts don’t burst

  
• (IMAGE) EXPLAIN WHY THE STUDENT SET UP TUBE 1

  1. To show that light doesn’t affect DCPIP
  2. to show that chloroplasts are required

  
• (IMAGE) EXPLAIN THE RESULTS IN TUBE 3

  1. Reduction of DCPIP by electrons
  2. from Chlorophyll / light dependent reaction

  
• (IMAGE) EXPLAIN THE ADVANTAGE OF THE STUDENT USING THE IC50 IN THIS INVESTIGATION

  • so that different chemicals or weed killers can be compared
  • so that different concentration of chemicals or weed killers can be compared

  
• (IMAGE) EXPLAIN HOW CHEMICALS WHICH INHIBIT THE DECOLOURISATION OF DCPIP COULD SLOW THE GROWTH OF WEEDS

  1. Less ATP produced
  2. less reduced NADP produced
  3. less GP reduced to be converted to TP

  
• (IMAGE) EXPLAIN WHY THE SCIENTISTS MEASURED THE RATE OF PRODUCTION OF OXYGEN IN THIS INVESTIGATION

  1. Oxygen is produced in the light-dependent reaction
  2. And the faster oxygen is produced, the faster the light dependent reaction