PRE 2016

• GIVE TWO WAYS IN WHICH THE STRUCTURE OF STARCH IS SIMILAR TO CELLULOSE

  1. are polymers/polysaccharides
  2. are made of monomers/monosaccharides
  3. contain glucose/carbon, hydrogen and oxygen
  4. contain glycosidic bonds
  5. have 1-4 links
  6. hydrogen bonding
• GIVE TWO WAYS IN WHICH THE STRUCTURE OF STARCH IS DIFFERENT FROM CELLULOSE

  (starch)

  1. Contains alpha glucose
  2. helical, coiled, compact, branched, not straight
  3. 1-6 bonds
  4. no H-bonds between molecules
  5. no micro fibrils
• (IMAGE) USING THE DIAGRAM, SUGGEST AND EXPLAIN ONE OTHER WAY IN WHICH SIEVE CELLS ARE ADAPTED FOR MASS TRANSPORT

  1. few organelles
  2. so easier flow

  
• USING THE DIAGRAM, SUGGEST AND EXPLAIN ONE OTHER WAY IN WHICH COMPANION CELLS ARE ADAPTED FOR THE TRANSPORT OF SUGARS BETWEEN CELLS

  1. Mitochondria release ATP
  2. For active transport against a concentration gradient

  OR

  1. ribosomes produce proteins
  2. proteins linked to transport such as carrier proteins or enzymes

  
• ORGANIC COMPOUNDS SYNTHESISED IN THE LEAVES OF A PLANT CAN BE TRANSPORTED TO THE PLANTS ROOTS. THIS TRANSPORT IS CALLED TRANSLOCATION AND OCCURS IN THE PHLOEM TISSUE OF THE PLANT. ONE THEORY OF TRANSLOCATION STATES THAT ORGANIC SUBSTANCES ARE PUSHED FROM A HIGH PRESSURE IN THE LEAVES TO A LOWER PRESSURE IN THE ROOTS.

  DESCRIBE HOW A HIGH PRESSURE IS PRODUCED IN THE LEAVES.

  1. Water potential becomes lower and negative as sugar enters the phloem
  2. water enters the phloem by osmosis
  3. increased volume of water creates a higher pressure
• (IMAGE) DURING THEIR EXPERIMENT, THE SCIENTISTS ENSURED THAT THE RATE OF PHOTOSYNTHESIS OF THEIR PLANTS REMAINED CONSTANT. EXPLAIN WHY THIS WAS IMPORTANT.

  1. Rate of photosynthesis related to rate of sucrose production
  2. rate of translocation higher when sucrose concentration is higher

  
• (IMAGE) THE SCIENTISTS CONCLUDED THAT SOME TRANSLOCATION MUST OCCUR IN THE SPACES IN THE CELL WALLS. EXPLAIN HOW THE INFORMATION IN THE FIGURE SUPPORTS THIS CONCLUSION.

  1. Rate of translocation doesn’t fall to zero, it still offers after 120 minutes
  2. but sucrose is no longer able to enter cytoplasm of phloem cells

  
• CONTRAST THE PROCESSES IF FACILITATED DIFFUSION AND ACTIVE TRANSPORT

  1. facilitated diffusion involves channel or carrier proteins whereas active transport only involves carrier proteins
  2. facilitated diffusion doesn’t use ATP as its passive whereas active transport does
  3. facilitated diffusion takes place down a concentration gradient whereas active transport can occur against a concentration gradient
• (IMAGE) EXPLAIN THE RESULTS SHOWN IN THE IMAGE

  1. Group A - initial uptake slower because of diffusion
  2. group A - levels off because same concentrations inside cells and outside cells/reached equillibrium
  3. Group B - uptake faster because by diffusion plus active transport
  4. Group B - fails to level off because uptake against gradient/no equilibrium to be reached
  5. group B - rate slows because fewer chloride ions in external solution

  
• DESCRIBE THE MASS FLOW HYPOTHESIS FOR THE MECHANISM OF TRANSLOCATION IN PLANTS

  1. In the leaf, the source, sugars are actively transported into phloem
  2. by companion cells
  3. which lowers the water potential of sieve cell and water enters by osmosis
  4. increase in pressure causes mass movement towards the roots, the sink
  5. sugars converted in root for respiration for storage
• BREATHING OUT AS HARD AS YOU CAN IS CALLED FORCED EXPIRATION.

  DESCRIBE AND EXPLAIN THE MECHANISM THAT CAUSES FORCED EXPIRATION.

  1. contraction of internal intercostal muscles
  2. Relaxation of diaphragm muscles / of external intercostal muscles
  3. causes decrease in volume of chest / thoracic cavity
  4. air pushed down pressure gradient
• (IMAGE) THE PEOPLE IN GROUP B WERE RECOVERING FROM AN ASTHMA ATTACK.

  EXPLAIN HOW AN ASTHMA ATTACK CAUSED THE DROP IN THE MEAN FEV SHOWN IN THE FIGURE,

  1. muscle walls of bronchi / bronchioles contract
  2. walls of bronchi / bronchioles secrete more mucus
  3. diameter of airways reduced
  4. therefore flow of air reduced

  
• (IMAGE) DESCRIBE HOW A HIGH PRESSURE IS PRODUCED IN THE LEAVES

  1. Water potential becomes lower / more negative as sugar enters the phloem
  2. water enters the phloem by osmosis
  3. increased volume of water causes increased pressure

  
• DESCRIBE HOW OXYGEN IN THE AIR REACHES CAPILLARIES SURROUNDING ALVEOLI IN THE LUNGS.

  DETAILS OF BREATHING ARENT REQUIRED

  1. trachea and bronchi and bronchioles
  2. down pressure gradient
  3. down diffusion gradient
  4. lacrosse alveolar epithelium
  5. across capillary endothelium / epithelium
• (IMAGE) ASTHMA AFFECTS BRONCHIOLES AND REDUCES FLOW OF AIR IN AND OUT OF THE LUNGS. FIBROSIS DOES NOT AFFECT BRONCHIOLES; IT REDUCES THE VOLUME OF THE LUNGS.

  WHICH GROUP, B OR C, WAS THE ONE CONTAINING PEOPLE WITH FIBROSIS OF THEIR LUNGS? EXPLAIN.

  1. group B because breathe out as quickly as healthy / have similar FEV to group A
  2. so bronchioles not affected
  3. FVC reduced / total volume breathed out reduced

  
• THE OXYGEN DISSOCIATION CURVE FOR HAEMOGLOBIN SHIFTS TO THE RIGHT DURING VIGOROUS EXERCISE.

  EXPLAIN THE ADVANTAGE OF THIS SHIFT.

  1. lower affinity for xylem / releases more oxygen / oxygen is released quicker / oxygen dissociates/ unloads more readily
  2. To muscles / tissues / cells
  3. for rapid respiration
• (IMAGE) EXPLAIN HOW THE BODY SHAPE OF A WEDDELL SEAL IS AN ADAPTATION TO LIVING IN A COLD ENVIRONMENT.

  1. Small SA:V
  2. so reduces heat loss

  
• (IMAGE) DESCRIBE AND EXPLAIN THE CHANGES IN THE RATE OF BLOOD FLOW TO THE DIFFERENT ORGANS DURING A LONG DIVE.

  1. brain is the same, others fall
  2. brain controls other organs and needs constant supply of oxygen
  3. lungs are used less and heart rate decreases as blood is diverted to muscles

  
• (IMAGE) HOW DID THE SCIENTISTS ENSURE THEY COULD MAKE A VALID COMPARISON BETWEEN LEAVES FROM DIFFERENT SPECIES

  scientists used fully grown leaves / five plants of each species

  
• DESCRIBE A METHOD YOU COULD USE TO FIND THE SURFACE AREA OF A LEAF

  1. draw around the leaf on graph paper
  2. count squares
  3. multiply by 2 for upper and lower leaf surface
• (IMAGE) WHICH SPECIES, A OR B, WOULD YOU PREDICT GREW IN A DRIER ENVIRONMENT? EXPLAIN ONE FEATURE THAT CAUSED YOU TO CHOOSE THIS SPECIES.

  1. species B
  2. smaller surface area so less evaporation
  3. thicker leaves so greater diffusion distance
  4. fewer stomata so less diffusion / evaporation
  5. smaller surface area to volume ratio so less evaporation

  
• OTHER THAN THE FEATURES OF LEAVES IN THE TABLE, GIVE TWO FEATURES OF LEAVES OF XEROPHYTES. FOR EACH FEATURE EXPLAIN HOW IT REDUCES WATER LOSS.

  1. Thicker cuticle so increase in diffusion distance
  2. hairs on leaves so
  3. curled leaves so
  4. sunken stomata so
  5. reduction in air movements / increase in humidity / decrease in water potential gradient

  
• SPECIES C HAS A HIGH NUMBER OF STOMATA PER MM3. DESPITE THIS IT LOSES A SMALL AMOUNT OF WATER.

  USE THE DATA TO EXPLAIN WHY.

  small leaves / surface area so number of stomata is low

  
• (IMAGE) THE PERCENTAGE OF OXYGEN AND CARBON DIOXIDE IN COLUMN A DO NOT ADD UP TO 100% BUT IN COLUMNS C AND D THEY DO.

  SUGGEST TWO REASONS FOR THIS DIFFERENCE.

  1. other gases in the atmosphere
  2. only oxygen and carbon dioxide in gas mixtures
  3. composition of gases in A not controlled and in B and C gas mixtures are controlled

  
• (IMAGE) USE ALL THE DATA TO DESCRIBE THE EFFECT OF CONCENTRATION OF CARBON DIOXIDE ON THE BREATHING RATE OF GRASSHOPPERS.

  1. breathing rate lowest when no carbon dioxide
  2. generally presence of carbon dioxide increase breathing rate - there is a positive correlation
  3. breathing rate increases when carbon dioxide high than 0.1%
  4. breathing rate of grasshopper 3 falls in D whereas others increase

  
• (IMAGE) THE ESTIMATE DOES NOT PROVIDE A RELIABLE VALUE FOR THE MEAN BREATHING RATE OF ALL INSECT SPECIES IN THE MEADOW.

  OTHER THAN BEING AN ESTIMATE, SUGGEST AND EXPLAIN THREE REASONS WHY THIS VALUE WOULD NOT BE RELIABLE

  1. small sample so may not be representative of all grasshoppers
  2. grasshoppers aren’t the only insects so genetic differences
  3. insects arent all mature - different sizes so different metabolic rates
  4. movement not restricted so rate of respiration higher
  5. Carbon dioxide concentration lower in meadow so breathing rate lower

  
• (IMAGE) SUGGEST A DIFFICULTY OF COUNTING MOVEMENTS OF GILL COVERS AS A METHOD OF MEASURING RATE OF VENTILATION IN FISH.

  fish keep moving so movement of gill covers too Fast to count

  
• (IMAGE) THE BIOLOGIST CONCLUDED THAT THERE WAS A CORRELATION BETWEEN RATE OF VENTILATION OF THE GILLS AND TEMPERATURE OF THE WATER. A SCATTER DIAGRAM CAN BE USED TO LOOK FOR A CORRELATION, BUT IN THIS INVESTIGATION, IT WAS NOT THE APPROPRIATE GRAPH FOR HER DATA. EXPLAIN WHY.

  1. there is only one dependent variable, water temperature is the Independent variable and breathing rate is depends on water temperature
  2. water temperature plus breathing rate aren’t both properties of fish

  
• (IMAGE) DESCRIBE THE RELATIONSHIP BETWEEN TEMPERATURE OF WATER, OXYGEN IN WATER AND RATE OF VENTILATION.

  as water temperature increases, oxygen concentration falls and ventilation rate increases

  
• (IMAGE) USE FIGURE 1 AND FIGURE 2 TO EXPLAIN THE ADVANTAGE TO THE FISH OF THE CHANGE IN ITS RATE OF VENTILATION.

  1. as concentration of oxygen falls less oxygen flows over gills / enters fish / enters gills
  2. as a result blood oxygen concentration falls
  3. an increase in ventilation rate increases the flow of oxygen / carbon dioxide across gills into or out if fish
  4. maintains concentration gradient in gills
  5. to maintain oxygen supply to cells to maintain respiration

  
• DESCRIBE AND EXPLAIN HOW THE COUNTERCURRENT SYSTEM LEADS TO EFFICIENT GAS EXCHANGE ACROSS THE GILLS OF A FISH

  1. Water and blood flow in opposite directions
  2. maintains concentration gradient / equilibrium not reached / water always next to blood with a lower concentration of oxygen
  3. along whole length of gill / lamellae
• AMOEBIC GILL DISEASE (AGD) IS CAUSED BY A PARASITE THAT LIVES ON THE GILLS OF SOME SPECIES OF FISH. THE DISEASE CAUSES THE LAMELLAE TO BECOME THICKER AND TO FUSE TOGETHER.

  AGD REDUCES THE EFFICIENCY OF GAS EXCHANGE IN FISH. GIVE TWO REASONS WHY.

  1. thicker lamellae so longer diffusion distance
  2. lamellae fuse so reduced surface area
• THE VOLUME OF WATER PASSING OVER THE GILLS INCREASES IF THE TEMPERATURE OF THE WATER INCREASES. SUGGEST WHY.

  1. increased metabolism / respiration / enzyme activity
  2. less oxygen dissolved in water
• NAME THE PROCESS BY WHICH OXYGEN PASSES FROM AN ALVEOLUS IN THE LUNGS INTO THE BLOOD

  simple diffusion
• DESCRIBE TWO ADAPTATIONS OF THE STRUCTURE OF ALVEOLI FOR EFFICIENT GAS EXCHANGE

  1. thin walls
  2. total surface area is large
• (IMAGE) PEOPLE WHO HAVE BEEN FIRE BREATHERS FOR MANY YEARS OFTEN FIND THEY CANNIT BREATHE OUT PROPERLY. EXPLAIN WHY.

  1. Loss of elasticity
  2. less recoil

  
• (IMAGE) GIVE THE LETTER OF THE ORGAN THAT PRODUCES AMYLASE AND THE ORGAN THAT PRODUCES MALTASE

  AMYLASE = C

  MALTASE = E

  
• MALTOSE IS HYDROLYSED BY THE ENZYME MALTASE.

  EXPLAIN WHY MALTASE CATALYSES ONLY THIS REACTION.

  1. active site of enzyme has active site complementary to substrate
  2. only maltose can fit
  3. to form enzyme substrate complex
• DESCRIBE HOW YOU WOULD TEST A SAMPLE OF FOOD FOR THE PRESENCE OF STARCH

  1. add iodine solution to food sample
  2. if starch present, it should turn blue / black
• THE CONCENTRATION OF GLUCOSE IN THE BLOOD RISES AFTER EATING A MEAL CONTAINING CARBOHYDRATES.

  THE RISE IS SLOWER IF THE CARBOHYDRATES IS STARCH RATHER THAN SUCROSE. EXPLAIN WHY.

  1. starch is digested to maltose by amylase
  2. maltose digested to glucose by maltase
  3. digestion of sucrose is only one enzyme, sucrase