EXAM QUESTIONS

• USING THE DATA IN FIGURE 3 WHAT CAN YOU CONCLUDE ABOUT THE UPTAKE OF OXYGEN OVER THE ENTIRE BODY OF THE LUGWORM?

  1. enters by diffusion down a concentration gradient.
  2. mostly across parts of the body with gills.
  3. this is because gills increase surface area to volume ratio for Absorption.
  4. 8.8kpa over gills
  5. 2.4kpa rest of body surface, 1.9kpa front end before gills, 0.5kpa rear end after gills.

  
• DESCRIBE HOW THE SCIENTIST WILL USE INFORMATION FROM THE COLORIMETER AND HER CALIBRATION CURVE TO DETERMINE THE PO2 IN A SAMPLE OF LUGWORM BLOOD.

  1. Measure light absorption.
  2. draw line to pO2.

  
• DESCRIBE THE ADVANTAGE OF THE BOHR EFFECT DURING INTENSE EXERCISE

  1. Increases the dissociation of oxygen
  2. for aerobic respiration at this tissues/muscles OR anaerobic respiration delayed at the tissues/muscles/cells OR less lactate at the tissues/muscles/cells.
• IMAGE

  1. increase in breathing rate
  2. similar partial pressure of carbon dioxide per breath, but more breaths

  OR

  1. Increase in tidal volume
  2. similar partial pressure of carbon dioxide per breath, but increased volume per breath

  
• DESCRIBE AND EXPLAIN THE EFFECT OF INCREASING CARBON DIOXIDE CONCENTRATION ON THE DISSOCIATION OF OXYHAEMOGLOBIN

  1. Increases oxygen dissociation OR decreases haemoglobin's affinity for oxygen
  2. by decreasing blood pH which increases the blood acidity
• IMAGE

  1. Higher affinity for oxygen than haemoglobin OR dissociates oxygen less readily OR associates more readily
  2. allows aerobic respiration when diving at lower partial pressure of oxygen OR provides oxygen when haemoglobin unloaded OR delays anaerobic respiration/lactate production

  
• BINDING OF ONE MOLECULE OF OXYGEN TO HAEMOGLOBIN MAKES IT EASIER FOR A SECOND OXYGEN MOLECULE TO BIND. EXPLAIN WHY.

  1. Binding of first oxygen allows conformational change to occur, which changes the tertiary/quaternary structure of the haemoglobin.
  2. this creates another binding site
• (IMAGE) EXPLAIN HOW CHANGES IN THE SHAPE OF HAEMOGLOBIN RESULT IN THE S-SHAPED (SIGMOID) OXYHAEMOGLOBIN DISSOCIATION CURVE FOR HBA

  the first oxygen bind to Hb causing the Hb to change shape which allows oxygen to bind more easily.

  
• (IMAGE) AT BIRTH 98% OF THE HAEMOGLOBIN IS HbF. BY THE AGE OF 6 MONTHS, THE HbF HAS USUALLY COMPLETELY DISAPPEARED FROM THE BABYS BLOOD AND BEEN REPLACED BY HbA. USE THE GRAOH TO EXPLAIN WHY THIS CHANGE IS AN ADVANTAGE FOR THE BABY.

  1. HbA has a lower affinity for oxygen at low partial pressures OR lower affinity for oxygen at pp found in tissues
  2. so there’s easier unloading of oxygen for aerobic respiration

  
• THE MASS FLOW HYPOTHESIS IS USED TO EXPLAIN THE MOVEMENT OF SUBSTANCES THROUGH PHLOEM.

  USE YOUR UNDERSTANDING OF THE MASS FLOW HYPOTHESIS TO EXPLAIN HOW PRESSURE IS GENERATED INSIDE THIS PHLOEM TUBE.

  1. sucrose is actively transported into phloem
  2. lowering the water potential
  3. which allows water to move into the phloem by osmosis from the xylem
• (IMAGE) DESCRIBE THE RELATIONSHIP BETWEEN PHLOEM PRESSURE AND THE RATE OF WATER MOVEMENT IN XYLEM IN THIS PLANT

  the phloem pressure falls as rate of water movement in the xylem increases

  
• PHLOEM PRESSURE IS REDUCED DURING THE HOTTEST OASRT OF THE DAY. USE INFORMATION IN THE GRAPH ALONG WITH YOUR UNDERSTANDING OF TRANSPIRATION AND MASS FLOW TO EXPLAIN WHY.

  1. High rate of transpiration
  2. as water is lost through stomata OR high tension in xylem
  3. which causes less water movement from the xylem to phloem OR insufficient water potential in phloem to draw water from xylem

  
• (IMAGE) THE STUDENT MEASURED THE TIME TAKEN FOR WATER MOVEMENT. GIVE TWO OTHER MEASUREMENTS HE MADE TO CALCULATE THE RATE OF WATER MOVEMENT.

  1. Initial and final mass
  2. number of xylem vessels

  
• GIVE THE REASON FOR ADDING A LAYER OF OIL TO THE WATER IN THE BEAKER

  to Prevent evaporation/water loss OR so that evaporation/water loss/transpiration only from celery

  
• (IMAGE) EXPLAIN WHY COLOURED WATER MOVED UP THE STALKS

  1. water is transpired from the leaves
  2. so a lower water potential creates negative pressure which pulls up the water
  3. and hydrogen bonds/cohesion maintains a column

  
• (IMAGE) THE STUDENT USED A SHARP SCALPEL TO CUT THE CELERY. DESCRIBE HOW SHE SHOULD ENSURE SHE HANDLED THE SCALPEL SAFELY DURING THIS PROCEDURE.

  1. Cut away from the body
  2. against a hard surface

  
• (IMAGE)

  median - due to small sample size or anomalies such as 80. ANSWER - 41

  
• SUGGEST HOW ULVA LACTUCA IS ABLE TO SURVIVE WITHOUT XYLEM TISSUE

  there’s a short diffusion pathway to cells OR it has a surface permeable to water/ions into cells

  
• (IMAGE) TREATMENT D IS A CONTROL. EXPLAIN HOW THE MEASUREMENT OBTAINED FROM THIS CONTROL IS USED BY THE SCIENTIST.

  1. used to compare effect of other treatments
  2. shows effect of substance X

  
• USING THE DIAGRAM AND TABLE, WHAT CAN YOU CONCLUDE FROM TREATMENTS D AND E ABOUT ROOT GROWTH?

  1. D shows substance x isn’t required for root growth OR substances were already present in stem because theres some root growth
  2. substance x moves through the plant
  3. e shows substance x increases root growth.

  
• THE MASS FLOW HYPOTHESIS IS USED TO EXPLAIN THE MOVEMENT OF SUBSTANCES THROUGH PHLOEM. EVALUATE WHETHER THE INFORMATION FROM THIS INVESTIGATION SUPPORTS THIS HYPOTHESIS.DO NOT CONSIDER STATISTICAL ANALYSIS IN THE ANSWER.

  SUPPORT

  1. f shows phloem is involved
  2. g shows active transport is involved
  3. because 4 degrees Celsius cooling slows movement
  4. the agar block is the source
  5. Roots are the sink

  AGAINST

  1. no bulge above ringing (in F)
  2. no osmosis
  3. movement could be due to gravity
  4. roots Still growth without functioning phloem
  5. No leaves to act as a source

  
• (IMAGE) THE SCIENTISTS CONCLUDED THAT THIS HEAT TREATMENT DAMAGED THE PHLOEM.

  EXPLAIN HOW THE RESULTS IN FIGURE 1 SUPPORT THIS CONCLUSION.

  EITHER

  1. The radioactively labelled carbon is converted into sugar/organic substances during photosynthesis
  2. mass flow/translocation in the phloem throughout the plant only in plants that were untreated/B/control.

  OR

  1. Movement in phloem requires living cells/respiration/active transport/ATP
  2. heat treatment damages living cells so transport in the phloem throughout the plant only in plants that were untreated/B/control

  
• (IMAGE) THE SCIENTISTS ALSO CONCLUDED THAT THIS HEAT TREATMENT DID NOT AFFECT THE XYLEM.

  EXPLAIN HOW THE RESULTS IN THE TABLE SUPPORT THIS CONCLUSION.

  1. the water content of the leaves wasn’t different because standard deviations overlap.
  2. water is therefore still being transported in the xylem to the leaf OR movement in xylem is passive so unaffected by heat treatment.

  
• (IMAGE)

  1. Heat treatment has greater effect on young leaves than old.
  2. heat treatment damages phloem
  3. Fe^3+ moves up leaf/plant
  4. suggests Fe^3+ is transported in xylem in older leaf
  5. in young leaf, some in xylem, as some still reaches top part of leaf
  6. suggests Fe^3+ mostly transported in phloem of young leaf OR xylem is damaged in young leaf OR xylem is alive in young leaf.
  7. Higher ratio of Fe^3+ in old leaves than young
  8. all ratios show less Fe^3+ in the top than lower part of leaves
  9. but there’s no statistical test to show if difference is significant

  
• (IMAGE) THE STUDENT WANTED TO DETERMINE THE RATE OF WATER LOSS PER MM^2 OF SURFACE AREA OF THE LEAVES OF THE SHOOT IN FIGURE 1.

  OUTLINE A METHOD SHE COULD HAVE USED TO FIND THIS RATE.YOU SHOULD ASSUME THAT ALL WTER LOSS FROM THE SHOOT IS FROM THE LEAVES.

  1. draw around each leaf on graph paper and count the squares
  2. of both sides of each leaf
  3. divide rate of uptake from POTOMETER by total surface area of leaves

  
• THE RATE OF MOVEMENT THROUGH A SHOOT IN A POTOMETER MAY NIT BE THE SAME A THE RATE OF WATER MOVEMENT THROUGH THE SHOOT OF A WHOLE PLANT.

  SUGGEST ONE REASON WHY,

  plant has roots OR xylem cells very narrow
• (IMAGE)NAME THE PROCESS THAT PRODUCED ^14CO2 RELEASED FROM THE TRUNK

  respiration

  
• (IMAGE) HOW LONG DID IT TAKE THE 14C LABEL TO GET FROM THE TOP OF THE TRUNK TO THE BOTTOM OF THE TRUNK? EXPLAIN.

  1. Around 30 hours
  2. time between peak 14C at top of trunk and bottom

  
• WHAT OTHER INFORMATION IS REQUIRED IN ORDER TO CALCULATE THE MEAN RATE OF MOVEMENT OF THE 14C DOWN THE TRUNK?

  length of trunk (between top and bottom)

  
• (IMAGE) USE THIS INFORMATION AND KNOWLEDGE OF SURFACE AREA TO VOLUME RATIOS TO SUGGEST AN EXPLANATION FOR THE POSITION OF MITOCHONDRIA IN LARGE U. MARINUM CELLS.

  1. larger cells have smaller surface area to volume ratios
  2. which means it’s takes longer for oxygen to diffuse to mitochondria OR less oxygen diffuses OR diffusion distance is longer

  
• EXPLAIN THE ADVANTAGE FOR LARGER ANMALS HAVING A SPECIALISED SYSTEM THAT FACILITATES OXYGEN UPTAKE

  1. larger organisms have a smaller surface area;volume ratio
  2. overcomes long diffusion pathway OR faster diffusion
• (IMAGE) SUGGEST HOW THE ENVIRONMENTAL CONDITIONS HAVE RESULTED IN ADAPTATIONS OF SYSTEMS USING MODEL A RATHER THAN MODEL B

  1. water has lower oxygen partial pressure than air
  2. so system on the outside gives larger surface area in contact with water

  OR

  1. Water is denser than air
  2. so water supports the systems / gills

  
• MAMMALS SUCH AS A MOUSE AND A HORSE ARE ABLE TO MAINTAIN A CONSTANT BODY TEMPERATURE.

  USE YOUR KNOWLEDGE OF SURFACE AREA TO VOLUME RATIO TO EXPLAIN THE HIGHER METABOLIC RATE OF A MOUSE COMPARED TO A HORSE

  MOUSE

  1. Smaller so larger surface to volume ratio
  2. more / faster heat loss
  3. faster rate of respiration / metabolism replaces heat.
• DESCRIBE THE RELATIONSHIP BETWEEN SIZE AND SURFACE AREA TO VOLUME RATIO OF ORGANISMS

  as size increases, the surface area to volume ratio decreases
• (IMAGE)

  measures uptake of oxygen

  OR

  avoids standard form

  
• (IMAGE) THE SCIENTIST DECIDED TO USE THE RATIO OF SURFACE AREA TO MASS RATHER THAN THE RATIO OF SURFACE AREA TO VOLUME. HE MADE THIS DECISION FOR PRACTICAL REASONS.

  SUGGEST ONE PRACTICAL ADVANTAGE OF MEASURING THE MASSES OF FROG EGGS, TADPOLES AND ADULTS COMPARED WITH MEASURING THEIR VOLUMES.

  more accurate / less error in measuring mass

  easier because irregular shapes

  fewer measurements

  
• EXPLAIN WHY OXYGEN UPTAKE IS A MEASURE OF METABOLIC RATE IN ORGANISMS.

  oxygen is used in respiration which provides energy OR oxygen used in respiration which is a metabolic process
• (IMAGE) A STUDENT WHO LOOKED AT THESE RESULTS SAID THAT THEY COULD NOT MAKE A CONCLUSION ABOUT THE RELATIONSHIP BETWEEN STAGE OF DEVELOPMENT AND METABOLIC RATE. USE INFORMATION IN THE TABLE TO EXPLAIN REASONS WHY THEY WERE UNABLE TO MAKE A CONCLUSION.

  1. no information about egg
  2. so can’t compare all stages
  3. no statistical information or comparison of standard deviations

  
• (IMAGE) NAME THE PROCESS BY WHICH OXYGEN REACHES THE CELLS INSIDE THE BODY OF A TUBIFEX WORM

  simple diffusion

  
• (IMAGE) USING THE INFORMATION PROVIDED, EXPLAIN HOW TWO FEATURES OF THE BODY OF THE TUBIFEX WORM ALLOW EFFICIENT GAS EXCHANGE.

  1. thin / small so short diffusion pathway
  2. Flat / long so large surface area to volume ratio