A level Biology 13.1 Photosynthesis as an energy transfer process – Exam style question – Paper 4

Question

 (a) The unicellular green alga, Chlorella, a photosynthetic protoctist, was originally studied for its potential as a food source. Although large-scale production proved to be uneconomic, the many health benefits provided by Chlorella mean that it is now mass
           produced and harvested for use as a health food supplement.

           Fig. 1.1 shows cells of Chlorella.

           In one study into the productivity of Chlorella, carbon dioxide concentration was altered to investigate its effects on the light-independent stage of photosynthesis.

    • A cell suspension of Chlorella was illuminated using a bench lamp.
    • The suspension was supplied with carbon dioxide at a concentration of 1% for 200 seconds.
    • The concentration of carbon dioxide was then reduced to 0.03% for a further 200 seconds.
    • The concentrations of RuBP and GP (PGA) were measured at regular intervals.
    • Throughout the investigation the temperature of the suspension was maintained at 25 °C.

           The results are shown in Fig. 1.2.

           (i) State precisely where in the chloroplast RuBP and GP are located.[1]

           (ii) Explain why the concentration of RuBP changed between 200 and 275 seconds.[2]

           (iii) Calculate the rate of decrease per second in the concentration of GP between 200 and 350 seconds.

                     Show your working and give your answer to two decimal places.

                     answer  arbitrary units per second [2]
    (b) Explain how the decrease in the concentration of GP leads to a decreased harvest for commercial suppliers of Chlorella.[2] [Total: 7]

Answer/Explanation

Ans:

1 (a) (i) stroma ; [1]

          (ii) lower CO2 concentration ;
                   less, carbon fixation/CO2 combining with RuBP/RuBP converted to GP ; RuBP reformed from TP ; [max 2]

          (iii) 0.01 ;;

                  A 0.012 or 1.8 ÷ 150 or \(\frac{2.0-0.2}{150}\) or \(\frac{2.0-0.2}{350-200}\) for 1 mark

    (b) less TP ;

          (so less) conversion to, (other) carbohydrates / lipids / amino acids / proteins ; A named examples, e.g. glucose/ hexose/ cellulose/ starch

          AVP ; e.g. 1 – (amino acids) used to make proteins for, growth/ cell division
                     e.g. 2 – (carbohydrate/ lipid) for respiration for, growth/ cell division

Question

 (a) Fig. 8.1 shows some of the reactions that take place inside a palisade mesophyll cell.

    (i) Identify substances A, B and C.[3] 

          A

          B

          C

  (ii) Name precisely the process that produces reduced NADP.[1]

  (iii) Name the type of reaction that takes place to produce starch from hexose sugars and name the type of bonds formed.[2]

          reaction

          bond

  (iv) Describe how carbon dioxide reaches the inside of a palisade mesophyll cell from the external atmosphere.[3] 

(b) The optimum pH for the activity of rubisco is pH8.

         Explain why the illumination of chloroplasts leads to optimum pH conditions for rubisco.[3] [Total: 12]

Answer/Explanation

Ans:

8 (a) (i) A – RuBP/ribulose bisphosphate ;
                 B – fatty acid ;
                 C – nitrates ; A suitable nitrogenous substance e.g. ammonium ions
                 I nitrogen/ammonia 
           (ii) non-cyclic photophosphorylation ; 
           (iii) condensation/polymerisation ; A anabolic
                    glycosidic ; 
           (iv) 1 enters via stoma(ta) ;
                    2 by diffusion/down a concentration gradient ;
                    3 passes through air spaces ;
                    4 dissolves in film of water (on cell surface) ;
                    5 (diffuses) through cell, wall/ surface membrane (of palisade cells) ;

     (b) 1 excited electrons leave, chlorophyll a/photosystem ;
            2 pass along ETC ;
            3 protons present from photolysis ;
            4 protons (pumped) into intermembrane space ;
            5 rubisco is in stroma ;
            6 idea that protons leaving stroma raises pH ;

Question

 (a) Fig. 1.1 shows a section through part of a dicotyledonous leaf of the tea plant Camellia sinensis.

        On Fig. 1.1, use label lines and letters to label each of the following parts:[2]

        X – xylem tissue

        P – palisade mesophyll tissue. 

  (b) The leaves of C. sinensis have a large surface area and are thin.

        Explain how each of these two features help the leaf to carry out photosynthesis.[2]

  (c) The lower epidermis contains stomata.

        (i) State one structural difference between a guard cell and other lower epidermal cells.[1]

        (ii) Abscisic acid has an important role in the closure of a stoma. It promotes the loss of potassium ions from guard cells.

              Outline how the loss of potassium ions from guard cells will lead to the closure of a stoma.[3] [Total: 8]

Answer/Explanation

Ans:

1 (a)

   (b) large surface area
          (to get) more, light/ carbon dioxide ; A gas exchange I oxygen

          thinness
          small(er)/ short(er)/reduced, diffusion distance for gases OR
          fast(er) diffusion of gases ; A named gas, either CO2 or O2

          1 mark only if both points made but not related to features in italics

   (c) (i) have chloroplasts / varying thickness of (cell) walls / no plasmodesmata ;
         (ii) water potential/Ψ, of (guard) cell(s), increases /becomes less negative ; water leaves cell(s) ;
                 (by) osmosis / down a water potential gradient ; I diffuses
                 (guard cell) becomes, flaccid/ less turgid/AW ;

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