AS Papers

• This fat substitute cannot be digested in the gut by lipase. Suggest why. [2 marks]

  1. Fat substitute is a different/wrong shape/not COMPLEMENTARY; OR Bond between glycerol/fatty acid and propylene glycol different (to that between glycerol and fatty acid)/no ester bond
  2. Unable to fit/bind to active site of lipase/no Enzyme substrate complex formed;

  
• This fat substitute is a lipid. Despite being a lipid, it cannot cross the cell-surface membranes of cells lining the gut. Suggest why it cannot cross cell-surface membranes. [1 mark]

  • It is hydrophilic/is polar/is too large/is too big;
• Give two ways in which the hydrolysis of ATP is used in cells. [2 marks

  1. To provide energy for other reactions/named process
  2. To add phosphate to other substances and make them more reactive/change their shape;
• Describe how ATP is resynthesised in cells. [2 marks]

  1. From ADP and phosphate
  2. By ATP synthase
  3. During respiration/photosynthesis;
• Y is a protein. One function of Y is to transport cellulose molecules across the phospholipid bilayer. Using information from Figure 3, describe the other function of Y. [2 marks]

  1. (Y is) an enzyme/has active site/forms ES complex
  2. That makes cellulose/attaches substrate to cellulose/joins β glucose; OR
  3. Makes cellulose/forms glycosidic bonds
  4. From β glucose

  
• What is the evidence in Figure 3 that the phospholipid bilayer shown is part of the cell-surface membrane? [1 mark]

  • Cell wall forms outside cell-surface membrane/has cellulose on it (on the outside);

  
• Describe the induced-fit model of enzyme action. [2 marks]

  1. (before reaction) active site not complementary to/does not fit substrate
  2. Shape of active site changes as substrate binds/as enzymesubstrate complex forms
  3. Stressing/distorting/bending bonds (in substrate leading to reaction)
• Explain the results shown in Figure 4

  1. (Rate of) increase in concentration of maltose slows as substrate/starch is used up OR High initial rate as plenty of starch/substrate/more E-S complexes
  2. No increase after 25 minutes/at end/levels off because no substrate/starch left;

  
• Describe how the scientist would have produced the calibration curve and used it to obtain the results in Figure 4.

  1. Make/use maltose solutions of known/different concentrations (and carry out quantitative Benedict’s test on each)
  2. Use colorimeter to measure colour/colorimeter value of each solution and plot calibration curve/graph described
  3. Find concentration of sample from calibration curve;
• Human papilloma virus (HPV) is the main cause of cervical cancer. A vaccine has been developed to protect girls and women from HPV. Describe how giving this vaccine leads to production of antibody against HPV. [4 marks]

  1. Vaccine/it contains antigen (from HPV)
  2. Displayed on antigen-presenting cells
  3. Specific HELPER T CELL (detects antigen and) stimulates specific B cell
  4. B cell divides/goes through mitosis/forms clone to give PLASMA CELLS
  5. B cell/plasma cell produces antibody;
• 1. Two (doses) because got more antibody
  2. With three doses, second dose/dose at 1 month doesn’t lead to production of any more antibody (than the twodose group)/get same/similar response
  3. Three doses would be more expensive/less popular with parents/girls (and serves no purpose)

  
• There is genetic diversity within HPV. Give two ways doctors could use base sequences to compare different types of HPV. [2 marks]

  1. Compare (base sequences of) DNA
  2. Look for mutations/named mutations (that change the base sequence)
  3. Compare (base sequences of) mRNA
• Suggest why the development of a monopolar mitotic spindle would prevent successful mitosis. [2 marks]

  1. No separation of chromatids/chromosomes/centromeres
  2. Chromatids/chromosomes all go to one pole/end/sides of cell/not pulled to opposite poles
  3. Doubles chromosome number in cell/one daughter cell gets no chromosomes or chromatids;

  
• Scientists investigated the effect of different concentrations of a kinesin inhibitor (KI) on mitosis of human bone-cancer cells grown in a culture. [4 marks]

  • 2. Cancer not completely destroyed/may continue to grow/spread/form tumours
  • Best concentration may be between 100 and 1000/need trials between 100 and 1000;
  • This research in culture, don’t know effect of KI on people;
  • Above 100 may be harmful (to body)
  • Higher concentrations more expensive
  • (above 100) will have more effect on (rapidly dividing) cancer cells

  
• Suggest how amyloid-precursor protein can be the substrate of two different enzymes, α-secretase and β-secretase (lines 3–5). [2 marks]

  1. Different parts/areas/amino acid sequences (of amyloid-precursor) protein
  2. Each enzyme is specific /fits/binds/ complementary to a different part of the APP;
• Suggest and explain why the combined actions of endopeptidases and exopeptidases are more efficient than exopeptidases on their own. [2 marks]

  1. Endopeptidases hydrolyse internal (peptide bonds) OR Exopeptidases remove amino acids/hydrolyse (bonds) at end(s)
  2. More ends or increase in surface area (for exopeptidases);
• Suggest appropriate units the student should use to compare the distribution of stomata on leaves. [1 mark]

  • Stomata per mm2 or cm2
  • Accept mm ^ -2
  • Reject the use of solidus being equivalent of per
• Give two reasons why it was important that the student counted the number of stomata in several parts of each piece of leaf tissue. [2 marks]

  1. Distribution may not be uniform
  2. OR So it is a representative sample
  3. ACCEPT more / fewer stomata in different areas
  4. To obtain a (reliable) mean;
• One of the two plant species used by the student in this investigation was a xerophyte.

  Other than the distribution of stomata, suggest and explain two xerophytic features the leaves of this plant might have. [2 marks]

  1. Hairs so ‘trap’ water vapour and water potential gradient decreased
  2. Stomata in pits/grooves so ‘trap’ water vapour and water potential gradient decreased
  3. Thick (cuticle/waxy) layer so increases diffusion distance;
• One of the two plant species used by the student in this investigation was a xerophyte.

  Other than the distribution of stomata, suggest and explain two xerophytic features the leaves of this plant might have. [2 marks]

  • 4. Waxy layer/cuticle so reduces evaporation/transpiration
  • Rolled/folded/curled leaves so ‘trap’ water vapour and water potential gradient decreased
  • Spines/needles so reduces surface area to volume ratio;
• Suggest two reasons why the rate of water uptake by a plant might not be the same as the rate of transpiration. [2 marks]

  1. Water used for support/turgidity
  2. Water used in photosynthesis
  3. Water used in hydrolysis
  4. Water produced during respiration;
• The scientists concluded that an increase in phosphate in the embryo was linked to growth of the embryo. Suggest two reasons why an increase in phosphate can be linked to growth of the embryo. [2 marks]

  1. (Phosphate required) to make RNA
  2. (Phosphate required) to make DNA
  3. (Phosphate required) to make ATP/ADP
  4. (Phosphate required) to make membranes
  5. (Phosphates required) for phosphorylation;
• 0 6 . 1 Scientists determined the mean FEV1 value of 25-year-olds in the population. Suggest two precautions that should have been taken to ensure that this mean FEV1 value was reliable. [2 marks]

  1. Large sample size
  2. Individuals chosen at random
  3. Are healthy
  4. Equal number of males and females
  5. Repeat readings;
• Explain the importance of determining a mean FEV1 value of 25-year-olds in this investigation. [2 marks]

  1. (For) comparison
  2. To see effect of age/emphysema/smoking OR Takes into account outliers/anomalous results;
• Forced expiration volume (FEV1) is the volume of air a person can breathe out in 1 second.

  The mean FEV1 value of non-smokers decreases after the age of 30. Use your knowledge of ventilation to suggest why. [1 mark]

  1. Internal intercostal muscle(s) less effective
  2. OR Less elasticity (of lung tissue);
• HIV attaches to a specific protein receptor on helper T cells. A low percentage of people have a mutation of the CCR5 gene which codes for this protein receptor. This mutation results in a non-functional protein receptor. Explain how this mutation can result in the production of a non-functional protein receptor. [4 marks]

  1. Change in DNA! base/nucleotide (sequence)
  2. Change in amino acid (sequence)/primary structure
  3. Alters (position of) hydrogen/ionic/disulfide bonds
  4. Change in TERTIARY! structure (of receptor);
• People with the CCR5 mutation show a greater resistance to developing AIDS. Explain why. [2 marks]

  1. (Receptor) is not complementary OR (HIV) cannot bind/attach and enter/infect (helper) T cell
  2. 2. No replication (of virus) OR No destruction of (helper) T cell;
• 1. Low/lower exposure to HIV (in Europe) OR Low/lower number of HIV/AIDS (infections/cases)
  2. (HIV) has only been present for a short time period OR (HIV relatively) recently evolved
  3. Mutation/CCR5 has been around for many years
  4. Mutation/CCR5 is advantageous (for something else)

  
• REQUIRED PRACTICAL 6 =

  Suggest two variables the student should control in using the filter paper discs in this investigation. [2 marks]

  1. (Same) size;
  2. Same material/absorbency
  3. In solution for same time period;
• Some people produce a much higher ventricular blood pressure than normal. This can cause tissue fluid to build up outside the blood capillaries of these people. Explain why. [2 marks]

  1. More fluid forced/filtered out of capillary/blood (due to high pressure)
  2. Less return of fluid (into capillary/blood) due to pressure OR Lymph(atic) (system) cannot drain away all excess fluid;