2019 a level paper 1

• Describe how a non-competitive inhibitor can reduce the rate of an enzyme-controlled reaction.

  non competitive inhibitor attaches at enzyme inhibitor site and changes the shape of the active site. so enzyme substrate complexes wont form as they're no longer complementary.
• The scientist concluded that pectin is a non-competitive inhibitor of the lipase enzyme. Use Figure 1 to explain why the scientist concluded that pectin is a non-competitive inhibitor.

  increased substrate concentration doesn't increase rate of reaction
• No large lipid droplets are visible with the optical microscope in the samples from suspension A. Explain why

  emulsification can't be seen due to resolution of optical microscope.
• A student concluded from Figure 3 that eating an extra 10 g of fibre per day would significantly lower his risk of cardiovascular disease. Evaluate his conclusion.

  the correlation doesn't mean causation. there is little evidence present for higher mass of fibre per day.
• Suggest one advantage of using the FFQ method and one disadvantage of using the FFQ method compared with the alternative method.

  advantage = diet over 24hrs may not be as representative as a year

  disadvantage = heavily reliant on memory so individual may not recall everything
• What data would the students need to collect to calculate their index of diversity in each habitat? Do not include apparatus used for species sampling in your answer.

  number of species and number of individuals in each species in each habitat.
• Give two ways the students would have ensured their index of diversity was representative of each habitat.

  random, large number of samples
• Modern farming techniques have led to larger fields and the removal of hedges between fields. Use Figure 4 to suggest why biodiversity decreases when farmers use larger fields

  it decreases as there's a larger centre
• Farmers are now being encouraged to replant hedges on their land. Suggest and explain one advantage and one disadvantage to a farmer of replanting hedges on her farmland.

  advantage = increase of biodiversity so increase in predators of pests

  disadvantage = reduced land so less crop income
• Use Figure 5 to explain how human mass at birth is affected by stabilising selection


  individuals under 2800g are likely to be transferred to a special care.

  extreme mass babies are not likely to survive and reproduce and therefore unlikely to pass on alleles for extreme mass at birth. therefore, extreme mass decreases in proportion.
• The scientists studied the effect of one form, KIR2DS1, of the human KIR gene on mass at birth. In the following passage the numbered spaces can be filled with biological terms.

  KIR2DS1 is an allele of the KIR gene, found at a loci on chromosome 19. KIR2DS1 is 14 021 bases long and is transcribed into mRNA that is 1101 bases long. This mRNA is then translated into a polypeptide 304 amino acids long. The polypeptide is then modified in the organelle, golgi apparatus , before forming its functional tertiary protein structure.
• The scientists calculated a P value of 0.03 when testing their null hypothesis. What can you conclude from this result? Explain your answer.

  probability that difference is due to chance is 0.03. so you would reject the null hypothesis.
• Describe the structure of the human immunodeficiency virus (HIV)

  contains RNA, reverse transcriptase, capsid and attachment proteins.
• Use the data in Table 3 and your knowledge of the immune response to suggest why HIV controllers do not develop symptoms of AIDS.

  all of them have more t helper cells.

  lower viral load to infect t helper cells

  so more activation of b cells which continuously produce plasma cells and there for are able to destroy other pathogens
• Describe and explain the data in Table 4

  slower growth from before birth to 21 days, so heart grows slowly until fully developed
• The scientists determined the percentage of heart cells undergoing DNA replication by using a chemical called BrdU. Cells use BrdU instead of nucleotides containing thymine during DNA replication. 0 6 . 2 Describe how BrdU would be incorporated into new DNA during semi-conservative replication.


  DNA helicase breaks hydrogen bonds betweens 2 dna strands, BrdU forms hydrogen bonds with adenine.

  DNA polymerase joins nucleotides and phosphodiester bonds form.
• Cells with BrdU in their DNA are detected using an anti-BrdU antibody with an enzyme attached. Use your knowledge of the ELISA test to suggest and explain how the scientists identified the cells that have BrdU in their DNA.


  add antibody to the cells, wash the cells to remove any excess antibodies and add substrate to cause colour change.
• Unlike plants, Ulva lactuca does not have xylem tissue. Suggest how Ulva lactuca is able to survive without xylem tissue.

  it has a shorter pathway for diffusion into cells
• Ulva prolifera also produces haploid, mobile single cells that can fuse to form a zygote. Suggest and explain one reason why successful reproduction between Ulva prolifera and Ulva lactuca does not happen

  they're of different species so if fused together they would form infertile offspring.
• Sunflowers are not xerophytic plants. The scientists repeated the experiment with xerophytic plants. Suggest and explain one way the leaf growth of xerophytic plants would be different from the leaf growth of sunflowers in Figure 9.

  they would grow slower due to the smaller number of stomata decreasing area for gas exchange
• Use your knowledge of gas exchange in leaves to explain why plants grown in soil with very little water grow only slowly.

  stomata close soless carbon dioxide uptake for photosynthesis
• The data in Table 7 show differences between the oxyhaemoglobin dissociation curve for a mouse and the oxyhaemoglobin dissociation curve for a horse. Suggest how these differences allow the mouse to have a higher metabolic rate than the horse

  mouse haemoglobin as lower affinity for oxygen so more oxygen can be unloaded
• 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.


  mouses have a larger surface area to volume ratio which allows for faster heat loss and a faster rate of respiration
• Explain five properties that make water important for organisms.

  metabolite in photosynthesis

  a solvent so reactions can occur

  high heat capacity so buffers change in temperatures

  cohesion between water molecules so supports columns of water in plants
• Describe the biochemical tests you would use to confirm the presence of lipid, non-reducing sugar and amylase in a sample.

  lipid, add ethanol then water and mix this should present a milky white emulsion

  non reducing sugars, add benedict's and stays blue. get new sample and boil with acid first and then neutralise with alkali, heat again with benedict's reagent and becomes red or orange,

  amylase, add biuret and becomes purple.
• Describe the chemical reactions involved in the conversion of polymers to monomers and monomers to polymers. Give two named examples of polymers and their associated monomers to illustrate your answer

  condensation reaction joins monomers together and forms a chemical bond and releases water.

  hydrolysis reaction breaks bonds between monomers using a molecule of water.

  polymers to monomers is DNA hydrolysed to form nucleotide, its monomer.

  monomers to polymers is beta glucose bonding to form a cellulose.