A-level biology paper 1 [2019], Exams of Advanced Education

Download A-level biology paper 1 [2019] and more Exams Advanced Education in PDF only on Docsity! A-level biology paper 1 [2019] Describe how a non-competitive inhibitor can reduce the rate of an enzyme-controlled reaction. - • Attaches to the enzyme at an allosteric site • Changes (shape of) the active site/tertiary structure (of enzyme) • So active site and substrate no longer complementary so less substrate can bind; 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. - With inhibitor increase substrate/lipid concentration does not change rate of reaction Calculate the maximum length of the large lipid droplet marked X in Figure 2. - 4mm = 4000um 4000/500 = 8 Maximum length) 8-10 (μm); (Uncertainty) (±) 2 (μm); No large lipid droplets are visible with the optical microscope in the samples from suspension A. Explain why. - • Emulsification • Cannot 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. - Plants: cellulose Algae: cellulose Fungi: chitin Prokaryotes: murein 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. - • Negative correlation between fibre eaten per day and risk of cardiovascular disease • Original/current fibre intake of student not known • Idea of significance linked to 2x standard deviation overlap (at 10 g day-1 change) • If current intake between 5 and 30g day-1 then eating 10g more results in a significant decrease in risk • Correlation does not mean causation • Little evidence/data for higher mass of fibre per day; • Large (2x) standard deviation at high/low mass of fibre makes mean less precise (less certainty) • No statistical test to show if differences are significant Suggest one advantage of using the FFQ method and one disadvantage of using the FFQ method compared with the alternative method. - (Advantage) Over longer period so more representative (Disadvantage) Relies on (long term) memory so may not be accurate 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 samples • 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. - Larger fields have relatively less hedge and fewer species Suggest and explain one advantage and one disadvantage to a farmer of replanting hedges on her farmland. - Advantage: • Greater (bio)diversity so increase in predators of pests Disadvantage: • Wash cells to remove unattached antibody • 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. - Short diffusion pathway to cells OR It has a surface permeable to water/ions into cells On Figure 7 complete each box with an appropriate letter to show the type of cell division happening between each stage in the life cycle. Use 'T' to represent mitosis and 'E' to represent meiosis. - • E in top right box • 3 x T in top and bottom left and bottom right boxes 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 are different species; • So if fused together they would not produce fertile offspring Complete Table 5 by giving all headings, units and volumes required to make 20 cm3 of this sodium chloride solution. - cm3 0.8 Volume of water cm3 19.2 Use the data in table 6 to calculate the concentration of sodium chloride solution with a water potential of -3.41 MPa - Use Figure 8 (on page 28) and Figure 9 to calculate by how much this leaf increased in area. Give your answer in cm2 - 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. - • Low/slow growth • Due to smaller number/area of stomata 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 • Less carbon dioxide uptake for less photosynthesis/glucose production Plot the haemoglobin saturation data from Table 7 and use these points to sketch the full oxyhaemoglobin dissociation curves for a horse and a mouse. - • y axis 0 - 100 in linear scale and x axis minimum 1 to 8 in linear scale and both axes use at least half size of grid; • Correct plots for 50% and 25% for both animals • Both curves levelling off at higher partial pressures and at percentage saturations ≤100% The following equation can be used to estimate the metabolic rate of an animal. Metabolic rate = 63 × BM-0.27 BM = body mass in grams Use this equation to calculate how many times faster the metabolic rate of a mouse is than the metabolic rate of a horse - 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/Hb has a lower affinity for oxygen • More oxygen can be dissociated/released/unloaded for metabolic reactions/respiration 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 • Smaller so larger surface area to volume ratio • More/faster heat loss per gram/in relation to body size • Faster rate of respiration/metabolism releases heat Explain five properties that make water important for organisms. - • A metabolite in condensation/hydrolysis/ photosynthesis/respiration • A solvent so (metabolic) reactions can occur OR a solvent so allowing transport of substances • High heat capacity so buffers changes in temperature • Large latent heat of vaporisation so provides a cooling effect through evaporation • Cohesion between water molecules so supports columns of water in plants Cohesion between water molecules so produces surface tension supporting small organisms Describe the biochemical tests you would use to confirm the presence of lipid, non-reducing sugar and amylase in a sample - ✅Lipid • Add ethanol/alcohol then add water and shake/mix • White/milky emulsion ✅Non-reducing sugar • Do Benedict's test and stays blue/negative • Boil with acid then neutralise with alkali • Heat with Benedict's and becomes red/orange precipitate ✅Amylase • Add biuret (reagent) and becomes purple/violet/mauve/lilac; • Add starch, (leave for a time), test for reducing sugar/absence of starch; 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. - • A condensation reaction joins monomers together and forms a (chemical) bond and releases water; • e.g. cellulose is made from multiple beta glucose molecules joined by glycosidic bonds • A hydrolysis reaction breaks a (chemical) bond between monomers and uses water;