AQA A-Level Biology Year 1 And AS Student Book Answers .
AQA A-Level Biology Year 1 and ASStudent Book AnswersChapter 1: Water and carbohydratesASSIGNMENT 1A1. There should be a straight line graph.A2.a. 87.5 mg cm–3A2.b. 4 mg cm–3A3. 50 secondsA4. It is very difficult to determine the end point; you may not stop the clock fast enough.A5. Fructose is also present as wine is made from fruit.ASSIGNMENT 2A1. Lactose is a sugar, so water enters by osmosis.A2. Watery faeces or diarrhoea.ASSIGNMENT 3A1.a. Add potassium iodide and look for a colour change from red/brown to blue/black.b. Add Benedict’s solution and heat and look for a colour change from blue to yellow/green/brick-red.c. First, carry out the reducing sugar test. If this is negative, hydrolyse the sample with dilute hydrochloricacid, neutralise with sodium hydroxide, and then add Benedict’s solution and heat. Look for a colour changefrom blue to yellow/green/brick-red.A2.a. Sample C had no carbohydrate as it was negative with all tests.b. Samples B and F were likely to contain sucrose as they were negative for a reducing sugar and positivefor a non-reducing sugar.c. Samples C and E would be suitable for someone with lactose intolerance as they were negative for thefermentation test.PRACTICE QUESTIONS1.a. Alpha, because the H on carbon 1 is above the ring, and OH is below.b. C6H12O6c.d. CondensationA-Level Biology Year 1 and AS Student Book Answers Page 1 HarperCollinsPublishers Limited 2015
2.a. Add Benedict's solution and heat to about 80 C. Glucose is a reducing sugar, and reduces the Cu2 ionsin the copper sulphate in the Benedict's solution to Cu ions, so the colour changes from blue to brick-red.b. Use an electronic balance to measure 10 g of glucose. Put this into a 1 dm 3 (1000 cm3) volumetric flaskand add a small amount of distilled water. Dissolve completely. Add more distilled water to exactly the 1000cm3 mark.c. For example, you could make an 8% solution by taking 80 cm 3 of the 10% solution, and then adding 20cm3 of distilled water.Use a similar method to make up solutions of a range of concentrations, say 6%, 4%, 2%, 1% and 0.5%.d. Measure a known volume of the 10% solution – say 10 cm3. Add a known volume of Benedict's solution– say 10 cm3. Note: it is important that there is enough Benedict's solution to react with all of the glucose. Inother words, the Benedict's must be in excess.Stand the mixture in a water bath at a steady temperature of 80 C for a standard length of time, say 10minutes.Repeat with each of your other standard solutions. Ensure that each tube is labelled, and stand them in arow so that you can compare their colours with the results from your unknown sample.e. Repeat the procedure with your unknown solution, using the same volume of solution, the same volumeof Benedict's solution, the same temperature and the same time. Compare the results with your standards.3.a. You could include these points in your answer: a polymer made of repeating monomers of α-glucose the glucose monomers are linked by 1–4 glycosidic bonds there are also 1–6 branches the molecule coils into a spiral.b. These large molecules are insoluble in water, so starch is a good storage substance. The spiral formmeans that they are compact, again good for storage inside a cell. The α-1–4 glycosidic bonds are easy tobreak by enzymes, so the starch can be converted to glucose when required for energy.4.i. Total loss after five days at 22 C is 18 g. The mean rate of loss is therefore 18 5 3.6 g day–1.a.ii. Total loss after five days at 28 C is 28 g. The mean rate of loss is therefore 28 5 5.6 g day–1.b. As temperature rises, the kinetic energy of water molecules increases. This means that water evaporatesfaster at higher temperatures, and the water vapour will also diffuse more quickly out of the leaf.c. Water has a high latent heat of evaporation. As the most fast-moving water molecules in the liquid waterin the walls of the mesophyll cells escape as gas, the average kinetic energy of the water molecules that areleft behind decreases. This makes the cell walls cooler.Chapter 2: Lipids and proteinsASSIGNMENT 1A1. The fatty acids in their molecules do not have charges, so are hydrophobic and so do not dissolve in water; thismeans they do not affect the osmotic properties of the cell in which they are stored. They have a high proportion ofhydrogen and carbon atoms in their molecules, which provide a lot of energy when the molecules are respired.A2. For example: the thickness of the blubber samples; the freshness of the samples; the site of the body from whichthey were taken; the temperature at which heat transfer was measured; the temperature at which they were keptbefore the measurements were made.A3. They had readings for two variables for each animal sampled. They did not choose the values for any of thereadings, so there was no 'independent variable'.A4.a. The thermal conductivity of the pilot whale blubber is generally greater than that of the sperm whaleblubber. This suggests that triglycerides have greater thermal conductivity than long chain alcohol lipids –the triglycerides are less good as thermal insulators.b. For the pygmy sperm whales, there appears to be a trend that the higher the lipid content, the lower thethermal conductivity. For the pilot whales, there could possibly be a trend in the other direction: the higherthe lipid content, the higher the thermal conductivity.A-Level Biology Year 1 and AS Student Book Answers Page 2 HarperCollinsPublishers Limited 2015
c. More samples could be taken and measured, so that there are more points on the scattergraph.Spearman's rank correlation test or Pearson's linear correlation test could be done to find out if there is asignificant correlation between one variable and the other.ASSIGNMENT 2A1. FiveA2. Spot A is leucine; B is valine; C is methionine; D is cysteine and E is lysine.A3. Spots may overlap or merge, e.g. glutamine and lysine.A4. (Research task) Students should find and present information about photosynthetic pigments.PRACTICE QUESTIONS1.a. Add a small amount of biuret solution to a solution containing the suspected protein. A purple or violetcolour indicates the presence of protein.b.i. Nitrogenii. Condensationiii.2.a.b. A polymer is a large molecule made up of a chain of repeating units, called monomers. Triglyceridemolecules do not have repeating units, nor are they made up of a long chain.3.a. The answer could include some or all of the following points: at least one named globular protein (e.g.haemoglobin, an enzyme, insulin) and one named fibrous protein (e.g. collagen, keratin, fibrin); a descriptionof the shape of the molecules of globular and fibrous proteins; reference to the solubility of many globularproteins, and an explanation of why they are soluble, referring to the presence of hydrophilic R groups onthe outside of the molecule; reference to why this solubility is important (e.g. allowing the molecules to betransported in blood, allowing them to take part in or catalyse chemical reactions); reference to the structuralroles of fibrous proteins, and how the shapes of their molecules suit them to these roles.b. Reference to proteins being made up of combinations of 20 building blocks/amino acids, giving rise tomany more combinations than carbohydrates which are made solely from sugar/glucosemolecules/monomers; and proteins having more complex structures and diverse range of roles within thebody.4.a. P hydrogen bond, Q peptide bond, R ionic bond, S disulfide bridge.b. Primary structure: only peptide bonds are involved. They link amino acids together into a chain;Secondary structure: hydrogen bonds link the R groups of different amino acids together, forming a regularalpha helix or beta sheet; Tertiary structure: hydrogen bonds, ionic bonds and disulfide bridges between Rgroups in different parts of the chain hold the whole chain in a complex 3D shape; Quaternary structure:hydrogen bonds, ionic bonds and especially disulfide bridges between R groups in different polypeptidechains hold the whole molecule in a complex 3D shape.A-Level Biology Year 1 and AS Student Book Answers Page 3 HarperCollinsPublishers Limited 2015
Chapter 3: EnzymesASSIGNMENT 1A1. Protease enzymes will digest proteins including those in the cell membranes of skin cells.A2. Enzymes are specific and only a complementary substrate will fit the active site.A3. Many bacteria will not be killed at this temperature so washed items may not be completely clean although theymay look clean.A4.a. and b.Factor to controlWhy it should be controlledHow it could be controlledSize of gelatine blocksDetergent and enzymes willdiffuse into blocks at differentrates if different sizesCut cubes from a solid block/usea cork borer to produce cylindersVolume of waterTo ensure equal concentrations ofwashing powderMeasure using measuringcylinderMass of washing powderTo ensure equal concentrations ofwashing powderWeigh using accurate scaleTemperature of waterHigher temperature increaseskinetic energy and will speed upreactionsUse a thermostatic water bath ata set temperatureMethod of measuring ‘digestedblocks’Need to measure both blocksaccuratelyWeigh blocks before and afterAmount of agitation / no agitationAgitation would increase rate ofcollisionsStir a given number of times ornot at allc. Use more than one block; repeat the experiment; carry out the experiment at a range of differenttemperatures.A5. No. The investigation only used blocks of gelatine, and not stains on clothes. It could only demonstrate theability of each washing powder to digest protein, and did not investigate the general cleaning ability of eachpowder.ASSIGNMENT 2A1. Enzymes are proteins and are made up of amino acids that contain an amino group with nitrogen.A2. Cellulase will break down cellulose in plant cell walls. Wood contains plant cells so the cellulose will start todigest the wood.A3. Student should plot suitable graph.A4. The protease is effective over a much wider range of pH, and will work well in alkaline solutions above pH 7.A5. The protease inhibitor stops the action of enzymes (proteases) that break up proteins. If a researcher wants atrue picture of what proteins are present in the cell, they need to stop the proteins being broken up. If not, theywould gain a false picture of the proteins in cells.REQUIRED PRACTICAL 1P1. Volumes of solutions used (enzyme and substrate), temperature and pH.P2. The diameter of the neck of the burette means that the volume is measured more accurately. A gas syringe isdesigned to accurately measure gas evolved.P3.a. Use a range of buffer solutions, keeping the enzyme and substrate concentrations constant.A-Level Biology Year 1 and AS Student Book Answers Page 4 HarperCollinsPublishers Limited 2015
b. Use water baths at a range of temperatures, keeping the enzyme and substrateconcentrations constant.P4. It is difficult to determine the end point accurately. The end point may occur between taking samples.P5. When using photographic, film the end point may occur between taking samples; it can be problematicdetermining the end point when using an indicator to observe colour change.P6. Keep a colour standard to compare with your samples.ASSIGNMENT 3A1. COOHA2. The graphs should be a similar shape to that in Figure 5, but with an optimum temperature of 55 C for glucosedehydrogenase and 85 C for hydrogenase.A3. 60 C, since glucose dehydrogenase is denatured above this temperature, so the whole process would cease.A4. Molecules have more kinetic energy so there are more frequent collisions / more enzyme substrate complexesformed.A5. By using the enzyme cellulase, which is produced by some species of bacteria.A6.a. 1300%b. 28 000 000 000c. Other costs need to be considered, such as fees for waste disposal – getting rid of anindustrial waste product.d. Lower percentage profit since start-up costs would be the same even though not all the gluconic acidcould be sold.e. Energy costs; cost of enzymes.ASSIGNMENT 4A1.a. Area of A 21.21 mm2, B 71.48 mm2, C 0.00 mm2 and D 43.20 mm2b. Data should be presented as a bar chart.c. B and D have a higher concentration of amylase than the standard and C has no amylase.d. Maltose (and amylase).A2. Place equal-sized samples of each strain of fungus on starch agar plates. Leave under controlled conditions fora set amount of time. Add iodine solution and measure the size of the clear areas around each strain of fungus.PRACTICE QUESTIONS1.a. Hydrogen peroxide was broken down to oxygen and water, and the oxygen was lost as a gas.b.i. Using a piece of boiled liver rather than fresh. All enzymes would be denatured by boiling,and so would be inactive.ii. This would be a check that the hydrogen peroxide does not decompose by itself, and wouldshow that it is the enzymes in the liver that are causing the reaction to take place.c. The rate of reaction is greatest at the start of the reaction, and then gradually decreases until it stopsat around five minutes.d.i. This is the most rapid stage of the reaction, when the concentration of substrate is greatest,A-Level Biology Year 1 and AS Student Book Answers Page 5 HarperCollinsPublishers Limited 2015
and the main limitation on the rate is the speed at which the catalase can break down andrelease each hydrogen peroxide molecule with which it forms a temporary enzyme–substratecomplex.ii. There is no change in mass during this time, because all of the hydrogen peroxide hasalready been broken down.e. The energy level of the hydrogen peroxide has to be raised before it will decompose. At roomtemperature, the energy level is not sufficient for most molecules to reach their activation energy.The presence of catalase lowers the activation energy, so this can be achieved even at roomtemperature.2.a. In order for a reaction to happen, energy often has to be supplied. The quantity of energy required iscalled activation energy. Enzymes lower activation energy, enabling reactions to take place at muchlower temperatures than is otherwise possible.b.i. Add iodine in potassium iodide solution. If starch is present, a blue-black colour is produced.ii. Add Benedict’s solution to a solution of the suspected reducing sugar, and heat to about 80 C. If reducing sugar is present, the blue Benedict’s solution will change to green, yellow or redbrown, as a coloured precipitate is produced.c.i.pHOptimum temperature / C468485754ii. As temperature increases from 40 to 48 C, the rate of reaction increases, as shown by theincrease in the percentage of starch converted to maltose in 10 minutes. This happens becausean increase in temperature means an increase in kinetic energy of the amylase and starchmolecules. They are therefore moving more quickly, and collide more frequently.As temperature increases above 48 C, hydrogen bonds in the enzyme molecule begin to break, sothe enzyme loses its three-dimensional shape. As the active site becomes distorted, it is no longerable to bond with starch. This happens progressively as temperature increases to 67 C, and at thattemperature all enzyme molecules are so denatured that none can bind with starch and no reactionoccurs at all.3.a. Enzyme has active site; Only substrate fits the active site.b. Allopurinol is a similar shape to xanthine; Allopurinol enters active site / is a competitive inhibitor; Lessxanthine binds / fewer enzyme–substrate complexes / fewer uric acid crystals formed / less uric acidformed.4.a. Phosphate changes shape of TK / changes shape of enzyme / changes the active site; Active siteforms/becomes the right shape / can bind to substrate / complementary to substrate / enzyme–substratecomplex can form.b. Faulty TK has functional active site without phosphate; So, faulty TK functional all the time / TKnot controlled by phosphate.c. Non-competitive inhibitor / binds to site other than active site; Causes TK to be in non-functional form /active site not formed / wrong shape / enzyme–substrate complex not formed; So, uncontrolled cell divisionstopped / slowed / controlled.5.a. Product is oxygen so volume can be measured; substrate is hydrogen peroxide which cannot be seen/ measured.A-Level Biology Year 1 and AS Student Book Answers Page 6 HarperCollinsPublishers Limited 2015
b.i. Accept any range in which minimum is between 0 and 10 and maximum is between 80 and 90 C.ii. Any suitable justification for either end of the range.iii. e.g. Cannot use temperatures below freezing point / above boiling point, catalase could have anoptimum anywhere within this range.c.i. Use a bufferii. Any two of: concentration of catalase; concentration of hydrogen peroxide; source ofcatalase; time over which oxygen production measured.6. Reference to active site of enzyme; active site has complementary shape to substrate; so each enzyme isspecific for a particular substrate / reaction; reference to primary structure of protein / sequence of amino acids;idea that very large variety of different primary structures possible; because there are 20 amino acids that canbe organised in any sequence; idea that primary structure determines tertiary structure; reference to threedimensional shape of protein; idea that huge variety of primary structures means enzymes with different shapescan be produced.7.a. It breaks down (hydrogen) peroxide.b. It is made up of more than one polypeptide chain.c. Similarities: it is made of four polypeptide chains; it contains haem groups; difference: it has an activesite (where hydrogen peroxide can bind).d. Affected by temperature: hydrogen bonds; affected by pH: ionic bonds.e. The active site must be a similar shape in all of them, even if the rest of the molecules have differentshapes.f. Idea that it is loss of catalase activity in hair follicles that is responsible for greying of hair; catalase in a pillwill be digested / hydrolysed / broken down; to individual amino acids; it is probable that the catalase in thecells in hair follicles is made in situ.g. The reaction is exothermic / reaction increases temperature / reaction releases heat energy; heatcauses the liquid to expand, increasing pressure; generation of oxygen gas increases pressure.Chapter 4: NucleotidesASSIGNMENT 1A1.a. They were easy to obtain and contain a nucleus.b. It has more phosphorus and different properties.A2. Chargaff’s rule is the evidence for base pairing and was crucial to Watson and Crick.A3. There is no correct answer to this question. Your answer could look at both sides of the argument, and use thefacts you have found to support your final viewpoint. It is important to remember that all scientific discoveries build onwork done earlier, so it is never easy to decide on exactly who should be credited with a discovery, but many peopledo feel that Rosalind Franklin could also have been given a Nobel Prize.ASSIGNMENT 2A1. DNA is double stranded, RNA is single stranded; DNA has deoxyribose, RNA has ribose; DNA has thymine, RNAhas uracil.A2. There will be 31% thymine; total A & T 62%, C & G 100 – 62 38% so, C 38/2 19%, and G 19%.A3. DNA runs in opposite directions (structurally) so that one side of the DNA ladder runs in the 5′ to 3′ direction andthe other runs from 3′ to 5′. This occurs due to the complementary shape of the base pairs and how they are able tobond together. They can only bond in one direction and that is parallel but in the opposite direction. This is calledantiparallel.A4. When cytosine and guanine pair, they do so with three hydrogen bonds, one more hydrogen bond than adenineand thymine.A-Level Biology Year 1 and AS Student Book Answers Page 7 HarperCollinsPublishers Limited 2015
PRACTICE QUESTIONS1.a. A: deoxyribose B: phosphate C: basesb. 32.a. Phosphate, a baseb.c.Number of bases3.AdenineThymineCytosineGuanineStrand X6244Strand Y2644a. Double helix.b. By hydrogen bonds between the bases.c. It has bases in a fixed sequence, which code for the sequence of amino acids in a protein.Adenine can o
A-Level Biology Year 1 and AS Student Book Answers Page 2 HarperCollinsPublishers Limited 2015 2. a. Add Benedict's solution and heat to about 80 C. Glucose is a reducing sugar, and reduces the Cu2 ions in the copper sulphate in the Benedict's solution to Cu ions, so the colour changes from blue to brick-red. b.
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AQA Topics Kerboodle Chapters 1 Energy P1 Conservation and Dissipation of Energy P2 Energy Transfer by Heating P3 Energy Resources Electricity P4 Electrical Circuits P5 Electricity in the Home Particle Model of Matter P
