AQA® Please write clearly in block capitals, Centre number Candidate number Surname Forename(s) Candidate signature I declare this is my own work. GCSE BIOLOGY Higher Tier Paper 2H Time allowed: 1 hour 45 minutes Materials For this paper you must have: + aruler Question | Mark + ascientific calculator. 4 For Examiner's Use Instructions ‘+ Use black ink or black ball-point pen Pencil should only be used for drawing Fill in the boxes at the top of this page. ‘Answer all questions in the spaces provided. If you need extra space for your answer(s), use the lined pages at the end of this book. Write the question number against your answer(s).. 7 Do all rough work in this book. Cross through any work you do not want 3 to be marked. + Inall calculations, show clearly how you work out your answer. 2 3 4 5 @ TOTAL Information ‘+ The maximum mark for this paper is 100. ‘+ The marks for questions are shown in brackets, + You are expected to use a calculator where appropriate. + You are reminded of the need for good English and clear presentation in your answers.There are no questions printed on this page DO NOT WRITE/ON THIS PAGE ANSWER IN THE/SPACES PROVIDED ‘tie re be‘Answer all questions in the spaces provided, There are two types of reproduction: * sexual reproduction ‘+ asexual reproduction .[1] Complete Table 1 to compare sexual reproduction with asexual reproduction. Write a tick (V) in the box if the statement is true. The first row has been completed for you. [2 marks] Table 1 Sexual Asexual reproduction | reproduction Cell division occurs v ¥ Fertilisation occurs Genes are passed on from parent to offspring Offspring are genetically identical to each other .[2] Gametes are formed in sexual reproduction. Name the male gamete formed in flowering plants. [1 mark] Question 1 continues on the next page Turn over >Cell division by meiosis forms gametes. Figure 1 shows the mean mass of DNA per cell before, during and after meiosis. Figure 1 Mean mass of DNA per cell in / arbitrary units 4 bo o 1 2 3 4 5 6 7 Time in hours Use information from Figure 4 to answer questions 01.3 to 01.6. When is the DNA in the chromosomes being copied? Tick (v7) one box. Between 0 and 3 hours Between 3 and 4 hours Between 4 and 5 hours Between 5 and 6 hours [1 mark]Cells divide twice during meiosis. Which two times in igure 1 show one cell dividing into two cells? [2 marks] Tick (v) two boxes 3 hours 4 hours 5 hours 6 hours 8 hours What is the mean mass of DNA in arbitrary units in a sperm cell? [1 mark] Tick (V) one box. What is the mean mass of DNA in arbitrary units in each cell in an embryo? [1 mark] Tick (V) one box. Tum over for the next question Turn over >Earthworms: « live in soil * feed on dead and decaying plant matter ‘+ have soft, moist skin ‘+ exchange gases through their skin. Give two abiotic factors and two biotic factors that could affect the size of an earthworm population Abiotic factors 1 [4 marks] Biotic factors 1‘Students investigated the populations of earthworms in the soil in two different areas: «Area A: a grass lawn : a farmer's field. * Area Chemical X can be mixed with water and poured onto the soil. The mixture brings earthworms to the surface of the soil but does net harm. the earthworms. Plan an investigation using chemical X to compare the number of earthworms per m? in areas A and B. [6 marks] not wate ‘utile ne tow Turn over for the next question Turn over >3] Itis important to control the concentration of glucose in the blood. Figure 2 shows how the concentration of glucose in the blood of a person changed over 4 hours. Figure 2 6.0 ss EN Blood glucose 5.0 concentration inmmoldm? 4.5 40 35 3.0 0 t 1 2 3 4 Time in hours Meal eaten Give one time when the concentration of insulin in the person's blood would be high. Use Figure 2. [1 mark] Time hours torDo rot te ‘se tne 0[3].[2] Explain the effect a high concentration of insulin has on blood glucose concentration. box [3 marks] Effect Explanation n 3 continues on the next page Turn over >40 People with diabetes have difficulty controlling the concentration of glucose in their blood. Type 2 diabetes is linked to obesity. Figure 3 shows how to find if an adult's body mass is healthy for their height. Figure 3 2.0 Healthy /Over- Underweight / reignt | weight (Obese 19 18 Height inm 17 16 15}/ 40 50 60 70 80 90 100 110 120 130 140 150 160 Body mass in kg tor4 Person A: + is 1.75 min height + has a body mass of 52 kg. What is person A's weight category? Tick (¥) one box. Underweight Healthy weight Overweight Obese Person B is 1.9 m in height. Give the range of body masses that would put person B in the healthy weight category. Range from kg to Question 3 continues on the next page [1 mark] [1 mark] kg Turn over >12 3].[5] Person C is obese. toe ‘A doctor thinks that person € has Type 2 diabetes. The doctor tests a sample of blood from person C. Table 2 shows « the results of the blood test ‘+ the mean results for people who do not have diabetes. Table 2 Concentration in blood Mean for people who Person | do not have diabetes Cholesterol in mmol/dm? 6.21 5.20 Glucose in mmol/ém? 9.56 451 Insulin in arbitrary units 24.32 14.83 ‘Type 2 diabetes occurs when body cells have a reduced response to insulin. Give two ways the results of the blood test show that person might have Type 2 diabetes. [2 marks]13 Give two ways that a person can reduce the chance of developing Type 2 diabetes. [2 marks] ‘Turn over for the next question not wate ‘ute ne tow Turn over >14 The rapid growth in human population means that more waste substances are released into the environment. The release of substances into the environment can cause pollution. Name one harmful substance that could cause air pollution. [1 mark] Name three harmful substances that could cause water pollution. Do not refer to plastic or to litter in your answer. [3 marks]15 Do netwate ‘ute ne 4].[3] Describe how substances that pollute air and water could be harmful to humans and fox other living organisms. [6 marks] 10 ‘Tum over for the next question Turn over >16 tor Maple syrup urine disease (MSUD) is a rare inherited human condition. MSUD is usually diagnosed early in childhood and can be controlled by having a low-protein diet. Figure 4 shows the inheritance of MSUD in one family. Figure 4 O @ 1 2 3 4 Key [ Mate with msud e } Male without MSUD 7 7 r 7 7 @ Femaie with msuD CO Femate without MSUD 10 1 12 The allele for MSUD is recessive. 5 ].[4] Give one piece of evidence from Figure 4 which shows that MSUD is a recessive condition. [1 mark]7 .[2] Persons 7 and 8 in Figure 4 are expecting a fourth child. Determine the probability that the child will have MSUD. You should: + draw a Punnett square diagram * identify the phenotype of each offspring genotype + use the symbols: N =allele for not having MSUD n= allele for MSUD. [4 marks] Probability = Question 5 continues on the next page Turn over >18 Figure 5 shows chemical reactions involved in the normal breakdown of some types of amino acid inside body cells, Figure 5 ‘Some amino acids Enzyme 1 Toxic substance P ‘Ammonia Enzyme 2 Other enzymes, Harmless products Urea ‘A person with MSUD cannot make Enzyme 2. ‘One of the final products shown in Figure 5 is urea. ns shown in Where in the human body are the rea jure 5 most likely to occur? [1 mark] Tick (v) one box. Kidney Liver Pancreas ‘Small intestine19 not wnte ‘uti ne Scientists can analyse blood samples or urine samples to see if a person has MSUD. foe The test identifies high concentrations of toxic substance P, shown in Figure 5. 3].[4] Explain why the blood of a person with MSUD have a high concentration of toxic substance P. Use information from Figure 5. [3 marks] 5 ].[5] €xplain why the urine of a person with MSUD will have a high concentration of toxic substance P. [2 marks] Question 5 continues on the next page Turn over >20 Explain why a person with MSUD must have a low-protein diet. [3 marks] tor2 ole Energy flows through an ecosystem and materials are recycled. toe Figure 6 shows the water cycle. Figure 6 0[6].[4] Name process x. [1 mark] 0 [6].[2] Name the process by which water is absorbed into plant roots. [1 mark] 0[6].[3] Give two uses of water in plants. [2 marks] Turn over >22 Figure 7 shows the flow of energy through a food chain. The numbers are in kilojoules/m*tyear. Figure 7 Sunlight Respiration 1700000 1025 1 Growth Food 24 eaten] \ iW, 3050 L Grass Other herbivores 3550 Decomposers 15200 tor23 not wnte ‘uti ne 0[6].[4] The cow is more efficient than the grass at converting energy. for The energy conversion efficiency of the cow is 4,098%. Calculate how many times more efficient the cow is at converting energy than the grass. The equation for energy conversion efficiency is: energy used for growth 100 energy input energy conversion efficiency = Give your answer to 3 significant figures. [5 marks] Number of times (3 significant figures) Question 6 continues on the next page Turn over >24 0[6].[5] Itis more energy-efficient to rear cows indoors than to rear cows outdoors. tor Give two reasons why. [2 marks] 1 2 0[6].[6] Suggest two possible disadvantages of rearing cows indoors [2 marks] 1 225 not wnte ‘uti ne o[7 A scientist found a polluted pond which had a new type of blue algae in the water. box The blue colour of the algae was caused by a mutation, 0[7].[4] What is a mutation? [1 mark] Question 7 continues on the next page Turn over >26 The scientist measured the number of blue algal cells in a sample of the pond water. The scientist used a special slide which has a counting grid. This is the method used. 1. Dilute 2.5 cm® of pond water to a volume of 10 cm* with distilled water. 2. Place a drop of the diluted pond water on the special slide, as shown in Figure 8, 3. Place a thick coverslip over the diluted pond water to give a depth of 0.1 mm of pond water. 4. Use a microscope to count the number of algal cells in a 0.2 mm x 0.2 mm square ‘on the counting grid. Figure 8 shows a side view of the special slide. Figure 8 Diluted pond water, Position of counting grid depth = 0.1 mm Coversiip | Special slide Figure 9 shows the view of the counting grid through a microscope. Figure 9 & @ io ® 6 > Algal cells 0.2mm a6 a od @|_ © © lq | °° ic} __e Q ,° torar How many algal cells are in the 0.2 mm x 0.2 mm square in Figure 9? Use the following procedure: ‘+ Count all cells that are completely within the 0.2 mm x 0,2 mm square in the ‘counting grid ‘+ Count cells that are touching the left side or the lower side of the square, ‘+ Do not count cells that are touching the right side or the top side of the square. [1 mark] Number of algal cells in the 0.2 mm x 0.2 mm square = ‘One week later the scientist repeated the test and counted 14 cells on the 0.2 mm x 0.2 mm counting grid. Calculate the number of algal cells in 1.0 mm? of undiluted pond water. Use the scientist's second count of 14 cells. [6 marks] Number of algal cells in 1.0 mm® of undiluted pond water = Que: ion 7 continues on the next page Turn over >28 .[4] Suggest why the scientist diluted the pond water before placing it on the special slide, | ~~ t« [1 mark] .[5] A student repeated the scientist's method. The student used a thin coverslip over the diluted pond water instead of the thick coverslip. The liquid pulled the thin coverslip downwards slightly. Explain how the use of the thin coverslip would affect the results for the cell count. [2 marks]29 Turn over for the next question DO NOT WRITE/ON THIS PAGE ANSWER IN THE/SPACES PROVIDED Turn over >30 ole An echidna is a mammal that lives in Australia. tor Figure 10 shows an echidna, Figure 10 Figure 11 shows how the body temperature of the echidna varies in warm weather and in cold weather. Figure 11 40. T T Warm weather: AS i soh A Nd [| Mook elwratial Body tt i 1 i Ty temperature iT Hy it i Ha ofechidna 20} 1} iH I A i in°c it rN ih 1 f Py Peo ay ELE weamer tt tihoet 10 EES i i rh EN San PoE ttt HL rt HoH ° it 0 2 4 6 8 Time in weeks Figure 12 shows how human body temperature varies. Figure 12, 38 Body temperature 37 of human. in" en 0 2 4 6 8 10 Time in weeks31 0[8].[1] Compare the variation in body temperature of the echidna in warm weather with the ax variation in body temperature of the human Use data from Figure 11 and Figure 12. [2 marks] In the cold winter months, the echidna hibemates. During hibernation: «+ the echidna’s body temperature decreases to below 5 °C «+ the echidna sleeps for up to 17 days ata time * the echidna’s rate of metabolism slows down. 0[8].[2] Explain why the decrease in body temperature is an advantage to the echidna during hibernation. [2 marks] Question 8 continues on the next page Turn over >32 During hibernation the echidna wakes up several times. Each time the echidna wakes up it becomes active and its body temperature increases to over 30 °C. Explain why the echidna has a higher body temperature when itis active. [2 marks] An echidna can dilate and constrict blood vessels in its skin. Explain how the dilation of blood vessels in the skin can help to decrease body temperature. [3 marks] tor33 An athlete trained in a hot climate. The athlete lost a large volume of water each day in sweat. The athlete's energy intake each day from food was 20 000 ki. Evaporation of 1 cm? of sweat requires 2.5 kJ of energy. 40% of the athlete's daily energy intake was used to evaporate sweat Calculate the volume of sweat the athlete lost each day. Give your answer in dm? 1. dm? = 1.000 cm? [3 marks] Volume of sweat lost in one day = dm? ‘Suggest why the athlete was advised to take salt tablets each day. [1 mark] Turn over for the next question Turn over >34 OE ‘Students investigated the response of plant shoots to one-sided light. tor Figure 13 shows how the students set up three experiments. Figure 13, karts = ours Experimenta — Sh —* 3 — Shoot cotton wool Seed Light-proof box 42 hours | Experiment B —s Light + Experimentc Light > [} <—Light 12 hours35 ‘Suggest two control variables the students should have used in their investigation. [2 marks] Describe how experiment B and experiment C acted as controls for the investigation. [2 marks] Experiment B Experiment C Give two conclusions that the students could make from the ink marks on the shoot in experiment A. [2 marks] Name the type of response shown by the seedling in experiment A. [1 mark] Question 9 continues on the next page Turn over >36 ‘Auxin is a plant hormone. Auxin is made in the shoot tip, Scientists investigated the role of auxin in the response of shoot tips to light. This is the method used. 1. Grow four seedlings in the dark for a few days. 2. Cut the tip off the shoot of each seedling, 3, Place each shoot tip on a small block of agar jelly. 4. Place the shoot tips and agar in different conditions as shown in Figure 14. 5, After 24 hours, measure the mass of auxin in the agar blocks. Figure 14 /\ Shoat \ =. ‘e ‘Agar block 253 3 125 12.4 Class Kept in the dark Kept in the dark — — (/\ tine 7. CM + + ‘A 17278 Class 12.6 12.6 One-sided light One-sided light ‘The numbers under each block show the mass of auxin that diffused into the blocks from the shoot tips. The mass of auxin is given in arbitrary units.37 A scientist made a hypothesis: ‘Light causes auxin to move from the side of the shoot nearest to the light to the side furthest from the light.” Describe the evidence from Figure 14 which supports the hypothesis. [3 marks] Another scientist made a different hypothesis: ‘Light causes the breakdown of auxin.’ Give the evidence from Figure 14 that shows that auxin is not broken down by light. [1 mark] END OF QUESTIONS Do not wre ‘ute he bow38 There are no questions printed on this page DO NOT WRITE/ON THIS PAGE ANSWER IN THE/SPACES PROVIDED39 Do not wre ‘ute ne Question Additional page, if required. number Write the question numbers in the left-hand margin.Question number 40 ‘Additional page, if required. 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