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TASMANIAN QUALIFICATIONS AUTHORITY PLACE LABEL HERE Tasmanian Certificate of Education BIOLOGY Senior Secondary Subject Code: BIO315109 External Assessment 2013 Part 1 Time: 35 minutes On the basis of your performance in this examination, the examiners will provide a result on the following criterion taken from the course statement: Criterion 2 Develop interpret and evaluate biological experiments. Section Total /32 Pages: Questions: © 12 4 Copyright for part(s) of this examination may be held by individuals and/or organisations other than the Tasmanian Qualifications Authority. Biology – Part 1 BLANK PAGE Page 2 Biology – Part 1 CANDIDATE INSTRUCTIONS Candidates MUST ensure that they have addressed the externally assessed criterion on this examination paper. Answer ALL questions. Answers must be written in the spaces provided on the examination paper. You should make sure you answer all parts within each question so that the criterion can be assessed. This examination is 3 hours in length. It is recommended that you spend approximately 35 minutes in total answering the questions in this booklet. The 2013 External Examination Information Sheet for Biology can be used throughout the examination. All written responses must be in English. Page 3 Biology – Part 1 Question 1 The diagram below shows the small freshwater Crustacean called Daphnia. The heart of Daphnia is clearly visible when observed with a microscope. 5 mm A biology class of 20 students investigated the effect of ethanol (alcohol) on the heart rate of Daphnia. Each student placed a live Daphnia specimen mounted in a drop of water on a microscope slide, observed it under low power and recorded the heart rate as number of beats per minute. Each student then replaced the water with 1% ethanol and then replaced it with 10% ethanol, repeating the count of beats per minute each time. (a) (i) Identify the independent variable in this experiment. (1 mark) .................................................................................................................................... (ii) Identify the dependent variable in this experiment. (1 mark) .................................................................................................................................... (b) What is the control in this experiment and what is its purpose? (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. Question 1 continues opposite. Page 4 For Marker Use Only Biology – Part 1 Question 1 (continued) (c) (i) List four factors that should be kept constant during this experiment. (2 marks) For Marker Use Only .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... (ii) Why is it important that these factors are kept constant? (1 mark) .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... (d) The students pooled their results. What are the benefits of this? (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. /9 Page 5 Biology – Part 1 BLANK PAGE Page 6 Biology – Part 1 Question 2 Two gardeners living in the same neighbourhood shared a packet of pea seeds. They each planted the same number of seeds in their own garden. For Marker Use Only Robert had a high yield of peas from his plants but Mary’s yield of peas was poor. Suggest one hypothesis that could account for these results. (3 marks) ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... /3 Page 7 Biology – Part 1 Question 3 Springtails, as shown in the diagram below, are small invertebrates which inhabit leaf litter under woodland in very large numbers. They consume dead leaves and other detritus and are important in beginning the process of decomposition. 0.5 mm Following the 2013 bushfires in southern Tasmania, a biologist decided to investigate the effect of the fires on the abundance of Springtails in leaf litter, including the time taken for their re-establishment in burnt areas. (a) Outline an effective procedure for this investigation. Identify the important variables and indicate how experimental bias can be avoided. (6 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. Question 3 continues opposite. Page 8 For Marker Use Only Biology – Part 1 Question 3 (continued) (b) Identify and explain one difficulty (not referred to in your answer to part a), in investigating the effects of bushfire on Springtail numbers. (2 marks) For Marker Use Only ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. /8 Page 9 Biology – Part 1 Question 4 In February 2013 a group of Norwegian scientists published results of an investigation into the effects of the vitamin folic acid when taken in tablet form by women in the early stages of pregnancy. This research took place between 2002 and 2008. Information was collected using questionnaires given to the mothers of 85 176 children born during this period. The women were asked whether or not they had taken folic acid tablets during early pregnancy. The number of children in this group who were diagnosed with autism was also traced. Autism is a disorder which can cause problems with language development, communication and social interaction. The results of the investigation are summarised below. Mothers Total number of births Number of autistic children % of children with autism Who took folic acid tablets 61 042 64 0.10 Who did not take folic acid supplement 24 134 50 0.21 (a) Write a hypothesis that the Norwegian scientists were testing in this investigation. (3 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. (b) Describe one major shortcoming of this investigation and explain its influence on the results. (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. Question 4 continues opposite. Page 10 For Marker Use Only Biology – Part 1 Question 4 (continued) (c) Explain why such a large sample size is desirable in this investigation involving humans. (3 marks) For Marker Use Only ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. (d) In many medical investigations involving humans a placebo is used. (i) Why is a placebo used in such investigations? (2 marks) .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... (ii) Explain why the scientists investigating the relationship between folic acid and autism did not design an experiment which involved the use of a placebo. (2 marks) .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... Page 11 /12 Biology – Part 1 This question paper and any materials associated with this examination (including answer booklets, cover sheets, rough note paper, or information sheets) remain the property of the Tasmanian Qualifications Authority. Page 12 TASMANIAN QUALIFICATIONS AUTHORITY PLACE LABEL HERE Tasmanian Certificate of Education BIOLOGY Senior Secondary Subject Code: BIO315109 External Assessment 2013 Part 2 Time: 35 minutes On the basis of your performance in this examination, the examiners will provide a result on the following criterion taken from the course statement: Criterion 5 Demonstrate knowledge and understanding of the chemical basis of life. Section Total /35 Pages: Questions: © 12 5 Copyright for part(s) of this examination may be held by individuals and/or organisations other than the Tasmanian Qualifications Authority. Biology – Part 2 BLANK PAGE Page 2 Biology – Part 2 CANDIDATE INSTRUCTIONS Candidates MUST ensure that they have addressed the externally assessed criterion on this examination paper. Answer ALL questions. Answers must be written in the spaces provided on the examination paper. You should make sure you answer all parts within each question so that the criterion can be assessed. This examination is 3 hours in length. It is recommended that you spend approximately 35 minutes in total answering the questions in this booklet. The 2013 External Examination Information Sheet for Biology can be used throughout the examination. All written responses must be in English. Page 3 Biology – Part 2 Question 5 Choose one biological compound from the list below to match each of the statements in the table. Compounds may be used more than once or not used at all. (3 marks) • • • • • For Marker Use Only Carbohydrate Lipid Protein Nucleic acid ATP Statement Biological compound All enzymes are this type of compound. This is a compound generated by cellular respiration. This compound is a polymer of nucleotides. This compound can be broken down into fatty acids and glycerol. This compound contains cytosine. Cell walls are composed of this type of compound. /3 Page 4 Biology – Part 2 Question 6 A scientist analysed the number of each of the four bases present in a molecule of DNA, obtaining the following results. Coding strand Adenine Thymine 60 21 Complementary strand Cytosine Guanine 32 44 (a) Complete the table above to show the composition of the DNA. (b) The scientist then analysed a molecule of messenger RNA produced from the coding strand. State one way in which the composition of the mRNA molecule would differ from the DNA molecule. (1 mark) For Marker Use Only (2 marks) ............................................................................................................................................. ............................................................................................................................................. (c) Outline how a DNA molecule replicates, referring to the importance of complementary base pairing. (3 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. /6 Page 5 Biology – Part 2 For Marker Use Only Question 7 The diagrams below show the mode of action of an enzyme. A D B C (a) Name the following: (2 marks) A ......................................................................................................................................... B .......................................................................................................................................... C ......................................................................................................................................... D ......................................................................................................................................... (b) Describe in detail how enzymes catalyse biological reactions, with reference to the diagrams above. (3 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. (c) What aspect of the mode of action of enzymes is not shown in the diagrams? (1 mark) ............................................................................................................................................. ............................................................................................................................................. Question 7 continues opposite. Page 6 Biology – Part 2 Question 7 (continued) (d) Pectinase is an enzyme which helps to break down plant cell walls. It can be added to fruit, such as apples, to speed up the process of juice extraction. For Marker Use Only Describe how temperature would influence juice extraction with pectinase and explain why it does so. (3 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. /9 Page 7 Biology – Part 2 Question 8 The graph below shows the rate of production of oxygen by a tomato plant growing in a greenhouse, measured over one day. Rate of oxygen production (arbitary units) 70 60 50 40 30 20 10 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 -10 -20 (a) Time of day (hours) Explain fully why peak oxygen production occurs between 7 and 19 hours (7 am and 7 pm). (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. (b) Explain why the rate of oxygen production sometimes has a negative value. (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. Question 8 continues opposite. Page 8 For Marker Use Only Biology – Part 2 Question 8 (continued) (c) At 21 hours (9 pm) the plant is not growing. Explain why this is the case. (2 marks) For Marker Use Only ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. (d) Plant roots normally have access to oxygen from the small air spaces in between soil particles. After a flood, soils become waterlogged and it can be observed that plants survive for a short time and then die. Name the process that is taking place in the cells of the plant roots in waterlogged soil. Explain why it enables plants to survive only for a short time. (3 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. /9 Page 9 Biology – Part 2 Question 9 Sickle cell anaemia is a genetic disease in which faulty haemoglobin is produced, resulting in red blood cells which tend to collapse and are less efficient at carrying oxygen. The table below shows part of the normal DNA base sequence for haemoglobin and the mutant code of a sickle cell anaemia sufferer. Complete the table below (i) to show the base sequence of the messenger RNA in each case. (ii) to identify the amino acid produced in each case, using the genetic code provided. (1 mark) Part I DNA base sequence II Messenger RNA base sequence III Amino acid coded for (1 mark) Normal person Sickle cell anaemia sufferer CTC CAC Genetic code on the messenger RNA SECOND BASE U C A G Key U Phe Phe C Ser Ser A Tyr Tyr G Cys Cys U C Leu Leu Ser Ser Stop Stop Stop Trp A G Leu Leu Pro Pro His His Arg Arg U C Leu Leu Pro Pro Gln Gln Arg Arg A G Ile Ile Ile Thr Thr Thr Asn Asn Lys Ser Ser Arg U C A Met Val Thr Ala Lys Asp Arg Gly G U Val Val Ala Ala Asp Glu Gly Gly C A Val Ala Glu Gly G THIRD BASE FIRST BASE (a) Phe = Phenylalanine Leu = Leucine Ile = Isoleucine Met =Methionine Val = Valine Ser = Serine Pro = Proline Thr = Threonine Ala = Alanine Tyr = Tyrosine His = Histidine Gln = Glutamine Asn = Asparagine Lys = Lysine Asp = Aspartic acid Glu = Glutamic acid Cys = Cysteine Try = Tryptophan Arg = Arginine Gly = Glycine Question 9 continues opposite. Page 10 For Marker Use Only Biology – Part 2 Question 9 (continued) (b) For Marker Use Only Outline the following stages in the process of production of haemoglobin: (i) from Part I to Part II of the table (2 marks) .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... (ii) from Part II to Part III of the table. (3 marks) .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... (c) Explain why the substitution of a single amino acid molecule can result in sickle cell anaemia. (1 mark) ............................................................................................................................................. ............................................................................................................................................. /8 Page 11 Biology – Part 2 This question paper and any materials associated with this examination (including answer booklets, cover sheets, rough note paper, or information sheets) remain the property of the Tasmanian Qualifications Authority. Page 12 TASMANIAN QUALIFICATIONS AUTHORITY PLACE LABEL HERE Tasmanian Certificate of Education BIOLOGY Senior Secondary Subject Code: BIO315109 External Assessment 2013 Part 3 Time: 35 minutes On the basis of your performance in this examination, the examiners will provide a result on the following criterion taken from the course statement: Criterion 6 Demonstrate knowledge and understanding of cells. Section Total /33 Pages: Questions: © 12 5 Copyright for part(s) of this examination may be held by individuals and/or organisations other than the Tasmanian Qualifications Authority. Biology – Part 3 BLANK PAGE Page 2 Biology – Part 3 CANDIDATE INSTRUCTIONS Candidates MUST ensure that they have addressed the externally assessed criterion on this examination paper. Answer ALL questions. Answers must be written in the spaces provided on the examination paper. You should make sure you answer all parts within each question so that the criterion can be assessed. This examination is 3 hours in length. It is recommended that you spend approximately 35 minutes in total answering the questions in this booklet. The 2013 External Examination Information Sheet for Biology can be used throughout the examination. All written responses must be in English. Page 3 Biology – Part 3 Question 10 The images on this page show three cells viewed through microscopes. Question 10 continues opposite. Page 4 Biology – Part 3 Question 10 (continued) (a) Complete the table below by correctly inserting in each box either a to mean ‘yes’ or an ! to mean ‘no’. (3 marks) The cell Cell A Cell B For Marker Use Only Cell C has a cell membrane. is a plant cell. has mitochondria. has a cell wall. has cytoplasm. has a single circular chromosome. (b) Which of the cells is a prokaryote? Give one clear piece of evidence from the diagram to support your choice. (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. (c) Was Cell C viewed using a light microscope or an electron microscope? Give one convincing piece of evidence that supports your answer. (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. (d) Estimate the width of the nucleus of Cell C and give the answer in both µm and mm. (1 mark) ....................................................... µm ....................................................... mm /8 Page 5 Biology – Part 3 Question 11 A scientist collected 50 marine worms which normally live in sea water. The worms were weighed in groups of 10 and each group placed in a different concentration of salts found in sea water. After 12 hours each group was reweighed and the percentage change in mass calculated. The results are shown in the graph below. (a) Explain fully what has occurred in the worms placed in the 5% solution. (3 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. (b) If worms were placed in distilled water they would die. Explain what would cause this. (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. Question 11 continues opposite. Page 6 For Marker Use Only Biology – Part 3 Question 11 (continued) (c) Predict the concentration of salts in the normal habitat of the worms. (1 mark) For Marker Use Only ............................................................................................................................................. ............................................................................................................................................. /6 Page 7 Biology – Part 3 Question 12 During sexual reproduction two cells (P and Q in the diagram) combine to produce cell R. Cell R divides to produce an embryo. Cell P Cell Q Cell R Embryo (a) State whether each of the following cells is haploid or diploid. (2 marks) (i) Cell P ........................................................................................................................ (ii) Cell Q ....................................................................................................................... (iii) Cell R ....................................................................................................................... (iv) Cell of the embryo .................................................................................................... (b) Each cell of the embryo in the diagram has the potential to generate any human cell. Explain why this is possible. (1 mark) ............................................................................................................................................. ............................................................................................................................................. (c) As the embryo develops, cells become specialised to perform particular functions. Name one specialised human cell you have studied, state its function and describe one way in which it is adapted to that function. (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. Question 12 continues opposite. Page 8 For Marker Use Only Biology – Part 3 Question 12 (continued) (d) Two different types of cell division are involved in the formation of an embryo. For each type of cell division, name it, state where in the process it occurs and explain its significance. (5 marks) For Marker Use Only ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. /10 Page 9 Biology – Part 3 Question 13 The table below shows the concentration of certain ions in sea water and in a seaweed living in sea water. Ion (a) For Marker Use Only Concentration (%) SEA WATER SEAWEED Sodium 1.08 3.50 Magnesium 0.13 0.70 Potassium 0.04 2.80 Name the process by which these ions enter seaweed cells and use the data to explain your answer. (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. (b) If the concentration of dissolved oxygen in sea water decreased, would you expect the concentration of these ions to decrease, increase or remain unchanged? Explain your answer. (3 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. /5 Page 10 Biology – Part 3 For Marker Use Only Question 14 Some organisms, such as Amoeba, are unicellular and some, such as Humans, are multicellular. (a) List two advantages of being multicellular. (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. (b) State one disadvantage of being multicellular and explain how this problem can be overcome. (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. /4 Page 11 Biology – Part 3 This question paper and any materials associated with this examination (including answer booklets, cover sheets, rough note paper, or information sheets) remain the property of the Tasmanian Qualifications Authority. Page 12 TASMANIAN QUALIFICATIONS AUTHORITY PLACE LABEL HERE Tasmanian Certificate of Education BIOLOGY Senior Secondary Subject Code: BIO315109 External Assessment 2013 Part 4 Time: 35 minutes On the basis of your performance in this examination, the examiners will provide a result on the following criterion taken from the course statement: Criterion 7 Demonstrate knowledge and understanding of organisms. Section Total /34 Pages: Questions: © 16 5 Copyright for part(s) of this examination may be held by individuals and/or organisations other than the Tasmanian Qualifications Authority. Biology – Part 4 BLANK PAGE Page 2 Biology – Part 4 CANDIDATE INSTRUCTIONS Candidates MUST ensure that they have addressed the externally assessed criterion on this examination paper. Answer ALL questions. Answers must be written in the spaces provided on the examination paper. You should make sure you answer all parts within each question so that the criterion can be assessed. This examination is 3 hours in length. It is recommended that you spend approximately 35 minutes in total answering the questions in this booklet. The 2013 External Examination Information Sheet for Biology can be used throughout the examination. All written responses must be in English. Page 3 Biology – Part 4 Question 15 The pedigree below shows the inheritance of a certain human disease. The individuals who have the disease are shown shaded. Generation I II III IV (a) Give two pieces of evidence from the pedigree above that show this disease is not X–linked recessive. (3 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. (b) Deduce whether this disease is due to an autosomal dominant or an autosomal recessive allele, giving one conclusive piece of evidence. Autosomal means not carried on the X chromosome. (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. Question 15 continues opposite. Page 4 For Marker Use Only Biology – Part 4 Question 15 (continued) (c) On the copy of the pedigree below, mark with an H all the individuals in this family who are definitely heterozygous. (2 marks) For Marker Use Only Generation I II III IV (d) Researchers may have data for families over many generations and occasionally they find that this disease suddenly occurs, even though previous members of the family are disease-free. Explain what could be causing this sudden appearance of the disease in a family. (1 mark) ............................................................................................................................................. ............................................................................................................................................. /8 Page 5 Biology – Part 4 Question 16 ( ) Diagram 1 shows an experiment in which radioactive carbon dioxide 14 CO2 was given to one leaf (X) of a pot plant. The plant was given water and light for 24 hours. It was then removed from the pot and placed on special film to produce an autoradiograph, as shown in Diagram 2. The autoradiograph registers the presence of radioactive 14 C by turning black. Diagram 1 Diagram 2 leaf X (a) Using the correct terms, account for the presence of radioactive 14 C in leaf X. (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. Question 16 continues opposite. Page 6 For Marker Use Only Biology – Part 4 Question 16 (continued) (b) Deduce one conclusion about the direction of transport of compounds containing carbon that can be drawn from this experiment. (1 mark) For Marker Use Only ............................................................................................................................................. ............................................................................................................................................. (c) Explain the high level of radioactive 14 C in the roots of the plant. (3 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. (d) Describe two ways in which the movement of water in a plant differs from the movement of carbon compounds. (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. /8 Page 7 Biology – Part 4 BLANK PAGE Page 8 Biology – Part 4 For Marker Use Only Question 17 Animals produce nitrogenous waste products which must be excreted. (a) What is the origin of nitrogenous waste products in animals? (1 mark) ............................................................................................................................................. ............................................................................................................................................. (b) Suggest why plants do not excrete nitrogenous waste products. (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. (c) Amphibians, such as frogs and toads, excrete their nitrogenous waste as ammonia during the tadpole phase but switch to producing urea during the adult phase. Some adult desert frogs, however, excrete nitrogenous waste as uric acid. (i) Explain the difference in excretory products of tadpoles and adult Amphibians. (3 marks) .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... (ii) Explain the benefit to desert frogs of excreting uric acid. (1 mark) .................................................................................................................................... .................................................................................................................................... /7 Page 9 Biology – Part 4 Question 18 Rhizopus is a fungus that is often found growing as a mould on decaying food. It spreads fungal threads called hyphae over food, sending up vertical hyphae with swellings or spore cases at the tip. Spores develop inside the spore cases and are dispersed when the spore case bursts. Each spore can germinate to produce a new mass of fungal threads (or mycelium). This process is shown in Diagram 1 below. Diagram 1 (x 100) (a) What type of reproduction is occurring in Diagram 1 above and what could you predict about all the spores from one mycelium? (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. (b) Suggest one advantage of this form of reproduction to the survival of Rhizopus. (1 mark) ............................................................................................................................................. ............................................................................................................................................. Question 18 continues opposite. Page 10 For Marker Use Only Biology – Part 4 Question 18 (continued) Occasionally hyphae from two different Rhizopus organisms meet, fuse together and produce a zygospore which germinates and forms a new mycelium. This is shown in Diagram 2 below. For Marker Use Only Diagram 2 (x 200) (c) Explain the importance of the form of reproduction shown in Diagram 2 to the longterm survival of Rhizopus. (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. /5 Page 11 Biology – Part 4 Question 19 For The carbon dioxide concentration of human blood is controlled by homeostatic mechanisms. Part of this process is illustrated in the diagram below. Question 19 continues opposite. Page 12 Marker Use Only Biology – Part 4 Question 19 (continued) With reference to this example, identify and describe the key aspects of homeostatic mechanisms in organisms in general. (6 marks) For Marker Use Only ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... /6 Page 13 Biology – Part 4 BLANK PAGE Page 14 Biology – Part 4 BLANK PAGE Page 15 Biology – Part 4 This question paper and any materials associated with this examination (including answer booklets, cover sheets, rough note paper, or information sheets) remain the property of the Tasmanian Qualifications Authority. Page 16 TASMANIAN QUALIFICATIONS AUTHORITY PLACE LABEL HERE Tasmanian Certificate of Education BIOLOGY Senior Secondary Subject Code: BIO315109 External Assessment 2013 Part 5 Time: 35 minutes On the basis of your performance in this examination, the examiners will provide a result on the following criterion taken from the course statement: Criterion 8 Demonstrate knowledge and understanding of the interaction of organisms with their environment. Section Total /35 Pages: Questions: © 16 6 Copyright for part(s) of this examination may be held by individuals and/or organisations other than the Tasmanian Qualifications Authority. Biology – Part 5 BLANK PAGE Page 2 Biology – Part 5 CANDIDATE INSTRUCTIONS Candidates MUST ensure that they have addressed the externally assessed criterion on this examination paper. Answer ALL questions. Answers must be written in the spaces provided on the examination paper. You should make sure you answer all parts within each question so that the criterion can be assessed. This examination is 3 hours in length. It is recommended that you spend approximately 35 minutes in total answering the questions in this booklet. The 2013 External Examination Information Sheet for Biology can be used throughout the examination. All written responses must be in English. Page 3 Biology – Part 5 Question 20 (a) In 2010 a new species of jellyfish was discovered in the River Derwent near Hobart. It was given the scientific name Csiromedusa medeopolis. Imagine you are studying the ecology of Csiromedusa medeopolis. (i) For Marker Use Only What important abiotic factors would you need to investigate? Name two. (1 mark) .................................................................................................................................... .................................................................................................................................... (ii) Choose one significant biotic factor which should also be investigated, and explain your choice. (2 marks) .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... (b) The list below gives the scientific names of some other marine species found in Australian waters. (i) 1 2 3 Chironex fleckeri Carukia barnesi Carybdea xaymacana 4 5 Phlytenanthus australis Carybdea sivickisi 6 Notoplana australis Explain why it is preferable for biologists to use scientific rather than common names. (1 mark) .................................................................................................................................... .................................................................................................................................... (ii) Which two species in the list are most closely related? Explain your reasoning. (2 marks) .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... /6 Page 4 Biology – Part 5 For Marker Use Only Question 21 Many species of Acacia, or Wattle, grow throughout Australia. One of the species of Acacia shown below is found in Tasmanian forests and one grows in the desert of Central Australia. A (a) B Identify which is the desert species and give one reason for your choice. (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. (b) Predict one structural adaptation to habitat that is not shown in the relevant diagram but which you would still expect to find in the desert species. (1 mark) ............................................................................................................................................. ............................................................................................................................................. /3 Page 5 Biology – Part 5 Question 22 An aquarium was set up as shown in the diagram below and no further components were added to it. Micro-organisms were also present in the aquarium which was kept in a laboratory, subject to natural light. (a) Use the diagram below to show how carbon moves through this ecosystem by (i) drawing in at least four arrows to show the direction of movement of carbon (2 marks) (ii) labelling each arrow with the name of the process that is occurring. (2 marks) Carbon dioxide in water Carbon in plants Carbon in snails Carbon in micro-organisms Question 22 continues opposite. Page 6 For Marker Use Only Biology – Part 5 Question 22 (continued) (b) Nitrogen is also an essential element for all organisms. Describe fully how nitrogen would be made available to the freshwater plants in the aquarium (2 marks) For Marker Use Only ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. /6 Page 7 Biology – Part 5 Question 23 The information below lists some of the organisms from a natural community found in Tasmania. It also shows the main components of the diets of the organisms. Organism Diet Eastern barred bandicoot (mammal) Spotted tailed quoll (mammal) Insects, spiders Small birds, mammals, insects, spiders Brown goshawk (bird) Native hen (bird) Small birds, mammals Grass Grasshopper (insect) Flower spider Grass Insects (a) Construct a food chain for this community that includes four of these organisms and indicate the trophic level of each of these four organisms. (2 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. (b) Give a detailed explanation of what happens to energy as it moves along a food chain. (4 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. Question 23 continues opposite. Page 8 For Marker Use Only Biology – Part 5 Question 23 (continued) PCBs (polychlorinated biphenyls) are chemicals which were used in electrical goods in the past. They are non-biodegradable and persistent in the environment. (c) For Marker Use Only PCBs are present in this ecosystem. If the Native hen and Brown goshawk were tested for PCBs, how would the levels in the birds compare? Explain your answer. (3 marks) ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. /9 Page 9 Biology – Part 5 Question 24 Paramecium is a unicellular organism which can easily be fed on yeast cells and cultured in a flask in the laboratory. In a famous experiment two different species of Paramecium were grown, either alone or mixed together. The sizes of the populations of Paramecium were estimated over two weeks and the results presented in the graphs below. Population size Time in days Population size Time in days (a) Consider Paramecium aurelia when grown alone and explain what is happening to the population at: (i) Day 6 (1 mark) .................................................................................................................................... .................................................................................................................................... (ii) Day 12 (1 mark) .................................................................................................................................... .................................................................................................................................... Question 24 continues opposite. Page 10 For Marker Use Only Biology – Part 5 Question 24 (continued) (b) Using the data, explain what causes the difference in results for the mixed cultures and those for single cultures. (2 marks) For Marker Use Only ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. (c) Predict what will happen to both single cultures if the experiment were continued for an extended period of time and explain your answer. (1 mark) ............................................................................................................................................. ............................................................................................................................................. /5 Page 11 Biology – Part 5 Question 25 Nothofagus (Southern Beech) is a genus of 36 different species of trees and shrubs found in the Southern Hemisphere. There are three Australian species of Nothofagus and their distribution is shown in the table below. Species Nothofagus gunnii Distribution Tasmania Nothofagus cunninghamii Nothofagus moorei Tasmania, Victoria New South Wales, Queensland Other species of Nothofagus are found in South America, New Zealand, New Caledonia and New Guinea. Fossils of Nothofagus species have also been found in Antarctica. Nothofagus cunninghamii leaves (half actual size) Question 25 continues opposite. Page 12 For Marker Use Only Biology – Part 5 Question 25 (continued) Give a scientific explanation, in as much detail as you can, of how all these different species of Nothofagus may have arisen. (6 marks) For Marker Use Only ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... /6 Page 13 Biology – Part 5 BLANK PAGE Page 14 Biology – Part 5 BLANK PAGE Page 15 Biology – Part 5 This question paper and any materials associated with this examination (including answer booklets, cover sheets, rough note paper, or information sheets) remain the property of the Tasmanian Qualifications Authority. Page 16 Biology Subject Code: BIO315109 2013 External Examination Information Sheet SOME COMMON TERMS EXPLAINED: Account explain Analyse examine data mathematically to gain understanding of the data Calculate to find the answer using mathematics Compare give an account of similarities and difference between two factors Construct represent information in a graphical form Deduce reach a conclusion from the information given Describe give a detailed account including all relevant information Design produce a plan Discuss give an account including a range of arguments, assessment of the importance of various factors or a comparison of alternatives Distinguish give the difference between two or more different items Draw represent by means of pencil lines – include labels (unless told not to) - not to be confused with Draw a Conclusion Estimate find an approximate value Evaluate assess the limitations and implications Explain give a clear account including causes and reasons List give a sequence of names or brief answers Identify find an answer for a quantity Measure Use a measuring instrument to find a value for a quantity. Always include the units of the number (value) Outline give a brief account or summary Predict give an expected result Page 1 of 19 “Cause” Criterion 2 – Design & evaluate experiments Formulating a Hypothesis “Effect” Your hypothesis should 1. be feasible (i.e. be sensible and based on scientific concepts) 2. be a statement (not a question) 3. be based on observations 4. involve one independent and one dependent variable in a cause-‐and-‐effect relationship 5. be testable and measurable in a way that demonstrates cause and effect Experimental Design State the hypothesis to be tested (unless already stated) 1. State the independent variable and how it is manipulated 2. State the dependent variable and how it is measured 3. Describe the procedure clearly in a step-‐by-‐step fashion which could be easily followed in a laboratory or the field. Indicate sample sizes, quantities of materials and time involved Indicate how many replicas there are (if needed). Describe which variables are controlled and how. Say why fixed variables are needed. Include factors relevant for the organism – biotic and abiotic relevant in the environment What is the control group in the experiment and why is it needed. 4. How are the results analysed. 5. Indicate any repetitions of the experiment If required: 6. State what results would support the hypothesis and what results would not support the hypothesis 7. Discuss any foreseeable problems in conducting the experiment. variables difficult to control, sample size issues, ethical issues, animal vs human experimentation/environmental impacts etc. INDEPENDENT VARIABLE (IV) the thing that the experimenter deliberately varies DEPENDENT VARIABLE (DV) the thing that the experimenter measures Designing Experiments with Humans Designing experiments with humans as subjects is not straightforward because there are a number of areas over which the experimenter has no control. The impact of these unfixed/uncontrolled variables can be reduced by a number of factors including: • use of placebos and double blind experiments/trials Also ethical aspects need to be considered. Trial experiments conducted in a confined situation (laboratory/greenhouse) need to precede field investigations. Distinction between a few terms: A replica is multiple identical groups within the one experiment. A repeat is doing the experiment again at a future time. A follow up experiment is one which builds on the original experiment in some way. Page 2 of 19 • Criterion 5 – Chemical Basis of Life • BIOLOGICAL MOLECULES A. • • • • CARBOHYDRATES contain carbon, hydrogen, oxygen general formula: CH2O e.g. glucose = C6H1206 main function -‐ energy source some used in growth/repair -‐ to build new cell components etc. • excess carbohydrates are stored in the body as: -‐ starch (in plants) -‐ glycogen (in animals) -‐ in liver, muscles Types: (i) Monosaccharides -‐ "single sugar" -‐ e.g. glucose, fructose -‐ 3 -‐ 6 C atoms, small – can fit through pores in cell membrane. -‐ soluble in water (ii) Disaccharides - "double sugars" -‐ e.g. sucrose (cane sugar), lactose (milk sugar) - too large to fit through cell membranes (iii) Polysaccharides - "many sugars", large number (1,000’s) of monosaccharide (monomer) units joined together – form a polymer. - too large to fit through cell membranes - insoluble in water (mostly) e.g. starch, glycogen, cellulose B. PROTEINS • contain carbon, hydrogen, oxygen and nitrogen (sometimes sulphur ) • large, complex • are polymers -‐ made up of a large number of individual units (monomers) called amino acids (AA) single AA -‐ (20 different types of AAs) dipeptide -‐ 2 amino acids joined polypeptide -‐ many amino acids protein -‐ one or more polypeptide chains, folded up to produce a specific shape • • Need all 20 kinds of amino acids in order to make all the proteins needed by the body. - Of the 20 different AA’s, 9 (8 in adults and 1 extra in infants) AA’s are called essential amino acids. These must be included in the diet, as they cannot be made by the body. - If non-‐essential AA’s are missing in the diet, they can be made by the liver, from glucose and nitrogen-‐containing compounds. Excess amino acids cannot be stored in the body. • C. LIPIDS • fats, oils • contain carbon, hydrogen, oxygen (less O than C) • consist of a glycerol molecule and usually 3 fatty acid molecules Same in ALL lipids The number and type of these determine the type of lipid Animal lipids -‐contain saturated fatty acids. Plant lipids -‐contain unsaturated fatty acids. Main function of lipids – high energy source -‐ (also used to build structural components -‐ e.g. cell membranes) - Main function of proteins -‐ to provide "building blocks"/raw materials for tissue growth, repair, function and metabolism. • different proteins have different AA sequences, and therefore fold up to produce different shapes. proteins in the diet need to be broken down (digested) into individual AAs so they can be absorbed into the blood. Inside cells (at the ribosomes) these AA’s are synthesised into new proteins. These proteins may be: - structural proteins – skin, muscles, hair - specialised proteins – have specific functions: e.g. haemoglobin – in red blood cells, enzymes, hormones (e.g. adrenalin, insulin), antibodies. Page 3 of 19 CELL RESPIRATION D. NUCLEIC ACIDS e.g. DNA (deoxyribonucleic acid) RNA (ribonucleic acid) an energy-‐releasing process (a) AEROBIC RESPIRATION SugarNitrogen → Polymers of nucleotides phosphate base pairs Sugarphosphate backbone backbone SUG AR 2 different types Ribose in R NA or deoxyribose in DN A uses/requires O2 -‐ occurs in mitochondria Overall Equation: PHO SPH ATE → carbon dioxide + water C6H12O6 + 6O2 → ENERGY + 6CO2 + 6H2O glucose + oxygen NITRO GENO US BASES A = adenine T = thymine (DNA) C = cytosine G = guanine U = uracil (RNA) lost as heat stored as ATP in mitochondria When 1 molecule of glucose is aerobically respired → 36 molecules of ATP ATP -‐ Adenosine triphosphate “high energy” molecule – “ like a charged battery” ADP -‐ Adenosine diphosphate CELL REACTIONS “low energy” molecule – “like a flat battery” (a) Decomposition /catabolic reactions - Breaking large molecules into smaller ones - produces/releases energy (b) Synthesis /anabolic reactions Joining together small molecules to make larger ones Needs energy to occur e.g. photosynthesis, DNA replication, protein synthesis - occurs without O2 -‐ occurs in cytoplasm Overall Equation: e.g. cell respiration - Storing Energy: ADP + energy + phosphate → ATP ATP → energy + ADP (+ phosphate) Releasing Energy: (b) ANAEROBIC RESPIRATION -‐ -‐ C6H12O6 → ENERGY + 2C3H6O3 -‐ in animal cells glucose lactic acid C6H12O6 → ENERGY + 2C2H5OH + 2CO2 – in plant & yeast cells glucose alcohol When 1 molecule of glucose is anaerobically respired → 2 molecules of ATP Brief pathway of glucose in cellular respiration Glucose Pyruvic acid (cytoplasm) Coenzyme A (mitochondria) Acetyl Carbon dioxide and water Page 4 of 19 Glycolysis takes place in cytoplasm This reaction takes place in the mitochondria of cells, only if O2 is present and releases large amounts of ATP ROLE OF ENZYMES PHOTOSYNTHESIS (P/S) an energy-‐storing process - light energy is stored as “chemical” energy Overall Equation: carbondioxide + water 6 CO2 + 6 H2O Enzymes are proteins which speed up (catalyse) the chemical reactions which occur in organisms. They are “biological catalysts”. Types of enzymes Enzymes fall into two types of categories: Stored as starch molecules of glucose glucose + oxygen light energy C6H12O6 + 6 O2 chlorophyll Respired to release energy 1) Intracellular enzymes: these occur inside cell and catalyse metabolic reactions 2) Extracellular enzymes: these occur outside cells and catalyse reactions involved in digestion Used to build new tissue (= growth) Enzymes are also classified according to the type of substance on which they act. Comparison of respiration and photosynthesis in plants and animals DARK LIGHT respiration & PLANT respiration photosynthesis (autotroph) ANIMAL (heterotroph) respiration Naming Enzymes -‐ -‐ respiration Coenzyme –a non-‐protein molecule that has to be present before an enzyme can stimulate its specific chemical reaction. The coenzyme takes part in the reaction and is changed by it. Examples include vitamins and ATP. Factors affecting the rate of photosynthesis • Temperature • Concentration of CO2 • Light intensity • Wavelength/colour of light Compensation Point-‐ the light intensity at which the rate of photosynthesis equals the rate of respiration. Cofactor – is a mineral ion of some kind that has to be present to help activate the enzyme. Cofactors influence reactions, but do not take part in them, nor are they changed. Examples include Mg2+ and K+. Inhibitors – poisons such as heavy metals and insecticides can prevent enzymes functioning by slowing down or stopping enzyme action by competing for or changing the shape of the active site. most end in -‐ASE first part relates to the substrate the enzyme usually works on: e.g. Lactase is the enzyme which works on the substrate Lactose Page 5 of 19 Mutations – A change in the sequence of nucleotide bases of the genetic material (DNA or RNA) Factors affecting enzyme activity • Concentration of substrate • Sensitivity to temperature changes • Sensitivity to pH environments • Presence of cofactors and coenzymes are required by some enzymes • Chemical energy in the form of ATP is required by a variety of enzymes before they can act. Changes may be: -‐spontaneous (no discernible reason) -‐ "base" mutation rate = 1 gene in every 100,000 per generation or -‐induced -‐ caused by mutagens (e.g. radiation, chemicals, high temperature) PROTEIN SYNTHESIS Process of making proteins in the cell Point (gene) mutation-‐ a mutation in the nucleotide sequence of a gene in which only one base is altered. Ribosome bases may be: Some of these changes may result in overall change to the protein being produced. Transcription –the synthesis of a mRNA (messenger RNA) molecule from a template strand of DNA in the nucleus. A tRNA (transfer RNA) molecule Translation-‐ involves the reading of the mRNA molecule by the ribosomes and the ordered sequential joining of the amino acids to form a protein. Codon-‐ A group of three nucleotides (or triplet) of DNA or RNA. A codon codes for a specific amino acid. Anticodon-‐ A sequence of three nucleotides on a molecule of tRNA that is complementary to the base sequence on a codon of mRNA. added (inserted) deleted substituted Page 6 of 19 Criterion 6 – Cells STRUCTURE Organelles visible with a light microscope: • • • • • Nucleus and nuclear membrane Nucleolus Cell membrane Chloroplasts Vacuoles !"#$%ℎ !" !"## = 1 mm = 1000 µm !"#$% !" !"#$ !"#$%& !" !"#$% !"## !"#$ !"#$%% ANIMAL CELL VIEWED WITH ELECTRON MICROSCOPE Endoplasmic reticulum – transport system Golgi Body – refines and packages proteins Lysosomes – contain digestive enzymes Centriole – involved in cell reproduction Contractile vacuole-‐fills with water and empties periodically to the cell environment Cilium (plural: cilia)-‐used for locomotion or movement of liquid over cell surface Flagellum-‐ also used for movement EUKARYOTIC CELLS • distinct nucleus surrounded by a membrane • have organelles that are surrounded by a membrane (e.g. mitochondria, ER, lysosomes) • e.g. cells of most plants and animals PROKARYOTIC CELLS • no definite nucleus – DNA spread through cytoplasm. • no membrane-‐bound organelles • eg bacteria, blue-‐green bacteria EXCHANGE OF MATERIALS PLANT CELL VIEWED WITH ELECTRON MICROSCOPE Organelles Mitochondria – site of aerobic respiration Chloroplasts – site of photosynthesis Ribosomes – site of protein synthesis Nucleolus – make ribosomes Diffusion -‐ movement of solute Osmosis -‐ movement of water from an area of lower concentration (“weak” solution) to an area of higher concentration (“strong” solution), through a semi-‐permeable membrane Active transport -‐ movement of solute from an area of lower concentration to any area of higher concentration with the use of energy Endocytosis -‐ movement of LARGE substances (too large to fit through pores in cell membrane) INTO a cell. e.g. phagocytosis – engulfing solid particles; pinocytosis – engulfing liquid droplets Exocytosis -‐ movement of LARGE substances OUT OF a cell. e.g. secretion of useful substances made in the cell (e.g. enzymes); excretion of wastes. Page 7 of 19 Diffusion and osmosis illustrated Factors affecting the rate of diffusion • • • • concentration gradient (conc. difference) temperature SA : Vol ratio medium of transport Surface area: Volume ratio (SA : Vol ratio) Important for cell functioning. Relates to cell size and shape. Adaptation to changing salinity Organisms that live in environments where the concentration changes (e.g. estuaries) respond to these changes in two main ways: Osmoconformers -‐ allow internal concentration to change Hypertonic -‐ more concentrated than … (e.g. seawater is hypertonic to freshwater). Hypotonic -‐ less concentrated than … (e.g. freshwater is hypotonic to seawater) Isotonic -‐ the same concentration as … Osmoregulators -‐ maintain a relatively constant internal environment Relative Sizes of Molecules The cell membrane has pores of a particular size. Only molecules which are small enough can pass through these pores (i.e. the cell membrane is semi-‐ permeable). SMALL ENOUGH TOO LARGE Oxygen (O2) Carbon dioxide (CO2) Cell organelles Other dissolved gases: e.g. N2 Amino acids Proteins Urea, ammonia CARBOHYDRATES CARBOHYDRATES Monosaccharides e.g. glucose Disaccharides e.g. sucrose, lactose, maltose Polysaccharides e.g. starch, glycogen, cellulose Lipids e.g. fats, oils Glycerol Fatty acids Dissolved ions e.g. Ca2+, Na+, Cl -‐ Water (H2O) Contractile vacuoles can be used to maintain equilibrium in some protists Plasmolysis -‐ contraction of cytoplasm of a plant cell as a result of osmosis out of the cell. Page 8 of 19 Meiosis CELL DIVISON There are two types of division: Mitosis – process of cell division in which daughter cells are identical to parent cells. Process used in growth, repair and replacement of cells. Meiosis – process of gamete (egg and sperm) production in most organisms. Gametes contain half the number of chromosomes as parent cells. Mitosis Diploid 2n -‐ Full number of chromosomes in the nuclei Haploid n-‐ Half the number of chromosomes in the nuclei. Usually found in the gametes. Page 9 of 19 Criterion 7 -‐ Organisms RESPIRATORY SYTEM IN HUMANS DIGESTION SYSTEM IN HUMANS Gas exchange at the alveoli Cross section through villi TRANSPORT SYSTEM IN HUMANS Stages in acquisition of nutrients Page 10 of 19 EXCRETION IN HUMANS Urinary system of a female PLANT TRANSPORT SYSTEM Vertical section through a kidney Cross section of the three vessels Structure of a nephron Components of blood Plasma (55%) Water Dissolved substances Plasma proteins Cells (45%) Red blood cells Lymphocytes Phagocytes Platelets Page 11 of 19 PLANT TRANSPORT SYSTEM Fate of excess amino acids Energy, or Carbohydrate stored as fat Transpiration – Loss of water vapour by a plant through xylem and stomata in the leaves. Translocation –the transfer of soluble food materials (eg sugars, vitamins etc) from the leaves where they are made by photosynthesis down and up through the phloem. amino acids Ammonia Urea Uric acid VASCULAR BUNDLES Comparison of nitrogenous waste products Ammonia Urea - Uric acid Solubility Soluble Soluble Insoluble Toxicity Toxic Less toxic Least toxic Energy required for conversion None Some Most energy demanding - - HOMEOSTASIS Optimal status for factor Negative feedback – a change which reverses a particular trend Positive feedback – a change in some variable triggers variables which amplify the change Comparison of control systems Nervous Nature of message Electrochemical impulses Route of message Specific nerve cells Types of effects Rapid, but usually short term Endocrine Chemical compounds (hormones) General blood system Usually slower, but generally longer lasting Page 12 of 19 xylem and phloem tissue occur together and form vascular bundles. vascular bundles in leaves form veins → branch into finer units → no cell is far from a vascular bundle in herbaceous dicotyledons, vascular bundles are arranged around edge of stem/root (xylem is on inside) Xylem vessels Phloem vessels Surface view of stomata Internal section of a leaf Page 13 of 19 ASEXUAL AND SEXUAL REPRODUCTION Transpiration system The path of water from soil across the root to the xylem and into the leaf 1. 2. ASEXUAL REPRODUCTION production of daughter cells (offspring) which are genetically identical to the parent cell. SEXUAL REPRODUCTION involves fertilization – the joining together of 2 gametes (sex cells – sperm and ovum) to produce a zygote. ovum + sperm → zygote Fertilisation requires moisture. • may be external (e.g. fish, frog) internal (e.g. mammals, birds) or • May involve separate “male” and “female” individuals ( cross fertilisation) or • “male” and “female” parts of the same individual (self-‐fertilisation) [hermaphrodites – have both ♀ and ♂ reproductive organs – e.g. snails, earthworms → can self-‐ or cross-‐fertilise] The path of water and particles from soil across the root Page 14 of 19 GENETICS Types of chromosomes Gene -‐the basic unit of inheritance for a given characteristic Allele-‐ Alternate forms of the same gene responsible for determining contrasting characteristics (T or t) Homozygous-‐ individuals have identical alleles for a particular characteristic – e.g. TT or tt Heterozygous individuals have different alleles for a particular characteristic – e.g. Tt Phenotype -‐ The physical or chemical expression of a characteristic Genotype -‐The genetic expression of a characteristic Monohybrid cross -‐A cross where only one gene or characteristic is being considered Dominant allele -‐ If present, hides the effect of the recessive allele. e.g. T is dominant over t → a Tt individual will have the characteristic for which T is dominant Recessive allele-‐ Only expressed in the phenotype if no dominant allele present. e.g. must be tt to be dwarf F1 generation -‐ the generation produced by crossing two parental stocks. F2 generation -‐ the generation produced by crossing two F1 organisms Incomplete (Partial) dominance -‐The condition where the alleles do not fully express themselves when in the heterozygous genotype and the offspring may be an “average” of the two characteristics Co dominance – the condition where both alleles express themselves in the heterozygous genotype. Usually expressed as CR CW Pedigree -‐ chart showing the family history of a particular condition Autosome Chromosome not associated with sex. In humans there are 22 autosomes Sex linked Chromosome that determines the sex of an individual Types of inheritance Page 15 of 19 Criterion 8 – Interaction of organisms with their environment Ecology The study of living organisms in the natural environment. How they interact with one another and how the interact with their nonliving environment. FOOD WEBS Ecosystem Community + Abiotic environment, interacting (i) (ii) (iii) (iv) Community All the populations of the different species living and interacting in the same ecosystem. Species Decomposers Heterotrophic organisms who secrete digestive enzymes onto dead organism matter and absorb the digested material. (e.g. fungi, bacteria) -‐completely breakdown the organic molecules into inorganic molecules → make matter available to producers -‐ i.e. responsible for the recycling of matter in the ecosystem A group of organisms that can breed to produce fully fertile offspring. Population A group of organism of the same species that live in the same habitat at the same time where they can freely interbreed. Biodiversity The total number of different species in an ecosystem and their relative abundance. Habitat The characteristics of the type environment where an organism normally lives. (e.g. a stony stream, a temperate woodland) Niche Habitat + role + tolerance limits to all limiting factors First Order (Primary) Consumers – Second Order (Secondary) Consumers Third Order (Tertiary) Consumers – Top Consumer/Carnivore – not usually eaten by other organisms Trophic level 4 – carnivore (tertiary consumer) Trophic level 3 – carnivore (secondary consumers) Trophic level 2 – herbivore (primary consumers) Trophic level 1 – producer ECOLOGICAL PYRAMIDS As you go up a food chain, the number of individuals at each level (usually) decreases. Due to energy losses at each level, each individual needs to consume a large number of the individuals below. Pyramid of Biomass uses total weight (dry mass) of organisms at each level and takes into account both numbers and mass Energy and organisms Autotrophs / Organisms which can synthesise their Producers own complex, energy rich, organic molecules from simple inorganic molecules (e.g. green plants synthesis sugars from CO2 and H2O) Heterotrophs Organisms who must obtain / Consumers complex, energy rich, organic compounds form the bodies of other organisms (dead or alive). Detritivores Heterotrophic organisms who ingest dead organic matter. (e.g. earthworms, woodlice, millipedes) but do not decompose it into inorganic matter Pyramid of Energy considers total energy at each level 1C P Each level must be no more than 10% of previous Page 16 of 19 BIOLOGICAL MAGNIFICATION Some poisons (those that are not biodegradable) are not completely broken down and excreted by organisms → instead; they accumulate in tissues and are passed along a food chain INTERACTIONS AMONGST ORGANISMS Type of interaction Predator Prey Herbivory Mutualism Commensalism Competition Parastism Species A Species B + + + + -‐ + -‐ -‐ + 0 -‐ -‐ BIOGEOCHEMICAL CYCLES • • • cannot create/destroy atoms → there is a finite (fixed) amount of matter in our biosphere (assume that Earth is a closed system – with no losses/gains of matter) recycling of matter by decomposers nutrient matter cycles Carbon cycle Key + benefits , -‐ harmed (may not mean death) 0 not affected ENERGY FLOW • is a one-‐way flow • energy can’t be created or destroyed, it can only be changed to a different form • source sun → light energy SUN reflected 97% Energy lost as heat and respiration PRODUCERS 10% ST DETRITIVORES AND 1 ORDER Energy lost DECOMPSERS CONSUMERS as heat and respiration 10% ND 2 ORDER CONSUMERS Nitrogen cycle Page 17 of 19 FACTORS WHICH AFFECT CARRYING CAPACITY / POPULATION SIZE ABIOTIC FACTORS – DENSITY INDEPENDENT e.g. temperature, humidity, rainfall, light intensity, sunlight hours, size of area, presence of trace elements, soil type may have a direct effect on the population – e.g. sunlight hours effects amount of photosynthesis and ∴plant growth OR an indirect effect – e.g. sunlight hours → P/S → plant growth → kangaroo population BIOTIC FACTORS –DENSITY DEPENDENT The impact of other organisms can be identified in two types of competition: • INTRASPECIFIC COMPETITION – competition between members of the same species POPULATION GROWTH measured as a rate – e.g. no. per 100,000 r = ( b -‐ d ) + ( i -‐ e ) ↓ ↓ ↓ ↓ ↓ growth birth death immigration emigration rate rate rate rate rate In Ideal conditions – J curve • • a “doubling” of numbers in a set time period e.g. bacteria, human world population. • not sustainable in the long term -‐ population either “crashes” OR external factors regulate the growth rate In reality – S curve Carrying capacity -‐ maximum no. of individuals which can be supported by the environment Environmental resistance Abiotic and biotic factors acting to limit population growth. Page 18 of 19 INTERSPECIFIC INTERACTIONS – competition between members of different species NATURAL SELECTION AND SPECIATION A species is a population of organisms that can potentially interbreed under natural conditions to provide fertile offspring. An extension of the species concept is a cline. Clines A cline is a gradual change in the structures among members of a species due to ecological of geographical distribution. E.g. in the Snow Gum the length of leaves is shorter at high altitude that at low altitude. Usually the trees next to each other are capable of interbreeding even though they look different and a common gene pool exists in the cline. Hybrids are offspring produced from parents of different species. They are sterile (can’t reproduce). Examples include zebroids (zebra and horse), mules (donkey and horse) Speciation begins when gene flow is prevented between populations of a species. Thereafter, mutation, natural selection, and genetic drift operate independently in each population and lead to irreversible genetic divergence of one from the other. - Individuals remain members of the same species even if they are geographically isolated provided that gene flow still occurs - Gene flow is the physical movement of alleles into and out of a population. These alleles counter the differences in populations that may occur through mutation, natural selection and genetic drift. To produce a new species, evolution must generate large enough genetic changes between populations so that mating cannot occur or the offspring produced are sterile. Isolating mechanisms: 1. Geographical Isolation • a single population is split into two separate populations by a physical barrier such as a river or mountain range. This prevents the organisms from mating and therefore each new population may develop large genetic differences and become separate species 2. Reproductive Isolation • Behavioural Isolation – the mating rituals of many species stops interbreeding e.g. female frogs will only respond to the correct call of the male of their species • Mechanical Incompatibility – the reproductive parts of the organism simply won’t “fit” together • Seasonal Isolation – interbreeding cannot occur if mating happens at different times of the year • Developmental Isolation – fertilisation may occur but the embryo does not survive Page 19 of 19
