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1 June 2017 Paper 1B (Higher Tier) Model Answers Level Edexcel Subject Biology Exam Board IGCSE Year June 2017 Paper Paper 1B (Double Award) Tier Higher Tier Booklet Model Answers Time Allowed: 60 minutes Score: /60 Percentage: /100 Model Mo del Answer Answer Key Red Re d = Answer Answer - This is what you need need to write to get get the marks Blue Blu e = Explanation Explanation - This is here here to to help help you unde unde nders rsttand the rs the answer - You DON T need to to writ writ itee this to get get the marks

2 Answer ALL questions. 1 The diagram shows part of the human breathing system. A B C (a) (i) Name the structures labelled A, B, and C. (3) A trachea B bronchus C bronchioles (ii) Suggest why the lung structure is sometimes referred to as a tree. the bronchus branches into lots of smaller bronchiole (1)

3 (b) Describe how ventilation of the lungs occurs when a person breathes in. (4) diaphragm muscle contracts causing the diaphragm to move downwards / flatten intercostal muscles contract causing the ribcage to move up and out this increases the volume inside the chest cavity / thorax decreasing the pressure inside the chest cavity (to below atmospheric pressure) The increase in volume and decrease in pressure causes air to be forced in from outside. (c) Smoking cigarettes inside public buildings has been banned in many countries. (i) Suggest why governments have banned smoking cigarettes inside public buildings. (3) non smokers will inhale second hand smoke (passive smoking) some chemicals in cigarettes smoke cause cancer inhaling smoke increases the risk of bronchitis, emphysema and can lead to COPD carbon monoxide in smoke reduces the oxygen carrying capacity of the blood as it binds irreversibly to haemoglobin smoking increases the risk of developing heart disease banning smoking outside public buildings discourages smoking and may encourage people who smoke to give up (ii) Suggest why children are particularly at risk from breathing in smoke from other people s cigarettes. (1) children s lungs are smaller so inhaling small amounts of smoke damages them more children are still growing and inhaling smoke can slow their development (Total for Question 1 = 12 marks)

4 2 A student investigates the movement of substances from the gut into the blood using a model. The Visking tubing bag represents the gut and water represents the blood. Visking tubing is permeable to small molecules. plastic cover substances Visking tubing bag measuring cylinder water The student puts different solutions of various substances into the bag. After 20 minutes he carries out food tests on the water..

5 (a) (i) Complete the table to show the results obtained by the student. One has been done for you. Solutions in bag (3) Starch present in water Glucose present in water Starch only No No Starch and glucose No Yes Starch and maltase No No Starch and boiled amylase No No Starch is a large molecule and cannot diffuse through the semi permeable membrane of the visking tubing into the water, therefore it will remain inside the bag. Glucose is a small molecule which will diffuse through the visking tubing into the water surrounding the bag. Maltase will not have any effect on the starch as enzymes are specific, meaning they only digest the molecules that fit into their active site. Maltase digests the sugar maltose, not starch. Amylase is the enzyme that digests starch however it will not do so in the fourth tube because it has been boiled. Enzymes are proteins and the shape of their active site is held in place by bonds. These bonds are destroyed by heating and the shape of the active site is lost, meaning substrates can no longer fit into them. We describe enzymes that have lost the shape of their active site as being denatured.

6 (ii) Explain how the volume of the bag would change when it only contains starch solution. (3) the volume of the bag would increase as water will enter from outside due to osmosis as there is a higher concentration of water outside the bag than inside (iii) Describe the test the student should use to find out if glucose is present in the water. (3) add a sample of the water to some Benedict s solution in a test tube boil the contents of the test tube if the Benedict s remains blue no glucose is present, if it turns brick red glucose is present

7 (b) Explain how you could use this apparatus to investigate the effect of bile on the digestion of lipid. (3) set up 3 tubes: tube 1 containing lipid + bile + lipase in the bag tube 2 containing lipid + lipase in the bag tube 3 containing lipid + bile in the bag leave for 20 minutes then test the water outside the bags in each tube for the presence of fatty acids by testing the ph the lower it is, the more fatty acids have been produced The apparatus referred to is the one at the start of the question, not the apparatus in a) iii). As the whole point of digestion is to produce molecules that are small enough to move through membranes, the idea is the same as with starch and glucose adding the lipase will digest the fats into fatty acids and glycerol which will diffuse through the partially permeable visking tubing into the water surrounding it. The expected results would be that the tube containing lipid (fat) + bile + lipase would show the greatest decrease in ph as the bile emulsifies the lipids into smaller droplets, providing a larger surface area for the lipase enzymes to work on and therefore speeding up the rate of reaction. The lipid + lipase would also show a reduction in ph but not as much. The lipid + bile would not show a reduction in ph as the bile does not have the ability to chemically digest the lipids into smaller molecules. (Total for Question 2 = 12 marks)

8 3 Galactosaemia is an inherited condition. It is caused by a mutation to a gene on chromosome 9. This gene codes for an enzyme that removes the sugar galactose from the body. The normal allele, G, can produce the enzyme and is dominant to the recessive allele, g. The pedigree shows the inheritance of galactosaemia in a family. Key 1 2 unaffected male unaffected female male with galactosaemia female with galactosaemia (a) (i) How many individuals in the family pedigree have an X and a Y chromosome in each of their body cells? (1) 8 All males, whether affected or unaffected, will contain an X and a Y chromosome. (ii) How many individuals in the family pedigree are homozygous recessive? (1) 4 The recessive allele, g, causes the condition so in order to have galactosaemia, an individual must be homozygous recessive (gg). There are four individuals with galactosaemia, therefore there are four homozygous recessive individuals.

9 (iii) Give the genotype of individual 1. (1) Gg He does not have galactosaemia so he cannot be homozygous recessive (gg). However, he has a daughter who has galactosaemia (so her genotype must be gg) and therefore he must have passed on a recessive allele (g) to her. This means that he must be heterozygous (Gg), as if he was homozygous dominant (GG) he would not have any offspring with galactosaemia as they would all have inherited a dominant allele, G, from him. (iv) Individuals 8 and 9 are going to have another child. Give the probability of this child having galactosaemia. (1) 1 / 4 or 25% G g G GG Gg g Gg gg Individuals 8 & 9 have produced offspring with galactosaemia even though neither of them have the condition themselves. This tells you that they must both be heterozygous. Each time they have a child, they have a 25% chance of producing a homozygous recessive child with galactosaemia as shown by the Punnett square above.

10 (b) (i) What is meant by the term mutation? (2) a random change in DNA (ii) The mutation for galactosaemia is harmful but some mutations can be beneficial. Describe one example of a beneficial mutation. (1) antibiotic resistance in bacteria fur colour enabling better camouflage Any beneficial characteristic described here would be awarded the mark. (c) Some inherited conditions can be fatal but medical treatment is often available. Explain what would happen to the frequency of alleles for these inherited conditions if medical treatment was not available. (3) the frequency of the alleles would decrease if no treatment was available as affected individuals would die sooner and therefore not have as many children so the allele would not be passed on Note that the term gene would not be accepted here. The gene codes for a protein that produces and enzyme that removes galactose from the body. There are different alleles for the gene one that codes correctly for the production of the enzyme and one that does not. It is the allele that will not be passed on, not the gene. (Total for Question 3 = 10 marks)

11 4 (a) A student reads the following statement. the father determines the sex of a baby Explain why this statement is true. (2) males produce sperm with either an X chromosome or a Y chromosome females produce eggs which all contain an X chromosome to produce a female (XX), the egg has to join with a sperm carrying an X chromosome to produce a male (XY), the egg has to join with a perm carrying a Y chromosome

12 (b) The diagram shows sperm cells, viewed using a microscope, from two different men. Man A Man B Use the information in the diagram to give two reasons why man A is more fertile than man B. (2) man A has more sperm than man B some of man B s sperm have two heads or two tails and wouldn t function properly

13 (c) The graph shows how smoking may affect the movement of sperm percentage of sperm seen moving (%) smokers non-smokers (i) The average (mean) sperm count in the semen from non-smokers is 55 million in 1 cm3. Calculate the number of moving sperm in 1 cm3 of semen from the non-smokers. (2) / 100 = x 45 = Exam tip:showing YOUR WORKING In the example above, there are two marks available. Only one is for the correct answer. The other mark is for showing your working as asked to in the question. This means writing down the calculation you did to get the answer. Make sure you follow the instruction and show your working to gain maximum marks as every mark counts! number of moving sperm = (ii) Use the information from the graph to explain why smoking could affect male fertility. (2) smokers have fewer moving sperm meaning there is less chance of a sperm reaching the egg and so less chance of fertilization occurring (Total for Question 4 = 8 marks)

14 5 In a survey, people of different ages were asked if they thought that animal cloning is a good idea or a bad idea. The table shows the results of the survey. Age group Percentage (%) of age group Good idea Bad idea under to to to over (5) (a) Plot a bar graph to show the data in the table percentage good idea 40 bad idea under age in years Scale (even) and size (over half of the graph paper used) Having an even scale means that you haven t just copied down all of the numbers from the table and put them onto the side of your graph. Look at the biggest number you have to plot for each axis and the figure out how to evenly space out the scale back to zero so that you have used as much of the graph paper as possible. If the numbers on your axes only reach half way up or across the graph paper you have been given, your graph is TOO SMALL and you will lose a mark.

15 Axes labelled Without labels and units your graph doesn t mean much. Use the column headings from the table to form your axes labels. Bars plotted accurately All bars have to be plotted accurately to get the mark. Line Neat lines drawn for bars. If you draw a line graph instead of a bar chart, you lose this mark. Key Key to show which columns represent good idea and which represent bad idea. Exam tip:graphs The independent variable (the factor being changed) always goes along the bottom and is always the first column in the table of data. The dependent variable (the factor being recorded to get the results) always goes up the side and is always the second column in the table of data. Use the column headings in the table to come up with your axis labels. Don t forget the UNITS!!! Always make your scale big enough to fill as much of the graph paper as possible. Plot your graph in pencil so you can easily erase any mistakes!

16 (b) Describe the relationship between age and what people think about animal cloning. (1) the older the individuals, the more think cloning animals is a good idea (c) In the 45 to 54 age group, 18 people think that cloning is a good idea. Calculate the total number of people surveyed in this age group. Show your working. 18 / 15 = (2) 1.2 x 100 = 1 0 Exam tip:showing YOUR WORKING In the example above, there are two marks available. Only one is for the correct answer. The other mark is for showing your working as asked to in the question. This means writing down the calculation you did to get the answer. Make sure you follow the instruction and show your working to gain maximum marks as every mark counts! total number of people.=

17 (d) Some people in the survey did not know anything about cloning. The process had to be described to them before they made a decision. Describe the process of cloning an adult animal using a named example. (6) Dolly the sheep was the first mammal successfully cloned the nucleus from the body cell of an adult sheep was placed into an enucleated egg cell in the lab a small jolt of electricity was applied to fuse the nucleus and the egg cell together the egg cell started dividing by mitosis when several cycles of cell division had formed an embryo it was placed into the uterus of a surrogate mother (where it continued to grow and divide until birth) (Total for Question 5 = 14 marks)

18 6 The amount of light falling on plant leaves is an abiotic (non-living) factor that affects photosynthesis. (a) Give the balanced chemical equation for photosynthesis. (2) 6 CO H 2 O C 6 H 12 O O 2 One mark is available for the correct formulae in the right places, the other mark is for the correct balancing of the equation. (b) The leaves of plants that live in the shade (low light) are different to the leaves of plants that live in full sunlight. (i) Suggest why leaves from plants that live in the shade are darker green than leaves from plants that live in full sunlight. (2) they contain more chloroplasts to try and absorb as much light as possible (ii) Explain why leaves from plants that live in the shade are thinner than leaves from plants that live in full sunlight. (2) if the leaf is thinner there is a shorter distance for the light to penetrate the leaf to reach the chloroplasts

19 (c) The rate of photosynthesis changes during the day. (i) Explain the factors that affect the rate of photosynthesis in the early morning. (3) in the early morning there is less light so there is less photosynthesis and stomata are not fully open so less gas exchange the temperature is also lower so enzymes controlling photosynthesis work more slowly (ii) Explain the factors that affect the rate of photosynthesis in the early afternoon. (3) there is more light in the afternoon so the rate of photosynthesis is faster and stomata are fully open so more gas exchange the temperature is higher so enzymes work faster the limiting factor on the rate of photosynthesis is likely to be carbon dioxide concentration (as there is not a huge amount in the air)

20 (d) Describe how you could compare the rate of photosynthesis in two different plant species. (4) by measuring the amount of oxygen produced by counting the number of bubble of gas produced or measuring the volume of gas produced using a syringe over a 5 minute period for each plant species ensuring both plants have the same amount of light and carbon dioxide and are at the same temperature and that the plants tested have the same number of leaves / are the same mass Other experiments described could also have been awarded marks here, for example using hydrogen carbonate indicator to test the levels of carbon dioxide in the water, or testing a leaf for starch. (Total for Question 6 = 16 marks)

21 7 A student uses this apparatus to investigate the effect of nitrate ions on the growth of plants. young plant air glass jar cotton wool wooden lid to cover the jar cardboard surrounding the jar solution A young plant is grown in a sterile solution containing all the mineral ions needed for growth. The student repeats the experiment with other young plants. The student also carries out the experiment with young plants grown in a sterile solution that contains all the mineral ions except nitrate. The student measures the length of the stem of each plant every five days. Some of the student s results are shown in the table. Time in days Average (mean) length of stem in mm Solution containing all mineral ions Solution without nitrate ions

22 (a) Describe the growth of the plants in each solution. (2) there is more growth of the plant in the solution containing all the mineral ions growth levels off at 45 days for the plant with all minerals and at 25 days for the plant grown without nitrates Exam tip:comparing DATA Unless the question asks you to compare AND explain or describe AND explain you don t need to give any reasons why the data looks like it does. Look at the number of marks for the question. You need to give an equivalent number of points in your answer to get all of the marks. So: in this question there are two marks therefore: - You must first say something aboutthe general difference between the two sets of data The second mark is awarded for quoting some actual figures from the data but again make sure you are comparing the data as you do so, not just writing down any numbers from the table! (b) Explain why young plants absorb more mineral ions when air is bubbled through the solutions. mineral ions are taken up by active transport this process requires energy from respiration bubbling oxygen through the water increases respiration rate as more oxygen is available to the root cells (3)

23 (c) (i) Suggest why each solution is sterilised at the start of this investigation. (2) to kill any microorganisms that might infect the plant or use the mineral ions in the water for themselves The dependent variable is the factor which is measured to get the results in an experiment. (ii) Suggest why the glass jar is surrounded by cardboard during this investigation. (2) to block light from the water preventing the growth of algae in it

24 (d) (i) Identify the dependent variable in this investigation. (1) length of stem The dependent variable is the factor which is measured to get the results in an experiment. (ii) Name one biotic (living) variable that should be controlled in this investigation. (1) age of plants species of plants mass of plants surface area of leaves (Total for Question 7 = 11 marks)

25 8 The diagram shows the distribution of two plant species in a small area of a field. 1 m Key bunchgrass plantain A student uses a square metal frame to help count all the plants in the area. (a) What is the name given to the metal apparatus that the student uses? (1) quadrat

26 (b) (i) He counts the number of plants in all of the squares he marked out in the field. Complete the table to show the number of plants of each species and the average (mean) number of plants per m2. Species Number of plants Average number of plants per m2 Bunchgrass Plantain (2) The total area of the quadrat is 36m2. 8 plants were found in the total area, so the average number per m2 is 8/36 = (ii) Frequency is another measure that can be used to study distribution. Frequency is the number of squares that contain at least one plant of the species being counted. This value can also be expressed as a percentage of the total number of squares sampled. Complete the table by giving the frequency and percentage of the total number of squares sampled for bunchgrass. Species Frequency Percentage (%) Bunchgrass 4 11 Plantain 8 22 (2) Although there are 7 bunchgrass plants, they are only found in 4 squares so the frequency is 4. To work out the percentage of the total number of squares sampled: 4 / 36 = 0.11 x 100 = 11

27 (c) Describe how the student could estimate the population size of plantain in a very large field. (4) place quadrats in several places at random using a number generator to produce coordinates for where to place the quadrats count the number of plantain plants in each quadrat calculate the mean number of plants per quadrat and multiply this by the total area of the field (eg if you had covered an area of 5m 2 with your quadrat throws but the total area was 50m 2 then you would multiple each of your mean values by 10 to get an estimate of the total number of plantain plants within the area) (Total for Question 8 = 9 marks)

28 9 The diagram shows a cell from the human nervous system. A B (a) (i) Name the structure labelled A. nucleus (1) (ii) Name the structure labelled B. (1) axon As the label line is not pointing at the entire axon but a part inside it, where the cytoplasm would be found, cytoplasm would also be marked correct.

29 (iii) Draw an arrow on the diagram to show the direction of a nerve impulse. (1) Nerve impulses always travel from the cell body towards the dendrites. (b) Describe the role of this neurone in a simple reflex arc. (2) to transmit nerve impulses from the central nervous system to the effector The drawing shown is of a motor neurone, identifiable as the cell body is at the end of the neurone (as opposed to a sensory neurone where it is found off the middle of the axon). Motor neurones transmit impulses from the CNS to the muscles or nerves which act as effectors.

30 (c) Nervous communication differs from hormonal communication. State three ways that nervous communication differs from hormonal communication. (3) nervous control is faster than hormonal control nervous control is electrical whereas hormonal is chemical nervous impulses travel in nerve cells unlike hormones which travel in blood the effects of nervous impulses tend to be short lasting however hormones tend to have longer lasting effects nervous control targets specific cells, on the other hand hormonal control can target several areas of the body nervous control has an all or nothing effect unlike hormonal control which can have an incremental effect Exam tip: ANSWERING A COMPARISON QUESTION When you have to answer a question by giving differences between two things, there is usually 1 mark awarded for each comparison. Often, students don t make a direct comparison and therefore they miss out on the marks. For example, this answer would not be awarded a mark: In nervous control the response is fast. In hormonal control the signals are chemicals. These are two true statements so why no marks? The reason is because this is not a direct comparison comparing like with like. To help you to do this, it can be useful to use a connective to link the two statements together (in the answers above the connectives are in italics). This helps because you can see that the sentence doesn t make sense unless the two parts are about the same thing. Take the example above again, adding the connective however : In nervous control the response is fast however in hormonal control the signals are chemicals. It doesn t make sense! So use connectives to help you structure your comparison questions better and get maximum marks! (Total for Question 9 = 8 marks)

31 10 The passage describes how different organisms are classified into groups. Complete the passage by writing a suitable word or words in each of the spaces. (10) Plants are multicellular organisms. They have chloroplasts to carry out cellulose photosynthesis and cell walls made of.... They store starch carbohydrate as or as sucrose. Animals are also multicellular but do not carry out photosynthesis. They are able consumers to move from place to place and are always described as in glycogen food chains. They store carbohydrate as.... Bacteria are single-celled organisms. They do not have a nucleus. chromosome Instead, they contain a circular and smaller circles plasmids of DNA called Most bacteria feed off other living or dead photosynthesis organisms but some bacteria can make their own food by yoghurt Examples of bacteria include Lactobacillus, used in the production of pathogen from milk, and Pneumococcus, that acts as a... pneumonia causing the disease (Total for Question 10 = 10 marks)

32 11 The photograph shows the logs left behind after an area of forest has been cut down. Calibas These logs are decomposed by fungi. (a) Describe how fungi decompose tree logs. fungi are saprophytic feeders meaning they produce extracellular enzymes which will digest the wood outside the body of the fungi into carbon dioxide and water (4)

33 (b) Some of the logs removed from the forest are used to make garden benches. The photograph shows a garden bench. James F. Perry The garden bench is painted with a fungicide solution. This prevents the wood being decomposed because fungicide kills fungi. There are different fungicides that can be used. Design an investigation to find out which fungicide is best at preventing the decomposition of wooden logs. (6) Your answer should include experimental details and be written in full sentences. take two logs from the same tree paint each with a different fungicidal solution and leave for one month under the same conditions amount of light, oxygen, temperature, humidity they should also be exposed to the same type and mass of fungi weigh the logs at the start and end to determine the change in mass throughout the experiment repeat with several logs for each fungicidal solution Exam tip: DESIGNING AN INVESTIGATION Mark schemes for designing investigations are often very general because the examiners want to see the same type of things mentioned regardless of what the actual experiment is. These things are: 1. That you have set up an experiment that is testing what you are being asked to 2. That you have identified how to get your results 3. That you identify some variables that should be kept constant to make the experiment fair (aim for at least 2 or 3) 4. That you make your results reliable by repeating the experiment or carry it out with lots of individuals 5. That you identify how long to do the experiment for