Biology 0610/52 February/March 2025: question paper and mark scheme

Question

Benedict’s solution is used to test for reducing sugars. Glucose is a reducing sugar. You will be estimating the concentration of glucose in a solution. Read all the instructions but DO NOT DO THEM until you have drawn a table for your results in the space provided in 1(a)(ii). You should use the safety equipment provided while you are doing the practical work. Step 1 Label six test-tubes 0.0%, 0.5%, 1.0%, 1.5%, 2.0% and U. Put the test-tubes in the test-tube rack. (a) (i) Decide the volumes of 2% glucose solution and distilled water needed to make 4 cm3 of a 1.5% glucose solution. Complete Table 1.1 by writing in the volumes of 2% glucose solution and distilled water you will use to make the 1.5% glucose solution. Table 1.1 percentage concentration volume of 2% glucose volume of distilled of glucose solution / cm3 water / cm3 0.0 0 4 0.5 1 3 1.0 2 2 1.5 2.0 4 0 [1] Step 2 Use a syringe to put the volumes of 2% glucose solution shown in Table 1.1 into the test-tubes labelled 0.5%, 1.0%, 1.5% and 2.0%. Step 3 Use the same syringe to put 4 cm3 of the unknown glucose solution U into the test-tube labelled U. Step 4 Use a clean syringe to put the volumes of distilled water shown in Table 1.1 into the test-tubes labelled 0.0%, 0.5%, 1.0%, and 1.5%. Step 5 Use a clean syringe to put 4 cm3 of Benedict’s solution into each of the test-tubes labelled 0.0%, 0.5%, 1.0%, 1.5%, 2.0% and U. Step 6 Raise your hand when you are ready for hot water to be added to the beaker labelled water-bath. Step 7 Put the test-tubes into the water-bath and start the stop-clock. Step 8 Wait for 5 minutes. Remove the test-tubes from the water-bath and put them back into the test-tube rack. Step 9 Record the colour of the liquid in the test-tubes labelled 0.0%, 0.5%, 1.0%, 1.5%, 2.0% and U in your table in 1(a)(ii). (ii) Prepare a table and record your results. [4] (iii) Test-tube U contains an unknown concentration of glucose solution. Use your results to estimate the percentage concentration of glucose in test-tube U. concentration of glucose in test-tube U …… % [1] (iv) State two variables that were kept constant in this investigation. 1 …… 2 …… [2] (v) Identify the possible source of error present in steps 2 and 3. State the effect of this error on the results of the investigation. error …… effect on the results …… [2] (b) Eggs contain a protein called albumen. Albumen will turn from cloudy to clear when it is digested by pepsin, a protease enzyme. Plan an investigation to determine the effect of pH on the digestion of albumen by protease. …… [6] (c) The emulsion test can be used to test for fat. State the result of a positive test. …… [1] [Total: 17]

Question

(a) Fig. 2.1 is a photograph of a leaf from an oak tree, Quercus sp. P Q magnification ×1.4 Fig. 2.1 (i) Draw a large diagram of the whole oak leaf shown in Fig. 2.1. [4] (ii) Line PQ on Fig. 2.1 represents the width of the oak leaf. Measure the length of line PQ on Fig. 2.1. length of PQ …… mm Calculate the actual width of the oak leaf using the formula and your measurement. length of line PQ in Fig. 2.1 magnification = actual width of the oak leaf Give your answer to two significant figures. Space for working. …… mm [3] (iii) Fig. 2.2 shows photographs of the oak leaf and a hollyhock (Alcea rosea) leaf. The magnification of both photographs is not the same. oak hollyhock Fig. 2.2 Table 2.1 gives one visible difference between the oak leaf and the hollyhock leaf shown in Fig. 2.2. Complete Table 2.1 to give two other visible differences between the leaves shown in Fig. 2.2. Do not include references to size in your answer. Table 2.1 difference oak leaf hollyhock leaf 1 The oak leaf is longer than it is The length and width of the hollyhock wide. leaf are similar. 2 3 [2] (b) Students investigated the effect of light intensity on the surface area of leaves of soybean plants. • 100 soybean seeds were planted in pots and put into the shade (low light). • 100 soybean seeds were planted in pots and put into full sun. • The soybean seeds were allowed to germinate and grow for 30 days. • After 30 days, three of the oldest leaves and three of the youngest leaves were removed from each plant. • The surface area of each of the removed leaves was measured. (i) State the dependent variable in this investigation. …… [1] (ii) State why the students used a large number of soybean plants. …… [1] (iii) The students estimated the surface area of each leaf using graph paper as shown in Fig. 2.3. Fig. 2.3 Suggest how the students used the graph paper to measure the surface area of each leaf. …… [2] (c) The results of the investigation are shown in Table 2.2. Table 2.2 mean surface area of lighting condition age of leaves leaves / cm2 full sun young 52 shade young 45 full sun old 78 shade old 58 (i) On the grid, using the data in Table 2.2, plot a bar chart to show the results of the investigation. [4] (ii) The students calculated that there was a 15.6% increase in the size of the young leaves when they were grown in full sun compared to in the shade. Calculate the percentage increase in the surface area of the old leaves that had been grown in full sun conditions compared to the old leaves that had been grown in the shade. Give your answer to one decimal place. …… % [3] (iii) State two conclusions from this investigation. 1 …… 2 …… [2] (d) Hydrogencarbonate indicator is used to test for the presence of carbon dioxide. An aquatic plant was placed in red hydrogencarbonate indicator and put under bright light. The plant takes in carbon dioxide as it photosynthesises. State the final colour of the hydrogencarbonate indicator. …… [1] [Total: 23]