AS BIOLOGY Paper 1 2 *02* IB/M/Jun22/7401/1

AS BIOLOGY Paper 1 2 *02* IB/M/Jun22/7401/1 Do not write outside the Answer all questions in the spaces provided. box 0 1 . 1 Figure 1 shows the structure of some biological molecules. Figure 1 3 *03* Turn ...

AS BIOLOGY Paper 1 2 *02* IB/M/Jun22/7401/1 Do not write outside the Answer all questions in the spaces provided. box 0 1 . 1 Figure 1 shows the structure of some biological molecules. Figure 1 3 *03* Turn over ► IB/M/Jun22/7401/1 Do not write outside the Comp box lete Table 1 by writing the correct letter, A, B, C, D or E, in the box next to each statement. Each letter may be used once, more than once, or not at all. [4 marks] Table 1 Letter Statement is hydrolysed in the ileum and a product of this hydrolysis is found in micelles is formed by a condensation reaction between two α-glucose molecules is formed by the action of DNA polymerase gives a positive result in an emulsion test 0 1 . 2 Describe the mechanism for the absorption of amino acids in the ileum. [4 marks] Turn over for the next question 8 4 *04* IB/M/Jun22/7401/1 Do not write outside the 0 2 box . 1 Give the two types of molecule from which a ribosome is made. [2 marks] 1 2 0 2 . 2 Complete Table 2 to give four structural differences between a DNA molecule and an mRNA molecule. [4 marks] Table 2 DNA structure mRNA structure 1 2 3 4 6 5 *05* Turn over ► IB/M/Jun22/7401/1 Do not write outside the 0 3 box . 1 Figure 2 is an image of a bacterium obtained using a scanning electron microscope. Figure 2 Name the structure labelled X. [1 mark] 0 3 . 2 Figure 2 is different from an image of this bacterium obtained using a transmission electron microscope. Describe and explain one difference between these images. [2 marks] Description Explanation 0 3 . 3 The resolution of an image obtained using an electron microscope is higher than the resolution of an image obtained using an optical microscope. Explain why. [1 mark] Question 3 continues on the next page 6 *06* IB/M/Jun22/7401/1 Do not write outside the 0 3 box . 4 A student determined the size of a cell structure from a photograph obtained using a microscope. He used a ruler and a calculator and gave the answer in μm Describe how the student determined the size of the structure. [2 marks] 0 3 . 5 Name two structures found in all bacteria that are not found in plant cells. [2 marks] 1 2 0 3 . 6 Name two features of HIV particles that are not found in bacteria. Do not include attachment protein in your answer. [2 marks] 1 2 10 7 *07* Turn over ► IB/M/Jun22/7401/1 Do not write outside the 0 4 box . 1 Lugworms create tubes in the sand on seashores. The tubes are filled with seawater. A scientist measured the partial pressure of dissolved oxygen (pO2) in seawater at different places in a tube with a lugworm inside. Figure 3 shows her results. Figure 3 The pO2 of dissolved oxygen in lugworm blood is < 2.7 kPa Using the data in Figure 3, what can you conclude about the uptake of oxygen over the entire body of the lugworm? [4 marks] Question 4 continues on the next page 8 *08* IB/M/Jun22/7401/1 Do not write outside the 0 4 box . 2 Figure 4 shows the oxyhaemoglobin dissociation curve for a lugworm. Figure 4 9 *09* Turn over ► IB/M/Jun22/7401/1 Do not write outside the box The oxygen saturation in the blood of a lugworm is 92% The lugworm has 0.2 cm3 of blood. Calculate the volume of dissolved oxygen in the blood of this lugworm using this equation pO2 = CdO2 0.000 031 CdO2 is the concentration of dissolved oxygen in the blood, with units cm3 oxygen per cm3 of blood. Show your working. [3 marks] Answer cm3 0 4 . 3 The intensity of the red colour in blood is affected by the pO2 of the blood. The intensity of the colour in a solution is measured using a colorimeter. The scientist used a colorimeter to measure the intensity of red colour in samples of lugworm blood with different pO2 values. She prepared a calibration curve with this information. Describe how the scientist will use information from the colorimeter and her calibration curve to determine the pO2 in a sample of lugworm blood. [2 marks] 9 10 *10* IB/M/Jun22/7401/1 Do not write outside the 0 5 box . 1 Describe how monomers join to form the primary structure of a protein. [3 marks] 0 5 . 2 Many proteins are enzymes. In 1894, a scientist suggested the lock and key model of enzyme action. Figure 5 shows the lock and key model. Figure 5 Describe one similarity and one difference between the induced-fit model of enzyme action and the lock and key model of enzyme action. [2 marks] Similarity Difference 11 *11* Turn over ► IB/M/Jun22/7401/1 Do not write outside the 0 5 box . 3 State how enzymes help reactions to proceed quickly at lower temperatures. Do not write about active sites in your answer. [1 mark] 0 5 . 4 The enzyme maltase catalyses the hydrolysis of maltose to glucose. A scientist investigated maltase activity in two different maltose solutions, G and H. For each solution, he measured: • the total number of glucose molecules produced by complete hydrolysis of the maltose • the time taken for the complete hydrolysis of the maltose. Table 3 shows his results. Table 3 Solution Total number of glucose molecules produced Time taken for complete hydrolysis of maltose / s G 4 × 107 20 H 6 × 108 Complete Table 3 by calculating the time taken for the complete hydrolysis of the maltose in solution H. Assume the rate of maltase activity is the same in solution G and in solution H. Show your working. [2 marks] Question 5 continues on the next page 12 *12* IB/M/Jun22/7401/1 Do not write outside the 0 5 box . 5 Figure 6 shows the scientist’s results for solution G. Curve I shows the results of a similar investigation in which he changed one independent variable. Figure 6 Tick () one box next to the statement that describes the independent variable that the scientist changed to give the results shown by curve I in Figure 6. [1 mark] Addition of a competitive inhibitor Increased maltase concentration Increased maltose concentration Reduced temperature 9 13 *13* Turn over ► IB/M/Jun22/7401/1 Do not write outside the 0 6 box . 1 Explain a property of iron ions that enables these ions to carry out their role in red blood cells. [2 marks] 0 6 . 2 The hormone hepcidin controls the iron ion concentration in blood plasma. Hepcidin affects ferroportin, the iron ion channel protein in cell-surface membranes. Figure 7 shows how hepcidin controls the iron ion concentration in plasma. Figure 7 People with the disease haemochromatosis do not produce hepcidin. Use information in Figure 7 to explain why the iron ion concentration is higher in the plasma of people with haemochromatosis. [3 marks] 14 *14* IB/M/Jun22/7401/1 Do not write outside the 0 6 box . 3 The mass of iron ions in the plasma of a person with haemochromatosis is 6104 μg The iron ion concentration in the plasma of a healthy person is 50 μg dm–3 The volume of blood in each of these people is 4000 cm3 Calculate the ratio of the mass of iron ions in the