4.2· 259 questions · 259 marks · 311 min · 2004–2025· Multiple choice
Every Cambridge A Level Biology Paper 1 question on movement into and out of cells, laid out as 94 A4 pages with the mark scheme below. Nothing is left out. Free to read, no account.




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94 / 94Answers below. Sit the paper first if you are practising.
Pastlit
Biology 9700 · Movement into and out of cells — Paper 1
A Level · topical answer key — answer key (teacher use)
Question
Answer
Marks
Pastlit
Biology 9700 · Movement into and out of cells — Paper 1
A Level · topical answer key — answer key (teacher use)
Question
Answer
Marks
Pastlit
Biology 9700 · Movement into and out of cells — Paper 1
A Level · topical answer key — answer key (teacher use)
Question
Answer
Marks
Pastlit
Biology 9700 · Movement into and out of cells — Paper 1
A Level · topical answer key — answer key (teacher use)
Question
Answer
Marks
Pastlit
Biology 9700 · Movement into and out of cells — Paper 1
A Level · topical answer key — answer key (teacher use)
Question
Answer
Marks
Pastlit
Biology 9700 · Movement into and out of cells — Paper 1
A Level · topical answer key — answer key (teacher use)
Question
Answer
Marks
| Question | Answer | Marks | From |
|---|---|---|---|
| 1 | C | 1 | 9700/11 Oct/Nov 2004 |
| 2 | C | 1 | 9700/11 Oct/Nov 2004 |
| 3 | C | 1 | 9700/11 May/June 2006 |
| 4 | B | 1 | 9700/11 May/June 2006 |
| 5 | C | 1 | 9700/11 May/June 2006 |
| 6 | A | 1 | 9700/11 May/June 2006 |
| 7 | D | 1 | 9700/11 Oct/Nov 2006 |
| 8 | C | 1 | 9700/11 Oct/Nov 2006 |
| 9 | B | 1 | 9700/11 Oct/Nov 2006 |
| 10 | B | 1 | 9700/11 Oct/Nov 2007 |
| 11 | A | 1 | 9700/11 Oct/Nov 2007 |
| 12 | D | 1 | 9700/11 May/June 2008 |
| 13 | C | 1 | 9700/11 May/June 2008 |
| 14 | B | 1 | 9700/11 May/June 2008 |
| 15 | B | 1 | 9700/11 Oct/Nov 2008 |
| 16 | C | 1 | 9700/11 Oct/Nov 2008 |
| 17 | D | 1 | 9700/11 Oct/Nov 2008 |
| 18 | C | 1 | 9700/11 Oct/Nov 2008 |
| 19 | B | 1 | 9700/12 Oct/Nov 2009 |
| 20 | A | 1 | 9700/12 Oct/Nov 2009 |
| 21 | C | 1 | 9700/12 Oct/Nov 2009 |
| 22 | D | 1 | 9700/12 Oct/Nov 2009 |
| 23 | D | 1 | 9700/11 May/June 2010 |
| 24 | D | 1 | 9700/11 May/June 2010 |
| 25 | D | 1 | 9700/12 May/June 2010 |
| 26 | D | 1 | 9700/12 May/June 2010 |
| 27 | D | 1 | 9700/13 May/June 2010 |
| 28 | D | 1 | 9700/13 May/June 2010 |
| 29 | A | 1 | 9700/11 Oct/Nov 2010 |
| 30 | B | 1 | 9700/11 Oct/Nov 2010 |
| 31 | A | 1 | 9700/12 Oct/Nov 2010 |
| 32 | C | 1 | 9700/12 Oct/Nov 2010 |
| 33 | B | 1 | 9700/11 May/June 2011 |
| 34 | A | 1 | 9700/11 May/June 2011 |
| 35 | D | 1 | 9700/12 May/June 2011 |
| 36 | A | 1 | 9700/12 May/June 2011 |
| 37 | B | 1 | 9700/13 May/June 2011 |
| 38 | A | 1 | 9700/13 May/June 2011 |
| 39 | D | 1 | 9700/11 Oct/Nov 2011 |
| 40 | B | 1 | 9700/11 Oct/Nov 2011 |
| 41 | C | 1 | 9700/11 Oct/Nov 2011 |
| 42 | B | 1 | 9700/13 Oct/Nov 2011 |
| 43 | D | 1 | 9700/13 Oct/Nov 2011 |
| 44 | C | 1 | 9700/13 Oct/Nov 2011 |
| 45 | D | 1 | 9700/11 May/June 2012 |
| 46 | C | 1 | 9700/11 May/June 2012 |
| 47 | C | 1 | 9700/12 May/June 2012 |
| 48 | D | 1 | 9700/12 May/June 2012 |
| 49 | D | 1 | 9700/13 May/June 2012 |
| 50 | C | 1 | 9700/13 May/June 2012 |
| 51 | A | 1 | 9700/11 Oct/Nov 2012 |
| 52 | C | 1 | 9700/11 Oct/Nov 2012 |
| 53 | B | 1 | 9700/11 Oct/Nov 2012 |
| 54 | D | 1 | 9700/12 Oct/Nov 2012 |
| 55 | D | 1 | 9700/12 Oct/Nov 2012 |
| 56 | A | 1 | 9700/12 Oct/Nov 2012 |
| 57 | A | 1 | 9700/13 Oct/Nov 2012 |
| 58 | C | 1 | 9700/13 Oct/Nov 2012 |
| 59 | B | 1 | 9700/11 May/June 2013 |
| 60 | B | 1 | 9700/11 May/June 2013 |
| 61 | B | 1 | 9700/11 May/June 2013 |
| 62 | A | 1 | 9700/11 May/June 2013 |
| 63 | A | 1 | 9700/12 May/June 2013 |
| 64 | D | 1 | 9700/12 May/June 2013 |
| 65 | C | 1 | 9700/12 May/June 2013 |
| 66 | B | 1 | 9700/12 May/June 2013 |
| 67 | B | 1 | 9700/13 May/June 2013 |
| 68 | A | 1 | 9700/13 May/June 2013 |
| 69 | B | 1 | 9700/13 May/June 2013 |
| 70 | C | 1 | 9700/11 Oct/Nov 2013 |
| 71 | A | 1 | 9700/11 Oct/Nov 2013 |
| 72 | B | 1 | 9700/11 Oct/Nov 2013 |
| 73 | C | 1 | 9700/12 Oct/Nov 2013 |
| 74 | D | 1 | 9700/12 Oct/Nov 2013 |
| 75 | C | 1 | 9700/13 Oct/Nov 2013 |
| 76 | A | 1 | 9700/13 Oct/Nov 2013 |
| 77 | B | 1 | 9700/13 Oct/Nov 2013 |
| 78 | C | 1 | 9700/13 Oct/Nov 2013 |
| 79 | C | 1 | 9700/11 May/June 2014 |
| 80 | A | 1 | 9700/11 May/June 2014 |
| 81 | D | 1 | 9700/11 May/June 2014 |
| 82 | B | 1 | 9700/11 May/June 2014 |
| 83 | B | 1 | 9700/12 May/June 2014 |
| 84 | C | 1 | 9700/12 May/June 2014 |
| 85 | D | 1 | 9700/12 May/June 2014 |
| 86 | C | 1 | 9700/13 May/June 2014 |
| 87 | D | 1 | 9700/13 May/June 2014 |
| 88 | C | 1 | 9700/11 Oct/Nov 2014 |
| 89 | D | 1 | 9700/11 Oct/Nov 2014 |
| 90 | A | 1 | 9700/12 Oct/Nov 2014 |
| 91 | D | 1 | 9700/13 Oct/Nov 2014 |
| 92 | A | 1 | 9700/11 May/June 2015 |
| 93 | D | 1 | 9700/12 May/June 2015 |
| 94 | C | 1 | 9700/12 May/June 2015 |
| 95 | B | 1 | 9700/13 May/June 2015 |
| 96 | B | 1 | 9700/13 May/June 2015 |
| 97 | A | 1 | 9700/11 Oct/Nov 2015 |
| 98 | D | 1 | 9700/11 Oct/Nov 2015 |
| 99 | B | 1 | 9700/11 Oct/Nov 2015 |
| 100 | B | 1 | 9700/12 Oct/Nov 2015 |
| 101 | B | 1 | 9700/12 Oct/Nov 2015 |
| 102 | B | 1 | 9700/12 Oct/Nov 2015 |
| 103 | C | 1 | 9700/13 Oct/Nov 2015 |
| 104 | C | 1 | 9700/13 Oct/Nov 2015 |
| 105 | C | 1 | 9700/12 Feb/March 2016 |
| 106 | D | 1 | 9700/11 May/June 2016 |
| 107 | C | 1 | 9700/11 May/June 2016 |
| 108 | A | 1 | 9700/11 May/June 2016 |
| 109 | D | 1 | 9700/12 May/June 2016 |
| 110 | D | 1 | 9700/12 May/June 2016 |
| 111 | C | 1 | 9700/12 May/June 2016 |
| 112 | see sheet | 1 | 9700/13 May/June 2016 |
| 113 | see sheet | 1 | 9700/13 May/June 2016 |
| 114 | B | 1 | 9700/11 Oct/Nov 2016 |
| 115 | B | 1 | 9700/11 Oct/Nov 2016 |
| 116 | C | 1 | 9700/13 Oct/Nov 2016 |
| 117 | D | 1 | 9700/13 Oct/Nov 2016 |
| 118 | B | 1 | 9700/12 Feb/March 2017 |
| 119 | A | 1 | 9700/12 Feb/March 2017 |
| 120 | B | 1 | 9700/11 May/June 2017 |
| 121 | D | 1 | 9700/11 May/June 2017 |
| 122 | C | 1 | 9700/11 May/June 2017 |
| 123 | C | 1 | 9700/12 May/June 2017 |
| 124 | D | 1 | 9700/13 May/June 2017 |
| 125 | B | 1 | 9700/13 May/June 2017 |
| 126 | A | 1 | 9700/12 Oct/Nov 2017 |
| 127 | A | 1 | 9700/13 Oct/Nov 2017 |
| 128 | C | 1 | 9700/13 Oct/Nov 2017 |
| 129 | D | 1 | 9700/12 Feb/March 2018 |
| 130 | B | 1 | 9700/12 Feb/March 2018 |
| 131 | A | 1 | 9700/11 May/June 2018 |
| 132 | B | 1 | 9700/11 May/June 2018 |
| 133 | C | 1 | 9700/11 May/June 2018 |
| 134 | B | 1 | 9700/12 May/June 2018 |
| 135 | C | 1 | 9700/12 May/June 2018 |
| 136 | B | 1 | 9700/12 May/June 2018 |
| 137 | B | 1 | 9700/13 May/June 2018 |
| 138 | C | 1 | 9700/13 May/June 2018 |
| 139 | A | 1 | 9700/11 Oct/Nov 2018 |
| 140 | A | 1 | 9700/11 Oct/Nov 2018 |
| 141 | C | 1 | 9700/11 Oct/Nov 2018 |
| 142 | D | 1 | 9700/12 Oct/Nov 2018 |
| 143 | A | 1 | 9700/12 Oct/Nov 2018 |
| 144 | C | 1 | 9700/13 Oct/Nov 2018 |
| 145 | C | 1 | 9700/13 Oct/Nov 2018 |
| 146 | D | 1 | 9700/12 Feb/March 2019 |
| 147 | A | 1 | 9700/11 May/June 2019 |
| 148 | B | 1 | 9700/11 May/June 2019 |
| 149 | B | 1 | 9700/11 May/June 2019 |
| 150 | B | 1 | 9700/13 May/June 2019 |
| 151 | A | 1 | 9700/13 May/June 2019 |
| 152 | C | 1 | 9700/13 May/June 2019 |
| 153 | B | 1 | 9700/11 Oct/Nov 2019 |
| 154 | D | 1 | 9700/11 Oct/Nov 2019 |
| 155 | A | 1 | 9700/11 Oct/Nov 2019 |
| 156 | C | 1 | 9700/12 Oct/Nov 2019 |
| 157 | D | 1 | 9700/12 Oct/Nov 2019 |
| 158 | B | 1 | 9700/13 Oct/Nov 2019 |
| 159 | C | 1 | 9700/13 Oct/Nov 2019 |
| 160 | D | 1 | 9700/12 Feb/March 2020 |
| 161 | B | 1 | 9700/12 Feb/March 2020 |
| 162 | A | 1 | 9700/12 Feb/March 2020 |
| 163 | A | 1 | 9700/12 Feb/March 2020 |
| 164 | D | 1 | 9700/11 May/June 2020 |
| 165 | B | 1 | 9700/11 May/June 2020 |
| 166 | D | 1 | 9700/12 May/June 2020 |
| 167 | B | 1 | 9700/13 May/June 2020 |
| 168 | A | 1 | 9700/13 May/June 2020 |
| 169 | A | 1 | 9700/11 Oct/Nov 2020 |
| 170 | A | 1 | 9700/11 Oct/Nov 2020 |
| 171 | A | 1 | 9700/13 Oct/Nov 2020 |
| 172 | A | 1 | 9700/12 Feb/March 2021 |
| 173 | D | 1 | 9700/12 Feb/March 2021 |
| 174 | B | 1 | 9700/12 Feb/March 2021 |
| 175 | B | 1 | 9700/11 May/June 2021 |
| 176 | A | 1 | 9700/12 May/June 2021 |
| 177 | A | 1 | 9700/12 May/June 2021 |
| 178 | D | 1 | 9700/12 May/June 2021 |
| 179 | D | 1 | 9700/12 May/June 2021 |
| 180 | A | 1 | 9700/13 May/June 2021 |
| 181 | D | 1 | 9700/13 May/June 2021 |
| 182 | D | 1 | 9700/11 Oct/Nov 2021 |
| 183 | D | 1 | 9700/11 Oct/Nov 2021 |
| 184 | D | 1 | 9700/11 Oct/Nov 2021 |
| 185 | D | 1 | 9700/12 Oct/Nov 2021 |
| 186 | B | 1 | 9700/12 Oct/Nov 2021 |
| 187 | C | 1 | 9700/13 Oct/Nov 2021 |
| 188 | D | 1 | 9700/13 Oct/Nov 2021 |
| 189 | A | 1 | 9700/12 Feb/March 2022 |
| 190 | B | 1 | 9700/12 Feb/March 2022 |
| 191 | B | 1 | 9700/11 May/June 2022 |
| 192 | B | 1 | 9700/11 May/June 2022 |
| 193 | A | 1 | 9700/11 May/June 2022 |
| 194 | C | 1 | 9700/12 May/June 2022 |
| 195 | A | 1 | 9700/12 May/June 2022 |
| 196 | A | 1 | 9700/12 May/June 2022 |
| 197 | A | 1 | 9700/13 May/June 2022 |
| 198 | B | 1 | 9700/13 May/June 2022 |
| 199 | C | 1 | 9700/13 May/June 2022 |
| 200 | C | 1 | 9700/11 Oct/Nov 2022 |
| 201 | D | 1 | 9700/11 Oct/Nov 2022 |
| 202 | C | 1 | 9700/12 Oct/Nov 2022 |
| 203 | C | 1 | 9700/12 Oct/Nov 2022 |
| 204 | C | 1 | 9700/12 Oct/Nov 2022 |
| 205 | C | 1 | 9700/13 Oct/Nov 2022 |
| 206 | B | 1 | 9700/13 Oct/Nov 2022 |
| 207 | A | 1 | 9700/13 Oct/Nov 2022 |
| 208 | D | 1 | 9700/13 Oct/Nov 2022 |
| 209 | D | 1 | 9700/12 Feb/March 2023 |
| 210 | D | 1 | 9700/12 Feb/March 2023 |
| 211 | B | 1 | 9700/11 May/June 2023 |
| 212 | C | 1 | 9700/11 May/June 2023 |
| 213 | C | 1 | 9700/12 May/June 2023 |
| 214 | A | 1 | 9700/12 May/June 2023 |
| 215 | A | 1 | 9700/12 May/June 2023 |
| 216 | B | 1 | 9700/13 May/June 2023 |
| 217 | B | 1 | 9700/13 May/June 2023 |
| 218 | A | 1 | 9700/13 May/June 2023 |
| 219 | C | 1 | 9700/12 Oct/Nov 2023 |
| 220 | C | 1 | 9700/12 Oct/Nov 2023 |
| 221 | D | 1 | 9700/13 Oct/Nov 2023 |
| 222 | D | 1 | 9700/13 Oct/Nov 2023 |
| 223 | B | 1 | 9700/12 Feb/March 2024 |
| 224 | A | 1 | 9700/12 Feb/March 2024 |
| 225 | C | 1 | 9700/12 Feb/March 2024 |
| 226 | B | 1 | 9700/11 May/June 2024 |
| 227 | B | 1 | 9700/11 May/June 2024 |
| 228 | B | 1 | 9700/11 May/June 2024 |
| 229 | D | 1 | 9700/12 May/June 2024 |
| 230 | D | 1 | 9700/12 May/June 2024 |
| 231 | C | 1 | 9700/12 May/June 2024 |
| 232 | C | 1 | 9700/12 May/June 2024 |
| 233 | C | 1 | 9700/13 May/June 2024 |
| 234 | B | 1 | 9700/13 May/June 2024 |
| 235 | C | 1 | 9700/13 May/June 2024 |
| 236 | A | 1 | 9700/13 May/June 2024 |
| 237 | A | 1 | 9700/11 Oct/Nov 2024 |
| 238 | B | 1 | 9700/11 Oct/Nov 2024 |
| 239 | B | 1 | 9700/12 Oct/Nov 2024 |
| 240 | D | 1 | 9700/12 Oct/Nov 2024 |
| 241 | C | 1 | 9700/13 Oct/Nov 2024 |
| 242 | C | 1 | 9700/13 Oct/Nov 2024 |
| 243 | C | 1 | 9700/13 Oct/Nov 2024 |
| 244 | B | 1 | 9700/12 Feb/March 2025 |
| 245 | A | 1 | 9700/12 Feb/March 2025 |
| 246 | D | 1 | 9700/12 May/June 2025 |
| 247 | A | 1 | 9700/12 May/June 2025 |
| 248 | D | 1 | 9700/12 May/June 2025 |
| 249 | C | 1 | 9700/13 May/June 2025 |
| 250 | D | 1 | 9700/13 May/June 2025 |
| 251 | D | 1 | 9700/14 May/June 2025 |
| 252 | C | 1 | 9700/14 May/June 2025 |
| 253 | C | 1 | 9700/14 May/June 2025 |
| 254 | D | 1 | 9700/14 May/June 2025 |
| 255 | B | 1 | 9700/11 Oct/Nov 2025 |
| 256 | B | 1 | 9700/11 Oct/Nov 2025 |
| 257 | D | 1 | 9700/11 Oct/Nov 2025 |
| 258 | A | 1 | 9700/13 Oct/Nov 2025 |
| 259 | C | 1 | 9700/13 Oct/Nov 2025 |
17 The diagram shows a plant cell. The plant cell is put into a solution with a water potential less negative (higher) than the cell contents. What will happen to the appearance of the cell? A B C D
1 marks
Answer: C
18 Which structures are present in large numbers at sites of active transport? A Golgi bodies B lysosomes C mitochondria D rough endoplasmic reticulum
1 marks
Answer: C
4 Which adaptation would increase the efficiency of active transport of carbohydrates from a plant cell? A areas where the cell wall is thin B increased permeability of the cell wall C large surface area of the cell surface membrane D selective permeability of the vacuole membrane
1 marks
Answer: C
15 Which statement defines active transport? A movement of large molecules through the cell surface membrane into the cytoplasm of a cell B movement of molecules or ions from where they are in a low concentration to where they are in a higher concentration C movement of molecules or ions from where they are in a high concentration to where they are in a lower concentration D net movement of water molecules across a partially permeable membrane from a region of higher water potential to one of lower water potential
1 marks
Answer: B
16 Which pair of factors is inversely proportional to the rate of diffusion? A concentration gradient and size of diffusing molecule B distance over which diffusion occurs and surface area over which diffusion occurs C size of diffusing molecule and distance over which diffusion occurs D surface area over which diffusion occurs and concentration gradient
1 marks
Answer: C
17 When cylinders of potato tissue were immersed in a 0.35 mol dm–3 sucrose solution, they showed no change in mass. What will happen when cylinders are immersed in a 0.1 mol dm–3 sucrose solution? A The pressure potential of the cells will become more positive. B The solute potential of the cell will become more negative. C The water potential of the cells will become more negative. D The water potential of the solution will become less negative.
1 marks
Answer: A
14 The diagram shows part of a cell surface membrane. P Q R What is the correct function for each of the structures labelled? regulates forms hydrogen bonds with transports ions and membrane fluidity water to stabilise membrane large polar molecules A R R Q B P Q R C Q R P D R P Q
1 marks
Answer: D
15 Strips of potato tuber tissue were immersed in distilled water or in sucrose solutions of different concentrations. The graph shows the percentage change in length of the strips. 20 distilled water 0.1 mol dm _3 10 change in length 0 of strips / % 0.4 mol dm _3 _10 _20 0 1 2 3 4 5 time / hours Which statement explains the change that occurred in the potato strips immersed in 0.1 mol dm–3 sucrose solution? A Sucrose molecules diffused into the potato cells. B Sucrose molecules were actively transported into the potato cells. C The water potential of the sucrose solution was less negative than the water potential inside the cells. D The water potential of the sucrose solution was more negative than the water potential inside the cells.
1 marks
Answer: C
16 Which process would allow the movement of large protein molecules out of the cell? A active transport B exocytosis C facilitated diffusion D phagocytosis
1 marks
Answer: B
15 Strips of plant tissue were immersed in a range of sucrose solutions of different concentrations. Their lengths were measured before immersion and after 30 minutes. The graph shows the ratio of initial length to final length. 1.4 1.2 1.0 initial length final length 0.8 0.6 0.4 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 sucrose concentration / mol dm–3 What is a correct description of the change in the cells and in their water potential as the sucrose concentration increases? change in the cells change in the water potential A more turgid less negative B less turgid more negative C more turgid more negative D less turgid less negative
1 marks
Answer: B
16 The graph shows rates of simple diffusion and facilitated diffusion, of substance X across a cell surface membrane, as the concentration of substance X increases. simple diffusion rate of facilitated diffusion diffusion / arbitrary units concentration of substance X Why does the rate of facilitated diffusion level off whereas the rate of simple diffusion does not? A Facilitated diffusion is limited by the number of protein channels in the membrane. B Facilitated diffusion is limited by the number of protein pumps in the membrane. C Facilitated diffusion requires ATP which will eventually be used up. D Only facilitated diffusion is affected by the kinetic energy of the molecules that are diffusing.
1 marks
Answer: A
14 The diagram shows a cell from the gut. The cell produces protease enzymes. What is correct? enzymes released by ATP needed A endocytosis no B endocytosis yes C exocytosis no D exocytosis yes
1 marks
Answer: D
16 Strips of plant tissue were immersed in a range of sucrose solutions of different concentrations. Their lengths were measured before immersion and after 30 minutes in the different solutions. The graph shows the ratio of initial length to final length. 1.04 1.02 1.00 initial length final length 0.98 0.96 0.94 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 sucrose concentration / mol dm–3 Which concentration of sucrose solution, in mol dm–3, has the same water potential as the cell sap before immersion? A 0.1 B 0.25 C 0.45 D 0.8
1 marks
Answer: C
17 The diagram shows three routes through which substances can pass across a cell membrane. X Y Z Which correctly shows the routes for vitamin D, which is fat soluble, and vitamin C, which is water soluble? vitamin D vitamin C A Y X B X Z C X Y D Z Y
1 marks
Answer: B
13 The water potential of three adjacent plant cells is shown. X Y –250 kPa –1000 kPa Z –4000 kPa In which direction will water move? A from cell X to cell Y and then cell Z only B from cell X to both cells Y and Z C from cell Z to cell Y and then cell X only D from cell Z to both cells Y and X
1 marks
Answer: B
14 The table shows three processes that contribute to transport across cell surface membranes. Which processes are the result of random movement of molecules? diffusion endocytosis osmosis A x x x B x v v Cc v x v D v v x < I key qj I random non random
1 marks
Answer: C
15 The epithelial cells of people with cystic fibrosis have a defect in the structure of the cell surface membrane. The ability of the cell to transport chloride ions out of the cell is affected. Which membrane component is involved? A cholesterol B glycolipid C phospholipid D protein
1 marks
Answer: D
23 Four solutions, with different water potentials are listed. 1 endodermal cell solution 2 root hair cell solution 3 soil water solution 4 solution in a xylem vessel Which list has the solutions in order from the highest (least negative) water potential to the lowest (most negative) water potential? highest lowest A 1 2 3 4 B 2 4 1 3 C 3 2 1 4 D 4 1 3 2
1 marks
Answer: C
14 The diagram shows a red blood cell and the concentrations of ions, in mmol dm−3, in the plasma and in the cell. Na+ 15 K+ 150 Cl 73 Na+ 144 K+ 5 Cl 111 Which ions are actively transported into and out of the cell? into cell out of cell A Cl − K+ B K+ Na+ C Na+ Cl − D Na+ K+
1 marks
Answer: B
15 Diagrams 1 and 2 show how the transverse section through a leaf changes when moved from one solution W to a different solution Y. X X W Y diagram 1 diagram 2 How has the water potential changed in diagram 2? difference in cells at X in solution Y compared to difference in solution Y the same cells in compared to solution W solution W A less negative less negative B less negative more negative C more negative less negative D more negative more negative
1 marks
Answer: A
17 Cystic fibrosis is a disease where Cl – ions are unable to be transported into cells. Which structure in the cell surface membrane is faulty? A B C D
1 marks
Answer: C
27 The diagram shows the changes in pressure potential (ΨP), solute potential (ΨS) and water potential (Ψ) when a plasmolysed plant cell is placed in pure water. + X kPa 0 time Y Z – Which shows the correct curves for each potential? X Y Z A Ψ ΨP ΨS B ΨP ΨS Ψ C ΨS Ψ ΨP D ΨP Ψ ΨS
1 marks
Answer: D
17 Which processes allow movement into and out of a cell? 1 active transport 2 diffusion 3 facilitated diffusion 4 osmosis A 2 and 4 only B 1, 2 and 3 only C 1, 3 and 4 only D 1, 2, 3 and 4
1 marks
Answer: D
18 Some plant and animal cells were placed in different solutions and the results are shown. 1 2 3 4 5 Which cells were placed in which solution? 1.0 mol dm–3 sucrose 0.1 mol dm–3 salt solution A 1 and 2 3 and 5 B 1 and 4 3 C 2 and 4 1 and 3 D 3 and 5 2 and 4
1 marks
Answer: D
4 Which processes allow movement into and out of a cell? 1 active transport 2 diffusion 3 facilitated diffusion 4 osmosis A 2 and 4 only B 1, 2 and 3 only C 1, 3 and 4 only D 1, 2, 3 and 4
1 marks
Answer: D
5 Some plant and animal cells were placed in different solutions and the results are shown. 1 2 3 4 5 Which cells were placed in which solution? 1.0 mol dm–3 sucrose 0.1 mol dm–3 salt solution A 1 and 2 3 and 5 B 1 and 4 3 C 2 and 4 1 and 3 D 3 and 5 2 and 4
1 marks
Answer: D
6 Some plant and animal cells were placed in different solutions and the results are shown. 1 2 3 4 5 Which cells were placed in which solution? 1.0 mol dm–3 sucrose 0.1 mol dm–3 salt solution A 1 and 2 3 and 5 B 1 and 4 3 C 2 and 4 1 and 3 D 3 and 5 2 and 4
1 marks
Answer: D
33 Which processes allow movement into and out of a cell? 1 active transport 2 diffusion 3 facilitated diffusion 4 osmosis A 2 and 4 only B 1, 2 and 3 only C 1, 3 and 4 only D 1, 2, 3 and 4
1 marks
Answer: D
16 The stalk of a dandelion is a hollow tube. Pieces of the stalk are cut as shown and placed in sucrose solutions of different water potentials. thick walled outer cells hollow centre of stalk cuts thin walled inner cells Which diagram shows the piece that is placed in the sucrose solution with the highest water potential? A B C D
1 marks
Answer: A
17 In an investigation, four sucrose solutions were separated from each other by partially permeable membranes. 1 1.1 mol dm–3 2 0.8 mol dm–3 3 0.5 mol dm–3 4 0.1 mol dm–3 Which shows the direction in which water will move between the solutions? A from 1 and 2 to 3 and 4 B from 2 and 3 to 1 C from 1 to 3 D from 2 to 4
1 marks
Answer: B
13 The diagram shows the transport of ions across the cell surface membrane. Inside the cell there is a low concentration of sodium ions (Na+) and a high concentration of potassium ions (K+). Outside the cell there is a low concentration of K+ and a high concentration of Na+. The carrier molecule is a pump which exchanges Na+ for K+ ions. inside cell 2 1 cell surface membrane 4 carrier 3 molecule outside cell Which ionic movements are represented by the arrows? active transport active transport diffusion of Na+ diffusion of K+ of K+ of Na+ A 2 3 1 4 B 2 3 4 1 C 3 2 1 4 D 3 2 4 1
1 marks
Answer: A
14 Plant cells were immersed in solutions of different water potential and left for one hour. Which row shows the effect of the different solutions on the plant cells? water potential of solution compared to plant cells less negative equal more negative A flaccid turgid unchanged B flaccid unchanged turgid C turgid unchanged flaccid D unchanged flaccid turgid
1 marks
Answer: C
16 What supports the view that a membrane protein is involved in active transport? A It allows movement of molecules across a membrane if concentration differences exist. B It can only function if mitochondria are supplied with sufficient oxygen. C It has a tertiary structure with a binding site with a specific shape. D It is found in the cell surface membranes and the mitochondrial membranes.
1 marks
Answer: B
17 Strips of plant tissue were immersed in a range of sucrose solutions of different concentrations. Their lengths were measured before immersion and after 30 minutes. The graph shows the ratio of initial length to final length. 1.4 1.2 1.0 initial length final length 0.8 0.6 0.4 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 sucrose concentration / mol dm–3 What is a correct description of the change in the cells and in their water potential as the sucrose concentration increases? change in the cells change in the water potential A less turgid more negative B less turgid less negative C more turgid less negative D more turgid more negative
1 marks
Answer: A
3 Visking tubing is an artificial partially permeable membrane used to demonstrate diffusion. Glucose molecules can pass through the pores in the membrane which are approximately 2.4 nm in diameter. Which of the following could pass through the pores? 1 bacteria 2 haemoglobin 3 ribosomes 4 glycogen A 2 only B 1 and 3 only C 2 and 4 only D none of these
1 marks
Answer: D
17 When cylinders of potato tissue were immersed in a 0.35 mol dm–3 sucrose solution, they showed no change in mass. What will happen when cylinders are immersed in a 0.1 mol dm–3 sucrose solution? A The pressure potential of the cells will become more positive. B The solute potential of the cell will become more negative. C The water potential of the cells will become more negative. D The water potential of the solution will become less negative.
1 marks
Answer: A
12 What supports the view that a membrane protein is involved in active transport? A It allows movement of molecules across a membrane if concentration differences exist. B It can only function if mitochondria are supplied with sufficient oxygen. C It has a tertiary structure with a binding site with a specific shape. D It is found in the cell surface membranes and the mitochondrial membranes.
1 marks
Answer: B
14 Strips of plant tissue were immersed in a range of sucrose solutions of different concentrations. Their lengths were measured before immersion and after 30 minutes. The graph shows the ratio of initial length to final length. 1.4 1.2 1.0 initial length final length 0.8 0.6 0.4 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 sucrose concentration / mol dm–3 What is a correct description of the change in the cells and in their water potential as the sucrose concentration increases? change in the cells change in the water potential A less turgid more negative B less turgid less negative C more turgid less negative D more turgid more negative
1 marks
Answer: A
14 Single-celled animals that live in fresh water have a vacuole that contracts regularly to remove excess water. Single-celled plants that live in fresh water do not have a similar vacuole. Which statement explains why only these animals need this vacuole? A Plant cell cytoplasm and animal cell cytoplasm both have a lower water potential than fresh water. B Plant cell sap has the same water potential as fresh water, animal cytoplasm has a lower water potential than fresh water. C Plant cell walls are impermeable to water, animal cell surface membranes are permeable to water. D Plant cell walls restrict the entry of water, animal cell membranes allow the free entry of water.
1 marks
Answer: D
15 Which statement defines active transport? A movement of large molecules through the cell surface membrane into the cytoplasm of a cell B movement of molecules or ions from where they are in a low concentration to where they are in a higher concentration C movement of molecules or ions from where they are in a high concentration to where they are in a lower concentration D net movement of water molecules across a partially permeable membrane from a region of higher water potential to one of lower water potential
1 marks
Answer: B
16 Which statements about the components of the cell surface membrane are correct? 1 Diffusion can take place through lipids and protein pores. 2 Endocytosis only involves lipids. 3 Facilitated diffusion only involves proteins. 4 Osmosis only involves proteins. A 1, 2, 3 and 4 B 1, 3 and 4 only C 1 and 3 only D 2 and 4 only
1 marks
Answer: C
24 Which statement defines active transport? A movement of large molecules through the cell surface membrane into the cytoplasm of a cell B movement of molecules or ions from where they are in a low concentration to where they are in a higher concentration C movement of molecules or ions from where they are in a high concentration to where they are in a lower concentration D net movement of water molecules across a partially permeable membrane from a region of higher water potential to one of lower water potential
1 marks
Answer: B
25 Single-celled animals that live in fresh water have a vacuole that contracts regularly to remove excess water. Single-celled plants that live in fresh water do not have a similar vacuole. Which statement explains why only these animals need this vacuole? A Plant cell cytoplasm and animal cell cytoplasm both have a lower water potential than fresh water. B Plant cell sap has the same water potential as fresh water, animal cytoplasm has a lower water potential than fresh water. C Plant cell walls are impermeable to water, animal cell surface membranes are permeable to water. D Plant cell walls restrict the entry of water, animal cell membranes allow the free entry of water.
1 marks
Answer: D
26 Which statements about the components of the cell surface membrane are correct? 1 Diffusion can take place through lipids and protein pores. 2 Endocytosis only involves lipids. 3 Facilitated diffusion only involves proteins. 4 Osmosis only involves proteins. A 1, 2, 3 and 4 B 1, 3 and 4 only C 1 and 3 only D 2 and 4 only
1 marks
Answer: C
15 Which of the following ways of moving substances across cell surface membranes allows movement in both directions? 1 active transport 2 diffusion 3 facilitated diffusion 4 osmosis A 2 only B 1 and 4 only C 2 and 3 only D 1, 2, 3 and 4
1 marks
Answer: D
16 The graphs show the rate of uptake of sugars by a culture of animal cells, under different conditions. air bubbled through the culture 3-carbon sugar rate of uptake 6-carbon sugar 0 0 10 20 30 temperature / °C nitrogen gas bubbled through the culture 3-carbon sugar rate of uptake 0 6-carbon sugar 0 10 20 30 temperature / °C How are the sugars taken up by the cells when air is bubbled through the culture? 3-carbon sugar 6-carbon sugar A active transport active transport B active transport diffusion C diffusion active transport D diffusion diffusion
1 marks
Answer: C
15 The following are all processes by which substances can enter a cell. 1 endocytosis 2 facilitated diffusion 3 osmosis Which processes are passive? A 2 only B 3 only C 2 and 3 only D 1, 2 and 3
1 marks
Answer: C
16 The diagram shows the water potential of three adjacent plant cells. –1200 kPa –800 kPa –1000 kPa In which directions will there be net movement of water by osmosis? A B C D
1 marks
Answer: D
10 Which of the following ways of moving substances across cell surface membranes allows movement in both directions? 1 active transport 2 diffusion 3 facilitated diffusion 4 osmosis A 2 only B 1 and 4 only C 2 and 3 only D 1, 2, 3 and 4
1 marks
Answer: D
11 The graphs show the rate of uptake of sugars by a culture of animal cells, under different conditions. air bubbled through the culture 3-carbon sugar rate of uptake 6-carbon sugar 0 0 10 20 30 temperature / °C nitrogen gas bubbled through the culture 3-carbon sugar rate of uptake 0 6-carbon sugar 0 10 20 30 temperature / °C How are the sugars taken up by the cells when air is bubbled through the culture? 3-carbon sugar 6-carbon sugar A active transport active transport B active transport diffusion C diffusion active transport D diffusion diffusion
1 marks
Answer: C
16 The graph shows how the rate of entry of glucose into a cell changes as the concentration of glucose outside the cell changes. X rate of entry of glucose into the cell concentration of glucose outside the cell What is the cause of the plateau at X? A All the carrier proteins are saturated with glucose. B The carrier proteins are denatured and no longer able to function. C The cell has used up its supply of ATP. D The concentrations of glucose inside and outside the cell are equal.
1 marks
Answer: A
17 A molecule can enter a cell by two different passive processes. Which process would increase the rate at which this molecule enters the cells? A diffusion B endocytosis C facilitated diffusion D osmosis
1 marks
Answer: C
30 The photograph shows a type of blood cell. Which statements about these cells are correct? 1 Oxygen diffuses through the phospholipid bilayer. 2 Sodium ions diffuse through the phospholipid bilayer. 3 Water passes in and out of these cells by osmosis. A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2 and 3
1 marks
Answer: B
15 Which role of the cell surface membrane is not a result of the properties of the phospholipids? A to allow cytokinesis to occur in mitotic cell division B to allow entry and exit of the water-soluble gases, oxygen and carbon dioxide C to allow phagocytosis of a bacterium into cells D to allow surface membranes to stabilise by binding with water molecules
1 marks
Answer: D
17 Which processes that move substances across cell surface membranes result in an equilibrium? 1 active transport 2 diffusion 3 facilitated diffusion 4 osmosis A 1, 2 and 3 only B 1, 2 and 4 only C 1, 3 and 4 only D 2, 3 and 4 only
1 marks
Answer: D
18 Diagrams 1 and 2 show how the transverse section through a leaf changes when moved from one solution W to a different solution Y. X X W Y diagram 1 diagram 2 How has the water potential changed in diagram 2? difference in cells at X in solution Y compared to difference in solution Y the same cells in compared to solution W solution W A less negative less negative B less negative more negative C more negative less negative D more negative more negative
1 marks
Answer: A
15 The graph shows how the rate of entry of glucose into a cell changes as the concentration of glucose outside the cell changes. X rate of entry of glucose into the cell concentration of glucose outside the cell What is the cause of the plateau at X? 1 All the carrier proteins are saturated with glucose. 2 The cell has used up its supply of ATP. 3 The concentrations of glucose inside and outside the cell are equal. A 1 only B 3 only C 1 and 2 only D 2 and 3 only
1 marks
Answer: A
16 The following are all processes by which substances can enter cells. 1 phagocytosis 2 active transport 3 facilitated diffusion Which processes require ATP? A 1 only B 2 only C 1 and 2 only D 2 and 3 only
1 marks
Answer: C
17 Which process allows the movement of molecules that are too large to enter through a cell surface membrane? A active transport B endocytosis C exocytosis D facilitated diffusion
1 marks
Answer: B
18 The diagram shows the water potential of three cells. cell X –700 kPa –800 kPa –1000 kPa In which directions will there be net movement of water by osmosis to or from cell X? A B C D
1 marks
Answer: B
25 The diagram shows a xerophytic leaf in different conditions, P and Q. X X P Q Which statements about the cells in layer X of the leaf in each of the conditions P and Q are correct? 1 less negative water potential in P than Q 2 cells may be turgid in P and plasmolysed in Q 3 cells less turgid in P than Q 4 no net diffusion of water into X in either P or Q A 1, 2, 3 and 4 B 1, 2 and 4 only C 1 and 4 only D 2 and 3 only
1 marks
Answer: B
28 Halophytes are plants that can survive in regions where they are regularly exposed to sea water. Sea water has a water potential of approximately –2500 kPa. What adaptations would you expect halophytes to show? 1 root hair cells that maintain a more negative water potential than sea water 2 root hair cells that accumulate salts and other solutes 3 stomata that are open most of the time A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2 and 3
1 marks
Answer: A
6 An animal cell and a plant cell are placed in distilled water. The animal cell swells and bursts, while the plant cell swells but does not burst. What accounts for this difference? A Animal cells have no cell wall. B Animal cells have no vacuole. C Plant cell surface membranes are partially permeable. D Plant cell walls are freely permeable.
1 marks
Answer: A
17 The diagram shows a cell that produces protease enzymes. Which row is correct? enzymes released by ATP needed A endocytosis no B endocytosis yes C exocytosis no D exocytosis yes
1 marks
Answer: D
18 The diagram shows two identical plant cells. One plant cell is put into a solution with a water potential less negative than the cell contents. The other is put into a solution with a water potential more negative than the cell contents. What will happen to the appearance of each cell? water potential of solution water potential of solution surrounding cell less surrounding cell more negative than cell contents negative than cell contents A B C D
1 marks
Answer: C
27 The diagram shows how sucrose is loaded into a sieve tube element. companion cell sieve tube element 1 hydrogen ions move out of the companion cell 3 sucrose molecules move into sucrose molecules move into sucrose molecules move into the sieve tube element the sieve tube element the sieve tube element hydrogen ions move into the companion cell 2 sucrose molecules move into the companion cell What type of transport is used to move the substances in steps 1, 2 and 3? 1 2 3 A active transport active transport diffusion B active transport facilitated diffusion diffusion C facilitated diffusion active transport active transport D facilitated diffusion facilitated diffusion active transport
1 marks
Answer: B
17 Which process would allow the movement of large protein molecules out of the cell? A active transport B exocytosis C facilitated diffusion D phagocytosis
1 marks
Answer: B
18 Strips of potato tissue were immersed in distilled water or in sucrose solutions of different concentrations. The graph shows the percentage change in length of the potato tissue over time. 20 distilled water 10 0.1 mol dm _3 sucrose solution percentage change in 0 length _10 _3 0.5 mol dm sucrose solution _20 0 1 2 3 4 5 time / hours Which row correctly shows how the water potentials of the distilled water and sucrose solutions differ from the initial water potential of the potato tissue? distilled 0.1 mol dm–3 0.5 mol dm–3 water sucrose solution sucrose solution A less negative less negative more negative B less negative more negative more negative C more negative less negative less negative D more negative more negative less negative
1 marks
Answer: A
26 The diagram shows a xerophytic leaf in different conditions, P and Q. X X P Q Which statements describe the difference between the cells in layer X in conditions P and Q? 1 More negative water potential in P than Q. 2 More cells plasmolysed in P. 3 Cells less turgid in Q. 4 Water potential becomes zero in Q. A 1, 2 and 3 only B 1 and 2 only C 2 and 4 only D 3 and 4 only
1 marks
Answer: B
14 What are the features of facilitated diffusion? uses proteins molecules move down a in membrane uses ATP concentration gradient A v v v B x v v Cc v x v D v v x key: JY correct X incorrect
1 marks
Answer: C
15 The stalk of a dandelion flower is a hollow tube. Pieces of the stalk are cut as shown and placed in sucrose solutions of different water potentials. thick walled outer cells hollow centre of stalk cuts thin walled inner cells Which diagram shows the piece that is placed in the sucrose solution with the highest water potential? A B C D
1 marks
Answer: A
16 What happens to an animal cell when it is placed in a solution with a more negative water potential? A It loses solutes to the solution and swells. B It loses water by osmosis and shrinks. C It takes in solutes and swells. D It takes in water by osmosis and bursts.
1 marks
Answer: B
5 Many single-celled animals, living in fresh water, possess vacuoles which contract regularly, expelling excess water. Why do the cells of plants living in fresh water not require such vacuoles? A Plant cells have a higher concentration of dissolved solutes than animal cells. B Plant cell walls are impermeable to water. C Plant cell walls limit cell size. D Water movement into plants is controlled by their roots.
1 marks
Answer: C
15 Which statements are descriptions of both facilitated diffusion and active transport? 1 moves substances against a concentration gradient 2 requires ATP 3 transports charged ions across the cell surface membrane 4 uses proteins A 1, 2, 3 and 4 B 1, 2 and 4 only C 2 and 3 only D 3 and 4 only
1 marks
Answer: D
5 When mitochondria are extracted from cells for biochemical study, they are usually kept in a 0.25 mol dm–3 sucrose solution. Why is the sucrose solution used? A to act as a solvent B to enable the rate of reaction of the mitochondria to be determined C to prevent the mitochondria from changing in structure D to provide a source of energy
1 marks
Answer: C
14 Which processes that contribute to transport across cell surface membranes are active or passive? endocytosis | exocytosis facilitated osmosis diffusion A v v x x key B v x v x ¥ =active Cc Xx v Xx v X = passive D x x v v
1 marks
Answer: A
15 In an investigation, four sucrose solutions were separated from each other by partially permeable membranes. solution 1 1.1 mol dm–3 solution 2 0.8 mol dm–3 solution 3 0.5 mol dm–3 solution 4 0.1 mol dm–3 Which shows the direction in which water will move between the solutions? A from 1 and 2 to 3 and 4 B from 2, 3 and 4 to 1 C from 1 to 3 to 2 and 4 D from 1, 2 and 3 to 4
1 marks
Answer: B
16 The diagram shows a partially plasmolysed plant cell. solution X Z solution Y What is found at Z? A air B solution X C solution Y D water
1 marks
Answer: C
3 Which adaptation would increase active transport of carbohydrates from a plant cell? A areas where the cell wall is thin B increased permeability of the cell wall C large surface area of the cell surface membrane D selective permeability of the vacuole membrane
1 marks
Answer: C
14 The fluidity of the cell surface membrane can be changed by a number of factors. As the fluidity of cell surface membranes decreases, which process would be least changed? A active transport B diffusion C endocytosis D osmosis
1 marks
Answer: A
16 Which is correct for facilitated diffusion and active transport? A both depend on the solubility of the transported molecule in the lipid bilayer B both increase as the concentration of the transported molecule increases C both require the use of ATP D both require the use of membrane proteins
1 marks
Answer: D
17 Which features increase the efficiency of ion uptake by a root hair cell? 1 many mitochondria in the cell 2 high concentration of ions in the vacuole 3 protein carriers in the cell surface membrane A 1, 2 and 3 B 1 and 3 only C 2 and 3 only D 1 only
1 marks
Answer: B
13 The drug ritonavir is sometimes used in the treatment of HIV / AIDS. Ritonavir consists of three amino acids and is a competitive inhibitor of HIV protease. HIV causes this protease to be made inside human cells. Ritonavir produces many side effects as it interferes with many metabolic processes in human cells. Which statements about ritonavir are correct? 1 Ritonavir has a shape complementary to the active site of HIV protease. 2 Ritonavir will enter human cells directly through the lipid bilayer and not require any transport proteins. 3 Ritonavir is likely to inhibit many of the enzymes of human cells. 4 Complete hydrolysis of ritonavir would require the addition of three water molecules. A 1, 2 and 3 B 1 and 3 only C 2 and 4 D 3 and 4
1 marks
Answer: B
16 The diagram shows a plant cell. The plant cell is put into a solution with a water potential less negative than the cell contents. What will happen to the appearance of the cell? A B C D
1 marks
Answer: C
30 In the lungs, oxygen and carbon dioxide pass through cell membranes by diffusion. Which row is correct? number of cell membranes diffused through by oxygen from air carbon dioxide to air A 3 2 B 3 2 or 3 C 5 4 D 5 4 or 5
1 marks
Answer: D
17 What are the features of facilitated diffusion? 1 It uses protein channels in the membrane and is driven by the energy from ATP. 2 It moves molecules from regions of higher concentration to lower concentration and is driven by the kinetic energy of the molecules which are diffusing. 3 It uses protein channels in the membrane, and the maximum rate of diffusion depends on the number of these channels. A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2 and 3
1 marks
Answer: C
18 Single-celled animals that live in fresh water have a vacuole that contracts regularly to remove excess water. Single-celled plants that live in fresh water do not have a similar vacuole. Which statement explains why these animals need this vacuole but plants do not? A Plant cell cytoplasm and animal cell cytoplasm both have a lower water potential than fresh water. B Plant cell sap has the same water potential as fresh water, animal cytoplasm has a lower water potential than fresh water. C Plant cell walls are impermeable to water, animal cell surface membranes are permeable to water. D Plant cell walls restrict the entry of water, animal cell membranes allow the entry of water.
1 marks
Answer: D
14 The diagram shows three routes, X, Y and Z, through which substances can pass across a cell surface membrane. X Y Z Which correctly shows the routes for vitamin D, which is fat soluble, and vitamin C, which is water soluble? vitamin D vitamin C A Y X B X Y C X Z D Z Y
1 marks
Answer: C
15 In plants adapted to cold conditions, their cell surface membranes change as the weather gets colder, allowing the plants to carry out exocytosis. Which change occurs in their cell surface membranes? A a decrease in the ratio of proteins to saturated phospholipids B a decrease in the ratio of unsaturated phospholipids to saturated phospholipids C an increase in the ratio of proteins to unsaturated phospholipids D an increase in the ratio of unsaturated phospholipids to saturated phospholipids
1 marks
Answer: D
26 The diagram shows four plant cells. 1 4 2 3 In which direction could there be net movement of water by osmosis? A 1 to 2 and 1 to 4 B 1 to 3 and 1 to 2 C 2 to 1 and 1 to 4 D 4 to 1 and 2 to 3
1 marks
Answer: A
16 The diagram shows apparatus set up to investigate the effect of changing the concentration of glucose in the surrounding solution on the movement of molecules through a selectively permeable membrane (Visking tubing) in 15 minutes. Visking tubing surrounding solution 10% glucose different concentrations solution from 1% to 10% glucose solution As the concentration of glucose solution in the surrounding solution increases, which statements are correct? 1 Net diffusion of water increases. 2 Glucose molecules reach an equilibrium quicker. 3 There is less change in the volume of surrounding solution. 4 Net diffusion of glucose increases. A 1, 2, 3 and 4 B 1, 2 and 4 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: D
16 The diagram shows the transport of ions across the cell surface membrane. Inside the cell there is a low concentration of sodium ions (Na+) and a high concentration of potassium ions (K+). Outside the cell there is a low concentration of K+ and a high concentration of Na+. The carrier molecule is a pump which exchanges Na+ for K+. inside cell 2 1 cell surface membrane 4 carrier 3 molecule outside cell Which ionic movements are represented by the arrows? active transport active transport diffusion of Na+ diffusion of K+ of K+ of Na+ A 2 3 1 4 B 2 3 4 1 C 3 2 1 4 D 3 2 4 1
1 marks
Answer: A
15 Which description of cell surface membrane permeability is correct? A An increase in the concentration of cholesterol molecules in the cell surface membrane can increase its permeability to hydrophilic substances. B Cell surface membrane permeability to large hydrophilic molecules is high and can be increased by membrane transport proteins involved in facilitated diffusion. C The permeability of the cell surface membrane to ions is increased with an increase in the proportion of saturated fatty acids in the phospholipids. D Without the presence of carrier and channel membrane proteins, the cell surface membrane has a low permeability to large polar molecules.
1 marks
Answer: D
16 The diagram represents a cell surface membrane. high concentration of substance 1 2 3 low concentration of substance Three pathways through the membrane are shown. Which process is represented by each arrow? 1 2 3 A active transport diffusion facilitated diffusion B diffusion active transport facilitated diffusion C diffusion facilitated diffusion active transport D facilitated diffusion diffusion active transport
1 marks
Answer: C
15 What supports the view that a membrane protein is involved in active transport? A It allows movement of molecules across a membrane if concentration differences exist. B It can only function if mitochondria are supplied with sufficient oxygen. C It has a tertiary structure with a binding site with a specific shape. D It is found in the cell surface membranes and the mitochondrial membranes.
1 marks
Answer: B
16 The graph shows the effect of increasing the concentration of glucose in a solution on the rate of entry of glucose into a cell. X rate of entry of glucose into the cell 00 concentration of glucose in solution What are not causes of the plateau at X? 1 All the carrier proteins are saturated with glucose. 2 The carrier proteins are denatured and no longer able to function. 3 The cell has used up its supply of ATP. 4 The concentrations of glucose inside and outside the cell are equal. A 1, 2 and 4 B 2, 3 and 4 C 2 and 3 only D 1 only
1 marks
Answer: B
15 Which descriptions are correct about transport across cell surface membranes? active processes passive processes A endocytosis and exocytosis diffusion and osmosis B exocytosis and facilitated diffusion osmosis and endocytosis C facilitated diffusion exocytosis and osmosis D facilitated diffusion and exocytosis endocytosis and diffusion
1 marks
Answer: A
16 The statements are comparisons of endocytosis and exocytosis. • Both are mechanisms that involve vesicles or vacuoles and the transport of materials across the cell surface membrane. • Both mechanisms occur to allow bulk transport across the cell surface membrane. • Endocytosis involves taking materials into the cell whereas exocytosis involves the release of materials from the cell. • Some of the cell surface membrane is lost when endocytosis occurs and there is an increase in the cell surface membrane when exocytosis occurs. How many of the statements are correct? A 1 B 2 C 3 D 4
1 marks
Answer: D
17 The diagram shows apparatus set up to investigate the effect of putting a 10 % glucose solution in a selectively permeable bag (Visking tubing) and surrounding it with water. water 10% glucose solution Samples from the surrounding water were tested with Benedict’s solution after 10 minutes and after 20 minutes. The change in volume of glucose solution was observed after 20 minutes. Which row is correct? result of Benedict’s test volume of glucose solution in Visking tubing / cm3 after 10 minutes after 20 minutes A green blue increased B green orange increased C red orange decreased D yellow green decreased
1 marks
Answer: B
15 Which descriptions are correct for transport across cell surface membranes? active processes passive processes A active transport exocytosis and osmosis B endocytosis and exocytosis facilitated diffusion and osmosis C exocytosis and active transport osmosis and endocytosis D facilitated diffusion and exocytosis endocytosis and diffusion
1 marks
Answer: B
16 The stages of an investigation using plant tissue are listed below. ● A freshly cut slice of plant tissue was rinsed in distilled water, dried and weighed. ● This slice was placed in a solution with a water potential of – 4 arbitrary units for thirty minutes. ● The slice was removed from the solution, dried and reweighed. ● The mass of the slice was the same as its original mass. Which conclusions can be drawn from this investigation? 1 The water potential of the cells of the plant tissue is – 4 arbitrary units. 2 The cell sap of the plant tissue has a lower water potential than the surrounding solution. 3 There has been no net movement of water. 4 The cell wall of the plant tissue will not be in contact with the cell membrane. A 1 and 2 B 1 and 3 C 2 and 4 D 3 and 4
1 marks
Answer: B
17 What explains the effect on a red blood cell of being placed into pure water? A Less water leaves the cell than enters it, so the cell shrinks. B More water enters the cell than leaves it, so the cell swells and bursts. C Water enters the cell and none leaves it, so the cell swells and bursts. D Water enters the cell and more leaves it, so the cell shrinks.
1 marks
Answer: B
15 Which descriptions are correct about transport across cell surface membranes? active processes passive processes A active transport exocytosis and osmosis B active transport and exocytosis endocytosis and diffusion C endocytosis and exocytosis diffusion and osmosis D exocytosis and active transport osmosis and endocytosis
1 marks
Answer: C
17 The diagram shows apparatus set up to investigate diffusion. selectively 20% glucose permeable solution membrane Y 10% glucose solution in V What shows net diffusion of glucose and water molecules? A glucose and water into V B glucose and water into Y C glucose into V and water into Y D glucose into Y and water into V
1 marks
Answer: C
15 Antimycin is a chemical that inhibits the function of mitochondria. Which methods of transport across the cell surface membrane would be inhibited by antimycin? 1 active transport 2 facilitated diffusion 3 endocytosis A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: C
4 Visking tubing is an artificial partially permeable membrane used to demonstrate diffusion. Glucose molecules can pass through the pores in the membrane which are approximately 2.4 nm in diameter. Which of the following could pass through the pores? 1 bacteria 2 haemoglobin 3 ribosomes 4 glycogen A 1 and 3 B 2 and 4 C 2 only D none of the above
1 marks
Answer: D
15 The diagram represents a cell surface membrane of a metabolically active cell and the direction of movement of some molecules through the membrane. 1 2 3 4 Which row shows a process by which the molecules may be moving through the membrane at each of the points 1, 2, 3 and 4? 1 2 3 4 A carbon dioxide water by osmosis glucose by sodium ions by by diffusion diffusion active transport B fatty acids oxygen by carbon dioxide water by osmosis by diffusion diffusion by diffusion C sodium ions by carbon dioxide water by osmosis glucose by active transport by diffusion facilitated diffusion D water by osmosis oxygen by fatty acids glucose by diffusion by diffusion active transport
1 marks
Answer: C
16 Batrachotoxin is a poison found in frogs in the Columbian jungle. The poison is used by Native Indians to produce poison darts. The poison works by increasing the permeability of some cell surface membranes to sodium ions, which move out of the cells. Which statements are correct for cells affected by batrachotoxin? 1 The intracellular fluid has a less negative water potential than the extracellular fluid. 2 The extracellular fluid has a less negative water potential than the intracellular fluid. 3 Water leaves the cells by osmosis, causing the cells to shrink. 4 Water enters the cells by osmosis, causing the cells to swell. A 1 and 3 B 1 and 4 C 2 and 3 D 2 and 4
1 marks
Answer: A
14 Which statements about active transport are always correct? 1 It does not require a membrane. 2 It occurs against the concentration gradient. 3 It moves oxygen molecules. A 1, 2 and 3 B 1 and 3 only C 1 only D 2 only
1 marks
Answer: D
15 There is a high concentration of molecule X outside the cell which enters the cell by facilitated diffusion. The results of measuring the concentration of X inside the cell at 30 s intervals are shown by the graph. concentration of X inside cell 0 30 60 90 120 150 180 time / s Why does the concentration of X inside the cell remain constant after 150 s? 1 There is no more of X outside the cell. 2 The number of carrier proteins is limiting. 3 There is no net movement of X. A 1 and 2 only B 1 and 3 only C 2 and 3 only D 3 only
1 marks
Answer: D
31 The photograph shows a type of blood cell. Which statements about these cells are correct? 1 Oxygen diffuses through the phospholipid bilayer. 2 Sodium ions diffuse through the phospholipid bilayer. 3 Water passes in and out of these cells by osmosis. A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: C
15 Which substances can pass directly through cell surface membranes without using a carrier protein or channel protein? 1 Ca2+ and Na+ 2 O2 3 C6H12O6 A 1 and 2 B 1 and 3 C 2 and 3 D 2 only
1 marks
16 Which statements about endocytosis are correct? 1 It is a process requiring energy in the form of ATP. 2 Phagocytosis is a form of endocytosis. 3 Substances brought into a cell by endocytosis are enclosed in a small vacuole. A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
1 marks
15 Which substances can pass directly through cell surface membranes without using a carrier protein or channel protein? 1 CO2 2 Ca2+ and Na+ 3 H2O A 1 and 2 B 1 and 3 C 2 and 3 D 2 only
1 marks
Answer: B
16 Which statements about the proteins in cell surface membranes are correct? 1 They can be involved in active transport and facilitated diffusion. 2 They can be involved in antigen recognition. 3 They have hydrophilic R groups to interact with the inner portion of the membrane. A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: B
15 The following are all processes by which substances can enter a cell. 1 endocytosis 2 facilitated diffusion 3 osmosis Which processes are passive? A 1 and 2 B 1 and 3 C 2 and 3 D 3 only
1 marks
Answer: C
16 Equal volumes of five concentrations of sodium chloride solution were placed into five containers. An identical piece of plant tissue was placed into each container and left for 48 hours. The plant tissues were removed and the volumes of the sodium chloride solution were accurately measured. The results are shown below. 1.5 1.0 0.5 change in volume of 0 solution / cm3 0.20 0.20 0.20 0.40 0.40 0.40 0.60 0.60 0.60 0.80 0.80 0.80 1.00 1.00 1.00 –0.5 –1.0 –1.5 concentration of sodium chloride / mol dm–3 Which statements explain the results from 0.80 to 1.00 mol dm–3 sodium chloride? 1 There was no net movement of water into or out of the plant tissues. 2 The plant root tissues had a water potential of zero. 3 The plant tissues were fully plasmolysed. A 1 and 2 B 1 and 3 C 2 and 3 D 3 only
1 marks
Answer: D
17 The cells in the roots of beetroot plants contain a red pigment. When pieces of root tissue are soaked in cold water, some of the red pigment leaks out of the cells into the water. An experiment was carried out to investigate the effect of temperature on the loss of red pigment from the root cells. It was found that the higher the temperature of the water, the higher the rate of loss of red pigment from the root cells. Which of these statements could explain this trend? 1 Enzymes in the cells denature as the temperature increases, so the pigment can no longer be used for reactions inside the cells and diffuses out. 2 As the temperature increases, the tertiary structure of protein molecules in the cell surface membrane changes, increasing the permeability of the membrane. 3 Phospholipid molecules gain kinetic energy as temperature rises, increasing the fluidity of the phospholipid bilayer and allowing pigment molecules to diffuse out more easily. A 1 and 2 B 2 and 3 C 2 only D 3 only
1 marks
Answer: B
19 The diagram shows two pathways, X and Y, through which molecules can diffuse across a cell surface membrane. X Y Which row correctly shows possible pathways for lipids, water and glucose? lipids water glucose A X only X and Y Y only B X only Y only Y only C X and Y X only X and Y D X and Y X and Y X only
1 marks
Answer: A
17 What describes a carrier protein in cell surface membranes? A a glycoprotein that is found on the outer surface of the membranes allowing cell recognition B a glycoprotein that is involved in moving substances through the membranes by both active and passive transport C a protein that allows the attachment of signalling molecules which brings about changes within the cell D a protein that is involved in moving substances through the membranes by passive transport through water-filled pores
1 marks
Answer: B
18 What could happen to a typical bacterium when it is placed in surroundings which have a less negative water potential than that inside the cell? A The bacterium will burst because the cell wall has no structural function. B The bacterium will die since water leaves the cell by osmosis. C There is no change because the cell wall is impermeable to water. D There will be a net movement of water into the bacterium.
1 marks
Answer: D
19 By which process do hydrogencarbonate ions leave red blood cells? A active transport B endocytosis C facilitated diffusion D phagocytosis
1 marks
Answer: C
14 The diagram shows a plant cell. The plant cell is put into a solution with a water potential less negative than the cell contents. What will happen to the appearance of the cell? A B C D
1 marks
Answer: C
13 Which row is correct for facilitated diffusion of molecules or ions into a cell? A ATP required movement against the membrane protein concentration gradient required B ATP required movement down the membrane protein concentration gradient not required C ATP not required movement against the membrane protein concentration gradient not required D ATP not required movement down the membrane protein concentration gradient required
1 marks
Answer: D
14 The diagram represents an experiment where two solutions, P and Q, were separated by a partially permeable membrane. partially permeable membrane solution P solution Q What is correct about the initial movement of the molecules , and between the two solutions, P and Q? net movement net movement no net from Q to P from P to Q movement A B C D
1 marks
Answer: B
18 The graph shows how the rate of entry of substance X into a cell changes as the concentration of substance X outside the cell increases. rate of entry of substance X concentration of substance X outside the cell The diagram shows part of a cell surface membrane. 1 2 3 Which pathways could substance X use to enter the cell? A 1 and 2 B 1 only C 2 and 3 D 2 only
1 marks
Answer: A
17 Which of these processes allow movement in both directions across cell surface membranes? 1 active transport 2 diffusion 3 facilitated diffusion 4 osmosis A 1, 2, 3 and 4 B 1 and 4 only C 2 and 3 only D 2 only
1 marks
Answer: A
19 Which row describes osmosis across a cell surface membrane? moves molecule molecule uses energy down a concentration moved from ATP gradient A solute Jv Jv key B solute v x JY = correct Cc solvent x v X = incorrect D solvent x x
1 marks
Answer: C
18 Which of these substances can pass directly through cell surface membranes without using a carrier protein or a channel protein? 1 Ca2+ 2 CO2 3 C6H12O6 A 1 and 2 B 1 and 3 C 2 and 3 D 2 only
1 marks
Answer: D
19 Companion cells use ATP to move hydrogen ions out of the cell and co-transporter proteins to allow hydrogen ions to return with sucrose molecules. Which two processes are involved in this movement? A active transport and diffusion B active transport and facilitated diffusion C exocytosis and diffusion D exocytosis and facilitated diffusion
1 marks
Answer: B
14 Which statements about the cell surface membrane are correct? 1 Channel proteins allow water soluble ions and molecules across the membrane. 2 Glucose can pass into the cell via carrier proteins. 3 Oxygen passes freely through the membrane as it is soluble in lipids. 4 Some glycoproteins act as antigens. A 1, 2, 3 and 4 B 1, 3 and 4 only C 1 and 2 only D 2, 3 and 4 only
1 marks
Answer: A
15 Which of these features increase the efficiency of ion uptake by a root hair cell? 1 many mitochondria in the cell 2 high concentration of ions in the vacuole 3 protein carriers in the cell surface membrane A 1, 2 and 3 B 1 and 3 only C 1 only D 2 and 3 only
1 marks
Answer: B
17 The diagrams show the shape and size of two types of cell. 50 μm 45 μm 10 μm 5 μm 10 μm 10 μm palisade mesophyll cell columnar epithelial cell surface area = 2200 μm2 volume = 2250 μm3 Which statement is correct about the palisade cell and epithelial cell shown in the diagrams? A An increase in surface area reduces the distance for gases to reach the centre of the cell. B The surface area of the palisade mesophyll cell is 500 µm2 greater than the columnar epithelial cell. C The surface area to volume ratio is greater in the columnar epithelial cell than the palisade mesophyll cell. D The volume of the palisade mesophyll cell is 2500 µm3 greater than that of the columnar epithelial cell.
1 marks
Answer: C
15 The formula shows how the rate of diffusion across a cell surface membrane can be calculated. surface area × difference in concentrat ion thickness of membrane Which row shows how the fastest rate of diffusion can be achieved? surface area difference in thickness of available concentration membrane A high high high B high high low C low high low D low low high
1 marks
Answer: B
16 The diagram shows part of the cell surface membrane of an active animal cell. more negative water potential X L M less negative water potential Y Which statements correctly describe the net movement of molecules across this membrane? 1 Oxygen diffuses through molecules M from X to Y. 2 Carbon dioxide diffuses through molecules M from X to Y. 3 Water moves from Y to X through molecule L. A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: C
17 Some epidermal cells from a well-watered plant are placed in three solutions which have different water potentials. Which row correctly shows the state of the plant cells in each of the solutions? water potential of solution lower than cells equal to cells higher than cells A plasmolysed turgid plasmolysed B plasmolysed turgid turgid C turgid plasmolysed plasmolysed D turgid plasmolysed turgid
1 marks
Answer: B
17 Which processes use energy in the form of ATP? 1 endocytosis 2 exocytosis 3 facilitated diffusion A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: B
18 The diagram shows a partially plasmolysed plant cell. solution X Z solution Y What is found at Z? A air B solution X C solution Y D water
1 marks
Answer: C
16 The diagram shows the transport of ions across the cell surface membrane. Inside the cell there is a low concentration of sodium ions (Na+) and a high concentration of potassium ions (K+). Outside the cell there is a low concentration of K+ and a high concentration of Na+. The carrier molecule is a pump which exchanges Na+ for K+. inside cell 2 1 cell surface membrane 4 carrier 3 molecule outside cell Which ionic movements are represented by the arrows? active transport active transport diffusion diffusion of K+ of Na+ of Na+ of K+ A 2 3 1 4 B 2 3 4 1 C 3 2 1 4 D 3 2 4 1
1 marks
Answer: A
17 The indicator cresol red changes from red to yellow when put into an acid. Some blocks of agar containing cresol red were cut to different sizes and put in an acid. All other variables were kept constant. The blocks were measured in mm. Which block became completely yellow most quickly? A 3 × 30 × 30 B 6 × 6 × 6 C 6 × 12 × 12 D 12 × 12 × 12
1 marks
Answer: A
18 When red blood cells are put into pure water they burst (haemolysis). Which statements explain this haemolysis? 1 The water potential of the surrounding liquid is lower than the water potential of the contents of the red blood cell. 2 The cell surface membranes of red blood cells are not supported by cell walls. 3 More water moves into the red blood cells by osmosis than leaves the cells. 4 Water enters the red blood cells by osmosis but does not leave the cells. A 1 and 3 B 1 and 4 C 2 and 3 D 2 and 4
1 marks
Answer: C
17 How could water molecules cross the cell surface membrane of animal cells? 1 carrier proteins 2 channel proteins 3 cholesterol molecules A 1 and 2 B 1 and 3 C 2 and 3 D 2 only
1 marks
Answer: D
19 Agar cubes can be used to demonstrate the effect of changing surface area to volume ratio on diffusion. Three different agar cubes made using a dilute acid were placed into an indicator solution that diffused into the cubes. When the indicator came into contact with the acid it changed colour. The cubes were 1 cm3, 2 cm3 and 3 cm3 and were left in the indicator solution for 10 minutes. All other variables were kept the same. The results were recorded as diagrams. The results for the 2 cm3 cube are shown. cube changed colour not to original colour to a depth of 0.5 cm scale Which diagrams represent the results for the 1 cm3 and the 3 cm3 cubes? not to scale cube changed colour cube changed colour A completely to a depth of 0.5 cm 1 cm3 3 cm3 cube changed colour cube changed colour B completely to a depth of 1 cm 1 cm3 3 cm3 cube changed colour cube changed colour C to a depth of 0.25 cm to a depth of 0.5 cm 1 cm3 3 cm3 cube changed colour cube changed colour D to a depth of 0.25 cm to a depth of 1 cm 1 cm3 3 cm3
1 marks
Answer: A
18 Which statement describes endocytosis? A movement across a membrane against the concentration gradient and requiring energy B movement across a membrane down the concentration gradient using a carrier molecule C movement across a membrane into a cell using a vesicle and requiring energy D movement across a membrane using a vesicle and requiring no energy
1 marks
Answer: C
19 An indicator is colourless in acid and pink in alkali. In an experiment a petri dish of agar was prepared using an acidic solution of this indicator. A disc of agar 1 cm in diameter was removed from the centre to create a well. A white card showing circular marker lines 1 cm apart was placed underneath the petri dish. 1 cm3 alkali solution was put into the well in the agar and a stop-watch was started. A circular disc of pink colour appeared and spread through the agar. It reached the first marker line in a short time but took longer to reach the second marker line and a very long time to reach the third marker line. What explains these observations? A facilitated diffusion of alkali solution B facilitated diffusion of the indicator C simple diffusion of alkali solution D simple diffusion of the indicator
1 marks
Answer: C
17 Which statements about the movement of water in and out of cells are correct? 1 Water moves from regions of more negative water potential to regions of less negative water potential. 2 Water can cross cell membranes by passing through channel proteins. 3 Water can pass through cellulose cell walls. A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: D
15 The graph shows how the rate of entry of glucose into a cell changes as the concentration of glucose outside the cell changes. X rate of entry of glucose into the cell concentration of glucose outside the cell What is the cause of the plateau at X? A All the carrier proteins are saturated with glucose. B The carrier proteins are denatured and no longer able to function. C The cell has used up its supply of ATP. D The concentrations of glucose inside and outside the cell are equal.
1 marks
Answer: A
16 The diagram shows the movement of substance Z across a cell surface membrane. Z outside cell membrane Z Z inside cell time Which process is involved in this movement? A endocytosis B exocytosis C phagocytosis D pinocytosis
1 marks
Answer: B
17 Visking tubing is often used as a model during experiments to investigate osmosis in plants. What could Visking tubing be used to represent? cell surface cell wall tonoplast membrane v v v v v x x v key / = represents X = does not represent
1 marks
Answer: B
18 The cell surface membranes of some cells are largely made up of phospholipids and cholesterol, with few proteins. Which transport mechanisms will be reduced across these membranes? 1 facilitated diffusion 2 active transport 3 diffusion A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: B
20 A student weighed a cylinder of potato and then put it into a test-tube containing a salt solution. The potato cylinder was removed from the salt solution after one hour. It was blotted dry and then reweighed. The student recorded that the potato had lost mass. Which row shows the correct explanation for the results the student collected? water potential condition of the of the potato cells potato cells before soaking after soaking A higher plasmolysed B higher turgid C lower plasmolysed D lower turgid
1 marks
Answer: A
30 Irrigating crop plants with water containing low concentrations of salt causes an increase in the concentration of salt in the soil. What explains why the increase in salt concentration eventually kills the crop? water potential water potential direction of water in roots in soil movement A ↓ into the roots key B ↑ into the soil solution ↑ = water potential increases C ↓ out of the roots ↓ = water potential decreases D ↑ into the roots
1 marks
Answer: C
15 Which statement suggests that a membrane protein is involved in active transport? A It allows movement of molecules across a membrane if concentration differences exist. B It can only function if mitochondria are supplied with sufficient oxygen. C It has a tertiary structure with a binding site with a specific shape. D It is found in the cell surface membranes and the mitochondrial membranes.
1 marks
Answer: B
17 Which substances can pass directly through cell surface membranes and do not use a carrier protein or channel protein? 1 K+ and Cl – 2 CO2 3 C6H12O6 A 1 and 2 B 1 and 3 C 2 and 3 D 2 only
1 marks
Answer: D
18 A student put a layer of plant epidermal cells on a microscope slide. The student put a drop of potassium nitrate solution on the layer of cells and observed that: • the cell surface membrane of many of the cells had separated from the cell wall • the cytoplasm and cell contents had shrunk. What explains these observations? direction of net water water potential of cells water potential of movement at start / kPa solution at start / kPa A cells to solution –100 –500 B cells to solution –500 –100 C solution to cells –100 –500 D solution to cells –500 –100
1 marks
Answer: A
18 The photomicrograph shows the appearance of onion epidermal cells after they have been soaked in solution X for one hour. Y What fills the space labelled Y? A air B cytoplasm C solution X D water
1 marks
Answer: C
19 Equal sized potato pieces were placed into a test-tube and covered with a sucrose solution. The test tube was left for 30 minutes. All other variables were controlled. After 30 minutes, the potato piece had not changed in size. What can be concluded from this result? A The concentration of sucrose is the same in the potato and in the solution and there is no more movement of water into or out of the potato. B The concentration of sucrose is the same in the potato and in the solution and there is no net movement of water into the potato. C The water potential is the same in the potato and in the sucrose solution and there is no more movement of water into or out of the potato. D The water potential is the same in the potato and in the sucrose solution and there is no net movement of water into or out of the potato.
1 marks
Answer: D
15 Which pair of factors is inversely proportional to the rate of diffusion? A concentration gradient and surface area over which diffusion occurs B distance over which diffusion occurs and size of diffusing molecule C size of diffusing molecule and concentration gradient D surface area over which diffusion occurs and distance over which diffusion occurs
1 marks
Answer: B
16 Raisins are dried fruit that contain high concentrations of sugar. Which row is correct when raisins are first put into water? water potential in the raisin direction of water movement compared to surrounding water A less negative into the raisin B less negative out of the raisin C more negative into the raisin D more negative out of the raisin
1 marks
Answer: C
15 Which description of cell surface membrane permeability is correct? A An increase in the concentration of cholesterol molecules in the cell surface membrane can increase its permeability to hydrophilic substances. B Cell surface membrane permeability to large hydrophilic molecules is high and can be increased by membrane transport proteins involved in facilitated diffusion. C The permeability of the cell surface membrane to ions increases as the proportion of saturated fatty acid chains in the phospholipids increases. D Without the presence of carrier and channel membrane proteins, the cell surface membrane has a low permeability to large polar molecules.
1 marks
Answer: D
16 Which transport mechanism within a cell can occur in the absence of membranes? A active transport B diffusion C facilitated diffusion D osmosis
1 marks
Answer: B
17 A student measured the time taken for complete diffusion of a dye into agar blocks of different sizes. The results are shown in the table. size of agar block time for / mm × mm × mm diffusion / s 5 × 5 × 5 6.2 10 × 10 × 10 16.1 15 × 15 × 15 34.5 5 × 10 × 15 What is the predicted time for complete diffusion of the dye into the agar block measuring 5 mm × 10 mm × 15 mm? A 6.2 s B 16.1 s C 34.5 s D more than 34.5 s
1 marks
Answer: A
18 A plant cell with a water potential of –600 kPa was placed in a solution with a water potential of –410 kPa for 10 minutes. Which row is correct? net movement water potential effect on cell of water of cell A into cell becomes higher becomes turgid B into cell becomes lower bursts C out of cell becomes higher swells D out of cell becomes lower becomes plasmolysed
1 marks
Answer: A
15 The diagram shows a cell surface membrane. 2 3 1 4 Which statements about the labelled molecules in the membrane are correct? ● 1 is involved in the diffusion of ions. ● 2 is involved in facilitated diffusion. ● 3 is involved in the recognition of antigens. ● 4 is involved in membrane fluidity. A 1, 2 and 3 B 1 and 3 only C 1 and 4 D 2 and 4 only
1 marks
Answer: D
16 Equal sized potato pieces were placed into test-tubes containing equal volumes of different concentrations of sucrose solution and left for 30 minutes. All other variables were controlled. After 30 minutes, the potato piece in one of the concentrations of sucrose solution had not changed in size. What can be concluded from this result? 1 There is no net movement of water into or out of the potato. 2 The water potential of the potato is the same as the water potential of the sucrose solution. 3 The concentration of sucrose in the potato is the same as the concentration of the sucrose solution. A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 only
1 marks
Answer: B
18 What are the features of facilitated diffusion? 1 It uses protein channels in the membrane and is driven by the energy from ATP. 2 It moves molecules from regions of higher concentration to lower concentration and is driven by the kinetic energy of the molecules which are diffusing. 3 It uses protein channels in the membrane, and the maximum rate of diffusion depends on the number of these channels. A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: D
15 Blocks of agar are stained with a pH indicator and used to investigate the diffusion of an acid solution. Which block would completely change colour the fastest? A 1mm × 1mm × 1 mm B 1mm × 0.25mm × 0.25 mm C 1mm × 0.5mm × 0.5 mm D 2mm × 0.5mm × 0.5 mm
1 marks
Answer: B
16 The diagram shows the water potential of three adjacent plant cells, P, Q and R. P –198 kPa Q –224 kPa R –212 kPa Which shows the net movement of water between cells P, Q and R? A P → Q and P → R and R → Q B P → Q and P → R C Q → P and Q → R and R → P D Q → P and R → P
1 marks
Answer: A
16 Which processes can allow transport into or out of a cell? 1 active transport 2 facilitated diffusion 3 osmosis A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: A
17 Plant cells were put into one of three different concentrations of sugar solution, 10%, 5% and 2.5%. The cells were left for 50 minutes and then observed using a light microscope. cell X cell Y cell Z vacuole Which statements are correct? 1 Cell Y had a lower water potential than the sugar solution it was put into. 2 Cell Z was put into the 10% sugar solution. 3 Cell Z had a less negative water potential than the sugar solution it was put into. A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: A
16 Solute X is at a higher concentration outside a cell than in the cytoplasm. Which processes may allow solute X to be moved through the cell surface membrane? A active transport, diffusion and facilitated diffusion B active transport, diffusion and osmosis C diffusion, facilitated diffusion and osmosis D exocytosis, facilitated diffusion and osmosis
1 marks
Answer: A
15 Diagram 1 and diagram 2 show how the transverse section through a leaf changes when the leaf is moved from solution X to solution Y. leaf solution X solution Y diagram 1 diagram 2 Which row describes the water potential of the leaf cells and surrounding solutions in diagram 2 compared with diagram 1? water potential of leaf cells water potential of solution Y in diagram 2 compared with compared with water water potential of leaf cells potential of solution X in diagram 1 A less negative less negative B less negative more negative C more negative less negative D more negative more negative
1 marks
Answer: A
16 Which features must always be present for water to move between two solutions by osmosis? 1 carrier proteins 2 cell surface membrane 3 selectively permeable membrane 4 water potential gradient A 1, 2 and 3 B 1, 3 and 4 C 2 and 4 D 3 and 4 only
1 marks
Answer: D
17 When living pancreatic cells were placed in a solution of a red stain called neutral red, the cytoplasm became red. The cells were then removed from the solution of neutral red. The red stain in the cytoplasm moved into vesicles, which were exported from the cell, eventually leaving the cell colourless. Which transport mechanisms could explain how the red stain entered and left the cells? A active transport and facilitated diffusion B diffusion and exocytosis C facilitated diffusion and endocytosis D osmosis and exocytosis
1 marks
Answer: B
26 The diagram shows a xerophytic leaf in different conditions, P and Q. Y Y P Q Which statements describe the difference between the cells in layer Y in conditions P and Q? 1 more negative water potential in P than Q 2 more cells plasmolysed in P 3 cells less turgid in Q 4 water potential becomes zero in Q A 1, 2 and 3 B 1 and 2 only C 2 and 4 D 3 and 4
1 marks
Answer: B
14 Which roles of the cell surface membrane result from the properties of the phospholipids? 1 to allow cytokinesis to occur in mitotic cell division 2 to allow entry and exit of oxygen and carbon dioxide 3 to allow the phagocytosis of a bacterium into a cell A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: A
15 Which factors can be changed to affect the rate of facilitated diffusion across a cell surface membrane? 1 the surface area of the membrane 2 the concentration gradient 3 the number of specific protein channels A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 3 only
1 marks
Answer: A
16 The diagram shows an experiment using a model cell to investigate the movement of substances. membrane permeable model cell to monosaccharides and water solution containing 0.03 mol dm–3 sucrose 0.02 mol dm–3 glucose solution containing 0.01 mol dm–3 sucrose 0.01 mol dm–3 glucose Which statements are correct? 1 There is net movement of sucrose out of the model cell. 2 There is net movement of glucose out of the model cell. 3 There is net movement of water into the model cell. A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: D
34 How many times must a molecule of carbon dioxide pass through a cell surface membrane as it diffuses from the plasma, through a cell in the capillary wall, into an alveolus? A 1 B 2 C 3 D 4
1 marks
Answer: D
16 An indicator mixed with agar forms a pink colour. The pink-coloured agar becomes colourless when put in acid. Blocks of pink-coloured agar are cut to different sizes and put in acid. All other variables are kept constant. Which block becomes colourless most quickly? A 3 mm 30 mm 30 mm B 6 mm 6 mm 6 mm C 6 mm 12 mm 12 mm D 12 mm 12 mm 12 mm
1 marks
Answer: A
17 Three identical plant cells were put into one of three different concentrations of sugar solution, 10%, 5% and 2.5%. The cells were left for 50 minutes and then observed using a light microscope. cell X cell Y cell Z vacuole Which statement is not correct? A Cell X has the same water potential as the sugar solution it was put into. B Cell Y is turgid and cell Z is plasmolysed. C Cell Y was put into the 2.5% sugar solution. D Cell Z had a more negative water potential than the sugar solution it was put into.
1 marks
Answer: D
15 Some enzymes are produced in the cells of the pancreas. The enzymes are secreted when required. Which process is used to transport these enzymes out of the cells of the pancreas? A active transport B facilitated diffusion C endocytosis D exocytosis
1 marks
Answer: D
16 Plant cells were submerged in a solution with a water potential less negative than that found inside the cells. What describes the condition of the plant cells after 20 minutes? A burst B incipient plasmolysis C plasmolysed D turgid
1 marks
Answer: D
23 An antibiotic enters bacterial cells through a membrane channel protein, P. Some bacterial cells have shown resistance to this antibiotic by acquiring a mutation which alters P. This mutation prevents the entry of the antibiotic into the cell. Which conclusions can be drawn about how resistance to this antibiotic developed in these bacteria? 1 The mutation changed the order of the amino acids in the gene coding for P. 2 The mutation resulted in the production of P with an altered tertiary structure. 3 The antibiotic is a hydrophobic molecule and so cannot cross the phospholipid bilayer to enter the cell. A 1, 2 and 3 B 1 and 3 only C 2 and 3 only D 2 only
1 marks
Answer: D
16 Liver cells contain vesicles that have proteins in their membranes which are specific for the transport of glucose. When these cells need to take up glucose, the vesicles fuse with the cell surface membrane. How does the uptake of glucose occur? A exocytosis B diffusion C endocytosis D facilitated diffusion
1 marks
Answer: D
17 A student set up an experiment to investigate diffusion. A block of agar, 1.0 cm 1.0 cm 1.0 cm, was stained uniformly with a water-soluble blue dye. The block of agar was put into a test-tube containing 10 cm3 of distilled water at 20 C. The intensity of the blue colour of the water after five minutes was measured. Four other experiments, A, B, C and D, were then carried out using different numbers of agar blocks, different sizes of agar blocks and different temperatures. All other variables were standardised. Which experiment would give a lighter blue colour in the water after five minutes compared to the first experiment? size of each number of temperature of agar block / cm agar blocks distilled water / C A 0.5 0.5 0.5 8 30 B 0.5 0.5 0.5 2 20 C 0.5 1.0 1.0 2 30 D 2.0 1.0 1.0 1 20
1 marks
Answer: B
16 The diagram shows a partially plasmolysed plant cell. solution X Z solution Y What is found at Z? A air B solution X C solution Y D water
1 marks
Answer: C
17 A single-celled organism lives in freshwater. Water that enters the cytoplasm of the cell by osmosis is collected into a structure called the contractile vacuole. To remove the water the contractile vacuole fuses with the cell surface membrane. A student counted the number of times that the contractile vacuole filled and emptied when the cell was placed in solutions with different water potentials. The results are shown in the table. water potential of rate of contractile external solution vacuole emptying / kPa / min–1 0 31 –100 20 –200 13 –300 8 –400 6 –500 0 Which statement explains the pattern observed as the water potential of the external solution decreased? A The water potential gradient between the cell and the solution increased, causing water to move into the cell more rapidly and the contractile vacuole to empty more frequently. B The water potential gradient between the cell and the solution increased, causing water to move into the cell less rapidly and the contractile vacuole to empty less frequently. C The water potential gradient between the cell and the solution decreased, causing water to move into the cell more rapidly and the contractile vacuole to empty more frequently. D The water potential gradient between the cell and the solution decreased, causing water to move into the cell less rapidly and the contractile vacuole to empty less frequently.
1 marks
Answer: D
17 Which transport mechanism does not require a concentration gradient to be present in order to take place? A exocytosis B facilitated diffusion C osmosis D transpiration
1 marks
Answer: A
19 In an investigation, a plant cell was placed in pure water. The initial rate at which water molecules entered the cell, R, was greater than the initial rate at which water molecules left the cell. In a second investigation, a plant cell of the same type was placed in a solution with a water potential equal to that of the cell contents. What will happen in the second investigation over a period of five minutes? A Water molecules will not enter or leave the cell because the water potential of the cell contents is equal to that of the solution. B Water molecules will enter and leave the cell in equal amounts, both at an initial rate that is less than R in the first investigation. C Water molecules will enter and leave the cell in equal amounts, both at an initial rate that is greater than R in the first investigation. D Water molecules will enter and leave the cell in equal amounts, both at an initial rate that is equal to R in the first investigation.
1 marks
Answer: B
15 Some processes occurring in cells are listed. 1 endocytosis of water into cells 2 exocytosis of enzymes from cells 3 facilitated diffusion of glucose into red blood cells 4 phagocytosis of dead cells by macrophages Which processes use ATP? A 1, 2 and 3 B 1, 2 and 4 C 1, 3 and 4 D 2, 3 and 4
1 marks
Answer: B
16 The graph shows changes in the concentration of a solute inside a cell. concentration of solute inside cell 0 0 time What explains this change in concentration? 1 diffusion 2 endocytosis 3 exocytosis 4 osmosis A 1, 2 and 3 B 1, 3 and 4 C 1 and 4 only D 2 and 4
1 marks
Answer: B
17 The indicator cresol red, changes from red to yellow when put into acid. Four blocks of agar containing cresol red were cut to different sizes measured in millimetres. The blocks were submerged in acid. All other variables were kept constant. The time taken for each of the blocks to completely turn yellow was recorded. Which of the four blocks became completely yellow most quickly? A 3 30 30 B 6 6 6 C 6 12 12 D 12 12 12
1 marks
Answer: A
14 When animal cells are cultured, salt solution is added to keep the cells alive. What is the purpose of the salt solution? A to allow facilitated diffusion of salts into the cells B to prevent diffusion of other ions in or out of the cells C to prevent net movement of water into or out of the cells D to provide a source of energy for active transport
1 marks
Answer: C
15 The following are all processes that allow movement into cells. 1 phagocytosis 2 active transport 3 facilitated diffusion Which processes require ATP? A 1 and 2 B 2 and 3 C 1 only D 2 only
1 marks
Answer: A
16 Which features are required to allow for efficient diffusion? 1 a large surface area 2 a short diffusion pathway 3 maintenance of a constant diffusion gradient A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: A
14 Batrachotoxin is a poison found in frogs in the Colombian jungle. The poison is used to produce poison darts. The poison works by increasing the permeability of the cell surface membrane of nerve and muscle cells to sodium ions, which move out of the cells. Four students made statements about how the poison affects the cells. 1 Water leaves the cells by osmosis, causing the cells to shrink. 2 Water enters the cells by osmosis, causing the cells to burst. 3 When the sodium ions move out of the cells the intracellular fluid has a more positive water potential than the extracellular fluid. 4 When the sodium ions move out of the cells the extracellular fluid has a more positive water potential than the intracellular fluid. Which statements are correct for the cells affected by batrachotoxin? A 1 and 3 B 1 and 4 C 2 and 3 D 2 and 4
1 marks
Answer: A
15 Which processes use energy in the form of ATP? 1 endocytosis 2 exocytosis 3 facilitated diffusion A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: B
16 An indicator is colourless in acid and pink in alkali. In an experiment a petri dish of agar was prepared using an acidic solution of this indicator. A disc of agar 1 cm in diameter was removed from the centre to create a well. A white card showing circular marker lines 1 cm apart was placed underneath the petri dish. 1 cm3 alkali solution was put into the well in the agar and a stop-watch was started. A circular disc of pink colour appeared and spread through the agar. It reached the first marker line in a short time but took longer to reach the second marker line and a very long time to reach the third marker line. What explains these observations? A facilitated diffusion of alkali solution B facilitated diffusion of the indicator C simple diffusion of alkali solution D simple diffusion of the indicator
1 marks
Answer: C
16 Which statement is correct for facilitated diffusion and active transport? A As the direction of the concentration gradient changes so does the direction of movement of the molecules. B Molecules always move at the same rate as simple diffusion. C Specific molecules are transported across a membrane. D The molecule ATP is required to move specific molecules quickly through proteins in the membrane.
1 marks
Answer: C
17 Equal sized potato pieces were placed into a test-tube and covered with a sucrose solution. The test-tube was left for 30 minutes. All other variables were standardised. After 30 minutes, the potato piece had not changed in size. What can be concluded from this result? A The concentration of sucrose is the same in the potato and in the solution and there is no more movement of water into or out of the potato. B The concentration of sucrose is the same in the potato and in the solution and there is no net movement of water into the potato. C The water potential is the same in the potato and in the sucrose solution and there is no more movement of water into or out of the potato. D The water potential is the same in the potato and in the sucrose solution and there is no net movement of water into or out of the potato.
1 marks
Answer: D
16 The diagram represents a process by which molecules move out of a cell through a cell surface membrane. Which process does this represent? A exocytosis B diffusion C facilitated diffusion D osmosis
1 marks
Answer: C
17 The photomicrograph shows the appearance of onion epidermal cells after they have been soaked in solution X for one hour. Y What fills the space labelled Y? A air B cytoplasm C solution X D water
1 marks
Answer: C
27 Sodium chloride is added to a culture solution containing freshwater single-celled plant cells. What happens to the water potential of the culture solution when the sodium chloride is added and what may happen to the plant cells after 5 minutes? water potential of the single-celled plant cells culture solution when after 5 minutes sodium chloride added A becomes less negative become plasmolysed B becomes less negative become turgid C becomes more negative become plasmolysed D becomes more negative become turgid
1 marks
Answer: C
15 Which features are correct for active transport and facilitated diffusion? 1 The movement of molecules and ions depends on ATP. 2 They are specific for one type of molecule or ion. 3 They use membrane proteins. A 1 and 2 B 1 and 3 C 2 and 3 D 3 only
1 marks
Answer: C
16 The diagram represents two solutions, P and Q, that were separated by a partially permeable dialysis tubing. partially permeable dialysis tubing solution P solution Q key molecule 1 molecule 2 molecule 3 partially permeable dialysis tubing What will be the initial movement of the molecules, 1, 2 and 3, between solution P and solution Q? net movement net movement no net from Q to P from P to Q movement A molecule 1 molecule 2 molecule 3 B molecule 1 molecule 3 molecule 2 C molecule 2 molecule 3 molecule 1 D molecule 3 molecule 1 molecule 2
1 marks
Answer: B
17 A student weighed a cylinder of potato and then put it into a test-tube containing a salt solution. The potato cylinder was removed from the salt solution after one hour. It was blotted dry and then reweighed. The student recorded that the potato had lost mass. Which row shows the correct explanation for the results the student collected? water potential of the condition of the potato cells before soaking potato cells compared to the after soaking water potential after soaking A higher plasmolysed B higher turgid C lower plasmolysed D lower turgid
1 marks
Answer: A
38 Certain bacteria develop resistance to an antibiotic by actively transporting the antibiotic out of the cell. This prevents the antibiotic building up to toxic concentrations. What must be found in a bacterial cell to allow it to develop this form of antibiotic resistance? gene coding for mitochondria a channel protein A B C D source of chemical energy
1 marks
Answer: D
5 Dialysis (Visking) tubing is an artificial partially permeable membrane with pore sizes of approximately 2.5 nm. Glucose molecules have a diameter of about 1.5 nm and can pass through the pores in the membrane. What else can pass through the pores? 1 bacteria 2 haemoglobin 3 ribosomes 4 fructose A 1 and 3 B 2 and 4 C 2 only D 4 only
1 marks
Answer: D
17 Which of these substances can pass directly through cell surface membranes without using a carrier protein or a channel protein? 1 Ca2+ 2 CO2 3 C6H12O6 A 1 and 2 B 1 and 3 C 2 and 3 D 2 only
1 marks
Answer: D
15 Four students, A, B, C and D, observed plant epidermal cells that had been placed in a concentrated sucrose solution for 30 minutes. They were asked to identify the partially permeable layer and to explain the appearance of the cells in terms of water potential and movement of water. Which student is correct? partially water potential movement of water permeable layer at start of experiment during experiment A cell surface cell contents have a lower water moved out of the cell membrane water potential than the and no water moved in sucrose solution B cell surface cell contents have a higher more water moved out membrane water potential than the of the cell than moved in sucrose solution C cell wall cell contents have a lower more water moved out water potential than the of the cell than moved in sucrose solution D cell wall cell contents have a higher water moved out of the cell water potential than the and no water moved in sucrose solution
1 marks
Answer: B
16 The table compares the surface area to volume ratios of five agar blocks that differ in dimensions but which all have the same volume. The agar blocks can be used to measure the efficiency of diffusion, where efficiency is measured as the time taken for a dye to reach all parts of the block. length width height surface volume surface area : / mm / mm / mm area / mm2 / mm3 volume ratio 1 8 8 8 384 512 0.75 2 16 16 2 640 512 1.3 3 32 4 4 544 512 1.1 4 32 32 0.5 2112 512 4.1 5 64 4 2 784 512 1.5 Which prediction can be made about the way in which size and dimensions of these blocks affect the efficiency of diffusion? A The efficiency of diffusion will decrease as the width of a block increases. B The efficiency of diffusion will increase as the height of a block increases. C The efficiency of diffusion will increase as a block of fixed volume is flattened. D The efficiency of diffusion will decrease as a block of fixed volume is elongated.
1 marks
Answer: C
17 The three main factors that affect the rate of diffusion across a membrane can be expressed by the relationship shown. surface area × concentration difference rate of diffusion is proportional to thickness of membrane Which changes in the factors would result in the rate of diffusion doubling? 1 Surface area has doubled. 2 Concentration difference has halved. 3 Thickness of membrane has doubled. 4 Thickness of membrane has halved. A 1, 2 and 4 B 1 and 3 C 1 and 4 only D 2 and 3
1 marks
Answer: C
18 A student measured the time taken for complete diffusion of a dye into agar blocks of different sizes which were suspended in the dye. The results are shown. size of agar block time for / mm × mm × mm diffusion / s 5 × 5 × 5 6.2 10 × 10 × 10 16.1 15 × 15 × 15 34.5 5 × 10 × 15 What is the predicted time for complete diffusion of the dye into the agar block measuring 5 mm × 10 mm × 15 mm? A 6.2 s B 16.1 s C 34.5 s D more than 34.5 s
1 marks
Answer: A
19 An experiment was carried out to investigate the effect of concentration of sucrose solution on cells in a plant tissue. A sample of plant tissue was cut into seven cylinders of equal length and diameter. The mass of each cylinder was recorded. Each of the seven cylinders was put into a different sucrose solution concentration. bung test-tube sucrose solution plant tissue cylinder After two hours, the cylinders were removed, blotted dry and reweighed. The percentage change in mass of each cylinder was recorded. The graph shows the results of this investigation. 20 15 10 5 percentage change in mass of plant 0 tissue cylinder 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 –5 –10 –15 –20 concentration of sucrose solution / mol dm–3 Which row explains the results if plant tissue cells were put in a sucrose solution of 0.45 mol dm–3? water potential of the cytoplasm change in volume of the of the cells at the start of the vacuoles of the cells at the end experiment compared with the of the experiment, that were water potential of 0.45 mol dm–3 initially placed in 0.45 mol dm–3 sucrose solution sucrose solution A less negative decreased B less negative increased C more negative decreased D more negative increased
1 marks
Answer: A
18 Which row correctly describes all the possible relative concentrations of a substance when the substance is moved by endocytosis or exocytosis? endocytosis exocytosis A concentrations equal concentrations equal B concentrations equal, concentrations equal, greater inside or greater inside or greater outside greater outside C concentrations equal concentrations equal or greater outside or lower outside D concentrations equal concentrations equal or lower outside or greater outside
1 marks
Answer: B
19 Which statement about simple diffusion is correct? A It requires specific molecules in the cell surface membrane. B It is a passive mode of transporting substances. C It always requires a membrane for transport of substances. D It only happens in the cells of prokaryotes and eukaryotes.
1 marks
Answer: B
20 Plant cells with the same water potential in their cytoplasm were each put into one of three different concentrations of sugar solution, 10%, 5% and 2.5%. The cells were left for 50 minutes and then observed using a light microscope. cell X cell Y cell Z vacuole Which statements are correct? 1 Cell Y had a lower water potential than the sugar solution it was put into. 2 Cell Z was put into the 10% sugar solution. 3 Cell Z had a less negative water potential than the sugar solution it was put into. A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: A
16 Plant cells were left for 50 minutes in three different sugar solutions, 10%, 5% and 1%. The water potential in the cytoplasm of the three cells was the same at the start of the experiment. The diagrams show the appearance of the cells after 50 minutes, using a light microscope. cell P cell Q cell R Which conclusion is correct? A Cell P has the same concentration of sugar inside and outside the cell. B Cell Q is flaccid and cell P is plasmolysed. C Cell Q was placed in the 1% solution. D The sugar solution outside cell R has a less negative water potential than inside cell R.
1 marks
Answer: C
17 The graph shows the effect of increasing the side length of agar cubes on the surface area and the volume of the cubes. surface 1 area or volume 2 0 0 side length Which row correctly identifies line 1, line 2 and the effect of increasing side length on the surface area : volume ratio of the cubes? surface area : line 1 line 2 volume ratio A surface area volume decreases B surface area volume increases C volume surface area decreases D volume surface area increases
1 marks
Answer: C
18 The statements are comparisons of endocytosis and exocytosis. ● Both are mechanisms that involve vesicles or vacuoles and the transport of materials across the cell surface membrane. ● Both mechanisms occur to allow bulk transport across the cell surface membrane. ● Endocytosis involves taking materials into the cell, whereas exocytosis involves the release of materials from the cell. ● Some of the cell surface membrane is lost when endocytosis occurs and there is an increase in the cell surface membrane when exocytosis occurs. How many statements are correct? A 1 B 2 C 3 D 4
1 marks
Answer: D
19 A student was asked to calculate the surface area : volume ratio for an agar cube with a side length of 5.5 mm. Which surface area : volume ratio is correct? A 0.2 : 1 B 0.9 : 1 C 1.0 : 1 D 1.1 : 1
1 marks
Answer: D
17 Sodium ions can enter cells across the cell surface membrane. Which methods could be used by sodium ions to cross a cell surface membrane and enter a cell? A active transport only B active transport and facilitated diffusion C facilitated diffusion and simple diffusion D simple diffusion only
1 marks
Answer: B
18 The diagram shows how an artificial partially permeable membrane was used to separate a 5% sodium chloride solution and a 10% sodium chloride solution in a beaker. The two sides of the beaker were labelled R and S. partially permeable membrane R S 5% 10% sodium chloride sodium chloride solution solution Which row correctly describes and explains what will happen in the half of the beaker labelled S? description of S explanation A volume of solution increases net movement of water from a higher water potential to a lower water potential B volume of solution increases net movement of water from a lower water potential to a higher water potential C volume of solution net movement of water from a higher water decreases potential to a lower water potential D volume of solution net movement of water from a lower water decreases potential to a higher water potential
1 marks
Answer: A
19 Agar cubes can be used to demonstrate the effect on diffusion of changing the surface area to volume ratio. Three different agar cubes made using a coloured indicator solution were placed into a dilute acid that diffused into the cubes. As the acid diffused into the agar cubes, the colour of the indicator solution changed. The cubes had volumes of 1 cm3, 2 cm3 and 3 cm3 and were left in the dilute acid for 10 minutes. All other variables were kept the same. After 10 minutes, the agar cubes were removed from the dilute acid and cut in half. The cut surfaces were observed and the results were recorded as diagrams. All diagrams were drawn to the same scale. The results for the 2 cm3 cube are shown. original colour Which diagrams show the results for the 1 cm3 and the 3 cm3 cubes? A 1 cm3 3 cm3 B 1 cm3 3 cm3 C 1 cm3 3 cm3 D 1 cm3 3 cm3
1 marks
Answer: C
13 The diagram shows how nicotine is transported from the blood plasma into a cell using a type of cotransporter mechanism. protons move down their concentration gradient blood out of the cell plasma cell surface membrane cytoplasm of the cell nicotine is transported cotransporter protein against its concentration gradient into the cell In the phloem tissue, there is a cotransporter mechanism that moves sucrose into the cytoplasm of a companion cell. Which statement correctly describes a similarity between the cotransport of nicotine and the cotransport of sucrose? A The cotransporter proteins generate a proton gradient by moving protons out of the cell by active transport. B The protons are transported through the cotransporter proteins by facilitated diffusion. C The protons move through the cotransporter proteins in the opposite direction to the movement of nicotine and sucrose. D The cotransporter proteins use energy from ATP to transport protons with nicotine and sucrose.
1 marks
Answer: B
15 A student observed the effect of two different concentrations of salt solution on blood cells. The student added each concentration of salt solution to one of two microscope slides, and then a small drop of fresh blood was added. Each slide was viewed using the high power lens of a microscope and the student’s observations were recorded. slide 1 No red blood cells were visible. slide 2 The red blood cells were visible but looked slightly crinkled. Which row correctly explains the results obtained? slide 1 slide 2 A Swelling of the cells The of the cell was caused them all to burst. more negative than the of the external solution. B The of the cell was The of the external solution key more negative than the was more negative than the = water potential of the external solution. of the cell. C The of the cell was There is a net movement of less negative than the water out of the cell by osmosis. of the external solution. D The of the external solution The of the cell was very was less negative than the similar to, but slightly of the cell. more negative than, the of the external solution.
1 marks
Answer: B
16 Which statement correctly describes facilitated diffusion? A The process only occurs using channel proteins that change shape and that use energy provided by the cell. B The process occurs using channel proteins or carrier proteins that may or may not change shape. C The process occurs using channel proteins or carrier proteins that use energy provided by the cell. D The process only occurs using carrier proteins that create a gradient to move ions in opposite directions.
1 marks
Answer: B
14 Which process always takes place without the involvement of energy from ATP? A active transport B endocytosis C exocytosis D facilitated diffusion
1 marks
Answer: D
15 The diagram shows the entry of molecule X into a cell. molecule X Which row shows a property of molecule X and the effect of the concentration of ATP in the cytoplasm on the rate of entry of molecule X? property of molecule X concentration of ATP in the cytoplasm A non-polar affects rate of entry of molecule X B non-polar has no effect on rate of entry of molecule X C polar affects rate of entry of molecule X D polar has no effect on rate of entry of molecule X
1 marks
Answer: D
16 The electron micrograph shows some human blood cells. X Which row correctly shows the net movement of water by osmosis and the water potential of the cytoplasm of cell X compared with the solution surrounding the cells? water potential of net movement of cytoplasm of cell X water by osmosis compared with the solution A into the cell higher B into the cell lower C out of the cell higher D out of the cell lower
1 marks
Answer: C
17 A red indicator solution was mixed with agar and the resulting solid was cut into small cylindrical blocks. The blocks were placed in an acid which turns the indicator yellow and all other variables were kept constant. The dimensions of the blocks are shown. block 1 height 3 mm diameter 6 mm block 2 height 6 mm diameter 12 mm block 3 height 8 mm diameter 16 mm The formula for calculating the surface area of a cylinder is 2rh + 2r 2. The formula for calculating the volume of a cylinder is r 2h. Which row shows the correct surface area (SA) to volume (V) ratio for each block and the time taken for the block to turn yellow? block 1 block 2 block 3 SA to V time to turn SA to V time to turn SA to V time to turn ratio yellow / mins ratio yellow / mins ratio yellow / mins A 0.75 : 1.0 4 1.5 : 1.0 5 2.0 : 1.0 11 B 0.75 : 1.0 11 1.5 : 1.0 5 2.0 : 1.0 4 C 1.33 : 1.0 4 0.67 : 1.0 5 0.5 : 1.0 11 D 1.33 : 1.0 11 0.67 : 1.0 5 0.5 : 1.0 4
1 marks
Answer: C
14 A red indicator solution was mixed with agar, and the resulting solid was cut into small cuboid blocks. The blocks were placed in an acid which turns the indicator yellow, and all other variables were kept constant. The dimensions of the three blocks used are shown. block 1 3 mm 3 mm 3 mm block 2 8 mm 8 mm 8 mm block 3 11 mm 11 mm 11 mm Which row shows the correct surface area (SA) to volume (V) ratio for each block, and the time taken for the block to turn yellow? block 1 block 2 block 3 SA to V time to turn SA to V time to turn SA to V time to turn ratio yellow / mins ratio yellow / mins ratio yellow / mins A 0.5 : 1.0 4 1.33 : 1.0 11 1.83 : 1.0 13 B 0.5 : 1.0 13 1.33 : 1.0 11 1.83 : 1.0 4 C 2.0 : 1.0 4 0.75 : 1.0 11 0.55 : 1.0 13 D 2.0 : 1.0 13 0.75 : 1.0 11 0.55 : 1.0 4
1 marks
Answer: C
15 The statements describe some events in the process of exocytosis of glycoprotein molecules. 1 Membrane of the Golgi body folds around glycoprotein molecules. 2 Vesicle binds to and fuses with the cell surface membrane. 3 Vesicle attached to microtubules moves through the cytoplasm. 4 Secretory vesicle forms. What is the correct order of events for exocytosis? A 1 4 2 3 B 1 4 3 2 C 2 3 1 4 D 4 1 3 2
1 marks
Answer: B
16 Four cylinders that were identical in size, A, B, C and D, were cut from potatoes that had been stored for different lengths of time. The cylinders were weighed, immersed in 10% salt solution for 45 minutes and then reweighed. The percentage change in mass was then calculated. Which cylinder had a water potential similar to the 10% salt solution? percentage change in mass A –7.2 B –2.5 C –0.9 D +3.4
1 marks
Answer: C
28 Carrier proteins in the cell surface membranes of companion cells are involved in the transfer of assimilates to phloem sieve tubes. The diagram represents the use of two types of carrier protein in this process. key X X X X Y X X pump X Y companion X X cell cytoplasm cotransporter cell surface membrane What are the substances labelled X and Y? X Y A H+ ions sucrose B H+ ions glucose C sucrose H+ ions D glucose H+ ions
1 marks
Answer: A
18 Which diagram shows the correct direction of net water movement between the four cells due to osmosis? key = water potential A B Ψ = – 40 kPa Ψ = –22 kPa Ψ = –22 kPa Ψ = –32 kPa Ψ = –36 kPa Ψ = –33 kPa Ψ = –16 kPa Ψ = –38 kPa C D Ψ = –33 kPa Ψ = –25 kPa Ψ = – 42 kPa Ψ = –33 kPa Ψ = –33 kPa Ψ = – 42 kPa Ψ = –24 kPa Ψ = –27 kPa
1 marks
Answer: A
19 An investigation was carried out into the effect of four different treatments on the permeability of the cell surface membranes and tonoplasts of beetroot cells. Beetroot cell vacuoles contain a red pigment. This pigment cannot diffuse through the tonoplasts or cell surface membranes. 1 cm3 cubes were cut from beetroot tissue and washed in running water for 20 minutes to remove any pigment released from damaged cells. Two cubes were then placed in each of the four test-tubes containing different contents and observed for five minutes. Which row shows a correct explanation for the observation recorded for one of the treatments? treatment observation explanation A dilute contents of membrane proteins hydrochloric acid test-tube stay clear have been denatured B ethanol contents of lipids, including membrane test-tube turn red phospholipids, have dissolved C water at 20 C contents of membrane proteins test-tube stay clear have been denatured D water at 80 C contents of lipids, including membrane test-tube turn red phospholipids, have dissolved
1 marks
Answer: B
10 A mixture of glucose and starch solutions was placed in a length of dialysis (Visking) tubing and the tubing sealed. The tubing was then placed in a boiling tube containing distilled water. Two samples were immediately removed from this water (time 0 minutes) and tested with either iodine solution or Benedict’s solution. This was repeated at 10 minute intervals for 30 minutes. The iodine solution gave an orange-brown colour each time. The table shows the results of the Benedict’s test. time / minutes 0 10 20 30 colour produced blue green yellow red by Benedict’s test What may be concluded from these results? 1 The pores in the Visking tubing are too small for a starch molecule to pass through. 2 Glucose diffuses through the Visking tubing down a diffusion gradient. 3 Water diffuses into the Visking tubing. A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: B
16 In an experiment, pieces of onion epidermis are put into three different concentrations of sucrose solutions, P, Q and R. The pieces of onion are left for an hour and then examined using the low power of a light microscope. Each diagram shows one cell from the epidermis that was placed in each of the sucrose concentrations. P Q R What explains the appearance of cells in solution Q? A The concentration of solution Q is equal to the concentration of the solutes in the cell sap. B The cytoplasm has the same concentration of sucrose as solution Q. C The water potential of the cytoplasm is equal to the water potential of the vacuole. D The water potential of the cell sap is equal to the water potential of solution Q.
1 marks
Answer: D
14 Before mitochondria are extracted from cells for microscopy, they are usually kept in a 0.25 mol dm–3 sucrose solution. Why is the sucrose solution used? A to act as a solvent B to enable the rate of reaction of the mitochondria to be determined C to prevent the mitochondria from changing in dimension D to provide a source of energy
1 marks
Answer: C
15 The photomicrograph shows a type of blood cell. Which statements about these cells are correct? 1 Oxygen diffuses through the phospholipid bilayer. 2 Sodium ions diffuse through the phospholipid bilayer. 3 Water passes in and out of these cells by osmosis. A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: C
17 A student filled dialysis tubing with a sucrose solution and knotted both ends. This formed a cylinder with a length of 5.0 cm and a radius of 2.0 cm. What is the surface area to volume ratio for this cylinder of dialysis tubing? A 0.7 : 1.0 B 1.0 : 1.0 C 1.4 : 1.0 D 2.5 : 1.0
1 marks
Answer: C
16 The diagram shows the dimensions of two blocks of agar. The diagram has been drawn to scale. width 30 mm width 10 mm length 20 mm length 15 mm height 4 mm height 8 mm The blocks of agar were stained pink with a pH indicator. In acidic conditions, the pink pH indicator becomes colourless. The two blocks of agar were placed in a beaker of acid at the same time. As the acid diffused into the blocks, the blocks became colourless. What is the surface area to volume ratio of the block that became completely colourless first? A 0.58 : 1 B 0.67 : 1 C 1.50 : 1 D 1.71 : 1
1 marks
Answer: B
17 The graph shows how the rate of facilitated diffusion of substance X across a cell surface membrane changed as the concentration of substance X increased. All conditions, except for the concentration of substance X, were kept constant. Temperature was maintained at 15 °C. Q P rate of facilitated diffusion concentration of substance X Which statement about the rate of facilitated diffusion is correct? A The rate of facilitated diffusion of substance X at Q will increase if the temperature is increased to 20 °C. B The rate of facilitated diffusion of substance X at P will increase if the concentration of ATP is increased. C The rate of facilitated diffusion of substance X at Q will increase if the concentration of substance X is increased. D The rate of facilitated diffusion of substance X at P will increase if the length of time over which the rate is measured is increased.
1 marks
Answer: A
18 Samples X, Y and Z are epidermal tissues cut from an onion. Each epidermal tissue was immersed in one of three different concentrations of a salt solution for 30 minutes. A student observed each tissue sample with a light microscope. Then the student estimated the concentration of each salt solution using a scale. The diagram shows the scale. P, Q and R represent the estimated concentration of the three salt solutions. distilled concentrated water salt solution P Q R The photomicrographs show the appearance of the tissues in samples X, Y and Z. sample X sample Y sample Z Which row shows the estimated concentrations of the salt solutions in which samples Y and Z were immersed? sample Y sample Z A P Q B P R C Q R D R P
1 marks
Answer: D
19 The graph shows the results of an osmosis investigation using potato tissue. 20 15 10 5 percentage change 0 in mass 0 0.1 0.2 0.3 0.4 0.4 0.4 0.5 0.6 0.7 0.8 0.9 –5 concentration of sodium chloride / mol dm–3 concentration of sodium chloride concentration of sodium chloride / mol mol dm dm–3 –3 –10 –15 –20 What is the concentration of sodium chloride solution that has the equivalent water potential to this potato tissue and what is correct about the movement of water at that point? A 0.37 mol dm–3 and no net movement of water B 17 mol dm–3 and net movement of water out of the potato tissue C 0 mol dm–3 and no net movement of water D 0.9 mol dm–3 and net movement of water into the potato tissue
1 marks
Answer: A
20 Some agar was coloured pink using a pH indicator. The pink agar was then used to make three agar cubes with different dimensions. 2 cm × 2 cm × 2 cm 4 cm × 4 cm × 4 cm 5 cm × 5 cm × 5 cm The cubes were placed in a beaker and covered with 0.1 mol dm–3 hydrochloric acid. The temperature of the experiment was standardised at 20 °C. Hydrochloric acid diffused into the agar cubes causing the cubes to become colourless. What is the surface area to volume ratio of the agar cube that became colourless in the least amount of time? A 0.33 : 1 B 0.83 : 1 C 1.2 : 1 D 3.0 : 1
1 marks
Answer: D
19 Antimycin is a chemical that inhibits the function of mitochondria. Which methods of transport across the cell surface membrane could be directly affected by antimycin? 1 active transport 2 facilitated diffusion 3 endocytosis A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: C
20 A student half filled a beaker with solution X. They placed a sealed Visking tubing bag containing solution Y into the beaker. X Y At 30 minutes, the solution in the beaker was orange and the solution inside the Visking tubing was blue-black. What did solutions X and Y contain at the start to give these results? solution X solution Y A starch amylase and iodine B iodine amylase C starch and amylase iodine D iodine starch
1 marks
Answer: D
17 Which of these substances can pass directly through cell surface membranes without using a carrier protein or channel protein? 1 K+ and Cl – 2 CO2 3 C6H12O6 A 1 and 2 B 1 and 3 C 2 and 3 D 2 only
1 marks
Answer: D
18 Which of these statements about facilitated diffusion are correct? 1 It is limited by the number of transport proteins. 2 It transports molecules against their concentration gradient. 3 It requires a source of ATP. A 1, 2 and 3 B 1 and 3 only C 1 only D 2 and 3 only
1 marks
Answer: C
19 A cell absorbs amino acids. This cell then synthesises and exports a digestive enzyme. Different cell structures are involved with different stages of this process. Which row shows a possible sequence of cell structures that the amino acids pass through? rough cell surface secretory Golgi body endoplasmic membrane vesicle reticulum A 1st 2nd 3rd 4th B 1st 3rd 4th 2nd C 4th 2nd 1st 3rd D 4th 1st 2nd 3rd
1 marks
Answer: C
20 The diagram shows apparatus set up to investigate the effect of changing the initial concentration of glucose in the surrounding solution on the movement of molecules through a selectively permeable membrane (Visking tubing) in 15 minutes. Visking tubing surrounding solution (different concentrations 10% glucose from 1% to 10% solution glucose solution) Which statements are correct as the initial concentration of glucose solution in the surrounding solution increases? 1 Net diffusion of water increases. 2 Glucose molecules reach an equilibrium quicker. 3 There is less change in the volume of the surrounding solution. 4 Net diffusion of glucose increases. A 1, 2, 3 and 4 B 1, 2 and 4 only C 1 and 3 only D 2 and 3 only
1 marks
Answer: D
16 A sample of healthy plant cells taken from the same tissue is placed in a beaker containing distilled water. The cells change in size. What would explain this change in size? A Water will leave the cells by active transport. B Water will enter the cells by osmosis. C Solutes will enter the cells by active transport. D Solutes will leave the cells by osmosis.
1 marks
Answer: B
17 A student studying surface area to volume ratio and diffusion made a cuboid, S1, using agar stained blue with a pH indicator. The dimensions of S1 are shown in the diagram. The student made a second agar cuboid, S2. Each dimension of S2, (the length, the width and the height), was half that of S1. S1 S2 30 mm not to scale 15 mm 5 mm 2.5 mm 8 mm 4 mm The student placed each cuboid in a test-tube and covered it in acid. The time taken for each cuboid to completely change colour was recorded. All variables other than the size of the cuboids were standardised. Which row shows the surface area to volume ratio of S1 and the time taken for S1 to change colour completely in acid compared to the time taken for S2 to change colour completely? time taken for S1 to change surface area to colour completely in acid volume ratio of S1 compared to S2 A 0.72 : 1 S1 takes less time than S2 B 0.72 : 1 S1 takes more time than S2 C 1.4 : 1 S1 takes less time than S2 D 1.4 : 1 S1 takes more time than S2
1 marks
Answer: B
18 Two test-tubes, labelled X and Y, were set up containing equal volumes of solution X or solution Y respectively. A large number of type P cells and a large number of type Q cells were added into test-tube X and also into test-tube Y. After a few minutes, samples of the solutions were taken and the cells were observed with a microscope. ● In solution X, all of cell type P had burst and cell type Q had not burst. ● In solution Y, no cells had burst. Which row correctly identifies cell type Q and solutions X and Y? cell type Q solution X solution Y A red blood cells 5% NaCl solution distilled water B goblet cells 5% glucose solution distilled water C liver cells distilled water 5% glucose solution D root hair cells distilled water 5% NaCl solution
1 marks
Answer: D
19 Which statements about the cell surface membrane are correct? 1 Channel proteins allow water soluble ions and molecules across the membrane. 2 Glucose can pass into the cell via carrier proteins. 3 Oxygen passes freely through the membrane as it is soluble in lipids. 4 Some glycoproteins act as antigens. A 1, 2, 3 and 4 B 1, 3 and 4 only C 1 and 2 only D 2, 3 and 4 only
1 marks
Answer: A
20 The diagram shows two cylinders of agar. The agar cylinders are placed into a solution of dye at the same time. The time taken for the dye to diffuse into the centre of each cylinder is recorded. 5 cm 3 cm 2 cm 1 cm X Y Which statement explains why the dye reaches the centre of one agar cylinder faster than the other agar cylinder? A The dye reaches the centre of X in a shorter time than the dye reaches the centre of Y because X has a larger surface area to volume ratio. B The dye reaches the centre of X in a shorter time than the dye reaches the centre of Y because X has a larger surface area. C The dye reaches the centre of Y in a shorter time than the dye reaches the centre of X because the dye has a shorter distance to diffuse to the centre. D The dye reaches the centre of Y in a shorter time than the dye reaches the centre of X because Y has a smaller volume.
1 marks
Answer: C