1. Unit 3 – Lecture 3

2. Diffusion Diffusion – movement of substances from an area of high concentration to an area of low concentration works to balance the concentration gradients of substances ex: perfume in a room, dye in a glass of water…

3. Diffusion – cont’d Concentration Gradient – difference in concentration of a substance over a certain area “with/down” the gradient = particles move from area of high conc. to low conc. typical diffusion; natural movement “against” the gradient = particles move from area of low conc. to high conc. non-natural movement; requires E

4. Diffusion – cont’d Dynamic Equilibrium – a balance of the distribution of particles in an area acknowledging the constant movement of the particles dynamic – opposite of sta(tic/sis) indicates constant change

5. Diffusion – cont’d Diffusion often occurs through one of two types of membranes: permeable membrane allows all molecules [solute, solvent, or other particle] to pass through semi/selectively permeable membrane allows only certain substances to pass through which type of membrane do cells have?

6. Diffusion Drawings Problem: In a membrane selectively permeable to X molecules, there are: 4 particles of X [3 left, 1 right] 12 particles of O [2 left, 10 right] Based on the given information, identify & understand the type of membrane draw before & after beaker with membrane

7. Problem: In a membrane selectively permeable to X molecules, there are: 4 particles of X [3 left, 1 right]; 12 particles of O [2 left, 10 right] Based on the given information, identify & understand the type of membrane draw before & after beakers with membrane

8. Problem: In a membrane selectively permeable to X molecules, there are: 4 particles of X [3 left, 1 right]; 12 particles of O [2 left, 10 right] before beaker: under beaker, record # of particles per side draw # particles in beaker X = 3X= 1 O = 2O = 10

9. Problem: In a membrane selectively permeable to X molecules, there are: 4 particles of X [3 left, 1 right]; 12 particles of O [2 left, 10 right] before beaker: determine directions of particle movement X = 3X= 1 O = 2O = 10

10. Problem: In a membrane selectively permeable to X molecules, there are: 4 particles of X [3 left, 1 right]; 12 particles of O [2 left, 10 right] before beaker: above, draw arrows specifying how particles will move X = 3X= 1 O = 2O = 10

11. Problem: In a membrane selectively permeable to X molecules, there are: 4 particles of X [3 left, 1 right]; 12 particles of O [2 left, 10 right] before beaker: large arrow = larger movement [specify type of particle] X = 3X= 1 O = 2O = 10 X

12. Problem: In a membrane selectively permeable to X molecules, there are: 4 particles of X [3 left, 1 right]; 12 particles of O [2 left, 10 right] before beaker: small arrow = smaller movement [specify type of particle] X = 3X= 1 O = 2O = 10 X

13. Problem: In a membrane selectively permeable to X molecules, there are: 4 particles of X [3 left, 1 right]; 12 particles of O [2 left, 10 right] after beaker: under beaker, record new # of particles per side X = 3X= 1 O = 2O = 10 X X = 2X= 2 O = 2O = 10

14. Problem: In a membrane selectively permeable to X molecules, there are: 4 particles of X [3 left, 1 right]; 12 particles of O [2 left, 10 right] after beaker: draw # of particles in beakers X = 3X= 1 O = 2O = 10 X X = 2X= 2 O = 2O = 10

15. Problem: In a membrane selectively permeable to X molecules, there are: 4 particles of X [3 left, 1 right]; 12 particles of O [2 left, 10 right] after beaker: draw appropriate arrows [specify type of particle] X = 3X= 1 O = 2O = 10 X X = 2X= 2 O = 2O = 10 X

16. Osmosis Osmosis – diffusion of water across a selectively permeable membrane membrane allows ONLY water to go through solute particles NEVER move water must be used to reach dynamic equilibrium more water moves to where there is more solute

17. Osmosis – cont’d Ex: which has the higher concentration of solute? how can you make the concentrations equal? 5 scoops of lemonade in a pitcher filled with 8 ounces of water 5 scoops of lemonade in a pitcher filled with 1 gallon of water

18. Osmosis Drawings Problem: With a membrane selectively permeable to water, there are: 4 particles of X [3 left, 1 right] Based on the given information, identify & understand the type of membrane draw before & after tubes with membrane

19. Problem: With a membrane selectively permeable to water, there are: 4 particles of X [3 left, 1 right] Based on the given information, identify & understand the type of membrane draw before & after tubes with membrane

20. Problem: With a membrane selectively permeable to water, there are: 4 particles of X [3 left, 1 right] before tube: under tube, record # of particles per side draw # particles in tube X = 3X= 1

21. Problem: With a membrane selectively permeable to water, there are: 4 particles of X [3 left, 1 right] before tube: determine directions of particle movement X = 3X= 1

22. Problem: With a membrane selectively permeable to water, there are: 4 particles of X [3 left, 1 right] before tube: above, draw arrows specifying how particles will move X = 3X= 1

23. Problem: With a membrane selectively permeable to water, there are: 4 particles of X [3 left, 1 right] before tube: large arrow = larger movement [specify type of particle] X = 3X= 1 H2OH2O

24. Problem: With a membrane selectively permeable to water, there are: 4 particles of X [3 left, 1 right] before tube: small arrow = smaller movement [specify type of particle] X = 3X= 1 H2OH2O

25. Problem: With a membrane selectively permeable to water, there are: 4 particles of X [3 left, 1 right] after tube: under tube, record # of particles per side X = 3X= 1 H2OH2O

26. Problem: With a membrane selectively permeable to water, there are: 4 particles of X [3 left, 1 right] after tube: draw # of particles in beakers AND water level X = 3X= 1 H2OH2O

27. Problem: With a membrane selectively permeable to water, there are: 4 particles of X [3 left, 1 right] after tube: draw appropriate arrows [specify type of particle] X = 3X= 1 H2OH2O H2OH2O

28. What happens to cells? Hypertonic Solution – solution outside of the cell has a higher concentration of solute than inside of the cell. “hyper” = more than normal water moves out of cell cell shrinks Why is this bad for the cell?

29. What happens to cells? – cont’d Isotonic Solution – solution outside of the cell has the same concentration as the solution inside of the cell. concentrations are equal cell remains the same size

30. What happens to cells? – cont’d Hypotonic Solution – solution outside of the cell has a lower concentration of solute than inside of the cell. “hypo” = less than normal water moves into the cell cell swells [possibly bursts]