Transport Across Cell Membranes

Question 1

Simple Diffusion Definition

What type of molecule?

Answer 1

Net movement of particles from high to low concentration down its concentration gradient until an equilibrium is reached. Doesn’t require ATP.

Small and lipid soluble.

Question 2

Facilitated diffusion definition

Channel Proteins:

Carrier Proteins:

Answer 2

Movement of larger, polar molecules through channel or carrier proteins- passive process.

Form tubes filled with water- water soluble ions can pass through.

Carrier: substrate binds with proteins: causes a change in shape.

Question 3

Osmosis definition

Answer 3

movement of water from a higher water potential to a lower water potential through a selectively permeable membrane.

Question 4

Active Transport definition:

ATP hydrolysis:

Answer 4

Movement of molecules or ions against their concentration gradient through protein carriers: requires ATP hydrolysis

ATP bins to protein, is hydrolysed to ATD+Pi
Causes protein to change shape- can transport substrate
As Pi leaves the protein returns to its original shape

Question 5

Co-transport: Na and glucose

Answer 5

Na+ actively transported into blood from epithelial cells- creates a conc gradient for Na+ to diffuse in
Na+ move down is concentration gradient through co transport protein carriers with glucose against its concentration gradient into the epithelial cells
glucose facilitated diffusion into the capillaries

Question 6

Adaptations to membranes:

Answer 6

Increased SA: more transport

Increased number of carrier/protein channels- more transport

Question 7

RP3:(1)
Steps:

Answer 7

1) Make a collection of samples with known concentrations- 1, .8, .6, .4, .2, 0.0
2) Use a cork borer- cut identically sized chips from the potato
3) blot to remove XS water
4) weigh each chip
5) place each chip in solution for 20 minutes
6) remove chips from solution and dab with paper towel to remove XS water
7) take final mass
8) calculate % change
9) plot graph x axis: conc of sucrose solution, y axis: % change
10) where line crosses X axis= water potential- water potential of sucrose solution is equal to water potential of Potato

Question 8

RP3) 2
1) Make a collection of samples with known concentrations- 1, .8, .6, .4, .2, 0.0- WHAT EQUATION?
2) Use a cork borer- cut identically sized chips from the same potato WHY IDENTICAL SIZE? WHY REMOVE SKIN? WHY SAME POTATO?
3)blot to remove XS water- WHY?
weigh each chip
place each chip in solution for 20 minutes
4) remove chips from solution and dab with paper towel to remove XS water-WHY?
take final mass
5) calculate % change- WHY? EQUATION?
plot graph x axis: conc of sucrose solution, y axis: % change
where line crosses X axis= water potential- water potential of sucrose solution is equal to water potential of Potato

Answer 8

1) conc of initial x volume of initial = conc of desired x volume of final solution
2) identical size- equal surface area for osmosis, remove skin- it is impermeable, same potato, natural variation in water potential among potatoes
3) blot- XS water would be present in variant amounts in initial mass of potato
4) XS water would be present in variant amounts in initial mass of potato- water on outside of potato varies
5) to allow comparison between slightly different sized chips,( change in size/initial size) x 100