2.4a Cells: Membranes (p. 27, 31, 32, 33) – September 19, 2012
«p. 27»
1) Explain what conclusion can be drawn from each of these pieces of experimental
evidence
a. When the cells were kept at normal body temperatures for mouse and
human cells, the red and the green markers became mixed.
It can be concluded that normal body temperature is the one at which the
rate of diffusion is the fastest in human and mouse cells.
b. Blocking ATP synthesis in the cells did not prevent the mixing of the red and
green markers.
The diffusion that occurred was passive, which does not require any ATP.
This means that no energy was required for the proteins to diffuse over the
membrane.
c. Inhibition of protein synthesis in the cells did not prevent mixing of the
markers.
The type of diffusion that occurred was simple diffusion, which does not
need proteins in order to be executed. This means that one of the cells (either the
mouse cell or the human cell) had a lower concentration of solutes than the other one.
2) Explain the trends shown in the graph
a. for temperatures between 15 and 35˚C.
Between 15 and 35˚C, the rate of diffusion increased exponentially.
b. for temperatures below 15˚C.
For temperatures below 15˚C, the rate of diffusion remained relatively
constant.
3) Predict, with reasons, the results of the experiment if it was repeated using cells from
Arctic fish rather than from mice or humans.
If the experiment was repeated using cells from Arctic fish, the results would be
different because the normal body temperature of Arctic fish is different from that of
mice and of humans. Because Arctic fish have a lower body temperature, their cells
would diffuse rapidly at a lower temperature than the mouse and human cells.
«p. 31»
1) Looking at Trace (i):
a. State the size of current flowing through the membrane when the ion
channels are closed.
0 pA.
b. State how many picoamps there are in one amp.
1.0 x 1012
2) Trace (i) shows five occasions when an ion channel opens and then closes.
a. State the maximum length of time that a channel remains open.
The maximum length of time that a channel remains open is approximately
54.5 milliseconds.
b. State the size of the current flowing when when an ion channel is open.
Approximately -2.75 pA.
3) ACh binds to the ion channels in this membrane at special binding sites, causing the
channels to open. The resulting flow of ions leads to the contraction of the muscle fibre.
Suggest what the natural source of ACh is in living muscle fibers.
The natural source of ACh in living muscle fibers could be amino acids
(which form a structure to fit into hormone binding sites).
4) Look at Trace (i). Explain why higher ACh concentrations:
a. increased the frequency of opening of channels
ACh is what caused the causes the channels to open, so higher ACh
concentrations mean that there would be more ACh present to open the channels.
Consequently, this would increase the frequency of opening of the channels.
b. did not increase the average time that the channels remained open.
Although higher frequencies of ACh mean that there is an increase in the
frequency of opening of the channels, it does not affect the average time that the
channels remain open. This is because ACh simply opens the channel; it doesn’t hold it
open.
5)
a. Compare Trace (ii) with Trace (i).
In Trace (i), there are two main sizes of current that are shown, whereas in
Trace (ii), there are four main sizes of current that are shown.
b. Suggest reasons for the differences between the two traces.
Reasons for the differences between the two traces could be that the two
membrane samples would react differently to different currents because they were
taken from two different locations.
«p. 32»
1)
a. Compare the levels of albumin in the three groups of children.
Children with Kwashiorkor have lower levels of albumin (15 g dm-3) than
children with marasmus (28 g dm-3), and healthy children have higher levels of
albumin (37 g dm-3) than children with marasmus.
b. Explain the differences in the levels of albumin.
Lack of food or proteins means that the children with marasmus or
kwashiorkor aren’t consuming the necessary nutrients in order to be healthy. This
means that their tissues won’t be healthy, so the reabsorption of water wouldn’t occur
properly – this means that there is more water present in the blood. This causes the
blood plasma albumin concentration levels to be lower in those children than in the
healthy children.
2) Explain how albumin dissolved in blood plasma causes reabsorption of water into
blood.
Albumin dissolved in blood plasma raises the concentration of the solute of the water,
triggering osmosis.
3) When water is not reabsorbed into blood capillaries, tissues become swollen by fluid
retention. This condition is called edema. Deduce which of the groups of children were
most at risk of edema.
The groups of children with marasmus or kwashiorkor are most at risk of edema.
«p. 33»
1)
a. State whether water moved into or out of the tissues at 0.0 mol dm-3 sodium
chloride solution.
Water moved into the tissues of all the plants.
b. State whether water moved into or out of the tissues at 1.0 mol dm-3 sodium
chloride solution.
Water moved into of the tissues of the pine kernel, butternut squash, and
sweet potato at 1.0 mol dm-3 sodium chloride solution, but moved out of the tissue of
the cactus.
2) Deduce which tissue had the lowest solute concentration in its cytoplasm. Include how
you reached your conclusion in your answer.
The tissue of the cactus had the lowest solute concentration in its cytoplasm because
it was the first one to have water moving out of it. Because water moves from an area
of low solute concentration to a high one, this would indicate that the solute
concentration of the cactus tissue in its cytoplasm was the lowest (out of the four
plants).
3) Suggest reasons for the differences in solute concentration between the tissues.
The reasons for the differences in solute concentration could be due to different
conditions that the plants live in and the fruit of the plant.
4) Explain the reasons for using percentage mass change rather than the actual mass
change in grams in this type of experiment.
The reasons for using percentage mass change rather than the actual mass change in
grams in this type of experiment is that it removes the problem of variation between
the masses of the samples used.