Cell Transport Mechanisms Essay Example

spread through the 200 MWCO while albumen could non.

5. Put the followers in order from smallest to largest molecular weight: glucose. Na chloride. albumen. and urea.

Sodium chloride & lt ; urea & lt ; glucose & lt ; albumen.

Activity 2: Simulated Facilitated Diffusion Lab Report

Review Sheet Results

1. Explain one manner in which facilitated diffusion is the same as simple diffusion and one manner in which it is different from simple diffusion.

One manner in which facilitated diffusion is the same as simple diffusion is that they both transport solutes with ( or along or down ) their concentration gradient and they do non necessitate the energy of ATP to back up the conveyance.

One manner in which facilitated diffusion is different from simple diffusion is that facilitated diffusion requires the aid of bearer proteins bound in the membrane to transport molecules across the membrane. On the other manus. simple diffusion does non necessitate the aid of membrane proteins.

2. The larger value obtained when more glucose bearers were present corresponds to an addition in the rate of glucose conveyance. Explain why the rate increased. How good did the consequence comparison with your anticipations?

When more glucose bearers were present. the rate of glucose conveyance increased. It was because with the addition in glucose bearers. more glucose molecules could adhere to the bearers and that would ensue in an addition figure of glucose molecules get across the membrane.

I obtained the consequences from the experiment supported my anticipations because when there was an addition in figure of glucose bearers. the rate of glucose conveyance would besides increase with the aid of facilitated diffusion.

3. Explain your anticipation of the consequence of Na+

Cl- might hold on glucose conveyance. In other words. explicate why you picked the pick you did. How good did the consequences compare with your anticipation?

Na+ Cl- will hold no consequence on the rate on glucose conveyance since Na+ Cl- is transported through a simple diffusion without the aid of a bearer protein. Adding Na+ Cl- will non impact the rate of glucose conveyance because the Na+ Cl- conveyance is wholly independent from the glucose conveyance.

I obtained the consequences from the experiment supported my anticipations because from tally figure 6. the consequence of Na+ Cl- did non impact the rate of glucose conveyance.

Activity 3: Simulating Osmotic Pressure Lab Report

1. Explain the consequence that increasing the Na+ Cl- concentration has on osmotic force per unit area and why it has this consequence. How good did the consequences compare with your anticipation?

Increasing the Na+ Cl- concentration in the left beaker while maintaining the size of MWCO at 20 would ensue in an addition in osmotic force per unit area ( Run No. 2 ) . This was because the high concentration of Na+ Cl- in the right side of membrane gives a increased force to H2O ( in left beaker ) to travel towards the solution with the highest concentration of solutes. Therefore. there was an addition in osmotic force per unit area.

However. when the membrane was changed to from 20 MWCO to 50 MWCO. the Na+ Cl- molecules were able to spread through the membrane. the equilibrium would be reached and no osmotic force per unit area was generated. If the concentration of Na+ Cl- of both size are equal. the osmotic force per unit area

would be zero.

I obtained the consequences from the experiment supported my anticipations because as the the concentration Na+ Cl- was increased from 5 millimeters to 10 millimeters ( by adding more Na+ Cl- ) . the osmotic force per unit area besides increased. However. after the membrane was changed to 50 MWCO. the Na+ Cl- molecules could spread easy through the membrane and did non caused an addition in osmotic force per unit area.

2. Describe on manner in which osmosis is similar to simple diffusion and on manner in which is different.

One manner in which osmosis is similar to simple diffusion is that both mechanisms are inactive that involve motion of a substance from an country of its high concentration to an country of its lower concentration. that is. with ( or along or down ) its concentration gradient without the input of energy.

On manner in which osmosis is different from simple diffusion is that ions and molecules are transported through the membrane while in osmosis H2O molecules are transported through a selectively permeable membrane.

3. Solutes are sometimes measured in millisomoles. Explain the statement. “Water chases millionsoles. ”

It means when the concentration of solutes addition. the concentration of H2O lessening. Osmosis is the procedure of diffusion of H2O. It happens when there is difference of H2O concentration between two topographic points which is separated by a membrane. Water will spread from a low concentration of solutes from one size to a higher concentration of solutes of the other size.

4. The conditions of 9mM albumen in the left beaker and 10mM glucose in the right beaker with 200 MWCO membrane in topographic point. Explain the consequences.

How did the consequences compared with your anticipation.

The molecules of 10mM glucose could go through through the 200 MWCO membrane from the right beaker to go forth beaker until the equilibrium was reached. At this point. the concentration of glucose between two sides is equal. On the other manus. the molecules of 9 Mm albumen could non go through through the 200 MWCO membrane that kept in the left beaker and a osmotic force per unit area of 15m mmHg was built up.

I obtained the consequences from the experiment supported my anticipations because glucose could spread through the 200 MWCO while albumen could non make so. Thus. a osmotic force per unit area was built up in the left size of beaker.

Activity 5: Stimulating Active Transport

1. Describe the significance of utilizing 9 mM Na chloride inside the cell and 6mM K chloride outside the cell alternatively of other concentration ratios.

The significance of utilizing 9 mM Na chloride inside the cell and 6 millimeter K chloride outside the cell is that for every three a+ ions transported out of the cell. two K+ ions are transported into the cell. The ratios of ions being out and in is 2: 3 is the same as the concentration ratio of 6 millimeters: 9 millimeter.

2. Explain why there was no Na transport even though ATP was present. How good did the consequences compare with your anticipation?

The ground there was no Na transport even though ATP was present because in order to do Na+/Ka+ pump perform successful. both Na+ ions and Ka+ ions must be present. . In Stimulation Run Number 3. there was no Ka+ nowadays. therefore the pump

could non work.

I obtained the consequences from the experiment supported my anticipations because no Na conveyance could be observed even with the presence of ATP.

3. Explain why the add-on of glucose bearers had no consequence on Na or K conveyance. How good did the consequences compared with your anticipation?

The add-on of glucose bearers had no consequence on Na or K conveyance because the bearers were used to transport the glucose molecules by facilitated diffusion without the input of ATP.

On the other manus. Na+ and Ka+ ions are transported actively by Na+/Ka+ pump across their concentration gradient. if glucose bearers were added. they had no effects on the conveyance of Na+and Ka+ ions

I obtained the consequences from the experiment supported my anticipations because the add-on of bearers had no consequence on the Na+and Ka+ ion conveyance.

4. Make you believe glucose is being actively transported or transported by facilitated diffusion in this experiment? Explain your reply.

From Run Number 5. Glucose molecules were transported passively by facilitated diffusion down their concentration gradient without the input of ATP. The presence of bearers were used to assist the glucose molecules to acquire across the membrane.