11.2a Muscles & Movement

09/12/2013 § Leave a comment

I don’t want a lot for Christmas, I just wanna get outta here.



Page 266, factors affecting muscle growth

1. Compare the changes in body mass in control group 2 and the exercise group. —> Control group 1 has very little to no change in increase in body mass that remains steady throughout the 6 weeks but control group 2’s change in body mass suddenly increases after the 2 weeks at a far greater rate than the other groups. The exercise group is the only group that experiences a decrease (after 4 weeks). The pectoralis mass of control 2 ended up being higher than that of control 1 or the exercise group.

2. Evaluate the claim that preventing exercise increases pectorals muscle mass. —> The claim seems valid enough because in both control groups (of which exercise was prevented), the pectoralis mass increased more than the exercise (control) group. The only setback could be the error bars but none of them overlap each other, indicating that the data is significant.

3. Suggest how the mass of the birds’ pectoralis muscle could be determined. —> Measuring the birds’ pectoralis muscle (ruler, tape measure) and calculating the volume of the weight then, or by simply weighing the bird before and after the experiment. Or if you’re really desperate, you could cut off the bird’s wing.

4. … Discuss the ethics of designing and carrying out experiments to test [this] hypothesis. —> By raising birds that don’t get to exercise or move as freely as they’d like, farmers are directly using birds for their own need, although that would be the birds’ purpose anyway. Animal rights activists might find a problem with this: the restriction of birds from living freely. However, the point of raising birds in a farm is to make them as big as possible before they are slaughtered and sold. While ethically it could feel like the exploitation of animals, the need to produce poultry still stands.

6.5b Synapses

05/12/2013 § Leave a comment




Page 257, Parkinson’s disease


1. Explain how symptoms of Parkinson’s disease are relieved by giving the following drugs:

  • a) L-dopa: Dopa decarboxylase decarboxylates L-DOPA to become Dopamine.
  • b) selegeline, which is an inhibitor of monoamine oxidase-B (MAO-B): Selegeline inhibits dopamine.
  • c) tolcapone, which is an inhibitor of catechol-O-methyl transferase (COMT): Tolcapone inhibits L-DOPA and a product of dopamine.
  • d) ropinirole, which is an agonist of dopamine: Instead of inhibiting dopamine, ropinirole promotes the production of dopamine.
  • e) safinamide, which inhibits reuptake of dopamine by presynaptic neurons: Safinamide inhibits the reuptake of dopamine and extends the duration of its effect.

2. Discuss how a cure for Parkinson’s disease might in the future be developed by:

  • a) stem cell therapy: Neurons that don’t function properly in Parkinson’s disease can be replaced by neurons produced through stem cell therapy. The biggest problem with this is mainly ethical in that some people question whether some stem cell procedures are ethical or not.
  • b) gene therapy: Through gene therapy, the actual DNA of an individual with Parkinson’s disease might be able to be altered in such a way that lets the individual’s neurons function again. This might mean retrieving healthier genes from other organisms, which could, again, bring up ethical issues as well as damage the DNA as a result of human error.

6.5a Neurons

03/12/2013 § Leave a comment

The other day I asked Santa if he could give me something for the circles around my eyes. I don’t think he’s going to reply.



page 255, abnormal action potentials

1. Using only the data in Figure 13, outline the effect of reduced Na+ concentration on:

a) the magnitude of depolarisation –> When the sea water is at the lowest (at 1 second), the magnitude of depolarisation decreased.

b) the duration of the action potential –>The reduced sodium levels caused a longer duration of the action potential.

2. Explain the effects of reduced Na+ concentration on the action potential. –> If the levels of sodium are low, that means that the resting potential cannot be low. The concentration gradient across the membrane isn’t high due to the low sodium concentration therefore there would be less active transport thus a smaller magnitude.

3. Discuss the effect of reduced Na+ concentration on the time taken to return to the resting potential. –> Reduced sodium levels increases the duration of action potential and increases the duration of time needed to return to the resting potential. But if there’s a low concentration of sodium ions, that could also mean that the membrane can actively transport sodium faster (because there’s less of them) and thus reduce the duration of time until the resting potential.

4. Compare the action potentials of shaker and normal fruit flies. –> Wild type flies have a higher action potential than mutant flies.  Wild type flies have about a 40 mV action potential peak while mutant only have about 25 mV.

5. Explain the differences between the action potentials. –> Because the K-channels aren’t working properly, it’s harder for the mutant flies to repolarise. Nonfunctioning K-channels limit the body from diffusing potassium across the membrane for repolarisation. A normal wild type fly can still repolarise normally because its K-channels do work.

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