## advanced higher biology 2008

### anna123

Posted 20 May 2008 - 05:46 PM

so how is everyone getting on with their revision for next tuesday?? i hate unit 1! hate it hate it dont understand DNA profile or chromosome walking at all. so...more studying!! ahhhhhh

### Chas

Posted 21 May 2008 - 09:24 PM

Revision has been ok for me. The DNA technology stuff in Unit 1 is annoying. Unit 2 i think is ok and Physiology for Unit 3 i found pretty straightforward.

### anna123

Posted 22 May 2008 - 11:52 AM

lol omg same here! the enviroment stuff is ok it's just learning all the names which i still get confused with niches and aestivation! physiology is simple as i did human biology last year so it's sort of similar but i just dont understand the DNA stuff and i hope the essays are based on unit 2 if its about the dna profile then i'm just gonna cry!

### anna123

Posted 22 May 2008 - 01:26 PM

i keep getting the wrong answer, does anyone know how to do this?

Energy expenditure can also be measured by indirect calorimetry.

The following information was gathered as an individual ran on a treadmill for 20 minutes:

Volume of air breathed out in 20 minutes = 600 litres

Inspired air contained 20% oxygen, while expired air contained 16% oxygen.

4.8 kcals of energy are released for every litre of oxygen used up.

How many kilocalories of energy were expended during the 20 minutes?

### Chas

Posted 22 May 2008 - 10:29 PM

Indirect calorimetry involves estimating an individual's oxygen consumption over a given time as oxygen is needed in the body to release energy for respiration. In fact 4.8 kcal of energy are released for each litre of oxygen used by an individual.

Volume of air breathed out=600 litres
Composition of oxygen in inspired air=20%x600=120 litres
Composition of oxygen in expired air=16%x600=96 litres

The amount of oxygen absorbed by an individual is equal to the amount of oxygen in the inspired air minus the amount of oxygen in the expired air

Volume of oxygen absorbed=120-96=24 litres

4.8 kcal of energy is released for every litre of oxygen used so:

Energy expended in 20 minutes=4.8x24=115.2 kcals of energy

Hope that helps

### anna123

Posted 23 May 2008 - 02:13 PM

thank you!

### anna123

Posted 23 May 2008 - 05:45 PM

i really don't understand the 3 and 5 end carbon chain stuff! how are you supposed to recognise it? q4 in the m.p part of 2006 is really confusing me. also how do you work out the mitotic index? i dont seem to remember doing that in class or reading about it in my notes? q5 2004 mp, q2 mp 2005 and q2 2007. can someone please explain? thank you!

### Chas

Posted 23 May 2008 - 07:46 PM

The mitotic index is a measure of how many cells are undergoing cell division at any one time. In Q2 of 2004(similar to the other mitotic index questions) it gives you the values for interphase and the four stages of mitosis. To work out the mitotic index you add up the number of cells in the four stages of mitosis and then divide this by the number of cells in interphase. The answer is then multiplied by 100 to give you the percentage.

So for Q2 2004 there are in total 50 cells in mitosis and 450 cells in interphase so the mitotic index=50/450x100=11.1% so the answer is C

Hope this helps

### anna123

Posted 24 May 2008 - 08:34 PM

well actually the marking scheme says that the answer is B , 10% and I've figured out how to do it.
interphase is not involved in mitosis ( 2 stages interphase and the mitotic phase) so what you do is you add the number of cells in total in this case there are 500 cells in total including interphase. then you divide the total of the other phases excluding interphase by the total in this case 50/500 x 100 = 10%. This also works for the other 2 questions.

### Chas

Posted 25 May 2008 - 08:54 PM

There are 270 cells in interphase and in total there are 90 cells in mitosis so the mitotic index=90/270x100=33%.

### anna123

Posted 26 May 2008 - 10:32 AM

QUOTE(Chas @ May 25 2008, 09:54 PM)
There are 270 cells in interphase and in total there are 90 cells in mitosis so the mitotic index=90/270x100=33%.

The answer is C, 25% ( marking scheme)

remember interphase isn't involved in mitosis so for the 2007 paper there are 360 cells in total including interphase. Then divide the total of the other phases except interphase by the total so in this case it's 90/360 x 100 = 25%.