Final Exam Study Guide
Supplement
- Bumble bee: Sample
calculations/Answers
1.
If the cycling provides heat to warm the insect,
estimate the time required for a bee to raise its muscle temperature from 20.0
°C to 30.0 °C if the substrate cycling takes place at an average rate of 20.0
micromoles/min/g of muscle and all of
the heat generated remains in the flight muscle.
- Assumptions:
- 1 ATP is hydrolyzed/turn of cycle
- All hydrolysis energy of ATP goes
to heat
- G_{ATP
hydrolysis} = 30.0 kJ*mol^{-1}
- the heat
capacity of bee muscle is the same as that for water = 1.00 cal*g^{-1}*°C^{-1} = 4.184 J*g^{-1}*°C^{-1}
- Calculation
- First calculate the energy
released for each turn of the cycle:
(20.0
x 10^{-6} mol*min^{-1}*g^{-1})*(30.0 x 10^{3} J*mol^{-1})
=
0.600 J*min^{-1}*g^{-1}
- Next calculate the energy
required:
(30.0
°C - 20.0°C)*(4.184 J*°C^{-1}*g^{-1})
=
41.84 J*g^{-1}
- Finally, dividing the total energy
required by the energy released/cycle give the time:
(41.84
J*3^{-1}) / (0.600 J*min^{-1}*g^{-1})
=
69.73 min
= 69.7 min
2.
Try a second, perhaps more realistic
calculation, by assuming a cycling rate of 20 micromoles/min/g of muscle and an
aerobic glycolytic rate of 16 micromoles/min/ g of
tissue.
3.
Assumptions:
- 1 ATP is hydrolyzed/turn of cycle
- All hydrolysis energy of ATP goes to heat
- all of the heat generated
remains in the flight muscle (the bee is perfectly insulated)
- G_{ATP
hydrolysis} = 30.0 kJ*mol^{-1}
- the heat capacity of bee
muscle is the same as that for water = 1.00 cal*g^{-1}*°C^{-1} = 4.184 J*g^{-1}*°C^{-1}
- For each glucose going through glycolysis the bee will generate 30 ATP (not 32, since bee flight muscle uses the
glycerol-P shuttle)
- All ATP's generated in glycolysis will be hydrolyzed to yield heat
(probably via muscle contractions due to shivering)
4.
Calculation
- First calculate the energy released/time interval:
Energy
from cycle + energy from glycolysis =
(20.0
x 10^{-6} mol*min^{-1}*g^{-1})*(30.0 x 10^{3} J*mol^{-1}) + (16.0 x 10^{-6} mol*min^{-1}*g^{-1})*(30
ATP)*(30.0 x 10^{3} J*mol^{-1})
=
0.600 J*min^{-1}*g^{-1} + 14.4 J*min^{-1}*g^{-1}
=
15.0 J*min^{-1}*g^{-1}
- Next calculate the energy required:
(30.0
°C - 20.0°C)*(4.184 J*°C^{-1}*g^{-1})
=
41.84 J*mol^{-1}
- Finally, dividing the total energy required by the
energy released/cycle give the time:
(41.84
J*mol^{-1}) / (15.0 J*min^{-1}*g^{-1})
Last modified 3 May 2011