3. Two animals of similar weight (150 kg) begin prep arations for winter hibernation. To
survive, the animals need to store enough energY n without losing any of their original weight
over a 90 day period. Prior to to allow them to burn 1000 kcal/day of hibernatio hibernation,
animal A added 20 kg of weight all as carbohydrate (-4 kcal/s), whereas animale kcal/g). added
10 kg of weight all as fat (-9 n? (Assume each animal does not gain or lose weight for any reason
How much will each animal weigh after 90 days of hibernatio other than energy storage noted
above (e.s., no water weight gain/loss). Will they both survive? (Show all your work). (5 pts)
Solution
3.During 90 days peroid of hibernation they require (1000×90)=90000 kcal
1gm carbohydrate gives 4 kcal
20Kg=( 20×1000 )gm carbohydrate gives (20000×4)=80000kcal
So, it needs extra 10000kcal .
Animal A will not survive
Animal B: 1gm fat gives 9 kcal
10kg=10000 gm fat gives 90000kcal
Required kcal during hybernation peroid is 90000 kcal
It will survive and weight after 90 days will be 150kg.
survive, the animals need to store enough energY n without losing any of their original weight
over a 90 day period. Prior to to allow them to burn 1000 kcal/day of hibernatio hibernation,
animal A added 20 kg of weight all as carbohydrate (-4 kcal/s), whereas animale kcal/g). added
10 kg of weight all as fat (-9 n? (Assume each animal does not gain or lose weight for any reason
How much will each animal weigh after 90 days of hibernatio other than energy storage noted
above (e.s., no water weight gain/loss). Will they both survive? (Show all your work). (5 pts)
Solution
3.During 90 days peroid of hibernation they require (1000×90)=90000 kcal
1gm carbohydrate gives 4 kcal
20Kg=( 20×1000 )gm carbohydrate gives (20000×4)=80000kcal
So, it needs extra 10000kcal .
Animal A will not survive
Animal B: 1gm fat gives 9 kcal
10kg=10000 gm fat gives 90000kcal
Required kcal during hybernation peroid is 90000 kcal
It will survive and weight after 90 days will be 150kg.
