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Ad libitum [116K]
3 years ago
10

Andrew snacks on high-carbohydrate and high-fat foods during his afternoons on Mondays through Thursdays. In fact, when he total

ed his daily caloric intake, he was surprised to find that he had eaten an average of 2,400 kilocalories per day! If his total daily energy requirement is 2,200 kilocalories, what is the metabolic fate of the extra carbohydrate and fat kilocalories he has consumed?
Biology
1 answer:
yulyashka [42]3 years ago
7 0

Answer:

  • Andrew is consuming extra 200 kilocalories every day. These calories are not required by the body so it will store them for further use.
  • Most of the extra carbohydrate will enter glycogenesis pathway to form glycogen which will be stored in liver and skeletal muscles.
  • Remaining carbohydrate and fat will enter the lipogenesis pathway will get converted into triglycerides which will be stored as fat in adipose tissue.
  • Excess storage of calories can make Andrew overweight in future.
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Answer:

0.153

Explanation:

We know the up-thrust on the fish, U = weight of water displaced = weight of fish + weight of air in air sacs.

So ρVg = ρ'V'g + ρ'V"g where ρ = density of water = 1 g/cm³, V = volume of water displaced, g = acceleration due to gravity, ρ'= density of fish = 1.18 g/cm³, V' = initial volume of fish, ρ"= density of air = 0.0012 g/cm³ and V" = volume of expanded air sac.

ρVg = ρ'V'g + ρ"V"g

ρV = ρ'V'g + ρ"V"

Its new body volume = volume of water displaced, V = V' + V"

ρ(V' + V") = ρ'V' + ρ"V"

ρV' + ρV" = ρ'V' + ρ"V"

ρV' - ρ"V'  = ρ'V" - ρV"

(ρ - ρ")V'  = (ρ' - ρ)V"

V'/V" = (ρ - ρ")/(ρ' - ρ)

= (1 g/cm³ - 0.0012 g/cm³)/(1.18 g/cm³ - 1 g/cm³)

= (0.9988 g/cm³ ÷ 0.18 g/cm³)

V'/V" = 5.55

Since V = V' + V"

V' = V - V"

(V - V")/V" = 5.55

V/V" - V"/V" = 5.55

V/V" - 1 = 5.55

V/V" = 5.55 + 1

V/V" = 6.55

V"/V = 1/6.55

V"/V = 0.153

So, the fish must inflate its air sacs to 0.153 of its expanded body volume

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