Daily Tissue Growth of a Pig from its Metabolizable-Energy Partition: 22-Agric-A1 Applied Plant, Animal or Human Physiology, May 2018, Q5

Question 5: Daily Tissue Growth of a Pig from its Metabolizable-Energy Partition (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Check: the source gives no separate energetic efficiency for NFVC deposition. Because NFVC shares protein's energy content (23.7 MJ/kg) and is compositionally mostly protein-based visceral tissue, this solution assumes NFVC deposition shares protein's energetic efficiency (0.511) — the standard assumption in this energy-partition model when a value is not given separately.

Given.

Quantity	Value
Feed offered	1.36 kg/day (5% wastage)
Metabolizable energy content of feed	13.8 MJ/kg
Energy partition: growth / maintenance	43% / 57% of ME intake
Energetic efficiency: protein / lipid / NFVC	0.511 / 0.919 / 0.511*
Energy content: lipid	39.6 MJ/kg DM
Energy content: protein & NFVC	23.7 MJ/kg DM
DM content: lean / fat / NFVC	22% / 90% / 22%
Tissue mass ratio, lean : fat : NFVC	1.00 : 0.29 : 0.35

Find. The daily fresh-mass growth rate of lean tissue, fat, and NFVC (g/day), and their total.

Approach. Convert the ME intake to ME available for growth (ME_g). Express the fat and NFVC daily growth rates as fixed multiples of the unknown lean-tissue growth rate $x$ using the given mass ratio, write each tissue's ME cost in terms of $x$ (fresh mass × DM fraction × energy content, divided by that tissue's efficiency), then solve the single equation "sum of the three ME costs = ME_g" for $x$.

Metabolizable energy intake and its growth share. Only 95% of the offered feed is actually consumed: $$\dot{m}_{feed} = 1.36 \times 0.95 = 1.292\ \text{kg/day}, \qquad \text{ME} = 1.292 \times 13.8 = \boxed{17.83\ \text{MJ/day}}$$ $$\text{ME}_g = 0.43 \times 17.83 = \boxed{7.667\ \text{MJ/day}}, \qquad \text{ME}_m = 0.57 \times 17.83 = 10.163\ \text{MJ/day}$$
Express each tissue's daily growth rate in terms of $x$ (lean tissue, kg/day). Using the given ratio: $$m_{lean} = x, \qquad m_{fat} = 0.29x, \qquad m_{nfvc} = 0.35x$$
ME cost per unit of $x$, tissue by tissue. Convert fresh mass to dry matter, multiply by the tissue's energy content to get deposited energy, then divide by that tissue's efficiency to get ME actually spent: $$\frac{\text{ME}_{lean}}{x} = \frac{0.22 \times 23.7}{0.511} = 10.204\ \tfrac{\text{MJ/day}}{\text{kg/day}}$$ $$\frac{\text{ME}_{fat}}{x} = \frac{0.29 \times 0.90 \times 39.6}{0.919} = 11.247\ \tfrac{\text{MJ/day}}{\text{kg/day}}$$ $$\frac{\text{ME}_{nfvc}}{x} = \frac{0.35 \times 0.22 \times 23.7}{0.511} = 3.571\ \tfrac{\text{MJ/day}}{\text{kg/day}}$$ Summing, $\text{ME}_g/x = 10.204 + 11.247 + 3.571 = \boxed{25.02}$ MJ/day per kg/day of lean tissue growth.
Solve for $x$ and the three tissue growth rates. $$x = \frac{\text{ME}_g}{25.02} = \frac{7.667}{25.02} = \boxed{0.3064\ \text{kg/day}} \;\; (306.4\ \text{g/day lean tissue})$$ $$m_{fat} = 0.29x = 0.0889\ \text{kg/day} = \boxed{88.9\ \text{g/day}}, \qquad m_{nfvc} = 0.35x = 0.1073\ \text{kg/day} = \boxed{107.2\ \text{g/day}}$$
Total daily mass gain. $$m_{total} = 306.4 + 88.9 + 107.2 = \boxed{502.5\ \text{g/day}}$$ Fat contributes the largest single ME cost per unit mass gained despite being the smallest mass fraction of the three, because it is both the most energy-dense tissue (39.6 vs 23.7 MJ/kg DM) and the driest (90% DM), so almost every kilogram of fresh fat mass counts as deposited energy.

Quantity	Result
ME intake	17.83 MJ/day
ME for growth, ME_g / maintenance, ME_m	7.667 / 10.163 MJ/day
Lean tissue growth	306.4 g/day
Fat growth	88.9 g/day
NFVC growth	107.2 g/day
Total daily mass gain	502.5 g/day

Concept & Theory

Topic: Solving a fixed-ratio, multi-tissue metabolizable-energy partition "backwards" — from a known ME_g to the unknown daily mass gains.

Key relations: ME = ME_m + ME_g; for each tissue, ME_spent = (fresh growth rate × DM fraction × energy content) / efficiency; with a fixed mass ratio between tissues, ME_g becomes a single linear equation in one unknown.

Why this works: Because the problem fixes the ratio of lean : fat : NFVC growth (rather than giving each mass directly), every tissue's ME cost can be written as a constant multiple of one unknown growth rate $x$; summing those three constants collapses the whole system to $\text{ME}_g = x \times (\text{sum of per-tissue ME coefficients})$, which is solved by one division.

Common pitfall: Applying the tissue mass ratio to the ME figures directly instead of to the fresh-mass growth rates that feed into the DM-then-energy-then- efficiency chain for each tissue — the ratio is a mass ratio, and each tissue's ME cost per unit mass differs (10.20 / 11.25 / 3.57 MJ per kg/day here), so the ME split is NOT 1.00 : 0.29 : 0.35.

Source: McDonald et al., Animal Nutrition, 7th ed., Ch. 12 (Energy metabolism and requirements).

Question 5 of 6: Daily Tissue Growth of a Pig from its Metabolizable-Energy Partition

Question 5: Daily Tissue Growth of a Pig from its Metabolizable-Energy Partition (20 marks)