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All Formulas

By Vieui Nov 17, 2013 817 Words

Fiber Content:

FC = Fiber Value / Dietary Fiber
--------------------------------------- X (n= E.P. wt; how many grams where consumed) 100

For Example:

FC = 70 /100 x 200 = 35grams of fiber.

% RENI:

Actual Nutrient Intake
%RENI = -----------------------------------
RENI

For Example:

VitA:

450 / 350 x 100 = 128.6%

Tannhauser’s (Broca) Method:
For example:
An adult sedentary male who stands 5 feet 5 inches.
Ht: = 5 feet 5 inches = 65.0 inches
= 65.0 inches x 2.54 cm
= 165.1 cm

DBW = 165.1 – 100 = 65.1 kg.
= 65.1 – 6.51 (10% of 65.1)
= 58.59 kg or 59 kg

NDAP Method:
DBW for men: 5 feet tall is 112lbs. Add(subtract) 4lbs for every inch above(below) 5 feet. DBW for women: 5 feetl tall is 106 lbs. Add(subtract) 4lbs for every inch above(below) 5 feet.

For Example:
Male: 5 feet and 5 inches
5 feet= 112 lbs
5 in= 20 lbs
= 132 lbs (60kg)

Adopted Method:
Male and Female (DBW)
5 feet = 106 lbs; for every inch above 5 ft, add 5 lbs.

For Example:
5 feet & 5 inches
5 feet= 106 lbs
5 in= 25 lbs
= 131 lbs (59.5kg)

Desirable BMI:
For men: 22
For women: 20.8
BMI = wt(kg)/H(m)2
Note: A BMI range of 19.0 to 24.9 is considered normal.

For Example:

= 61kg / (1.65m)2
= 61kg / 2.7225
= 22.4

Total Energy Allowance (TEA) (Krause Method):

Activity:kcal/kg DBW/day
Bed rest but mobile (hospital patients)27.5
Sedentary (mostly sitting)30.0
Light (tailor, nurse, physician, jeepney driver)35.0
Moderate (carpenter, painter, heavy housework)40.0
Very Active (swimming, lumberman)45.0

For Example:
Total Energy Allowance (TEA) of a Moderate person weighing 59kg.
= 59 x 40 = 2360 kcal or 9875 kj.

Determining CHO, PRO and Fat:
A. Percentage Distribution:
Carbs- 50-70% of TEA
Pro- 10-15% of TEA
Fat- 20-30% of TEA

For Example:
CHO - 65%
PRO - 15%
FAT - 20%

CHO= 2360 x 0.65 =1534 kcal
PRO= 2360 x 0.15 =354 kcal
FAT= 2360 x 0.20 =372 kcal

B. Kcal to Grams
CHO = 4kcal/gram; PRO = 4 kcal/gram; FAT = 9kcal/gram

For Example:
CHO= 1534 / 4 = 383.5 or 385g
PRO= 354 / 4 = 88.5 or 90g
FAT = 372 / 9 = 41.3 or 40g

Diet Rx: Kcal 2350; CHO 385g; PRO 90g; Fat 40g

Deriving Nutrient Values:

1. Carbohydrates
Formula:
% Carbohydrate = 100 – [%Water + %Protein + %Fat + %Ash]

For Example:
Given: Rice, well-milled, boiled (bigas, maputi, sinaing) A020 Water = 67.6g per 100g E.P.
Protein = 2.1g per 100g E.P.
Fat = 0.2g per 100g E.P.
Ash = 0.4g per 100g E.P.

Solution:
Carbohydrate= 100g – (67.6 + 2.1g + 0.2g + 0.4g)
= 100 – 70.3
= 29.7g per 100g

2. Energy
Formula:
Energy (kcal) = [(4xg protein) + (9 x g fat) + (4 x g carbs)] :4,9,4 are Atwater factors

For Example:
Given:Protein = 2.1g per 100g E.P.
Fat = 0.2g per 100g E.P.
Carbohydrate = 29.7g per 100g E.P.

Solution:
Energy (kcal)= [(4 x 2.1g) + (9 x 0.2g) + (4x 29.7g)]
= 8.4 + 1.8 + 118.8
= 129 kcal per 100g E.P.

3. Kcal to kj conversion
1Kcal = 4.184 kj.
For Example:
= 2350kcal x 4.184 kj = 9832.4 or 9850kj.

4. Retinol Equivalent (RE)
Formula: µ

1 µg retinol = 1 RE

1 µg ß-carotene
------------------------ = 1 RE
6
For Example:

Total RE = 9205 (Chicken Liver) + 135 µg per 100g
-----------------------------
6
= 9205 + 23 = 9228 µg per 100g or 9228 RE per 100g.

5. Niacin Equivalent (NE)
Formula:
Mg Tryptophan
Mg niacin = ------------------------------= Niacin + Niacin values from Tryptophan
60= Niacin Equivalent.

6. Niacin Equivalent from Protein
Formula:
Protein / 6 = Additional Niacin.

For Example:

Niacin from soybean seed, dried (C044)= 2.0mg per 100g.
Protein of soybean seed, dried (C044)= 35.8g per 100g.
35.8g / 6 = 5.97 or 6.0mg
= 2.0mg + 6.0mg
= 8.0mg per 100g

Computing Nutrient Content of Foods/Translating Foods of Different Forms and/or Units into Nutrients

7. Convert As Purchased (AP) to Edible Portion (EP) Weight.
Formula: You need FEL manual

E.P. weight(g) = % E.P.
-------------- x A.P. weight (g)

For Example:
Given:
1 pc. Milkfish= 250g A.P. Weight
E.P. Weight of milkfish= 65% (Get it from FCT)
= 250 x .65 = 162.5 or 162g E.P.

8. Convert Cooked A.P./E.P. weight to its raw E.P. weight equivalent. Formulas:

Raw A.P. weight equivalent= A.P/E.P weight of X “Cooked-to-raw” weight CF of the cooked food cooked food

Raw E.P. weight equivalent= A.P. weight X % E.P. / 100
of the cooked food equivalent

For Example:
Given: Cooked weight of scad, round, fried (G105)= 75g
“Cooked-to-raw” weight CF for scad, round, fried= 1.523
E.P. of scad, round = 49%
No FCT data for scad, round, fried

Solution:

Raw A.P. wt equivalent of 75g scad, round, fried= 75 x 1.523
= 114g raw A.P. wt
Raw E.P. wt equivalent of 114g scad, round, fried = 114g x 49/100
= 56g raw E.P. wt

Another Formula:

Raw weight
--------------------- = “Cooked-to-raw” weight CF
Cooked weight

Cooked weight
--------------------- = “Raw-to-Cooked” weight CF
Raw weight

For Example:
“Cooked-to-raw” and “Raw-to-cooked” weight in CF of mungbean
Given:
Wt (g) of raw mungbean= 50g
Wt (g) of cooked mungbean= 120g

Solution:
“Cooked-to-raw” weight CF = 50 / 120 = 0.42
“Raw-to-cooked” weight CF= 120 / 50 = 2.4

Calculate Nutrient Content of Food/Diet

9. Formulas:

FORMULA 10A:
Nutrient per gram E.P. = nutrient per 100g EP (FCT Value) X gram EP Hh measure Hh Measure ------------------------------------------------
100

FORMULA 10B:

Nutrient per 100 EP = X nutrient per gram E.P. Hh measure
---------------------------- -----------------------------------------------------
100g gram E.P. Hh measure

Example: Protein content of 1 pc Banana, lacatan.
Given: 1 pc banana, lacatan = 60g AP
E.P. of banana, lacatan (E011) = 69%
Protein = 1.4g per 100g E.P.

Solution 1 – Using FORMULA 7

E.P. wt of 60g A.P. Banana, lacatan = 69 X 60g
____
100=41g E.P.

Then using FORMULA 10A:

Protein content per 41g E.P. banana, lacatan = 1.4 / 100 x 41g E.P. = 0.6g

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