Name:____________________

Exercise #1

ANSC 3613 - Refer to Chapter Two 

Table of Equivalents and Calculations

Because nutrient composition of feeds are not always based on the same units, it is important that the student be able to make conversions from one unit to another. Therefore, this laboratory exercise is designed to familiarize students with the composition of feedstuffs and basic calculations used in ration formulation.

There are certain constants that should eventually become second nature to one involved with ration formulation and feeding of livestock. Of course, conversion tables do exist in books, but books are not always available. Below are a list of unit equivalents and conversions one should commit to memory.

1 kilogram (kg) = 1000 grams (g)

1 kg = 2.2 pounds (lb)

454 g = 1 lb

1 kg = 1000 g

1 g = 1000 milligrams (mg)

1 mg = 1000 micrograms (µg)

1 kilocalorie (kcal) = 1000 calories^*

1 megacalorie (Mcal) = 1000 kcal

1 ton = 2000 lb

percent (%) = parts per one-hundred

ppm = parts per million

ppm / 10,000 = %

ppm = 1 µg/g, 1 mg/kg,  or 1 mg/L

^*A calorie defined as the amount of heat energy needed to raise 1 gram of water 1^oC.

Students should always express numbers in terms of their units. A number without a unit is meaningless and answers without the correct units are incorrect.! Sometimes nutrients are expressed as amounts or as a concentration. True animal requirements are expressed as amounts, but often for simplicity, requirements are expressed based on the concentration of the nutrient that must be included in a feed to meet a specific animal's needs. In this case, one must assume the animal will consume enough of the feed to get its required amounts of nutrients.  For example, a cow may require 1.2 lb of crude protein (CP) per day.  In this case, her CP requirements are expressed as an amount.   Also, a cow may require 6% CP in her feed. In this instance, the nutrient requirement is expressed as a concentration but cow will have to eat at least 20 lb (1.2 lb of CP/.06) of a 6% CP ration to get an adequate amount (1.2 lb) of CP. However, if the cow became full with only 15 pounds of the 6% CP feed, then she would not have consumed an adequate amount (.9 lbs) of CP. A practical example of this concept may involve a 1000-lb beef cow consuming rice straw. Rice straw in a poor-quality forage and intake is usually limited to 1.5% of the cow's body weight, which would be 15 lb (1000 lb cow X .015) in our example. Here, it was assumed that rice straw contains 6% CP, which is probably too high in practice.

Pounds, kg, g, Mcal, kcal, etc. are all amounts. Concentrations are expressed as an amount per unit weight of feed. Common expression are: % (parts/100), ppm (parts/million), Mcal/kg, Mcal/lb, kcal/kg, IU/lb., etc. A feed that has a net energy value of 2.3 kcal/lb means that each pound of that feed contains 2.3 kcal of net energy. Below are a few examples of how you should learn to make conversions:

A. What do we know?

(1) 1 kg = 2.2 lb, we can also express this equality as  1 kg/2.2 lb or 2.2 lb/1kg

B. What do we need?

(1) pounds of feed

(2) so our answer will be X lb of feed. RIGHT!

C. So lets work towards X lb of feed.

10 kg of feed X 2.2 lb of feed = 22 lb of feed 

                              1 kg feed

* Notice like units is the numerator and the denominator (highlighted in yellow  or in italics), can be cancelled leaving the desired unit  in the numerator. If you will remember that the units you want your answer in must be in the numerator after all other units are cancelled, barring any careless errors, your answer will be correct!

A. What do we know?

(1) that pounds is an amount and % is a concentration

(2) that % equals to parts per 100.

(3) In this example, 15% means there are 15 lb of protein per 100 lb of feed.  An amount divided by another amount is a concentration, in this case %.

B. What do we want?

(1) to know how many pounds of protein are in the feed

C. So lets work towards X pounds of protein

10 lb of feed X 15 lb of protein = 1.5 lb of protein 

                     100 lb of feed

* Notice the like units in the numerator and denominator cancelled to leave us with pounds of protein in the numerator.

* When working with percentages a short cut is to change the % to its decimal equivalent by dividing the number by 100%. Notice that this number is unitless.

15% = .15 

100%

10 lbs of feed X .15 = 1.5 pounds of protein

Example 3: How many Mcal of net energy (NE) will a cow get if she eats 10 kg of hay that has 1.4 Mcal NE/kg.

A. What do we know?

(1) Mcal is an amount

(2) Mcal/kg is a concentration

(3) the cow eats 10 kg of hay  (which is an amount)

(1) Mcal of NE the cow eats

C. Let's work towards X Mcal of NE.

10 kg of hay X 1.4 Mcal of NE = 14 Mcal of NE 

                        kg of hay

1. 2.2 lbs of corn =       1         kg of corn

2. 30 mg/kg Zn =                      ppm Zn or                      % Zn ( refer back to the list of equivalents)

3. 1000 Kcal of DM =                       Mcal of DE

4. 1 ton of corn =                     bushels of  shelled corn ( assume there are 56 lb/bu)

5.  0.05% Ca =                  ppm Cu

B. Calculate the amount of each nutrient or energy in the following feeds. Assume weights and content to be on same dry matter basis.

2. 100 lb hay at 55% TDN =                   lb TDN

3. 10 kg soybean meal at 44% CP                     kg CP  (CP stands for crude protein)

4. 10 kg alfalfa hay at 25 ppm Zn =                       mg Zn (Hint: remember what ppm is)

5. 500 lb wheat at 3920 Kcal/kg =                         Mcal DE

Converting nutrients compositions from an as-fed to a dry matter basis. (refer to page 76 (60) in your textbook).

Nutrient requirements of an animal and nutrient composition of feeds may be expressed on dry matter (moisture-free) or as-fed basis. A dry matter basis is often used with cattle because cattle diets may contain ingredients that are substantially different based on the amount of moisture (water) they contain. Therefore, in order to more accurately formulate a diet that includes ingredients with very different moisture levels, it is better to formulate on a dry matter basis. However, once the percentages of each ingredient is determined, values should be converted back to an as-fed basis. This is necessary because naturally all feeds will have some moisture and final analysis of nutrients in a diet based on dry matter can be very different from those based on an as-fed basis. Furthermore, feed mill operators weigh out 25 lb of corn not 25 lb of corn dry matter.

Let's see how we can convert feed nutrient concentrations from an as-fed basis to a dry matter basis.

Concentration of feed nutrient (as-fed)=X concentration (DM basis)

DM percentage (as-fed)                                     100% DM

25 lb of corn DM X 100 lb corn               =  28.4%

                            88 lb of corn DM

or use the short method below where you divide the amount of dry matter  by the as-fed dry matter

25 lb of corn DM   =          28.4%                  

        .88

Assume alfalfa silage analyzed 6.5% crude protein on an as-fed basis and contained 40% dry matter.  What percent crude protein would the alfalfa silage contain when expressed on a dry matter basis?

Example 2. 

Another reason for expressing nutrients on a dry matter basis is for comparison of feeds such as the example 5b on page 77 (61) of your text, which is also illustrated below.  In this example, the crude protein is compared between a "dry" corn samples and a "high-moisture" corn sample.  By expressing the CP content  on dry-matter basis, we have now a "fair" (because we have removed the variable, water) comparison and find that the CP content is virtually the same.

Two samples of shelled corn were sent to a laboratory for analysis of crude protein (CP).   One sample was "dry" corn and the other "high-moisture" corn.  The laboratory sent back the following analysis:

Calculate the protein content (%) of each corn type on a dry matter basis.

What is your conclusion about the protein content of these corn types on a dry matter basis?

Problem 1.
 A feed has the following composition on an as-fed basis.

1. Calculate the corresponding values on a dry matter basis  (dry matter = moisture free) and insert them into the table above.

2. What is the % nitrogen on a DM basis? _________

3. What is the % total carbohydrate on an as-fed basis? ____

4. What is the % crude fiber on an air-dry basis?_____

* Sometimes it may be necessary to convert feed nutrients from a dry matter to an as-fed basis such as illustrated in 6a on page 77 (61) of your text.

Problem 2.

How many  kilograms of ration dry matter are consumed daily if a steer is being fed the following amounts of as-fed feeds? Refer to example 7 on page  78 (62) of your text. Complete the table below.

Problem 3

The following concentrate mix is being fed to yearling horses.   Feeds are presented as pounds of dry matter.  Calculate the pounds of as-fed feeds in this diet.  This is example where the feed formula was determined on a dry matter basis, but to actually be able to weight out the ingredients we must first convert the dry matter amounts to as-fed amounts.  Refer to example 8 on page 78 (62) of your textbook and complete the table below

Problem 4

The following two laboratory analyses of two feeds are provided for your use. These data illustrate that different laboratories may report nutrient (CP, ADF, TDN, etc.) values on differing bases. Lab 1 always expresses its feed analyses on a 100% DM basis while Lab 2 expresses theirs on an as-fed basis.  Remember that all feed contains some moisture some just more than others.  This problem will require conversion of nutrient values from as-fed to DM or DM to as-fed depending on the question asked below.   Study this section carefully making sure you understand the concepts and calculations presented because you will be performing these calculations from time-to-time through out this course.

2. What is the % Zn in each feed on an as-fed basis?

3. If 40 lbs of each feed were fed (fed implies as-fed basis) how much of each nutrient or energy would be provided?

Problem 5                                        

 The following information was obtained from a feed tag on each of three feeds. Always assume DM from feed tags to be 90%.

1. What is the Ca:P ratio of each feed on an as-fed and DM basis?  The ratio is the parts of Ca to the part of P.  It can be calculated by dividing in each feed the % Ca by % P and %P divided by itself.  (Hint: When changing values from as-fed to dry matter, every values is changed by the same proportion.)

What is your conclusion about these ratios?

2. Which one is the best buy in relation to cost per pound of CP? (Refer to page 75 of your text, section VI. Evaluating Feeds on the Basis of Cost per Unit of Nutrient.

Problem 6.

 A ration formula calls for the following ingredients on an as-fed basis.

Ingredients (see pg  105 (89)) . You will need to find each ingredient  in the table and record their dry matter and crude protein values.  The number in parenthesis corresponds to the number to the left of the ingredient name in the tables. Note that there are two  rows of values for each ingredient - nutrients are expressed on either an as-fed basis or dry matter basis.  In this exercise, you will be reported the % crude protein on an as-fed basis.  An example has been provided for grain sorghum. 

Use that values in the table above to answer the following questions.                               

2. What is the DM and CP content of the final mixture?  __________% DM ____________% CP (Tip: This is not calculated by merely summing the %DM and %CP values because the ingredients are all included in the diet at different proportions.  For example, grain sorghum contributes 32.7 lb. (37 lb X 88.5%)  dry matter 3.7 lb. (37 lb X 10% CP)  of protein to the diet.  You must calculate the amounts the rest of the ingredients contribute, dry matter and crude protein, and then sum the values to arrive at the answers to this questions.

Note the "Thumb rules" presented in G-9 on page 79 (63).