Animal Science 320-18

al Science 320 Fall 2023 Problem Set 1 50 points EXERCISE L Youw < many conversions throughout the semester. Here are the most common ones we will use with the factors that vou can utilize throughout the semester: 1 pound = 454 grams 1 ounce = 28 grams 1 pound = 0.45 kilograms 1 kg = 2.2 pounds 1,000 gram = 1 kilogram 1,000 milligram = 1 gram 1 ton = 2,000 pounds 1 bushel corn = 56 pounds Read Chapter 1 of the E Text and complete the Chapter 1 Pre-Lab Assessment for 4 points (1/2 point per question). EXERCISE 2 — CRITICAL SKILL: Percent. Decimal & Algebra Review 1. Ingredients in diets are typically calculated as a % in the diet. A percent (%) is always out of 100 of something. Therefore. 20 out of 100 = 20% and that could be 20 pounds, 20 grams, 20 crackers, 20 candies etc. As a basic rule “part/whole x 100 = %" For example: if we have 10 out of a total of 300 crackers, the math is completed as (10/300) x 100 =3.3% 20 Lbcom out of 100 Lb=__ L0 % Zjos X /°° Biboasomofssb=_ 9.5 % 25’/45 x 100 H 30 Lb barley out 450 Lb= ; % 3”/450 x (D0 10 tons alfalfa hay out of 20 tons = _ S % 19/, x 100 2. Diet formulations are represented as weights (or mass) and also as concentrations (typically %) and can be sed either way if you know the formulation. Take for example the following formulation represented and note the 2 columns on the right are the exact same formulation using 100 pounds and 100 Kg: [ You could also say. You could also say this: | |25 pounds com 25 Kg com | 20 pounds distillers grains 20 Ky distillers grains 130 pounds soybean meal 30 Ky soybean meal | | 20 pounds alfalfa hay 20 Ky alfalfa h | |5 pounds vitamin/mineral premix 5 Ky vi al premix Total - 100 pounds Total 100 Ky

I£1 wish to make 10.000 pounds (#ions) of the diet represented in the table above, how many pounds of cach ingredient do I need: = 286D Poundscom /0,000 X .25 - 2060 pounds distillers grains )0, 600 X%+ 2.0 2000 Pounds ofalfalfatay /0, co0x . 22,20 = 300D Poundsofsoybeanmeal " x 2O = o me 500 pounds of vitamin/mineral premix sp , p00< . OS = /10 600)) 4 / 3. Now assume you have the following diet in units of mass (kg in this case). Convert the diet to percent units. 200 kilograms corn —_ w kilograms soybean meal (SBM) Jotol = ROSO K p 300 kilograms of oats 150 kilograms of soybean hulls 50 kilograms of vitamin/mineral premix (VMP) S8.5 %com /Z°°/zoso x lo0 [Z. ] %sBM ZSD/ZDSD x (00 9.l %oas 3045/2050 X o0 7.3 % soybean hulls ’So/zoso %100 2.4 o o YME ;V,’bso Xxr00 = (99.9) 100% (diet formulations should always total 100% 4. We will also use algebra when formulating diets. These types of algebra problems should be review from previous courses and we will be using this level of algebra throughout the course. If needed, use 1 decimal == place in g 4 a6x+3=7x—1(0lveforx) [y 071;/ b.x+y=100 (x = 10; solve for y) /gr@ o0 C.2x+2(x+5)= 16x-7 4’}(/—/0: /[;X % d. 100 —x=25 z =/ 0-x=2 1#212x [x= 1Y) | 180

EXERCISE 3 — Proximat alysis: Dry Matter Concepts Chapter 2 in the E text and viewed the lectures in Module 2. The “Diagram™ in the E Text should be reviewed as well. 1. E Text: Review Chapter 2 and the “Diagram”. Complete the Pre-Lab Assessment for chapter 2 (4 points) . Zn Lab 2. Labexercise 1 : answer the following questions. e a. What was the weight of the dry beet pulp in (g) This is called dry matter intake (DMI). b. What was the weight of water added (g)? This is called moisture (water): ¢ What is the total weight in the cup including water and beet pulp? This is called the as fed intake: (g):. d. What is the % water (AKA: moisture) (%): ¢. What is the % dry matter (%): £ Which weighs more dry matter intake or as fed intake? Lab Exercise 2: Follow instructions provided in lab then fill in the table. Nutrients on an As-Fed Basis Nutrient Weight of Nutrient (grams) % of Nutrient (As-fed) Blue (water) Yellow (fat) | Red (protein) White (carbohydrates) Green (vitamins) Total 100% Now remove the water/moisture from your diet (subtract out the blue ones) Nutrients on an Dry Matter Basis " Nutrient ‘Weight of Nutrient (grams) % of Nutrient (DMB — dry matter basis) [Yellow (fat) | Red (protein) White (carbohydrates) Green (vitamins) _Total 100% Notice the number of colored beads DID NOT CHANGE but now you have fewer total dots. What does that do to your nutrient concentrations?

4. Notice that dry matter intake is a smaller number and dry matter basis of nutrient concentrations is a larger number even though both are considered “dry matter”. Explain this difference and the difference in how you complete the math in your own words. You can use examples. lohen Yhe water i's remowed , +he mass (intatke) Oje‘t’f:;es- W t [ " f concentrates +he nuirients So the nubtber is /Mjer'. 5. This table includes the Dry Matter (DM) concentration and the crude protein concentration of 5 different feed ingredients. Complete the table and show your calculations on either an as-fed or DM basis as indicated. Hint: Look at the units and notice it’s a concentration (%). Use 2 decimals in answers. Crude protein, % Feedstuff DM, % As-fed | DM basis Calculations basis 1 10.00 6.50 (S.00 ©.5/.70 2 85. . 5.00 BW | 55 9 |28/-85 3 91.00 0.00 00 ap099 |3/ 4 45.00 13.05 2900 [29 x.4S 5 22.00 4500 |4S x.22 9.9 Using the data below on 5 feedstuffs, fill in the table below with the INTAKE of each feedstuff on either an as-fed or DM basis as indicated. Hint: Look at the units and notice it’s a unit of mass (Lb) [ Intake, Lb Feedstuff’ DM, % As-fed DM basis Calculations | basis 1 1 75.00 Q?,l} 20.50 20.5/75‘ | 2 90.00 10.00 9.00 Jox .90 3 91.00 Zfi%’ 35.00 35/9/ 4 34.50 11.25 3 gg 1. 25)( } ?t/{ s 35.00 050 g a3 | /25 .95

Protein Generally, most samples use a 6. from the lab. Calculate % protein in the feed samples factor for calculation of protein. Below are your % nitrogen values Feed sample Nitrogen, % Crude protein, % A 170 x (.25 10.643 B 580 x .74 3 .25 C .90 4000 /6.25 D ). 2l 850 / 4.25 2. Explain what the 6.25 factor means (why is it used)? Amine acids in most feds contain /(a%n/%ojm ”’% = 4.25 3. Nitrogen Free Extract The following lab values came back on your feed samples. Notice the first table indicates % as fed: therefore, the water is included. Meaning, you need to subtract the water and other nutrients on an as fed basis to get NFE on an as fed basis. Round to 1 decimal. Example I: Calculate nitrogen free extract (NFE) on an as fed basis. %, As Fed Feedstuff | Moisture cP Ash CF EE NFE A —_ 100 250 L5 80 10.0 9L, 5 B 80.0 40 05 10 25 250 Now consider the next table. For these feed samples, the nutrients are already on a dry matter basis indicated by the % dry matter heading. When all the nutrients are already on a dry matter basis, then you still start with 100 and do not need to worry about subtracting out water since it is already “dry”. Example 2: Calculate nitrogen free extract (NFE) on a dry matter basis (DMB). - % of Dry Matter Feedstuff CP__ | Ash CF EE NFE [ ¢c 1 350 | 35 50 200 20.5 D 25.0 25 15.0 7.0 50.5