Protein Supply - RDP - RUP - Rate of Passage - Intestinal Digestibility of RUP
Background: The following activity is designed to demonstrate the use of some of the equations presented in the reading. In particular we will determine the proportion of dietary crude protein (CP) that will become rumen degraded protein (RDP) and rumen undegraded protein (RUP). Hopefully, the concept of degradable protein vs. degraded protein will become clear. One of the steps will include the calculation of the rate of passage from the rumen.
Instructions Part I:
For each cow described in scenario 1 and scenario 2, below, complete Table 1 indicating the percentage and amount of the SBM CP that will be degraded in the rumen (RDP) as well as the percentage and amount that will escape ruminal degradation and flow to the small intestine as (RUP). The characteristics of the SBM CP (A, B and C fractions and Kd, the rate of degradation of the B fraction) are indicated in the Table.
| Scenario | CP % of DM |
A % of CP |
B % of CP |
C % of CP |
Kd of B %/h |
RUP int. Dig. % |
Kp %/h |
RDP % of CP |
RUP % of CP |
CP Lb/d |
RDP Lb/d |
RUP Lb/d |
Dig. RUP Lb/d |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Scenario 1: 6.27 lbs of SBM DM fed to a high producing cow | 49.9 | 22 | 77 | 1 | 9.14 | 93 | 8.06 | 62.9 | 37.1 | 3.13 | 1.97 | 1.16 | 1.08 |
| Scenario 2: 1.4 lbs of SBM DM fed to a dry cow | 49.9 | 22 | 77 | 1 | 9.14 | 93 | 5.91 | 71.1 | 28.8 | 0.70 | 0.50 | 0.20 | 0.19 |
Scenario 1: The cow weighs 1433 lbs. She is consuming 62.7 lb/d of dry matter (DM). Her days in milk (DIM) is 100 and she produces 110 lbs of 3.5% fat milk. Her diet includes 6.27 lbs of soybean meal (SBM) DM (i.e., her diet comprise 10% of SBM).
Step 1: Calculation of rate of passage of SBM. The equation for rate of passage of a concentrate feed is:
Kp Conc., %/hr = 2.904 + 1.375 × DMI, %BW - 0.020 × (% Conc. in diet DM).
DMI, % BW = 62.7*100/1433 = 4.37
% conc. in diet DM = 100 - % forage in diet DM = 100 – 57.3 = 42.7%
Thus,
Kp Conc., %/hr = 2.904 + 1.375 × 4.37 - 0.020 × 42.7 = 8.06 %/hr.
Step 2: Calculate RDP and RUP (% of CP) with eq. 10 and eq 11 from the reading:
RDP (% of CP) = A + B × [Kd / (Kd + Kp)] and RUP (% of CP) = C + B × [Kp / (Kd + Kp)]
RDP (% of CP) = A + B × [Kd / (Kd + Kp)]
= 22 + 77 × [9.14 / (9.14 + 8.06)]
= 22 + 77 × 0.531
= 22 + 40.92
= 62.9
RUP (% of CP) = C + B × [Kp / (Kd + Kp)]
= 1 + 77 × [8.06 / (9.14 + 8.06)]
= 1 + 77 × 0.469
= 1 + 36.08
= 37.1
Note that RDP (% of CP) + RUP (% of CP) = 62.9 + 37.1 = 100%
Step 3: Calculation of the supply of CP, RDP and RUP (lb/d)
SBM CP (lb/d) = SBM DM (lb/d) × CP % of DM / 100 = 6.27 × 49.9 / 100 = 3.13 lb/d
RDP (lb/d) = SBM CP (lb/d) x RDP (% of CP) / 100 = 3.13 × 62.9 / 100 = 1.97 lb/d
RUP (lb/d) = SBM CP (lb/d) x RUP (% of CP) / 100 = 3.13 × 37.1 / 100 = 1.16 lb/d
Note that RDP + RUP = CP : 1.97 lb/d + 1.16 lb/d = 3.13 lb/d
Scenario 2: The cow weighs 1433 lbs. She is dry and her intake of DM is 28.0 lb/d. Her diet includes 1.4 lbs of (SBM) DM (i.e., her diet comprise 5% of SBM).
Step 1: Calculation of rate of passage of SBM. The equation for rate of passage of a concentrate feed is:
Kp Conc., %/hr = 2.904 + 1.375 × DMI, %BW - 0.020 × (% Conc. in diet DM).
DMI, % BW = 28.0*100/1433 = 1.95
% conc. in diet DM = 100 - % forage in diet DM = 100 – 80 = 20%
Thus,
Kp Conc., %/hr = 2.904 + 1.375 × 1.95 - 0.020 × 20 = 5.19 %/hr.
Step 2: Calculate RDP and RUP (% of CP) with eq. 10 and eq 11 from the reading:
RDP (% of CP) = A + B × [Kd / (Kd + Kp)] and RUP (% of CP) = C + B × [Kp / (Kd + Kp)]
RDP (% of CP) = A + B × [Kd / (Kd + Kp)]
= 22 + 77 × [9.14 / (5.19 + 9.14)]
= 22 + 77 × 0.638
= 22 + 49.11
= 71.1
RUP (% of CP) = C + B × [Kp / (Kd + Kp)]
= 1 + 77 × [5.19 / (5.19 + 9.14)]
= 1 + 77 × 0.362
= 1 + 27.89
= 28.8
Note that RDP (% of CP) + RUP (% of CP) = 71.1 + 28.8 = 100%
Step 3: Calculation of the supply of CP, RDP and RUP (lb/d)
SBM CP (lb/d) = SBM DM (lb/d) × CP % of DM / 100 = 1.4 × 49.9 / 100 = 0.70 lb/d
RDP (lb/d) = SBM CP (lb/d) x RDP (% of CP) / 100 = 0.70 × 71.1 / 100 = 0.50 lb/d
RUP (lb/d) = SBM CP (lb/d) x RUP (% of CP) / 100 = 0.70 × 28.8 / 100 = 0.20 lb/d
Note that RDP + RUP = CP : 0.50 lb/d + 0.20 lb/d = 0.50 lb/d
Instructions Part II:
Analyze the data presented in Table below. First, describe under which condition the RUP of a feed is high and under which conditions the RDP of a feed is high. Second, Explain the impact of the changing rate of passage (Kp; see table footnote) associated with a high DMI (lactating cow) or a low DMI (dry cow) on the differential breakdown of CP into RUP and RDP.
| Feed | CP (% of DM) | A (% of CP) | B (% of CP) | C (% of CP) | Kd of B (% /h) | RUP (% of CP) Dry cow1 |
RUP (% of CP) Lac. cow2 |
|---|---|---|---|---|---|---|---|
| Alfalfa Pasture | 26.5 | 31 | 62 | 7 | 12.3 | 23 | 26 |
| Alfalfa silage (<40% NDF) | 23.2 | 62 | 29 | 9 | 13.1 | 16 | 18 |
| Corn silage (32-38%DM) | 8.8 | 51 | 30 | 19 | 4.4 | 34 | 36 |
| Corn grain (dry rolled) | 9.2 | 28 | 71 | 1 | 5.1 | 37 | 43 |
| Soy Bean Meal (44% CP) | 49.9 | 22 | 77 | 1 | 9.1 | 28 | 35 |
| Meat and bone meal | 54.2 | 18 | 48 | 34 | 7.2 | 54 | 58 |
Feed ingredients that have a CP characterized by a high C fraction and/or high B fraction with a low Kd tend to have high RUP value. In contrast, feed ingredients that have a CP characterized by a high A fraction and/or high B fraction with a high Kd tend to have high RDP value.
In essence increasing Kp (with increasing DMI) leads to variable increase in the RUP value of a feed. The increase in RUP value is large when the B or the C fractions are a large fraction of CP (see corn grain for large B fraction and meat an bone for large C fractions). Conversely, the increase in RUP with increasing intake is small when the A fraction is a large fraction of the CP (see for example alfalfa silage)