Nutrient Requirements of Beef Cattle: Seventh Revised Edition: Update 2000

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Nutrient Requirements of Beef Cattle: Seventh Revised Edition, 1996

TABLE 3–5 Empty Body (EB) Chemical Composition at Different Condition Scores (CS)

CS	Percent in EB				SBW, percent of CS 5^a
CS	Fat	Protein	Ash	Water	SBW, percent of CS 5^a
1	3.77	19.42	7.46	69.35	77
2	7.54	18.75	7.02	66.69	81
3	11.30	18.09	6.58	64.03	87
4	15.07	17.04	6.15	61.36	93
5	18.84	16.75	5.71	58.70	100
6	22.61	16.08	5.27	56.04	108
7	26.38	15.42	4.83	53.37	118
8	30.15	14.75	4.39	50.71	130
9	33.91	14.08	3.96	48.05	144
^aWeight change from CS5 weight can be estimated from the difference between CS5 weight and CS5 weight * percent of CS5 weight for the CS in question. Net energy reserves provided, or required to change CS, is kg weight change * 5.82.

81.3, 86.7, 92.9, 108.3, 118.1, 129.9, and 144.3 percent of a CS 5 cow for CS 1, 2, 3, 4, 6, 7, 8, and 9, respectively. A 500 kg cow is predicted to weigh 465, 434, 407, and 383 kg at CS 4, 3, 2, and 1 with weight losses of 35, 31, 27, and 24 kg for CS 5, 4, 3 and 2, respectively. Corresponding values for a 650 kg cow are 604, 564, 528, and 497 kg SBW at CS 4, 3, 2 and 1 with weight losses per CS of 46, 40, 35 and 31 kg for CS 5, 4, 3 and 2, respectively.

Table 3–6 gives Mcal mobilized in moving to the next lower score, or required to move from the next lower score, to the one being considered for cows with different mature sizes. These cows are within the range included in the data base used to develop the regression equations (433 to 887 kg SBW). Diet NE_m replaced by mobilized reserves, or required to replenish reserves, are computed by assuming 1 Mcal of mobilized tissue will replace 0.8 Mcal of diet N_Em, and 1 Mcal of diet N_Em will provide 1 Mcal of tissue NE, based on Moe (1981) and NRC (1989). For example, a 500 kg cow at CS 5 will mobilize 207 Mcal in declining to a CS 4. If NE_m intake is deficient 3 Mcal/day, this cow will lose 1 CS in (207 * 0.8)/3=55 days. If consuming 3 Mcal NE_m above daily requirements, this cow will move back to a CS 5 in 207/3=69 days.

The weakest link in this model is the prediction of body

TABLE 3–6 Energy Reserves for Cows with Different Body Sizes and Condition Scores

CS	Mcal NE Required or Provided for Each CS^a at CS 5 Mature Weight
CS	400	450	500	550	600	650	700	750	800
2	112	126	140	154	168	182	196	210	223
3	126	141	157	173	189	204	220	236	251
4	144	162	180	198	217	235	253	271	289
5	165	186	207	227	248	269	289	310	331
6	193	217	242	266	290	314	338	362	386
7	228	267	285	314	342	371	399	428	456
8	275	309	343	378	412	446	481	515	549
9	335	377	419	461	503	545	587	629	670
^aRepresents the energy mobilized in moving to the next lower score, or required to move from the next lower score to this one. Each kg of SBW change contains 5.82 Mcal, and SBW at CS 1, 2, 3, 6, 1, 8, and 9 are 76.5, 81.3, 86.7, 92.9, 108.3, 118.1, 129.9, and 144.3 percent of CS 5 weight, respectively.

weight change associated with each CS change. This is a critical step because it is used to compute total energy reserves available and energy required to replenish reserves. In this model, this calculation is based on the assumption that ash mass is constant. The weights and weight changes appear to agree well with other data at CS 5 and below, but appear to be high above CS 7. A reasonable alternative would be to use the weight change and energy reserves per CS computed for CS 5 for CS categories above a 5. Additional research is needed to be able to predict more accurately the body weights and weight changes associated with each condition score on diverse cattle types.

REFERENCES

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