Lab 2 Flashcards

1
Q

what are the steps in silage making

A
  1. harvest and chop silage crop into smaller particles
  2. compaction - use rollers to compact material in silo (helps expel O2 and stops respiration process)
  3. phase 1 - increase cell respiration which makes CO2 and heat
  4. phase 2 - fermentation begins and acetic acid is made, pH drops from 6 to 5, temperature increases from 70 to 95F
  5. phase 3 - lactic acid production begins, pH drops to 4, temperature decreases to 80 to 85F
  6. phase 4 - lactic acid production continues, pH drops to 4, temperature maintains at 80-85F
  7. phases 1-4 last 21 days, which means silage is ready to use after day 21
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2
Q

what are the steps to hay making

A
  1. monitor the weather to find a window of 3-4 sunny days without rain in the forecast
  2. cutting/ moving - use mower or similar equipment to cut hay crop
  3. tedding - use equipment to lift and turn cut hay to promote even drying
  4. raking/windrowing - use appropriate machinery to rake/ windrow or gather dried hay into rows to prepare for baling, ideal moisture may not exceed 15%
  5. baling - baler gathers dried hay along the rows and convert this into rectangular/ round bales
  6. storage - baled hay is next transported to hay barn for dry storage or wrapped into plastic bad to enable storage of hay in open fields
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3
Q

the average yield of corn silage crop is 30 tons/acre with moisture (DM=35%). How many acres will be planted with silage corn to last 1000 dairy cows for a period of 120 days if silage comprises 40% of total mixed ration and daily feed intake per cow is 18 kg DM?

A
  1. determine cow needs: 1000 cows x 18kg DM/day x 0.4 x 120 days = 864,000 kg DM
  2. determine yield of silage crop: (30 tons x 2000 lb/ton x 0.35)/ 2.2 = 9,545.45 kg DM/acre
  3. determine acreage needed: 864,000/ 9,545.45 = 90.5 acres
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4
Q

although evalution of body condition can be looked at as an ongoing process, state 3 key ideal times when and why body condition scoring should be considered in a cow calf herd

A
  1. late summer - plan for management strategies eg: early weaning or supplementation for cows grazing warm season pastures or pasture quality is decreasing
  2. fall - plan for economical winter supplementation program to get females back to target BCS
  3. weaning time - cows weaning their first calves, most likely to be thin at this time
  4. 45 days after weaning - provides good information on how fast the cows are bouncing back after weaning
  5. 90 days before calving - last opportunity to get condition back on cows economically, separate thin cows from cows in good condition and feed them separately, particularly benefits young cows
  6. calving - indicates how well the pre calving feeding program went and the BCS score helps to plan for an appropriate weaning date
  7. breeding - thin cows at this time may indicate a poor match of calving season to feed resources eg: cows calving too early in the spring
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5
Q
  1. if the target bodyweight of sheep with BCS of 3.5 is 70 kg, calculate the current body weight of the sheep if their present BCS is 2.25
A

bodyweight change expected: 0.13 x 1.25 x 70 = 11.375 kg
current bodyweight: 70 kg - 11.375 kg = 58.625 kg

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6
Q

if the target body weight of a sheep with BCS of 3.5 is 70 kg, calculate the desired average daily gain (grams) if the animals would require 50 days to reach the target bodyweight (70 kg)

A

average daily gain expected: 11.375 kg x 1000 g = 11,375 g/50 days = 227.5 day

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7
Q

if the target bodyweight of sheep with BCS of 3.5 is 70 kg, use NRC guidelines to calculate the daily amount kg of supplement (ME= 3.00 Mcal/kg DM) to be fed to this sheep, the sheep is mainly fed oat hay (ME = 1.99 Mcal/kg DM)

A
  1. using NRC tables, you determine that sheep expected to gain 227.5 g/day needs 5.03 Mcal/day and consume 2.4 kg DM/day
  2. determine ME content of feed: 5.03 Mcal/day/ 2.4 kg/day = 2.10 Mcal/kg feed
  3. apply pearson square:
    supplement 3.00 0.11/1.01 = 0.1089
    2.10
    hay 1.99 0.90/1.01 =0.8911
    1.01 1.01
  4. total amount of supplement DM: 0.1089 x 2.4 kg/day x 50 sheep = 13.07 kg/day
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