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Physiology of Rum Digestion > Exam 3 > Flashcards

Exam 3 Flashcards

1
Q

What is te difference in starch vs. cellulose?

A
  • Starch (amylose, amylopectin)
    • amylose is not branched
    • amylopectin is branched
  • structure vs. physical properties
  • rate of hydration
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2
Q

What affects ruminant digestion?

A
  • Microbial enzymes
    • Rumen and Hind gut
    • Amylase, proteases, Lipase: slight differences from mammalian enzymes
    • Cellulase, other enzymes for fiber digestion: Species differences
  • Rate of digestion
    • Cellulose vs Starch/sugars
    • Processing
    • Rate of glucose entry into fermentation
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3
Q

Formula for Acetate

A

C2H4O2

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4
Q

Total Acetates synthesized

A

2

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5
Q

ATP generate by Acetate

A

4

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6
Q

Total carbons in acetate product

A

4

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7
Q

total gas lost by acetate

A

2

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8
Q

oxidation equivalent of acetate

A

0

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9
Q

formula for propionate

A

C3H6O2

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10
Q

Total propionate synthesized via the randomizing pathway

A

2

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11
Q

total ATP generated by propionate via the randomizing pathway

A

4

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12
Q

total carbons in propionate via the randomizing pathway

A

6

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13
Q

total gas loss by propionate via the randomizing pathway

A

0

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14
Q

oxidation equivalent of propionate via the randomizing pathway

A

-1

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15
Q

total propionate synthesized via acrylate pathway

A

2

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16
Q

total atp generated by propionate via the acrylate pathway

A

2

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17
Q

total carbons in propionate via the acrylate pathway

A

6

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18
Q

total gas loss by propionate via the acrylate pathway

A

0

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19
Q

oxidation equivalent of propionate via the acrylate pathway

A

-1

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20
Q

formula for butyrate

A

C4H8O2

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21
Q

total butyrate synthesized

A

1

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22
Q

total atp generated by butyrate

A

3

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23
Q

total carbons in butyrate

A

4

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24
Q

gas loss by butyrate

A

2

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25
Q

oxidation equivalent of butyrate

A

-2

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26
Q

formula for pyruvate

A

C3H4O3

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27
Q

total pyruvate synthsized

A

2

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28
Q

total atp generated by pyruvate

A

2

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29
Q

total carbons in pyruvate

A

6

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30
Q

gas loss by pyruvate

A

0

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31
Q

oxidation equivalent of pyruvate

A

1

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32
Q

total acetyl CoA synthesized

A

2

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33
Q

total ATP generated by Acetyl CoA

A

2

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34
Q

total carbons in Acetyl CoA

A

4

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35
Q

gas loss by Acetyl CoA

A

2

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36
Q

formula for succinate

A

C4H6O4

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37
Q

total succinate synthesized

A

2

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38
Q

total atp generated by succinate

A

4

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39
Q

total carbons in succinate

A

8

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40
Q

total gas loss by succinate

A

-2

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41
Q

oxidation equivalent of succinate

A

1

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42
Q

formula for lactate

A

C3H6O3

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43
Q

total lactate synthesized

A

2

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44
Q

total atp generated by lactate

A

2

45
Q

total carbons in lactate

A

6

46
Q

gas loss by lactate

A

0

47
Q

oxidation equivalent of lactate

A

0

48
Q

total CO2 synthesized

A

2

49
Q

total atp generated from CO2

A

2

50
Q

total carbons in CO2

A

2

51
Q

gas loss by CO2

A

2

52
Q

oxidation equivalent of CO2

A

2

53
Q

total CH4 synthesized

A

2

54
Q

total atp generated by CH4

A

4

55
Q

total carbons in CH4

A

2

56
Q

gas loss by CH4

A

2

57
Q

oxidation equivalent of CH4

A

-2

58
Q

Metabolism of glucose under anaerobic conditions causes what?

A
  • no oxygen so no oxidative phosphorylation
    • no conversion of NADH to ATP
    • less ATP production
  • prevents the metabolism of hexose
    • glucose –> VFA instead of Glucose –> CO2+H2O
    • VFA used for energy by animal instead of glucose
59
Q

Oxidation equivalent formula

A

of O atoms - (# of H atoms/2)

60
Q

Outcomes of oxidation equivalents indicate what?

A

More positive = more H produced

More negative = more H consumed

61
Q

Net Oxidation equivalent per mole

  1. Glucose
  2. Acetic Acid
  3. Propionic Acid
  4. Butyric Acid
  5. Carbon Dioxide
  6. Methane
A
  1. Glucose = 0
  2. Acetic Acid = 0
  3. Propionic Acid = -1
  4. Butyric Acid = -2
  5. Carbon Dioxide = +2
  6. Methane = -2
62
Q

Fermentation Balance

Step One

A

Step 1: Oxidation Equivalent Balance

Hexose (0) = Acetate (0) + Propionate (-1) + Butyrate (-2) + CO2 (+2) + CH4 (-2)

63
Q

Fermentation Balance

Step 2

A

Step 2: Total Gas production

CO2 + CH4 = Acetate + 2(Butyrate)

64
Q

Fermentation Balance

Step 3

A

Step 3: Solve simultaneous equations

65
Q

Fermentation Balance

Step 4

A

Step 4: Substitute and solve for methane

66
Q

Fermentation Balance

Step 5

A

Step 5: calculate amount of hexose fermented

67
Q

Fermentation Balance

Step 6

A

Step 6: balanced equation

68
Q

Fermentation balance in VITRO can determine…

A

gas production (or VFA from gas production)

missing products

69
Q

Fermentation balance in VIVO can determine…

A
  • total productio of each product can be calculated if any 1 product is known
    • difficult to measure VFA production (absorption)
    • can measure gas production
      • enclosed chamber
      • head mask
70
Q

Why is fermentation considered inefficient?

A

Up to 20% energy loss via methanogenesis

up to 10% energy used by microbes

71
Q

What are some goals for manipulating fermentation

A
  • enhance a beneficial process
    • propionate pathway over acetate pathway
  • minimize, alter, or eleminate inefficient or harmful processes
    • gas production
      • any reaction that consumes hydrogen instead of producing it is beneficial
72
Q

How can you manipulate CHO digestion?

A
  • type of CHO
  • manipulating rate and extent of digestion
    • manipulate forage maturity (harvest)
    • moisture content
    • processing
    • acid/alkali treatment
    • protein and/or CHO supplementation
    • control rate of intake, pattern
73
Q

How does the rate of glucose entry into fermentation affect digestion?

A
  • changes fermentation pathways
    • slower digestion –> increased acetate
      • higher ATP yield per glucose
      • make most of limited glucose supply
    • Faster digestion –> increased propionate
      • lower ATP yield per glucose
      • greater overal ATP production (surplus of glucose)
      • leads to more energy being available to the animal and not used by the microbes
74
Q

What are some advantages to feeding ionophores?

A
  • inhibit methanogenesis
    • less gaseous energy loss
    • removal of primary H sink
  • promotes use of alternative H sinks
    • propionate, butyrate, ethanol
  • greater energy capture in a usable form (VFA)
  • reduce incidence of acidosis
75
Q

What are some disadvantages of feeding ionophores?

A
  • general decrease in Gram + bacteria, protozoa
    • decreased cellulolytics
    • depressed fiber digestion (esp. short term)
  • negative influence on feed intake
  • public fears of antibiotic resistance
76
Q

What is an advantage of lipid supplementation?

A
  • physically coats the fiber particles, decreasing attachment
    • negative effects on fiber degradation
77
Q

How is feeding a lipid supplementation advantagous?

A
  • USFA are toxic to Gram + bacteria, protozoa
    • double bonds alter cell membrane permeability
    • similar effect to that seen with ionophores
78
Q

How does providing alternative H sinks affect the ruminant?

A
  • Unsaturated Fatty Acids
    • biohydrogenation uses H
    • fat feeeding reduces methanogenesis
    • amount to replace methanogenesis is impractical
  • Ethanol Supplementation
    • conversion of ethanol to acetate uses H
    • stimulation of fiber digestibility
    • delivery is logistical problem
79
Q

How does feeding a nitrate, sulfate reduction work as a H sink?

A
  • higher affinity for H than methanogens
  • free energy change greater than methanogenesis
  • toxicity is a concern
    • nitrate poisoning, PEM
80
Q

What are some advantages to reductive acetogens?

A
  • VFA capture vs. loss via methane
  • promotes fiber degredation
  • increased acetate production (milk fat)
  • decreased greenhouse gas emmissions
  • decreased public scrutiny over antibiotic use
81
Q

What is a problem with reductive acetogens?

A
  • methanogens outcompete acetogens
    • lower affinity for H
82
Q

Where are some competitive acetogens located?

A

wood eating termites

human hindgut

kangaroos

83
Q

What is DFD?

A

Direct-Fed microbials

adding new or improved organisms to the rumen

84
Q

What are some possible advantages to DFDs?

A
  • increased acetate production
85
Q

What are some disadvantages to DFDs?

A
  • generally short lived
    • uncompetitive w/existing microbes
    • require chronic supplementation
    • expensive –> impractical
    • interaction with feed/nutrional mgt.
86
Q

Why do microbes attach?

A
  • avoid washout from rumen
    • need generation interval to exceed rumen turnover rate
  • substrate competition
    • allows bacteria to seek and sequester their preferred substrate
      • identify digestible substrate
      • minimize loss to secondary fermenters
    • dominance over substrate environment
87
Q

Who attaches??

A
  • attachment is more common in slower digested CHO
    • cellulolytics >50% attached
    • amylolytics <1% attached
  • attached pass out with particles, remain resident
  • unattached pass with fluid
    • fluid passage is typically 2x faster
    • shoer generation interval reqt.
88
Q

What do microbes attach to?

A
  • feed particles (substrate)
  • rumen wall
  • protozoa
89
Q

Where do microbes attach?

A
  • access through broken edges of plant surface
    • stomata
    • cut ends (importance of processing, mastication)
    • role of fungi
90
Q

Describe nonspecific attachment

A
  • No structural attachment involved
    • H bonding
    • electrostatic attraction, ion gradients
    • pH, temperature, hydrophobicity
  • weaker than specific attachment
    • most are reversible
91
Q

Describe specific attachment mechanisms

A
  • use cell membrane structure for attachment
    • majority are nonreversible
  • capsules
  • fimbrae or pili
    • surface appendages
  • glycocalyx
    • external glycoprotein filaments
      • conform to substrate surface
      • may aid in sequestering substrate
    • used by some cellulolytics
92
Q

What are the components of cellulosome?

A
  1. membrane anchor
  2. scaffolding
  3. CHO binding module
    1. glycosylated protein structure
    2. binds specifically to cellulose
    3. aids in penetration of cellulose fibers
  4. enzyme subunits
    1. cellulase
    2. metabolic enzymes
93
Q

What happens once microbes are attached?

A
  • colinize/reproduce
    • etching of “pits” on particle surface
      • follows patter of cellulose strands
    • competition for attachment sites
      • replacement of old with new cells
      • maintains active metabolism
    • proceeds until all attachment sites are taken
      • new cells then release for new substrate
94
Q

How do microbes use ATP from CHO fermentation?

A
  1. maintenance
    • repair of cellular structure
    • ion gradient across cell membrane
    • enzymes
    • attachment
  2. Growth
    • cellular division
    • increase in cell numbers, pop. size
95
Q

Describe a batch culture

A
  • classical culture for microbes
  • substrate added only at beginning of fermentation
  • population growth limited by supply of substrate added
  • endproducts, microbes only removed at the end
96
Q

Describe a bacterial growth curve. Label it completely

A
97
Q

Describe the lag phase

A
  • period of adjustment
    • locating substrate
    • attachment
    • enzyme synthesis
    • “young bugs”
  • little or no growth occurs
    • population remains stable
    • typically few minutes to several hours
    • ATP supply is about maintenance
98
Q

Describe the log phase

A
  • period of rapid, exponential growth
    • ATP supply surpasses maintenance
    • not limited by supply of substrate available
      • surplus of attachment sites
    • more replication than death
99
Q

describe the stationary phase

A
  • overall numbers remain constant
    • Rate of ATP production slows
      • less digestible substrate
      • fewer attachment sites
      • initial pop (“old bugs”) starved for substrate
        • energy insufficient for maintenance –> cell death
    • Accumulation of endproducts
      • VFA, H, intermediate acids, etc.
      • feedback inhibition
    • replication rate = death rate
100
Q

Describe the death phase

A
  • energy available is less than maintenance
    • lack of digestible substrate
    • insufficient energy to support normal cellular function
  • rapid cell death
  • overal decrease in pop. size
  • recycling of nutrients from lysed cells (intra-ruminal recycling)
101
Q

How is the rumen an example of a continuous culture?

A

repeated (continuous) addition of substrate

continual removal of endproducts/microbes

102
Q

Describe a continuous culture fermenter

A

It has four components. Solid, liquid, stirrer, and effluent.

solid–> substrate/feed into vat

liquid –> McDougals buffer into vat

Stirrer–> continous mixing of vat

Effluent–> removal of endproducts

103
Q

Why does the microbial growth curve still apply in a continous culture?

A
  • numberous simultaneous growth curves
    • each feeding/meal
    • each forage, grain particle
    • each site of attachment on particle
  • results in microbes at nearly all stages of growth
  • positon of average bug is more important
104
Q
A

A) Max MOEFF

B) Ideal

C) Dairy cow

D) HQ forages

E) Grain diet

F) LQ forages

105
Q

What is microbial efficiency and how can it be quantified?

A

yield of microbes per unit of substrate

  • Yatp
    • grams of microbial cells/mole of ATP
  • MOEFF
    • gram of microbial Nitrogen/Kg of organic matter fermented
    • standard used
106
Q

How does passage rateK effect MOEFF

A
  • Kp = dilution rate (D)
    • inverse of retention time (Kp = 1/RT)
    • typical range of 2-10%/hr
  • increased Kp
    • selection for faster growing species
    • reduced intraruminal recycling
      • wash out old bugs prior to death phase
      • maintain younger pop, more active bugs
    • shift average bug to the left in growth curve
107
Q

Where does max MOEFF occur?

A
  • Max MOEFF occurs at Kp higher than observed in vivo
    • junction of lag and log
      • generation interval exceeds retention time
    • Kp needed to achieve max MOEFF
      • FC–> 17.7%/hr
      • NFC–> 31.4%/hr
108
Q

What is the primary deterinant of kp?

A

Feed intake

  • forage
    • gut fill limits voluntary feed intake
      • relationship to forage NDF
    • higher quality –> greater DMI –> faster Kp
  • Concentrate
    • intake is limited by chemostatic regulation
      • VFA, pH
    • higher grain –> depression in DMI –> slower Kp
109
Q

What are the typical Kp values observed in:

beef cattle, sheep?

dairy cattle?

A
  • Beef cattle, sheep
    • 2-6%/hr
      • forage intake (gut fill), forage quality
  • Dairy cattle
    • 5-10%/hr
      • higher forage intake, quality

Physiology of Rum Digestion (3 decks)

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