the principals of assessing energy balance and metabolic rate in mice

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The principals of assessing energy balance and metabolic rate in mice. Energy balance- the game has changed. Tschop et al Nature methods Dec 2012. What is energy balance? And what is metabolic rate? How do they differ?. 30g. 30g of energy out. 31g of food in. 1 tonne of energy out. - PowerPoint PPT Presentation

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The principals of assessing energy balance and metabolic rate in mice

Energy balance- the game has changed.

Tschop et al Nature methods Dec 2012

What is energy balance? And what is metabolic rate? How do they differ?

1. ENERGY BALANCE

Energy balance describes whether an organism will lose or gain weight. The sizeOf the organism is actually irrelevent for considering energy balance

31g of food in 30g of energy out

1 tonne and 1gOf food in

1 tonne of energy out

30g31g

1 tonne1 tonne+ 1g

1. ENERGY BALANCE and rate of weight gain

31g of food in 30g of energy out

1 tonne and 1gOf food in

1 tonne of energy out

5g of food in 5g of energy out30g

7g of food in 7g of energy out30g

2 Metabolic rate With metabolic rate then the size of organism is very important, as we care aboutThe rate of energy expenditure per g of the organism

How do we measure food intake, energy balance and metabolic rate?

To assess metabolic rate and energy balance we need 3 pieces of information

1. The body weight of the animal2. The amount of food it has consumed*3. How much energy it has expended

*This correctly should be the amount it has assimilated. See point at the end.

How to measure food intake and body weight

How to measure energy expenditure

Direct Methods: Calorimetry

Indirect methods:Indirect calorimetry using gas exchange Doubly labelled water

Direct Calorimetry

Advantage – Most accurate method – does exactly what it says on the tin.Disadvantage – Sealed unit so (very) limited time scale for experiments.

Indirect calorimetry 1 - Doubly labelled water

Indirect calorimetry 2 – Gas exchange indirect calorimetry.

What is an indirect calorimeter?

Why does the volume of the chamber matter matter?

0

0.2

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1.2

0 20 40 60 80 100

6.2l chamber2.7l chamber10l chamber

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uncorrected for lag

corrected for lag

What is Energy expenditure and how does it relate to VO2 and VCO2?

EEJ = 15.818xV02 + 5.176*VCO2

Indirect calorimetery

Oxygen consumption gives an indication of howmuch energy an animal is using as an animal will produce 4.8 KJ of energy per litre of oxygen with an error of around 6%.

Carbon dioxide production is not a 1 to 1 ratio with oxygen consumption and the amount of CO2 produced varies dependenton the source of energy (Carbohydrate, Fat or Protein)

Combining the amount of O2 consumed and the amount of CO2produced gives us a value called the RQ which provides some information about fuel usage.

RQ varies from 0.7-1 dependent on metabolic status of the animal

What is Energy expenditure and how does it relate to VO2 and VCO2?

EEJ = 15.818xV02 + 5.176*VCO2

Varies in ratio to oxygen by about 30%

Therefore about 75% of the EE equation is driven by VO2

25% is driven by VCO2, which varies from equalling V02 to -30%.

Therefore 25% of 30 ~6% - the error based on using 02 alone

How about RQ

A bit more on RQ

RQ is VC02/VO2

Burning carbohydrate has an RQ of 1 because….

C

H

H

O

n

C

O

O

O

H H

Therefore1 molecule of C02

Requires 1 molecule of O2

Burning fat has an RQ of 0.7 because….

C

O

O

O

H H

Therefore1 molecule of C02

Requires 1.5 molecules of O2 (1/1.5 =0.7)

C

H

H n

What does RQ mean?

-Gives information about substrate utilisation

-Mice fed a constant diet, so in mice probably gives more information aboutEnergy balance

-Can exceed 1 during periods of lipogenesis as lipogenesis has an RQ of about 5

Important points about body weight and experimental design

Important point 1

A larger mouse WILL expend more energy than a smaller mouse.

y = 0.8886x + 9.4426

R2 = 0.3757

05

101520253035404550

0 10 20 30 40Bodyweight (g)

EE

J/m

in/m

ou

se P=0.0004

Important point 2

A larger mouse WILL eat more energy than a smaller mouse.

y = 22.081x - 143.72

R2 = 0.8604

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bodyweight (g)

Fo

od

inta

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in (

kj/m

ou

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/8 d

ay

s)

Important point 3

When you study your mouse is very important.

Time

Bod

ywei

ght

If a larger animal expends more energy, and my genotype of interest is obese andgetting more obese relative to a control when I measure it, will it have higher EE?

Probably yes

Therefore what I need to know is if my animal has a DISPROPORTIONATELY Low EE for its body weight.

How to normalise for body weight

Traditionally people have used different comparators to straight body weight.

A particularly common comparator is BW0.75

However this, along with BW0.72 and BW0.66 were developed for comparing across species, ie elephants and mice, rather than within species.

Evidence suggests that these are not valid comparators for comparing within species, as we do.

So then describing oxygen consumption as

Ml/min/kg0.75 is not a good idea

Is there a better method?

ANCOVA

Body weight

Oxy

gen

cons

umpt

ion

How ANCOVA works

A group with increased oxygen consumption per gram of animal

Body weight

Oxy

gen

cons

umpt

ion

A group with the same oxygen consumption per gram of animal

Body weight

Oxy

gen

cons

umpt

ion

Idealised regression

When regression lines are not equal

How about some real data?

O2 ml/min/mouse Body weight gWT 2.1425 34WT 2.3217 32WT 2.108415148 31.1WT 2.3652 29WT 2.240591387 34.3WT 2.2635 35WT 2.40299321 37KO 1.5767 29KO 2.0381 33KO 1.9817 29KO 2.110144446 33.3KO 2.031 33KO 1.979966916 37.7KO 2.318167748 37.6KO 2.220327835 35.4

Using traditional ml/min/kg0.75

NS

Now to do some ANCOVAring

Body weight

Oxy

gen

cons

umpt

ion

Idealised regression

When regression lines are not equal and cross in the area of interest

So the model is valid!

Oxygen consumption rates from 7 month old male mice fed a standard laboratory chow diet. Data collected from free living mice with ad libitum access to food over a period of 48 hours. N=8 per group. Chow diet. Oxygen consumption expressed as adjusted means based on a normalised mouse weight of 33.36g determined using ANCOVA. P<0.05

Other types of data from metabolic caging systems

Delta Body weight: weight out - weight inImportant for understanding RQ and energy balance. Mice in negative energybalance will have lower RQ values (they are oxidising fat) and potentiallydisproportionately low EE (as they try to maintain fat reserves)

ActivityA very poorly understood variable and would take about another hour to discuss fully. I would recommend entirely ignoring this for the time being.

Water intakeMay give very useful information regarding diabetic mice/kidney function. In generalnot a major variable in energy balance studies, however a dramatic loss in weight may be attributable to a failure to drink. Important to check this.

Other types of data

Gut assimilation efficiencyNot all food that is consumed will be absorbed into the body. Mice general operate inThe mid-80% range for assimilation efficiency. Assimilation efficiency can be assessed using a bomb calorimeter to measure fecal energy content.

Conclusions

Energy balance describes the processes by which animals get fat.

Metabolic rate describes the unitary mass energy expenditure.

Assessing these variables accurately is essential for determining why your mouse is obese, lean or has altered metabolic function.

Practical session

1. Testing that your data is valid

2. How to analyse your data.

Testing that your data is valid

Basic checks

An important point about how a calorimeter measures CO2 and O2

CO2 = Air out – air in to give VCO2O2 = Air in – Air out to give VO2

This is not a trivial difference!!!

Lets imagine that we have a lot of water in our cage…

The water dilutes the gases in the cage, reducing the CO2 and O2 measurementsComing out of the chamber

02in = 100% of normalCO2in = 100% of normal

02out = 90% of normalCO2out = 90% of normal

CO2 = Air out – air in to give VCO2O2 = Air in – Air out to give VO2

These errors do not cancel because:

CO2 = 90 -100% of correct values so error is negative O2 = 100 – 90% of correct values so error is positive

Therefore excessive moisture will drive down RQ as RQ = VCO2/VO2

Unusually low RQ values are indicative of an issue with moisture or drying…

If all cages drift down, as well as the room air value, it may suggest the Air drying on your system is not working.

If one animal has a drift in RER it may be a problem with a specific cage

Exercise 1 – checking the data

On your computer you will find the EXCEL sheet ‘training analysis’.

Please open this file and proceed to tab ‘training data 1’

On this data set please graph (scatter plot) the O2% for room air

Also place on a separate graph the RQ values for each mouse.

You can compare this to training data 2, which does not have any problems

Can we see any problems?

Exercise 2

Look at training data 1

Please graph the EE column for each mouse

Can we see a problem? What might have caused it?

Summary of checks:

Make sure zero gas = 0Make sure span gas gives same reading as the bottleMake sure room air gives stable readingsMake sure RQ average is sensible (average of 0.9 for high carb diet, nearer to 0.78 for HFD)

Make sure there are no freak out lying values (may happen occasionally for CO2).

Make sure data shows expected circadian rhythms (EE and RQ high at night, low in the day)

Check all cages have roughly similar SD for RQ and EE.

The minimum EE would normally be expected to be the first value (due to time constant of cage).

How to analyse the data

At this stage we will just take the average RER and average EE for the data setWe have

I have generated this data for you in the sheet called training analysis 1

Please calculate the average EE, body weight and perform a ttest between them

Formula:

=TTEST(G3:G9,G10:G16,2,2)

So the animals have a lower average energy expenditure, while they do not have A significant difference in body weight there is a tendency for it to be lower. We now want to know if they also have a lower metabolic rate (EE per KG)

To do so we need to normalise for body weight using ANCOVA.

First we can get a feel for this by plotting BW on the x axis and EE on the y axis ofA scatter plot. If the data fall on the same line, then the reduced EE is just caused by Reduced body weight.

Try and make a plot with two data series placing BW on the x axis and EE on the y axis

This is what you should get. As we can see the lines are parralel. We now need to seeIf they are statistically different…

ANCOVA.

EE vs Average BW

y = 0.4974x + 22.004

R2 = 0.5976

y = 0.5948x + 20.747

R2 = 0.6232

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25 27 29 31 33 35 37

Weight (g)

EE

J/m

in/m

ou

se

Series1

Series2

Linear (Series2)

Linear (Series1)

How to actually RUN the ANCOVA

SPSS data input screen – copy data into here

Select variable tab to show this screen – type in names and change measure type.

Select variable tab to show this screen – type in names and change measure type.

Select ‘data’ menu and select define variable properties

Select ‘genotype’ and press continue

Type ‘WT’ and KO into the label section. Press ‘ok’

Select analysis menu > general linear model > univariate

Make EE the dependent variable, genotype the fixed factor and body weight thecovariate

Select options and then ‘display means for:’ genotype. Tick compare mainEffects box. Hit ok! The analysis will run.

Test of between subject effects should be signficant for BW and genotype.

Estimated marginal means show mean and standard error based on a 32.56g mouse

Testing model validity…

Go back to: analysis menu > general linear model > univariateSelect model button.

Select ‘custom’ model. Move genotype and BW indidually, then select both and Move them across to test for an interaction. Press continue and run the model.

Ignore every thing but the genotype* BW value – this should be NON-significantIf your model is valid…

EE vs Average BW

y = 0.4974x + 22.004

R2 = 0.5976

y = 0.5948x + 20.747

R2 = 0.6232

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25 27 29 31 33 35 37

Weight (g)

EE

J/m

in/m

ou

se

Series1

Series2

Linear (Series2)

Linear (Series1)

As you can see the lines do not cross in the region of interest…

Exercise 3

Open the ‘formatted for SPSS tab’

Copy the data into SPSS and run the analysis.

Refer to the hand out for instructions on how to do this…

Compare the results of the ANCOVA analysis to the EE/BW and EE/BW^0.75 Data in the excel sheet ‘training analysis 1’ tab. What do you find?

Bonus exercises

If you have not already done so try plotting delta body weight against RER For the data on ‘training analysis 1’.

What do you find and why?

Secondly look at bonus training analysis.

Check the average and the significance (t-test) of all the different values.

Are any significant? If so what do you think this may tell us biologically?

So overall…

1. Analysis of energy expenditure for Energy balance should not be normalisedto body weight

2. Normalisation of energy expenditure or food intake should ideally be conductedUsing ANCOVA. If this is not possible, then stick with per mouse.

3. Activity data can simply be expressed as counts per mouse.

4. Metabolic chambers will give you VAST quantities of data but much of it ismeaningless unless you consider how you are analysing it carefully.

Answer to bonus question 1

RER vs delta body weight

y = -39.852x + 38.606

R2 = 0.7325

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4

0.88 0.9 0.92 0.94 0.96 0.98

RER

wei

gh

t lo

ss

Series1

Linear (Series1)

Answer to bonus question 2

Lipolysis at E max is impaired in basal and B3-adrenergic receptor stimulatedstates

WT KO

Lipid Clearance TG

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Time (Hours)

TG WT

PPARy2

Answer to bonus question 2

And in human obesity this is what we see…

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