Vanessa Goodman, Allison FisherZOOL 430Dr. Butler10/4/21

Comparative Digestive Techniques for the Massive Herbivorous Dinosaur: Borealopeltamarkmitchelli

Introduction● Vertebrate digestion follows a basic model of stomach-> small intestine -> large intestine

but with intense variation based on type of diet and digestive mechanisms

● Digestion tract models provide insight into how organisms process energy and can

therefore give clues into various other traits like feeding behaviors and adaptations

● Our animal ate only plant matter with majority of its diet coming from l eptosporangiate

ferns (Brown et al. 2020)

● Our animal had a gastrolith matrix for enhanced breakdown of plant material inside the

posterior stomach (Brown et al. 2020)

● No resources provide exact nutritional breakdown of these plants so approximate nutrient

ratios obtained through comparison of tree fern, a vegetable fern consumed by humans

● It’s unlikely that an animal of such massive size would be able to efficiently obtain

enough energy for growth and locomotion from plainly digesting plants with low

concentrations of available carbohydrates and proteins. Previous studies find that it is

more likely that giant herbivorous dinosaurs relied on fermentation as a means to utilize

the high volumes of cellulose present in the available flora (Farlow 1987). Specifically,

hindgut fermentation is most likely due to the animals enormous size which wouldn’t be

well supported with the delayed digestion involved in foregut fermentation (Farlow

1987).

● One study found the metabolic energy released by leptosporangiate ferns and other food

sources which are comparable to what would have been available to B. markmitchelli via

in vitro fermentation studies using symbiotic microorganisms (Hummel et al. 2008)

● The aim of this study is to create two digestion models: one of basic nutrient acquisition

via mechanical and chemical breakdown followed by small intestine absorption and one

with heavy reliance on fermenting micro symbionts for energy acquisition

● We hypothesize that hindgut fermentation was an essential mechanism for optimal

digestion of low-energy plant materials in B. markmitchelli. The confirmation of this

adaptation would yield strong supporting evidence for other behaviors such as selective

feeding which has been speculated by many researchers.

MethodsBasal Model

Digestion tract dimensions and structure were modeled using the nutritional content of a

tree fern, Diplazium esculentum. Using data from the USDA FoodData Central on percentage

composition of each nutrient coupled with energy density of each nutrient (carbohydrate, protein,

fat), we were able to calculate the net amount of energy per gram of food that the dinosaur would

obtain with consumption.

Percent of food mass*Energy density of nutrient (kJ/g)=Energy contribution (kJ/g)

The energy contributions were summed to find the total energy density of a tree fern. The

amount of tree fern required to assimilate per day was found using daily metabolic rate (Fisher

Goodman 2021).

Amount of food assimilated required per day (g/day)= daily metabolic rate

(kJ/day)/energy density of food (kJ/g)

Additionally, amounts of each nutrient assimilated per day were found and converted to

moles.

Moles of nutrient required per day (mol/day)=(Amount of food assimilated per day

(g/day)*nutrient percent of food mass)/molecular weight of nutrient (g/mol)

Dietary fiber was originally treated as an indigestible food source when building this

model absent of fermenting adaptations.

The amount of food that would need to be ingested and assimilated in order to meet the

energetic demands of daily metabolic rate was calculated using an assimilation efficiency value

of 50%. This number is comparable to previous estimates of other megaherbivore dinosaurs

based on the high fiber content of available forage (Weaver 1983).

Mass of food ingested (g/day)=mass of food assimilated per day(g/day)/assimilation

efficiency as a percentage

The mass of a single fern frond was calculated using dimensions of the plant to find its

volume. Volume of a rectangle divided by 2 may accurately describe the shape of the plant,

(Figure 1).

Figure 1. Approximating the volume of a ground fern.

V (m3)=(l*w*h)/2

The density of water was chosen as an accurate representation of the density of the plant

since the USDA reports tree fern having 88% water content (2021).

Density of water (kg/m3)*volume (m3)=mass of tree fern (kg)

Using this calculated mass we were able to deduce the number of food items that the

dinosaur would have to consume per day for both digestive tract models.

Number of items per day=Mass of food ingested (kg/day)/mass of tree fern (kg)

Since carbohydrates are the most abundant nutrient of the food source, it was considered

as the limiting factor when investigating the digestive tract surface area and dimensions. The rate

of glucose uptake in the small intestines of reptiles was corrected to the body temperature of our

animal during its daytime digestive phase (31.56oC) using a Q10 conversion (Fisher, Goodman

2021; Withers 1992).

Uptake rate at body temp=rate at 37oC(2.5)(Tb-37)/10

The required surface area of the gut was computed using the amount of carbohydrate

intake per day and the uptake rate per cm2.

Surface area (cm2)=mol carbohydrate (mol/day)/uptake rate (mol/day/cm2)

Gut diameter was estimated based on the dimensions of a cololite fossil (Fig. 2) which

was assumed to represent the approximate size of the posterior stomach (Brown et al, 2020).

Looking at the attachment sites of stomach and small intestine in other animals, 5 cm seems to be

an appropriate diameter of the small intestine tube.

Figure 2. Model of the ellipsoid cololite specimen with dimensions measured in cm (Brown et

al. 2020).

The surface area of the gut is represented by that of a cylinder. Accordingly, we were able

to speculate gut length.

Length (cm)=Surface area/pi/diameter

Fermentation Model

As an alternative model for digestion, we speculated that B. Markmitchelli utilized gut

fermentation via symbiotic microorganisms to increase the nutritional yield of its foods.

Using the gas production of fermenting microbes in vitro, Hummel et al. found the gross

energy of dry matter from a variety of leptosporangiate ferns, representing those which would

have been consumed by B. markmitchelli (2008). This was corrected to metabolizable energy

using previously established coefficients (Weaver 1983).

The amount of dry weight needed to sustain metabolic needs of B. markmitchelli was

found using:

Daily metabolic rate (kJ/day)/energy density of various ferns (kJ/kg)

The weight of fresh ferns required was calculated using a ratio of 0.27 for average dry

mass:fresh mass (Weaver 1983). The number of ferns required was found using the same mass

model method used in the basal model.

Given that crude protein represents 11.5% of dry mass of leptosporangiate ferns, proline

uptake was used as a limiting factor in building the small intestine model for this scenario

(Hummel et al. 2008). Moles of protein needing to be absorbed from daily food intake was found

using:

(Protein percentage of dry mass*amount of dry mass required(g))/molecular weight of

protein (g/mol)

The rate of proline uptake for a lizard was assumed to be the limiting factor in this

scenario (Withers 1992). Body temperature correction, surface area, and length of the small

intestine were calculated using the same methods as the basal model.

Results

Basal Model

Table 1. Energy values and contributions of macronutrients in the tree fern.

Nutrient Percent of totalmass (USDA)

Energy density(kJ/g) (Withers1992)

Energycontribution(kJ/g)

Moles of foodassimilated(mol/day)

Carbohydrate 11% (-3.7%fiber)=7.3% net

15.9 1.1607 1.81

Protein 0.29% 18.15 0.05264 0. 099

Lipid 0.07% 39.2 0.02744 0.012

Water 88.6% 0 0

Total 1.24

The tree fern was found to provide 1.24kJ of energy per gram of weight (1.1607kJ/g of

carbohydrate, 0.05264kJ/g of protein, and 0.02744kJ/g of lipid) (Table 1). To support the daily

metabolic rate of B. markmitchelli, it must have assimilated 4647.09g of tree-fern equivalent

food. To assimilate this quantity at 50% efficiency, it must have consumed 9294g or 20.12 full

fern fronds per day. Given the need to uptake 1.81 moles of carbohydrate daily, the surface area

of the small intestine would have been 3201.7cm2 with a length of 2.04 meters.

Fermentation Model

Using the microbial fermentation of leptosporangiate ferns, 0.874 kg of dry fern matter

would have supplied enough metabolizable energy to support daily metabolic rate. This amount

is equivalent to 3236 grams of fresh fern or 7 full fronds. The protein content in this amount of

food is 100.5g or 0.74 moles. In order to uptake this amount of protein, the small intestine would

have had a surface area of 3121.66cm2 and a length of 1.99 meters.

Table 2. Comparison of parameters between 2 digestion models.

Parameter Non-Fermentation Model Fermentation Model

Mass of food assimilated (g/day) 4647.09 3237

Mass of food ingested (g/day) 9294 6474

Number of food items 200.07 139.37

Required surface area of intestine(cm2)

3201.7 3121.6

Gut length (m) 2.04 1.99

Figure 2. Gastrointestinal model for B. Markmitchelli.

Discussion

Number of Food Items

The number of food items required when digestion was modeled without fermentation

was over 20 full tree fern fronds. With fermentation, this number decreased to only 7 full fronds.

The narrow shape and size of B. markmitchelli’s snout provides evidence that it was a selective

feeder and likely spent long periods browsing for forage (Brown et al. 2020). This means it was

not eating an entire frond at once but grazing on different plant individuals throughout the day.

With the fermentation model, it needed to consume less plant volume and therefore could be

more selective when browsing. This is supported by evidence that a very small ratio of its

stomach contents was represented by less attractive plant material like twigs (Brown et al. 2020).

This type of material may have been more present if less discretion was used when feeding,

which would have been necessary to obtain the food volume sufficient for non-fermenting

digestion.

Small Intestine Size

The resultant small intestine surface area and length from both models were strikingly

similar, though seemingly small. For an herbivore of such size, a small intestine of about 2

meters, spanning less than half of the body length (5.5 meters) is likely an underestimation

(Brown et al. 2017). For the basal model, this could have been the result of the food source

referenced. Because no available sources give the macronutrient breakdown of a more accurate

fern type, the USDA databank supplied data on a human consumable tree fern (USDA 2021).

Use of this human food source likely overestimates the amount of available energy and

carbohydrates and underestimates non-digestible material as humans don’t possess the cellulose

digesting capabilities that most herbivores utilize (Withers 1992). In reality, the ferns consumed

by B. markmitchelli likely contained a much higher concentration of cellulose which wouldn’t

have been broken down into digestible material by mechanical and chemical processes alone.

The amount of usable carbohydrates would likely represent a lower fraction in the overall mass

of food consumed. Therefore, the energy density of the food source would be lower, the amount

consumed would be higher, and the small intestine would be longer to increase absorption

opportunity.

The high ratio of net carbohydrates in the human-suitable food source chosen makes

glucose absorption the limiting factor when calculating small intestine surface area and length.

However, the carbohydrate content of the more accurate food source analyzed in the

fermentation scenario was not provided. Crude protein, as a fraction of dry food mass, was the

only macronutrient ratio given for this scenario. In absence of other nutrient amounts, it was

assumed that absorption of protein was the limiting parameter when calculating surface area and

length of the small intestine. This may have caused the resulting small intestine size to be

obscured because herbivore digestion is normally not dependent on protein uptake (Withers

1992). Additionally, crude protein amounts present in the food source does not necessarily

represent the amount of protein effectively absorbed, which could be measured if protein content

of excrement was known.

Though it is clear that some confounding factors have contributed to a smaller small

intestine than expected based on body size, there is some biological evidence that may

rationalize the function of this decreased size. Primarily, the presence of gastroliths found in the

cololite fossil of the posterior stomach (Brown et al. 2020). Gastroliths are stones held in the

stomach for grinding food to a much smaller particle size than could be obtained by biting or

shearing alone. B. markmitchelli was not known to have an intensive food milling apparatus in

the mouth so it likely relied on gastric grinding to increase nutrient availability of food

consumed. This type of processing is advantageous because it decreases the time needed to attain

adequate digestibility in the gastrointestinal tract (Farlow 1987). For animals without this type of

breakdown, digestibility is compensated with a longer small intestine and a longer duration of

digestion. The presence of gastroliths accounted for 50% of the stomach composition in the

cololite fossil (Brown et al. 2020). In order to accommodate this spatial demand, the stomach

was probably enlarged and other organs, like the small intestine, were consequently decreased.

Conclusion

Based on the comparative models for digestion with and without fermentation, we can

speculate that fermentation by symbiotic microorganisms in the gut played an important role in

digestion. Fermentation requires a decreased load of food mass consumed daily which

accommodates more selective feeding behaviors. Additionally, food particle reduction by gastric

grinding made digestion more efficient and nutrient uptake more available even with a reduced

small intestine length. The combination of these digestive feeding mechanisms meant that B.

markmitchelli was able to grow and maintain its massive body size even when available food

sources had generally low energy density.

To support and improve these findings, a more accurate reference for nutrient

composition of the food actually consumed by B. markmitchelli would provide a better

comparison between these two digestive techniques. Better understanding of what ratio of

nutrients were available in the food supply could give insight to which nutrient uptake rate

limited digestion in the small intestine to produce a more accurate size model.

ReferencesBrown, C. M., Henderson, D. M., Vinther, J., Fletcher, I., Sistiaga, A., Herrera, J., & Summons,

R. E. (2017). An Exceptionally Preserved Three-Dimensional Armored Dinosaur RevealsInsights into Coloration and Cretaceous Predator-Prey Dynamics. Current biology : CB,27(16), 2514–2521.e3. https://doi.org/10.1016/j.cub.2017.06.071

Brown, C. M., Greenwood, D. R., Kalyniuk, J. E., Braman, D. R., Henderson, D. M.,Greenwood, C. L., & Basinger, J. F. (2020). Dietary palaeoecology of an EarlyCretaceous armoured dinosaur (Ornithischia; Nodosauridae) based on floral analysis ofstomach contents. Royal Society open science, 7(6), 200305. Retrieved from:Dietary palaeoecology of an Early Cretaceous armoured dinosaur (Ornithischia;Nodosauridae) based on floral analysis of stomach contents

Farlow, J. O. (1987). Speculations About the Diet and Digestive Physiology of HerbivorousDinosaurs. Paleobiology, 13(1), 60–72. http://www.jstor.org/stable/2400838

Fooddata Central Search Results. FoodData Central. (n.d.). Retrieved October 1, 2021, fromhttps://fdc.nal.usda.gov/fdc-app.html#/food-details/170464/nutrients.

Hummel, J., Gee, C. T., Südekum, K. H., Sander, P. M., Nogge, G., & Clauss, M. (2008). In vitrodigestibility of fern and gymnosperm foliage: implications for sauropod feeding ecologyand diet selection. Proceedings. Biological sciences, 275(1638), 1015–1021.https://doi.org/10.1098/rspb.2007.1728

Stewart, R. (n.d.). U.S. Forest Service. Forest Service Shield. Retrieved October 3, 2021, fromhttps://www.fs.fed.us/wildflowers/plant-of-the-week/athyrium_filix-femina.shtml.

Weaver J.C. The improbable endotherm: the energetics of the sauropod dinosaur Brachiosaurus.Paleobiology. 1983;9:173–182.

Outline (used for calculations, will be made into appendix):● Daily metabolic rate?

○ 5766 kJ/day

● Anaerobic metabolism? Byproducts?● What did it eat?

○ Stomach contents of suncor nodosaur were analyzed and shown to have 88% leafmatter of which 85% was leptosporangiate fern remnants (Brown et al. 2020)

● How did it obtain this food?○ Selective browsing - narrow snout

● Photosynthetic or chemosynthetic symbionts?● Describe and draw GI tract. Include: storage, mechanical breakdown, chemical digestion,

secretion of digestive enzymes, absorption, and storage of fecal material○ Gastrolith is a stone held in the stomach for grinding food for animals that lack

grinding teeth (Brown et al., 2020)○

● Did it use cellulase symbiotic microbes? Include fermentation region● Dietary model

○ For each food type: percentages of carbohydrates, fats, protein, water, andindigestible material (USDA, 2021)

■ Tree Fern (USDA, 2021):● 11% Carbs (3.7% fiber), 0.07 % lipid, 0.29% protein, 88.6% water● 167 kJ/100g

○ Use these percentages to calculate the overall energy density of the food (kJ g-1)■ Molecular weights: Carbohydrates=180g/mol, lipids=256g/mol,

protein=135g/mol■ Energy: Carbohydrates=15.9 kJ/g, Lipids= 39.2kJ/g, Protein=18.15kJ/g

(Withers, year)Without fermentationCarbohydrates

● 11% - 3.7% of which is fiber● Net 7.3% Carbohydrates● .073 * 15.9 kJ/g = 1.1607 kJ/g

Lipids● .0007 * 39.2 kJ/g = .02744 kJ/g

Protein● .0029 * 18.15 kJ/g = .05264 kJ/g

Non-Digestible material● 3.7% fiber

Water● 88.6%

Net energy from ferns● 1.1607 kJ/g + .02744 kJ/g + .05264 kJ/g = 1.24 kJ/g

Food mass assimilated to support DMR of 5766 kJ/day

● 5766 kJ/day / 1.24 kJ/g = 4647.09 g/day● Mass of Carbs

○ 4647.09 g/day * .073 = 325.29 g Carbohydrates/day● Mass of Lipids

○ 4647.09 g/day * .0007 = 3.25 g Lipids/day● Mass of Protein

○ 4647.09 g/day * .029 = 13.47 g Protein/day● Convert grams to moles

○ 325.29 g / 180 g/mol = 1.81 mol Carbohydrates/day○ 3.25 g / 256 g/mol = .012 mol Lipids/day○ 13.47 g / 135 g/mol = .099 mol Protein/day

Food mass ingested to support DMR of 5766 kJ/day● Assume 50% digestive efficiency based on high fiber diet (Weaver 1983)● 4647.09 g/day / 0.5 = 9294 g/day

Number of food items per day● Average ground fern grows to be about 2ft tall and 1 ft wide with a thickness of 5 mm

(Stewart, 2021)● Volume of a rectangle with these dimensions divided by 2 may accurately describe the

shape of the plant, see figure 1.● V =(l*w*h)/2 = (.6096 m *.3048 m*.0005m)/2● V = 4.645*10^-5 m^3● Since the plant is around 90% water we assume its density to be equivalent to water

(1000 kg/m^3)● 1000kg/m3*(4.645*10^-5)=0.04645● M = 0.04645 kg or 46.45 g● 9294 g/day / 46.45 g/fern = 200.07 full fern fronds● Is the required foraging effort reasonable?

○ This sounds like a reasonable amount of food foraging effortSupporting metabolism of the brain

● How many grams of glucose per day are required to support the metabolism of the brain ?(The metabolic rate of an average mammal brain is about 8% of BMR, of a reptile or fishis about 2% of SMR)

● SMR=116.747 kJ/hr=2802kj/day

● 2% of SMR=2802*0.02=56kJ/day of glucose

○ 325.29 g Carbohydrates assimilated/day / 15.9 kJ/g = 20.45 kJ

○ This is not enough glucose absorption

● If it did not absorb enough glucose through digestion, where did the glucose come from?

○ VFAs of fermentation?Surface area of the intestine without fermentationNominal surface area of intestine that your animal required to absorb the carbohydrates andproteins in its diet (don’t forget to correct absorption rates for Q10 effects – assume Q10 = 2.5)

● Limiting nutrient: carbohydrates● Rate of glucose uptake in reptiles 357nmole/min/cm2 at 37oC (Withers 1992)● Daytime body temp: 31.56oC● Q10 correction for glucose uptake: 357(2.5)(31.56-37)/10=216.9nmole/min/cm2 at daytime

body temp● 216.9nmole=0.0000002169mole/min/cm2=0.00031mol/day/cm2

● 1cm2/0.00031mol/day*1.81 mol Carbohydrates/day=3201.7cm2

Calculate intestine length and intestine volume

●● Fig 2. Model of the ellipsoid cololite specimen with dimensions measured in cm (Brown

et al. 2020).● The cololite fossil represents an approximate representation of posterior stomach

dimension so small intestine diameter can be estimated based on where the two organsshould attach.

● 5cm diameter, SA=pi*d*l● 3201.7cm2=pi*5*l● Length=203.8cm long=2.04meters

● Are the length and volume appropriate for the dimensions of your animal? If not, changethe absorption rate and/or change the internal diameter to change the length of theintestine.

Fermentation model● DMR=5766 kJ/day● Metabolizable energy of various ferns = 6,600 kJ/kg of dry matter (Hummel et al. 2008)

○ 11.5% of dry mass is crude protein (Hummel et al. 2008)Dry fern matter required to be assimilated per day

● 5766kJ/day*1kg dry matter/6600kJ=0.874 kgDry fern matter needed to be consumed per day

● Assimilation efficiency of 50%● 0.874 kg/0.5 = 1.748 kg

Converting dry fern matter to fresh fern matter● Average dry weight/wet weight ratio of ferns = 0.27 (Weaver 1983)● 1.748kg dry/0.27=6.47kg 6474 grams fresh ferns required per day

Number of food items per day● 6474 grams/46.45g/fern=139.37 full fern fronds required

Finding surface area and length of the gut● 11.5% of dry mass is protein -> 0.115*874g dry mass=100.5 grams of protein in a day's

worth of fern mass● 100.5 g protein / 135 g protein/mol=0.74 mol protein/day● Rate of proline uptake for reptiles is 271nm/min/cm2 at 37oC (Withers 1992)● Daytime body temp: 31.56oC● Q10 correction for glucose uptake: 271(2.5)(31.56-37)/10=164.62nmole/min/cm2 at daytime

body temp● 164.62nmole=0.00000016462mole/min/cm2=0.000237mol/day/cm2

● 1cm2/0.000237mol/day*0.74 mol protein/day=3121.66cm2

● SA=pi*d*l, 3121.66=pi*5*l● Length of small intestine=198.73cm=1.99m