Supporting Information
Mahmoudabadi et al. 10.1073/pnas.1701670114
I. Introduction: Viral Rewiring of the Host Metabolism
Viruses rely entirely on their host as an energy source. Instead of
passively exploiting the host
’
s metabolic energy, some viruses
appear to augment it (2). A particularly compelling example is
demonstrated by phages that infect cyanobacteria. Cyanophages
carry genes for photosystem II, high-light inducible protein,
transaldolase, and ribonucleotide reductase, which are all tran-
scribed during an infection (3). Given the unprecedented pres-
ence of photosynthetic genes in viral genomes and the active
expression of these genes during an infection, it is proposed that
cyanophages carry these genes to increase the host energy supply
and deoxynucleotide production for their own replication (3). An
analogous finding is the presence of sulfite reductase genes in
genomes of phages that infect deep-sea bacteria, which use sul-
fur as their energy source (36). Here, too, these phages are hy-
pothesized to add to the host
’
s metabolic output. It was also
shown that large DNA algal viruses encode deoxynucleotide
synthesis enzymes. For example, PBCV-1 encodes 13 nucleotide
metabolism enzymes and EsV-1 encodes an ATPase and both
subunits of ribonucleotide reductase (34). The most recent study
on this topic identified more than 200 virus-encoded auxiliary
metabolic genes such as those used in nitrogen and sulfur me-
tabolism in marine viral metagenomes (6).
Adenoviruses have been shown to reprogram the host
’
s gluta-
mine metabolism by up-regulating glutamine transporters and
glutamine catabolism enzymes. Glutamine is a critical amino acid
used in the synthesis and import of other amino acids. In-
terestingly, this viral rewiring of glutamine metabolism is shown to
boost the concentration of certain amino acids as well as increase
glutamine reductive carboxylation. Together, these effects are
required for optimal viral production not only during an adeno-
virus infection but also during herpes and influenza viral infections
(4). In addition to virus-infected cells, which have high demand for
molecular building blocks and energy, cancer cells, immune cells,
and other proliferating cells similarly rewire glutamine metabolism
for energy production and biosynthesis (44).
Moreover, several examples from the early years of virology have
shown that viruses such as Rous sarcoma virus and feline leukemia
virus increase their hosts
’
glycolytic rate upon infection (35).
Similarly, the vaccinia virus was shown to up-regulate mitochon-
drial genes involved in the electron transport chain (ETC),
thereby increasing the ATP production within its host (5).
II. Energetic Cost Definitions and Assumptions
Viral synthesis requires the expenditure of the host
’
s ATP-equivalent
molecules as well as the usurpation of the host
’
s monomeric building
blocks such as nucleotides, amino acids, and in the case of some
viruses, lipids. To synthesize these monomeric building blocks and
generate ATP-equivalent molecules, heterotrophic cells, such as the
hosts of T4 or influenza, rely on reduced carbon sources from their
environment. In calculating the ener
geticcostofaviralinfectionand
its impact on the host energy supply, it is critical that we state several
assumptions about the host growt
h conditions. First, we assume that
the host is growing aerobically at 37 °C, with glucose as the sole
carbon source. Second, we assume that sources of nitrogen, sulfur,
phosphorus, and other trace elements are in excess, which is
typical of culture conditions in the laboratory, and from which
burst size measurements are commonly obtained. Third, we as-
sume that the growth media contain only inorganic sources of
nitrogen, requiring cells to synthesize amino acids rather than
salvaging them from a growth medium supplemented with pep-
tides (although this assumption should be modified in the case of a
mammalian host cell that cannot synthesize all amino acids).
As the sole carbon source in the growth media, glucose will serve
both as a source of energy and biomass. As glucose molecules enter
the cell, some are fully metabolized through glycolysis and the
tricarboxylic acid (TCA) cycle, and result in the formation of
carbon dioxide, water, and ATPs (Fig. S1). Other glucose mole-
cules, however, are not fully metabolized. Instead, they serve as
precursor metabolites for the synthesis of building blocks. In fact,
all building blocks, coenzymes, and prosthetic groups are synthe-
sized from just 12 precursor metabolites, which serve as links
between fueling and biosynthesis reactions (21). The diversion of
these precursor metabolites away from energy-producing path-
ways toward the synthesis of building blocks (Fig. S1) results in an
“
opportunity cost.
”
We define the opportunity cost of a precursor
metabolite as the number of ATP molecules that could have been
generated had the precursor metabolite not been diverted away
from energy-producing pathways (Fig. S1, step 1). Fig. S2
A
shows
the placement of precursor metabolites along metabolic pathways,
and summarizes the net energetic gains and losses of converting
one precursor metabolite to another. Fig. S2
B
builds upon Fig.
S2
A
and the detailed estimates provided in SI sections III and IV
to delineate the opportunity costs of precursor metabolites in
heterotrophic bacteria and eukaryotes.
In the synthesis of building blocks from precursor metabolites,
there is an additional source of opportunity cost, namely, the
oxidation of electron carrier molecules (Fig. S1, steps 2). Had
these electron carriers been preserved for energy production
rather than being oxidized during biosynthesis pathways, 2 ATP
molecules could have been generated from each NAD(P)H
molecule using the bacterial ETC (ref. 21, pp. 158
–
159) and 2.5
ATP molecules from the eukaryotic ETC (45). The combination
of the opportunity cost of precursor metabolites and the op-
portunity cost of building-block synthesis from precursor me-
tabolites will be referred to as the opportunity cost of a building
block, or
E
O
(Fig. S1, steps 1 and 2). To simplify, we will gen-
erally refer to the opportunity cost of a building block in our
calculations.
In addition to an opportunity cost, a monomeric building block
incurs a direct expenditure of ATP-equivalent molecules. To
convert a precursor metabolite into a monomeric building block
such as an amino acid, nucleotide, or lipid, the cell must pay a
synthesis cost (Fig. S1, step 3). The cell must also pay in ATP-
equivalent molecules to polymerize or assemble building blocks
into genomes, proteins, and membranes (Fig. S1, step 4).
III. Opportunity Costs of Precursor Metabolites for
Heterotrophic Bacteria
In this section, we will use the units of P, without any subscripts, to
refer to ATP (and ATP-equivalent) hydrolysis events (using the
subscripts without having first derived the opportunity cost of a
metabolite is meaningless). Once we have derived the opportunity
cost of a precursor metabolite here in this section, we will ac-
company it with the symbol P
O
in later figures and tables to
clearly mark these costs as opportunity costs. We will estimate
the opportunity cost of precursor metabolites,
C
opportunity
, by the
following:
C
opportunity
=
E
glu
−
E
partial
,
[S1]
where
E
glu
represents the net energetic gain from the complete
metabolism of a glucose molecule into water and carbon dioxide,
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and
E
partial
is the net energetic gain from the partial metabolism
of a glucose molecule into a precursor metabolite. Under aerobic
respiration,
E
glu
is
≈
26 P in
E. coli
(24). In the event that there is
a net energetic cost from the conversion of glucose into a pre-
cursor metabolite,
E
partial
will be a negative value.
In the synthesis of lipids, dihydroxyacetone-phosphate (dhap) is
used as a precursor. This precursor is generated during glycolysis
(Fig. S2
A
). Each glucose molecule results in the production of
two dihydroxyacetone-phosphate molecules, with this process
having a net energetic cost of 2 P (
E
partial
=
−
2) (Fig. S2
A
). The
opportunity cost of two dihydroxyacetone-phosphate molecules
is 2 P greater than that of glucose, or 28 P (Eq.
S1
). The op-
portunity cost of one dihydroxyacetone-phosphate molecule is
therefore
≈
14 P (Fig. S2
B
).
To simplify the opportunity cost estimates further, we could use
a shortcut. Rather than estimating the opportunity cost of each
precursor metabolite by calculating the
E
partial
from glucose as the
starting point, we could obtain the opportunity cost of metabolite
j
,
C
j
opportunity
, from the opportunity cost of metabolite
i
,
C
i
opportunity
,
by the following:
C
j
opportunity
=
C
i
opportunity
−
E
i
→
j
partial
,
[S2]
where
E
i
→
j
partial
represents the net energetic gain from the conver-
sion of metabolite
i
to metabolite
j
.
For example, during glycolysis each molecule of dihydroxyacetone-
phosphate is converted to a molecule of 3-phosphoglycerate (3pg),
resulting in the production of 1 NADH molecule and 1 ATP. In
E. coli
, each NADH molecule results in the production of
≈
2
ATP molecules under aerobic conditions (ref. 21, pp. 158
–
159);
therefore, the net energetic gain from this conversion is
≈
3P.
Hence, the opportunity cost of each 3-phosphoglycerate molecule
would be
≈
11 P, which is 3 P less than the opportunity cost of a
dihydroxyacetone-phosphate molecule (Eq.
S2
) (Fig. S2
B
).
If not used as a precursor, 3-phosphoglycerate is converted to
phosphoenolpyruvate (pep) in glycolysis (Fig. S2
A
). In this process,
however, there is zero energy expenditure or gain. Hence, the
opportunity cost of a phosphoenolpyruvate molecule is the same
as that of a 3-phosphoglycerate molecule, which is
≈
11 P (Fig.
S2
B
). Both of these precursors come before pyruvate (pyr) in
glycolysis. In converting a phosphoenolpyruvate molecule into a
pyruvate molecule, there is a net energy gain of 1 P (Fig. S2
A
).
The opportunity cost of a pyruvate molecule is therefore
≈
10 P
(1 P less than a phosphoenolpyruvate opportunity cost) (Fig. S2
B
).
Pyruvate can be converted to oxaloacetate (oaa) with the expen-
diture of 1 ATP. The opportunity cost of oxaloacetate is
≈
11 P
(10 P
+
1 P) (Fig. S2
B
).
Pyruvate is further converted to acetyl-CoA (acCoA), and in the
process one molecule of NADH is generated, which is equivalent to
2 P (Fig. S2
A
). The opportunity cost of acetyl-CoA is therefore
≈
8P(10P
−
2P)(Fig.S2
B
). One molecule of acetyl-CoA and
one molecule of oxaloacetate are then eventually converted to
α
-ketoglutarate (
α
kg) (Fig. S2
A
). The sum of the opportunity
costs of acetyl-CoA (8 P) and oxaloacetate (11 P) is 19 P, and
because 1 molecule of NADH is generated in their conversion
to
α
-ketoglutarate, the opportunity cost of
α
-ketoglutarate is
≈
17
P(or19P
–
2P)(Fig.S2
B
). Similarly,
α
-ketoglutarate is eventually
converted to oxaloacetate, and 2 NADH, 1 GTP, and 1 FADH
2
molecules are generated (Fig. S2
A
). This is a net gain of
≈
6
(assuming 1 P from each FADH
2
), reducing the opportunity cost
of oxaloacetate to
≈
11 P (17 P
–
6 P). Note that this is consistent
with the opportunity cost of oxaloacetate derived from the ana-
plerotic pathway described ear
lier (the conversion of pyruvate to
oxaloacetate via the pyruva
te decarboxylase enzyme).
Glucose can also be converted to ribose-5-phosphate (r5p) in
the pentose phosphate pathway, and in the process 2 NADPH
molecules are generated, which is equivalent to 4 P (Fig. S2
A
).
We subtract 4 P from the possible 26 P that glucose would be
converted to under respiratory conditions, and we arrive at 22 P as
the opportunity cost of ribose-5-phosphate (Eq.
S1
) (Fig. S2
B
).
The same calculation can be used for erythrose-4-phosphate
(e4p), resulting in 22 P as its opportunity cost (Fig. S2
B
).
IV. Opportunity Costs of Precursor Metabolites for
Heterotrophic Eukaryotes
To estimate the opportunity costs of precursor metabolites in
heterotrophic eukaryotes, we can carry out very similar calcula-
tions to those performed for heterotrophic bacteria (Fig. 1
B
and
C
). Similar to SI section III, we will refrain from using subscripts
for P in this section. For eukaryotes,
E
glu
, or the total energetic
gain from the complete metabolism of a glucose molecule into
carbon dioxide and water is higher. This is because each NAD(P)H
and FADH
2
molecule results in a higher number of ATPs within
the mitochondrial ETC compared with the bacterial ETC. Spe-
cifically, each NAD(P)H molecule is equivalent to
≈
2.5 P and
each FADH
2
molecule corresponds to
≈
1.5 P (ref. 45 and ref.
46, pp. 517
–
518), resulting in 30
–
32 P per glucose molecule.
Note that the theoretical yield of 38 P per glucose molecule has
been shown to be an overestimate due to the outdated as-
sumptions that each NAD(P)H molecule is equivalent to 3 P and
that each FADH
2
molecule generates 2 P (45). We will therefore
use 32 P as
E
glu
.
Each glucose molecule results in the production of two
dihydroxyacetone-phosphate molecules, with this process having a net
energetic cost of 2 P. The opportunity cost of two dihydroxyacetone-
phosphate molecules is 2 P greater than that of glucose, or 34 P
(Eq.
S1
). The opportunity cost of one dihydroxyacetone-
phosphate molecule is therefore 17 P (Fig. S2
B
). As described
earlier, in the conversion of dihydroxyacetone-phosphate mole-
cule into a 3-phosphoglycerate molecule, 1 NADH and 1 ATP
molecules are produced. This is equivalent to a net energetic gain
of
≈
3.5 P. The opportunity cost of a 3-phosphoglycerate mole-
cule is therefore
≈
13.5 P (17 P
–
3.5 P) (Eq.
S2
) (Fig. S2
B
).
The opportunity cost of a phosphoenolpyruvate is the same as
that of a 3-phosphoglycerate, which is
≈
13.5 P. In converting
phosphoenolpyruvate into pyruvate, there is a net energy gain of
1 P (1 ATP molecule is formed). The opportunity cost of pyru-
vate is therefore
≈
12.5 P (Fig. S2
B
). Pyruvate can be converted
to oxaloacetate with the expenditure of 1 ATP. Hence, the op-
portunity cost of oxaloacetate will be
≈
13.5 P (Fig. S2
B
).
Pyruvate is converted to acetyl-CoA, and in the process one
molecule of NADH is generated. As a result, the opportunity cost of
acetyl-CoA is
≈
10 P (12.5 P
–
2.5P).Onemoleculeofacetyl-CoA
and one molecule of oxaloacetate are converted to
α
-ketoglutarate
in the TCA cycle. The sum of the opportunity costs of acetyl-CoA
(10 P) and oxaloacetate (13.5 P) is 23.5 P, and because 1 molecule of
NADH is generated in their conversion to
α
-ketoglutarate, the op-
portunity cost of
α
-ketoglutarate is
≈
21 P (23.5 P
–
2.5 P) (Fig. S2
B
).
α
-Ketoglutarate is eventually co
nverted to oxaloacetate, and 2
NADH, 1 GTP, and 1 FADH
2
molecules are generated. This is a
net gain of
≈
7.5 P (assuming 1.5 P from each FADH
2
), reducing the
opportunity cost of oxaloacetate to
≈
13.5 P (21 P
–
7.5 P). Note that
this is again consistent with the opportunity cost of oxaloacetate
derived from the anaplerotic pathway.
Glucose can also be converted to ribose-5-phosphate in the
pentose phosphate pathway, and in the process 2 NADPH
molecules are generated, which is equivalent to 5 P. The op-
portunity cost of ribose-5-phosphate is thus
≈
27 P (Fig. S2
B
).
The same calculation can be used for erythrose-4-phosphate,
resulting in
≈
27 P as its opportunity cost (Fig. S2
B
).
V. Viral Entry Cost
The process of viral entry varies extensively across different viral
groups. Although many animal viruses enter their host cell through
clathrin-mediated endocytosis or fusion with the cell membrane
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(47), most phages inject their genetic material with the capsid
remaining outside of the cell. In both cases, however, entry is
mediated by the interaction between viral entry proteins and host
receptors (47).
For T4, it is the interactions between a minimum of three long
tail fibers and cellular receptors that initiates a cascade of con-
formational changes (48) (Fig. 1, step 1). After this preliminary
interaction, the base plate is subsequently brought closer to the
cell membrane, allowing the short tail fibers to interact with their
host receptors. The tail sheath contracts, resulting in the tail tube
puncturing the outer cell membrane (Fig. 1, step 2). Then,
conformational changes in gp5 phage protein activate its lysozyme
domains, resulting in the digestion of the peptidoglycan layer (Fig.
1, step 2). Effectively, the tail tube passes through the periplasm.
At this point, the phage DNA is passed through the inner
membrane via the tail tube and is then exposed to the intracellular
environment (Fig. 1, step 2). In general, viral entry proceeds
through protein conformational changes and does not rely on
ATP expenditure (49). In the case of the T5 phage, this point has
been explicitly demonstrated (50).
The influenza virus is composed of a capsid that is enveloped by a
lipid membrane (Fig. 2). Inside the capsid reside the ribonucleo-
protein complexes, which are composed of the segmented viral
genome encapsidated by proteins. The viral membrane is deco-
rated with hemagglutinin (HA) proteins, which bind to the host
sialic acid receptors, thereby initiating clathrin-mediated endocy-
tosis (Fig. 2, step 1). During endocytosis, a self-assembled protein
cage composed of clathrin triskelions forms around the inward
budding vesicle. Once the clathrin cage has formed, dynamins
perform the last stage of endocytosis (Fig. 2, step 2).
Dynamin is a mechanochemical GTPase that self-assembles
into multimeric spirals at the necks of clathrin-coated endo-
cytic vesicles to catalyze membrane fission. The dynamin helix
wrapped around the neck of an endocytic vesicle forms a protein
–
lipid tube with an inner diameter of 20 nm. In the presence of
GTP, the dynamin helix undergoes structural changes that result
in the reduction of the inner diameter (51). Because the dynamin
helix is composed of at least two turns (37), each turn is com-
posed of 13 dynamins, and each dynamin consumes 1 GTP, the
energy requirement for vesicle fission can be approximated as
30
–
100 P
D
(Fig. 2, step 2).
Once the vesicle is released from the cell membrane, the
clathrin cage has to be disassembled. This process requires the
expenditure of 3 ATPs per triskelion (52). For a clathrin cage
composed of 36 triskelions, this is equivalent to about 100 P
D
(Fig.
2, step 3). As the endosome matures, the endosomal lumen
becomes more acidic. The influenza virus has exploited this
feature for the uncoating of its lipid membrane as well as the
disassembly of its capsid. The endosomal pH drop activates the
viral transmembrane proton channel, M2. The influx of H
+
ions
from the endosome through M2 leads to the dissociation of the
viral capsid proteins. The acidic environment also triggers con-
formational changes that expose the viral HA2 fusion peptide,
which subsequently fuses the viral and endosomal membranes
(53). These two events together enable the release of the viral
ribonucleoprotein complexes into the host
’
s cytoplasm (54). To
summarize, the cost of entry for influenza per virion,
E
Entry
, falls
within the range of 10
2
to 10
3
P
D
. It is important to note that, in
our estimates, we will generally not include the cost of host
protein production, unless a protein is exclusively produced for
viral synthesis. For example, the cost of producing the clathrin
cage and dynamin proteins are not included due to the fact that
these proteins are recycled and produced for the host
’
s own
functions.
T4 efficiently exposes its genomic material to the host cytoplasm
where it can be readily transcribed and translated. The influenza
virus, however, due to the much larger volume and extensive
compartmentalization of its eukaryotic host, faces additional ob-
stacles in the way of reaching sites of transcription and translation.
We will describe these obstacles in the next section.
VI. Intracellular Transport Cost
Replication and transcription of the influenza virus occur inside
the nucleus. Like other cargo destined toward the nucleus from
the plasma membrane, the endosome carrying the influenza virus
is transported via the action of the dynein motor proteins along
microtubule tracks (Fig. 2, step 4). Unlike the kinesin motor
protein, dynein is one that takes variable step sizes along the
microtubule. As a result of having a hexameric ring of AAA
+
ATPases (similar to Vps4, described later in SI section XI),
dynein has multiple ATP binding sites. For the purposes of our
estimates, we assume a step size of 8 nm and the expenditure of 1
ATP per step (55). If we assume that the nucleus resides roughly
in the center of a cell
≈
10
μ
m in diameter (18), this will require
≈
10
3
dynein steps in carrying the endosome. As a result, the cost
of transport for the viral genome from the vicinity of the cell
membrane to the nucleus is
≈
10
3
P
D
(Fig. 2, step 4).
The ribonucleoproteins are imported to and exported from the
nucleus via nuclear localization signals (Fig. 2, step 5). The nu-
clear import of influenza ribonucleoproteins has been experi-
mentally demonstrated to be an energy-consuming process (56).
To get access to the nucleus, these ribonucleoproteins have to
pass through the nuclear pore complex. They do so by binding to
importins using nuclear localization signals. Each cargo imported
into the nucleus will require the hydrolysis of one GTP. Whether
ribonucleoproteins travel in and out of the nucleus separately or
together is not definitely known, but there is recent evidence
suggesting that the eight ribonucleoproteins travel as one cargo
(57). Hence, we estimate that the import of the viral genome will
cost 1 P
D
per virus.
Once inside the nucleus, the influenza genome is transcribed
and replicated, the costs of which will be discussed in later sec-
tions. The production of the full ribonucleoprotein complex is a
convoluted process, requiring the transcription of the viral ge-
nome (Fig. 2, step 6), the nuclear export of the resulting tran-
scripts (Fig. 2, step 5), as well as the translation of viral mRNA
transcripts (Fig. 2, steps 7). Once the viral proteins are generated,
some proteins travel back into the nucleus to encapsidate the
viral
−
ssRNA genome and form the next generation of ribonu-
cleoprotein complexes (PB1, PB2, NP, and PA proteins) (Fig. 2,
steps 5 and 8). Other viral proteins are transported to the cell
membrane, and together with ribonucleoprotein complexes,
which are destined to the same site, bud off.
Similar to the nuclear entrance, cargo exiting the nucleus must
also pay a price. The nuclear export cost through the CRM1 export
pathway, which is the one used by influenza, incurs 1 P
D
per cargo,
similar to the import pathway (57). Considering that on average
6,000 influenza genomes have to leave the nucleus to eventually
give rise to 6,000 influenza virions, we estimate the export of
ribonucleoproteins from the nucleus will cost
≈
6
×
10
3
P
D
.The
mechanism and energetics of viral mRNA export from the nucleus
are not well understood (58), although there is a growing body of
evidence implicating the involvement of the nuclear RNA export
factor 1. We suspect the cost of mRNA nuclear export is on par
with the cost of ribonucleoprotein nuclear export and estimate the
cost of nuclear import and export to be
≈
10
4
P
D
.
Upon exiting the nucleus, ribonucleoproteins have to be
transported toward the cell membrane where they can coassemble
with other viral proteins and bud off. Their path to the cytoplasm
starts at the microtubule organizing center and proceeds via
recycling endosomes bound to the RAB-11 GTPase (57). This
endocytic transport is powered by kinesin motor proteins trekking
along microtubules with a step size of 8 nm and the hydrolysis of 1
ATP per step (59) (Fig. 2, step 9). Assuming the eight ribonu-
cleoproteins travel as one cargo
—
consistent with the reported
typical size of an endosome (200 nm in diameter) (60)
—
a5-
μ
m
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transport of 6,000 ribonucleoproteins from the nucleus to the
cytoplasm costs
≈
3
×
10
6
P
D
. Under a similar assumption (i.e.,
each endosome carries the proteins required to build a single
virus), the cost of protein transport to the apical cell surface is
similar to the cost of ribonucleoprotein transport. Considering
the sum of the costs from nuclear import and export as well as
travel along the microtubules, the cost of intracellular transport
for an influenza infection,
E
Transit
=
i
, can be approximated as 6
×
10
6
P
D
.
The cost of GTPases mediating endosome docking and fusion
is likely negligible compared with the heavy cost of motor proteins
mobilizing the endosomes. This is because, during endosome
trafficking, there are far more steps taken by motor proteins than
there are endosome fusion or docking events. However, each
fusion or docking event incurs a similar cost to a step taken by a
motor protein. More detailed estimates of endosomal trafficking
are exceedingly difficult due to the unavailability of studies that
shed light on (
i
) the average number of proteins carried per
endosome, (
ii
) the average number of GTPases needed per
endosome, and (
iii
) the average length that an endosome travels
within the cell, among other topics.
VII. Genome Replication Cost
The Direct Cost.
Genome replication is a complex phenomenon
requiring many different steps, such as the unwinding of the
parent helix, RNA primer synthesis, Okazaki fragment ligation,
and proofreading. Although there are many facets of genome
replication in cells that require energy expenditure, the direct cost
of replication lies primarily in the direct synthesis cost of nu-
cleotides from precursor metabolites as well as the polymerization
of individual nucleotides. The energetic cost of genome repli-
cation for a virus with a dsDNA genome can be approximated as
follows:
E
REP
ð
dsDNA
Þ
=
v
≈
2
L
g
e
d
+
e
p
.
[S3]
Here,
L
g
corresponds to the genome length and is multiplied by
2 to account for T4
’
s double-stranded genome. The cost of each
DNA nucleotide can be stated as the sum of
e
d
, which represents
the average direct cost of DNA synthesis from precursor metab-
olites, and
e
p
, which denotes the cost of chain elongation per
base [equivalent to 2 P
D
(20)] (Fig. 1, step 5). The energetic cost
of replicating a
−
ssRNA genome is similarly as follows:
E
REP
ð
−
ssRNA
Þ
=
v
≈
2
L
g
e
r
+
e
p
,
[S4]
where
e
r
represents the average direct cost of RNA synthesis
from precursor metabolites. The factor of 2 stems from the fact
that a
−
ssRNA has to first be converted to
+
ssRNA before a
second copy of
−
ssRNA can be synthesized (Fig. 2, step 8).
e
d
in
the context of bacterial metabolism is equivalent to 11 P
D
(Fig.
S3
A
and Dataset S1). The average synthesis cost of an RNA base
from a precursor metabolite in the context of eukaryotic metab-
olism is 10 P
D
(Fig. S3
B
and Dataset S1). With these values in
hand, we estimate the direct cost of T4 (
L
g
≈
2
×
10
5
) and in-
fluenza (
L
g
≈
1
×
10
4
) genome replication to be
≈
4
×
10
6
P
D
and
≈
3
×
10
5
P
D
, respectively.
The Opportunity Cost.
The opportunity cost of T4 phage replication
can be estimated as 2
L
g
e
od
, where
e
od
corresponds to the average
opportunity cost of a DNA base synthesized in bacteria and is
≈
34 P
O
(Fig. S3
A
and Dataset S1). Note that
e
od
represents the
sum of the average opportunity cost of precursor metabolites
required for DNA synthesis (33 P
O
) and the average opportunity
cost of DNA synthesis from precursor metabolites (1 P
O
). Under
this estimate, the opportunity cost of genome replication for a
T4 phage is
≈
1
×
10
7
P
O
. Similarly, the opportunity cost of
genome replication for an influenza virus can be estimated as
2
L
g
e
or
, where
e
or
refers to the average opportunity cost of an
RNA nucleotide synthesized in a eukaryotic organism, which is
≈
39 P
O
(Fig. S3
B
and Dataset S1). Hence, the opportunity cost
of influenza genome replication is
≈
8
×
10
5
P
O
.
The Total Cost.
The total cost of T4 phage replication is the sum of
opportunity and direct costs: 2
L
g
ð
e
d
+
e
p
+
e
od
Þ
. Under this esti-
mate, the cost of genome replication for a T4 phage is
≈
2
×
10
7
P
T
.
The same logic follows for influenza, where the total cost of its
genome replication can be estimated as 2
L
g
ð
e
r
+
e
p
+
e
or
Þ
.Hence,
the total cost of influenza genome replication is
≈
1
×
10
6
P
T
.
The Replication Cost Component of an Infection.
Furthermore, the
replication cost per infection is the cost of replication for a virus
multiplied by its burst size,
B
, resulting in the following:
E
REP
=
i
≈
BE
REP
=
v
.
[S5]
The direct, opportunity, and total costs of genome replication dur-
ing a T4 infection with an average burst size of 200 are
≈
9
×
10
8
P
D
,
≈
2
×
10
9
P
O
,and
≈
3
×
10
9
P
T
(Fig. 1, step 5). The direct,
opportunity, and total costs of genome replication during an in-
fluenza infection with an average burst size of 6,000 are
≈
2
×
10
9
P
D
,
≈
5
×
10
9
P
O
,and
≈
6
×
10
9
P
T
(Fig. 2, step 8).
VIII. Transcriptional Cost
As in the case of replication, transcription involves various
energy-consuming processes, such as transcriptional activation,
initiation and termination, as well as proofreading and mRNA
splicing. However, the dominant cost in transcription is similar to
that in replication, namely, the investment in the synthesis of the
nucleotides themselves (20). See Fig. 1 (step 3) and Fig. 2 (step 6).
The Direct Cost.
The direct cost of transcription is approximately
the following:
E
TX
=
v
≈
ð
e
er
+
e
r
Þ
X
m
i
=
1
L
ri
N
ri
,
[S6]
where
L
ri
and
N
ri
represent the length and the copy number of
each viral mRNA transcript, respectively. The average direct cost
of synthesizing an RNA base from precursor metabolites is
denoted as
e
r
, which is 10 P
D
(Fig. S3 and Dataset S1). The
direct cost of polymerizing an RNA base is symbolized by
e
er
,
which is equal to 2 P
D
(20). Because the mRNA copy number for
each viral gene is largely an unknown parameter, we approxi-
mated the viral genome as one long gene. This allowed us to
eliminate the
L
r
and
N
r
dependence, and replace them with
constants,
l
r
and
n
r
, such that
E
TX
=
v
≈
ð
e
er
+
e
r
Þ
l
r
n
r
.
[S7]
The constant
l
r
corresponds to the length of the mRNA tran-
script, and thus is equal to the length of the genome. The average
copy number of this transcript,
n
r
, can be approximated by the
observed ratio of 1 mRNA transcript per 1,000 resulting proteins
both in prokaryotes (18) as well as in mammalian cells (61). The
average protein copy number of a virus,
n
p
, can be related to its
average transcript number according to
n
r
≈
n
p
=
1,000.
To obtain
n
p
, we have used available data on viral protein copy
numbers. For a T4 phage, with a total estimated protein count of
5,000 representing 35 structural and 3 lysis genes (Table S1), the
average protein copy number per virus is
≈
130. Influenza
’
s
total protein count is also
≈
5,000, representing the products of
nine proteins (Table S2). This results in an estimate for the av-
erage protein copy number for an influenza virus of
≈
550. With
Mahmoudabadi et al.
www.pnas.org/cgi/content/short/1701670114
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these numbers in hand, we estimate
n
r
to be
≈
0.13 for a single
T4 phage and 0.55 for an influenza virus. We note that, gener-
ally, mRNAs are relatively short-lived and each individual
mRNA on average produces between 10 and 100 proteins.
However, the pool of such proteins is then used to synthesize all
of the virus particles that make up a given burst.
Although it is sufficient to consider only the synthesis and
polymerization costs of nucleotides in the direct cost estimates for
genome replication, the transcriptional cost should additionally
encompass the cost of mRNA repolymerization (20). This is
because mRNA transcripts have a short life span and must be
regenerated throughout the duration of the infection. This cost is
perhaps more prominent in bacteria where the lifetime of a
transcript is about 3 min (18), whereas the lifetime of a mam-
malian cellular transcript is
≈
10 h (18), and therefore compa-
rable to the lifetime of the influenza infection itself. The first
step in calculating the mRNA repolymerization cost is to mul-
tiply the lifetime of the infection,
t
, by the mRNA degradation
rate,
δ
r
.
The second step is to take into account the average transcript
copy number and its length to determine the number of RNA
bases that have to be repolymerized during the course of an
infection. Effectively, the assumption is that the RNA nucleotides
are only being repolymerized, not resynthesized. The repoly-
merization cost of transcription can be stated as the number of
nucleotides to be repolymerized,
l
r
n
r
δ
r
t
, multiplied by
e
er
. The
direct cost of transcription per virus can then be revised as such:
E
TX
=
v
≈
l
r
n
r
ð
e
r
+
e
er
δ
r
t
Þ
.
[S8]
The lifetime of a T4 infection is
≈
30 min (62), and the lifetime
of an influenza infection is roughly 12 h (14). With these param-
eters in hand, the direct cost of transcription (per virus) for
T4 and influenza are
≈
7
×
10
5
P
D
and
≈
7
×
10
4
P
D
, respec-
tively (Table 1).
The Opportunity Cost.
The opportunity cost of transcription can be
obtained by
l
r
n
r
e
or
, where
e
or
represents the opportunity cost of
an RNA nucleotide.
e
or
is
≈
31 P
O
in bacteria and 39 P
O
in eu-
karyotes (Dataset S1 and Fig. S3). Note that, in this expression, we
do not account for repolymerization events as RNA nucleotides
are assumed to be recycled rather than resynthesized. The op-
portunity costs of transcription for a single T4 phage and
an influenza virus are thus
≈
7
×
10
5
P
O
and
≈
2
×
10
5
P
O
,re-
spectively.
The Total Cost.
The total cost of transcription can be obtained by
l
r
n
r
ð
e
or
+
e
r
+
e
er
δ
r
t
Þ
, which represents the sum of opportunity
and direct costs of transcription. According to this estimate, the
total costs of transcription for a single T4 phage and an influenza
virus are
≈
1
×
10
6
P
T
and
≈
3
×
10
5
P
T
, respectively (Table 1).
The Transcriptional Cost of an Infection.
The transcriptional cost of
an infection is the transcriptional cost of a virus multiplied by its
burst size, namely, the following:
E
TX
=
i
≈
BE
TX
=
v
.
[S9]
For T4 with an average burst size of 200 and for influenza with an
average burst size of 6,000, the direct cost of transcription at the
level of an infection is
≈
1
×
10
8
P
D
(Fig. 1, step 3) and
≈
4
×
10
8
P
D
, respectively (Fig. 2, step 6). The opportunity cost of tran-
scription for these two infections are
≈
1
×
10
8
P
O
(T4) and
≈
1
×
10
9
P
O
(influenza).
Their total costs are
≈
3
×
10
8
P
T
(T4, Fig. 1, step 3) and
≈
2
×
10
9
P
T
(influenza,Fig.2,step6)(Table1).
IX. Translational Cost
There are important biosynthetic costs associated with proteins
just as there are with nucleic acids. Here, we attempt to capture
the most significant costs in the protein synthesis pathway while
making some simplifying assumptions that neglect substantially
smaller cost components such as the costs of translational initi-
ation and termination and posttranslational modifications (20).
The estimate for translational cost follows the same rationale as
the cost calculation for transcription.
The Direct Cost.
The direct cost of translating the viral proteome
can be estimated as follows:
E
TL
=
v
≈
ð
e
a
+
e
ea
Þ
X
k
j
=
1
L
pj
N
pj
.
[S10]
Here, we have multiplied the total number of amino acids by the
per-amino acid costs of synthesis from precursor metabolites,
e
a
,
and polymerization,
e
ea
. The arrays
L
p
and
N
p
hold values for the
length of each protein and its copy number per virus, respectively
(Tables S1 and S2). We show that
P
k
j
=
1
L
pj
N
pj
for the influenza
virus and the T4 phage are about 1.7 and 1.2 million amino acids,
respectively (Tables S1 and S2). In both bacteria and eukaryotes,
e
a
, is on average equal to 2 P
D
(Fig. S3 and Dataset S1) and
e
ea
is
4P
D
(25). Due to the relatively slow protein degradation rates for
bacteria (0.4/h) and human cells (0.08/h) (20, 26) compared with
infection lifetimes, we have neglected costs associated with this
process. Using this information, the direct cost of translation for a
T4 phage and an influenza virus are
≈
7
×
10
6
P
D
and
≈
1
×
10
7
P
D
, respectively (Table 1). This finding stems mainly from the fact
that both viruses are based on
≈
10
6
aa.
The Opportunity Cost.
The opportunity cost of viral translation is
approximately
e
oa
P
k
j
=
1
L
pj
N
pj
, where
e
oa
denotes the average
opportunity cost of an amino acid and corresponds to 25 P
O
in
bacteria and 30 P
O
in eukaryotes (Fig. S3 and Dataset S1). The
opportunity cost of viral translation for a T4 phage and an in-
fluenza virus are therefore
≈
3
×
10
7
P
O
and
≈
5
×
10
7
P
O
,
respectively.
The Total Cost.
The total cost of viral translation is the sum of
direct and opportunity costs of translation. The total cost of viral
translation for a T4 phage and an influenza virus are therefore
≈
4
×
10
7
P
T
and
≈
6
×
10
7
P
T
, respectively (Table 1).
The Translational Cost Component of an Infection.
The translational
cost of an infection is simply the cost of translation per virion
multiplied by its burst size, namely, the following:
E
TL
=
i
≈
BE
TL
=
v
.
[S11]
The direct, opportunity, and total translational costs for a
T4 infection with a burst size of 200 are
≈
1
×
10
9
P
D
,6
×
10
9
P
O
, and 8
×
10
9
P
T
(Fig. 1, step 4). For an influenza infection
with a burst size of 6,000, these costs are
≈
6
×
10
10
P
D
,3
×
10
11
P
O
, and 4
×
10
11
P
T
(Fig. 2, step 7) (Table 1).
Protein Folding and Quality Control.
Just as in any other biological
process, protein folding is subject to errors. To correct for such
errors and prevent the aggregation of misfolded proteins, cells from
all three domains of life have evolved elaborate mechanisms for the
detection of misfolded proteins. Through various ATP-dependent
(e.g., Hsp90, Hsp70, and Hsp60) and ATP-independent processes,
a triage is carried out in which some proteins are refolded and
others are degraded (63
–
65). From an energetic standpoint, pro-
tein quality control mechanisms are likely to cost substantially less
than the cost of translation. As shown above, a protein with an
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average length of 300 aa (26) will have a direct translational cost
of 1,800 P
D
and a total cost of 9,300 P
T
in bacteria and 10,800 P
T
in eukaryotes, respectively. On the other hand, the energetic cost
of various protein quality control pathways can range from a few
ATPs (66) to a few hundred ATPs per protein (67), which is re-
quired for protein degradation. Thus, it is likely that protein
quality control will be a fractional cost compared with the trans-
lational cost of a protein. A similar conclusion was drawn in the
context of cellular protein cost (20). Because we were unable to
ascertain the fraction of viral proteins that may be degraded, and
because different proteins require different quality control path-
ways (68), any more detailed estimates are difficult to make. Fu-
ture experimental studies would be needed to determine any
substantial costs incurred by protein quality control or translation
at large that may be missing from our estimates.
Another very interesting experimental avenue would be to
explore the consequences of the quality control mechanisms as
they are being partially recruited toward maintaining the viral
proteome. Particularly in the context of a host cell that survives
the infection, how the cell responds to the additional burden from
viral protein production and maintenance would be a fascinating
topic of study. Constructed from roughly 5,000 proteins (Table
S2), and with an average burst size of about 6,000, the influenza
infection will produce about 3
×
10
7
viral proteins. Considering
that a human cell will harbor more than 10
9
proteins (69), we
would expect the extra load from viral proteins on the quality
control machinery to be minimal. For a T4 phage, composed of
roughly 5,000 proteins (Table S1) and with a burst size of roughly
200, the total number of viral proteins during an infection would
be
≈
10
6
. This is comparable to the number of bacterial proteins,
which is estimated to be 10
6
per cubic micrometer (70), or the
approximate volume of an
E. coli
cell. Hence, in the case of a
T4 infection, the viral quality control pathways are likely to more
heavily affected than in the case of an influenza infection.
X. Assembly and Genome Packaging Cost
Upon translation, the influenza viral proteins and ribonucleo-
proteins travel toward the cell membrane via Rab-bound endo-
somes that are carried by kinesins on microtubules. Interactions
of the matrix protein M1 with ribonucleoproteins and the viral
transmembrane proteins, namely HA, NA, and M2, result in the
assembly of the influenza virus (71). Although the kinetics of the
assembly steps remains to be delineated, influenza virus assembly
and genome packaging are not regarded as energy-consuming
processes. In general, virus assembly is described as an ener-
getically favorable process, typically driven by the burial of hy-
drophobic surfaces (72, 73), and therefore independent of host
energy expenditure. As an example, the assembly of hepatitis
B virus is shown to occur spontaneously through weak protein
–
protein interactions (74).
Although the assembly of the T4 capsid is spontaneous, the
packaging of the genome inside the capsid is not (Fig. 1, step 7).
The cost of genome packaging for a T4 phage is as follows:
E
Pack
=
v
=
e
p
L
g
,
[S12]
where the cost to package a base pair,
e
p
,is2P
D
(75). For the
169-kb genome of T4, this cost is as follows:
≈
3
×
10
5
P
D
.
The packaging cost of a T4 infection is simply the cost of
packaging for a single T4 phage multiplied by the T4 burst size:
E
Pack
=
i
=
BE
Pack
=
v
.
[S13]
For T4, with a burst size of 200, the contribution of packaging to the
total cost of infection is
≈
7
×
10
7
P
D
(Fig. 1, step 7) (Table 1).
XI. Viral Exit Cost
Viruses use two primary exit strategies. Generally, enveloped
viruses such as influenza and HIV bud off from the host mem-
brane. Phages, on the other hand, generally lyse their host cells.
For T4, the cost of exit is primarily the production cost of proteins
that together break down the cell wall. We have already included
the cost of lysis proteins in our translational cost estimates. The
lysis proteins include holin, endolysin, and spanin proteins.
The holins create holes in the host inner membrane, enabling
the endolysins to reach the peptidoglycan layer. The spanins fuse
the inner and the outer membrane as a requirement for lysis of
Gram-negative bacteria (Fig. 1, step 9, only holins and endolysins
are shown). Considering that T4 holin, endolysin, and spanin
proteins are 218, 164, and 216 aa in length, respectively, and each
has a copy number of about 4,000 per infection (or
≈
20 per virus
considering a burst size of 200) (76), the contribution of lysis
proteins to the translational cost of an infection is negligible.
In the case of influenza, the exocytosis of virions is energy-
consuming. However, the exact mechanism remains a mystery,
with influenza M2 protein so far serving as the most likely agent
for mediating exocytosis (77). Three separate costs exist: (
i
) the
cost to locally bend the membrane outward, (
ii
) the cost to scisse
the budding virion from the cell membrane, and (
iii
) the cost of
the cellular membrane that becomes part of the viral membrane
(Fig. 2, step 10). The cost to bend the membrane into the shape
of a sphere of any size, is equivalent to 25 P
D
(25). One way to
estimate the cost of scission is to assume that it incurs a com-
parable cost to the HIV scission process, which, similar to several
other enveloped viruses, is mediated through the ESCRT-III
(endosomal sorting complexes required for transport) assem-
bly. ESCRT-III complex self-assembles into filaments around
the neck of a budding vesicle (similar to dynamin), and its dis-
assembly requires the expenditure of 6 P
D
via the Vps4 ATPase
(78). The sum of these two costs results in the use of 31 P
D
as
one influenza virus leaves the cell. Because 6,000 virions exit the
cell on average, exocytosis costs
≈
2
×
10
5
P
D
.
An alternative, order-of-magnitude estimate could be made
with the assumption that the cost of membrane scission during
endocytosis equals the cost of membrane scission during exo-
cytosis. In estimating the influenza entry cost, we showed that the
cost of membrane scission during influenza endocytosis via the
dynamin polymer is
≈
30 P
D
. Together with the cost of mem-
brane bending,
≈
25 P
D
, this exit estimate is slightly higher (55
P
D
per virus) than the previous (31 P
D
per virus). Under this
estimate, the cost of exocytosis for all 6,000 influenza virions
amounts to
≈
3
×
10
5
P
D
.
The primary cost of viral exit for influenza, however, is the cost
of lipids that are taken from the host cell to form the viral
membrane. The cost of lipids per virus can be estimated by the
number of lipid molecules needed per virion multiplied by the
cost of a lipid molecule,
e
l
:
E
Exit
=
v
=
8
π
r
2
s
e
l
.
[S14]
In Eq.
S14
, the numerator represents twice the viral surface area
(accounting for the bilayer) and the denominator,
s
, denotes the
surface area of a lipid head group, which is
≈
0.5 nm
2
(26, 79).
With an average radius of 50 nm, the influenza virus is composed
of
≈
1
×
10
5
lipid molecules. The direct, opportunity, and total
costs of a lipid molecule in a eukaryotic cell are 18 P
D
, 264 P
O
,
and 282 P
T
, respectively (Dataset S1 and Fig. S3
B
). As a result,
the direct, opportunity, and total costs of lipids per influenza
virus are 2
×
10
6
P
D
,3
×
10
7
P
O
, and 4
×
10
7
P
T
, respectively.
The exit costs of an influenza infection can be derived by the
following:
Mahmoudabadi et al.
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E
Exit
=
i
=
BE
Exit
=
v
,
[S15]
and are
≈
1
×
10
10
P
D
,2
×
10
11
P
O
, and 2
×
10
11
P
T
for an
infection with a burst size of 6,000 (Table 1).
XII. Estimating the Total Host Energy Budget from Growth
Experiments in Chemostats
The total basal and growth metabolic requirements of various
organisms have been shown to correlate with the cellular volume
(20). We have used Eqs.
S16
–
S18
presented by Lynch and
Marinov (20) to estimate the energetic budget of hosts consid-
ered in this study. Basal metabolic requirement of a cell scales
with the cell volume according to the following:
E
M
≈
0.39
V
0.88
,
[S16]
where
V
, represents the cell volume in units of cubic microme-
ters, and
E
M
is in the units of 10
9
ATP per cell per hour. The
growth metabolic requirement of a cell similarly scales with the
cell volume according to the following:
E
G
≈
27
V
0.97
,
[S17]
where
E
G
is in the units of 10
9
ATP per cell. The total energy
requirement of a cell is simply the sum of the basal and growth
energy requirements,
E
T
=
E
G
+
tE
M
,
[S18]
where
t
is the cell division time in the units of hours.
For a bacterial cell with a volume of 1
μ
m
3
, the maintenance
metabolic cost is
≈
10
8
P
T
in the duration of a T4 phage in-
fection that lasts about 30 min. A mammalian cell, on the other
hand, with a characteristic volume of 2,000
μ
m
3
, has a basal
metabolic cost of
≈
10
12
P
T
over the course of a 12-h influenza
infection. The total energetic cost of a cell should also encom-
pass the cost of cellular growth. Hence, the total energetic cost of
a bacterium and a mammalian cell with the dimensions high-
lighted above are
≈
3
×
10
10
P
T
and
≈
5
×
10
13
P
T
, respectively,
during the course of their viral infection. The correlation
between cellular volume and metabolic capacity is supported by
the observation that larger
E. coli
cells produce higher T4 burst
sizes (80).
XIII. The Heat Production and Power Consumption of a Viral
Infection
In our estimates for heat production and power consumption of a
viral infection, we will not consider the total cost of an infection as it
contains the opportunity costs; by definition, these opportunity
costs do not represent direct use of ATP (and ATP-equivalent)
molecules; rather, they represent the ATP (and ATP-equivalent)
molecules that could have been generated in the absence of a
viral infection (SI section II). For these estimates, we will rely on
the direct costs.
To estimate the amount of heat that is generated due to a viral
infection, we have to consider the inefficiency of aerobic me-
tabolism. The burning of glucose results in the production of
≈
2,800 kJ/mol of heat (81). The same reaction takes place inside
our cells, with the difference being that cells are capable of
harnessing a fraction of the free energy that would otherwise be
liberated as heat. When glucose is aerobically metabolized into
water and carbon dioxide, a fraction of the free energy is used to
convert ADP into ATP, whereas the remaining free energy is dis-
sipated as heat. By assuming physiological conditions, the free-
energy change of ATP hydrolysis can be approximated as
−
50 kJ/mol
(16). In bacterial metabolism, 26 ATPs are generated from each
glucose molecule; hence the free energy captured by the conversion
of ADP into ATP is approximately
−
1,300 kJ/mol of glucose. As a
result, in this simple estimate, we consider that about 50% of the
energy from the aerobic metabolism of glucose is dissipated as heat:
ð
1
−
ð
1,300
kJ
=
mol
=
2,800
kJ
=
mol
ÞÞ
×
100%. For eukaryotic cells,
with 32 ATPs generated per glucose molecule, about 40% of the
energy stored in glucose is dissipated as heat.
The T4 infection has a direct cost of 3
×
10
9
P
D
(Table 1).
Because each glucose molecule results in 26 ATPs in the bacterial
metabolism, T4 infection
’
s direct cost would require the complete
metabolism of 10
8
glucose molecules. The influenza infection
’
s
direct cost is about an order of magnitude higher than that of T4
(Table 1) and is equivalent to the aerobic metabolism of
≈
10
9
glucose molecules. Based on the number of glucose molecules
required to cover the direct costs of each infection, the free energy
stored in glucose (
−
2,800 kJ/mol), and the percentage of the en-
ergy released as heat during the aerobic metabolism of glucose
(
≈
40
–
50%), we can conclude that the heat generated during
T4 and influenza infections are
≈
0.2 and 2 nJ, respectively.
In half an hour, the T4 infection results in the hydrolysis of ATP-
equivalent molecules at an average rate of 2
×
10
6
P
D
per second. In
half a day, an influenza infection a
lso has an average ATP-hydrolysis
rate of 2
×
10
6
P
D
per second. Put in terms of the more familiar units
of watts (by assuming
−
50 kJ/mol of free energy change per P
D
), the
power of both viral infections is on the order of 200 fW.
XIV. Generalizing Viral Energetics for dsDNA Phages
Fig. 5
A
shows how we can generalize the estimates presented
here by thinking of dsDNA phages as approximately spherical
objects with an outer layer of thickness,
t
. In this model, the inner
volume of a virus containing the viral genome is given by 4
π
r
3
=
3,
which can be used to estimate the viral genome length (28). The
total cost of genome replication for a dsDNA genome can be
obtained from Eq.
S3
(Fig. 4). However, instead of using the
viral genome length directly, we can divide the capsid inner
volume by the volume of a base pair,
v
d
(
≈
1nm
3
) (26):
E
REP
ð
dsDNA
Þ
=
v
≈
4
π
r
3
3
v
d
e
d
+
e
p
+
e
od
.
[S19]
Because for many dsDNA phages only about one-half of the cap-
sid is filled with the viral genome (28), the cost of a DNA base was
not multiplied by a factor of 2 (even though the genome is double
stranded) as the two multipliers cancel each other out. The direct
cost of replication can be obtained similarly by the exclusion of the
opportunity cost of a DNA base or
e
od
(in bacteria) from Eq.
S19
.
Moreover, the translational cost of a virus can be obtained from
a modification of
ð
e
a
+
e
ea
+
e
oa
Þ
P
k
j
=
1
L
pj
N
pj
, derived previously
(SI section IX). The total cost of viral translation,
E
TL
=
v
≈
4
π
R
3
−
r
3
3
v
a
ð
e
a
+
e
ea
+
e
oa
Þ
,
[S20]
can be obtained from multiplying the total number of amino acids
by the total cost of an amino acid. The number of amino acids is
estimated by dividing the outer capsid volume (denoted by the
shaded blue region in Fig. 5
A
), 4
π
ð
R
3
−
r
3
Þ
=
3, by the volume of
an amino acid,
v
a
, which can be approximate as 0.1 nm
3
(82).
This expression can be further simplified by replacing the outer
radius,
R
, with
r
+
t
(Fig. 4):
E
TL
=
v
≈
4
π
ð
r
+
t
Þ
3
−
r
3
3
v
a
ð
e
a
+
e
ea
+
e
oa
Þ
≈
4
π
r
2
t
v
a
ð
e
a
+
e
ea
+
e
oa
Þ
.
[S21]
The direct translation cost of a virus can be similarly obtained
from Eq.
S21
by excluding the average opportunity cost of an
Mahmoudabadi et al.
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amino acid,
e
oa
. The critical radius,
r
crit
,atwhichtranslationand
replication will have equal cost can be obtained by setting Eqs.
S19
and
S21
equal and solving for
r
(Fig. 4). Because capsid shell
thickness is relatively conserved across icosahedral viruses studied
to date, it can be treated as a constant equal to 3 nm (83). The
critical radius for total cost estimates,
r
crit
−
Tot
,is59nm.Forthe
direct cost estimates, the critical radius,
r
crit
−
Dir
,is42nm(Fig.4).
XV. Discussion
In addition to the energetics of viral synthesis, another burst-
limiting metric to consider is a volumetric one, namely the frac-
tion of the host volume occupied by viruses during an infection.
Taking influenza and T4 as our representative viruses, it is clear
that they both occupy a relatively small percentage of the total host
volume (84). A T4 infection takes up less than 5% of its host
’
s total
volume. An influenza infection takes up even less space (
≈
0.2% of
its host volume). These estimates suggest that (
i
) for T4, the en-
ergy requirement is more likely a burst-limiting factor than the
volumetric requirement, and (
ii
) for influenza, neither energetic
nor volumetric factors seem to be limiting the burst size.
We have already considered possible causes for the inefficiencies
of an influenza infection, which have experimentally been shown to
result in only 1 infectious virus out of every 10 produced (40).
Accounting for this inefficiency boosts the total cost of an influenza
infection to
≈
10% of its host
’
s. A second consideration that
could explain the relatively low energy uptake of an influenza
infection is the growth state of its host cell. Our current estimate
assumes that the host cell is under maximal growth conditions.
When the host cell is not dividing, however, its energy supply
could be as low as 10
12
P
T
(estimated for a 12-h infection period;
SI section VIII). In considering both the 10% infection efficiency
and assuming a slow-growing host cell, the influenza infection
’
s
total cost could also be a significant fraction of its host
’
s total cost.
A final consideration is that implicit in our original question is
the assumption that all viruses must conform to producing the
maximum burst size allowed by their host energetic supply. This
assumption, although it appears compatible with our findings for
phages, may simply not be true for viruses such as influenza. It may
be that a eukaryotic virus within a multicellular setting has an
entirely different growth strategy than a prokaryotic virus infecting
a single-celled organism. The influenza virus burst size is not only
under selection pressure within its host cell but also within the
multicellular organism that serves as its secondary host.
Mahmoudabadi et al.
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glucose
precursor metabolites
ATPs
building blocks
proteins, genome,
membranes
1
synthesis
post-synthesis
Opportunity Cost of precursor
metabolites
Opportunity Cost of precursor metabolites
Opporunity Cost of synthesis from precursor metabolites
Direct Cost of synthesis from precursor metabolites
Direct Cost of polymerization
Opportunity Costs of a building block (total)
Direct Costs of a building block (total)
Total Cost of a building block
2
3
4
Fig. S1.
Defining the opportunity and the direct costs of monomeric building blocks. During aerobic conditions, glucose is converted to precursor metabolit
es
and eventually results in the production of 26 ATPs (in bacteria) when fully metabolized into water and carbon dioxide. If a precursor metabolite is in
stead
channeled toward the synthesis of building blocks (step 2 and 3), the number of ATPs that could have been generated from its full metabolism minus the
number of ATPs that have already been generated from its partial metabolism serves as its opportunity cost (step 1). The synthesis of building blocks f
rom
precursor metabolites incurs both opportunity (step 2) and direct (step 3) costs. The creation of macromolecular structures such as proteins from bu
ilding blocks
invokes postsynthesis costs such as the cost of polymerization (step 4). Direct costs are denoted in shades of orange, and opportunity costs are denot
ed in
shades of green. The direct costs of a building block can be attained by the sum of steps 3 and 4, whereas the opportunity costs of a building block is the su
mof
steps 1 and 2. The total cost of a building block is the sum of steps 1
–
4.
glucose
penP
eryP
αkg
dhap
3pg
oaa
pep
pyr
acCoA
Opporunity Cost of Precursor Metabolites in
Eukaryotes
and
Bacteria
glucose
αkg
3pg
oaa
pep
pyr
acCoA
*dhap
Glycolysis
TCA
2 NADPH
-1 ATP
r5p
NADPH
Pentose Phosphate
Pathway
-1 ATP
ATP
ATP
g6p
FADH
ATP
2 NADH
A.
B.
0
5
10
15
20
25
30
35
NADH
NADH
(P
O
)
Fig. S2.
The metabolic pathways and opportunity cost associated with each precursor metabolites. (
A
) Metabolic pathways and precursor metabolites in-
volved in the synthesis of building blocks. Metabolites that do not serve as precursors are denoted as gray circles. The energetic gain or loss from the
conversion
of one precursor metabolite to another is depicted (the energetic loss is denoted by a negative sign). The asterisk signifies that, from dhap onward, t
wo
molecules of each precursor metabolite are generated; however, the energetic gains or losses are reported on a per-molecule basis. (
B
) The opportunity cost of
each precursor metabolite in the context of bacterial and eukaryotic aerobic metabolism (using glucose as the sole carbon source) in units of P
O
. See SI sections
II
–
IV for a full description of these opportunity cost estimates.
Mahmoudabadi et al.
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alanine
arginine
asparagine
aspartate
cysteine
glutamate
glutamine
glycine
histidine
isoleucine
leucine
lysine
methionine
phenylalanine
proline
serine
threonine
tryptophan
tyrosine
valine
dATP
dGTP
dCTP
dTTP
ATP
GTP
CTP
UTP
Amino Acids
DNA
RNA
Total Cost:
Opportunity Cost of Precursor Metabolites:
Opporunity Cost of Synthesis from Precursor Metabolites:
Direct Cost of Synthesis from Precurosr Metabolites:
Direct Cost of Polymerization:
-10
0
10
20
30
40
50
60
70
80
36
25
5
2
4
55
41
1
11
2
51
41
-2
10
2
0
50
100
150
200
250
300
Lipid
282
194
70
18
N/A
lipid
Costs of building blocks in heterotrophic eukaryotes
alanine
arginine
asparagine
aspartate
cysteine
glutamate
glutamine
glycine
histidine
isoleucine
leucine
lysine
methionine
phenylalanine
proline
serine
threonine
tryptophan
tyrosine
valine
dATP
dGTP
dCTP
dTTP
ATP
GTP
CTP
UTP
Amino Acids
DNA
RNA
-10
0
10
20
30
40
50
60
70
80
Total Cost:
Opportunity Cost of precursor metabolites:
Opporunity Cost of synthesis from precursor metabolites:
Direct Cost of synthesis from precursor metabolites:
Direct Cost of polymerization:
31
21
4
2
4
47
33
1
11
2
44
33
-2
10
2
0
50
100
150
200
250
Lipid
230
156
56
18
N/A
lipid
Costs of building blocks in heterotrophic bacteria
A.
B.
Fig. S3.
The breakdown of direct and opportunity costs associated with amino acids, DNA, RNA, and lipids in the context of (
A
) bacterial and (
B
) eukaryotic
metabolism. Average cost values are reported in the table above the chart. See Dataset S1 for a detailed derivation of these costs.
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Table S1. T4 bacteriophage structural proteins and their average copy numbers per virion
T4 bacteriophage
Protein
Protein length
ð
L
pj
Þ
Average protein copy
number
ð
N
pj
Þ
Total no. of amino acids in
protein
j
ð
L
pj
N
pj
Þ
23*
521
930
484,530
20*
524
12
6,288
24*
427
55
23,485
soc*
80
840
67,200
hoc*
376
160
60,160
22*
269
576
154,944
21*
212
72
15,264
IPIII*
194
370
71,780
IPI*
95
360
34,200
IPII*
100
360
36,000
alt*
682
40
27,280
68*
141
240
33,840
67*
80
341
27,280
3
†
176
6
1,056
53
†
196
6
1,176
5
†
575
3
1,725
6
†
660
12
7,920
7
†
1,032
6
6,192
8
†
334
12
4,008
9
†
288
18
5,184
10
†
602
18
10,836
11
†
219
18
3,942
12
†
527
18
9,486
15
†
272
6
1,632
18
†
659
144
94,896
19
†
163
144
23,472
25
†
132
6
792
26
†
208
Assumed 1
208
27
†
391
3
1,173
28
†
177
Assumed 1
177
29
†
590
6
3,540
48
†
364
6
2,184
54
†
320
6
1,920
td
†
286
3
858
frd
†
193
6
1,158
holin
‡
218
20
4,360
endolysin
‡
164
20
3,280
spanin
‡
216
20
4,320
Totals
P
k
j
=
1
N
pj
=
4,805 proteins
P
k
j
=
1
L
pj
N
pj
=
1,225,786 aa
The number of amino acids comprising each virion is calculated by the product of the average protein copy
number and the length of the corresponding protein.
*These genes together compose the phage head.
†
These genes are those that make up the tail tube, the tail sheath, and the base plate (table modified from ref.
85).
‡
These genes are those that are involved in lysis (76).
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Table S2. Influenza A virus proteins and their average copy numbers per virion
Influenza A virus
RNA segment lengths
(no. of nucleotides)
Protein product
Protein
length
ð
L
pj
Þ
Average protein
copy number
ð
N
pj
Þ
Total no. of amino acids
in protein
j
ð
L
pj
N
pj
Þ
1 (2,341)
Polymerase PB2
759
45
34,155
2 (2,341)
Polymerase PB1
757
45
34,065
3 (2,233)
Polymerase PA
716
45
32,220
4 (1,778)
Hemagglutinin
566
500
283,000
5 (1,565)
Nucleoprotein
498
1,000
498,000
6 (1,413)
Neuraminidase
454
100
45,400
7 (1,027)
Matrix protein M1
252
3,000
756,000
Matrix protein M2
97
40
3,880
8 (890)
NS1
230
0
0
NS2
121
165
19,965
Totals
P
k
j
=
1
N
pj
=
4,940 proteins
P
k
j
=
1
L
pj
N
pj
=
1,706,685 aa
The number of amino acids comprising each virion is calculated by the product of the average protein copy number and the length
of the corresponding protein (table modified from ref. 86).
Dataset S1. The detailed breakdown of opportunity and direct costs of building blocks across heterotrophic bacterial and eukaryotic
metabolisms (using glucose as the sole carbon source). All references are provided as notes in the Excel sheets
Dataset S1
Dataset S2. A list of viruses and their associated costs used to estimate replication to translation cost ratios shown in Fig. 4
Dataset S2
Dataset S3. A list of direct and total fractional cost estimates,
E
g
, for genetic elements of lengths 1, 10, 100, 1,000, and 10,000 bp across
30 dsDNA viruses (Fig. 5). Genetic elements are assumed to have no functional benefit to the virus and to be nontranscribed. Viruses A, B,
and C correspond to hypothetical viruses
Dataset S3
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