Asako Isogawa

Robert P Fuchs

Shingo Fujii
email: shingo.fujii@inserm.fr

Enrichment of Plasmid Pool from E. coli for Mass Sequencing to Detect Untargeted Mutagenic Events

Translesion synthesis (TLS) is an event to cope with DNA damages. During TLS, the responsible TLS polymerase frequently elicits untargeted mutagenesis as potentially a source of genetic diversity. Identifying such untargeted mutations in vivo is challenging due to the bulk of DNA that does not undergo TLS. Here, we present a protocol to enrich a plasmid pool that underwent Pol V-mediated TLS in Escherichia coli for mass sequencing. The concept of this protocol could be applied into any species.

For complete details on the use and execution of this protocol, please refer to Isogawa et al. (2018).

Before you Begin

To enrich a plasmid pool that underwent TLS for mass sequencing, we utilized the characteristic mutagenic feature of Pol V at a TT pyrimidine-pyrimidone (6-4) photoproduct (TT (6-4)) [START_REF] Isogawa | Pol V-Mediated Translesion Synthesis Elicits Localized Untargeted Mutagenesis during Post-replicative Gap Repair[END_REF]. Pol V frequently misinserts a guanine opposite the 3'-thymine at TT (6-4) [START_REF] Tang | Roles of E. coli DNA polymerases IV and V in lesion-targeted and untargeted SOS mutagenesis[END_REF]. Therefore, when a single TT (6-4)-containing plasmid is introduced into E. coli, a subfraction of the plasmid progeny harbors the mutagenic signature induced by Pol V. The sequence context in which the TT (6-4) lesion is located belongs to a restriction enzyme recognition sequence that is altered by the Pol V-induced targeted mutation. Thus, the plasmid pool that experienced mutagenic Pol V bypass can be isolated by virtue of its resistance to restriction enzyme cleavage (Figure 1). Mass sequencing of that plasmid pool allows untargeted mutagenic events associated with the TLS events to be identified in comparison with control samples (see Figure 2). In principle, this protocol will be applicable into any species of interest provided a targeted mutagenic signature by a TLS polymerase of interest is known and if there is an available shuttle vector.

Design of a plasmid to detect untargeted mutagenic events Timing: 1 day 1. A plasmid to detect untargeted mutagenesis is a heteroduplex construct prepared through inserting a 13-mer oligo with a single TT (6-4) photoproduct into a gapped plasmid [START_REF] Becherel | SOS mutagenesis results from up-regulation of translesion synthesis[END_REF][START_REF] Koffel-Schwartz | Cellular strategies for accommodating replication-hindering adducts in DNA: control by the SOS response in Escherichia coli[END_REF]. This construct is designed to inactivate the lacZ gene in both strands: the lesion-containing strand includes a stop codon, while the complementary strand carries a frameshift mutation. If a TLS event across the TT (6-4) induces a mutation (i.e., targeted mutagenesis), the stop codon is reverted into an amino acid codon, leading to the reversion from lacZ -to lacZ + , and thus the formation of blue colonies as visualized on X-gal-containing indicator plates. Such a mutagenic plasmid progeny can be specifically isolated from the plasmid pool by virtue of restriction enzyme digestion as depicted in Figure 1 [START_REF] Isogawa | Pol V-Mediated Translesion Synthesis Elicits Localized Untargeted Mutagenesis during Post-replicative Gap Repair[END_REF].

Note: A lesion-free control plasmid, which is the same heteroduplex construct mentioned above while not containing TT (6-4), is also prepared (Figure 1).

Note: Depending on a TLS polymerase of interest, a type of lesion and/or a surrounding sequence context should be appropriately designed.

Competent cells prepared under SOS-induced conditions

Timing: 1 week Since a TLS polymerase, Pol V, functions only under SOS-induced conditions, competent cells to monitor Pol V-mediated TLS events are treated by UV irradiation. As in the "Primary transformation" step (Figure 2), mutant strains should be used as a host strain in order to circumvent repair of the heteroduplex region on plasmid (Figure 1). The heteroduplex regions on the lesion-free and lesioncontaining plasmid are substrates to mismatch repair (MMR) (repair of the mismatched base-pair)

and nucleotide excision repair (NER) (removal of the TT (6-4)), respectively. In our assay system, an NER-defective strain (ΔuvrA) with SOS induction (for Pol V activation) is chosen as a standard strain in the "Primary transformation" step when dealing with the lesion-containing plasmid. In the case of the lesion-free plasmid, an NER / MMR-defective strain (ΔuvrAΔmutS) without SOS induction is chosen. Thus, the minimum set of assays during the "Primary transformation" step requires two sorts of competent cells (Figure 2). In this protocol, we basically describe methodologies in the case of the minimum set. Depending on genes of interest to investigate their interplay, the number of strains to prepare competent cells varies. As an example in our case for the lesion-containing plasmid, we opted to test three mutants as host strains [START_REF] Isogawa | Pol V-Mediated Translesion Synthesis Elicits Localized Untargeted Mutagenesis during Post-replicative Gap Repair[END_REF] derived from MGZ (Tc r ) [START_REF] Napolitano | All three SOS-inducible DNA polymerases (Pol II, Pol IV and Pol V) are involved in induced mutagenesis[END_REF]: uvrA (Tc r Cm r ), uvrAdinB (Tc r Cm r, Km r ), uvrAdinBmutS (Tc r Cm r, Km r Spc r ), as indicated in the Key Resources Table . The dinB mutant is defective for another TLS polymerase (Pol IV).

Note: With respect to the control assay using a lesion-free plasmid: in the minimum assay set (Figure 2), we choose a ΔuvrAΔmutS strain without SOS induction for the lesion-free plasmid differently from a ΔuvrA strain with SOS induction for the lesion-containing plasmid. Due to the absence of replication blocking lesion in the lesion-free plasmid, this plasmid is normally replicated in host strains and potential mutations happened on the plasmid rely on the replication errors. Practically, such mutations irrespective of SOS-induction in cells are rare events and are not meaningfully detected by mass sequencing. Therefore, any arbitrary strain could be suitable as a host strain for the lesion-free plasmid.

LB

Reagent Final Concentration Amount

Bacto tryptone 1% 10 g Bacto yeast extract 0.5% 5 g NaCl 0.5% 5 g ddH2O n/a Fill to 1 L Total n/a 1 L Autoclave. If required, add an antibiotic(s) as follows: tetracycline (Tc), 10 ug/ml; chloramphenicol (Cm), 20 ug/ml; kanamycin (Km), 40 ug/ml; spectinomycin (Spc), 20 ug/ml; ampicillin (Ap), 100 ug/ml. LB-Agar for 16 plates (use ~25 ml per plate)

Reagent

Final Concentration Amount Bacto agar 1.5% 6 g LB n/a Fill to 400 ml Total n/a 400 ml Autoclave. When melted LB-Agar cools down (~55°C) in a water bath, add an antibiotic(s) as follows: Tc, 10 ug/ml; Cm, 20 ug/ml; Km, 40 ug/ml; Spc, 20 ug/ml; Ap, 100 ug/ml.

Tetracycline stock solution

Reagent

Final Concentration Amount Tetracycline 10 mg/ml 50 mg Ethanol (50%) 50% Fill to 5 ml e. Inoculate 4 ml of the culture into pre-warmed 200 ml LB in a sterilized 2 L flask (final 50-fold dilution) f. Incubate the flask until OD600 = 0.4-0.5 (it takes ~80 min) at 37°C, with shaking (200 rpm)

Note: In our assays, all of the mutant strains (uvrA, uvrAdinB, uvrAmutS) except for a strain (uvrAdinBmutS) exhibit similar growth rates in rich media (e.g., LB). The ~80 min reaching OD = 0.4-0.5 are an approximate incubation time in a condition, LB at 37°C, for our mutant strains. For the uvrAdinBmutS strain, this exhibits ~50% slower growth rates compared with the other strains under the same growth conditions. If choosing mutant strains having different genetic backgrounds, their growth rates may be different depending on types of mutants. In addition, if using a minimal medium or incubating at a lower temperature, a longer incubation time will be required to reach the desired OD.

3. Competent cells under SOS induction a. Transfer the culture into a sterilized 250 ml centrifuge tube b. Centrifuge the tube (3,000g, 15 min, 15°C)

Note: We normally choose a relatively mild centrifugation condition (3,000g) during competent cells preparation. This is to avoid potential risks might be caused by high centrifugation forces (e.g., 15,000g) such as damages to cells. In addition, cell pellets prepared by a mild centrifugation condition are highly compatible to homogeneous resuspension in the subsequent step. If the centrifugation time (15 min) seems to be inadequate to make cell pellets, we recommend to prolong the centrifugation time rather than increasing the centrifugation force.

c. Discard the supernatant d. Resuspend the pellet with 0.6 volumes (120 ml) of 10 mM MgSO4

Note: Withdraw 100 ul of the suspension to estimate cell viability by comparison between before and after UV irradiation: in our strains, their colony forming units (cfu) per ml before and after UV irradiation are expected to be ~10 8 and ~10 7 , respectively. i. Dilute the suspension (the factors are 10 -4 , 10 -5 and 10 -6 ) with LB (each final volume is 1 ml) ii.

Inoculate 200 ul of the 1 ml diluted suspensions onto LB agar plates with antibiotics iii.

Incubate the plates for ~16 h at 37°C iv.

Count colonies on the plates and calculate cfu/ml (in this "before UV irradiation", hundreds of colonies will appear per plate in the case of 10 -5 dilution)

e. Pour 10 ml of the suspension into each 10 cm dish, total 12 dishes f. Irradiate UV (254 nm; 8 J/m 2 ): When using a UV lamp (see Key Resources Table ), the lamp is set on two (or more) appropriate boxes (see Figure 3) in a dark room. If there is no dark room, any kind of shielding space can be utilized. Subsequently, a desired UV irradiation time is determined via monitoring strength of UV by a UV detector (see Key Resources Table ). In our case, the distance between the UV lamp and the dish is ~25 cm, and the irradiation time is ~8 s to reach 8 J/m 2 .

Note: In our UV-irradiation setting, one dish is irradiated at once. Therefore, 12 turns are required to irradiate all 12 dishes. g. Collect the suspension from all 10 cm dishes into a sterilized 250 ml centrifuge tube Note: Withdraw 100 ul of the suspension to estimate cell viability by comparison between before and after UV irradiation (Implement the same steps i-iv as mentioned above)

h. Centrifuge the tube (3,000g, 15 min, 15°C) i. Discard the supernatant from the tube j. Resuspend the pellet with a small amount (~20 ml) of 37°C pre-warmed LB k. Transfer the suspension into a sterilized 2 L flask l. Rinse the same centrifuge tube collected cells with 37°C pre-warmed LB (~20 ml) to collect residual cells in the tube, transfer it into the same 2 L flask (cells are totally resuspended with 1 volume (200 ml) of LB)

Note: If residual cells/suspension in the tube are not visible, this "rinse" step can be omitted. In this case, the suspension in the flask is simply filled up to final 200 ml.

m. Incubate the flask for 30 min at 37°C, with shaking (200 rpm) n. Put the flask in wet ice, and incubate it for 10 min o. Transfer the suspension into a sterilized 250 ml centrifuge tube and centrifuge (3,000g, 15 min, 4°C) p. Wash 1: Discard the supernatant, and resuspend the pellet with a small amount (~20 ml) of ice cold sterilized MilliQ water, subsequently add more MilliQ water (total 1 volume (200 ml)), centrifuge (3,000g, 15 min, 4°C) and discard the supernatant q. Wash 2 & 3: Repeat two more washes with the same centrifuge conditions as Wash 1, but using 1/2 volumes (100 ml) of MilliQ water in the same tube (Wash 2) and using 1/5 volumes (40 ml) of ice cold 10% glycerol in a sterilized 50 ml centrifuge tube (Wash 3) r. Resuspend the pellet with 1/250 volumes (~800 ul) of ice cold 10% glycerol s. Aliquot the suspension in 0.5 ml tubes (40 ul per tube). Store the aliquots at -80°C CRITICAL: UV-irradiated competent cells should exhibit around 10% of cell survival compared with non-irradiated cells. Otherwise, competent cells should be newly prepared with adjusted settings of UV-irradiation (see Troubleshooting 2).

CRITICAL: Concentration of UV-irradiated competent cells should be around 10 8 cells per 40 ul (see Troubleshooting 3). Estimation of the cell number: i. Dilute the competent cells by the factor of 10 -5 with LB (final volume is 1 ml) ii. Inoculate 200 ul of the 1 ml diluted suspension onto LB agar plates with antibiotics. iii. Incubate the plates for ~16 h at 37°C iv. If competent cells are properly prepared, about 200 colonies should appear per plate

Competent cells prepared under normal growth conditions

Timing: 1 week At the stages of "Primary transformation" for the lesion-free plasmid and "Secondary transformation" for both plasmids (Figure 2), we use competent cells without SOS-induction. The way to prepare the competent cells is essentially the same as the way described above section, "Competent cells prepared under SOS-induced conditions", except for omitting the steps related to UV-irradiation. In order to do the minimum set of assays, competent cells need to be prepared from two strains, a ΔuvrAΔmutS strain and a wild-type strain.

Measuring transformation efficiency of the competent cells

Timing: 2-3 days

In order to estimate transformation efficiencies of the prepared competent cells, various amounts of the lesion-containing plasmid are used as an input sample.

Note: For selection of transformants harboring the plasmid on LB-Agar plates, ampicillin (100 ug/ml) is used in addition to antibiotics required to select for chromosomal markers.

SOB

Reagent Final Concentration Amount

Bacto water) on ice d. Add 1 ul (1, 2, 4, and 8 ng) of the plasmid into the competent cells (make sure that the competent cells are thawed) e. Agitate the competent cells by pipetting. Transfer the mixtures into the electroporation cuvettes f. Electroporate the cells according to cuvette manufacture's instructions for bacteria (Bio-Rad): our electroporation apparatus is Gene Pulser Xcell (Bio-Rad); the settings, Bacterial 2 (for 0.2 cm gap cuvettes) = 25 uF, 200 ohm, 2500 V. g. Add 960 ul of SOC into each cuvette immediately (e.g., < 5 s) after the electroporation h. Transfer the suspensions from the cuvettes into 15 ml tubes i. Incubate with shaking (180 rpm) at 37°C for 1 h j. Dilute the suspensions (the factors are 10 -2 and 10 -3 ) with LB (each final volume is 1 ml) k. Spread 200 ul of the 1 ml diluted suspensions onto LB agar plates with antibiotics l. Incubate the plates for ~16 h at 37°C m. Count colonies on the plates Note: If transformation efficiency is 4 x 10 8 per ug of plasmid, in the case of 1 ng input plasmid, expected number of colonies per plate will be ~800 and ~80 in the dilution factors of 5 x 10 -2 and 5 x 10 -3 , respectively.

CRITICAL: Transformation (electroporation) efficiency of competent cells should reach ~4

x 10 8 per ug of plasmid and its efficiency will not be affected up to ~8 ng plasmid. If the efficiency of plasmid uptake is significantly low (e.g., < 5 x 10 7 ), competent cells should be newly prepared (see Troubleshooting 4). Total n/a ~400 ml Add also appropriate antibiotics; X-gal is dissolved by dimethylformamide (DMF) and needs protection from light; IPTG and X-gal are stored at -20°C; When autoclaved LB-Agar is cooled down (~55°C) in a water bath, add the supplements (IPTG, X-gal, antibiotics). Stored at 4°C; Protect the prepared plates from light and use within one week.

Materials and Equipment

Step-by-Step Method Details Primary transformation

Timing: 2 days

In this step, a subfraction of the single lesion-containing plasmid in cells undergoes TLS events. The following protocol indicates the minimum set of this assay system exemplified in Figure 2: the combination is to transform a ΔuvrAΔmutS strain without SOS induction with the lesion-free plasmid and a ΔuvrA strain with SOS induction with the lesion-containing plasmid. Depending on experimental purposes, the number of combinations between a plasmid and a strain will vary. In this case, perform the following steps for each additional combination.

1. Plasmid recovery from the transformants a. Electroporate 40 µl of the competent mutant strain of interest with 8 ng (1 µl) of the lesion containing plasmid (total two times: finally, 16 ng of plasmid are used for 80 ul of competent cells), and the control strain with 2.5 ng (1 µl) of the lesion free plasmid. Follow steps a-i of the previous section, "4. Transformation".

Note: 16 ng of plasmid (~2.7 kbp) contains ~3.9 x 10 9 plasmid molecules.

Note:

In order to check efficiency of TLS events, also perform the following steps: i.

The way is the same as previously mentioned "check of cell viability" on LB plates containing X-gal and the dilution factors are 5 x 10 -2 and 5 x 10 -3 . Incubate the plates for ~16 h at 37°C ii.

Count blue and white colonies on the plates Note: As depicted in Figure 1, appearance of blue colonies basically relies on Pol Vmediated targeted mutagenesis for the lesion-containing plasmid. On the other hand, there is no appearance of blue colony for the lesion-free plasmid (although a few blue colonies may appear due to pre-existing errors in the lesion-free plasmid construct (see the section of "Limitations")).

iii.

Calculate mutation frequency (blue / (blue + white))

Note: Proportion of blue colonies relative to all colonies (blue + white) should be ~10%. For example, if the total number of viable cells (i.e., blue + white) is 10 7 , the number of plasmid molecules that underwent independent TLS events would be around 10 6 . The value (~10% blue) is specific to the present assay conditions (i.e., the bypass of the TT (6-4) lesion in the SOS-induced strains) and may be varying when assaying a different TLS polymerase, mutant background or organism.

CRITICAL: As error rates of TLS polymerases are around 10 -4 (a range of 10 -3 to 10 -5 ) per base [START_REF] Fujii | A Comprehensive View of Translesion Synthesis in Escherichia coli[END_REF] and the length of template DNA filled by a Pol V may be up to 57 nt in vivo as well as in vitro [START_REF] Fujii | Biochemical basis for the essential genetic requirements of RecA and the beta-clamp in Pol V activation[END_REF], the detection of one untargeted mutagenic event will require around 10 2 to 10 3 plasmid molecules. Since we wished to detect hundreds to thousands of untargeted mutagenic events by mass sequencing, we aimed to prepare plasmid pools containing around 10 6 plasmid molecules that underwent independent TLS events. If a plasmid pool is not a proper size (i.e., < 10 6 ), this transformation step should repeat until reaching the proper size.

Concentrate the plasmid underwent TLS events

Timing: 1 week

As the plasmid pool in the "1st prep" derived from the lesion-containing plasmid contains largely undesired plasmid molecules (~90% of total plasmid) not undergoing TLS events in cells, this step aims to increase proportion of desired closed circular plasmid molecules (~10% of total plasmid) that underwent TLS events from the plasmid pool. The following steps aim to digest the undesired plasmid molecules into linear DNA, while plasmids that were mutated by TLS events have become resistant to the specific endonucleases. Linear and closed circular DNA can then be separated by CsCl density gradient centrifugation. In addition, the transformation efficiency of linear DNA is negligible compared with closed circular DNA in E. coli.

CsCl saturated isopropanol with TE

Reagent Final Concentration Amount

Tris-Cl (1 M) (pH 8.0 at 25°C) 10 mM 500 ul EDTA-Na (0.5 M) (pH 8.0) 1 mM 100 ul ddH2O n/a 49.4 ml Cesium chloride (CsCl) saturation > 100 g Isopropanol < 50% 50 ml

Total n/a > 100 ml

Add and mix CsCl until reaching saturation in TE (Tris + EDTA + ddH2O), then add more CsCl (~10 g). Subsequently, add and mix isopropanol. The resultant mixed solution appears as two separated, aqueous (bottom) and organic (top), layers.

The following steps are specific to the case of "1st prep" derived from the lesion-containing plasmid.

With respect to the lesion-free control plasmid, we describe the protocol later on.

2. Double digestion by restriction enzymes, KpnI / HpaI a. Transfer 20 ug equivalent volume of the "1st prep" derived from the lesioncontaining plasmid into a 1.5 ml tube b. Adjust the volume of DNA solution to 291 ul with MilliQ water c. Add 36 ul of 10x Cut Smart buffer (NEB) d. Mix well e. Add 26 ul of 5 u/ul HpaI and 6.8 ul of 100 u/ul KpnI-HF f. Incubate the tube (~360 ul of reaction mixture) at 37°C for 2 h g. Purify DNA through phenol/chloroform treatment and ethanol precipitation h. Transfer the reaction mixture into a 15 ml tube Note: Check if the digestion is efficient by agarose gel electrophoresis. In our assay conditions, the plasmid pool in the "1st prep" derived from the lesion-containing plasmid is composed of ~90% (not associated to TLS; indicated by appearance of white colonies) and ~10% (associated to TLS; indicated by appearance of blue colonies) of plasmid. These outcomes indicate the ~90% of plasmid molecules are progenies derived from the KpnI strand, leading to KpnI sensitivity and HpaI resistance. The remaining ~10% are progenies derived from the HpaI strand, leading to resistance to both KpnI and HpaI (see Figure 1). With respect to HpaI digestion in addition to KpnI digestion, there are two sources for generation of HpaI sensitive progenies: 1) derived from the HpaI strand in the absence of the TT (6-4) lesion (i.e., this is the same as the lesion-free control plasmid) (such a contamination is normally inevitable during construction of the lesion-containing plasmid due to the presence of lesion-free oligo as a minor contamination (e.g., ~0.5% in our case) in the lesion-containing oligo pool; 2) as a minor event, Pol V faithfully bypasses the TT (6-4) lesion, resulting in HpaI sensitive progenies and appearance of white colonies. Whereas the proportion of HpaI sensitive plasmids in the plasmid pool would be minor, we exclude such progenies via HpaI digestion in order to simplify interpretation of obtained data. As shown in Figure 4, the vast majority of plasmid is indeed sensitive to KpnI (the double digestion, KpnI/HpaI, exhibits visibly similar pattern as the KpnI single digestion as expected).

Note: If the digestion is inefficient compared with an expected outcome based on proportion of blue and white colonies on X-gal-containing LB plates, repeat the digestion process again.

CRITICAL:

When SOS-induced host strains are transformed with the plasmid after the double digestion (see the lane of KpnI/HpaI digestion in Figure 4), around 70-80% of blue colonies appear on X-gal-containing LB plates in contrast to ~10% of blue colonies by the plasmid of the "1st prep" (see the lanes of no digestion in Figure 4).

CRITICAL:

If the proportion of blue colonies reach ~90% by this double digestion step, the following "Ultracentrifugation" and "Second double digestion" steps can be skipped because the value is high enough and a further increase of the proportion of blue colonies will be difficult. In this case, the next step becomes "Secondary transformation" step.

Ultracentrifugation and fractionation

a. Add 3.6 ml of TE to the reaction mixture in the 15 ml tube (total 4.3 ml) b. Add 100 ul of 10 mg/ml ethidium bromide (EtBr) c. Add 4 g of cesium chloride (CsCl) (total volume becomes around 5.4 ml). Mix well d. Transfer ~5 ml of the mixture into a 5.1 ml quick seal ultracentrifuge tube with a Pasteur pipette e. Seal the tube by heat sealer f. Ultracentrifuge the tube in a Beckman NVT 65.2 rotor (50,000 rpm, 16 h, 20°C) g. The tube is carefully fixed with an appropriate support rod. Subsequently, cut an upper side of the disposable ultracentrifuge tube carefully to avoid disturbing the gradient formed in the tube.

Note: Presence of DNA is visibly observable by different contrast in the tube. However, in our sample preparation conditions (i.e., the presence of excess amounts of linear DNA), we cannot see a single band composed of closed circular DNA because significant amounts of linear DNA are also contaminated in the same local area of the closed circular DNA (see Figure 5).

h. Withdraw aliquots (~170 ul each) from the upper phase of the sample using a tube connected to a pump (total ~20 fractions).

Note:

The total recovery volume is ~3.4 ml and the remaining ~1.6 ml from the bottom side is discarded.

Note: If there is not an appropriate pump, withdraw aliquots (~170 ul each) manually from the upper phase by using a pipette i. Each sample is adjusted to 0.3 ml with MilliQ water j. Add 1 ml of the organic isopropanol phase (top layer) of "CsCl saturated isopropanol with TE". Mix well Note: In the tube, two phases visibly appear, an aqueous phase containing DNA sample at the bottom side and an organic phase containing isopropanol and EtBr on top side. As solution including EtBr exhibits a color of red just under normal light, conversion of red to clear color indicates removal of EtBr.

k. Take off and discard supernatant (the top layer). Repeat steps j and k 2 times more in order to thoroughly remove EtBr from the samples (total 3 times) l. Add 3 volumes of MilliQ water m. Ethanol precipitate Note: Whereas the sample already contains high concentration of CsCl as a salt, we implement a standard ethanol precipitation (adding 1/5 volumes of 3 M NaOAc and 3 volumes of ethanol).

n. Resuspend the pellet with 25 ul of 1/10 TE (10-fold dilution of TE by MilliQ water) o. Measure DNA concentration by Nanodrop p. Choose fractions containing closed circular DNA via checking migration pattern of DNA in agarose gel electrophoresis (Figure 5)

Note: Amounts of the recovered DNA will be 3-6 ug.

Note: Owing to the presence of excess amounts of linear DNA in the input sample, it is not easy task to clearly separate closed circular DNA (ccDNA) from linear DNA during the ultracentrifugation. Instead, this step aims to moderately increase relative concentration of ccDNA by choosing fractions containing relatively high amounts of ccDNA. (e.g., choose fractions containing >20% of ccDNA relative to the total amounts of ccDNA).

Note: Although this ultracentrifugation step will moderately increase the relative proportion of ccDNA in the total DNA (and also exclude minor genomic DNA contaminants in principle), the percentage of blue colonies is not changed because the sources of blue and white colonies rely on ccDNA.

Second double digestion

Note: In order to obtain a high-quality DNA substrate (e.g., result in around 90% of blue colonies) consisting of the desired plasmid molecules that underwent TLS events, DNA samples are digested by endonucleases again to linearize residual undesired plasmid molecules that do not contain the TLS mutation signature and thus remain sensitive to KpnI or HpaI digestion.

Note: When assaying a different TLS polymerase, mutant background or organism, proportion of blue colonies may never reach ~90% due to generation of untargeted secondary mutations leading to inactivation of lacZ (see Expected Outcomes).

a. Repeat steps a-f of the section "Double digestion by restriction enzymes, KpnI / HpaI", but adjust the experimental settings (reaction volume, amounts of restriction enzymes) to the amounts of DNA (will be 3-6 ug). b. Purify DNA through phenol/chloroform treatment and ethanol precipitation c. Resuspend the pellet in 10 ul of 1/10 TE. This sample is termed "treated 1st prep" d. Measure DNA concentration by Nanodrop e. Estimate amounts of ccDNA by agarose gel electrophoresis CRITICAL: Check the quality of plasmid in the "treated 1st prep" via transformation to measure percentage of blue colonies. By this second double digestion, proportion of blue colonies will slightly increase to ~90% from 70-80% in the first double digestion (and also in the sample following the ultracentrifugation). Although DNA profiles on agarose gel are indistinguishable before and after the second double digestion (Figure 6), a subfraction of ccDNA leading to appearance of white colonies would be digested. If the second double digestion does not slightly increase proportion of the blue colonies, sequencing a few white colonies may be useful to judge if re-digestion is meaningful (see Troubleshooting 5).

Restriction digestion of the lesion-free control plasmid

Note: As shown in Figure 1, when a host strain is transformed by the lesion-free plasmid, two different plasmid progenies (i.e., KpnI strand derived and HpaI strand derived) are equally amplified in the strain, and both of which just lead to appearance of white colonies on X-gal-containing LB plates. Our experimental purpose is to detect untargeted mutagenic events associated to Pol V-induced targeted mutagenesis occurring on the HpaI strand in the lesion-containing plasmid. Therefore, as a control, we aim to enrich the HpaI strand-derived plasmid progenies from the lesion-free plasmid pool through digestion of the KpnI strand-derived plasmid progenies with the restriction enzyme, KpnI. a. Digest the "1st prep" derived from the lesion-free plasmid as previously mentioned in the section, "Double digestion by restriction enzymes, KpnI / HpaI", but omit HpaI. This sample is termed "treated 1st prep" Note: As shown in Figure 2, there is no requirement to do further processing for the lesion-free prep (i.e., ultracentrifugation and second double digestion) that were implemented in the preparation of "treated 1st prep" of the lesion-containing plasmid pool.

Secondary transformation

Timing: 2 days

This step aims to amplify the plasmid pool in the "treated 1st prep" (prepared from both lesion-free and lesion-containing plasmid pools) in order to obtain sufficient amounts of plasmid for mass sequencing (Figure 2). 6. Transformation and recovery of plasmid are implemented as the same way described in the section, "Primary transformation" except that the host strain is a wild-type strain (MGZ) without SOS-induction; input DNA is 2 ng of ccDNA in "treated 1st prep". The recovered plasmid preparation is termed "2nd prep"

Note: As the purpose is to amplify the plasmid pool in cells, any wild type cells can be used as competent cells.

Note: 2 ng of plasmid (~2.7 kbp) contains ~4.9 x 10 8 plasmid molecules. Around 10 6 of transformants will appear and cover most of independent TLS events occurred on plasmid in the "1st prep".

Note: This "2nd prep" is subjected to mass sequencing such as Plasmid SMRT sequencing (Pacific Biosciences) that requires 1-2 ug of input DNA.

Expected Outcomes

This protocol generates a plasmid pool containing ~10 6 independent TLS events. When sequencing ~10 5 molecules chosen randomly from the pool, the vast majority (>90%) of sequenced molecules could stochastically reflect progenies derived from plasmid molecules that underwent independent TLS events in the pool [START_REF] Isogawa | Pol V-Mediated Translesion Synthesis Elicits Localized Untargeted Mutagenesis during Post-replicative Gap Repair[END_REF]. In order to obtain intelligible mass sequencing data, it is important to prepare high-quality plasmid pool consisting of plasmid molecules underwent TLS events as much as possible. Indeed, sequencing data using plasmid pools following this protocol revealed that ~90% of sequenced molecules contain the mutagenic signature of Pol V at the TT (6-4) site, demonstrating that most plasmid molecules in the pool underwent TLS in vivo [START_REF] Isogawa | Pol V-Mediated Translesion Synthesis Elicits Localized Untargeted Mutagenesis during Post-replicative Gap Repair[END_REF].

In principle, as untargeted mutagenic events (especially frameshift mutagenesis) associated with TLS events may disrupt a functional gene such as lacZ, plasmids possessing Pol V-induced targeted mutagenic signature would induce not only blue colonies but also white colonies on X-gal-containing LB plates. On the other hand, as untargeted mutagenic events per se are infrequent events, such white colonies derived from the untargeted mutagenic events are unlikely to contribute significantly to decrease of proportion of blue colonies.

Limitations

When constructing a single lesion-containing plasmid (Figure 1), a short synthetic oligo is inserted into a short gap-containing plasmid. Any synthetic oligo typically exhibits a high error frequency introduced during its chemical synthesis. In our 13-mer oligo, the mutation frequency is ~0.25% per nt [START_REF] Isogawa | Pol V-Mediated Translesion Synthesis Elicits Localized Untargeted Mutagenesis during Post-replicative Gap Repair[END_REF] that is higher than the mutation rates of TLS polymerases in E. coli [START_REF] Fujii | A Comprehensive View of Translesion Synthesis in Escherichia coli[END_REF]. Thus, detection of untargeted mutagenic events is unreliable within the region of oligo.

On the other hand, as the targeted mutagenic event at the TT (6-4) site is ~90%, such frequent events can be readily detected even in the region of the oligo. When applying this assay system in any other species, if an average size of TLS patch synthesized by TLS polymerases is shorter than the length of the inserted oligo that contains the lesion, detection of untargeted mutagenic events will be unsuccessful. Thus, a prerequisite to detect untargeted mutagenic events associated with TLS events is that a TLS patch size extends beyond the length of the inserted synthetic oligo. In the case of Pol Vmediated TLS events in E. coli, a DNA gap appears triggered by a lesion that blocks progress of the replicative DNA polymerase (this event is essential to induce the SOS response via RecA nucleoprotein filament formation), subsequently Pol V bypasses the lesion and produces a TLS patch. As the TLS patch size is much shorter than the DNA gap size, normal gap-filling events occur following the TLS patch formation to thoroughly fill in the gap [START_REF] Fujii | A Comprehensive View of Translesion Synthesis in Escherichia coli[END_REF]. During the normal gap-filling events, Pol V participates frequently in the events despite the absence of lesion [START_REF] Isogawa | Pol V-Mediated Translesion Synthesis Elicits Localized Untargeted Mutagenesis during Post-replicative Gap Repair[END_REF]. Thus, Pol V-induced untargeted mutations occur in the whole region of DNA gap not restricted within the TLS patch. This feature is highly compatible to this protocol by which the untargeted mutations occur outside of the synthetic oligo region. Similarly, even if a TLS patch size is expected to be short in any other species, if a TLS polymerase of interest participates also in the normal gap-filling events, the untargeted mutations could be detected by this protocol. If untargeted mutations are not detected, this would indicate either the DNA gap size is short or the TLS polymerase does not participate in gap filling.

Troubleshooting

Problem 1:

A host strain using in "Primary transformation" is genetically unstable Potential Solution:

Readers should seek for a better growth condition (types of media, growing temperatures, etc.) or test other strains possessing a mutation of interest and a different genetic background. At least, overgrowth conditions should be avoided to maintain the genetic stability.

Problem 2:

UV-irradiated competent cells exhibit lower or higher cell survival Potential Solution:

For fully inducing the SOS response by UV irradiation, ~10% cell survival compared to non-irradiated cells is appropriate. If cell survival is significantly lower (e.g. < 4%) or higher (e.g. > 20%), competent cells should be prepared again. Since nucleotide excision repair (NER) defective strains (e.g., uvrA) show hyper sensitivity to UV irradiation, setting of a UV lamp needs to be carefully adjusted and the strength of UV should always be checked by a UV detector. In addition, the depth of cell suspension in a plate should be shallow (e.g. ~2.5 mm in the case of 10 ml suspension in a 10 cm dish) in order to uniformly irradiate cells.

Problem 3:

Concentration of competent cells does not reach around 10 8 cells per 40 ul Potential Solution:

In the MilliQ wash steps, as cell pellets following centrifugation are soft, significant amounts of cells may be lost during the steps. Following centrifugation, the supernatant should be carefully removed and a small portion of the supernatant should be left rather than thoroughly removing it Problem 4:

Competent cells do not show enough transformation efficiency Potential Solution:

Always keep the centrifuge tubes with cells in ice water during manipulation at the wash steps to avoid an increase in temperature of the cells. In addition, the number of viable cells in the tube of competent cells may need to be checked. If the cell number is too low, preparing new competent cells.

Problem 5:

Transformation of "treated 1st prep" does not increase the ratio of blue colonies Potential Solution:

In order to get insight for quality of the plasmid pool, sequence plasmid preps prepared from blue and white colonies (e.g., each 10 colonies) obtained after transformation with the "treated 1st prep". Depending on the sequencing data, e.g. detection of the KpnI site from one or more white colonies, the endonuclease treatments should be repeated. Fractions 8-17 are analyzed by a 0.7% agarose gel with EtBr (0.5 ug/ml): 3 ul of each fraction are loaded. We choose fractions 12-13 as a ccDNA enriched fraction. The lane control is loaded 10 ng of plasmid (no treatment of restriction endonuclease)

Figure 6: DNA profiles before and after "Second double digestion" Each 30 ng of DNA is analyzed by a 0.7% agarose gel with EtBr (0.5 ug/ml).

KEY RESOURCES

  b. Transfer the suspension into a 2 L flask containing 200 ml of LB with antibiotics c. Cultivate the culture at 37°C with shaking (200 rpm) until OD = 1 (it takes ~7 h) d. Collect the cells by centrifugation (3,000 g, 15 min, 15°C) e. Purify plasmid from the collected cells according to Qiagen's MAXI prep plasmid preparation protocol. The plasmid preparation is termed "1st prep" f. Measure DNA concentration by Nanodrop

Figure 1 :Figure 2 :Figure 3 :Figure 4 :

 1234 Figure1: Characteristic features of plasmid to detect untargeted mutagenic events associated to TLS events

  Figure 1

Figure 5 Figure 6

 56 Figure 5

  Place 4 tubes of competent cells (40 ul) on ice b. Place 4 electroporation cuvettes (0.2 cm) on ice c. Place 4 tubes of the lesion-containing plasmid (1, 2, 4, and 8 ng/ul, diluted by MilliQ

	a.
	tryptone	2%	20 g
	Bacto yeast extract	0.5%	5 g
	NaCl	0.05%	0.5 g
	KCl (250 mM)	2.5 mM	10 ml
	ddH2O	n/a	Fill to 1 L
	Total	n/a	1 L
	Autoclave
	SOC
	Reagent	Final Concentration	Amount
	SOB	n/a	10 ml
	MgCl2 (1 M)	10 mM	100 ul
	MgSO4 (1 M)	10 mM	100 ul
	Glucose (1 M)	20 mM	200 ul
	Total	n/a	~10 ml
	Filtered by 0.22um filter
	4. Transformation

TABLE REAGENT

 REAGENT

	Plasmid
	Single TT (6-4) containing plasmid	Koffel-Schwartz et al.,	N/A
		1996; Becherel and
	or RESOURCE	SOURCE Fuchs, 1999	IDENTIFIER
	E. coli Strains Other
	MGZ (wild type, MG1655 derived) NanoDrop™ 2000	Napolitano et al., 2000 N/A ThermoFisher ND-2000
	uvrA (MGZ derived) Density meter, Ultrospec™ 10 Classic	Napolitano et al., 2000 N/A VWR 634-0882
	uvrAdinB (MGZ derived) Disposable cells, PS, 1,5 ml, 10 mm	Napolitano et al., 2000 N/A VWR SCLI80-2084-11
	uvrAdinBmutS (MGZ derived) UVP Blak-Ray Lamp	Isogawa et al., 2018 VWR	N/A XX-15S
	uvrAmutS (MGZ derived) UVP UVX Radiometer	Napolitano et al., 2000 N/A Fisherscientific 11881563
	Chemicals, Peptides, and Recombinant Proteins Gene Pulser	Bio-Rad	165-2660
	Trizma base, bioxtra Gene Pulser Electroporation Cuvettes 0.2cm gap	Sigma-Aldrich Bio-Rad	T6791-100G 165-2086
	EDTA Centrifuge 5910 R	Sigma-Aldrich Eppendorf	E5134-50G 5942000315
	Agarose Centrifuge 5427 R	Sigma-Aldrich Eppendorf	A9539-100G 5409000535
	TAE Buffer Beckman NVT65.2 rotor	Fisherscientific Beckman	10490264 361073
	Smart ladder Beckman polyallomer quick seal centrifuge tube	Eurogentec Beckman	MW-1700-10 342412
	Bromophenol Blue (13x51mm) Xylene Cyanol FF Beckman TA-10-250 Fixed-Angle Aluminum Rotor Ethidium bromide Beckman 250ml Polycarbonate Bottle with Screw on cap Beckman Sigma-Aldrich Sigma-Aldrich Beckman Sigma-Aldrich 2-Propanol Sigma-Aldrich 2.0 mL Microcentrifuge Tubes (Eppendorf Safe-Lock , Sigma-Aldrich Cesium chloride Euromedex natural) 1.5 mL Microcentrifuge Tubes (Eppendorf Safe-Lock , Sigma-Aldrich Phenol:Chloroform:Isoamyl Alcohol (25:24:1) Sigma-Aldrich natural) Chloroform:Isoamyl alcohol (24:1) Sigma-Aldrich Apparatus Mupid®-One Electrophoresis System Complete Eurogentec	114391-5G 368293 X4126-10G 356013 E1510-10ML EP0030123620-I9516-1L 500EA EU0770-EP0030123611-B 500EA P2069-100ML C0549-1PT MU-0041-
	Ethanol	Sigma-Aldrich	51976-500ML-F
	Glycogen (Roche, 20mg/ml solution)	Sigma-Aldrich	10901393001
	Sodium acetate	Sigma-Aldrich	71183-250G
	KpnI HF (100,000 units/ml)	NEB	R3142M
	HpaI (5,000 units/ml)	NEB	R0105S
	Magnesium chloride	Sigma-Aldrich	M2670-100G
	Magnesium sulfate heptahydrate	Sigma-Aldrich	M2773-500G
	Potassium chloride	Sigma-Aldrich	P9333-500G
	Sodium chloride	Sigma-Aldrich	S7653-250G
	D-(+)-Glucose	Sigma-Aldrich	G7528-250G
	Gibco Bacto tryptone	ThermoFisher	211705
	Gibco Difco Bacto yeast extract	ThermoFisher	212750
	BD Bacto™ Dehydrated Agar	Fisherscientific	10455513
	Glycerol	Sigma-Aldrich	G2025-500ML
	N,N-Dimethylformamide	Sigma-Aldrich	227056-100ML
	IPTG	Sigma-Aldrich	I5502-1G
	X-gal	Sigma-Aldrich	B9146-10MG
	Tetracycline hydrochloride	Sigma-Aldrich	T7660-5G
	Ampicillin sodium salt	Sigma-Aldrich	A9518-5G
	Kanamycin sulfate	Sigma-Aldrich	60615-5G
	Spectinomycin dihydrochloride pentahydrate	Sigma-Aldrich	S4014-5G
	Chloramphenicol	Sigma-Aldrich	C0378-5G
	Critical Commercial Assays
	Qiagen Plasmid Maxi kit	Qiagen	12163

Acknowledgments

This work did not receive any specific grant.

Resource Availability

Lead Contact

Further information and requests for resources and reagents should be directed to and will be fulfilled by the Lead Contact, Shingo Fujii (shingo.fujii@inserm.fr).

Materials Availability

This study did not generate any new unique reagents.

Data and Code Availability

This protocol does not include any datasets or code.

Author Contributions

Conceptualization, R.P.F. and S.F.; Investigation, A.I.; Writing -Original Draft, A.I. and S.F.; Writing -Review & Editing, A.I., R.P.F. and S.F.

Declaration of Interests

The authors declare no competing interests.


Jamil A Nasser
email: jamil.nasser@uvsq.fr

A microscopic model for the phase diagramm of a pur substance

Recently, a two-dimensional model that leads to …rst order phase transitions has been proposed. This model also leads to phase diagrams which present the characteristics observed in phase diagrams of the substances : three thermodynamic phases, three coexistence curves, a triple point and a critical point.

In the present study, the calculation of the latent heats of the di¤erent …rst order phase transitions as well as the experimental data concerning the phase diagram of a pure substance, here carbon dioxyde, are introduced into the two-dimensional model.

By assuming that the pressure does not modify the chemical bond between two nearest-neighbors atoms but modi…es that betwen two next nearest-neighbors atoms we can calculated the phase diagram of the compound in (bar; Kelvin) plane and the Joule values of the di¤erent latent heats of transition. Moreover, the energy value of the chemical bond between two nearest-atoms is obtained.

Introduction

Let us consider a cristal of N identical atoms. In the adiabatic approximation, it can be considered as a set of electronic bonds ( or chemical bonds) and elastic springs between the atoms, each elastic spring being related to a bond [ 1 2]. So any modi…cation of the quantum states of the electrons involved in a chemical bond can modify the elastic force constant of the spring associated to it. We consider harmonic vibrations.

Recently [START_REF] Nasser | [END_REF], a two-dimensionnal model taking into account this kind of atoms-phonon coupling leads to a phase diagram which displays, at least qualitatively, the features observed in the phase diagram of a pure substance: …rst order phase transitions between three thermodynamic phases, three coexistence curves, one triple point and one critical point. In this model it is assumed that, for each atom, the electrons involved in the bond between this atom and its nearest-neighbors are di¤erent from those involved in the bond between this atom and its next nearest-neighbors. It is assumed that each kind of electrons has two energy levels. These assumptions lead to a coupling between a chemical bond and the spring associated to it. This coupling is described below.

Let us consider the bond between two atoms nearest neighbors. The potential energy between those atoms can have three values depending on the quantum states of electrons involved in the bond. Let us call V 0 (r) this potential energy when the electrons of both atoms are in their fundamental level, where r is the distance between both atoms. It is assumed that V 0 (r) has a minimun for r = r 0 and a second derivative equal to at the minimun. Let us call V 1 (r) this potential energy when the electrons of both atoms are in their excited level. It is assumed that V 1 (r) has a minimun for r = r 1 and a second derivative equal to at the minimun. We neglect the di¤erence between r 0 and r 1 and we assume that is smaller than . As < , the vibrational energy of both atoms is smaller when the electrons are in their excited level than when they are in their fundamental level. Consequently, when the electrons are in their fundamental level the electronic energy is small but the vibrational energy is large, while, when they are in their excited level, the electronic energy is large but the vibrational energy is small. So there is a competition between the electronic interactions which favour the fundamental levels and the vibration interactions which favour the excited one. This competition can lead to a …rst order phase transition [4].

In the present study we add to the two-dimensional model the calculation of the latent heats of the di¤erent …rst order phase transitions. Moreover, using the experimental data concerning the phase diagram of the carbon dioxide, CO 2 , we show how those data can be introduced in the model and which results can be obtained. In Section 2, we present the crystal Hamiltonian and the method used to study it. In Section 3 we give the results obtained by the numerical study, and the last Section is devoted to discussion and conclusion.

2 Theoretical study

Lattice Hamiltonian

Let us consider a square lattice of N identical atoms. Each atom l moves around a lattice point l. The position vector of the lattice point l is

! R l = l 1 ! a 1 + l 2 ! a 2 (1)
where ! a 1 and ! a 2 , the basis vectors of the lattice, are orthogonal and have the same length a 0 , and where l 1 = 0; 1; 2; :::N 1 1 and l 2 = 0; 1; 2; :::N 2 1. So the lattice point l is de…ned by the set (l 1 ; l 2 ), and we can write l = (l 1 ; l 2 ). It is clear that

N = N 1 N 2 .
For studying the atoms movements, we introduce the orthogonal axes ! Ox and ! Oy which are parallel to ! a 1 and ! a 2 , respectively, the origin, O; being at the lattice point (0; 0). We assume that each atom is linked to its four nearest neighbors and to its four next-nearest neighbors by harmonic springs.

The four nearest neighbors of the atom l = (l 1 ; l 2 ), are the atoms :

l (1) = (l 1 + 1; l 2 ), l (2) = (l 1 ; l 2 + 1), l (3) = (l 1 1; l 2
) and l (4) = (l 1 ; l 2 1). We call e ll ( ) , the elastic force constant of the spring linking the atoms l and l ( ) , with = 1; 2; 3; 4. The potential energy of the elastic interaction between the atom l and its four nearest neighbors is E p1 (l) given by

E p1 (l) = 1 2 (e ll (1) (u l u l (1) ) 2 + e ll (3) (u l u l (3) ) 2 + e ll (2) (v l v l (2) ) 2 + e ll (4) (v l v l (4) ) 2 ) (2)
where u l and v l are the components, on the respective axes ! Ox and ! Oy, of the displacement of the atom l around the lattice point l. So, the total elastic interaction between pairs of atoms nearest neighbors is

E p1tot = 1 2 X (l) E p1 (l) (3)
P (l)
is the sum over the N lattice points. Periodic boundary conditions are assumed.

The four next-nearest neighbors of the atom l = (l 1 ; l 2 ) are the atoms:

L (1) = (l 1 + 1; l 2 + 1), L (2) = (l 1 1; l 2 + 1), L (3) = (l 1 1; l 2 1
) and L (4) = (l 1 + 1; l 2 1). We call e e l;L ( ) , the elastic force constant of the spring which links the atoms l and L ( ) , with = 1; 2; 3; 4. The potential energy of the elastic interaction between the atom l and its four next-nearest neighbors is

E p2 (l) = 1 2
(e e lL

with b lL (1) = 1 2 u L (1) u l + v L (1) v l 2 (5) b lL (2) = 1 2 (u l u L (2) + v L (2) v l ) 2 (6) b lL (3) = 1 2 (u L (3) u l + v L (3) v l ) 2 (7) b lL (4) = 1 2 (u l u L (4) + v L (4) v l ) 2 (4)
So, the total elastic interaction between pairs of atoms next-nearest neighbors is

E p2tot = 1 2 X (l) E p2 (l) (9)
The Hamiltonian of the crystal vibrations is

H ph = E c + E p1tot + E p2tot (10)
where E c is the kinetic energy of the atoms. Now we introduce the following assumptions : i) Each atom l contains two electronic subsystems (1) and (2). The electronic subsystem (1) contains the electrons which are involved in the bond between the atom l and its four nearest neighbors and the electronic subsystem (2) contains the electrons which are involved in the bond between the atom l and its four next-nearest neighbors. The electronic subsystem (i) (i = 1; 2) has two energy levels separated by i : the fundamental level, called (a i ) is not degenerated while the excited one, called (b i ) has the degeneracy r i . To the subsystem (i) is associated the …ctitious spin b il which has two eigenvalues il = 1. In others words, each subsystem (i) has its own space of quantum states and its physical observable b il . And we work in the tensorial ( or direct) product of both spaces.

The electronic Hamiltonian of the atom l can be written

H e (l) = 1 2 b 1l + 2 2 b 2l (11)
and the electronic Hamiltonian of the crystal is

H e = X (l) 1 2 b 1l + 2 2 b 2l (12)
ii) The value of the spring constant e ll ( ) , with = 1; ::4, depends only on the quantum states of the electronic subsystems (1) of the atoms l and l ( ) following the relation

e ll ( ) = + 2 + 4 + 4 ( 1l + 1l ( ) ) + 2 + 4 1l 1l ( ) (13)
We can verify that the above formula gives e ll ( ) = when 1l = 1l ( ) = 1, e ll ( ) = when 1l = 1l ( ) = +1 and e ll ( ) = when 1l 6 = 1l ( ) . Moreover, we assume that (14)

iii) The value of the spring constant e e lL ( ) , with = 1; ::4, depends only on the quantum states of the electronic subsystems (2) of the atoms l and L

We can verify that the above formula gives e e l;L ( ) = e when 2l = 2L ( ) = 1, e e l;L ( ) = e when 2l = 2L ( ) = +1 and e e l;L ( ) = when 2l 6 = 2L ( ) . Moreover, we assume that e e e

The Hamiltonian of the crystal, H, is the sum of H ph and H e . So,

H = E c + E p1tot + E p2tot + X (l) 1 2 b 1l + 2 2 b 2l (17)

Reduced parameters

It is interesting to take and e as the units of elastic force constant and to introduce the sets of reduced parameters (x; y) and (e x; e y) de…ned by

x = (18) e x = e e ( 19
)
2 = ( + ) + y ( ) (20)
2e = e + e + e y e e (21)

From the relations (14) and (16), the parameters x and e x are comprised between 0 and 1, and the parameters y and e y must be comprised between 1 and +1.

With these new parameters, e ll ( ) and e e lL ( ) can be written

e ll ( ) = (a + b ( 1l + 1l ( ) ) + c 1l 1l ( ) ) (22)
It is worth to notice that the parameters b and e b are negative or equal to zero.

Phonon -atom interaction

By inserting relation (22) in the expression of E p1 (l), equation (2), the potential energy E p1tot can be writen

E p1tot = 1 4 a X (l) A l + V Z + V Ex (30) with A l = (u l u l (1) ) 2 + (u l u l (3) ) 2 + (v l v l (2) ) 2 + (v l v l (4) ) 2 (31)
V Z = 1 2 b X (l) A l 1l (32)
and

V Ex = 1 2 c X ((u l u l (1) ) 2 1l 1l (1) + (u l u l (3) ) 2 1l 1l (3)
(33)

+ (v l v l (2) ) 2 1l 1l (2) + (v l v l (4) ) 2 1l 1l (4) )
The energy term V Z is a Zeeman-like interaction which can be written

V Z = X (l) h l b 1l (34)
with

h l = 1 2 bA l (35)
As the parameter b is negative, the …eld-like h l which acts on the …ctitious spins b 1l favours the eigen value 1l = +1.

The energy term V Ex is an exchange-like interaction between the …ctitious spin b 1l and its four nearest neighbors b 1l ( ) ( = 1; ::4). When the parameter c is positive, this interaction favours the case 1l = 1l ( ) when this parameter is negative this interaction favours the case 1l = 1l ( ) and when it is equal to zero, the exchange-like interaction is equal to zero.

For the folIowing we assume y = 0 (36) which implies that the exchange-like interaction V Ex is equal to zero. By inserting relation (23) in the expression of E p2 (l), equation ( 4), the potential energy E p2tot can be written

E p2tot = 1 4 e e a X (l) B l + e V Z + e V Ex ( 37
) with B l = b lL (1) + b lL (2) + b lL (3) + b lL (4) (38) e V Z = 1 2 e e b X (l) B l b 2l (39)
and

e V Ex = e e c 2
X (l) (b lL (1) 2l 2L (1) + b lL (2) 2l 2L (2) + b lL (3) 2l 2L (3) + b lL (4) 2l 2L (4) )
(40) As previously, the energy term e V Z is a Zeeman-like interaction. As the parameter e b is negative, the …eld-like which acts on the …ctitious spin b 2l favours the eigenvalue 2l = +1. The energy term e V Ex is an exchange-like interaction between the …ctitious spin b 2l and its next-nearest neighbors b 2L ( ) ( = 1; ::4).

When the parameter e c is positive, this interaction favours the case 2l = 2L ( ) when this parameter is negative this interaction favours the case 2l = 1L ( ) and when it is equal to zero, the exchange-like interaction is equal to zero.

For the following we assume that e y = 0 (41) which implies that the exchange-like interaction e V Ex is equal to zero.

Variational method: e¤ective parameters

For the variational Hamiltonian H 0 [5; 6], we take the sum of a phonon Hamiltonian H 0ph and a spin Hamiltonian H 0sp

H 0 = H 0ph + H 0sp (42)

Variational phonon Hamiltonian H 0ph

We introduce two parameters E and e E. The …rst one, E, is an e¤ective spring constant that replaces the spring constants e ll ( ) , and the second one, e E, is an e¤ective spring constant that replaces the spring constants e e lL ( ) . Those E and e E do not depend on the electronic states of the subsystems (1) and (2). The phonon Hamiltonian H 0ph E; e E is then

H 0ph E; e E = E c + E p10 (E) + E p20 e E (43)
with

E p10 (E) = 1 2 X (l) 1 2 EA l (44)
E p20 e E = 1 2
X (l) 1 2 e EB l (45)
The Hamiltonian H 0ph E; e E is the Hamiltonian of vibrations of a square lattice of atoms linked by springs between atoms nearest neighbors and atoms next-nearest neighbors with the spring constants E and e E, respectively. It is known that there are two dispersion relations

! = ! 1 ! k and ! = ! 2 ! k
which correspond to two acoustic branches of the crystal. In the previous relations the vector ! k is the phonons wave vector. The previous dispersion relations are in agreement with those obtained by Blackman [START_REF] Blackman | [END_REF] and de Launay [START_REF] De Launay | Solid State Physics[END_REF]. They are given in the Appendix.

The free energy associated to the Hamiltonian

H 0ph E; e E is F 0ph = k B T X ! k 0 ln 0 @ 2 sinh 0 @ ~! ! k 2 1 A 1 A (46)
where k B is the Boltzmann constant, = 1 k B T and

X ! k 0
is the sum over the two phonons branches. Using the matrix density of H 0ph E; e E , the thermal mean values hA l i 0 and hB l i 0 of the parameters A l and B l , respectively, are

hA l i 0 = 4 N X ! k 0 coth 0 @ ~! ! k 2 1 A 2 @! @E (47)
and

hB l i 0 = 4 N X ! k 0 coth 0 @ ~! ! k 2 1 A 2 @! @ e E ( 48
)
where @! @E and @! @ e E are the partial derivatives of the dispersion relations. The expressions of @! @E and @! @ e E are given in the Appendix.

Variational spin Hamiltonian H 0sp

We consider the case where the exchange-like interactions V Ex and e V Ex are equal to zero. We introduce two …elds-like h 1 and h 2 which act over the …ctitious spin b 1l and b 2l , respectively. Those …elds-like are uniform.

The spin Hamiltonian H 0sp is

H 0sp = H 0sp1 + H 0sp2 (49)
with

H 0sp1 = X (l) h 1 b 1l (50)
and

H 0sp2 = X (l) h 2 b 2l (51)
With this spin Hamiltonians, all the …ctitious spins of the electronic subsystem i, (i = 1; 2), have the same thermal mean value m i given by

m i = exp( h i ) + r i exp( h i ) exp( h i ) + r i exp( h i ) (52)
The equations ( 52) are called self-consistent equations.

The free energy related to H 0sp is

F 0sp = N k B T (ln z 0sp1 + ln z 0sp2 ) (53)
where the partition function z 0spi is given by

z 0spi = exp( h i ) + r i exp( h i ) (54)

Variational Hamiltonian H 0

So, the variational Hamiltonian is given by

H 0 = E c + E p10 (E) + E p20 e E + X (l) h 1 b 1l + X (l) h 2 b 2l (55)
The free energy associated with H 0 is given by

F 0 = F 0ph + F 0sp (56)
that is

F 0 = k B T X ! k 0 ln 0 @ 2 sinh 0 @ ~! ! k 2 1 A 1 A N k B T ln z 0sp1 N k B T ln z 0sp2 (57)

Thermodynamic parameters

At the …rst order of a perturbation calculation we obtain the following results. (59)

E = a + 2bm 1 + cm 2 1 ( 58
h 1 = 1 2 1 2 hA l i 0 (b + cm 1 ) (60)
h 2 = 2 2 1 2 hB l i 0 e e b + e cm 2 (61)
In the present study the parameters c and e c are equal to zero. The crystal free energy, at the thermodynamic equilibrium, is

F = F 0 + N 1 2 + h 1 m 1 + N 2 2 + h 2 m 2 (62)
The fraction of atoms in the excited level (b 1 ) is

n 1 = 1 + m 1 2 (63)
and the fraction of atoms in the excited level (b 2 ) is

n 2 = 1 + m 2 2 (64)
When the parameter m i (i = 1; 2) is equal to 1, the fraction n i is equal to zero and the electronic subsystems (i) are in the fundamental level (a i ) and when m i = +1, they are in the excited level (b i ).

We can calculate the crystal entropy by using the relation

S = @F @T (65)
So we obtain

S = S ph + S sp (66)
with

S ph = 1 T X ! k 0 coth 0 @ ~! ! k 2 1 A ~! ! k 2 k B X ! k 0 ln 0 @ 2 sinh 0 @ ~! ! k 2 1 A 1 A (67)
and

S sp = N k B (ln z 0sp1 + ln z 0sp2 ) N 1 T (m 1 h 1 + m 2 h 2 ) (68)
From the relation

F = U T S (69)
we deduce for the internal energy

U U = X ! k 0 coth 0 @ ~! ! k 2 1 A ~! ! k 2 + N 1 2 m 1 + N 2 2 m 2 (70)
We recall that in a …rst order phase transition there is not a discontinuity in F , but there are a discontinuity in the extensive parameters N n 1 , N n 2 , S, and U . At the transition, S, the discontinuity in S, and U , the discontinuity in U , are related by

U = T S ( 71
)
where T is the transition temperature. The latent heat of the transition L is given by

L = T S = U (72)

Numerical study

The numerical study consists essentially in solving the self-consistent equations (52) by taking into account the relations (58) to (61) . Both self-consistent equations are coupled.

New reduced parameters

It is interesting to introduce new reduced parameters by taking ~!M ( ) as the unit of energy with

! M ( ) = 2 r m a (73)
where m a is the mass of the atoms. The value of ~!M ( ) is roughly estimated to 1000K or 695 cm 1 .

With this unit of energy, we introduce the following reduced parameters: the reduced temperature

t = k B T ~!M ( ) (74)
the reduced energy gaps

1 = 1 ~!M ( ) (75)
and

2 = 2 ~!M ( ) (76)
the reduced free energy per atom that we call reduced free energy

f = 1 N ~!M ( ) F (77)
the reduced internal energie per atom that we call reduced internal energy

u = U N ~!M ( ) (78)
The quantities t, 1 , 2 , f and u have no unit.

We call reduced entropy, which is denoted s, the entropy per atom in the unit k B ( Boltzmann constant). So,

s = 1 k B S N (79)
The quantity s has no unit.

From (79), a variation of s, s, is related to an entropy variation, S, by the relation

s = 1 k B S N (80)
Using the de…nition of the reduced temperature t, we have

t s = T S N ~!M ( ) (81)
In a …rst order transition T S is the latent of the transition, S being the discontinuity in entropy at the transition. Then, t s is the reduced latent heat per atom at the transition that we call reduced latent heat. We can caculate f , u and s.

Study of the self-consistent equations

We …x the values of the temperature and that of all the model parameters and we look for the values of m 1 and m 2 which satisfy the self-consistent equations (52). A solution is a set (m 1 ; m 2 ) or (n 1 ; n 2 ). For each solution we can calculate the values of the thermodynamic parameters f , u and s.

If there is one solution, this solution is the stable solution ( or the stable state or the thermodynamic state of the crystal). If there are several solutions, the reduced free energy values of those solutions must be compared. The solution which has the lowest reduced free energy value is the stable state. When the reduced free energy values of two solutions (n 1 ; n 2 ) and (n 0 1 ; n 0 2 ) are equal and are the lowest, both solutions are the thermodynamic states of the crystal which then displays a …rst order phase transition between those thermodynamic states. At the transition, the parameters N n i (i = 1; 2), the crystal entropy and the internal energy display a discontinuity. In some cases, the magnitude of those discontinuities decrease when the transition temperature increases and are equal to zero when the temperature is higher or equal to a temperature value T C ( or t C ). We have used this property for …nding t C . The value t C is called reduced critical temperature.

It is worth to count the number of independent parameters. In this study

N 1 = 40, N 2 = 50, r 1 = r 2 = 4, y = e y = 0, z = e = 0:2, 1 = 10 and x = 10 5 .
So, a thermodynamic state of the system, that is a stable solution, depends on the three parameters : 2 , e x and t. A …rst order phase transition implies the existence of one relation between those parameters and a triple point implies the existence of another relation between those parameters. So, at the phase transition, the transition temperature, the magnitude of the discontinuities and the latent heat values depend on the two parameters 2 and e

x. At a triple point, the triple point temperature, the magnitude of the discontinuities and the latent heats values, depend on one parameter e

x or 2 . We take e x as the independent parameter.

One can expect that the stable state is (0:000; 0:000) at very low temperature and (1:; 1:) at high temperature. Between the brackets, the …rst number corresponds to the value of n 1 and the second one to that of n 2 .

Phase diagram for e

x constant

Phases and discontinuities

A phase diagram of the crystal in the ( 2 ; t) plane for e x = 0:0263 is shown in Fig. 1. The coordinates of the point T are ( 2T = 14:84919; t T = 3:032334). The method for obtening those coordinates is explained in [START_REF] Nasser | [END_REF] and in the following.

When the value of 2 is …xed the thermodynamic state of the crystal depends only on the temperature. As shown in Fig. 1, when we …x the value of 2 , the study of the thermal variation of the stable solution shows the presence of one or two …rst order phase transitions depending on whether 2 is smaller or larger than 2T .

Case 2 = 14 At very low temperature, t = 0:001, the thermodynamic state is (0:000; 0:000). Increasing the temperature, the values of n 1 , n 2 , f , s and u vary but there is only one stable state. At t = 2:917849, which corresponds to point A 3 in Fig. 1, there are two stable states (1; 1) and (0:249; 0:043) : So, there is a …rst order transition between both states.

For the state (0:249; 0:043), as for the state (0:000; 0:000), the majority of the electronic subsystems (1) and (2) are in their fundamental level (a 1 ) and (a 2 ), respectively. We say that those states belong to the thermodynamic phase (a; a). For the state (1; 1) all the electronic subsystems (1) and (2) are in their excited level (b 1 ) and (b 2 ), respectively. We say that this state belongs to the (b; b) phase. So, at t = 2:917849, we say that there is a …rst order phase transition between the (a; a) and (b; b) phases.

For t lower than the transition temperature, 2:917849, the stable state belongs to the (a; a) phase because n 1 and n 2 are lower than 0:5, and for t higher than 2:917849 the stable state belongs to the (b; b) phase because n 1 and n 2 are equal to 1:. We denote t 3 the temperature of the transition (a; a) (b; b). So for 2 = 14, t 3 = 2:917849. In Fig. 1, the curve with black circles represents the values of t 3 for di¤erent values of 2 lower than 2T . This curve is the (a; a) (b; b) coexistence curve.

At the transition, the parameters n i (i = 1; 2), s and u display a discontinuity. We call ( n i ) 3 the discontinuity in n i , ( s) 3 that in s and ( u) 3 that in u. The value of the discontinuity in n i is the value of n i in the (1; 1) state minus its value in the (0:249; 0:043) state. So, for 2 = 14, at t 3 = 2:917849, ( n 1 ) 3 = 0:751 and ( n 2 ) 3 = 0:957.

For 2 = 14, at t 3 = 2:917849, the values of s and u for the (0:249; 0:043) state are 6:325423 and 3:050334, respectively, and they are 13:482498 and 17:832931, respectively, for the (1; 1) state. So, the discontinuity s is ( s) 3 = 7:157075 and that in u is ( u) 3 = 20:883265.

We denote l 3 the reduced latent heat for the (a; a) (b; b) transition . We have

l 3 = t 3 ( s) 3 (82)
So, for 2 = 14, at t 3 = 2:917849, l 3 = 20:883264. We we can verify that l 3 = ( u) 3 .

Case 2 = 17 At very low temperature, t = 0:001, the thermodynamic state is (0:000; 0:000). As the temperature increases, the values of n 1 , n 2 , f , s and u but there is only one stable state. At t = 3:056919, which corresponds to point A 1 in Fig. 1, there are two stable solutions which are (1; 0:040) and (0:286; 0:020).

The state (0:286; 0:020) belongs to the (a; a) phase. In the state (1; 0:040) all the electronic subsystems (1) are in their excited level (b 1 ) while the majority of the electronic subsystems (2) are in their fundamental level (a 2 ). We say that this state belongs to the (b; a) phase. So, at t = 3:056919, there is a …rst order phase transition between the (a; a) and (b; a) phases.

Above the temperature values 3:056919 and up to the temperature value 3:471697, which corresponds to point A 2 in Fig. 1, there is only one stable state (1; n 2 ), where n 2 varies from the value 0:040 to the value 0:079. It is clear that this state (1; n 2 ) belongs to the (b; a). At the temperature value t = 3:471697 there are two stable states which are (1; 1) and (1; 0:079). So, at t = 3:471697, there is another …rst order phase transition between the (b; a) and (b; b) phases. In some case the value of n 2 is higher than 0:5 for the state (1; n 2 ). We consider that this state (1; n 2 ) belongs to the (b; a) phase even if the value of n 2 is higher than 0:5.

In summary for temperature value lower than 3:056919 the stable state belongs to the (a; a) phase, between the temperature values 3:056919 and 3:471697 it belongs to the (b; a) phase and for temperature value higher than 3:471697 it belongs to the (b; b) phase because n 1 and n 2 are equal to 1:. We denote t 1 the temperature of the transition (a; a) (b; a) and t 2 that of the transition (b; a) (b; b). So, for 2 = 17, t 1 = 3:056919 and t 2 = 3:471697. In Fig. 1, the curve with open circles and that with crosses represent the values t 1 and t 2 , respectively, for di¤erent values of 2 higher than 2T . The …rst curve is the (a; a) (b; a) coexistence curve and the second one the (b; a) (b; b) coexistence curve.

At each of the two phase transitions, the parameters n i (i = 1; 2), s and u display a discontinuity. We call ( n i ) 1 , ( s) 1 and ( u) 1 the values of this discontinuity at t 1 and ( n i ) 2 , ( s) 2 and ( u) 2 those values at t 2 . It is worth to notice that ( n 1 ) 2 is always equal to zero.

We denote l 1 the reduced the latent heat for the (a; a) (b; a) transition and l 2 that for the (b; a) (b; b) transition. We have

l 1 = t 1 ( s) 1 (83)
and x. For given value to 2 , there are one or two …rst order phase transitions depending on the value of 2 compared to that of 2T . We recall that in all this study 1 = 10, x = 10 5 and z = 0:2.

l 2 = t 2 ( s) 2 (84)

Triple point T

In Fig. 1, the value of t 2 minus that of t 1 , t 2 t 1 , is equal to 0:414778 for 2 = 17 and decreases when the value of 2 decreases. There is a value of 2 such as t 2 is equal to t 1 . Let us call 2T this value. Similarly let us call t T the value of t 1 when t 1 and t 2 are equal. The set ( 2T ; t T ) are the coordinates of the point T in Fig. 1.

When t 1 and t 2 are equal, there are three stable solutions. In Fig. 1, those solutions are (0:280; 0:039), (1; 0:079) and (1; 1). They belong to the (a; a), (b; a), and (b; b) phases, respectively. The values of f for those solutions are 22:630346, 22:630347 and 22:630340, respectively. So we can say that the state T is a triple point. Indeed, in thermodynamics, << a state of three-phase comptability is a"triple point" >> [START_REF] Callen | Thermodynamics and an introduction to Thermostatistics[END_REF]. We introduce the following notations: the values of l 1 and l 2 at a triple point are denoted l 1T and l 2T , respectively. So, for e x = 0:0263 the triple point is characterized by the values 2T = 14:84919, t T = 3:032334, l 1T = 7:760728 and l 2T = 13:678070. We recall that those values are obtained for 1 = 10, z = 0:2 and x = 10 5 .

In fact, it is di¢ cult to obtain the condition t 1 = t 2 . We consider that this condition is reached when t 2 t 1 5 10 6 . We then take for t T the value of t 1 and for 2T the corresponding value of 2 . When t 2 t 1 5 10 6 the di¤erence between the three values of f is lower than 10 10 6 .

For each value of e x there is a triple point characterized by the values 2T , t T , l 1T and l 2T . The variation of 2T with the value of e

x is shown in Fig. 2, those of l 1T and l 2T in Fig. 3. As for t T , it decreases from 3:081895 to 2:877612 when the value of e

x increases from 5 10 4 to 10 1 . Fig. 2. Variation of 2T with e x. The parameter 2T is the value of 2 at the Triple point. The numbers between brackets are the values of the ratio l 2T l 1T . It is interesting to notice that for a value of e

x corresponds only one value of this ratio. x increases and is equal to zero when e

x is higher than the critical value e x c which is near 0:166.

Critical value e

x C

In the phase diagram of The value e x C is lower than 0:18. For e x equal to 0:166 and 0:169, ( n 2 ) 2 is equal to 0:159 and 0:078, respectively. Taking into account the di¢ culty for working around the value e

x C , we consider that the (b; a) (b; b) phase transition does not exist when e

x > 0:166 and we consider that the critical value e x C is equal to 0:166.

In this study, we have verify that the values of ( s) 2 , ( u) 2 , and that of the latent heat l 2 decrease when e

x increases and are nearly equal to zero when e x is equal to 0:166.

The phase diagram of the crystal in the ( 2 ; t) plane for e x = e x C = 0:166 looks like that of Fig. 1. So, for each value of 2 , larger than 2T , there is a (b; a) (b; b) phase transition and a transition temperature t 2 . This value of t 2 is the critical temperature value that we denote t C . The variation of t C with 2 is shown in Fig. 4.

So, in this study, the critical point is characterised by e x = e x C = 0:166 and by the values of 2 and t. Fig. 4. Variation of the critical temperature t C with 2 . The critical tem-perature is the transition temperature t 2 when the parameter e

x is equal to e x C = 0:166. The value 2 must be greater than the triple point value 2T for x = e

x C = 0:166.

Phase diagram for e x and 2 varying

We look for a phase diagram where z, 1 and x are constant while 2 and e x can vary. We consider the case where the variations of 2 and e

x are controled by a parameter P following the relations

e x = e x 0 + e cP (85)
and

2 = 20 + c 2 P ( 86
)
where the coe¢ cients e x 0 ,e c, 20 and c 2 are constants. In order to give values to those coe¢ cients, we use the experimental data concerning the phase diagram of carbon dioxyde CO 2 . So, we must assume that the phases (a; a), (b; a) and (b; b) of the model correspond to the solid, liquid and gas states, respectively and that P is the applied pressure. Consequently, l 1 corresponds to the latent heat of fusion L f , l 2 to the latent heat of vaporization L v and l 3 to the latent heat of sublimation L s . We recall that z = 0:2, 1 = 10 and x = 10 5 .

Experimental data concerning CO 2

For CO 2 , the coordinates of the triple point in the (P; T ) diagram are P T = 5:19bar and T T = 216:55K and those of the critical point are P C = 73:8bar and T C = 304:25K : . Moreover, for P = 1atm = 1:013bar the temperature of sublimation is T s (1atm) = 194:67K. Concerning the latent heats : the values of the latent heat of fusion and of vaporization at the triple point are L f T = 199kJkg 1 and L vT = 350:83kJkg 1 , respectively. The latent heat of sublimation at 1atm is L s (1atm) = 573kJkg 1 . The latent heat L vT has been obtained by extrapolating data found in gazechim website . All the other data are obtained from wikipedia website. We accept those experimental data and we shall show how they can be introduced in the model.

Determination of the constants e

x 0 , e c, 20 and c 2

Triple point data At the triple point of CO 2 the ratio of the latent heats of vaporization and fusion, L vT L f T , is equal to 1:7630.

We look for, in the model, the triple point for which the ratio l 2T l 1T is equal to 1:7630. This triple point correspons to e x = 0:0263 (see Fig. 2.). For this value of e

x, the latent heats l 1 and l 2 are equal to l 1T = 7:760728 and l 2T = 13:678108, respectively, and then the ratio l 2T l 1T is equal to 1:7625. Moreover for this value of e

x, the value of the parameter 2 is 2T = 14:84919 and the value of the triple point temperature is t T = 3:032334.

As the pressure value at the triple point of CO 2 is 5:19 bar, we can write the following equations 0:0263 = e

x 0 + e c 5:19 (1a)

14:84919 = 20 + c 2 5:19 (2a)
Critical point data For CO 2 , the ratio of the values of the critical temperature and the triple point temperature, T C T T , is equal to 1:4050. We look for , in the model, the value t C such as t C t T = 1:4050. As we have obtained for t T the value 3:032334, the reduced temperature t C should be equal to 4:260429. As shown in Fig. 4 the critical temperature value t C depends on that of 2 . We look for the value of 2 such as t C = 4:260429. This value is 2 = 10:6115.

As the pressure value at the critical point of CO 2 is 73:8 bar, we can write the following equations 0:166 = e

x 0 + 73:8 e c (1b)

10:6115 = 20 + 73:8 c 2 (2b)
The solution of the system of equations (1a) and ( 1b) is e x = 15:732414 10 3 + 2:036146 10 3 P (I)

The solution of the system of equations (2a) and (2b) is 2 = 15:16975 61:764903 10 3 P (II)

The relations (I) and (II) can be used above the triple point of CO 2 , that is for P 5:19 bar.

Data at P = 1atm For CO 2 , the ratio of the values of the sublimation temperature at the atmospheric pressure and the triple point temperature, Ts(1atm) T T , is equal to 0:8990. Moreover, the ratio of the values of the latent heat of sublimation at the atmospheric pressure and that of vaporization at the triple point,

Ls(1atm) L vT
, is equal to 1:6333. We look for, in the model, a state on the (a; a) (b; b) coexistence curve where t 3 satis…es the relation t3 t T = 0:8990 and where l 3 satis…es the relation l3 l 2T = 1:6333. As t T is equal to 3:032334, the transition temperature t 3 (1atm) should be equal to t 3 (1atm) = 2:726068. As l 2T is equal to 13:678108, the reduced latent heat l 3 (1atm) should be equal to l 3 (1atm) = 22:340454. The state looking for is de…ned by a value for e x and a value for 2 . Let us use the relations (I) and (II). For P = 1atm = 1:013bar, those relations lead to the values e

x calc = 0:017795 and 2calc = 15:107182. Using these values of e

x and 2 in the self-consistent equations we obtain for t 3 and l 3 the values t 3calc = 2:912993 and l 3calc = 22:163028. We see that the set (l 3calc ; t 3calc ) is di¤erent from the set (l 3 (1atm) ; t 3 (1atm)). This disagreement means that the relations (I) and (II) are not valid for P P T = 5:3 bar.

Finally, with e x = 0:011625 and 2 = 14:720, the self-consistent equations lead to l 3 = 22:359920 and t 3 = 2:724432. Those values are very near the set (l 3 (1atm) ; t 3 (1atm)).

We then can write for P = 1atm = 1:013 bar the following equations 0:011625 = e x 0 + 1:013 e c (1c) and 14:720 = 20 + 1:013 c 2 (2c)

The solution of the system of equations (1a) and ( 1c) is e x = 8:06604 10 3 + 3:51329 10 3 P (III)

The solution of the system of equations (2a) and (2c) is

Use of the relations (I) to (IV)

The relations (I) to (IV) contain more informations than the set of reduced values obtained above in relations (87) and (88). Indeed, for each value of the applied pressure they allow to calculate the values of e x and 2 . Then using the self-consistent equations we can calculate the values of the transition temperature t 1 , t 2 and t 3 and the values of the corresponding latent heat l 1 , l 2 and l 3 .

First, we have veri…ed that using the relations (I) to (IV) we obtain for P = 5:19bar the values t T , l 1T and l 2T contained in relation ( 87) and (88), for P = 73:8bar the value t C contained in relation (87) and for P = 1atm the values t 3 (1atm) and l 3 (1atm) contained in relation ( 87) and (88). Then, we have calculated the phase diagram of the crystal in the (P; t) plane and the variations with the applied pressure of the di¤erent latent heats.

The phase diagram in the (P; t) plane is shown in Fig. 5. For P = 0bar the value of reduced temperature of sublimation is t 3 (0bar) = 2:606381. As shown in Fig. 5, the temperature of vaporization and that of sublimation increase when the applied pressure increases, which is in agreement with the experimental results. But the temperature of fusion decreases when the applied pressure increases. In this study, we have not analyse this last result which has been observed only for water, H 2 O. x and 2 can vary with the applied pressure.

The variations with pressure of the di¤erent latent heats are shown in Fig. 6. For P = 0bar the value of the latent heat of sublimation is l 3 (0bar) = 22:685849.

For a given atom, when the electrons involved in the bonds are in their excited level, the increase of electronic energy is 1 + 2 , in reduced unit. But the thermal energy necessary to move these electrons from their fundamental level to their excited one is l 3 , the reduced latent heat of sublimation. As shown in Fig. 6, this latent heat is lower that the sum 1 + . The values of the ratio l 3 = ( 1 + 2 ) is equal to 0:92 and to 0:91 for 0bar and 1atm, respectivement. Likewise, for pressure values greater than P T the reduced latent of fusion l 1 is lower than the energy gap 1 which is equal to 10. For all substances the latent heat of vaporization decreases when the pressure increases and vanishes at the critical pressure. This result is veri…ed in this study as seen in Fig. 7. Fig. 8. Variation with pressure of the latent heat of fusion (l 1 ) and vaporization (l 2 ). The pressure varies from P T to P C .

Determination of the energy unit ~!M ( )

For the following, we use those numerical values : Boltzmann's constant k B = 1:381 10 23 J K 1 , elementary charge e = 1:602 10 19 C, Avogadro's number N A = 6:023 10 23 mol 1 , Planck's constant h = 6:626 10 34 J s and ~= 1:055 10 34 J s, Gas constant R = 8:314 J mol 1 K 1 , molar mass of CO 2 M = 44 g.

If we assume that t T is the reduced value of the experimental transition temperature T T we can write

k B T T ~!M ( ) = t T (89) So ~!M ( ) = k B T T t T (

Conclusion and discussion

To apply the model to the study of the phase diagram of a substance, we have assumed that the phases (a; a), (b; a) and (b; b) of the model correspond to the solid, liquid and gas states, respectively. And consequently, l 1 corresponds to the latent heat of fusion L f , l 2 to the latent heat of vaporization L v and l 3 to the latent heat of sublimation L s . Moreover, we have assumed that the parameters z, 1 and x are constant while the parameters 2 and e x vary linearly with the applied pressure P . The parameters 1 and x refer to the chemical bond between two atoms nearest-neighbors and the parameters 2 and e

x refer to the chemical bond between two atoms next nearest-neighbors. Taking into account the experimental data of CO 2 we get the reduced values contained in the relations (87) and (88) and the equations (I) to (IV). By using the self-consistents equations, these equations allow to determine the phase diagram of CO 2 in the plane (P; t) and the reduced values of the latent heat in the di¤erent …rstorder transitions (fusion, vaporization and sublimation), t being the reduced temperature, We obtain the results in Kelvin and Joule by multiplying the reduced values by the energy unit ~!M ( ). From equation (89) the energy unit is chosen so that the calculated value in Kelvin of the triple point temperature is equal to the experimental value.

Meaning of the results

The parameter x is the ratio , where is the value in the solid state of elastic constant of the elastic spring which links two atoms nearest-neighbors and is its value in the liquid and gas states. The value x = 10 5 means that the value of is zero. So, in the liquid and gas phases the spring does not exist neither does the chemical bond. Moreover this result means that the value of the parameter 1 is that of the energy of the chemical bond between two nearest-neighbors atoms.

As for the elastic constant of the spring which links two atoms next nearestneighbors its value is never zero. Consequently, the chemical bond between two atoms next nearest-neighbors exists in the three phases. Concerning the gas phase, this result is contained in the van der Waals equation.

As shown in Fig. 6 the latent heat of sublimation l 3 is smaller than the electronic energy 1 + 2 . This result is contained in the expression of the internal energy U and is independent of relations (I) to (IV) as explained below.

As the values of the elastic force constants are lower in the (b; b) phase than in the (a; a) phase, the value of the vibration energy in the (b; b) phase, U vib(b;b) , is lower than its value in the (a; a) phase, U vib(a;a) . So the variation of internal energy is less than the variation of electronic energy, N ( 1 + 2 ).

Comments on the results

The method used to introduce the experimental data of the substance into the model causes that if the Kelvin value of one of the three reduced temperatures t T , t C and t 3 (1atm) ( relations 87) is equal to the corresponding experimental value it is the same for the other two values. The use of relation (89) imposes that the Kelvin value of t T is equal to the experimental value of the triple point temperature of CO 2 . So three points of the phase diagram of the compound in the plane (bar; Kelvin) agree with the experimental data whatever the values of the coe¢ cients of equations (I) to (IV), that is whatever the values of 1 and z.

Likewise, if the Joule value of one of the three reduced latent heats l 1T , l 2T and l 3 (1atm) (relations 88) is equal to the corresponding experimental value it is the same for the other two values. With the energy unit given by relation (89), the Joule value of l 2T is equal to 185kJ kg 1 while the experimental value of the latent heat of vaporization at the triple point is equal to 350; 83kJ kg 1 . We think that the disagreement between these two values comes from the fact the values of 1 and z used in this study are arbitrary values. It would be interesting to see how vary all the results obtained when one varies the values of 1 and z = e .

The study of speed of sound in the crystal at low temperature should make it possible to determine the values of the elastic constants and e . We then can deduce the value of the report z = e and that of the energy unit ~!M ( ).

Let us denote by (~! M ( )) sound this value. Using relation (89) we write

k B T T t T = (~! M ( )) sound (96)
As the value of t T depends on those of 1 and z, we seek the value of 1 so that this relation is satis…ed.

For some substances, the ratio L vT =L f T is equal to 10. In Fig. 3, the highest value for the ratio l 2T l 1T is around 4. Preliminary calculations show that the value of the ratio l 2T l 1T increases when z increases, the value of 1 being …xed. Moreover, we think that the slope of the (a; a) (b; a) coexistence curve in Fig. 5 depends on the values of the coe¢ cients of relation (I) to (IV) and then it depends on the values of 1 and z.

Conclusion

Having found the values of 1 and z for the substance, we can carry out the study describe here. This study is based on the fact that when atoms ( or molecules) have two energy levels an atom-vibrations coupling can exist. Spin conversion compounds are an example of such coupling.

The author has done alone this study and has written alone this manuscript. 4.5 Calculation of the partial derivatives @! @E and @! @ e E

We introduce r by

! = p r with r = 1 2 (c 1 + c 2 ) q (c 1 c 2 ) 2 + 4c

( 1 )

 1 b lL (1) + e e lL (2) b lL (2) + e e lL (3) b lL (3) + e e lL (4) b lL (4) )

  and e e lL ( ) = e e a + e b ( 2l + 2L ( ) ) + e c 2l 2L ( )

  ) e E = e e a + 2 e bm 2 + e c m 2 2

Fig. 1 .

 1 Fig. 1. Phase diagram for ex constant. The symbol xtd stands for e x. As the parameters 1 , x and z are constant In this study a thermodynamic state depends on the three parameters t, 2 and ex. For given value to 2 , there are one or two …rst order phase transitions depending on the value of 2 compared to that of 2T . We recall that in all this study 1 = 10, x = 10 5 and z = 0:2.

For

  each of the three stable solutions we can calculate the values of s and u. So, for the stable solutions (0:280; 0:039), (1; 0:079), and (1; 1) the values of s are 6:489363, 9:048688, and 13:559420, respectively. Consequently, for the transition (a; a) (b; a), ( s) 1 = 2:559325 and l 1 = 7:760728, for the transition (b; a) (b; b), ( s) 2 = 4:510740 and l 2 = 13:678070 and for the transition (a; a) (b; b), ( s) 3 = 7:070065 and l 3 = 21:438799.

Fig. 3 .

 3 Fig. 3. Variation of l 1T and l 2T with e x. The parameter l 1T is the value at the Triple point of the latent heat of the (a; a) (b; a) phase transition and l 2T is that of the (b; a) (b; b) phase transition. The value l 2T decreases when ex increases and is equal to zero when ex is higher than the critical value e x c which is near 0:166.

Fig. 1 ,

 1 ( n 2 ) 2 , the value of the discontinuity in n 2 along the (b; a) (b; b) coexistence curve is constant and equal to 0:921. In a previous study[START_REF] Nasser | [END_REF], we have shown that ( n 2 ) 2 does not depend on the values of 1 and 2 , but depends only on that of ex. The discontinuity value ( n 2 ) 2 decreases when ex increases and is equal to zero when e x is higher than a value called critical value of ex and denoted e x C . For e x higher than e x C , the …rst order (b; a) (b; b) phase transition does not exist and the crystal passes from the (b; a) phase to the (b; b) phase continuously.

2

  = 14:688669 + 30:928896 10 3 P (IV) The relations (III) and (IV) are valid for P P T . In summary, for CO 2 we have obtained the following values t T = 3:032334, t C = 4:260429, t 3 (1atm) = 2:726068 (87) l 1T = 7:760728, l 2T = 13:678108, l 3 (1atm) = 22:340454 (88) and the relations (I) to (IV). Those results are obtained for 1 = 10, z = 0:2 and x = 10 5 .

Fig. 5 .

 5 Fig. 5. Phase diagram obtained by using the relations (I) to (IV). It is assume that only the parameters ex and 2 can vary with the applied pressure.

Fig. 6 .

 6 Fig.[START_REF] Boccara | Symétries Brisées -Théorie des Transitions avec Paramètre d'Ordre[END_REF]. Variation with pressure of the latent heat of fusion (l 1 ), vaporization (l 2 ) and sublimation ( 3 ). The unit of energy is ~!M ( ).

  90) With T T = 216:55K and t T = 3:032334 the energy unit is equal to ~!M ( ) = 98:622 10 23 J = 61:562 10 4 eV (91)

  In the (a; a) phase , m 1 and m 2 are equal to 1 and they are equal to 1 in the (b; b) phase. Calling U (a;a) and U (b;b) the values of the internal energy in the (a; a) and (b; b) phases, respectively, we have from relation (70) U (b;b) U (a;a) = N ( 1 + 2 ) U vib(a;a) U vib(b;b)

APPENDIX 4 . 4 k x a cos k y a) c 3 = 2 e E m a sin k x a sin k y a! k are k x = x 2 N 1 a and k y = y 2 2 :

 4422 Phonon dispersion relations for the square latticeThe dispersion relations for the two acoustics branches are k x a cos k y a)In the above relations, m a is the atoms mass, a is the distance of the equilibrium positions of the atoms, E and e E are the elastic force constants for atoms pairs …rst neighbors and second neighbors, respectively.With the boundary periodic conditions, the components k x and k y of the wave vectors The wave vector ! k (0; 0) corresponds to a translation mode. So, there are N 1 N 2 1 di¤erent wave vectors which correspond to phonon frequency values.The parameters N 1 ; N 2 and a are de…ned in the text.

  = 2 cos k x a cos k y a d 2 = 1 cos k x a cos k y a d 3 = cos k y a cos k x a d 4 = sin k x a sin k y aFor some values of wave vector components (k x ; k y ) the parameter d 4 is equal to zero. In that case the above relations are modi…ed. So, r becomes

	with								0					1
		@r @ e E	=	1 m a		@ 2d 2		4 e E m a	q	E 2 m 2 a	d 2 4 d 2 3 + 4 e E 2 m 2 a	d 2 4	A
					r =		E m a	d 1 +	2 e E m a	d 2	E m a	jd 3 j
	the parameters @r @E and @r @ e E become
									@r @E	=	d 1 m a	jd 3 j m a
	r = We can write 1 2 and	0 B @	2E m a	d 1 +	4 e E m a	d 2	v u u t 2E m a @r @ e E 2d 2 = m	2	d 2 3 + 4	2 3 2 e m a E	! 2	d 2 4	C A 1
	or		r =	E m a	d 1 +	2 e E m a	d 2	s	E 2 m 2 a	d 2 3 +	4 e E 2 a m 2	d 2 4
	where
	d 1 So we have				@! @E	=	1 2!	@r @E
	with								0					1
		@r @E		=	1 m a	@ d 1			E m a	q	E 2 m 2 a	d 2 3 d 2 3 + 4 e E 2 m 2 a	d 2 4	A
	and										@! @ e E	=	1 2!	@r @ e E

a

where jd 3 j is the absolute value of d 3

References

(0bar) = 90:426 10

eV

Using this value for ~!M ( ) we obtain for the critical temperature

and for temperature of sublimation at the atmospheric pressure

Both calculated values are equal to the experimental values. So, if we pass from the phase diagram in the (P; t) plane to the phase diagram in the (bar; Kelvin) plane by using this value for the energy unit, the values of the triple point temperature, the critical temperature and the sublimation temperature at the atmospheric pressure are equal to the respective experimental values. The transition temperature values for other values of the pressure depend on the relation (I) to (IV).

Using l 2T = 13:678070 and this energy unit we can calculate the latent heat of vaporization at the triple point. So

In relation (94) 1000 44 is the number of mole per kg. This value, 185kJ kg 1 , is di¤erent from the experimental value L vT = 350:83kJ kg 1 . For the latent heat of fusion at the triple point and that of sublimation at the atmospheric pressure the calculated values are 104:8kJ kg 1 and 301:6kJ kg 1 . Both calculated values are di¤erent from the experimental values which are equal to 199kJ kg 1 and 573kJ kg 1 , respectively. These disagreements between the calculated values for the latent heats and the experimental one are discussed in the next section. We think that a part of the disagreement comes from the fact that the sum P ! k 0 is made on two phonon branches and not three.

The parameter 1 , which is constant, is equal to

The parameter 2 varies with the pressure. Its value for P = 0 is 14:688669. So, for P = 0 the value of 2 is 2 (0bar) with