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Part 2 The use of biochemical information to find the metabolic basis for pink fruit colour

Part 2 The use of biochemical information to find the metabolic basis for pink fruit colour

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Last updated: sep. 05, 2022 by Michelle Pijpers (Deactivated)
7 min read

The principles of analysis of metabolic composition in plants

Cells that constitute plant tissue, e.g. fruits, produce and accumulate a few hundred different secondary metabolites, such as pigments, antioxidants, vitamins etc. All these different metabolites may have very different concentrations in fruits. To study the metabolic composition of fruits researchers perform metabolic profiling, or metabolomic analysis, which consists of several essential steps:

  1. Extraction of metabolites from fruit material

  2. Instrumental chemical analysis of extracts

  3. Identification of metabolites

  4. Determination of concentrations of identified metabolites

 

Extraction of metabolites

To extract metabolites from fruit (or any other plant tissue), fruit tissue is homogenized with addition of an organic solvent, e.g. methanol, chloroform, hexane etc. The choice of an organic solvent (or their combination) depends on the chemical properties of metabolites a researcher is interested in. In particular, a chemical property called ‘polarity’ of the solvent has to match the polarity of the metabolites we aim to study. Here is a sequence of common solvents used for extraction of metabolites in the order of decreasing polarity: water>acetonitrile>methanol>acetone>chloroform>hexane. The polarity of chemicals depends on their chemical structure, e.g. hydroxyl groups (-OH) make chemicals more polar and methyl groups (-CH3) make chemicals more a-polar. In the case of the present example study, we are interested in semi-polar secondary metabolites and methanol seems, therefore, the appropriate choice. After homogenization of fruit tissue with methanol the insoluble cell parts (cell walls, proteins etc.) are precipitated by centrifugation and the resulting clear metabolic extract is subjected to an instrumental chemical analysis. 

 

Instrumental chemical analysis of extracts

A metabolic extract is a mixture of a few hundred different metabolites. But the identity (chemical name) and concentration of each metabolite can only be determined individually. To achieve this, the mixture is first separated for individual metabolites and the number of molecules of each metabolite is counted and will then represent its concentration. Liquid metabolic extracts are often separated using Liquid Chromatography (LC). Metabolites of an extract first enter a tube filled with an absorbent material – a chromatographic column - and start to move through this tube with different speed, which depends on the strength of their interaction with the absorbent material of the column. So metabolites reach the end of the column at a different time, which is called retention time (in other words it tells for how long metabolites are retained in the column). A life example of chromatographic separation is marathon competition, when all runners start altogether at the start and arrive to the finish at different times depending on how fast they can move.

At the end of the chromatographic column the metabolites arrive to a detector – a mass spectrometer (MS) – which measures the weight (mass) of the arrived molecules and count their number. But to be able to perform these two tasks the molecules must become visible for the detector. For this the molecules are converted into ions – molecules with a negative (-) or a positive (+) charge. This can be achieved either by removing one hydrogen atom (H) from a molecule to produce a negatively charged ion or by adding one H to produce a positively charged ion. For example, the chemical formula of a neutral (non-charged) molecule of salicylic acid is C7H6O3 which gives an exact mass of this molecule [M]=138.031694 (try to calculate this yourself at https://www.sisweb.com/referenc/tools/exactmass.htm). If your mass spectrometer is set up for negative ionization one H atom will be subtracted from the molecule which will give C7H5O3- with an exact mass [M-H]-=137.0239, since the exact mass of H=1.007825 (check on the same website). This mass of salicylic acid will be recorded by the mass spectrometer. What will be the formula of salicylic acid and the exact mass of a detected molecule if the mass spectrometer is set up for positive ionization?

A typical result of LC-MS analysis of a metabolic extract is shown in the chromatogram below (Figure 3).



Figure 3. Click to enlarge. A typical chromatogram of a tomato fruit secondary metabolite extract obtained by LC-MS analysis. Individual metabolites of the extract are separated in a chromatographic column and detected by a mass spectrometer at the end of the column. This makes metabolites to look like peaks along the retention time scale. The height (or area) of each peak is proportional to the number of molecules detected by the mass spectrometer for a corresponding metabolite, or in other words it shows the concentration of this metabolite in the extract.  For each metabolite peak the mass spectrometer detects a mass of the corresponding molecules. E.g. a peak at 24.26 min consists of 26,030 molecules with an average mass 609.1342 Da (Dalton).

 

Identification of metabolites

The simplest way to identify metabolites is to compare the two properties provided by your LC-MS analysis – 1) retention time of a metabolite and 2) its mass – with a reference metabolite library in which many metabolites were characterized for these properties by analysing the corresponding pure chemical standards. In the example of salicylic acid, you would first purchase pure salicylic acid from one of the chemical distributors, dissolve a small amount in a solvent that you want to use for your plant material (e.g. methanol), analyse this sample with LC-MS using settings that you will use to analyse your plant extracts and record the data into your reference library, e.g. mass [M-H]-=137.0239 (if you set up the mass spectrometer for negative ionization) at retention time 29.71 minutes. Now to identify salicylic acid in your fruit metabolic extract you would search for a metabolite with a mass CLOSE to 137.0239 that has a retention time CLOSE to 29.71. Why CLOSE? Because all analytical measurements always produce a small error which causes small deviations from the reference library (theoretical) values. For the mass this deviation should not exceed a certain threshold often expressed in ppm (parts per million) – typically 5 ppm. Mass deviation can be calculated as follows:

ppm = ((ME – ML)/ML)*1,000,000,

where, ME is an experimental mass of a metabolite from your extract, that you want to identify, ML – the mass of a metabolite present in the reference library and calculated based on the chemical formula of this metabolite.

Example. You would like to identify a metabolite which in a chromatogram of your fruit extract has a retention time of 29.64 min. and [M-H]-=137.0244. In your reference library you have 200 metabolites whose chemical standards you have previously analysed. In this library you find two metabolites: one has a formula C3H6O3, exact mass [M] = 138.0164 and a retention time at 31.52 minutes, and the other one is salicylic acid, C7H6O3, [M] = 138.0317 at 29.71 minutes. Which of the two fits best to the metabolite in your extract? First you calculate masses of negative ions [M-H]- for both the library metabolites C3H6O3 –H =C3H5O3- =137.0086 and C7H6O3 – H = C7H5O3-= 137.0239. Thus, both masses are quite close to 137.0216 in the chromatogram, however, we have to choose the correct one by calculating the ppm deviations. Ppm of (C3H5O3-) = ((137.0244 – 137.0086)/137.0086)*1000000 = 115.32 and ppm of C7H5O3- = ((137.0244 – 137.0239)/137.0239)*1000000 =3.65. Thus the ppm deviation of salicylic acid of the reference library is much smaller than the C3H6O3 compound. In addition the reference retention time of salicylic acid is closer to or compound – 0.07 min compared to 1.81 minutes of C3H6O3. Thus, we can conclude that our compound with negative exact mass of 137.0244 at 29.64 minutes is most likely salicylic acid.



LC-MS analysis of pink tomato fruits

The population of 49 tomato introgression lines (ILs) and the wild type (WT) cv. Moneymaker was grown in a greenhouse with three plants per introgression line. Ripe fruits were harvested from individual plants of each IL and the WT. Thus for each IL and the three WT fruit samples – biological replicates – were obtained. As mentioned in Part 1, fruits of IL01b had a pink colour. In the worksheet Metabolic data you can find concentrations of 12 metabolite peaks detected by LC-MS in the three replicates of the WT with red fruits and three replicates of IL01b. You have to find out which of the 12 metabolites are significantly different between the WT fruits and fruits of IL01b by calculating:

1)    For each metabolite make average concentrations of the three biological replicates separately for WT and IL01b in columns P and R of the Metabolic Data file using AVERAGE function of Excel.

2)    Calculate the standard deviations of biological replicates in WT and IL01b in columns Q and S using STDEV function of Excel

3)    What metabolites have different concentrations between WT and IL01b?

4)    In biology we normally consider statistically significant differences only. Student’s t-test is a statistical method used to find reliable differences. Perform a t-test between the WT and IL01b groups using TTEST function of Excel. Which metabolites accumulate at different levels in the WT compared to IL1B: put ‘yes’ or ‘no’ in column U? A hint: look for metabolites with a t-test value <0.05

5)    Identify all differentially accumulating compounds, using the information available in the Library spreadsheet:

  1. convert the negative masses present in the Metabolic data file into neutral masses in column D of the Metabolic data of the Excel data file;

  2. now for the calculated neutral masses try to find the closest hits from the Metabolic library by calculation of mass deviation (ppm) and retention time deviations in columns G and H.

  3. In column I put the name of the identified (closest hits) metabolites from the library . What is the name of the metabolite(s) which is/are significantly different between WT and IL01b?

6)    Search for literature about this metabolite

  1. to which class of metabolites does it belong and via which biochemical pathway is this class of metabolites produced in plants?

  2. what colour has this metabolite?

  3. how can this metabolic difference explain the pink colour of the IL01b fruits

  4. present the data of the significantly different metabolite(s) as a bar chart of WT and IL01b averages, use error (SD) bars to show the variation between biological replicates.