1. The Critical Requirement for Linolenic Acid Is Pollen Development, Not Photosynthesis, in an Arabidopsis Mutant
Michele McConn and John Browse (1996)
- Presented by: Christina, Eric and Gurkirat

2. Background What are lipids and glycerolipids? What is desaturation?
How are desaturation reactions performed on glycerolipids?
What are trienoic fatty acids?
What are free radicals?
Why are free radicals detrimental to plant cells?
Useful things to know before we begin the actual research

3. Central portion of thylakoid lamella bilayer
Lipids are a major group of biological macromolecules defined by their insolubility in water
Fatty acids are a type of lipid - carboxylic acids with long alkane or alkene groups
A glycerolipid is a lipid whose structure consists of a glycerol molecule with 2 fatty acids esterified to C1 and C2, and a variable head group molecule esterified to C3
The example shows MDGD whose head group is a single galactose

4. Desaturation Oxidation
C-C single bond -> C=C double bond

5. What Desaturates Glycerolipids?
Glycerolipid desaturation catalyzed by 7 FAD (fatty acid desaturase) enzymes
FADs are integral membrane proteins and mostly use glycerolipid as substrate
Desaturation of 16:0 and 18:1 occur in membrane lipids
Membrane desaturases cannot retain function after solubilization or purification
FAD mutants must be used instead to study desaturation

6. What are trienoic fatty acids?
Fatty acids with three cis double bonds
16:3, 18:3 most abundant fatty acids in membrane glycerolipids of plants – particularly chloroplast membranes
Dienoic fatty acid desaturation catalyzed by 3 ω-3 fatty acid desaturases - FAD3, FAD7, FAD8
Hexadecatrienoic acid 16:3
α-linolenic acid (ω-3) 18:3

7. Pay attention to checkpoints that introduce a third double bond to dienoic fatty acids
Source: Haugn G., Kunst L., Song L.

8. Dienoic Fatty Acid Desaturase (FAD) Enzymes
FAD-coding genes
Location
Substrate
Activity
Known mutants
FAD7
Chloroplast
Any chloroplast glycerolipid
16:2→16:3, 18:2→18:3
fad7-1 (leaky), fad7-2 (negligible function)
FAD8 (isozyme of FAD7)
fad8 (negligible function)
FAD3 ER
Phosphatidylcholine
18:2→18:3
fad3-1(leaky), fad3-2 (negligible function)

9. Free Radicals in Photosynthesis (Krieger-Liszkay 2004)
Free Radicals in Photosynthesis (Krieger-Liszkay 2004)
Highly oxidative compounds with an unpaired electron
Photosynthesis - biological process using light energy to convert CO2 and H2O to carbohydrates and O2
occurs in plant plastids
Chlorophylls absorb light energy and can convert O2 into a free radical O2-
O2- reacts with nearby molecules
You can see that PSII and PSI, which is where chlorophyll is typically found, are embedded in the thylakoid lumen membrane… lipids

10. Lipid Peroxidation
Free radicals cause peroxidation of unsaturated lipids
Lipid peroxides:
Changes cell membrane properties
Associated with cell death
Oxidative stress
Free radicals can react with unsaturated fatty acids to cause negative effects
Oxidative stress can cause damage in a multitude of macromolecules
Source: Young and McEneny (2001)

11. Motivation for the study
Photosynthesis produces free radicals
Polyunsaturated lipids are peroxidized more readily
Plastid membranes have a lot of trienoic acids (16:3, 18:3) via MGDG
Peroxidation is unhealthy
Question: Why might trienoic acids be important for photosynthesis?
The rationale behind why people believed trienoic fatty acids were important for photosynthesis follows some indirect reasoning based knowledge that photosynthesis makes free radicals, knowledge on the deleterious effects of free radicals on polyunsaturated fatty acids, the high levels of polyunsaturated fatty acids in plastid membranes, and evolutionary principles arguing such a combination would be impossible if there wasn’t some positive tradeoff that outweighs the risks

12. We expect that fewer trienoic acids in the thylakoid membrane would give higher fitness against oxidative stress
However, the trienoic acids remain prevalent in plastid membranes despite the risk
Therefore, trienoic acids might be important for the plastid (eg. photosynthesis)
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13. Method and Results Big Question
Is having a very high proportion of trienoic fatty acids in chloroplast membranes essential for photosynthesis?

14. Parallel pathways of glycerolipid desaturation
Alternate desaturation pathways introduce backdoors to alleviate dienoic desaturase loss
18:3 can be imported to plastids from ER (red)
alleviates fad7,fad8 double mutation
18:2 can be imported to plastids, desaturated, then exported to ER (blue)
alleviates fad3 mutation

15. How did the presence of two parallel pathways of glycerolipid synthesis impact the researchers’ capacity to observe the effects of trienoic FA deficiency?
Transport mechanism between Chloroplast and ER does not completely revert the membrane fatty acid compositions to WT
If very small levels of trienoic acids are necessary for normal function and parallel pathways allow plastids/ER to have low trienoic acid levels, then none of the double mutants will reduce trienoic acids enough

16. Question: How can you work around the presence of alternate dienoic desaturation pathways?

17. Question: How can you work around the presence of alternate dienoic desaturation pathways?
Make a triple mutant
You can’t use parallel pathways if all pathways are blocked

18. Mutant Lines … Leaky alleles – retain small desaturase activity
Double mutant 1: a cross between fad7-2 and fad8
fad3-1
fad7-1
fad8
fad7-2
fad3-2
Leaky alleles – retain small desaturase activity
Double mutant 2: a cross between fad3-2 and fad7-2
5 dienoic desaturase mutant lines available to make higher order mutants from
The authors chose to cross fad8 with fad7-2 and fad7-2 with fad3-2 for reasons we’ll see further on

19. Production of healthy triple mutant plants
Question: Why did the authors cross a fad3-2;fad7-2 double mutant with a fad7-2;fad8 double mutant?
Hint: Which is more efficient cross to screen the F2?
A. a cross between fad3;fad7-2 and fad7-2;fad8
B. a cross between fad7-2;fad8 and fad3-2
Hint 2: Think about the ratios

20. Production of healthy triple mutant plants
Question: Why did the authors cross a fad3-2;fad7-2 double mutant with a fad7-2;fad8 double mutant? Why not fad7-2;fad8 and fad3-2?
fad3-2(-/-);fad7-2(-/-)
fad7-2(-/-);fad8(-/-)
fad3-2(-/-)
fad7-2(-/-);fad8(-/-) VS F1
fad3-2(+/-);fad7-2(-/-);fad8(+/-) F1
fad3-2(+/-);fad7-2(+/-);fad8(+/-) F2
The double mutant cross makes fad3-2(+/-)fad7-2(-/-)fad8(+/-) which self crosses and produces triple mutant at 1/16 probability. Crossing fad7-2;fad8 and fad3-2 makes a triple heterozygote which produces triple mutant at 1/64 probability.
fad3-2(-/-);fad7-2(-/-);fad8(-/-)
1/16 F2
fad3-2(-/-);fad7-2(-/-);fad8(-/-)
1/64

21. Healthy Triple Mutant
F1 obtained from a cross between fad7-2 fad-8 and fad3-2 fad7-2 double mutants
F1 self pollinate – seeds germinated
F2 screened – leaf fatty acid examined with GC
17/240 F2 plants with no detectable trienoic FAs
expected = 240 ÷ 16 = 15
homozygous triple mutants were not selected against
F2 – screened using GC
Double mutants had lower trienoic levels
Heterozygous at fad8 locus – may produce more homozygous triple mutants when needed

22. Healthy Triple Mutant fad3-2;fad7-2;fad8 Wild Type
The wild type plant looks healthy, grows to the same size, looks similar below the flowers
There is something odd about the flowers though, covered later
Source: McConn and Browse (1996)

23. Potential quantum efficiency (Fv/Fm): sensitivity to light
Source: McConn and Browse (1996)
Potential quantum efficiency (Fv/Fm):
sensitivity to light
Steady state quantum efficiency (Seaten and Walker 1990):
relates to how much oxygen is produced at different light intensities
Both metrics for photosynthesis effectiveness
Neither reflect significant differences at 22° C
Possible difference at low temperature
How can you be certain?
Additional quantification required
Lets do a fluorescence analysis
Resposnses temperature dependent
Fluoresence quenching done as well

24. Triple mutant and WT have similar fresh weight over time:
No significant difference between the triple mutant and WT in growth rate.
No overall effect of trienoic acid deficiency on growth
What is the phenotype of triple mutants? <0.1% trienoic FA, growth development similar to WT at varied temperatures (look at discussion for a change), photosynthesis not affected, male plants are sterile!
Relative growth rates are the same as WT
Increase in shoot weight measured for plants grown under continuous light
But, can you determine growth rate by other means? - this is where they did fluorescence
Source: McConn and Browse (1996)

25. Partial Dienoic Desaturation Knockout
How does each dienoic desaturase (fad3, fad7, fad8) contribute to fatty acid levels?
Single + double mutants available from early experiments
Made new triple mutants with fad7-1(leaky) and fad7-2
Displays codominance with fad7-2
Crossed mutants to produce leaky triple mutants
homozygous for fad3, fad8-2
fad7-1/fad7-1 homozygous or fad7-1/fad7-2 heterozygous
Performed GLC on leaves from samples from each genotype
In order to determine contribution of each desaturase to trienoic acid synthesis, we need to look at how much each desaturase works in isolation
Hence, we need to examine our batches of different higher order mutants
Note: To make the fad7-1/fad7-1 homozygous triple mutant, they did something similar to how they made the fad7-2/fad7-2 homozygous triple mutant
To make the fad7-1/fad7-2 heterozygous triple mutant, they crossed the two homozygous triple mutants so all F1 were our genotype of interest

26. Relevance of Partial Dienoic Desaturation Knockout
Shows contribution of individual fads to dienoic desaturation
Generates intermediate phenotypes to investigate partial trienoic deficiency
Confirms that there are only 3 major dienoic desaturases
other tissues (eg. roots) agree with leaf data that this is the case (data unpublished)
desaturation of dienoic acids has been essentially completely knocked out without fad3, fad7, fad8
fad7 + fad3 active
fad7 + fad8 active
fad8 + fad3 active
fad8 active
fad3 active
2 leaky fad7 active; ~5% trienoic
1 leaky fad7 active; ~2% trienoic
~0.1% trienoic
Authors still acknowledge there might be other contributors as trace amounts of trienoic acids still remain but the 3 fads are the main workers
This slide tells you why making all these intermediate genotypes was useful
Source: McConn and Browse (1996)

27. In Depth Leaf Fatty Acid Analysis
For 3 individuals of wild type, leaky triple mutant, and triple mutant:
Isolate individual glycerolipids by Thin Layer Chromatography
For each purified glycerolipid:
Do GLC on the glycerolipid’s fatty acids
Calculate average mole % for each fatty acid within a glycerolipid
more precise and reliable
Calculate average mole % for each glycerolipid within an individual
Visual pipeline on the next page
The researchers also did this because this process was more precise at finding exact fatty acid composition levels which allowed them to validate their observation that the 3 fads were the primary contributors to dienoic desaturation

28. Leaf Fatty Acid Analysis Pipeline
MGDG
GLC
Pick a genotype
Process tissue for
TLC
DGDG
GLC PG
GLC
Thin Layer Chromatography SL
GLC PC
GLC PE
GLC
eg. fad3;fad7-2;fad8 PI
GLC
Allows separation of each individual glycerolipid by how polar their molecules are
Visual pipeline
TLC is basically just a molecular isolation method
Sources:

29. Question
Given that all major dienoic -> trienoic desaturases have fully/partially lost function in the triple mutants:
What changes (increase, decrease, no change) do you expect will happen to overall glycerolipid composition (eg. MGDG, DGDG)?
Block out the products of each of the enzymes
What changes?

30. Overall Glycerolipid Composition
Just the raw numbers, skip at your leisure
Source: McConn and Browse (1996)

32. Overall Glycerolipid Composition
Expectation: glycerolipid composition doesn’t change
although desaturation is affected, each glycerolipid should still be synthesized normally
In reality, relative to wild type, MGDG decreased in mutants and DGDG + eukaryotic pathway phospholipids increased
Unclear why

33. Question
Given that all major dienoic -> trienoic desaturases have fully/partially lost function in the triple mutants:
What changes (increase, decrease, no change) do you expect will happen to saturated, monosaturated, dienoic, and trienoic fatty acids?
Source: Haugn G., Kunst L., Song L.

34. More raw numbers

35. Prokaryotic Pathway
MGDG = monogalactosyldiacylglycerol
DGDG = digalactosyldiacylglycerol
PG = phosphatidylglycerol
SL = sulfoquinovosyldiacylglycerol
The key thing to note here is just that for any of the glycerolipids, N:3 fatty acids drop in relative concentration whereas N:2 fatty acids rise in triple mutants, demonstrating a blockage in dienoic desaturation

36. Eukaryotic Pathway
PC = phosphatidylcholine
PI = phosphatidylinositol
PE = phosphatidylethanolamine
Note that because we’re dealing with the Eukaryotic pathway glycerolipids, there are no desaturated C16 fatty acids
However, we still see the trend of mutants having no 18:3 and a buildup of 18:2 (or a partial version of this with the leaky triple mutant)

37. Glycerolipid Fatty Acid Content
Triple mutant:
Trace levels of trienoic acids compared to wild type
Higher levels of dienoic acids compared to wild type
Leaky mutant heterozygote:
Similar pattern as triple mutant but not as pronounced
Some noticeable fad7 function still remains
As expected

38. Trienoics make up ~1.5-2X more of MGDG and DGDG than SL and PG
Confirmation of FAD7/FAD8 Substrate Specificity
Belief: FAD7 prefers MGDG and DGDG and FAD8 prefers SL and PG (McConn et al. 1994)
Trienoics make up ~1.5-2X more of MGDG and DGDG than SL and PG
However, leaky fad7 desaturates ~4X more dienoics in MGDG and DGDG
Conclusion: fad7 has more activity in MGDG and DGDG
The researchers seemed to have used this as an opportunity to analyze substrate specificity since their data allowed them to do so for free
Because fad7 works more on MGDG and DGDG, fad8 must make up for fad7 on the other 2 glycerolipids

39. If you don’t have trienoic acids, then you don’t have photosynthesis
At most minimally

40. Ethylene Small organic gas Developmental hormone
Produced after fertilization
Involved in processes such as ripening of fruits and plant senescence (Schaller 2012)
Exposure to ethylene also causes these effects (eg. bananas)
Source: American Chemical Society

41. Wild type self fertilizes Develops siliques Petals senesce
4 weeks
Source: McConn and Browse (1996)
Wild type self fertilizes
Develops siliques
Petals senesce
Triple null mutant retains petals and doesn’t develop siliques
Ethylene probably not released
Infertility
Wild Type
Triple Mutant

42. Who is sterile?
Given that you can fertilize an Arabidopsis flower by placing anthers (male reproductive organ) near its stigma (female reproductive organ):
How can you test whether the triple mutant is male sterile or female sterile?

43. Who is sterile?
Wild type anthers could always pollinate triple mutants
Triple null mutant (fad3-2;fad7-2;fad8) anthers could not pollinate wild type flowers without male organs
=> Male sterility
Why?

44. Pollen development
The development of the pollen grain is dependent on maternal tissue and signalling as well as gamete- sourced tissues and signalling.
Prod. by maternal tissue and gamete tissue
Prod. by gamete tissue
Gilbert, S. F. Developmental Biology, 6e.Sunderland (MA): Sinauer Associates, 2000.

45. Pollen Release Mechanism (Goldberg et al. 1993)
Stomium - thin wall that undergoes apoptosis when pollen is mature
“splits open”
Pollen is released - dehiscence

46. Triple mutant anthers did not dehisce (SEM micrographs verify this)
Manual dehiscence of anthers also failed to fertilize
Wild Type Anthers (2 days after petals)
Triple Mutant Anthers (4 days after petals)
Note that the anthers in the mutant look intact rather than split
The pollen in the SEM after eventually dehiscence look normal but they’re trapped in the anther of the triple mutant
Wild Type Anthers (2 days after petals)
Triple Mutant Anthers (4 days after petals)
Source: McConn and Browse (1996)

47. However...
fad3-2(-/-) fad7-2(-/-) fad8(+/-) could be self crossed to to generate a 3:1 Mendelian ratio of triple mutant:double mutant
Recall: fad3-2(-/-) fad7-2(-/-) fad7(-/-) could not be self-crossed
What does this demonstrate about the role of trienoic acids in fertility?
Even though the triple mutant (not the triple leaky mutant) could not be self crossed… a double mutant could be self crossed
Source: McConn and Browse (1996)

48. However...
fad3-2(-/-) fad7-2(-/-) fad8(+/-) could be self crossed to to generate a 3:1 Mendelian ratio of triple mutant:double mutant
Recall: fad3-2(-/-) fad7-2(-/-) fad7(-/-) could not be self-crossed
What does this demonstrate about the role of trienoic acids in fertility?
Low levels of trienoic acids are sufficient for fertility
Source: McConn and Browse (1996)

49. Question
Recall that the triple null mutant fad3(-/-)fad7-2(-/-)fad8(-/-) is male sterile and cannot self-cross, but the fad3(-/-)fad7-2(-/-)fad8(+/-) is fertile and can self-cross.
Is male sterility due to the diploid genotype of the parent plant, or the haploid genotype of the pollen?
NICE!

50. Hypothesis 1: Haploid genotype of the pollen causes pollen sterility
= Viable gamete
= Non-viable gamete
Expectation:
fad3(-/-)fad7-2(-/-)fad8(-/-)
fad3(-/-)fad7-2(-/-)fad8(+/-)
Male gamete
Female gamete
Male gamete
Female gamete
fad3(-)fad7-2(-)fad8(-)
fad3(-)fad7-2(-)fad8(-)
fad3(-)fad7-2(-)fad8(-)
fad3(-)fad7-2(-)fad8(-)
fad3(-)fad7-2(-)fad8(+)
fad3(-)fad7-2(-)fad8(+)
Cannot self-cross
Self crosses with 100% double mutant
If the haploid genotype of the pollen causes pollen sterility, then we can infer pollen viability straight from the haploid genotype of the pollen

51. Hypothesis 2: Diploid genotype of the parent causes pollen sterility
= Viable gamete
= Non-viable gamete
Expectation:
fad3(-/-)fad7-2(-/-)fad8(-/-)
fad3(-/-)fad7-2(-/-)fad8(+/-)
Male gamete
Female gamete
Male gamete
Female gamete
fad3(-)fad7-2(-)fad8(-)
fad3(-)fad7-2(-)fad8(-)
fad3(-)fad7-2(-)fad8(-)
fad3(-)fad7-2(-)fad8(-)
fad3(-)fad7-2(-)fad8(+)
fad3(-)fad7-2(-)fad8(+)
Cannot self-cross
Self crosses with 3:1 double mutant:triple mutant
If the diploid genotype of the parent causes pollen sterility, then we can infer pollen sterility from the parent genotype, not the pollen genotype
This outcome was observed

52. Pollen development Haploid cell, arose through meiosis
Generative cell and vegetative cell arise from the microspore through mitosis
Tricellular “mature” phase
Img: DOI:

53. What do we take from this figure?
Source: McConn and Browse (1996)
What do we take from this figure?

54. Did the triple mutants’ pollen develop to the tricellular stage?
QUESTION:
Did the triple mutants’ pollen develop to the tricellular stage?
Wild type
TEST:
DNA fluorescent stain.
Observed 3 fluorescent “spots” per pollen grain.
Triple mutant
CONCLUSION:
Yes, the pollen matured to the tricellular stage.

55. QUESTION: Is the pollen viable?
Wild type
TEST:
Pollen placed on sugar-containing medium, then mixed with orange fluorescent dye that only stains dead cells, and blue that only stains living cells. Observed
11% alive in triple mutants compared to 84% in WT
Triple mutant
CONCLUSION:
The triple mutants’ pollen is much less viable than the wild type’s.

56. Is the viable pollen capable of producing seeds?
QUESTION:
Is the viable pollen capable of producing seeds?
Wild type
TEST:
Initiate germination by placing on germination medium and incubating. Observed 0.06% germination in triple mutants compared to 82% in WT, with pollen tubes <⅓ the length of WT
Triple mutant
CONCLUSION:
Many observable barriers contributing to the triple mutant’s male infertility.
(Adjusted colour on Fig. 6)

57. Triple mutant plants still produced a tiny amount of viable (but less effective) pollen. What, if anything, is important about this? How could that have taken place?

58. At most minimally
Yes
If you don’t have trienoic acids, then you don’t have male fertility

59. Exogenous linolenate complements the male-sterile phenotype
QUESTION: Could the small requirement for trienoic acids be fulfilled with supplementation with linolenate?
TEST: Spray plants with fatty acid salts onto their flower buds at roughly 4 weeks of age. Fatty acid salt treatments were as follows:
The normal Arabidopsis isomer of 18:3 (α-Linolenic acid)
Another isomer of 18:3 (γ-Linolenic acid)
18:2
Did the fatty acid get incorporated into plant tissues? Rosette leaf analysis (GLC) after 10 days of treatment shows that it did, making up approx. 5% of acyl groups.

60. Why were the fatty acids applied as a salt?

61. Why were the fatty acids applied as a salt?
Sodium salts are more soluble than free fatty acids
Linolenate

62. Exogenous linolenate complements the male-sterile phenotype
Plants left to self-fertilize throughout treatment applications
RESULTS:
Only plants that had received the α-Linolenic acid (normal plant isomer) produced seeds. Each seed, after being planted, presented with the triple mutant phenotype.

63. fad3-2 fad7-1 fad8 (leaky) mutant:
Significantly smaller amount of linolenic acid in the floral organs compared to WT, produced seeds
fad3-2 fad7-2 fad8 (triple) mutant:
Even less linolenic acid than the leaky mutant in anthers, still produced seeds
Thus, extremely low quantities of exogenous linolenate in the anthers are sufficient for normal fertility.

64. Exogenous linolenate complements the male-sterile phenotype
QUESTION: In the previous experiment, did the 18:3 penetrate the unopened buds and incorporate into the anther tissue in order to restore male fertility?
TEST: GC on floral organs.
RESULT: Floral organs had taken up the applied 18:3, but the anthers did not contain detectable levels.
CONCLUSION: α-Linolenic acid is likely not required as a structural component of the anthers and/or pollen, but may instead be used as a precursor for another molecule that regulates pollen maturation and release...

65. An essential role for jasmonate in pollen development
What is jasmonate?
Plant hormone (with derivatives that perform related functions) synthesized from 18:3 (postulated to be released from membranes). Oxylipin, made from oxygenated fatty acid.
Acts in minute quantities because of its cascading effect, a characteristic of hormones in general.
Jasmonic acid
Methyl jasmonate
Fun fact: This was likely the first study that investigated the effect of jasmonate deprivation in Arabidopsis.
Images:

66. An essential role for jasmonate in pollen development
Is 18:3 required for male fertility because it is the precursor for jasmonate?
QUESTION:
Does supplementation with any of the the derivatives of 18:3, including jasmonate, complement the male sterile phenotype?
TEST:
Applied solutions of 18:3 derivative molecules jasmonic acid, trans-3-hexenol, trans-2-hexenal, cis-3-hexen-1-ol, and traumatic acid to flower buds on triple mutant plants and allowed them to self-fertilize.
RESULT:
Only the plants treated with jasmonic acid produced seed.

67. An essential role for jasmonate in pollen development
Further viability testing showed that mutant plants treated with jasmonic acid had their pollen restored to 80% of the viability of WT plants.
CONCLUSION:
Strong evidence that jasmonates play a critical role in pollen development, maturation, and release.

68. Review of male sterility experiments
Discovered male sterility phenotype
Extremely small amounts of 18:3 are sufficient for revival of male fertility
Application of 18:3 can restore fertility, but doesn’t integrate into the membranes of the anther tissue
Jasmonate, a derivative of 18:3, also restores fertility
Therefore, the critical role of 18:3 is very likely in its being a precursor for jasmonate. Jasmonate is necessary and sufficient to complement the male sterile phenotype in fad3-2 fad7-2 fad8 plants.

69. Results Timeline summary
Their action-plan
Additional lines with very low levels of 16:3 and 18:3
Production of Triple Mutants
Exogenous linolenate complements the male sterile phenotype
Growth rate and photosynthesis are normal in triple mutants
The triple mutant is male sterile
WT and triple mutant pollen development
The essential role for Jasmonate in Pollen Development

70. Importance
Thoughts?

71. Importance Disproved a common belief about trienoic acids
however, trienoic acids may be necessary for photosynthesis at very low temperatures
18:3’s function as a signal molecule precursor was demonstrated
jasmonate’s importance in pollen development shown
High order mutants can be used to investigate intermediate phenotypes
Chemical complementation is a useful tool

72. Further Investigation
Thoughts?

73. Further Investigation
Why do the triple mutants have lower MGDG and higher levels of eukaryotic phospholipids?
What is the exact mechanism of jasmonate with respect to pollen development?
What about 16:3?
also a precursor to jasmonate in A. thaliana (Liechti and Farmer 2003)
Maybe 16:3 doesn’t have as clearly defined a function because 18:3 plants work fine

74. Questions?
“It is not the answer that enlightens, but the question.” – Eugene Ionesco

75. References
McConn, M., Hugly, S., Browse, J., and Sommerville, C. (1994). A mutation at the fad8 locus of Arabidopsis identifies a second chloroplast ω-3 desaturase. Plant Physiol. 106,
McConn, M., and Browse, J. (1996). The critical requirement for linolenic acid is pollen development, not photosynthesis, in an Arabidopsis mutant. Plant Cell. 8,
Schaller G. (2012). Ethylene and the regulation of plant development. BMC Biol. 10:9.
Goldberg, R. B., Peals, T. P., Sanders, P. M. (1993). Anther Development: Basic Principles and Practical Applications. Plant Cell. 5,
Liechti, R., Farmer, E.E. (2003). The jasmonate biochemical pathway. Sci STKE. 2003(203):CM18.
Young, I. S. and McEneny, J. (2001). Lipoprotein oxidation and atherosclerosis. Biochem Soc Trans. 29(2):
Gaschler M. M., and Stockwell B. R. (2017). Lipid peroxidation in cell death. Biochem Biophys Res Commun. 482(3):
Quilichini, T. D., Quilichini, T. D., Douglas, C. J., & Samuels, A. L. (2014). New views of tapetum ultrastructure and pollen exine development in arabidopsis thaliana. Academic Press [etc. doi: /aob/mcu042
Seaton, G. G. R., and Walker, D. A. (1990). Chlorophyll fluorescence as a measure of photosynthetic carbon assimilation. Proc R Soc Lond Ser B. 242:19-35.
Krieger-Liszkay, A. (2005). Singlet oxygen production in photosynthesis. J Exp Biol. 411(1):

76. Thank You