Fatty Acid Desaturation

Objectives:

1. Isolation of desaturase mutants

2. Substrates for fatty acid desaturation

3. Cellular localization of desaturases

Fatty Acid Desaturation

References:

Buchanan et al. 2000. Biochemistry and Molecular Biology of Plants. American Society of Plant Physiologists, Rockville MD. Chapter 10.

Wallis and Browse (2002) Progress in Lipid Research 41, 254-278. .

Glycerolipid Synthesis “16:3” plants –Arabidopsis, spinach

ER

Plastid

PA

18:1 18:1 (16:0)

18:1

18:1 18:1

18:2

18:1 16:0

18:2 16:0

18:1 18:1 (16:0)

18:2 18:2 (16:0)

18:2

18:1 (16:0)

18:1

Where does 18:1 come from?

Fatty Acid Synthesis

FAS

16:0-ACP

18:0-ACP

18:1-ACP

FATB

FATA

LACS

LACS

16:0-CoA

18:1-CoA

AT

AT

16:0

18:1 Chl

orop

last

lipi

ds

DES

FAS = fatty acid synthesis

FAT = acyl-ACP thioesterase

AT = acyl transferase

LACS = acyl-CoA synthetase

DES = stearoyl-ACP desaturase

Main products of FAS are 16:0 and 18:1

Fatty Acid Desaturation Where does 18:1 come from?

It is made in the plastid stroma by the only known soluble desaturase enzyme in eukaryotes

Fig. 10-18

Plant Membrane Lipids are Highly Unsaturated

The fatty acid compositions of the major leaf glycerolipids from wild type Arabidopsis

PC PE PI SL DGD PG MGD Lipid

Fatty acid

Plant Membrane Lipids are Highly Unsaturated

In all plant tissues the first synthesized glycerolipids PA and DAG contain only 16:0 and 18:1 fatty acyl groups.

Where does desaturation beyond 18:1 occur?

Prediction:

Desaturase enzymes must exist in both the plastid and the ER to carry out desaturation of 18:1 to 18:2 and 18:3

18:1 is made in the plastid stroma by the stearoyl-ACP desaturase

Unique desaturase enzymes must exist in the plastid to carry out desaturation of 16:0 to 16:1, 16:2 and 16:3

Where does desaturation of 16:0 to 16:1, 16:2 and 16:3 occur?

Approaches to Identification and Characterization of Membrane Bound Plant Desaturase Enzymes

1.  Traditional biochemical approach:

Detergent solubilization results in loss of enzyme activity -> isolation and characterization were unsuccessful!

2. Genetic approach:

Requires isolation of mutants

Can we isolate mutants deficient in fatty acid unsaturation?

What kind of phenotype would such mutants have?


ER

Plastid

PA

18:1 18:1 (16:0)

18:1

18:1 18:1

18:2

18:1 16:0

18:2 16:0

18:1 18:1 (16:0)

18:2 18:2 (16:0)

18:2

18:1 (16:0)

18:1

Marker for prokaryotic pathway

Identification of Plant Fatty Acid Unsaturation Mutants

How many mutants would you expect to find?

16:0 t16:1

18:0 18:1

16:0 c16:1

16:1 16:2

16:2 16:3

18:1 18:2

18:2 18:3

Plastid 18:1 18:2

18:2 18:3 ER


What types of mutants have we isolated?

Fatty acid WT JB60 JB27 JB67 LK3 JB1 LK9 JB12 JB25 LK8

16:0 15 18 17 24 15 15 13 15 13 12

16:1 cis 0 0 0 0 11 2 2 0 0 0

16:1 trans 2 0 0 2 2 2 3 4 2 2

16:2 0 0 0 0 0 12 12 0 0 0

16:3 12 12 13 0 0 3 4 10 0 0

18:0 1 1 1 1 1 1 2 1 1 1

18:1 3 3 3 2 21 3 3 27 8 9

18:2 18 19 19 14 14 26 32 4 23 23

18:3 48 47 47 56 36 36 29 36 53 53



16:0 t16:1 on sn-2 of PG JB60 and JB27 FAD4

16:0 c16:1 on sn-2 of MGD JB67 FAD5

16:1 16:2 and 18:1 18:2 LK3 FAD6

16:2 16:3 and 18:2 18:3 JB1 and LK9 FAD7

PLASTID

18:1 18:2 on sn-1 and sn-2 of PC JB12 FAD2 ER

Mutations in FAD4, FAD5, FAD6 and FAD7 affect plastid lipids

Mutations in FAD2 affect extraplastidial lipids and plastid lipids

SUBSTRATE MUTANT GENE


ER

Plastid

PA

18:1 18:1 (16:0)

18:1

18:1 18:1

18:2

18:1 16:0

18:2 16:0

18:1 18:1 (16:0)

18:2 18:2 (16:0)

18:2

18:1 (16:0)

18:1

fab1

fab2

act1

fad4

fad6

fad7 fad8

fad5

fad6 fad6 fad6

fad7 fad8

fad7 fad8

fad7 fad8

fad2 fad3

Marker for prokaryotic pathway





16:1 16:2 and 18:1 18:2 LK3 FAD6

16:2 16:3 and 18:2 18:3 JB1 and LK9 FAD7

PLASTID

18:1 18:2 on sn-1 and sn-2 of PC JB12 FAD2

ER

Mutations in FAD4, FAD5, FAD6 and FAD7 affect plastid lipids

Mutations in FAD2 and FAD3 affect extraplastidial lipids and plastid lipids


SUBSTRATE SPECIFIC

SUBSTRATE NON-SPECIFIC

18:2 18:3 on sn-1 and sn-2 of PC FAD3


ER

Plastid

PA

18:1 18:1 (16:0)

18:1

18:1 18:1

18:2

18:1 16:0

18:2 16:0

18:1 18:1 (16:0)

18:2 18:2 (16:0)

18:2

18:1 (16:0)

18:1

fab1

fab2

act1

fad4

fad6

fad7 fad8

fad5

fad6 fad6 fad6

fad7 fad8

fad7 fad8

fad7 fad8

fad2 fad3

JB1 (Fad7) Story

Fatty acid WT JB60 fad4

JB27 fad4

JB67 fad5

LK3 fad6

JB1 fad7

LK9 fad7

JB12 fad2

JB25 act1

LK8 act1

16:0 15 18 17 24 15 15 13 15 13 12

16:1 cis 0 0 0 0 11 2 2 0 0 0

16:1 trans 2 0 0 2 2 2 3 4 2 2

16:2 0 0 0 0 0 12 12 0 0 0

16:3 12 12 13 0 0 3 4 10 0 0

18:0 1 1 1 1 1 1 2 1 1 1

18:1 3 3 3 2 21 3 3 27 8 9

18:2 18 19 19 14 14 26 32 4 23 23

18:3 48 47 47 56 36 36 29 36 53 53

JB1 (Fad7) Phenotype

The Fad7 phenotype is apparent only when plants are grown at temperatures greater than 18oC

JB1 (Fad7) Phenotype Effect of temperature on the proportion of trienoic fatty acids (16:3 + 18:3)

in leaves

WT

JB1

Hypothesis: This effect is due to a temperature sensitive mutation

JB1 (Fad7) Phenotype

Several additional alleles of fad7 were isolated. They were all temperature sensitive!!!

What does that suggest about our hypothesis?

It is likely not correct. Why?

Temperature sensitive mutations are very rare!

Hypothesis: This effect is due to a temperature sensitive mutation

Alternate hypothesis: There must be a second plastidial 16:2, 18:2 desaturase in addition to FAD7 that functions at low temperature

Isolation of the Low Temperature Desaturase Mutant (Fad8)

How?

1. Mutagenize the Fad7 mutant

2. Grow M2 population at low temperature

3. Screen for alterations in leaf fatty acid composition by gas chromatography

4. Identify mutants with lower 16:3 and 18:3 content than that found in the Fad 7 mutant

Which phenotype would you look for?

Isolation of the FAD8 gene

FAD8 gene was cloned by heterologous hybridization using FAD3 ER Δ15 desaturase gene as a probe.

FAD8 gene is not linked to the FAD7 gene.

FAD7 and FAD8 genes share about 75% nucleotide identity. The FAD8 gene functionally complements the fad7 mutation when expressed using the FAD7 promoter. This demonstrates that FAD7 and FAD8 gene products are functionally equivalent.

fad8-1 mutation created a premature stop codon 149 amino acids from the amino-terminal end of the 435 amino acid long predicted polypeptide, suggesting that this mutation results in a complete loss of FAD8 activity.

Summary of Plant Fatty Acid Unsaturation Mutants





16:1 16:2 and 18:1 18:2 LK3 FAD6

16:2 16:3 and 18:2 18:3 JB1 and LK9 FAD7

PLASTID

Mutations in FAD4, FAD5, FAD6, FAD7 and FAD8 affect plastid lipids

SUBSTRATE SPECIFIC



ER 18:2 18:3 on sn-1 and sn-2 of PC FAD3

16:2 16:3 and 18:2 18:3 FAD8 (Low temperature inducible)

Isolation of the other FAD genes

All seven genes identified by mutation in the seven classes of fad mutants have been cloned and all encode desaturase enzymes:

FAD2 – cloned by two research teams by T-DNA tagging and map-based cloning

FAD3 – cloned by map-based cloning

FAD6 – cloned by heterologous hybridization using FAD2 ER Δ12 desaturase gene as a probe

FAD7,8 – cloned by heterologous hybridization using FAD3 ER Δ15 desaturase gene as a probe

FAD4 – cloned by map-based cloning

FAD5– cloned by map-based cloning

Contributions of Plant FAD Mutants to Understanding of the Fatty Acid Unsaturation

Process 1.  Substrates of all membrane-bound FAD enzymes are lipid-bound

fatty acids

2. Fatty acid unsaturation is a sequential process – insertion of the first double bond is required (16:1 or 18:1) before the next desaturase can use this fatty acid chain as a substrate for insertion of the second double bond

Fatty Acid Unsaturation

Where does 18:1 come from?

It is made in the plastid stroma by the only known soluble desaturase enzyme in eukaryotes which acts on a soluble 18:0-ACP substrate

Fig. 10-18


ER

Plastid

PA

18:1 18:1 (16:0)

18:1

18:1 18:1

18:2

18:1 16:0

18:2 16:0

18:1 18:1 (16:0)

18:2 18:2 (16:0)

18:2

18:1 (16:0)

18:1

fab1

fab2

act1

fad4

fad6

fad7 fad8

fad5

fad6 fad6 fad6

fad7 fad8

fad7 fad8

fad7 fad8

fad2 fad3



fatty acids


3. Because some desaturases can use both C16 and C18 fatty acid substrates, they must determine the site of double bond insertion relative to an existing double bond or relative to the methyl end of the fatty acyl chain

18:1Δ9 -> 18:2Δ9, 12 -> 18:3 Δ9, 12, 15

->16:3 Δ7, 10, 13 -> 16:2Δ7, 10

Carboxyl end

Methyl end -

H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

O

HO C - C

H

- C H

- H C H

- H C H -

H C H

- H C H

- H C H

- H C H

- H C H

- H C H

H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

O

HO C - C

H

- C H

- H C H

- H C H -

H C H

- H C H

- H C H

- H C H

- H C H

- H C H

H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

O

HO C - C

H -

H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

O

HO C - C

H C H

- C H

- H C H

- H C H -

H C H

- H C H

- H C H

- H C H

- H C H

- H C H

H

- C H - C

H

- H C H

- H C H -

H C H

- H C H

- H C H

- H C H

- H C H

- H C H

H

- H C H

- H C H -

H C H

- H C H

- H C H

- H C H -

H C H

- H C H -

H C H

- H C H

- H C H

- H C H

- H C H

- H C H -

H C H

- H C H

- H C H

- H

H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

16:1Δ7

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

O

HO C - C

H

- C H

- H C H

- H C H -

H C H

- H C H

- H C H

- H C H

- H C H

- H C H

H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

O

HO C - C

H

- C H

- H C H

- H C H -

H C H

- H C H

- H C H

- H C H

- H C H

- H C H

H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- C H C H

- C H

- H C H

- H C H -

H C H

- H C H

- H C H

- H C H

- H C H

- H C H

H

- C H - C

H

- H C H

- H C H -

H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H

- H C H -

H C H

- H C H

- H C H

- H C H -

H C H

- H C H -

H C H

- H C H

- H C H

- H C H

- H C H

- H C H -

H C H

- H C H

- H C H

- H

H

- H C H

- H C H

- H C H

- H C H

- H

H

- H C H

- H

- H C H

C HO

O



fatty acids


3. Because some desaturases can use both C16 and C18 fatty acid substrates, they must determine the site of double bond insertion relative to an existing double bond or relative to the methyl end of the fatty acyl chain

4. fad mutants made it possible to clone the FAD genes and characterize FAD enzymes biochemically

Objectives:

1. Analysis of the role of Δ-3 trans16:1 (Tutorial)

2. Determining the role of trienoic fatty acids (16:3 and 18:3) using the Fad3, Fad7, Fad8 triple mutant (Tutorial)

3. The importance of polyunsaturated fatty acids for photosynthesis

Biological Roles of Fatty Acid Unsaturation

References:

Lightner et al. (1994) Altered body morphology is caused by increased stearate levels in a mutant of Arabidopsis. Plant J. 6, 401-412.

McConn and Browse (1998) Polyunsaturated membranes are required for photosynthetic competence in a mutant of Arabidopsis. Plant J. 15, 521-530. .

Thylakoid Membranes are the Most Highly Unsaturated Membranes in Eukaryotes

Chloroplast

Thylakoid

Lipid Lipid SL DGD PG MGD Lipid

Thylakoid membrane glycerolipids from wild type Arabidopsis

Fatty acid

Are such high levels of thylakoid membrane unsaturation critical for photosynthesis?

Thylakoid membranes are 75-80% polyunsaturated

None of the Isolated Unsaturation Mutants Displays a Visible Phenotype





16:1 16:2 and 18:1 18:2 LK3 FAD6

16:2 16:3 and 18:2 18:3 JB1 and LK9 FAD7

PLASTID

Mutations in FAD4, FAD5, FAD6, FAD7 and FAD8 affect plastid lipids

SUBSTRATE SPECIFIC



ER 18:2 18:3 on sn-1 and sn-2 of PC FAD3

16:2 16:3 and 18:2 18:3 FAD8 (Low temperature inducible)

Hypothesis: More substantial changes in membrane unsaturation are probably required to affect membrane functions.

Tutorial: Fad3 Fad7 Fad8 triple mutant is not impaired in photosynthesis Under normal growth conditions indicating that 16:3 and 18:3 are not essential for this process

Which double/triple mutant would you try to make?

Can we conclude that unsaturation is irrelevant to membrane functions? Why or why not?

ER

Plastid

PA

18:1 18:1 (16:0)

18:1

18:1 18:1

18:2

18:1 16:0

18:2 16:0

18:1 18:1 (16:0)

18:2 18:2 (16:0)

18:2

18:1 (16:0)

18:1

fab1

fab2

act1

fad4

fad6

fad7 fad8

fad5

fad6 fad6 fad6

fad7 fad8

fad7 fad8

fad7 fad8

fad2 fad3

The Fad2Fad6 Story

Generation of the Fad2Fad6 Double Mutant

Fad2 x Fad6

F1 x F1 (selfing)

F2 No Fad2Fad6 double mutant was found

Hypotheses:

1.  Double mutant failed to germinate

2.  Double mutant could not get established autotrophically

Seed Germination and Seedling Establishment

Fad2Fad6 Double Mutant Grows on Sucrose

F2 population Fad2Fad6 double mutants are chlorotic and contain only 10% of WT chlorophyll levels

When maintained on sucrose medium Fad2Fad6 double mutants develop relatively normal shoots and roots, but no flowers

WT Fad2Fad6

When transplanted into soil, Fad2Fad6 double mutants die within 7 days

Conclusion: Fad2Fad6 double mutant is not capable of autotrophic growth (photosynthesis)

77% polyunsaturated fatty acids

6% polyunsaturated fatty acids

Fad2Fad6 Phenotype

The observation that growth and organ development of Fad2Fad6 plants is almost normal on sucrose medium indicates that:

1. The majority of membrane functions required for these processes is not compromised;

2. Photosynthesis is the only function that requires high levels of membrane polyunsaturation.

Fatty Acid Desaturation

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