Chapter 20 – ‘The Calvin Cycle and the Pentose Phosphate Pathway’ PDF

Title Chapter 20 – ‘The Calvin Cycle and the Pentose Phosphate Pathway’
Course Biochemistry
Institution Rutgers University
Pages 3
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Chapter 20 – ‘The Calvin Cycle and the Pentose Phosphate Pathway’ •



The pentose phosphate pathway (PPP) provides a mean by which glucose can be oxidized to generate NADPH (the currency of readily available reducing power in the cell) o Note the difference between NADH vs. NADPH §฀ NADH isoxidized by respiratory chain to generate ATP §฀ NADPH serves as reductant in biosynthesis o PPP can also be used for catabolizing pentose sugars, synthesis of pentose sugars for nucleotide biosynthesis, and catabolism/synthesis of less common 4 and 7 sugars You should be able to write the PPP (oxidative and non-oxidative branches) with all structures, enzymes, and cofactors o The PPP consists of two phases: (all take place cyto) §฀ (1) Oxidative generation of NADPH • NADPH is generated when glucose-6-phosphate is oxidized to ribulose 5phosphate à฀ ribose 5-phosphate §฀





(2) Nonoxidative interconversion of sugars (‘carbon shuffling’) • The pathway catalyzes the interconversion of 3,4,5, 6, 7—C sugars in series nonoxy rxns PPP – OXIDATIVE Phase

o The rxn yields 2 molecules of NADPH and 1 molecule of ribulose 5-phosphate for per G6P PPP – NONOXIDATIVE Phase o The PPP and glycolysis are linked by the enzymes transketolase and transaldolase §฀ Ribose 5-phosphate can be converted à฀ G3P and F,6-P by these enzymes o These enzymes create a reversible link b/w PPP and glycolysis by catalyzing 3 successive rxns: §฀ The net result of these rxns is the formation of 2 hexoses and 1 triose from 3 pentoses o (1) Formation of GA3P/sedoheptulose 7-phosphate from 2 pentose (5)

§฀

*Note that a ketose is substrate oftransketolase ONLY if –OH group at C-3 has xylulose configuration (not ribulose)

o (2) G3P and sedoheptulose 7-phosphate à฀ F,6-P and erythrose 4-phosphate

o (3) Xylulose 5-phosphate and erythrose 4-phosphate à฀ G3P and F,6-P





o Note that excess ribose 5-phosphate formed by PPP can be completely converted to glycolytic intermediates The rate of PPP is controlled by level of NADP+ o First reaction of oxy branch (irreversible) is rate limiting o Low levels of NADP+ reduce the dehydrogenation of glucose 6-phospate because it is needed as electron acceptor §฀ These effects are intensifies since NADPH competes with NADP to bind w/enzyme o Thus NADPH is not generated unless the supply needed for reduc biosyn is low o Non-oxy branch is controlled by availability of substrates The PPP functions to provide NADPH and ribose o You should understand why these products are important and how both branches of the PP can interact to provide the products which are needed by a cell (modes – Fig. 20.23)

o Mode 1 – Much more ribose 5-phosphate than NADPH is required. §฀ Utilize thenon-oxidative branch to make R5P o Mode 2 – The needs for NADPH and for ribose 5-phosphate are balanced. §฀ Utilize theoxidative branch to make both NADPH and R5P o Mode 3 – Much more NADPH than ribose 5-phosphate is required. §฀ First run oxidative branch to form 2 NADPH and ribose 5-P §฀ Then ribose 5-P à฀ F6P and G3P (transketolase, transaldolase) §฀ G6P resynthesized by from F6P and G3P by gluconeogenic pathway §฀ R5P produced by PPP is recycled to G6P by transketolase, transaldolase, and some enzymes of gluconeogenic pathway. o Mode 4 – Both NADPH and ATP are required.

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In which tissues is the PPP most active? o Adipose tissue requires a high level of NADPH for synthesis of fatty acids o Other tissues include mammary glands, adrenal, liver, ovary, testes, RBCs Understand the mechanism of transketolase and transaldolase (parallel to PYR DH complex and aldolase respectively) (Fig. 20.20 and 20.21) o Transketolase – transfers 2-C unit Transaldolase – transfers 3-C unit Glutathione acts as a ‘redox’ buffer to protect cells against oxidation o NADPH plays important role in protecting cells from ROS §฀ Reduced glutathione (GSH) a tripeptide with a free sulfhydryl group combats oxidative stress by reducing ROS to harmless forms §฀ Afterwards it becomes oxidized as GSSG and must be reduced to regenerate GSH • NADPH from PPP is used for reduction Understand why cells with decreased G6PDH activity are more sensitive to oxidative stress §฀ RBCs are more sensitive because they lack mito and have no means reducing power

Why does the drug pamaquine cause hemolytic anemia? o Antimalarial drug; some ppl given drug had severe symptoms and reactions §฀ Black urine, jaundice, sharp decrease Hb content in blood o Makes membrane RBCs brittle leading to break down o Drug-induced hemolytic anemia caused by deficiency of G6PDH à฀ low NADPH in cells §฀ NADHP is used to reduce GSH o Pamaquine is a pure glycoside of fava bean à฀ ppl who ingest fava beans à฀ favism §฀ Pamaquine is oxidative reagent à฀ ROS (peroxides) §฀ Peroxides are not eliminated à฀ glutathione peroxidase uses GSH as reducing agent o W/o enough GSH, Hb –SH groups can no longer remain reduced form à฀ leading to Hb linkage that form Heinz bodies (which damages membrane RBC leading to lysis) o *G6PDH deficiency protects against falciparum malaria Wernicke-Korsakoff syndrome (weak binding of TPP to transketolase) o Neuropsychiatric disorder caused by a lack of thiamine in diet §฀ Vitamin free diet (alcoholism) §฀ Paralysis of eye movement; impaired memory, gait, abnormal stance o May be caused by genetic factors of thiamine deficiency o Transketolase (High KM) from ppl with syndrome binds TPP less than enzyme dos for normal person §฀ Abnormality of transketolase à฀ level of TPP too low to saturate enzyme The Calvin Cycle is very important in supporting life on Earth o Basically you should understand that is it made up almost entirely of reactions from gluconeogenesis and PPP. (plus carbon fixation) In photosynthesis the ‘Light Reaction’ uses energy from photons to provide the cell with ATP and NADH. o The ‘Dark Reaction’ –the Calvin Cycle ---utilizes these products to form the whole panoply of organic compounds in cell starting with CO2 o This is pure standard biochem and not dependent on light in anyway o It is the light reaction, which produces O2, by removing e- from water and promoting them to higher energies....


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