Ðåôåðàòû

Ó÷åáíîå ïîñîáèå: Îáì³í åíåð㳿 â îðãàí³çì³

Ó÷åáíîå ïîñîáèå: Îáì³í åíåð㳿 â îðãàí³çì³

Ïëàí

1. Îáì³í ðå÷îâèí ôóíêö³ÿ æèâîãî

2. Ïåðåòâîðåííÿ õ³ì³÷íî¿ åíåð㳿 â îðãàí³çì³:

2.1. Ïåðåêèñíà òåîð³ÿ

2.2. Òåîð³ÿ Ïàëàä³íà

2.3. Ñó÷àñí óÿâëåííÿ ïðî ìåõàí³çì ïðîöåñó á³îëîã³÷íîãî îêèñëåííÿ

3. Ìåõàí³çì ïåðåíîñó ïðîòîí³â ³ åëåêòðîí³â ïî äèõàëüíîìó ëàíöþãó:

3.1. Àåðîáíå òà àíàåðîáíå îêèñëåííÿ

3.2. Îêèñëþâàëüíå ôîñôîðèëþâàííÿ

3.3. ³ëüíå òà ñïðÿæåíå îêèñëåííÿ

4. Ìàêðîåðã³÷í çâ‘ÿçêè òà ìàêðîåðã³÷í³ ñïîëóêè


Ïî÷èíàþ÷è âèâ÷åííÿ êóðñó á³îõ³ì³¿ ìè âèçíà÷èëè îñíîâí³ õàðàêòåðèñòèêè æèâîãî îðãàí³çìó. Íà â³äì³íó â³ä ïðåäìåò³â íåæèâî¿ ïðèðîäè æèâ³ îᑺêòè çäàòí äî ñàìîâ³äòâîðåííÿ ç ïåðåäà÷åþ ãåíåòè÷íî¿ ³íôîðìàö³¿. ßâëÿþ÷èñü â³äêðèòèìè ñèñòåìàìè – àêòèâíî âçàºìîä³þòü ç íàâêîëèøí³ì ñåðåäîâèùåì, ï³äòðèìóþ÷è òèì ñàìèì âëàñíó âíóòð³øíþ ð³âíîâàãó (ãîìåîñòàç).  îñíîâ îñòàííüîãî ëåæèòü íàäõîäæåííÿ åíåð㳿 õàð÷îâèõ ñóáñòðàò³â ç îòî÷óþ÷îãî ñåðåäîâèùà â îðãàí³çì ³ âèâåäåííÿ ç íüîãî ê³íöåâèõ ïðîäóêò³â âíóòð³øíüîãî îáì³íó. Ñóêóïí³ñòü îáì³ííèõ ïðîöåñ³â ì³æ æèâîþ ñèñòåìîþ òà îòî÷óþ÷èì ñåðåäîâèùåì îòðèìàëè íàçâó ìåòàáîë³çìó, à ðåàêö³¿ ùî ïðèâîäÿòü äî óòâîðåííÿ ê³íöåâèõ ïðîäóêò³â æèòòºä³ÿëüíîñò³ îðãàí³çìó íàçèâàþòü âíóòð³øí³ì îáì³íîì.

Ðåàêö³¿ âíóòð³øíüîãî îáì³íó çà íàïðÿìêàìè ¿õ ïðîò³êàííÿ òà õàðàêòåðîì ê³íöåâèõ ïðîäóêò³â ïîä³ëÿþòü íà:

Ïðîì³æíèé îáì³í – êîìïëåêñ õ³ì³÷íèõ ïåðåòâîðåíü ðå÷îâèí ÿê³ ùîéíî ïîòðàïè-ëè â îðãàí³çì ççîâí³.

Ïëàñòè÷íèé îáì³í – ñóêóïí³ñòü õ³ì³÷íèõ ïðîöåñ³â ùî ïðèâîäÿòü äî ñèíòåçó ñïåöèô³÷íèõ äëÿ îðãàí³çìó ðå÷îâèí: ôåðìåíò³â, ãîðìîí³â, åíåðãîâì³ñíèõ ñïîëóê, ñòðóêòóðíèõ åëåìåíò³â êë³òèíè.

Ôóíêö³îíàëüíèé îáì³í – âêëþ÷ຠá³îõ³ì³÷í³ ðåàêö³¿ ÿê³ çàáåçïå÷óþòü ôóíêö³îíàëüíó àêòèâí³ñòü êë³òèí, òêàíèí òà îðãàí³â.

Îáì³í ðå÷îâèíàìè â æèâ³é ñèñòåì³ çàáåçïå÷óºòüñÿ âçàºìîïîâ‘ÿçàíèìè àëå ïðîòèëåæíèìè ïðîöåñàìè àñèì³ëÿö³¿ òà äèñèì³ëÿö³¿.

Àñèì³ëÿö³ÿ – çàñâîºííÿ òà ïåðåòâîðåííÿ ðå÷îâèí îòî÷óþ÷îãî ñåðåäîâèùà ó ñïåöèô³÷í³ äëÿ îðãàí³çìó ñïîëóêè.

Äèñèì³ëÿö³ÿ – ðîçïàä ðå÷îâèí îðãàí³çìó äî ê³íöåâèõ ïðîäóêò³â òà âèâåäåííÿ ¿õ ç îðãàí³çìó.

Ñîíÿ÷íà åíåðã³ÿ, ó ãëîáàëüíîìó ïîíÿòò³, ºäèíå äæåðåëî åíåð㳿 ïëàíåòè Çåìëÿ. Ñîíöå âèä³ëÿº 13 1023 Êêàëîð³é åíåð㳿 íà ð³ê. ʳëüê³ñòü åíåð㳿 ùî ïàäຠíà Çåëþ ùîäåííî, åêâ³âàëåíòíà 1000000 àòîìíèõ çàðÿä³â ùî âïàëè íà Õ³ðîñ³ìó. ×àñòèíà åíåð㳿 çàòðèìóºòüñÿ ó âåðõí³õ øàðàõ àòìîñôåðè, ÷àñòèíà, â³äáèâàþ÷èñü â³ä ïîâåðõí³, ðîçñ³þºòüñÿ ó âèãëÿä³ òåïëà ³ ëèøå 1 % ïåðåòâîðþºòüñÿ â åíåðã³þ õ³ì³÷íèõ çâ‘ÿçê³â îðãàí³÷íèõ ñïîëóê çàâäÿêè ïðîöåñàì ôîòîñèíòåçó. Ëþäèíà ÿê íå ôîòîñèíòåçóþ÷à ñèñòåìà íå ìຠìîæëèâîñò áåçïîñåðåäíüî çàñâîþâàòè cîíÿ÷íó åíåðã³þ äëÿ ï³äòðèìàííÿ ïðîöåñ³â æèòòÿ. Ôóíêö³îíàëüíà àêòèâí³ñòü îðãàí³çìó çàáåçïå÷óºòüñÿ çîâí³øí³ì íàäõîäæåííÿì åíåð㳿 ó âèãëÿä³ õ³ì³÷íî¿ åíåð㳿 õàð÷îâèõ ïðîäóêò³â.


2. Ïåðåòâîðåííÿ õ³ì³÷íî¿ åíåð㳿 â îðãàí³çì³

Ïèòàííÿ ïðî òðàíñôîðìàö³þ åíåð㳿 õ³ì³÷íèõ çâ‘ÿçê³â ó ô³ç³îëîã³÷í ôóíêö³¿ äàâíî õâèëþâàëî â÷åíèõ. Ùå ó 18 ñòîð³÷÷³ Ëàâóàçüº ïîì³òèâ ïîä³áí³ñòü ïðîöåñ³â äèõàííÿ îðãàí³çìó òà ãîð³ííÿ îðãàí³÷íèõ ðå÷îâèí. Äëÿ ï³äòðèìàííÿ îáîõ ïðîöåñ³â íåîáõ³äíèì º êèñåíü, à ê³íöåâèìè ïðîäóêòàìè ÿâëÿþòüñÿ âîäà âóãëåêèñëèé ãàç.

ϳñëÿ â³äêðèòòÿ Ëàâóàçüº çà êîðîòêèé ïðîì³æîê ÷àñó âèíèêëî áàãàòî òåîð³é á³îëîã³÷íîãî îêèñëåííÿ îðãàí³÷íèõ ðå÷îâèí.

2.1 Ïåðåêèñíà òåîð³ÿ Øåíáàéíà–Áàõà

Ïåðåäóìîâîþ ñòâîðåííÿ ïîñëóæèëî â³äêðèòòÿ Øåíáàéíîì îçîíó, ÿêèé çäàòíèé îêèñëþâàòè äåÿê³ ðå÷îâèíè ïðè çâè÷àéí³é òåìïåðàòóð³ óòâîðþþ÷è ïåðîêñèäè.

Ñóòü:

1. Êèñåíü ïîâ³òðÿ ïîïàäàþ÷è â îðãàí³çì àêòèâóºòüñÿ ï³ä 䳺þ ëåãêîîêèñíèõ ðå÷îâèí ( îêñèãåíàç) ç óòâîðåííÿì ïåðîêñèä³â.    

                                                                                                Î

îêñèãåíàçà           +       Î2               ÎÊÑÈÃÅÍÀÇÀ            

Î

2. Êèñåíü ïåðîêñèäó îêñèãåíàçè, ï³ä âïëèâîì ôåðìåíòó ïåðîêñèäàçè, ïåðåíîñèòüñÿ íà á³ëüø âàæêîîêèñí³ ñóáñòðàòè îêèñëþþ÷è ¿õ.       

          Π      Π                                   Î
ÎÊÑÈÃÅÍÀÇÀ   +       S                 S                 +       îêñèãåíàçà.

Π                                  


2.2 Òåîð³ÿ Ïàëàä³íà–³ëàíäà (îêèñëåííÿ ðå÷îâèí øëÿõîì ¿õ äåã³äðóâàííÿ)

Ïåðåäóìîâîþ âèíèêíåííÿ áóëî â³äêðèòòÿ â ðîñëèíàõ îñîáëèâèõ ï³ãìåíò³â, ùî çäàòí îêèñëþâàòè ñóáñòðàòè áåç êèñíþ.

Ñóòü:

1. Îêèñëåííÿ ñóáñòðàò³â ïðîõîäèòü çàâäÿêè ¿õ äåã³äðóâàííþ ( â³äùåï-ëåííþ â³ä ìîëåêóë ñóáñòðàòó àòîì³â âîäíþ), ùî çä³éñíþºòüñÿ çà äîïîìîãîþ äå-ã³äðîãåíàçíèõ ôåðìåíòíèõ ñèñòåì:

H                                                                          H

          S       +       D                                    S       +       D               

           H                                                                         H

          Red.            Ox                                  Ox              Red

2. Êèñåíü áåçïîñåðåäíüî ç ñóáñòðàòîì íå âçàºìî䳺, à îêèñëþþ÷è â³äíîâëåíó ôîðìó äåã³äðîãåíàç çàïîá³ãຠçóïèíö³ ïðîöåñ³â á³îëîã³÷íîãî îêèñëåííÿ:

                             H                                                                          H               

                   D       +                 O                          D       +       O                + Å   

                             H                                                                          H               

                   Red                       Ox                        Ox              Red

Ïîäàëüøèé ðîçâèòîê íàóêè äîâ³â ñïðàâåäëèâ³ñòü îáîõ òåîð³é ÿê³ ³ ëÿãëè â îñíîâó ñó÷àñíèõ óÿâëåíü ïðî ìåõàí³çì á³îëîã³÷íîãî îêèñëåííÿ.


2.3 Ñó÷àñíà òåîð³ÿ á³îëîã³÷íîãî îêèñëåííÿ

Á³îëîã³÷íå îêèñëåííÿ – ñóêóïí³ñòü õ³ì³÷íèõ ðåàêö³é ó ÿêèõ ïðîõîäèòü ïåðåíîñ åëåêòðîí³â ³ ïðîòîí³â â³ä îêèñëþâàëüíî¿ ðå÷îâèíè ( äîíîðà ) äî ðå÷îâèíè ùî â³äíîâëþºòüñÿ (àêöåïòîðà). Ïðîöåñ â³äáóâàºòüñÿ ç âèä³ëåííÿì åíåð㳿.

Ïåðåíîñ åëåêòðîí³â ³ ïðîòîí³â âîäíþ ì³æ ðå÷îâèíàìè (êîìïîíåíòàìè äèõàëüíîãî ëàíöþãà ì³òîõîíäð³é) â³äáóâàºòüñÿ çã³äíî ¿õ âåëè÷èí ñïîð³äíåíîñò³ äî åëåêòðîíó (ñèëà ïðèòÿãàííÿ åëåêòðîíà äî ÿäðà). ʳëüê³ñíîþ ì³ðîþ îñòàííüî¿ º âåëè÷èíà îêèñíî-â³äíîâíîãî ïîòåíö³àëó Å.î.: – â³ä‘ºìíà âåëè÷èíà Å.î. õàðàêòåðèçóº íèçüêó ñïîð³äíåí³ñòü äî åëåêòðîíà, ðå÷îâèíà äîáðå â³ääຠåëåêòðîíè ³ ïðîòîíè, òîáòî º ¿õ äîíîðîì; – ïîçèòèâíà âåëè÷èíà Å.î. õàðàêòåðèçóº âèñîêó ñïîð³äíåí³ñòü äî åëåêòðîíà, ðå÷îâèíà êðàùå ïðèºäíó åëåêòðîíè ³ ïðîòîíè í³æ â³ääຠ¿õ, òîáòî º àêöåïòîðîì ïðîòîí³â ³ åëåêòðîí³â.

Ïåðåíîñ åëåêòðîí³â ³ ïðîòîí³â â³äáóâàºòüñÿ çà äîïîìîãîþ ð³çíîìàí³òíèõ ôåðìåíòíèõ ñèñòåì.

Äåã³äðîãåíàçè – ñêëàäí³ á³ëêè ÿê³ êàòàë³çóþòü â³äùåïëåííÿ àòîì³â âîäíþ òà åëåêòðîí³â â³ä îêèñëþâàëüíèõ ñóáñòðàò³â. Ñêëàäàþòüñÿ ç á³ëêîâî ÷àñòèíè – àïîôåðìåíòó, òà íåá³ëêîâî¿ – êîôåðìåíòó ÷è ïðîñòåòè÷íî¿ ãðóïè.  îðãàí³çì³ íàë³÷óºòüñÿ á³ëÿ 150 âèä³â äåã³äðîãåíàçíèõ ôåðìåíòíèõ ñèñòåì.

 ïðîöåñàõ îêèñëåííÿ íàé÷àñò³øå áåðóòü ó÷àñòü òðè òèïè ñèñòåì ÿê â³äð³çíÿþòüñÿ áóäîâîþ íåá³ëêîâî¿ ÷àñòèíè.

1). ßêùî êîôåðìåíòîì äåã³äðîãåíàçíî¿ ñèñòåìè º â³òàì³í ÐÐ (àì³ä í³êîòèíîâî¿ êèñëîòè), òî ãîâîðÿòü ïðî ÍÀÄ– òà ÍÀÄÔ-çàëåæí³ äåã³äðîãåíàçè.

ÍÀÄ ³ ÍÀÄÔ–çàëåæí³ äåã³äðîãåíàçè: ëåãêî îêèñëþþòü âóãëåâîäíåâ³, æèðîâ³ òà á³ëêîâ³ ñóáñòðàòè; âèñîêîðóõëèâ³ íîñ³¿ ïðîòîí³â âîäíþ ì³æ ð³çíèìè äåã³äðîãåíàçíèìè ñèñòåìàìè; Å.î. = –0.3 Â.

ͳêîòèíàì³äàäåí³íäèíóêëåîòèä ͳêîòèíàì³äàäåí³íäèíóêëåîòèäôîñôàò

                              CONH2                                                                CONH2      

                                                                                                                            

                    N                                                               N                        

 O–CH2 O                                         O–CH2 O                               

           H H                                                  H H                              

                                                                                                                            

 H      OH   OH H                                               H OH OH H                          

O=P–OH                                                    O=P–OH              NH2                                  

                              NH2                                                                              

 O               N               N                         O       N                         N               

                                                                                                                            

O=P–OH                                                    O=P–OH                                          

                    N      N                                                      N      N                        

 O–CH2                                                                                           O–CH2                                                                            

                                                                                                                            

           O                                                      O                                           

           H      H                                                      H H                              

                                                                                                                            

H OH          HO H                                      H O HO H                             

                                                                   HO–P=O                                          

                                                                             OH                      
Àêòèâíîþ ÷àñòèíîþ ¿õ º ï³ðèäèíîâå ê³ëüöå í³êîòèíàì³äó:

H                H                          H H                                       

          S       +       H      CONH2                H      CONH2                                  

                   H       H      H                         H      H               +       H+ + S        

                                      N+                                         N                                                                         

                                                                                                         

2) ßêùî ïðîñòåòè÷íîþ ãðóïîþ º â³òàì³í Â2 (ðèáîôëàâ³í) òî ãîâîðÿòü ïðî ôëàâ³íçàëåæí³ äåã³äðîãåíàçè ÔÌÍ òà ÔÀÄ. ÔÀÄ–çàëåæí³ äåã³äðîãåíàçè, ÿê³ ïåðåâàæíî îêèñëþþòü æèðîâ³ ñóáñòðàòè òà ÍÀÄ-çàëåæí³ äåã³äðîãåíàçè ìàþ÷è îêèñíî-â³äíîâíèé ïîòåíö³àë Å.î. = - 0.1 Â.

Àêòèâíîþ ÷àñòèíîþ ¿õ ìîëåêóëè º ³çîàëëîêñàçèíîâå ê³ëüöå ðèáîôëàâ³íó:

                                       O                                   O     

          H       CH3  N       NH             CH3            NH    NH

S                 +                                                                                    +       S

          H       CH3                     O      CH3                     O     

                                      N+     N                                    N NH         


Ôëàâ³íìîíîíóêëåîòèä        Ôëàâ³íàäåí³íäèíóêëåîòèä 

                                      O                                    O                        

                   CH3   N       NH             CH3            N       NH                     

                                                                                                                            

                    CH3                     O      CH3                     O      NH2 

                                      N+     N                                    N N N                           

                                      CH2                                         CH2                                                    

                                      CH–OH                                   CH–OH      N              

                                      CH–OH                                   CH–OH      N     

                                      CH–OH      OH                       CH–OH OH         OH             

                                      CH2–O–P–OH                         CH2-O-P-O-P-O-CH2    

                                                 O                                            O O           

                                                                                                                   O      

                                                                                                           H H 

                                                                                                                            

                                                                                                          H OH OH  

                                                                                                                   H

3) ßêùî êîôåðìåíòîì º ïîõ³äíå â³òàì³íó Ê, òî â³äïîâ³äí äåã³äðîãåíàçè íàçèâàþòü óá³õ³íîíçàëåæíèìè ÷è êîôåðìåíòîì Q.

                              O              

                   CH3O                   CH3

                                                 CH3            CH3

                   CH3O                   (CH2–CH=C–CH2)9–CH2–CH=C–CH3

                                                         

                              O

Àêòèâíîþ ÷àñòèíîþ ìîëåêóëè êîôåðìåíòà Q º óá³õ³íîí:

                              O                                            OH                     

          H       CH3O                   CH3                                     CH3O                   CH3                  

 S      +                                                                                    +       S      

          H       CH3O                   R                          CH3O                   R                        

                                                                                                                  

                             O                                             OH                     

Óá³õ³íîíçàëåæí³ äåã³äðîãåíàçè âîëîä³þ÷è Å.î. = – 0.05  îêèñëþþòü ò³ëüêè ÍÀÄ- ³ ÔÀÄ-çàëåæí³ äåã³äðîãåíàçè. ßâëÿþ÷èñü íàäçâè÷àéíî âèñîêîðóõëèâèìè, âóçëîâèìè ïóíêòàìè çáîðó ïðîòîí³â âîäíþ îêèñëåíèõ ñóáñòðàò³â.

Îñîáëèâèì òèïîì ïåðåíîñ÷èê³â º öèòîõðîìè. Íà â³äì³íó â³ä äåã³äðîãå-íàç, ÿê³ áåçïîñåðåäíüî îêèñëþþòü ñóáñòðàòè çàáèðàþ÷è â³ä íèõ ïðîòîíè âîäíþ òà åëåêòðîíè, öèòîõðîìè ëèøå ïåðåíîñÿòü åëåêòðîíè îêèñëåíèõ ñóáñòðàò³â íà ê³íöåâèé àêöåïòîð.

Öèòîõðîìè – ñêëàäí³ á³ëêè, ùî ñêëàäàþòüñÿ ³ç çàë³çîâì³ñíî¿ ïðîñòåòè÷íî¿ ãðóïè òà á³ëêà íîñ³ÿ. Ïåðåíîñ åëåêòðîí³â â³äáóâàºòüñÿ çà ðàõóíîê çì³íè îêèñíî-â³äíîâíèõ ïîòåíö³àë³â ³îíó çàë³çà:

Fe3+ + e                 Fe2+

Fe2+ – å                 Fe3+

Ïî áóäîâ³ á³ëêà íîñ³ÿ òà çíà÷åííÿì îêèñíî–â³äíîâíèõ ïîòåíö³àë³â öèòîõðîìè ïîä³ëÿþòü íà ñë³äóþ÷³ òèïè :

Öèò Â < öèò C < öèò Ñ < öèò À1 < öèò À3

+0.7ì + 212ì + 210ì + 210ì + 385ìÂ.

Îòæå:

1. Îêèñëåííÿ ñóáñòðàò³â îðãàí³çìó ïðîõîäèòü øëÿõîì ¿õ äåã³äðóâàííÿ çà äîïîìîãîþ äåã³äðîãåíàçíèõ ôåðìåíòíèõ ñèñòåì;

2. Ïåðåíîñ ïðîòîí³â âîäíþ òà åëåêòðîí³â îêèñëåíèõ ñóáñòðàò³â â³äáó-âàºòüñÿ çà äîïîìîãîþ äåã³äðîãåíàçíèõ ôåðìåíòíèõ ñèñòåì çã³äíî âåëè-÷èí ¿õ îêèñíî-â³äíîâíèõ ïîòåíö³àë³â (â³ä ðå÷îâèíè ç íèæ÷èì éîãî çíà÷åííÿì äî ðå÷îâèíè ç á³ëüø âèñîêèì çíà÷åííÿì);

3. ÍÀÄ, ÔÀÄ, ÓÕ – çàëåæí³ äåã³äðîãåíàçè òà öèòîõðîìè çíàõîäÿòüñÿ â ì³òîõîíäð³àëüí³é ìåìáðàí³, ÿâëÿþ÷è ñîáîþ ºäèíó ñèñòåìó á³îëîã³÷íîãî îêèñëåííÿ, òàê çâàíèé «äèõàëüíèé ëàíöþã»;

4. Äåã³äðîãåíàçíà ñèñòåìà ôåðìåíò³â çàáåçïå÷óº îêèñëåííÿ åíåðãåòè÷íèõ ñóáñòðàò³â ³ òðàíñïîðò ïðîòîí³â òà åëåêòðîí³â à öèòîõðîìíà, òðàíñïîðòó-þ÷è åëåêòðîíè, àêòèâóº ê³íöåâèé àêöåïòîð.


3. Ìåõàí³çì ïåðåíîñó ïðîòîí³â ³ åëåêòðîí³â ïî äèõàëüíîìó ëàíöþãó

3.1. Àåðîáíå òà àíàåðîáíå îêèñëåííÿ

 çàëåæíîñò³ â³ä ê³íöåâîãî àêöåïòîðó ïðîòîí³â âîäíþ òà åëåêòðîí³â, ùî ïðîéøëè êð³çü äèõàëüíèé ëàíöþã, ðåàêö³¿ á³îëîã³÷íîãî îêèñëåííÿ ïîä³ëÿþòü íà àåðîáí³ òà àíàåðîáí³.

Àåðîáíèìè – íàçèâàþòü ðåàêö³¿ â ÿêèõ ê³íöåâèì àêöåïòîðîì ïðîòîí³â âîäíþ òà åëåêòðîí³â, ùî ïðîéøëè ïî äèõàëüíîìó ëàíöþãó, º êèñåíü.

          H                                                       2Í+                                                  

S                 ÍÀÄ+                   ÔÀÄÍÍ     ÓÕ+            â++     ñ+++    à++               Π     

H                                                                öèò    öèò    öèò                      

                                                                             â+++    ñ++     à+++             Î2–   

S                 ÍÀÄÍÍ     ÔÀÄ+                   ÓÕÍÍ                                                       

                                                                                                                   H2O 

ʳíöåâèì ïðîäóêòîì àåðîáíèõ ðåàêö³é á³îëîã³÷íîãî îêèñëåííÿ º âîäà ³ âóãëå-êèñëèé ãàç.

3.2. Îêèñëþâàëüíå ôîñôîðèëþâàííÿ

ßêùî ê³íöåâèì àêöåïòîðîì åëåêòðîí³â òà ïðîòîí³â âîäíþ äèõàëüíîãî ëàíöþãà íø³ ðå÷îâèíè êð³ì êèñíþ (ìîëî÷íà ÷è ï³ðîâèíîãðàäíà êèñëîòè) òî ãîâîðÿòü ïðî àíàåðîáíèé òèï á³îëîã³÷íîãî îêèñëåííÿ. Ïðîò³êຠïðè âåëèêèõ ô³çè÷íèõ íàâàíòàæåííÿõ ÷è îòðóºííÿõ îðãàí³çìó. Îáóìîâëþºòüñÿ äåô³öèòîì òêàíèííîãî êèñíþ. Öå ïðèâîäèòü äî «çàòîâàðåííÿ åëåêòðîíàìè» ñèñòåìè öèòîõðîì, çíèæåííÿ àêòèâíîñò³ äåã³äðîãåíàç âíàñë³äîê íàêîïè÷åííÿ ¿õ â³äíîâëåíèõ ôîðì. Ïðîòå, çóïèíêè ïðîöåñ³â îêèñëåííÿ íå â³äáóâàºòüñÿ çàâäÿêè îêèñëåííþ â³ä-íîâëåíèõ ôîðì ÍÀÄ-çàëåæíèõ äåã³äðîãåíàç ï³ðîâèíîãðàäíîþ êèñëîòîþ. Îñ-òàííÿ â³äíîâëþþ÷èñü äî ìîëî÷íî¿ êèñëîòè ïîíîâëþº ¿õ àêòèâí³ñòü, ùî ðîáèòü ìîæëèâèì ïîäàëüøå ïðîò³êàííÿ ïðîöåñ³â îêèñëåííÿ ñóáñòðàò³â.

          H                         

S                 ÍÀÄ+                   CH3–CH–COOH (ìîëî÷íà êèñëîòà)

H                          OH

                                     

                   ÍÀÄÍÍ     CH3–C–COOH (ï³ðîâèíîãðàäíà êèñëîòà)

                                                O

Ìîëî÷íà êèñëîòà – ê³íöåâèé ïðîäóêò àíàåðîáíîãî (áåçêèñíåâîãî) îêèñëåííÿ ñóáñòðàò³â. Çà íîðìàëüíèõ ô³ç³îëîã³÷íèõ óìîâ äîîêèñëþºòüñÿ äî âóãëåêèñëîãî ãàçó òà âîäè.

Åíåðã³ÿ ùî âèä³ëÿºòüñÿ â ðåçóëüòàò³ îêèñëåííÿ îðãàí³÷íèõ ñóáñòðàò³â ìîæå ïî ð³çíîìó âèêîðèñòîâóâàòèñÿ îðãàí³çìîì. ßêùî âñÿ åíåðã³ÿ âèä³ëÿºòüñÿ âèêëþ÷íî ó âèãëÿä³ òåïëà, òî ãîâîðÿòü ïðî â³ëüíå îêèñëåííÿ. Ñïîñòåð³ãàºòüñÿ ïðè îòðóºííÿõ îðãàí³çìó õ³ì³÷íèìè ðå÷îâèíàìè ÷è áàêòåð³àëüíèìè òîêñèíàìè, âíàñë³äîê áëîêóâàííÿ ôàêòîð³â ñïðÿæåííÿ – á³ëê³â ì³òîõîíäð³àëüíî¿ ìåìáðàíè ÿê íåîáõ³äíèìè äëÿ ñèíòåçó ìîëåêóë ÀÒÔ.

ßêùî æ åíåðã³ÿ âèêîðèñòîâóºòüñÿ äëÿ îäíî÷àñíèõ ðåàêö³é ñèíòåçó á³îîðãàí³÷íèõ ñïîëóê, òî ãîâîðÿòü ïðî ñïðÿæåíå îêèñëåííÿ. Ñïðÿæåíå îêèñëåííÿ îñíîâíèé òèï îêèñíî-â³äíîâíèõ ïðîöåñ³â îðãàí³çìó.Ó 1940 ðîö³ ðàäÿíñüêèé â÷åíèé Áåë³öåð ïîì³òèâ ñë³äóþ÷èé ôàêò: îêèñëåííÿ ð³çíèõ ñóáñòðàò³â ñóïðîâîäèòüñÿ ñïîæèâàííÿì íåîðãàí³÷íîãî ôîñôàòó.

ϳçí³øå áóëî âñòàíîâëåíî, ùî ñïîæèòèé ôîñôàò âèêîðèñòîâóºòüñÿ íà ñèíòåç ÀÒÔ. Òàêèì ÷èíîì â³ëüíà åíåðã³ÿ îêèñëåííÿ îðãàí³÷íèõ ñóáñòðàò³â ïåðåòâîðþºòüñÿ ó äîñòóïíó äëÿ îðãàí³çìó åíåðã³þ õ³ì³÷íèõ çâ‘ÿçê³â.

Ïîñê³ëüêè ïðîöåñ óòâîðåííÿ ÀÒÔ ïðîõîäèòü ñïðÿæåíî (çâ‘ÿçàíî) ç ïðîöåñàìè îêèñëåííÿ ñóáñòðàò³â òî â³í íîñèòü íàçâó ñïðÿæåíîãî îêèñëþâàëüíîãî ôîñôîðèëþâàííÿ.Äëÿ óòâîðåííÿ ìàêðîåðã³÷íîãî çâ‘ÿçêó ìîëåêóëè ÀÒÔ íåîáõ³äíà ð³çíèöÿ ïîòåí-ö³àë³â ì³æ êîìïîíåíòàìè äèõàëüíîãî ëàíöþãà ïîâèííà ñêëàäàòè 0.15Â. Âîíà ñïîñòåð³ãàºòüñÿ ó ñë³äóþ÷èõ ïóíêòàõ ñïðÿæåííÿ (ì³ñöÿõ ñèíòåçó ìîëåêóë ÀÒÔ).

²-é – çàáåçïå÷óºòüñÿ åíåð㳺þ ïåðåíîñó ïðîòîí³â âîäíþ â³ä ÍÀÄ äî ÔÀÄ-çà-ëåæíèõ äåã³äðîãåíàçíèõ ôåðìåíòíèõ ñèñòåì. гçíèöÿ îêèñíî-â³äíîâíèõ ïîòåí-ö³àë³â –0.26 Â. Ñèíòåçóºòüñÿ îäíà ìîëåêóëà ÀÒÔ.

²²-é – çàáåçïå÷óºòüñÿ åíåð㳺þ ïåðåíîñó ïàðè åëåêòðîí³â â³ä êîôåðìåíòà Q íà öèòîõðîì Ñ ÷åðåç öèòîõðîì Â. гçíèöÿ îêèñíî-â³äíîâíèõ ïîòåíö³àë³â – 0.29 Â. Ñèíòåçóºòüñÿ îäíà ìîëåêóëà ÀÒÔ.

²²²-é – çàáåçïå÷óºòüñÿ åíåð㳺þ ïåðåíîñó ïàðè åëåêòðîí³â ç öèòîõðîìà À íà êè-ñåíü. гçíèöÿ îêèñíî-â³äíîâíèõ ïîòåíö³àë³â –0.29 Â. Ñèíòåçóºòüñÿ îäíà ìîëå-êóëà ÀÒÔ.

Íàäëèøêîâ³ ê³ëüêîñò³ åíåð㳿 ùî âèä³ëÿþòüñÿ â ïóíêòàõ ñïðÿæåííÿ, âèêîðèñòî-âóþòüñÿ äëÿ ï³äòðèìàííÿ ôóíêö³îíàëüíî àêòèâíîñò³ öèòîõðîìîêñèäàçíî¿ ñè-ñòåìè (êîìïîíåíò³â «äèõàëüíîãî ëàíöþãà»):

          H                                                       2Í+                                                  

S                 ÍÀÄ+                   ÔÀÄÍÍ     ÓÕ+            â++     ñ+++    à++               Π     

H                                                                öèò    öèò    öèò                      

                                                                             â+++    ñ++     à+++             Î2–   

S                 ÍÀÄÍÍ     ÔÀÄ+                   ÓÕÍÍ                                                       

Å î-â.          (Â)     –0.32                    –0.06          0                 +0.26 +0.29 +0.53           +0.82

                             ÀÒÔ                                        ÀÒÔ ÀÒÔ

Âèñíîâîê:

1.  ïðîöåñ³ ðåàêö³é ñïðÿæåíîãî îêèñëþâàëüíîãî ôîñôîðèëþâàííÿ ïåðåíîñ ïàðè åëåêòðîí³â òà ïðîòîí³â âîäíþ ÷åðåç ñèñòåìó äèõàëüíîãî ëàíöþãà, çà àåðîáíèõ óìîâ, ïðèâîäèòü äî óòâîðåííÿ òðüîõ ìîëåêóë ÀÒÔ, à ïåðåíîñ ïàðè åëåêòðîí³â òà ïðîòîí³â çà àíàåðîáíèõ óìîâ ïðèâäèòü äî óòâîðåííÿ îäí ìîëåêóëè ÀÒÔ.         
2.Çàâäÿêè ïðîöåñàì îêèñëþâàëüíîãî ôîñôîðèëþâàííÿ åíåðã³ÿ îêèñíèõ ïðîöå-ñ³â òðàíñôîðìóºòüñÿ ó åíåðã³þ çâ‘ÿçê³â ìàêðîåðã³÷íèõ ñïîëóê.


4. Ìàêðîåðã³÷í³ çâ‘ÿçêè òà ìàêðîåðã³÷í³ ñïîëóêè

Ìàêðîåðã³÷íèìè – íàçèâàþòüñÿ ñïîëóêè â ñêëàä³ ìîëåêóë ÿêèõ áàãàò³ åíåð㳺þ (ìàêðîåðã³÷í³) çâ‘ÿçêè.

Çàëåæíî â³ä âåëè÷èíè åíåð㳿 ùî àêóìóëüîâàíà ó çâ‘ÿçêàõ, ìàêðîåðã³÷í³ ñïî-ëóêè ïîä³ëÿþòü íà íèçüêîåíåðãåòè÷í³ (2-3 êÊàë/çâ.) òà âèñîêîåíåðãåòè÷í³ (10-16 êÊàë /çâ.) ìàêðîåðãè.

Äî íèçüêîåíåðãåòè÷íèõ ìàêðîåðã³â â³äíîñÿòü ïðîñò³ åô³ðè ñïèðò³â òà ôîñ-ôîðíî¿ êèñëîòè, ôîñôîãë³öåðèíîâà êèñëîòà, ñêëàäí³ ìîíî- òà äèôîñôîðí³ åô³ðè ãëþêîçè, ðèáîçè, ôðóêòîçè.

Äî âèñîêîåíåðãåòè÷íèõ ìàêðîåðã³â â³äíîñÿòü ï³ðîôîñôàò, êðåàòèíôîñôàò, àöåòèëôîñôàò, 1,3-äèôîñôîãë³öåðèíîâó êèñëîòó , 2­ôîñôîåíîëï³ðîâèíîãðàäíó êèñëîòó.

Îñîáëèâå ì³ñöå ñåðåä ìàêðîåðã³â êë³òèíè íàëåæèòü ñèñòåì³ ÀÒÔ-ÀÄÔ, ùî ñêëàäàºòüñÿ ç àäåí³íó, ðèáîçè, òà òðüîõ ÷è äâîõ çàëèøê³â ôîñôîðíî¿ êèñëîòè â³äïîâ³äíî. Ìîëåêóëà ÀÒÔ ì³ñòèòü äâà ìàêðîåðã³÷í³ çâ‘ÿçêè, ìîëåêóëà ÀÄÔ – îäèí. Ïðè ðîçðèâ îäíîãî ìàêðîåðã³÷íîãî çâ‘ÿçêó âèä³ëÿºòüñÿ 6-8 êÊàë. åíåð㳿.

Öåíòðàëüíå ì³ñöå ñèñòåìè ÀÒÔ-ÀÄÔ ñåðåä ìàêðîåðã³â êë³òèíè çàáåçïå÷óºòüñÿ ñë³äóþ÷èìè ôàêòîðàìè: ïðîì³æíèìè çíà÷åííÿìè åíåð㳿 ìàêðîåðã³÷íîãî çâ‘ÿç-êó, âåëèêîþ ìîá³ëüí³ñòþ òà âèñîêîþ êîíöåíòðàö³ºþ ó êë³òèí³.

Ö³ ôàêòîðè äàþòü çìîãó ñèñòåì³ ÀÒÔ-ÀÄÔ áóòè îñíîâíèìè åíåðãåòè÷íèìè ñóáñòðàòàìè êë³òèíè.

         
HO              O

          P       OH                                 NH2                                                          

O                                              N               N                                                     

HO    P       O                                                                                           

          O                                             N      N                                                      

O=P                                                                                                                     

OH    O                                                                                    + H2O                 

           CH2 O                                                                                                      

                   H       H                          ÀÒÔ                                                          

           H                         H                                                                                   

           OH   OH                                                                                        

                                                           NH2                                                          

ÍO                                           N               N                                                     

HO    P       O                                                                                                     

          O                                             N      N                                                      

O=P                                                                                                                     

OH    O                                                                                    + H3ÐÎ4 + Å       

           CH2 O                                                                                                      

                   H       H                          ÀÄÔ                                                          

           H                         H                                                                                   

           OH   OH                               

                   ìàêðîåðã³÷íèé çâ‘ÿçîê.


Ïèòàííÿ äëÿ ñàìîêîíòðîëþ.

1. Îõàðàêòåðèçóéòå îñíîâí ïîëîæåííÿ ïåðåêèñíî¿ òåîð³¿.

2. Îõàðàêòåðèçóéòå îñíîâí³ ïîëîæåííÿ òåîð Ïàëàä³íà.

3. Ïîíÿòòÿ «äèõàëüíîãî» ëàíöþãà. ̳òîõîíäð îñíîâí³ åíåðãåòè÷í³ ñòàíö³¿ îðãàí³çìó.

4. Õàðàêòåðèñòèêà ÍÀÄ–çàëåæíèõ ôåðìåíòíèõ ñèñòåì.

5. Õàðàêòåðèñòèêà ÔÀÄ–çàëåæíèõ äåã³äðîãåíàçíèõ ñèñòåì.

6. Õàðàêòåðèñòèêà êîôåðìåíòó Q òà ñèñòåìè öèòîõðîì.

7. Ïîíÿòòÿ ïðîöåñ³â àåðîáíîãî òà àíàåðîáíîãî îêèñëåííÿ õàð÷îâèõ ñóáñòðàò³â, â³ëüíîãî òà ñïðÿæåíîãî îêèñëåííÿ. Åíåðãåòè÷í åôåêòè ïðîöåñ³â. Ïóíêòè ñïðÿæåííÿ.

8. Õàðàêòåðèñòèêà ìàêðîåðã³÷íèõ ñïîëóê êë³òèíè. ÀÒÔ – êëþ÷îâà ñïîëóêà åíåðãåòè÷íîãî îáì³íó.



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