NORGES TEKNISK-NATURVITENSKAPELIGE UNIVERSITET INSTITUTT FOR BIOTEKNOLOGI 1 Faglig kontakt under ekasmen: Institutt for bioteknologi, Gløshaugen Førsteamanuensis Sergey B. Zotchev, tlf. 98679. EKSAMEN I EMNE TBT4100 BIOKJEMI GRUNNKURS Tirsdag 14. desember 2006 Tid: kl. 09:00 14:00 Hjelpemidler: B1-Typegodkjent kalkulator (med tomt minne) i henhold til liste utarbeidet av NTNU tillatt. Ingen trykte eller håndskrevne hjelpemidler tillatt. NORSK VERSjON LØSNINGSFORSLAG Problem 3 a) Du har spist 5 glukose molekyler. Regn ut hvor mye ATP som vil bli produsert ved fullstendig oksydering av disse molekylene gjennom glykolyse, sitronsyresyklus og oksydativ fosforylering; Glycolysis: conversion of one glucose molecule to 2 pyruvate molecules gives 2 ATPs + 2 NADH molecules, which will donate their electrons to the oxidative phosphorylation pathway, producing 3 or 4.5 ATPs (depending on the shuttle system used to transport electrons from cytosole to the mitochondrial matrix); Pyrivate dehydrogenase complex converts 2 pyruvate molecules to 2 acetyl CoA molecules, producing 2 NADH molecules, which will donate electrons to the electron transport chain in oxidative phosphorylation, resulting in production of 5 ATPs; Citric acid cycle: 2 acetyl CoAs while oxidized in the cytric acid cycle will give 2 ATPs, 2 FADH 2, and 6 NADH molecules. The two latter molecules will yield 3 ATPs (for FADH 2 ) and 15 ATPs (for NADH) while donating their electrons to the oxidative phosphorylation; Therefore, total amount of ATP per molecule of glucose = 30 31.5, depending on the shuttle system used. 5 glucose molecules will give 150-157.5 ATPs.
b) Beskriv funksjon og hovedkomponentene av fotosyntese system II (PSII). Skriv netto reaksjon katalysert av PSII; 2 PSII oxidizes water molecule and donates energized electrons to electron carriers that transfer electrons to Photosystem I. Main components of PSII: - Light-harvesting complex II (LHCII) containing antenna pigments (chlorophylls + carotenoids); - Reaction center composed of D1/D2 dimer, 2 P680 chlorophyll a, and electron acceptors. Net reaction catalyzed by PSII: 2 Q + 2 H 2 O --> O 2 + 2 QH 2 (dependent on light). c) Hva er de to hovedfunksjonene av pentose fosfat sporet (pentose phosphate pathway, PPP)? Skriv (med strukturformler) netto reaksjon for PPP, og den første reaksjon I den oksydative delen av PPP; Major functions of the PPP are generation of NADPH and synthesis of five-carbon sugars (pentoses). Net reaction: + 2 NADP + + H 2 O - + 2 NADPH + 2H + + CO 2 First reaction on the oxidative phase of PPP:
3 d) Skriv ned (med strukturformler, navner til intermediater, enzymer og ko-faktorer) oksydering-hydrering-oksidering (oxidation-hydration-oxidation) trinnene i fettsyre nedbrytningen. Acyl CoA dehydrogenase Enoyl CoA hydratase L-3-hydroxyacyl CoA dehydrogenase Problem 4 a) Her er nukleotidsekvensen til et RNA molekyl: 5 - UUUGGACAACGUCCAGCGAUC - 3, som var transkribert fra følgende dobbeltrådige DNA molekyl: 5 - TTTGGACAACGTCCAGCGATC - 3 3 - AAACCTGTTGCAGGTCGCTAG - 5 Vis både kodende tråd og templat tråd på denne DNA molekyl;
4 Coding strand 5 - TTTGGACAACGTCCAGCGATC - 3 3 - AAACCTGTTGCAGGTCGCTAG - 5 Template strand b) Beskriv hovedfunksjonen av DNA polymerase og de 3 forskjellige enzymaktiviteter som dette enzymet har; DNA polymerase is a large enzyme complex responsible for the synthesis of new DNA strands during replication. It has three activities crucial for DNA replication: 5' to 3' elongation (polymerase activity), 3' to 5' exonuclease (proof-reading activity), 5' to 3' exonuclease (repair activity). c) Sammenlign prosessene for transkripsjon i prokaryoter og eukaryoter (gjerne med en tegning), og vis hovedforskjellene. Prøv å analysere både fordelene og ulempene i transkripsjonsmåter i disse organismetypene;
5 Prokaryotes: Advantages (i) transcription very fast and efficient, ribosomes usually start to synthesize proteins from mrna template almost immediately after the message has left the RNA polymerase-dna cpmplex; (ii) polycistronic mrna provides possibilities for expressing several genes for proteins involved in the same biochemical process simulteneously. Disadvantages - (i) transcription is regulated mostly by a limited number of proteins, fewer levels of regulation compared to eukaryotes; (ii) mrna is unstable and is rapidly degraded because of lack of modifications. Eukaryotes: Advantages (i) mrna is modified by capping and polyadenylation, which prevent degradation by RNAses; (ii) several proteins can be synthesized based on one original transcipt after alternative splicing; (iii) many more levels of regulation of gene expression due to a larger number of regulatory proteins involved and spacial separation of transcription and translation; Disadvantages considerably slower process compared to prokaryotes. d) Beskriv I detalj det terminerende trinn i proteinsyntesen. Navngi faktorer some er involvert og mekanismen bak frigjørelse av polypeptidkjeden fra ribosomet. Stop codons in th enormal cells are not recognized by any aminoacyl trnas. Instead, these stop codons are recognized by protein release factors (RFs). RF1 and RF2 recognize different stop codons, while RF3 mediates interaction between RF1 or RF2 and the ribosome. RF1 and RF2 resemble in their structure aminoacyl trna, but unlike the latter they bind a water molecule. RFs bring this water molecule into the peptidyl transferase center of the ribosome, promoting water molecule s attack on the ester linkage between the peptide chain and the last trna. This leads to hydrolysis, which releases the mature peptide chain from the ribosome.