Exam 2 Flashcards

Question

Electron transport chain

Answer 1

-electron transport through chain of protein complexes generates a proton (H+) gradient across inner mitochondrial membrane -aka oxidative phosphorylation -High [H+] (low pH) in intermembrane space, low [H+] (high pH) in matrix -Energy from proton gradient is used to generate ATP using an enzyme complex called ATP synthase -Protons pumped into complexes I, III, and IV generate high [H+] in the intermembrane space

Answer 2

-Complex I: NADH donates electrons here (Protons pumped into intermembrane space) -Complex II: FADH2 donates electrons here (No protons pumped) -Complex III: protons pumped into intermembrane space -Complex IV: protons pumped into intermembrane space (Reduction of O2 to H2O (oxygen is final electron acceptor)) -ATP synthase: protons move from intermembrane space to drive ATP production

Answer 3

-Electrons are transferred from high to low free energy -NADH is best electron donor (high free energy) -O2 is best electron acceptor (low free energy)

Answer 4

-proton gradient generated by ETC provides energy needed to make ATP -Protons move down gradient, from intermembrane space into matrix --Movement of H+ ions across inner mitochondrial membrane down the gradient -Drives ATP production --Explained how ETC provided energy that could be used to make ATP --Made connection between making high energy NADH and FADH2 molecules and converting that energy into ATP -Proposed by Mitchell in 1961

Answer 5

-molecular machine, enzyme that catalyzes the formation of ATP using ADP and Pi by using energy from proton gradient -Subunits group to form two main components, F0 and F1

Answer 6

-Gamma y (1) -Alpha a (3) -Beta B (3) -C (10-14) -A (1) -B2 (1)

Answer 7

-spans membrane to form a rotating proton channel -Protons (H+) cause F0 to rotate -Subunit a contains half-channels in which protons enter and exit through, causing c subunits to rotate clockwise -Subunit c has aspartic acid side chain (negatively charged when not protonated, neutral when protonated, allowing it to better interact with membrane)

Answer 8

-includes an enzyme “catalytic wheel” in the mitochondrial matrix, catalyzes ATP synthesis from ADP and Pi -Rotation of the c ring rotates the gamma y subunit -Irregular shape of y subunit forces conformational changes of the beta (B) subunits -Open (O), Loose (L), and Tight (T) --Open: contains ADP + Pi --Loose: contains ADP + Pi --Tight: contains transfer of ADP to ATP -Chemical energy of proton gradient is converted to mechanical energy of rotating c ring, which is converted to chemical energy in ATP -Every 360 rotation of y results in 3 ATP --Depends on number of c subunits (# c subunits / 3 is number of H+/ATP)

Answer 9

-reduction of carbon (CO2) to make sugars and oxidation of carbon (sugar) to obtain energy, electrons to do work -6CO2 + 12H2O + light energy → C6H12O6 + H2O + 6O2 -Carbon dioxide = oxidized (4 C-O, 0 C-H bond per C) -Glucose= reduced (1 C-O, 1 C-H bond per C) -Takes place in two steps: --Light-dependent reactions --Light independent reactions

Answer 10

-where photosynthesis takes place -Has 3 membranes (outer membrane, inner membrane, and thylakoid membrane)

Answer 11

membrane-bound compartments within chloroplasts and cyanobacteria that are crucial for photosynthesis

Answer 12

-Uses light energy (photons to excite electrons (from H2O)) -Electrons travel through a series of electron carriers and pump protons (H+) into thylakoid lumen to generate a proton motive force (PMF) -H+ pass through ATP synthase to produce ATP --Electrons passed through ETC reduced NADP+ to produced NADPH --ATP and NADPH used to fix CO2 to sugar

Answer 13

-electron transfer and ATP synthesis -Takes place in thylakoid membrane -Light energy is captured by chlorophyll in the photosystems -After absorbing light energy, a photosystem’s chlorophyll reaction center is oxidized (loses electron) -electrons travel from PSII to PSI, and transfer of electrons creates proton gradient to power ATP synthase -Products: ATP, NADPH, and O2

Answer 14

-Electron hole in PSII is filled by H2O (water is oxidized, produces O2) -The electron is raised to an excited state (possesses more energy) after absorption of photon and is highly unstable -Electron and H+ from H2O are transferred to electron acceptor (electron transport system), travels down electron transport chain (loses energy) --Transferred to Pq then to Cyt then Pc --Some released energy drives pumping of H+ ions and builds gradient, driving ATP production (stroma has low H+ concentration, lumen has high H+) -Gets electrons from water

Answer 15

-electron from PSII arrives at PSI and joins at pair of chlorophylls in reaction center -After light energy is absorbed by pigments and passed to reaction center, electron is boosted to very high energy level and transferred to acceptor molecule -High energy electron travels down second leg of electron transport chain (Fd) and is passed to NADP+ (along with second electron from same pathway) to make NADPH --Electron is passed to Fd then to NADP+ reductase -Gets electrons from flow down electron transport chain from PSII

Answer 16

-uses proton gradient produced from transfer of electron through PSII and PSI to create ATP -3 ATP are produced for every 2 water molecules used -1 NADPH is produced per water molecule

Answer 17

-light independent reaction of photosynthesis, chemical energy generated during light reactions used to reduce CO2 into sugars -Occurs in the stroma -3 stages: --Carboxylation --Reduction --Regeneration

Answer 18

-first step of calvin cycle -carbon fixation, CO2 combines with 5C intermediate (RuBP) -CO2 + RuBP produces unstable 6C intermediate -Splits into 2 x 3C molecules (3PGA)

Answer 19

-catalyzes the first step of Calvin Cycle -Most abundant enzyme on earth -30-50% of total leaf protein

Answer 20

-second step of calvin cycle -3PGA is reduced to Glyceraldehyde-3P -NADPH is used to reduce 3-PGA to make glyceraldehyde-3P (G3P/GAP) --G3P: intermediate in pathway and intermediate in glycolysis -ATP is used for coupling reactions and to drive unfavorable reactions in synthesizing glucose and regenerating RuBP

Answer 21

-third step of Calvin Cycle -Takes 6CO2 to form 1 glucose (takes 6 turns of calvin cycle to form 1 6C sugar) -30 C stay in cycle

Answer 22

-sugars, the products of photosynthesis, are transported throughout the entire plant -Used as energy source and as carbon building blocks in roots, leaves, flowers, fruits, seeds, and storage in starch -Excess sugars are stored in starch during the day --Broken down at night into glucose for metabolism --Glucose oxidation (glycolysis, CAC → ETC, ATP generation)

Answer 23

-a pore found in the epidermis of leaves -facilitate gas exchange -CO2 is reduced in leaves to form sugars

Answer 24

-During dry, hot days, stomata close to prevent excess water loss --This limits gas exchange and O2 accumulates in leaf --Relative concentration of O2 to CO2 goes up in leaf interior -Rubisco can react with either CO2 or O2 --Rubisco can function as a carboxylase or an oxygenase --Rubisco reacts with O2 instead of CO2 in reaction called photorespiration ---Results in less CO2 ending up as sugar

Answer 25

-alternative carbon fixation pathway used by many plants that grow well in hot climates to reduce photorespiration -Prevents exposure of Rubisco to high [O2] -C4 plants include corn and sugar cane -Have different leaf structure that separates CO2 capture and CO2 reduction in different cells --Concentrates CO2 and delivers it to Rubisco -Rubisco is in bundle sheathes (separate from where light reaction w/ O2 occurs)

Answer 26

-do not have specialized leaf structure and do more photorespiration in hot, dry weather -Rubisco is in mesophyll cells (where light reaction occurs and O2 is produced) -Includes wheat, rice, 95% of plants

Answer 27

-metabolic pathways are regulated by availability of carbon resources, activity level, energy status of the cell -Often regulated at level of enzyme activity -Allosteric feedback regulation plays an important role in metabolic pathways --Accumulation of some products can inhibit their own synthesis or can stimulate other pathways that require the same raw materials

Answer 28

-a product of the pathway regulates enzyme activity earlier in pathway, or in a related pathway -can be positive feedback or negative feedback

Answer 29

-the initial stimulus is amplified, leading to an increased response -when a compound G reduces its own synthesis by activating enzyme catalyzing conversion of substrate into alternate compound, pullign substrate away from making G

Answer 30

-a stimulus triggers a response that counteracts the original stimulus -compound G inhibits its own synthesis by inhibiting the enzyme catalyzing conversion of substrate into compound G producing enzyme

Answer 31

-Several calvin cycle enzymes are regulated by oxidation/reduction -If there is plenty of light → electron transport, there is excess reduced Fd to activate these enzymes --Ferredoxin (Fd) part of ETC in light reactions ---Reduced Fd activates enzymes in Calvin Cycle --Light reactions stimulate Calvin Cycle ---By reducing (breaking) disulfide bridges, electrons from the light reactions activate Calvin cycle enzymes

Answer 32

-PFK (phosphofructiokinase) inhibited by high ATP -Highly regulated -Reaction is irreversible under cellular conditions -ATP is both substrate and an allosteric regulator --Under low ATP concentrations, ATP only binds to the active site → phosphorylation --Under high ATP concentrations, ATP also binds to inhibitor site, inhibits phosphorylation -AMP and ADP can increase PFK activity -High levels of citrate: inhibit PFK

Answer 33

-convert R (non-virulent) → S (virulent) Streptococcus pneumoniae -When S strain was given to mouse, mouse died -When R strain was given to mouse, mouse lived -When heat-killed S strain was given to mouse, it live -When R Strain and heat-killed S strain was given to mouse, mous died --Something from heat-killed S strain caused R strain to become virulent

Answer 34

-designed experiment to find out why heat-killed S strain caused virulent R strain (Griffith experiment) -Prepared a whole cell extract of killed S strain bacteria -Destroyed each known chemical type one by one, mixed with R strain bacteria, and tested in mice for lethality -When DNA was destroyed, the mouse lived

Answer 35

-chemical unit (monomer) that is repeated in the DNA polymer -Double stranded DNA is composed of two strands (or chains) of nucleotide subuntis -Each nucleotide has three components: --Sugar --Nitrogenous base --Phosphate

Answer 36

-Deoxyribose: found in DNA, has -H at 2'C -Ribose: found in RNA, has -OH at 2'C

Answer 37

-Flat molecules with 1 or 2 rings -Carbon and nitrogen ring backbone -5 different bases gound in nucleic acids (4 in DNA, 5 in RNA) -2 kinds of nitrogenous bases: --pyrimidines --purines -Base is always attached to C1’ of the sugar by B-glycosidic linkage

Answer 38

-2 carbon rings -includes adenine (A) and guanine (G) -Guanine has O, Adenine has NH2

Answer 39

-1 ring -includes cytosine (C), uracil (U), and thymine (T) -Thymine has CH3, cytosine has NH2

Answer 40

-formed by 3’-5’ phosphodiester bonds -Polarity: 3’(bottom) → 5’ (top) --3’ and 5; based on carbon numbering of ribose sugars ---oxygen in ribose sugar points down when 3’ is at top, points up when 5’ is at top ---Phosphate group is at 5’ end, -OH group is at 3’ end -Negative charges on the phosphates -Two DNA strands are put in anti-parallel fashion (opposite directions) --Two strands are held together by H-bonding between bases -Hydrophilic, interacts with water one outside of ladder --Bases inside molecule are hydrophobic --Twisting of ladder eliminates space between bases

Answer 41

-are the same width and form the steps of the DNA ladder -Each Watson-Crick base-pair contains one purine and one pyrimidine -Chargraff rules: #A = #T, #C = #G -Guanine and Cytosine have 3 H-bonds, Adenine and Thymine have 2 H-bonds

Answer 42

-Prokaryotic cells often contain one or fewer circular DNA molecules --e.g. E. coli contain a single circular chromosome of 4.6 million bp -Most eukaryotic cells contain several linear chromosomes

Answer 43

-Information content must be maintained and passed on to progeny -Faithful replication = high fidelity (faithfulness) --New copy must have essentially no mistakes -Rapidly, within one cycle -Double helix suggests a mechanism that could facilitate this --Watson-Crick base-pairing leads to a “complementary” sequence of bases on each strand --Polarity is based on the carbons of the ribose sugars

Answer 44

-both replicated DNA helixes has one strand of original DNA and one strand of new DNA -accurate DNA structure -showed as one DNA line, in middle of tube (between high and low) in meselson-stahl experiment after first round, middle and high after second

Answer 45

-one replicated DNA helix is completely original DNA, one replicated helix is completely new DNA -showed as low and high in meselson-stahl

Answer 46

-each replicated DNA helix have mix of both original and new on each strand -showed as middle in both rounds of meselson-stahl experiment

Answer 47

-based on density labeling of DNA -Grow a bacterial culture (E. coli) for many cell generations in a liquid medium containing “heavy” isotope of nitrogen (N15) -Collect sample of culture at time = 0 -Transfer the rest of the culture to a new liquid medium containing the common “light” isotope of nitrogen (N14) -At various times after transfer, remove sample of bacteria --Purify DNA from samples and measure density -Density of DNA was measured by equilibrium gradient centrifugation --Low density = low height, vice versa

Answer 48

-proved that semiconservative model was accurate -one DNA line, in middle of tube (between high and low) after first round -two DNA lines (one middle, one high) after second round -matched with results of semiconservative

Answer 49

-class of enzymes that catalyzes the polymerization of deoxyribonucleotide triphosphates (dNTPs) -Requires a single-stranded DNA template -Order of nucleotides determined by the template and Watson-Crick base pairing --Newly synthesized strand must be antiparallel to template strand -Polymerization proceeds in 5’ → 3’ direction --Template strand is read in 3’ → 5’ direction -Requires a primer to donate a 3’ -OH group

Answer 50

-separates DNA strands, allowing DNA polymerase access to single-stranded templates -Requires ATP hydrolysis

Answer 51

binds ssDNA after helicase, holds strands apart

Answer 52

-synthesizes a short (~10 nucleotides) RNA primer -Provides a 3’-OH group for DNA polymerase to begin synthesis

Answer 53

-a Y-shaped structure that forms during DNA replication when the parental DNA double helix separates into two single strands

Answer 54

strand is replicated continuously, follows replication fork, requires only one RNA primer, replicated by one DNA polymerase

Answer 55

-synthesized discontinuously in Okazaki fragments -synthesized in opposite direction of fork (but still in 5’ → 3’ direction) -requires multiple RNA primers -Okazaki fragments are joined by ligase -replicated by more than one DNA polymerase

Answer 56

-fills in the gap (5’ → 3’ polymerase) and digests RNA primer (5’ → 3’ exonuclease) -leaves a nick (break in sugar-phosphate backbone) -DNA ligase seals nick -Also does proof-reading

Answer 57

-synthesizes DNA of both template and lagging strand -Also does proof reading using 3’ to 5’ exonuclease activity (incorrect base is detected and removed)

Answer 58

creates a phosphodiester bond between adjacent 5’-phosphate and 3’-OH

Answer 59

-E. coli chromosome is a circular double-stranded DNA molecule -DNA replication starts at a specific site (oriC) -A bubble forms at oriC with two replication forks -DNA replication will proceed in both directions from origin of replication

Answer 60

enzymes that relieve torsional stress in front of fork and regulate DNA topology by temporarily breaking and rejoining DNA strands

Answer 61

-small, circular “Extrachromosomal” DNA molecules that replicate independently of the bacterial host chromosome also have fixed origins of replication -These plasmid ori sequences are different from oriC and are characteristic for each plasmid type

Answer 62

-Prokaryotic: small genomes (viral, plasmid, chromosomal) with a single origin of replication -Eukaryotic: extremely large nuclear genomes with multiple replication origins

Answer 63

-no way to prime DNA synthesis at the very 5’ end of the lagging strand. Predicts that linear chromosomal DNA will become shorter with each round of DNA replication -Does not apply to prokaryotic DNA (circular, no end)

Answer 64

-protective caps located at the ends of chromosomes, composed of variable numbers of 5’-TTAGGG-3’ repeats -telomere length is determined by a balance between mechanisms: -Telomerase-mediated telomere repeat addition -Loss of telomere repeats due to the end-replication problem

Answer 65

-a ribonucleoprotein complex that polymerizes a single-stranded 3’ DNA extension using an internal RNA template -Reverse transcriptase (RNA-directed DNA polymerase, converts RNA to DNA): synthesizes DNA in 5’--> 3’ direction starting from 3’-OH of leading strand -Short length of telomere repeat reflects the limited coding capacity of the short telomerase RNA

Answer 66

any alteration in the genetic material that can be inherited

Answer 67

alteration in the nucleotide sequence of a gene

Answer 68

alteration of chromosome structure or chromosome number

Answer 69

-an organism or cell carrying one or more mutations -Mutations are rare

Answer 70

-DNA replication is a high fidelity process (error frequency of the 5’--> 3’ polymerase activity of a typical DNA polymerase is ~10^-5 -Many DNA polymerases contain proof-reading 3’-->5’ exonuclease activity that reduces the error frequency an additional 100-1000 fold by removing wrongly incorporated nucleotides -Post-replicative DNA repair of errors in replication and damaged DNA

Answer 71

-Spontaneous mutations (examples: errors during replication) -Mutations from DNA damage (examples: chemicals like base analogs, intercalating agents, UV light, X-rays) --Ionizing radiation: causes single and double strand DNA breaks --UV radiation: results in pyrimidine dimers --Chemicals: results in damaged bases or cross-links

Answer 72

the insertion or deletion of a small number of base pairs due to slippage during replication

Answer 73

-coupled to replication -MutS protein recognizes the mismatch or small deletion and binds -recruits MutL and MutH to form complex -an exonuclease excises a section of the new DNA strand including mismatch -DNA pol III and ligase repair the gap and produce correct base pair

Answer 74

-1. Direct reversal of DNA damage (e.g. in E.Coli, photo-reactivated repair of pyrimidine dimers caused by UV light) -2. Excision repair pathways (nucleotide excision repair via recognition, removal, and replacement) --Exclusion nuclease recognizes bulky distortion of helix, removes fragment containing damage via DNA helicase, DNA pol replaces it with new synthesis of DNA -3. Error-prone repair pathways (damage is not removed but copied) -4. Strand-break repair (double-strand break repair)

Answer 75

-has two versions: -1. Direct rejoining of broken ends (non-homologous end-joining (NHEJ)), confined mostly to eukaryotic cells --Single strands are cut by exonuclease, DNA strands are brought together and strands are filled in joined by ligation, double helix is reconstructed --Imprecise, as some nucleotides are lost and several base pairs in original wild-type sequence are missing (results in mutant sequence) -2. Copying of “broken” sequence from “sister chromatid” in a cell that has a “good” copy of the “broken” sequence following DNA replication (homologous recombination) --DNA is broken by exonuclease, base-pairing occurs with unwound sister chromatids, strand is extended, sister chromatids are disengaged and paired, gaps are restored

Answer 76

-reiterative DNA synthesis in vitro using specific oligonucleotide primer pairs -Exponential amplification of a specific genomic DNA fragment --Number of double-stranded DNA segments = 2^n (n = # of cycles) -Components: template DNA, DNA polymerase, dNTPs, Buffer, pair of single-stranded oligonucleotide primers (hybridize to a specific genomic position, oriented in 5’ → 3’ direction at 3’ ends of template DNA) -Starts with 1 molecule of double-stranded DNA (template DNA) -Allows isolation and amplification of specific DNA fragments from a large complex genome

Answer 77

-Reaction is run for many cycles, and each cycle has three steps: -1. Denature --Strands are separated using heat -2. Anneal --Primers combine to DNA at 3’ end by cooling -3. Polymerization --Taq pol adds dNTP’s from 3’-OH of primer, polymerizes DNA in 5’-->3’ direction

Answer 78

-determination of linear order of nucleotides -Different techniques based on DNA replication: --Sanger sequencing and next-generation sequencing

Answer 79

-based on DNA polymerization and chain termination, in vitro DNA synthesis on a DNA template from a fixed site, which is determined by position of a single oligonucleotide primer -DNA synthesis reactions “poisoned” by a chain terminating “dideoxy” nucleotides (eg ddATP), each associated with a different fluorescent dye -Identical to PCR, except ddCTP is also present which terminates chain at certain nucleotide -Leads to separation of molecules based on size via gel electrophoresis -Each ddNTP has its own fluorescent label

Answer 80

-2’,3’ dideoxyribonucleotide -No 2’-OH on sugar, no 3’-OH on sugar, cannot form phosphodiester bond due to lack of 3’-OH substrate for DNA polymerase -Once this base has been added by DNA polymerase, no other nucleotides can be added to chain

Answer 81

-sequencing the DNA recovered directly from (a mixed) environmental sample -used to study microbial community composition -Two approaches (targeted and shotgun)

Answer 82

sequence single genetic locus (eg 16S rRNA gene) within community DNA

Answer 83

-sequence community DNA in unbiased manner (eg whole genomes) -Allows analysis of various functions in community

Answer 84

-acts as a ‘molecular fingerprint’ that can be used to identify different bacteria -Ribosomes in prokaryotic and eukaryotic cells are similar, but not identical -Eukaryotic ribosomes do not have a 16S rRNA molecule -Presence of both conserved and variable regions is critical

Exam 2 Flashcards

(108 cards)