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57 Cards in this Set
- Front
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Enzymes for eukaryotic replication |
> DNA polymerase delta =5'>3' =3'>5' exonuclease > DNase I = 3'>5' exonuclease > RNA primase > DNA ligase (fill in 'nicks') > HELICase |
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Enzymes for prokaryotic replication |
RMT > DNA polymerase 1 (rRNA) > DNA polymerase 2 (mRNA) > DNA polymerase 3 (tRNA) = 5'>3' =3'>5' exonuclease --- > helicase > DNase I Relieving supercoiling, stress > Topoisomerase IV (gram +) (- quinolone) > DNA gyrase (gram -) (quinolone, i.e: ciprofloxacin) |
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Types DNA repair mechanisms |
Nucleotide excision repair - Disorder causes xeroderma pigmentosum steps > DNA endonuclease > DNA polymerase, ligase Base excision repair > DNA glycosylase > DNA endonuclease > DNA polymerase, ligase DNA mistmatch repair - Disorder causes Hereditary nonpolyposis colorectal cancer (aka, Lynch syndrome) Mechanism: > Nonmethylated DNA strand (new strand) is worked on |
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DNA disorders |
Thymine dimer > cause: UV light Guanine adduct > cause: oxidation of Cigarette smokg Xeroderma pigmentosum > cause: faulty nucleotide excision repair Tx: dimericine Hereditary nonpolyposis colorectal cancer > cause: faulty DNA mismatch repair |
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Dimericine is used to treat what? a. thymine dimer b. xeroderm pigmentosum c. heredity nonpolyposis colorectal cancer d. guanine adduct e. ATM mutation f. ATR mutation |
b. xeroderma pigmentosum Expl: restores the nucleotide excision repair mechanism that is faulty. in xeroderm pigmentosum |
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A defect in nucleotide excision repair causes what? a. thymine dimer b. xeroderm pigmentosum c. heredity nonpolyposis colorectal cancer d. guanine adduct e. ATM mutation f. ATR mutation |
a. thymine dimer b xeroderm pigmentosum d. guanine adduct |
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A defect in mismatch repair causes which of the following? a. thymine dimer b. xeroderm pigmentosum c. heredity nonpolyposis colorectal cancer d. guanine adduct e. ATM mutation f. ATR mutation |
c. heredity nonpolyposis colorectal cancer Expl: A defect in mismatch repair causes hereditary nonpolyposis colorectal cancer. |
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Structure nucleic acid |
Nucleotide polymers Nucleoside = N2 base + sugar Nucletide = Nucleoside + PO43- = N2 base + sugar + PO43- 1' attached to base VIA N-glycosidic bond 2': -H in DNA, -OH in RNA (Makes it less stable) 3': attached to PO43- 5': attached to PO43- Phosphodieste rbackbone |
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Bases connect to sugar via what? |
N-glycosidic bond |
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Charge of DNA? |
- due to negative charges of pO43- in the sugar phosphate backbone |
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In DNA, th bond between deoxyribose sugar + phosphate best described by what? a. polar bond b. ionic bond c. hydrogen bond d. covalent bond e. van der waals bond |
d. covalent bond expl: phopshodiester bond is a type of covalent bond |
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Explain base pairing |
A (2 H2 bonds) T G (3 H2 bonds) C Charfaff's rules: [A] = [T], [G] = C] or purines = pyrimidines |
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Explain Tm |
(denaturation temp or temp at which 50% strands denatured) = 69.3 + 0.41 (%G-C) the %G-C is a WHOLE number, not a decimal |
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Impact of adding basic solution to nucleic acid. |
Only denature DNA Breaks apart RNA, b/c ribose has a 2-OH that deprotonates in basic solution and then breaks apart phopshodiester bond. |
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Explain nucleic acid hybridization |
Single strands with complementary sequences can hybridize, which can be useful in molecular biology research. |
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Explain stability between bases, why are bases stable in DNA? |
Pi stacking interactions between bases to increase stability |
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Match. 1. Dominates in living systems and is right handed 2. Dominates in DNA-RNA hybrids, "tightly wrapped." 3. Dehydration form of B-DNA 4. Line connecting phosphate zags a. A DNA b. B DNA c. Z DNA |
1. B-DNA 2. A-DNA 3. A-DNA 4. Z-DNA |
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Describe chromatin and its structure |
Chromatin = DNA + various proteins Beads on a string essentially: complexes of histone proteins + DNA = centrosomes |
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What are centrosomes? |
Complexes of histone proteins (2 H2A, 2 H2B, 2 H3, 2 H4) octamer w/ DNA |
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Describe histones -structure -types |
Structure: High % of lysine and arginine that are able to interact w/ - charged sugar phosphate DNA backbone ~ 140 base pairs of DNA wrap around each nucleosome core Types histones: 8 on nucleosome > 2 H2A, 2 H2B, 2 H3, 2 H4 H1 outside between the nucleosomes 'beads' |
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How many base pairs of DNA wrap around nucleosome core? |
About 140 base pairs. |
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What is a gene? |
Specific point on chromosome that has a functional role in producing a protein |
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Explain 5-Fluouracil (5-FU) - Use - Mechanism |
Normally uracil - methylation at C-5 --> 5-methyl uracil But the 5-FU inhibits thimidylate synthase, inhibiting production of deoxythymidine. Use: slowing progression of tumor cell |
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Azithromycin Ciprofloxazin |
Azithro > binds 50S ribosome subunit, preventing protien synthesis Ciprofloxacin > targets bacterial DNA gyrase |
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Explain retroviruses |
Have RNA genomes Use enzyme, reverse transcriptase (RT) tomake DNA copy to insert into nucleus of host cell |
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Compare ribosomes + rRNA vs pro + euks |
Pros/euks: ribosome = 30s, 50s, 70s / 40S, 60S, 80S rRNA = 3 types rRNA (23, 16, and 5S) / 4 types rRNA (28, 18, 5.8, 5S) |
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Explain proteins in prokaryotic replication |
Helicase > unwind helix to separate pairs Topoisomerase, gyrase > rleive supercoiling Single stranded binding protien > maintains single strands |
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Match. 1. What is involved in relieving supercoiling? 2. What unwinds helix to separate pairs before DNA replication? 3. What maintains single strands? 4. Ciprofloxacin inhibits what? a. DNA gyrase b. helicase c. single stranded binding protein (SSB) d. topoisomerase |
1. DNA gyrase, topoisomerase 2. DNA helicase 3. single stranded binding DNA 4. ciprofloxacin inhibits DNA gyrase |
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Describe DNA polymerase activity |
> Primes adds RNA primer = Reason: DNA polymerase needs 3' OH to function to perform nucleophilic attack > Works 5'>3' directions > Adds new deoxynucleotide triphosphate to 3' OH group of chain while releasing PPi, (very exergonic) > Primers then removed by RNase H and DNA polymerase I (via DNA polymerase 1's 5'>3' exonuclease activity) > Okazaki fragments ( on lagging strand) , joined by DNA ligase |
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1. How are primers removing during DNA synthesis? 2. Okazaki fragments are joined by what? a. DNA polymerase I b. DNA P 2 c. DNA P 3 d. DNA gyrase e. RNase H f. DNA ligase g. DNA polmyerase delta |
1. a. DNA polymerase 1 > via its 5'>3' exonuclease activity e. RNase H 2. f. DNA ligase |
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Proofreading activities in DNA replication |
DNA p 1 > 5'>3' exonuclease (removes RNA priemrs) DNA p 3 > 3'>5' exonuclease activity |
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What stage of mitosis does synthesis occur? |
S phase |
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Difference between eukaryotic and prokaryotic DNA replication? |
Prokaryotic: > DNA Pol I, Replication, 5'>3' exonuclease activity > Polmyerase III, Major replicative polymerase, 3'>5' exonuclease for proofreading Eukaryotes > DNA polymerase delta: replication, 3'>5' exonuclease for proofreading Diferences: (pro/euk) > single origin of replication/multiple origins of replication > similar enzymes (topoisomerase, primes, polymerase, leading lagging strands) > circular/linear |
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1. What is the major replicative enzyme in prokaryotes? 2. What is the replicative enzyme in prokaryotes with 5'>3' exonuclease activity? 3. What is the replicative enzyme in eukaryotes with 5'>3' exonuclease activity? 4. What is the replicative enzyme in prokaryotes with 3'>5' exonuclease activity? a. DNA polymerase I b. DNA P 2 c. DNA P 3 d. DNA gyrase e. RNase H f. DNA ligase g. DNA polmyerase delta |
1. DNA P 3 > Also has 3'>5' activity 2. DNA P 1 3. None (not listed on slide) 4. DNA P delta |
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What are the ends of chromosomes? |
Teloemres |
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What enzyme maintains telomere length during replication? |
Telomerase |
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Teloemrase > function > mechanism > properties |
Function: Enzyme that maintains telomere length during replication Mechanism: > RNA reverse transcriptase mechanism Properties: > Importance in senesence, aging, cancer > Telomerase may increase in some cancer cells |
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DNA methylation -result -mechanism -importance |
Result: > Inhibits gene expression Mechanism > Adenine, cytosine methylated Importance > gene regulation > epigenetics |
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1. DNA mutations due to cigarette smoking that are uncorrected is a result in a defect of which of the following? a. nucleotide excision repair b. DNA mismatch repair c. DNA base excisino repair d. ligation 2. The product resulting from the effects of cigarette smoking on DNA reacts with what to distort DNA helix? a. adenine b. guanine c. thymine d. cytosine e. uracil 3. What is a possible outcome of this? a. malignant melanoma b. brain cancer c. adenocarcinoma d. pyrmidine dimer e. hereditary nonpolyposis colorectal cancer |
1. a. nucleotide excision repair 2. b. guanine > guanine adduct is formed 3. c. adenocarcinoma |
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DNA mutations due to UV light exposure that are uncorrected can result in what? a. malignant melanoma b. brain cancer c. adenocarcinoma d. pyrmidine dimer e. hereditary nonpolyposis colorectal cancer |
a. malignant melanoma |
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DNA mutations due to UV light exposure result in what usually? a. malignant melanoma b. brain cancer c. adenocarcinoma d. pyrmidine dimer e. hereditary nonpolyposis colorectal cancer |
d. poyrimidine dimer > later can become malignant melanoma if not treated |
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A person has xeroderma pigmentosum. This is due to a defect in which of the following? a. nucleotide excision repair b. DNA mismatch repair c. DNA base excisino repair d. ligation |
a. nucleotide excisionr epair |
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A person has a defect in DNA mismatch repair mechanisms. This leads to what? a. malignant melanoma b. brain cancer c. adenocarcinoma d. pyrmidine dimer e. hereditary nonpolyposis colorectal cancer |
e. hereditary nonpolyposis colorectal cancer |
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Matching 1. Which occurs during meiosis? 2. What are 'jumping genes' or mobile genetic elements? 3. Which results in Holliday structure formation? 4. What is caused by breaks in 2 non homologous chromosomes? 5. What uses reverse transcription to make a DNA copy? a. homologous recombination b. retroviral insertion c. translocation d. transposons |
1. a. homologous recombination 2. d. transposons 3. a. homologous recombination 4. translocation 5. retroviral insertion |
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Deaminations and significance |
Deamination of cytosine to uracil can be repaired Deamination of 5-methylcytosine cannot be repaired. |
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1. Defect in one of several mismatch repair genes result in what? 2. Which of the following may progress to malignant melanomas? 3. Which of the following can cause an adenocarcinoma? 4. Defect in nucleotide excision repair results in what? 5. Dimericine is a treatment for what? a. xeroderma pigmentosum b. hereditary nonpolyposis colorectal cancer c. thymine dimer d. guanine adduct |
1. hereditary nonpolyposis colorectal cancer 2. thymine dimers 3. guanine adduct 4. xeroderma pigmentosum, thymine dimer, guanine adduct 5. Xeroderma pigmentosum |
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How is RNA polymerase difference from DNA polymerase? |
No 3'>5' exonuclease (as in DNA polymerase 1), but both 5'>3' ok |
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1. What produces tRNA? 2. What produces mRNA? 3. What produces rRNA? 4. What produces 5S rRNA? 5. Alpha amantin inhibits what? a. RNA polymerase I b. RNA polymerase II c. RNA polymerase III |
rmt/5S rRNA 1. RNA P 3 2. RNA P 2 3. RNA P 1 4. RNA P 3 5. RNA P 3 |
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Pro vs. euk transcription |
Pro/euk: type mRNA: polycystronic mRNA (1 mRNA many proteins)/ monocystronic mRNA (1 mRNA, 1 protein) starting factors: sigma factor/TFIID connections: transcription, translation together/translation, transcription separated location transcription: cytosol/nucleus |
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1. Redunancy and degeneracy occurs during what? a. DNA replication b. transcription c. translation |
1. translation |
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1. Putting in the wrong AA is an example of what type of mutation? 2. Prematurely ending translation is due to what type of mutation? 3. Cystic fibrosis is an example of what type of mutation? 4. Progeria is due to what type of mutation? 5. Tay sachs disease is due to what type of mutation? a. silent b. missence c. nonsense d. insertion e. deletion |
1. missence 2. nonsense 3. missence mutation 4. Silent mutation 5. Insertion (Biochem dan , slide 171) |
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What is the first AA in translation? |
Euks: methionine proks: formyl-methionine |
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1. Transcription of what gene results in the breakdown of lactose into glucose and galactose 2. Transcription of what gene results in transport of of lactose into cells? 3. Transcription of what gene result in acetylation of B-galactosides? a. lac a gene b. lac x gene c. lac y gene d. lac z gene |
1. lac z gene (B-galactosidase) 2. lac y gene (lac permease) 2. lac a gene (transacetylase) |
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Explain positive and negative control of lac operon? |
Negative: lactose present, lactose binds to receptor and allows RNA polymerase to bind Positive: no glucose, CAP (bound activator) protein binds to activator site for binding protein Net result: transcription of lac operon with no glucose and lactose present |
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Explain tryptophan and attenuation |
Without tryptophan Result: Transcription occurs of tryptophan receptor Mechanism: segments 2, 3 bind allowing slower transcription With tryptophan (attenuation results) Result: Transcription of tryptophan receptor doe snot occur Mechanism: segments 3, 4 bind allowing faster transcription and thus limited production tryptophan receptors |
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Heterochromatin to eucharomatin |
Acetylation allows activation of DNA (heterochromatin to euchromatin) |
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DNA methyulation experiments on RAT |
DNA methyulation: gene deactivation > high rat anxiety > no care of young No DNA methyulation: gene activation remains > low rat anxiety > grooming + care of young |