Midterm 1 Flashcards
(92 cards)
1
Q
Genetics
A
Scientific study of genes and heredity of how certain qualities or traits are passed from parents to offspring as a result of changes in DNA sequence
2
Q
Molecular genetics
A
Scientific discipline concerned with the structure and function of genes at a molecular level and includes technique of genetic engineering, direct manipulation of an organism’s genome
3
Q
Gene/cistron
A
Basic unit of hereditary material
4
Q
Genome
A
All chromosomal genes in an organism
-only in the chromosome, not plasmid
5
Q
Locus
A
Physical location of a gene within a genome
6
Q
Allele
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Alternate form of a gene
7
Q
Genotype
A
allelic composition of an organism
8
Q
Phenotype
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Outcome of a given genotype
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Q
Haploid
A
One complete set of genes
10
Q
Diploid
A
2 complete sets of genes
11
Q
Merozygote (merodiploid)
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A partial diploid; has 2 copies of some genes
-gene can be in chromosome and in plasmid
12
Q
Wild type
A
Allelic form of a gene that is most prevalent in the “wild” population
-most common
13
Q
Mutation
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Genetic alteration (usually observed by a change in the phenotype)
14
Q
Mutant
A
An organism that had one or more mutations; terms may also be applied to genes containing mutations and proteins encoded by mutant genes
15
Q
Reverse mutation “reversion”
A
A second mutation that restores a mutant cell to the wild type phenotype
16
Q
Nomenclature of genes and proteins
A
-gene->lacZ (italicized)
-wild type-> lacZ^+
-mutant-> lacZ^-
-drug resistance-> amp^r, amp^s
-r=resistance, s=susceptible
-protein->LacZ (not italicized, 1st letter capitalized)
17
Q
Methods of bacterial culture
A
1) suspension of bacterial cells
2)suspension spread on petri plate with agar gel
3)incubate for couple days
4)bacteria colonies will be visible, each a clone of the corresponding single cell)
18
Q
Differential media
A
Reveals the presence or absence of metabolites/changes in pH
19
Q
Selective media
A
-allow growth for specific microorganisms
-antibiotic resistant
-synthesis metabolites
20
Q
Classical characterization of genes
A
-based on gene expression (phenotype)
-utilizes mutations to ID functions of unknown genes
21
Q
Molecular characterization of genes
A
-gene defined as an “open reading frame” in a DNA sequence
- functions of unknown genes inferred from sequence similarity to known genes or biochemical characterization of gene products
22
Q
Prototrophs
A
wild type strains that can synthesize all essential nutrient
-grow on minimal media and rich media
23
Q
auxotrophs
A
-mutations in metabolic pathways lead to autotrophic mutations
-they are unable to synthesize essential nutrients
-require rich or supplemented minimal media
24
Q
What is the difference between minimal and rich media?
A
-minimal:NH4+, PO4-, SO4-, Na+, K+, Cl-, Mg2+, Cu2+, trace minerals
-rich: proteolytic digest of casein (amino acids), yeast extract (vitamins, growth factors, nucleic acids)
25
Identifying autotrophic mutations in neurospora
1) gametes are exposed to x-rays that damages molecule in random places
2) produces spores where the specific genotype that either is or isn’t damaged from x-rays
3) spores are grown in tubes in rich media
4) spores are transferred to minimal media, if the gene is damaged, then it can’t survive in the minimal medium
5) mutant genes are transferred to minimal media with amino acids, vitamins to figure out where the gene was defective
26
identifying auxotrophic mutants in bacteria
1)plate cells on rich media
2)replica plate to minimal media
3)look for colonies that only grow on rich media (auxotrophic mutants)
4) plate mutants on minimal media + individual supplements to identify the mutant gene
-mutant is missing on replica plate and present on the original
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conditional mutations
mutation that has the wild-type phenotype under certain (permissive) environmental conditions and a mutant phenotype under other (restrictive conditions)
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locus 1 in determining T2 plaque morphology
-h+ allele (wild type) = B
-h allele = B and B/2
-strain B = permissive bc it allows for plaque formation by h and h+
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locus 2 in determining T2 plaque morphology
r+ (wild type) = slow lysis
r = rapid lysis
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plaque morphology with h and r
h+ = turbid
h = clear
r+ = small
r = large
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Cis-trans test
measures the relative distance btwn mutations and determines if 2 mutations are located in the same or different genes
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Cis - test
recombination:
-exchange of DNA
-btwn 2 molecules
-within the same molecule
homology:
-50 base pairs (BP)
-95% the same
-recomb is inversely proportional to distance
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what does a cis test tell you?
-distance btwn mutations
-relative mutation order
-DOESN'T tell you if the mutations are in the same or different genes
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Recombination frequency of T2 system
(# of recombinants (mix plate of B and B/2))/(total plaques (mix plate of B and B/2))
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Recombination frequency of T4 system
(# of recombinants (non permissive K12))/ (total plaques(permissive B))x2
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trans test
1)2 mutations will complement each other only if each is located in a different gene
2) the trans test tells you if 2 mutations are located in the same or different genes
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how do you look for complementation btwn RII mutants
1)coinfect non-permissive K12 strain, grown in liquid media, with 2 independently derrived rII mutants. incubate mixture for sufficient time to allow phage to attach to the surface of the cells
2)wash cells and incubate in new media for duration of phage life cycle
3) centrifuge sampe to pellet unlysed cells. Add the liquid phase to E. coli B and spread on solid media
4)presence of plaques indicates complementation btwn 2 phage mutants
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when are plaques formed?
complement protein, different proteins
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what happens when no plaques form?
the products come from the same gene, not complementary
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complementation rII region of T4
-bacterium is infected with 2 different phage genomes (rIIA and rIIB)
-rIIA: nonfunctional A product and functional B product
-rIIB: functional A product and nonfunctional B product
-plaques are formed
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no complementation of rII region of T4
-bacterium is infected with 2 phage genomes, with a different mutation on the same gene, rIIA
-both mutants produce a functional B product, neither make a functional A product
-no plaques are formed
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dominant mutations
-mutant alleles that suppress the phenotype of the wild type allele
-most often found in gene for proteins that are associated with a multi-subunit complexes
-a dominant and recessive mutation will appear to be in the same gene even when located on DIFFERENT genes
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alpha complementation
-proteins with domains that function independently
-mutations in different domains of a gene that exhibit a complementation may complement each other and appear to be on a different gene even though they are on the SAME gene
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dominant vs recessive mutations
-phenotype of a dominant mutation will be visible in the presence of either a wild type or recessive mutation when the latter are present in a 2nd copy of the same gene
-phenotype of a recessive mutation will not be visible if a 2nd gene copy is present in the form of a wild type or dominant mutant allele
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griffith and bacterial transformation
s-strain (encapsulated): lethal to mice and higher virulence in humans
-shiny, smooth colonies
r-strain: nonpathogenic
-rough colonies
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IIR vs IIIS
IIR=mice lived
IIIS=mice died (IIIS bacteria was recovered)
IIIS + heat=mice lived (heat killed IIIS bacteria)
-bacteria needs to be alive and have polysaccharide coat to be virulent and kill host
IIR + IIIS + heat=mice died
-IIIS bacteria was recovered
-IIR had been converted into virulent IIIS by interacting with dead IIIS bacteria
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Avery, Macleod, and McCarty (1944)
1)heat kill S-strain
2)detergent lyse bacteria and separate cytosol from cellular debris
3)treat cell extract with enzymes (RNase and DNase) to degrade proteins, lipids and carbs
RNase: only DNA remains
DNase: only RNA remains
4)mix enzyme treated cytosol with R-strain
Results:
-IIIS transformants were produced on RNase
-no IIIS transformants were produced on DNase
-DNA is the transforming principle, not protein
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Hershey-Chase Experiment
1)label phage with 35S or 32 P
2)mix phage with E. coli
3)separate phage from bacteria with a Waring blender
4)pellet bacteria in centrifuge
5)determine how much radioactivity associated with each isotope is in pellet vs supernatant
results:
-35S: majority of radioactivity was in the supernatant
-32P: majority of radioactivity was in the pellet
-DNA is genetic material in T2 phages
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underblending (protein)
separation of phage and bacteria isn't gonna happen
-radioactivity will appear in pellet when its supposed to be in supernatant
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overblending (DNA)
breakdown of DNA that was in the bacteria
-radioactivity will appear in supernatant when its supposed to be in the pellet
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post-translational phosphorylation (DNA)
-add a radioactive phosphate group after translation
-why radioactivity might appear in supernatant and the pellet
-unlabeled protein in supernatant of DNA part of experiment becomes radioactively labeled
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liquid trapping (protein)
-pulling bacteria by gravity/centrifugal force
-traps droplets of liquid btwn bacteria
-as you pull it down, you can bring some of the supernatant into the pellet
-more radioactivity in the pellet
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how do deoxyribose and ribose sugars differ?
absence of hydroxyl group at 2' position on deoxyribose
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what are the purines?
adenine and guanine
-pure As Gold
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what are the pyrimidines?
cytosine, uracil, thymine
-CUTie pie
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Chargaff's rules
A=T
G=C
purines=pyrimidines
50% purines
50% pyrimidines
(A+T)/(C+G)=1
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Watson and Crick
1)right-handed double helix
2)antiparallel
3)backbone is on outside of double helix
4)nucleotide bases are bound together by H-bonds
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Rosalind Franklin
-DNA models are helical
-each molecule contains 2 repeated structures, one with 34 A (3.4 nm) repeat (360 turn) and the other with 3.4 (0.34nm) repeat (space in between rungs)
-the most x-ray dense parts of the molecule (bases) are stacked perpendicular to the long axis of the molecule)
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what did x-ray data suggest about pyrimidines and purines?
pyrimidine+pyrimidine=DNA is too thin
purine+purine=DNA too thick
purine+pyrimidine=thickness compatible with x-ray data
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how many H-bonds does A-T have?
2
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how many H-bonds does C-G have?
3
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what bond does origins of replication have the most of?
A-T because they are easier to break due to less H-bonds, more energy is needed to break G-C bonds
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A-DNA
-right handed
-short and wide
-dehydration state
-increase in salt changes shape of double helix
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B-DNA
-right handed
-long and thin
-most abundant
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Z-DNA
-left handed
-co-exist with B-DNA
-phosphodiester binds go in zig-zag cuts
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how does DNA exist in prokaryotes
relaxed and supercoiled
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L(linking #)=T+W
Twist (T)=# of helix turns
Writhe (W)=# supercoils
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positive supercoiling
-superhelical winding of a DNA molecule in a direction opposite to the helix twist
-positive supercoils are generated by DNA and RNA polymerase movement along a DNA molecule
-inhibits strand separation
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negative supercoiling
superhelical winding of a DNA molecule in the same direction as the helical twist
-promotes DNA strand separation
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who uses plectonemic supercoiling
prokaryotes
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who uses solenoidal supercoiling
eukaryotes
-has histones
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topoisomerase I
-cuts one strand and passes another segment of dsDNA molecule through the break
-alters linking number by 1
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Topoisomerase II
-cuts both strands and passes another segment of the dsDNA molecule through the break
-alters linking number by 2
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Meselson and Stahl
1)grow E. coli for many generations in 15N media
2)transfer to 14N for successive generations
3)following each round of replication, separate DNA with CsCl gradient
4)determine whether DNA molecules are heavy, light or hybrid
results: proved that DNA rep is semi-conservative
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dna synthesis
read 3' to 5'
writes 5' to 3'
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DNA initiator complex
-dnaA and C
-denatures strands
-opens replication bubble
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DNA helicase
-dnaB
-extends replication fork
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DNA gyrase
-gyrA and B
-relieves supercoiling ahead of replication fork
-a type II topoisomerase
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single-stranded binding protein
-ssb
-stabilizes ssDNA, prevents intrastrand pairing of ssDNA
-prevents palindromic DNA sequences (they cant be replicated
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DNA pol III
-10 genes
-synthesizes leading strand
-elongates lagging strand
-needs an -OH
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primase
-dnaG
-synthesizes RNA primers
-compared to DNA template with OH group
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DNA pol I
replaces RNA primer with DNA
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DNA ligase
-lig
-joins Okazaki fragment and the ends of the leading strand fragment
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DNA replication (initiation)
1)DNA-A protein binds to origin of replication (oriC), denaturing DNA strands
2)DNA-C protein binds to DNA-A and deliveres helicase to DNA molecule
3)helicase extends replication forks beyond oriC
4)SSB binds to ssDNA, preventing formation of secondary DNA structures
5)primase binds to DNA behind helicase and polymerizes a short 10-12 nt RNA primer complementary to the DNA sequence
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DNA replication (elongation)
1)DNA pol III elongates the primer, using dNTPs
2)DNA pol I removes the primer and replaces it with dNTPs
3) DNA ligase joins the end of the new leading strand sequence to the lagging strand initiated at the opposite replication fork
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replisome
molecular machine that allows replication to take place simultaneously on both strands
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alpha subunits
polymerase
-makes phosphodiester bonds
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beta subunits
processivity
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epsilon subunits
proofreading
-senses wrong base pair, removes it and substitute it for the correct one
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theta and tau subunits
linking
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theta replication
-prokaryotic
-2 origins of replication
-DNA T produces protein T aids in separation of replisome
-DNA pol stalls when it reaches site of termination (180 degrees away from origin of rep)
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rolling circle replication
1)nick is made in + strand of the parental duplex (O=origin_
2)5' end is displaced and covered by SSBs (lagging strand)
3)polymerization at 3' end adds new deoxyribonucleotides (leading strand)
4)attachment of replisome and formation of okazaki fragments
-produces tandem copies of the plasmid or phage genome
-has endonuclease cut site that corresponds to the end of the genome
