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Genetics Biology test 3 Flashcards

1
Q

linkage

A

genes show linkage if they are on the same chromosome
linked genes are not free to undergo independent assortment

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2
Q

crossing over

A

results in reshuffling (recombination) of alleles between homologs

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3
Q

Frequency of crossing over

A

proportional to distance between any two loci on single chromosome
farther away more crossover frequency
closer together less crossover frequency

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4
Q

independent assortment (unlinked)

A

different chromosomes each with own gene

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5
Q

linkage without crossing over

A

two genetically different gamete produced
complete linkage produces parental or no crossover gamete in equal proportions

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6
Q

linkage with crossing over

A

involves two non-sister chromatids
generates two new allele combinations called recombinant or crossover gametes

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7
Q

Alfred sturtevant

A

first chromosome map
reliazed morgans proposal could be used to map sequence of linked genes
recombination frequencies between linked genes are additive

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8
Q

map units

A

frequency of exchange used to estimate distance two genes along chromosome
firmly established chromosomal theory of inheritance

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9
Q

single crossovers

A

the closer two loci reside - less likely of single crossover event
two loci farther apart along chromosome - random crossover more likely
occurs between two non-sister chromatids

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10
Q

multiple crossovers

A

between chromatids of tetrad facilitate more extensive chromosome maps

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11
Q

double crossovers

A

results from double exchanges of genetic material
three loci must be used for determination - each pair must be heterozygous for two alleles

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12
Q

“duck test” for heritable material

A

stable and contain the information
able to reproduce itself faithfully
able to change so that there is variation on which selection can occur so that populations can evolve

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13
Q

friedrich miescher

A

discovers nucleic acids
collected cells from pus in discarded bandages from a local health clinic
isolated molecules from nuclei composed of nitrogen, hydrogen, carbon, oxygen, and phosphorus
believed protiens were the molecules of heredity

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14
Q

sutton-boveri

A

chromosomes in nuclei carry hereditary information

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15
Q

Albrecht Kossel

A

chromosomes are composed of protien and nuclei acids
protiens are composed of 20 amino acids and DNA is composed of 4 nucleotides

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16
Q

Frederick Griffith

A

there are two strains of Pneumonia but never change into each other
Either strain can change into an R strain spontaneously

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17
Q

Avery’s experiments

A

first strong evidence that DNA was the transforming factor

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18
Q

trypsin

A

destroyes protiens

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19
Q

chymotrypsin

A

destroys protien

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20
Q

RNase

A

destroys RNA

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21
Q

DNase

A

destroyed DNA

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22
Q

deoxyribose

A

Deoxy= having less oxygen”
stable
DNA

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23
Q

Ribose

A

more reactive
RNA

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24
Q

Purine

A

2 rings
adenine
guanine

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25
Pymidine
1 ring cytosine uracil thymine
26
DNA nucleotide - adenine
nucleoside (sugar + base) = deoxyadenosine Nucleotide (sugar + base + phosphate group) = deoxyadenosine 5' - monophosphate
27
RNA nucleotide - uracil
nucleoside (sugar + base) = uridine Nucleotide (Sugar + base + phosphate grou) = uridine 5' - monophosphate or uridylic acid
28
reaction between two nucleotides
Condensation reaction - H2O creates phosphodiester linkage
29
watson and crick
correctly modeling the sruture of DNA
30
Erwin Chargaff
determined that A=T and G=C in DNA, but the ratio of (A+T)/(G+C) varied among species
31
Rosalind Franklin and Maurice Wilkins
independently analyzed DNA by X-ray diffraction cyrstallography
32
Crystallography
revealed a helical molecule gave spacing values for repaeting features revealed that the backbone was alternating sugar and phosphate groups, placing the bases on the inside
33
how many bonds between the bases
g-c = 3 a-t = 2
34
what are the three models of DNA
semi-conservative - half of new DNA would be old; proposed 1953 watson and crick conservative - new DNA would be all new dispersive - 1/2 old and 1/2 new; old DNA gets diluted over many generations
35
Cesium chloride gradient centrifugation
Done by meselson and stahl a substance/molecule into the cesium chloride gradient spins for a long time at high speeds DNA of different sizes spearate by density
36
meselson and stahl experiments
Mixed DNA made of 14N and 15 N F1 generation disposed of the conservative model F2 generation disposed of the dispersive model
37
Taylor-woods-hughes experiment
demonstrated that semi-conservative replication is the mechanism in Eukaryotes too
38
Kornberg
isolates DNA polymerase 1 Achieves DNA replication in Vitro Demostrates that DNA precipitate includes radiolabel
39
DNA polymerases
extend existing strands of DNA using a complementary strand as a template requires single strand/double strand interface diverse in prokaryotes and eukaryotes addition of nucleotides to existing strand creates new phosphodiester bond; released two of three phosphates of the nucleotide only extend DNA strands in the 5' to 3' direction
40
oriC
a sequence region has 4 * 9nt units 3 * AT rich 13nt units
41
why does one of the DNA strands loops
looping of the lagging strand permits the replication complex to move together in one direction
42
what are the discontinuous strands on the lagging strands called
okazaki fragments
43
phases of DNA replication
initiation elongation maturation of okazaki fragments replication of telomeres (eukaryotes only)
44
elongation
DNA replication is semidiscontinuous
45
Step 1 in replication of DNA in prokaryotes
- replicator motif at the orgin of replication binds initiator protein (DNAa): denaturation of DNA, producing replication bubble
46
Step 2 in replication of DNA in prokaryotes
- Helicases elongate the replication fork by untwisting, denaturing DNA
47
Step 3 in replication of DNA in prokaryotes
- DNA primase binds each helicase forming the primosome. Primase initiates RNA primers complementary to strands or template DNA
48
Step 4 in replication of DNA in prokaryotes
- DNA polymerase III extends primers in 5' - 3' direction
49
Step 6 in replication of DNA in prokaryotes
- replication proceeds toward the replication fork on the leading strand but away from it on the lagging strand
50
Step 5 in replication of DNA in prokaryotes
- Single stranded binding protines bind ssDNA; stabilize ssDNA and prevent renaturation
51
Step 7 in replication of DNA in prokaryotes
- okazaki fragments on lagging strand
52
Step 8 in replication of DNA in prokaryotes
- leading strand replication is continous
53
Step 9 in replication of DNA in prokaryotes
- lagging strand replication is discontinuous: polymerase falls off when reaching the next 3' okazaki fragment
54
Step 10 in replication of DNA in prokaryotes
- as replication fork open, new RNA primers form and new polymerase binds creating a new upstream Okazaki fragment. Gaps or 'nicks' occur between okazaki fragments
55
Step 11 in replication of DNA in prokaryotes
- Dna polymerase I replaces RNA in Okazaki fragment with DNA
56
Step 12 in replication of DNA in prokaryotes
- DNA ligase patches Okazaki fragments together
57
replicon
unit (amount/lenght) of DNA replicated from one ORI
58
Initiation of replication in eukaryotes
ORC = Origin recognition complex - protien complex ORC binds ORI in G1 Once the DNA is melted and stabilized - SSBs come in primase and polymerase are recruited to start replication
59
DNA replication only occurs in
S phase of the cell cycle
60
similarities with prokaryotic replication
- there are lagging strands and leading strands - lagging strands created a series of okazaki fragments - a helicase, gyrase, SSBs, polymerase, primase, all required - replication is bidirectional
61
differences with prokaryotic replication
- there are many ORI's per chromosome - there are many more kinds of polymerase, and some have more specialized jobs - the chromosomes have ends - chromatin is packed arounf histones - multiple replicons per chromosome
62
what is the problem at the ends of DNA
after semiconservative replication, new DNA strands have RNA primers at their 5' ends RNA primers removed, leaving single stranded overhangs at telomers because DNA polymerase cannot fill them the sequence can be lost
63
How do we prevent the telomers from shortening
binding of telemores to the overhanging 3' end of the chromosome synthesis of new telomere DNA using telomerase RNA as template telomerase movement to 3' end of newly synthesized telomere DNA synthesis of new telomere DNA chromosome end after telomerase leaves new end of the chromosome after replication and primer removal
64
packing new chromatin
number of histones must doublew, so many new subunits must be made during S phase nucleosomes are disassembled as the replication fork approaches Daughter histones are typically a mix of 'old' and 'new' histones components

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