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College > Genetics Biology test 3 > Flashcards

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
Q

Pymidine

A

1 ring
cytosine
uracil
thymine

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26
Q

DNA nucleotide - adenine

A

nucleoside (sugar + base) = deoxyadenosine
Nucleotide (sugar + base + phosphate group) = deoxyadenosine 5’ - monophosphate

27
Q

RNA nucleotide - uracil

A

nucleoside (sugar + base) = uridine
Nucleotide (Sugar + base + phosphate grou) = uridine 5’ - monophosphate or uridylic acid

28
Q

reaction between two nucleotides

A

Condensation reaction - H2O
creates phosphodiester linkage

29
Q

watson and crick

A

correctly modeling the sruture of DNA

30
Q

Erwin Chargaff

A

determined that A=T and G=C in DNA, but the ratio of (A+T)/(G+C) varied among species

31
Q

Rosalind Franklin and Maurice Wilkins

A

independently analyzed DNA by X-ray diffraction cyrstallography

32
Q

Crystallography

A

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
Q

how many bonds between the bases

A

g-c = 3
a-t = 2

34
Q

what are the three models of DNA

A

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
Q

Cesium chloride gradient centrifugation

A

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
Q

meselson and stahl experiments

A

Mixed DNA made of 14N and 15 N
F1 generation disposed of the conservative model
F2 generation disposed of the dispersive model

37
Q

Taylor-woods-hughes experiment

A

demonstrated that semi-conservative replication is the mechanism in Eukaryotes too

38
Q

Kornberg

A

isolates DNA polymerase 1
Achieves DNA replication in Vitro
Demostrates that DNA precipitate includes radiolabel

39
Q

DNA polymerases

A

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
Q

oriC

A

a sequence region
has 4 * 9nt units
3 * AT rich 13nt units

41
Q

why does one of the DNA strands loops

A

looping of the lagging strand permits the replication complex to move together in one direction

42
Q

what are the discontinuous strands on the lagging strands called

A

okazaki fragments

43
Q

phases of DNA replication

A

initiation
elongation
maturation of okazaki fragments
replication of telomeres (eukaryotes only)

44
Q

elongation

A

DNA replication is semidiscontinuous

45
Q

Step 1 in replication of DNA in prokaryotes

A
  • replicator motif at the orgin of replication binds initiator protein (DNAa): denaturation of DNA, producing replication bubble
46
Q

Step 2 in replication of DNA in prokaryotes

A
  • Helicases elongate the replication fork by untwisting, denaturing DNA
47
Q

Step 3 in replication of DNA in prokaryotes

A
  • DNA primase binds each helicase forming the primosome. Primase initiates RNA primers complementary to strands or template DNA
48
Q

Step 4 in replication of DNA in prokaryotes

A
  • DNA polymerase III extends primers in 5’ - 3’ direction
49
Q

Step 6 in replication of DNA in prokaryotes

A
  • replication proceeds toward the replication fork on the leading strand but away from it on the lagging strand
50
Q

Step 5 in replication of DNA in prokaryotes

A
  • Single stranded binding protines bind ssDNA; stabilize ssDNA and prevent renaturation
51
Q

Step 7 in replication of DNA in prokaryotes

A
  • okazaki fragments on lagging strand
52
Q

Step 8 in replication of DNA in prokaryotes

A
  • leading strand replication is continous
53
Q

Step 9 in replication of DNA in prokaryotes

A
  • lagging strand replication is discontinuous: polymerase falls off when reaching the next 3’ okazaki fragment
54
Q

Step 10 in replication of DNA in prokaryotes

A
  • 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
Q

Step 11 in replication of DNA in prokaryotes

A
  • Dna polymerase I replaces RNA in Okazaki fragment with DNA
56
Q

Step 12 in replication of DNA in prokaryotes

A
  • DNA ligase patches Okazaki fragments together
57
Q

replicon

A

unit (amount/lenght) of DNA replicated from one ORI

58
Q

Initiation of replication in eukaryotes

A

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
Q

DNA replication only occurs in

A

S phase of the cell cycle

60
Q

similarities with prokaryotic replication

A
  • 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
Q

differences with prokaryotic replication

A
  • 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
Q

what is the problem at the ends of DNA

A

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
Q

How do we prevent the telomers from shortening

A

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
Q

packing new chromatin

A

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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