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Genetics > Week 5 > Flashcards

Week 5 Flashcards

1
Q

Interrupted “mating” and time of entry mapping

A
  • ## Samples plated on selective media to evaluate the phenotypes of exconjugants
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2
Q

F factor excision form Hfr strains

A
  • F factor separated from the bacterial chromosome of a Hfr strain
  • Excision can include genes from the bacterial chromosome in the F plasmid
  • Becomes F’ plasmid
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3
Q

Effect of F’ plasmids on ploidy in recipient cells

A

If the F’ plasmid that has a gene from a donor cell is transferred to a recipient cell, that recipient cell will have two copies of a gene - one from the doner and one from the original gene in the recipient

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4
Q

Steps of Transformation

A
  1. Donor DNA binds at the receptor site. One strand is degraded as it enters the recipient cell
  2. The transforming strand pairs with the homologous region of the recipient chromosome
  3. The transforming strand displaces a recipient strand, forming complementary heteroduplex DNA. The excess strand degrades
  4. DNA replication and cell division produces one transformant and one nontransformant
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5
Q

What does the term competent mean when discussing transformation

A

Refers to a cell capable of being transformed

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6
Q

Co-transformation

A

Transformation of multiple genes - indicates they are closely linked

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7
Q

Lytic cycle of bacteriophage

A
  • The phage genome does not get incorporated into the bacterial genome
  • Donor genes can be incorporated in the phage; this DNA can be inserted into recipient cell to produce a transductant when recombination integrates the DNA into the recipient genome
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8
Q

Steps of the lytic cycle of a bacteriophage

A
  1. Phage attaches to host cell
  2. Phage injects DNA through hollow tail
  3. Replication of phage chromosome occurs; host DNA breaks down
  4. Under the direction of phage genes, transcription and translation produce new phage components
  5. DNA and proteins are assembled into progeny phages
  6. Progeny phage particles are released by lysis from host bacteria
    * Repeat Cycle *
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9
Q

Lysogenic cycle of bacteriophage

A
  • Phage DNA gets incorporated
  • The resulting prophage may undergo multiple cell divisions before the phage genome is excised and the lytic cycle continues
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10
Q

Stages of the Lysogenic cycle in bacteriophage

A
  1. Phage attaches to host cell
  2. Phage injects DNA through hollow tail
  3. Integration of phage DNA into the host chromosome
  4. Generation(s) of cell divisions occur
  5. Excision of prophage from the host chromosome
  6. The lytic cycle resumes
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11
Q

Effect of bacteriophage on plaque formation

A

When bacteriophages infect bacteria, a spot with dead bacteria forms call a “plaque”
- If genes in the bacteriophage are defective, plaques either don’t form or don’t look like wildtype plaques

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12
Q

How can recombination effect plaque formation with defective bacteriophages

A
  • Bacteria can be coinfected by phage strains with different mutations that prevent normal plaque development
  • Strains with different mutations in the same gene fail to complement meaning plaques either don’t form or they look weird
  • Recombination could occur within defective genes to generate a functional wildtype allele resulting in normal plaque formation
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13
Q

How do we know that DNA and not protein is the heredity material

A
  • Protein contains large amounts of sulfur but almost no phosphorus
  • DNA is the opposite
  • When bacteriophages attach to the outside of bacterial cell and inject DNA, the empty phages can be separated by shaking and centrifugation
  • Using radioactive tracers, can determine the function of protein vs DNA
  • When protein traced; found non-radioactive plaques formed
  • When DNA traced; plaques seen through radiation
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14
Q

DNA 3D structure

A

The most common 3D form of DNA - the B form - has a major and minor groove that is recognized by proteins
- Minor groove: less exposed promiscuous protein binging (non-specific)
- Major Groove: Interior Portion exposed, sequence specific protein binding

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15
Q

What are the three proposed models for how DNA is copied during cell division?

A

CONSERVATION REPLICATION
- one daughter strand made completely of parental strands and other made of synthesized strands
SEMICONSERVATIVE REPLICATION
- Daughter DNA made of one parent strand and one synthesized strand each
DISPERSIVE REPLICATION
- Daughter DNA stands made up each of a combination of parental DNA and newly synthesized DNA

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16
Q

The Meselson-Stahl Experiment

A

Grow generation 0 E. coli cultures in 15N growth medium resulting in heavy DNA
- After transferring to a medium with 14N only, all hybrid DNA was of intermediate weight, rejecting the CONSERVATION THEORY
- Following generation result in light and intermediate weights, rejects dispersive replication

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17
Q

of Origin of replications

A
  • Bacteria and archaea typically have one origin of replication (ori)
  • Eukaryotes have multiple ori per chromosome (human genome contains more than 50,000 ori)
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18
Q

Replisome

A

Complex of proteins at each replication fork
Includes;
- DNA topoisomerase
- Helicase
- SSB
- Primase
- DNA polymerase III
- DNA polymerase I
- DNA ligase

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19
Q

DNA topoisomerase

A

Relaxes supercoiling

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20
Q

Helicase

A

Unwinds the double helix

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21
Q

SSB

A

Prevents reannealing of separated strands

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22
Q

Primase

A

Synthesizes RNA primers

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23
Q

DNA Polymerase III

A

Synthesizes DNA

24
Q

DNA polymerase I

A

Removes and replaces RNA primer with DNA

25
Q

DNA Ligase

A

Joins DNA segments

26
Q

Proteins role in DNA replication

A
  1. Helicase breaks hydrogen bonds. Topoisomerase relaxes supercoiling
  2. Single-stranded binding (SSB) protien precents reannealing
  3. Primase synthesizes RNA primers
  4. DNA polymerase III synthesizes daughter strand
  5. DNA polymerase III elongates the leading strand continuously and the lagging strand discontinuously
  6. DNA polymerase I removes and replaces nucleotides of the RNA primer
  7. DNA ligase joins okazaki fragments
27
Q

Why does DNA polymerase need an RNA primer in order to begin synthesizing DNA

A
  • DNA polymerase is unable to initiate synthesis - it can only extend the strand
  • The essential role of RNA in DNA replication is consistent with the “RNA world” hypothesis
28
Q

DNA Proofreading

A
  1. DNA polymerase error
    - An incorrect nucleotide is incorporated in the extending strand - rare
  2. Exonuclease removal of mismatched base pair
    - Many DNA polymerases have a 3’ to 5’ exonuclease activity
  3. Daughter stand resumes DNA synthesis
    - The incorrect nucleotide is excised and replication continues
29
Q

Mutations

A
  • Due to errors in replication
  • Most are form fathers
  • some parts of the genome mutate more than others
  • Some types of mutations happen more frequently
30
Q

Mutation rate

A
  • Typically about 1 out of every 1 billion base pairs but this rate varies among species
  • Humans inherit about 60 new mutations each generation
31
Q

Replication at the ends of linear chromosomes

A
  • The replication process is unable to replicate linear chromosomes to the end due to the requirement of an RNA primer
  • Chromosomes become slightly shorter with each replication cycle
  • Repetitive sequences at the ends of linear chromosomes called telomeres ensure that portion of the chromosome can be safely lost without serious consequences to the organism
  • Telomeres do not contain coding sequences
32
Q

Hayflick limit

A

The number of times that a normal, differentiated somatic cell can undergo cell division before cell division stops
- One reason for the cessation of cell division is that each cycle results in chromosome shortening
- For many cells, limit is 50-70 cell divisions
- After this cell undergoes apoptosis

33
Q

Cells that do not exhibit the Hayflick limit

A
  • Some cells have an extended or non-existent limit
  • These exceptional cells typically have telomerase activity that maintains the chromosome ends
  • E.g. stem sperm progenitor cells and cancer cells
34
Q

What is the purpose of sex?

A

Recombination

35
Q

What are the pros of sexual reporduction

A
  • Male parental care
  • Sexual selection removes bad alleles
  • Recombination
36
Q

What are the cons of sexual reproduction

A
  • Have to make males (2 fold cost)
  • May disrupt advantageous combinations of alleles
  • Could be risk (predation, STDs)
37
Q

Why have separate sexes instead of hermaphrodites

A
  • Cost: you have to find someone to have sex with
  • Sexual selection only possible when you have separate sexes - advantageous by purging deleterious alleles carried by males
  • may better exploit a complex fitness landscape
  • inbreeding in hermaphrodites may increase homozygosity
38
Q

Advantages of recombination

A
  • Leads to offspring that have combinations of traits that differ from parents
  • Increases rand of phenotypic value for polygenic traits
  • more options for natural selection to choose from
39
Q

Muller’s ratchet

A

A decline in fitness due to stochastic loss of the least deleterious allele

40
Q

Hill-Robertson effect

A

Interference between linked beneficial and deleterious mutation slows fixation of beneficial mutations and slows removal of deleterious ones

41
Q

“Genetic Hitchhiking”

A

Some deleterious mutations rise to high frequency or fixation because they are linked to advantageous mutations

42
Q

“Background selection”

A

Some beneficial mutations are removed from a population because they are linked to deleterious mutations

43
Q

Why is genetic recombination useful

A
  1. Recombination can separate good mutations for bad mutations
  2. Recombination can unite multiple advantageous mutations that appeared in different individuals
44
Q

Process of DNA replication

A
  1. Helicase breaks hydrogen topoisomerase relaxes supercoiling
  2. Single-stranded binding protein prevents reannealing
  3. Primase synthesizes RNA primers
  4. DNA polymerase III synthesizes daughter strand
  5. DNA polymerase III elongates the leading strand continuously and the lagging strand discontinuously
  6. DNA polymerase I removes and replaces nucleotides of the RNA primer
  7. DNA ligase joins Okazaki fragments
45
Q

What is the mutation rate

A

typically 1 out of every 1 billion base pairs but varies among species

46
Q

Mutations in humans

A
  • about 60 new mutations each generation
  • most are form our fathers
  • some parts of the genome mutate more than others (mtDNA)
  • Some types of mutations happen more frequently
47
Q

Replication at the ends of linear chromosomes

A
  • The replication process is unable to replicate linear chromosomes to the end due to the requirement of an RNA primer
  • Repetitive sequences at the ends of linear chromosomes called telomeres ensure that a portion of the chromosome can be safely lost without serious consequences to the organism
48
Q

How does telomerase maintain chromosomal integrity

A
  1. Attachment of telomerase
  2. Elongation of DNA
    3.Translocation of telomerase: moves over after bit of DNA synthesis
  3. Elongation of DNA
  4. Telomere completion (by DNA polymerase)
49
Q

Telomeres linked to aging

A
  • At least a few hundred nucleotides of telomere repeats must cap the end of each chromosome in order to avoid apoptosis or cellular senescence
  • Telomere length declines with age
  • Dysfunctional telomeres are associated with genome instability and aneuploidy
  • Patients with low telomerase activity have a variety of disorders
50
Q

What is a ribonucleoprotein

A

An enzymatic complex that includes protein and ribonucleic acid
- E.g. telomerase, ribosomes, and components of the splicesome

51
Q

Where is telomerase active

A

germ cells and stem cells

52
Q

Polymerase Chain Reaction

A
  • in vitro DNA replication
  • uses special type of DNA polymerase - taq polymerase
  • uses DNA primer instead of RNA
  • uses temperature to unwind
  • still semiconservative
  • done in small volume of fluid
53
Q

Thermus Aquaticus

A
  • A bacteria that can tolerate high temperatures
  • it lives and undergoes cell division in hot springs
  • it does not denature at high temperatures
  • allows for high temperature to ce used to denature DNA without denaturing the polymerase
54
Q

What does PCR require

A
  • DNA template
  • A supply of the four nucleotides
  • a heat-stable DNA polymerase
  • Two different single stranded DNA primers
  • a buffer solution
55
Q

Process of PCR

A
  1. Denaturation of DNA by heating 95 degrees
  2. Primer annealing (45-68)
  3. Primer extension at 72 degrees
56
Q

Limitations of PCR

A
  • You need to know something about the sequence you want to amplify in advance in order to design primers
  • PCR only works for small fragments
57
Q

Sanger Sequencing

A
  • DNA sequencing using fluorescently labeled DDCTP
  • Sequencing occurs with ddCTP terminating elongation at various lengths
  • Organizing strands by length allows the fluorescent colors to be read to determine sequence

Genetics (18 decks)

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