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GEGE2901 > Mutations > Flashcards

Mutations Flashcards

1
Q

Is DNA replication 100% accurate? What is its error rate if it isn’t?

A

• DNA replication is accurate, but not 100% accurate
o DNA replication has a template, which is why it is precise
o Reasonably high error rate-10-9 is error rate of incorporating new mutation through each DNA replication cycle

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2
Q

Is genetic variation per generation predictable or random? Include an example

A

• Predictable rate of genetic variation per generation
o Incorporated through DNA replication and other mutation events
o Humans have about 70 new base pairs different from parents’ genomes due to changes in germline through mutation

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3
Q

Describe Darwin’s variation in finch beaks, what he obesrved from that and what gene is responsible for this variation

A

• Darwin- variation in finch beaks
o Some finches with blunt beaks
o Some finches with sharp pointed beaks
o Inferred that there was a difference in way finches developed their beaks
o Direct relationship to physical changes and DNA sequence
 Gene responsible for beak shape is ALX1 (a gene in region D expressed in craniofacial region)
• In mammals with ALX1 gene, mutation causes craniofacial defects

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4
Q

Are bases chemically inert (cannot change)?

A

• The bases can change in chemical forms

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5
Q

What two forms can bases have and which is more common/rarer and why?

A

• The bases can change in chemical forms
o Switch between keto and enol form
 Normal -keto form which are more energetically stable than enol forms
 Rare- enol form called tautomers

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6
Q

What are tautomers?

A

Enol forms of nucleotide bases

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7
Q

How are enol forms different from base keto forms?

A

o Different types of hydrogen bonds form between bases when they’re in their enol form vs their keto form
 Tautomeric forms change position of the double bonds and proton shifts within structure of the bases, forming less stable pairings
o Normal keto (common) form: makes TA (2 hydrogen bonds) and CG (3 hydrogen bonds) base pairing (Watson-Ccrick base pairs)

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8
Q

Describe how mispairings in DNA replication relating to different base forms is possible and what base pairs this result in

A

• This can lead to mispairings in DNA replication-Tautomeric shift of nucleotide bases can cause mispairings of bases and can result in newly synthesised DNA pair having a mutation
o When have Cytosine in enol form and adenine in common form, these two bases can base pair with 2 hydrogen bonds (slightly energy stable, but not as much as TA pair)
o When have thymine in common form and guanine in rare form, these two bases can base pair with 3 hydrogen bonds

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9
Q

Describe how frequently enol-keto mispairing occurs and what can be done to reduce this

A

o There is mispairing 1 in 100,000 bases -> tens of thousands of errors in 1 replication of mammalian genome
 100 fold reduction of these mutations by error correction in DNA replication

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10
Q

Describe the process of mispairing in DNA replication and why it is a problem

A

• Process of DNA replication-tautomers
o Standard nucleotides and base pairing
o Tautomeric shift of nucleotide from keto form to enol form
o Replication occurs
o The enol tautomer mispairs with a keto form nucleotide that it normally doesn’t pair with
o Strands separated
o Tautomeric shift of enol form back to keto form
o Replication occurs
o There is a mutation in the newly synthesised strand in second step of replication-mutant strand of DNA in ½ the progeny

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11
Q

Where do most mutations take place?

A

In DNA replication

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12
Q

What is the mechanism of repairing mispaired bases during replication? Is this mechanism slow or fast and why?

A

• Polymerase exonuclease
o Polymerase removes most mispaired bases during replication by measuring distance between nucleotides (mispaired nucleotides have different distances between each other than normal ones)
o Removes wrong ones via 3’ exonuclease activity
o New opportunity for pairing and inserts right base pairs
o Proofreading slows down replication but makes it more accurate
o Fast replication done by error-prone polymerases with no exonuclease function so make it less accurate (sometimes these are used in PCR)

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13
Q

Describe the purpose of base excision repair and its mechanism

A

• Base excision repair
o Breakage of bond connecting the base to the sugar-phosphate backbone
o Loss of base creates an apurinic or apyrimidinic site (AP)
o AP is recognised by AP endonuclease, which cleaves the backbone on each side of the base
o DNA polymerases can fill in this gap

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14
Q

How was the first bacterial repair system identified by?

A

• First bacterial repair system identified by:
o Phenotype- high rate of mutation in E.Coli
o Genotype- mut genes were mutant
 Mut genes wild type- DNA repair is normal
 Mut genes are mutant- DNA repair is less efficient

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15
Q

Describe the purpose of mismatch repair and how it occurs

A

o Catches mismatches missed by DNA replication proofreading or that have occurred at another time
o Mismatch is detected with proteins
o The mismatch repair proteins remove the mispaired base and surrounding bases on the new strand
 Removes about 10 nucleotides or so around the mismatch and does it on newly synthesised strand
o DNA polymerase fills the gap and ligase seals the nicks

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16
Q

What is the difference in the detection step in mismatch repair between bacteria and other organisms?

A

 In bacteria, template strand is more highly modified than newly synthesised strand which is how it knows what to use as a template. Doesn’t happen in all organisms
• Old strand- has methylated adenines
• New strand- unmethylated adenines

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17
Q

Describe how direct repair occurs

A

• Direct repair
o Many environmental agents attach methyl groups to bases, especially guanine which can cause guanine to mispair with thymine
o Chemically modified bases can be directly repaired by removal of methyl groups by methyltransferases

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18
Q

Describe how break repair occurs (purpose and process)

A

• Break repair
o Double-stranded breaks to DNA are repaired by reattaching the broken end to another DNA molecule
o Two different processes, with many variations, are used
o Commonly, the DNA molecules are not similar in sequence, so the breaks are repaired by non-homologous end joining
 Less commonly, the DNA molecules are highly similar in sequence, and homologous recombination occurs

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19
Q

What is the danger of UV light?

A

• UV light causes lots of mutations

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20
Q

Describe the mutation induced by UV light and how it triggers repair

A

• UV light has high energy and causes pyrimidine dimer
o Mutation event that isn’t normal-physically joins two adjacent pyrimidines (usually Ts)
• This leads to a bulge in the helix which triggers repair

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21
Q

What is photoreactivation and in what organisms does it occur?

A

• Photoreactivation- Bacteria, single-celled eukaryotes, plants and some animals (not humans)
o Pyrimidine dimers repaired by photoreactivation
o Photolyase binds the thymine dimer and enzyme photolyase uses energy from visible light to break the bonds between pyrimidine dimers
o Photolyase is encoded by the E.coli phr (photoreactive repair) gene

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22
Q

Describe how UV repair proceeds in humans

A

• UV repair in humans-
o Nucleotide excision repair
 Changes to multiple adjacent bases or more extensive damage
 When damage to the nucleotides is greater than a single base, the damaged region is recognised: the strands are separated, and oen strand is degraded to create a gap of 10-20 nucleotides
 This gap is filled in by DNA polymerases
 While the overall process is similar in bacteria and eukaryotes, the proteins are not highly conserved

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23
Q

What is the consequence of people who have defects in process of nucleotide excision repair?

A

o People who have defects in process of nucleotide excision repair are affected by xeroderma pigmentosa and are extremely sensitive to UV light, often have basal cell cacinomas and have reduced life expectancy
 People have mutations in 1 of 7 genes of encoding repair proteins

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24
Q

What are 3 different types of mutations?

A

• Indels
o Small insertions or deletions (generally 1,2,3 or 4 base pairs)
• Transitions
o A purine (A,G) is replaced by a purine or a pyrimidine (C,T) is replaced by a pyrimidine
• Transversions
o A purine is replaced by a pyrimidine or vice versa

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25
Q

Does the location of a mutation in the genetic code matter?

A

• Location of small mutations will have different consequences

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26
Q

Describe where mutations can occur in a genome

A

o Coding region
o Regulatory region
o Splice sites
o Between genes

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27
Q

What are the 4 different types of mutation that can occur in a coding region of a gene?

A

 Will have great impact from base change
 Silent mutation: base pair change, no amino acid sequence change
 Missense mutation: base pair change, amino acid change
 Nonsense mutation: base pair change, early stop codon
 Frameshift: 1 base pair or more, insertion or deletion which alters the reading frame and makes a change in the string of codons. Can have an early stop

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28
Q

What happens to a cell when a mutation affects an essential function?

A

o Mutation affects an essential function

 Cell undergoes cell death

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29
Q

What happens to a cell when the mutation affects a non-essential function?

A

o Mutation affects another function

 Cell divides normally despite the mutation

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30
Q

What are mutations that can occur in genes associated with cell division and what are the consequence of these?

A

o Growth and cell division
 Tumour suppressor genes keep cell division in check
 Mutation in gene suppressor gene may remove regulation and cell division will occur extremely rapidly (dysregulation)

o Promote cell division
 Proto-oncogene (wildtype allele) becomes oncogene (mutant allele)
 Proto-oncogene regulate and promote normal cell growth and division
 Oncogene overexpress this regulation- promote too much
 Overexpression mutations

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31
Q

Describe the 3 types of mutation that are important in evolution

A

• Mutations in evolution
o Looking at a population
o Advantageous change- positive selection
 Mutation allele gives positive advantage and frequency greatly increases as generations go
o Neutral change- no selection
 Mutation allele gives no advantage and frequency doesn’t change as generations go
o Deleterious change- purifying selection
 Mutation allele negatively impacts survival and frequency decreases as generations go

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32
Q

What was the purpose of the fluctuation test?

A

o To address whether mutation is always there or only present when it is needed

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33
Q

Who invented the fluctuation test?

A

Luria and Delbruck 1943

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34
Q

Describe the method of the fluctuation test and its results

A

o Process-
 Flask of bacteria in medium
 A resistant (to a virus) mutation arises in growing culture
 The resistant cells increase in frequency as they grow
 Cells are exposed to virus and plated onto petri dish
 Found that different plates have similar numbers of resistant cells
 Had a control:
• Started with 4 independent tasks
• Each inoculated with population of bacteria and allowed to grow
o Know now that some of these flasks would have contained resistant bacteria but mutagenic events would take place at different stages (randomly)
• Plated each flask culture on one petri dish
• Found that resistant cells arise at different times in different cultures- different plates have very different numbers of resistant cells
o Found that mutations were always there, not simply induced through exposure of selective pressure
 Exposure to virus is test to identify mutagenic events that had arisen beforehand

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35
Q

What are sources of phenotypic and genomic variation in individuals of the same species (relatively)?

A 
•	Individuals of the same species
o	Gene number
o	Regulatory sequences
o	Variation in amino acid sequences
	Largest portion
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36
Q

What are sources of phenotypic and genomic variation in closely related species

A 
•	Closely related species- all variations smaller than distantly related species 
o	Variation in amino acid sequences
o	Regulatory sequences
o	Variation in gene number
o	Variation in splicing 
o	Novel genes
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37
Q

What are sources of phenotypic and genomic variation in distantly related species and their different impacts

A 
•	Distantly related species 
o	Variation in amino acid sequences (coding differences)
	Large portion
o	Regulatory sequences
	Large portion
o	Variation in gene number
	Medium portion
o	Variation in splicing
	Medium portion 
o	Novel genes 
	Small portion
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38
Q

How can sequences be added to a genome?

A
  • Polyploidization
  • Horizontal gene transfer
  • Copy number variation in blocks of DNA
  • Expansion of gene families
  • Novel genes
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39
Q

What is polyoploidization?

A

o Extra copy of genome is added

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40
Q

What is ploidy?

A

the number of sets of chromosomes in a cell, or in the cells of an organism.

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41
Q

Describe examples of haploid organisms

A

 Bacteria and other organisms are haploid

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42
Q

Describe examples of diploid organisms

A

 Humans and many eukaryotes are diploid

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43
Q

Describe examples of triploid organisms and why we want commercial food products to be triploid

A

 Triploids are usually infertile
• Bananas have been bred to be triploid-those that are bought in shops are triploid to make them infertile and reduce seed size
o Propagated through cuttings
• Seedless watermelons-process of triploidy (seeds are infertile)

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44
Q

Describe examples of tetraploid/hexaploid organisms

A

 Plants can be tetraploids and hexaploids

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45
Q

Describe why some plants are not diploid, and what plants these are

A

• ½ of flowering plants and crops are not diploid. Can occur through:
o Mitosis-replication occurs but not division
o Meiosis-replication occurs but only one division occurs

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46
Q

Have there been ploidy events in humans? Can they easily be seen now?

A

 There is evidence of ploidy events through evolution
• Evidence of 2 past polyploidisation events in humans
o But cannot be seen now because of change in structure in chromosomes over time

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47
Q

What is copy number variation?

A

o Copy number variation- a block of DNA sequence (1-25kb) on the chromosome becomes duplicated and repeated multiple times

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48
Q

How can copy number variation occur?

A

o Can occur when comparing individuals within a species and comparing between species

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49
Q

Are copy number variations random?

A

No

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50
Q

Can copy number variation in blocks of DNA occur in twins?How?

A

o Can vary between identical twins, thus during mitosis

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51
Q

How many cases of copy number variation in blocks of DNA occur in humans?

A

 1400 cases in humans (1/2% of human genome varies because of this)

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52
Q

What phenotypes does copy number variation usually influence?

A

 Often copy number variation influences cognitive or personality differences

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53
Q

What variation is the major differences between humans and chimps

A

 One of the major differences between humans and chimps

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54
Q

What is a gene family?

A

 Gene family-a set of several similar genes, formed by duplication of a single original gene, and generally with similar biochemical functions
 Gene family members carry out highly related functions
 Similar DNA sequences, paralogs

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55
Q

How many genes in the human genome are members of a gene family?

A

 Half of genes in human genome are members of a gene family

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56
Q

Are gene families the same within species? Are there exceptions (if so, talk about it)

A

 Usually a set number of members of gene family for a species
• Exception is gene encoding amylase which is encoded by a variable number of a gene family in humans
o People with high-starch diets have a median gene number of 7
o People with low-starch diets have a median gene number of 5

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57
Q

What is the effect of a greater number of genes in the family?

A

 A greater number of genes in the family gives higher gene product levels

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58
Q

How do gene families arise?

A

 Arising of gene families
• Ancestral gene is duplicated
• Further duplication of ancestral gene and there is divergence in function: similar versions of same gene
• Continue to duplicate and diverge in function and expression

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59
Q

What is a pseudogene?

A

o Pseudogene- Nonfunctional copies of a gene with a mutation that prevents its expression. Indication that it used to be a functional gene but has now accumulated so many mutations that it is no longer functional

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60
Q

How can expansion or contraction of gene families occur?

A

o Expansion or contraction of gene families along a specific lineage can be due to change, or can be the result of natural selection

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61
Q

Describe the arising of the myoglobin gene family (and the use of that gene). Include which chromosomes each resultant gene is in

A

o E.g. Myoglobin gene (carries oxygen and carbon dioxide) in humans found in muscle
 Myoglobin gene maintained in single copy of chromosome 22 in humans
 But there is a gene family with myoglobin as its ancestor
• Myoglobin duplicated and gave rise to alpha-beta globin gene (which was expressed in both blood and muscles)
• Alpha-beta globin gene duplicated and diverged to become the alpha globin and beta globin gene
• Alpha globin gene further duplicated and diverged in times of expression into 3 genes on chromosome 16 in humans
• Beta globin gene further duplicated and diverged in times of expression into 5 genes on chromosome 11 in humans

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62
Q

Describe the role, composition and turnover rate of haemoglobin

A

 Haemoglobin transports oxygen and is composed for 4 globin protein molecules
• 2 alpha and 2 beta chains
• Within each globin is an iron-containing heme group
• Has high turnover rate (replaced every 120 days) so high need for globin production

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63
Q

What is thalassemia and in what populations is it common?

A

 Thalassemia- imbalance in the relative amounts of alpha and beta globins produced
• Common in some populations (originated from Mediterranean and south-east Asia)

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64
Q

What are two types of thalassemia?

A

o Alpha thalassemia

o Beta thalassemia

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65
Q

Describe alpha thalassemia in terms of possible alleles and diffferent resultant phenotypes possible

A

 Chromosome 16
• Normal chromosome has 2 copies of alpha globin gene
• Alpha-thal-2 chromosome has deletion of 1 copy of alpha-globin gene
• Alpha-thal-1 has deletion of both copies of alpha-globin gene
 Situations-
• When 4 copies of alpha globin genes in genome, genotype is normal and there is no anemia
• When 3 (alpha- Thal-2/wild type heterozygote), or 2 (alpha-thal-1/wild type heterozygote or alpha-thal-2 homozygote) copies of alpha globin genes, anemia is mild
• When 1 copy of alpha globin gene (alpha thal-2/alpha-thal-1 heterozygote), there is moderate aenemia
• When 0 copies of alpha globin gene (alpha-thal-1 homozygote), anemia level is lethal

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66
Q

What are the current aims of thalassemia treatments?

A

• Treatments of thalassemia aim to (if foetal forms of gene are intact its just adult forms which are missing) continue to express foetal forms to compensate for adult form damage

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67
Q

Describe the alpha globin genes and what chromosome they’re on as well as when they’re expressed

A

 On chromosome 16, there is evidence for 3 different type of alpha globin genes and 2 pseudo alpha related genes
• Zeta form- expressed in embryo until switched off in later in development
• Pseudo-zeta and pseudo-alpha form- no longer have function but show remnant sequences
• Alpha 2 and alpha 1- expressed in foetal stage and adult stage

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68
Q

Describe the beta globin genes and what chromosome they’re on as well as when they’re expressed

A

 On chromosome 11, there is evidence for 5 different types of beta globin genes and 1 pseudo beta related gene
• Epsilon- expressed in the embryo
• G gamma and A gamma- expressed in the foetus then stops after birth
• Delta and Beta- expressed in the adult

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69
Q

Is pattern of globin gene expression constant?What is the effect of this?
Sketch the graph.

A

 Pattern of globin gene expression changes with age
• Expression of gene family members can change during life cycle
o This is a way that members of gene family can accumulate to have a specific function
• Can give rise to a trait having varying severity
Timestamp: 4:40 pm at 4/09/2019

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70
Q

Describe the FAD2 gene family in coffee and where in each tissue each gene is expressed

A

o FAD2 gene family in coffee- encodes for linoleic acid to give coffee its aroma
 Duplication events and selective breeding of coffee plants have fiven rise to 6 different FAD2 gene family copies
• Gene version 1 and 2 are a little expressed in stamen (gene 1 and 2), pistil (gene 2) and leaf (gene 1 and 2)
• Gene version 3 is expressed a little at every life cycle, but a lot in the root, stamen and endosperm
• Gene version 4 and 5 are only a little expressed in the leaf
• Gene version 6 is expressed throughout entire life, especially in pistil, lead and perisperm
 Has tissue specific expression in gene family

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71
Q

Do novel genes have a small or large effect on variation?

A

Large

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72
Q

What genes are novel in humans but not in chimps and what phenotypes do these genes influenced

A

o 200 novel genes in humans, not in chimps

 Genes give rise to gene expression in brain, skin and immune system

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73
Q

How are novel genes theorised to be accumulated?

A

o Don’t quite understand how novel genes are accumulated, but theorise might be horizontal gene transfer

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74
Q

What is CYP1A2 gene product and what does it encode/its role? Can the CYP1A2 gene be swapped between species?

A

 Caffeine breakdown rate by CYP1A2 gene product
• Gene encodes cytochrome C oxidase subunit VII2
o Related to breakdown of medical drugs and other compounds
• Many alleles-highly variable in humans

o Genes have many differences between species
o Some have the same function but not swappable
 Gene is not swappable between species- tried to place E.Coli. Gene in human culture cell to see if it could replace the human gene (knocked out the human gene) and it didn’t. Vice versa didn’t work either.
 So divergent that one cannot swap for the other one
 But in closely related organisms this can be done

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75
Q

Are all genes swappable between all species?

A

o Many amino acid differences between individuals
o Genes have many differences between species
o Some have the same function but not swappable
 So divergent that one cannot swap for the other one
 But in closely related organisms this can be done

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76
Q

What are variations in gene regulation?

A

o Mutations that later amount of protein product by changing recruitment efficiency of RNA polymerase to promotor site of gene
o No amino acid sequence change
o Can be in promoters, introns, 5’-UTR, 3’-UTR (sequences that influence regulation)
 Where is a gene expressed
 When is the gene expressed
 How much of a gene is expressed

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77
Q

Describe the hair colour gene as a variation in gene regulation

A

o Hair colour gene (kit ligand gene product)
 Level of expression varies between individuals
• Low kit ligand gene expression which results in reduced differentiation of melanocytes (which reduces pigment production) in blonde people
• High expression and differentiation of melanocytes in brown haired people
 Same amino acid coding sequence

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78
Q

What is alternative splicing and is it the same between species and within species?

A

• Alternative splicing
o Transcript spliced in different ways
 Different introns removed and different exons included in different versions of the transcript
o Different between species
o Should not vary from individual to individual within a species (expect genes to be spliced identically with species) but could vary between tissues in one individual (tissue specific expression which would be constant in the entire species)

79
Q

Describe the effect of the TYR gene missense mutation where glycine changed to an arginine on cat phenotypes

A

• In cats
o TYR gene codes tyrosinase, which is involved in melanin production (when melanin produced, dark pigment)
o Missense mutation where glycine changed to arginine which produces a tyrosinase which is active at low temperatures but is inactive at high temperatures (temperature sensitive)
 So in warm parts of cat, melanin is not produced but in cold parts of cat, melanin is produced
 Same protein everywhere but temperature gives different appearance

80
Q

Describe the effect of serotonin transporter genotype on people with major depression

A

• Major depression
o People with SS and SL genotypes for serotonin transporter are much more likely than LL genotypes to develop severe depression in response to 3 or more stressful life events

81
Q

What is vertical gene transfer and in what organisms can it be done?

A
  • Transmission of DNA from one generation to the next generation
  • Sexually or asexually (such as binary fission) reproducing organisms
  • New alleles now and then
82
Q

Describe what horizontal gene transfer is and between what organisms it occurs

  • How does the transfer take place
  • What kind of genetic information
  • What happens if integration is stable
A

• All organisms
• Plasmid transferred from donor to recipient
o Donor cell was lysed and dsDNA was ejected in external environment
o dsDNA is taken up by recipient as single stranded DNA
o It is then copied to double stranded DNA and is integrated into genome
• Transfer between cells within a generation
o Donor transfers dsDNA to recipient
o Acquired DNA is integrated into genome
• Uptake of DNA from environment
• RNA and single stranded DNA can also be taken up
• Can be stably integrated and maintained to subsequent generations

83
Q

Can horizontal gene transfer happen in humans? How?

A

• Humans do this-
o Viral infection can cause horizontal gene transfer in humans-Virus can become integrated in chromosome, and genotype has changed in particular cell, and can be passed on to next generation if germline cells are affected

84
Q

What are importation vectors in horizontal gene transfer and how do they function?

A

o Viruses
 Virus can infect a cell and transmit its nucleic acid in cell- horizontal gene transfer
o Transposable elements
 Type of DNA that has functional components that give it motility and allow it to move from a genome to another genome, or within a genome

85
Q

Is horizontal gene transfer integrated or not? what happens if it is/is not integrated? In which organisms is it most stable?

A

o Stable integration into the genome
 Only if stable will it be maintained in population of cells- otherwise it will be lost
o Infrequent in many organisms
 A lot more uptake occurs than stable integration occurs- DNA will be lost in many cases
 May get mutagenic events-integration get lead to problems
o More frequent in bacteria

86
Q

Describe Escherichia Coli in terms of:

  • chromosome shape
  • kb length
  • gene length
  • DNA components
A 
	Escherichia coli
•	Closed circular chromosome
•	4600kb in length
•	4300 genes
•	Extrachromosomal DNA
o	DNA outside the chromosome
87
Q

What are plasmids? How are they transmitted? Describe their size and replication when compared to chromosomes, and what they may encode for

A

• Plasmids
o Replicate independently of chromosome
o Relatively small compared to chromosomal DNA
o Transient or stably transmitted between generations
o 2-2000kb of double stranded DNA
o Single or multiple copies
o May encode antibiotic resistance, pathogenicity, substrate utilisation and DNA mobility

88
Q

What is transformation?

A

• Transformation-a bacterial cell taking up free DNA in the environment

89
Q

What are the cell requirements for transformation?

A

• Cell is competent when it is able to take up free DNA
o Cell must be competent for transformation
o Takes about 14 proteins to be expressed at the right time for it to be competent
o Can also do it artificially in the lab

90
Q

What is uptaked in transformation?

A

• Direct uptake of DNA molecules
o Plasmid DNA can also be taken if free in environment, but will probably be maintained in the cell as a free plasmid in recipient (won’t be integrated)

91
Q

Does transformation require energy?

A

• Requires energy

92
Q

Is transformation highly efficient?

A

• Low efficiency

o Lot of it happens but not stable

93
Q

Is the DNA often transmitted or lost in transformation?

A

• DNA often degraded and lost and not passed on to subsequent cell lineages

94
Q

Describe the entire process of transformation?

A

• Process-
o One strand of free DNA is digested/degraded, while the other strand is taken up (single strand of DNA taken up)
o The DNA is replicated
o The double stranded DNA is stably incorporated in the transformant cell
 Incorporation occurs through double recombination
 Incorporation happens as crossovers (double recombination events) between transforming DNA and host chromosome occurs, and the transforming DNA replaces the corresponding region in the host chromosome

95
Q

What is the main use of recombinant DNA technology and how is this done?

A

• Recombinant DNA technology
o Plasmids manipulated
 Isolate and purify plasmid to be used to be vectors of horizontal gene transfers
o Transformed into host bacteria
o Stable as plasmids
o Transformants identified by selective gene e.g. antibiotic resistance

96
Q

What is the name of transformation in eukaryotes?

A

• Transformation is transfection in eukaryotes

97
Q

What is the F plasmid?

A

o F plasmid- stands for fertility factor and gives cell ability to donate DNA to neighbour

98
Q

Is F plasmid conjugation unidirectional or bidirectional?

A

o One way transfer-no transfer from recipient back to donor

99
Q

Describe which F plasmid cell is a donor and which one is a recipient

A

o F+ is donor (with F plasmid)

o F- is recipient (without F plasmid)

100
Q

Describe F+ x F- cell conjugation (as well as its results)

A

• Process-F+ and F- cells
o The pili from an F+ cell make contact with an F- cell and a channel forms between them
o As the F plasmid is replicated, it is transferred to the F- cell
 Plasmid is nicked at the oriT (origin of Transfer) sequence on one strand of the plasmid
 Nicked strand peels off and moves to recipient cell through pili
 Intact strand remains in donor and is replicated
o Pili will break, cells will be separated and both cells have the F plasmid and are F+ (recipient also becomes F+)
o Exconjugant (F-) is converted to donor state (F+)
o Donor bacterial genes are not transferred to exconjugant
 Recombinants is not part of process- irrelevant in this time of mating

101
Q

Is the F plasmid always free in cell cytoplasm?

A

o The F plasmid can be free in the cytoplasm or can integrate in the chromosome

102
Q

What is an Hfr cell?

A

o If plasmid integrated in the chromosome, it is an Hfr cell (high frequency of recombination cell)

103
Q

Where does integration in an Hfr cell occur and why?

A

 Integration is primarily at one site but can occur somewhere else
• Crossing over between the plasmid and chromosome integrates the plasmid into the chromosome
• Through single recombination event
• Specific sites that are similar between the F plasmid and the chromosome- integration does not happen randomly

104
Q

Describe HfrxF- conjugation and its process and results

A

o Transfer begins at oriT
o First transfers some F plasmid genes, then some chromosomal genes
o Conjugation and partial T strand transfer due to interrupted mating
 Conjugation doesn’t stay stable long enough for entire chromosome to be transferred so usually just a few genes will come in with portion of F plasmid
o Incomplete oriT means can’t recircularize
 Incomplete F factor segment is degraded by enzymes
o Chromosomal genes may recombine into recipient chromosome
o Replaces recipient allele
 Only if recombined in do genes maintain their stable nature in recipient chromosome
o Recipient remains F-
o Exconjugant (F-) is not converted to donor state (Hfr)
o Donor bacterial genes are transferred to exconjugant
 High frequency of transfer
 Recombination is part of process- can bring bacterial alleles from donor to recipient

105
Q

What is a timed map?

A

 Timed map-plasmid map in minutes: distance in genes mapped in minutes using process of conjugation

106
Q

What are necessary components for mobilisation plasmids in conjugation and what is their function?

A 
•	Components for mobilisation plasmids in conjugation
o	oriT
o	Pilus protein T4SS
	To make a pilus
o	Relaxase
	To make the oriT nick
o	Coupling protein T4CP 
	Involved in maintaining connection with relaxase and pilus
107
Q

What is transduction?

A

• Transduction- transfer of DNA through horizontal mechanism using viruses as vectors through which transfer occurs

108
Q

What do phages affect and what is their composition?

A

• Bacteriophages or phages affect bacteria

o Contain nucleic acid and protective shell around it

109
Q

What is a naked virus made of?

A

• Nucleic acid + capsid protein-naked virus

110
Q

What is an enveloped virus made of?

A

• Nucleic acid + capsid protein + membrane- enveloped virus

111
Q

What is a T4 phage made of?

A

• Head (with capsid protein), tail and tail fibres-T4 phage

112
Q

Describe the life cycle of lytic phage

A

• Docks on surface
• Injects linear DNA
• DNA recircularises
• Production of phage components, degradation of host DNA
o Tricks host into producing many copies of phage genome and phage components
o Phage heads can sometimes acquire fragments of host DNA-packaging process is sensitive to size but not origin
 When the phage infects a new host, DNA from the previous host is transduced into the new host cell-> recombination can potentially insert DNA into chromosome of recipient
• Assembly of phage particles and packaging of DNA
• Lysis of host and release of new phages in neighbouring environments
• Next infection

113
Q

Describe the lysogenic phage life cycle

A

• Can follow lytic or lysogenic cycles
• Lambda phage (lysogenic phage) can integrate into host genome
o Inject linear DNA
o DNA recircularises
o Integrates in chromosome
• Can remain indefinitely throughout generations
• Through some kind of signal, excision of that phage genome can occur and can be excised out of chromosome
• Then goes on to rest of lytic cycle
• Phage now called prophage

114
Q

Describe how integration of viral DNA in bacterial DNA occurs in lysogenic life cycle

  • Components
  • Process
  • Use
A

 Integration occurs at the att (attachement) sites
• 15bp sequence in phage and bacterial genomes
o attP-phage att site
o attB-bacterium att site
• Site specific recombination at att site catalysed by integrase
o AttL or AttR- recombination of attP and attB
• Used in recombinant DNA technology

115
Q

Describe how phage genome is excised out of bacterial chromosome during lysogenesis

A

o When trigger occurs and virus genome recombines out to go through lytic cycle, normally separates into attP and attB through precise excision and this attP continues through life cycle
o Specialised transduction-Rare
 Imprecise excision occurs and attL and attR separate on different chromosomes
 Requires helper phage for re-infection
 Transduction of gal gene to new host

116
Q

What are the three phages of life cycles and their lethality to the invaded cell?

A 
•	Lytic
o	Kill the cell
•	Lysogenic
o	Kill the cell
•	Temperate- no immediate death
o	Can enter the cell and go through life cycle of can integrate into genome, but DO NOT kill the cell
117
Q

Describe the difference between recombination in meiosis and recombination in transduction (phages)

A

• Recombination in meiosis requires 1000s bp homologous DNA
o 15 base pairs (recombination site length) not normally enough to get recombination-need to be catalysed by enzyme (integrase)
 Site-specific recombination requires short homologous sequence plus enzyme
• Can integrate new genes into genome
• Great way to insert genes into genome

118
Q

What is the impact of integration?

A

• Mechanism of integration can cause rapid adaptation and antibiotic resistance in hospital bacterial populations

119
Q

What is a retrovirus?

A

• Retrovirus- viruses with RNA genome
o Infective particles
o Exist outside cells for a period of time

120
Q

Describe how human immunodeficiency virus integrates its DNA into the host cell

A

• Human immunodeficiency virus is a retrovirus
o To integrate into genome of host cell, reverse transcriptase is needed
 RNA copied to single stranded DNA which is made into double stranded DNA
o DNA then able to integrate into chromosome of host cell using integrase

121
Q

Describe how insertional mutagenesis can occur

A

o As integration event occurs, insertional mutagenesis can occur

122
Q

Have integration events been happening over time in the human and mammalian genome? Give examples

A

o Many infections over time and indications in human and mammalian genomes of integration events happening over time
 For example, amylase gene has different type of integration and evidence in regulatory region of amylase gene of upregulation through past viral infection through integration

123
Q

What is the percentage of human genome showing remnants of past retroviral infections?

A

8%

124
Q

Describe the retrovirus life cycle

A

• Retrovirus life cycle
o Fusion to and entry of RNA into susceptible cell
o Virus uncoats
o RNA is a template for reverse transcription
o RNA degrades, dsDNA synthesised
o Integrase carries DNA into nucleus, insertion into genome
o Transcription by host
o Viral RNA and proteins packaged, bud off from cell without lysing the cell

125
Q

What is the general composition of transposable elements?

A 
•	Transposable elements
o	Can move within a genome
	Make copies of themselves or jump 
o	Can encode their own movements 
o	Have defined ends
o	They are small: 300bp-10kbp
126
Q

What are type I retrotransposons?

-Composition

A

o Long terminal repeats at ends- same sequence at ends
o Gag and pol genes- these genes can also be seen in retroviruses
 In the middle of the terminal repeats
• Pol- encodes reverse transcriptase
• Gag- part of the capsid protein

127
Q

How do type I retrotransposons move and where can they be found?

A

o Can move from place to place in a genome by reverse transcription of an RNA transposition intermediate
o In eukaryotes
o Copy through transcription and paste (put themselves in new location) mechanism adds copies to the genome
 Increase in number over time of retrotransposons as original copy stays where it is
o Common in mammals

128
Q

What percentage of the human genome is type I retrotransposons?

A

o 40% of human genome

129
Q

Describe insertion sequences

-Composition

A

o Inverted repeat at ends (5’-3’ on top strand on left is same as 5’-3’ on bottom strand on right)
o In the middle, encodes transposase
 Encodes for movement
 Indications that small bits of homology used for transport mechanisms

130
Q

Describe where insertion sequences can be found and how they move

A

o In bacteria
o Move via stem-loop cut and paste mechanism
 Enzyme cleaves target site in sticky end manner
 Ligation occurs-insertion sequence pops in middle
 Gaps are formed as a result
 These gaps are filled in by polymerases
 Result is direct repeat outside of genome sequence

131
Q

What are type II transposable elements in terms of composition and how do they move, and where are they found?

A 
•	Type II transposable elements
o	In eukaryotes
o	DNA intermediate- transpose directly without an RNA intermediate 
o	Similar to insertion sequence elements in bacteria
	Transposase middle
	Terminal inverted repeat at ends
	Cut and paste mechanism 
o	A bit bigger than IS
132
Q

What are transposon Tn5’s, what are they made of, where they can move and where they can be found?

A

• Transposon Tn5
o In bacteria
o Ends are insertion sequence elements, and one encodes transposase
o Insertion sequences at end often mutated and can’t move themselves as an IS
o One or more antibiotic resistance genes between the IS elements
o Can insert into plasmids
o Can insert into other transposons, makes multi-resistant Tn

133
Q

Describe what makes up the human genome and what the impact of this is

A 
•	Exons
•	DNA transposons
•	Retrotransposons
o	LINE
o	LTR
o	SINE
•	Satellite short repeats
•	Introns 
•	Other 
•	Can get recombination between chromosomes because there are similarities between these elements
134
Q

What are advantages and disadvantages of horizontal gene transfer?

A

Advantages-
Possible selective advantage
Neutral variation that may confer a future selective advantage

Disadvantages-
Mutation via insertion of rearrangement
Metabolic toll of replicating extra DNA
Possible harmful transcripts encoded

135
Q

What are spacers and how and when are they made?

A

• Phage goes through lytic cycle and cell will sometimes survive
• Bacteria that survive incorporate genome into a new spacer
• Invading DNA fragmented to less than 30 bp
• Adjacent to 3bp motif, PAM (protospacer-adjacent motif)
o Not random DNA fragment fragmentation
• Added to 5’ position of array of repeats
• Serves as reference library of previous invasions, which will be passed on to daughter cells
• When a new infection occurs, can get transcription of CRISPR array (which includes spacers)

136
Q

What is CRISPR and in what species does it happen?

A

o CRISPR-Clustered Regularly Interspaced Short Palindromic Repeats
 Adjacent to Cas genes (CRISPIR associated genes) which encode for nucleases, helicases and DNA-binding proteins- involved in degradation and recognition of incoming infective viral DNA
 Direct repeats have internal palindrome
 Forms stem and loop hairpin structures once they’re transcribed and when single stranded
 Variable spacer DNA even if same species as is library of past infections (which can vary)
• Specific to isolates, different within a species
 CRISPR array reference library to identify invading DNA
 Bacteria, archaea

137
Q

Describe how CRISPR works

A

• RNA repeat regions can form stem and loop structure (form hydrogen bonds between them)
• Cleaved to small RNAs (guide RNAs)
• Guide RNA hybridises to invading genome and signals degradation mechanism
o Guide RNA is homologous to new viral infection and hence able to hybridise to viral DNA
o Cas9 finds guide RNA and protein catalyses the breakage of the double stranded DNA and fragments the viral DNA
• Degrades invading DNA and makes it non-functional inside cell
• Double stranded cut repairs in 2 different mechanisms
o Error-prone non-homologous end joining (NHEJ)
 Used for gene knockout-some DNA is lost at the cut site
o Homology-directed repair (HDR)
 Uses an oligo template-piece of DNA available to be incorporated in that site and has homology with that site
 Used for gene replacement or knock-in

138
Q

What are key useful properties and components of CRISPR/Cas

A 
•	Key useful properties
o	Searches for a specific DNA sequence
o	Produces double stranded cut
•	Components
o	Manufactured guide RNA 
o	Case9 protein (endonuclease function)
o	Donor DNA, if you want to add it in
139
Q

Describe the process of genetic analysis

A

• Genetic analysis
o Complicated biological process to investigate
o Disrupt process through mutagenesis
o Identify defective genes
o Understand how the process works through understanding details of gene expression

140
Q

Describe the outcomes of screening and genetic analysis

A

• Outcomes of screening and genetic analysis
o Multiple genes identified
o Ordered into a pathway by supplementing mutants with pathway intermediates
o Detailed analysis of each gene can follow

141
Q

What are the key requirements for genetic analysis?

A

• Key ingredients for genetic analysis
o Question to answer
o Suitable organism
 Ethical to mutate
 Easy to keep in the lab
 Simple close relative to model
o Unbiased approach
 No prejudice/preconceived idea of which genes are involved-answer could be all genes
o Clear assay
 Quickly look at them- screen a lot of mutants
o Scale and efficiency
 Need a large scale and large efficiency
o Assigning defects to genes
 Have a mutagenesis protocol that tags the gene in some way
o Follow up with detailed analysis of each gene

142
Q

Describe Beadle’s screening process

A

• Screen-Beadle
o Neurospora crassa wildtype synthesises its own vitamins and amino acids (including tryptophan and arginine)-only simple ingredients needed
 Fungus is haploid- if mutation occurs in a gene, it shows straight away
o Beadle randomly mutated organism through X-rays or UV light and grew it in a nutrient rich medium
 Supplied all amino acids/vitamins so that mutated organisms could still survive
o He replicated isolates or fungus to test media each lacking a single vitamin or amino acid to see what the neurospora crassa could not produce – if it grew in a single amino acid and didn’t grow in the others, means that it could no longer synthesis that single amino acid

143
Q

Describe the Beadle and Tatum experiment

A

• Experiment-Beadle and Tatum
o Isolated genes that contained mutation that in their wildtype form would synthesise amino acid
 E.g. arginine synthesis pathway inspection
o Tested intermediates in amino acid production pathway to see where pathway is disrupted
 Test intermediates of arginine synthesis pathway (ornithine, citrulline) to see where disruption in the pathway occurs

144
Q

Describe how arginine is made from its precursor

A

 E.g. arginine synthesis pathway inspection
• Enzyme A from gene arg4 catalyses precursor production into ornithine
• Enzyme B from gene arg2 catalyses precursor production into citrulline
• Enzyme C from gene arg1 catalyses precursor production into arginine

145
Q

What are unbiased mutant screening strategies?

A

• Unbiased
o Forward genetic screen
 Organism mutated and mutants are screened. Mutants are isolated and mutant genes are identified
o Genome-wide search
 Make mutations in every gene through technology such as CRISPR/Cas9
 Library of organisms

146
Q

What is a biased mutant screening strategy

A

• Biased
o Reverse genetic screen
 Know what gene is involved and deliberately target gene to mutate it
• Doesn’t tell you if other genes are involved

147
Q

Describe the main components of a Drosophila fly embryo

A 
•	Biology
o	Mouthparts- at the front 
o	Thorax-middle: 3 main thoraxic segments 
o	Abdomen- back: 8 segments of abdomen 
	Bristles at each segments
148
Q

Describe the development of a drosophila

A

o After fertilisation of fruit fly, get a single nucleus
o Single nucleus divides into multiple nuclei
 Rapid mitosis
o Nuclei migrate to edges of egg
o Nuclei become individual cells and membranes form
o Pole cell down one end of organism will become germ line later in life cycle
o Gastrolation
o Segmentation
o Completion of embryosis development and hatching
o Three larval stages
o Pupation
o Metamorphisis into adult fruit fly

149
Q

What are the complications and process requirements required due to the fact that drosophila is diploid

A

• Diploid requirements- crosses to find mutations affecting segmentation
o Segmentation homozygous mutants likely to be lethal
 Will not live into adulthood
o Need to keep the mutations in heterozygotes that will live
 Most mutations recessive- heterozygotes wild type
o Could cross heterozygotes to get homozygote mutants (for examination for phenotype)
o Keep track of chromosome carrying the mutation

150
Q

Describe how the drosophila development example fulfilled the requirement of a suitable organism for genetic analysis

A 
o	Suitable organism
	Known embryology
	Small genome
	Rapid generation time
	Eggs laid externally
151
Q

Describe how the drosophila development example fulfilled the requirement of an unbiased approach for genetic analysis

A

 Experiment:
• Male received mutagenesis at a dosage- one gene per genome should be mutated
• Mutants crossed with female that had different visible markers but also carried dominant mutation that gives heat sensitivity (if heated to a specific temperature, will die)
o Female had chromosome inversion to prevent recombination
• Ended up with male flies that contained the mutation (all other flies were discarded) and inverted chromosome (recombinants would not occur)-heterozygote mutants
• Heterozygote mutants crossed with parent female
• Got males and females with mutation (from dad) and inverted chromosome (from mum)-stock 2
o Any unwanted flies are heat killed (due to dominant mutation)
• Females and males carrying mutation (stock 2) crossed together so could get homozygotes with mutation
• Homozygotes can be examined in embryonic or larval stages before they die
o Due to cinnabar marker, knew if flies homozygous for mutation

152
Q

Describe how the drosophila development example fulfilled the requirement of a clear assay for genetic analysis

A

o Clear assay

 Denticle patterns on thoracic and abdominal segments identify what is normal and what is not normal development

153
Q

What is the gap gene in drosophila and what is its effect in wild type vs mutant? Give a specific example

A

 Gap genes
• Wild type-establish pattern of segments in broad area of embryo
• Mutants- loss of broad region that would cover several segments later in development
o Kruppel mutant is missing 8 segments

154
Q

What are pair-rule genes in drosophila d what is its effect in wild type vs mutant? Give a specific example

A

 Pair-rule genes
• Wild type- establish greater definition in embryo and work after the action of gap genes
• Mutants-missing every second segment
o Eve mutant-mutant in eve gene. Wild type product is even-skipped expressed in 7 (out of 14) segments- alternative segments fail to differentiate appropriately

155
Q

What are segment polarity genes in drosophila and what is its effect in wild type vs mutant? Give a specific example

A

 Segment polarity genes
• Wild type- divide pairs of segments into segments and define pattern within segments- expressed at boundary of segments
• Mutants- altered regions within a segment
o Engrailed mutant- mutant in en gene: wild type engrailed expressed in segment boundaries , which is absent in the mutants

156
Q

What are hox genein drosophila and what is its effect in wild type vs mutant? Give two specific examples

A

 Hox gene mutations-encode regulatory molecules in development
• Wild type-establishes identity of a region of the embryo
• Mutants- a region of an embryo changes identity
o Antennapedia mutant- mutation in upstream regulatory region : normally expressed in mid-thoracic segment
 In mutant, regulation of gene is changed
o Haltere mutant-ultrabithorax homeotic gene in flies is normally expressed from 3rd thoracic segments to abdomen and gives rise to halteres and the 3rd pair of legs
 In mutant, have second pair of wings instead of halteres due to different regulation pattern

157
Q

Describe the gene order in the development of drosophila embryos

A

 Gap genes -> pair rule genes -> segment polarity genes and hox genes is the order of development

158
Q

Describe how the drosophila development example fulfilled the requirement of assigning defects to genes for genetic analysis

A

 Assigning mutations to genes
• Isolated 641 different mutant lines
• When did complementation test, found mutants related to 121 different genes
• Mapping techniques and complementation studies to work out how many genes are affected in the process
o Saturation in genetics screens
• Screening all genes

159
Q

Describe genetic screens in terms of features and saturation

A

o Saturation in genetics screens
 Number of mutants and number of genes isolated on 2 different y axes
 Number of chromosomes tested on x axis
 Efficiency of finding additional genes decreases as the screen continues
o Efficiency of screen depends on mutagenesis method and ability to identify mutants

160
Q

What is the difference in animals vs plants in terms of determination of basic body plan?

A

• Animals
o Basic animal body plant set down in embryogenesis and then growth occurs
• Plants
o New organs are added through lifespan

161
Q

Describe normal flower structure

A 
o	Flower structure
	4 concentric whorls of organs
•	Sepal-encloses the bud before it opens up
•	Petal 
•	Stamens (pollens) 
•	Carpels (ovules)
162
Q

Which flower is most commonly used in the lab?

A

o Use of flower

 Arabidopsis-used in lab due to easy growth

163
Q

Describe genes which control flowering in plants

A 
•	Genes-
o	AP2- Apetala2 gene
o	Pi-pistillata gene
o	AP3-apetala3
o	AG-agamous
164
Q

What genes are responsible for septal development in plants?

A

o Septal: AP2 expression

165
Q

What genes are responsible for petal development in plants?

A

o Petal: AP2, Pi and AP3 expression

166
Q

What genes are responsible for stamen development in plants?

A

o Stamen: AG, Pi, AP3 expression

167
Q

What genes are responsible for carpel development in plants?

A

o Carpels: AG expression

168
Q

Describe the ABC model of gene expression in terms off flower whorls

A

o 3 areas of gene expression
o Different combinations of homeotic gene activity
o Each whorl has a different combination of gene activity
 Whorl 1-A class gene activity
 Whorl 2- A+B class gene activity
 Whorl 3—B+C class gene activity
 Whorl 4- C class gene activity
 A and C-class gene activity is mutually antagonistic- when one is absent, the other takes over

169
Q

Describe the appearance of A-class flower mutants and the cause

A 
•	A-class mutant (apetala2 no longer normally expressed)
o	2 outer whorls affected:
	Sepals changed to carpels
	Petals changed to stamens
170
Q

Describe the appearance of B-class flower mutants and the cause

A 
•	B-class mutant (apetala3 or pistillata no longer normally expressed)
o	2 middle whorls affected
	Petals changed to sepals
	Stamens changed to carpels
171
Q

Describe the appearance of C-class flower mutants and the cause

A

• C-class mutant (agamous no longer normally expressed)
o 2 inner whorls affected
 Stamens changed to petals
 Carpels changed to sepals (repeat structure)
• Multiple level of petals/sepals in flowers

172
Q

Describe the appearance of BC-class flower mutants and the cause

A

• BC double mutants (apetal3 agamous no longer normally expressed)
o Additive effect-two mutations considered together in order to predict organs in flower
o Sepals in entire flower

173
Q

Describe the appearance of AC-class flower mutants and the cause

A 
•	AC double mutants (apetala2 agamous no longer normally expressed)
o	Novel organs
	1st whorl- leaf-like carpels
	2nd whorl- petal-like stamens
	3rd whorls- petal-like stamens
	4th whorl-leaf-like carpels
o	Repeat 
o	Suggests that basal organs are leaves
174
Q

Describe the appearance of ABC-class flower mutants and the cause

A

• Floral triple mutant (ABC)
o All leaf-like carpels
o Flower organs may be derived from leaves
o A,B,C gene activity modifies leaves into specialised organs

175
Q

Is gene position expression related to structure?

A

• Different positions in which the genes are expressed has got a lot to do with the overall structure

176
Q

How can genes be ordered in a pathway?

A

• Order of genes in a pathway
 Used nutritional intermediates to bypass the mutations in the pathway to discover the order of the steps
o Conditional mutants

177
Q

Describe the cell cycle of yeast saccharomyces cerevisiae and why it is a good organism to study eukaryotic mutants

A 
•	Yeast saccharomyces cerevisiae used 
o	Single celled organism
o	Spends most of its life as haploid, so each mutation can affect phenotype
o	Cell cycle is vital process so mutants might be lethal 
	So need to view mutants but switch off mutagenic lethality in organism 
•	Cell cycle-
o	G1 phase-growth
o	S- DNA synthesis phase
	Bud emergence
	DNA replication
o	G2-second growth phase
	Nuclear migration
o	Mitosis/meiosis- cell division
	Spindle formation 
	Chromosome segregation and nuclear fission 
	Cytokinesis
178
Q

What is a conditional mutation?

A

o A type of conditional mutation

 Effect is seen only in some conditions and not others

179
Q

Describe the use of temperature sensitive mutations in yeast and their specific temperatures

A

o Protein product has full function at regular (permissive) temperatures
 Yeast growing temperature- 22 degrees: permissive temperature
 Yeast will make a daughter cell and will go through entire cell cycle
o Protein product destabilises with heat and no functional ability (restrictive temperature)
 At 36 degrees in yeast, protein product is no longer stable and cannot function
 Cell cycle will stop at specific point-can identify mutants

180
Q

Describe how a temperature shift experiment with temperature sensitive mutants is performed

A

o Temperature shift experiment-
 A single isolate was used to inoculate the flask and a population of yeast cells all with the same mutation grow and divided at the permissive temperature
 They are all at different cell cycle stages
 Move flask to restrictive temperature and over time, all mutants will go through cycle they are able to do but will eventually get stuck due to the mutation
• Overtime, they will all arrest at the same position of the cell cycle -this can then be identified as the phenotype of the mutation

181
Q

Describe the results and uses of temperature shift experiment with temperature sensitive mutants, and how many yeast mutants and genes were isolated in this experiment

A

o Results of temperature shift experiment-
 Cells eventually all arrest at the same stage of cycle
 Mutation is in a gene encoding a product that is needed for that stage of the cell cycle
 Gene product is not needed for other stages
 Isolated 150 mutants that arrested at same position of cell cycle
 Using mapping and complementation studies, isolated 32 genes associated with these mutations

182
Q

What are cdc mutants?

A

cell division cycle mutants

183
Q

Describe gene order in the yeast cycle

A 
	Cdc28 at the start of G1
	Cdc4 and cdc6 for bud emergence
	Cdc9 and cdc2 for nuclear migration 
	Cdc13 for beginning of mitosis
	Cdc14 and cdc15 for chromosome segregation and nuclear division
	Cdc3, cdc10 and cdc11 for cytokinesis
184
Q

Describe the role of cdc28

A

• Cdc28-cyclin-dependent kinase responsible for start

185
Q

Describe the role of cdc6

A

• Cdc6-part of pre-replication complex

186
Q

Describe the role of cdc4

A

• Cdc4-part of ubiquitin ligase complex that targets proteins for degradation

187
Q

Describe the role of cdc2

A

• Cdc2-DNA polymerase δ evolved in lagging strand synthesis

188
Q

Describe the role of cdc9

A

• Cdc9-DNA ligase

189
Q

Describe the role of cdc13

A

• Cdc13- Telomerase regulation

190
Q

Describe the role of cdc14

A

• Cdc14- phosphatase needed for mitotic exit

191
Q

Describe the role of cdc15

A

• Cdc15- kinase needed for mitotic exit

192
Q

Describe the role of cdc3

A

• Cdc3-septin ring component

193
Q

Describe the role of cdc10

A

• Cdc10-septin ring component

194
Q

Describe the role of cdc11

A

• Cdc11- septin ring component

GEGE2901 (4 decks)

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