Love

Question 0

Synonymous mutation

Answer 0

No change in amino acid.

Question 1

Transition and trans versions?

Answer 1

Transition is purine to purine or pyramidine to pyramidine. Occur more frequently.
Transversions are purine to pyrimadine or vice versa.

Question 2

Missense mutation?

Answer 2

Change in amino acid coding

Question 3

Conservation amino acid change is when?

Answer 3

When there is a change in aa coding but the both aa are chemically similar.

Question 4

Non conservative change is?

Answer 4

When there is a change in aa and both aa are chemically dissimilar.

Question 5

Silent mutations?

Answer 5

No effect on the aa.

Question 6

Two types of missence mutation?

Answer 6

Conservative and non conservative

Question 7

Nonsense mutation?

Answer 7

Stop codon at a spot which abruptly stops translation.

Question 8

Base deletion causes?

Answer 8

Frame shift mutation causing a downstream change.

Question 9

Spontaneous mutation rate?

Answer 9

All the time everywhere and only sometimes they are corrected. Each gene has a basal mutation rate no external influence is required.
Errors in dna replication and spontaneous lesions such as depurination/deamination.
G-c can be converted to A-T = transition mutation events occur.

Question 10

Induced mutations done how?

Answer 10

Replace a base
Base alteration
Damage a base

Question 11

What are the variations in base pairing and how does it happen?

Answer 11

Guanine can take the enol form and bond with thymine
Thymine can take the enol form and bond with guanine
Cytosine can take the imino form and bond with adenine
Adenine can take the imino form and bond with cytosine.
These forms are possible but are rare so the other dominant forms are seen.
The change is spontaneous and the rare structures are protonated.

Question 12

Frame shifts occur normally where in dna?

Answer 12

At repetitive sequences. Loops occur to add or delete bases this is called slipped pair mis pairing. The 2 products of which one will be normal and other will be a mutant.

Question 13

What is a diagnostic test for huntingtons?

Answer 13

If a sequence is repeated over 39. Autosomal dominant disease.

Question 14

Where do most mutations occur?

Answer 14

At repetitive sequences

Question 15

Spontaneous lesion example

Answer 15

Depurination. Removing a base but which can be repaired and it is more efficient when there is a high fidelity.
Deamination another example. Cytosine to uracil.

Question 16

How does repairing of a spontaneous lesion occur?

Answer 16

Dna glycosylase cleaves the base-sugar bond and the ap endonuclease makes cut. Drpase removes stretch of dna. Polymerises synthesises new dna. Ligase seals nick.
High fidelity you done see a consequence.

Question 17

Transition mutation events increase using ?

Answer 17

Base analogues cause an increase mutation rate and are used to induce mutation and cause a fast mutation than spontaneous mutation. Alternative pairing using 5-BU increase transition mutation rate. At to gc and vice versa. Occurs both ways.

Question 18

Induced mutations

Answer 18

Base analogues 2 amino purine. When protonated change the thing it is bonded to and so it is replicated differently.

Question 19

Induced mutations - specific mispairing uses which agent?

Answer 19

Alkylationg agents - ethylmethane sulphonate (ems) which adds an ethyl group which nitrosoguanidine (ng) adds a methyl group. These convert ONLY gc to at transition mutation events in prokaryotes.

Question 20

In induced mutations what do intercalating agents do?

Answer 20

Intercalators are frame shifting mutations I.e addition or deletion which are inserted into the double stranded dna and is fluorescent. E.g proflavin. It doesn’t revert because it only adds and deletes.

Question 21

Induced mutation - base damage done by?

Answer 21

Uv radiation can give rise to pyramidine dimers. The repair of these dimers result in : transitions, trans versions, frame shift mutations (duplications, addition and deletion).

Question 22

What are the minimal medium or a defined medium requirements for an environment for prototrophic bacterial growth?

Answer 22

Carbon source- glucose, lactose, glycerol
Nitrogen- nh4
Phosphorus source -po4
Sulfur source
Cations
Ph buffer
Concentration of everything is controlled by the scientist and nothing extra

Question 23

Autotrophs can’t survive in a defined medium because?

Answer 23

They are unable to make certain amino acids and so must have the required amino acids added to the medium.
The alternative to using a defined medium is to use a complex medium (= broth); cells usually grow more quickly because many compounds in medium are ready-made and do not have to be made by the cell. A broth is usually made from the acid hydrolysis of protein rich material, e.g. yeast.

Question 24

Optimum temperature for E. coli is?

Answer 24

Optimum for E. coli is 37oC but range is 10°C-45oC. Wide differences in temperature optima among bacteria – from

Question 25

psychrophiles and hyper thermophiles grow in temperatures that range from?

Answer 25

psychrophiles (grow best in cold, <15oC) to hyper thermophiles (grow above 90oC).

Question 26

Optimum ph for acidophile, alkaphile and E. coli is?

Answer 26

Optimum pH for E. coli and for most organisms is ~7, but acidophiles (pH ~2) and alkalophiles (pH ~9-10) also exist.

Question 27

What do you need to enumerate bacteria?

Answer 27

You need a solid surface to enumerate bacteria. This is done by microwaving agar and putting it on the plate. Agar is the nutrient medium required by the bacteria.
1.​enumeration of cells in liquid – sample diluted and spread, agar plate incubated, and colonies counted = number of viable cells on plate (as for phage).

Question 28

How do you know what is the number of colonies after the night?

Answer 28

Number of colonies/volume plated X dilution.

Question 29

selection of minority types (e.g. mutants) from majority type is done by?

Answer 29

2.​selection of minority types (e.g. mutants) from majority type, by colony appearance or ability to grow.

Question 30

How do you purify a strain?

Answer 30

3.​strain purification – sample streaked on surface of agar, incubated, cells in single colony picked to determine characteristics or to start an experiment. In this way one starts with progeny of one cell, i.e. pure strain.
“Streaking for single colonies” is a crucial step at one or more stages in nearly all molecular genetic experiments.

Question 31

How can E. coli be resistant to a phage?

Answer 31

Mutation on a membrane protein so the phage cannot adsorb

Question 32

What is the life cycle of a virulent bacteriophage?

Answer 32

Free phages adsorb to host cell. Entry of phage nucleus acid. Phage proteins synthesised and genetic material replicated and the host chromosomes then degraded. Assembly of phages within Host cell.

Question 33

What are the major structural components of bacteriophage t4?

Answer 33

Head, neck and collar, core, sheath, end plate and fibres. ​Early measurements employed ultrafiltration and ultracentrifugation, suggesting sizes of 0.01-0.1m and masses of 1-4 x 10-16g. Early electron microscope observations confirmed these measurements and revealed a complex structure.

Structure
·​Basically head plus tail, but there is considerable variation – some B. cereus phages have small heads and long, thick tails; coliphage T3 have a head and a very short tail. Later analyses of phages revealed heads with no tails (fX174), curly tails (P1), a head only = filament (M13), a head with a lipid layer (PM2).

Composition
·​All early investigations were of phages (e.g. T phages) having approximately 50% mass = protein, 50% = double-stranded DNA.
·​Subsequently, other nucleic acids, e.g. ssDNA (M13, fX174) or ssRNA

Question 34

What is the life cycle of T4 bacteriophage?

Answer 34

Transcription, replication, recombination, translation, maturation, cell lysis.

Question 35

What are plaques?

Answer 35

Zones of lysis on the agar plate

Question 36

Ellis and Delbrück carried out the one-step growth experiment (1939). Describe the steps and what would the results be?

Answer 36

1. ​Log-phase E. coli cells mixed with T4 phage for adsorption.
2. ​Mixture diluted into fresh medium – growth restarts, further adsorption prevented.
3. ​Samples taken at various times, plated on bacteria in soft agar to enumerate infective centres (= infected cells + free phage).

Results
·​Titre of infective centres stays constant for ~25 min. = latent period – time between infection and lysis.
·​Titre rises rapidly over next 5-10 min = rise period – period of lysis.
·​Titre reaches plateau level, which is the ratio of final titre to initial titre = burst size – average number of progeny phage per infected cell.

·​During latent period, each plaque is due to an infected cell – lyses on plate.
·​After cell lysis, each plaque is due to a free phage particle.
·​During the rise period, plaques are due to both. It is important to remember that , plating an infected cell yields a single plaque no matter how many phages are contained within it at the time of plating.

Question 37

What is happening inside the cell during infection?

Answer 37

J

Question 38

What are the stages of cell growth on a liquid medium?

Answer 38

Initially, cells do not grow – lag phase – nutrients are absorbed and energy metabolism gradually restarts. Then growth begins, and soon reaches its maximum rate – exponential (log) phase. Usually manipulations are carried out on cells at this stage.

When the culture broth is thick, nutrients are exhausted, O2 is limiting or toxic products accumulate, and cell growth slows then stops – stationary phase – (culture is saturated). Cells usually start to die after a long time in stationary phase.

Growth can be followed (measured) by counting the cells or, more conveniently, by measuring turbidity of culture samples in a spectrophotometer.

Question 39

What are the stages of cell growth on a solid medium

Answer 39

Growth of cells on a solid support involves a medium exactly as for liquid, but with agar added to gel (solidify) it, in Petri dishes.
The advantage over liquid cultures is that every cell forms its own culture (colony) where the cell happened to land on the plate, i.e. every cell is a colony: the progeny of a single parent cell.
Used for:
1.​enumeration of cells in liquid – sample diluted and spread, agar plate incubated, and colonies counted = number of viable cells on plate (as for phage).
2.​selection of minority types (e.g. mutants) from majority type, by colony appearance or ability to grow.
3.​strain purification – sample streaked on surface of agar, incubated, cells in single colony picked to determine characteristics or to start an experiment. In this way one starts with progeny of one cell, i.e. pure strain.
“Streaking for single colonies” is a crucial step at one or more stages in nearly all molecular genetic experiments.

Question 40

How much dna is in the head of phage?

Answer 40

130 x 106 Daltons of DNA in each particle of phage T2. 200000 base pairs

Question 41

The formation of plaques is useful in the same way as the formation of bacterial colonies

Answer 41

·​for enumeration.
·​for selecting or recognising mutants.
·​for purifying phage strains.

Question 42

What is an eclipse period?

Answer 42

The time during which an infected cell does not contain any mature infectious phage particles, but rather the component parts of these particles, corresponds to the eclipse period.Hence, for the first 10 minutes after infection, no infective phages are present = eclipse. After eclipse, the progeny of the infecting phage appear and increase linearly with time until host cell lysis.

Question 43

What are the two paths for temperate bacteriophage?

Answer 43

Lytic and lysogenic paths

Question 44

Process of the temperate bacteriophage for the lytic pathway?

Answer 44

Chromosomes expressed then viral assembly and then cell lysis

Question 45

Process of the temperate bacteriophage for the lysogenic pathway?

Answer 45

Infection, recombination and integration and the cell divides, progeny of the cells have the dna of the phage. Lastly prophase induction and the cell lysis.

Question 46

G-ve

Answer 46

·​flagellum, common pilus (fimbria), sex pilus.
·​cell envelope ~20% dry weight; ~½ protein, ~100 different proteins.
·​includes: lipopolysaccharide (antigenic determinants), outer membrane (uptake of glucose, vitamins, metal ions, etc.) and inner membrane (energy generation on internal surface). Membranes sandwich peptidoglycan (murein) rigid layer – responsible for cell shape, target of lysozyme, penicillin, and periplasmic space.
·​cytoplasm – contains ribosomes, many enzymes, and the chromosome (= nucleoid). Each cell has only a single chromosome, and frequently has other small, dispensable DNA molecules called plasmids.
readily exchange DNA with each other, but not with G +ve cells. Don’t take up stain.

Question 47

G+ve

Answer 47

spores, secretion of proteins, express G +ve genes only.

​G +ve differs from G –ve in having a single (inner) membrane containing teichoic acid. In both types, chromosome attached to the cell envelope – most clearly seen in G +ve cells (mesosome).

Typical G –ve​=​E. coli
Typical G +ve​=​Bacillus subtilis

A more detailed means of classification is based on enzyme activities.
Rationale: bacteria occupying specific niches tend to have particular growth requirements and hence particular enzyme activities – detection of these in the lab can be used for classification.

Other classification criteria:
·​morphology and staining
·​immunological – surface antigens
• phage sensitivity – cell wall for adsorption, cytoplasm for replication.

Question 48

Conjugation

Answer 48

transfer of DNA via cell-to-cell contact

Question 49

fertility factor, or F

Answer 49

small circular mini chromosome.
The ability to transfer DNA by conjugation is dependent on the presence of a cytoplasmic factor termed the fertility factor, or F (small circular mini chromosome), which carries approximately 100 genes.

​cells carrying F in the cytoplasm are designated F+
​cells not carrying F are designated F-

Question 50

Conjugation mapping requires even or odd crossing over events?

Answer 50

Even

Question 51

Conjugation

Answer 51

The ability to transfer DNA by conjugation is dependent on the presence of a cytoplasmic factor termed the fertility factor, or F (small circular mini chromosome), which carries approximately 100 genes. there is integration of incoming DNA into the recipient chromosome.

Question 52

Transformation

Answer 52

uptake of naked DNA from the environment. there is integration of incoming DNA into the recipient chromosome.

Question 53

Transduction

Answer 53

transfer of DNA from one cell to another via bacteriophage. there is integration of incoming DNA into the recipient chromosome.

Question 54

What facilitates conjugation

Answer 54

(1)​F replicates and segregates into daughter cells.
(2)​F+ cells produce pili, allowing attachment to other cells and to maintain contact with them.
(3)​F+ cells can transfer a newly synthesised copy of the F genome to F- cells, thereby converting them to F+.
(4)​F+/F+ cells are contact inhibited.
(5)​F can integrate into a host chromosome so that upon conjugation host chromosome markers can be transferred to a recipient cell. The level of transfer is small but detectable because genetic recombination can occur in the recipient cell between the incoming DNA and the recipient chromosome.
​Only a small number of F+ cells have F integrated into the chromosome. These special cells can be isolated to give pure populations.

Question 55

genetic transfer occurred between strains is in what direction?

Answer 55

unidirectional

Question 56

Treated strain A culture with streptomycin (prevents bacterial protein synthesis, but marker transfer can still take place), washed out the streptomycin, then mixed with strain B and plated samples to minimal medium. Prototrophs were obtained. The reciprocal experiment led to no prototrophs! Transfer is not reciprocal.
Therefore all detected genetic recombinants took place in strain B. why?

Answer 56

Reason: Strain A was the donor strain (F+) in that it carried F, while strain B was the recipient F- cell. Approximately 1/1000 F+ cells are in fact Hfr, which accounts for the low frequency of gene transfer we see when using an F+ population of cells.

Question 57

What is the f- factor, f + factor and Hfr

Answer 57

F- no f factor and is a recipient
F+ - has f factor and is seperate from the cell chromosome
Hfr- f factor integrated into cells chromosome. So dna of cell is dragged along when recombination takes place.
F ‘ - f factor removed

Question 58

modified F, called F’, can be transferred upon conjugation to a recipient cell so that the recipient becomes a

Answer 58

Partial diploid

Question 59

The recipient cells are scored for the presence of marker alleles from the donor cell. These cells are called

Answer 59

exconjugants

Question 60

interrupted mating. Why is streptomycin added?

Answer 60

At specific time intervals after establishing the mating, samples are removed and blended to disrupt mating then plated to media containing streptomycin to prevent the growth of donor Hfr cells. streptomycin stops the growth of Hfr cells.

Question 61

What can be created from interrupted mating?

Answer 61

A linkage map can be constructed from this interrupted mating experiment using the time at which donor alleles first appear after mating. The distance between markers can be translated into a time scale which is a function of the transfer process.

Question 62

What are the features of the transfer via the transfer process?

Answer 62

(1) ​Donor alleles appear in recipient cells at a defined time.
(2) ​Donor alleles appear in a specific time sequence.
(3) ​The maximum yield of cells containing a specific donor allele is smaller for markers that enter later.

Question 63

transfer of f continues for 2hrs after mating then what happens to the f-?

Answer 63

then some of the F- cells can be converted to Hfr as it requires this long for the terminal portion of F to be transferred.

Question 64

By looking at interrupted mating using a series of Hfr strains the realisation was made that:

Answer 64

​•​F is circular.

​•​the orientation of integration of F into a host chromosome determines the polarity of the F chromosome.

​•​the integration of F involves crossing-over between two rings.

Question 65

endogenote

Answer 65

complete recipient cell genome

Question 66

Rapid lysis mutants

Answer 66

r+ ​E. coli B​ small plaque

​r-​ E. coli B​ large plaque