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Flashcards in Exan Dos Deck (45)
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1

Proof of crossing over

Creighton and McClintock

2

What did creighton and McClintock conclude

Crossing over does involve a physical exchange between homologous chromosomes

3

Advantages of using bacteria for genetic studies

Lots of progeny
Can be genetically engineered
Reproduction is rapid
Growth is easy with little space
Easy to isolate

4

Prototroph

Can grow on minimal media

5

Auxotroph

Mutant that requires additional nutrients

6

Lederberg and Tatum

Conducted an experiment to determine if bacteria can transfer genetic material

7

Purpose of U tube experiment

To understand if bacteria must touch in order to exchange genetic information

8

Davis

U tube experience for to understand if cell contact is required for transfer of genetic information

9

Conjugation

Temporary fusion of two single cell organisms for the sexual exchange of genetic material

10

Bacteria cells that can transfer genetic material through fusion

F+ and F- cells

11

F-

Recipient cell

12

F+

Contains F factor and donates through pili hat contact F-

13

Process of Conjugation

Pilus of F+ come in contact with F-
Conjugation tube forms
One strand of F factor is nicked and the 5’ end leads the way to the F- cell
Replication of the F factor occurs

14

Benzer experiment

Wondered if mutations were all in same gene or if several genes were controlling lysis
Coinfext bacterial with two strains of bacteriophage and see if normal plaques are produced

15

Population genetics

Study of inherited variation within and between populations over time and space

16

CIs

Mutation one same chromosome

17

Trans

Mutation on different chromosomes

18

Cistron

Place where complementation cannot occur (gene)

19

Two mutations in same gene

Alleles

20

Two mutations on different genes

Nonalleles

21

Lytic cycle

Phage infects
Injects genetic material
Takes over
Replicates
Creates new phages
Lysis of cell

22

Plaque

Seen on lawn of bacteria when lysis occurs

23

Lysogenic cell

Bacterial cell with a prophage

24

Prophage

A bacterial cell with a phage integrated into the chromosome

25

Life cycle of a temperate phage

Lytic or lysogenic

26

How do we use conjugation to map?

Combine several Hfr x F- makings

27

Interrupted Mating

Combining several HFR x F- pairs to map circular bacteria because different Hfr strands integrate in different places if the bacterial chromosome and will begin transfer af different places and sometimes different directions

28

F’ cells

F factor within an HFR pops out of the bacterial chromosome to produce an F’ cell

29

F’ xF-

F’ merezygote (partial diploid)

30

Transformation

Exogenous DNA transfers genes to competent bacterial cell and brings about heritable change in the cell

31

Result of transformation

Only one daughter cell is transduced

32

How can we use translation for genetic mapping

Cotransformation to determine if genres are located near each other

33

Generalized transduction

Bacteria are infected with phage
Bacterial chromosome is fragmented and some of the genes become incorporated into a few phages
The cell lysis releases the phages to attach and release bacterial genetics instead of the phage genetics and create a transduced bacterial cell

34

How can we use generalized transduction for mapping?

Two genes co transduce if they are close together

35

Specialized transduction

Prophage pops out of bacterial chromosome and takes one or more genes with it
The progeny virus transfer these genes when infecting other cells

36

How can we use specialized transduction for genetic mapping

Not useful since the only genes are affects are those next to the integration site

37

Cell contact required in conjugation?

Yes

38

Cell contact required in transformation?

No

39

Cell contact required in transduction?

No

40

Conjugation sensitive to DNase?

No

41

Transformation sensitive to DNase?

Yes

42

Transduction sensitive to DNase?

Yes

43

Hardy Weinberg law

Allele and genotypes frequencies will arrive at and remain at equilibrium frequencies after one generation of random mating of all sssumotions are met

44

Assumptions of hardy Weinberg

Infinitely large population
Random mating
No selection (all genotypes equally fit)
No migration
No mutation

45

At equilibrium.. what equation represents genotypic frequency?

P2AA + 2pq Aa + q2 aa
p and q are the frequencies of the A and a alleles respectively