首页 > > 详细

辅导 BIO2101 Comprehensive Biology Laboratory Exercise 4讲解 留学生SQL语言

BIO2101 Comprehensive Biology Laboratory

Exercise 4:  Activation of the b-galactosidase gene in transformed E. coli

Pre-lab exercise:

1. What is transformation?

2. What is competent cells? What is the function of CaCl2 in the preparation of competent cells?

3. Please describe what is a-complementation?

4. What is the function of IPTG and X-Gal in exercise 4?

Purpose: To transform. E. coli with plasmids containing the b-galactosidase gene

I. Introduction:

Lactose metabolism in E. coli is controlled by a set of enzymes consisting of b-galactosidase, permease and transacetylase. The structural genes (lac Z, Y and A) encoding these enzymes, as well as the gene (lac I) that codes for a repressor protein and the promoter-operator elements constitute a regulatory unit called the lac operon (see appendix).

The coordinated expression of the Z, Y and A genes can be induced by the substrate lactose or its synthetic analog, IPTG (Isopropyl-β-D-thiogalactoside, a lactose analog, see Figure 1 & Figure 2). A chromogenic substrate used in this experiment is X-Gal (5-bromo-4-chloro-3-indolyl-β-D-galactopyranoside). In a cell line that has a full LacZ gene and produces a functional β-gal, when they are plated on media containing X-Gal the β-gal enzyme will cleave the colorless X-Gal and form. an insoluble bright blue precipitate, which will turn the bacterial colony blue. If the β-gal enzyme is not functional the colony will remain white on the plate.

In this experiment, you will study the expression of the b-galactosidase gene in a mutant strain of E. coliDH5α. β-galactosidase enzyme (or β-gal for short) is a naturally occurring enzyme in E. coli cells to cut lactose molecules into glucose and galactose, so the cells can utilize lactose as food source. In DH5α, part of the lac Z gene coding for the amino terminal of b-galactosidase has been deleted and it does not exhibit any b-galactosidase activity (Lac-) because the enzyme is defective. However, when a plasmid (e.g. pUC19) that carries the lac promoter and the missing part of the Z gene (lac Z’) is introduced into DH5α by chemical transformation, active b-galactosidase is regenerated by a-complementation (see Supplementary Information) and the bacteria becomes Lac+.

Transformation is one way to introduce foreign genes into bacteria. When pUC19 is taken up by DH5α, the plasmid becomes part of the extra-chromosomal genetic material and brings new features to the host bacteria.

1) Presence of the lac promoter and the lac Z’ gene will allow gene activation by lactose or IPTG and the expression of the b-galactosidase activity.

2) The product of the ampicillin resistant gene (a b-lactamase that degrades ampicillin) will protect the bacteria in the presence of ampicillin (an antibiotic that kills the bacteria) so that those bacteria carrying the plasmid will survive while those bacteria lacking the plasmid will not (thus, ampicillin selection).

3) The origin of replication in pUC19 will allow the plasmid to duplicate itself during mitosis and the information it carries will be passed onto future generations of DH5α.

In the first part of this exercise, you will be provided with pUC19 plasmid to transform. DH5α. The transformed bacteria will be plated onto agar plates containing X-gal (a chromogenic substrate for b-galactosidase), IPTG (an inducer of lac operon) and ampicillin (an antibiotic, for selecting plasmid containing bacteria).

 

Figure 1

 

Figure 2

Supplementary Information

a-Complementation

Many of the vectors (plasmids constructed for cloning DNA) in current use (e.g. pUC19) carry a short segment of E. coli DNA that contains the regulatory sequences and the coding information for the first 146 amino acids (the operator-proximal region) of the b-galactosidase gene (lacZ).  Embedded in this cloning region is a polycloning site that does not disrupt the reading frame. but results in the harmless interpolation of a small number of amino acids into the amino-terminal fragment of b-galactosidase. Vectors of this type are used in host cells mutated to carry only the carboxy-terminal portion of the enzyme.  In such a system (a-complementation), the deletion mutants of the operator-proximal segment of the lacZ gene are complemented by the plasmid that contains an intact operator-proximal region of the lacZ gene.  Although neither the host-encoded nor the plasmid encoded fragments of the b-galactosidase are themselves active, they can associate to form. an enzymatically active protein.

The bacteria that result from a-complementation are b-galactosidase active (Lac+) and the enzymatic activity can be increased greatly in the presence of an inducer, such as IPTG. The Lac+ bacteria are easily recognized because they form. blue colonies in the presence of the chromogenic substrate X-gal.  Insertion of a fragment of foreign DNA into the polycloning site of the plasmid, however, almost invariably results in the production of an amino-terminal fragment that is not capable of a-complementation. Bacteria carrying recombinant plasmids are therefore Lac- and form. white colonies. The development of this simple colour test has greatly simplified the identification of recombinants constructed in plasmid vectors of this type.  It is easily possible to screen many thousands of colonies visually and to recognize colonies that carry putative recombinant plasmids. 

 

Supplies and Equipment for Experiment

Supplies:

§ pUC19 plasmid, concentration is 50 ng/ml in TE buffer

§ DH5α competent bacteria

§ dd H2O

§ Luria-Bertani (LB) broth media (bacterial media which contain Tryptone 10 g, NaCl 10 g, Yeast Extract 5 g in 1 L H2O)

§ LB plates with 20 mg/ml X-gal and 50 mg/ml IPTG

§ LB plates with 100 mg/ml ampicillin, 20 mg/ml X-gal and 50 mg/ml IPTG

§ PCR Mix

§ PCR Forward Primer

§ PCR Reverse Primer

Note: LB broth and LB plate, is liquid and solidified media, respectively, for culturing bacteria.

Equipment:

§ Eppendorf refrigerated table-top centrifuge

§ Sterile plastic centrifuge tubes, 15 ml and 50 ml

§ Microfuge and sterile 1.5 ml plastic centrifuge tubes

§ 200ul PCR plastic tubes

§ Incubator and metal baths set at 37℃

§ Metal bathes set at 42℃

§ Racks for 1.5 ml tubes

§ Ice buckets

§ Pipetman P 20, 200 and 1000

§ Coating bar

II. Procedure: Transformation of DH5α with pUC19

1. Before the practical session, DH5α cells cultured in LB-broth are grown to mid-log phase and chilled on ice.

2. Place about 40 ml of the bacterial culture in a sterile 50 ml centrifuge tube and spin 4000 r.p.m. (revolution per minute) for 5 minutes at 4oC in an Eppendorf table top centrifuge.

3. Pour out as much as of the supernatant as possible into a waste container provided. Use Pipetman P 1000 to remove the media remained. Do not discard the cell pellet.  

4. Add 4 ml of ice-cold sterile 0.1M CaCl2 to the tube. Gently pipet to resuspend the pellet.

5. Incubate the bacterial suspension on ice for 10 minutes.

6. Centrifuge as Step 2 to collect the bacteria. Then Use Pipetman P 1000 to remove the supernatant.

7. Add 0.4 ml of ice-cold sterile 0.1 M CaCl2 to the tube. Gently pipet to resuspend the pellet. Keep the competent bacteria on ice.

Note: at this point, the CaCl2-treated bacteria are said to be competent.

Note: Step 1 – 7 is done by TA. Following procedures (Step 8 – 13) should be performed on ice for maximal transformation rate.

8. Prepare three 1.5 ml microfuge tubes and have them pre-chilled in ice. Set up the following reactions (a, b, c) and label the tubes accordingly:

(a) 30μl DH5α competent bacteria

(b) 30μl of 0.1 M CaCl2 + 2.5 ml plasmid solution

(c) 30μl DH5α competent bacteria + 2.5 ml plasmid solution

Tap the tubes for several times to mix the content.

9. Incubate the tubes for 20 minutes on ice.

10. Place the tubes in water bath at 42℃ for 45 seconds (heat shock induction).  

11. Immediately put tubes on ice for at least 2 minutes.  

12. Then add 500ul room temperature LB liquid media to each tube. Shake tubes at 225 rpm at 37℃ for at least 45 minutes to recover the bacteria.

13. During the incubation period, prepare PCR replication system in the PCR microtube with plasmid remaining as follows:

Template DNA (Plasmid remaining in PCR tube)

1μl

PCR Mix

10μl

Forward Primer, 10 μM each

0.5μl

Reverse Primer, 10 μM each

0.5μl

ddH2O

8μl

Total volume

20μl

Transfer PCR tubes from ice to a PCR machine and begin thermocycling:

Step

Temperature

Time

Initial denaturation

95°C

2 min

 

35 cycles

95°C

30 s

60°C

30 s

72°C

10 s

Final Extension

72°C

10 min

Hold

4°C

-

Note: TA will collect the samples later and store them until next week.

14.  Gently spread the bacterial suspensions after incubation on the pre-made using cell spreader. After drying the plate surface, close the dish and label every plate lid with group number and sample number.

a-XI-LB: 50 ml of (a) on LB plate with X-gal and IPTG

a-XI-A: 50 ml of (a) on LB plate with ampicillin, X-gal and IPTG

b-XI-A: 50 ml of (b) on LB plate with ampicillin, X-gal and IPTG

c-XI-A: 50ml of (c) on LB plate with ampicillin, X-gal and IPTG

Note: The work of bacterial spreading on LB plates requires aseptic techniques to minimize the air-borne contamination.

15. Place the plates upside down in an incubator at 37℃ overnight.

16. Give tube (c) with transformed bacteria back to TA.

Note: Bacteria remained should be given back to TA for sterilization before disposal. The antibiotic-resisting gene might cause genetic pollution in environment due to “plasmid transfer” phenomenon between different bacteria.

17. TA will collect the plates on the subsequent day and store them at 4oC until next week for your observation.

18. One week later, check the results of your sets of transformation.  Count the number of colonies in each plate and mark down their color on the experimental datasheet.

Experimental Datasheet of Exercise 4

1. Use the following table to report the outcomes of transformations:

Plate

Inoculants

No. of colonies

Color of colonies

a-XI-LB

(a)

 

 

 

 

a-XI-A

(a)

 

 

 

 

b-XI-A

(b)

 

 

 

 

c-XI-A

(c)

 

 

 

 

Key: A = Ampicillin, X = X-gal, I = IPTG

2. Is the plasmid received for transformation “pUC19” or “pUC19 with insert”? Why?

3. Are the outcomes of transformation on every plate the same as your expectation? Try to give possible explaination for every single plate.

 

 



联系我们
  • QQ:99515681
  • 邮箱:99515681@qq.com
  • 工作时间:8:00-21:00
  • 微信:codinghelp
热点标签

联系我们 - QQ: 99515681 微信:codinghelp
程序辅导网!