Lab report Microbial Physiology

Group members

1. Miss Kanoksachinut Ngeonsudja. ID: 5231105036

2. Miss Phratchaya Seeladlao. ID: 5231105037

3. Miss Phatamaporn Sooksai. ID: 5231105039

Objectives

1. To screening & isolation of protease producing from bacteria in Thua nao.2. To understand the method and technique for prepare the pure culture, effect of inducer in SM.3. To practice and understand the relationship between bacteria growth and enzyme protease of OD measurement, protease casein assay and Bradford’s method.

Preparation of sample (Thua nao) & pre-screening

Material & Method

- Sample (Thua nao) 10 g- 1% of peptone solution- Sterilized pipette- Stomacher- Plastic bag- Bunsen burner- 70% of ethanol for cleaning- Test tubes- Spreader- Petri-dish or skin milk agar- Rack

1. Weight 10 g of sample, then rip or break down Thua nao sample in 10 ml of 1% peptonein plastic bag.

2. Spinning in stomacher 200 rpm 2 min.

3. Transfer in bottle and prepared 10 -2 until 10 -8dilution technique

4. Pipette 0.5 ml of 10-6, 10-7 and 10-8 dilution and then drop on skin milk agar by using spread plate technique respectively.5. Incubate at 37˚C and invers plates overnight, then observation.

ResultsSample : Thua nao from FAH THAI market (MAE FAH LUANG University)

Observation: The sample is dry product of fermented soybean, orange and brown colour that show in this figure:1

figure1: The sample (fermented soybean from FAH THAI market)

Table1: Morphology & Characteristic of colonies in Thua nao and select from 10-6 dilution series on skim milk agar.

Morphology & Characteristic Description & Total plates count (TPC)

- Total colonies around 30 colonies , variety of shape filamentous, cocci, mucus and dry and produces pretease enzyme (observed from clear zone around colonies).

- Mainly colonies in this plate filamentous shape and dull , white color and produces pretease enzyme (observed from clear zone around colonies).

Isotate No. θ Clear zone(cm)

θ of Colonies (cm)

Statistic values

Supernatant 1 2 3 1 2 3 R1 R2 R3 R average

SD

NO.1 2.2 2.6 1.7 2.2 1.294118 1.181818 1.237968 0.079408NO.2* 1.5 1.35 1.6 1 0.9 1.15 1.5 1.5 1.391304 1.463768 0.062755NO.3NO.4 1.2 1.45 1.2 0.95 1.2 1.1 1.263158 1.208333 1.090909 1.187467 0.077622NO.5 2.5 1.55 2.25 1.5 1.111111 1.033333 1.072222 0.054997NO.6NO.7 1 1.05 0.9 0.9 0.8 0.8 1.111111 1.3125 1.125 1.18287 0.11247NO.8NO.9NO.10 1.5 1.4 1.4 1.25 1.25 1.25 1.2 1.12 1.12 1.146667 0.035777

Table2: The RI value (Relatives index) of protease activities used supernatant from 10 isolates colonies on Skim milk agar plates

Table3: The RI value (Relatives index) of protease activities used cell from 10 isolates colonies on Skim milk agar plates

Isotate No. Cell

θ Clear zone(cm)

θ of Colonies

(cm)

Statistic values

1 2 3 1 2 3 R1 R2 R3 R average SD

NO.1NO.2 2.1 2.65 1.8 1.5 1.8 1.3 1.4 1.472222 1.384615 1.41894586

90.04677551

7NO.3 0.45 1 0.7 0.3 0.7 0.9 1.5 1.428571 0.777778 1.23545 0.370644NO.4NO.5NO.6NO.7 1.2 1.1 1.2 0.85 0.8 0.7 1.411765 1.375 0.602083 1.129616 0.430318NO.8NO.9NO.10

* From table 2 and 3 showed the efficient of protease enzyme and collect the superior isolate is NO.2 from supernatant (select from the highest of R average value).

Table 4: Showed the morphology of 3 isolates in sample NO.2* from supernatant and picture under microscope used Gram’s staining technique.

Picture of isolates Description

NO. 2 from supernatant

NO.2* NO.1

From the table 4 used Gram’s staining technique under microscope for make sure pure culture that show the structure must be the same and the morphology isolates in triplicate the colonies must be similar.

Relationship between bacteria growth and enzyme protease

- They are 4 analyses for measurements

1. Bacteria growth by measuring OD (optical density) values & pH at 3 days every 24 hr or (0 hr, 24 hr, 48 hr and 72 hr) that showed in Table:5

Table 5: Bacteria growth by measuring OD (optical density) values & pH at 3 days every 24 hr

Time of measure Absorbance values pH

0 hour

NB 0.007 6.0

Skim milk (SM) 0.041 6.0

24 hour

NB 0.288 6.0

Skim milk (SM) 0.328 6.0

48 hour

NB 1.71 8.0

Skim milk (SM) 0.077 8.0

72 hour

NB 1.81 8.0

Skim milk (SM) 0.0064 8.0

2. Protease Casein assay - On skim milk agar drop 10 µl of supernatant every 24 hour in 3 days form NB and SM culture that show in Table 6

Table 6: Showed the activity of protease enzyme on SM agar at 48 and 72 hour

SM at 48 hour & RI values SM at 72 hour & RI values

R1 = 0.8 R1 = 0.95

R2 = 0.75 R2 = 0.6

R3 = 0.9 R3 = 0.65

R v = 0.82 R v = 0.73

- At 48 hour the RI values more than 72 hour that showed the optical growth condition of culture in SM is suitable for produce maximize of enzyme.

3. Enzyme activity- The method used to Tyrosine is standard protein

Reagent

- 800 µl of Tris-buffer, pH =7 (for keep pH constant)

- Enzyme 200 µl incubate at 37 C 5 min * in blank use H ̊� 2O

- 1,000 µl of casein (substrate) , then incubate 37 C in water bath, 30 min ̊�

- 3 ml of TCA for stop reaction

- Centrifuge 10,000 rpm 10 min * collect supernatant

- Prepare the STD tyrosine (100 µg/ml) dilute to 0, 10,20,30, 40, 50 until 100 µg/ml, then measure A 275 and record

Graph 1: The calibration curve between A 275 versus concentration of standard Tyrosine

- The result of R2 = 0.991 that show the trend of each points on the line is high reliable, because

R2 value of Tyrosine standard must be nearly or equal 1.

Thus we can find the concentration of samples NB and SM from calibration curve y = 0.06x - 0.013

Table 7: A 275 of standard Tyrosine [10 – 100 µg/ml] and sample (NB and SM culture)

A 275

valuesSTD tyrosine

µg/mlNB Absorbanceℷ 275 nm

Conc.µg/ml

SM Absorbanceℷ 275 nmConc.µg/ml

0.058 10 µg/ml 0 hour 0.17 30.5 0 hour 0.26 45.50.112 20 µg/ml 24 0.167 30.0 24 0.34 58.830.176 30 µg/ml 48 0.092 17.5 48 0.193 34.330.278 40 µg/ml 72 0.136 24.83 72 0.27 47.70.305 50 µg/ml - - - - - -0.366 60 µg/ml - - - - - -

0.408 70 µg/ml - - - - - -0.514 80 µg/ml - - - - - -0.593 90 µg/ml - - - - - -0.639 100 µg/ml - - - - - -

Find the unit of enzyme

Total volume = 5 ml, reaction of time = 30 min

Sample 1 is NB (0, 24, 48, 72 hour)

In 0 hour A 275 = 0.17 and concentration = 30.5 µg/ml

1 ml 30.5 µg

5 ml 30.5 µg * 5 ml / 1ml = 152.5 µg, that mean in 200 µl of enzyme contain 152.5 µg of product (tyrosine)

S0, we put 1,000 µl of tyrosine = 1,000 µl * 152.5 µg / 200 µl = 762.5 µg of enzyme in 30 min

In 1 min = 762.5 µg * 1 min / 30 min = 25.42 µg/min

Enzyme 1 ml = 25.42 µg/min

MW of tyrosine = 181.19 µg/µmol (25.42 µg / 181.19 µg/µmol) / min = 0.14 µmol/min (U)

The total volume is 1 ml so the unit of enzyme is 0.14 U /ml (in 0 hour of NB culture) and fine in all condition of sample that showed in table below

Table 8: The unit of enzyme in NB and SM culture every 24 hour in 3 day 1 ml of enzyme in 1 min activated

Time of NB/Conc. µg/ml

Enzyme activity (U/ml) Time of SM /Conc. µg/ml

Enzyme activity (U/min)

0 hour, 30.5 0.140 0 hour, 45.5 0.209

24 hour, 30.0 0.138 24 hour, 58.83 0.271

48 hour, 17.5 0.08 48 hour, 34.33 0.158

72 hour, 24.83 0.114 72 hour, 47.7 0.219

Protein assay- Used Bradford’s method

Table 9: Bradford’s reagent

Reagent of Bradford’s method

Reagent (ml)

1 2 3 4 5 6 7 8

0.2 mg/ml of BSA

0 0.1 0.2 0.3 0.4 0.5 - -

Coomessie Blue G

5 5 5 5 5 5 5 5

H2O 0.5 0.4 0.3 0.2 0.1 - 0.3 -

Sample - - - - - - 0.2 0.5

- Wait 10 min for dye complex of protein, then measure A 595 nm

Table 10: A 595 of standard BSA and concentration of BSA (µg/ml)

STD BSA Concentration of BSA

A 595 (µg/ml)

0.146 3.64

0.334 7.27

0.568 1 0.9

0.56 14.5

0.669 18.18

Calculation the concentration standard of BSA and plot graph

Test tube 2 Test tube 3

CV initial = CV final CV initial = CV final

C final = (0.2 mg/ml) (0.1 ml)/5.5 ml C final = (0.2 mg/ml) (0.2 ml)/5.5 ml

C final = 0.00364 mg/ml C final = 0.00727 mg/ml

= 0.00364mg/ml*10-3mg/10-6 μg = 0.00727mg/ml*10-3mg/10-6 μg

=3.64μg/ml =7.27μg/ml

Test tube 4 Test tube 5

CV initial = CV final CV initial = CV final

C final = (0.2 mg/ml) (0.3 ml)/5.5 ml C final = (0.2 mg/ml) (0.4 ml)/5.5 ml

C final = 0.0109mg/ml C final = 0.0145 mg/ml

= 0.0109mg/ml*10-3mg/10-6 μg = 0.0145mg/ml*10-3mg/10-6 μg

=10.9μg/ml =14.5μg/ml

Test tube 6

CV initial = CV final

C final = (0.2 mg/ml) (0.5 ml)/5.5 ml

C final = 0.01818 mg/ml

= 0.01818mg/ml*10-3mg/10-6 μg

=18.18μg/ml

Graph 2: The calibration curve between A 595 versus standard BSA

- The result of R2 = 0.936 that show the trend of each points on the line is high reliable,

because

R2 value of Tyrosine standard must be nearly or equal 1.

Thus we can find the concentration of samples NB and SM from calibration curve y = 0.044x - 0.012

Table 11: The concentration of protein contain in 0.2 of samples at A 595

* 0.2 ml of sample contain X µg 1 ml contain 1ml * X µg / 0.2 ml

tube7NB A ̊595 ̊ Concentrati

on ̊(0.2 ̊µg/ml)

SM A595 Concentration ̊(0.2 ̊µg/ml)

0 hour 0.004 1.8 0 hour 0.532 61.824

hour0.01 2.5 24

hour0.69 79.75

48 hour

0.08 10.45 48 hour

0.323 38.05

72 hour

0.006 2.05 72hour 0.209 21.1

tube8NB A595 Concentrati

on ̊(0.5 ̊µg/ml)

SM A595 Concentration ̊(0.5 ̊µg/ml)

0 hour 0.064 8.635 0 hour 0.684 79.0924

hour 0.069 9.2 24

hour 1.212 139.09

48hour

0.063 8.5 48hour 0.723 83.5

72 hour

0.068 5.9 72 hour

0.435 50.8

Table 12: The concentration of protein contain in 0.5 ml of samples at A 595

* 0.5 ml of sample contain X µg 1 ml contain 1ml * X µg / 0.5 ml

Discussion