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The Steps of the Scientific Method for Kids – Science for Children: FreeSchool.

The hypothesis–prediction approach
traditionally has five steps:
1. Making observations
2. Asking questions
3. Forming hypotheses, or tentative
answers
4. Making predictions based on the
hypotheses
5. Testing the predictions
3. Develop hypothesis
The project will isolate bacteria from SSU campus and conduct transformation using the isolate
bacteria and Escherichia coli (E. coli). Bacterial transformation is a process of horizontal gene
transfer by which some bacteria take up foreign genetic material (naked DNA) from the
environment. It was first reported in Streptococcus pneumoniae by Griffith in 1928. The
prerequisite for bacteria to undergo transformation is its ability to take up free, extracellular genetic
material. Such bacteria are termed as competent cells.
In general, isolating bacteria from soil is an important first step in many microbiology experiments.
Once they are isolated, bacteria can be further analyzed to determine things, such as their species
and their function in the soil environment. Even a tiny amount of soil can contain millions of
bacteria, which makes it necessary to dilute a soil sample before isolating bacteria from the sample.
Activity 1: Students will get samples from soil on SSU campus (Day 2). Develop your hypothesis
for the following questions and provide the reasons how your hypothesis was formulated.
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Q1. Where can you find bacteria on campus?
Hypothesis:
Reasons:
Q2. What types of bacteria could be found from your samples?
Hypothesis:
Reasons:

Q3. How many bacteria could be found in 1 gram of soil?
Hypothesis:
Reasons:

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4. Proposing a set of steps to isolate bacteria from the soil sample.
For isolating bacteria, we could use standard plate count. This technique allows microbiologists
to estimate the population of that samples by plating a very dilute portion of that sample and
counting the number of colonies it produces. The samples will be prepared through serial dilution
procedure.
For preventing contamination, bacterial isolation needs distilled (DI) water and the sterile
tubes/bottles.
Serial Dilution and Plate Counts.

Activity 2: Watch the movie “Serial Dilution and Plate Counts”. Propose a set of steps to isolate
bacteria from the soil sample. Students can use DI water, the sterile tubes/bottles and the plates
containing culture media that contain the nutrients needed to sustain microbes. Consider how to
dilute samples (e.g. How many ml of DI water will be given to how many gram of soil? How many
ml of water and samples should be used for serial dilution?)
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Describe procedure:
5. Nucleic Acids
Nucleic acids are molecules that consist of repeating units called nucleotides. Each nucleotide
contains a pentose sugar (deoxyribose in DNA or ribose in RNA), a phosphate group, and a
nitrogenous (nitrogen-containing) base. There are four nitrogenous bases in DNA: adenine (A),
cytosine (C), guanine (G) and thymine (T). In DNA, adenine always binds with thymine (A-T)
and cytosine always binds with guanine (C-G). In RNA, uracil (U) replaces thymine. Adenosine
triphosphate (ATP) is the energy-carrying molecule in cells. ATP is composed of adenosine
(adenine + ribose) modified by the addition of three phosphate groups. When ATP is hydrolyzed
to form ADP (adenosine diphosphate) and a molecule of phosphate, energy is released for cellular
metabolism.
Basic questions about isolation of genomic DNA
Watch the following two movies and answer the questions
• Nucleic acids – DNA and RNA structure

• DNA isolation from strawberries:

1. Describe structure of DNA at molecular level?
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2. Why is it important for scientists to be able to remove DNA from cells? In your own words,
describe the structure and function of DNA.
3. What does mashing the strawberries do to their cells?
4. DNA extraction buffer contains detergent and salt, what are the purposes of detergent and
salt, respectively?
5. What is the purpose of cold ethanol used in this lab?
6. If you had extracted DNA from animal cells instead of plant cells, what cell barrier would
have been different? What do plant cells have that animal cells do not?
7. Do you think that DNA from animal cells would look the same as DNA from plant cells?
Why or why not?
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Name: ____________________________ Last 4-digits of Student ID number: ________
Project: Bacterial Isolation and DNA Transformation (Day 2) / Lab Note 4/5
Day 2
1. Sampling bacteria
2. Serial dilution and spreading bacteria on plates
3. Kirby-Bauer test (preparation)
0. Discussion about sampling. Discuss the following items with the lab Note 4
• Where will you get samples?
• Types of sample (soil, water, etc)?
• Expected results after inoculation (bacterial growth on plates)
• Confirmation of experimental procedures
1. Sampling bacteria
Activity 3: Students need to obtain one soil sample on campus (within 15 minutes) and record
the location where the samples are collected.
Location:
2. Serial dilution and spreading bacteria on plates
Activity 4: Weigh out 1 g of the soil sample and add it to the sterile tube containing 50 ml of
distilled water. Tightly cap the bottle and shake it to thoroughly mix the solution (Tube A).
Transfer 100 ul of the soil sample to the conical tube containing 9.9 ml of DI water (Tube B).
Repeat the steps to make serial dilution samples (Tubes C, D and E; see below)
Tube/Sample Soil sample DI water Dilution
A 1 g of soil 10 ml Original (1 g soil in 10 ml)
B 1 ml of sample A 9 ml 10-1
(101 dilution)
C 1 ml of sample B 9 ml 10-2
(102 dilution)
D 1 ml of sample C 9 ml 10-3
(103 dilution)
E 1 ml of sample D 9 ml 10-4
(104 dilution)

When your tubes A-E are ready, give 1 ml of each samples to the plate containing culture media.
Label the plates (Your group name and plate names A-E) and transfer the plates to the instructor.
The instructor will incubate the samples at 37Co
for 24 hrs – 5 days (depends on bacteria).
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How to use micropipette

How to use Power pipette

3. Kirby-Bauer test
The Kirby-Bauer test for antibiotic susceptibility, called the disc diffusion test, is a standard that
has been used for years. First developed in the 1950s, it was refined and by W. Kirby and A. Bauer,
then standardized by the World Health Organization in 1961. This test is used to determine the
resistance or sensitivity of aerobes or facultative anaerobes to specific chemicals, which can
then be used by the clinician for treatment of patients with bacterial infections. The presence or
absence of an inhibitory area around the disc identifies the bacterial sensitivity to the drug.
The basics are easy: The bacterium is swabbed on the agar and the antibiotic discs are placed on
top. The antibiotic diffuses from the disc into the agar in decreasing amounts the further it is away
from the disc. If the organism is killed or inhibited by the concentration of the antibiotic, there will
be NO growth in the immediate area around the disc: This is called the zone of inhibition. The
zone sizes are looked up on a standardized chart to give a result of sensitive, resistant, or
intermediate. Many charts have a corresponding column that also gives the MIC (minimal
inhibitory concentration) for that drug. The MIC is currently the standard test run for antibiotic
sensitivity testing because it produces more pertinent information on minimal dosages.
Kirby-Bauer test

Zone of inhibition
Chlorine
Hexachlorophene
O-phenylphenol
Quat
Chlorine Chlorine
Quat Quat
Hexachlorophene
O-phenylphenol
O-phenylphenol Hexachlorophene
Staphylococcus
aureus
(gram-positive)
Escherichia coli
(gram-negative)
Pseudomonas
aeruginosa
(gram-negative)
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Activity 5: Students will isolate/subclone samples from soil on SSU campus (Day 3) and
determine the resistance or sensitivity of the bacteria to specific chemicals. In this Lab, we will
use the following antibiotics: (use 20 ul of each antibiotic)
(1) Kanamycin sulfate (0.5 mg/ml – 10 ug)
(2) Chloramphenicol (0.1 mg/ml – 2 ug)
(3) Cefoxitin sodium (0.5 mg/ml – 10 ug)
(4) Gentamicin sulfate (0.1 mg/ml – 2 ug)
(5) Ampicillin sodium salt (0.5 mg/ml – 10 ug)
(6) Ceftriaxone sodium salt (0.1 mg/ml – 2 ug)
Research characteristics of the antibiotics online or textbooks, develop your hypothesis for the
following questions and provide the reasons how your hypothesis was formulated.
Q4. Describe characteristics of the six antibiotics. Hypothesize your results (response of your
bacteria / zone of inhibition) against the six antibiotics listed above when you will conduct
the Kirby-Bauer test.
Characteristics:
(1) Kanamycin sulfate
(2) Chloramphenicol
(3) Cefoxitin sodium
(4) Gentamicin sulfate
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(5) Ampicillin sodium salt
(6) Ceftriaxone sodium salt
Hypothesis(bacterial growth with antibiotics):
Reasons:
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Name: ____________________________ Last 4-digits of Student ID number: ________
Project: Bacterial Isolation and DNA Transformation (Day 3) / Lab Note 4/5
Day 3
1. Isolation of bacteria (NA plate) and Kirby-Bauer test
2. Introduction to Transformation
1. Isolation and Kirby-Bauer test
Identify one big colony on the plate, pick up a portion of bacteria and spread bacteria to a new
plate with aseptic technique.
Materials
• 70% ethanol
• Bunsen burner
• Forceps
• Micropipette
• Tips
• Disk (4 disks in total)
• Kanamycin sulfate (0.5 mg/ml – 10 ug)
• Chloramphenicol (0.1 mg/ml – 2 ug)
• Cefoxitin sodium (0.5 mg/ml – 10 ug)
• Gentamicin sulfate (0.1 mg/ml – 2 ug)
• Ampicillin sodium salt (0.5 mg/ml – 10 ug)
• Ceftriaxone sodium salt (0.1 mg/ml – 2 ug)
Activity 6
Bacterial isolation
1. Disinfect the workspace with 70% ethanol
2. Sterilize an inoculation loop with flame (Bunsen burner)
3. Transfer bacteria to a new test tube (sterile) containing 2 ml of sterile saline
4. Vortex the saline tube to make a smooth suspension.
5. Dip a sterile swab into the inoculum tube
6. Spread bacteria on a new NA plate by streaking the swab. (Inoculate the plate with the test
organism by streaking the swab in a back-and-forth motion very close together as you move
across and down the plate. Rotate the plate 60° and repeat this action. Rotate the plate once
more and repeat the streaking action. This ensures an even distribution of inoculum that will
result in a confluent lawn of growth.
7. Repeat streaking and make another NA plate with bacteria
8. Give your name and the information of the sample (where did you get your soil)
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Kirby-Bauer test
9. Steps 1-8 is the same as bacterial isolation
10. Sterilize forceps with 70% ethanol and flame
11. Place four disks at even intervals (see figure).
12. Select 4 antibiotics and give 20 ul of antibiotics with micropipette.
13. Store the plates you inoculated bacteria (with and without disks) in the 37Co
incubator.
2. Introduction to Transformation
DNA CAN BE TRANSFERRED BETWEEN BACTERIA
In nature, DNA is transferred between bacteria using two main methods, transformation and
conjugation. In transformation, a bacterium takes up exogenous DNA from the surrounding
environment (Figure 1). In contrast, conjugation relies upon direct contact between two bacterial
cells. A piece of DNA is copied in one cell (the donor) and then is transferred into the other
(recipient).
Frederick Griffith first discovered transformation in 1928
when he observed that living cultures of a normally nonpathogenic strain of Streptococcus pneumonia were able
to kill mice, but only after being mixed with a heat-killed
pathogenic strain. Because the non-pathogenic strain had
been “transformed” into a pathogenic strain, he named
this transfer of virulence “transformation”. In 1944,
Oswald Avery and his colleagues purified DNA, RNA
and protein from a virulent strain of S. pneumonia to
determine which was responsible for transformation.
Each component was mixed with a non-pathogenic strain
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3
4 2
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of bacteria. Only those recipient cells exposed to DNA became pathogenic. These transformation
experiments not only revealed this virulence is transferred but also led to the recognition of DNA
as the genetic material.
The process of gene transfer by transformation does not require a living donor cell but only requires
the presence of persistent DNA in the environment. The prerequisite for bacteria to undergo
transformation is its ability to take up free, extracellular genetic material. Such bacteria are
termed as competent cells. The exact mode of transformation can differ between bacteria species.
For example, Haemophilus influenzae uses membrane-bound vesicles to capture double-stranded
DNA from the environment. In contrast, S. pneumoniae expresses competency factors that allow
the cells to take in single-stranded DNA molecules. In the laboratory, scientists can induce cells—
even those that are not naturally competent—to take up DNA and become transformed. To
accomplish this, DNA is added to the cells in the presence of specific chemicals (like calcium,
rubidium, or magnesium chloride), and the suspension is “heat shocked”—moved quickly between
widely different temperatures. It is believed that a combination of chemical ions and the rapid
change in temperature alters the permeability of the cell wall and membrane, allowing the DNA
molecules to enter the cell.
The factors that regulate natural competence vary between various genera. Once the transforming
factor (DNA) enters the cytoplasm, it may be degraded by nucleases if it is different from the
bacterial DNA. If the exogenous genetic material is similar to bacterial DNA, it may integrate into
the chromosome. Sometimes the exogenous genetic material may co-exist as a plasmid with
chromosomal DNA.
The Mechanism of Transformation with Competent Cells

Plasmids.
A plasmid is a small, extrachromosomal DNA
molecule within a cell that is physically separated
from chromosomal DNA and can replicate
independently. They are most commonly found as
small circular, double-stranded DNA molecules in
bacteria.
Calculation of Transformation Efficiency.
The transformation efficiency is defined as the number of transformants generated per µg of
supercoiled plasmid DNA used in the transformation reaction.
Transformation efficiency is calculated using the formula below:
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Transformation efficiency = Number of colonies on plate / Amount of DNA plated (ng) x
1000
If you gained 250 bacterial colonies on plate after giving 1 ug (=1000 ng) of DNA, transformation
efficiency of this competent cells is 250 / 1000 x 1000 = 250
Required: The more information about transformation, plasmids, control of gene expression
and GFP is available from https://www.edvotek.com/site/pdf/223.pdf
Q5. Describe applications of transformation.
Q6. What is Required in a Typical Transformation Reaction?
Q7. If you gained 100 bacterial colonies on plate after giving 10 ng of DNA, what is
transformation efficiency of this competent cells? Provide your calculation and answer.
Q8. See the figure on the left. A typical plasmid had some unique
genes/elements such as amp, ori and lacZ. Research and describe roles
of the elements.
ori:
lacZ:
amp:
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Read procedures for Transformation of E.coli with Green Fluorescence Protein
(https://www.edvotek.com/site/pdf/223.pdf) and answer the following questions.
Q9: How many microtubes are required for this experiment?
Q10: Each group needs 4 plates. What names should be given to each plate?
Q11: The competent cells will be incubated in the water bath twice. Identify temperature and
time for the first and second incubation
First:
Second:
Q12: On which plates would you expect to find bacteria most like the E. coli on the source
plate? Explain.
Q13: On which plates would you find only genetically transformed bacterial cells? Why?
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Q14. What is the purpose of the control plates? Explain the difference between the controls and
why each one is necessary.
Q15. Why would one compare the -DNA/+Amp and +DNA/+Amp plates?
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Name: ____________________________ Last 4-digits of Student ID number: ________
Project: Bacterial Isolation and DNA Transformation (Day 4) / Lab Note 4/5
Day 4
1. Record zone of inhibition in Kirby-Bauer test
2. Introduction to Green Fluorescent Protein (GFP)
3. Transformation
1. Record zone of inhibition in Kirby-Bauer test
Activity 7: Record the data of zone of inhibition
1. Check mechanisms of Kirby-Bauer test (if required) and
measure zone of inhibition (ZOI) of each antibiotic you
selected on Day 3.
2. Take pictures with your cellphone (for making a report
later).
3. Record your data and complete the Table 1 below.
4. Find two groups that use the same antibiotics, gather
information and compare sizes of ZOI. Antibiotics among the groups do not have to match
completely (e.g., Your group had antibiotics A, B, C and D)
Source of your bacteria
Table 1
Antibiotic ZOI (mm) ZOI (mm)/
Antibiotics (ug)
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Table 2
ZOI (mm) / antibiotics (ug)
Antibiotics Your group
Source __________
Group _____
Source __________
Group _____
Source __________
Kanamycin
(10 ug)
Chloramphenicol
(2 ug)
Cefoxitin
(10 ug)
Gentamicin
(10 ug)
Ampicillin
(10 ug)
Ceftriaxone
(2 ug)
2. Introduction to Green Fluorescent Protein (GFP)
FLUORESCENT PROTEINS
The plasmid that we will be using to transform our E. coli has been engi neered to contain the
DNA sequence the Green Fluorescent Protein (GFP). GFP are small proteins, approximately
27 kilodaltons in size. GFP possesses the ability to absorb blue light and emit green light in
response. This activity, known as fluorescence, does not require any additional special substrates,
gene products or cofactors to produce visible light.
GFP was first isolated from the jellyfish Ae quorea
victoria in the 1970’s. Once scientists identified its
DNA sequence, they were able to use genetic
engineering to introduce fluorescent proteins into
other organisms, such as E. coli and the nematode
Caenorhabditis elegans. Scientists also identified
particular amino acid substitutions in GFP that
altered the behavior of its ‘chromophore’, a special
structure within the protein that is responsible for
light production (Figure 4a). Different changes bring
about different patterns of light absorption and
emission, allowing scientists to develop a rainbow of
fluorescent proteins (Figure 4b). For example, GFP can be converted to BFP by making two amino
acid substitutions, one of which is in the chromophore (His-Tyr). For their discovery and
development of GFP and other fluorescent proteins, Osamu Shimomura, Martin Chalfie and Roger
Tsien were awarded the Nobel Prize in Chemistry in 2008.
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Fluorescent proteins have become an essential tool in cell and
molecular biology. Using DNA cloning strategies, proteins can be
“tagged” with fluorescent proteins and then expressed in cells.
These tags simplify purification because fluorescently labeled
proteins can be tracked using UV light. The most useful
application of fluorescent proteins is as a visualization tool during
fluorescent microscopy studies. By tagging other proteins with
GFP, researchers can determine where those proteins are normally
found in the cell. Similarly, using a fluorescent protein as a
reporter, scientists can observe biological processes as they occur
within living cells. For example, in the model organism zebrafish
(Danio rerio), scientists use GFP to fluorescently label blood vessel proteins so they can track
blood vessel growth patterns and networks. Scientists also tag regulatory DNA sequences with the
GFP coding sequence so they can observe patterns of when and where the gene is expressed. In
this way, GFP can reveal the role these regulatory sequences might normally play in a cell. In
summary, fluorescent proteins, including GFP and BFP, and fluorescent micros copy have
enhanced our understanding of many biological processes by allowing scientists to watch
biological processes in real-time.
A movie: GFP Overview

3. Transformation
Activity 8: Transformation with GFP
Read experimental overview and procedures in the manual “Transformation with Green
Fluorescence Protein”(https://www.edvotek.com/site/pdf/223.pdf)” and complete the experiments
(steps 1-17).
When your 4 plates receive the bacteria with or without DNA and store the plates with inverted
position in the 37oC incubator for overnight.
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Name: ____________________________ Last 4-digits of Student ID number: ________
Project: Bacterial Isolation and DNA Transformation (Day 5) / Lab Note 8
Activity 9: Transformation with GFP / Record results and analyze the data
*1. observe the transformation and control plates using long wave UV light. For each of the plates,
record the following:
-DNA -DNA/+Amp +DNA/+AMP +DNA/+AMP/
+IPTG
The number of
colonies on the plate
What color of the
bacteria without UV
light?
The color of the
bacteria under UV
light / the number of
colonies showing GFP
*Corresponding to step 18 in the manual “Transformation with Green Fluorescence
Protein”(https://www.edvotek.com/site/pdf/223.pdf)” and complete the experiments.
Q16. Calculate your transformation efficiency.
• Calculate transformation efficiency (check Day 3)
Transformation efficiency =
Number of transformants (colonies on +Amp plate) final volume (ml) at recovery
X ug DNA volume (ml) plated
Quick reference of this experiments
Number of transformants (colonies on +Amp plate) 0.5 ml
0.05 ug 0.25 ml
X
X
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Activity 10. Record the number of colonies showing GFP under UV light. Gather data of other
groups
The number of colonies on the plate Your group Group ________ Group ________
+DNA/+AMP
+DNA/+AMP/+IPTG
Student’s t-test using Excel
The t-test is any statistical hypothesis test in which the test statistic follows a Student’s tdistribution under the null hypothesis. A t-test is the most commonly applied when the test statistic
would follow a normal distribution if the value of a scaling term in the test statistic were known.
When the scaling term is unknown and is replaced by an estimate based on the data, the test
statistics (under certain conditions) follow a Student’s t distribution. The t-test can be used, for
example, to determine if the means of two sets of data are significantly different from each other.
How To Perform T-Tests In Microsoft Excel

Activity 11. Regarding the number of colonies showing GFP under UV light, compare the results
of two teams (your team and Team X-Y-Z) with Excel (Student’s t-test)
The number of colonies on the plate Group X Group Y Group Z
+DNA/+AMP 0 0 0
+DNA/+AMP/+IPTG 225 215 190
p value = .
The numbers in the two teams are significantly different Yes or No
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Activity 12. (Homework) Select one antibiotic from your Table 2 – Day 4. Using the Data set B
below and compare two groups.
Data set B
ZOI (mm) / antibiotics (ug)
Antibiotics Group A Group B Group C
Kanamycin
(10 ug)
10 8 5
Chloramphenicol
(2 ug)
8 3 6
Cefoxitin
(10 ug)
2 0 1
Gentamicin
(10 ug)
5 2 10
Ampicillin
(10 ug)
20 12 15
Ceftriaxone
(2 ug)
3 5 9
Exmple. Your classroom data of ZOI (mm) / Ampicillin (ug) are 10, 7 and 11. The
corresponding data in the Data set B are 20, 12 and 15.
Claaroom Data set B
10 20
7 12
11 15
Student’s
t-test 0.076188387
Provide p value = .
The numbers in the two teams are significantly different Yes or No
Q17. Exogenous DNA does not passively enter E. coli that are not competent. What treatment do
cells require to be competent?
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Q18. Why doesn’t the recovery broth used in this experiment contain ampicillin?
Q19. What is the difference in the amino acid structure of Green and Blue fluorescence?
Q20. What evidence do you have the transformation is successful?
Q21. What are some reasons why transformation may not be successful?
Q22. What is the source of fluorescence? What are cells on the +DNA/+AMP/ +IPTG plate
fluorescence while cells on the +DNA/+AMP plate not fluoresce